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Research Article On the Bishop-Phelps-Bollobás Property for Numerical Radius Sun Kwang Kim, 1 Han Ju Lee, 2 and Miguel Martín 3 1 Department of Mathematics, Kyonggi University, Suwon 443-760, Republic of Korea 2 Department of Mathematics Education, Dongguk University-Seoul, Seoul 100-715, Republic of Korea 3 Departamento de An´ alisis M´ atematico, Facultad de Ciencias, Universidad de Granada, E-18071 Granada, Spain Correspondence should be addressed to Han Ju Lee; [email protected] Received 30 December 2013; Accepted 14 February 2014; Published 6 April 2014 Academic Editor: Manuel Maestre Copyright © 2014 Sun Kwang Kim et al. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We study the Bishop-Phelps-Bollob´ as property for numerical radius (in short, BPBp-nu) and find sufficient conditions for Banach spaces to ensure the BPBp-nu. Among other results, we show that 1 ()-spaces have this property for every measure . On the other hand, we show that every infinite-dimensional separable Banach space can be renormed to fail the BPBp-nu. In particular, this shows that the Radon-Nikod´ ym property (even reflexivity) is not enough to get BPBp-nu. 1. Introduction Let be a (real or complex) Banach space and its dual space. e unit sphere of will be denoted by . We write L() for the space of all bounded linear operators on . For L(), its numerical radius is defined by V () = sup { : (, ) ∈ Π ()} , (1) where Π() = {(, )∈ × : () = 1}. It is clear that V is a seminorm on L(). We refer the reader to the monographs [1, 2] for background. An operator L() attains its numerical radius if there exists ( 0 , 0 ) ∈ Π() such that V() = | 0 0 |. In this paper we will discuss the density of numerical radius attaining operators, actually on a stronger property called Bishop-Phelps-Bollob´ as property for numerical radius. Let us present first a short account on the known results about numerical radius attaining operators. Motivated by the study of norm attaining operators initiated by J. Lindenstrauss in the 1960s, Sims [3] asked in 1972 whether the numerical radius attaining operators are dense in the space of all bounded linear operators on a Banach space. Berg and Sims [4] gave a positive answer for uniformly convex spaces and Cardassi showed that the answer is positive for 1 , 0 , () (where is a metrizable compact), 1 (), and uniformly smooth spaces [57]. Acosta showed that the numerical radius attaining operators are dense in () for every compact Hausdorff space [8]. Acosta and Pay´ a showed that numerical radius attaining operators are dense in L() if has the Radon-Nikod´ ym property [9]. On the other hand, Pay´ a[10] showed in 1992 that there is a Banach space such that the numerical radius attaining operators are not dense in L(), which gave a negative answer to Sims’ question. Some authors also paid attention to the study of denseness of numerical radius attaining nonlinear mappings [1114]. Motivated by the work [15] of Acosta et al. on the Bishop-Phelps-Bollob´ as property for operators, Guirao and Kozhushkina [16] introduced very recently the notion of Bishop-Phelps-Bollob´ as property for numerical radius. Definition 1 (see [16]). A Banach space is said to have the Bishop-Phelps-Bollob´ as property for numerical radius (in short, BPBp-nu) if, for every 0<<1, there exists () > 0 such that, whenever L() and (, ) ∈ Π() satisfy V() = 1 and | | > 1 − (), there exit L() and (, ) ∈ Π() such that V () = = 1, ‖ − ‖ < , < , < . (2) Notice that if a Banach space has the BPBp-nu, then the numerical radius attaining operators are dense in L(). One of the main results of this paper is to show that the converse result is no longer true (Section 5). Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 479208, 15 pages http://dx.doi.org/10.1155/2014/479208

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Page 1: Research Article On the Bishop-Phelps-Bollobás Property ...downloads.hindawi.com/journals/aaa/2014/479208.pdf · Research Article On the Bishop-Phelps-Bollobás Property for Numerical

Research ArticleOn the Bishop-Phelps-Bollobaacutes Property for Numerical Radius

Sun Kwang Kim1 Han Ju Lee2 and Miguel Martiacuten3

1 Department of Mathematics Kyonggi University Suwon 443-760 Republic of Korea2Department of Mathematics Education Dongguk University-Seoul Seoul 100-715 Republic of Korea3 Departamento de Analisis Matematico Facultad de Ciencias Universidad de Granada E-18071 Granada Spain

Correspondence should be addressed to Han Ju Lee hanjuleedonggukedu

Received 30 December 2013 Accepted 14 February 2014 Published 6 April 2014

Academic Editor Manuel Maestre

Copyright copy 2014 Sun Kwang Kim et alThis is an open access article distributed under theCreativeCommonsAttribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We study the Bishop-Phelps-Bollobas property for numerical radius (in short BPBp-nu) and find sufficient conditions for Banachspaces to ensure the BPBp-nu Among other results we show that 119871

1(120583)-spaces have this property for every measure 120583 On the

other hand we show that every infinite-dimensional separable Banach space can be renormed to fail the BPBp-nu In particularthis shows that the Radon-Nikodym property (even reflexivity) is not enough to get BPBp-nu

1 Introduction

Let 119883 be a (real or complex) Banach space and 119883lowast its dualspace The unit sphere of 119883 will be denoted by 119878

119883 We write

L(119883) for the space of all bounded linear operators on119883 For119879 isinL(119883) its numerical radius is defined by

V (119879) = sup 1003816100381610038161003816119909lowast1198791199091003816100381610038161003816 (119909 119909

lowast) isin Π (119883) (1)

where Π(119883) = (119909 119909lowast) isin 119878119883times 119878

119883lowast 119909

lowast(119909) = 1 It is clear

that V is a seminorm on L(119883) We refer the reader to themonographs [1 2] for background An operator 119879 isin L(119883)attains its numerical radius if there exists (119909

0 119909

lowast

0) isin Π(119883)

such that V(119879) = |119909lowast0119879119909

0|

In this paper we will discuss the density of numericalradius attaining operators actually on a stronger propertycalled Bishop-Phelps-Bollobas property for numerical radiusLet us present first a short account on the known resultsabout numerical radius attaining operators Motivated by thestudy of normattaining operators initiated by J Lindenstraussin the 1960s Sims [3] asked in 1972 whether the numericalradius attaining operators are dense in the space of allbounded linear operators on a Banach space Berg and Sims[4] gave a positive answer for uniformly convex spaces andCardassi showed that the answer is positive for ℓ

1 119888

0 119862(119870)

(where 119870 is a metrizable compact) 1198711(120583) and uniformly

smooth spaces [5ndash7] Acosta showed that the numericalradius attaining operators are dense in 119862(119870) for every

compact Hausdorff space119870 [8] Acosta and Paya showed thatnumerical radius attaining operators are dense inL(119883) if 119883has the Radon-Nikodym property [9] On the other handPaya [10] showed in 1992 that there is a Banach space119883 suchthat the numerical radius attaining operators are not densein L(119883) which gave a negative answer to Simsrsquo questionSome authors also paid attention to the study of densenessof numerical radius attaining nonlinear mappings [11ndash14]

Motivated by the work [15] of Acosta et al on theBishop-Phelps-Bollobas property for operators Guirao andKozhushkina [16] introduced very recently the notion ofBishop-Phelps-Bollobas property for numerical radius

Definition 1 (see [16]) A Banach space 119883 is said to havethe Bishop-Phelps-Bollobas property for numerical radius (inshort BPBp-nu) if for every 0 lt 120576 lt 1 there exists 120578(120576) gt 0such that whenever 119879 isin L(119883) and (119909 119909lowast) isin Π(119883) satisfyV(119879) = 1 and |119909lowast119879119909| gt 1 minus 120578(120576) there exit 119878 isin L(119883) and(119910 119910

lowast) isin Π(119883) such that

V (119878) = 1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 = 1 119879 minus 119878 lt 120576

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowastminus 119910

lowast1003817100381710038171003817 lt 120576

(2)

Notice that if a Banach space119883 has the BPBp-nu then thenumerical radius attaining operators are dense inL(119883) Oneof the main results of this paper is to show that the converseresult is no longer true (Section 5)

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 479208 15 pageshttpdxdoiorg1011552014479208

2 Abstract and Applied Analysis

It is shown in [16] that the real or complex spaces 1198880and

ℓ1have the BPBp-nu This result has been extended to the

real space 1198711(R) by Falco [17] Aviles et al [18] give sufficient

conditions on a compact space 119870 for the real space 119862(119870) tohave the BPBp-nu which in particular include all metrizablecompact spaces

The content of this paper is the following First weintroduce in Section 2 a modulus of the BPBp-nu analogousto the one introduced in [19] for the Bishop-Phelps-Bollobasproperty for the operator norm and we will use it as a toolin the rest of the paper As easy applications we prove thatfinite-dimensional spaces always have the BPBp-nu and thata reflexive space has the BPBp-nu if and only if its dual doesNext Section 3 is devoted to prove that Banach spaces whichare both uniformly convex and uniformly smooth satisfy aweaker version of the BPBp-nu and to discuss such weakerversion In particular it is shown that 119871

119901(120583) spaces have the

BPBp-nu for every measure 120583 when 1 lt 119901 lt infin 119901 = 2We show in Section 4 that given any measure 120583 the real orcomplex space 119871

1(120583) has the BPBp-nu Finally we prove in

Section 5 that every separable infinite-dimensional Banachspace can be equivalently renormed to fail the BPBp-nu(actually to fail the weaker version) In particular this showsthat reflexivity (or even superreflexivity) is not enough for theBPBp-nu while the Radon-Nikodym property was knownto be sufficient for the density of numerical radius attainingoperators

Let us introduce some notations for later use The 119899-dimensional space with the ℓ

1norm is denoted by ℓ(119899)

1

Given a family 119883119896infin

119896=1of Banach spaces [⨁infin

119896=1119883

119896]1198880

(resp[⨁

infin

119896=1119883

119896]ℓ1

) is the Banach space consisting of all sequences(119909

119896)infin

119896=1such that each 119909

119896is in 119883

119896and lim

119896rarrinfin119909

119896 = 0

(resp suminfin

119896=1119909

119896 lt infin) equipped with the norm (119909

119896)infin

119896=1 =

sup119896119909

119896 (resp (119909

119896)infin

119896=1 = sum

infin

119896=1119909

119896)

2 Modulus of the Bishop-Phelps-Bollobaacutes forNumerical Radius

Analogously to what is done in [19] for the BPBp for theoperator norm we introduce here a modulus to quan-tify the Bishop-Phelps-Bollobas property for numericalradius

Notation 1 Let119883 be a Banach space Consider the set

Πnu (119883) = (119909 119909lowast 119879) (119909 119909

lowast) isin Π (119883)

119879 isinL (119883) V (119879) = 1 = 1003816100381610038161003816119909lowast1198791199091003816100381610038161003816

(3)

which is closed in 119878119883times 119878

119883lowast times L(119883) with respect to the

following metric

dist ((119909 119909lowast 119879) (119910 119910lowast 119878))

= max 1003817100381710038171003817119909 minus 1199101003817100381710038171003817 1003817100381710038171003817119909

lowastminus 119910

lowast1003817100381710038171003817 119879 minus 119878

(4)

The modulus of the Bishop-Phelps-Bollobas property fornumerical radius is the function defined by

120578nu (119883) (120576)

= inf 1 minus 1003816100381610038161003816119909lowast1198791199091003816100381610038161003816 (119909 119909

lowast) isin Π (119883) 119879 isinL (119883)

V (119879) = 1 dist ((119909 119909lowast 119879) Πnu (119883)) ⩾ 120576

(5)

for every 120576 isin (0 1) Equivalently 120578nu(119883)(120576) is the supremumof those scalars 120578 gt 0 such that whenever 119879 isin L(119883) and(119909 119909

lowast) isin Π(119883) satisfy V(119879) = 1 and |119909lowast119879119909| gt 1 minus 120578 there

exist 119878 isinL(119883) and (119910 119910lowast) isin Π(119883) such that

V (119878) = 1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 = 1 119879 minus 119878 lt 120576

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowastminus 119910

lowast1003817100381710038171003817 lt 120576

(6)

It is immediate that a Banach space119883 has the BPBp-nu ifand only if 120578nu(120576) gt 0 for every 0 lt 120576 lt 1 By construction ifa function 120576 997891rarr 120578(120576) is valid in the definition of the BPBp-nuthen 120578nu(120576) ⩾ 120578(120576)

An immediate consequence of the compactness of theunit ball of a finite-dimensional space is the following resultIt was previously known to A Guirao (private communica-tion)

Proposition 2 Let 119883 be a finite-dimensional Banach spaceThen119883 has the Bishop-Phelps-Bollobas property for numericalradius

Proof Let 119870 = 119878 isin L(119883) V(119878) = 0 Then 119870 is anorm-closed subspace of L(119883) Hence L(119883)119870 is a finite-dimensional space with two norms

V ([119879]) = inf V (119879 minus 119878) 119878 isin 119870 = V (119879)

[119879] = inf 119879 minus 119878 119878 isin 119870 (7)

where [119879] is the class of 119879 in the quotient space L(119883)119870Hence there is a constant 0 lt 119888 ⩽ 1 such that

119888 [119879] ⩽ V (119879) ⩽ [119879] (8)

Suppose that119883 does not have the BPBp-nuThen there is0 lt 120576 lt 1 such that 120578nu(119883)(120576) = 0That is there are sequences(119909

119899 119909

lowast

119899) isin Π(119883) and (119879

119899) isinL(119883) with V(119879

119899) = 1 such that

dist ((119909119899 119909

lowast

119899 119879

119899) Πnu (119883)) ⩾ 120576 (119899 isin N)

lim119899

1003816100381610038161003816119909lowast

119899119879119899119909119899

1003816100381610038161003816 = 1

(9)

By compactness wemay assume that lim119899[119879

119899]minus[119879

0] = 0 for

some1198790isinL(119883) and V(119879

0) = 1 Hence there exists a sequence

119878119899119899in 119870 such that lim

119899119879

119899minus (119879

0+ 119878

119899) = 0 Observe that

V(1198790+ 119878

119899) = V(119879

0) = 1 for every 119899 isin N

By compactness again we may assume that (119909119899 119909

lowast

119899)

converges to (1199090 119909

lowast

0) isin 119883 times 119883

lowast This implies that (1199090 119909

lowast

0) isin

Π(119883) and |119909lowast0(119879

0+ 119878

119899)119909

0| = V(119879

0+ 119878

119899) = 1 that is

(1199090 119909

lowast

0 119879

0+ 119878

119899) isin Πnu(119883) for all 119899 This is a contradiction

with the fact that0 = lim

119899dist ((119909

119899 119909

lowast

119899 119879

119899) (119909

0 119909

lowast

0 119879

0+ 119878

119899))

⩾ lim119899

dist ((119909119899 119909

lowast

119899 119879

119899) Πnu (119883)) ⩾ 120576

(10)

Abstract and Applied Analysis 3

We may also give the following easy result concerningduality

Proposition 3 Let119883 be a reflexive space Then

120578nu (119883) (120576) = 120578nu (119883lowast) (120576) (11)

for every 120576 isin (0 1) In particular 119883 has the BPBp-nu if andonly if 119883lowast has the BPBp-nu

Wewill use that V(119879lowast) = V(119879) for all 119879 isinL(119883) where 119879lowastdenotes the adjoint operator of 119879 This result can be found in[1] but it is obvious if119883 is reflexive

Proof By reflexivity it is enough to show that 120578nu(119883)(120576) ⩽120578nu(119883

lowast)(120576) Let 120576 isin (0 1) be fixed If 120578nu(119883)(120576) = 0 there is

nothing to prove Otherwise consider 0 lt 120578 lt 120578nu(119883)(120576)Suppose that 119879

1isinL(119883lowast

) and (119909lowast1 119909

1) isin Π(119883

lowast) satisfy

V (1198791) = 1

100381610038161003816100381611990911198791119909lowast

1

1003816100381610038161003816 gt V (1198791) minus 120578 (12)

By considering 119879lowast1isin L(119883) we may find 119878

1isin L(119883) and

(1199101 119910

lowast

1) isin Π(119883) such that

1003816100381610038161003816119910lowast

111987811199101

1003816100381610038161003816 = V (1198781) = 1

10038171003817100381710038171199101 minus 11990911003817100381710038171003817 lt 120576

1003817100381710038171003817119910lowast

1minus 119909

lowast

1

1003817100381710038171003817 lt 1205761003817100381710038171003817119879

lowast

1minus 119878

1

1003817100381710038171003817 lt 120576

(13)

Then 119878lowast1isinL(119883lowast

) and (119910lowast1 119910

1) isin Π(119883

lowast) satisfy

1003816100381610038161003816⟨1199101 119878lowast

1119910lowast

1⟩1003816100381610038161003816 = V (119878

1) = 1

1003817100381710038171003817119910lowast

1minus 119909

lowast

1

1003817100381710038171003817 lt 120576

10038171003817100381710038171199101 minus 11990911003817100381710038171003817 lt 120576

10038171003817100381710038171198791 minus 119878lowast

1

1003817100381710038171003817 lt 120576

(14)

This implies that 120578nu(119883lowast)(120576) ⩾ 120578 We finish by just taking

supremum on 120578

We do not know whether the result above is valid in thenonreflexive case

3 Spaces Which Are Both Uniformly Convexand Uniformly Smooth

For a Banach space which is both uniformly convex anduniformly smooth we get a property which is weaker thanBPBp-nu This result was known to A Guirao (privatecommunication)

Proposition 4 Let 119883 be a uniformly convex and uniformlysmooth Banach space Then given 120576 gt 0 there exists 120578(120576) gt 0such that whenever 119879

0isinL(119883) with V(119879

0) = 1 and (119909

0 119909

lowast

0) isin

Π(119883) satisfy |119909lowast011987901199090| gt 1 minus 120578(120576) there exist 119878 isin L(119883) and

(119910 119910lowast) isin Π(119883) such that

V (119878) = 1003816100381610038161003816119910lowast1198781199101003816100381610038161003816

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowastminus 119910

lowast1003817100381710038171003817 lt 1205761003817100381710038171003817119878 minus 1198790

1003817100381710038171003817 lt 120576

(15)

Proof Notice that the uniform smoothness of119883 is equivalentto the uniform convexity of 119883lowast Let 120575

119883(120576) and 120575

119883lowast(120576) be the

moduli of convexity119883 and119883lowast respectively Given 0 lt 120576 lt 1consider

120578 (120576) =120576

4min 120575

119883(120576

4) 120575

119883lowast (120576

4) gt 0 (16)

Consider 1198790isin L(119883) with V(119879

0) = 1 and (119909

0 119909

lowast

0) isin Π(119883)

satisfying |119909lowast011987901199090| gt 1 minus 120578(120576) Define 119879

1isinL(119883) by

1198791119909 = 119879

0119909 + 120582

1(120576

4) 119909

lowast

0(119909) 119909

0(17)

for all 119909 isin 119883 where 1205821is the scalar satisfying |120582

1| = 1 and

|119909lowast

011987901199090+ 120582

1(1205764)| = |119909

lowast

011987901199090| + 1205764 Now choose 119909

1isin 119878

119883

and 119909lowast1isin 119878

119883lowast such that |119909lowast

1(119909

1)| = 1 119909lowast

1(119909

0) = |119909

lowast

1(119909

0)| and

1003816100381610038161003816119909lowast

111987911199091

1003816100381610038161003816 ⩾ V (1198791) minus 120578(

1205762

42) (18)

Now we define a sequence (119909119899 119909

lowast

119899 119879

119899) in 119878

119883times 119878

119883lowast times L(119883)

inductively Indeed suppose that we have a defined sequence(119909

119895 119909

lowast

119895 119879

119895) for 0 ⩽ 119895 ⩽ 119899 and let

119879119899+1119909 = 119879

119899119909 + 120582

119899+1

120576119899+1

4119899+1119909lowast

119899(119909) 119909

119899 (19)

Then choose 119909119899+1

isin 119878119883

and 119909lowast119899+1

isin 119878119883lowast such that

|119909lowast

119899+1(119909

119899+1)| = 1 and |119909lowast

119899+1(119909

119899)| = 119909

lowast

119899+1(119909

119899)

1003816100381610038161003816119909lowast

119899+1119879119899+1119909119899+1

1003816100381610038161003816 ⩾ V (119879119899+1) minus 120578(

120576119899+2

4119899+2) (20)

Notice that for all 119899 ⩾ 0 we have

1003817100381710038171003817119879119899+1 minus 1198791198991003817100381710038171003817 ⩽120576119899+1

4119899+1

1003816100381610038161003816V (119879119899+1) minus V (119879119899)1003816100381610038161003816 ⩽120576119899+1

4119899+1

(21)

This implies that (119879119899) is a Cauchy sequence and assume that

it converges to 119878 isinL(119883) Then we have

lim119899119879119899= 119878

10038171003817100381710038171198790 minus 1198781003817100381710038171003817 lt 120576

lim119899

1003816100381610038161003816119909lowast

119899119879119899119909119899

1003816100381610038161003816 = lim119899V (119879

119899) = V (119878)

(22)

We will show that both sequences (119909119899) and (119909lowast

119899) are Cauchy

From the definition we have

V (119879119899+1) minus 120578(

120576119899+2

4119899+2)

⩽1003816100381610038161003816119909

lowast

119899+1119879119899+1119909119899+1

1003816100381610038161003816

4 Abstract and Applied Analysis

100381610038161003816100381610038161003816100381610038161003816

119909lowast

119899+1119879119899119909119899+1+ 120582

119899+1

120576119899+1

4119899+1119909lowast

119899(119909

119899+1) 119909

lowast

119899+1(119909

119899)

100381610038161003816100381610038161003816100381610038161003816

⩽ V (119879119899) +120576119899+1

4119899+1119909lowast

119899+1(119909

119899)

V (119879119899+1)

⩾1003816100381610038161003816119909

lowast

119899119879119899+1119909119899

1003816100381610038161003816 =

100381610038161003816100381610038161003816100381610038161003816

119909lowast

119899119879119899119909119899+ 120582

119899+1

120576119899+1

4119899+1

100381610038161003816100381610038161003816100381610038161003816

=1003816100381610038161003816119909

lowast

119899119879119899119909119899

1003816100381610038161003816 +120576119899+1

4119899+1⩾ V (119879

119899) minus 120578(

120576119899+1

4119899+1) +120576119899+1

4119899+1

(23)

In summary we have

V (119879119899) +120576119899+1

4119899+1119909lowast

119899+1(119909

119899)

⩾ V (119879119899) minus 120578(

120576119899+1

4119899+1) +120576119899+1

4119899+1minus 120578(

120576119899+2

4119899+2)

(24)

Hence

119909lowast

119899+1(119909

119899) ⩾ 1 minus 2

4119899+1

120576119899+1120578(120576119899+1

4119899+1)

= 1 minus1

2min120575

119883(120576119899+1

4119899+2) 120575

119883lowast (120576119899+1

4119899+2)

1003817100381710038171003817100381710038171003817

119909119899+ 119909

119899+1

2

1003817100381710038171003817100381710038171003817⩾ 119909

lowast

119899+1(119909119899+ 119909

119899+1

2) ⩾ 1 minus 120575

119883(120576119899+1

4119899+2)

100381710038171003817100381710038171003817100381710038171003817

119909lowast

119899+ 119909

lowast

119899+1

2

100381710038171003817100381710038171003817100381710038171003817

⩾119909lowast

119899+ 119909

lowast

119899+1

2(119909

119899) ⩾ 1 minus 120575

119883lowast (120576119899+1

4119899+2)

(25)

This means that

1003817100381710038171003817119909119899 minus 119909119899+11003817100381710038171003817 ⩽ 120576

119899+14

119899+2

1003817100381710038171003817119909lowast

119899minus 119909

lowast

119899+1

1003817100381710038171003817 ⩽ 120576119899+14

119899+2

(26)

for all 119899 So (119909119899) and (119909lowast

119899) are Cauchy Let 119909

infin= lim

119899119909119899and

119909lowast

infin= lim

119899119909lowast

119899 Then 119909

0minus 119909

infin lt 1205764 and 119909lowast

0minus 119909

lowast

infin lt 1205764

Hence |119909lowastinfin(119909

infin)| = lim

119899|119909

lowast

119899(119909

119899)| = 1 and

V (119878) = lim119899V (119879

119899) = lim

119899

1003816100381610038161003816119909lowast

119899119879119899119909119899

1003816100381610038161003816 =1003816100381610038161003816119909

lowast

infin119878119909

infin

1003816100381610038161003816 (27)

Let 120572 = 119909lowastinfin(119909

infin) 119910lowast = 119909lowast

infin and 119910 = 119909

infin Then we have

119910lowast(119910) = 1 V(119878) = |119910lowast119878119910| and 119910 minus 119909

0 lt 120576 Notice that

|120572 minus 1| =1003816100381610038161003816119909

lowast

infin(119909

infin) minus 119909

lowast

0(119909

0)1003816100381610038161003816

⩽1003816100381610038161003816(119909

lowast

infinminus 119909

lowast

0) (119909

infin)1003816100381610038161003816 +1003816100381610038161003816119909

lowast

0(119909

infin) minus 119909

lowast

0(119909

0)1003816100381610038161003816 lt120576

2

(28)

Therefore

1003817100381710038171003817119910lowastminus 119909

lowast1003817100381710038171003817 ⩽1003817100381710038171003817119910

lowastminus 119910

lowast1003817100381710038171003817 +1003817100381710038171003817119910

lowastminus 119909

lowast1003817100381710038171003817 lt120576

2+120576

4lt 120576 (29)

This completes the proof

Let us discuss a little bit about the equivalence betweenthe property in the result above and the BPBp-nu Forconvenience let us introduce the following definition

Definition 5 A Banach space 119883 has the weak Bishop-Phelps-Bollobas property for the numerical radius (in short weak-BPBp-nu) if given 120576 gt 0 there exists 120578(120576) gt 0 such thatwhenever 119879

0isin L(119883) with V(119879

0) = 1 and (119909

0 119909

lowast

0) isin Π(119883)

satisfy |119909lowast011987901199090| gt 1 minus 120578(120576) there exist 119878 isin L(119883) and

(119910 119910lowast) isin Π(119883) such that

V (119878) = 1003816100381610038161003816119910lowast1198781199101003816100381610038161003816

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowastminus 119910

lowast1003817100381710038171003817 lt 120576 119878 minus 119879 lt 120576

(30)

Notice that the only difference between this concept andthe BPBp-nu is the normalization of the operator 119878 by thenumerical radius Of course if the numerical radius andthe operator norm are equivalent these two properties arethe same This equivalence is measured by the so-callednumerical index of the Banach space as follows For a Banachspace119883 the numerical index of119883 is defined by

119899 (119883) = inf V (119879) 119879 isinL (119883) 119879 = 1 (31)

It is clear that 0 ⩽ 119899(119883) ⩽ 1 and 119899(119883)119879 ⩽ V(119879) ⩽ 119879for all 119879 isin L(119883) The value 119899(119883) = 1 means that V equalsthe usual operator norm This is the case of 119883 = 119871

1(120583) and

119883 = 119862(119870) among many others On the other hand 119899(119883) gt 0if and only if the numerical radius is equivalent to the normof L(119883) We refer the reader to [20] for more informationand background

The following result is immediate We include a proof forthe sake of completeness

Proposition 6 Let119883 be a Banach space with 119899(119883) gt 0 Then119883 has the BPBp-nu if and only if119883 has the weak-BPBp-nu

Proof Thenecessity is clear For the converse assume that wehave 120578(120576) gt 0 satisfying the conditions of the weak-BPBp-nufor all 0 lt 120576 lt 1 If 119879 isin L(119883) with V(119879) = 1 and (119909

0 119909

lowast

0) isin

Π(119883) satisfy |119909lowast0119879119909

0| gt 1 minus 120578(120576) for 0 lt 120576 lt 1 then there exist

119878 isinL(119883) and (119910 119910lowast) isin Π(119883) such that

V (119878) = 1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 119878 minus 119879 lt 120576

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowastminus 119910

lowast1003817100381710038171003817 lt 120576

(32)

As V(119878) gt 0 by the above let 1198781= (1V(119878))119878 Then we have

1 = V (1198781) =1003816100381610038161003816119910

lowast11987811199101003816100381610038161003816

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowastminus 119910

lowast1003817100381710038171003817 lt 120576

(33)

Abstract and Applied Analysis 5

Finally we have

10038171003817100381710038171198781 minus 1198791003817100381710038171003817 ⩽

10038171003817100381710038171003817100381710038171003817

1

V (119878)119878 minus 119878

10038171003817100381710038171003817100381710038171003817

+ 119878 minus 119879

=119878

V (119878)|V (119878) minus 1| + 119878 minus 119879

⩽1

119899 (119883)|V (119878) minus V (119879)| + 119878 minus 119879

⩽ (1

119899 (119883)+ 1) 119878 minus 119879 lt

119899 (119883) + 1

119899 (119883)120576

(34)

An obvious change of parameters finishes the proof

We do not know whether the hypothesis of 119899(119883) gt 0 canbe omitted in the above result

Putting together Propositions 4 and 6 we get the follow-ing

Corollary 7 Let 119883 be a uniformly convex and uniformlysmooth Banach space with 119899(119883) gt 0 Then 119883 has the BPBp-nu

Let us comment that every complex Banach space 119883satisfies 119899(119883) ⩾ 1119890 so the above corollary automaticallyapplies in the complex case In the real case this is no longertrue as the numerical index of a Hilbert space of dimensiongreater than or equal to two is 0 On the other hand it isproved in [21] that real 119871

119901(120583) spaces have nonzero numerical

index for every measure 120583when 119901 = 2Therefore we have thefollowing examples

Example 8 (a) Complex Banach spaces which are uniformlysmooth and uniformly convex satisfy the BPBp-nu

(b) In particular for every measure 120583 the complex spaces119871119901(120583) have the BPBp-nu for 1 lt 119901 lt infin(c) For every measure 120583 the real spaces 119871

119901(120583) have the

BPBp-nu for 1 lt 119901 lt infin 119901 = 2

Note Added in Revision Very recently H J Lee M Martınand J Merı have proved that Proposition 6 can be extendedto some Banach spaces with numerical index zero as forinstance real Hilbert spaces Hence they have shown thatHilbert spaces have the BPBp-nu These results will appearelsewhere

4 L1

Spaces

In this section we will show that 1198711(120583) has the BPBp-nu for

every measure 120583 In the proof we are dealing with complexintegrable functions since the real case is followed easilyby applying the same proof Our main result here is thefollowing

Theorem 9 Let 120583 be a measure Then 1198711(120583) has the Bishop-

Phelps-Bollobas property for numerical radius More preciselygiven 120576 gt 0 there exists 120578(120576) gt 0 (which does not depend on120583) such that whenever 119879

0isin L(119871

1(120583)) with V(119879

0) = 1 and

(1198910 119892

0) isin Π(119871

1(120583)) satisfy |⟨119879

01198910 119892

0⟩| gt 1 minus 120578(120576) then there

exist 119879 isinL(1198711(120583)) (119891

1 119892

1) isin Π(119871

1(120583)) such that

1003816100381610038161003816⟨1198791198911 1198921⟩1003816100381610038161003816 = V (119879) = 1 10038171003817100381710038171198910 minus 1198911

1003817100381710038171003817 lt 120576

10038171003817100381710038171198920 minus 11989211003817100381710038171003817 lt 120576

1003817100381710038171003817119879 minus 11987901003817100381710038171003817 lt 120576

(35)

As a first step we have to start dealing with finite regularpositive Borel measures for which a representation theoremfor operators exists

Proposition 10 Let 119898 be a finite regular positive Borelmeasure on a compact Hausdorff spaceΩ Then 119871

1(119898) has the

Bishop-Phelps-Bollobas property for numerical radius Moreprecisely given 120576 gt 0 there is 120578(120576) gt 0 (which is independent ofthemeasure119898) such that if a norm-one element119879 inL(119871

1(119898))

and (1198910 119892

0) isin Π(119871

1(119898)) satisfy |⟨119879119891

0 119892

0⟩| gt 1 minus 120578(120576) then

there exist an operator 119878 isin L(1198711(119898)) (119891

1 119892

1) isin Π(119871

1(119898))

such that1003816100381610038161003816⟨1198781198911 1198921⟩

1003816100381610038161003816 = 119878 = 110038171003817100381710038171198910 minus 1198911

1003817100381710038171003817 ⩽ 120576

10038171003817100381710038171198920 minus 11989211003817100381710038171003817 ⩽ 120576 119879 minus 119878 ⩽ 120576

(36)

To prove this proposition we need some backgroundon representation of operators on Lebesgue spaces on finiteregular positive Borel measures and several preliminarylemmas

Let 119898 be a finite regular positive Borel measure on acompact Hausdorff space Ω If 120583 is a complex-valued Borelmeasure on the product space Ω times Ω then define theirmarginal measures 120583119894 on Ω (119894 = 1 2) as follows 1205831(119860) =120583(119860 times Ω) and 1205832(119861) = 120583(Ω times 119861) where 119860 and 119861 are Borelmeasurable subsets of Ω

Let 119872(119898) be the complex Banach lattice of measuresconsisting of all complex-valued Borel measures 120583 on theproduct spaceΩ timesΩ such that |120583|119894 are absolutely continuouswith respect to119898 for 119894 = 1 2 endowed with the norm

1003817100381710038171003817100381710038171003817100381710038171003817

11988910038161003816100381610038161205831003816100381610038161003816

1

119889119898

1003817100381710038171003817100381710038171003817100381710038171003817infin

(37)

Each 120583 isin 119872(119898) defines a bounded linear operator 119879120583from

1198711(119898) to itself by

⟨119879120583(119891) 119892⟩ = int

ΩtimesΩ

119891 (119909) 119892 (119910) 119889120583 (119909 119910) (38)

where 119891 isin 1198711(119898) and 119892 isin 119871

infin(119898) Iwanik [22] showed that

the mapping 120583 997891rarr 119879120583is a lattice isometric isomorphism from

119872(119898) onto L(1198711(119898)) Even though he showed this for the

real case it can be easily generalized to the complex caseFor details see [22 Theorem 1] and [23 IV Theorem 15(ii)Corollary 2]

We will also use that given an arbitrary measure 120583 every119879 isin L(119871

1(120583)) satisfies V(119879) = 119879 [24] (that is the space

1198711(120583) has numerical index 1)

Lemma 11 (see [15 Lemma 33]) Let 119888119899 be a sequence of

complex numbers with |119888119899| ⩽ 1 for every 119899 and let 120578 gt 0 such

6 Abstract and Applied Analysis

that for a convex series sum120572119899 Resuminfin

119899=1120572119899119888119899gt 1 minus 120578 Then for

every 0 lt 119903 lt 1 the set 119860 = 119894 isin N Re 119888119894gt 119903 satisfies the

estimate

sum

119894isin119860

120572119894⩾ 1 minus

120578

1 minus 119903 (39)

From now on 119898 will be a finite regular positive Borelmeasure on the compact Hausdorff space Ω

Lemma 12 Suppose that there exist a nonnegative simplefunction 119891 isin 119878

1198711(119898)

and a function 119892 isin 119878119871infin(119898)

such that

Re ⟨119891 119892⟩ gt 1 minus 1205763

16 (40)

Then there exist a nonnegative simple function 1198911isin 119878

1198711(119898)

anda function 119892

1isin 119878

119871infin(119898)

such that

1198921(119909) = 120594supp(119891

1)(119909) + 119892 (119909) 120594

Ωsupp(1198911)(119909)

⟨1198911 119892

1⟩ = 1

1003817100381710038171003817119891 minus 119891110038171003817100381710038171lt 120576

1003817100381710038171003817119892 minus 11989211003817100381710038171003817infinlt radic120576 supp (119891

1) sub supp (119891)

(41)

Proof Let119891 = sum119898

119895=1(120573

119895119898(119861

119895))120594

119861119895

for some (120573119895) such that

120573119895⩾ 0 for all 119895 andsum119898

119895=1120573119895= 1 and119861

119895rsquos aremutually disjoint

By the assumption we have

Re ⟨119891 119892⟩ =119899

sum

119895=1

120573119895

1

119898 (119861119895)

int119861119895

Re119892 (119909) 119889119898 (119909) gt 1 minus 1205763

16

(42)

and letting

119869 = 119895 1 ⩽ 119895 ⩽ 1198991

119898 (119861119895)

int119861119895

Re119892 (119909) 119889119898 (119909) gt 1 minus 1205762

4

(43)

we have by Lemma 11

sum

119895isin119869

120573119895gt 1 minus

120576

4 (44)

For each 119895 isin 119869 we have

1 minus1205762

4lt

1

119898 (119861119895)

int119861119895

Re119892 (119909) 119889119898 (119909)

=1

119898 (119861119895)

int119861119895capRe119892⩽1minus120576

Re119892 (119909) 119889119898 (119909)

+ int119861119895capRe119892gt1minus120576

Re119892 (119909) 119889119898 (119909)

⩽1

119898 (119861119895)

((1 minus 120576)119898 (119861119895cap Re119892 ⩽ 1 minus 120576)

+ 119898 (119861119895cap Re119892 gt 1 minus 120576))

= 1 minus 120576

119898 (119861119895cap Re119892 ⩽ 1 minus 120576)

119898 (119861119895)

(45)

This implies that

119898(119861119895cap Re119892 ⩽ 1 minus 120576)

119898 (119861119895)

lt120576

4 (46)

Define 119861119895= 119861

119895cap Re119892 gt 1 minus 120576 for all 119895 isin 119869

1198911= (1sum

119895isin119869

120573119895)sum

119895isin119869

120573119895(120594

119861119895

119898 (119861119895)) (47)

and 1198921(119909) = 1 on supp(119891

1) and 119892

1(119909) = 119892(119909) elsewhereThen

it is clear that supp(1198911) sub supp(119891) 119892 minus 119892

1infinlt radic120576 and

⟨1198911 119892

1⟩ = 1 Finally we will show that 119891 minus 119891

1 lt 120576 Notice

first that10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

= 2sum

119895isin119869

120573119895

119898(119861119895 119861

119895)

119898 (119861119895)

lt120576

2

(48)

Hence1003817100381710038171003817119891 minus 1198911

1003817100381710038171003817

10038171003817100381710038171003817100381710038171003817100381710038171003817

1

sum119895isin119869120573119895

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus 119891

10038171003817100381710038171003817100381710038171003817100381710038171003817

1 minus sum119895isin119869120573119895

sum119895isin119869120573119895

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+120576

4

= (1 minussum

119895isin119869

120573119895) +

120576

2+120576

4

⩽120576

4+120576

2+120576

4= 120576

(49)

Lemma 13 (see [25 Lemma 33]) Suppose that 119879120583is a norm-

one element in L(1198711(119898)) for some 120583 isin 119872(119898) and there is

a nonnegative simple function 1198910such that 119891

0is a norm-one

element of 1198711(119898) and 119879

1205831198910 ⩾ 1 minus 120576

32

6 for some 0 lt 120576 lt 1

Abstract and Applied Analysis 7

Then there exist a norm-one bounded linear operator 119879] forsome ] isin 119872(119898119898) and a nonnegative simple function 119891

1

in 1198781198711(119898)

such that 119879120583minus 119879] ⩽ 120576 1198911 minus 1198910 ⩽ 3120576 and

(119889|]|1119889119898)(119909) = 1 for all 119909 isin supp(1198911)

Lemma 14 Suppose that 119879] isin L(1198711(119898)) is a norm-one

operator 119891 = sum119899

119894=1120573119894(120594

119861119894

119898(119861119894)) where 119898(119861

119895) gt 0 for all

1 ⩽ 119895 ⩽ 119899 and 119861119895119899

119895=1are mutually disjoint Borel subsets ofΩ

is a norm-one nonnegative simple function and 119892 is an elementof 119878

119871infin(119898)

such that

Re ⟨119892 119879]119891⟩ ⩾ 1 minus1205766

27(50)

for some 0 lt 120576 lt 1 and

119889|]|1

119889119898(119909) = 1 119892 (119909) = 1 (51)

for all x in the support of 119891Then there exist a nonnegative simple function 119891 isin 119878

1198711(119898)

a function119892 isin 119878

119871infin(119898)

and an operator119879] inL(1198711(119898) 1198711(119898))such that

⟨119892 119879]119891⟩ =1003817100381710038171003817119879]1003817100381710038171003817 = 1

1003817100381710038171003817119879] minus 119879]1003817100381710038171003817 ⩽ 2120576

10038171003817100381710038171003817119891 minus 119891

10038171003817100381710038171003817⩽ 3120576

1003817100381710038171003817119892 minus 1198921003817100381710038171003817 ⩽radic120576 ⟨119891 119892⟩ = 1

(52)

Proof Since

Re ⟨119892 119879]119891⟩ ⩾ 1 minus1205766

27 (53)

we have

1 minus1205766

27lt Re ⟨119892 119879]119891⟩

= intΩtimesΩ

119891 (119909)Re119892 (119910) 119889120583 (119909 119910)

=

119899

sum

119895=1

120573119895intΩtimesΩ

120594119861119895

(119909)

119898 (119861119895)

Re119892 (119910) 119889] (119909 119910)

(54)

Let

119869 = 119895 intΩtimesΩ

(

120594119861119895

(119909)

119898 (119861119895)

)Re119892 (119910) 119889] (119909 119910) gt 1 minus 1205763

26

(55)

Then from Lemma 11 we have sum119895isin119869120573119895gt 1 minus 120576

32 Let 119891

1=

sum119895isin119869120573119895(120594

119861119895

119898(119861119895)) where 120573

119895= 120573

119895(sum

119895isin119869120573119895) for all 119895 isin 119869

Then

10038171003817100381710038171198911 minus 1198911003817100381710038171003817 ⩽

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

(120573119895minus 120573

119895)

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+ sum

119895isin119869

120573119895⩽ 120576

3⩽ 120576 (56)

Note that there is a Borel measurable function ℎ on Ω times Ωsuch that 119889](119909 119910) = ℎ(119909 119910) 119889|]|(119909 119910) and |ℎ(119909 119910)| = 1 forall (119909 119910) isin Ω times Ω Let

119862 = (119909 119910) 1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 1

1003816100381610038161003816 ltradic120576

232 (57)

Define two measures ]119891and ]

119888as follows

]119891(119860) = ] (119860 119862) ]

119888(119860) = ] (119860 cap 119862) (58)

for every Borel subset 119860 ofΩ times Ω It is clear that

119889] = 119889]119891+ 119889]

119888 119889

10038161003816100381610038161003816]119891

10038161003816100381610038161003816= ℎ 119889]

119891

1198891003816100381610038161003816]1198881003816100381610038161003816 = ℎ 119889]119888 119889 |]| = 119889

10038161003816100381610038161003816]119891

10038161003816100381610038161003816+ 1198891003816100381610038161003816]1198881003816100381610038161003816

(59)

Since (119889|]|11198891198981)(119909) = 1 for all 119909 isin ⋃119899

119895=1119861119895 we have

1 =119889|]|1

1198891198981

(119909) =

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) +1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909) (60)

for all 119909 isin 119861 = ⋃119899

119895=1119861119895 and we deduce that |]|1(119861

119895) = 119898

1(119861

119895)

for all 1 ⩽ 119895 ⩽ 119899We claim that |]

119891|1(119861

119895)119898

1(119861

119895) ⩽ 120576

22

2 for all 119895 isin 119869Indeed if |119892(119910)ℎ(119909 119910) minus 1| ⩾ radic120576232 then

Re (119892 (119910) ℎ (119909 119910)) ⩽ 1 minus 12057624 (61)

So we have

1 minus1205763

26

⩽1

1198981(119861

119895)

ReintΩtimesΩ

120594119861119895(119909)119892 (119910) 119889] (119909 119910)

=1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 |]| (119909 119910)

=1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 10038161003816100381610038161003816]11989110038161003816100381610038161003816(119909 119910)

+1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 1003816100381610038161003816]1198881003816100381610038161003816 (119909 119910)

⩽1

1198981(119861

119895)

((1 minus120576

24)10038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

(119861119895) +1003816100381610038161003816]1198881003816100381610038161003816

1

(119861119895))

= 1 minus120576

24

10038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

(119861119895)

1198981(119861

119895)

(62)

This proves our claimWe also claim that for each 119895 isin 119869 there exists a Borel

subset 119861119895of 119861

119895such that

(1 minus120576

2)119898

1(119861

119895) ⩽ 119898

1(119861

119895) ⩽ 119898

1(119861

119895)

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) ⩽120576

2

(63)

8 Abstract and Applied Analysis

for all 119909 isin 119861119895 Indeed set 119861

119895= 119861

119895cap 119909 isin Ω (119889|]

119891|1119889119898

1)

(119909) ⩽ 1205762 Then

int119861119895119861119895

120576

2119889119898

1(119909) ⩽ int

119861119895

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) 1198891198981(119909)

=10038161003816100381610038161003816]1119891

10038161003816100381610038161003816(119861

119895) ⩽1205762

22119898

1(119861

119895)

(64)

This shows that 1198981(119861

119895 119861

119895) ⩽ (1205762)119898

1(119861

119895) This proves our

second claimNow we define 119892 by 119892(119910) = 119892(119910)|119892(119910)| if |119892(119910)| ⩾ 1 minus

radic120576232 and 119892(119910) = 119892(119910) if |119892(119910)| lt 1 minus radic120576232 and we

write 119891 = sum119895isin119869120573119895(120594

119861119895

1198981(119861

119895)) It is clear that 119892 isin 119878

119871infin(119898)

119892 minus 119892 lt radic120576 and 119892(119910) = 1 for all 119909 isin supp119891

Finally we define the measure

119889] (119909 119910)

= sum

119895isin119869

120594119861119895

(119909) 119892 (119910)ℎ (119909 119910)119889]119888(119909 119910)(

1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+ 1205941198691119861(119909) 119889] (119909 119910)

(65)

where 119861 = ⋃119895isin119869119861119895 It is easy to see that (119889|]|1119889119898

1)(119909) = 1

on 119861 and (119889|]|11198891198981)(119909) ⩽ 1 elsewhere Note that

119889 (] minus ]) (119909 119910)

= sum

119895isin119869

120594119861119895

(119909) [

[

119892 (119910)ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1]

]

119889]119888(119909 119910)

minus sum

119895isin119869

120594119861119895

(119909) 119889]119891(119909 119910)

(66)

If (119909 119910) isin 119862 then |119892(119910)| ⩾ 1 minus radic120576232 ⩾ 1 minus 1232 and

100381610038161003816100381610038161003816119892 (119910)ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

119892 (119910)

1003816100381610038161003816119892 (119910)1003816100381610038161003816

ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816100381610038161003816

1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 11003816100381610038161003816

1003816100381610038161003816119892 (119910)1003816100381610038161003816

+

10038161003816100381610038161 minus1003816100381610038161003816119892 (119910)

1003816100381610038161003816

10038161003816100381610038161003816100381610038161003816119892 (119910)

1003816100381610038161003816

⩽ 2

1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 11003816100381610038161003816

1003816100381610038161003816119892 (119910)1003816100381610038161003816

⩽ 2radic120576

232

232

232 minus 1⩽ 2radic120576

(67)

Hence for all (119909 119910) isin 119862 we have

10038161003816100381610038161003816100381610038161003816100381610038161003816

119892 (119910)ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

⩽100381610038161003816100381610038161003816119892 (119910)ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

⩽ 2radic120576(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

(68)

So we have for all 119909 isin 1198691

119889|] minus ]|1

1198891198981

(119909)

⩽ sum

119895isin119869

120594119861119895

(119909) [

[

2radic120576(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

]

]

1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909)

+ sum

119895isin119869

120594119861119895

(119909)

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909)

⩽ sum

119895isin119869

120594119861119895

(119909)(2radic120576 + (1 minus1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909)))

+sum

119895isin119869

120594119861119895

(119909)(

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909))

⩽ 2radic120576 + 120576 lt 3radic120576

(69)

This gives that 119879] minus 119879] lt 3radic120576 Note also that for all 119895 isin 119869

⟨119879]

120594119861119895

1198981(119861

119895)

119892⟩

= intΩtimesΩ

120594119861119895

(119909)

1198981(119861

119895)

119892 (119910) 119889] (119909 119910)

= intΩtimesΩ

120594119861119895

(119909)

1198981(119861

119895)

ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

119889]119888(119909 119910)

Abstract and Applied Analysis 9

= intΩ

120594119861119895

(119909)

1198981(119861

119895)

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

1198891003816100381610038161003816]1198881003816100381610038161003816

1

(119909)

= intΩ

120594119861119895

(119909)

1198981(119861

119895)

1198891198981(119909) = 1

(70)

Hence we get ⟨119879]119891 119892⟩ = 1 which implies that 119879]119891 =119879] = 1 Finally10038171003817100381710038171003817119891 minus 119891

10038171003817100381710038171003817

⩽10038171003817100381710038171003817119891 minus 119891

1

10038171003817100381710038171003817+10038171003817100381710038171198911 minus 119891

1003817100381710038171003817

=

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

1198981(119861

119895)

minus sum

119895isin119869

120573119895

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+ 120576

⩽ sum

119895isin119869

120573119895(

10038171003817100381710038171003817100381710038171003817100381710038171003817

120594119861119895

1198981(119861

119895)

minus

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

120594119861119895

1198981(119861

119895)

minus

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

) + 120576

= 2sum

119895isin119869

120573119895

1198981(119861

119895 119861

119895)

1198981(119861

119895)

+ 120576

⩽ 2sum

119895isin119869

120573119895

(1205762)1198981(119861

119895)

1198981(119861

119895)

+ 120576 ⩽120576

1 minus 1205762+ 120576 lt 3120576

(71)

We are now ready to present the proof of the main resultin the case of finite regular positive Borel measures

Proof of Proposition 10 Let 1205751= 120575

3

2(5sdot2

4) 120575

2= 120575

12

3(3

2sdot2

14)

and 1205753= (12057610)

2 for some 0 lt 120576 lt 1 Suppose that 119879 isinL(119871

1(119898)) with 119879 = 1 and that there is an 119891

0isin 119878

1198711(119898)

and1198920isin 119878

119871infin(119898)

such that ⟨1198910 119892

0⟩ = 1 and |⟨119879119891

0 119892

0⟩| gt 1minus120575

3

12

6Then there is an isometric isomorphism Ψ from 119871

1(119898) onto

itself such thatΨ(1198910) = |119891

0| and there is a scalar number 120572 in

119878R such that |⟨1198791198910 119892

0⟩| = ⟨120572119879119891

0 119892

0⟩ Then letting 119891

1= Ψ119891

0

1198921= (Ψ

minus1)lowast1198920 and 119879

1= 120572Ψ119879Ψ

minus1 we have

⟨1198781198911 119892

1⟩ = ⟨120572Ψ119879

minus1Ψ119891

0 (Ψ

minus1)lowast

1198920⟩

= ⟨1205721198791198910 119892

0⟩ gt 1 minus

1205753

1

26

⟨1198911 119892

1⟩ = ⟨Ψ119891

0 (Ψ

minus1)lowast

1198920⟩ = 1

(72)

Since 11987911198911 gt 1 minus 120575(120575

3

12

6) by Lemma 13 there exists a

norm-one bounded operator 119879] and a nonnegative simplefunction119891

2isin 119878

1198711(119898)

such that 1198791minus119879] ⩽ 1205751 1198912minus1198911 ⩽ 31205751

and (119889|]|11198891198981)(119909) = 1 for all 119909 isin supp(119891

2) Then

⟨119879]1198912 1198921⟩

= ⟨11987911198911 119892

1⟩ minus ⟨119879

11198911minus 119879

11198912 119892

1⟩ minus ⟨119879

11198912minus 119879]1198912 1198921⟩

⩾ ⟨11987911198911 119892

1⟩ minus10038171003817100381710038171198911 minus 1198912

1003817100381710038171003817 minus10038171003817100381710038171198791 minus 119879]

1003817100381710038171003817

⩾ 1 minus1205753

1

26minus 3120575

1minus 120575

1⩾ 1 minus 5120575

1= 1 minus

1205753

2

16

(73)

Notice also that

⟨1198912 119892

1⟩ = ⟨119891

1 119892

1⟩ minus ⟨119891

1minus 119891

2 119892

1⟩ ⩾ 1 minus

10038171003817100381710038171198911 minus 11989121003817100381710038171003817

⩾ 1 minus 31205751⩾ 1 minus 5120575

1= 1 minus

1205753

2

16

(74)

By Lemma 12 there are a nonnegative simple function 1198913isin

1198781198711(119898)

and a function 1198923isin 119878

119871infin(119898)

such that

1198923(119909) = 120594supp119891

3

(119909) + 1198922(119909) 120594

Ωsupp1198913

(119909)

10038171003817100381710038171198912 minus 11989131003817100381710038171003817 ⩽ 1205752

10038171003817100381710038171198923 minus 11989211003817100381710038171003817 ⩽radic120575

2 ⟨119891

3 119892

3⟩ = 1

(75)

So we have

⟨119879]1198913 1198923⟩ = ⟨119879]1198912 1198921⟩ minus ⟨119879]1198912 minus 119879]1198913 1198921⟩

minus ⟨119879]1198913 1198921 minus 1198923⟩

⩾ 1 minus1205753

2

16minus 2radic120575

2⩾ 1 minus 3radic120575

2= 1 minus

1205756

3

27

(76)

By Lemma 14 there exist 1198914isin 119878

1198711(119898)

and 1198924isin 119878

119871infin(119898)

and anoperator 119879

4such that

⟨1198924 119879

41198914⟩ = 1 =

100381710038171003817100381711987941003817100381710038171003817

10038171003817100381710038171198794 minus 119879]1003817100381710038171003817 ⩽ 21205753

10038171003817100381710038171198914 minus 11989131003817100381710038171003817 ⩽ 31205753

10038171003817100381710038171198924 minus 11989231003817100381710038171003817 ⩽radic120575

3 ⟨119891

4 119892

4⟩ = 1

(77)

So we have

10038171003817100381710038171198794 minus 11987911003817100381710038171003817 ⩽10038171003817100381710038171198794 minus 119879]

1003817100381710038171003817 +1003817100381710038171003817119879] minus 1198791

1003817100381710038171003817

⩽ 1205751+ 2120575

3⩽ 3120575

3

10038171003817100381710038171198911 minus 11989141003817100381710038171003817 ⩽10038171003817100381710038171198911 minus 1198912

1003817100381710038171003817 +10038171003817100381710038171198912 minus 1198913

1003817100381710038171003817 +10038171003817100381710038171198913 minus 1198914

1003817100381710038171003817

⩽ 31205751+ 120575

2+ 3120575

3⩽ 10120575

3

10038171003817100381710038171198921 minus 11989241003817100381710038171003817 ⩽10038171003817100381710038171198921 minus 1198923

1003817100381710038171003817 + 1198923 minus 1198924

⩽ 1205752+ radic120575

3⩽ 2radic120575

3

(78)

10 Abstract and Applied Analysis

Let 119878 = 120572Ψminus11198794Ψ 119891 = Ψminus1

1198914 and 119892 = Ψlowast

1198924 then we have

119879 minus 119878 =10038171003817100381710038171003817119879 minus 120572Ψ

minus11198794Ψ10038171003817100381710038171003817=10038171003817100381710038171003817120572Ψ119879Ψ

minus1minus 119879

4

10038171003817100381710038171003817

=10038171003817100381710038171198791 minus 1198794

1003817100381710038171003817 ⩽ 31205753

100381710038171003817100381710038171198910minus 11989110038171003817100381710038171003817=100381710038171003817100381710038171198910minus Ψ

minus11198914

10038171003817100381710038171003817=10038171003817100381710038171198911 minus 1198914

1003817100381710038171003817 ⩽ 101205753

100381710038171003817100381710038171198920minus 11989110038171003817100381710038171003817=10038171003817100381710038171198920 minus Ψ

lowast1198924

1003817100381710038171003817 =100381710038171003817100381710038171003817(Ψ

minus1)lowast

1198920minus 119892

4

100381710038171003817100381710038171003817

=10038171003817100381710038171198921 minus 1198924

1003817100381710038171003817 ⩽ 2radic120575

3

⟨119891 119892⟩ = ⟨Ψminus11198914 Ψ

lowast1198924⟩ = ⟨119891

4 119892

4⟩ = 1

10038161003816100381610038161003816⟨119878119891 119892⟩

10038161003816100381610038161003816=10038161003816100381610038161003816⟨120572Ψ

minus11198794ΨΨ

minus11198914 Ψ

lowast1198924⟩10038161003816100381610038161003816= |120572| = 1

(79)

This completes the proof

Finally we may give the proof of the main result in fullgenerality

Proof of Theorem 9 Notice that the Kakutani representationtheorem (see [26] for a reference) says that for every 120590-finitemeasure ] the space 119871

1(]) is isometrically isomorphic to

1198711(119898) for some positive Borel regular measure on a compact

Hausdorff space Then by Proposition 10 there is a universalfunction 120576 997891rarr 120578(120576) gt 0 which gives the BPBp-nu for 119871

1(]) for

every 120590-finite measure ]Fix 120576 gt 0 Suppose that 119879

0isin L(119871

1(120583)) with V(119879

0) = 1

and (1198910 119891

lowast

0) isin Π(119871

1(120583)) satisfy

1003816100381610038161003816⟨119891lowast

0 119879

01198910⟩1003816100381610038161003816 gt 1 minus 120578 (120576) (80)

Choose a sequence 119891119899 in 119871

1(120583) such that sup

119899119879

0119891119899 = 1

and let 119866 be the closed linear span of

119879119899

0119891119898 119899 119898 isin N cup 0 (81)

As 119866 is separable there is a dense subset 119892119899 119899 isin N of 119866

and let 119864 = ⋃infin

119899=1supp119892

119899 where supp119892

119899is the support of 119892

119899

Then the measure 120583|119864is 120590-finite Let

119884 = 119891 isin 1198711(120583) supp (119891) sub 119864 (82)

be a closed subspace of 1198711(120583) It is clear that 119871

1(120583) = 119884oplus

1119885

and 119884 is isometrically isomorphic to 1198711(120583|

119864) So 119884 has the

BPBp-nu with 120578(120576)Now write 119878

0= 119879

0|119884 119884 rarr 119884 consider 119910

0= 119891

0isin

119878119884 119910lowast

0= 119891

lowast

0|119884isin 119878

119884lowast and observe that 119910lowast

0(119910

0) = 1 and

|119910lowast

0(119878

01199100)| = |119891

lowast

0(119879

01198910)| gt 1 minus 120578(120576) Hence there exist 119878 isin

L(119884) and (1199100 119910

lowast

0) isin Π(119884) such that

1003816100381610038161003816119910lowast

0(119878119910

0)1003816100381610038161003816 = 1 = V (119878) 1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

10038171003817100381710038171199100 minus 11991001003817100381710038171003817 lt 120576

1003817100381710038171003817119910lowast

0minus 119910

lowast

0

1003817100381710038171003817 lt 120576

(83)

Finally consider the operator 119879 isinL(1198711(120583)) given by

119879 (119910 119911)

= (119878119910 0) + 1198790(0 119911) ((119910 119911) isin 119871

1(120583) equiv 119884oplus

1119885)

(84)

We have 119879 = 1 (and so V(119879) = 1) Indeed1003817100381710038171003817119879 (119910 119911)

1003817100381710038171003817 =1003817100381710038171003817(119878119910 0)

1003817100381710038171003817 +10038171003817100381710038171198790 (0 119911)

1003817100381710038171003817 ⩽10038171003817100381710038171199101003817100381710038171003817 + 119911 =

1003817100381710038171003817(119910 119911)1003817100381710038171003817

(85)

for all (119910 119911) isin 1198711(120583) and 119879(119910

0 0) = (119878119910

0 0) = 119878119910

0 = 1

Let 119909 = (1199100 0) and 119909lowast = (119910lowast

0 119891

0|119885) Then (119909 119909lowast) isin Π(119871

1(120583))

Moreover we have1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119910

lowast

0119878119910

0

1003816100381610038161003816 = 1 = V (119879) 1003817100381710038171003817119909 minus 1198910

1003817100381710038171003817 =1003817100381710038171003817119910 minus 1199100

1003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowast

0minus 119891

lowast

0

1003817100381710038171003817

= max 1003817100381710038171003817119910 minus 119891lowast

0|119884

1003817100381710038171003817 1003817100381710038171003817119891

lowast

0|119885minus 119891

lowast

0|119885

1003817100381710038171003817

=1003817100381710038171003817119910

lowastminus 119910

lowast

0

1003817100381710038171003817 lt 120576

1003817100381710038171003817119879 minus 11987901003817100381710038171003817 = sup119910+119911⩽1

1003817100381710038171003817119879 (119910 119911) minus 1198790 (119910 119911)1003817100381710038171003817

= sup119910⩽1

1003817100381710038171003817119878119910 minus 11987801199101003817100381710038171003817 =1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

(86)

This completes the proof

5 Examples of Spaces Failing theBishop-Phelps-Bollobaacutes Property forNumerical Radius

Our goal here is to prove that the density of numerical radiusattaining operators does not imply the BPBp-nu Actuallywe will show that among separable spaces there is noisomorphic property implying the BPBp-nu other than finite-dimensionality

We need to relate the BPBp-nu to the Bishop-Phelps-Bollobas property for operators which as mentioned in theintroduction was introduced in [15] A pair (119883 119884) of Banachspaces has the Bishop-Phelps-Bollobas property for operators(in short BPBp) if given 120576 gt 0 there exists 120578(120576) gt 0 suchthat given 119879 isin L(119883 119884) with 119879 = 1 and 119909 isin 119878

119883such that

119879119909 gt 1 minus 120578(120576) then there exist 119911 isin 119878119883and 119878 isin L(119883 119884)

satisfying

119878 = 119878119911 = 1 119909 minus 119911 lt 120576

119879 minus 119878 lt 120576

(87)

We refer the reader to [15 19 25] and references therein formore information and background Among the interestingresults on the BPBp we emphasize that a pair (119883 119884) when119883is finite-dimensional does not necessarily have the BPBp Forinstance if119884 is a strictly convex space which is not uniformlyconvex then the pair (ℓ(2)

1 119884) fails to have the BPBp (this is

contained in [15] see [19 Section 3])

Abstract and Applied Analysis 11

The next result relates the BPBp-nu to the BPBp foroperators in a particular case We will deduce our examplefrom it

Theorem 15 If 1198711(120583)oplus

1119883 has the BPBp-119899119906 then the pair

(1198711(120583) 119883) has the BPBp for operators

Before proving this proposition we will use it to get themain examples of this section The first example shows thatthe density of numerical radius attaining operators does notimply the BPBp-nu

Example 16 There is a reflexive space (and so numerical radiusattaining operators on it are dense) which fails to have theBPBp-nu Indeed let 119884 be a reflexive separable space whichis not superreflexive and we may suppose that 119884 is strictlyconvex Observe that 119884 cannot be uniformly convex since itis not superreflexive Now 119883 = ℓ(2)

1oplus1119884 is reflexive but the

pair (ℓ(2)1 119884) fails the BPBp since 119884 is strictly convex but not

uniformly convex [19 Corollary 33] Therefore Theorem 15gives us that119883 does not have the BPBp-nu

The example above can be extended to get the result thatevery infinite-dimensional separable Banach space can berenormed to fail the BPBp-nu This follows from the factthat every infinite-dimensional separable Banach space canbe renormed to be strictly convex but not uniformly convex(this result can be proved ldquoby handrdquo an alternative categoricalargument for it can be found in [27] and references therein)With a little more of effort we may get the main result of thesection

Theorem 17 Every infinite-dimensional separable Banachspace can be renormed to fail the weak-BPBp-nu (and so inparticular to fail the BPBp-nu)

We need the following result which is surely well knownAs we have not found a reference we include a nice and easyproof kindly given to us by Vladimir Kadets We recall thatgiven a Banach space 119884 the set of all equivalent norms on119884 can be viewed as a metric space using the Banach-Mazurdistance

Lemma 18 Let119884 be an infinite-dimensional separable Banachspace Then the set of equivalent norms on 119884 which are strictlyconvex and are not (locally) uniformly convex is dense in the setof all equivalent norms on119884 (with respect to the Banach-Mazurdistance)

Proof Fix 119890 isin 119878119884and 119890lowast

1isin 119878

119884lowast such that 119890lowast

1(119890) = 1 For a fixed

120576 isin (0 12) denote

119902 (119910) = max (1 minus 120576) 10038171003817100381710038171199101003817100381710038171003817 1003816100381610038161003816119890lowast

1(119910)1003816100381610038161003816 (119910 isin 119884) (88)

Evidently (1 minus 120576)119910 ⩽ 119902(119910) ⩽ 119910 for every 119910 isin 119884 Fix asequence 119890lowast

119896 119896 ⩾ 2 of norm-one functionals separating

the points of 119884 and denote

119901 (119910) = radic

infin

sum

119896=1

1

2119896

1003816100381610038161003816119890lowast

119896(119910)1003816100381610038161003816

2

(119910 isin 119884) (89)

Then 119901 is a strictly convex norm on 119884 119901(119890) ⩾ 1radic2 and119901(119910) ⩽ 119910 for all 119910 isin 119883 Finally write

100381710038171003817100381711991010038171003817100381710038171= (1 minus 120576) 119902 (119910) + 120576

119901 (119910)

119901 (119890)(119910 isin 119884) (90)

Then sdot 1is a strictly convex norm on 119884 and

(1 minus 120576)2 10038171003817100381710038171199101003817100381710038171003817 ⩽100381710038171003817100381711991010038171003817100381710038171⩽ (1 + 120576)

10038171003817100381710038171199101003817100381710038171003817 (119910 isin 119884) (91)

Wewill finish the proof by showing that sdot 1is not uniformly

convex (actually it is not locally uniformly convex) Indeedfor each 119899 isin N we select 119910

119899isin ⋂

119899

119896=1ker 119890lowast

119896with 119910

119899 = 1 and

consider 119890119899= 119890 + (1205764)119910

119899 Then 119902(119890) = 1 119902(119890

119899) = 1 and

119902(119890+119890119899) = 2 At the same time 119901(119910

119899) rarr 0 so 119901(119890

119899) rarr 119901(119890)

and 119901(119890 + 119890119899) rarr 2119901(119890) Consequently

1198901 = 1100381710038171003817100381711989011989910038171003817100381710038171997888rarr 1

1003817100381710038171003817119890 + 11989011989910038171003817100381710038171997888rarr 2 (92)

but 119890 minus 1198901198991= (1205764)119910

1198991⩾ (1 minus 120576)

2(1205764) which means the

absence of local uniform convexity at 119890

Proof of Theorem 17 Let 119883 be an infinite-dimensional sepa-rable Banach space Take a closed subspace 119884 of 119883 of codi-mension two By [28 Proposition 2] the map carrying everyequivalent norm on 119884 to its numerical index is continuousand so the set of values of the numerical index of 119884 upto reforming is a nontrivial interval [28 Theorem 9] ThenLemma 18 allows us to find an equivalent norm | sdot | on 119884 insuch a way that (119884 | sdot |) is strictly convex and is not uniformlyconvex and 119899(119884 | sdot |) gt 0 Now the space 119883 = ℓ(2)

1oplus1(119884 | sdot |)

is an equivalent renorming of 119883 which does not have theBPBp-119899119906 (indeed otherwise the pair (ℓ(2)

1 (119884 | sdot |)) would

have the BPBp for the operator norm and so (119884 | sdot |) wouldbe uniformly convex by [19 Corollary 33] a contradiction)Moreover as

119899 (119883) = min 119899 (ℓ(2)1) 119899 (119884 |sdot|) gt 0 (93)

(see [20 Proposition 2] for instance) 119883 also fails the weak-BPBp-nu by Proposition 6

To finish the section with the promised proof ofTheorem 15 we first see the following stability result

Lemma 19 Let119883 = [⨁infin

119896=1119883

119896]1198880

or [⨁infin

119896=1119883

119896]ℓ1

If119883 has theBishop-Phelps-Bollobas property for numerical radius with afunction 120578 then each Banach space 119883

119894has the Bishop-Phelps-

Bollobas property for numerical radius with 120578nu(119883119894) ⩾ 120578 That

is inf119894120578nu(119883119894

)(120576) ⩾ 120578nu(119883)(120576) for all 0 lt 120576 lt 1

Proof Let 119875119894 119883 rarr 119883

119894and 1198751015840

119894 119883

lowastrarr 119883

lowast

119894be the natural

projections and let 119876119894 119883

119894rarr 119883 and 1198761015840

119894 119883

lowast

119894rarr 119883

lowast be thenatural embeddings

Assume that an operator 119879119894 119883

119894rarr 119883

119894and a pair

(119909119894 119909

lowast

119894) isin Π(119883

119894) satisfy that

V (119879119894) = 1

1003816100381610038161003816119909lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (94)

We define an operator 119879 119883 rarr 119883 and (119909 119909lowast) isin Π(119883) by

119879 = 119876119894∘ 119879

119894∘ 119875

119894 (119909 119909

lowast) = (119876

119894119909119894 119876

1015840

119894119909lowast

119894) (95)

12 Abstract and Applied Analysis

then clearly we see that1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119909

lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (96)

From the assumption there exist 119878 119883 rarr 119883 and a pair(119910 119910

lowast) isin Π(119883) such that

1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 = 1 = V (119878) 119878 minus 119879 lt 120576

1003817100381710038171003817119910lowastminus 119909

lowast1003817100381710038171003817 lt 1205761003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576

(97)

Since this clearly shows that1003817100381710038171003817119875119894 ∘ 119878 ∘ 119876119894

minus 119879119894

1003817100381710038171003817 lt 120576100381710038171003817100381710038171198751015840

119894119910lowastminus 119909

lowast

119894

10038171003817100381710038171003817lt 120576

1003817100381710038171003817119875119894119910 minus 1199091198941003817100381710038171003817 lt 120576

(98)

we only need to show that |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = 1

We first show the case of 1198880sum Since

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817=10038171003817100381710038171003817119875119895119910 minus 119875

11989511990910038171003817100381710038171003817⩽1003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576 (99)

for every 119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽100381710038171003817100381710038171198751015840

119894119910lowast10038171003817100381710038171003817+ 120576 sum

119895isinN119895 = 119894

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817⩽1003817100381710038171003817119910

lowast1003817100381710038171003817 = 1

(100)

This shows that 1198751015840119894119910lowast = 1 and 1198751015840

119895119910lowast= 0 for every 119895 = 119894

So 119910lowast = 1198761015840

1198941198751015840

119894119910lowast and 1198751015840

119894119910lowast(119875

119894119910) = 1 This and the fact that

119910 minus 119876119894119875119894119910 lt 120576 imply that

(119876119894119875119894119910 + (

1

120576) (119910 minus 119876

119894119875119894119910) 119876

1015840

1198941198751015840

119894119910lowast) isin Π (119883) (101)

So we get that (1198761015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910 + (1120576)(119910 minus 119876

119894119875119894119910))V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =10038161003816100381610038161003816(119876

1015840

1198941198751015840

119894119910lowast) 11987811991010038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910 +1

120576(119910 minus 119876

119894119875119894119910))

1003816100381610038161003816100381610038161003816⩽ 1

(102)

and so we get |1198751015840119894119910lowast(119875

119894∘119878 ∘119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

We next show the case of ℓ1sum The proof is almost

the same as that of the 1198880case However for the sake of

completeness we provide it hereSince 1198751015840

119895119910lowast = 119875

1015840

119895119910lowastminus 119875

1015840

119895119909lowast ⩽ 119910

lowastminus 119909

lowast lt 120576 for every

119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽10038171003817100381710038171198751198941199101003817100381710038171003817 + 120576 sum

119895isinN119895 = 119894

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817⩽10038171003817100381710038171199101003817100381710038171003817 = 1

(103)

which shows 119875119894119910 = 1 and 119875

119895119910 = 0 for every 119895 = 119894 Since this

implies (119876119894119875119894119910119876

1015840

1198941198751015840

119894119910lowast+(1120576)(119910

lowastminus119876

1015840

1198941198751015840

119894119910lowast)) isin Π(119883) we get

that |(1198761015840

1198941198751015840

119894119910lowast+ (1120576)(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast))119878(119876

119894119875119894119910)| ⩽ V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =1003816100381610038161003816119910

lowast119878 (119876

119894119875119894119910)1003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast+1

120576(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast)) 119878 (119876

119894119875119894119910)

1003816100381610038161003816100381610038161003816⩽ 1

(104)

and so |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

Proof of Theorem 15 Note that 120578nu(1198711(120583)oplus1119883)(120576) rarr 0 as120576 rarr 0 Fix 0 lt 120576

0lt 1 and choose 0 lt 120576 lt 1 such that

6120576 + 120578nu(1198711(120583)oplus1119883)(120576) lt 1205760 Let 120578(1205760) = 120578nu(1198711(120583)oplus1119883)(120576)Suppose that 119879

0isin L(119871

1(120583) 119883) with 119879

0 = 1 and 119891

0isin

1198781198711(120583)

satisfy

1003817100381710038171003817119879011989101003817100381710038171003817 gt 1 minus 120578 (1205760) (105)

For any measurable subset 119861 let

1198711(120583|

119861) = 119891|

119861 119891 isin 119871

1(120583) (106)

with the norm 119891|119861 = 119891120594

1198611 Then it is easy to see

that 1198711(120583|

119861) is isometrically isomorphic to a complemented

subspace of 1198711(120583) Let 119875

119861 119871

1(120583) rarr 119871

1(120583|

119861) be the

restriction defined by 119875119861(119891) = 119891|

119861for all 119891 isin 119871

1(120583) and

let 119869119861 119871

1(120583|

119861) rarr 119871

1(120583) be the extension defined by

119869119861(119891)(120596) = 119891(120596) if 120596 isin 119861 and 119869

119861(119891)(120596) = 0 otherwise

It is clear that 119875119861119869119861= Id

1198711(120583|119861)and 119869

119861119875119861(119891) = 119891120594

119861for all

119891 isin 1198711(120583) Notice also that 119871

1(120583) is isometrically isomorphic

to 1198711(120583|

119861)oplus

11198711(120583|

119861119888)

Let 119860 = supp1198910and 119892

0= 119875

1198601198910 Then

1003817100381710038171003817119879011986911986011989201003817100381710038171003817 =100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) gt 0 (107)

and define the operator 119879119860 119871

1(120583|

119860) rarr 119883 by 119879

119860119891 =

1198790119869119860119891119879

0119869119860 for every 119891 isin 119871

1(120583|

119860) Then

100381710038171003817100381711987911986011989201003817100381710038171003817 ⩾100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) (108)

Since 120583|119860is 120590-finite 119871

1(120583|

119860)lowast= 119871

infin(120583|

119860) Let 119892lowast

0isin 119878

119871infin(120583|119860)

be a function such that ⟨119892lowast0 119892

0⟩ = 1 and choose 119909lowast

0isin 119878

119883lowast

such that 119909lowast0(119879

1198601198920) = 119879

1198601198920 Define the operator 119878

0isin

L(1198711(120583|

119860)oplus

1119883) by

1198780(119891 119909) = (0 119879

119860119891) ((119891 119909) isin 119871

1(120583|

119860) oplus

1119883) (109)

and observe that 1198780 = V(119878

0) = 1 Indeed

100381710038171003817100381711987801003817100381710038171003817 ⩽ 1 =

10038171003817100381710038171198791198601003817100381710038171003817

= sup 1003816100381610038161003816119909lowast1198791198601198911003816100381610038161003816 119909

lowastisin 119878

119883lowast 119891 isin 119878

1198711(120583|119860)

= sup 1003816100381610038161003816(119891lowast 119909

lowast) 119878

0(119891 119909)

1003816100381610038161003816

((119891lowast 119909

lowast) (119891 119909)) isin Π (119871

1(120583|

119860) oplus

1119883)

= V (1198780) ⩽100381710038171003817100381711987801003817100381710038171003817

(110)

Abstract and Applied Analysis 13

It is immediate that

(119892lowast

0 119909

lowast

0) 119878

0(119892

0 0) = 119909

lowast

0(119879

1198601198920) =10038171003817100381710038171198791198601198920

1003817100381710038171003817 gt 1 minus 120578 (1205760)

(111)

By Lemma 19 1198711(120583|

119860)oplus

1119883 has the BPBp-nu with the func-

tion 120578 Therefore there exist 1198781isin L(119871

1(120583|

119860)oplus

1119883) (119892

1 119909

1) isin

1198781198711(120583|119860)oplus1119883 and (119892lowast

1 119909

lowast

1) isin 119878

119871infin(120583|119860)oplusinfin119883lowast such that

1003817100381710038171003817(1198921 1199091) minus (1198920 0)1003817100381710038171003817 lt 120576

1003817100381710038171003817(119892lowast

1 119909

lowast

1) minus (119891

lowast

0 119909

lowast

0)1003817100381710038171003817 lt 120576

10038171003817100381710038171198781 minus 11987801003817100381710038171003817 lt 120576 ⟨(119892

lowast

1 119909

lowast

1) (119892

1 119909

1)⟩ = 1

1003816100381610038161003816(119892lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 = V (119878

1) = 1

(112)

Claim 1 We claim that 1199091= 0

Otherwise

1 = Re ⟨(119892lowast1 119909

lowast

1) (119892

1 119909

1)⟩

=100381710038171003817100381711989211003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

+100381710038171003817100381711990911003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

⟩ ⩽ 1

(113)

We deduce that

((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

(119892lowast

1 119909

lowast

1)) (

(0 1199091)

100381710038171003817100381711990911003817100381710038171003817

(119892lowast

1 119909

lowast

1))

isin Π (1198711(120583|

119860) oplus

1119883)

(114)

Since 1198781((0 119909

1)119909

1) = (119878

1minus 119878

0)((0 119909

1)119909

1) lt 120576 we

get that

1 =1003816100381610038161003816⟨(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)⟩1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

100381710038171003817100381711989211003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

)⟩

+100381710038171003817100381711990911003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

)⟩

100381610038161003816100381610038161003816100381610038161003816

⩽100381710038171003817100381711989211003817100381710038171003817 V (1198781) + 120576

100381710038171003817100381711990911003817100381710038171003817 lt100381710038171003817100381711989211003817100381710038171003817 +100381710038171003817100381711990911003817100381710038171003817 = 1

(115)

a contradiction This proves the claimWe define the operator 119878

2 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883

by 1198782(119891 119909) = 119878

1(119891 0) for every 119891 isin 119871

1(120583|

119860) and for every

119909 isin 119883 Then we have

V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 = 1

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ 120576 (116)

Indeed from Claim 1 we have

V (1198781) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 ⩽ V (119878

2)

(117)

On the other hand we see that1003816100381610038161003816(119891

lowast 119909

lowast) 119878

2(119891 119909)

1003816100381610038161003816

=1003816100381610038161003816(119891

lowast 119909

lowast) 119878

1(119891 0)

1003816100381610038161003816 ⩽10038171003817100381710038171198911003817100381710038171003817 V (1198781) ⩽ V (119878

1)

(118)

for every ((119891lowast 119909lowast) (119891 119909)) isin Π(1198711(120583|

119860)oplus

1119883) Therefore

V(1198782) ⩽ V(119878

1) Also

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ sup

119909isin119878119883

10038171003817100381710038171198781 (0 119909)1003817100381710038171003817

= sup119909isin119878119883

10038171003817100381710038171198781 (0 119909) minus 1198780 (0 119909)1003817100381710038171003817 ⩽ 120576

(119)

Claim 2 There exists 1198783 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883 such

that 1198783(119892

1 0) = 119878

3 = 1 119878

3(0 119909) = 0 119878

3(119891 119909) isin 0oplus

1119883

for every (119891 119909) isin 1198711(120583|

119860)oplus

1119883 and 119878

3minus 119878

2 lt 4120576

Indeed write 1198781= (119863

1 119863

2) where 119863

1 119871

1(120583|

119860)oplus

1119883 rarr

1198711(120583|

119860) and119863

2 119871

1(120583|

119860)oplus

1119883 rarr 119883 We have that

sup 1003816100381610038161003816119892lowast119863

1(119892

1 0) + 119909

lowast119863

2(119892

1 0)1003816100381610038161003816

119909lowastisin 119878

119883lowast ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860) +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⩽ V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119892

lowast

1119863

1(119892

1 0) + 119909

lowast

1119863

2(119892

1 0)1003816100381610038161003816

(120)

This implies that

1003816100381610038161003816119909lowast

1119863

2(119892

1 0)1003816100381610038161003816 =10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

1003816100381610038161003816119892lowast

1119863

1(119892

1 0)1003816100381610038161003816

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119871

infin(120583|

119860)

(121)

Therefore |119892lowast1| equals 1 on the support of 119863

1(119892

1 0) As

|⟨119892lowast

1 119892

1⟩| = 1 we also have that |119892lowast

1| equals 1 on the support

of 1198921 Changing the values of 119892lowast

1by the ones of 119891lowast

0on 119860

(supp(1198631(119892

1 0)) cup supp(g

1)) we may and do suppose that

|119892lowast

1| = 1 on the whole 119860We also have 119863

2(119892

1 0) gt 0 Indeed

10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)1003817100381710038171003817

⩽10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198921 0)

1003817100381710038171003817 +10038171003817100381710038171198780 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817

lt 2120576 + 120576 = 3120576

(122)

So we have

10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

⩽10038171003817100381710038171198631(119892

1 0)1003817100381710038171003817 +10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

14 Abstract and Applied Analysis

=1003817100381710038171003817(1198631

(1198921 0) 119863

2(119892

1 0)) minus (0 119879

1198601198920)1003817100381710038171003817

=10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817 lt 3120576

(123)

and 1198632(119892

1 0) gt 119879

1198601198920 minus 3120576 ⩾ 1 minus 120578(120576

0) minus 3120576 gt 0

Finally define the operator 1198783by

1198783(119891 119909) = (0119863

2(119891 0) + 119892

lowast

1(119863

1(119891 0))

1198632(119892

1 0)

119909lowast

1119863

2(119892

1 0))

(124)

for (119891 119909) isin 1198711(120583|

119860)oplus

1119883 It is clear that

100381710038171003817100381711987831003817100381710038171003817 ⩽ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) (125)

Notice also that

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817

⩽10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 +10038171003817100381710038171198632(119891 0) minus 119879

1198601198911003817100381710038171003817

=1003817100381710038171003817(1198631

(119891 0) 1198632(119891 0)) minus (0 119879

119860119891)1003817100381710038171003817

=10038171003817100381710038171198782 (119891 119909) minus 1198780 (119891 119909)

1003817100381710038171003817

(126)

for all (119891 119909) isin 1198711(120583|

119860)oplus

1119883 Hence we have

10038171003817100381710038171198783 minus 11987821003817100381710038171003817

= 2 sup119891isin1198781198711(120583|119860)

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 ⩽ 210038171003817100381710038171198782 minus 1198780

1003817100381710038171003817 lt 4120576(127)

On the other hand let 119866 1198711(120583|

119860) rarr 119871

1(120583|

119860) be defined by

119866(119891) = 119892lowast

1119891 for every 119891 isin 119871

1(120583|

119860) Then we have

V (1198782)

= sup 1003816100381610038161003816119911lowast11987821199111003816100381610038161003816 (119911 119911

lowast) isin Π (119871

1(120583|

119860) oplus

1119883)

⩾ sup10038161003816100381610038161003816100381610038161003816

119909lowast119863

2(119866(

1

120583 (119862)120594119862) 0)

+119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119909lowastisin 119878

119883lowast 119862 isin Σ

119860 120583 (119862) gt 0

= sup10038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119862)120594119862) 0)

10038171003817100381710038171003817100381710038171003817

+

10038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119862 isin Σ119860 120583 (119862) gt 0

(128)

where Σ119860is the family of measurable subsets of 119860

Hence for any simple function 119904 = sum119899

119894=1((120572

119894120583(119860

119894))120594

119860119894

) isin

1198781198711(120583|119860) where 119860

119894119894is a family of disjoint measurable subsets

with strictly positive measure we have

V (1198782)

119899

sum

119894=1

10038161003816100381610038161205721198941003816100381610038161003816 (

100381710038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381710038171003817100381710038171003817100381710038171003817

+

100381610038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381610038161003816100381610038161003816100381610038161003816

)

⩾10038171003817100381710038171198632(119866 (119904) 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119866 (119904) 0)

1003816100381610038161003816

(129)

Since |119892lowast1| = 1119866 is an isometric isomorphism so for each119891 isin

1198781198711(120583|119860)there exists a sequence of norm-one simple functions

(119904119896) such that 119866(119904

119896) converges to 119891 Therefore

V (1198782) ⩾ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) ⩾100381710038171003817100381711987831003817100381710038171003817

(130)

On the other hand we see that100381710038171003817100381711987831003817100381710038171003817 ⩾1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

3(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119909

lowast

1119863

2(119892

1 0) + 119892

lowast

1119863

1(119892

1 0)1003816100381610038161003816 = V (119878

2) = 1

(131)

Therefore 1 = 1198783 = 119878

3(119892

1 0) which proves Claim 2

Finally set 1198783= (0 ) for a suitable 119871

1(120583|

119860)oplus

1119883 rarr

119883 and define the operator 1198791 119871

1(120583) rarr 119883 by

1198791(119891) = 119879

0(119891120594

119860119888) + (119875

119860119891 0) (132)

for every 119891 isin 1198711(120583) Then we have

10038171003817100381710038171198791 (119891)1003817100381710038171003817 ⩽100381710038171003817100381711987901003817100381710038171003817

10038171003817100381710038171198911205941198601198881003817100381710038171003817 +1003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171198911205941198601003817100381710038171003817 =10038171003817100381710038171198911003817100381710038171003817 (133)

for every 119891 isin 1198711(120583) so 119879

1 ⩽ 1 Also

10038171003817100381710038171198791 (1198691198601198921)1003817100381710038171003817 =10038171003817100381710038171198783 (1198921 0)

1003817100381710038171003817 =100381710038171003817100381711987831003817100381710038171003817 = 1 (134)

so 1198791attains its norm at 119869

1198601198921isin 119871

1(120583) and

10038171003817100381710038171198691198601198921 minus 11989101003817100381710038171003817 =10038171003817100381710038171198921 minus 1198920

1003817100381710038171003817 lt 120576 (135)

We also have that for any 119891 isin 1198781198711(120583)

10038171003817100381710038171198790 (119891) minus 1198791 (119891)1003817100381710038171003817 =100381710038171003817100381710038171198790(119891120594

119860) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790 (119869119860119875119860119891) minus 119879119860 (119875119860119891)

1003817100381710038171003817

+10038171003817100381710038171003817119879119860(119875

119860119891) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790119869119860 minus 119879119860

1003817100381710038171003817 +10038171003817100381710038171198780 minus 1198783

1003817100381710038171003817

lt 120578 (1205760) + 6120576

(136)

Hence 1198790minus 119879

1 ⩽ 120578(120576

0) + 6120576 lt 120576

0

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Abstract and Applied Analysis 15

Acknowledgments

The authors thank Gilles Godefroy and Rafael Paya forfruitful conversations about the content of this paper andVladimir Kadets for providing Lemma 18 They also appre-ciate anonymous referees for careful reading and fruitfulsuggestions about revision The second author was sup-ported by Basic Science Research Program through theNational Research Foundation of Korea (NRF) fundedby the Ministry of Education Science and Technology(NRF-2012R1A1A1006869) The third author was partiallysupported by Spanish MICINN and FEDER project noMTM2012-31755 and by Junta de Andalucıa and FEDERGrants FQM-185 and P09-FQM-4911

References

[1] F F Bonsall and J Duncan Numerical Ranges of Operators onNormed Spaces and of Elements of Normed Algebras vol 2 ofLondon Mathematical Society Lecture Note Series 2 CambridgeUniversity Press London UK 1971

[2] F F Bonsall and J Duncan Numerical Ranges II CambridgeUniversity Press London UK 1973 London MathematicalSociety Lecture Notes Series 10

[3] B Sims On numerical range and its applications to Banachalgebras [PhD thesis] University of Newcastle New SouthWales Australia 1972

[4] I D Berg and B Sims ldquoDenseness of operators which attaintheir numerical radiusrdquoAustralian Mathematical Society A vol36 no 1 pp 130ndash133 1984

[5] C S Cardassi ldquoDensity of numerical radius attaining operatorson some reflexive spacesrdquo Bulletin of the Australian Mathemat-ical Society vol 31 no 1 pp 1ndash3 1985

[6] C S Cardassi ldquoNumerical radius attaining operatorsrdquo inBanach Spaces (Columbia Mo 1984) vol 1166 of Lecture Notesin Mathematics pp 11ndash14 Springer Berlin Germany 1985

[7] C S Cardassi ldquoNumerical radius-attaining operators on119862(119870)rdquoProceedings of the AmericanMathematical Society vol 95 no 4pp 537ndash543 1985

[8] MDAcostaOperadores que alcanzan su radio numerico [PhDthesis] Universidad de Granada Granada Spain 1990

[9] M D Acosta and R Paya ldquoNumerical radius attaining oper-ators and the Radon-Nikodym propertyrdquo The Bulletin of theLondon Mathematical Society vol 25 no 1 pp 67ndash73 1993

[10] R Paya ldquoA counterexample on numerical radius attainingoperatorsrdquo Israel Journal of Mathematics vol 79 no 1 pp 83ndash101 1992

[11] Y S Choi and S G Kim ldquoNorm or numerical radius attainingmultilinear mappings and polynomialsrdquo Journal of the LondonMathematical Society vol 54 no 1 pp 135ndash147 1996

[12] M D Acosta J Becerra Guerrero and M Ruiz GalanldquoNumerical-radius-attaining polynomialsrdquo The Quarterly Jour-nal of Mathematics vol 54 no 1 pp 1ndash10 2003

[13] M D Acosta and S G Kim ldquoDenseness of holomorphicfunctions attaining their numerical radiirdquo Israel Journal ofMathematics vol 161 pp 373ndash386 2007

[14] S G Kim and H J Lee ldquoNumerical peak holomorphicfunctions on Banach spacesrdquo Journal of Mathematical Analysisand Applications vol 364 no 2 pp 437ndash452 2010

[15] M D Acosta R M Aron D Garcıa and M MaestreldquoThe Bishop-Phelps-Bollobas theorem for operatorsrdquo Journal ofFunctional Analysis vol 254 no 11 pp 2780ndash2799 2008

[16] A J Guirao and O Kozhushkina ldquoThe Bishop-Phelps-Bollobasproperty for numerical radius in ℓ

1(119862)rdquo Studia Mathematica

vol 218 no 1 pp 41ndash54 2013[17] J Falco ldquoThe Bishop-Phelps-Bollobas property for numerical

radius on 1198711rdquo Journal of Mathematical Analysis and Applica-

tions vol 414 no 1 pp 125ndash133 2014[18] A Aviles A J Guirao and J Rodriguez ldquoOn the Bishop-

Phelps-Bollobas property for numerical radius in119862(119870)-spacesrdquoPreprint

[19] RM Aron Y S Choi S K Kim H J Lee andMMartin ldquoTheBishop-Phelps-Bollobas version of Lindenstrauss properties Aand Brdquo Preprint

[20] V Kadets M Martın and R Paya ldquoRecent progress and openquestions on the numerical index of Banach spacesrdquo Revista dela Real Academia de Ciencias Exactas Fısicas y Naturales A vol100 no 1-2 pp 155ndash182 2006

[21] M Martın J Merı and M Popov ldquoOn the numerical index ofreal 119871

119901(120583)-spacesrdquo Israel Journal of Mathematics vol 184 pp

183ndash192 2011[22] A Iwanik ldquoNorm attaining operators on Lebesgue spacesrdquo

Pacific Journal of Mathematics vol 83 no 2 pp 381ndash386 1979[23] HH SchaeferBanach Lattices and Positive Operators Springer

New York NY USA 1974[24] J Duncan C M McGregor J D Pryce and A J White ldquoThe

numerical index of a normed spacerdquo Journal of the LondonMathematical Society Second Series vol 2 pp 481ndash488 1970

[25] Y S Choi S K Kim H J Lee and M Martin ldquoThe Bishop-Phelps-Bollobas theorem for operators on 119871

1(120583)rdquo Preprint

[26] H E Lacey The Isometric Theory of Classical Banach SpacesSpringer New York NY USA 1974

[27] G Godefroy ldquoUniversal spaces for strictly convex Banachspacesrdquo Revista de la Real Academia de Ciencias Exactas Fısicasy Naturales A vol 100 no 1-2 pp 137ndash146 2006

[28] C Finet M Martın and R Paya ldquoNumerical index andrenormingrdquo Proceedings of the American Mathematical Societyvol 131 no 3 pp 871ndash877 2003

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Stochastic AnalysisInternational Journal of

Page 2: Research Article On the Bishop-Phelps-Bollobás Property ...downloads.hindawi.com/journals/aaa/2014/479208.pdf · Research Article On the Bishop-Phelps-Bollobás Property for Numerical

2 Abstract and Applied Analysis

It is shown in [16] that the real or complex spaces 1198880and

ℓ1have the BPBp-nu This result has been extended to the

real space 1198711(R) by Falco [17] Aviles et al [18] give sufficient

conditions on a compact space 119870 for the real space 119862(119870) tohave the BPBp-nu which in particular include all metrizablecompact spaces

The content of this paper is the following First weintroduce in Section 2 a modulus of the BPBp-nu analogousto the one introduced in [19] for the Bishop-Phelps-Bollobasproperty for the operator norm and we will use it as a toolin the rest of the paper As easy applications we prove thatfinite-dimensional spaces always have the BPBp-nu and thata reflexive space has the BPBp-nu if and only if its dual doesNext Section 3 is devoted to prove that Banach spaces whichare both uniformly convex and uniformly smooth satisfy aweaker version of the BPBp-nu and to discuss such weakerversion In particular it is shown that 119871

119901(120583) spaces have the

BPBp-nu for every measure 120583 when 1 lt 119901 lt infin 119901 = 2We show in Section 4 that given any measure 120583 the real orcomplex space 119871

1(120583) has the BPBp-nu Finally we prove in

Section 5 that every separable infinite-dimensional Banachspace can be equivalently renormed to fail the BPBp-nu(actually to fail the weaker version) In particular this showsthat reflexivity (or even superreflexivity) is not enough for theBPBp-nu while the Radon-Nikodym property was knownto be sufficient for the density of numerical radius attainingoperators

Let us introduce some notations for later use The 119899-dimensional space with the ℓ

1norm is denoted by ℓ(119899)

1

Given a family 119883119896infin

119896=1of Banach spaces [⨁infin

119896=1119883

119896]1198880

(resp[⨁

infin

119896=1119883

119896]ℓ1

) is the Banach space consisting of all sequences(119909

119896)infin

119896=1such that each 119909

119896is in 119883

119896and lim

119896rarrinfin119909

119896 = 0

(resp suminfin

119896=1119909

119896 lt infin) equipped with the norm (119909

119896)infin

119896=1 =

sup119896119909

119896 (resp (119909

119896)infin

119896=1 = sum

infin

119896=1119909

119896)

2 Modulus of the Bishop-Phelps-Bollobaacutes forNumerical Radius

Analogously to what is done in [19] for the BPBp for theoperator norm we introduce here a modulus to quan-tify the Bishop-Phelps-Bollobas property for numericalradius

Notation 1 Let119883 be a Banach space Consider the set

Πnu (119883) = (119909 119909lowast 119879) (119909 119909

lowast) isin Π (119883)

119879 isinL (119883) V (119879) = 1 = 1003816100381610038161003816119909lowast1198791199091003816100381610038161003816

(3)

which is closed in 119878119883times 119878

119883lowast times L(119883) with respect to the

following metric

dist ((119909 119909lowast 119879) (119910 119910lowast 119878))

= max 1003817100381710038171003817119909 minus 1199101003817100381710038171003817 1003817100381710038171003817119909

lowastminus 119910

lowast1003817100381710038171003817 119879 minus 119878

(4)

The modulus of the Bishop-Phelps-Bollobas property fornumerical radius is the function defined by

120578nu (119883) (120576)

= inf 1 minus 1003816100381610038161003816119909lowast1198791199091003816100381610038161003816 (119909 119909

lowast) isin Π (119883) 119879 isinL (119883)

V (119879) = 1 dist ((119909 119909lowast 119879) Πnu (119883)) ⩾ 120576

(5)

for every 120576 isin (0 1) Equivalently 120578nu(119883)(120576) is the supremumof those scalars 120578 gt 0 such that whenever 119879 isin L(119883) and(119909 119909

lowast) isin Π(119883) satisfy V(119879) = 1 and |119909lowast119879119909| gt 1 minus 120578 there

exist 119878 isinL(119883) and (119910 119910lowast) isin Π(119883) such that

V (119878) = 1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 = 1 119879 minus 119878 lt 120576

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowastminus 119910

lowast1003817100381710038171003817 lt 120576

(6)

It is immediate that a Banach space119883 has the BPBp-nu ifand only if 120578nu(120576) gt 0 for every 0 lt 120576 lt 1 By construction ifa function 120576 997891rarr 120578(120576) is valid in the definition of the BPBp-nuthen 120578nu(120576) ⩾ 120578(120576)

An immediate consequence of the compactness of theunit ball of a finite-dimensional space is the following resultIt was previously known to A Guirao (private communica-tion)

Proposition 2 Let 119883 be a finite-dimensional Banach spaceThen119883 has the Bishop-Phelps-Bollobas property for numericalradius

Proof Let 119870 = 119878 isin L(119883) V(119878) = 0 Then 119870 is anorm-closed subspace of L(119883) Hence L(119883)119870 is a finite-dimensional space with two norms

V ([119879]) = inf V (119879 minus 119878) 119878 isin 119870 = V (119879)

[119879] = inf 119879 minus 119878 119878 isin 119870 (7)

where [119879] is the class of 119879 in the quotient space L(119883)119870Hence there is a constant 0 lt 119888 ⩽ 1 such that

119888 [119879] ⩽ V (119879) ⩽ [119879] (8)

Suppose that119883 does not have the BPBp-nuThen there is0 lt 120576 lt 1 such that 120578nu(119883)(120576) = 0That is there are sequences(119909

119899 119909

lowast

119899) isin Π(119883) and (119879

119899) isinL(119883) with V(119879

119899) = 1 such that

dist ((119909119899 119909

lowast

119899 119879

119899) Πnu (119883)) ⩾ 120576 (119899 isin N)

lim119899

1003816100381610038161003816119909lowast

119899119879119899119909119899

1003816100381610038161003816 = 1

(9)

By compactness wemay assume that lim119899[119879

119899]minus[119879

0] = 0 for

some1198790isinL(119883) and V(119879

0) = 1 Hence there exists a sequence

119878119899119899in 119870 such that lim

119899119879

119899minus (119879

0+ 119878

119899) = 0 Observe that

V(1198790+ 119878

119899) = V(119879

0) = 1 for every 119899 isin N

By compactness again we may assume that (119909119899 119909

lowast

119899)

converges to (1199090 119909

lowast

0) isin 119883 times 119883

lowast This implies that (1199090 119909

lowast

0) isin

Π(119883) and |119909lowast0(119879

0+ 119878

119899)119909

0| = V(119879

0+ 119878

119899) = 1 that is

(1199090 119909

lowast

0 119879

0+ 119878

119899) isin Πnu(119883) for all 119899 This is a contradiction

with the fact that0 = lim

119899dist ((119909

119899 119909

lowast

119899 119879

119899) (119909

0 119909

lowast

0 119879

0+ 119878

119899))

⩾ lim119899

dist ((119909119899 119909

lowast

119899 119879

119899) Πnu (119883)) ⩾ 120576

(10)

Abstract and Applied Analysis 3

We may also give the following easy result concerningduality

Proposition 3 Let119883 be a reflexive space Then

120578nu (119883) (120576) = 120578nu (119883lowast) (120576) (11)

for every 120576 isin (0 1) In particular 119883 has the BPBp-nu if andonly if 119883lowast has the BPBp-nu

Wewill use that V(119879lowast) = V(119879) for all 119879 isinL(119883) where 119879lowastdenotes the adjoint operator of 119879 This result can be found in[1] but it is obvious if119883 is reflexive

Proof By reflexivity it is enough to show that 120578nu(119883)(120576) ⩽120578nu(119883

lowast)(120576) Let 120576 isin (0 1) be fixed If 120578nu(119883)(120576) = 0 there is

nothing to prove Otherwise consider 0 lt 120578 lt 120578nu(119883)(120576)Suppose that 119879

1isinL(119883lowast

) and (119909lowast1 119909

1) isin Π(119883

lowast) satisfy

V (1198791) = 1

100381610038161003816100381611990911198791119909lowast

1

1003816100381610038161003816 gt V (1198791) minus 120578 (12)

By considering 119879lowast1isin L(119883) we may find 119878

1isin L(119883) and

(1199101 119910

lowast

1) isin Π(119883) such that

1003816100381610038161003816119910lowast

111987811199101

1003816100381610038161003816 = V (1198781) = 1

10038171003817100381710038171199101 minus 11990911003817100381710038171003817 lt 120576

1003817100381710038171003817119910lowast

1minus 119909

lowast

1

1003817100381710038171003817 lt 1205761003817100381710038171003817119879

lowast

1minus 119878

1

1003817100381710038171003817 lt 120576

(13)

Then 119878lowast1isinL(119883lowast

) and (119910lowast1 119910

1) isin Π(119883

lowast) satisfy

1003816100381610038161003816⟨1199101 119878lowast

1119910lowast

1⟩1003816100381610038161003816 = V (119878

1) = 1

1003817100381710038171003817119910lowast

1minus 119909

lowast

1

1003817100381710038171003817 lt 120576

10038171003817100381710038171199101 minus 11990911003817100381710038171003817 lt 120576

10038171003817100381710038171198791 minus 119878lowast

1

1003817100381710038171003817 lt 120576

(14)

This implies that 120578nu(119883lowast)(120576) ⩾ 120578 We finish by just taking

supremum on 120578

We do not know whether the result above is valid in thenonreflexive case

3 Spaces Which Are Both Uniformly Convexand Uniformly Smooth

For a Banach space which is both uniformly convex anduniformly smooth we get a property which is weaker thanBPBp-nu This result was known to A Guirao (privatecommunication)

Proposition 4 Let 119883 be a uniformly convex and uniformlysmooth Banach space Then given 120576 gt 0 there exists 120578(120576) gt 0such that whenever 119879

0isinL(119883) with V(119879

0) = 1 and (119909

0 119909

lowast

0) isin

Π(119883) satisfy |119909lowast011987901199090| gt 1 minus 120578(120576) there exist 119878 isin L(119883) and

(119910 119910lowast) isin Π(119883) such that

V (119878) = 1003816100381610038161003816119910lowast1198781199101003816100381610038161003816

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowastminus 119910

lowast1003817100381710038171003817 lt 1205761003817100381710038171003817119878 minus 1198790

1003817100381710038171003817 lt 120576

(15)

Proof Notice that the uniform smoothness of119883 is equivalentto the uniform convexity of 119883lowast Let 120575

119883(120576) and 120575

119883lowast(120576) be the

moduli of convexity119883 and119883lowast respectively Given 0 lt 120576 lt 1consider

120578 (120576) =120576

4min 120575

119883(120576

4) 120575

119883lowast (120576

4) gt 0 (16)

Consider 1198790isin L(119883) with V(119879

0) = 1 and (119909

0 119909

lowast

0) isin Π(119883)

satisfying |119909lowast011987901199090| gt 1 minus 120578(120576) Define 119879

1isinL(119883) by

1198791119909 = 119879

0119909 + 120582

1(120576

4) 119909

lowast

0(119909) 119909

0(17)

for all 119909 isin 119883 where 1205821is the scalar satisfying |120582

1| = 1 and

|119909lowast

011987901199090+ 120582

1(1205764)| = |119909

lowast

011987901199090| + 1205764 Now choose 119909

1isin 119878

119883

and 119909lowast1isin 119878

119883lowast such that |119909lowast

1(119909

1)| = 1 119909lowast

1(119909

0) = |119909

lowast

1(119909

0)| and

1003816100381610038161003816119909lowast

111987911199091

1003816100381610038161003816 ⩾ V (1198791) minus 120578(

1205762

42) (18)

Now we define a sequence (119909119899 119909

lowast

119899 119879

119899) in 119878

119883times 119878

119883lowast times L(119883)

inductively Indeed suppose that we have a defined sequence(119909

119895 119909

lowast

119895 119879

119895) for 0 ⩽ 119895 ⩽ 119899 and let

119879119899+1119909 = 119879

119899119909 + 120582

119899+1

120576119899+1

4119899+1119909lowast

119899(119909) 119909

119899 (19)

Then choose 119909119899+1

isin 119878119883

and 119909lowast119899+1

isin 119878119883lowast such that

|119909lowast

119899+1(119909

119899+1)| = 1 and |119909lowast

119899+1(119909

119899)| = 119909

lowast

119899+1(119909

119899)

1003816100381610038161003816119909lowast

119899+1119879119899+1119909119899+1

1003816100381610038161003816 ⩾ V (119879119899+1) minus 120578(

120576119899+2

4119899+2) (20)

Notice that for all 119899 ⩾ 0 we have

1003817100381710038171003817119879119899+1 minus 1198791198991003817100381710038171003817 ⩽120576119899+1

4119899+1

1003816100381610038161003816V (119879119899+1) minus V (119879119899)1003816100381610038161003816 ⩽120576119899+1

4119899+1

(21)

This implies that (119879119899) is a Cauchy sequence and assume that

it converges to 119878 isinL(119883) Then we have

lim119899119879119899= 119878

10038171003817100381710038171198790 minus 1198781003817100381710038171003817 lt 120576

lim119899

1003816100381610038161003816119909lowast

119899119879119899119909119899

1003816100381610038161003816 = lim119899V (119879

119899) = V (119878)

(22)

We will show that both sequences (119909119899) and (119909lowast

119899) are Cauchy

From the definition we have

V (119879119899+1) minus 120578(

120576119899+2

4119899+2)

⩽1003816100381610038161003816119909

lowast

119899+1119879119899+1119909119899+1

1003816100381610038161003816

4 Abstract and Applied Analysis

100381610038161003816100381610038161003816100381610038161003816

119909lowast

119899+1119879119899119909119899+1+ 120582

119899+1

120576119899+1

4119899+1119909lowast

119899(119909

119899+1) 119909

lowast

119899+1(119909

119899)

100381610038161003816100381610038161003816100381610038161003816

⩽ V (119879119899) +120576119899+1

4119899+1119909lowast

119899+1(119909

119899)

V (119879119899+1)

⩾1003816100381610038161003816119909

lowast

119899119879119899+1119909119899

1003816100381610038161003816 =

100381610038161003816100381610038161003816100381610038161003816

119909lowast

119899119879119899119909119899+ 120582

119899+1

120576119899+1

4119899+1

100381610038161003816100381610038161003816100381610038161003816

=1003816100381610038161003816119909

lowast

119899119879119899119909119899

1003816100381610038161003816 +120576119899+1

4119899+1⩾ V (119879

119899) minus 120578(

120576119899+1

4119899+1) +120576119899+1

4119899+1

(23)

In summary we have

V (119879119899) +120576119899+1

4119899+1119909lowast

119899+1(119909

119899)

⩾ V (119879119899) minus 120578(

120576119899+1

4119899+1) +120576119899+1

4119899+1minus 120578(

120576119899+2

4119899+2)

(24)

Hence

119909lowast

119899+1(119909

119899) ⩾ 1 minus 2

4119899+1

120576119899+1120578(120576119899+1

4119899+1)

= 1 minus1

2min120575

119883(120576119899+1

4119899+2) 120575

119883lowast (120576119899+1

4119899+2)

1003817100381710038171003817100381710038171003817

119909119899+ 119909

119899+1

2

1003817100381710038171003817100381710038171003817⩾ 119909

lowast

119899+1(119909119899+ 119909

119899+1

2) ⩾ 1 minus 120575

119883(120576119899+1

4119899+2)

100381710038171003817100381710038171003817100381710038171003817

119909lowast

119899+ 119909

lowast

119899+1

2

100381710038171003817100381710038171003817100381710038171003817

⩾119909lowast

119899+ 119909

lowast

119899+1

2(119909

119899) ⩾ 1 minus 120575

119883lowast (120576119899+1

4119899+2)

(25)

This means that

1003817100381710038171003817119909119899 minus 119909119899+11003817100381710038171003817 ⩽ 120576

119899+14

119899+2

1003817100381710038171003817119909lowast

119899minus 119909

lowast

119899+1

1003817100381710038171003817 ⩽ 120576119899+14

119899+2

(26)

for all 119899 So (119909119899) and (119909lowast

119899) are Cauchy Let 119909

infin= lim

119899119909119899and

119909lowast

infin= lim

119899119909lowast

119899 Then 119909

0minus 119909

infin lt 1205764 and 119909lowast

0minus 119909

lowast

infin lt 1205764

Hence |119909lowastinfin(119909

infin)| = lim

119899|119909

lowast

119899(119909

119899)| = 1 and

V (119878) = lim119899V (119879

119899) = lim

119899

1003816100381610038161003816119909lowast

119899119879119899119909119899

1003816100381610038161003816 =1003816100381610038161003816119909

lowast

infin119878119909

infin

1003816100381610038161003816 (27)

Let 120572 = 119909lowastinfin(119909

infin) 119910lowast = 119909lowast

infin and 119910 = 119909

infin Then we have

119910lowast(119910) = 1 V(119878) = |119910lowast119878119910| and 119910 minus 119909

0 lt 120576 Notice that

|120572 minus 1| =1003816100381610038161003816119909

lowast

infin(119909

infin) minus 119909

lowast

0(119909

0)1003816100381610038161003816

⩽1003816100381610038161003816(119909

lowast

infinminus 119909

lowast

0) (119909

infin)1003816100381610038161003816 +1003816100381610038161003816119909

lowast

0(119909

infin) minus 119909

lowast

0(119909

0)1003816100381610038161003816 lt120576

2

(28)

Therefore

1003817100381710038171003817119910lowastminus 119909

lowast1003817100381710038171003817 ⩽1003817100381710038171003817119910

lowastminus 119910

lowast1003817100381710038171003817 +1003817100381710038171003817119910

lowastminus 119909

lowast1003817100381710038171003817 lt120576

2+120576

4lt 120576 (29)

This completes the proof

Let us discuss a little bit about the equivalence betweenthe property in the result above and the BPBp-nu Forconvenience let us introduce the following definition

Definition 5 A Banach space 119883 has the weak Bishop-Phelps-Bollobas property for the numerical radius (in short weak-BPBp-nu) if given 120576 gt 0 there exists 120578(120576) gt 0 such thatwhenever 119879

0isin L(119883) with V(119879

0) = 1 and (119909

0 119909

lowast

0) isin Π(119883)

satisfy |119909lowast011987901199090| gt 1 minus 120578(120576) there exist 119878 isin L(119883) and

(119910 119910lowast) isin Π(119883) such that

V (119878) = 1003816100381610038161003816119910lowast1198781199101003816100381610038161003816

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowastminus 119910

lowast1003817100381710038171003817 lt 120576 119878 minus 119879 lt 120576

(30)

Notice that the only difference between this concept andthe BPBp-nu is the normalization of the operator 119878 by thenumerical radius Of course if the numerical radius andthe operator norm are equivalent these two properties arethe same This equivalence is measured by the so-callednumerical index of the Banach space as follows For a Banachspace119883 the numerical index of119883 is defined by

119899 (119883) = inf V (119879) 119879 isinL (119883) 119879 = 1 (31)

It is clear that 0 ⩽ 119899(119883) ⩽ 1 and 119899(119883)119879 ⩽ V(119879) ⩽ 119879for all 119879 isin L(119883) The value 119899(119883) = 1 means that V equalsthe usual operator norm This is the case of 119883 = 119871

1(120583) and

119883 = 119862(119870) among many others On the other hand 119899(119883) gt 0if and only if the numerical radius is equivalent to the normof L(119883) We refer the reader to [20] for more informationand background

The following result is immediate We include a proof forthe sake of completeness

Proposition 6 Let119883 be a Banach space with 119899(119883) gt 0 Then119883 has the BPBp-nu if and only if119883 has the weak-BPBp-nu

Proof Thenecessity is clear For the converse assume that wehave 120578(120576) gt 0 satisfying the conditions of the weak-BPBp-nufor all 0 lt 120576 lt 1 If 119879 isin L(119883) with V(119879) = 1 and (119909

0 119909

lowast

0) isin

Π(119883) satisfy |119909lowast0119879119909

0| gt 1 minus 120578(120576) for 0 lt 120576 lt 1 then there exist

119878 isinL(119883) and (119910 119910lowast) isin Π(119883) such that

V (119878) = 1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 119878 minus 119879 lt 120576

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowastminus 119910

lowast1003817100381710038171003817 lt 120576

(32)

As V(119878) gt 0 by the above let 1198781= (1V(119878))119878 Then we have

1 = V (1198781) =1003816100381610038161003816119910

lowast11987811199101003816100381610038161003816

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowastminus 119910

lowast1003817100381710038171003817 lt 120576

(33)

Abstract and Applied Analysis 5

Finally we have

10038171003817100381710038171198781 minus 1198791003817100381710038171003817 ⩽

10038171003817100381710038171003817100381710038171003817

1

V (119878)119878 minus 119878

10038171003817100381710038171003817100381710038171003817

+ 119878 minus 119879

=119878

V (119878)|V (119878) minus 1| + 119878 minus 119879

⩽1

119899 (119883)|V (119878) minus V (119879)| + 119878 minus 119879

⩽ (1

119899 (119883)+ 1) 119878 minus 119879 lt

119899 (119883) + 1

119899 (119883)120576

(34)

An obvious change of parameters finishes the proof

We do not know whether the hypothesis of 119899(119883) gt 0 canbe omitted in the above result

Putting together Propositions 4 and 6 we get the follow-ing

Corollary 7 Let 119883 be a uniformly convex and uniformlysmooth Banach space with 119899(119883) gt 0 Then 119883 has the BPBp-nu

Let us comment that every complex Banach space 119883satisfies 119899(119883) ⩾ 1119890 so the above corollary automaticallyapplies in the complex case In the real case this is no longertrue as the numerical index of a Hilbert space of dimensiongreater than or equal to two is 0 On the other hand it isproved in [21] that real 119871

119901(120583) spaces have nonzero numerical

index for every measure 120583when 119901 = 2Therefore we have thefollowing examples

Example 8 (a) Complex Banach spaces which are uniformlysmooth and uniformly convex satisfy the BPBp-nu

(b) In particular for every measure 120583 the complex spaces119871119901(120583) have the BPBp-nu for 1 lt 119901 lt infin(c) For every measure 120583 the real spaces 119871

119901(120583) have the

BPBp-nu for 1 lt 119901 lt infin 119901 = 2

Note Added in Revision Very recently H J Lee M Martınand J Merı have proved that Proposition 6 can be extendedto some Banach spaces with numerical index zero as forinstance real Hilbert spaces Hence they have shown thatHilbert spaces have the BPBp-nu These results will appearelsewhere

4 L1

Spaces

In this section we will show that 1198711(120583) has the BPBp-nu for

every measure 120583 In the proof we are dealing with complexintegrable functions since the real case is followed easilyby applying the same proof Our main result here is thefollowing

Theorem 9 Let 120583 be a measure Then 1198711(120583) has the Bishop-

Phelps-Bollobas property for numerical radius More preciselygiven 120576 gt 0 there exists 120578(120576) gt 0 (which does not depend on120583) such that whenever 119879

0isin L(119871

1(120583)) with V(119879

0) = 1 and

(1198910 119892

0) isin Π(119871

1(120583)) satisfy |⟨119879

01198910 119892

0⟩| gt 1 minus 120578(120576) then there

exist 119879 isinL(1198711(120583)) (119891

1 119892

1) isin Π(119871

1(120583)) such that

1003816100381610038161003816⟨1198791198911 1198921⟩1003816100381610038161003816 = V (119879) = 1 10038171003817100381710038171198910 minus 1198911

1003817100381710038171003817 lt 120576

10038171003817100381710038171198920 minus 11989211003817100381710038171003817 lt 120576

1003817100381710038171003817119879 minus 11987901003817100381710038171003817 lt 120576

(35)

As a first step we have to start dealing with finite regularpositive Borel measures for which a representation theoremfor operators exists

Proposition 10 Let 119898 be a finite regular positive Borelmeasure on a compact Hausdorff spaceΩ Then 119871

1(119898) has the

Bishop-Phelps-Bollobas property for numerical radius Moreprecisely given 120576 gt 0 there is 120578(120576) gt 0 (which is independent ofthemeasure119898) such that if a norm-one element119879 inL(119871

1(119898))

and (1198910 119892

0) isin Π(119871

1(119898)) satisfy |⟨119879119891

0 119892

0⟩| gt 1 minus 120578(120576) then

there exist an operator 119878 isin L(1198711(119898)) (119891

1 119892

1) isin Π(119871

1(119898))

such that1003816100381610038161003816⟨1198781198911 1198921⟩

1003816100381610038161003816 = 119878 = 110038171003817100381710038171198910 minus 1198911

1003817100381710038171003817 ⩽ 120576

10038171003817100381710038171198920 minus 11989211003817100381710038171003817 ⩽ 120576 119879 minus 119878 ⩽ 120576

(36)

To prove this proposition we need some backgroundon representation of operators on Lebesgue spaces on finiteregular positive Borel measures and several preliminarylemmas

Let 119898 be a finite regular positive Borel measure on acompact Hausdorff space Ω If 120583 is a complex-valued Borelmeasure on the product space Ω times Ω then define theirmarginal measures 120583119894 on Ω (119894 = 1 2) as follows 1205831(119860) =120583(119860 times Ω) and 1205832(119861) = 120583(Ω times 119861) where 119860 and 119861 are Borelmeasurable subsets of Ω

Let 119872(119898) be the complex Banach lattice of measuresconsisting of all complex-valued Borel measures 120583 on theproduct spaceΩ timesΩ such that |120583|119894 are absolutely continuouswith respect to119898 for 119894 = 1 2 endowed with the norm

1003817100381710038171003817100381710038171003817100381710038171003817

11988910038161003816100381610038161205831003816100381610038161003816

1

119889119898

1003817100381710038171003817100381710038171003817100381710038171003817infin

(37)

Each 120583 isin 119872(119898) defines a bounded linear operator 119879120583from

1198711(119898) to itself by

⟨119879120583(119891) 119892⟩ = int

ΩtimesΩ

119891 (119909) 119892 (119910) 119889120583 (119909 119910) (38)

where 119891 isin 1198711(119898) and 119892 isin 119871

infin(119898) Iwanik [22] showed that

the mapping 120583 997891rarr 119879120583is a lattice isometric isomorphism from

119872(119898) onto L(1198711(119898)) Even though he showed this for the

real case it can be easily generalized to the complex caseFor details see [22 Theorem 1] and [23 IV Theorem 15(ii)Corollary 2]

We will also use that given an arbitrary measure 120583 every119879 isin L(119871

1(120583)) satisfies V(119879) = 119879 [24] (that is the space

1198711(120583) has numerical index 1)

Lemma 11 (see [15 Lemma 33]) Let 119888119899 be a sequence of

complex numbers with |119888119899| ⩽ 1 for every 119899 and let 120578 gt 0 such

6 Abstract and Applied Analysis

that for a convex series sum120572119899 Resuminfin

119899=1120572119899119888119899gt 1 minus 120578 Then for

every 0 lt 119903 lt 1 the set 119860 = 119894 isin N Re 119888119894gt 119903 satisfies the

estimate

sum

119894isin119860

120572119894⩾ 1 minus

120578

1 minus 119903 (39)

From now on 119898 will be a finite regular positive Borelmeasure on the compact Hausdorff space Ω

Lemma 12 Suppose that there exist a nonnegative simplefunction 119891 isin 119878

1198711(119898)

and a function 119892 isin 119878119871infin(119898)

such that

Re ⟨119891 119892⟩ gt 1 minus 1205763

16 (40)

Then there exist a nonnegative simple function 1198911isin 119878

1198711(119898)

anda function 119892

1isin 119878

119871infin(119898)

such that

1198921(119909) = 120594supp(119891

1)(119909) + 119892 (119909) 120594

Ωsupp(1198911)(119909)

⟨1198911 119892

1⟩ = 1

1003817100381710038171003817119891 minus 119891110038171003817100381710038171lt 120576

1003817100381710038171003817119892 minus 11989211003817100381710038171003817infinlt radic120576 supp (119891

1) sub supp (119891)

(41)

Proof Let119891 = sum119898

119895=1(120573

119895119898(119861

119895))120594

119861119895

for some (120573119895) such that

120573119895⩾ 0 for all 119895 andsum119898

119895=1120573119895= 1 and119861

119895rsquos aremutually disjoint

By the assumption we have

Re ⟨119891 119892⟩ =119899

sum

119895=1

120573119895

1

119898 (119861119895)

int119861119895

Re119892 (119909) 119889119898 (119909) gt 1 minus 1205763

16

(42)

and letting

119869 = 119895 1 ⩽ 119895 ⩽ 1198991

119898 (119861119895)

int119861119895

Re119892 (119909) 119889119898 (119909) gt 1 minus 1205762

4

(43)

we have by Lemma 11

sum

119895isin119869

120573119895gt 1 minus

120576

4 (44)

For each 119895 isin 119869 we have

1 minus1205762

4lt

1

119898 (119861119895)

int119861119895

Re119892 (119909) 119889119898 (119909)

=1

119898 (119861119895)

int119861119895capRe119892⩽1minus120576

Re119892 (119909) 119889119898 (119909)

+ int119861119895capRe119892gt1minus120576

Re119892 (119909) 119889119898 (119909)

⩽1

119898 (119861119895)

((1 minus 120576)119898 (119861119895cap Re119892 ⩽ 1 minus 120576)

+ 119898 (119861119895cap Re119892 gt 1 minus 120576))

= 1 minus 120576

119898 (119861119895cap Re119892 ⩽ 1 minus 120576)

119898 (119861119895)

(45)

This implies that

119898(119861119895cap Re119892 ⩽ 1 minus 120576)

119898 (119861119895)

lt120576

4 (46)

Define 119861119895= 119861

119895cap Re119892 gt 1 minus 120576 for all 119895 isin 119869

1198911= (1sum

119895isin119869

120573119895)sum

119895isin119869

120573119895(120594

119861119895

119898 (119861119895)) (47)

and 1198921(119909) = 1 on supp(119891

1) and 119892

1(119909) = 119892(119909) elsewhereThen

it is clear that supp(1198911) sub supp(119891) 119892 minus 119892

1infinlt radic120576 and

⟨1198911 119892

1⟩ = 1 Finally we will show that 119891 minus 119891

1 lt 120576 Notice

first that10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

= 2sum

119895isin119869

120573119895

119898(119861119895 119861

119895)

119898 (119861119895)

lt120576

2

(48)

Hence1003817100381710038171003817119891 minus 1198911

1003817100381710038171003817

10038171003817100381710038171003817100381710038171003817100381710038171003817

1

sum119895isin119869120573119895

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus 119891

10038171003817100381710038171003817100381710038171003817100381710038171003817

1 minus sum119895isin119869120573119895

sum119895isin119869120573119895

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+120576

4

= (1 minussum

119895isin119869

120573119895) +

120576

2+120576

4

⩽120576

4+120576

2+120576

4= 120576

(49)

Lemma 13 (see [25 Lemma 33]) Suppose that 119879120583is a norm-

one element in L(1198711(119898)) for some 120583 isin 119872(119898) and there is

a nonnegative simple function 1198910such that 119891

0is a norm-one

element of 1198711(119898) and 119879

1205831198910 ⩾ 1 minus 120576

32

6 for some 0 lt 120576 lt 1

Abstract and Applied Analysis 7

Then there exist a norm-one bounded linear operator 119879] forsome ] isin 119872(119898119898) and a nonnegative simple function 119891

1

in 1198781198711(119898)

such that 119879120583minus 119879] ⩽ 120576 1198911 minus 1198910 ⩽ 3120576 and

(119889|]|1119889119898)(119909) = 1 for all 119909 isin supp(1198911)

Lemma 14 Suppose that 119879] isin L(1198711(119898)) is a norm-one

operator 119891 = sum119899

119894=1120573119894(120594

119861119894

119898(119861119894)) where 119898(119861

119895) gt 0 for all

1 ⩽ 119895 ⩽ 119899 and 119861119895119899

119895=1are mutually disjoint Borel subsets ofΩ

is a norm-one nonnegative simple function and 119892 is an elementof 119878

119871infin(119898)

such that

Re ⟨119892 119879]119891⟩ ⩾ 1 minus1205766

27(50)

for some 0 lt 120576 lt 1 and

119889|]|1

119889119898(119909) = 1 119892 (119909) = 1 (51)

for all x in the support of 119891Then there exist a nonnegative simple function 119891 isin 119878

1198711(119898)

a function119892 isin 119878

119871infin(119898)

and an operator119879] inL(1198711(119898) 1198711(119898))such that

⟨119892 119879]119891⟩ =1003817100381710038171003817119879]1003817100381710038171003817 = 1

1003817100381710038171003817119879] minus 119879]1003817100381710038171003817 ⩽ 2120576

10038171003817100381710038171003817119891 minus 119891

10038171003817100381710038171003817⩽ 3120576

1003817100381710038171003817119892 minus 1198921003817100381710038171003817 ⩽radic120576 ⟨119891 119892⟩ = 1

(52)

Proof Since

Re ⟨119892 119879]119891⟩ ⩾ 1 minus1205766

27 (53)

we have

1 minus1205766

27lt Re ⟨119892 119879]119891⟩

= intΩtimesΩ

119891 (119909)Re119892 (119910) 119889120583 (119909 119910)

=

119899

sum

119895=1

120573119895intΩtimesΩ

120594119861119895

(119909)

119898 (119861119895)

Re119892 (119910) 119889] (119909 119910)

(54)

Let

119869 = 119895 intΩtimesΩ

(

120594119861119895

(119909)

119898 (119861119895)

)Re119892 (119910) 119889] (119909 119910) gt 1 minus 1205763

26

(55)

Then from Lemma 11 we have sum119895isin119869120573119895gt 1 minus 120576

32 Let 119891

1=

sum119895isin119869120573119895(120594

119861119895

119898(119861119895)) where 120573

119895= 120573

119895(sum

119895isin119869120573119895) for all 119895 isin 119869

Then

10038171003817100381710038171198911 minus 1198911003817100381710038171003817 ⩽

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

(120573119895minus 120573

119895)

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+ sum

119895isin119869

120573119895⩽ 120576

3⩽ 120576 (56)

Note that there is a Borel measurable function ℎ on Ω times Ωsuch that 119889](119909 119910) = ℎ(119909 119910) 119889|]|(119909 119910) and |ℎ(119909 119910)| = 1 forall (119909 119910) isin Ω times Ω Let

119862 = (119909 119910) 1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 1

1003816100381610038161003816 ltradic120576

232 (57)

Define two measures ]119891and ]

119888as follows

]119891(119860) = ] (119860 119862) ]

119888(119860) = ] (119860 cap 119862) (58)

for every Borel subset 119860 ofΩ times Ω It is clear that

119889] = 119889]119891+ 119889]

119888 119889

10038161003816100381610038161003816]119891

10038161003816100381610038161003816= ℎ 119889]

119891

1198891003816100381610038161003816]1198881003816100381610038161003816 = ℎ 119889]119888 119889 |]| = 119889

10038161003816100381610038161003816]119891

10038161003816100381610038161003816+ 1198891003816100381610038161003816]1198881003816100381610038161003816

(59)

Since (119889|]|11198891198981)(119909) = 1 for all 119909 isin ⋃119899

119895=1119861119895 we have

1 =119889|]|1

1198891198981

(119909) =

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) +1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909) (60)

for all 119909 isin 119861 = ⋃119899

119895=1119861119895 and we deduce that |]|1(119861

119895) = 119898

1(119861

119895)

for all 1 ⩽ 119895 ⩽ 119899We claim that |]

119891|1(119861

119895)119898

1(119861

119895) ⩽ 120576

22

2 for all 119895 isin 119869Indeed if |119892(119910)ℎ(119909 119910) minus 1| ⩾ radic120576232 then

Re (119892 (119910) ℎ (119909 119910)) ⩽ 1 minus 12057624 (61)

So we have

1 minus1205763

26

⩽1

1198981(119861

119895)

ReintΩtimesΩ

120594119861119895(119909)119892 (119910) 119889] (119909 119910)

=1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 |]| (119909 119910)

=1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 10038161003816100381610038161003816]11989110038161003816100381610038161003816(119909 119910)

+1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 1003816100381610038161003816]1198881003816100381610038161003816 (119909 119910)

⩽1

1198981(119861

119895)

((1 minus120576

24)10038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

(119861119895) +1003816100381610038161003816]1198881003816100381610038161003816

1

(119861119895))

= 1 minus120576

24

10038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

(119861119895)

1198981(119861

119895)

(62)

This proves our claimWe also claim that for each 119895 isin 119869 there exists a Borel

subset 119861119895of 119861

119895such that

(1 minus120576

2)119898

1(119861

119895) ⩽ 119898

1(119861

119895) ⩽ 119898

1(119861

119895)

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) ⩽120576

2

(63)

8 Abstract and Applied Analysis

for all 119909 isin 119861119895 Indeed set 119861

119895= 119861

119895cap 119909 isin Ω (119889|]

119891|1119889119898

1)

(119909) ⩽ 1205762 Then

int119861119895119861119895

120576

2119889119898

1(119909) ⩽ int

119861119895

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) 1198891198981(119909)

=10038161003816100381610038161003816]1119891

10038161003816100381610038161003816(119861

119895) ⩽1205762

22119898

1(119861

119895)

(64)

This shows that 1198981(119861

119895 119861

119895) ⩽ (1205762)119898

1(119861

119895) This proves our

second claimNow we define 119892 by 119892(119910) = 119892(119910)|119892(119910)| if |119892(119910)| ⩾ 1 minus

radic120576232 and 119892(119910) = 119892(119910) if |119892(119910)| lt 1 minus radic120576232 and we

write 119891 = sum119895isin119869120573119895(120594

119861119895

1198981(119861

119895)) It is clear that 119892 isin 119878

119871infin(119898)

119892 minus 119892 lt radic120576 and 119892(119910) = 1 for all 119909 isin supp119891

Finally we define the measure

119889] (119909 119910)

= sum

119895isin119869

120594119861119895

(119909) 119892 (119910)ℎ (119909 119910)119889]119888(119909 119910)(

1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+ 1205941198691119861(119909) 119889] (119909 119910)

(65)

where 119861 = ⋃119895isin119869119861119895 It is easy to see that (119889|]|1119889119898

1)(119909) = 1

on 119861 and (119889|]|11198891198981)(119909) ⩽ 1 elsewhere Note that

119889 (] minus ]) (119909 119910)

= sum

119895isin119869

120594119861119895

(119909) [

[

119892 (119910)ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1]

]

119889]119888(119909 119910)

minus sum

119895isin119869

120594119861119895

(119909) 119889]119891(119909 119910)

(66)

If (119909 119910) isin 119862 then |119892(119910)| ⩾ 1 minus radic120576232 ⩾ 1 minus 1232 and

100381610038161003816100381610038161003816119892 (119910)ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

119892 (119910)

1003816100381610038161003816119892 (119910)1003816100381610038161003816

ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816100381610038161003816

1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 11003816100381610038161003816

1003816100381610038161003816119892 (119910)1003816100381610038161003816

+

10038161003816100381610038161 minus1003816100381610038161003816119892 (119910)

1003816100381610038161003816

10038161003816100381610038161003816100381610038161003816119892 (119910)

1003816100381610038161003816

⩽ 2

1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 11003816100381610038161003816

1003816100381610038161003816119892 (119910)1003816100381610038161003816

⩽ 2radic120576

232

232

232 minus 1⩽ 2radic120576

(67)

Hence for all (119909 119910) isin 119862 we have

10038161003816100381610038161003816100381610038161003816100381610038161003816

119892 (119910)ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

⩽100381610038161003816100381610038161003816119892 (119910)ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

⩽ 2radic120576(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

(68)

So we have for all 119909 isin 1198691

119889|] minus ]|1

1198891198981

(119909)

⩽ sum

119895isin119869

120594119861119895

(119909) [

[

2radic120576(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

]

]

1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909)

+ sum

119895isin119869

120594119861119895

(119909)

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909)

⩽ sum

119895isin119869

120594119861119895

(119909)(2radic120576 + (1 minus1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909)))

+sum

119895isin119869

120594119861119895

(119909)(

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909))

⩽ 2radic120576 + 120576 lt 3radic120576

(69)

This gives that 119879] minus 119879] lt 3radic120576 Note also that for all 119895 isin 119869

⟨119879]

120594119861119895

1198981(119861

119895)

119892⟩

= intΩtimesΩ

120594119861119895

(119909)

1198981(119861

119895)

119892 (119910) 119889] (119909 119910)

= intΩtimesΩ

120594119861119895

(119909)

1198981(119861

119895)

ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

119889]119888(119909 119910)

Abstract and Applied Analysis 9

= intΩ

120594119861119895

(119909)

1198981(119861

119895)

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

1198891003816100381610038161003816]1198881003816100381610038161003816

1

(119909)

= intΩ

120594119861119895

(119909)

1198981(119861

119895)

1198891198981(119909) = 1

(70)

Hence we get ⟨119879]119891 119892⟩ = 1 which implies that 119879]119891 =119879] = 1 Finally10038171003817100381710038171003817119891 minus 119891

10038171003817100381710038171003817

⩽10038171003817100381710038171003817119891 minus 119891

1

10038171003817100381710038171003817+10038171003817100381710038171198911 minus 119891

1003817100381710038171003817

=

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

1198981(119861

119895)

minus sum

119895isin119869

120573119895

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+ 120576

⩽ sum

119895isin119869

120573119895(

10038171003817100381710038171003817100381710038171003817100381710038171003817

120594119861119895

1198981(119861

119895)

minus

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

120594119861119895

1198981(119861

119895)

minus

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

) + 120576

= 2sum

119895isin119869

120573119895

1198981(119861

119895 119861

119895)

1198981(119861

119895)

+ 120576

⩽ 2sum

119895isin119869

120573119895

(1205762)1198981(119861

119895)

1198981(119861

119895)

+ 120576 ⩽120576

1 minus 1205762+ 120576 lt 3120576

(71)

We are now ready to present the proof of the main resultin the case of finite regular positive Borel measures

Proof of Proposition 10 Let 1205751= 120575

3

2(5sdot2

4) 120575

2= 120575

12

3(3

2sdot2

14)

and 1205753= (12057610)

2 for some 0 lt 120576 lt 1 Suppose that 119879 isinL(119871

1(119898)) with 119879 = 1 and that there is an 119891

0isin 119878

1198711(119898)

and1198920isin 119878

119871infin(119898)

such that ⟨1198910 119892

0⟩ = 1 and |⟨119879119891

0 119892

0⟩| gt 1minus120575

3

12

6Then there is an isometric isomorphism Ψ from 119871

1(119898) onto

itself such thatΨ(1198910) = |119891

0| and there is a scalar number 120572 in

119878R such that |⟨1198791198910 119892

0⟩| = ⟨120572119879119891

0 119892

0⟩ Then letting 119891

1= Ψ119891

0

1198921= (Ψ

minus1)lowast1198920 and 119879

1= 120572Ψ119879Ψ

minus1 we have

⟨1198781198911 119892

1⟩ = ⟨120572Ψ119879

minus1Ψ119891

0 (Ψ

minus1)lowast

1198920⟩

= ⟨1205721198791198910 119892

0⟩ gt 1 minus

1205753

1

26

⟨1198911 119892

1⟩ = ⟨Ψ119891

0 (Ψ

minus1)lowast

1198920⟩ = 1

(72)

Since 11987911198911 gt 1 minus 120575(120575

3

12

6) by Lemma 13 there exists a

norm-one bounded operator 119879] and a nonnegative simplefunction119891

2isin 119878

1198711(119898)

such that 1198791minus119879] ⩽ 1205751 1198912minus1198911 ⩽ 31205751

and (119889|]|11198891198981)(119909) = 1 for all 119909 isin supp(119891

2) Then

⟨119879]1198912 1198921⟩

= ⟨11987911198911 119892

1⟩ minus ⟨119879

11198911minus 119879

11198912 119892

1⟩ minus ⟨119879

11198912minus 119879]1198912 1198921⟩

⩾ ⟨11987911198911 119892

1⟩ minus10038171003817100381710038171198911 minus 1198912

1003817100381710038171003817 minus10038171003817100381710038171198791 minus 119879]

1003817100381710038171003817

⩾ 1 minus1205753

1

26minus 3120575

1minus 120575

1⩾ 1 minus 5120575

1= 1 minus

1205753

2

16

(73)

Notice also that

⟨1198912 119892

1⟩ = ⟨119891

1 119892

1⟩ minus ⟨119891

1minus 119891

2 119892

1⟩ ⩾ 1 minus

10038171003817100381710038171198911 minus 11989121003817100381710038171003817

⩾ 1 minus 31205751⩾ 1 minus 5120575

1= 1 minus

1205753

2

16

(74)

By Lemma 12 there are a nonnegative simple function 1198913isin

1198781198711(119898)

and a function 1198923isin 119878

119871infin(119898)

such that

1198923(119909) = 120594supp119891

3

(119909) + 1198922(119909) 120594

Ωsupp1198913

(119909)

10038171003817100381710038171198912 minus 11989131003817100381710038171003817 ⩽ 1205752

10038171003817100381710038171198923 minus 11989211003817100381710038171003817 ⩽radic120575

2 ⟨119891

3 119892

3⟩ = 1

(75)

So we have

⟨119879]1198913 1198923⟩ = ⟨119879]1198912 1198921⟩ minus ⟨119879]1198912 minus 119879]1198913 1198921⟩

minus ⟨119879]1198913 1198921 minus 1198923⟩

⩾ 1 minus1205753

2

16minus 2radic120575

2⩾ 1 minus 3radic120575

2= 1 minus

1205756

3

27

(76)

By Lemma 14 there exist 1198914isin 119878

1198711(119898)

and 1198924isin 119878

119871infin(119898)

and anoperator 119879

4such that

⟨1198924 119879

41198914⟩ = 1 =

100381710038171003817100381711987941003817100381710038171003817

10038171003817100381710038171198794 minus 119879]1003817100381710038171003817 ⩽ 21205753

10038171003817100381710038171198914 minus 11989131003817100381710038171003817 ⩽ 31205753

10038171003817100381710038171198924 minus 11989231003817100381710038171003817 ⩽radic120575

3 ⟨119891

4 119892

4⟩ = 1

(77)

So we have

10038171003817100381710038171198794 minus 11987911003817100381710038171003817 ⩽10038171003817100381710038171198794 minus 119879]

1003817100381710038171003817 +1003817100381710038171003817119879] minus 1198791

1003817100381710038171003817

⩽ 1205751+ 2120575

3⩽ 3120575

3

10038171003817100381710038171198911 minus 11989141003817100381710038171003817 ⩽10038171003817100381710038171198911 minus 1198912

1003817100381710038171003817 +10038171003817100381710038171198912 minus 1198913

1003817100381710038171003817 +10038171003817100381710038171198913 minus 1198914

1003817100381710038171003817

⩽ 31205751+ 120575

2+ 3120575

3⩽ 10120575

3

10038171003817100381710038171198921 minus 11989241003817100381710038171003817 ⩽10038171003817100381710038171198921 minus 1198923

1003817100381710038171003817 + 1198923 minus 1198924

⩽ 1205752+ radic120575

3⩽ 2radic120575

3

(78)

10 Abstract and Applied Analysis

Let 119878 = 120572Ψminus11198794Ψ 119891 = Ψminus1

1198914 and 119892 = Ψlowast

1198924 then we have

119879 minus 119878 =10038171003817100381710038171003817119879 minus 120572Ψ

minus11198794Ψ10038171003817100381710038171003817=10038171003817100381710038171003817120572Ψ119879Ψ

minus1minus 119879

4

10038171003817100381710038171003817

=10038171003817100381710038171198791 minus 1198794

1003817100381710038171003817 ⩽ 31205753

100381710038171003817100381710038171198910minus 11989110038171003817100381710038171003817=100381710038171003817100381710038171198910minus Ψ

minus11198914

10038171003817100381710038171003817=10038171003817100381710038171198911 minus 1198914

1003817100381710038171003817 ⩽ 101205753

100381710038171003817100381710038171198920minus 11989110038171003817100381710038171003817=10038171003817100381710038171198920 minus Ψ

lowast1198924

1003817100381710038171003817 =100381710038171003817100381710038171003817(Ψ

minus1)lowast

1198920minus 119892

4

100381710038171003817100381710038171003817

=10038171003817100381710038171198921 minus 1198924

1003817100381710038171003817 ⩽ 2radic120575

3

⟨119891 119892⟩ = ⟨Ψminus11198914 Ψ

lowast1198924⟩ = ⟨119891

4 119892

4⟩ = 1

10038161003816100381610038161003816⟨119878119891 119892⟩

10038161003816100381610038161003816=10038161003816100381610038161003816⟨120572Ψ

minus11198794ΨΨ

minus11198914 Ψ

lowast1198924⟩10038161003816100381610038161003816= |120572| = 1

(79)

This completes the proof

Finally we may give the proof of the main result in fullgenerality

Proof of Theorem 9 Notice that the Kakutani representationtheorem (see [26] for a reference) says that for every 120590-finitemeasure ] the space 119871

1(]) is isometrically isomorphic to

1198711(119898) for some positive Borel regular measure on a compact

Hausdorff space Then by Proposition 10 there is a universalfunction 120576 997891rarr 120578(120576) gt 0 which gives the BPBp-nu for 119871

1(]) for

every 120590-finite measure ]Fix 120576 gt 0 Suppose that 119879

0isin L(119871

1(120583)) with V(119879

0) = 1

and (1198910 119891

lowast

0) isin Π(119871

1(120583)) satisfy

1003816100381610038161003816⟨119891lowast

0 119879

01198910⟩1003816100381610038161003816 gt 1 minus 120578 (120576) (80)

Choose a sequence 119891119899 in 119871

1(120583) such that sup

119899119879

0119891119899 = 1

and let 119866 be the closed linear span of

119879119899

0119891119898 119899 119898 isin N cup 0 (81)

As 119866 is separable there is a dense subset 119892119899 119899 isin N of 119866

and let 119864 = ⋃infin

119899=1supp119892

119899 where supp119892

119899is the support of 119892

119899

Then the measure 120583|119864is 120590-finite Let

119884 = 119891 isin 1198711(120583) supp (119891) sub 119864 (82)

be a closed subspace of 1198711(120583) It is clear that 119871

1(120583) = 119884oplus

1119885

and 119884 is isometrically isomorphic to 1198711(120583|

119864) So 119884 has the

BPBp-nu with 120578(120576)Now write 119878

0= 119879

0|119884 119884 rarr 119884 consider 119910

0= 119891

0isin

119878119884 119910lowast

0= 119891

lowast

0|119884isin 119878

119884lowast and observe that 119910lowast

0(119910

0) = 1 and

|119910lowast

0(119878

01199100)| = |119891

lowast

0(119879

01198910)| gt 1 minus 120578(120576) Hence there exist 119878 isin

L(119884) and (1199100 119910

lowast

0) isin Π(119884) such that

1003816100381610038161003816119910lowast

0(119878119910

0)1003816100381610038161003816 = 1 = V (119878) 1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

10038171003817100381710038171199100 minus 11991001003817100381710038171003817 lt 120576

1003817100381710038171003817119910lowast

0minus 119910

lowast

0

1003817100381710038171003817 lt 120576

(83)

Finally consider the operator 119879 isinL(1198711(120583)) given by

119879 (119910 119911)

= (119878119910 0) + 1198790(0 119911) ((119910 119911) isin 119871

1(120583) equiv 119884oplus

1119885)

(84)

We have 119879 = 1 (and so V(119879) = 1) Indeed1003817100381710038171003817119879 (119910 119911)

1003817100381710038171003817 =1003817100381710038171003817(119878119910 0)

1003817100381710038171003817 +10038171003817100381710038171198790 (0 119911)

1003817100381710038171003817 ⩽10038171003817100381710038171199101003817100381710038171003817 + 119911 =

1003817100381710038171003817(119910 119911)1003817100381710038171003817

(85)

for all (119910 119911) isin 1198711(120583) and 119879(119910

0 0) = (119878119910

0 0) = 119878119910

0 = 1

Let 119909 = (1199100 0) and 119909lowast = (119910lowast

0 119891

0|119885) Then (119909 119909lowast) isin Π(119871

1(120583))

Moreover we have1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119910

lowast

0119878119910

0

1003816100381610038161003816 = 1 = V (119879) 1003817100381710038171003817119909 minus 1198910

1003817100381710038171003817 =1003817100381710038171003817119910 minus 1199100

1003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowast

0minus 119891

lowast

0

1003817100381710038171003817

= max 1003817100381710038171003817119910 minus 119891lowast

0|119884

1003817100381710038171003817 1003817100381710038171003817119891

lowast

0|119885minus 119891

lowast

0|119885

1003817100381710038171003817

=1003817100381710038171003817119910

lowastminus 119910

lowast

0

1003817100381710038171003817 lt 120576

1003817100381710038171003817119879 minus 11987901003817100381710038171003817 = sup119910+119911⩽1

1003817100381710038171003817119879 (119910 119911) minus 1198790 (119910 119911)1003817100381710038171003817

= sup119910⩽1

1003817100381710038171003817119878119910 minus 11987801199101003817100381710038171003817 =1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

(86)

This completes the proof

5 Examples of Spaces Failing theBishop-Phelps-Bollobaacutes Property forNumerical Radius

Our goal here is to prove that the density of numerical radiusattaining operators does not imply the BPBp-nu Actuallywe will show that among separable spaces there is noisomorphic property implying the BPBp-nu other than finite-dimensionality

We need to relate the BPBp-nu to the Bishop-Phelps-Bollobas property for operators which as mentioned in theintroduction was introduced in [15] A pair (119883 119884) of Banachspaces has the Bishop-Phelps-Bollobas property for operators(in short BPBp) if given 120576 gt 0 there exists 120578(120576) gt 0 suchthat given 119879 isin L(119883 119884) with 119879 = 1 and 119909 isin 119878

119883such that

119879119909 gt 1 minus 120578(120576) then there exist 119911 isin 119878119883and 119878 isin L(119883 119884)

satisfying

119878 = 119878119911 = 1 119909 minus 119911 lt 120576

119879 minus 119878 lt 120576

(87)

We refer the reader to [15 19 25] and references therein formore information and background Among the interestingresults on the BPBp we emphasize that a pair (119883 119884) when119883is finite-dimensional does not necessarily have the BPBp Forinstance if119884 is a strictly convex space which is not uniformlyconvex then the pair (ℓ(2)

1 119884) fails to have the BPBp (this is

contained in [15] see [19 Section 3])

Abstract and Applied Analysis 11

The next result relates the BPBp-nu to the BPBp foroperators in a particular case We will deduce our examplefrom it

Theorem 15 If 1198711(120583)oplus

1119883 has the BPBp-119899119906 then the pair

(1198711(120583) 119883) has the BPBp for operators

Before proving this proposition we will use it to get themain examples of this section The first example shows thatthe density of numerical radius attaining operators does notimply the BPBp-nu

Example 16 There is a reflexive space (and so numerical radiusattaining operators on it are dense) which fails to have theBPBp-nu Indeed let 119884 be a reflexive separable space whichis not superreflexive and we may suppose that 119884 is strictlyconvex Observe that 119884 cannot be uniformly convex since itis not superreflexive Now 119883 = ℓ(2)

1oplus1119884 is reflexive but the

pair (ℓ(2)1 119884) fails the BPBp since 119884 is strictly convex but not

uniformly convex [19 Corollary 33] Therefore Theorem 15gives us that119883 does not have the BPBp-nu

The example above can be extended to get the result thatevery infinite-dimensional separable Banach space can berenormed to fail the BPBp-nu This follows from the factthat every infinite-dimensional separable Banach space canbe renormed to be strictly convex but not uniformly convex(this result can be proved ldquoby handrdquo an alternative categoricalargument for it can be found in [27] and references therein)With a little more of effort we may get the main result of thesection

Theorem 17 Every infinite-dimensional separable Banachspace can be renormed to fail the weak-BPBp-nu (and so inparticular to fail the BPBp-nu)

We need the following result which is surely well knownAs we have not found a reference we include a nice and easyproof kindly given to us by Vladimir Kadets We recall thatgiven a Banach space 119884 the set of all equivalent norms on119884 can be viewed as a metric space using the Banach-Mazurdistance

Lemma 18 Let119884 be an infinite-dimensional separable Banachspace Then the set of equivalent norms on 119884 which are strictlyconvex and are not (locally) uniformly convex is dense in the setof all equivalent norms on119884 (with respect to the Banach-Mazurdistance)

Proof Fix 119890 isin 119878119884and 119890lowast

1isin 119878

119884lowast such that 119890lowast

1(119890) = 1 For a fixed

120576 isin (0 12) denote

119902 (119910) = max (1 minus 120576) 10038171003817100381710038171199101003817100381710038171003817 1003816100381610038161003816119890lowast

1(119910)1003816100381610038161003816 (119910 isin 119884) (88)

Evidently (1 minus 120576)119910 ⩽ 119902(119910) ⩽ 119910 for every 119910 isin 119884 Fix asequence 119890lowast

119896 119896 ⩾ 2 of norm-one functionals separating

the points of 119884 and denote

119901 (119910) = radic

infin

sum

119896=1

1

2119896

1003816100381610038161003816119890lowast

119896(119910)1003816100381610038161003816

2

(119910 isin 119884) (89)

Then 119901 is a strictly convex norm on 119884 119901(119890) ⩾ 1radic2 and119901(119910) ⩽ 119910 for all 119910 isin 119883 Finally write

100381710038171003817100381711991010038171003817100381710038171= (1 minus 120576) 119902 (119910) + 120576

119901 (119910)

119901 (119890)(119910 isin 119884) (90)

Then sdot 1is a strictly convex norm on 119884 and

(1 minus 120576)2 10038171003817100381710038171199101003817100381710038171003817 ⩽100381710038171003817100381711991010038171003817100381710038171⩽ (1 + 120576)

10038171003817100381710038171199101003817100381710038171003817 (119910 isin 119884) (91)

Wewill finish the proof by showing that sdot 1is not uniformly

convex (actually it is not locally uniformly convex) Indeedfor each 119899 isin N we select 119910

119899isin ⋂

119899

119896=1ker 119890lowast

119896with 119910

119899 = 1 and

consider 119890119899= 119890 + (1205764)119910

119899 Then 119902(119890) = 1 119902(119890

119899) = 1 and

119902(119890+119890119899) = 2 At the same time 119901(119910

119899) rarr 0 so 119901(119890

119899) rarr 119901(119890)

and 119901(119890 + 119890119899) rarr 2119901(119890) Consequently

1198901 = 1100381710038171003817100381711989011989910038171003817100381710038171997888rarr 1

1003817100381710038171003817119890 + 11989011989910038171003817100381710038171997888rarr 2 (92)

but 119890 minus 1198901198991= (1205764)119910

1198991⩾ (1 minus 120576)

2(1205764) which means the

absence of local uniform convexity at 119890

Proof of Theorem 17 Let 119883 be an infinite-dimensional sepa-rable Banach space Take a closed subspace 119884 of 119883 of codi-mension two By [28 Proposition 2] the map carrying everyequivalent norm on 119884 to its numerical index is continuousand so the set of values of the numerical index of 119884 upto reforming is a nontrivial interval [28 Theorem 9] ThenLemma 18 allows us to find an equivalent norm | sdot | on 119884 insuch a way that (119884 | sdot |) is strictly convex and is not uniformlyconvex and 119899(119884 | sdot |) gt 0 Now the space 119883 = ℓ(2)

1oplus1(119884 | sdot |)

is an equivalent renorming of 119883 which does not have theBPBp-119899119906 (indeed otherwise the pair (ℓ(2)

1 (119884 | sdot |)) would

have the BPBp for the operator norm and so (119884 | sdot |) wouldbe uniformly convex by [19 Corollary 33] a contradiction)Moreover as

119899 (119883) = min 119899 (ℓ(2)1) 119899 (119884 |sdot|) gt 0 (93)

(see [20 Proposition 2] for instance) 119883 also fails the weak-BPBp-nu by Proposition 6

To finish the section with the promised proof ofTheorem 15 we first see the following stability result

Lemma 19 Let119883 = [⨁infin

119896=1119883

119896]1198880

or [⨁infin

119896=1119883

119896]ℓ1

If119883 has theBishop-Phelps-Bollobas property for numerical radius with afunction 120578 then each Banach space 119883

119894has the Bishop-Phelps-

Bollobas property for numerical radius with 120578nu(119883119894) ⩾ 120578 That

is inf119894120578nu(119883119894

)(120576) ⩾ 120578nu(119883)(120576) for all 0 lt 120576 lt 1

Proof Let 119875119894 119883 rarr 119883

119894and 1198751015840

119894 119883

lowastrarr 119883

lowast

119894be the natural

projections and let 119876119894 119883

119894rarr 119883 and 1198761015840

119894 119883

lowast

119894rarr 119883

lowast be thenatural embeddings

Assume that an operator 119879119894 119883

119894rarr 119883

119894and a pair

(119909119894 119909

lowast

119894) isin Π(119883

119894) satisfy that

V (119879119894) = 1

1003816100381610038161003816119909lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (94)

We define an operator 119879 119883 rarr 119883 and (119909 119909lowast) isin Π(119883) by

119879 = 119876119894∘ 119879

119894∘ 119875

119894 (119909 119909

lowast) = (119876

119894119909119894 119876

1015840

119894119909lowast

119894) (95)

12 Abstract and Applied Analysis

then clearly we see that1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119909

lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (96)

From the assumption there exist 119878 119883 rarr 119883 and a pair(119910 119910

lowast) isin Π(119883) such that

1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 = 1 = V (119878) 119878 minus 119879 lt 120576

1003817100381710038171003817119910lowastminus 119909

lowast1003817100381710038171003817 lt 1205761003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576

(97)

Since this clearly shows that1003817100381710038171003817119875119894 ∘ 119878 ∘ 119876119894

minus 119879119894

1003817100381710038171003817 lt 120576100381710038171003817100381710038171198751015840

119894119910lowastminus 119909

lowast

119894

10038171003817100381710038171003817lt 120576

1003817100381710038171003817119875119894119910 minus 1199091198941003817100381710038171003817 lt 120576

(98)

we only need to show that |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = 1

We first show the case of 1198880sum Since

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817=10038171003817100381710038171003817119875119895119910 minus 119875

11989511990910038171003817100381710038171003817⩽1003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576 (99)

for every 119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽100381710038171003817100381710038171198751015840

119894119910lowast10038171003817100381710038171003817+ 120576 sum

119895isinN119895 = 119894

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817⩽1003817100381710038171003817119910

lowast1003817100381710038171003817 = 1

(100)

This shows that 1198751015840119894119910lowast = 1 and 1198751015840

119895119910lowast= 0 for every 119895 = 119894

So 119910lowast = 1198761015840

1198941198751015840

119894119910lowast and 1198751015840

119894119910lowast(119875

119894119910) = 1 This and the fact that

119910 minus 119876119894119875119894119910 lt 120576 imply that

(119876119894119875119894119910 + (

1

120576) (119910 minus 119876

119894119875119894119910) 119876

1015840

1198941198751015840

119894119910lowast) isin Π (119883) (101)

So we get that (1198761015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910 + (1120576)(119910 minus 119876

119894119875119894119910))V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =10038161003816100381610038161003816(119876

1015840

1198941198751015840

119894119910lowast) 11987811991010038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910 +1

120576(119910 minus 119876

119894119875119894119910))

1003816100381610038161003816100381610038161003816⩽ 1

(102)

and so we get |1198751015840119894119910lowast(119875

119894∘119878 ∘119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

We next show the case of ℓ1sum The proof is almost

the same as that of the 1198880case However for the sake of

completeness we provide it hereSince 1198751015840

119895119910lowast = 119875

1015840

119895119910lowastminus 119875

1015840

119895119909lowast ⩽ 119910

lowastminus 119909

lowast lt 120576 for every

119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽10038171003817100381710038171198751198941199101003817100381710038171003817 + 120576 sum

119895isinN119895 = 119894

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817⩽10038171003817100381710038171199101003817100381710038171003817 = 1

(103)

which shows 119875119894119910 = 1 and 119875

119895119910 = 0 for every 119895 = 119894 Since this

implies (119876119894119875119894119910119876

1015840

1198941198751015840

119894119910lowast+(1120576)(119910

lowastminus119876

1015840

1198941198751015840

119894119910lowast)) isin Π(119883) we get

that |(1198761015840

1198941198751015840

119894119910lowast+ (1120576)(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast))119878(119876

119894119875119894119910)| ⩽ V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =1003816100381610038161003816119910

lowast119878 (119876

119894119875119894119910)1003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast+1

120576(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast)) 119878 (119876

119894119875119894119910)

1003816100381610038161003816100381610038161003816⩽ 1

(104)

and so |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

Proof of Theorem 15 Note that 120578nu(1198711(120583)oplus1119883)(120576) rarr 0 as120576 rarr 0 Fix 0 lt 120576

0lt 1 and choose 0 lt 120576 lt 1 such that

6120576 + 120578nu(1198711(120583)oplus1119883)(120576) lt 1205760 Let 120578(1205760) = 120578nu(1198711(120583)oplus1119883)(120576)Suppose that 119879

0isin L(119871

1(120583) 119883) with 119879

0 = 1 and 119891

0isin

1198781198711(120583)

satisfy

1003817100381710038171003817119879011989101003817100381710038171003817 gt 1 minus 120578 (1205760) (105)

For any measurable subset 119861 let

1198711(120583|

119861) = 119891|

119861 119891 isin 119871

1(120583) (106)

with the norm 119891|119861 = 119891120594

1198611 Then it is easy to see

that 1198711(120583|

119861) is isometrically isomorphic to a complemented

subspace of 1198711(120583) Let 119875

119861 119871

1(120583) rarr 119871

1(120583|

119861) be the

restriction defined by 119875119861(119891) = 119891|

119861for all 119891 isin 119871

1(120583) and

let 119869119861 119871

1(120583|

119861) rarr 119871

1(120583) be the extension defined by

119869119861(119891)(120596) = 119891(120596) if 120596 isin 119861 and 119869

119861(119891)(120596) = 0 otherwise

It is clear that 119875119861119869119861= Id

1198711(120583|119861)and 119869

119861119875119861(119891) = 119891120594

119861for all

119891 isin 1198711(120583) Notice also that 119871

1(120583) is isometrically isomorphic

to 1198711(120583|

119861)oplus

11198711(120583|

119861119888)

Let 119860 = supp1198910and 119892

0= 119875

1198601198910 Then

1003817100381710038171003817119879011986911986011989201003817100381710038171003817 =100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) gt 0 (107)

and define the operator 119879119860 119871

1(120583|

119860) rarr 119883 by 119879

119860119891 =

1198790119869119860119891119879

0119869119860 for every 119891 isin 119871

1(120583|

119860) Then

100381710038171003817100381711987911986011989201003817100381710038171003817 ⩾100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) (108)

Since 120583|119860is 120590-finite 119871

1(120583|

119860)lowast= 119871

infin(120583|

119860) Let 119892lowast

0isin 119878

119871infin(120583|119860)

be a function such that ⟨119892lowast0 119892

0⟩ = 1 and choose 119909lowast

0isin 119878

119883lowast

such that 119909lowast0(119879

1198601198920) = 119879

1198601198920 Define the operator 119878

0isin

L(1198711(120583|

119860)oplus

1119883) by

1198780(119891 119909) = (0 119879

119860119891) ((119891 119909) isin 119871

1(120583|

119860) oplus

1119883) (109)

and observe that 1198780 = V(119878

0) = 1 Indeed

100381710038171003817100381711987801003817100381710038171003817 ⩽ 1 =

10038171003817100381710038171198791198601003817100381710038171003817

= sup 1003816100381610038161003816119909lowast1198791198601198911003816100381610038161003816 119909

lowastisin 119878

119883lowast 119891 isin 119878

1198711(120583|119860)

= sup 1003816100381610038161003816(119891lowast 119909

lowast) 119878

0(119891 119909)

1003816100381610038161003816

((119891lowast 119909

lowast) (119891 119909)) isin Π (119871

1(120583|

119860) oplus

1119883)

= V (1198780) ⩽100381710038171003817100381711987801003817100381710038171003817

(110)

Abstract and Applied Analysis 13

It is immediate that

(119892lowast

0 119909

lowast

0) 119878

0(119892

0 0) = 119909

lowast

0(119879

1198601198920) =10038171003817100381710038171198791198601198920

1003817100381710038171003817 gt 1 minus 120578 (1205760)

(111)

By Lemma 19 1198711(120583|

119860)oplus

1119883 has the BPBp-nu with the func-

tion 120578 Therefore there exist 1198781isin L(119871

1(120583|

119860)oplus

1119883) (119892

1 119909

1) isin

1198781198711(120583|119860)oplus1119883 and (119892lowast

1 119909

lowast

1) isin 119878

119871infin(120583|119860)oplusinfin119883lowast such that

1003817100381710038171003817(1198921 1199091) minus (1198920 0)1003817100381710038171003817 lt 120576

1003817100381710038171003817(119892lowast

1 119909

lowast

1) minus (119891

lowast

0 119909

lowast

0)1003817100381710038171003817 lt 120576

10038171003817100381710038171198781 minus 11987801003817100381710038171003817 lt 120576 ⟨(119892

lowast

1 119909

lowast

1) (119892

1 119909

1)⟩ = 1

1003816100381610038161003816(119892lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 = V (119878

1) = 1

(112)

Claim 1 We claim that 1199091= 0

Otherwise

1 = Re ⟨(119892lowast1 119909

lowast

1) (119892

1 119909

1)⟩

=100381710038171003817100381711989211003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

+100381710038171003817100381711990911003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

⟩ ⩽ 1

(113)

We deduce that

((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

(119892lowast

1 119909

lowast

1)) (

(0 1199091)

100381710038171003817100381711990911003817100381710038171003817

(119892lowast

1 119909

lowast

1))

isin Π (1198711(120583|

119860) oplus

1119883)

(114)

Since 1198781((0 119909

1)119909

1) = (119878

1minus 119878

0)((0 119909

1)119909

1) lt 120576 we

get that

1 =1003816100381610038161003816⟨(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)⟩1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

100381710038171003817100381711989211003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

)⟩

+100381710038171003817100381711990911003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

)⟩

100381610038161003816100381610038161003816100381610038161003816

⩽100381710038171003817100381711989211003817100381710038171003817 V (1198781) + 120576

100381710038171003817100381711990911003817100381710038171003817 lt100381710038171003817100381711989211003817100381710038171003817 +100381710038171003817100381711990911003817100381710038171003817 = 1

(115)

a contradiction This proves the claimWe define the operator 119878

2 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883

by 1198782(119891 119909) = 119878

1(119891 0) for every 119891 isin 119871

1(120583|

119860) and for every

119909 isin 119883 Then we have

V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 = 1

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ 120576 (116)

Indeed from Claim 1 we have

V (1198781) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 ⩽ V (119878

2)

(117)

On the other hand we see that1003816100381610038161003816(119891

lowast 119909

lowast) 119878

2(119891 119909)

1003816100381610038161003816

=1003816100381610038161003816(119891

lowast 119909

lowast) 119878

1(119891 0)

1003816100381610038161003816 ⩽10038171003817100381710038171198911003817100381710038171003817 V (1198781) ⩽ V (119878

1)

(118)

for every ((119891lowast 119909lowast) (119891 119909)) isin Π(1198711(120583|

119860)oplus

1119883) Therefore

V(1198782) ⩽ V(119878

1) Also

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ sup

119909isin119878119883

10038171003817100381710038171198781 (0 119909)1003817100381710038171003817

= sup119909isin119878119883

10038171003817100381710038171198781 (0 119909) minus 1198780 (0 119909)1003817100381710038171003817 ⩽ 120576

(119)

Claim 2 There exists 1198783 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883 such

that 1198783(119892

1 0) = 119878

3 = 1 119878

3(0 119909) = 0 119878

3(119891 119909) isin 0oplus

1119883

for every (119891 119909) isin 1198711(120583|

119860)oplus

1119883 and 119878

3minus 119878

2 lt 4120576

Indeed write 1198781= (119863

1 119863

2) where 119863

1 119871

1(120583|

119860)oplus

1119883 rarr

1198711(120583|

119860) and119863

2 119871

1(120583|

119860)oplus

1119883 rarr 119883 We have that

sup 1003816100381610038161003816119892lowast119863

1(119892

1 0) + 119909

lowast119863

2(119892

1 0)1003816100381610038161003816

119909lowastisin 119878

119883lowast ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860) +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⩽ V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119892

lowast

1119863

1(119892

1 0) + 119909

lowast

1119863

2(119892

1 0)1003816100381610038161003816

(120)

This implies that

1003816100381610038161003816119909lowast

1119863

2(119892

1 0)1003816100381610038161003816 =10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

1003816100381610038161003816119892lowast

1119863

1(119892

1 0)1003816100381610038161003816

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119871

infin(120583|

119860)

(121)

Therefore |119892lowast1| equals 1 on the support of 119863

1(119892

1 0) As

|⟨119892lowast

1 119892

1⟩| = 1 we also have that |119892lowast

1| equals 1 on the support

of 1198921 Changing the values of 119892lowast

1by the ones of 119891lowast

0on 119860

(supp(1198631(119892

1 0)) cup supp(g

1)) we may and do suppose that

|119892lowast

1| = 1 on the whole 119860We also have 119863

2(119892

1 0) gt 0 Indeed

10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)1003817100381710038171003817

⩽10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198921 0)

1003817100381710038171003817 +10038171003817100381710038171198780 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817

lt 2120576 + 120576 = 3120576

(122)

So we have

10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

⩽10038171003817100381710038171198631(119892

1 0)1003817100381710038171003817 +10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

14 Abstract and Applied Analysis

=1003817100381710038171003817(1198631

(1198921 0) 119863

2(119892

1 0)) minus (0 119879

1198601198920)1003817100381710038171003817

=10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817 lt 3120576

(123)

and 1198632(119892

1 0) gt 119879

1198601198920 minus 3120576 ⩾ 1 minus 120578(120576

0) minus 3120576 gt 0

Finally define the operator 1198783by

1198783(119891 119909) = (0119863

2(119891 0) + 119892

lowast

1(119863

1(119891 0))

1198632(119892

1 0)

119909lowast

1119863

2(119892

1 0))

(124)

for (119891 119909) isin 1198711(120583|

119860)oplus

1119883 It is clear that

100381710038171003817100381711987831003817100381710038171003817 ⩽ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) (125)

Notice also that

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817

⩽10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 +10038171003817100381710038171198632(119891 0) minus 119879

1198601198911003817100381710038171003817

=1003817100381710038171003817(1198631

(119891 0) 1198632(119891 0)) minus (0 119879

119860119891)1003817100381710038171003817

=10038171003817100381710038171198782 (119891 119909) minus 1198780 (119891 119909)

1003817100381710038171003817

(126)

for all (119891 119909) isin 1198711(120583|

119860)oplus

1119883 Hence we have

10038171003817100381710038171198783 minus 11987821003817100381710038171003817

= 2 sup119891isin1198781198711(120583|119860)

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 ⩽ 210038171003817100381710038171198782 minus 1198780

1003817100381710038171003817 lt 4120576(127)

On the other hand let 119866 1198711(120583|

119860) rarr 119871

1(120583|

119860) be defined by

119866(119891) = 119892lowast

1119891 for every 119891 isin 119871

1(120583|

119860) Then we have

V (1198782)

= sup 1003816100381610038161003816119911lowast11987821199111003816100381610038161003816 (119911 119911

lowast) isin Π (119871

1(120583|

119860) oplus

1119883)

⩾ sup10038161003816100381610038161003816100381610038161003816

119909lowast119863

2(119866(

1

120583 (119862)120594119862) 0)

+119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119909lowastisin 119878

119883lowast 119862 isin Σ

119860 120583 (119862) gt 0

= sup10038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119862)120594119862) 0)

10038171003817100381710038171003817100381710038171003817

+

10038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119862 isin Σ119860 120583 (119862) gt 0

(128)

where Σ119860is the family of measurable subsets of 119860

Hence for any simple function 119904 = sum119899

119894=1((120572

119894120583(119860

119894))120594

119860119894

) isin

1198781198711(120583|119860) where 119860

119894119894is a family of disjoint measurable subsets

with strictly positive measure we have

V (1198782)

119899

sum

119894=1

10038161003816100381610038161205721198941003816100381610038161003816 (

100381710038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381710038171003817100381710038171003817100381710038171003817

+

100381610038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381610038161003816100381610038161003816100381610038161003816

)

⩾10038171003817100381710038171198632(119866 (119904) 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119866 (119904) 0)

1003816100381610038161003816

(129)

Since |119892lowast1| = 1119866 is an isometric isomorphism so for each119891 isin

1198781198711(120583|119860)there exists a sequence of norm-one simple functions

(119904119896) such that 119866(119904

119896) converges to 119891 Therefore

V (1198782) ⩾ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) ⩾100381710038171003817100381711987831003817100381710038171003817

(130)

On the other hand we see that100381710038171003817100381711987831003817100381710038171003817 ⩾1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

3(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119909

lowast

1119863

2(119892

1 0) + 119892

lowast

1119863

1(119892

1 0)1003816100381610038161003816 = V (119878

2) = 1

(131)

Therefore 1 = 1198783 = 119878

3(119892

1 0) which proves Claim 2

Finally set 1198783= (0 ) for a suitable 119871

1(120583|

119860)oplus

1119883 rarr

119883 and define the operator 1198791 119871

1(120583) rarr 119883 by

1198791(119891) = 119879

0(119891120594

119860119888) + (119875

119860119891 0) (132)

for every 119891 isin 1198711(120583) Then we have

10038171003817100381710038171198791 (119891)1003817100381710038171003817 ⩽100381710038171003817100381711987901003817100381710038171003817

10038171003817100381710038171198911205941198601198881003817100381710038171003817 +1003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171198911205941198601003817100381710038171003817 =10038171003817100381710038171198911003817100381710038171003817 (133)

for every 119891 isin 1198711(120583) so 119879

1 ⩽ 1 Also

10038171003817100381710038171198791 (1198691198601198921)1003817100381710038171003817 =10038171003817100381710038171198783 (1198921 0)

1003817100381710038171003817 =100381710038171003817100381711987831003817100381710038171003817 = 1 (134)

so 1198791attains its norm at 119869

1198601198921isin 119871

1(120583) and

10038171003817100381710038171198691198601198921 minus 11989101003817100381710038171003817 =10038171003817100381710038171198921 minus 1198920

1003817100381710038171003817 lt 120576 (135)

We also have that for any 119891 isin 1198781198711(120583)

10038171003817100381710038171198790 (119891) minus 1198791 (119891)1003817100381710038171003817 =100381710038171003817100381710038171198790(119891120594

119860) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790 (119869119860119875119860119891) minus 119879119860 (119875119860119891)

1003817100381710038171003817

+10038171003817100381710038171003817119879119860(119875

119860119891) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790119869119860 minus 119879119860

1003817100381710038171003817 +10038171003817100381710038171198780 minus 1198783

1003817100381710038171003817

lt 120578 (1205760) + 6120576

(136)

Hence 1198790minus 119879

1 ⩽ 120578(120576

0) + 6120576 lt 120576

0

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Abstract and Applied Analysis 15

Acknowledgments

The authors thank Gilles Godefroy and Rafael Paya forfruitful conversations about the content of this paper andVladimir Kadets for providing Lemma 18 They also appre-ciate anonymous referees for careful reading and fruitfulsuggestions about revision The second author was sup-ported by Basic Science Research Program through theNational Research Foundation of Korea (NRF) fundedby the Ministry of Education Science and Technology(NRF-2012R1A1A1006869) The third author was partiallysupported by Spanish MICINN and FEDER project noMTM2012-31755 and by Junta de Andalucıa and FEDERGrants FQM-185 and P09-FQM-4911

References

[1] F F Bonsall and J Duncan Numerical Ranges of Operators onNormed Spaces and of Elements of Normed Algebras vol 2 ofLondon Mathematical Society Lecture Note Series 2 CambridgeUniversity Press London UK 1971

[2] F F Bonsall and J Duncan Numerical Ranges II CambridgeUniversity Press London UK 1973 London MathematicalSociety Lecture Notes Series 10

[3] B Sims On numerical range and its applications to Banachalgebras [PhD thesis] University of Newcastle New SouthWales Australia 1972

[4] I D Berg and B Sims ldquoDenseness of operators which attaintheir numerical radiusrdquoAustralian Mathematical Society A vol36 no 1 pp 130ndash133 1984

[5] C S Cardassi ldquoDensity of numerical radius attaining operatorson some reflexive spacesrdquo Bulletin of the Australian Mathemat-ical Society vol 31 no 1 pp 1ndash3 1985

[6] C S Cardassi ldquoNumerical radius attaining operatorsrdquo inBanach Spaces (Columbia Mo 1984) vol 1166 of Lecture Notesin Mathematics pp 11ndash14 Springer Berlin Germany 1985

[7] C S Cardassi ldquoNumerical radius-attaining operators on119862(119870)rdquoProceedings of the AmericanMathematical Society vol 95 no 4pp 537ndash543 1985

[8] MDAcostaOperadores que alcanzan su radio numerico [PhDthesis] Universidad de Granada Granada Spain 1990

[9] M D Acosta and R Paya ldquoNumerical radius attaining oper-ators and the Radon-Nikodym propertyrdquo The Bulletin of theLondon Mathematical Society vol 25 no 1 pp 67ndash73 1993

[10] R Paya ldquoA counterexample on numerical radius attainingoperatorsrdquo Israel Journal of Mathematics vol 79 no 1 pp 83ndash101 1992

[11] Y S Choi and S G Kim ldquoNorm or numerical radius attainingmultilinear mappings and polynomialsrdquo Journal of the LondonMathematical Society vol 54 no 1 pp 135ndash147 1996

[12] M D Acosta J Becerra Guerrero and M Ruiz GalanldquoNumerical-radius-attaining polynomialsrdquo The Quarterly Jour-nal of Mathematics vol 54 no 1 pp 1ndash10 2003

[13] M D Acosta and S G Kim ldquoDenseness of holomorphicfunctions attaining their numerical radiirdquo Israel Journal ofMathematics vol 161 pp 373ndash386 2007

[14] S G Kim and H J Lee ldquoNumerical peak holomorphicfunctions on Banach spacesrdquo Journal of Mathematical Analysisand Applications vol 364 no 2 pp 437ndash452 2010

[15] M D Acosta R M Aron D Garcıa and M MaestreldquoThe Bishop-Phelps-Bollobas theorem for operatorsrdquo Journal ofFunctional Analysis vol 254 no 11 pp 2780ndash2799 2008

[16] A J Guirao and O Kozhushkina ldquoThe Bishop-Phelps-Bollobasproperty for numerical radius in ℓ

1(119862)rdquo Studia Mathematica

vol 218 no 1 pp 41ndash54 2013[17] J Falco ldquoThe Bishop-Phelps-Bollobas property for numerical

radius on 1198711rdquo Journal of Mathematical Analysis and Applica-

tions vol 414 no 1 pp 125ndash133 2014[18] A Aviles A J Guirao and J Rodriguez ldquoOn the Bishop-

Phelps-Bollobas property for numerical radius in119862(119870)-spacesrdquoPreprint

[19] RM Aron Y S Choi S K Kim H J Lee andMMartin ldquoTheBishop-Phelps-Bollobas version of Lindenstrauss properties Aand Brdquo Preprint

[20] V Kadets M Martın and R Paya ldquoRecent progress and openquestions on the numerical index of Banach spacesrdquo Revista dela Real Academia de Ciencias Exactas Fısicas y Naturales A vol100 no 1-2 pp 155ndash182 2006

[21] M Martın J Merı and M Popov ldquoOn the numerical index ofreal 119871

119901(120583)-spacesrdquo Israel Journal of Mathematics vol 184 pp

183ndash192 2011[22] A Iwanik ldquoNorm attaining operators on Lebesgue spacesrdquo

Pacific Journal of Mathematics vol 83 no 2 pp 381ndash386 1979[23] HH SchaeferBanach Lattices and Positive Operators Springer

New York NY USA 1974[24] J Duncan C M McGregor J D Pryce and A J White ldquoThe

numerical index of a normed spacerdquo Journal of the LondonMathematical Society Second Series vol 2 pp 481ndash488 1970

[25] Y S Choi S K Kim H J Lee and M Martin ldquoThe Bishop-Phelps-Bollobas theorem for operators on 119871

1(120583)rdquo Preprint

[26] H E Lacey The Isometric Theory of Classical Banach SpacesSpringer New York NY USA 1974

[27] G Godefroy ldquoUniversal spaces for strictly convex Banachspacesrdquo Revista de la Real Academia de Ciencias Exactas Fısicasy Naturales A vol 100 no 1-2 pp 137ndash146 2006

[28] C Finet M Martın and R Paya ldquoNumerical index andrenormingrdquo Proceedings of the American Mathematical Societyvol 131 no 3 pp 871ndash877 2003

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 3: Research Article On the Bishop-Phelps-Bollobás Property ...downloads.hindawi.com/journals/aaa/2014/479208.pdf · Research Article On the Bishop-Phelps-Bollobás Property for Numerical

Abstract and Applied Analysis 3

We may also give the following easy result concerningduality

Proposition 3 Let119883 be a reflexive space Then

120578nu (119883) (120576) = 120578nu (119883lowast) (120576) (11)

for every 120576 isin (0 1) In particular 119883 has the BPBp-nu if andonly if 119883lowast has the BPBp-nu

Wewill use that V(119879lowast) = V(119879) for all 119879 isinL(119883) where 119879lowastdenotes the adjoint operator of 119879 This result can be found in[1] but it is obvious if119883 is reflexive

Proof By reflexivity it is enough to show that 120578nu(119883)(120576) ⩽120578nu(119883

lowast)(120576) Let 120576 isin (0 1) be fixed If 120578nu(119883)(120576) = 0 there is

nothing to prove Otherwise consider 0 lt 120578 lt 120578nu(119883)(120576)Suppose that 119879

1isinL(119883lowast

) and (119909lowast1 119909

1) isin Π(119883

lowast) satisfy

V (1198791) = 1

100381610038161003816100381611990911198791119909lowast

1

1003816100381610038161003816 gt V (1198791) minus 120578 (12)

By considering 119879lowast1isin L(119883) we may find 119878

1isin L(119883) and

(1199101 119910

lowast

1) isin Π(119883) such that

1003816100381610038161003816119910lowast

111987811199101

1003816100381610038161003816 = V (1198781) = 1

10038171003817100381710038171199101 minus 11990911003817100381710038171003817 lt 120576

1003817100381710038171003817119910lowast

1minus 119909

lowast

1

1003817100381710038171003817 lt 1205761003817100381710038171003817119879

lowast

1minus 119878

1

1003817100381710038171003817 lt 120576

(13)

Then 119878lowast1isinL(119883lowast

) and (119910lowast1 119910

1) isin Π(119883

lowast) satisfy

1003816100381610038161003816⟨1199101 119878lowast

1119910lowast

1⟩1003816100381610038161003816 = V (119878

1) = 1

1003817100381710038171003817119910lowast

1minus 119909

lowast

1

1003817100381710038171003817 lt 120576

10038171003817100381710038171199101 minus 11990911003817100381710038171003817 lt 120576

10038171003817100381710038171198791 minus 119878lowast

1

1003817100381710038171003817 lt 120576

(14)

This implies that 120578nu(119883lowast)(120576) ⩾ 120578 We finish by just taking

supremum on 120578

We do not know whether the result above is valid in thenonreflexive case

3 Spaces Which Are Both Uniformly Convexand Uniformly Smooth

For a Banach space which is both uniformly convex anduniformly smooth we get a property which is weaker thanBPBp-nu This result was known to A Guirao (privatecommunication)

Proposition 4 Let 119883 be a uniformly convex and uniformlysmooth Banach space Then given 120576 gt 0 there exists 120578(120576) gt 0such that whenever 119879

0isinL(119883) with V(119879

0) = 1 and (119909

0 119909

lowast

0) isin

Π(119883) satisfy |119909lowast011987901199090| gt 1 minus 120578(120576) there exist 119878 isin L(119883) and

(119910 119910lowast) isin Π(119883) such that

V (119878) = 1003816100381610038161003816119910lowast1198781199101003816100381610038161003816

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowastminus 119910

lowast1003817100381710038171003817 lt 1205761003817100381710038171003817119878 minus 1198790

1003817100381710038171003817 lt 120576

(15)

Proof Notice that the uniform smoothness of119883 is equivalentto the uniform convexity of 119883lowast Let 120575

119883(120576) and 120575

119883lowast(120576) be the

moduli of convexity119883 and119883lowast respectively Given 0 lt 120576 lt 1consider

120578 (120576) =120576

4min 120575

119883(120576

4) 120575

119883lowast (120576

4) gt 0 (16)

Consider 1198790isin L(119883) with V(119879

0) = 1 and (119909

0 119909

lowast

0) isin Π(119883)

satisfying |119909lowast011987901199090| gt 1 minus 120578(120576) Define 119879

1isinL(119883) by

1198791119909 = 119879

0119909 + 120582

1(120576

4) 119909

lowast

0(119909) 119909

0(17)

for all 119909 isin 119883 where 1205821is the scalar satisfying |120582

1| = 1 and

|119909lowast

011987901199090+ 120582

1(1205764)| = |119909

lowast

011987901199090| + 1205764 Now choose 119909

1isin 119878

119883

and 119909lowast1isin 119878

119883lowast such that |119909lowast

1(119909

1)| = 1 119909lowast

1(119909

0) = |119909

lowast

1(119909

0)| and

1003816100381610038161003816119909lowast

111987911199091

1003816100381610038161003816 ⩾ V (1198791) minus 120578(

1205762

42) (18)

Now we define a sequence (119909119899 119909

lowast

119899 119879

119899) in 119878

119883times 119878

119883lowast times L(119883)

inductively Indeed suppose that we have a defined sequence(119909

119895 119909

lowast

119895 119879

119895) for 0 ⩽ 119895 ⩽ 119899 and let

119879119899+1119909 = 119879

119899119909 + 120582

119899+1

120576119899+1

4119899+1119909lowast

119899(119909) 119909

119899 (19)

Then choose 119909119899+1

isin 119878119883

and 119909lowast119899+1

isin 119878119883lowast such that

|119909lowast

119899+1(119909

119899+1)| = 1 and |119909lowast

119899+1(119909

119899)| = 119909

lowast

119899+1(119909

119899)

1003816100381610038161003816119909lowast

119899+1119879119899+1119909119899+1

1003816100381610038161003816 ⩾ V (119879119899+1) minus 120578(

120576119899+2

4119899+2) (20)

Notice that for all 119899 ⩾ 0 we have

1003817100381710038171003817119879119899+1 minus 1198791198991003817100381710038171003817 ⩽120576119899+1

4119899+1

1003816100381610038161003816V (119879119899+1) minus V (119879119899)1003816100381610038161003816 ⩽120576119899+1

4119899+1

(21)

This implies that (119879119899) is a Cauchy sequence and assume that

it converges to 119878 isinL(119883) Then we have

lim119899119879119899= 119878

10038171003817100381710038171198790 minus 1198781003817100381710038171003817 lt 120576

lim119899

1003816100381610038161003816119909lowast

119899119879119899119909119899

1003816100381610038161003816 = lim119899V (119879

119899) = V (119878)

(22)

We will show that both sequences (119909119899) and (119909lowast

119899) are Cauchy

From the definition we have

V (119879119899+1) minus 120578(

120576119899+2

4119899+2)

⩽1003816100381610038161003816119909

lowast

119899+1119879119899+1119909119899+1

1003816100381610038161003816

4 Abstract and Applied Analysis

100381610038161003816100381610038161003816100381610038161003816

119909lowast

119899+1119879119899119909119899+1+ 120582

119899+1

120576119899+1

4119899+1119909lowast

119899(119909

119899+1) 119909

lowast

119899+1(119909

119899)

100381610038161003816100381610038161003816100381610038161003816

⩽ V (119879119899) +120576119899+1

4119899+1119909lowast

119899+1(119909

119899)

V (119879119899+1)

⩾1003816100381610038161003816119909

lowast

119899119879119899+1119909119899

1003816100381610038161003816 =

100381610038161003816100381610038161003816100381610038161003816

119909lowast

119899119879119899119909119899+ 120582

119899+1

120576119899+1

4119899+1

100381610038161003816100381610038161003816100381610038161003816

=1003816100381610038161003816119909

lowast

119899119879119899119909119899

1003816100381610038161003816 +120576119899+1

4119899+1⩾ V (119879

119899) minus 120578(

120576119899+1

4119899+1) +120576119899+1

4119899+1

(23)

In summary we have

V (119879119899) +120576119899+1

4119899+1119909lowast

119899+1(119909

119899)

⩾ V (119879119899) minus 120578(

120576119899+1

4119899+1) +120576119899+1

4119899+1minus 120578(

120576119899+2

4119899+2)

(24)

Hence

119909lowast

119899+1(119909

119899) ⩾ 1 minus 2

4119899+1

120576119899+1120578(120576119899+1

4119899+1)

= 1 minus1

2min120575

119883(120576119899+1

4119899+2) 120575

119883lowast (120576119899+1

4119899+2)

1003817100381710038171003817100381710038171003817

119909119899+ 119909

119899+1

2

1003817100381710038171003817100381710038171003817⩾ 119909

lowast

119899+1(119909119899+ 119909

119899+1

2) ⩾ 1 minus 120575

119883(120576119899+1

4119899+2)

100381710038171003817100381710038171003817100381710038171003817

119909lowast

119899+ 119909

lowast

119899+1

2

100381710038171003817100381710038171003817100381710038171003817

⩾119909lowast

119899+ 119909

lowast

119899+1

2(119909

119899) ⩾ 1 minus 120575

119883lowast (120576119899+1

4119899+2)

(25)

This means that

1003817100381710038171003817119909119899 minus 119909119899+11003817100381710038171003817 ⩽ 120576

119899+14

119899+2

1003817100381710038171003817119909lowast

119899minus 119909

lowast

119899+1

1003817100381710038171003817 ⩽ 120576119899+14

119899+2

(26)

for all 119899 So (119909119899) and (119909lowast

119899) are Cauchy Let 119909

infin= lim

119899119909119899and

119909lowast

infin= lim

119899119909lowast

119899 Then 119909

0minus 119909

infin lt 1205764 and 119909lowast

0minus 119909

lowast

infin lt 1205764

Hence |119909lowastinfin(119909

infin)| = lim

119899|119909

lowast

119899(119909

119899)| = 1 and

V (119878) = lim119899V (119879

119899) = lim

119899

1003816100381610038161003816119909lowast

119899119879119899119909119899

1003816100381610038161003816 =1003816100381610038161003816119909

lowast

infin119878119909

infin

1003816100381610038161003816 (27)

Let 120572 = 119909lowastinfin(119909

infin) 119910lowast = 119909lowast

infin and 119910 = 119909

infin Then we have

119910lowast(119910) = 1 V(119878) = |119910lowast119878119910| and 119910 minus 119909

0 lt 120576 Notice that

|120572 minus 1| =1003816100381610038161003816119909

lowast

infin(119909

infin) minus 119909

lowast

0(119909

0)1003816100381610038161003816

⩽1003816100381610038161003816(119909

lowast

infinminus 119909

lowast

0) (119909

infin)1003816100381610038161003816 +1003816100381610038161003816119909

lowast

0(119909

infin) minus 119909

lowast

0(119909

0)1003816100381610038161003816 lt120576

2

(28)

Therefore

1003817100381710038171003817119910lowastminus 119909

lowast1003817100381710038171003817 ⩽1003817100381710038171003817119910

lowastminus 119910

lowast1003817100381710038171003817 +1003817100381710038171003817119910

lowastminus 119909

lowast1003817100381710038171003817 lt120576

2+120576

4lt 120576 (29)

This completes the proof

Let us discuss a little bit about the equivalence betweenthe property in the result above and the BPBp-nu Forconvenience let us introduce the following definition

Definition 5 A Banach space 119883 has the weak Bishop-Phelps-Bollobas property for the numerical radius (in short weak-BPBp-nu) if given 120576 gt 0 there exists 120578(120576) gt 0 such thatwhenever 119879

0isin L(119883) with V(119879

0) = 1 and (119909

0 119909

lowast

0) isin Π(119883)

satisfy |119909lowast011987901199090| gt 1 minus 120578(120576) there exist 119878 isin L(119883) and

(119910 119910lowast) isin Π(119883) such that

V (119878) = 1003816100381610038161003816119910lowast1198781199101003816100381610038161003816

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowastminus 119910

lowast1003817100381710038171003817 lt 120576 119878 minus 119879 lt 120576

(30)

Notice that the only difference between this concept andthe BPBp-nu is the normalization of the operator 119878 by thenumerical radius Of course if the numerical radius andthe operator norm are equivalent these two properties arethe same This equivalence is measured by the so-callednumerical index of the Banach space as follows For a Banachspace119883 the numerical index of119883 is defined by

119899 (119883) = inf V (119879) 119879 isinL (119883) 119879 = 1 (31)

It is clear that 0 ⩽ 119899(119883) ⩽ 1 and 119899(119883)119879 ⩽ V(119879) ⩽ 119879for all 119879 isin L(119883) The value 119899(119883) = 1 means that V equalsthe usual operator norm This is the case of 119883 = 119871

1(120583) and

119883 = 119862(119870) among many others On the other hand 119899(119883) gt 0if and only if the numerical radius is equivalent to the normof L(119883) We refer the reader to [20] for more informationand background

The following result is immediate We include a proof forthe sake of completeness

Proposition 6 Let119883 be a Banach space with 119899(119883) gt 0 Then119883 has the BPBp-nu if and only if119883 has the weak-BPBp-nu

Proof Thenecessity is clear For the converse assume that wehave 120578(120576) gt 0 satisfying the conditions of the weak-BPBp-nufor all 0 lt 120576 lt 1 If 119879 isin L(119883) with V(119879) = 1 and (119909

0 119909

lowast

0) isin

Π(119883) satisfy |119909lowast0119879119909

0| gt 1 minus 120578(120576) for 0 lt 120576 lt 1 then there exist

119878 isinL(119883) and (119910 119910lowast) isin Π(119883) such that

V (119878) = 1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 119878 minus 119879 lt 120576

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowastminus 119910

lowast1003817100381710038171003817 lt 120576

(32)

As V(119878) gt 0 by the above let 1198781= (1V(119878))119878 Then we have

1 = V (1198781) =1003816100381610038161003816119910

lowast11987811199101003816100381610038161003816

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowastminus 119910

lowast1003817100381710038171003817 lt 120576

(33)

Abstract and Applied Analysis 5

Finally we have

10038171003817100381710038171198781 minus 1198791003817100381710038171003817 ⩽

10038171003817100381710038171003817100381710038171003817

1

V (119878)119878 minus 119878

10038171003817100381710038171003817100381710038171003817

+ 119878 minus 119879

=119878

V (119878)|V (119878) minus 1| + 119878 minus 119879

⩽1

119899 (119883)|V (119878) minus V (119879)| + 119878 minus 119879

⩽ (1

119899 (119883)+ 1) 119878 minus 119879 lt

119899 (119883) + 1

119899 (119883)120576

(34)

An obvious change of parameters finishes the proof

We do not know whether the hypothesis of 119899(119883) gt 0 canbe omitted in the above result

Putting together Propositions 4 and 6 we get the follow-ing

Corollary 7 Let 119883 be a uniformly convex and uniformlysmooth Banach space with 119899(119883) gt 0 Then 119883 has the BPBp-nu

Let us comment that every complex Banach space 119883satisfies 119899(119883) ⩾ 1119890 so the above corollary automaticallyapplies in the complex case In the real case this is no longertrue as the numerical index of a Hilbert space of dimensiongreater than or equal to two is 0 On the other hand it isproved in [21] that real 119871

119901(120583) spaces have nonzero numerical

index for every measure 120583when 119901 = 2Therefore we have thefollowing examples

Example 8 (a) Complex Banach spaces which are uniformlysmooth and uniformly convex satisfy the BPBp-nu

(b) In particular for every measure 120583 the complex spaces119871119901(120583) have the BPBp-nu for 1 lt 119901 lt infin(c) For every measure 120583 the real spaces 119871

119901(120583) have the

BPBp-nu for 1 lt 119901 lt infin 119901 = 2

Note Added in Revision Very recently H J Lee M Martınand J Merı have proved that Proposition 6 can be extendedto some Banach spaces with numerical index zero as forinstance real Hilbert spaces Hence they have shown thatHilbert spaces have the BPBp-nu These results will appearelsewhere

4 L1

Spaces

In this section we will show that 1198711(120583) has the BPBp-nu for

every measure 120583 In the proof we are dealing with complexintegrable functions since the real case is followed easilyby applying the same proof Our main result here is thefollowing

Theorem 9 Let 120583 be a measure Then 1198711(120583) has the Bishop-

Phelps-Bollobas property for numerical radius More preciselygiven 120576 gt 0 there exists 120578(120576) gt 0 (which does not depend on120583) such that whenever 119879

0isin L(119871

1(120583)) with V(119879

0) = 1 and

(1198910 119892

0) isin Π(119871

1(120583)) satisfy |⟨119879

01198910 119892

0⟩| gt 1 minus 120578(120576) then there

exist 119879 isinL(1198711(120583)) (119891

1 119892

1) isin Π(119871

1(120583)) such that

1003816100381610038161003816⟨1198791198911 1198921⟩1003816100381610038161003816 = V (119879) = 1 10038171003817100381710038171198910 minus 1198911

1003817100381710038171003817 lt 120576

10038171003817100381710038171198920 minus 11989211003817100381710038171003817 lt 120576

1003817100381710038171003817119879 minus 11987901003817100381710038171003817 lt 120576

(35)

As a first step we have to start dealing with finite regularpositive Borel measures for which a representation theoremfor operators exists

Proposition 10 Let 119898 be a finite regular positive Borelmeasure on a compact Hausdorff spaceΩ Then 119871

1(119898) has the

Bishop-Phelps-Bollobas property for numerical radius Moreprecisely given 120576 gt 0 there is 120578(120576) gt 0 (which is independent ofthemeasure119898) such that if a norm-one element119879 inL(119871

1(119898))

and (1198910 119892

0) isin Π(119871

1(119898)) satisfy |⟨119879119891

0 119892

0⟩| gt 1 minus 120578(120576) then

there exist an operator 119878 isin L(1198711(119898)) (119891

1 119892

1) isin Π(119871

1(119898))

such that1003816100381610038161003816⟨1198781198911 1198921⟩

1003816100381610038161003816 = 119878 = 110038171003817100381710038171198910 minus 1198911

1003817100381710038171003817 ⩽ 120576

10038171003817100381710038171198920 minus 11989211003817100381710038171003817 ⩽ 120576 119879 minus 119878 ⩽ 120576

(36)

To prove this proposition we need some backgroundon representation of operators on Lebesgue spaces on finiteregular positive Borel measures and several preliminarylemmas

Let 119898 be a finite regular positive Borel measure on acompact Hausdorff space Ω If 120583 is a complex-valued Borelmeasure on the product space Ω times Ω then define theirmarginal measures 120583119894 on Ω (119894 = 1 2) as follows 1205831(119860) =120583(119860 times Ω) and 1205832(119861) = 120583(Ω times 119861) where 119860 and 119861 are Borelmeasurable subsets of Ω

Let 119872(119898) be the complex Banach lattice of measuresconsisting of all complex-valued Borel measures 120583 on theproduct spaceΩ timesΩ such that |120583|119894 are absolutely continuouswith respect to119898 for 119894 = 1 2 endowed with the norm

1003817100381710038171003817100381710038171003817100381710038171003817

11988910038161003816100381610038161205831003816100381610038161003816

1

119889119898

1003817100381710038171003817100381710038171003817100381710038171003817infin

(37)

Each 120583 isin 119872(119898) defines a bounded linear operator 119879120583from

1198711(119898) to itself by

⟨119879120583(119891) 119892⟩ = int

ΩtimesΩ

119891 (119909) 119892 (119910) 119889120583 (119909 119910) (38)

where 119891 isin 1198711(119898) and 119892 isin 119871

infin(119898) Iwanik [22] showed that

the mapping 120583 997891rarr 119879120583is a lattice isometric isomorphism from

119872(119898) onto L(1198711(119898)) Even though he showed this for the

real case it can be easily generalized to the complex caseFor details see [22 Theorem 1] and [23 IV Theorem 15(ii)Corollary 2]

We will also use that given an arbitrary measure 120583 every119879 isin L(119871

1(120583)) satisfies V(119879) = 119879 [24] (that is the space

1198711(120583) has numerical index 1)

Lemma 11 (see [15 Lemma 33]) Let 119888119899 be a sequence of

complex numbers with |119888119899| ⩽ 1 for every 119899 and let 120578 gt 0 such

6 Abstract and Applied Analysis

that for a convex series sum120572119899 Resuminfin

119899=1120572119899119888119899gt 1 minus 120578 Then for

every 0 lt 119903 lt 1 the set 119860 = 119894 isin N Re 119888119894gt 119903 satisfies the

estimate

sum

119894isin119860

120572119894⩾ 1 minus

120578

1 minus 119903 (39)

From now on 119898 will be a finite regular positive Borelmeasure on the compact Hausdorff space Ω

Lemma 12 Suppose that there exist a nonnegative simplefunction 119891 isin 119878

1198711(119898)

and a function 119892 isin 119878119871infin(119898)

such that

Re ⟨119891 119892⟩ gt 1 minus 1205763

16 (40)

Then there exist a nonnegative simple function 1198911isin 119878

1198711(119898)

anda function 119892

1isin 119878

119871infin(119898)

such that

1198921(119909) = 120594supp(119891

1)(119909) + 119892 (119909) 120594

Ωsupp(1198911)(119909)

⟨1198911 119892

1⟩ = 1

1003817100381710038171003817119891 minus 119891110038171003817100381710038171lt 120576

1003817100381710038171003817119892 minus 11989211003817100381710038171003817infinlt radic120576 supp (119891

1) sub supp (119891)

(41)

Proof Let119891 = sum119898

119895=1(120573

119895119898(119861

119895))120594

119861119895

for some (120573119895) such that

120573119895⩾ 0 for all 119895 andsum119898

119895=1120573119895= 1 and119861

119895rsquos aremutually disjoint

By the assumption we have

Re ⟨119891 119892⟩ =119899

sum

119895=1

120573119895

1

119898 (119861119895)

int119861119895

Re119892 (119909) 119889119898 (119909) gt 1 minus 1205763

16

(42)

and letting

119869 = 119895 1 ⩽ 119895 ⩽ 1198991

119898 (119861119895)

int119861119895

Re119892 (119909) 119889119898 (119909) gt 1 minus 1205762

4

(43)

we have by Lemma 11

sum

119895isin119869

120573119895gt 1 minus

120576

4 (44)

For each 119895 isin 119869 we have

1 minus1205762

4lt

1

119898 (119861119895)

int119861119895

Re119892 (119909) 119889119898 (119909)

=1

119898 (119861119895)

int119861119895capRe119892⩽1minus120576

Re119892 (119909) 119889119898 (119909)

+ int119861119895capRe119892gt1minus120576

Re119892 (119909) 119889119898 (119909)

⩽1

119898 (119861119895)

((1 minus 120576)119898 (119861119895cap Re119892 ⩽ 1 minus 120576)

+ 119898 (119861119895cap Re119892 gt 1 minus 120576))

= 1 minus 120576

119898 (119861119895cap Re119892 ⩽ 1 minus 120576)

119898 (119861119895)

(45)

This implies that

119898(119861119895cap Re119892 ⩽ 1 minus 120576)

119898 (119861119895)

lt120576

4 (46)

Define 119861119895= 119861

119895cap Re119892 gt 1 minus 120576 for all 119895 isin 119869

1198911= (1sum

119895isin119869

120573119895)sum

119895isin119869

120573119895(120594

119861119895

119898 (119861119895)) (47)

and 1198921(119909) = 1 on supp(119891

1) and 119892

1(119909) = 119892(119909) elsewhereThen

it is clear that supp(1198911) sub supp(119891) 119892 minus 119892

1infinlt radic120576 and

⟨1198911 119892

1⟩ = 1 Finally we will show that 119891 minus 119891

1 lt 120576 Notice

first that10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

= 2sum

119895isin119869

120573119895

119898(119861119895 119861

119895)

119898 (119861119895)

lt120576

2

(48)

Hence1003817100381710038171003817119891 minus 1198911

1003817100381710038171003817

10038171003817100381710038171003817100381710038171003817100381710038171003817

1

sum119895isin119869120573119895

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus 119891

10038171003817100381710038171003817100381710038171003817100381710038171003817

1 minus sum119895isin119869120573119895

sum119895isin119869120573119895

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+120576

4

= (1 minussum

119895isin119869

120573119895) +

120576

2+120576

4

⩽120576

4+120576

2+120576

4= 120576

(49)

Lemma 13 (see [25 Lemma 33]) Suppose that 119879120583is a norm-

one element in L(1198711(119898)) for some 120583 isin 119872(119898) and there is

a nonnegative simple function 1198910such that 119891

0is a norm-one

element of 1198711(119898) and 119879

1205831198910 ⩾ 1 minus 120576

32

6 for some 0 lt 120576 lt 1

Abstract and Applied Analysis 7

Then there exist a norm-one bounded linear operator 119879] forsome ] isin 119872(119898119898) and a nonnegative simple function 119891

1

in 1198781198711(119898)

such that 119879120583minus 119879] ⩽ 120576 1198911 minus 1198910 ⩽ 3120576 and

(119889|]|1119889119898)(119909) = 1 for all 119909 isin supp(1198911)

Lemma 14 Suppose that 119879] isin L(1198711(119898)) is a norm-one

operator 119891 = sum119899

119894=1120573119894(120594

119861119894

119898(119861119894)) where 119898(119861

119895) gt 0 for all

1 ⩽ 119895 ⩽ 119899 and 119861119895119899

119895=1are mutually disjoint Borel subsets ofΩ

is a norm-one nonnegative simple function and 119892 is an elementof 119878

119871infin(119898)

such that

Re ⟨119892 119879]119891⟩ ⩾ 1 minus1205766

27(50)

for some 0 lt 120576 lt 1 and

119889|]|1

119889119898(119909) = 1 119892 (119909) = 1 (51)

for all x in the support of 119891Then there exist a nonnegative simple function 119891 isin 119878

1198711(119898)

a function119892 isin 119878

119871infin(119898)

and an operator119879] inL(1198711(119898) 1198711(119898))such that

⟨119892 119879]119891⟩ =1003817100381710038171003817119879]1003817100381710038171003817 = 1

1003817100381710038171003817119879] minus 119879]1003817100381710038171003817 ⩽ 2120576

10038171003817100381710038171003817119891 minus 119891

10038171003817100381710038171003817⩽ 3120576

1003817100381710038171003817119892 minus 1198921003817100381710038171003817 ⩽radic120576 ⟨119891 119892⟩ = 1

(52)

Proof Since

Re ⟨119892 119879]119891⟩ ⩾ 1 minus1205766

27 (53)

we have

1 minus1205766

27lt Re ⟨119892 119879]119891⟩

= intΩtimesΩ

119891 (119909)Re119892 (119910) 119889120583 (119909 119910)

=

119899

sum

119895=1

120573119895intΩtimesΩ

120594119861119895

(119909)

119898 (119861119895)

Re119892 (119910) 119889] (119909 119910)

(54)

Let

119869 = 119895 intΩtimesΩ

(

120594119861119895

(119909)

119898 (119861119895)

)Re119892 (119910) 119889] (119909 119910) gt 1 minus 1205763

26

(55)

Then from Lemma 11 we have sum119895isin119869120573119895gt 1 minus 120576

32 Let 119891

1=

sum119895isin119869120573119895(120594

119861119895

119898(119861119895)) where 120573

119895= 120573

119895(sum

119895isin119869120573119895) for all 119895 isin 119869

Then

10038171003817100381710038171198911 minus 1198911003817100381710038171003817 ⩽

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

(120573119895minus 120573

119895)

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+ sum

119895isin119869

120573119895⩽ 120576

3⩽ 120576 (56)

Note that there is a Borel measurable function ℎ on Ω times Ωsuch that 119889](119909 119910) = ℎ(119909 119910) 119889|]|(119909 119910) and |ℎ(119909 119910)| = 1 forall (119909 119910) isin Ω times Ω Let

119862 = (119909 119910) 1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 1

1003816100381610038161003816 ltradic120576

232 (57)

Define two measures ]119891and ]

119888as follows

]119891(119860) = ] (119860 119862) ]

119888(119860) = ] (119860 cap 119862) (58)

for every Borel subset 119860 ofΩ times Ω It is clear that

119889] = 119889]119891+ 119889]

119888 119889

10038161003816100381610038161003816]119891

10038161003816100381610038161003816= ℎ 119889]

119891

1198891003816100381610038161003816]1198881003816100381610038161003816 = ℎ 119889]119888 119889 |]| = 119889

10038161003816100381610038161003816]119891

10038161003816100381610038161003816+ 1198891003816100381610038161003816]1198881003816100381610038161003816

(59)

Since (119889|]|11198891198981)(119909) = 1 for all 119909 isin ⋃119899

119895=1119861119895 we have

1 =119889|]|1

1198891198981

(119909) =

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) +1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909) (60)

for all 119909 isin 119861 = ⋃119899

119895=1119861119895 and we deduce that |]|1(119861

119895) = 119898

1(119861

119895)

for all 1 ⩽ 119895 ⩽ 119899We claim that |]

119891|1(119861

119895)119898

1(119861

119895) ⩽ 120576

22

2 for all 119895 isin 119869Indeed if |119892(119910)ℎ(119909 119910) minus 1| ⩾ radic120576232 then

Re (119892 (119910) ℎ (119909 119910)) ⩽ 1 minus 12057624 (61)

So we have

1 minus1205763

26

⩽1

1198981(119861

119895)

ReintΩtimesΩ

120594119861119895(119909)119892 (119910) 119889] (119909 119910)

=1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 |]| (119909 119910)

=1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 10038161003816100381610038161003816]11989110038161003816100381610038161003816(119909 119910)

+1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 1003816100381610038161003816]1198881003816100381610038161003816 (119909 119910)

⩽1

1198981(119861

119895)

((1 minus120576

24)10038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

(119861119895) +1003816100381610038161003816]1198881003816100381610038161003816

1

(119861119895))

= 1 minus120576

24

10038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

(119861119895)

1198981(119861

119895)

(62)

This proves our claimWe also claim that for each 119895 isin 119869 there exists a Borel

subset 119861119895of 119861

119895such that

(1 minus120576

2)119898

1(119861

119895) ⩽ 119898

1(119861

119895) ⩽ 119898

1(119861

119895)

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) ⩽120576

2

(63)

8 Abstract and Applied Analysis

for all 119909 isin 119861119895 Indeed set 119861

119895= 119861

119895cap 119909 isin Ω (119889|]

119891|1119889119898

1)

(119909) ⩽ 1205762 Then

int119861119895119861119895

120576

2119889119898

1(119909) ⩽ int

119861119895

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) 1198891198981(119909)

=10038161003816100381610038161003816]1119891

10038161003816100381610038161003816(119861

119895) ⩽1205762

22119898

1(119861

119895)

(64)

This shows that 1198981(119861

119895 119861

119895) ⩽ (1205762)119898

1(119861

119895) This proves our

second claimNow we define 119892 by 119892(119910) = 119892(119910)|119892(119910)| if |119892(119910)| ⩾ 1 minus

radic120576232 and 119892(119910) = 119892(119910) if |119892(119910)| lt 1 minus radic120576232 and we

write 119891 = sum119895isin119869120573119895(120594

119861119895

1198981(119861

119895)) It is clear that 119892 isin 119878

119871infin(119898)

119892 minus 119892 lt radic120576 and 119892(119910) = 1 for all 119909 isin supp119891

Finally we define the measure

119889] (119909 119910)

= sum

119895isin119869

120594119861119895

(119909) 119892 (119910)ℎ (119909 119910)119889]119888(119909 119910)(

1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+ 1205941198691119861(119909) 119889] (119909 119910)

(65)

where 119861 = ⋃119895isin119869119861119895 It is easy to see that (119889|]|1119889119898

1)(119909) = 1

on 119861 and (119889|]|11198891198981)(119909) ⩽ 1 elsewhere Note that

119889 (] minus ]) (119909 119910)

= sum

119895isin119869

120594119861119895

(119909) [

[

119892 (119910)ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1]

]

119889]119888(119909 119910)

minus sum

119895isin119869

120594119861119895

(119909) 119889]119891(119909 119910)

(66)

If (119909 119910) isin 119862 then |119892(119910)| ⩾ 1 minus radic120576232 ⩾ 1 minus 1232 and

100381610038161003816100381610038161003816119892 (119910)ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

119892 (119910)

1003816100381610038161003816119892 (119910)1003816100381610038161003816

ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816100381610038161003816

1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 11003816100381610038161003816

1003816100381610038161003816119892 (119910)1003816100381610038161003816

+

10038161003816100381610038161 minus1003816100381610038161003816119892 (119910)

1003816100381610038161003816

10038161003816100381610038161003816100381610038161003816119892 (119910)

1003816100381610038161003816

⩽ 2

1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 11003816100381610038161003816

1003816100381610038161003816119892 (119910)1003816100381610038161003816

⩽ 2radic120576

232

232

232 minus 1⩽ 2radic120576

(67)

Hence for all (119909 119910) isin 119862 we have

10038161003816100381610038161003816100381610038161003816100381610038161003816

119892 (119910)ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

⩽100381610038161003816100381610038161003816119892 (119910)ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

⩽ 2radic120576(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

(68)

So we have for all 119909 isin 1198691

119889|] minus ]|1

1198891198981

(119909)

⩽ sum

119895isin119869

120594119861119895

(119909) [

[

2radic120576(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

]

]

1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909)

+ sum

119895isin119869

120594119861119895

(119909)

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909)

⩽ sum

119895isin119869

120594119861119895

(119909)(2radic120576 + (1 minus1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909)))

+sum

119895isin119869

120594119861119895

(119909)(

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909))

⩽ 2radic120576 + 120576 lt 3radic120576

(69)

This gives that 119879] minus 119879] lt 3radic120576 Note also that for all 119895 isin 119869

⟨119879]

120594119861119895

1198981(119861

119895)

119892⟩

= intΩtimesΩ

120594119861119895

(119909)

1198981(119861

119895)

119892 (119910) 119889] (119909 119910)

= intΩtimesΩ

120594119861119895

(119909)

1198981(119861

119895)

ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

119889]119888(119909 119910)

Abstract and Applied Analysis 9

= intΩ

120594119861119895

(119909)

1198981(119861

119895)

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

1198891003816100381610038161003816]1198881003816100381610038161003816

1

(119909)

= intΩ

120594119861119895

(119909)

1198981(119861

119895)

1198891198981(119909) = 1

(70)

Hence we get ⟨119879]119891 119892⟩ = 1 which implies that 119879]119891 =119879] = 1 Finally10038171003817100381710038171003817119891 minus 119891

10038171003817100381710038171003817

⩽10038171003817100381710038171003817119891 minus 119891

1

10038171003817100381710038171003817+10038171003817100381710038171198911 minus 119891

1003817100381710038171003817

=

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

1198981(119861

119895)

minus sum

119895isin119869

120573119895

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+ 120576

⩽ sum

119895isin119869

120573119895(

10038171003817100381710038171003817100381710038171003817100381710038171003817

120594119861119895

1198981(119861

119895)

minus

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

120594119861119895

1198981(119861

119895)

minus

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

) + 120576

= 2sum

119895isin119869

120573119895

1198981(119861

119895 119861

119895)

1198981(119861

119895)

+ 120576

⩽ 2sum

119895isin119869

120573119895

(1205762)1198981(119861

119895)

1198981(119861

119895)

+ 120576 ⩽120576

1 minus 1205762+ 120576 lt 3120576

(71)

We are now ready to present the proof of the main resultin the case of finite regular positive Borel measures

Proof of Proposition 10 Let 1205751= 120575

3

2(5sdot2

4) 120575

2= 120575

12

3(3

2sdot2

14)

and 1205753= (12057610)

2 for some 0 lt 120576 lt 1 Suppose that 119879 isinL(119871

1(119898)) with 119879 = 1 and that there is an 119891

0isin 119878

1198711(119898)

and1198920isin 119878

119871infin(119898)

such that ⟨1198910 119892

0⟩ = 1 and |⟨119879119891

0 119892

0⟩| gt 1minus120575

3

12

6Then there is an isometric isomorphism Ψ from 119871

1(119898) onto

itself such thatΨ(1198910) = |119891

0| and there is a scalar number 120572 in

119878R such that |⟨1198791198910 119892

0⟩| = ⟨120572119879119891

0 119892

0⟩ Then letting 119891

1= Ψ119891

0

1198921= (Ψ

minus1)lowast1198920 and 119879

1= 120572Ψ119879Ψ

minus1 we have

⟨1198781198911 119892

1⟩ = ⟨120572Ψ119879

minus1Ψ119891

0 (Ψ

minus1)lowast

1198920⟩

= ⟨1205721198791198910 119892

0⟩ gt 1 minus

1205753

1

26

⟨1198911 119892

1⟩ = ⟨Ψ119891

0 (Ψ

minus1)lowast

1198920⟩ = 1

(72)

Since 11987911198911 gt 1 minus 120575(120575

3

12

6) by Lemma 13 there exists a

norm-one bounded operator 119879] and a nonnegative simplefunction119891

2isin 119878

1198711(119898)

such that 1198791minus119879] ⩽ 1205751 1198912minus1198911 ⩽ 31205751

and (119889|]|11198891198981)(119909) = 1 for all 119909 isin supp(119891

2) Then

⟨119879]1198912 1198921⟩

= ⟨11987911198911 119892

1⟩ minus ⟨119879

11198911minus 119879

11198912 119892

1⟩ minus ⟨119879

11198912minus 119879]1198912 1198921⟩

⩾ ⟨11987911198911 119892

1⟩ minus10038171003817100381710038171198911 minus 1198912

1003817100381710038171003817 minus10038171003817100381710038171198791 minus 119879]

1003817100381710038171003817

⩾ 1 minus1205753

1

26minus 3120575

1minus 120575

1⩾ 1 minus 5120575

1= 1 minus

1205753

2

16

(73)

Notice also that

⟨1198912 119892

1⟩ = ⟨119891

1 119892

1⟩ minus ⟨119891

1minus 119891

2 119892

1⟩ ⩾ 1 minus

10038171003817100381710038171198911 minus 11989121003817100381710038171003817

⩾ 1 minus 31205751⩾ 1 minus 5120575

1= 1 minus

1205753

2

16

(74)

By Lemma 12 there are a nonnegative simple function 1198913isin

1198781198711(119898)

and a function 1198923isin 119878

119871infin(119898)

such that

1198923(119909) = 120594supp119891

3

(119909) + 1198922(119909) 120594

Ωsupp1198913

(119909)

10038171003817100381710038171198912 minus 11989131003817100381710038171003817 ⩽ 1205752

10038171003817100381710038171198923 minus 11989211003817100381710038171003817 ⩽radic120575

2 ⟨119891

3 119892

3⟩ = 1

(75)

So we have

⟨119879]1198913 1198923⟩ = ⟨119879]1198912 1198921⟩ minus ⟨119879]1198912 minus 119879]1198913 1198921⟩

minus ⟨119879]1198913 1198921 minus 1198923⟩

⩾ 1 minus1205753

2

16minus 2radic120575

2⩾ 1 minus 3radic120575

2= 1 minus

1205756

3

27

(76)

By Lemma 14 there exist 1198914isin 119878

1198711(119898)

and 1198924isin 119878

119871infin(119898)

and anoperator 119879

4such that

⟨1198924 119879

41198914⟩ = 1 =

100381710038171003817100381711987941003817100381710038171003817

10038171003817100381710038171198794 minus 119879]1003817100381710038171003817 ⩽ 21205753

10038171003817100381710038171198914 minus 11989131003817100381710038171003817 ⩽ 31205753

10038171003817100381710038171198924 minus 11989231003817100381710038171003817 ⩽radic120575

3 ⟨119891

4 119892

4⟩ = 1

(77)

So we have

10038171003817100381710038171198794 minus 11987911003817100381710038171003817 ⩽10038171003817100381710038171198794 minus 119879]

1003817100381710038171003817 +1003817100381710038171003817119879] minus 1198791

1003817100381710038171003817

⩽ 1205751+ 2120575

3⩽ 3120575

3

10038171003817100381710038171198911 minus 11989141003817100381710038171003817 ⩽10038171003817100381710038171198911 minus 1198912

1003817100381710038171003817 +10038171003817100381710038171198912 minus 1198913

1003817100381710038171003817 +10038171003817100381710038171198913 minus 1198914

1003817100381710038171003817

⩽ 31205751+ 120575

2+ 3120575

3⩽ 10120575

3

10038171003817100381710038171198921 minus 11989241003817100381710038171003817 ⩽10038171003817100381710038171198921 minus 1198923

1003817100381710038171003817 + 1198923 minus 1198924

⩽ 1205752+ radic120575

3⩽ 2radic120575

3

(78)

10 Abstract and Applied Analysis

Let 119878 = 120572Ψminus11198794Ψ 119891 = Ψminus1

1198914 and 119892 = Ψlowast

1198924 then we have

119879 minus 119878 =10038171003817100381710038171003817119879 minus 120572Ψ

minus11198794Ψ10038171003817100381710038171003817=10038171003817100381710038171003817120572Ψ119879Ψ

minus1minus 119879

4

10038171003817100381710038171003817

=10038171003817100381710038171198791 minus 1198794

1003817100381710038171003817 ⩽ 31205753

100381710038171003817100381710038171198910minus 11989110038171003817100381710038171003817=100381710038171003817100381710038171198910minus Ψ

minus11198914

10038171003817100381710038171003817=10038171003817100381710038171198911 minus 1198914

1003817100381710038171003817 ⩽ 101205753

100381710038171003817100381710038171198920minus 11989110038171003817100381710038171003817=10038171003817100381710038171198920 minus Ψ

lowast1198924

1003817100381710038171003817 =100381710038171003817100381710038171003817(Ψ

minus1)lowast

1198920minus 119892

4

100381710038171003817100381710038171003817

=10038171003817100381710038171198921 minus 1198924

1003817100381710038171003817 ⩽ 2radic120575

3

⟨119891 119892⟩ = ⟨Ψminus11198914 Ψ

lowast1198924⟩ = ⟨119891

4 119892

4⟩ = 1

10038161003816100381610038161003816⟨119878119891 119892⟩

10038161003816100381610038161003816=10038161003816100381610038161003816⟨120572Ψ

minus11198794ΨΨ

minus11198914 Ψ

lowast1198924⟩10038161003816100381610038161003816= |120572| = 1

(79)

This completes the proof

Finally we may give the proof of the main result in fullgenerality

Proof of Theorem 9 Notice that the Kakutani representationtheorem (see [26] for a reference) says that for every 120590-finitemeasure ] the space 119871

1(]) is isometrically isomorphic to

1198711(119898) for some positive Borel regular measure on a compact

Hausdorff space Then by Proposition 10 there is a universalfunction 120576 997891rarr 120578(120576) gt 0 which gives the BPBp-nu for 119871

1(]) for

every 120590-finite measure ]Fix 120576 gt 0 Suppose that 119879

0isin L(119871

1(120583)) with V(119879

0) = 1

and (1198910 119891

lowast

0) isin Π(119871

1(120583)) satisfy

1003816100381610038161003816⟨119891lowast

0 119879

01198910⟩1003816100381610038161003816 gt 1 minus 120578 (120576) (80)

Choose a sequence 119891119899 in 119871

1(120583) such that sup

119899119879

0119891119899 = 1

and let 119866 be the closed linear span of

119879119899

0119891119898 119899 119898 isin N cup 0 (81)

As 119866 is separable there is a dense subset 119892119899 119899 isin N of 119866

and let 119864 = ⋃infin

119899=1supp119892

119899 where supp119892

119899is the support of 119892

119899

Then the measure 120583|119864is 120590-finite Let

119884 = 119891 isin 1198711(120583) supp (119891) sub 119864 (82)

be a closed subspace of 1198711(120583) It is clear that 119871

1(120583) = 119884oplus

1119885

and 119884 is isometrically isomorphic to 1198711(120583|

119864) So 119884 has the

BPBp-nu with 120578(120576)Now write 119878

0= 119879

0|119884 119884 rarr 119884 consider 119910

0= 119891

0isin

119878119884 119910lowast

0= 119891

lowast

0|119884isin 119878

119884lowast and observe that 119910lowast

0(119910

0) = 1 and

|119910lowast

0(119878

01199100)| = |119891

lowast

0(119879

01198910)| gt 1 minus 120578(120576) Hence there exist 119878 isin

L(119884) and (1199100 119910

lowast

0) isin Π(119884) such that

1003816100381610038161003816119910lowast

0(119878119910

0)1003816100381610038161003816 = 1 = V (119878) 1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

10038171003817100381710038171199100 minus 11991001003817100381710038171003817 lt 120576

1003817100381710038171003817119910lowast

0minus 119910

lowast

0

1003817100381710038171003817 lt 120576

(83)

Finally consider the operator 119879 isinL(1198711(120583)) given by

119879 (119910 119911)

= (119878119910 0) + 1198790(0 119911) ((119910 119911) isin 119871

1(120583) equiv 119884oplus

1119885)

(84)

We have 119879 = 1 (and so V(119879) = 1) Indeed1003817100381710038171003817119879 (119910 119911)

1003817100381710038171003817 =1003817100381710038171003817(119878119910 0)

1003817100381710038171003817 +10038171003817100381710038171198790 (0 119911)

1003817100381710038171003817 ⩽10038171003817100381710038171199101003817100381710038171003817 + 119911 =

1003817100381710038171003817(119910 119911)1003817100381710038171003817

(85)

for all (119910 119911) isin 1198711(120583) and 119879(119910

0 0) = (119878119910

0 0) = 119878119910

0 = 1

Let 119909 = (1199100 0) and 119909lowast = (119910lowast

0 119891

0|119885) Then (119909 119909lowast) isin Π(119871

1(120583))

Moreover we have1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119910

lowast

0119878119910

0

1003816100381610038161003816 = 1 = V (119879) 1003817100381710038171003817119909 minus 1198910

1003817100381710038171003817 =1003817100381710038171003817119910 minus 1199100

1003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowast

0minus 119891

lowast

0

1003817100381710038171003817

= max 1003817100381710038171003817119910 minus 119891lowast

0|119884

1003817100381710038171003817 1003817100381710038171003817119891

lowast

0|119885minus 119891

lowast

0|119885

1003817100381710038171003817

=1003817100381710038171003817119910

lowastminus 119910

lowast

0

1003817100381710038171003817 lt 120576

1003817100381710038171003817119879 minus 11987901003817100381710038171003817 = sup119910+119911⩽1

1003817100381710038171003817119879 (119910 119911) minus 1198790 (119910 119911)1003817100381710038171003817

= sup119910⩽1

1003817100381710038171003817119878119910 minus 11987801199101003817100381710038171003817 =1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

(86)

This completes the proof

5 Examples of Spaces Failing theBishop-Phelps-Bollobaacutes Property forNumerical Radius

Our goal here is to prove that the density of numerical radiusattaining operators does not imply the BPBp-nu Actuallywe will show that among separable spaces there is noisomorphic property implying the BPBp-nu other than finite-dimensionality

We need to relate the BPBp-nu to the Bishop-Phelps-Bollobas property for operators which as mentioned in theintroduction was introduced in [15] A pair (119883 119884) of Banachspaces has the Bishop-Phelps-Bollobas property for operators(in short BPBp) if given 120576 gt 0 there exists 120578(120576) gt 0 suchthat given 119879 isin L(119883 119884) with 119879 = 1 and 119909 isin 119878

119883such that

119879119909 gt 1 minus 120578(120576) then there exist 119911 isin 119878119883and 119878 isin L(119883 119884)

satisfying

119878 = 119878119911 = 1 119909 minus 119911 lt 120576

119879 minus 119878 lt 120576

(87)

We refer the reader to [15 19 25] and references therein formore information and background Among the interestingresults on the BPBp we emphasize that a pair (119883 119884) when119883is finite-dimensional does not necessarily have the BPBp Forinstance if119884 is a strictly convex space which is not uniformlyconvex then the pair (ℓ(2)

1 119884) fails to have the BPBp (this is

contained in [15] see [19 Section 3])

Abstract and Applied Analysis 11

The next result relates the BPBp-nu to the BPBp foroperators in a particular case We will deduce our examplefrom it

Theorem 15 If 1198711(120583)oplus

1119883 has the BPBp-119899119906 then the pair

(1198711(120583) 119883) has the BPBp for operators

Before proving this proposition we will use it to get themain examples of this section The first example shows thatthe density of numerical radius attaining operators does notimply the BPBp-nu

Example 16 There is a reflexive space (and so numerical radiusattaining operators on it are dense) which fails to have theBPBp-nu Indeed let 119884 be a reflexive separable space whichis not superreflexive and we may suppose that 119884 is strictlyconvex Observe that 119884 cannot be uniformly convex since itis not superreflexive Now 119883 = ℓ(2)

1oplus1119884 is reflexive but the

pair (ℓ(2)1 119884) fails the BPBp since 119884 is strictly convex but not

uniformly convex [19 Corollary 33] Therefore Theorem 15gives us that119883 does not have the BPBp-nu

The example above can be extended to get the result thatevery infinite-dimensional separable Banach space can berenormed to fail the BPBp-nu This follows from the factthat every infinite-dimensional separable Banach space canbe renormed to be strictly convex but not uniformly convex(this result can be proved ldquoby handrdquo an alternative categoricalargument for it can be found in [27] and references therein)With a little more of effort we may get the main result of thesection

Theorem 17 Every infinite-dimensional separable Banachspace can be renormed to fail the weak-BPBp-nu (and so inparticular to fail the BPBp-nu)

We need the following result which is surely well knownAs we have not found a reference we include a nice and easyproof kindly given to us by Vladimir Kadets We recall thatgiven a Banach space 119884 the set of all equivalent norms on119884 can be viewed as a metric space using the Banach-Mazurdistance

Lemma 18 Let119884 be an infinite-dimensional separable Banachspace Then the set of equivalent norms on 119884 which are strictlyconvex and are not (locally) uniformly convex is dense in the setof all equivalent norms on119884 (with respect to the Banach-Mazurdistance)

Proof Fix 119890 isin 119878119884and 119890lowast

1isin 119878

119884lowast such that 119890lowast

1(119890) = 1 For a fixed

120576 isin (0 12) denote

119902 (119910) = max (1 minus 120576) 10038171003817100381710038171199101003817100381710038171003817 1003816100381610038161003816119890lowast

1(119910)1003816100381610038161003816 (119910 isin 119884) (88)

Evidently (1 minus 120576)119910 ⩽ 119902(119910) ⩽ 119910 for every 119910 isin 119884 Fix asequence 119890lowast

119896 119896 ⩾ 2 of norm-one functionals separating

the points of 119884 and denote

119901 (119910) = radic

infin

sum

119896=1

1

2119896

1003816100381610038161003816119890lowast

119896(119910)1003816100381610038161003816

2

(119910 isin 119884) (89)

Then 119901 is a strictly convex norm on 119884 119901(119890) ⩾ 1radic2 and119901(119910) ⩽ 119910 for all 119910 isin 119883 Finally write

100381710038171003817100381711991010038171003817100381710038171= (1 minus 120576) 119902 (119910) + 120576

119901 (119910)

119901 (119890)(119910 isin 119884) (90)

Then sdot 1is a strictly convex norm on 119884 and

(1 minus 120576)2 10038171003817100381710038171199101003817100381710038171003817 ⩽100381710038171003817100381711991010038171003817100381710038171⩽ (1 + 120576)

10038171003817100381710038171199101003817100381710038171003817 (119910 isin 119884) (91)

Wewill finish the proof by showing that sdot 1is not uniformly

convex (actually it is not locally uniformly convex) Indeedfor each 119899 isin N we select 119910

119899isin ⋂

119899

119896=1ker 119890lowast

119896with 119910

119899 = 1 and

consider 119890119899= 119890 + (1205764)119910

119899 Then 119902(119890) = 1 119902(119890

119899) = 1 and

119902(119890+119890119899) = 2 At the same time 119901(119910

119899) rarr 0 so 119901(119890

119899) rarr 119901(119890)

and 119901(119890 + 119890119899) rarr 2119901(119890) Consequently

1198901 = 1100381710038171003817100381711989011989910038171003817100381710038171997888rarr 1

1003817100381710038171003817119890 + 11989011989910038171003817100381710038171997888rarr 2 (92)

but 119890 minus 1198901198991= (1205764)119910

1198991⩾ (1 minus 120576)

2(1205764) which means the

absence of local uniform convexity at 119890

Proof of Theorem 17 Let 119883 be an infinite-dimensional sepa-rable Banach space Take a closed subspace 119884 of 119883 of codi-mension two By [28 Proposition 2] the map carrying everyequivalent norm on 119884 to its numerical index is continuousand so the set of values of the numerical index of 119884 upto reforming is a nontrivial interval [28 Theorem 9] ThenLemma 18 allows us to find an equivalent norm | sdot | on 119884 insuch a way that (119884 | sdot |) is strictly convex and is not uniformlyconvex and 119899(119884 | sdot |) gt 0 Now the space 119883 = ℓ(2)

1oplus1(119884 | sdot |)

is an equivalent renorming of 119883 which does not have theBPBp-119899119906 (indeed otherwise the pair (ℓ(2)

1 (119884 | sdot |)) would

have the BPBp for the operator norm and so (119884 | sdot |) wouldbe uniformly convex by [19 Corollary 33] a contradiction)Moreover as

119899 (119883) = min 119899 (ℓ(2)1) 119899 (119884 |sdot|) gt 0 (93)

(see [20 Proposition 2] for instance) 119883 also fails the weak-BPBp-nu by Proposition 6

To finish the section with the promised proof ofTheorem 15 we first see the following stability result

Lemma 19 Let119883 = [⨁infin

119896=1119883

119896]1198880

or [⨁infin

119896=1119883

119896]ℓ1

If119883 has theBishop-Phelps-Bollobas property for numerical radius with afunction 120578 then each Banach space 119883

119894has the Bishop-Phelps-

Bollobas property for numerical radius with 120578nu(119883119894) ⩾ 120578 That

is inf119894120578nu(119883119894

)(120576) ⩾ 120578nu(119883)(120576) for all 0 lt 120576 lt 1

Proof Let 119875119894 119883 rarr 119883

119894and 1198751015840

119894 119883

lowastrarr 119883

lowast

119894be the natural

projections and let 119876119894 119883

119894rarr 119883 and 1198761015840

119894 119883

lowast

119894rarr 119883

lowast be thenatural embeddings

Assume that an operator 119879119894 119883

119894rarr 119883

119894and a pair

(119909119894 119909

lowast

119894) isin Π(119883

119894) satisfy that

V (119879119894) = 1

1003816100381610038161003816119909lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (94)

We define an operator 119879 119883 rarr 119883 and (119909 119909lowast) isin Π(119883) by

119879 = 119876119894∘ 119879

119894∘ 119875

119894 (119909 119909

lowast) = (119876

119894119909119894 119876

1015840

119894119909lowast

119894) (95)

12 Abstract and Applied Analysis

then clearly we see that1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119909

lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (96)

From the assumption there exist 119878 119883 rarr 119883 and a pair(119910 119910

lowast) isin Π(119883) such that

1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 = 1 = V (119878) 119878 minus 119879 lt 120576

1003817100381710038171003817119910lowastminus 119909

lowast1003817100381710038171003817 lt 1205761003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576

(97)

Since this clearly shows that1003817100381710038171003817119875119894 ∘ 119878 ∘ 119876119894

minus 119879119894

1003817100381710038171003817 lt 120576100381710038171003817100381710038171198751015840

119894119910lowastminus 119909

lowast

119894

10038171003817100381710038171003817lt 120576

1003817100381710038171003817119875119894119910 minus 1199091198941003817100381710038171003817 lt 120576

(98)

we only need to show that |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = 1

We first show the case of 1198880sum Since

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817=10038171003817100381710038171003817119875119895119910 minus 119875

11989511990910038171003817100381710038171003817⩽1003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576 (99)

for every 119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽100381710038171003817100381710038171198751015840

119894119910lowast10038171003817100381710038171003817+ 120576 sum

119895isinN119895 = 119894

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817⩽1003817100381710038171003817119910

lowast1003817100381710038171003817 = 1

(100)

This shows that 1198751015840119894119910lowast = 1 and 1198751015840

119895119910lowast= 0 for every 119895 = 119894

So 119910lowast = 1198761015840

1198941198751015840

119894119910lowast and 1198751015840

119894119910lowast(119875

119894119910) = 1 This and the fact that

119910 minus 119876119894119875119894119910 lt 120576 imply that

(119876119894119875119894119910 + (

1

120576) (119910 minus 119876

119894119875119894119910) 119876

1015840

1198941198751015840

119894119910lowast) isin Π (119883) (101)

So we get that (1198761015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910 + (1120576)(119910 minus 119876

119894119875119894119910))V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =10038161003816100381610038161003816(119876

1015840

1198941198751015840

119894119910lowast) 11987811991010038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910 +1

120576(119910 minus 119876

119894119875119894119910))

1003816100381610038161003816100381610038161003816⩽ 1

(102)

and so we get |1198751015840119894119910lowast(119875

119894∘119878 ∘119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

We next show the case of ℓ1sum The proof is almost

the same as that of the 1198880case However for the sake of

completeness we provide it hereSince 1198751015840

119895119910lowast = 119875

1015840

119895119910lowastminus 119875

1015840

119895119909lowast ⩽ 119910

lowastminus 119909

lowast lt 120576 for every

119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽10038171003817100381710038171198751198941199101003817100381710038171003817 + 120576 sum

119895isinN119895 = 119894

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817⩽10038171003817100381710038171199101003817100381710038171003817 = 1

(103)

which shows 119875119894119910 = 1 and 119875

119895119910 = 0 for every 119895 = 119894 Since this

implies (119876119894119875119894119910119876

1015840

1198941198751015840

119894119910lowast+(1120576)(119910

lowastminus119876

1015840

1198941198751015840

119894119910lowast)) isin Π(119883) we get

that |(1198761015840

1198941198751015840

119894119910lowast+ (1120576)(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast))119878(119876

119894119875119894119910)| ⩽ V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =1003816100381610038161003816119910

lowast119878 (119876

119894119875119894119910)1003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast+1

120576(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast)) 119878 (119876

119894119875119894119910)

1003816100381610038161003816100381610038161003816⩽ 1

(104)

and so |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

Proof of Theorem 15 Note that 120578nu(1198711(120583)oplus1119883)(120576) rarr 0 as120576 rarr 0 Fix 0 lt 120576

0lt 1 and choose 0 lt 120576 lt 1 such that

6120576 + 120578nu(1198711(120583)oplus1119883)(120576) lt 1205760 Let 120578(1205760) = 120578nu(1198711(120583)oplus1119883)(120576)Suppose that 119879

0isin L(119871

1(120583) 119883) with 119879

0 = 1 and 119891

0isin

1198781198711(120583)

satisfy

1003817100381710038171003817119879011989101003817100381710038171003817 gt 1 minus 120578 (1205760) (105)

For any measurable subset 119861 let

1198711(120583|

119861) = 119891|

119861 119891 isin 119871

1(120583) (106)

with the norm 119891|119861 = 119891120594

1198611 Then it is easy to see

that 1198711(120583|

119861) is isometrically isomorphic to a complemented

subspace of 1198711(120583) Let 119875

119861 119871

1(120583) rarr 119871

1(120583|

119861) be the

restriction defined by 119875119861(119891) = 119891|

119861for all 119891 isin 119871

1(120583) and

let 119869119861 119871

1(120583|

119861) rarr 119871

1(120583) be the extension defined by

119869119861(119891)(120596) = 119891(120596) if 120596 isin 119861 and 119869

119861(119891)(120596) = 0 otherwise

It is clear that 119875119861119869119861= Id

1198711(120583|119861)and 119869

119861119875119861(119891) = 119891120594

119861for all

119891 isin 1198711(120583) Notice also that 119871

1(120583) is isometrically isomorphic

to 1198711(120583|

119861)oplus

11198711(120583|

119861119888)

Let 119860 = supp1198910and 119892

0= 119875

1198601198910 Then

1003817100381710038171003817119879011986911986011989201003817100381710038171003817 =100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) gt 0 (107)

and define the operator 119879119860 119871

1(120583|

119860) rarr 119883 by 119879

119860119891 =

1198790119869119860119891119879

0119869119860 for every 119891 isin 119871

1(120583|

119860) Then

100381710038171003817100381711987911986011989201003817100381710038171003817 ⩾100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) (108)

Since 120583|119860is 120590-finite 119871

1(120583|

119860)lowast= 119871

infin(120583|

119860) Let 119892lowast

0isin 119878

119871infin(120583|119860)

be a function such that ⟨119892lowast0 119892

0⟩ = 1 and choose 119909lowast

0isin 119878

119883lowast

such that 119909lowast0(119879

1198601198920) = 119879

1198601198920 Define the operator 119878

0isin

L(1198711(120583|

119860)oplus

1119883) by

1198780(119891 119909) = (0 119879

119860119891) ((119891 119909) isin 119871

1(120583|

119860) oplus

1119883) (109)

and observe that 1198780 = V(119878

0) = 1 Indeed

100381710038171003817100381711987801003817100381710038171003817 ⩽ 1 =

10038171003817100381710038171198791198601003817100381710038171003817

= sup 1003816100381610038161003816119909lowast1198791198601198911003816100381610038161003816 119909

lowastisin 119878

119883lowast 119891 isin 119878

1198711(120583|119860)

= sup 1003816100381610038161003816(119891lowast 119909

lowast) 119878

0(119891 119909)

1003816100381610038161003816

((119891lowast 119909

lowast) (119891 119909)) isin Π (119871

1(120583|

119860) oplus

1119883)

= V (1198780) ⩽100381710038171003817100381711987801003817100381710038171003817

(110)

Abstract and Applied Analysis 13

It is immediate that

(119892lowast

0 119909

lowast

0) 119878

0(119892

0 0) = 119909

lowast

0(119879

1198601198920) =10038171003817100381710038171198791198601198920

1003817100381710038171003817 gt 1 minus 120578 (1205760)

(111)

By Lemma 19 1198711(120583|

119860)oplus

1119883 has the BPBp-nu with the func-

tion 120578 Therefore there exist 1198781isin L(119871

1(120583|

119860)oplus

1119883) (119892

1 119909

1) isin

1198781198711(120583|119860)oplus1119883 and (119892lowast

1 119909

lowast

1) isin 119878

119871infin(120583|119860)oplusinfin119883lowast such that

1003817100381710038171003817(1198921 1199091) minus (1198920 0)1003817100381710038171003817 lt 120576

1003817100381710038171003817(119892lowast

1 119909

lowast

1) minus (119891

lowast

0 119909

lowast

0)1003817100381710038171003817 lt 120576

10038171003817100381710038171198781 minus 11987801003817100381710038171003817 lt 120576 ⟨(119892

lowast

1 119909

lowast

1) (119892

1 119909

1)⟩ = 1

1003816100381610038161003816(119892lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 = V (119878

1) = 1

(112)

Claim 1 We claim that 1199091= 0

Otherwise

1 = Re ⟨(119892lowast1 119909

lowast

1) (119892

1 119909

1)⟩

=100381710038171003817100381711989211003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

+100381710038171003817100381711990911003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

⟩ ⩽ 1

(113)

We deduce that

((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

(119892lowast

1 119909

lowast

1)) (

(0 1199091)

100381710038171003817100381711990911003817100381710038171003817

(119892lowast

1 119909

lowast

1))

isin Π (1198711(120583|

119860) oplus

1119883)

(114)

Since 1198781((0 119909

1)119909

1) = (119878

1minus 119878

0)((0 119909

1)119909

1) lt 120576 we

get that

1 =1003816100381610038161003816⟨(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)⟩1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

100381710038171003817100381711989211003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

)⟩

+100381710038171003817100381711990911003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

)⟩

100381610038161003816100381610038161003816100381610038161003816

⩽100381710038171003817100381711989211003817100381710038171003817 V (1198781) + 120576

100381710038171003817100381711990911003817100381710038171003817 lt100381710038171003817100381711989211003817100381710038171003817 +100381710038171003817100381711990911003817100381710038171003817 = 1

(115)

a contradiction This proves the claimWe define the operator 119878

2 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883

by 1198782(119891 119909) = 119878

1(119891 0) for every 119891 isin 119871

1(120583|

119860) and for every

119909 isin 119883 Then we have

V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 = 1

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ 120576 (116)

Indeed from Claim 1 we have

V (1198781) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 ⩽ V (119878

2)

(117)

On the other hand we see that1003816100381610038161003816(119891

lowast 119909

lowast) 119878

2(119891 119909)

1003816100381610038161003816

=1003816100381610038161003816(119891

lowast 119909

lowast) 119878

1(119891 0)

1003816100381610038161003816 ⩽10038171003817100381710038171198911003817100381710038171003817 V (1198781) ⩽ V (119878

1)

(118)

for every ((119891lowast 119909lowast) (119891 119909)) isin Π(1198711(120583|

119860)oplus

1119883) Therefore

V(1198782) ⩽ V(119878

1) Also

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ sup

119909isin119878119883

10038171003817100381710038171198781 (0 119909)1003817100381710038171003817

= sup119909isin119878119883

10038171003817100381710038171198781 (0 119909) minus 1198780 (0 119909)1003817100381710038171003817 ⩽ 120576

(119)

Claim 2 There exists 1198783 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883 such

that 1198783(119892

1 0) = 119878

3 = 1 119878

3(0 119909) = 0 119878

3(119891 119909) isin 0oplus

1119883

for every (119891 119909) isin 1198711(120583|

119860)oplus

1119883 and 119878

3minus 119878

2 lt 4120576

Indeed write 1198781= (119863

1 119863

2) where 119863

1 119871

1(120583|

119860)oplus

1119883 rarr

1198711(120583|

119860) and119863

2 119871

1(120583|

119860)oplus

1119883 rarr 119883 We have that

sup 1003816100381610038161003816119892lowast119863

1(119892

1 0) + 119909

lowast119863

2(119892

1 0)1003816100381610038161003816

119909lowastisin 119878

119883lowast ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860) +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⩽ V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119892

lowast

1119863

1(119892

1 0) + 119909

lowast

1119863

2(119892

1 0)1003816100381610038161003816

(120)

This implies that

1003816100381610038161003816119909lowast

1119863

2(119892

1 0)1003816100381610038161003816 =10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

1003816100381610038161003816119892lowast

1119863

1(119892

1 0)1003816100381610038161003816

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119871

infin(120583|

119860)

(121)

Therefore |119892lowast1| equals 1 on the support of 119863

1(119892

1 0) As

|⟨119892lowast

1 119892

1⟩| = 1 we also have that |119892lowast

1| equals 1 on the support

of 1198921 Changing the values of 119892lowast

1by the ones of 119891lowast

0on 119860

(supp(1198631(119892

1 0)) cup supp(g

1)) we may and do suppose that

|119892lowast

1| = 1 on the whole 119860We also have 119863

2(119892

1 0) gt 0 Indeed

10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)1003817100381710038171003817

⩽10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198921 0)

1003817100381710038171003817 +10038171003817100381710038171198780 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817

lt 2120576 + 120576 = 3120576

(122)

So we have

10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

⩽10038171003817100381710038171198631(119892

1 0)1003817100381710038171003817 +10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

14 Abstract and Applied Analysis

=1003817100381710038171003817(1198631

(1198921 0) 119863

2(119892

1 0)) minus (0 119879

1198601198920)1003817100381710038171003817

=10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817 lt 3120576

(123)

and 1198632(119892

1 0) gt 119879

1198601198920 minus 3120576 ⩾ 1 minus 120578(120576

0) minus 3120576 gt 0

Finally define the operator 1198783by

1198783(119891 119909) = (0119863

2(119891 0) + 119892

lowast

1(119863

1(119891 0))

1198632(119892

1 0)

119909lowast

1119863

2(119892

1 0))

(124)

for (119891 119909) isin 1198711(120583|

119860)oplus

1119883 It is clear that

100381710038171003817100381711987831003817100381710038171003817 ⩽ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) (125)

Notice also that

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817

⩽10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 +10038171003817100381710038171198632(119891 0) minus 119879

1198601198911003817100381710038171003817

=1003817100381710038171003817(1198631

(119891 0) 1198632(119891 0)) minus (0 119879

119860119891)1003817100381710038171003817

=10038171003817100381710038171198782 (119891 119909) minus 1198780 (119891 119909)

1003817100381710038171003817

(126)

for all (119891 119909) isin 1198711(120583|

119860)oplus

1119883 Hence we have

10038171003817100381710038171198783 minus 11987821003817100381710038171003817

= 2 sup119891isin1198781198711(120583|119860)

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 ⩽ 210038171003817100381710038171198782 minus 1198780

1003817100381710038171003817 lt 4120576(127)

On the other hand let 119866 1198711(120583|

119860) rarr 119871

1(120583|

119860) be defined by

119866(119891) = 119892lowast

1119891 for every 119891 isin 119871

1(120583|

119860) Then we have

V (1198782)

= sup 1003816100381610038161003816119911lowast11987821199111003816100381610038161003816 (119911 119911

lowast) isin Π (119871

1(120583|

119860) oplus

1119883)

⩾ sup10038161003816100381610038161003816100381610038161003816

119909lowast119863

2(119866(

1

120583 (119862)120594119862) 0)

+119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119909lowastisin 119878

119883lowast 119862 isin Σ

119860 120583 (119862) gt 0

= sup10038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119862)120594119862) 0)

10038171003817100381710038171003817100381710038171003817

+

10038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119862 isin Σ119860 120583 (119862) gt 0

(128)

where Σ119860is the family of measurable subsets of 119860

Hence for any simple function 119904 = sum119899

119894=1((120572

119894120583(119860

119894))120594

119860119894

) isin

1198781198711(120583|119860) where 119860

119894119894is a family of disjoint measurable subsets

with strictly positive measure we have

V (1198782)

119899

sum

119894=1

10038161003816100381610038161205721198941003816100381610038161003816 (

100381710038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381710038171003817100381710038171003817100381710038171003817

+

100381610038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381610038161003816100381610038161003816100381610038161003816

)

⩾10038171003817100381710038171198632(119866 (119904) 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119866 (119904) 0)

1003816100381610038161003816

(129)

Since |119892lowast1| = 1119866 is an isometric isomorphism so for each119891 isin

1198781198711(120583|119860)there exists a sequence of norm-one simple functions

(119904119896) such that 119866(119904

119896) converges to 119891 Therefore

V (1198782) ⩾ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) ⩾100381710038171003817100381711987831003817100381710038171003817

(130)

On the other hand we see that100381710038171003817100381711987831003817100381710038171003817 ⩾1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

3(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119909

lowast

1119863

2(119892

1 0) + 119892

lowast

1119863

1(119892

1 0)1003816100381610038161003816 = V (119878

2) = 1

(131)

Therefore 1 = 1198783 = 119878

3(119892

1 0) which proves Claim 2

Finally set 1198783= (0 ) for a suitable 119871

1(120583|

119860)oplus

1119883 rarr

119883 and define the operator 1198791 119871

1(120583) rarr 119883 by

1198791(119891) = 119879

0(119891120594

119860119888) + (119875

119860119891 0) (132)

for every 119891 isin 1198711(120583) Then we have

10038171003817100381710038171198791 (119891)1003817100381710038171003817 ⩽100381710038171003817100381711987901003817100381710038171003817

10038171003817100381710038171198911205941198601198881003817100381710038171003817 +1003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171198911205941198601003817100381710038171003817 =10038171003817100381710038171198911003817100381710038171003817 (133)

for every 119891 isin 1198711(120583) so 119879

1 ⩽ 1 Also

10038171003817100381710038171198791 (1198691198601198921)1003817100381710038171003817 =10038171003817100381710038171198783 (1198921 0)

1003817100381710038171003817 =100381710038171003817100381711987831003817100381710038171003817 = 1 (134)

so 1198791attains its norm at 119869

1198601198921isin 119871

1(120583) and

10038171003817100381710038171198691198601198921 minus 11989101003817100381710038171003817 =10038171003817100381710038171198921 minus 1198920

1003817100381710038171003817 lt 120576 (135)

We also have that for any 119891 isin 1198781198711(120583)

10038171003817100381710038171198790 (119891) minus 1198791 (119891)1003817100381710038171003817 =100381710038171003817100381710038171198790(119891120594

119860) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790 (119869119860119875119860119891) minus 119879119860 (119875119860119891)

1003817100381710038171003817

+10038171003817100381710038171003817119879119860(119875

119860119891) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790119869119860 minus 119879119860

1003817100381710038171003817 +10038171003817100381710038171198780 minus 1198783

1003817100381710038171003817

lt 120578 (1205760) + 6120576

(136)

Hence 1198790minus 119879

1 ⩽ 120578(120576

0) + 6120576 lt 120576

0

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Abstract and Applied Analysis 15

Acknowledgments

The authors thank Gilles Godefroy and Rafael Paya forfruitful conversations about the content of this paper andVladimir Kadets for providing Lemma 18 They also appre-ciate anonymous referees for careful reading and fruitfulsuggestions about revision The second author was sup-ported by Basic Science Research Program through theNational Research Foundation of Korea (NRF) fundedby the Ministry of Education Science and Technology(NRF-2012R1A1A1006869) The third author was partiallysupported by Spanish MICINN and FEDER project noMTM2012-31755 and by Junta de Andalucıa and FEDERGrants FQM-185 and P09-FQM-4911

References

[1] F F Bonsall and J Duncan Numerical Ranges of Operators onNormed Spaces and of Elements of Normed Algebras vol 2 ofLondon Mathematical Society Lecture Note Series 2 CambridgeUniversity Press London UK 1971

[2] F F Bonsall and J Duncan Numerical Ranges II CambridgeUniversity Press London UK 1973 London MathematicalSociety Lecture Notes Series 10

[3] B Sims On numerical range and its applications to Banachalgebras [PhD thesis] University of Newcastle New SouthWales Australia 1972

[4] I D Berg and B Sims ldquoDenseness of operators which attaintheir numerical radiusrdquoAustralian Mathematical Society A vol36 no 1 pp 130ndash133 1984

[5] C S Cardassi ldquoDensity of numerical radius attaining operatorson some reflexive spacesrdquo Bulletin of the Australian Mathemat-ical Society vol 31 no 1 pp 1ndash3 1985

[6] C S Cardassi ldquoNumerical radius attaining operatorsrdquo inBanach Spaces (Columbia Mo 1984) vol 1166 of Lecture Notesin Mathematics pp 11ndash14 Springer Berlin Germany 1985

[7] C S Cardassi ldquoNumerical radius-attaining operators on119862(119870)rdquoProceedings of the AmericanMathematical Society vol 95 no 4pp 537ndash543 1985

[8] MDAcostaOperadores que alcanzan su radio numerico [PhDthesis] Universidad de Granada Granada Spain 1990

[9] M D Acosta and R Paya ldquoNumerical radius attaining oper-ators and the Radon-Nikodym propertyrdquo The Bulletin of theLondon Mathematical Society vol 25 no 1 pp 67ndash73 1993

[10] R Paya ldquoA counterexample on numerical radius attainingoperatorsrdquo Israel Journal of Mathematics vol 79 no 1 pp 83ndash101 1992

[11] Y S Choi and S G Kim ldquoNorm or numerical radius attainingmultilinear mappings and polynomialsrdquo Journal of the LondonMathematical Society vol 54 no 1 pp 135ndash147 1996

[12] M D Acosta J Becerra Guerrero and M Ruiz GalanldquoNumerical-radius-attaining polynomialsrdquo The Quarterly Jour-nal of Mathematics vol 54 no 1 pp 1ndash10 2003

[13] M D Acosta and S G Kim ldquoDenseness of holomorphicfunctions attaining their numerical radiirdquo Israel Journal ofMathematics vol 161 pp 373ndash386 2007

[14] S G Kim and H J Lee ldquoNumerical peak holomorphicfunctions on Banach spacesrdquo Journal of Mathematical Analysisand Applications vol 364 no 2 pp 437ndash452 2010

[15] M D Acosta R M Aron D Garcıa and M MaestreldquoThe Bishop-Phelps-Bollobas theorem for operatorsrdquo Journal ofFunctional Analysis vol 254 no 11 pp 2780ndash2799 2008

[16] A J Guirao and O Kozhushkina ldquoThe Bishop-Phelps-Bollobasproperty for numerical radius in ℓ

1(119862)rdquo Studia Mathematica

vol 218 no 1 pp 41ndash54 2013[17] J Falco ldquoThe Bishop-Phelps-Bollobas property for numerical

radius on 1198711rdquo Journal of Mathematical Analysis and Applica-

tions vol 414 no 1 pp 125ndash133 2014[18] A Aviles A J Guirao and J Rodriguez ldquoOn the Bishop-

Phelps-Bollobas property for numerical radius in119862(119870)-spacesrdquoPreprint

[19] RM Aron Y S Choi S K Kim H J Lee andMMartin ldquoTheBishop-Phelps-Bollobas version of Lindenstrauss properties Aand Brdquo Preprint

[20] V Kadets M Martın and R Paya ldquoRecent progress and openquestions on the numerical index of Banach spacesrdquo Revista dela Real Academia de Ciencias Exactas Fısicas y Naturales A vol100 no 1-2 pp 155ndash182 2006

[21] M Martın J Merı and M Popov ldquoOn the numerical index ofreal 119871

119901(120583)-spacesrdquo Israel Journal of Mathematics vol 184 pp

183ndash192 2011[22] A Iwanik ldquoNorm attaining operators on Lebesgue spacesrdquo

Pacific Journal of Mathematics vol 83 no 2 pp 381ndash386 1979[23] HH SchaeferBanach Lattices and Positive Operators Springer

New York NY USA 1974[24] J Duncan C M McGregor J D Pryce and A J White ldquoThe

numerical index of a normed spacerdquo Journal of the LondonMathematical Society Second Series vol 2 pp 481ndash488 1970

[25] Y S Choi S K Kim H J Lee and M Martin ldquoThe Bishop-Phelps-Bollobas theorem for operators on 119871

1(120583)rdquo Preprint

[26] H E Lacey The Isometric Theory of Classical Banach SpacesSpringer New York NY USA 1974

[27] G Godefroy ldquoUniversal spaces for strictly convex Banachspacesrdquo Revista de la Real Academia de Ciencias Exactas Fısicasy Naturales A vol 100 no 1-2 pp 137ndash146 2006

[28] C Finet M Martın and R Paya ldquoNumerical index andrenormingrdquo Proceedings of the American Mathematical Societyvol 131 no 3 pp 871ndash877 2003

Submit your manuscripts athttpwwwhindawicom

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 4: Research Article On the Bishop-Phelps-Bollobás Property ...downloads.hindawi.com/journals/aaa/2014/479208.pdf · Research Article On the Bishop-Phelps-Bollobás Property for Numerical

4 Abstract and Applied Analysis

100381610038161003816100381610038161003816100381610038161003816

119909lowast

119899+1119879119899119909119899+1+ 120582

119899+1

120576119899+1

4119899+1119909lowast

119899(119909

119899+1) 119909

lowast

119899+1(119909

119899)

100381610038161003816100381610038161003816100381610038161003816

⩽ V (119879119899) +120576119899+1

4119899+1119909lowast

119899+1(119909

119899)

V (119879119899+1)

⩾1003816100381610038161003816119909

lowast

119899119879119899+1119909119899

1003816100381610038161003816 =

100381610038161003816100381610038161003816100381610038161003816

119909lowast

119899119879119899119909119899+ 120582

119899+1

120576119899+1

4119899+1

100381610038161003816100381610038161003816100381610038161003816

=1003816100381610038161003816119909

lowast

119899119879119899119909119899

1003816100381610038161003816 +120576119899+1

4119899+1⩾ V (119879

119899) minus 120578(

120576119899+1

4119899+1) +120576119899+1

4119899+1

(23)

In summary we have

V (119879119899) +120576119899+1

4119899+1119909lowast

119899+1(119909

119899)

⩾ V (119879119899) minus 120578(

120576119899+1

4119899+1) +120576119899+1

4119899+1minus 120578(

120576119899+2

4119899+2)

(24)

Hence

119909lowast

119899+1(119909

119899) ⩾ 1 minus 2

4119899+1

120576119899+1120578(120576119899+1

4119899+1)

= 1 minus1

2min120575

119883(120576119899+1

4119899+2) 120575

119883lowast (120576119899+1

4119899+2)

1003817100381710038171003817100381710038171003817

119909119899+ 119909

119899+1

2

1003817100381710038171003817100381710038171003817⩾ 119909

lowast

119899+1(119909119899+ 119909

119899+1

2) ⩾ 1 minus 120575

119883(120576119899+1

4119899+2)

100381710038171003817100381710038171003817100381710038171003817

119909lowast

119899+ 119909

lowast

119899+1

2

100381710038171003817100381710038171003817100381710038171003817

⩾119909lowast

119899+ 119909

lowast

119899+1

2(119909

119899) ⩾ 1 minus 120575

119883lowast (120576119899+1

4119899+2)

(25)

This means that

1003817100381710038171003817119909119899 minus 119909119899+11003817100381710038171003817 ⩽ 120576

119899+14

119899+2

1003817100381710038171003817119909lowast

119899minus 119909

lowast

119899+1

1003817100381710038171003817 ⩽ 120576119899+14

119899+2

(26)

for all 119899 So (119909119899) and (119909lowast

119899) are Cauchy Let 119909

infin= lim

119899119909119899and

119909lowast

infin= lim

119899119909lowast

119899 Then 119909

0minus 119909

infin lt 1205764 and 119909lowast

0minus 119909

lowast

infin lt 1205764

Hence |119909lowastinfin(119909

infin)| = lim

119899|119909

lowast

119899(119909

119899)| = 1 and

V (119878) = lim119899V (119879

119899) = lim

119899

1003816100381610038161003816119909lowast

119899119879119899119909119899

1003816100381610038161003816 =1003816100381610038161003816119909

lowast

infin119878119909

infin

1003816100381610038161003816 (27)

Let 120572 = 119909lowastinfin(119909

infin) 119910lowast = 119909lowast

infin and 119910 = 119909

infin Then we have

119910lowast(119910) = 1 V(119878) = |119910lowast119878119910| and 119910 minus 119909

0 lt 120576 Notice that

|120572 minus 1| =1003816100381610038161003816119909

lowast

infin(119909

infin) minus 119909

lowast

0(119909

0)1003816100381610038161003816

⩽1003816100381610038161003816(119909

lowast

infinminus 119909

lowast

0) (119909

infin)1003816100381610038161003816 +1003816100381610038161003816119909

lowast

0(119909

infin) minus 119909

lowast

0(119909

0)1003816100381610038161003816 lt120576

2

(28)

Therefore

1003817100381710038171003817119910lowastminus 119909

lowast1003817100381710038171003817 ⩽1003817100381710038171003817119910

lowastminus 119910

lowast1003817100381710038171003817 +1003817100381710038171003817119910

lowastminus 119909

lowast1003817100381710038171003817 lt120576

2+120576

4lt 120576 (29)

This completes the proof

Let us discuss a little bit about the equivalence betweenthe property in the result above and the BPBp-nu Forconvenience let us introduce the following definition

Definition 5 A Banach space 119883 has the weak Bishop-Phelps-Bollobas property for the numerical radius (in short weak-BPBp-nu) if given 120576 gt 0 there exists 120578(120576) gt 0 such thatwhenever 119879

0isin L(119883) with V(119879

0) = 1 and (119909

0 119909

lowast

0) isin Π(119883)

satisfy |119909lowast011987901199090| gt 1 minus 120578(120576) there exist 119878 isin L(119883) and

(119910 119910lowast) isin Π(119883) such that

V (119878) = 1003816100381610038161003816119910lowast1198781199101003816100381610038161003816

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowastminus 119910

lowast1003817100381710038171003817 lt 120576 119878 minus 119879 lt 120576

(30)

Notice that the only difference between this concept andthe BPBp-nu is the normalization of the operator 119878 by thenumerical radius Of course if the numerical radius andthe operator norm are equivalent these two properties arethe same This equivalence is measured by the so-callednumerical index of the Banach space as follows For a Banachspace119883 the numerical index of119883 is defined by

119899 (119883) = inf V (119879) 119879 isinL (119883) 119879 = 1 (31)

It is clear that 0 ⩽ 119899(119883) ⩽ 1 and 119899(119883)119879 ⩽ V(119879) ⩽ 119879for all 119879 isin L(119883) The value 119899(119883) = 1 means that V equalsthe usual operator norm This is the case of 119883 = 119871

1(120583) and

119883 = 119862(119870) among many others On the other hand 119899(119883) gt 0if and only if the numerical radius is equivalent to the normof L(119883) We refer the reader to [20] for more informationand background

The following result is immediate We include a proof forthe sake of completeness

Proposition 6 Let119883 be a Banach space with 119899(119883) gt 0 Then119883 has the BPBp-nu if and only if119883 has the weak-BPBp-nu

Proof Thenecessity is clear For the converse assume that wehave 120578(120576) gt 0 satisfying the conditions of the weak-BPBp-nufor all 0 lt 120576 lt 1 If 119879 isin L(119883) with V(119879) = 1 and (119909

0 119909

lowast

0) isin

Π(119883) satisfy |119909lowast0119879119909

0| gt 1 minus 120578(120576) for 0 lt 120576 lt 1 then there exist

119878 isinL(119883) and (119910 119910lowast) isin Π(119883) such that

V (119878) = 1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 119878 minus 119879 lt 120576

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowastminus 119910

lowast1003817100381710038171003817 lt 120576

(32)

As V(119878) gt 0 by the above let 1198781= (1V(119878))119878 Then we have

1 = V (1198781) =1003816100381610038161003816119910

lowast11987811199101003816100381610038161003816

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowastminus 119910

lowast1003817100381710038171003817 lt 120576

(33)

Abstract and Applied Analysis 5

Finally we have

10038171003817100381710038171198781 minus 1198791003817100381710038171003817 ⩽

10038171003817100381710038171003817100381710038171003817

1

V (119878)119878 minus 119878

10038171003817100381710038171003817100381710038171003817

+ 119878 minus 119879

=119878

V (119878)|V (119878) minus 1| + 119878 minus 119879

⩽1

119899 (119883)|V (119878) minus V (119879)| + 119878 minus 119879

⩽ (1

119899 (119883)+ 1) 119878 minus 119879 lt

119899 (119883) + 1

119899 (119883)120576

(34)

An obvious change of parameters finishes the proof

We do not know whether the hypothesis of 119899(119883) gt 0 canbe omitted in the above result

Putting together Propositions 4 and 6 we get the follow-ing

Corollary 7 Let 119883 be a uniformly convex and uniformlysmooth Banach space with 119899(119883) gt 0 Then 119883 has the BPBp-nu

Let us comment that every complex Banach space 119883satisfies 119899(119883) ⩾ 1119890 so the above corollary automaticallyapplies in the complex case In the real case this is no longertrue as the numerical index of a Hilbert space of dimensiongreater than or equal to two is 0 On the other hand it isproved in [21] that real 119871

119901(120583) spaces have nonzero numerical

index for every measure 120583when 119901 = 2Therefore we have thefollowing examples

Example 8 (a) Complex Banach spaces which are uniformlysmooth and uniformly convex satisfy the BPBp-nu

(b) In particular for every measure 120583 the complex spaces119871119901(120583) have the BPBp-nu for 1 lt 119901 lt infin(c) For every measure 120583 the real spaces 119871

119901(120583) have the

BPBp-nu for 1 lt 119901 lt infin 119901 = 2

Note Added in Revision Very recently H J Lee M Martınand J Merı have proved that Proposition 6 can be extendedto some Banach spaces with numerical index zero as forinstance real Hilbert spaces Hence they have shown thatHilbert spaces have the BPBp-nu These results will appearelsewhere

4 L1

Spaces

In this section we will show that 1198711(120583) has the BPBp-nu for

every measure 120583 In the proof we are dealing with complexintegrable functions since the real case is followed easilyby applying the same proof Our main result here is thefollowing

Theorem 9 Let 120583 be a measure Then 1198711(120583) has the Bishop-

Phelps-Bollobas property for numerical radius More preciselygiven 120576 gt 0 there exists 120578(120576) gt 0 (which does not depend on120583) such that whenever 119879

0isin L(119871

1(120583)) with V(119879

0) = 1 and

(1198910 119892

0) isin Π(119871

1(120583)) satisfy |⟨119879

01198910 119892

0⟩| gt 1 minus 120578(120576) then there

exist 119879 isinL(1198711(120583)) (119891

1 119892

1) isin Π(119871

1(120583)) such that

1003816100381610038161003816⟨1198791198911 1198921⟩1003816100381610038161003816 = V (119879) = 1 10038171003817100381710038171198910 minus 1198911

1003817100381710038171003817 lt 120576

10038171003817100381710038171198920 minus 11989211003817100381710038171003817 lt 120576

1003817100381710038171003817119879 minus 11987901003817100381710038171003817 lt 120576

(35)

As a first step we have to start dealing with finite regularpositive Borel measures for which a representation theoremfor operators exists

Proposition 10 Let 119898 be a finite regular positive Borelmeasure on a compact Hausdorff spaceΩ Then 119871

1(119898) has the

Bishop-Phelps-Bollobas property for numerical radius Moreprecisely given 120576 gt 0 there is 120578(120576) gt 0 (which is independent ofthemeasure119898) such that if a norm-one element119879 inL(119871

1(119898))

and (1198910 119892

0) isin Π(119871

1(119898)) satisfy |⟨119879119891

0 119892

0⟩| gt 1 minus 120578(120576) then

there exist an operator 119878 isin L(1198711(119898)) (119891

1 119892

1) isin Π(119871

1(119898))

such that1003816100381610038161003816⟨1198781198911 1198921⟩

1003816100381610038161003816 = 119878 = 110038171003817100381710038171198910 minus 1198911

1003817100381710038171003817 ⩽ 120576

10038171003817100381710038171198920 minus 11989211003817100381710038171003817 ⩽ 120576 119879 minus 119878 ⩽ 120576

(36)

To prove this proposition we need some backgroundon representation of operators on Lebesgue spaces on finiteregular positive Borel measures and several preliminarylemmas

Let 119898 be a finite regular positive Borel measure on acompact Hausdorff space Ω If 120583 is a complex-valued Borelmeasure on the product space Ω times Ω then define theirmarginal measures 120583119894 on Ω (119894 = 1 2) as follows 1205831(119860) =120583(119860 times Ω) and 1205832(119861) = 120583(Ω times 119861) where 119860 and 119861 are Borelmeasurable subsets of Ω

Let 119872(119898) be the complex Banach lattice of measuresconsisting of all complex-valued Borel measures 120583 on theproduct spaceΩ timesΩ such that |120583|119894 are absolutely continuouswith respect to119898 for 119894 = 1 2 endowed with the norm

1003817100381710038171003817100381710038171003817100381710038171003817

11988910038161003816100381610038161205831003816100381610038161003816

1

119889119898

1003817100381710038171003817100381710038171003817100381710038171003817infin

(37)

Each 120583 isin 119872(119898) defines a bounded linear operator 119879120583from

1198711(119898) to itself by

⟨119879120583(119891) 119892⟩ = int

ΩtimesΩ

119891 (119909) 119892 (119910) 119889120583 (119909 119910) (38)

where 119891 isin 1198711(119898) and 119892 isin 119871

infin(119898) Iwanik [22] showed that

the mapping 120583 997891rarr 119879120583is a lattice isometric isomorphism from

119872(119898) onto L(1198711(119898)) Even though he showed this for the

real case it can be easily generalized to the complex caseFor details see [22 Theorem 1] and [23 IV Theorem 15(ii)Corollary 2]

We will also use that given an arbitrary measure 120583 every119879 isin L(119871

1(120583)) satisfies V(119879) = 119879 [24] (that is the space

1198711(120583) has numerical index 1)

Lemma 11 (see [15 Lemma 33]) Let 119888119899 be a sequence of

complex numbers with |119888119899| ⩽ 1 for every 119899 and let 120578 gt 0 such

6 Abstract and Applied Analysis

that for a convex series sum120572119899 Resuminfin

119899=1120572119899119888119899gt 1 minus 120578 Then for

every 0 lt 119903 lt 1 the set 119860 = 119894 isin N Re 119888119894gt 119903 satisfies the

estimate

sum

119894isin119860

120572119894⩾ 1 minus

120578

1 minus 119903 (39)

From now on 119898 will be a finite regular positive Borelmeasure on the compact Hausdorff space Ω

Lemma 12 Suppose that there exist a nonnegative simplefunction 119891 isin 119878

1198711(119898)

and a function 119892 isin 119878119871infin(119898)

such that

Re ⟨119891 119892⟩ gt 1 minus 1205763

16 (40)

Then there exist a nonnegative simple function 1198911isin 119878

1198711(119898)

anda function 119892

1isin 119878

119871infin(119898)

such that

1198921(119909) = 120594supp(119891

1)(119909) + 119892 (119909) 120594

Ωsupp(1198911)(119909)

⟨1198911 119892

1⟩ = 1

1003817100381710038171003817119891 minus 119891110038171003817100381710038171lt 120576

1003817100381710038171003817119892 minus 11989211003817100381710038171003817infinlt radic120576 supp (119891

1) sub supp (119891)

(41)

Proof Let119891 = sum119898

119895=1(120573

119895119898(119861

119895))120594

119861119895

for some (120573119895) such that

120573119895⩾ 0 for all 119895 andsum119898

119895=1120573119895= 1 and119861

119895rsquos aremutually disjoint

By the assumption we have

Re ⟨119891 119892⟩ =119899

sum

119895=1

120573119895

1

119898 (119861119895)

int119861119895

Re119892 (119909) 119889119898 (119909) gt 1 minus 1205763

16

(42)

and letting

119869 = 119895 1 ⩽ 119895 ⩽ 1198991

119898 (119861119895)

int119861119895

Re119892 (119909) 119889119898 (119909) gt 1 minus 1205762

4

(43)

we have by Lemma 11

sum

119895isin119869

120573119895gt 1 minus

120576

4 (44)

For each 119895 isin 119869 we have

1 minus1205762

4lt

1

119898 (119861119895)

int119861119895

Re119892 (119909) 119889119898 (119909)

=1

119898 (119861119895)

int119861119895capRe119892⩽1minus120576

Re119892 (119909) 119889119898 (119909)

+ int119861119895capRe119892gt1minus120576

Re119892 (119909) 119889119898 (119909)

⩽1

119898 (119861119895)

((1 minus 120576)119898 (119861119895cap Re119892 ⩽ 1 minus 120576)

+ 119898 (119861119895cap Re119892 gt 1 minus 120576))

= 1 minus 120576

119898 (119861119895cap Re119892 ⩽ 1 minus 120576)

119898 (119861119895)

(45)

This implies that

119898(119861119895cap Re119892 ⩽ 1 minus 120576)

119898 (119861119895)

lt120576

4 (46)

Define 119861119895= 119861

119895cap Re119892 gt 1 minus 120576 for all 119895 isin 119869

1198911= (1sum

119895isin119869

120573119895)sum

119895isin119869

120573119895(120594

119861119895

119898 (119861119895)) (47)

and 1198921(119909) = 1 on supp(119891

1) and 119892

1(119909) = 119892(119909) elsewhereThen

it is clear that supp(1198911) sub supp(119891) 119892 minus 119892

1infinlt radic120576 and

⟨1198911 119892

1⟩ = 1 Finally we will show that 119891 minus 119891

1 lt 120576 Notice

first that10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

= 2sum

119895isin119869

120573119895

119898(119861119895 119861

119895)

119898 (119861119895)

lt120576

2

(48)

Hence1003817100381710038171003817119891 minus 1198911

1003817100381710038171003817

10038171003817100381710038171003817100381710038171003817100381710038171003817

1

sum119895isin119869120573119895

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus 119891

10038171003817100381710038171003817100381710038171003817100381710038171003817

1 minus sum119895isin119869120573119895

sum119895isin119869120573119895

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+120576

4

= (1 minussum

119895isin119869

120573119895) +

120576

2+120576

4

⩽120576

4+120576

2+120576

4= 120576

(49)

Lemma 13 (see [25 Lemma 33]) Suppose that 119879120583is a norm-

one element in L(1198711(119898)) for some 120583 isin 119872(119898) and there is

a nonnegative simple function 1198910such that 119891

0is a norm-one

element of 1198711(119898) and 119879

1205831198910 ⩾ 1 minus 120576

32

6 for some 0 lt 120576 lt 1

Abstract and Applied Analysis 7

Then there exist a norm-one bounded linear operator 119879] forsome ] isin 119872(119898119898) and a nonnegative simple function 119891

1

in 1198781198711(119898)

such that 119879120583minus 119879] ⩽ 120576 1198911 minus 1198910 ⩽ 3120576 and

(119889|]|1119889119898)(119909) = 1 for all 119909 isin supp(1198911)

Lemma 14 Suppose that 119879] isin L(1198711(119898)) is a norm-one

operator 119891 = sum119899

119894=1120573119894(120594

119861119894

119898(119861119894)) where 119898(119861

119895) gt 0 for all

1 ⩽ 119895 ⩽ 119899 and 119861119895119899

119895=1are mutually disjoint Borel subsets ofΩ

is a norm-one nonnegative simple function and 119892 is an elementof 119878

119871infin(119898)

such that

Re ⟨119892 119879]119891⟩ ⩾ 1 minus1205766

27(50)

for some 0 lt 120576 lt 1 and

119889|]|1

119889119898(119909) = 1 119892 (119909) = 1 (51)

for all x in the support of 119891Then there exist a nonnegative simple function 119891 isin 119878

1198711(119898)

a function119892 isin 119878

119871infin(119898)

and an operator119879] inL(1198711(119898) 1198711(119898))such that

⟨119892 119879]119891⟩ =1003817100381710038171003817119879]1003817100381710038171003817 = 1

1003817100381710038171003817119879] minus 119879]1003817100381710038171003817 ⩽ 2120576

10038171003817100381710038171003817119891 minus 119891

10038171003817100381710038171003817⩽ 3120576

1003817100381710038171003817119892 minus 1198921003817100381710038171003817 ⩽radic120576 ⟨119891 119892⟩ = 1

(52)

Proof Since

Re ⟨119892 119879]119891⟩ ⩾ 1 minus1205766

27 (53)

we have

1 minus1205766

27lt Re ⟨119892 119879]119891⟩

= intΩtimesΩ

119891 (119909)Re119892 (119910) 119889120583 (119909 119910)

=

119899

sum

119895=1

120573119895intΩtimesΩ

120594119861119895

(119909)

119898 (119861119895)

Re119892 (119910) 119889] (119909 119910)

(54)

Let

119869 = 119895 intΩtimesΩ

(

120594119861119895

(119909)

119898 (119861119895)

)Re119892 (119910) 119889] (119909 119910) gt 1 minus 1205763

26

(55)

Then from Lemma 11 we have sum119895isin119869120573119895gt 1 minus 120576

32 Let 119891

1=

sum119895isin119869120573119895(120594

119861119895

119898(119861119895)) where 120573

119895= 120573

119895(sum

119895isin119869120573119895) for all 119895 isin 119869

Then

10038171003817100381710038171198911 minus 1198911003817100381710038171003817 ⩽

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

(120573119895minus 120573

119895)

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+ sum

119895isin119869

120573119895⩽ 120576

3⩽ 120576 (56)

Note that there is a Borel measurable function ℎ on Ω times Ωsuch that 119889](119909 119910) = ℎ(119909 119910) 119889|]|(119909 119910) and |ℎ(119909 119910)| = 1 forall (119909 119910) isin Ω times Ω Let

119862 = (119909 119910) 1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 1

1003816100381610038161003816 ltradic120576

232 (57)

Define two measures ]119891and ]

119888as follows

]119891(119860) = ] (119860 119862) ]

119888(119860) = ] (119860 cap 119862) (58)

for every Borel subset 119860 ofΩ times Ω It is clear that

119889] = 119889]119891+ 119889]

119888 119889

10038161003816100381610038161003816]119891

10038161003816100381610038161003816= ℎ 119889]

119891

1198891003816100381610038161003816]1198881003816100381610038161003816 = ℎ 119889]119888 119889 |]| = 119889

10038161003816100381610038161003816]119891

10038161003816100381610038161003816+ 1198891003816100381610038161003816]1198881003816100381610038161003816

(59)

Since (119889|]|11198891198981)(119909) = 1 for all 119909 isin ⋃119899

119895=1119861119895 we have

1 =119889|]|1

1198891198981

(119909) =

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) +1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909) (60)

for all 119909 isin 119861 = ⋃119899

119895=1119861119895 and we deduce that |]|1(119861

119895) = 119898

1(119861

119895)

for all 1 ⩽ 119895 ⩽ 119899We claim that |]

119891|1(119861

119895)119898

1(119861

119895) ⩽ 120576

22

2 for all 119895 isin 119869Indeed if |119892(119910)ℎ(119909 119910) minus 1| ⩾ radic120576232 then

Re (119892 (119910) ℎ (119909 119910)) ⩽ 1 minus 12057624 (61)

So we have

1 minus1205763

26

⩽1

1198981(119861

119895)

ReintΩtimesΩ

120594119861119895(119909)119892 (119910) 119889] (119909 119910)

=1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 |]| (119909 119910)

=1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 10038161003816100381610038161003816]11989110038161003816100381610038161003816(119909 119910)

+1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 1003816100381610038161003816]1198881003816100381610038161003816 (119909 119910)

⩽1

1198981(119861

119895)

((1 minus120576

24)10038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

(119861119895) +1003816100381610038161003816]1198881003816100381610038161003816

1

(119861119895))

= 1 minus120576

24

10038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

(119861119895)

1198981(119861

119895)

(62)

This proves our claimWe also claim that for each 119895 isin 119869 there exists a Borel

subset 119861119895of 119861

119895such that

(1 minus120576

2)119898

1(119861

119895) ⩽ 119898

1(119861

119895) ⩽ 119898

1(119861

119895)

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) ⩽120576

2

(63)

8 Abstract and Applied Analysis

for all 119909 isin 119861119895 Indeed set 119861

119895= 119861

119895cap 119909 isin Ω (119889|]

119891|1119889119898

1)

(119909) ⩽ 1205762 Then

int119861119895119861119895

120576

2119889119898

1(119909) ⩽ int

119861119895

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) 1198891198981(119909)

=10038161003816100381610038161003816]1119891

10038161003816100381610038161003816(119861

119895) ⩽1205762

22119898

1(119861

119895)

(64)

This shows that 1198981(119861

119895 119861

119895) ⩽ (1205762)119898

1(119861

119895) This proves our

second claimNow we define 119892 by 119892(119910) = 119892(119910)|119892(119910)| if |119892(119910)| ⩾ 1 minus

radic120576232 and 119892(119910) = 119892(119910) if |119892(119910)| lt 1 minus radic120576232 and we

write 119891 = sum119895isin119869120573119895(120594

119861119895

1198981(119861

119895)) It is clear that 119892 isin 119878

119871infin(119898)

119892 minus 119892 lt radic120576 and 119892(119910) = 1 for all 119909 isin supp119891

Finally we define the measure

119889] (119909 119910)

= sum

119895isin119869

120594119861119895

(119909) 119892 (119910)ℎ (119909 119910)119889]119888(119909 119910)(

1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+ 1205941198691119861(119909) 119889] (119909 119910)

(65)

where 119861 = ⋃119895isin119869119861119895 It is easy to see that (119889|]|1119889119898

1)(119909) = 1

on 119861 and (119889|]|11198891198981)(119909) ⩽ 1 elsewhere Note that

119889 (] minus ]) (119909 119910)

= sum

119895isin119869

120594119861119895

(119909) [

[

119892 (119910)ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1]

]

119889]119888(119909 119910)

minus sum

119895isin119869

120594119861119895

(119909) 119889]119891(119909 119910)

(66)

If (119909 119910) isin 119862 then |119892(119910)| ⩾ 1 minus radic120576232 ⩾ 1 minus 1232 and

100381610038161003816100381610038161003816119892 (119910)ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

119892 (119910)

1003816100381610038161003816119892 (119910)1003816100381610038161003816

ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816100381610038161003816

1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 11003816100381610038161003816

1003816100381610038161003816119892 (119910)1003816100381610038161003816

+

10038161003816100381610038161 minus1003816100381610038161003816119892 (119910)

1003816100381610038161003816

10038161003816100381610038161003816100381610038161003816119892 (119910)

1003816100381610038161003816

⩽ 2

1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 11003816100381610038161003816

1003816100381610038161003816119892 (119910)1003816100381610038161003816

⩽ 2radic120576

232

232

232 minus 1⩽ 2radic120576

(67)

Hence for all (119909 119910) isin 119862 we have

10038161003816100381610038161003816100381610038161003816100381610038161003816

119892 (119910)ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

⩽100381610038161003816100381610038161003816119892 (119910)ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

⩽ 2radic120576(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

(68)

So we have for all 119909 isin 1198691

119889|] minus ]|1

1198891198981

(119909)

⩽ sum

119895isin119869

120594119861119895

(119909) [

[

2radic120576(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

]

]

1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909)

+ sum

119895isin119869

120594119861119895

(119909)

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909)

⩽ sum

119895isin119869

120594119861119895

(119909)(2radic120576 + (1 minus1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909)))

+sum

119895isin119869

120594119861119895

(119909)(

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909))

⩽ 2radic120576 + 120576 lt 3radic120576

(69)

This gives that 119879] minus 119879] lt 3radic120576 Note also that for all 119895 isin 119869

⟨119879]

120594119861119895

1198981(119861

119895)

119892⟩

= intΩtimesΩ

120594119861119895

(119909)

1198981(119861

119895)

119892 (119910) 119889] (119909 119910)

= intΩtimesΩ

120594119861119895

(119909)

1198981(119861

119895)

ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

119889]119888(119909 119910)

Abstract and Applied Analysis 9

= intΩ

120594119861119895

(119909)

1198981(119861

119895)

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

1198891003816100381610038161003816]1198881003816100381610038161003816

1

(119909)

= intΩ

120594119861119895

(119909)

1198981(119861

119895)

1198891198981(119909) = 1

(70)

Hence we get ⟨119879]119891 119892⟩ = 1 which implies that 119879]119891 =119879] = 1 Finally10038171003817100381710038171003817119891 minus 119891

10038171003817100381710038171003817

⩽10038171003817100381710038171003817119891 minus 119891

1

10038171003817100381710038171003817+10038171003817100381710038171198911 minus 119891

1003817100381710038171003817

=

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

1198981(119861

119895)

minus sum

119895isin119869

120573119895

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+ 120576

⩽ sum

119895isin119869

120573119895(

10038171003817100381710038171003817100381710038171003817100381710038171003817

120594119861119895

1198981(119861

119895)

minus

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

120594119861119895

1198981(119861

119895)

minus

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

) + 120576

= 2sum

119895isin119869

120573119895

1198981(119861

119895 119861

119895)

1198981(119861

119895)

+ 120576

⩽ 2sum

119895isin119869

120573119895

(1205762)1198981(119861

119895)

1198981(119861

119895)

+ 120576 ⩽120576

1 minus 1205762+ 120576 lt 3120576

(71)

We are now ready to present the proof of the main resultin the case of finite regular positive Borel measures

Proof of Proposition 10 Let 1205751= 120575

3

2(5sdot2

4) 120575

2= 120575

12

3(3

2sdot2

14)

and 1205753= (12057610)

2 for some 0 lt 120576 lt 1 Suppose that 119879 isinL(119871

1(119898)) with 119879 = 1 and that there is an 119891

0isin 119878

1198711(119898)

and1198920isin 119878

119871infin(119898)

such that ⟨1198910 119892

0⟩ = 1 and |⟨119879119891

0 119892

0⟩| gt 1minus120575

3

12

6Then there is an isometric isomorphism Ψ from 119871

1(119898) onto

itself such thatΨ(1198910) = |119891

0| and there is a scalar number 120572 in

119878R such that |⟨1198791198910 119892

0⟩| = ⟨120572119879119891

0 119892

0⟩ Then letting 119891

1= Ψ119891

0

1198921= (Ψ

minus1)lowast1198920 and 119879

1= 120572Ψ119879Ψ

minus1 we have

⟨1198781198911 119892

1⟩ = ⟨120572Ψ119879

minus1Ψ119891

0 (Ψ

minus1)lowast

1198920⟩

= ⟨1205721198791198910 119892

0⟩ gt 1 minus

1205753

1

26

⟨1198911 119892

1⟩ = ⟨Ψ119891

0 (Ψ

minus1)lowast

1198920⟩ = 1

(72)

Since 11987911198911 gt 1 minus 120575(120575

3

12

6) by Lemma 13 there exists a

norm-one bounded operator 119879] and a nonnegative simplefunction119891

2isin 119878

1198711(119898)

such that 1198791minus119879] ⩽ 1205751 1198912minus1198911 ⩽ 31205751

and (119889|]|11198891198981)(119909) = 1 for all 119909 isin supp(119891

2) Then

⟨119879]1198912 1198921⟩

= ⟨11987911198911 119892

1⟩ minus ⟨119879

11198911minus 119879

11198912 119892

1⟩ minus ⟨119879

11198912minus 119879]1198912 1198921⟩

⩾ ⟨11987911198911 119892

1⟩ minus10038171003817100381710038171198911 minus 1198912

1003817100381710038171003817 minus10038171003817100381710038171198791 minus 119879]

1003817100381710038171003817

⩾ 1 minus1205753

1

26minus 3120575

1minus 120575

1⩾ 1 minus 5120575

1= 1 minus

1205753

2

16

(73)

Notice also that

⟨1198912 119892

1⟩ = ⟨119891

1 119892

1⟩ minus ⟨119891

1minus 119891

2 119892

1⟩ ⩾ 1 minus

10038171003817100381710038171198911 minus 11989121003817100381710038171003817

⩾ 1 minus 31205751⩾ 1 minus 5120575

1= 1 minus

1205753

2

16

(74)

By Lemma 12 there are a nonnegative simple function 1198913isin

1198781198711(119898)

and a function 1198923isin 119878

119871infin(119898)

such that

1198923(119909) = 120594supp119891

3

(119909) + 1198922(119909) 120594

Ωsupp1198913

(119909)

10038171003817100381710038171198912 minus 11989131003817100381710038171003817 ⩽ 1205752

10038171003817100381710038171198923 minus 11989211003817100381710038171003817 ⩽radic120575

2 ⟨119891

3 119892

3⟩ = 1

(75)

So we have

⟨119879]1198913 1198923⟩ = ⟨119879]1198912 1198921⟩ minus ⟨119879]1198912 minus 119879]1198913 1198921⟩

minus ⟨119879]1198913 1198921 minus 1198923⟩

⩾ 1 minus1205753

2

16minus 2radic120575

2⩾ 1 minus 3radic120575

2= 1 minus

1205756

3

27

(76)

By Lemma 14 there exist 1198914isin 119878

1198711(119898)

and 1198924isin 119878

119871infin(119898)

and anoperator 119879

4such that

⟨1198924 119879

41198914⟩ = 1 =

100381710038171003817100381711987941003817100381710038171003817

10038171003817100381710038171198794 minus 119879]1003817100381710038171003817 ⩽ 21205753

10038171003817100381710038171198914 minus 11989131003817100381710038171003817 ⩽ 31205753

10038171003817100381710038171198924 minus 11989231003817100381710038171003817 ⩽radic120575

3 ⟨119891

4 119892

4⟩ = 1

(77)

So we have

10038171003817100381710038171198794 minus 11987911003817100381710038171003817 ⩽10038171003817100381710038171198794 minus 119879]

1003817100381710038171003817 +1003817100381710038171003817119879] minus 1198791

1003817100381710038171003817

⩽ 1205751+ 2120575

3⩽ 3120575

3

10038171003817100381710038171198911 minus 11989141003817100381710038171003817 ⩽10038171003817100381710038171198911 minus 1198912

1003817100381710038171003817 +10038171003817100381710038171198912 minus 1198913

1003817100381710038171003817 +10038171003817100381710038171198913 minus 1198914

1003817100381710038171003817

⩽ 31205751+ 120575

2+ 3120575

3⩽ 10120575

3

10038171003817100381710038171198921 minus 11989241003817100381710038171003817 ⩽10038171003817100381710038171198921 minus 1198923

1003817100381710038171003817 + 1198923 minus 1198924

⩽ 1205752+ radic120575

3⩽ 2radic120575

3

(78)

10 Abstract and Applied Analysis

Let 119878 = 120572Ψminus11198794Ψ 119891 = Ψminus1

1198914 and 119892 = Ψlowast

1198924 then we have

119879 minus 119878 =10038171003817100381710038171003817119879 minus 120572Ψ

minus11198794Ψ10038171003817100381710038171003817=10038171003817100381710038171003817120572Ψ119879Ψ

minus1minus 119879

4

10038171003817100381710038171003817

=10038171003817100381710038171198791 minus 1198794

1003817100381710038171003817 ⩽ 31205753

100381710038171003817100381710038171198910minus 11989110038171003817100381710038171003817=100381710038171003817100381710038171198910minus Ψ

minus11198914

10038171003817100381710038171003817=10038171003817100381710038171198911 minus 1198914

1003817100381710038171003817 ⩽ 101205753

100381710038171003817100381710038171198920minus 11989110038171003817100381710038171003817=10038171003817100381710038171198920 minus Ψ

lowast1198924

1003817100381710038171003817 =100381710038171003817100381710038171003817(Ψ

minus1)lowast

1198920minus 119892

4

100381710038171003817100381710038171003817

=10038171003817100381710038171198921 minus 1198924

1003817100381710038171003817 ⩽ 2radic120575

3

⟨119891 119892⟩ = ⟨Ψminus11198914 Ψ

lowast1198924⟩ = ⟨119891

4 119892

4⟩ = 1

10038161003816100381610038161003816⟨119878119891 119892⟩

10038161003816100381610038161003816=10038161003816100381610038161003816⟨120572Ψ

minus11198794ΨΨ

minus11198914 Ψ

lowast1198924⟩10038161003816100381610038161003816= |120572| = 1

(79)

This completes the proof

Finally we may give the proof of the main result in fullgenerality

Proof of Theorem 9 Notice that the Kakutani representationtheorem (see [26] for a reference) says that for every 120590-finitemeasure ] the space 119871

1(]) is isometrically isomorphic to

1198711(119898) for some positive Borel regular measure on a compact

Hausdorff space Then by Proposition 10 there is a universalfunction 120576 997891rarr 120578(120576) gt 0 which gives the BPBp-nu for 119871

1(]) for

every 120590-finite measure ]Fix 120576 gt 0 Suppose that 119879

0isin L(119871

1(120583)) with V(119879

0) = 1

and (1198910 119891

lowast

0) isin Π(119871

1(120583)) satisfy

1003816100381610038161003816⟨119891lowast

0 119879

01198910⟩1003816100381610038161003816 gt 1 minus 120578 (120576) (80)

Choose a sequence 119891119899 in 119871

1(120583) such that sup

119899119879

0119891119899 = 1

and let 119866 be the closed linear span of

119879119899

0119891119898 119899 119898 isin N cup 0 (81)

As 119866 is separable there is a dense subset 119892119899 119899 isin N of 119866

and let 119864 = ⋃infin

119899=1supp119892

119899 where supp119892

119899is the support of 119892

119899

Then the measure 120583|119864is 120590-finite Let

119884 = 119891 isin 1198711(120583) supp (119891) sub 119864 (82)

be a closed subspace of 1198711(120583) It is clear that 119871

1(120583) = 119884oplus

1119885

and 119884 is isometrically isomorphic to 1198711(120583|

119864) So 119884 has the

BPBp-nu with 120578(120576)Now write 119878

0= 119879

0|119884 119884 rarr 119884 consider 119910

0= 119891

0isin

119878119884 119910lowast

0= 119891

lowast

0|119884isin 119878

119884lowast and observe that 119910lowast

0(119910

0) = 1 and

|119910lowast

0(119878

01199100)| = |119891

lowast

0(119879

01198910)| gt 1 minus 120578(120576) Hence there exist 119878 isin

L(119884) and (1199100 119910

lowast

0) isin Π(119884) such that

1003816100381610038161003816119910lowast

0(119878119910

0)1003816100381610038161003816 = 1 = V (119878) 1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

10038171003817100381710038171199100 minus 11991001003817100381710038171003817 lt 120576

1003817100381710038171003817119910lowast

0minus 119910

lowast

0

1003817100381710038171003817 lt 120576

(83)

Finally consider the operator 119879 isinL(1198711(120583)) given by

119879 (119910 119911)

= (119878119910 0) + 1198790(0 119911) ((119910 119911) isin 119871

1(120583) equiv 119884oplus

1119885)

(84)

We have 119879 = 1 (and so V(119879) = 1) Indeed1003817100381710038171003817119879 (119910 119911)

1003817100381710038171003817 =1003817100381710038171003817(119878119910 0)

1003817100381710038171003817 +10038171003817100381710038171198790 (0 119911)

1003817100381710038171003817 ⩽10038171003817100381710038171199101003817100381710038171003817 + 119911 =

1003817100381710038171003817(119910 119911)1003817100381710038171003817

(85)

for all (119910 119911) isin 1198711(120583) and 119879(119910

0 0) = (119878119910

0 0) = 119878119910

0 = 1

Let 119909 = (1199100 0) and 119909lowast = (119910lowast

0 119891

0|119885) Then (119909 119909lowast) isin Π(119871

1(120583))

Moreover we have1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119910

lowast

0119878119910

0

1003816100381610038161003816 = 1 = V (119879) 1003817100381710038171003817119909 minus 1198910

1003817100381710038171003817 =1003817100381710038171003817119910 minus 1199100

1003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowast

0minus 119891

lowast

0

1003817100381710038171003817

= max 1003817100381710038171003817119910 minus 119891lowast

0|119884

1003817100381710038171003817 1003817100381710038171003817119891

lowast

0|119885minus 119891

lowast

0|119885

1003817100381710038171003817

=1003817100381710038171003817119910

lowastminus 119910

lowast

0

1003817100381710038171003817 lt 120576

1003817100381710038171003817119879 minus 11987901003817100381710038171003817 = sup119910+119911⩽1

1003817100381710038171003817119879 (119910 119911) minus 1198790 (119910 119911)1003817100381710038171003817

= sup119910⩽1

1003817100381710038171003817119878119910 minus 11987801199101003817100381710038171003817 =1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

(86)

This completes the proof

5 Examples of Spaces Failing theBishop-Phelps-Bollobaacutes Property forNumerical Radius

Our goal here is to prove that the density of numerical radiusattaining operators does not imply the BPBp-nu Actuallywe will show that among separable spaces there is noisomorphic property implying the BPBp-nu other than finite-dimensionality

We need to relate the BPBp-nu to the Bishop-Phelps-Bollobas property for operators which as mentioned in theintroduction was introduced in [15] A pair (119883 119884) of Banachspaces has the Bishop-Phelps-Bollobas property for operators(in short BPBp) if given 120576 gt 0 there exists 120578(120576) gt 0 suchthat given 119879 isin L(119883 119884) with 119879 = 1 and 119909 isin 119878

119883such that

119879119909 gt 1 minus 120578(120576) then there exist 119911 isin 119878119883and 119878 isin L(119883 119884)

satisfying

119878 = 119878119911 = 1 119909 minus 119911 lt 120576

119879 minus 119878 lt 120576

(87)

We refer the reader to [15 19 25] and references therein formore information and background Among the interestingresults on the BPBp we emphasize that a pair (119883 119884) when119883is finite-dimensional does not necessarily have the BPBp Forinstance if119884 is a strictly convex space which is not uniformlyconvex then the pair (ℓ(2)

1 119884) fails to have the BPBp (this is

contained in [15] see [19 Section 3])

Abstract and Applied Analysis 11

The next result relates the BPBp-nu to the BPBp foroperators in a particular case We will deduce our examplefrom it

Theorem 15 If 1198711(120583)oplus

1119883 has the BPBp-119899119906 then the pair

(1198711(120583) 119883) has the BPBp for operators

Before proving this proposition we will use it to get themain examples of this section The first example shows thatthe density of numerical radius attaining operators does notimply the BPBp-nu

Example 16 There is a reflexive space (and so numerical radiusattaining operators on it are dense) which fails to have theBPBp-nu Indeed let 119884 be a reflexive separable space whichis not superreflexive and we may suppose that 119884 is strictlyconvex Observe that 119884 cannot be uniformly convex since itis not superreflexive Now 119883 = ℓ(2)

1oplus1119884 is reflexive but the

pair (ℓ(2)1 119884) fails the BPBp since 119884 is strictly convex but not

uniformly convex [19 Corollary 33] Therefore Theorem 15gives us that119883 does not have the BPBp-nu

The example above can be extended to get the result thatevery infinite-dimensional separable Banach space can berenormed to fail the BPBp-nu This follows from the factthat every infinite-dimensional separable Banach space canbe renormed to be strictly convex but not uniformly convex(this result can be proved ldquoby handrdquo an alternative categoricalargument for it can be found in [27] and references therein)With a little more of effort we may get the main result of thesection

Theorem 17 Every infinite-dimensional separable Banachspace can be renormed to fail the weak-BPBp-nu (and so inparticular to fail the BPBp-nu)

We need the following result which is surely well knownAs we have not found a reference we include a nice and easyproof kindly given to us by Vladimir Kadets We recall thatgiven a Banach space 119884 the set of all equivalent norms on119884 can be viewed as a metric space using the Banach-Mazurdistance

Lemma 18 Let119884 be an infinite-dimensional separable Banachspace Then the set of equivalent norms on 119884 which are strictlyconvex and are not (locally) uniformly convex is dense in the setof all equivalent norms on119884 (with respect to the Banach-Mazurdistance)

Proof Fix 119890 isin 119878119884and 119890lowast

1isin 119878

119884lowast such that 119890lowast

1(119890) = 1 For a fixed

120576 isin (0 12) denote

119902 (119910) = max (1 minus 120576) 10038171003817100381710038171199101003817100381710038171003817 1003816100381610038161003816119890lowast

1(119910)1003816100381610038161003816 (119910 isin 119884) (88)

Evidently (1 minus 120576)119910 ⩽ 119902(119910) ⩽ 119910 for every 119910 isin 119884 Fix asequence 119890lowast

119896 119896 ⩾ 2 of norm-one functionals separating

the points of 119884 and denote

119901 (119910) = radic

infin

sum

119896=1

1

2119896

1003816100381610038161003816119890lowast

119896(119910)1003816100381610038161003816

2

(119910 isin 119884) (89)

Then 119901 is a strictly convex norm on 119884 119901(119890) ⩾ 1radic2 and119901(119910) ⩽ 119910 for all 119910 isin 119883 Finally write

100381710038171003817100381711991010038171003817100381710038171= (1 minus 120576) 119902 (119910) + 120576

119901 (119910)

119901 (119890)(119910 isin 119884) (90)

Then sdot 1is a strictly convex norm on 119884 and

(1 minus 120576)2 10038171003817100381710038171199101003817100381710038171003817 ⩽100381710038171003817100381711991010038171003817100381710038171⩽ (1 + 120576)

10038171003817100381710038171199101003817100381710038171003817 (119910 isin 119884) (91)

Wewill finish the proof by showing that sdot 1is not uniformly

convex (actually it is not locally uniformly convex) Indeedfor each 119899 isin N we select 119910

119899isin ⋂

119899

119896=1ker 119890lowast

119896with 119910

119899 = 1 and

consider 119890119899= 119890 + (1205764)119910

119899 Then 119902(119890) = 1 119902(119890

119899) = 1 and

119902(119890+119890119899) = 2 At the same time 119901(119910

119899) rarr 0 so 119901(119890

119899) rarr 119901(119890)

and 119901(119890 + 119890119899) rarr 2119901(119890) Consequently

1198901 = 1100381710038171003817100381711989011989910038171003817100381710038171997888rarr 1

1003817100381710038171003817119890 + 11989011989910038171003817100381710038171997888rarr 2 (92)

but 119890 minus 1198901198991= (1205764)119910

1198991⩾ (1 minus 120576)

2(1205764) which means the

absence of local uniform convexity at 119890

Proof of Theorem 17 Let 119883 be an infinite-dimensional sepa-rable Banach space Take a closed subspace 119884 of 119883 of codi-mension two By [28 Proposition 2] the map carrying everyequivalent norm on 119884 to its numerical index is continuousand so the set of values of the numerical index of 119884 upto reforming is a nontrivial interval [28 Theorem 9] ThenLemma 18 allows us to find an equivalent norm | sdot | on 119884 insuch a way that (119884 | sdot |) is strictly convex and is not uniformlyconvex and 119899(119884 | sdot |) gt 0 Now the space 119883 = ℓ(2)

1oplus1(119884 | sdot |)

is an equivalent renorming of 119883 which does not have theBPBp-119899119906 (indeed otherwise the pair (ℓ(2)

1 (119884 | sdot |)) would

have the BPBp for the operator norm and so (119884 | sdot |) wouldbe uniformly convex by [19 Corollary 33] a contradiction)Moreover as

119899 (119883) = min 119899 (ℓ(2)1) 119899 (119884 |sdot|) gt 0 (93)

(see [20 Proposition 2] for instance) 119883 also fails the weak-BPBp-nu by Proposition 6

To finish the section with the promised proof ofTheorem 15 we first see the following stability result

Lemma 19 Let119883 = [⨁infin

119896=1119883

119896]1198880

or [⨁infin

119896=1119883

119896]ℓ1

If119883 has theBishop-Phelps-Bollobas property for numerical radius with afunction 120578 then each Banach space 119883

119894has the Bishop-Phelps-

Bollobas property for numerical radius with 120578nu(119883119894) ⩾ 120578 That

is inf119894120578nu(119883119894

)(120576) ⩾ 120578nu(119883)(120576) for all 0 lt 120576 lt 1

Proof Let 119875119894 119883 rarr 119883

119894and 1198751015840

119894 119883

lowastrarr 119883

lowast

119894be the natural

projections and let 119876119894 119883

119894rarr 119883 and 1198761015840

119894 119883

lowast

119894rarr 119883

lowast be thenatural embeddings

Assume that an operator 119879119894 119883

119894rarr 119883

119894and a pair

(119909119894 119909

lowast

119894) isin Π(119883

119894) satisfy that

V (119879119894) = 1

1003816100381610038161003816119909lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (94)

We define an operator 119879 119883 rarr 119883 and (119909 119909lowast) isin Π(119883) by

119879 = 119876119894∘ 119879

119894∘ 119875

119894 (119909 119909

lowast) = (119876

119894119909119894 119876

1015840

119894119909lowast

119894) (95)

12 Abstract and Applied Analysis

then clearly we see that1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119909

lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (96)

From the assumption there exist 119878 119883 rarr 119883 and a pair(119910 119910

lowast) isin Π(119883) such that

1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 = 1 = V (119878) 119878 minus 119879 lt 120576

1003817100381710038171003817119910lowastminus 119909

lowast1003817100381710038171003817 lt 1205761003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576

(97)

Since this clearly shows that1003817100381710038171003817119875119894 ∘ 119878 ∘ 119876119894

minus 119879119894

1003817100381710038171003817 lt 120576100381710038171003817100381710038171198751015840

119894119910lowastminus 119909

lowast

119894

10038171003817100381710038171003817lt 120576

1003817100381710038171003817119875119894119910 minus 1199091198941003817100381710038171003817 lt 120576

(98)

we only need to show that |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = 1

We first show the case of 1198880sum Since

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817=10038171003817100381710038171003817119875119895119910 minus 119875

11989511990910038171003817100381710038171003817⩽1003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576 (99)

for every 119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽100381710038171003817100381710038171198751015840

119894119910lowast10038171003817100381710038171003817+ 120576 sum

119895isinN119895 = 119894

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817⩽1003817100381710038171003817119910

lowast1003817100381710038171003817 = 1

(100)

This shows that 1198751015840119894119910lowast = 1 and 1198751015840

119895119910lowast= 0 for every 119895 = 119894

So 119910lowast = 1198761015840

1198941198751015840

119894119910lowast and 1198751015840

119894119910lowast(119875

119894119910) = 1 This and the fact that

119910 minus 119876119894119875119894119910 lt 120576 imply that

(119876119894119875119894119910 + (

1

120576) (119910 minus 119876

119894119875119894119910) 119876

1015840

1198941198751015840

119894119910lowast) isin Π (119883) (101)

So we get that (1198761015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910 + (1120576)(119910 minus 119876

119894119875119894119910))V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =10038161003816100381610038161003816(119876

1015840

1198941198751015840

119894119910lowast) 11987811991010038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910 +1

120576(119910 minus 119876

119894119875119894119910))

1003816100381610038161003816100381610038161003816⩽ 1

(102)

and so we get |1198751015840119894119910lowast(119875

119894∘119878 ∘119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

We next show the case of ℓ1sum The proof is almost

the same as that of the 1198880case However for the sake of

completeness we provide it hereSince 1198751015840

119895119910lowast = 119875

1015840

119895119910lowastminus 119875

1015840

119895119909lowast ⩽ 119910

lowastminus 119909

lowast lt 120576 for every

119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽10038171003817100381710038171198751198941199101003817100381710038171003817 + 120576 sum

119895isinN119895 = 119894

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817⩽10038171003817100381710038171199101003817100381710038171003817 = 1

(103)

which shows 119875119894119910 = 1 and 119875

119895119910 = 0 for every 119895 = 119894 Since this

implies (119876119894119875119894119910119876

1015840

1198941198751015840

119894119910lowast+(1120576)(119910

lowastminus119876

1015840

1198941198751015840

119894119910lowast)) isin Π(119883) we get

that |(1198761015840

1198941198751015840

119894119910lowast+ (1120576)(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast))119878(119876

119894119875119894119910)| ⩽ V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =1003816100381610038161003816119910

lowast119878 (119876

119894119875119894119910)1003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast+1

120576(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast)) 119878 (119876

119894119875119894119910)

1003816100381610038161003816100381610038161003816⩽ 1

(104)

and so |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

Proof of Theorem 15 Note that 120578nu(1198711(120583)oplus1119883)(120576) rarr 0 as120576 rarr 0 Fix 0 lt 120576

0lt 1 and choose 0 lt 120576 lt 1 such that

6120576 + 120578nu(1198711(120583)oplus1119883)(120576) lt 1205760 Let 120578(1205760) = 120578nu(1198711(120583)oplus1119883)(120576)Suppose that 119879

0isin L(119871

1(120583) 119883) with 119879

0 = 1 and 119891

0isin

1198781198711(120583)

satisfy

1003817100381710038171003817119879011989101003817100381710038171003817 gt 1 minus 120578 (1205760) (105)

For any measurable subset 119861 let

1198711(120583|

119861) = 119891|

119861 119891 isin 119871

1(120583) (106)

with the norm 119891|119861 = 119891120594

1198611 Then it is easy to see

that 1198711(120583|

119861) is isometrically isomorphic to a complemented

subspace of 1198711(120583) Let 119875

119861 119871

1(120583) rarr 119871

1(120583|

119861) be the

restriction defined by 119875119861(119891) = 119891|

119861for all 119891 isin 119871

1(120583) and

let 119869119861 119871

1(120583|

119861) rarr 119871

1(120583) be the extension defined by

119869119861(119891)(120596) = 119891(120596) if 120596 isin 119861 and 119869

119861(119891)(120596) = 0 otherwise

It is clear that 119875119861119869119861= Id

1198711(120583|119861)and 119869

119861119875119861(119891) = 119891120594

119861for all

119891 isin 1198711(120583) Notice also that 119871

1(120583) is isometrically isomorphic

to 1198711(120583|

119861)oplus

11198711(120583|

119861119888)

Let 119860 = supp1198910and 119892

0= 119875

1198601198910 Then

1003817100381710038171003817119879011986911986011989201003817100381710038171003817 =100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) gt 0 (107)

and define the operator 119879119860 119871

1(120583|

119860) rarr 119883 by 119879

119860119891 =

1198790119869119860119891119879

0119869119860 for every 119891 isin 119871

1(120583|

119860) Then

100381710038171003817100381711987911986011989201003817100381710038171003817 ⩾100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) (108)

Since 120583|119860is 120590-finite 119871

1(120583|

119860)lowast= 119871

infin(120583|

119860) Let 119892lowast

0isin 119878

119871infin(120583|119860)

be a function such that ⟨119892lowast0 119892

0⟩ = 1 and choose 119909lowast

0isin 119878

119883lowast

such that 119909lowast0(119879

1198601198920) = 119879

1198601198920 Define the operator 119878

0isin

L(1198711(120583|

119860)oplus

1119883) by

1198780(119891 119909) = (0 119879

119860119891) ((119891 119909) isin 119871

1(120583|

119860) oplus

1119883) (109)

and observe that 1198780 = V(119878

0) = 1 Indeed

100381710038171003817100381711987801003817100381710038171003817 ⩽ 1 =

10038171003817100381710038171198791198601003817100381710038171003817

= sup 1003816100381610038161003816119909lowast1198791198601198911003816100381610038161003816 119909

lowastisin 119878

119883lowast 119891 isin 119878

1198711(120583|119860)

= sup 1003816100381610038161003816(119891lowast 119909

lowast) 119878

0(119891 119909)

1003816100381610038161003816

((119891lowast 119909

lowast) (119891 119909)) isin Π (119871

1(120583|

119860) oplus

1119883)

= V (1198780) ⩽100381710038171003817100381711987801003817100381710038171003817

(110)

Abstract and Applied Analysis 13

It is immediate that

(119892lowast

0 119909

lowast

0) 119878

0(119892

0 0) = 119909

lowast

0(119879

1198601198920) =10038171003817100381710038171198791198601198920

1003817100381710038171003817 gt 1 minus 120578 (1205760)

(111)

By Lemma 19 1198711(120583|

119860)oplus

1119883 has the BPBp-nu with the func-

tion 120578 Therefore there exist 1198781isin L(119871

1(120583|

119860)oplus

1119883) (119892

1 119909

1) isin

1198781198711(120583|119860)oplus1119883 and (119892lowast

1 119909

lowast

1) isin 119878

119871infin(120583|119860)oplusinfin119883lowast such that

1003817100381710038171003817(1198921 1199091) minus (1198920 0)1003817100381710038171003817 lt 120576

1003817100381710038171003817(119892lowast

1 119909

lowast

1) minus (119891

lowast

0 119909

lowast

0)1003817100381710038171003817 lt 120576

10038171003817100381710038171198781 minus 11987801003817100381710038171003817 lt 120576 ⟨(119892

lowast

1 119909

lowast

1) (119892

1 119909

1)⟩ = 1

1003816100381610038161003816(119892lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 = V (119878

1) = 1

(112)

Claim 1 We claim that 1199091= 0

Otherwise

1 = Re ⟨(119892lowast1 119909

lowast

1) (119892

1 119909

1)⟩

=100381710038171003817100381711989211003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

+100381710038171003817100381711990911003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

⟩ ⩽ 1

(113)

We deduce that

((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

(119892lowast

1 119909

lowast

1)) (

(0 1199091)

100381710038171003817100381711990911003817100381710038171003817

(119892lowast

1 119909

lowast

1))

isin Π (1198711(120583|

119860) oplus

1119883)

(114)

Since 1198781((0 119909

1)119909

1) = (119878

1minus 119878

0)((0 119909

1)119909

1) lt 120576 we

get that

1 =1003816100381610038161003816⟨(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)⟩1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

100381710038171003817100381711989211003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

)⟩

+100381710038171003817100381711990911003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

)⟩

100381610038161003816100381610038161003816100381610038161003816

⩽100381710038171003817100381711989211003817100381710038171003817 V (1198781) + 120576

100381710038171003817100381711990911003817100381710038171003817 lt100381710038171003817100381711989211003817100381710038171003817 +100381710038171003817100381711990911003817100381710038171003817 = 1

(115)

a contradiction This proves the claimWe define the operator 119878

2 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883

by 1198782(119891 119909) = 119878

1(119891 0) for every 119891 isin 119871

1(120583|

119860) and for every

119909 isin 119883 Then we have

V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 = 1

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ 120576 (116)

Indeed from Claim 1 we have

V (1198781) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 ⩽ V (119878

2)

(117)

On the other hand we see that1003816100381610038161003816(119891

lowast 119909

lowast) 119878

2(119891 119909)

1003816100381610038161003816

=1003816100381610038161003816(119891

lowast 119909

lowast) 119878

1(119891 0)

1003816100381610038161003816 ⩽10038171003817100381710038171198911003817100381710038171003817 V (1198781) ⩽ V (119878

1)

(118)

for every ((119891lowast 119909lowast) (119891 119909)) isin Π(1198711(120583|

119860)oplus

1119883) Therefore

V(1198782) ⩽ V(119878

1) Also

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ sup

119909isin119878119883

10038171003817100381710038171198781 (0 119909)1003817100381710038171003817

= sup119909isin119878119883

10038171003817100381710038171198781 (0 119909) minus 1198780 (0 119909)1003817100381710038171003817 ⩽ 120576

(119)

Claim 2 There exists 1198783 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883 such

that 1198783(119892

1 0) = 119878

3 = 1 119878

3(0 119909) = 0 119878

3(119891 119909) isin 0oplus

1119883

for every (119891 119909) isin 1198711(120583|

119860)oplus

1119883 and 119878

3minus 119878

2 lt 4120576

Indeed write 1198781= (119863

1 119863

2) where 119863

1 119871

1(120583|

119860)oplus

1119883 rarr

1198711(120583|

119860) and119863

2 119871

1(120583|

119860)oplus

1119883 rarr 119883 We have that

sup 1003816100381610038161003816119892lowast119863

1(119892

1 0) + 119909

lowast119863

2(119892

1 0)1003816100381610038161003816

119909lowastisin 119878

119883lowast ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860) +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⩽ V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119892

lowast

1119863

1(119892

1 0) + 119909

lowast

1119863

2(119892

1 0)1003816100381610038161003816

(120)

This implies that

1003816100381610038161003816119909lowast

1119863

2(119892

1 0)1003816100381610038161003816 =10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

1003816100381610038161003816119892lowast

1119863

1(119892

1 0)1003816100381610038161003816

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119871

infin(120583|

119860)

(121)

Therefore |119892lowast1| equals 1 on the support of 119863

1(119892

1 0) As

|⟨119892lowast

1 119892

1⟩| = 1 we also have that |119892lowast

1| equals 1 on the support

of 1198921 Changing the values of 119892lowast

1by the ones of 119891lowast

0on 119860

(supp(1198631(119892

1 0)) cup supp(g

1)) we may and do suppose that

|119892lowast

1| = 1 on the whole 119860We also have 119863

2(119892

1 0) gt 0 Indeed

10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)1003817100381710038171003817

⩽10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198921 0)

1003817100381710038171003817 +10038171003817100381710038171198780 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817

lt 2120576 + 120576 = 3120576

(122)

So we have

10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

⩽10038171003817100381710038171198631(119892

1 0)1003817100381710038171003817 +10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

14 Abstract and Applied Analysis

=1003817100381710038171003817(1198631

(1198921 0) 119863

2(119892

1 0)) minus (0 119879

1198601198920)1003817100381710038171003817

=10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817 lt 3120576

(123)

and 1198632(119892

1 0) gt 119879

1198601198920 minus 3120576 ⩾ 1 minus 120578(120576

0) minus 3120576 gt 0

Finally define the operator 1198783by

1198783(119891 119909) = (0119863

2(119891 0) + 119892

lowast

1(119863

1(119891 0))

1198632(119892

1 0)

119909lowast

1119863

2(119892

1 0))

(124)

for (119891 119909) isin 1198711(120583|

119860)oplus

1119883 It is clear that

100381710038171003817100381711987831003817100381710038171003817 ⩽ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) (125)

Notice also that

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817

⩽10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 +10038171003817100381710038171198632(119891 0) minus 119879

1198601198911003817100381710038171003817

=1003817100381710038171003817(1198631

(119891 0) 1198632(119891 0)) minus (0 119879

119860119891)1003817100381710038171003817

=10038171003817100381710038171198782 (119891 119909) minus 1198780 (119891 119909)

1003817100381710038171003817

(126)

for all (119891 119909) isin 1198711(120583|

119860)oplus

1119883 Hence we have

10038171003817100381710038171198783 minus 11987821003817100381710038171003817

= 2 sup119891isin1198781198711(120583|119860)

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 ⩽ 210038171003817100381710038171198782 minus 1198780

1003817100381710038171003817 lt 4120576(127)

On the other hand let 119866 1198711(120583|

119860) rarr 119871

1(120583|

119860) be defined by

119866(119891) = 119892lowast

1119891 for every 119891 isin 119871

1(120583|

119860) Then we have

V (1198782)

= sup 1003816100381610038161003816119911lowast11987821199111003816100381610038161003816 (119911 119911

lowast) isin Π (119871

1(120583|

119860) oplus

1119883)

⩾ sup10038161003816100381610038161003816100381610038161003816

119909lowast119863

2(119866(

1

120583 (119862)120594119862) 0)

+119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119909lowastisin 119878

119883lowast 119862 isin Σ

119860 120583 (119862) gt 0

= sup10038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119862)120594119862) 0)

10038171003817100381710038171003817100381710038171003817

+

10038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119862 isin Σ119860 120583 (119862) gt 0

(128)

where Σ119860is the family of measurable subsets of 119860

Hence for any simple function 119904 = sum119899

119894=1((120572

119894120583(119860

119894))120594

119860119894

) isin

1198781198711(120583|119860) where 119860

119894119894is a family of disjoint measurable subsets

with strictly positive measure we have

V (1198782)

119899

sum

119894=1

10038161003816100381610038161205721198941003816100381610038161003816 (

100381710038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381710038171003817100381710038171003817100381710038171003817

+

100381610038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381610038161003816100381610038161003816100381610038161003816

)

⩾10038171003817100381710038171198632(119866 (119904) 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119866 (119904) 0)

1003816100381610038161003816

(129)

Since |119892lowast1| = 1119866 is an isometric isomorphism so for each119891 isin

1198781198711(120583|119860)there exists a sequence of norm-one simple functions

(119904119896) such that 119866(119904

119896) converges to 119891 Therefore

V (1198782) ⩾ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) ⩾100381710038171003817100381711987831003817100381710038171003817

(130)

On the other hand we see that100381710038171003817100381711987831003817100381710038171003817 ⩾1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

3(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119909

lowast

1119863

2(119892

1 0) + 119892

lowast

1119863

1(119892

1 0)1003816100381610038161003816 = V (119878

2) = 1

(131)

Therefore 1 = 1198783 = 119878

3(119892

1 0) which proves Claim 2

Finally set 1198783= (0 ) for a suitable 119871

1(120583|

119860)oplus

1119883 rarr

119883 and define the operator 1198791 119871

1(120583) rarr 119883 by

1198791(119891) = 119879

0(119891120594

119860119888) + (119875

119860119891 0) (132)

for every 119891 isin 1198711(120583) Then we have

10038171003817100381710038171198791 (119891)1003817100381710038171003817 ⩽100381710038171003817100381711987901003817100381710038171003817

10038171003817100381710038171198911205941198601198881003817100381710038171003817 +1003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171198911205941198601003817100381710038171003817 =10038171003817100381710038171198911003817100381710038171003817 (133)

for every 119891 isin 1198711(120583) so 119879

1 ⩽ 1 Also

10038171003817100381710038171198791 (1198691198601198921)1003817100381710038171003817 =10038171003817100381710038171198783 (1198921 0)

1003817100381710038171003817 =100381710038171003817100381711987831003817100381710038171003817 = 1 (134)

so 1198791attains its norm at 119869

1198601198921isin 119871

1(120583) and

10038171003817100381710038171198691198601198921 minus 11989101003817100381710038171003817 =10038171003817100381710038171198921 minus 1198920

1003817100381710038171003817 lt 120576 (135)

We also have that for any 119891 isin 1198781198711(120583)

10038171003817100381710038171198790 (119891) minus 1198791 (119891)1003817100381710038171003817 =100381710038171003817100381710038171198790(119891120594

119860) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790 (119869119860119875119860119891) minus 119879119860 (119875119860119891)

1003817100381710038171003817

+10038171003817100381710038171003817119879119860(119875

119860119891) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790119869119860 minus 119879119860

1003817100381710038171003817 +10038171003817100381710038171198780 minus 1198783

1003817100381710038171003817

lt 120578 (1205760) + 6120576

(136)

Hence 1198790minus 119879

1 ⩽ 120578(120576

0) + 6120576 lt 120576

0

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Abstract and Applied Analysis 15

Acknowledgments

The authors thank Gilles Godefroy and Rafael Paya forfruitful conversations about the content of this paper andVladimir Kadets for providing Lemma 18 They also appre-ciate anonymous referees for careful reading and fruitfulsuggestions about revision The second author was sup-ported by Basic Science Research Program through theNational Research Foundation of Korea (NRF) fundedby the Ministry of Education Science and Technology(NRF-2012R1A1A1006869) The third author was partiallysupported by Spanish MICINN and FEDER project noMTM2012-31755 and by Junta de Andalucıa and FEDERGrants FQM-185 and P09-FQM-4911

References

[1] F F Bonsall and J Duncan Numerical Ranges of Operators onNormed Spaces and of Elements of Normed Algebras vol 2 ofLondon Mathematical Society Lecture Note Series 2 CambridgeUniversity Press London UK 1971

[2] F F Bonsall and J Duncan Numerical Ranges II CambridgeUniversity Press London UK 1973 London MathematicalSociety Lecture Notes Series 10

[3] B Sims On numerical range and its applications to Banachalgebras [PhD thesis] University of Newcastle New SouthWales Australia 1972

[4] I D Berg and B Sims ldquoDenseness of operators which attaintheir numerical radiusrdquoAustralian Mathematical Society A vol36 no 1 pp 130ndash133 1984

[5] C S Cardassi ldquoDensity of numerical radius attaining operatorson some reflexive spacesrdquo Bulletin of the Australian Mathemat-ical Society vol 31 no 1 pp 1ndash3 1985

[6] C S Cardassi ldquoNumerical radius attaining operatorsrdquo inBanach Spaces (Columbia Mo 1984) vol 1166 of Lecture Notesin Mathematics pp 11ndash14 Springer Berlin Germany 1985

[7] C S Cardassi ldquoNumerical radius-attaining operators on119862(119870)rdquoProceedings of the AmericanMathematical Society vol 95 no 4pp 537ndash543 1985

[8] MDAcostaOperadores que alcanzan su radio numerico [PhDthesis] Universidad de Granada Granada Spain 1990

[9] M D Acosta and R Paya ldquoNumerical radius attaining oper-ators and the Radon-Nikodym propertyrdquo The Bulletin of theLondon Mathematical Society vol 25 no 1 pp 67ndash73 1993

[10] R Paya ldquoA counterexample on numerical radius attainingoperatorsrdquo Israel Journal of Mathematics vol 79 no 1 pp 83ndash101 1992

[11] Y S Choi and S G Kim ldquoNorm or numerical radius attainingmultilinear mappings and polynomialsrdquo Journal of the LondonMathematical Society vol 54 no 1 pp 135ndash147 1996

[12] M D Acosta J Becerra Guerrero and M Ruiz GalanldquoNumerical-radius-attaining polynomialsrdquo The Quarterly Jour-nal of Mathematics vol 54 no 1 pp 1ndash10 2003

[13] M D Acosta and S G Kim ldquoDenseness of holomorphicfunctions attaining their numerical radiirdquo Israel Journal ofMathematics vol 161 pp 373ndash386 2007

[14] S G Kim and H J Lee ldquoNumerical peak holomorphicfunctions on Banach spacesrdquo Journal of Mathematical Analysisand Applications vol 364 no 2 pp 437ndash452 2010

[15] M D Acosta R M Aron D Garcıa and M MaestreldquoThe Bishop-Phelps-Bollobas theorem for operatorsrdquo Journal ofFunctional Analysis vol 254 no 11 pp 2780ndash2799 2008

[16] A J Guirao and O Kozhushkina ldquoThe Bishop-Phelps-Bollobasproperty for numerical radius in ℓ

1(119862)rdquo Studia Mathematica

vol 218 no 1 pp 41ndash54 2013[17] J Falco ldquoThe Bishop-Phelps-Bollobas property for numerical

radius on 1198711rdquo Journal of Mathematical Analysis and Applica-

tions vol 414 no 1 pp 125ndash133 2014[18] A Aviles A J Guirao and J Rodriguez ldquoOn the Bishop-

Phelps-Bollobas property for numerical radius in119862(119870)-spacesrdquoPreprint

[19] RM Aron Y S Choi S K Kim H J Lee andMMartin ldquoTheBishop-Phelps-Bollobas version of Lindenstrauss properties Aand Brdquo Preprint

[20] V Kadets M Martın and R Paya ldquoRecent progress and openquestions on the numerical index of Banach spacesrdquo Revista dela Real Academia de Ciencias Exactas Fısicas y Naturales A vol100 no 1-2 pp 155ndash182 2006

[21] M Martın J Merı and M Popov ldquoOn the numerical index ofreal 119871

119901(120583)-spacesrdquo Israel Journal of Mathematics vol 184 pp

183ndash192 2011[22] A Iwanik ldquoNorm attaining operators on Lebesgue spacesrdquo

Pacific Journal of Mathematics vol 83 no 2 pp 381ndash386 1979[23] HH SchaeferBanach Lattices and Positive Operators Springer

New York NY USA 1974[24] J Duncan C M McGregor J D Pryce and A J White ldquoThe

numerical index of a normed spacerdquo Journal of the LondonMathematical Society Second Series vol 2 pp 481ndash488 1970

[25] Y S Choi S K Kim H J Lee and M Martin ldquoThe Bishop-Phelps-Bollobas theorem for operators on 119871

1(120583)rdquo Preprint

[26] H E Lacey The Isometric Theory of Classical Banach SpacesSpringer New York NY USA 1974

[27] G Godefroy ldquoUniversal spaces for strictly convex Banachspacesrdquo Revista de la Real Academia de Ciencias Exactas Fısicasy Naturales A vol 100 no 1-2 pp 137ndash146 2006

[28] C Finet M Martın and R Paya ldquoNumerical index andrenormingrdquo Proceedings of the American Mathematical Societyvol 131 no 3 pp 871ndash877 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Algebra

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 5: Research Article On the Bishop-Phelps-Bollobás Property ...downloads.hindawi.com/journals/aaa/2014/479208.pdf · Research Article On the Bishop-Phelps-Bollobás Property for Numerical

Abstract and Applied Analysis 5

Finally we have

10038171003817100381710038171198781 minus 1198791003817100381710038171003817 ⩽

10038171003817100381710038171003817100381710038171003817

1

V (119878)119878 minus 119878

10038171003817100381710038171003817100381710038171003817

+ 119878 minus 119879

=119878

V (119878)|V (119878) minus 1| + 119878 minus 119879

⩽1

119899 (119883)|V (119878) minus V (119879)| + 119878 minus 119879

⩽ (1

119899 (119883)+ 1) 119878 minus 119879 lt

119899 (119883) + 1

119899 (119883)120576

(34)

An obvious change of parameters finishes the proof

We do not know whether the hypothesis of 119899(119883) gt 0 canbe omitted in the above result

Putting together Propositions 4 and 6 we get the follow-ing

Corollary 7 Let 119883 be a uniformly convex and uniformlysmooth Banach space with 119899(119883) gt 0 Then 119883 has the BPBp-nu

Let us comment that every complex Banach space 119883satisfies 119899(119883) ⩾ 1119890 so the above corollary automaticallyapplies in the complex case In the real case this is no longertrue as the numerical index of a Hilbert space of dimensiongreater than or equal to two is 0 On the other hand it isproved in [21] that real 119871

119901(120583) spaces have nonzero numerical

index for every measure 120583when 119901 = 2Therefore we have thefollowing examples

Example 8 (a) Complex Banach spaces which are uniformlysmooth and uniformly convex satisfy the BPBp-nu

(b) In particular for every measure 120583 the complex spaces119871119901(120583) have the BPBp-nu for 1 lt 119901 lt infin(c) For every measure 120583 the real spaces 119871

119901(120583) have the

BPBp-nu for 1 lt 119901 lt infin 119901 = 2

Note Added in Revision Very recently H J Lee M Martınand J Merı have proved that Proposition 6 can be extendedto some Banach spaces with numerical index zero as forinstance real Hilbert spaces Hence they have shown thatHilbert spaces have the BPBp-nu These results will appearelsewhere

4 L1

Spaces

In this section we will show that 1198711(120583) has the BPBp-nu for

every measure 120583 In the proof we are dealing with complexintegrable functions since the real case is followed easilyby applying the same proof Our main result here is thefollowing

Theorem 9 Let 120583 be a measure Then 1198711(120583) has the Bishop-

Phelps-Bollobas property for numerical radius More preciselygiven 120576 gt 0 there exists 120578(120576) gt 0 (which does not depend on120583) such that whenever 119879

0isin L(119871

1(120583)) with V(119879

0) = 1 and

(1198910 119892

0) isin Π(119871

1(120583)) satisfy |⟨119879

01198910 119892

0⟩| gt 1 minus 120578(120576) then there

exist 119879 isinL(1198711(120583)) (119891

1 119892

1) isin Π(119871

1(120583)) such that

1003816100381610038161003816⟨1198791198911 1198921⟩1003816100381610038161003816 = V (119879) = 1 10038171003817100381710038171198910 minus 1198911

1003817100381710038171003817 lt 120576

10038171003817100381710038171198920 minus 11989211003817100381710038171003817 lt 120576

1003817100381710038171003817119879 minus 11987901003817100381710038171003817 lt 120576

(35)

As a first step we have to start dealing with finite regularpositive Borel measures for which a representation theoremfor operators exists

Proposition 10 Let 119898 be a finite regular positive Borelmeasure on a compact Hausdorff spaceΩ Then 119871

1(119898) has the

Bishop-Phelps-Bollobas property for numerical radius Moreprecisely given 120576 gt 0 there is 120578(120576) gt 0 (which is independent ofthemeasure119898) such that if a norm-one element119879 inL(119871

1(119898))

and (1198910 119892

0) isin Π(119871

1(119898)) satisfy |⟨119879119891

0 119892

0⟩| gt 1 minus 120578(120576) then

there exist an operator 119878 isin L(1198711(119898)) (119891

1 119892

1) isin Π(119871

1(119898))

such that1003816100381610038161003816⟨1198781198911 1198921⟩

1003816100381610038161003816 = 119878 = 110038171003817100381710038171198910 minus 1198911

1003817100381710038171003817 ⩽ 120576

10038171003817100381710038171198920 minus 11989211003817100381710038171003817 ⩽ 120576 119879 minus 119878 ⩽ 120576

(36)

To prove this proposition we need some backgroundon representation of operators on Lebesgue spaces on finiteregular positive Borel measures and several preliminarylemmas

Let 119898 be a finite regular positive Borel measure on acompact Hausdorff space Ω If 120583 is a complex-valued Borelmeasure on the product space Ω times Ω then define theirmarginal measures 120583119894 on Ω (119894 = 1 2) as follows 1205831(119860) =120583(119860 times Ω) and 1205832(119861) = 120583(Ω times 119861) where 119860 and 119861 are Borelmeasurable subsets of Ω

Let 119872(119898) be the complex Banach lattice of measuresconsisting of all complex-valued Borel measures 120583 on theproduct spaceΩ timesΩ such that |120583|119894 are absolutely continuouswith respect to119898 for 119894 = 1 2 endowed with the norm

1003817100381710038171003817100381710038171003817100381710038171003817

11988910038161003816100381610038161205831003816100381610038161003816

1

119889119898

1003817100381710038171003817100381710038171003817100381710038171003817infin

(37)

Each 120583 isin 119872(119898) defines a bounded linear operator 119879120583from

1198711(119898) to itself by

⟨119879120583(119891) 119892⟩ = int

ΩtimesΩ

119891 (119909) 119892 (119910) 119889120583 (119909 119910) (38)

where 119891 isin 1198711(119898) and 119892 isin 119871

infin(119898) Iwanik [22] showed that

the mapping 120583 997891rarr 119879120583is a lattice isometric isomorphism from

119872(119898) onto L(1198711(119898)) Even though he showed this for the

real case it can be easily generalized to the complex caseFor details see [22 Theorem 1] and [23 IV Theorem 15(ii)Corollary 2]

We will also use that given an arbitrary measure 120583 every119879 isin L(119871

1(120583)) satisfies V(119879) = 119879 [24] (that is the space

1198711(120583) has numerical index 1)

Lemma 11 (see [15 Lemma 33]) Let 119888119899 be a sequence of

complex numbers with |119888119899| ⩽ 1 for every 119899 and let 120578 gt 0 such

6 Abstract and Applied Analysis

that for a convex series sum120572119899 Resuminfin

119899=1120572119899119888119899gt 1 minus 120578 Then for

every 0 lt 119903 lt 1 the set 119860 = 119894 isin N Re 119888119894gt 119903 satisfies the

estimate

sum

119894isin119860

120572119894⩾ 1 minus

120578

1 minus 119903 (39)

From now on 119898 will be a finite regular positive Borelmeasure on the compact Hausdorff space Ω

Lemma 12 Suppose that there exist a nonnegative simplefunction 119891 isin 119878

1198711(119898)

and a function 119892 isin 119878119871infin(119898)

such that

Re ⟨119891 119892⟩ gt 1 minus 1205763

16 (40)

Then there exist a nonnegative simple function 1198911isin 119878

1198711(119898)

anda function 119892

1isin 119878

119871infin(119898)

such that

1198921(119909) = 120594supp(119891

1)(119909) + 119892 (119909) 120594

Ωsupp(1198911)(119909)

⟨1198911 119892

1⟩ = 1

1003817100381710038171003817119891 minus 119891110038171003817100381710038171lt 120576

1003817100381710038171003817119892 minus 11989211003817100381710038171003817infinlt radic120576 supp (119891

1) sub supp (119891)

(41)

Proof Let119891 = sum119898

119895=1(120573

119895119898(119861

119895))120594

119861119895

for some (120573119895) such that

120573119895⩾ 0 for all 119895 andsum119898

119895=1120573119895= 1 and119861

119895rsquos aremutually disjoint

By the assumption we have

Re ⟨119891 119892⟩ =119899

sum

119895=1

120573119895

1

119898 (119861119895)

int119861119895

Re119892 (119909) 119889119898 (119909) gt 1 minus 1205763

16

(42)

and letting

119869 = 119895 1 ⩽ 119895 ⩽ 1198991

119898 (119861119895)

int119861119895

Re119892 (119909) 119889119898 (119909) gt 1 minus 1205762

4

(43)

we have by Lemma 11

sum

119895isin119869

120573119895gt 1 minus

120576

4 (44)

For each 119895 isin 119869 we have

1 minus1205762

4lt

1

119898 (119861119895)

int119861119895

Re119892 (119909) 119889119898 (119909)

=1

119898 (119861119895)

int119861119895capRe119892⩽1minus120576

Re119892 (119909) 119889119898 (119909)

+ int119861119895capRe119892gt1minus120576

Re119892 (119909) 119889119898 (119909)

⩽1

119898 (119861119895)

((1 minus 120576)119898 (119861119895cap Re119892 ⩽ 1 minus 120576)

+ 119898 (119861119895cap Re119892 gt 1 minus 120576))

= 1 minus 120576

119898 (119861119895cap Re119892 ⩽ 1 minus 120576)

119898 (119861119895)

(45)

This implies that

119898(119861119895cap Re119892 ⩽ 1 minus 120576)

119898 (119861119895)

lt120576

4 (46)

Define 119861119895= 119861

119895cap Re119892 gt 1 minus 120576 for all 119895 isin 119869

1198911= (1sum

119895isin119869

120573119895)sum

119895isin119869

120573119895(120594

119861119895

119898 (119861119895)) (47)

and 1198921(119909) = 1 on supp(119891

1) and 119892

1(119909) = 119892(119909) elsewhereThen

it is clear that supp(1198911) sub supp(119891) 119892 minus 119892

1infinlt radic120576 and

⟨1198911 119892

1⟩ = 1 Finally we will show that 119891 minus 119891

1 lt 120576 Notice

first that10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

= 2sum

119895isin119869

120573119895

119898(119861119895 119861

119895)

119898 (119861119895)

lt120576

2

(48)

Hence1003817100381710038171003817119891 minus 1198911

1003817100381710038171003817

10038171003817100381710038171003817100381710038171003817100381710038171003817

1

sum119895isin119869120573119895

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus 119891

10038171003817100381710038171003817100381710038171003817100381710038171003817

1 minus sum119895isin119869120573119895

sum119895isin119869120573119895

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+120576

4

= (1 minussum

119895isin119869

120573119895) +

120576

2+120576

4

⩽120576

4+120576

2+120576

4= 120576

(49)

Lemma 13 (see [25 Lemma 33]) Suppose that 119879120583is a norm-

one element in L(1198711(119898)) for some 120583 isin 119872(119898) and there is

a nonnegative simple function 1198910such that 119891

0is a norm-one

element of 1198711(119898) and 119879

1205831198910 ⩾ 1 minus 120576

32

6 for some 0 lt 120576 lt 1

Abstract and Applied Analysis 7

Then there exist a norm-one bounded linear operator 119879] forsome ] isin 119872(119898119898) and a nonnegative simple function 119891

1

in 1198781198711(119898)

such that 119879120583minus 119879] ⩽ 120576 1198911 minus 1198910 ⩽ 3120576 and

(119889|]|1119889119898)(119909) = 1 for all 119909 isin supp(1198911)

Lemma 14 Suppose that 119879] isin L(1198711(119898)) is a norm-one

operator 119891 = sum119899

119894=1120573119894(120594

119861119894

119898(119861119894)) where 119898(119861

119895) gt 0 for all

1 ⩽ 119895 ⩽ 119899 and 119861119895119899

119895=1are mutually disjoint Borel subsets ofΩ

is a norm-one nonnegative simple function and 119892 is an elementof 119878

119871infin(119898)

such that

Re ⟨119892 119879]119891⟩ ⩾ 1 minus1205766

27(50)

for some 0 lt 120576 lt 1 and

119889|]|1

119889119898(119909) = 1 119892 (119909) = 1 (51)

for all x in the support of 119891Then there exist a nonnegative simple function 119891 isin 119878

1198711(119898)

a function119892 isin 119878

119871infin(119898)

and an operator119879] inL(1198711(119898) 1198711(119898))such that

⟨119892 119879]119891⟩ =1003817100381710038171003817119879]1003817100381710038171003817 = 1

1003817100381710038171003817119879] minus 119879]1003817100381710038171003817 ⩽ 2120576

10038171003817100381710038171003817119891 minus 119891

10038171003817100381710038171003817⩽ 3120576

1003817100381710038171003817119892 minus 1198921003817100381710038171003817 ⩽radic120576 ⟨119891 119892⟩ = 1

(52)

Proof Since

Re ⟨119892 119879]119891⟩ ⩾ 1 minus1205766

27 (53)

we have

1 minus1205766

27lt Re ⟨119892 119879]119891⟩

= intΩtimesΩ

119891 (119909)Re119892 (119910) 119889120583 (119909 119910)

=

119899

sum

119895=1

120573119895intΩtimesΩ

120594119861119895

(119909)

119898 (119861119895)

Re119892 (119910) 119889] (119909 119910)

(54)

Let

119869 = 119895 intΩtimesΩ

(

120594119861119895

(119909)

119898 (119861119895)

)Re119892 (119910) 119889] (119909 119910) gt 1 minus 1205763

26

(55)

Then from Lemma 11 we have sum119895isin119869120573119895gt 1 minus 120576

32 Let 119891

1=

sum119895isin119869120573119895(120594

119861119895

119898(119861119895)) where 120573

119895= 120573

119895(sum

119895isin119869120573119895) for all 119895 isin 119869

Then

10038171003817100381710038171198911 minus 1198911003817100381710038171003817 ⩽

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

(120573119895minus 120573

119895)

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+ sum

119895isin119869

120573119895⩽ 120576

3⩽ 120576 (56)

Note that there is a Borel measurable function ℎ on Ω times Ωsuch that 119889](119909 119910) = ℎ(119909 119910) 119889|]|(119909 119910) and |ℎ(119909 119910)| = 1 forall (119909 119910) isin Ω times Ω Let

119862 = (119909 119910) 1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 1

1003816100381610038161003816 ltradic120576

232 (57)

Define two measures ]119891and ]

119888as follows

]119891(119860) = ] (119860 119862) ]

119888(119860) = ] (119860 cap 119862) (58)

for every Borel subset 119860 ofΩ times Ω It is clear that

119889] = 119889]119891+ 119889]

119888 119889

10038161003816100381610038161003816]119891

10038161003816100381610038161003816= ℎ 119889]

119891

1198891003816100381610038161003816]1198881003816100381610038161003816 = ℎ 119889]119888 119889 |]| = 119889

10038161003816100381610038161003816]119891

10038161003816100381610038161003816+ 1198891003816100381610038161003816]1198881003816100381610038161003816

(59)

Since (119889|]|11198891198981)(119909) = 1 for all 119909 isin ⋃119899

119895=1119861119895 we have

1 =119889|]|1

1198891198981

(119909) =

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) +1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909) (60)

for all 119909 isin 119861 = ⋃119899

119895=1119861119895 and we deduce that |]|1(119861

119895) = 119898

1(119861

119895)

for all 1 ⩽ 119895 ⩽ 119899We claim that |]

119891|1(119861

119895)119898

1(119861

119895) ⩽ 120576

22

2 for all 119895 isin 119869Indeed if |119892(119910)ℎ(119909 119910) minus 1| ⩾ radic120576232 then

Re (119892 (119910) ℎ (119909 119910)) ⩽ 1 minus 12057624 (61)

So we have

1 minus1205763

26

⩽1

1198981(119861

119895)

ReintΩtimesΩ

120594119861119895(119909)119892 (119910) 119889] (119909 119910)

=1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 |]| (119909 119910)

=1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 10038161003816100381610038161003816]11989110038161003816100381610038161003816(119909 119910)

+1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 1003816100381610038161003816]1198881003816100381610038161003816 (119909 119910)

⩽1

1198981(119861

119895)

((1 minus120576

24)10038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

(119861119895) +1003816100381610038161003816]1198881003816100381610038161003816

1

(119861119895))

= 1 minus120576

24

10038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

(119861119895)

1198981(119861

119895)

(62)

This proves our claimWe also claim that for each 119895 isin 119869 there exists a Borel

subset 119861119895of 119861

119895such that

(1 minus120576

2)119898

1(119861

119895) ⩽ 119898

1(119861

119895) ⩽ 119898

1(119861

119895)

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) ⩽120576

2

(63)

8 Abstract and Applied Analysis

for all 119909 isin 119861119895 Indeed set 119861

119895= 119861

119895cap 119909 isin Ω (119889|]

119891|1119889119898

1)

(119909) ⩽ 1205762 Then

int119861119895119861119895

120576

2119889119898

1(119909) ⩽ int

119861119895

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) 1198891198981(119909)

=10038161003816100381610038161003816]1119891

10038161003816100381610038161003816(119861

119895) ⩽1205762

22119898

1(119861

119895)

(64)

This shows that 1198981(119861

119895 119861

119895) ⩽ (1205762)119898

1(119861

119895) This proves our

second claimNow we define 119892 by 119892(119910) = 119892(119910)|119892(119910)| if |119892(119910)| ⩾ 1 minus

radic120576232 and 119892(119910) = 119892(119910) if |119892(119910)| lt 1 minus radic120576232 and we

write 119891 = sum119895isin119869120573119895(120594

119861119895

1198981(119861

119895)) It is clear that 119892 isin 119878

119871infin(119898)

119892 minus 119892 lt radic120576 and 119892(119910) = 1 for all 119909 isin supp119891

Finally we define the measure

119889] (119909 119910)

= sum

119895isin119869

120594119861119895

(119909) 119892 (119910)ℎ (119909 119910)119889]119888(119909 119910)(

1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+ 1205941198691119861(119909) 119889] (119909 119910)

(65)

where 119861 = ⋃119895isin119869119861119895 It is easy to see that (119889|]|1119889119898

1)(119909) = 1

on 119861 and (119889|]|11198891198981)(119909) ⩽ 1 elsewhere Note that

119889 (] minus ]) (119909 119910)

= sum

119895isin119869

120594119861119895

(119909) [

[

119892 (119910)ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1]

]

119889]119888(119909 119910)

minus sum

119895isin119869

120594119861119895

(119909) 119889]119891(119909 119910)

(66)

If (119909 119910) isin 119862 then |119892(119910)| ⩾ 1 minus radic120576232 ⩾ 1 minus 1232 and

100381610038161003816100381610038161003816119892 (119910)ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

119892 (119910)

1003816100381610038161003816119892 (119910)1003816100381610038161003816

ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816100381610038161003816

1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 11003816100381610038161003816

1003816100381610038161003816119892 (119910)1003816100381610038161003816

+

10038161003816100381610038161 minus1003816100381610038161003816119892 (119910)

1003816100381610038161003816

10038161003816100381610038161003816100381610038161003816119892 (119910)

1003816100381610038161003816

⩽ 2

1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 11003816100381610038161003816

1003816100381610038161003816119892 (119910)1003816100381610038161003816

⩽ 2radic120576

232

232

232 minus 1⩽ 2radic120576

(67)

Hence for all (119909 119910) isin 119862 we have

10038161003816100381610038161003816100381610038161003816100381610038161003816

119892 (119910)ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

⩽100381610038161003816100381610038161003816119892 (119910)ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

⩽ 2radic120576(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

(68)

So we have for all 119909 isin 1198691

119889|] minus ]|1

1198891198981

(119909)

⩽ sum

119895isin119869

120594119861119895

(119909) [

[

2radic120576(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

]

]

1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909)

+ sum

119895isin119869

120594119861119895

(119909)

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909)

⩽ sum

119895isin119869

120594119861119895

(119909)(2radic120576 + (1 minus1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909)))

+sum

119895isin119869

120594119861119895

(119909)(

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909))

⩽ 2radic120576 + 120576 lt 3radic120576

(69)

This gives that 119879] minus 119879] lt 3radic120576 Note also that for all 119895 isin 119869

⟨119879]

120594119861119895

1198981(119861

119895)

119892⟩

= intΩtimesΩ

120594119861119895

(119909)

1198981(119861

119895)

119892 (119910) 119889] (119909 119910)

= intΩtimesΩ

120594119861119895

(119909)

1198981(119861

119895)

ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

119889]119888(119909 119910)

Abstract and Applied Analysis 9

= intΩ

120594119861119895

(119909)

1198981(119861

119895)

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

1198891003816100381610038161003816]1198881003816100381610038161003816

1

(119909)

= intΩ

120594119861119895

(119909)

1198981(119861

119895)

1198891198981(119909) = 1

(70)

Hence we get ⟨119879]119891 119892⟩ = 1 which implies that 119879]119891 =119879] = 1 Finally10038171003817100381710038171003817119891 minus 119891

10038171003817100381710038171003817

⩽10038171003817100381710038171003817119891 minus 119891

1

10038171003817100381710038171003817+10038171003817100381710038171198911 minus 119891

1003817100381710038171003817

=

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

1198981(119861

119895)

minus sum

119895isin119869

120573119895

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+ 120576

⩽ sum

119895isin119869

120573119895(

10038171003817100381710038171003817100381710038171003817100381710038171003817

120594119861119895

1198981(119861

119895)

minus

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

120594119861119895

1198981(119861

119895)

minus

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

) + 120576

= 2sum

119895isin119869

120573119895

1198981(119861

119895 119861

119895)

1198981(119861

119895)

+ 120576

⩽ 2sum

119895isin119869

120573119895

(1205762)1198981(119861

119895)

1198981(119861

119895)

+ 120576 ⩽120576

1 minus 1205762+ 120576 lt 3120576

(71)

We are now ready to present the proof of the main resultin the case of finite regular positive Borel measures

Proof of Proposition 10 Let 1205751= 120575

3

2(5sdot2

4) 120575

2= 120575

12

3(3

2sdot2

14)

and 1205753= (12057610)

2 for some 0 lt 120576 lt 1 Suppose that 119879 isinL(119871

1(119898)) with 119879 = 1 and that there is an 119891

0isin 119878

1198711(119898)

and1198920isin 119878

119871infin(119898)

such that ⟨1198910 119892

0⟩ = 1 and |⟨119879119891

0 119892

0⟩| gt 1minus120575

3

12

6Then there is an isometric isomorphism Ψ from 119871

1(119898) onto

itself such thatΨ(1198910) = |119891

0| and there is a scalar number 120572 in

119878R such that |⟨1198791198910 119892

0⟩| = ⟨120572119879119891

0 119892

0⟩ Then letting 119891

1= Ψ119891

0

1198921= (Ψ

minus1)lowast1198920 and 119879

1= 120572Ψ119879Ψ

minus1 we have

⟨1198781198911 119892

1⟩ = ⟨120572Ψ119879

minus1Ψ119891

0 (Ψ

minus1)lowast

1198920⟩

= ⟨1205721198791198910 119892

0⟩ gt 1 minus

1205753

1

26

⟨1198911 119892

1⟩ = ⟨Ψ119891

0 (Ψ

minus1)lowast

1198920⟩ = 1

(72)

Since 11987911198911 gt 1 minus 120575(120575

3

12

6) by Lemma 13 there exists a

norm-one bounded operator 119879] and a nonnegative simplefunction119891

2isin 119878

1198711(119898)

such that 1198791minus119879] ⩽ 1205751 1198912minus1198911 ⩽ 31205751

and (119889|]|11198891198981)(119909) = 1 for all 119909 isin supp(119891

2) Then

⟨119879]1198912 1198921⟩

= ⟨11987911198911 119892

1⟩ minus ⟨119879

11198911minus 119879

11198912 119892

1⟩ minus ⟨119879

11198912minus 119879]1198912 1198921⟩

⩾ ⟨11987911198911 119892

1⟩ minus10038171003817100381710038171198911 minus 1198912

1003817100381710038171003817 minus10038171003817100381710038171198791 minus 119879]

1003817100381710038171003817

⩾ 1 minus1205753

1

26minus 3120575

1minus 120575

1⩾ 1 minus 5120575

1= 1 minus

1205753

2

16

(73)

Notice also that

⟨1198912 119892

1⟩ = ⟨119891

1 119892

1⟩ minus ⟨119891

1minus 119891

2 119892

1⟩ ⩾ 1 minus

10038171003817100381710038171198911 minus 11989121003817100381710038171003817

⩾ 1 minus 31205751⩾ 1 minus 5120575

1= 1 minus

1205753

2

16

(74)

By Lemma 12 there are a nonnegative simple function 1198913isin

1198781198711(119898)

and a function 1198923isin 119878

119871infin(119898)

such that

1198923(119909) = 120594supp119891

3

(119909) + 1198922(119909) 120594

Ωsupp1198913

(119909)

10038171003817100381710038171198912 minus 11989131003817100381710038171003817 ⩽ 1205752

10038171003817100381710038171198923 minus 11989211003817100381710038171003817 ⩽radic120575

2 ⟨119891

3 119892

3⟩ = 1

(75)

So we have

⟨119879]1198913 1198923⟩ = ⟨119879]1198912 1198921⟩ minus ⟨119879]1198912 minus 119879]1198913 1198921⟩

minus ⟨119879]1198913 1198921 minus 1198923⟩

⩾ 1 minus1205753

2

16minus 2radic120575

2⩾ 1 minus 3radic120575

2= 1 minus

1205756

3

27

(76)

By Lemma 14 there exist 1198914isin 119878

1198711(119898)

and 1198924isin 119878

119871infin(119898)

and anoperator 119879

4such that

⟨1198924 119879

41198914⟩ = 1 =

100381710038171003817100381711987941003817100381710038171003817

10038171003817100381710038171198794 minus 119879]1003817100381710038171003817 ⩽ 21205753

10038171003817100381710038171198914 minus 11989131003817100381710038171003817 ⩽ 31205753

10038171003817100381710038171198924 minus 11989231003817100381710038171003817 ⩽radic120575

3 ⟨119891

4 119892

4⟩ = 1

(77)

So we have

10038171003817100381710038171198794 minus 11987911003817100381710038171003817 ⩽10038171003817100381710038171198794 minus 119879]

1003817100381710038171003817 +1003817100381710038171003817119879] minus 1198791

1003817100381710038171003817

⩽ 1205751+ 2120575

3⩽ 3120575

3

10038171003817100381710038171198911 minus 11989141003817100381710038171003817 ⩽10038171003817100381710038171198911 minus 1198912

1003817100381710038171003817 +10038171003817100381710038171198912 minus 1198913

1003817100381710038171003817 +10038171003817100381710038171198913 minus 1198914

1003817100381710038171003817

⩽ 31205751+ 120575

2+ 3120575

3⩽ 10120575

3

10038171003817100381710038171198921 minus 11989241003817100381710038171003817 ⩽10038171003817100381710038171198921 minus 1198923

1003817100381710038171003817 + 1198923 minus 1198924

⩽ 1205752+ radic120575

3⩽ 2radic120575

3

(78)

10 Abstract and Applied Analysis

Let 119878 = 120572Ψminus11198794Ψ 119891 = Ψminus1

1198914 and 119892 = Ψlowast

1198924 then we have

119879 minus 119878 =10038171003817100381710038171003817119879 minus 120572Ψ

minus11198794Ψ10038171003817100381710038171003817=10038171003817100381710038171003817120572Ψ119879Ψ

minus1minus 119879

4

10038171003817100381710038171003817

=10038171003817100381710038171198791 minus 1198794

1003817100381710038171003817 ⩽ 31205753

100381710038171003817100381710038171198910minus 11989110038171003817100381710038171003817=100381710038171003817100381710038171198910minus Ψ

minus11198914

10038171003817100381710038171003817=10038171003817100381710038171198911 minus 1198914

1003817100381710038171003817 ⩽ 101205753

100381710038171003817100381710038171198920minus 11989110038171003817100381710038171003817=10038171003817100381710038171198920 minus Ψ

lowast1198924

1003817100381710038171003817 =100381710038171003817100381710038171003817(Ψ

minus1)lowast

1198920minus 119892

4

100381710038171003817100381710038171003817

=10038171003817100381710038171198921 minus 1198924

1003817100381710038171003817 ⩽ 2radic120575

3

⟨119891 119892⟩ = ⟨Ψminus11198914 Ψ

lowast1198924⟩ = ⟨119891

4 119892

4⟩ = 1

10038161003816100381610038161003816⟨119878119891 119892⟩

10038161003816100381610038161003816=10038161003816100381610038161003816⟨120572Ψ

minus11198794ΨΨ

minus11198914 Ψ

lowast1198924⟩10038161003816100381610038161003816= |120572| = 1

(79)

This completes the proof

Finally we may give the proof of the main result in fullgenerality

Proof of Theorem 9 Notice that the Kakutani representationtheorem (see [26] for a reference) says that for every 120590-finitemeasure ] the space 119871

1(]) is isometrically isomorphic to

1198711(119898) for some positive Borel regular measure on a compact

Hausdorff space Then by Proposition 10 there is a universalfunction 120576 997891rarr 120578(120576) gt 0 which gives the BPBp-nu for 119871

1(]) for

every 120590-finite measure ]Fix 120576 gt 0 Suppose that 119879

0isin L(119871

1(120583)) with V(119879

0) = 1

and (1198910 119891

lowast

0) isin Π(119871

1(120583)) satisfy

1003816100381610038161003816⟨119891lowast

0 119879

01198910⟩1003816100381610038161003816 gt 1 minus 120578 (120576) (80)

Choose a sequence 119891119899 in 119871

1(120583) such that sup

119899119879

0119891119899 = 1

and let 119866 be the closed linear span of

119879119899

0119891119898 119899 119898 isin N cup 0 (81)

As 119866 is separable there is a dense subset 119892119899 119899 isin N of 119866

and let 119864 = ⋃infin

119899=1supp119892

119899 where supp119892

119899is the support of 119892

119899

Then the measure 120583|119864is 120590-finite Let

119884 = 119891 isin 1198711(120583) supp (119891) sub 119864 (82)

be a closed subspace of 1198711(120583) It is clear that 119871

1(120583) = 119884oplus

1119885

and 119884 is isometrically isomorphic to 1198711(120583|

119864) So 119884 has the

BPBp-nu with 120578(120576)Now write 119878

0= 119879

0|119884 119884 rarr 119884 consider 119910

0= 119891

0isin

119878119884 119910lowast

0= 119891

lowast

0|119884isin 119878

119884lowast and observe that 119910lowast

0(119910

0) = 1 and

|119910lowast

0(119878

01199100)| = |119891

lowast

0(119879

01198910)| gt 1 minus 120578(120576) Hence there exist 119878 isin

L(119884) and (1199100 119910

lowast

0) isin Π(119884) such that

1003816100381610038161003816119910lowast

0(119878119910

0)1003816100381610038161003816 = 1 = V (119878) 1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

10038171003817100381710038171199100 minus 11991001003817100381710038171003817 lt 120576

1003817100381710038171003817119910lowast

0minus 119910

lowast

0

1003817100381710038171003817 lt 120576

(83)

Finally consider the operator 119879 isinL(1198711(120583)) given by

119879 (119910 119911)

= (119878119910 0) + 1198790(0 119911) ((119910 119911) isin 119871

1(120583) equiv 119884oplus

1119885)

(84)

We have 119879 = 1 (and so V(119879) = 1) Indeed1003817100381710038171003817119879 (119910 119911)

1003817100381710038171003817 =1003817100381710038171003817(119878119910 0)

1003817100381710038171003817 +10038171003817100381710038171198790 (0 119911)

1003817100381710038171003817 ⩽10038171003817100381710038171199101003817100381710038171003817 + 119911 =

1003817100381710038171003817(119910 119911)1003817100381710038171003817

(85)

for all (119910 119911) isin 1198711(120583) and 119879(119910

0 0) = (119878119910

0 0) = 119878119910

0 = 1

Let 119909 = (1199100 0) and 119909lowast = (119910lowast

0 119891

0|119885) Then (119909 119909lowast) isin Π(119871

1(120583))

Moreover we have1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119910

lowast

0119878119910

0

1003816100381610038161003816 = 1 = V (119879) 1003817100381710038171003817119909 minus 1198910

1003817100381710038171003817 =1003817100381710038171003817119910 minus 1199100

1003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowast

0minus 119891

lowast

0

1003817100381710038171003817

= max 1003817100381710038171003817119910 minus 119891lowast

0|119884

1003817100381710038171003817 1003817100381710038171003817119891

lowast

0|119885minus 119891

lowast

0|119885

1003817100381710038171003817

=1003817100381710038171003817119910

lowastminus 119910

lowast

0

1003817100381710038171003817 lt 120576

1003817100381710038171003817119879 minus 11987901003817100381710038171003817 = sup119910+119911⩽1

1003817100381710038171003817119879 (119910 119911) minus 1198790 (119910 119911)1003817100381710038171003817

= sup119910⩽1

1003817100381710038171003817119878119910 minus 11987801199101003817100381710038171003817 =1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

(86)

This completes the proof

5 Examples of Spaces Failing theBishop-Phelps-Bollobaacutes Property forNumerical Radius

Our goal here is to prove that the density of numerical radiusattaining operators does not imply the BPBp-nu Actuallywe will show that among separable spaces there is noisomorphic property implying the BPBp-nu other than finite-dimensionality

We need to relate the BPBp-nu to the Bishop-Phelps-Bollobas property for operators which as mentioned in theintroduction was introduced in [15] A pair (119883 119884) of Banachspaces has the Bishop-Phelps-Bollobas property for operators(in short BPBp) if given 120576 gt 0 there exists 120578(120576) gt 0 suchthat given 119879 isin L(119883 119884) with 119879 = 1 and 119909 isin 119878

119883such that

119879119909 gt 1 minus 120578(120576) then there exist 119911 isin 119878119883and 119878 isin L(119883 119884)

satisfying

119878 = 119878119911 = 1 119909 minus 119911 lt 120576

119879 minus 119878 lt 120576

(87)

We refer the reader to [15 19 25] and references therein formore information and background Among the interestingresults on the BPBp we emphasize that a pair (119883 119884) when119883is finite-dimensional does not necessarily have the BPBp Forinstance if119884 is a strictly convex space which is not uniformlyconvex then the pair (ℓ(2)

1 119884) fails to have the BPBp (this is

contained in [15] see [19 Section 3])

Abstract and Applied Analysis 11

The next result relates the BPBp-nu to the BPBp foroperators in a particular case We will deduce our examplefrom it

Theorem 15 If 1198711(120583)oplus

1119883 has the BPBp-119899119906 then the pair

(1198711(120583) 119883) has the BPBp for operators

Before proving this proposition we will use it to get themain examples of this section The first example shows thatthe density of numerical radius attaining operators does notimply the BPBp-nu

Example 16 There is a reflexive space (and so numerical radiusattaining operators on it are dense) which fails to have theBPBp-nu Indeed let 119884 be a reflexive separable space whichis not superreflexive and we may suppose that 119884 is strictlyconvex Observe that 119884 cannot be uniformly convex since itis not superreflexive Now 119883 = ℓ(2)

1oplus1119884 is reflexive but the

pair (ℓ(2)1 119884) fails the BPBp since 119884 is strictly convex but not

uniformly convex [19 Corollary 33] Therefore Theorem 15gives us that119883 does not have the BPBp-nu

The example above can be extended to get the result thatevery infinite-dimensional separable Banach space can berenormed to fail the BPBp-nu This follows from the factthat every infinite-dimensional separable Banach space canbe renormed to be strictly convex but not uniformly convex(this result can be proved ldquoby handrdquo an alternative categoricalargument for it can be found in [27] and references therein)With a little more of effort we may get the main result of thesection

Theorem 17 Every infinite-dimensional separable Banachspace can be renormed to fail the weak-BPBp-nu (and so inparticular to fail the BPBp-nu)

We need the following result which is surely well knownAs we have not found a reference we include a nice and easyproof kindly given to us by Vladimir Kadets We recall thatgiven a Banach space 119884 the set of all equivalent norms on119884 can be viewed as a metric space using the Banach-Mazurdistance

Lemma 18 Let119884 be an infinite-dimensional separable Banachspace Then the set of equivalent norms on 119884 which are strictlyconvex and are not (locally) uniformly convex is dense in the setof all equivalent norms on119884 (with respect to the Banach-Mazurdistance)

Proof Fix 119890 isin 119878119884and 119890lowast

1isin 119878

119884lowast such that 119890lowast

1(119890) = 1 For a fixed

120576 isin (0 12) denote

119902 (119910) = max (1 minus 120576) 10038171003817100381710038171199101003817100381710038171003817 1003816100381610038161003816119890lowast

1(119910)1003816100381610038161003816 (119910 isin 119884) (88)

Evidently (1 minus 120576)119910 ⩽ 119902(119910) ⩽ 119910 for every 119910 isin 119884 Fix asequence 119890lowast

119896 119896 ⩾ 2 of norm-one functionals separating

the points of 119884 and denote

119901 (119910) = radic

infin

sum

119896=1

1

2119896

1003816100381610038161003816119890lowast

119896(119910)1003816100381610038161003816

2

(119910 isin 119884) (89)

Then 119901 is a strictly convex norm on 119884 119901(119890) ⩾ 1radic2 and119901(119910) ⩽ 119910 for all 119910 isin 119883 Finally write

100381710038171003817100381711991010038171003817100381710038171= (1 minus 120576) 119902 (119910) + 120576

119901 (119910)

119901 (119890)(119910 isin 119884) (90)

Then sdot 1is a strictly convex norm on 119884 and

(1 minus 120576)2 10038171003817100381710038171199101003817100381710038171003817 ⩽100381710038171003817100381711991010038171003817100381710038171⩽ (1 + 120576)

10038171003817100381710038171199101003817100381710038171003817 (119910 isin 119884) (91)

Wewill finish the proof by showing that sdot 1is not uniformly

convex (actually it is not locally uniformly convex) Indeedfor each 119899 isin N we select 119910

119899isin ⋂

119899

119896=1ker 119890lowast

119896with 119910

119899 = 1 and

consider 119890119899= 119890 + (1205764)119910

119899 Then 119902(119890) = 1 119902(119890

119899) = 1 and

119902(119890+119890119899) = 2 At the same time 119901(119910

119899) rarr 0 so 119901(119890

119899) rarr 119901(119890)

and 119901(119890 + 119890119899) rarr 2119901(119890) Consequently

1198901 = 1100381710038171003817100381711989011989910038171003817100381710038171997888rarr 1

1003817100381710038171003817119890 + 11989011989910038171003817100381710038171997888rarr 2 (92)

but 119890 minus 1198901198991= (1205764)119910

1198991⩾ (1 minus 120576)

2(1205764) which means the

absence of local uniform convexity at 119890

Proof of Theorem 17 Let 119883 be an infinite-dimensional sepa-rable Banach space Take a closed subspace 119884 of 119883 of codi-mension two By [28 Proposition 2] the map carrying everyequivalent norm on 119884 to its numerical index is continuousand so the set of values of the numerical index of 119884 upto reforming is a nontrivial interval [28 Theorem 9] ThenLemma 18 allows us to find an equivalent norm | sdot | on 119884 insuch a way that (119884 | sdot |) is strictly convex and is not uniformlyconvex and 119899(119884 | sdot |) gt 0 Now the space 119883 = ℓ(2)

1oplus1(119884 | sdot |)

is an equivalent renorming of 119883 which does not have theBPBp-119899119906 (indeed otherwise the pair (ℓ(2)

1 (119884 | sdot |)) would

have the BPBp for the operator norm and so (119884 | sdot |) wouldbe uniformly convex by [19 Corollary 33] a contradiction)Moreover as

119899 (119883) = min 119899 (ℓ(2)1) 119899 (119884 |sdot|) gt 0 (93)

(see [20 Proposition 2] for instance) 119883 also fails the weak-BPBp-nu by Proposition 6

To finish the section with the promised proof ofTheorem 15 we first see the following stability result

Lemma 19 Let119883 = [⨁infin

119896=1119883

119896]1198880

or [⨁infin

119896=1119883

119896]ℓ1

If119883 has theBishop-Phelps-Bollobas property for numerical radius with afunction 120578 then each Banach space 119883

119894has the Bishop-Phelps-

Bollobas property for numerical radius with 120578nu(119883119894) ⩾ 120578 That

is inf119894120578nu(119883119894

)(120576) ⩾ 120578nu(119883)(120576) for all 0 lt 120576 lt 1

Proof Let 119875119894 119883 rarr 119883

119894and 1198751015840

119894 119883

lowastrarr 119883

lowast

119894be the natural

projections and let 119876119894 119883

119894rarr 119883 and 1198761015840

119894 119883

lowast

119894rarr 119883

lowast be thenatural embeddings

Assume that an operator 119879119894 119883

119894rarr 119883

119894and a pair

(119909119894 119909

lowast

119894) isin Π(119883

119894) satisfy that

V (119879119894) = 1

1003816100381610038161003816119909lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (94)

We define an operator 119879 119883 rarr 119883 and (119909 119909lowast) isin Π(119883) by

119879 = 119876119894∘ 119879

119894∘ 119875

119894 (119909 119909

lowast) = (119876

119894119909119894 119876

1015840

119894119909lowast

119894) (95)

12 Abstract and Applied Analysis

then clearly we see that1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119909

lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (96)

From the assumption there exist 119878 119883 rarr 119883 and a pair(119910 119910

lowast) isin Π(119883) such that

1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 = 1 = V (119878) 119878 minus 119879 lt 120576

1003817100381710038171003817119910lowastminus 119909

lowast1003817100381710038171003817 lt 1205761003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576

(97)

Since this clearly shows that1003817100381710038171003817119875119894 ∘ 119878 ∘ 119876119894

minus 119879119894

1003817100381710038171003817 lt 120576100381710038171003817100381710038171198751015840

119894119910lowastminus 119909

lowast

119894

10038171003817100381710038171003817lt 120576

1003817100381710038171003817119875119894119910 minus 1199091198941003817100381710038171003817 lt 120576

(98)

we only need to show that |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = 1

We first show the case of 1198880sum Since

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817=10038171003817100381710038171003817119875119895119910 minus 119875

11989511990910038171003817100381710038171003817⩽1003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576 (99)

for every 119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽100381710038171003817100381710038171198751015840

119894119910lowast10038171003817100381710038171003817+ 120576 sum

119895isinN119895 = 119894

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817⩽1003817100381710038171003817119910

lowast1003817100381710038171003817 = 1

(100)

This shows that 1198751015840119894119910lowast = 1 and 1198751015840

119895119910lowast= 0 for every 119895 = 119894

So 119910lowast = 1198761015840

1198941198751015840

119894119910lowast and 1198751015840

119894119910lowast(119875

119894119910) = 1 This and the fact that

119910 minus 119876119894119875119894119910 lt 120576 imply that

(119876119894119875119894119910 + (

1

120576) (119910 minus 119876

119894119875119894119910) 119876

1015840

1198941198751015840

119894119910lowast) isin Π (119883) (101)

So we get that (1198761015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910 + (1120576)(119910 minus 119876

119894119875119894119910))V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =10038161003816100381610038161003816(119876

1015840

1198941198751015840

119894119910lowast) 11987811991010038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910 +1

120576(119910 minus 119876

119894119875119894119910))

1003816100381610038161003816100381610038161003816⩽ 1

(102)

and so we get |1198751015840119894119910lowast(119875

119894∘119878 ∘119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

We next show the case of ℓ1sum The proof is almost

the same as that of the 1198880case However for the sake of

completeness we provide it hereSince 1198751015840

119895119910lowast = 119875

1015840

119895119910lowastminus 119875

1015840

119895119909lowast ⩽ 119910

lowastminus 119909

lowast lt 120576 for every

119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽10038171003817100381710038171198751198941199101003817100381710038171003817 + 120576 sum

119895isinN119895 = 119894

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817⩽10038171003817100381710038171199101003817100381710038171003817 = 1

(103)

which shows 119875119894119910 = 1 and 119875

119895119910 = 0 for every 119895 = 119894 Since this

implies (119876119894119875119894119910119876

1015840

1198941198751015840

119894119910lowast+(1120576)(119910

lowastminus119876

1015840

1198941198751015840

119894119910lowast)) isin Π(119883) we get

that |(1198761015840

1198941198751015840

119894119910lowast+ (1120576)(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast))119878(119876

119894119875119894119910)| ⩽ V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =1003816100381610038161003816119910

lowast119878 (119876

119894119875119894119910)1003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast+1

120576(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast)) 119878 (119876

119894119875119894119910)

1003816100381610038161003816100381610038161003816⩽ 1

(104)

and so |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

Proof of Theorem 15 Note that 120578nu(1198711(120583)oplus1119883)(120576) rarr 0 as120576 rarr 0 Fix 0 lt 120576

0lt 1 and choose 0 lt 120576 lt 1 such that

6120576 + 120578nu(1198711(120583)oplus1119883)(120576) lt 1205760 Let 120578(1205760) = 120578nu(1198711(120583)oplus1119883)(120576)Suppose that 119879

0isin L(119871

1(120583) 119883) with 119879

0 = 1 and 119891

0isin

1198781198711(120583)

satisfy

1003817100381710038171003817119879011989101003817100381710038171003817 gt 1 minus 120578 (1205760) (105)

For any measurable subset 119861 let

1198711(120583|

119861) = 119891|

119861 119891 isin 119871

1(120583) (106)

with the norm 119891|119861 = 119891120594

1198611 Then it is easy to see

that 1198711(120583|

119861) is isometrically isomorphic to a complemented

subspace of 1198711(120583) Let 119875

119861 119871

1(120583) rarr 119871

1(120583|

119861) be the

restriction defined by 119875119861(119891) = 119891|

119861for all 119891 isin 119871

1(120583) and

let 119869119861 119871

1(120583|

119861) rarr 119871

1(120583) be the extension defined by

119869119861(119891)(120596) = 119891(120596) if 120596 isin 119861 and 119869

119861(119891)(120596) = 0 otherwise

It is clear that 119875119861119869119861= Id

1198711(120583|119861)and 119869

119861119875119861(119891) = 119891120594

119861for all

119891 isin 1198711(120583) Notice also that 119871

1(120583) is isometrically isomorphic

to 1198711(120583|

119861)oplus

11198711(120583|

119861119888)

Let 119860 = supp1198910and 119892

0= 119875

1198601198910 Then

1003817100381710038171003817119879011986911986011989201003817100381710038171003817 =100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) gt 0 (107)

and define the operator 119879119860 119871

1(120583|

119860) rarr 119883 by 119879

119860119891 =

1198790119869119860119891119879

0119869119860 for every 119891 isin 119871

1(120583|

119860) Then

100381710038171003817100381711987911986011989201003817100381710038171003817 ⩾100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) (108)

Since 120583|119860is 120590-finite 119871

1(120583|

119860)lowast= 119871

infin(120583|

119860) Let 119892lowast

0isin 119878

119871infin(120583|119860)

be a function such that ⟨119892lowast0 119892

0⟩ = 1 and choose 119909lowast

0isin 119878

119883lowast

such that 119909lowast0(119879

1198601198920) = 119879

1198601198920 Define the operator 119878

0isin

L(1198711(120583|

119860)oplus

1119883) by

1198780(119891 119909) = (0 119879

119860119891) ((119891 119909) isin 119871

1(120583|

119860) oplus

1119883) (109)

and observe that 1198780 = V(119878

0) = 1 Indeed

100381710038171003817100381711987801003817100381710038171003817 ⩽ 1 =

10038171003817100381710038171198791198601003817100381710038171003817

= sup 1003816100381610038161003816119909lowast1198791198601198911003816100381610038161003816 119909

lowastisin 119878

119883lowast 119891 isin 119878

1198711(120583|119860)

= sup 1003816100381610038161003816(119891lowast 119909

lowast) 119878

0(119891 119909)

1003816100381610038161003816

((119891lowast 119909

lowast) (119891 119909)) isin Π (119871

1(120583|

119860) oplus

1119883)

= V (1198780) ⩽100381710038171003817100381711987801003817100381710038171003817

(110)

Abstract and Applied Analysis 13

It is immediate that

(119892lowast

0 119909

lowast

0) 119878

0(119892

0 0) = 119909

lowast

0(119879

1198601198920) =10038171003817100381710038171198791198601198920

1003817100381710038171003817 gt 1 minus 120578 (1205760)

(111)

By Lemma 19 1198711(120583|

119860)oplus

1119883 has the BPBp-nu with the func-

tion 120578 Therefore there exist 1198781isin L(119871

1(120583|

119860)oplus

1119883) (119892

1 119909

1) isin

1198781198711(120583|119860)oplus1119883 and (119892lowast

1 119909

lowast

1) isin 119878

119871infin(120583|119860)oplusinfin119883lowast such that

1003817100381710038171003817(1198921 1199091) minus (1198920 0)1003817100381710038171003817 lt 120576

1003817100381710038171003817(119892lowast

1 119909

lowast

1) minus (119891

lowast

0 119909

lowast

0)1003817100381710038171003817 lt 120576

10038171003817100381710038171198781 minus 11987801003817100381710038171003817 lt 120576 ⟨(119892

lowast

1 119909

lowast

1) (119892

1 119909

1)⟩ = 1

1003816100381610038161003816(119892lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 = V (119878

1) = 1

(112)

Claim 1 We claim that 1199091= 0

Otherwise

1 = Re ⟨(119892lowast1 119909

lowast

1) (119892

1 119909

1)⟩

=100381710038171003817100381711989211003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

+100381710038171003817100381711990911003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

⟩ ⩽ 1

(113)

We deduce that

((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

(119892lowast

1 119909

lowast

1)) (

(0 1199091)

100381710038171003817100381711990911003817100381710038171003817

(119892lowast

1 119909

lowast

1))

isin Π (1198711(120583|

119860) oplus

1119883)

(114)

Since 1198781((0 119909

1)119909

1) = (119878

1minus 119878

0)((0 119909

1)119909

1) lt 120576 we

get that

1 =1003816100381610038161003816⟨(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)⟩1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

100381710038171003817100381711989211003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

)⟩

+100381710038171003817100381711990911003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

)⟩

100381610038161003816100381610038161003816100381610038161003816

⩽100381710038171003817100381711989211003817100381710038171003817 V (1198781) + 120576

100381710038171003817100381711990911003817100381710038171003817 lt100381710038171003817100381711989211003817100381710038171003817 +100381710038171003817100381711990911003817100381710038171003817 = 1

(115)

a contradiction This proves the claimWe define the operator 119878

2 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883

by 1198782(119891 119909) = 119878

1(119891 0) for every 119891 isin 119871

1(120583|

119860) and for every

119909 isin 119883 Then we have

V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 = 1

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ 120576 (116)

Indeed from Claim 1 we have

V (1198781) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 ⩽ V (119878

2)

(117)

On the other hand we see that1003816100381610038161003816(119891

lowast 119909

lowast) 119878

2(119891 119909)

1003816100381610038161003816

=1003816100381610038161003816(119891

lowast 119909

lowast) 119878

1(119891 0)

1003816100381610038161003816 ⩽10038171003817100381710038171198911003817100381710038171003817 V (1198781) ⩽ V (119878

1)

(118)

for every ((119891lowast 119909lowast) (119891 119909)) isin Π(1198711(120583|

119860)oplus

1119883) Therefore

V(1198782) ⩽ V(119878

1) Also

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ sup

119909isin119878119883

10038171003817100381710038171198781 (0 119909)1003817100381710038171003817

= sup119909isin119878119883

10038171003817100381710038171198781 (0 119909) minus 1198780 (0 119909)1003817100381710038171003817 ⩽ 120576

(119)

Claim 2 There exists 1198783 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883 such

that 1198783(119892

1 0) = 119878

3 = 1 119878

3(0 119909) = 0 119878

3(119891 119909) isin 0oplus

1119883

for every (119891 119909) isin 1198711(120583|

119860)oplus

1119883 and 119878

3minus 119878

2 lt 4120576

Indeed write 1198781= (119863

1 119863

2) where 119863

1 119871

1(120583|

119860)oplus

1119883 rarr

1198711(120583|

119860) and119863

2 119871

1(120583|

119860)oplus

1119883 rarr 119883 We have that

sup 1003816100381610038161003816119892lowast119863

1(119892

1 0) + 119909

lowast119863

2(119892

1 0)1003816100381610038161003816

119909lowastisin 119878

119883lowast ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860) +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⩽ V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119892

lowast

1119863

1(119892

1 0) + 119909

lowast

1119863

2(119892

1 0)1003816100381610038161003816

(120)

This implies that

1003816100381610038161003816119909lowast

1119863

2(119892

1 0)1003816100381610038161003816 =10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

1003816100381610038161003816119892lowast

1119863

1(119892

1 0)1003816100381610038161003816

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119871

infin(120583|

119860)

(121)

Therefore |119892lowast1| equals 1 on the support of 119863

1(119892

1 0) As

|⟨119892lowast

1 119892

1⟩| = 1 we also have that |119892lowast

1| equals 1 on the support

of 1198921 Changing the values of 119892lowast

1by the ones of 119891lowast

0on 119860

(supp(1198631(119892

1 0)) cup supp(g

1)) we may and do suppose that

|119892lowast

1| = 1 on the whole 119860We also have 119863

2(119892

1 0) gt 0 Indeed

10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)1003817100381710038171003817

⩽10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198921 0)

1003817100381710038171003817 +10038171003817100381710038171198780 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817

lt 2120576 + 120576 = 3120576

(122)

So we have

10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

⩽10038171003817100381710038171198631(119892

1 0)1003817100381710038171003817 +10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

14 Abstract and Applied Analysis

=1003817100381710038171003817(1198631

(1198921 0) 119863

2(119892

1 0)) minus (0 119879

1198601198920)1003817100381710038171003817

=10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817 lt 3120576

(123)

and 1198632(119892

1 0) gt 119879

1198601198920 minus 3120576 ⩾ 1 minus 120578(120576

0) minus 3120576 gt 0

Finally define the operator 1198783by

1198783(119891 119909) = (0119863

2(119891 0) + 119892

lowast

1(119863

1(119891 0))

1198632(119892

1 0)

119909lowast

1119863

2(119892

1 0))

(124)

for (119891 119909) isin 1198711(120583|

119860)oplus

1119883 It is clear that

100381710038171003817100381711987831003817100381710038171003817 ⩽ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) (125)

Notice also that

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817

⩽10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 +10038171003817100381710038171198632(119891 0) minus 119879

1198601198911003817100381710038171003817

=1003817100381710038171003817(1198631

(119891 0) 1198632(119891 0)) minus (0 119879

119860119891)1003817100381710038171003817

=10038171003817100381710038171198782 (119891 119909) minus 1198780 (119891 119909)

1003817100381710038171003817

(126)

for all (119891 119909) isin 1198711(120583|

119860)oplus

1119883 Hence we have

10038171003817100381710038171198783 minus 11987821003817100381710038171003817

= 2 sup119891isin1198781198711(120583|119860)

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 ⩽ 210038171003817100381710038171198782 minus 1198780

1003817100381710038171003817 lt 4120576(127)

On the other hand let 119866 1198711(120583|

119860) rarr 119871

1(120583|

119860) be defined by

119866(119891) = 119892lowast

1119891 for every 119891 isin 119871

1(120583|

119860) Then we have

V (1198782)

= sup 1003816100381610038161003816119911lowast11987821199111003816100381610038161003816 (119911 119911

lowast) isin Π (119871

1(120583|

119860) oplus

1119883)

⩾ sup10038161003816100381610038161003816100381610038161003816

119909lowast119863

2(119866(

1

120583 (119862)120594119862) 0)

+119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119909lowastisin 119878

119883lowast 119862 isin Σ

119860 120583 (119862) gt 0

= sup10038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119862)120594119862) 0)

10038171003817100381710038171003817100381710038171003817

+

10038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119862 isin Σ119860 120583 (119862) gt 0

(128)

where Σ119860is the family of measurable subsets of 119860

Hence for any simple function 119904 = sum119899

119894=1((120572

119894120583(119860

119894))120594

119860119894

) isin

1198781198711(120583|119860) where 119860

119894119894is a family of disjoint measurable subsets

with strictly positive measure we have

V (1198782)

119899

sum

119894=1

10038161003816100381610038161205721198941003816100381610038161003816 (

100381710038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381710038171003817100381710038171003817100381710038171003817

+

100381610038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381610038161003816100381610038161003816100381610038161003816

)

⩾10038171003817100381710038171198632(119866 (119904) 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119866 (119904) 0)

1003816100381610038161003816

(129)

Since |119892lowast1| = 1119866 is an isometric isomorphism so for each119891 isin

1198781198711(120583|119860)there exists a sequence of norm-one simple functions

(119904119896) such that 119866(119904

119896) converges to 119891 Therefore

V (1198782) ⩾ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) ⩾100381710038171003817100381711987831003817100381710038171003817

(130)

On the other hand we see that100381710038171003817100381711987831003817100381710038171003817 ⩾1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

3(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119909

lowast

1119863

2(119892

1 0) + 119892

lowast

1119863

1(119892

1 0)1003816100381610038161003816 = V (119878

2) = 1

(131)

Therefore 1 = 1198783 = 119878

3(119892

1 0) which proves Claim 2

Finally set 1198783= (0 ) for a suitable 119871

1(120583|

119860)oplus

1119883 rarr

119883 and define the operator 1198791 119871

1(120583) rarr 119883 by

1198791(119891) = 119879

0(119891120594

119860119888) + (119875

119860119891 0) (132)

for every 119891 isin 1198711(120583) Then we have

10038171003817100381710038171198791 (119891)1003817100381710038171003817 ⩽100381710038171003817100381711987901003817100381710038171003817

10038171003817100381710038171198911205941198601198881003817100381710038171003817 +1003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171198911205941198601003817100381710038171003817 =10038171003817100381710038171198911003817100381710038171003817 (133)

for every 119891 isin 1198711(120583) so 119879

1 ⩽ 1 Also

10038171003817100381710038171198791 (1198691198601198921)1003817100381710038171003817 =10038171003817100381710038171198783 (1198921 0)

1003817100381710038171003817 =100381710038171003817100381711987831003817100381710038171003817 = 1 (134)

so 1198791attains its norm at 119869

1198601198921isin 119871

1(120583) and

10038171003817100381710038171198691198601198921 minus 11989101003817100381710038171003817 =10038171003817100381710038171198921 minus 1198920

1003817100381710038171003817 lt 120576 (135)

We also have that for any 119891 isin 1198781198711(120583)

10038171003817100381710038171198790 (119891) minus 1198791 (119891)1003817100381710038171003817 =100381710038171003817100381710038171198790(119891120594

119860) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790 (119869119860119875119860119891) minus 119879119860 (119875119860119891)

1003817100381710038171003817

+10038171003817100381710038171003817119879119860(119875

119860119891) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790119869119860 minus 119879119860

1003817100381710038171003817 +10038171003817100381710038171198780 minus 1198783

1003817100381710038171003817

lt 120578 (1205760) + 6120576

(136)

Hence 1198790minus 119879

1 ⩽ 120578(120576

0) + 6120576 lt 120576

0

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Abstract and Applied Analysis 15

Acknowledgments

The authors thank Gilles Godefroy and Rafael Paya forfruitful conversations about the content of this paper andVladimir Kadets for providing Lemma 18 They also appre-ciate anonymous referees for careful reading and fruitfulsuggestions about revision The second author was sup-ported by Basic Science Research Program through theNational Research Foundation of Korea (NRF) fundedby the Ministry of Education Science and Technology(NRF-2012R1A1A1006869) The third author was partiallysupported by Spanish MICINN and FEDER project noMTM2012-31755 and by Junta de Andalucıa and FEDERGrants FQM-185 and P09-FQM-4911

References

[1] F F Bonsall and J Duncan Numerical Ranges of Operators onNormed Spaces and of Elements of Normed Algebras vol 2 ofLondon Mathematical Society Lecture Note Series 2 CambridgeUniversity Press London UK 1971

[2] F F Bonsall and J Duncan Numerical Ranges II CambridgeUniversity Press London UK 1973 London MathematicalSociety Lecture Notes Series 10

[3] B Sims On numerical range and its applications to Banachalgebras [PhD thesis] University of Newcastle New SouthWales Australia 1972

[4] I D Berg and B Sims ldquoDenseness of operators which attaintheir numerical radiusrdquoAustralian Mathematical Society A vol36 no 1 pp 130ndash133 1984

[5] C S Cardassi ldquoDensity of numerical radius attaining operatorson some reflexive spacesrdquo Bulletin of the Australian Mathemat-ical Society vol 31 no 1 pp 1ndash3 1985

[6] C S Cardassi ldquoNumerical radius attaining operatorsrdquo inBanach Spaces (Columbia Mo 1984) vol 1166 of Lecture Notesin Mathematics pp 11ndash14 Springer Berlin Germany 1985

[7] C S Cardassi ldquoNumerical radius-attaining operators on119862(119870)rdquoProceedings of the AmericanMathematical Society vol 95 no 4pp 537ndash543 1985

[8] MDAcostaOperadores que alcanzan su radio numerico [PhDthesis] Universidad de Granada Granada Spain 1990

[9] M D Acosta and R Paya ldquoNumerical radius attaining oper-ators and the Radon-Nikodym propertyrdquo The Bulletin of theLondon Mathematical Society vol 25 no 1 pp 67ndash73 1993

[10] R Paya ldquoA counterexample on numerical radius attainingoperatorsrdquo Israel Journal of Mathematics vol 79 no 1 pp 83ndash101 1992

[11] Y S Choi and S G Kim ldquoNorm or numerical radius attainingmultilinear mappings and polynomialsrdquo Journal of the LondonMathematical Society vol 54 no 1 pp 135ndash147 1996

[12] M D Acosta J Becerra Guerrero and M Ruiz GalanldquoNumerical-radius-attaining polynomialsrdquo The Quarterly Jour-nal of Mathematics vol 54 no 1 pp 1ndash10 2003

[13] M D Acosta and S G Kim ldquoDenseness of holomorphicfunctions attaining their numerical radiirdquo Israel Journal ofMathematics vol 161 pp 373ndash386 2007

[14] S G Kim and H J Lee ldquoNumerical peak holomorphicfunctions on Banach spacesrdquo Journal of Mathematical Analysisand Applications vol 364 no 2 pp 437ndash452 2010

[15] M D Acosta R M Aron D Garcıa and M MaestreldquoThe Bishop-Phelps-Bollobas theorem for operatorsrdquo Journal ofFunctional Analysis vol 254 no 11 pp 2780ndash2799 2008

[16] A J Guirao and O Kozhushkina ldquoThe Bishop-Phelps-Bollobasproperty for numerical radius in ℓ

1(119862)rdquo Studia Mathematica

vol 218 no 1 pp 41ndash54 2013[17] J Falco ldquoThe Bishop-Phelps-Bollobas property for numerical

radius on 1198711rdquo Journal of Mathematical Analysis and Applica-

tions vol 414 no 1 pp 125ndash133 2014[18] A Aviles A J Guirao and J Rodriguez ldquoOn the Bishop-

Phelps-Bollobas property for numerical radius in119862(119870)-spacesrdquoPreprint

[19] RM Aron Y S Choi S K Kim H J Lee andMMartin ldquoTheBishop-Phelps-Bollobas version of Lindenstrauss properties Aand Brdquo Preprint

[20] V Kadets M Martın and R Paya ldquoRecent progress and openquestions on the numerical index of Banach spacesrdquo Revista dela Real Academia de Ciencias Exactas Fısicas y Naturales A vol100 no 1-2 pp 155ndash182 2006

[21] M Martın J Merı and M Popov ldquoOn the numerical index ofreal 119871

119901(120583)-spacesrdquo Israel Journal of Mathematics vol 184 pp

183ndash192 2011[22] A Iwanik ldquoNorm attaining operators on Lebesgue spacesrdquo

Pacific Journal of Mathematics vol 83 no 2 pp 381ndash386 1979[23] HH SchaeferBanach Lattices and Positive Operators Springer

New York NY USA 1974[24] J Duncan C M McGregor J D Pryce and A J White ldquoThe

numerical index of a normed spacerdquo Journal of the LondonMathematical Society Second Series vol 2 pp 481ndash488 1970

[25] Y S Choi S K Kim H J Lee and M Martin ldquoThe Bishop-Phelps-Bollobas theorem for operators on 119871

1(120583)rdquo Preprint

[26] H E Lacey The Isometric Theory of Classical Banach SpacesSpringer New York NY USA 1974

[27] G Godefroy ldquoUniversal spaces for strictly convex Banachspacesrdquo Revista de la Real Academia de Ciencias Exactas Fısicasy Naturales A vol 100 no 1-2 pp 137ndash146 2006

[28] C Finet M Martın and R Paya ldquoNumerical index andrenormingrdquo Proceedings of the American Mathematical Societyvol 131 no 3 pp 871ndash877 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 6: Research Article On the Bishop-Phelps-Bollobás Property ...downloads.hindawi.com/journals/aaa/2014/479208.pdf · Research Article On the Bishop-Phelps-Bollobás Property for Numerical

6 Abstract and Applied Analysis

that for a convex series sum120572119899 Resuminfin

119899=1120572119899119888119899gt 1 minus 120578 Then for

every 0 lt 119903 lt 1 the set 119860 = 119894 isin N Re 119888119894gt 119903 satisfies the

estimate

sum

119894isin119860

120572119894⩾ 1 minus

120578

1 minus 119903 (39)

From now on 119898 will be a finite regular positive Borelmeasure on the compact Hausdorff space Ω

Lemma 12 Suppose that there exist a nonnegative simplefunction 119891 isin 119878

1198711(119898)

and a function 119892 isin 119878119871infin(119898)

such that

Re ⟨119891 119892⟩ gt 1 minus 1205763

16 (40)

Then there exist a nonnegative simple function 1198911isin 119878

1198711(119898)

anda function 119892

1isin 119878

119871infin(119898)

such that

1198921(119909) = 120594supp(119891

1)(119909) + 119892 (119909) 120594

Ωsupp(1198911)(119909)

⟨1198911 119892

1⟩ = 1

1003817100381710038171003817119891 minus 119891110038171003817100381710038171lt 120576

1003817100381710038171003817119892 minus 11989211003817100381710038171003817infinlt radic120576 supp (119891

1) sub supp (119891)

(41)

Proof Let119891 = sum119898

119895=1(120573

119895119898(119861

119895))120594

119861119895

for some (120573119895) such that

120573119895⩾ 0 for all 119895 andsum119898

119895=1120573119895= 1 and119861

119895rsquos aremutually disjoint

By the assumption we have

Re ⟨119891 119892⟩ =119899

sum

119895=1

120573119895

1

119898 (119861119895)

int119861119895

Re119892 (119909) 119889119898 (119909) gt 1 minus 1205763

16

(42)

and letting

119869 = 119895 1 ⩽ 119895 ⩽ 1198991

119898 (119861119895)

int119861119895

Re119892 (119909) 119889119898 (119909) gt 1 minus 1205762

4

(43)

we have by Lemma 11

sum

119895isin119869

120573119895gt 1 minus

120576

4 (44)

For each 119895 isin 119869 we have

1 minus1205762

4lt

1

119898 (119861119895)

int119861119895

Re119892 (119909) 119889119898 (119909)

=1

119898 (119861119895)

int119861119895capRe119892⩽1minus120576

Re119892 (119909) 119889119898 (119909)

+ int119861119895capRe119892gt1minus120576

Re119892 (119909) 119889119898 (119909)

⩽1

119898 (119861119895)

((1 minus 120576)119898 (119861119895cap Re119892 ⩽ 1 minus 120576)

+ 119898 (119861119895cap Re119892 gt 1 minus 120576))

= 1 minus 120576

119898 (119861119895cap Re119892 ⩽ 1 minus 120576)

119898 (119861119895)

(45)

This implies that

119898(119861119895cap Re119892 ⩽ 1 minus 120576)

119898 (119861119895)

lt120576

4 (46)

Define 119861119895= 119861

119895cap Re119892 gt 1 minus 120576 for all 119895 isin 119869

1198911= (1sum

119895isin119869

120573119895)sum

119895isin119869

120573119895(120594

119861119895

119898 (119861119895)) (47)

and 1198921(119909) = 1 on supp(119891

1) and 119892

1(119909) = 119892(119909) elsewhereThen

it is clear that supp(1198911) sub supp(119891) 119892 minus 119892

1infinlt radic120576 and

⟨1198911 119892

1⟩ = 1 Finally we will show that 119891 minus 119891

1 lt 120576 Notice

first that10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

= 2sum

119895isin119869

120573119895

119898(119861119895 119861

119895)

119898 (119861119895)

lt120576

2

(48)

Hence1003817100381710038171003817119891 minus 1198911

1003817100381710038171003817

10038171003817100381710038171003817100381710038171003817100381710038171003817

1

sum119895isin119869120573119895

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus 119891

10038171003817100381710038171003817100381710038171003817100381710038171003817

1 minus sum119895isin119869120573119895

sum119895isin119869120573119895

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

minus sum

119895isin119869

120573119895

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+120576

4

= (1 minussum

119895isin119869

120573119895) +

120576

2+120576

4

⩽120576

4+120576

2+120576

4= 120576

(49)

Lemma 13 (see [25 Lemma 33]) Suppose that 119879120583is a norm-

one element in L(1198711(119898)) for some 120583 isin 119872(119898) and there is

a nonnegative simple function 1198910such that 119891

0is a norm-one

element of 1198711(119898) and 119879

1205831198910 ⩾ 1 minus 120576

32

6 for some 0 lt 120576 lt 1

Abstract and Applied Analysis 7

Then there exist a norm-one bounded linear operator 119879] forsome ] isin 119872(119898119898) and a nonnegative simple function 119891

1

in 1198781198711(119898)

such that 119879120583minus 119879] ⩽ 120576 1198911 minus 1198910 ⩽ 3120576 and

(119889|]|1119889119898)(119909) = 1 for all 119909 isin supp(1198911)

Lemma 14 Suppose that 119879] isin L(1198711(119898)) is a norm-one

operator 119891 = sum119899

119894=1120573119894(120594

119861119894

119898(119861119894)) where 119898(119861

119895) gt 0 for all

1 ⩽ 119895 ⩽ 119899 and 119861119895119899

119895=1are mutually disjoint Borel subsets ofΩ

is a norm-one nonnegative simple function and 119892 is an elementof 119878

119871infin(119898)

such that

Re ⟨119892 119879]119891⟩ ⩾ 1 minus1205766

27(50)

for some 0 lt 120576 lt 1 and

119889|]|1

119889119898(119909) = 1 119892 (119909) = 1 (51)

for all x in the support of 119891Then there exist a nonnegative simple function 119891 isin 119878

1198711(119898)

a function119892 isin 119878

119871infin(119898)

and an operator119879] inL(1198711(119898) 1198711(119898))such that

⟨119892 119879]119891⟩ =1003817100381710038171003817119879]1003817100381710038171003817 = 1

1003817100381710038171003817119879] minus 119879]1003817100381710038171003817 ⩽ 2120576

10038171003817100381710038171003817119891 minus 119891

10038171003817100381710038171003817⩽ 3120576

1003817100381710038171003817119892 minus 1198921003817100381710038171003817 ⩽radic120576 ⟨119891 119892⟩ = 1

(52)

Proof Since

Re ⟨119892 119879]119891⟩ ⩾ 1 minus1205766

27 (53)

we have

1 minus1205766

27lt Re ⟨119892 119879]119891⟩

= intΩtimesΩ

119891 (119909)Re119892 (119910) 119889120583 (119909 119910)

=

119899

sum

119895=1

120573119895intΩtimesΩ

120594119861119895

(119909)

119898 (119861119895)

Re119892 (119910) 119889] (119909 119910)

(54)

Let

119869 = 119895 intΩtimesΩ

(

120594119861119895

(119909)

119898 (119861119895)

)Re119892 (119910) 119889] (119909 119910) gt 1 minus 1205763

26

(55)

Then from Lemma 11 we have sum119895isin119869120573119895gt 1 minus 120576

32 Let 119891

1=

sum119895isin119869120573119895(120594

119861119895

119898(119861119895)) where 120573

119895= 120573

119895(sum

119895isin119869120573119895) for all 119895 isin 119869

Then

10038171003817100381710038171198911 minus 1198911003817100381710038171003817 ⩽

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

(120573119895minus 120573

119895)

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+ sum

119895isin119869

120573119895⩽ 120576

3⩽ 120576 (56)

Note that there is a Borel measurable function ℎ on Ω times Ωsuch that 119889](119909 119910) = ℎ(119909 119910) 119889|]|(119909 119910) and |ℎ(119909 119910)| = 1 forall (119909 119910) isin Ω times Ω Let

119862 = (119909 119910) 1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 1

1003816100381610038161003816 ltradic120576

232 (57)

Define two measures ]119891and ]

119888as follows

]119891(119860) = ] (119860 119862) ]

119888(119860) = ] (119860 cap 119862) (58)

for every Borel subset 119860 ofΩ times Ω It is clear that

119889] = 119889]119891+ 119889]

119888 119889

10038161003816100381610038161003816]119891

10038161003816100381610038161003816= ℎ 119889]

119891

1198891003816100381610038161003816]1198881003816100381610038161003816 = ℎ 119889]119888 119889 |]| = 119889

10038161003816100381610038161003816]119891

10038161003816100381610038161003816+ 1198891003816100381610038161003816]1198881003816100381610038161003816

(59)

Since (119889|]|11198891198981)(119909) = 1 for all 119909 isin ⋃119899

119895=1119861119895 we have

1 =119889|]|1

1198891198981

(119909) =

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) +1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909) (60)

for all 119909 isin 119861 = ⋃119899

119895=1119861119895 and we deduce that |]|1(119861

119895) = 119898

1(119861

119895)

for all 1 ⩽ 119895 ⩽ 119899We claim that |]

119891|1(119861

119895)119898

1(119861

119895) ⩽ 120576

22

2 for all 119895 isin 119869Indeed if |119892(119910)ℎ(119909 119910) minus 1| ⩾ radic120576232 then

Re (119892 (119910) ℎ (119909 119910)) ⩽ 1 minus 12057624 (61)

So we have

1 minus1205763

26

⩽1

1198981(119861

119895)

ReintΩtimesΩ

120594119861119895(119909)119892 (119910) 119889] (119909 119910)

=1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 |]| (119909 119910)

=1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 10038161003816100381610038161003816]11989110038161003816100381610038161003816(119909 119910)

+1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 1003816100381610038161003816]1198881003816100381610038161003816 (119909 119910)

⩽1

1198981(119861

119895)

((1 minus120576

24)10038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

(119861119895) +1003816100381610038161003816]1198881003816100381610038161003816

1

(119861119895))

= 1 minus120576

24

10038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

(119861119895)

1198981(119861

119895)

(62)

This proves our claimWe also claim that for each 119895 isin 119869 there exists a Borel

subset 119861119895of 119861

119895such that

(1 minus120576

2)119898

1(119861

119895) ⩽ 119898

1(119861

119895) ⩽ 119898

1(119861

119895)

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) ⩽120576

2

(63)

8 Abstract and Applied Analysis

for all 119909 isin 119861119895 Indeed set 119861

119895= 119861

119895cap 119909 isin Ω (119889|]

119891|1119889119898

1)

(119909) ⩽ 1205762 Then

int119861119895119861119895

120576

2119889119898

1(119909) ⩽ int

119861119895

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) 1198891198981(119909)

=10038161003816100381610038161003816]1119891

10038161003816100381610038161003816(119861

119895) ⩽1205762

22119898

1(119861

119895)

(64)

This shows that 1198981(119861

119895 119861

119895) ⩽ (1205762)119898

1(119861

119895) This proves our

second claimNow we define 119892 by 119892(119910) = 119892(119910)|119892(119910)| if |119892(119910)| ⩾ 1 minus

radic120576232 and 119892(119910) = 119892(119910) if |119892(119910)| lt 1 minus radic120576232 and we

write 119891 = sum119895isin119869120573119895(120594

119861119895

1198981(119861

119895)) It is clear that 119892 isin 119878

119871infin(119898)

119892 minus 119892 lt radic120576 and 119892(119910) = 1 for all 119909 isin supp119891

Finally we define the measure

119889] (119909 119910)

= sum

119895isin119869

120594119861119895

(119909) 119892 (119910)ℎ (119909 119910)119889]119888(119909 119910)(

1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+ 1205941198691119861(119909) 119889] (119909 119910)

(65)

where 119861 = ⋃119895isin119869119861119895 It is easy to see that (119889|]|1119889119898

1)(119909) = 1

on 119861 and (119889|]|11198891198981)(119909) ⩽ 1 elsewhere Note that

119889 (] minus ]) (119909 119910)

= sum

119895isin119869

120594119861119895

(119909) [

[

119892 (119910)ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1]

]

119889]119888(119909 119910)

minus sum

119895isin119869

120594119861119895

(119909) 119889]119891(119909 119910)

(66)

If (119909 119910) isin 119862 then |119892(119910)| ⩾ 1 minus radic120576232 ⩾ 1 minus 1232 and

100381610038161003816100381610038161003816119892 (119910)ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

119892 (119910)

1003816100381610038161003816119892 (119910)1003816100381610038161003816

ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816100381610038161003816

1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 11003816100381610038161003816

1003816100381610038161003816119892 (119910)1003816100381610038161003816

+

10038161003816100381610038161 minus1003816100381610038161003816119892 (119910)

1003816100381610038161003816

10038161003816100381610038161003816100381610038161003816119892 (119910)

1003816100381610038161003816

⩽ 2

1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 11003816100381610038161003816

1003816100381610038161003816119892 (119910)1003816100381610038161003816

⩽ 2radic120576

232

232

232 minus 1⩽ 2radic120576

(67)

Hence for all (119909 119910) isin 119862 we have

10038161003816100381610038161003816100381610038161003816100381610038161003816

119892 (119910)ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

⩽100381610038161003816100381610038161003816119892 (119910)ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

⩽ 2radic120576(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

(68)

So we have for all 119909 isin 1198691

119889|] minus ]|1

1198891198981

(119909)

⩽ sum

119895isin119869

120594119861119895

(119909) [

[

2radic120576(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

]

]

1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909)

+ sum

119895isin119869

120594119861119895

(119909)

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909)

⩽ sum

119895isin119869

120594119861119895

(119909)(2radic120576 + (1 minus1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909)))

+sum

119895isin119869

120594119861119895

(119909)(

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909))

⩽ 2radic120576 + 120576 lt 3radic120576

(69)

This gives that 119879] minus 119879] lt 3radic120576 Note also that for all 119895 isin 119869

⟨119879]

120594119861119895

1198981(119861

119895)

119892⟩

= intΩtimesΩ

120594119861119895

(119909)

1198981(119861

119895)

119892 (119910) 119889] (119909 119910)

= intΩtimesΩ

120594119861119895

(119909)

1198981(119861

119895)

ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

119889]119888(119909 119910)

Abstract and Applied Analysis 9

= intΩ

120594119861119895

(119909)

1198981(119861

119895)

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

1198891003816100381610038161003816]1198881003816100381610038161003816

1

(119909)

= intΩ

120594119861119895

(119909)

1198981(119861

119895)

1198891198981(119909) = 1

(70)

Hence we get ⟨119879]119891 119892⟩ = 1 which implies that 119879]119891 =119879] = 1 Finally10038171003817100381710038171003817119891 minus 119891

10038171003817100381710038171003817

⩽10038171003817100381710038171003817119891 minus 119891

1

10038171003817100381710038171003817+10038171003817100381710038171198911 minus 119891

1003817100381710038171003817

=

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

1198981(119861

119895)

minus sum

119895isin119869

120573119895

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+ 120576

⩽ sum

119895isin119869

120573119895(

10038171003817100381710038171003817100381710038171003817100381710038171003817

120594119861119895

1198981(119861

119895)

minus

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

120594119861119895

1198981(119861

119895)

minus

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

) + 120576

= 2sum

119895isin119869

120573119895

1198981(119861

119895 119861

119895)

1198981(119861

119895)

+ 120576

⩽ 2sum

119895isin119869

120573119895

(1205762)1198981(119861

119895)

1198981(119861

119895)

+ 120576 ⩽120576

1 minus 1205762+ 120576 lt 3120576

(71)

We are now ready to present the proof of the main resultin the case of finite regular positive Borel measures

Proof of Proposition 10 Let 1205751= 120575

3

2(5sdot2

4) 120575

2= 120575

12

3(3

2sdot2

14)

and 1205753= (12057610)

2 for some 0 lt 120576 lt 1 Suppose that 119879 isinL(119871

1(119898)) with 119879 = 1 and that there is an 119891

0isin 119878

1198711(119898)

and1198920isin 119878

119871infin(119898)

such that ⟨1198910 119892

0⟩ = 1 and |⟨119879119891

0 119892

0⟩| gt 1minus120575

3

12

6Then there is an isometric isomorphism Ψ from 119871

1(119898) onto

itself such thatΨ(1198910) = |119891

0| and there is a scalar number 120572 in

119878R such that |⟨1198791198910 119892

0⟩| = ⟨120572119879119891

0 119892

0⟩ Then letting 119891

1= Ψ119891

0

1198921= (Ψ

minus1)lowast1198920 and 119879

1= 120572Ψ119879Ψ

minus1 we have

⟨1198781198911 119892

1⟩ = ⟨120572Ψ119879

minus1Ψ119891

0 (Ψ

minus1)lowast

1198920⟩

= ⟨1205721198791198910 119892

0⟩ gt 1 minus

1205753

1

26

⟨1198911 119892

1⟩ = ⟨Ψ119891

0 (Ψ

minus1)lowast

1198920⟩ = 1

(72)

Since 11987911198911 gt 1 minus 120575(120575

3

12

6) by Lemma 13 there exists a

norm-one bounded operator 119879] and a nonnegative simplefunction119891

2isin 119878

1198711(119898)

such that 1198791minus119879] ⩽ 1205751 1198912minus1198911 ⩽ 31205751

and (119889|]|11198891198981)(119909) = 1 for all 119909 isin supp(119891

2) Then

⟨119879]1198912 1198921⟩

= ⟨11987911198911 119892

1⟩ minus ⟨119879

11198911minus 119879

11198912 119892

1⟩ minus ⟨119879

11198912minus 119879]1198912 1198921⟩

⩾ ⟨11987911198911 119892

1⟩ minus10038171003817100381710038171198911 minus 1198912

1003817100381710038171003817 minus10038171003817100381710038171198791 minus 119879]

1003817100381710038171003817

⩾ 1 minus1205753

1

26minus 3120575

1minus 120575

1⩾ 1 minus 5120575

1= 1 minus

1205753

2

16

(73)

Notice also that

⟨1198912 119892

1⟩ = ⟨119891

1 119892

1⟩ minus ⟨119891

1minus 119891

2 119892

1⟩ ⩾ 1 minus

10038171003817100381710038171198911 minus 11989121003817100381710038171003817

⩾ 1 minus 31205751⩾ 1 minus 5120575

1= 1 minus

1205753

2

16

(74)

By Lemma 12 there are a nonnegative simple function 1198913isin

1198781198711(119898)

and a function 1198923isin 119878

119871infin(119898)

such that

1198923(119909) = 120594supp119891

3

(119909) + 1198922(119909) 120594

Ωsupp1198913

(119909)

10038171003817100381710038171198912 minus 11989131003817100381710038171003817 ⩽ 1205752

10038171003817100381710038171198923 minus 11989211003817100381710038171003817 ⩽radic120575

2 ⟨119891

3 119892

3⟩ = 1

(75)

So we have

⟨119879]1198913 1198923⟩ = ⟨119879]1198912 1198921⟩ minus ⟨119879]1198912 minus 119879]1198913 1198921⟩

minus ⟨119879]1198913 1198921 minus 1198923⟩

⩾ 1 minus1205753

2

16minus 2radic120575

2⩾ 1 minus 3radic120575

2= 1 minus

1205756

3

27

(76)

By Lemma 14 there exist 1198914isin 119878

1198711(119898)

and 1198924isin 119878

119871infin(119898)

and anoperator 119879

4such that

⟨1198924 119879

41198914⟩ = 1 =

100381710038171003817100381711987941003817100381710038171003817

10038171003817100381710038171198794 minus 119879]1003817100381710038171003817 ⩽ 21205753

10038171003817100381710038171198914 minus 11989131003817100381710038171003817 ⩽ 31205753

10038171003817100381710038171198924 minus 11989231003817100381710038171003817 ⩽radic120575

3 ⟨119891

4 119892

4⟩ = 1

(77)

So we have

10038171003817100381710038171198794 minus 11987911003817100381710038171003817 ⩽10038171003817100381710038171198794 minus 119879]

1003817100381710038171003817 +1003817100381710038171003817119879] minus 1198791

1003817100381710038171003817

⩽ 1205751+ 2120575

3⩽ 3120575

3

10038171003817100381710038171198911 minus 11989141003817100381710038171003817 ⩽10038171003817100381710038171198911 minus 1198912

1003817100381710038171003817 +10038171003817100381710038171198912 minus 1198913

1003817100381710038171003817 +10038171003817100381710038171198913 minus 1198914

1003817100381710038171003817

⩽ 31205751+ 120575

2+ 3120575

3⩽ 10120575

3

10038171003817100381710038171198921 minus 11989241003817100381710038171003817 ⩽10038171003817100381710038171198921 minus 1198923

1003817100381710038171003817 + 1198923 minus 1198924

⩽ 1205752+ radic120575

3⩽ 2radic120575

3

(78)

10 Abstract and Applied Analysis

Let 119878 = 120572Ψminus11198794Ψ 119891 = Ψminus1

1198914 and 119892 = Ψlowast

1198924 then we have

119879 minus 119878 =10038171003817100381710038171003817119879 minus 120572Ψ

minus11198794Ψ10038171003817100381710038171003817=10038171003817100381710038171003817120572Ψ119879Ψ

minus1minus 119879

4

10038171003817100381710038171003817

=10038171003817100381710038171198791 minus 1198794

1003817100381710038171003817 ⩽ 31205753

100381710038171003817100381710038171198910minus 11989110038171003817100381710038171003817=100381710038171003817100381710038171198910minus Ψ

minus11198914

10038171003817100381710038171003817=10038171003817100381710038171198911 minus 1198914

1003817100381710038171003817 ⩽ 101205753

100381710038171003817100381710038171198920minus 11989110038171003817100381710038171003817=10038171003817100381710038171198920 minus Ψ

lowast1198924

1003817100381710038171003817 =100381710038171003817100381710038171003817(Ψ

minus1)lowast

1198920minus 119892

4

100381710038171003817100381710038171003817

=10038171003817100381710038171198921 minus 1198924

1003817100381710038171003817 ⩽ 2radic120575

3

⟨119891 119892⟩ = ⟨Ψminus11198914 Ψ

lowast1198924⟩ = ⟨119891

4 119892

4⟩ = 1

10038161003816100381610038161003816⟨119878119891 119892⟩

10038161003816100381610038161003816=10038161003816100381610038161003816⟨120572Ψ

minus11198794ΨΨ

minus11198914 Ψ

lowast1198924⟩10038161003816100381610038161003816= |120572| = 1

(79)

This completes the proof

Finally we may give the proof of the main result in fullgenerality

Proof of Theorem 9 Notice that the Kakutani representationtheorem (see [26] for a reference) says that for every 120590-finitemeasure ] the space 119871

1(]) is isometrically isomorphic to

1198711(119898) for some positive Borel regular measure on a compact

Hausdorff space Then by Proposition 10 there is a universalfunction 120576 997891rarr 120578(120576) gt 0 which gives the BPBp-nu for 119871

1(]) for

every 120590-finite measure ]Fix 120576 gt 0 Suppose that 119879

0isin L(119871

1(120583)) with V(119879

0) = 1

and (1198910 119891

lowast

0) isin Π(119871

1(120583)) satisfy

1003816100381610038161003816⟨119891lowast

0 119879

01198910⟩1003816100381610038161003816 gt 1 minus 120578 (120576) (80)

Choose a sequence 119891119899 in 119871

1(120583) such that sup

119899119879

0119891119899 = 1

and let 119866 be the closed linear span of

119879119899

0119891119898 119899 119898 isin N cup 0 (81)

As 119866 is separable there is a dense subset 119892119899 119899 isin N of 119866

and let 119864 = ⋃infin

119899=1supp119892

119899 where supp119892

119899is the support of 119892

119899

Then the measure 120583|119864is 120590-finite Let

119884 = 119891 isin 1198711(120583) supp (119891) sub 119864 (82)

be a closed subspace of 1198711(120583) It is clear that 119871

1(120583) = 119884oplus

1119885

and 119884 is isometrically isomorphic to 1198711(120583|

119864) So 119884 has the

BPBp-nu with 120578(120576)Now write 119878

0= 119879

0|119884 119884 rarr 119884 consider 119910

0= 119891

0isin

119878119884 119910lowast

0= 119891

lowast

0|119884isin 119878

119884lowast and observe that 119910lowast

0(119910

0) = 1 and

|119910lowast

0(119878

01199100)| = |119891

lowast

0(119879

01198910)| gt 1 minus 120578(120576) Hence there exist 119878 isin

L(119884) and (1199100 119910

lowast

0) isin Π(119884) such that

1003816100381610038161003816119910lowast

0(119878119910

0)1003816100381610038161003816 = 1 = V (119878) 1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

10038171003817100381710038171199100 minus 11991001003817100381710038171003817 lt 120576

1003817100381710038171003817119910lowast

0minus 119910

lowast

0

1003817100381710038171003817 lt 120576

(83)

Finally consider the operator 119879 isinL(1198711(120583)) given by

119879 (119910 119911)

= (119878119910 0) + 1198790(0 119911) ((119910 119911) isin 119871

1(120583) equiv 119884oplus

1119885)

(84)

We have 119879 = 1 (and so V(119879) = 1) Indeed1003817100381710038171003817119879 (119910 119911)

1003817100381710038171003817 =1003817100381710038171003817(119878119910 0)

1003817100381710038171003817 +10038171003817100381710038171198790 (0 119911)

1003817100381710038171003817 ⩽10038171003817100381710038171199101003817100381710038171003817 + 119911 =

1003817100381710038171003817(119910 119911)1003817100381710038171003817

(85)

for all (119910 119911) isin 1198711(120583) and 119879(119910

0 0) = (119878119910

0 0) = 119878119910

0 = 1

Let 119909 = (1199100 0) and 119909lowast = (119910lowast

0 119891

0|119885) Then (119909 119909lowast) isin Π(119871

1(120583))

Moreover we have1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119910

lowast

0119878119910

0

1003816100381610038161003816 = 1 = V (119879) 1003817100381710038171003817119909 minus 1198910

1003817100381710038171003817 =1003817100381710038171003817119910 minus 1199100

1003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowast

0minus 119891

lowast

0

1003817100381710038171003817

= max 1003817100381710038171003817119910 minus 119891lowast

0|119884

1003817100381710038171003817 1003817100381710038171003817119891

lowast

0|119885minus 119891

lowast

0|119885

1003817100381710038171003817

=1003817100381710038171003817119910

lowastminus 119910

lowast

0

1003817100381710038171003817 lt 120576

1003817100381710038171003817119879 minus 11987901003817100381710038171003817 = sup119910+119911⩽1

1003817100381710038171003817119879 (119910 119911) minus 1198790 (119910 119911)1003817100381710038171003817

= sup119910⩽1

1003817100381710038171003817119878119910 minus 11987801199101003817100381710038171003817 =1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

(86)

This completes the proof

5 Examples of Spaces Failing theBishop-Phelps-Bollobaacutes Property forNumerical Radius

Our goal here is to prove that the density of numerical radiusattaining operators does not imply the BPBp-nu Actuallywe will show that among separable spaces there is noisomorphic property implying the BPBp-nu other than finite-dimensionality

We need to relate the BPBp-nu to the Bishop-Phelps-Bollobas property for operators which as mentioned in theintroduction was introduced in [15] A pair (119883 119884) of Banachspaces has the Bishop-Phelps-Bollobas property for operators(in short BPBp) if given 120576 gt 0 there exists 120578(120576) gt 0 suchthat given 119879 isin L(119883 119884) with 119879 = 1 and 119909 isin 119878

119883such that

119879119909 gt 1 minus 120578(120576) then there exist 119911 isin 119878119883and 119878 isin L(119883 119884)

satisfying

119878 = 119878119911 = 1 119909 minus 119911 lt 120576

119879 minus 119878 lt 120576

(87)

We refer the reader to [15 19 25] and references therein formore information and background Among the interestingresults on the BPBp we emphasize that a pair (119883 119884) when119883is finite-dimensional does not necessarily have the BPBp Forinstance if119884 is a strictly convex space which is not uniformlyconvex then the pair (ℓ(2)

1 119884) fails to have the BPBp (this is

contained in [15] see [19 Section 3])

Abstract and Applied Analysis 11

The next result relates the BPBp-nu to the BPBp foroperators in a particular case We will deduce our examplefrom it

Theorem 15 If 1198711(120583)oplus

1119883 has the BPBp-119899119906 then the pair

(1198711(120583) 119883) has the BPBp for operators

Before proving this proposition we will use it to get themain examples of this section The first example shows thatthe density of numerical radius attaining operators does notimply the BPBp-nu

Example 16 There is a reflexive space (and so numerical radiusattaining operators on it are dense) which fails to have theBPBp-nu Indeed let 119884 be a reflexive separable space whichis not superreflexive and we may suppose that 119884 is strictlyconvex Observe that 119884 cannot be uniformly convex since itis not superreflexive Now 119883 = ℓ(2)

1oplus1119884 is reflexive but the

pair (ℓ(2)1 119884) fails the BPBp since 119884 is strictly convex but not

uniformly convex [19 Corollary 33] Therefore Theorem 15gives us that119883 does not have the BPBp-nu

The example above can be extended to get the result thatevery infinite-dimensional separable Banach space can berenormed to fail the BPBp-nu This follows from the factthat every infinite-dimensional separable Banach space canbe renormed to be strictly convex but not uniformly convex(this result can be proved ldquoby handrdquo an alternative categoricalargument for it can be found in [27] and references therein)With a little more of effort we may get the main result of thesection

Theorem 17 Every infinite-dimensional separable Banachspace can be renormed to fail the weak-BPBp-nu (and so inparticular to fail the BPBp-nu)

We need the following result which is surely well knownAs we have not found a reference we include a nice and easyproof kindly given to us by Vladimir Kadets We recall thatgiven a Banach space 119884 the set of all equivalent norms on119884 can be viewed as a metric space using the Banach-Mazurdistance

Lemma 18 Let119884 be an infinite-dimensional separable Banachspace Then the set of equivalent norms on 119884 which are strictlyconvex and are not (locally) uniformly convex is dense in the setof all equivalent norms on119884 (with respect to the Banach-Mazurdistance)

Proof Fix 119890 isin 119878119884and 119890lowast

1isin 119878

119884lowast such that 119890lowast

1(119890) = 1 For a fixed

120576 isin (0 12) denote

119902 (119910) = max (1 minus 120576) 10038171003817100381710038171199101003817100381710038171003817 1003816100381610038161003816119890lowast

1(119910)1003816100381610038161003816 (119910 isin 119884) (88)

Evidently (1 minus 120576)119910 ⩽ 119902(119910) ⩽ 119910 for every 119910 isin 119884 Fix asequence 119890lowast

119896 119896 ⩾ 2 of norm-one functionals separating

the points of 119884 and denote

119901 (119910) = radic

infin

sum

119896=1

1

2119896

1003816100381610038161003816119890lowast

119896(119910)1003816100381610038161003816

2

(119910 isin 119884) (89)

Then 119901 is a strictly convex norm on 119884 119901(119890) ⩾ 1radic2 and119901(119910) ⩽ 119910 for all 119910 isin 119883 Finally write

100381710038171003817100381711991010038171003817100381710038171= (1 minus 120576) 119902 (119910) + 120576

119901 (119910)

119901 (119890)(119910 isin 119884) (90)

Then sdot 1is a strictly convex norm on 119884 and

(1 minus 120576)2 10038171003817100381710038171199101003817100381710038171003817 ⩽100381710038171003817100381711991010038171003817100381710038171⩽ (1 + 120576)

10038171003817100381710038171199101003817100381710038171003817 (119910 isin 119884) (91)

Wewill finish the proof by showing that sdot 1is not uniformly

convex (actually it is not locally uniformly convex) Indeedfor each 119899 isin N we select 119910

119899isin ⋂

119899

119896=1ker 119890lowast

119896with 119910

119899 = 1 and

consider 119890119899= 119890 + (1205764)119910

119899 Then 119902(119890) = 1 119902(119890

119899) = 1 and

119902(119890+119890119899) = 2 At the same time 119901(119910

119899) rarr 0 so 119901(119890

119899) rarr 119901(119890)

and 119901(119890 + 119890119899) rarr 2119901(119890) Consequently

1198901 = 1100381710038171003817100381711989011989910038171003817100381710038171997888rarr 1

1003817100381710038171003817119890 + 11989011989910038171003817100381710038171997888rarr 2 (92)

but 119890 minus 1198901198991= (1205764)119910

1198991⩾ (1 minus 120576)

2(1205764) which means the

absence of local uniform convexity at 119890

Proof of Theorem 17 Let 119883 be an infinite-dimensional sepa-rable Banach space Take a closed subspace 119884 of 119883 of codi-mension two By [28 Proposition 2] the map carrying everyequivalent norm on 119884 to its numerical index is continuousand so the set of values of the numerical index of 119884 upto reforming is a nontrivial interval [28 Theorem 9] ThenLemma 18 allows us to find an equivalent norm | sdot | on 119884 insuch a way that (119884 | sdot |) is strictly convex and is not uniformlyconvex and 119899(119884 | sdot |) gt 0 Now the space 119883 = ℓ(2)

1oplus1(119884 | sdot |)

is an equivalent renorming of 119883 which does not have theBPBp-119899119906 (indeed otherwise the pair (ℓ(2)

1 (119884 | sdot |)) would

have the BPBp for the operator norm and so (119884 | sdot |) wouldbe uniformly convex by [19 Corollary 33] a contradiction)Moreover as

119899 (119883) = min 119899 (ℓ(2)1) 119899 (119884 |sdot|) gt 0 (93)

(see [20 Proposition 2] for instance) 119883 also fails the weak-BPBp-nu by Proposition 6

To finish the section with the promised proof ofTheorem 15 we first see the following stability result

Lemma 19 Let119883 = [⨁infin

119896=1119883

119896]1198880

or [⨁infin

119896=1119883

119896]ℓ1

If119883 has theBishop-Phelps-Bollobas property for numerical radius with afunction 120578 then each Banach space 119883

119894has the Bishop-Phelps-

Bollobas property for numerical radius with 120578nu(119883119894) ⩾ 120578 That

is inf119894120578nu(119883119894

)(120576) ⩾ 120578nu(119883)(120576) for all 0 lt 120576 lt 1

Proof Let 119875119894 119883 rarr 119883

119894and 1198751015840

119894 119883

lowastrarr 119883

lowast

119894be the natural

projections and let 119876119894 119883

119894rarr 119883 and 1198761015840

119894 119883

lowast

119894rarr 119883

lowast be thenatural embeddings

Assume that an operator 119879119894 119883

119894rarr 119883

119894and a pair

(119909119894 119909

lowast

119894) isin Π(119883

119894) satisfy that

V (119879119894) = 1

1003816100381610038161003816119909lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (94)

We define an operator 119879 119883 rarr 119883 and (119909 119909lowast) isin Π(119883) by

119879 = 119876119894∘ 119879

119894∘ 119875

119894 (119909 119909

lowast) = (119876

119894119909119894 119876

1015840

119894119909lowast

119894) (95)

12 Abstract and Applied Analysis

then clearly we see that1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119909

lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (96)

From the assumption there exist 119878 119883 rarr 119883 and a pair(119910 119910

lowast) isin Π(119883) such that

1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 = 1 = V (119878) 119878 minus 119879 lt 120576

1003817100381710038171003817119910lowastminus 119909

lowast1003817100381710038171003817 lt 1205761003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576

(97)

Since this clearly shows that1003817100381710038171003817119875119894 ∘ 119878 ∘ 119876119894

minus 119879119894

1003817100381710038171003817 lt 120576100381710038171003817100381710038171198751015840

119894119910lowastminus 119909

lowast

119894

10038171003817100381710038171003817lt 120576

1003817100381710038171003817119875119894119910 minus 1199091198941003817100381710038171003817 lt 120576

(98)

we only need to show that |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = 1

We first show the case of 1198880sum Since

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817=10038171003817100381710038171003817119875119895119910 minus 119875

11989511990910038171003817100381710038171003817⩽1003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576 (99)

for every 119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽100381710038171003817100381710038171198751015840

119894119910lowast10038171003817100381710038171003817+ 120576 sum

119895isinN119895 = 119894

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817⩽1003817100381710038171003817119910

lowast1003817100381710038171003817 = 1

(100)

This shows that 1198751015840119894119910lowast = 1 and 1198751015840

119895119910lowast= 0 for every 119895 = 119894

So 119910lowast = 1198761015840

1198941198751015840

119894119910lowast and 1198751015840

119894119910lowast(119875

119894119910) = 1 This and the fact that

119910 minus 119876119894119875119894119910 lt 120576 imply that

(119876119894119875119894119910 + (

1

120576) (119910 minus 119876

119894119875119894119910) 119876

1015840

1198941198751015840

119894119910lowast) isin Π (119883) (101)

So we get that (1198761015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910 + (1120576)(119910 minus 119876

119894119875119894119910))V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =10038161003816100381610038161003816(119876

1015840

1198941198751015840

119894119910lowast) 11987811991010038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910 +1

120576(119910 minus 119876

119894119875119894119910))

1003816100381610038161003816100381610038161003816⩽ 1

(102)

and so we get |1198751015840119894119910lowast(119875

119894∘119878 ∘119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

We next show the case of ℓ1sum The proof is almost

the same as that of the 1198880case However for the sake of

completeness we provide it hereSince 1198751015840

119895119910lowast = 119875

1015840

119895119910lowastminus 119875

1015840

119895119909lowast ⩽ 119910

lowastminus 119909

lowast lt 120576 for every

119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽10038171003817100381710038171198751198941199101003817100381710038171003817 + 120576 sum

119895isinN119895 = 119894

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817⩽10038171003817100381710038171199101003817100381710038171003817 = 1

(103)

which shows 119875119894119910 = 1 and 119875

119895119910 = 0 for every 119895 = 119894 Since this

implies (119876119894119875119894119910119876

1015840

1198941198751015840

119894119910lowast+(1120576)(119910

lowastminus119876

1015840

1198941198751015840

119894119910lowast)) isin Π(119883) we get

that |(1198761015840

1198941198751015840

119894119910lowast+ (1120576)(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast))119878(119876

119894119875119894119910)| ⩽ V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =1003816100381610038161003816119910

lowast119878 (119876

119894119875119894119910)1003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast+1

120576(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast)) 119878 (119876

119894119875119894119910)

1003816100381610038161003816100381610038161003816⩽ 1

(104)

and so |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

Proof of Theorem 15 Note that 120578nu(1198711(120583)oplus1119883)(120576) rarr 0 as120576 rarr 0 Fix 0 lt 120576

0lt 1 and choose 0 lt 120576 lt 1 such that

6120576 + 120578nu(1198711(120583)oplus1119883)(120576) lt 1205760 Let 120578(1205760) = 120578nu(1198711(120583)oplus1119883)(120576)Suppose that 119879

0isin L(119871

1(120583) 119883) with 119879

0 = 1 and 119891

0isin

1198781198711(120583)

satisfy

1003817100381710038171003817119879011989101003817100381710038171003817 gt 1 minus 120578 (1205760) (105)

For any measurable subset 119861 let

1198711(120583|

119861) = 119891|

119861 119891 isin 119871

1(120583) (106)

with the norm 119891|119861 = 119891120594

1198611 Then it is easy to see

that 1198711(120583|

119861) is isometrically isomorphic to a complemented

subspace of 1198711(120583) Let 119875

119861 119871

1(120583) rarr 119871

1(120583|

119861) be the

restriction defined by 119875119861(119891) = 119891|

119861for all 119891 isin 119871

1(120583) and

let 119869119861 119871

1(120583|

119861) rarr 119871

1(120583) be the extension defined by

119869119861(119891)(120596) = 119891(120596) if 120596 isin 119861 and 119869

119861(119891)(120596) = 0 otherwise

It is clear that 119875119861119869119861= Id

1198711(120583|119861)and 119869

119861119875119861(119891) = 119891120594

119861for all

119891 isin 1198711(120583) Notice also that 119871

1(120583) is isometrically isomorphic

to 1198711(120583|

119861)oplus

11198711(120583|

119861119888)

Let 119860 = supp1198910and 119892

0= 119875

1198601198910 Then

1003817100381710038171003817119879011986911986011989201003817100381710038171003817 =100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) gt 0 (107)

and define the operator 119879119860 119871

1(120583|

119860) rarr 119883 by 119879

119860119891 =

1198790119869119860119891119879

0119869119860 for every 119891 isin 119871

1(120583|

119860) Then

100381710038171003817100381711987911986011989201003817100381710038171003817 ⩾100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) (108)

Since 120583|119860is 120590-finite 119871

1(120583|

119860)lowast= 119871

infin(120583|

119860) Let 119892lowast

0isin 119878

119871infin(120583|119860)

be a function such that ⟨119892lowast0 119892

0⟩ = 1 and choose 119909lowast

0isin 119878

119883lowast

such that 119909lowast0(119879

1198601198920) = 119879

1198601198920 Define the operator 119878

0isin

L(1198711(120583|

119860)oplus

1119883) by

1198780(119891 119909) = (0 119879

119860119891) ((119891 119909) isin 119871

1(120583|

119860) oplus

1119883) (109)

and observe that 1198780 = V(119878

0) = 1 Indeed

100381710038171003817100381711987801003817100381710038171003817 ⩽ 1 =

10038171003817100381710038171198791198601003817100381710038171003817

= sup 1003816100381610038161003816119909lowast1198791198601198911003816100381610038161003816 119909

lowastisin 119878

119883lowast 119891 isin 119878

1198711(120583|119860)

= sup 1003816100381610038161003816(119891lowast 119909

lowast) 119878

0(119891 119909)

1003816100381610038161003816

((119891lowast 119909

lowast) (119891 119909)) isin Π (119871

1(120583|

119860) oplus

1119883)

= V (1198780) ⩽100381710038171003817100381711987801003817100381710038171003817

(110)

Abstract and Applied Analysis 13

It is immediate that

(119892lowast

0 119909

lowast

0) 119878

0(119892

0 0) = 119909

lowast

0(119879

1198601198920) =10038171003817100381710038171198791198601198920

1003817100381710038171003817 gt 1 minus 120578 (1205760)

(111)

By Lemma 19 1198711(120583|

119860)oplus

1119883 has the BPBp-nu with the func-

tion 120578 Therefore there exist 1198781isin L(119871

1(120583|

119860)oplus

1119883) (119892

1 119909

1) isin

1198781198711(120583|119860)oplus1119883 and (119892lowast

1 119909

lowast

1) isin 119878

119871infin(120583|119860)oplusinfin119883lowast such that

1003817100381710038171003817(1198921 1199091) minus (1198920 0)1003817100381710038171003817 lt 120576

1003817100381710038171003817(119892lowast

1 119909

lowast

1) minus (119891

lowast

0 119909

lowast

0)1003817100381710038171003817 lt 120576

10038171003817100381710038171198781 minus 11987801003817100381710038171003817 lt 120576 ⟨(119892

lowast

1 119909

lowast

1) (119892

1 119909

1)⟩ = 1

1003816100381610038161003816(119892lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 = V (119878

1) = 1

(112)

Claim 1 We claim that 1199091= 0

Otherwise

1 = Re ⟨(119892lowast1 119909

lowast

1) (119892

1 119909

1)⟩

=100381710038171003817100381711989211003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

+100381710038171003817100381711990911003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

⟩ ⩽ 1

(113)

We deduce that

((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

(119892lowast

1 119909

lowast

1)) (

(0 1199091)

100381710038171003817100381711990911003817100381710038171003817

(119892lowast

1 119909

lowast

1))

isin Π (1198711(120583|

119860) oplus

1119883)

(114)

Since 1198781((0 119909

1)119909

1) = (119878

1minus 119878

0)((0 119909

1)119909

1) lt 120576 we

get that

1 =1003816100381610038161003816⟨(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)⟩1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

100381710038171003817100381711989211003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

)⟩

+100381710038171003817100381711990911003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

)⟩

100381610038161003816100381610038161003816100381610038161003816

⩽100381710038171003817100381711989211003817100381710038171003817 V (1198781) + 120576

100381710038171003817100381711990911003817100381710038171003817 lt100381710038171003817100381711989211003817100381710038171003817 +100381710038171003817100381711990911003817100381710038171003817 = 1

(115)

a contradiction This proves the claimWe define the operator 119878

2 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883

by 1198782(119891 119909) = 119878

1(119891 0) for every 119891 isin 119871

1(120583|

119860) and for every

119909 isin 119883 Then we have

V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 = 1

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ 120576 (116)

Indeed from Claim 1 we have

V (1198781) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 ⩽ V (119878

2)

(117)

On the other hand we see that1003816100381610038161003816(119891

lowast 119909

lowast) 119878

2(119891 119909)

1003816100381610038161003816

=1003816100381610038161003816(119891

lowast 119909

lowast) 119878

1(119891 0)

1003816100381610038161003816 ⩽10038171003817100381710038171198911003817100381710038171003817 V (1198781) ⩽ V (119878

1)

(118)

for every ((119891lowast 119909lowast) (119891 119909)) isin Π(1198711(120583|

119860)oplus

1119883) Therefore

V(1198782) ⩽ V(119878

1) Also

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ sup

119909isin119878119883

10038171003817100381710038171198781 (0 119909)1003817100381710038171003817

= sup119909isin119878119883

10038171003817100381710038171198781 (0 119909) minus 1198780 (0 119909)1003817100381710038171003817 ⩽ 120576

(119)

Claim 2 There exists 1198783 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883 such

that 1198783(119892

1 0) = 119878

3 = 1 119878

3(0 119909) = 0 119878

3(119891 119909) isin 0oplus

1119883

for every (119891 119909) isin 1198711(120583|

119860)oplus

1119883 and 119878

3minus 119878

2 lt 4120576

Indeed write 1198781= (119863

1 119863

2) where 119863

1 119871

1(120583|

119860)oplus

1119883 rarr

1198711(120583|

119860) and119863

2 119871

1(120583|

119860)oplus

1119883 rarr 119883 We have that

sup 1003816100381610038161003816119892lowast119863

1(119892

1 0) + 119909

lowast119863

2(119892

1 0)1003816100381610038161003816

119909lowastisin 119878

119883lowast ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860) +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⩽ V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119892

lowast

1119863

1(119892

1 0) + 119909

lowast

1119863

2(119892

1 0)1003816100381610038161003816

(120)

This implies that

1003816100381610038161003816119909lowast

1119863

2(119892

1 0)1003816100381610038161003816 =10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

1003816100381610038161003816119892lowast

1119863

1(119892

1 0)1003816100381610038161003816

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119871

infin(120583|

119860)

(121)

Therefore |119892lowast1| equals 1 on the support of 119863

1(119892

1 0) As

|⟨119892lowast

1 119892

1⟩| = 1 we also have that |119892lowast

1| equals 1 on the support

of 1198921 Changing the values of 119892lowast

1by the ones of 119891lowast

0on 119860

(supp(1198631(119892

1 0)) cup supp(g

1)) we may and do suppose that

|119892lowast

1| = 1 on the whole 119860We also have 119863

2(119892

1 0) gt 0 Indeed

10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)1003817100381710038171003817

⩽10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198921 0)

1003817100381710038171003817 +10038171003817100381710038171198780 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817

lt 2120576 + 120576 = 3120576

(122)

So we have

10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

⩽10038171003817100381710038171198631(119892

1 0)1003817100381710038171003817 +10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

14 Abstract and Applied Analysis

=1003817100381710038171003817(1198631

(1198921 0) 119863

2(119892

1 0)) minus (0 119879

1198601198920)1003817100381710038171003817

=10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817 lt 3120576

(123)

and 1198632(119892

1 0) gt 119879

1198601198920 minus 3120576 ⩾ 1 minus 120578(120576

0) minus 3120576 gt 0

Finally define the operator 1198783by

1198783(119891 119909) = (0119863

2(119891 0) + 119892

lowast

1(119863

1(119891 0))

1198632(119892

1 0)

119909lowast

1119863

2(119892

1 0))

(124)

for (119891 119909) isin 1198711(120583|

119860)oplus

1119883 It is clear that

100381710038171003817100381711987831003817100381710038171003817 ⩽ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) (125)

Notice also that

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817

⩽10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 +10038171003817100381710038171198632(119891 0) minus 119879

1198601198911003817100381710038171003817

=1003817100381710038171003817(1198631

(119891 0) 1198632(119891 0)) minus (0 119879

119860119891)1003817100381710038171003817

=10038171003817100381710038171198782 (119891 119909) minus 1198780 (119891 119909)

1003817100381710038171003817

(126)

for all (119891 119909) isin 1198711(120583|

119860)oplus

1119883 Hence we have

10038171003817100381710038171198783 minus 11987821003817100381710038171003817

= 2 sup119891isin1198781198711(120583|119860)

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 ⩽ 210038171003817100381710038171198782 minus 1198780

1003817100381710038171003817 lt 4120576(127)

On the other hand let 119866 1198711(120583|

119860) rarr 119871

1(120583|

119860) be defined by

119866(119891) = 119892lowast

1119891 for every 119891 isin 119871

1(120583|

119860) Then we have

V (1198782)

= sup 1003816100381610038161003816119911lowast11987821199111003816100381610038161003816 (119911 119911

lowast) isin Π (119871

1(120583|

119860) oplus

1119883)

⩾ sup10038161003816100381610038161003816100381610038161003816

119909lowast119863

2(119866(

1

120583 (119862)120594119862) 0)

+119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119909lowastisin 119878

119883lowast 119862 isin Σ

119860 120583 (119862) gt 0

= sup10038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119862)120594119862) 0)

10038171003817100381710038171003817100381710038171003817

+

10038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119862 isin Σ119860 120583 (119862) gt 0

(128)

where Σ119860is the family of measurable subsets of 119860

Hence for any simple function 119904 = sum119899

119894=1((120572

119894120583(119860

119894))120594

119860119894

) isin

1198781198711(120583|119860) where 119860

119894119894is a family of disjoint measurable subsets

with strictly positive measure we have

V (1198782)

119899

sum

119894=1

10038161003816100381610038161205721198941003816100381610038161003816 (

100381710038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381710038171003817100381710038171003817100381710038171003817

+

100381610038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381610038161003816100381610038161003816100381610038161003816

)

⩾10038171003817100381710038171198632(119866 (119904) 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119866 (119904) 0)

1003816100381610038161003816

(129)

Since |119892lowast1| = 1119866 is an isometric isomorphism so for each119891 isin

1198781198711(120583|119860)there exists a sequence of norm-one simple functions

(119904119896) such that 119866(119904

119896) converges to 119891 Therefore

V (1198782) ⩾ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) ⩾100381710038171003817100381711987831003817100381710038171003817

(130)

On the other hand we see that100381710038171003817100381711987831003817100381710038171003817 ⩾1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

3(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119909

lowast

1119863

2(119892

1 0) + 119892

lowast

1119863

1(119892

1 0)1003816100381610038161003816 = V (119878

2) = 1

(131)

Therefore 1 = 1198783 = 119878

3(119892

1 0) which proves Claim 2

Finally set 1198783= (0 ) for a suitable 119871

1(120583|

119860)oplus

1119883 rarr

119883 and define the operator 1198791 119871

1(120583) rarr 119883 by

1198791(119891) = 119879

0(119891120594

119860119888) + (119875

119860119891 0) (132)

for every 119891 isin 1198711(120583) Then we have

10038171003817100381710038171198791 (119891)1003817100381710038171003817 ⩽100381710038171003817100381711987901003817100381710038171003817

10038171003817100381710038171198911205941198601198881003817100381710038171003817 +1003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171198911205941198601003817100381710038171003817 =10038171003817100381710038171198911003817100381710038171003817 (133)

for every 119891 isin 1198711(120583) so 119879

1 ⩽ 1 Also

10038171003817100381710038171198791 (1198691198601198921)1003817100381710038171003817 =10038171003817100381710038171198783 (1198921 0)

1003817100381710038171003817 =100381710038171003817100381711987831003817100381710038171003817 = 1 (134)

so 1198791attains its norm at 119869

1198601198921isin 119871

1(120583) and

10038171003817100381710038171198691198601198921 minus 11989101003817100381710038171003817 =10038171003817100381710038171198921 minus 1198920

1003817100381710038171003817 lt 120576 (135)

We also have that for any 119891 isin 1198781198711(120583)

10038171003817100381710038171198790 (119891) minus 1198791 (119891)1003817100381710038171003817 =100381710038171003817100381710038171198790(119891120594

119860) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790 (119869119860119875119860119891) minus 119879119860 (119875119860119891)

1003817100381710038171003817

+10038171003817100381710038171003817119879119860(119875

119860119891) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790119869119860 minus 119879119860

1003817100381710038171003817 +10038171003817100381710038171198780 minus 1198783

1003817100381710038171003817

lt 120578 (1205760) + 6120576

(136)

Hence 1198790minus 119879

1 ⩽ 120578(120576

0) + 6120576 lt 120576

0

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Abstract and Applied Analysis 15

Acknowledgments

The authors thank Gilles Godefroy and Rafael Paya forfruitful conversations about the content of this paper andVladimir Kadets for providing Lemma 18 They also appre-ciate anonymous referees for careful reading and fruitfulsuggestions about revision The second author was sup-ported by Basic Science Research Program through theNational Research Foundation of Korea (NRF) fundedby the Ministry of Education Science and Technology(NRF-2012R1A1A1006869) The third author was partiallysupported by Spanish MICINN and FEDER project noMTM2012-31755 and by Junta de Andalucıa and FEDERGrants FQM-185 and P09-FQM-4911

References

[1] F F Bonsall and J Duncan Numerical Ranges of Operators onNormed Spaces and of Elements of Normed Algebras vol 2 ofLondon Mathematical Society Lecture Note Series 2 CambridgeUniversity Press London UK 1971

[2] F F Bonsall and J Duncan Numerical Ranges II CambridgeUniversity Press London UK 1973 London MathematicalSociety Lecture Notes Series 10

[3] B Sims On numerical range and its applications to Banachalgebras [PhD thesis] University of Newcastle New SouthWales Australia 1972

[4] I D Berg and B Sims ldquoDenseness of operators which attaintheir numerical radiusrdquoAustralian Mathematical Society A vol36 no 1 pp 130ndash133 1984

[5] C S Cardassi ldquoDensity of numerical radius attaining operatorson some reflexive spacesrdquo Bulletin of the Australian Mathemat-ical Society vol 31 no 1 pp 1ndash3 1985

[6] C S Cardassi ldquoNumerical radius attaining operatorsrdquo inBanach Spaces (Columbia Mo 1984) vol 1166 of Lecture Notesin Mathematics pp 11ndash14 Springer Berlin Germany 1985

[7] C S Cardassi ldquoNumerical radius-attaining operators on119862(119870)rdquoProceedings of the AmericanMathematical Society vol 95 no 4pp 537ndash543 1985

[8] MDAcostaOperadores que alcanzan su radio numerico [PhDthesis] Universidad de Granada Granada Spain 1990

[9] M D Acosta and R Paya ldquoNumerical radius attaining oper-ators and the Radon-Nikodym propertyrdquo The Bulletin of theLondon Mathematical Society vol 25 no 1 pp 67ndash73 1993

[10] R Paya ldquoA counterexample on numerical radius attainingoperatorsrdquo Israel Journal of Mathematics vol 79 no 1 pp 83ndash101 1992

[11] Y S Choi and S G Kim ldquoNorm or numerical radius attainingmultilinear mappings and polynomialsrdquo Journal of the LondonMathematical Society vol 54 no 1 pp 135ndash147 1996

[12] M D Acosta J Becerra Guerrero and M Ruiz GalanldquoNumerical-radius-attaining polynomialsrdquo The Quarterly Jour-nal of Mathematics vol 54 no 1 pp 1ndash10 2003

[13] M D Acosta and S G Kim ldquoDenseness of holomorphicfunctions attaining their numerical radiirdquo Israel Journal ofMathematics vol 161 pp 373ndash386 2007

[14] S G Kim and H J Lee ldquoNumerical peak holomorphicfunctions on Banach spacesrdquo Journal of Mathematical Analysisand Applications vol 364 no 2 pp 437ndash452 2010

[15] M D Acosta R M Aron D Garcıa and M MaestreldquoThe Bishop-Phelps-Bollobas theorem for operatorsrdquo Journal ofFunctional Analysis vol 254 no 11 pp 2780ndash2799 2008

[16] A J Guirao and O Kozhushkina ldquoThe Bishop-Phelps-Bollobasproperty for numerical radius in ℓ

1(119862)rdquo Studia Mathematica

vol 218 no 1 pp 41ndash54 2013[17] J Falco ldquoThe Bishop-Phelps-Bollobas property for numerical

radius on 1198711rdquo Journal of Mathematical Analysis and Applica-

tions vol 414 no 1 pp 125ndash133 2014[18] A Aviles A J Guirao and J Rodriguez ldquoOn the Bishop-

Phelps-Bollobas property for numerical radius in119862(119870)-spacesrdquoPreprint

[19] RM Aron Y S Choi S K Kim H J Lee andMMartin ldquoTheBishop-Phelps-Bollobas version of Lindenstrauss properties Aand Brdquo Preprint

[20] V Kadets M Martın and R Paya ldquoRecent progress and openquestions on the numerical index of Banach spacesrdquo Revista dela Real Academia de Ciencias Exactas Fısicas y Naturales A vol100 no 1-2 pp 155ndash182 2006

[21] M Martın J Merı and M Popov ldquoOn the numerical index ofreal 119871

119901(120583)-spacesrdquo Israel Journal of Mathematics vol 184 pp

183ndash192 2011[22] A Iwanik ldquoNorm attaining operators on Lebesgue spacesrdquo

Pacific Journal of Mathematics vol 83 no 2 pp 381ndash386 1979[23] HH SchaeferBanach Lattices and Positive Operators Springer

New York NY USA 1974[24] J Duncan C M McGregor J D Pryce and A J White ldquoThe

numerical index of a normed spacerdquo Journal of the LondonMathematical Society Second Series vol 2 pp 481ndash488 1970

[25] Y S Choi S K Kim H J Lee and M Martin ldquoThe Bishop-Phelps-Bollobas theorem for operators on 119871

1(120583)rdquo Preprint

[26] H E Lacey The Isometric Theory of Classical Banach SpacesSpringer New York NY USA 1974

[27] G Godefroy ldquoUniversal spaces for strictly convex Banachspacesrdquo Revista de la Real Academia de Ciencias Exactas Fısicasy Naturales A vol 100 no 1-2 pp 137ndash146 2006

[28] C Finet M Martın and R Paya ldquoNumerical index andrenormingrdquo Proceedings of the American Mathematical Societyvol 131 no 3 pp 871ndash877 2003

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Stochastic AnalysisInternational Journal of

Page 7: Research Article On the Bishop-Phelps-Bollobás Property ...downloads.hindawi.com/journals/aaa/2014/479208.pdf · Research Article On the Bishop-Phelps-Bollobás Property for Numerical

Abstract and Applied Analysis 7

Then there exist a norm-one bounded linear operator 119879] forsome ] isin 119872(119898119898) and a nonnegative simple function 119891

1

in 1198781198711(119898)

such that 119879120583minus 119879] ⩽ 120576 1198911 minus 1198910 ⩽ 3120576 and

(119889|]|1119889119898)(119909) = 1 for all 119909 isin supp(1198911)

Lemma 14 Suppose that 119879] isin L(1198711(119898)) is a norm-one

operator 119891 = sum119899

119894=1120573119894(120594

119861119894

119898(119861119894)) where 119898(119861

119895) gt 0 for all

1 ⩽ 119895 ⩽ 119899 and 119861119895119899

119895=1are mutually disjoint Borel subsets ofΩ

is a norm-one nonnegative simple function and 119892 is an elementof 119878

119871infin(119898)

such that

Re ⟨119892 119879]119891⟩ ⩾ 1 minus1205766

27(50)

for some 0 lt 120576 lt 1 and

119889|]|1

119889119898(119909) = 1 119892 (119909) = 1 (51)

for all x in the support of 119891Then there exist a nonnegative simple function 119891 isin 119878

1198711(119898)

a function119892 isin 119878

119871infin(119898)

and an operator119879] inL(1198711(119898) 1198711(119898))such that

⟨119892 119879]119891⟩ =1003817100381710038171003817119879]1003817100381710038171003817 = 1

1003817100381710038171003817119879] minus 119879]1003817100381710038171003817 ⩽ 2120576

10038171003817100381710038171003817119891 minus 119891

10038171003817100381710038171003817⩽ 3120576

1003817100381710038171003817119892 minus 1198921003817100381710038171003817 ⩽radic120576 ⟨119891 119892⟩ = 1

(52)

Proof Since

Re ⟨119892 119879]119891⟩ ⩾ 1 minus1205766

27 (53)

we have

1 minus1205766

27lt Re ⟨119892 119879]119891⟩

= intΩtimesΩ

119891 (119909)Re119892 (119910) 119889120583 (119909 119910)

=

119899

sum

119895=1

120573119895intΩtimesΩ

120594119861119895

(119909)

119898 (119861119895)

Re119892 (119910) 119889] (119909 119910)

(54)

Let

119869 = 119895 intΩtimesΩ

(

120594119861119895

(119909)

119898 (119861119895)

)Re119892 (119910) 119889] (119909 119910) gt 1 minus 1205763

26

(55)

Then from Lemma 11 we have sum119895isin119869120573119895gt 1 minus 120576

32 Let 119891

1=

sum119895isin119869120573119895(120594

119861119895

119898(119861119895)) where 120573

119895= 120573

119895(sum

119895isin119869120573119895) for all 119895 isin 119869

Then

10038171003817100381710038171198911 minus 1198911003817100381710038171003817 ⩽

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

(120573119895minus 120573

119895)

120594119861119895

119898(119861119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+ sum

119895isin119869

120573119895⩽ 120576

3⩽ 120576 (56)

Note that there is a Borel measurable function ℎ on Ω times Ωsuch that 119889](119909 119910) = ℎ(119909 119910) 119889|]|(119909 119910) and |ℎ(119909 119910)| = 1 forall (119909 119910) isin Ω times Ω Let

119862 = (119909 119910) 1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 1

1003816100381610038161003816 ltradic120576

232 (57)

Define two measures ]119891and ]

119888as follows

]119891(119860) = ] (119860 119862) ]

119888(119860) = ] (119860 cap 119862) (58)

for every Borel subset 119860 ofΩ times Ω It is clear that

119889] = 119889]119891+ 119889]

119888 119889

10038161003816100381610038161003816]119891

10038161003816100381610038161003816= ℎ 119889]

119891

1198891003816100381610038161003816]1198881003816100381610038161003816 = ℎ 119889]119888 119889 |]| = 119889

10038161003816100381610038161003816]119891

10038161003816100381610038161003816+ 1198891003816100381610038161003816]1198881003816100381610038161003816

(59)

Since (119889|]|11198891198981)(119909) = 1 for all 119909 isin ⋃119899

119895=1119861119895 we have

1 =119889|]|1

1198891198981

(119909) =

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) +1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909) (60)

for all 119909 isin 119861 = ⋃119899

119895=1119861119895 and we deduce that |]|1(119861

119895) = 119898

1(119861

119895)

for all 1 ⩽ 119895 ⩽ 119899We claim that |]

119891|1(119861

119895)119898

1(119861

119895) ⩽ 120576

22

2 for all 119895 isin 119869Indeed if |119892(119910)ℎ(119909 119910) minus 1| ⩾ radic120576232 then

Re (119892 (119910) ℎ (119909 119910)) ⩽ 1 minus 12057624 (61)

So we have

1 minus1205763

26

⩽1

1198981(119861

119895)

ReintΩtimesΩ

120594119861119895(119909)119892 (119910) 119889] (119909 119910)

=1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 |]| (119909 119910)

=1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 10038161003816100381610038161003816]11989110038161003816100381610038161003816(119909 119910)

+1

1198981(119861

119895)

intΩtimesΩ

120594119861119895(119909)

Re (119892 (119910) ℎ (119909 119910)) 119889 1003816100381610038161003816]1198881003816100381610038161003816 (119909 119910)

⩽1

1198981(119861

119895)

((1 minus120576

24)10038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

(119861119895) +1003816100381610038161003816]1198881003816100381610038161003816

1

(119861119895))

= 1 minus120576

24

10038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

(119861119895)

1198981(119861

119895)

(62)

This proves our claimWe also claim that for each 119895 isin 119869 there exists a Borel

subset 119861119895of 119861

119895such that

(1 minus120576

2)119898

1(119861

119895) ⩽ 119898

1(119861

119895) ⩽ 119898

1(119861

119895)

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) ⩽120576

2

(63)

8 Abstract and Applied Analysis

for all 119909 isin 119861119895 Indeed set 119861

119895= 119861

119895cap 119909 isin Ω (119889|]

119891|1119889119898

1)

(119909) ⩽ 1205762 Then

int119861119895119861119895

120576

2119889119898

1(119909) ⩽ int

119861119895

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) 1198891198981(119909)

=10038161003816100381610038161003816]1119891

10038161003816100381610038161003816(119861

119895) ⩽1205762

22119898

1(119861

119895)

(64)

This shows that 1198981(119861

119895 119861

119895) ⩽ (1205762)119898

1(119861

119895) This proves our

second claimNow we define 119892 by 119892(119910) = 119892(119910)|119892(119910)| if |119892(119910)| ⩾ 1 minus

radic120576232 and 119892(119910) = 119892(119910) if |119892(119910)| lt 1 minus radic120576232 and we

write 119891 = sum119895isin119869120573119895(120594

119861119895

1198981(119861

119895)) It is clear that 119892 isin 119878

119871infin(119898)

119892 minus 119892 lt radic120576 and 119892(119910) = 1 for all 119909 isin supp119891

Finally we define the measure

119889] (119909 119910)

= sum

119895isin119869

120594119861119895

(119909) 119892 (119910)ℎ (119909 119910)119889]119888(119909 119910)(

1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+ 1205941198691119861(119909) 119889] (119909 119910)

(65)

where 119861 = ⋃119895isin119869119861119895 It is easy to see that (119889|]|1119889119898

1)(119909) = 1

on 119861 and (119889|]|11198891198981)(119909) ⩽ 1 elsewhere Note that

119889 (] minus ]) (119909 119910)

= sum

119895isin119869

120594119861119895

(119909) [

[

119892 (119910)ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1]

]

119889]119888(119909 119910)

minus sum

119895isin119869

120594119861119895

(119909) 119889]119891(119909 119910)

(66)

If (119909 119910) isin 119862 then |119892(119910)| ⩾ 1 minus radic120576232 ⩾ 1 minus 1232 and

100381610038161003816100381610038161003816119892 (119910)ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

119892 (119910)

1003816100381610038161003816119892 (119910)1003816100381610038161003816

ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816100381610038161003816

1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 11003816100381610038161003816

1003816100381610038161003816119892 (119910)1003816100381610038161003816

+

10038161003816100381610038161 minus1003816100381610038161003816119892 (119910)

1003816100381610038161003816

10038161003816100381610038161003816100381610038161003816119892 (119910)

1003816100381610038161003816

⩽ 2

1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 11003816100381610038161003816

1003816100381610038161003816119892 (119910)1003816100381610038161003816

⩽ 2radic120576

232

232

232 minus 1⩽ 2radic120576

(67)

Hence for all (119909 119910) isin 119862 we have

10038161003816100381610038161003816100381610038161003816100381610038161003816

119892 (119910)ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

⩽100381610038161003816100381610038161003816119892 (119910)ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

⩽ 2radic120576(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

(68)

So we have for all 119909 isin 1198691

119889|] minus ]|1

1198891198981

(119909)

⩽ sum

119895isin119869

120594119861119895

(119909) [

[

2radic120576(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

]

]

1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909)

+ sum

119895isin119869

120594119861119895

(119909)

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909)

⩽ sum

119895isin119869

120594119861119895

(119909)(2radic120576 + (1 minus1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909)))

+sum

119895isin119869

120594119861119895

(119909)(

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909))

⩽ 2radic120576 + 120576 lt 3radic120576

(69)

This gives that 119879] minus 119879] lt 3radic120576 Note also that for all 119895 isin 119869

⟨119879]

120594119861119895

1198981(119861

119895)

119892⟩

= intΩtimesΩ

120594119861119895

(119909)

1198981(119861

119895)

119892 (119910) 119889] (119909 119910)

= intΩtimesΩ

120594119861119895

(119909)

1198981(119861

119895)

ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

119889]119888(119909 119910)

Abstract and Applied Analysis 9

= intΩ

120594119861119895

(119909)

1198981(119861

119895)

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

1198891003816100381610038161003816]1198881003816100381610038161003816

1

(119909)

= intΩ

120594119861119895

(119909)

1198981(119861

119895)

1198891198981(119909) = 1

(70)

Hence we get ⟨119879]119891 119892⟩ = 1 which implies that 119879]119891 =119879] = 1 Finally10038171003817100381710038171003817119891 minus 119891

10038171003817100381710038171003817

⩽10038171003817100381710038171003817119891 minus 119891

1

10038171003817100381710038171003817+10038171003817100381710038171198911 minus 119891

1003817100381710038171003817

=

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

1198981(119861

119895)

minus sum

119895isin119869

120573119895

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+ 120576

⩽ sum

119895isin119869

120573119895(

10038171003817100381710038171003817100381710038171003817100381710038171003817

120594119861119895

1198981(119861

119895)

minus

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

120594119861119895

1198981(119861

119895)

minus

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

) + 120576

= 2sum

119895isin119869

120573119895

1198981(119861

119895 119861

119895)

1198981(119861

119895)

+ 120576

⩽ 2sum

119895isin119869

120573119895

(1205762)1198981(119861

119895)

1198981(119861

119895)

+ 120576 ⩽120576

1 minus 1205762+ 120576 lt 3120576

(71)

We are now ready to present the proof of the main resultin the case of finite regular positive Borel measures

Proof of Proposition 10 Let 1205751= 120575

3

2(5sdot2

4) 120575

2= 120575

12

3(3

2sdot2

14)

and 1205753= (12057610)

2 for some 0 lt 120576 lt 1 Suppose that 119879 isinL(119871

1(119898)) with 119879 = 1 and that there is an 119891

0isin 119878

1198711(119898)

and1198920isin 119878

119871infin(119898)

such that ⟨1198910 119892

0⟩ = 1 and |⟨119879119891

0 119892

0⟩| gt 1minus120575

3

12

6Then there is an isometric isomorphism Ψ from 119871

1(119898) onto

itself such thatΨ(1198910) = |119891

0| and there is a scalar number 120572 in

119878R such that |⟨1198791198910 119892

0⟩| = ⟨120572119879119891

0 119892

0⟩ Then letting 119891

1= Ψ119891

0

1198921= (Ψ

minus1)lowast1198920 and 119879

1= 120572Ψ119879Ψ

minus1 we have

⟨1198781198911 119892

1⟩ = ⟨120572Ψ119879

minus1Ψ119891

0 (Ψ

minus1)lowast

1198920⟩

= ⟨1205721198791198910 119892

0⟩ gt 1 minus

1205753

1

26

⟨1198911 119892

1⟩ = ⟨Ψ119891

0 (Ψ

minus1)lowast

1198920⟩ = 1

(72)

Since 11987911198911 gt 1 minus 120575(120575

3

12

6) by Lemma 13 there exists a

norm-one bounded operator 119879] and a nonnegative simplefunction119891

2isin 119878

1198711(119898)

such that 1198791minus119879] ⩽ 1205751 1198912minus1198911 ⩽ 31205751

and (119889|]|11198891198981)(119909) = 1 for all 119909 isin supp(119891

2) Then

⟨119879]1198912 1198921⟩

= ⟨11987911198911 119892

1⟩ minus ⟨119879

11198911minus 119879

11198912 119892

1⟩ minus ⟨119879

11198912minus 119879]1198912 1198921⟩

⩾ ⟨11987911198911 119892

1⟩ minus10038171003817100381710038171198911 minus 1198912

1003817100381710038171003817 minus10038171003817100381710038171198791 minus 119879]

1003817100381710038171003817

⩾ 1 minus1205753

1

26minus 3120575

1minus 120575

1⩾ 1 minus 5120575

1= 1 minus

1205753

2

16

(73)

Notice also that

⟨1198912 119892

1⟩ = ⟨119891

1 119892

1⟩ minus ⟨119891

1minus 119891

2 119892

1⟩ ⩾ 1 minus

10038171003817100381710038171198911 minus 11989121003817100381710038171003817

⩾ 1 minus 31205751⩾ 1 minus 5120575

1= 1 minus

1205753

2

16

(74)

By Lemma 12 there are a nonnegative simple function 1198913isin

1198781198711(119898)

and a function 1198923isin 119878

119871infin(119898)

such that

1198923(119909) = 120594supp119891

3

(119909) + 1198922(119909) 120594

Ωsupp1198913

(119909)

10038171003817100381710038171198912 minus 11989131003817100381710038171003817 ⩽ 1205752

10038171003817100381710038171198923 minus 11989211003817100381710038171003817 ⩽radic120575

2 ⟨119891

3 119892

3⟩ = 1

(75)

So we have

⟨119879]1198913 1198923⟩ = ⟨119879]1198912 1198921⟩ minus ⟨119879]1198912 minus 119879]1198913 1198921⟩

minus ⟨119879]1198913 1198921 minus 1198923⟩

⩾ 1 minus1205753

2

16minus 2radic120575

2⩾ 1 minus 3radic120575

2= 1 minus

1205756

3

27

(76)

By Lemma 14 there exist 1198914isin 119878

1198711(119898)

and 1198924isin 119878

119871infin(119898)

and anoperator 119879

4such that

⟨1198924 119879

41198914⟩ = 1 =

100381710038171003817100381711987941003817100381710038171003817

10038171003817100381710038171198794 minus 119879]1003817100381710038171003817 ⩽ 21205753

10038171003817100381710038171198914 minus 11989131003817100381710038171003817 ⩽ 31205753

10038171003817100381710038171198924 minus 11989231003817100381710038171003817 ⩽radic120575

3 ⟨119891

4 119892

4⟩ = 1

(77)

So we have

10038171003817100381710038171198794 minus 11987911003817100381710038171003817 ⩽10038171003817100381710038171198794 minus 119879]

1003817100381710038171003817 +1003817100381710038171003817119879] minus 1198791

1003817100381710038171003817

⩽ 1205751+ 2120575

3⩽ 3120575

3

10038171003817100381710038171198911 minus 11989141003817100381710038171003817 ⩽10038171003817100381710038171198911 minus 1198912

1003817100381710038171003817 +10038171003817100381710038171198912 minus 1198913

1003817100381710038171003817 +10038171003817100381710038171198913 minus 1198914

1003817100381710038171003817

⩽ 31205751+ 120575

2+ 3120575

3⩽ 10120575

3

10038171003817100381710038171198921 minus 11989241003817100381710038171003817 ⩽10038171003817100381710038171198921 minus 1198923

1003817100381710038171003817 + 1198923 minus 1198924

⩽ 1205752+ radic120575

3⩽ 2radic120575

3

(78)

10 Abstract and Applied Analysis

Let 119878 = 120572Ψminus11198794Ψ 119891 = Ψminus1

1198914 and 119892 = Ψlowast

1198924 then we have

119879 minus 119878 =10038171003817100381710038171003817119879 minus 120572Ψ

minus11198794Ψ10038171003817100381710038171003817=10038171003817100381710038171003817120572Ψ119879Ψ

minus1minus 119879

4

10038171003817100381710038171003817

=10038171003817100381710038171198791 minus 1198794

1003817100381710038171003817 ⩽ 31205753

100381710038171003817100381710038171198910minus 11989110038171003817100381710038171003817=100381710038171003817100381710038171198910minus Ψ

minus11198914

10038171003817100381710038171003817=10038171003817100381710038171198911 minus 1198914

1003817100381710038171003817 ⩽ 101205753

100381710038171003817100381710038171198920minus 11989110038171003817100381710038171003817=10038171003817100381710038171198920 minus Ψ

lowast1198924

1003817100381710038171003817 =100381710038171003817100381710038171003817(Ψ

minus1)lowast

1198920minus 119892

4

100381710038171003817100381710038171003817

=10038171003817100381710038171198921 minus 1198924

1003817100381710038171003817 ⩽ 2radic120575

3

⟨119891 119892⟩ = ⟨Ψminus11198914 Ψ

lowast1198924⟩ = ⟨119891

4 119892

4⟩ = 1

10038161003816100381610038161003816⟨119878119891 119892⟩

10038161003816100381610038161003816=10038161003816100381610038161003816⟨120572Ψ

minus11198794ΨΨ

minus11198914 Ψ

lowast1198924⟩10038161003816100381610038161003816= |120572| = 1

(79)

This completes the proof

Finally we may give the proof of the main result in fullgenerality

Proof of Theorem 9 Notice that the Kakutani representationtheorem (see [26] for a reference) says that for every 120590-finitemeasure ] the space 119871

1(]) is isometrically isomorphic to

1198711(119898) for some positive Borel regular measure on a compact

Hausdorff space Then by Proposition 10 there is a universalfunction 120576 997891rarr 120578(120576) gt 0 which gives the BPBp-nu for 119871

1(]) for

every 120590-finite measure ]Fix 120576 gt 0 Suppose that 119879

0isin L(119871

1(120583)) with V(119879

0) = 1

and (1198910 119891

lowast

0) isin Π(119871

1(120583)) satisfy

1003816100381610038161003816⟨119891lowast

0 119879

01198910⟩1003816100381610038161003816 gt 1 minus 120578 (120576) (80)

Choose a sequence 119891119899 in 119871

1(120583) such that sup

119899119879

0119891119899 = 1

and let 119866 be the closed linear span of

119879119899

0119891119898 119899 119898 isin N cup 0 (81)

As 119866 is separable there is a dense subset 119892119899 119899 isin N of 119866

and let 119864 = ⋃infin

119899=1supp119892

119899 where supp119892

119899is the support of 119892

119899

Then the measure 120583|119864is 120590-finite Let

119884 = 119891 isin 1198711(120583) supp (119891) sub 119864 (82)

be a closed subspace of 1198711(120583) It is clear that 119871

1(120583) = 119884oplus

1119885

and 119884 is isometrically isomorphic to 1198711(120583|

119864) So 119884 has the

BPBp-nu with 120578(120576)Now write 119878

0= 119879

0|119884 119884 rarr 119884 consider 119910

0= 119891

0isin

119878119884 119910lowast

0= 119891

lowast

0|119884isin 119878

119884lowast and observe that 119910lowast

0(119910

0) = 1 and

|119910lowast

0(119878

01199100)| = |119891

lowast

0(119879

01198910)| gt 1 minus 120578(120576) Hence there exist 119878 isin

L(119884) and (1199100 119910

lowast

0) isin Π(119884) such that

1003816100381610038161003816119910lowast

0(119878119910

0)1003816100381610038161003816 = 1 = V (119878) 1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

10038171003817100381710038171199100 minus 11991001003817100381710038171003817 lt 120576

1003817100381710038171003817119910lowast

0minus 119910

lowast

0

1003817100381710038171003817 lt 120576

(83)

Finally consider the operator 119879 isinL(1198711(120583)) given by

119879 (119910 119911)

= (119878119910 0) + 1198790(0 119911) ((119910 119911) isin 119871

1(120583) equiv 119884oplus

1119885)

(84)

We have 119879 = 1 (and so V(119879) = 1) Indeed1003817100381710038171003817119879 (119910 119911)

1003817100381710038171003817 =1003817100381710038171003817(119878119910 0)

1003817100381710038171003817 +10038171003817100381710038171198790 (0 119911)

1003817100381710038171003817 ⩽10038171003817100381710038171199101003817100381710038171003817 + 119911 =

1003817100381710038171003817(119910 119911)1003817100381710038171003817

(85)

for all (119910 119911) isin 1198711(120583) and 119879(119910

0 0) = (119878119910

0 0) = 119878119910

0 = 1

Let 119909 = (1199100 0) and 119909lowast = (119910lowast

0 119891

0|119885) Then (119909 119909lowast) isin Π(119871

1(120583))

Moreover we have1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119910

lowast

0119878119910

0

1003816100381610038161003816 = 1 = V (119879) 1003817100381710038171003817119909 minus 1198910

1003817100381710038171003817 =1003817100381710038171003817119910 minus 1199100

1003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowast

0minus 119891

lowast

0

1003817100381710038171003817

= max 1003817100381710038171003817119910 minus 119891lowast

0|119884

1003817100381710038171003817 1003817100381710038171003817119891

lowast

0|119885minus 119891

lowast

0|119885

1003817100381710038171003817

=1003817100381710038171003817119910

lowastminus 119910

lowast

0

1003817100381710038171003817 lt 120576

1003817100381710038171003817119879 minus 11987901003817100381710038171003817 = sup119910+119911⩽1

1003817100381710038171003817119879 (119910 119911) minus 1198790 (119910 119911)1003817100381710038171003817

= sup119910⩽1

1003817100381710038171003817119878119910 minus 11987801199101003817100381710038171003817 =1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

(86)

This completes the proof

5 Examples of Spaces Failing theBishop-Phelps-Bollobaacutes Property forNumerical Radius

Our goal here is to prove that the density of numerical radiusattaining operators does not imply the BPBp-nu Actuallywe will show that among separable spaces there is noisomorphic property implying the BPBp-nu other than finite-dimensionality

We need to relate the BPBp-nu to the Bishop-Phelps-Bollobas property for operators which as mentioned in theintroduction was introduced in [15] A pair (119883 119884) of Banachspaces has the Bishop-Phelps-Bollobas property for operators(in short BPBp) if given 120576 gt 0 there exists 120578(120576) gt 0 suchthat given 119879 isin L(119883 119884) with 119879 = 1 and 119909 isin 119878

119883such that

119879119909 gt 1 minus 120578(120576) then there exist 119911 isin 119878119883and 119878 isin L(119883 119884)

satisfying

119878 = 119878119911 = 1 119909 minus 119911 lt 120576

119879 minus 119878 lt 120576

(87)

We refer the reader to [15 19 25] and references therein formore information and background Among the interestingresults on the BPBp we emphasize that a pair (119883 119884) when119883is finite-dimensional does not necessarily have the BPBp Forinstance if119884 is a strictly convex space which is not uniformlyconvex then the pair (ℓ(2)

1 119884) fails to have the BPBp (this is

contained in [15] see [19 Section 3])

Abstract and Applied Analysis 11

The next result relates the BPBp-nu to the BPBp foroperators in a particular case We will deduce our examplefrom it

Theorem 15 If 1198711(120583)oplus

1119883 has the BPBp-119899119906 then the pair

(1198711(120583) 119883) has the BPBp for operators

Before proving this proposition we will use it to get themain examples of this section The first example shows thatthe density of numerical radius attaining operators does notimply the BPBp-nu

Example 16 There is a reflexive space (and so numerical radiusattaining operators on it are dense) which fails to have theBPBp-nu Indeed let 119884 be a reflexive separable space whichis not superreflexive and we may suppose that 119884 is strictlyconvex Observe that 119884 cannot be uniformly convex since itis not superreflexive Now 119883 = ℓ(2)

1oplus1119884 is reflexive but the

pair (ℓ(2)1 119884) fails the BPBp since 119884 is strictly convex but not

uniformly convex [19 Corollary 33] Therefore Theorem 15gives us that119883 does not have the BPBp-nu

The example above can be extended to get the result thatevery infinite-dimensional separable Banach space can berenormed to fail the BPBp-nu This follows from the factthat every infinite-dimensional separable Banach space canbe renormed to be strictly convex but not uniformly convex(this result can be proved ldquoby handrdquo an alternative categoricalargument for it can be found in [27] and references therein)With a little more of effort we may get the main result of thesection

Theorem 17 Every infinite-dimensional separable Banachspace can be renormed to fail the weak-BPBp-nu (and so inparticular to fail the BPBp-nu)

We need the following result which is surely well knownAs we have not found a reference we include a nice and easyproof kindly given to us by Vladimir Kadets We recall thatgiven a Banach space 119884 the set of all equivalent norms on119884 can be viewed as a metric space using the Banach-Mazurdistance

Lemma 18 Let119884 be an infinite-dimensional separable Banachspace Then the set of equivalent norms on 119884 which are strictlyconvex and are not (locally) uniformly convex is dense in the setof all equivalent norms on119884 (with respect to the Banach-Mazurdistance)

Proof Fix 119890 isin 119878119884and 119890lowast

1isin 119878

119884lowast such that 119890lowast

1(119890) = 1 For a fixed

120576 isin (0 12) denote

119902 (119910) = max (1 minus 120576) 10038171003817100381710038171199101003817100381710038171003817 1003816100381610038161003816119890lowast

1(119910)1003816100381610038161003816 (119910 isin 119884) (88)

Evidently (1 minus 120576)119910 ⩽ 119902(119910) ⩽ 119910 for every 119910 isin 119884 Fix asequence 119890lowast

119896 119896 ⩾ 2 of norm-one functionals separating

the points of 119884 and denote

119901 (119910) = radic

infin

sum

119896=1

1

2119896

1003816100381610038161003816119890lowast

119896(119910)1003816100381610038161003816

2

(119910 isin 119884) (89)

Then 119901 is a strictly convex norm on 119884 119901(119890) ⩾ 1radic2 and119901(119910) ⩽ 119910 for all 119910 isin 119883 Finally write

100381710038171003817100381711991010038171003817100381710038171= (1 minus 120576) 119902 (119910) + 120576

119901 (119910)

119901 (119890)(119910 isin 119884) (90)

Then sdot 1is a strictly convex norm on 119884 and

(1 minus 120576)2 10038171003817100381710038171199101003817100381710038171003817 ⩽100381710038171003817100381711991010038171003817100381710038171⩽ (1 + 120576)

10038171003817100381710038171199101003817100381710038171003817 (119910 isin 119884) (91)

Wewill finish the proof by showing that sdot 1is not uniformly

convex (actually it is not locally uniformly convex) Indeedfor each 119899 isin N we select 119910

119899isin ⋂

119899

119896=1ker 119890lowast

119896with 119910

119899 = 1 and

consider 119890119899= 119890 + (1205764)119910

119899 Then 119902(119890) = 1 119902(119890

119899) = 1 and

119902(119890+119890119899) = 2 At the same time 119901(119910

119899) rarr 0 so 119901(119890

119899) rarr 119901(119890)

and 119901(119890 + 119890119899) rarr 2119901(119890) Consequently

1198901 = 1100381710038171003817100381711989011989910038171003817100381710038171997888rarr 1

1003817100381710038171003817119890 + 11989011989910038171003817100381710038171997888rarr 2 (92)

but 119890 minus 1198901198991= (1205764)119910

1198991⩾ (1 minus 120576)

2(1205764) which means the

absence of local uniform convexity at 119890

Proof of Theorem 17 Let 119883 be an infinite-dimensional sepa-rable Banach space Take a closed subspace 119884 of 119883 of codi-mension two By [28 Proposition 2] the map carrying everyequivalent norm on 119884 to its numerical index is continuousand so the set of values of the numerical index of 119884 upto reforming is a nontrivial interval [28 Theorem 9] ThenLemma 18 allows us to find an equivalent norm | sdot | on 119884 insuch a way that (119884 | sdot |) is strictly convex and is not uniformlyconvex and 119899(119884 | sdot |) gt 0 Now the space 119883 = ℓ(2)

1oplus1(119884 | sdot |)

is an equivalent renorming of 119883 which does not have theBPBp-119899119906 (indeed otherwise the pair (ℓ(2)

1 (119884 | sdot |)) would

have the BPBp for the operator norm and so (119884 | sdot |) wouldbe uniformly convex by [19 Corollary 33] a contradiction)Moreover as

119899 (119883) = min 119899 (ℓ(2)1) 119899 (119884 |sdot|) gt 0 (93)

(see [20 Proposition 2] for instance) 119883 also fails the weak-BPBp-nu by Proposition 6

To finish the section with the promised proof ofTheorem 15 we first see the following stability result

Lemma 19 Let119883 = [⨁infin

119896=1119883

119896]1198880

or [⨁infin

119896=1119883

119896]ℓ1

If119883 has theBishop-Phelps-Bollobas property for numerical radius with afunction 120578 then each Banach space 119883

119894has the Bishop-Phelps-

Bollobas property for numerical radius with 120578nu(119883119894) ⩾ 120578 That

is inf119894120578nu(119883119894

)(120576) ⩾ 120578nu(119883)(120576) for all 0 lt 120576 lt 1

Proof Let 119875119894 119883 rarr 119883

119894and 1198751015840

119894 119883

lowastrarr 119883

lowast

119894be the natural

projections and let 119876119894 119883

119894rarr 119883 and 1198761015840

119894 119883

lowast

119894rarr 119883

lowast be thenatural embeddings

Assume that an operator 119879119894 119883

119894rarr 119883

119894and a pair

(119909119894 119909

lowast

119894) isin Π(119883

119894) satisfy that

V (119879119894) = 1

1003816100381610038161003816119909lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (94)

We define an operator 119879 119883 rarr 119883 and (119909 119909lowast) isin Π(119883) by

119879 = 119876119894∘ 119879

119894∘ 119875

119894 (119909 119909

lowast) = (119876

119894119909119894 119876

1015840

119894119909lowast

119894) (95)

12 Abstract and Applied Analysis

then clearly we see that1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119909

lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (96)

From the assumption there exist 119878 119883 rarr 119883 and a pair(119910 119910

lowast) isin Π(119883) such that

1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 = 1 = V (119878) 119878 minus 119879 lt 120576

1003817100381710038171003817119910lowastminus 119909

lowast1003817100381710038171003817 lt 1205761003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576

(97)

Since this clearly shows that1003817100381710038171003817119875119894 ∘ 119878 ∘ 119876119894

minus 119879119894

1003817100381710038171003817 lt 120576100381710038171003817100381710038171198751015840

119894119910lowastminus 119909

lowast

119894

10038171003817100381710038171003817lt 120576

1003817100381710038171003817119875119894119910 minus 1199091198941003817100381710038171003817 lt 120576

(98)

we only need to show that |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = 1

We first show the case of 1198880sum Since

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817=10038171003817100381710038171003817119875119895119910 minus 119875

11989511990910038171003817100381710038171003817⩽1003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576 (99)

for every 119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽100381710038171003817100381710038171198751015840

119894119910lowast10038171003817100381710038171003817+ 120576 sum

119895isinN119895 = 119894

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817⩽1003817100381710038171003817119910

lowast1003817100381710038171003817 = 1

(100)

This shows that 1198751015840119894119910lowast = 1 and 1198751015840

119895119910lowast= 0 for every 119895 = 119894

So 119910lowast = 1198761015840

1198941198751015840

119894119910lowast and 1198751015840

119894119910lowast(119875

119894119910) = 1 This and the fact that

119910 minus 119876119894119875119894119910 lt 120576 imply that

(119876119894119875119894119910 + (

1

120576) (119910 minus 119876

119894119875119894119910) 119876

1015840

1198941198751015840

119894119910lowast) isin Π (119883) (101)

So we get that (1198761015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910 + (1120576)(119910 minus 119876

119894119875119894119910))V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =10038161003816100381610038161003816(119876

1015840

1198941198751015840

119894119910lowast) 11987811991010038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910 +1

120576(119910 minus 119876

119894119875119894119910))

1003816100381610038161003816100381610038161003816⩽ 1

(102)

and so we get |1198751015840119894119910lowast(119875

119894∘119878 ∘119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

We next show the case of ℓ1sum The proof is almost

the same as that of the 1198880case However for the sake of

completeness we provide it hereSince 1198751015840

119895119910lowast = 119875

1015840

119895119910lowastminus 119875

1015840

119895119909lowast ⩽ 119910

lowastminus 119909

lowast lt 120576 for every

119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽10038171003817100381710038171198751198941199101003817100381710038171003817 + 120576 sum

119895isinN119895 = 119894

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817⩽10038171003817100381710038171199101003817100381710038171003817 = 1

(103)

which shows 119875119894119910 = 1 and 119875

119895119910 = 0 for every 119895 = 119894 Since this

implies (119876119894119875119894119910119876

1015840

1198941198751015840

119894119910lowast+(1120576)(119910

lowastminus119876

1015840

1198941198751015840

119894119910lowast)) isin Π(119883) we get

that |(1198761015840

1198941198751015840

119894119910lowast+ (1120576)(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast))119878(119876

119894119875119894119910)| ⩽ V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =1003816100381610038161003816119910

lowast119878 (119876

119894119875119894119910)1003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast+1

120576(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast)) 119878 (119876

119894119875119894119910)

1003816100381610038161003816100381610038161003816⩽ 1

(104)

and so |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

Proof of Theorem 15 Note that 120578nu(1198711(120583)oplus1119883)(120576) rarr 0 as120576 rarr 0 Fix 0 lt 120576

0lt 1 and choose 0 lt 120576 lt 1 such that

6120576 + 120578nu(1198711(120583)oplus1119883)(120576) lt 1205760 Let 120578(1205760) = 120578nu(1198711(120583)oplus1119883)(120576)Suppose that 119879

0isin L(119871

1(120583) 119883) with 119879

0 = 1 and 119891

0isin

1198781198711(120583)

satisfy

1003817100381710038171003817119879011989101003817100381710038171003817 gt 1 minus 120578 (1205760) (105)

For any measurable subset 119861 let

1198711(120583|

119861) = 119891|

119861 119891 isin 119871

1(120583) (106)

with the norm 119891|119861 = 119891120594

1198611 Then it is easy to see

that 1198711(120583|

119861) is isometrically isomorphic to a complemented

subspace of 1198711(120583) Let 119875

119861 119871

1(120583) rarr 119871

1(120583|

119861) be the

restriction defined by 119875119861(119891) = 119891|

119861for all 119891 isin 119871

1(120583) and

let 119869119861 119871

1(120583|

119861) rarr 119871

1(120583) be the extension defined by

119869119861(119891)(120596) = 119891(120596) if 120596 isin 119861 and 119869

119861(119891)(120596) = 0 otherwise

It is clear that 119875119861119869119861= Id

1198711(120583|119861)and 119869

119861119875119861(119891) = 119891120594

119861for all

119891 isin 1198711(120583) Notice also that 119871

1(120583) is isometrically isomorphic

to 1198711(120583|

119861)oplus

11198711(120583|

119861119888)

Let 119860 = supp1198910and 119892

0= 119875

1198601198910 Then

1003817100381710038171003817119879011986911986011989201003817100381710038171003817 =100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) gt 0 (107)

and define the operator 119879119860 119871

1(120583|

119860) rarr 119883 by 119879

119860119891 =

1198790119869119860119891119879

0119869119860 for every 119891 isin 119871

1(120583|

119860) Then

100381710038171003817100381711987911986011989201003817100381710038171003817 ⩾100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) (108)

Since 120583|119860is 120590-finite 119871

1(120583|

119860)lowast= 119871

infin(120583|

119860) Let 119892lowast

0isin 119878

119871infin(120583|119860)

be a function such that ⟨119892lowast0 119892

0⟩ = 1 and choose 119909lowast

0isin 119878

119883lowast

such that 119909lowast0(119879

1198601198920) = 119879

1198601198920 Define the operator 119878

0isin

L(1198711(120583|

119860)oplus

1119883) by

1198780(119891 119909) = (0 119879

119860119891) ((119891 119909) isin 119871

1(120583|

119860) oplus

1119883) (109)

and observe that 1198780 = V(119878

0) = 1 Indeed

100381710038171003817100381711987801003817100381710038171003817 ⩽ 1 =

10038171003817100381710038171198791198601003817100381710038171003817

= sup 1003816100381610038161003816119909lowast1198791198601198911003816100381610038161003816 119909

lowastisin 119878

119883lowast 119891 isin 119878

1198711(120583|119860)

= sup 1003816100381610038161003816(119891lowast 119909

lowast) 119878

0(119891 119909)

1003816100381610038161003816

((119891lowast 119909

lowast) (119891 119909)) isin Π (119871

1(120583|

119860) oplus

1119883)

= V (1198780) ⩽100381710038171003817100381711987801003817100381710038171003817

(110)

Abstract and Applied Analysis 13

It is immediate that

(119892lowast

0 119909

lowast

0) 119878

0(119892

0 0) = 119909

lowast

0(119879

1198601198920) =10038171003817100381710038171198791198601198920

1003817100381710038171003817 gt 1 minus 120578 (1205760)

(111)

By Lemma 19 1198711(120583|

119860)oplus

1119883 has the BPBp-nu with the func-

tion 120578 Therefore there exist 1198781isin L(119871

1(120583|

119860)oplus

1119883) (119892

1 119909

1) isin

1198781198711(120583|119860)oplus1119883 and (119892lowast

1 119909

lowast

1) isin 119878

119871infin(120583|119860)oplusinfin119883lowast such that

1003817100381710038171003817(1198921 1199091) minus (1198920 0)1003817100381710038171003817 lt 120576

1003817100381710038171003817(119892lowast

1 119909

lowast

1) minus (119891

lowast

0 119909

lowast

0)1003817100381710038171003817 lt 120576

10038171003817100381710038171198781 minus 11987801003817100381710038171003817 lt 120576 ⟨(119892

lowast

1 119909

lowast

1) (119892

1 119909

1)⟩ = 1

1003816100381610038161003816(119892lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 = V (119878

1) = 1

(112)

Claim 1 We claim that 1199091= 0

Otherwise

1 = Re ⟨(119892lowast1 119909

lowast

1) (119892

1 119909

1)⟩

=100381710038171003817100381711989211003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

+100381710038171003817100381711990911003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

⟩ ⩽ 1

(113)

We deduce that

((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

(119892lowast

1 119909

lowast

1)) (

(0 1199091)

100381710038171003817100381711990911003817100381710038171003817

(119892lowast

1 119909

lowast

1))

isin Π (1198711(120583|

119860) oplus

1119883)

(114)

Since 1198781((0 119909

1)119909

1) = (119878

1minus 119878

0)((0 119909

1)119909

1) lt 120576 we

get that

1 =1003816100381610038161003816⟨(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)⟩1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

100381710038171003817100381711989211003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

)⟩

+100381710038171003817100381711990911003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

)⟩

100381610038161003816100381610038161003816100381610038161003816

⩽100381710038171003817100381711989211003817100381710038171003817 V (1198781) + 120576

100381710038171003817100381711990911003817100381710038171003817 lt100381710038171003817100381711989211003817100381710038171003817 +100381710038171003817100381711990911003817100381710038171003817 = 1

(115)

a contradiction This proves the claimWe define the operator 119878

2 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883

by 1198782(119891 119909) = 119878

1(119891 0) for every 119891 isin 119871

1(120583|

119860) and for every

119909 isin 119883 Then we have

V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 = 1

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ 120576 (116)

Indeed from Claim 1 we have

V (1198781) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 ⩽ V (119878

2)

(117)

On the other hand we see that1003816100381610038161003816(119891

lowast 119909

lowast) 119878

2(119891 119909)

1003816100381610038161003816

=1003816100381610038161003816(119891

lowast 119909

lowast) 119878

1(119891 0)

1003816100381610038161003816 ⩽10038171003817100381710038171198911003817100381710038171003817 V (1198781) ⩽ V (119878

1)

(118)

for every ((119891lowast 119909lowast) (119891 119909)) isin Π(1198711(120583|

119860)oplus

1119883) Therefore

V(1198782) ⩽ V(119878

1) Also

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ sup

119909isin119878119883

10038171003817100381710038171198781 (0 119909)1003817100381710038171003817

= sup119909isin119878119883

10038171003817100381710038171198781 (0 119909) minus 1198780 (0 119909)1003817100381710038171003817 ⩽ 120576

(119)

Claim 2 There exists 1198783 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883 such

that 1198783(119892

1 0) = 119878

3 = 1 119878

3(0 119909) = 0 119878

3(119891 119909) isin 0oplus

1119883

for every (119891 119909) isin 1198711(120583|

119860)oplus

1119883 and 119878

3minus 119878

2 lt 4120576

Indeed write 1198781= (119863

1 119863

2) where 119863

1 119871

1(120583|

119860)oplus

1119883 rarr

1198711(120583|

119860) and119863

2 119871

1(120583|

119860)oplus

1119883 rarr 119883 We have that

sup 1003816100381610038161003816119892lowast119863

1(119892

1 0) + 119909

lowast119863

2(119892

1 0)1003816100381610038161003816

119909lowastisin 119878

119883lowast ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860) +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⩽ V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119892

lowast

1119863

1(119892

1 0) + 119909

lowast

1119863

2(119892

1 0)1003816100381610038161003816

(120)

This implies that

1003816100381610038161003816119909lowast

1119863

2(119892

1 0)1003816100381610038161003816 =10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

1003816100381610038161003816119892lowast

1119863

1(119892

1 0)1003816100381610038161003816

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119871

infin(120583|

119860)

(121)

Therefore |119892lowast1| equals 1 on the support of 119863

1(119892

1 0) As

|⟨119892lowast

1 119892

1⟩| = 1 we also have that |119892lowast

1| equals 1 on the support

of 1198921 Changing the values of 119892lowast

1by the ones of 119891lowast

0on 119860

(supp(1198631(119892

1 0)) cup supp(g

1)) we may and do suppose that

|119892lowast

1| = 1 on the whole 119860We also have 119863

2(119892

1 0) gt 0 Indeed

10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)1003817100381710038171003817

⩽10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198921 0)

1003817100381710038171003817 +10038171003817100381710038171198780 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817

lt 2120576 + 120576 = 3120576

(122)

So we have

10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

⩽10038171003817100381710038171198631(119892

1 0)1003817100381710038171003817 +10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

14 Abstract and Applied Analysis

=1003817100381710038171003817(1198631

(1198921 0) 119863

2(119892

1 0)) minus (0 119879

1198601198920)1003817100381710038171003817

=10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817 lt 3120576

(123)

and 1198632(119892

1 0) gt 119879

1198601198920 minus 3120576 ⩾ 1 minus 120578(120576

0) minus 3120576 gt 0

Finally define the operator 1198783by

1198783(119891 119909) = (0119863

2(119891 0) + 119892

lowast

1(119863

1(119891 0))

1198632(119892

1 0)

119909lowast

1119863

2(119892

1 0))

(124)

for (119891 119909) isin 1198711(120583|

119860)oplus

1119883 It is clear that

100381710038171003817100381711987831003817100381710038171003817 ⩽ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) (125)

Notice also that

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817

⩽10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 +10038171003817100381710038171198632(119891 0) minus 119879

1198601198911003817100381710038171003817

=1003817100381710038171003817(1198631

(119891 0) 1198632(119891 0)) minus (0 119879

119860119891)1003817100381710038171003817

=10038171003817100381710038171198782 (119891 119909) minus 1198780 (119891 119909)

1003817100381710038171003817

(126)

for all (119891 119909) isin 1198711(120583|

119860)oplus

1119883 Hence we have

10038171003817100381710038171198783 minus 11987821003817100381710038171003817

= 2 sup119891isin1198781198711(120583|119860)

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 ⩽ 210038171003817100381710038171198782 minus 1198780

1003817100381710038171003817 lt 4120576(127)

On the other hand let 119866 1198711(120583|

119860) rarr 119871

1(120583|

119860) be defined by

119866(119891) = 119892lowast

1119891 for every 119891 isin 119871

1(120583|

119860) Then we have

V (1198782)

= sup 1003816100381610038161003816119911lowast11987821199111003816100381610038161003816 (119911 119911

lowast) isin Π (119871

1(120583|

119860) oplus

1119883)

⩾ sup10038161003816100381610038161003816100381610038161003816

119909lowast119863

2(119866(

1

120583 (119862)120594119862) 0)

+119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119909lowastisin 119878

119883lowast 119862 isin Σ

119860 120583 (119862) gt 0

= sup10038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119862)120594119862) 0)

10038171003817100381710038171003817100381710038171003817

+

10038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119862 isin Σ119860 120583 (119862) gt 0

(128)

where Σ119860is the family of measurable subsets of 119860

Hence for any simple function 119904 = sum119899

119894=1((120572

119894120583(119860

119894))120594

119860119894

) isin

1198781198711(120583|119860) where 119860

119894119894is a family of disjoint measurable subsets

with strictly positive measure we have

V (1198782)

119899

sum

119894=1

10038161003816100381610038161205721198941003816100381610038161003816 (

100381710038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381710038171003817100381710038171003817100381710038171003817

+

100381610038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381610038161003816100381610038161003816100381610038161003816

)

⩾10038171003817100381710038171198632(119866 (119904) 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119866 (119904) 0)

1003816100381610038161003816

(129)

Since |119892lowast1| = 1119866 is an isometric isomorphism so for each119891 isin

1198781198711(120583|119860)there exists a sequence of norm-one simple functions

(119904119896) such that 119866(119904

119896) converges to 119891 Therefore

V (1198782) ⩾ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) ⩾100381710038171003817100381711987831003817100381710038171003817

(130)

On the other hand we see that100381710038171003817100381711987831003817100381710038171003817 ⩾1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

3(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119909

lowast

1119863

2(119892

1 0) + 119892

lowast

1119863

1(119892

1 0)1003816100381610038161003816 = V (119878

2) = 1

(131)

Therefore 1 = 1198783 = 119878

3(119892

1 0) which proves Claim 2

Finally set 1198783= (0 ) for a suitable 119871

1(120583|

119860)oplus

1119883 rarr

119883 and define the operator 1198791 119871

1(120583) rarr 119883 by

1198791(119891) = 119879

0(119891120594

119860119888) + (119875

119860119891 0) (132)

for every 119891 isin 1198711(120583) Then we have

10038171003817100381710038171198791 (119891)1003817100381710038171003817 ⩽100381710038171003817100381711987901003817100381710038171003817

10038171003817100381710038171198911205941198601198881003817100381710038171003817 +1003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171198911205941198601003817100381710038171003817 =10038171003817100381710038171198911003817100381710038171003817 (133)

for every 119891 isin 1198711(120583) so 119879

1 ⩽ 1 Also

10038171003817100381710038171198791 (1198691198601198921)1003817100381710038171003817 =10038171003817100381710038171198783 (1198921 0)

1003817100381710038171003817 =100381710038171003817100381711987831003817100381710038171003817 = 1 (134)

so 1198791attains its norm at 119869

1198601198921isin 119871

1(120583) and

10038171003817100381710038171198691198601198921 minus 11989101003817100381710038171003817 =10038171003817100381710038171198921 minus 1198920

1003817100381710038171003817 lt 120576 (135)

We also have that for any 119891 isin 1198781198711(120583)

10038171003817100381710038171198790 (119891) minus 1198791 (119891)1003817100381710038171003817 =100381710038171003817100381710038171198790(119891120594

119860) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790 (119869119860119875119860119891) minus 119879119860 (119875119860119891)

1003817100381710038171003817

+10038171003817100381710038171003817119879119860(119875

119860119891) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790119869119860 minus 119879119860

1003817100381710038171003817 +10038171003817100381710038171198780 minus 1198783

1003817100381710038171003817

lt 120578 (1205760) + 6120576

(136)

Hence 1198790minus 119879

1 ⩽ 120578(120576

0) + 6120576 lt 120576

0

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Abstract and Applied Analysis 15

Acknowledgments

The authors thank Gilles Godefroy and Rafael Paya forfruitful conversations about the content of this paper andVladimir Kadets for providing Lemma 18 They also appre-ciate anonymous referees for careful reading and fruitfulsuggestions about revision The second author was sup-ported by Basic Science Research Program through theNational Research Foundation of Korea (NRF) fundedby the Ministry of Education Science and Technology(NRF-2012R1A1A1006869) The third author was partiallysupported by Spanish MICINN and FEDER project noMTM2012-31755 and by Junta de Andalucıa and FEDERGrants FQM-185 and P09-FQM-4911

References

[1] F F Bonsall and J Duncan Numerical Ranges of Operators onNormed Spaces and of Elements of Normed Algebras vol 2 ofLondon Mathematical Society Lecture Note Series 2 CambridgeUniversity Press London UK 1971

[2] F F Bonsall and J Duncan Numerical Ranges II CambridgeUniversity Press London UK 1973 London MathematicalSociety Lecture Notes Series 10

[3] B Sims On numerical range and its applications to Banachalgebras [PhD thesis] University of Newcastle New SouthWales Australia 1972

[4] I D Berg and B Sims ldquoDenseness of operators which attaintheir numerical radiusrdquoAustralian Mathematical Society A vol36 no 1 pp 130ndash133 1984

[5] C S Cardassi ldquoDensity of numerical radius attaining operatorson some reflexive spacesrdquo Bulletin of the Australian Mathemat-ical Society vol 31 no 1 pp 1ndash3 1985

[6] C S Cardassi ldquoNumerical radius attaining operatorsrdquo inBanach Spaces (Columbia Mo 1984) vol 1166 of Lecture Notesin Mathematics pp 11ndash14 Springer Berlin Germany 1985

[7] C S Cardassi ldquoNumerical radius-attaining operators on119862(119870)rdquoProceedings of the AmericanMathematical Society vol 95 no 4pp 537ndash543 1985

[8] MDAcostaOperadores que alcanzan su radio numerico [PhDthesis] Universidad de Granada Granada Spain 1990

[9] M D Acosta and R Paya ldquoNumerical radius attaining oper-ators and the Radon-Nikodym propertyrdquo The Bulletin of theLondon Mathematical Society vol 25 no 1 pp 67ndash73 1993

[10] R Paya ldquoA counterexample on numerical radius attainingoperatorsrdquo Israel Journal of Mathematics vol 79 no 1 pp 83ndash101 1992

[11] Y S Choi and S G Kim ldquoNorm or numerical radius attainingmultilinear mappings and polynomialsrdquo Journal of the LondonMathematical Society vol 54 no 1 pp 135ndash147 1996

[12] M D Acosta J Becerra Guerrero and M Ruiz GalanldquoNumerical-radius-attaining polynomialsrdquo The Quarterly Jour-nal of Mathematics vol 54 no 1 pp 1ndash10 2003

[13] M D Acosta and S G Kim ldquoDenseness of holomorphicfunctions attaining their numerical radiirdquo Israel Journal ofMathematics vol 161 pp 373ndash386 2007

[14] S G Kim and H J Lee ldquoNumerical peak holomorphicfunctions on Banach spacesrdquo Journal of Mathematical Analysisand Applications vol 364 no 2 pp 437ndash452 2010

[15] M D Acosta R M Aron D Garcıa and M MaestreldquoThe Bishop-Phelps-Bollobas theorem for operatorsrdquo Journal ofFunctional Analysis vol 254 no 11 pp 2780ndash2799 2008

[16] A J Guirao and O Kozhushkina ldquoThe Bishop-Phelps-Bollobasproperty for numerical radius in ℓ

1(119862)rdquo Studia Mathematica

vol 218 no 1 pp 41ndash54 2013[17] J Falco ldquoThe Bishop-Phelps-Bollobas property for numerical

radius on 1198711rdquo Journal of Mathematical Analysis and Applica-

tions vol 414 no 1 pp 125ndash133 2014[18] A Aviles A J Guirao and J Rodriguez ldquoOn the Bishop-

Phelps-Bollobas property for numerical radius in119862(119870)-spacesrdquoPreprint

[19] RM Aron Y S Choi S K Kim H J Lee andMMartin ldquoTheBishop-Phelps-Bollobas version of Lindenstrauss properties Aand Brdquo Preprint

[20] V Kadets M Martın and R Paya ldquoRecent progress and openquestions on the numerical index of Banach spacesrdquo Revista dela Real Academia de Ciencias Exactas Fısicas y Naturales A vol100 no 1-2 pp 155ndash182 2006

[21] M Martın J Merı and M Popov ldquoOn the numerical index ofreal 119871

119901(120583)-spacesrdquo Israel Journal of Mathematics vol 184 pp

183ndash192 2011[22] A Iwanik ldquoNorm attaining operators on Lebesgue spacesrdquo

Pacific Journal of Mathematics vol 83 no 2 pp 381ndash386 1979[23] HH SchaeferBanach Lattices and Positive Operators Springer

New York NY USA 1974[24] J Duncan C M McGregor J D Pryce and A J White ldquoThe

numerical index of a normed spacerdquo Journal of the LondonMathematical Society Second Series vol 2 pp 481ndash488 1970

[25] Y S Choi S K Kim H J Lee and M Martin ldquoThe Bishop-Phelps-Bollobas theorem for operators on 119871

1(120583)rdquo Preprint

[26] H E Lacey The Isometric Theory of Classical Banach SpacesSpringer New York NY USA 1974

[27] G Godefroy ldquoUniversal spaces for strictly convex Banachspacesrdquo Revista de la Real Academia de Ciencias Exactas Fısicasy Naturales A vol 100 no 1-2 pp 137ndash146 2006

[28] C Finet M Martın and R Paya ldquoNumerical index andrenormingrdquo Proceedings of the American Mathematical Societyvol 131 no 3 pp 871ndash877 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 8: Research Article On the Bishop-Phelps-Bollobás Property ...downloads.hindawi.com/journals/aaa/2014/479208.pdf · Research Article On the Bishop-Phelps-Bollobás Property for Numerical

8 Abstract and Applied Analysis

for all 119909 isin 119861119895 Indeed set 119861

119895= 119861

119895cap 119909 isin Ω (119889|]

119891|1119889119898

1)

(119909) ⩽ 1205762 Then

int119861119895119861119895

120576

2119889119898

1(119909) ⩽ int

119861119895

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909) 1198891198981(119909)

=10038161003816100381610038161003816]1119891

10038161003816100381610038161003816(119861

119895) ⩽1205762

22119898

1(119861

119895)

(64)

This shows that 1198981(119861

119895 119861

119895) ⩽ (1205762)119898

1(119861

119895) This proves our

second claimNow we define 119892 by 119892(119910) = 119892(119910)|119892(119910)| if |119892(119910)| ⩾ 1 minus

radic120576232 and 119892(119910) = 119892(119910) if |119892(119910)| lt 1 minus radic120576232 and we

write 119891 = sum119895isin119869120573119895(120594

119861119895

1198981(119861

119895)) It is clear that 119892 isin 119878

119871infin(119898)

119892 minus 119892 lt radic120576 and 119892(119910) = 1 for all 119909 isin supp119891

Finally we define the measure

119889] (119909 119910)

= sum

119895isin119869

120594119861119895

(119909) 119892 (119910)ℎ (119909 119910)119889]119888(119909 119910)(

1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+ 1205941198691119861(119909) 119889] (119909 119910)

(65)

where 119861 = ⋃119895isin119869119861119895 It is easy to see that (119889|]|1119889119898

1)(119909) = 1

on 119861 and (119889|]|11198891198981)(119909) ⩽ 1 elsewhere Note that

119889 (] minus ]) (119909 119910)

= sum

119895isin119869

120594119861119895

(119909) [

[

119892 (119910)ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1]

]

119889]119888(119909 119910)

minus sum

119895isin119869

120594119861119895

(119909) 119889]119891(119909 119910)

(66)

If (119909 119910) isin 119862 then |119892(119910)| ⩾ 1 minus radic120576232 ⩾ 1 minus 1232 and

100381610038161003816100381610038161003816119892 (119910)ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

119892 (119910)

1003816100381610038161003816119892 (119910)1003816100381610038161003816

ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816100381610038161003816

1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 11003816100381610038161003816

1003816100381610038161003816119892 (119910)1003816100381610038161003816

+

10038161003816100381610038161 minus1003816100381610038161003816119892 (119910)

1003816100381610038161003816

10038161003816100381610038161003816100381610038161003816119892 (119910)

1003816100381610038161003816

⩽ 2

1003816100381610038161003816119892 (119910) ℎ (119909 119910) minus 11003816100381610038161003816

1003816100381610038161003816119892 (119910)1003816100381610038161003816

⩽ 2radic120576

232

232

232 minus 1⩽ 2radic120576

(67)

Hence for all (119909 119910) isin 119862 we have

10038161003816100381610038161003816100381610038161003816100381610038161003816

119892 (119910)ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

⩽100381610038161003816100381610038161003816119892 (119910)ℎ (119909 119910) minus 1

100381610038161003816100381610038161003816(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

⩽ 2radic120576(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

(68)

So we have for all 119909 isin 1198691

119889|] minus ]|1

1198891198981

(119909)

⩽ sum

119895isin119869

120594119861119895

(119909) [

[

2radic120576(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

+

10038161003816100381610038161003816100381610038161003816100381610038161003816

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816

]

]

1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909)

+ sum

119895isin119869

120594119861119895

(119909)

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909)

⩽ sum

119895isin119869

120594119861119895

(119909)(2radic120576 + (1 minus1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909)))

+sum

119895isin119869

120594119861119895

(119909)(

11988910038161003816100381610038161003816]119891

10038161003816100381610038161003816

1

1198891198981

(119909))

⩽ 2radic120576 + 120576 lt 3radic120576

(69)

This gives that 119879] minus 119879] lt 3radic120576 Note also that for all 119895 isin 119869

⟨119879]

120594119861119895

1198981(119861

119895)

119892⟩

= intΩtimesΩ

120594119861119895

(119909)

1198981(119861

119895)

119892 (119910) 119889] (119909 119910)

= intΩtimesΩ

120594119861119895

(119909)

1198981(119861

119895)

ℎ (119909 119910)(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

119889]119888(119909 119910)

Abstract and Applied Analysis 9

= intΩ

120594119861119895

(119909)

1198981(119861

119895)

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

1198891003816100381610038161003816]1198881003816100381610038161003816

1

(119909)

= intΩ

120594119861119895

(119909)

1198981(119861

119895)

1198891198981(119909) = 1

(70)

Hence we get ⟨119879]119891 119892⟩ = 1 which implies that 119879]119891 =119879] = 1 Finally10038171003817100381710038171003817119891 minus 119891

10038171003817100381710038171003817

⩽10038171003817100381710038171003817119891 minus 119891

1

10038171003817100381710038171003817+10038171003817100381710038171198911 minus 119891

1003817100381710038171003817

=

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

1198981(119861

119895)

minus sum

119895isin119869

120573119895

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+ 120576

⩽ sum

119895isin119869

120573119895(

10038171003817100381710038171003817100381710038171003817100381710038171003817

120594119861119895

1198981(119861

119895)

minus

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

120594119861119895

1198981(119861

119895)

minus

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

) + 120576

= 2sum

119895isin119869

120573119895

1198981(119861

119895 119861

119895)

1198981(119861

119895)

+ 120576

⩽ 2sum

119895isin119869

120573119895

(1205762)1198981(119861

119895)

1198981(119861

119895)

+ 120576 ⩽120576

1 minus 1205762+ 120576 lt 3120576

(71)

We are now ready to present the proof of the main resultin the case of finite regular positive Borel measures

Proof of Proposition 10 Let 1205751= 120575

3

2(5sdot2

4) 120575

2= 120575

12

3(3

2sdot2

14)

and 1205753= (12057610)

2 for some 0 lt 120576 lt 1 Suppose that 119879 isinL(119871

1(119898)) with 119879 = 1 and that there is an 119891

0isin 119878

1198711(119898)

and1198920isin 119878

119871infin(119898)

such that ⟨1198910 119892

0⟩ = 1 and |⟨119879119891

0 119892

0⟩| gt 1minus120575

3

12

6Then there is an isometric isomorphism Ψ from 119871

1(119898) onto

itself such thatΨ(1198910) = |119891

0| and there is a scalar number 120572 in

119878R such that |⟨1198791198910 119892

0⟩| = ⟨120572119879119891

0 119892

0⟩ Then letting 119891

1= Ψ119891

0

1198921= (Ψ

minus1)lowast1198920 and 119879

1= 120572Ψ119879Ψ

minus1 we have

⟨1198781198911 119892

1⟩ = ⟨120572Ψ119879

minus1Ψ119891

0 (Ψ

minus1)lowast

1198920⟩

= ⟨1205721198791198910 119892

0⟩ gt 1 minus

1205753

1

26

⟨1198911 119892

1⟩ = ⟨Ψ119891

0 (Ψ

minus1)lowast

1198920⟩ = 1

(72)

Since 11987911198911 gt 1 minus 120575(120575

3

12

6) by Lemma 13 there exists a

norm-one bounded operator 119879] and a nonnegative simplefunction119891

2isin 119878

1198711(119898)

such that 1198791minus119879] ⩽ 1205751 1198912minus1198911 ⩽ 31205751

and (119889|]|11198891198981)(119909) = 1 for all 119909 isin supp(119891

2) Then

⟨119879]1198912 1198921⟩

= ⟨11987911198911 119892

1⟩ minus ⟨119879

11198911minus 119879

11198912 119892

1⟩ minus ⟨119879

11198912minus 119879]1198912 1198921⟩

⩾ ⟨11987911198911 119892

1⟩ minus10038171003817100381710038171198911 minus 1198912

1003817100381710038171003817 minus10038171003817100381710038171198791 minus 119879]

1003817100381710038171003817

⩾ 1 minus1205753

1

26minus 3120575

1minus 120575

1⩾ 1 minus 5120575

1= 1 minus

1205753

2

16

(73)

Notice also that

⟨1198912 119892

1⟩ = ⟨119891

1 119892

1⟩ minus ⟨119891

1minus 119891

2 119892

1⟩ ⩾ 1 minus

10038171003817100381710038171198911 minus 11989121003817100381710038171003817

⩾ 1 minus 31205751⩾ 1 minus 5120575

1= 1 minus

1205753

2

16

(74)

By Lemma 12 there are a nonnegative simple function 1198913isin

1198781198711(119898)

and a function 1198923isin 119878

119871infin(119898)

such that

1198923(119909) = 120594supp119891

3

(119909) + 1198922(119909) 120594

Ωsupp1198913

(119909)

10038171003817100381710038171198912 minus 11989131003817100381710038171003817 ⩽ 1205752

10038171003817100381710038171198923 minus 11989211003817100381710038171003817 ⩽radic120575

2 ⟨119891

3 119892

3⟩ = 1

(75)

So we have

⟨119879]1198913 1198923⟩ = ⟨119879]1198912 1198921⟩ minus ⟨119879]1198912 minus 119879]1198913 1198921⟩

minus ⟨119879]1198913 1198921 minus 1198923⟩

⩾ 1 minus1205753

2

16minus 2radic120575

2⩾ 1 minus 3radic120575

2= 1 minus

1205756

3

27

(76)

By Lemma 14 there exist 1198914isin 119878

1198711(119898)

and 1198924isin 119878

119871infin(119898)

and anoperator 119879

4such that

⟨1198924 119879

41198914⟩ = 1 =

100381710038171003817100381711987941003817100381710038171003817

10038171003817100381710038171198794 minus 119879]1003817100381710038171003817 ⩽ 21205753

10038171003817100381710038171198914 minus 11989131003817100381710038171003817 ⩽ 31205753

10038171003817100381710038171198924 minus 11989231003817100381710038171003817 ⩽radic120575

3 ⟨119891

4 119892

4⟩ = 1

(77)

So we have

10038171003817100381710038171198794 minus 11987911003817100381710038171003817 ⩽10038171003817100381710038171198794 minus 119879]

1003817100381710038171003817 +1003817100381710038171003817119879] minus 1198791

1003817100381710038171003817

⩽ 1205751+ 2120575

3⩽ 3120575

3

10038171003817100381710038171198911 minus 11989141003817100381710038171003817 ⩽10038171003817100381710038171198911 minus 1198912

1003817100381710038171003817 +10038171003817100381710038171198912 minus 1198913

1003817100381710038171003817 +10038171003817100381710038171198913 minus 1198914

1003817100381710038171003817

⩽ 31205751+ 120575

2+ 3120575

3⩽ 10120575

3

10038171003817100381710038171198921 minus 11989241003817100381710038171003817 ⩽10038171003817100381710038171198921 minus 1198923

1003817100381710038171003817 + 1198923 minus 1198924

⩽ 1205752+ radic120575

3⩽ 2radic120575

3

(78)

10 Abstract and Applied Analysis

Let 119878 = 120572Ψminus11198794Ψ 119891 = Ψminus1

1198914 and 119892 = Ψlowast

1198924 then we have

119879 minus 119878 =10038171003817100381710038171003817119879 minus 120572Ψ

minus11198794Ψ10038171003817100381710038171003817=10038171003817100381710038171003817120572Ψ119879Ψ

minus1minus 119879

4

10038171003817100381710038171003817

=10038171003817100381710038171198791 minus 1198794

1003817100381710038171003817 ⩽ 31205753

100381710038171003817100381710038171198910minus 11989110038171003817100381710038171003817=100381710038171003817100381710038171198910minus Ψ

minus11198914

10038171003817100381710038171003817=10038171003817100381710038171198911 minus 1198914

1003817100381710038171003817 ⩽ 101205753

100381710038171003817100381710038171198920minus 11989110038171003817100381710038171003817=10038171003817100381710038171198920 minus Ψ

lowast1198924

1003817100381710038171003817 =100381710038171003817100381710038171003817(Ψ

minus1)lowast

1198920minus 119892

4

100381710038171003817100381710038171003817

=10038171003817100381710038171198921 minus 1198924

1003817100381710038171003817 ⩽ 2radic120575

3

⟨119891 119892⟩ = ⟨Ψminus11198914 Ψ

lowast1198924⟩ = ⟨119891

4 119892

4⟩ = 1

10038161003816100381610038161003816⟨119878119891 119892⟩

10038161003816100381610038161003816=10038161003816100381610038161003816⟨120572Ψ

minus11198794ΨΨ

minus11198914 Ψ

lowast1198924⟩10038161003816100381610038161003816= |120572| = 1

(79)

This completes the proof

Finally we may give the proof of the main result in fullgenerality

Proof of Theorem 9 Notice that the Kakutani representationtheorem (see [26] for a reference) says that for every 120590-finitemeasure ] the space 119871

1(]) is isometrically isomorphic to

1198711(119898) for some positive Borel regular measure on a compact

Hausdorff space Then by Proposition 10 there is a universalfunction 120576 997891rarr 120578(120576) gt 0 which gives the BPBp-nu for 119871

1(]) for

every 120590-finite measure ]Fix 120576 gt 0 Suppose that 119879

0isin L(119871

1(120583)) with V(119879

0) = 1

and (1198910 119891

lowast

0) isin Π(119871

1(120583)) satisfy

1003816100381610038161003816⟨119891lowast

0 119879

01198910⟩1003816100381610038161003816 gt 1 minus 120578 (120576) (80)

Choose a sequence 119891119899 in 119871

1(120583) such that sup

119899119879

0119891119899 = 1

and let 119866 be the closed linear span of

119879119899

0119891119898 119899 119898 isin N cup 0 (81)

As 119866 is separable there is a dense subset 119892119899 119899 isin N of 119866

and let 119864 = ⋃infin

119899=1supp119892

119899 where supp119892

119899is the support of 119892

119899

Then the measure 120583|119864is 120590-finite Let

119884 = 119891 isin 1198711(120583) supp (119891) sub 119864 (82)

be a closed subspace of 1198711(120583) It is clear that 119871

1(120583) = 119884oplus

1119885

and 119884 is isometrically isomorphic to 1198711(120583|

119864) So 119884 has the

BPBp-nu with 120578(120576)Now write 119878

0= 119879

0|119884 119884 rarr 119884 consider 119910

0= 119891

0isin

119878119884 119910lowast

0= 119891

lowast

0|119884isin 119878

119884lowast and observe that 119910lowast

0(119910

0) = 1 and

|119910lowast

0(119878

01199100)| = |119891

lowast

0(119879

01198910)| gt 1 minus 120578(120576) Hence there exist 119878 isin

L(119884) and (1199100 119910

lowast

0) isin Π(119884) such that

1003816100381610038161003816119910lowast

0(119878119910

0)1003816100381610038161003816 = 1 = V (119878) 1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

10038171003817100381710038171199100 minus 11991001003817100381710038171003817 lt 120576

1003817100381710038171003817119910lowast

0minus 119910

lowast

0

1003817100381710038171003817 lt 120576

(83)

Finally consider the operator 119879 isinL(1198711(120583)) given by

119879 (119910 119911)

= (119878119910 0) + 1198790(0 119911) ((119910 119911) isin 119871

1(120583) equiv 119884oplus

1119885)

(84)

We have 119879 = 1 (and so V(119879) = 1) Indeed1003817100381710038171003817119879 (119910 119911)

1003817100381710038171003817 =1003817100381710038171003817(119878119910 0)

1003817100381710038171003817 +10038171003817100381710038171198790 (0 119911)

1003817100381710038171003817 ⩽10038171003817100381710038171199101003817100381710038171003817 + 119911 =

1003817100381710038171003817(119910 119911)1003817100381710038171003817

(85)

for all (119910 119911) isin 1198711(120583) and 119879(119910

0 0) = (119878119910

0 0) = 119878119910

0 = 1

Let 119909 = (1199100 0) and 119909lowast = (119910lowast

0 119891

0|119885) Then (119909 119909lowast) isin Π(119871

1(120583))

Moreover we have1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119910

lowast

0119878119910

0

1003816100381610038161003816 = 1 = V (119879) 1003817100381710038171003817119909 minus 1198910

1003817100381710038171003817 =1003817100381710038171003817119910 minus 1199100

1003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowast

0minus 119891

lowast

0

1003817100381710038171003817

= max 1003817100381710038171003817119910 minus 119891lowast

0|119884

1003817100381710038171003817 1003817100381710038171003817119891

lowast

0|119885minus 119891

lowast

0|119885

1003817100381710038171003817

=1003817100381710038171003817119910

lowastminus 119910

lowast

0

1003817100381710038171003817 lt 120576

1003817100381710038171003817119879 minus 11987901003817100381710038171003817 = sup119910+119911⩽1

1003817100381710038171003817119879 (119910 119911) minus 1198790 (119910 119911)1003817100381710038171003817

= sup119910⩽1

1003817100381710038171003817119878119910 minus 11987801199101003817100381710038171003817 =1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

(86)

This completes the proof

5 Examples of Spaces Failing theBishop-Phelps-Bollobaacutes Property forNumerical Radius

Our goal here is to prove that the density of numerical radiusattaining operators does not imply the BPBp-nu Actuallywe will show that among separable spaces there is noisomorphic property implying the BPBp-nu other than finite-dimensionality

We need to relate the BPBp-nu to the Bishop-Phelps-Bollobas property for operators which as mentioned in theintroduction was introduced in [15] A pair (119883 119884) of Banachspaces has the Bishop-Phelps-Bollobas property for operators(in short BPBp) if given 120576 gt 0 there exists 120578(120576) gt 0 suchthat given 119879 isin L(119883 119884) with 119879 = 1 and 119909 isin 119878

119883such that

119879119909 gt 1 minus 120578(120576) then there exist 119911 isin 119878119883and 119878 isin L(119883 119884)

satisfying

119878 = 119878119911 = 1 119909 minus 119911 lt 120576

119879 minus 119878 lt 120576

(87)

We refer the reader to [15 19 25] and references therein formore information and background Among the interestingresults on the BPBp we emphasize that a pair (119883 119884) when119883is finite-dimensional does not necessarily have the BPBp Forinstance if119884 is a strictly convex space which is not uniformlyconvex then the pair (ℓ(2)

1 119884) fails to have the BPBp (this is

contained in [15] see [19 Section 3])

Abstract and Applied Analysis 11

The next result relates the BPBp-nu to the BPBp foroperators in a particular case We will deduce our examplefrom it

Theorem 15 If 1198711(120583)oplus

1119883 has the BPBp-119899119906 then the pair

(1198711(120583) 119883) has the BPBp for operators

Before proving this proposition we will use it to get themain examples of this section The first example shows thatthe density of numerical radius attaining operators does notimply the BPBp-nu

Example 16 There is a reflexive space (and so numerical radiusattaining operators on it are dense) which fails to have theBPBp-nu Indeed let 119884 be a reflexive separable space whichis not superreflexive and we may suppose that 119884 is strictlyconvex Observe that 119884 cannot be uniformly convex since itis not superreflexive Now 119883 = ℓ(2)

1oplus1119884 is reflexive but the

pair (ℓ(2)1 119884) fails the BPBp since 119884 is strictly convex but not

uniformly convex [19 Corollary 33] Therefore Theorem 15gives us that119883 does not have the BPBp-nu

The example above can be extended to get the result thatevery infinite-dimensional separable Banach space can berenormed to fail the BPBp-nu This follows from the factthat every infinite-dimensional separable Banach space canbe renormed to be strictly convex but not uniformly convex(this result can be proved ldquoby handrdquo an alternative categoricalargument for it can be found in [27] and references therein)With a little more of effort we may get the main result of thesection

Theorem 17 Every infinite-dimensional separable Banachspace can be renormed to fail the weak-BPBp-nu (and so inparticular to fail the BPBp-nu)

We need the following result which is surely well knownAs we have not found a reference we include a nice and easyproof kindly given to us by Vladimir Kadets We recall thatgiven a Banach space 119884 the set of all equivalent norms on119884 can be viewed as a metric space using the Banach-Mazurdistance

Lemma 18 Let119884 be an infinite-dimensional separable Banachspace Then the set of equivalent norms on 119884 which are strictlyconvex and are not (locally) uniformly convex is dense in the setof all equivalent norms on119884 (with respect to the Banach-Mazurdistance)

Proof Fix 119890 isin 119878119884and 119890lowast

1isin 119878

119884lowast such that 119890lowast

1(119890) = 1 For a fixed

120576 isin (0 12) denote

119902 (119910) = max (1 minus 120576) 10038171003817100381710038171199101003817100381710038171003817 1003816100381610038161003816119890lowast

1(119910)1003816100381610038161003816 (119910 isin 119884) (88)

Evidently (1 minus 120576)119910 ⩽ 119902(119910) ⩽ 119910 for every 119910 isin 119884 Fix asequence 119890lowast

119896 119896 ⩾ 2 of norm-one functionals separating

the points of 119884 and denote

119901 (119910) = radic

infin

sum

119896=1

1

2119896

1003816100381610038161003816119890lowast

119896(119910)1003816100381610038161003816

2

(119910 isin 119884) (89)

Then 119901 is a strictly convex norm on 119884 119901(119890) ⩾ 1radic2 and119901(119910) ⩽ 119910 for all 119910 isin 119883 Finally write

100381710038171003817100381711991010038171003817100381710038171= (1 minus 120576) 119902 (119910) + 120576

119901 (119910)

119901 (119890)(119910 isin 119884) (90)

Then sdot 1is a strictly convex norm on 119884 and

(1 minus 120576)2 10038171003817100381710038171199101003817100381710038171003817 ⩽100381710038171003817100381711991010038171003817100381710038171⩽ (1 + 120576)

10038171003817100381710038171199101003817100381710038171003817 (119910 isin 119884) (91)

Wewill finish the proof by showing that sdot 1is not uniformly

convex (actually it is not locally uniformly convex) Indeedfor each 119899 isin N we select 119910

119899isin ⋂

119899

119896=1ker 119890lowast

119896with 119910

119899 = 1 and

consider 119890119899= 119890 + (1205764)119910

119899 Then 119902(119890) = 1 119902(119890

119899) = 1 and

119902(119890+119890119899) = 2 At the same time 119901(119910

119899) rarr 0 so 119901(119890

119899) rarr 119901(119890)

and 119901(119890 + 119890119899) rarr 2119901(119890) Consequently

1198901 = 1100381710038171003817100381711989011989910038171003817100381710038171997888rarr 1

1003817100381710038171003817119890 + 11989011989910038171003817100381710038171997888rarr 2 (92)

but 119890 minus 1198901198991= (1205764)119910

1198991⩾ (1 minus 120576)

2(1205764) which means the

absence of local uniform convexity at 119890

Proof of Theorem 17 Let 119883 be an infinite-dimensional sepa-rable Banach space Take a closed subspace 119884 of 119883 of codi-mension two By [28 Proposition 2] the map carrying everyequivalent norm on 119884 to its numerical index is continuousand so the set of values of the numerical index of 119884 upto reforming is a nontrivial interval [28 Theorem 9] ThenLemma 18 allows us to find an equivalent norm | sdot | on 119884 insuch a way that (119884 | sdot |) is strictly convex and is not uniformlyconvex and 119899(119884 | sdot |) gt 0 Now the space 119883 = ℓ(2)

1oplus1(119884 | sdot |)

is an equivalent renorming of 119883 which does not have theBPBp-119899119906 (indeed otherwise the pair (ℓ(2)

1 (119884 | sdot |)) would

have the BPBp for the operator norm and so (119884 | sdot |) wouldbe uniformly convex by [19 Corollary 33] a contradiction)Moreover as

119899 (119883) = min 119899 (ℓ(2)1) 119899 (119884 |sdot|) gt 0 (93)

(see [20 Proposition 2] for instance) 119883 also fails the weak-BPBp-nu by Proposition 6

To finish the section with the promised proof ofTheorem 15 we first see the following stability result

Lemma 19 Let119883 = [⨁infin

119896=1119883

119896]1198880

or [⨁infin

119896=1119883

119896]ℓ1

If119883 has theBishop-Phelps-Bollobas property for numerical radius with afunction 120578 then each Banach space 119883

119894has the Bishop-Phelps-

Bollobas property for numerical radius with 120578nu(119883119894) ⩾ 120578 That

is inf119894120578nu(119883119894

)(120576) ⩾ 120578nu(119883)(120576) for all 0 lt 120576 lt 1

Proof Let 119875119894 119883 rarr 119883

119894and 1198751015840

119894 119883

lowastrarr 119883

lowast

119894be the natural

projections and let 119876119894 119883

119894rarr 119883 and 1198761015840

119894 119883

lowast

119894rarr 119883

lowast be thenatural embeddings

Assume that an operator 119879119894 119883

119894rarr 119883

119894and a pair

(119909119894 119909

lowast

119894) isin Π(119883

119894) satisfy that

V (119879119894) = 1

1003816100381610038161003816119909lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (94)

We define an operator 119879 119883 rarr 119883 and (119909 119909lowast) isin Π(119883) by

119879 = 119876119894∘ 119879

119894∘ 119875

119894 (119909 119909

lowast) = (119876

119894119909119894 119876

1015840

119894119909lowast

119894) (95)

12 Abstract and Applied Analysis

then clearly we see that1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119909

lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (96)

From the assumption there exist 119878 119883 rarr 119883 and a pair(119910 119910

lowast) isin Π(119883) such that

1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 = 1 = V (119878) 119878 minus 119879 lt 120576

1003817100381710038171003817119910lowastminus 119909

lowast1003817100381710038171003817 lt 1205761003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576

(97)

Since this clearly shows that1003817100381710038171003817119875119894 ∘ 119878 ∘ 119876119894

minus 119879119894

1003817100381710038171003817 lt 120576100381710038171003817100381710038171198751015840

119894119910lowastminus 119909

lowast

119894

10038171003817100381710038171003817lt 120576

1003817100381710038171003817119875119894119910 minus 1199091198941003817100381710038171003817 lt 120576

(98)

we only need to show that |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = 1

We first show the case of 1198880sum Since

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817=10038171003817100381710038171003817119875119895119910 minus 119875

11989511990910038171003817100381710038171003817⩽1003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576 (99)

for every 119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽100381710038171003817100381710038171198751015840

119894119910lowast10038171003817100381710038171003817+ 120576 sum

119895isinN119895 = 119894

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817⩽1003817100381710038171003817119910

lowast1003817100381710038171003817 = 1

(100)

This shows that 1198751015840119894119910lowast = 1 and 1198751015840

119895119910lowast= 0 for every 119895 = 119894

So 119910lowast = 1198761015840

1198941198751015840

119894119910lowast and 1198751015840

119894119910lowast(119875

119894119910) = 1 This and the fact that

119910 minus 119876119894119875119894119910 lt 120576 imply that

(119876119894119875119894119910 + (

1

120576) (119910 minus 119876

119894119875119894119910) 119876

1015840

1198941198751015840

119894119910lowast) isin Π (119883) (101)

So we get that (1198761015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910 + (1120576)(119910 minus 119876

119894119875119894119910))V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =10038161003816100381610038161003816(119876

1015840

1198941198751015840

119894119910lowast) 11987811991010038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910 +1

120576(119910 minus 119876

119894119875119894119910))

1003816100381610038161003816100381610038161003816⩽ 1

(102)

and so we get |1198751015840119894119910lowast(119875

119894∘119878 ∘119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

We next show the case of ℓ1sum The proof is almost

the same as that of the 1198880case However for the sake of

completeness we provide it hereSince 1198751015840

119895119910lowast = 119875

1015840

119895119910lowastminus 119875

1015840

119895119909lowast ⩽ 119910

lowastminus 119909

lowast lt 120576 for every

119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽10038171003817100381710038171198751198941199101003817100381710038171003817 + 120576 sum

119895isinN119895 = 119894

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817⩽10038171003817100381710038171199101003817100381710038171003817 = 1

(103)

which shows 119875119894119910 = 1 and 119875

119895119910 = 0 for every 119895 = 119894 Since this

implies (119876119894119875119894119910119876

1015840

1198941198751015840

119894119910lowast+(1120576)(119910

lowastminus119876

1015840

1198941198751015840

119894119910lowast)) isin Π(119883) we get

that |(1198761015840

1198941198751015840

119894119910lowast+ (1120576)(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast))119878(119876

119894119875119894119910)| ⩽ V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =1003816100381610038161003816119910

lowast119878 (119876

119894119875119894119910)1003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast+1

120576(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast)) 119878 (119876

119894119875119894119910)

1003816100381610038161003816100381610038161003816⩽ 1

(104)

and so |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

Proof of Theorem 15 Note that 120578nu(1198711(120583)oplus1119883)(120576) rarr 0 as120576 rarr 0 Fix 0 lt 120576

0lt 1 and choose 0 lt 120576 lt 1 such that

6120576 + 120578nu(1198711(120583)oplus1119883)(120576) lt 1205760 Let 120578(1205760) = 120578nu(1198711(120583)oplus1119883)(120576)Suppose that 119879

0isin L(119871

1(120583) 119883) with 119879

0 = 1 and 119891

0isin

1198781198711(120583)

satisfy

1003817100381710038171003817119879011989101003817100381710038171003817 gt 1 minus 120578 (1205760) (105)

For any measurable subset 119861 let

1198711(120583|

119861) = 119891|

119861 119891 isin 119871

1(120583) (106)

with the norm 119891|119861 = 119891120594

1198611 Then it is easy to see

that 1198711(120583|

119861) is isometrically isomorphic to a complemented

subspace of 1198711(120583) Let 119875

119861 119871

1(120583) rarr 119871

1(120583|

119861) be the

restriction defined by 119875119861(119891) = 119891|

119861for all 119891 isin 119871

1(120583) and

let 119869119861 119871

1(120583|

119861) rarr 119871

1(120583) be the extension defined by

119869119861(119891)(120596) = 119891(120596) if 120596 isin 119861 and 119869

119861(119891)(120596) = 0 otherwise

It is clear that 119875119861119869119861= Id

1198711(120583|119861)and 119869

119861119875119861(119891) = 119891120594

119861for all

119891 isin 1198711(120583) Notice also that 119871

1(120583) is isometrically isomorphic

to 1198711(120583|

119861)oplus

11198711(120583|

119861119888)

Let 119860 = supp1198910and 119892

0= 119875

1198601198910 Then

1003817100381710038171003817119879011986911986011989201003817100381710038171003817 =100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) gt 0 (107)

and define the operator 119879119860 119871

1(120583|

119860) rarr 119883 by 119879

119860119891 =

1198790119869119860119891119879

0119869119860 for every 119891 isin 119871

1(120583|

119860) Then

100381710038171003817100381711987911986011989201003817100381710038171003817 ⩾100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) (108)

Since 120583|119860is 120590-finite 119871

1(120583|

119860)lowast= 119871

infin(120583|

119860) Let 119892lowast

0isin 119878

119871infin(120583|119860)

be a function such that ⟨119892lowast0 119892

0⟩ = 1 and choose 119909lowast

0isin 119878

119883lowast

such that 119909lowast0(119879

1198601198920) = 119879

1198601198920 Define the operator 119878

0isin

L(1198711(120583|

119860)oplus

1119883) by

1198780(119891 119909) = (0 119879

119860119891) ((119891 119909) isin 119871

1(120583|

119860) oplus

1119883) (109)

and observe that 1198780 = V(119878

0) = 1 Indeed

100381710038171003817100381711987801003817100381710038171003817 ⩽ 1 =

10038171003817100381710038171198791198601003817100381710038171003817

= sup 1003816100381610038161003816119909lowast1198791198601198911003816100381610038161003816 119909

lowastisin 119878

119883lowast 119891 isin 119878

1198711(120583|119860)

= sup 1003816100381610038161003816(119891lowast 119909

lowast) 119878

0(119891 119909)

1003816100381610038161003816

((119891lowast 119909

lowast) (119891 119909)) isin Π (119871

1(120583|

119860) oplus

1119883)

= V (1198780) ⩽100381710038171003817100381711987801003817100381710038171003817

(110)

Abstract and Applied Analysis 13

It is immediate that

(119892lowast

0 119909

lowast

0) 119878

0(119892

0 0) = 119909

lowast

0(119879

1198601198920) =10038171003817100381710038171198791198601198920

1003817100381710038171003817 gt 1 minus 120578 (1205760)

(111)

By Lemma 19 1198711(120583|

119860)oplus

1119883 has the BPBp-nu with the func-

tion 120578 Therefore there exist 1198781isin L(119871

1(120583|

119860)oplus

1119883) (119892

1 119909

1) isin

1198781198711(120583|119860)oplus1119883 and (119892lowast

1 119909

lowast

1) isin 119878

119871infin(120583|119860)oplusinfin119883lowast such that

1003817100381710038171003817(1198921 1199091) minus (1198920 0)1003817100381710038171003817 lt 120576

1003817100381710038171003817(119892lowast

1 119909

lowast

1) minus (119891

lowast

0 119909

lowast

0)1003817100381710038171003817 lt 120576

10038171003817100381710038171198781 minus 11987801003817100381710038171003817 lt 120576 ⟨(119892

lowast

1 119909

lowast

1) (119892

1 119909

1)⟩ = 1

1003816100381610038161003816(119892lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 = V (119878

1) = 1

(112)

Claim 1 We claim that 1199091= 0

Otherwise

1 = Re ⟨(119892lowast1 119909

lowast

1) (119892

1 119909

1)⟩

=100381710038171003817100381711989211003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

+100381710038171003817100381711990911003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

⟩ ⩽ 1

(113)

We deduce that

((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

(119892lowast

1 119909

lowast

1)) (

(0 1199091)

100381710038171003817100381711990911003817100381710038171003817

(119892lowast

1 119909

lowast

1))

isin Π (1198711(120583|

119860) oplus

1119883)

(114)

Since 1198781((0 119909

1)119909

1) = (119878

1minus 119878

0)((0 119909

1)119909

1) lt 120576 we

get that

1 =1003816100381610038161003816⟨(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)⟩1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

100381710038171003817100381711989211003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

)⟩

+100381710038171003817100381711990911003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

)⟩

100381610038161003816100381610038161003816100381610038161003816

⩽100381710038171003817100381711989211003817100381710038171003817 V (1198781) + 120576

100381710038171003817100381711990911003817100381710038171003817 lt100381710038171003817100381711989211003817100381710038171003817 +100381710038171003817100381711990911003817100381710038171003817 = 1

(115)

a contradiction This proves the claimWe define the operator 119878

2 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883

by 1198782(119891 119909) = 119878

1(119891 0) for every 119891 isin 119871

1(120583|

119860) and for every

119909 isin 119883 Then we have

V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 = 1

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ 120576 (116)

Indeed from Claim 1 we have

V (1198781) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 ⩽ V (119878

2)

(117)

On the other hand we see that1003816100381610038161003816(119891

lowast 119909

lowast) 119878

2(119891 119909)

1003816100381610038161003816

=1003816100381610038161003816(119891

lowast 119909

lowast) 119878

1(119891 0)

1003816100381610038161003816 ⩽10038171003817100381710038171198911003817100381710038171003817 V (1198781) ⩽ V (119878

1)

(118)

for every ((119891lowast 119909lowast) (119891 119909)) isin Π(1198711(120583|

119860)oplus

1119883) Therefore

V(1198782) ⩽ V(119878

1) Also

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ sup

119909isin119878119883

10038171003817100381710038171198781 (0 119909)1003817100381710038171003817

= sup119909isin119878119883

10038171003817100381710038171198781 (0 119909) minus 1198780 (0 119909)1003817100381710038171003817 ⩽ 120576

(119)

Claim 2 There exists 1198783 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883 such

that 1198783(119892

1 0) = 119878

3 = 1 119878

3(0 119909) = 0 119878

3(119891 119909) isin 0oplus

1119883

for every (119891 119909) isin 1198711(120583|

119860)oplus

1119883 and 119878

3minus 119878

2 lt 4120576

Indeed write 1198781= (119863

1 119863

2) where 119863

1 119871

1(120583|

119860)oplus

1119883 rarr

1198711(120583|

119860) and119863

2 119871

1(120583|

119860)oplus

1119883 rarr 119883 We have that

sup 1003816100381610038161003816119892lowast119863

1(119892

1 0) + 119909

lowast119863

2(119892

1 0)1003816100381610038161003816

119909lowastisin 119878

119883lowast ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860) +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⩽ V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119892

lowast

1119863

1(119892

1 0) + 119909

lowast

1119863

2(119892

1 0)1003816100381610038161003816

(120)

This implies that

1003816100381610038161003816119909lowast

1119863

2(119892

1 0)1003816100381610038161003816 =10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

1003816100381610038161003816119892lowast

1119863

1(119892

1 0)1003816100381610038161003816

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119871

infin(120583|

119860)

(121)

Therefore |119892lowast1| equals 1 on the support of 119863

1(119892

1 0) As

|⟨119892lowast

1 119892

1⟩| = 1 we also have that |119892lowast

1| equals 1 on the support

of 1198921 Changing the values of 119892lowast

1by the ones of 119891lowast

0on 119860

(supp(1198631(119892

1 0)) cup supp(g

1)) we may and do suppose that

|119892lowast

1| = 1 on the whole 119860We also have 119863

2(119892

1 0) gt 0 Indeed

10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)1003817100381710038171003817

⩽10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198921 0)

1003817100381710038171003817 +10038171003817100381710038171198780 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817

lt 2120576 + 120576 = 3120576

(122)

So we have

10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

⩽10038171003817100381710038171198631(119892

1 0)1003817100381710038171003817 +10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

14 Abstract and Applied Analysis

=1003817100381710038171003817(1198631

(1198921 0) 119863

2(119892

1 0)) minus (0 119879

1198601198920)1003817100381710038171003817

=10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817 lt 3120576

(123)

and 1198632(119892

1 0) gt 119879

1198601198920 minus 3120576 ⩾ 1 minus 120578(120576

0) minus 3120576 gt 0

Finally define the operator 1198783by

1198783(119891 119909) = (0119863

2(119891 0) + 119892

lowast

1(119863

1(119891 0))

1198632(119892

1 0)

119909lowast

1119863

2(119892

1 0))

(124)

for (119891 119909) isin 1198711(120583|

119860)oplus

1119883 It is clear that

100381710038171003817100381711987831003817100381710038171003817 ⩽ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) (125)

Notice also that

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817

⩽10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 +10038171003817100381710038171198632(119891 0) minus 119879

1198601198911003817100381710038171003817

=1003817100381710038171003817(1198631

(119891 0) 1198632(119891 0)) minus (0 119879

119860119891)1003817100381710038171003817

=10038171003817100381710038171198782 (119891 119909) minus 1198780 (119891 119909)

1003817100381710038171003817

(126)

for all (119891 119909) isin 1198711(120583|

119860)oplus

1119883 Hence we have

10038171003817100381710038171198783 minus 11987821003817100381710038171003817

= 2 sup119891isin1198781198711(120583|119860)

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 ⩽ 210038171003817100381710038171198782 minus 1198780

1003817100381710038171003817 lt 4120576(127)

On the other hand let 119866 1198711(120583|

119860) rarr 119871

1(120583|

119860) be defined by

119866(119891) = 119892lowast

1119891 for every 119891 isin 119871

1(120583|

119860) Then we have

V (1198782)

= sup 1003816100381610038161003816119911lowast11987821199111003816100381610038161003816 (119911 119911

lowast) isin Π (119871

1(120583|

119860) oplus

1119883)

⩾ sup10038161003816100381610038161003816100381610038161003816

119909lowast119863

2(119866(

1

120583 (119862)120594119862) 0)

+119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119909lowastisin 119878

119883lowast 119862 isin Σ

119860 120583 (119862) gt 0

= sup10038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119862)120594119862) 0)

10038171003817100381710038171003817100381710038171003817

+

10038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119862 isin Σ119860 120583 (119862) gt 0

(128)

where Σ119860is the family of measurable subsets of 119860

Hence for any simple function 119904 = sum119899

119894=1((120572

119894120583(119860

119894))120594

119860119894

) isin

1198781198711(120583|119860) where 119860

119894119894is a family of disjoint measurable subsets

with strictly positive measure we have

V (1198782)

119899

sum

119894=1

10038161003816100381610038161205721198941003816100381610038161003816 (

100381710038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381710038171003817100381710038171003817100381710038171003817

+

100381610038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381610038161003816100381610038161003816100381610038161003816

)

⩾10038171003817100381710038171198632(119866 (119904) 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119866 (119904) 0)

1003816100381610038161003816

(129)

Since |119892lowast1| = 1119866 is an isometric isomorphism so for each119891 isin

1198781198711(120583|119860)there exists a sequence of norm-one simple functions

(119904119896) such that 119866(119904

119896) converges to 119891 Therefore

V (1198782) ⩾ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) ⩾100381710038171003817100381711987831003817100381710038171003817

(130)

On the other hand we see that100381710038171003817100381711987831003817100381710038171003817 ⩾1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

3(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119909

lowast

1119863

2(119892

1 0) + 119892

lowast

1119863

1(119892

1 0)1003816100381610038161003816 = V (119878

2) = 1

(131)

Therefore 1 = 1198783 = 119878

3(119892

1 0) which proves Claim 2

Finally set 1198783= (0 ) for a suitable 119871

1(120583|

119860)oplus

1119883 rarr

119883 and define the operator 1198791 119871

1(120583) rarr 119883 by

1198791(119891) = 119879

0(119891120594

119860119888) + (119875

119860119891 0) (132)

for every 119891 isin 1198711(120583) Then we have

10038171003817100381710038171198791 (119891)1003817100381710038171003817 ⩽100381710038171003817100381711987901003817100381710038171003817

10038171003817100381710038171198911205941198601198881003817100381710038171003817 +1003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171198911205941198601003817100381710038171003817 =10038171003817100381710038171198911003817100381710038171003817 (133)

for every 119891 isin 1198711(120583) so 119879

1 ⩽ 1 Also

10038171003817100381710038171198791 (1198691198601198921)1003817100381710038171003817 =10038171003817100381710038171198783 (1198921 0)

1003817100381710038171003817 =100381710038171003817100381711987831003817100381710038171003817 = 1 (134)

so 1198791attains its norm at 119869

1198601198921isin 119871

1(120583) and

10038171003817100381710038171198691198601198921 minus 11989101003817100381710038171003817 =10038171003817100381710038171198921 minus 1198920

1003817100381710038171003817 lt 120576 (135)

We also have that for any 119891 isin 1198781198711(120583)

10038171003817100381710038171198790 (119891) minus 1198791 (119891)1003817100381710038171003817 =100381710038171003817100381710038171198790(119891120594

119860) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790 (119869119860119875119860119891) minus 119879119860 (119875119860119891)

1003817100381710038171003817

+10038171003817100381710038171003817119879119860(119875

119860119891) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790119869119860 minus 119879119860

1003817100381710038171003817 +10038171003817100381710038171198780 minus 1198783

1003817100381710038171003817

lt 120578 (1205760) + 6120576

(136)

Hence 1198790minus 119879

1 ⩽ 120578(120576

0) + 6120576 lt 120576

0

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Abstract and Applied Analysis 15

Acknowledgments

The authors thank Gilles Godefroy and Rafael Paya forfruitful conversations about the content of this paper andVladimir Kadets for providing Lemma 18 They also appre-ciate anonymous referees for careful reading and fruitfulsuggestions about revision The second author was sup-ported by Basic Science Research Program through theNational Research Foundation of Korea (NRF) fundedby the Ministry of Education Science and Technology(NRF-2012R1A1A1006869) The third author was partiallysupported by Spanish MICINN and FEDER project noMTM2012-31755 and by Junta de Andalucıa and FEDERGrants FQM-185 and P09-FQM-4911

References

[1] F F Bonsall and J Duncan Numerical Ranges of Operators onNormed Spaces and of Elements of Normed Algebras vol 2 ofLondon Mathematical Society Lecture Note Series 2 CambridgeUniversity Press London UK 1971

[2] F F Bonsall and J Duncan Numerical Ranges II CambridgeUniversity Press London UK 1973 London MathematicalSociety Lecture Notes Series 10

[3] B Sims On numerical range and its applications to Banachalgebras [PhD thesis] University of Newcastle New SouthWales Australia 1972

[4] I D Berg and B Sims ldquoDenseness of operators which attaintheir numerical radiusrdquoAustralian Mathematical Society A vol36 no 1 pp 130ndash133 1984

[5] C S Cardassi ldquoDensity of numerical radius attaining operatorson some reflexive spacesrdquo Bulletin of the Australian Mathemat-ical Society vol 31 no 1 pp 1ndash3 1985

[6] C S Cardassi ldquoNumerical radius attaining operatorsrdquo inBanach Spaces (Columbia Mo 1984) vol 1166 of Lecture Notesin Mathematics pp 11ndash14 Springer Berlin Germany 1985

[7] C S Cardassi ldquoNumerical radius-attaining operators on119862(119870)rdquoProceedings of the AmericanMathematical Society vol 95 no 4pp 537ndash543 1985

[8] MDAcostaOperadores que alcanzan su radio numerico [PhDthesis] Universidad de Granada Granada Spain 1990

[9] M D Acosta and R Paya ldquoNumerical radius attaining oper-ators and the Radon-Nikodym propertyrdquo The Bulletin of theLondon Mathematical Society vol 25 no 1 pp 67ndash73 1993

[10] R Paya ldquoA counterexample on numerical radius attainingoperatorsrdquo Israel Journal of Mathematics vol 79 no 1 pp 83ndash101 1992

[11] Y S Choi and S G Kim ldquoNorm or numerical radius attainingmultilinear mappings and polynomialsrdquo Journal of the LondonMathematical Society vol 54 no 1 pp 135ndash147 1996

[12] M D Acosta J Becerra Guerrero and M Ruiz GalanldquoNumerical-radius-attaining polynomialsrdquo The Quarterly Jour-nal of Mathematics vol 54 no 1 pp 1ndash10 2003

[13] M D Acosta and S G Kim ldquoDenseness of holomorphicfunctions attaining their numerical radiirdquo Israel Journal ofMathematics vol 161 pp 373ndash386 2007

[14] S G Kim and H J Lee ldquoNumerical peak holomorphicfunctions on Banach spacesrdquo Journal of Mathematical Analysisand Applications vol 364 no 2 pp 437ndash452 2010

[15] M D Acosta R M Aron D Garcıa and M MaestreldquoThe Bishop-Phelps-Bollobas theorem for operatorsrdquo Journal ofFunctional Analysis vol 254 no 11 pp 2780ndash2799 2008

[16] A J Guirao and O Kozhushkina ldquoThe Bishop-Phelps-Bollobasproperty for numerical radius in ℓ

1(119862)rdquo Studia Mathematica

vol 218 no 1 pp 41ndash54 2013[17] J Falco ldquoThe Bishop-Phelps-Bollobas property for numerical

radius on 1198711rdquo Journal of Mathematical Analysis and Applica-

tions vol 414 no 1 pp 125ndash133 2014[18] A Aviles A J Guirao and J Rodriguez ldquoOn the Bishop-

Phelps-Bollobas property for numerical radius in119862(119870)-spacesrdquoPreprint

[19] RM Aron Y S Choi S K Kim H J Lee andMMartin ldquoTheBishop-Phelps-Bollobas version of Lindenstrauss properties Aand Brdquo Preprint

[20] V Kadets M Martın and R Paya ldquoRecent progress and openquestions on the numerical index of Banach spacesrdquo Revista dela Real Academia de Ciencias Exactas Fısicas y Naturales A vol100 no 1-2 pp 155ndash182 2006

[21] M Martın J Merı and M Popov ldquoOn the numerical index ofreal 119871

119901(120583)-spacesrdquo Israel Journal of Mathematics vol 184 pp

183ndash192 2011[22] A Iwanik ldquoNorm attaining operators on Lebesgue spacesrdquo

Pacific Journal of Mathematics vol 83 no 2 pp 381ndash386 1979[23] HH SchaeferBanach Lattices and Positive Operators Springer

New York NY USA 1974[24] J Duncan C M McGregor J D Pryce and A J White ldquoThe

numerical index of a normed spacerdquo Journal of the LondonMathematical Society Second Series vol 2 pp 481ndash488 1970

[25] Y S Choi S K Kim H J Lee and M Martin ldquoThe Bishop-Phelps-Bollobas theorem for operators on 119871

1(120583)rdquo Preprint

[26] H E Lacey The Isometric Theory of Classical Banach SpacesSpringer New York NY USA 1974

[27] G Godefroy ldquoUniversal spaces for strictly convex Banachspacesrdquo Revista de la Real Academia de Ciencias Exactas Fısicasy Naturales A vol 100 no 1-2 pp 137ndash146 2006

[28] C Finet M Martın and R Paya ldquoNumerical index andrenormingrdquo Proceedings of the American Mathematical Societyvol 131 no 3 pp 871ndash877 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 9: Research Article On the Bishop-Phelps-Bollobás Property ...downloads.hindawi.com/journals/aaa/2014/479208.pdf · Research Article On the Bishop-Phelps-Bollobás Property for Numerical

Abstract and Applied Analysis 9

= intΩ

120594119861119895

(119909)

1198981(119861

119895)

(1198891003816100381610038161003816]1198881003816100381610038161003816

1

1198891198981

(119909))

minus1

1198891003816100381610038161003816]1198881003816100381610038161003816

1

(119909)

= intΩ

120594119861119895

(119909)

1198981(119861

119895)

1198891198981(119909) = 1

(70)

Hence we get ⟨119879]119891 119892⟩ = 1 which implies that 119879]119891 =119879] = 1 Finally10038171003817100381710038171003817119891 minus 119891

10038171003817100381710038171003817

⩽10038171003817100381710038171003817119891 minus 119891

1

10038171003817100381710038171003817+10038171003817100381710038171198911 minus 119891

1003817100381710038171003817

=

10038171003817100381710038171003817100381710038171003817100381710038171003817

sum

119895isin119869

120573119895

120594119861119895

1198981(119861

119895)

minus sum

119895isin119869

120573119895

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+ 120576

⩽ sum

119895isin119869

120573119895(

10038171003817100381710038171003817100381710038171003817100381710038171003817

120594119861119895

1198981(119861

119895)

minus

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

+

10038171003817100381710038171003817100381710038171003817100381710038171003817

120594119861119895

1198981(119861

119895)

minus

120594119861119895

1198981(119861

119895)

10038171003817100381710038171003817100381710038171003817100381710038171003817

) + 120576

= 2sum

119895isin119869

120573119895

1198981(119861

119895 119861

119895)

1198981(119861

119895)

+ 120576

⩽ 2sum

119895isin119869

120573119895

(1205762)1198981(119861

119895)

1198981(119861

119895)

+ 120576 ⩽120576

1 minus 1205762+ 120576 lt 3120576

(71)

We are now ready to present the proof of the main resultin the case of finite regular positive Borel measures

Proof of Proposition 10 Let 1205751= 120575

3

2(5sdot2

4) 120575

2= 120575

12

3(3

2sdot2

14)

and 1205753= (12057610)

2 for some 0 lt 120576 lt 1 Suppose that 119879 isinL(119871

1(119898)) with 119879 = 1 and that there is an 119891

0isin 119878

1198711(119898)

and1198920isin 119878

119871infin(119898)

such that ⟨1198910 119892

0⟩ = 1 and |⟨119879119891

0 119892

0⟩| gt 1minus120575

3

12

6Then there is an isometric isomorphism Ψ from 119871

1(119898) onto

itself such thatΨ(1198910) = |119891

0| and there is a scalar number 120572 in

119878R such that |⟨1198791198910 119892

0⟩| = ⟨120572119879119891

0 119892

0⟩ Then letting 119891

1= Ψ119891

0

1198921= (Ψ

minus1)lowast1198920 and 119879

1= 120572Ψ119879Ψ

minus1 we have

⟨1198781198911 119892

1⟩ = ⟨120572Ψ119879

minus1Ψ119891

0 (Ψ

minus1)lowast

1198920⟩

= ⟨1205721198791198910 119892

0⟩ gt 1 minus

1205753

1

26

⟨1198911 119892

1⟩ = ⟨Ψ119891

0 (Ψ

minus1)lowast

1198920⟩ = 1

(72)

Since 11987911198911 gt 1 minus 120575(120575

3

12

6) by Lemma 13 there exists a

norm-one bounded operator 119879] and a nonnegative simplefunction119891

2isin 119878

1198711(119898)

such that 1198791minus119879] ⩽ 1205751 1198912minus1198911 ⩽ 31205751

and (119889|]|11198891198981)(119909) = 1 for all 119909 isin supp(119891

2) Then

⟨119879]1198912 1198921⟩

= ⟨11987911198911 119892

1⟩ minus ⟨119879

11198911minus 119879

11198912 119892

1⟩ minus ⟨119879

11198912minus 119879]1198912 1198921⟩

⩾ ⟨11987911198911 119892

1⟩ minus10038171003817100381710038171198911 minus 1198912

1003817100381710038171003817 minus10038171003817100381710038171198791 minus 119879]

1003817100381710038171003817

⩾ 1 minus1205753

1

26minus 3120575

1minus 120575

1⩾ 1 minus 5120575

1= 1 minus

1205753

2

16

(73)

Notice also that

⟨1198912 119892

1⟩ = ⟨119891

1 119892

1⟩ minus ⟨119891

1minus 119891

2 119892

1⟩ ⩾ 1 minus

10038171003817100381710038171198911 minus 11989121003817100381710038171003817

⩾ 1 minus 31205751⩾ 1 minus 5120575

1= 1 minus

1205753

2

16

(74)

By Lemma 12 there are a nonnegative simple function 1198913isin

1198781198711(119898)

and a function 1198923isin 119878

119871infin(119898)

such that

1198923(119909) = 120594supp119891

3

(119909) + 1198922(119909) 120594

Ωsupp1198913

(119909)

10038171003817100381710038171198912 minus 11989131003817100381710038171003817 ⩽ 1205752

10038171003817100381710038171198923 minus 11989211003817100381710038171003817 ⩽radic120575

2 ⟨119891

3 119892

3⟩ = 1

(75)

So we have

⟨119879]1198913 1198923⟩ = ⟨119879]1198912 1198921⟩ minus ⟨119879]1198912 minus 119879]1198913 1198921⟩

minus ⟨119879]1198913 1198921 minus 1198923⟩

⩾ 1 minus1205753

2

16minus 2radic120575

2⩾ 1 minus 3radic120575

2= 1 minus

1205756

3

27

(76)

By Lemma 14 there exist 1198914isin 119878

1198711(119898)

and 1198924isin 119878

119871infin(119898)

and anoperator 119879

4such that

⟨1198924 119879

41198914⟩ = 1 =

100381710038171003817100381711987941003817100381710038171003817

10038171003817100381710038171198794 minus 119879]1003817100381710038171003817 ⩽ 21205753

10038171003817100381710038171198914 minus 11989131003817100381710038171003817 ⩽ 31205753

10038171003817100381710038171198924 minus 11989231003817100381710038171003817 ⩽radic120575

3 ⟨119891

4 119892

4⟩ = 1

(77)

So we have

10038171003817100381710038171198794 minus 11987911003817100381710038171003817 ⩽10038171003817100381710038171198794 minus 119879]

1003817100381710038171003817 +1003817100381710038171003817119879] minus 1198791

1003817100381710038171003817

⩽ 1205751+ 2120575

3⩽ 3120575

3

10038171003817100381710038171198911 minus 11989141003817100381710038171003817 ⩽10038171003817100381710038171198911 minus 1198912

1003817100381710038171003817 +10038171003817100381710038171198912 minus 1198913

1003817100381710038171003817 +10038171003817100381710038171198913 minus 1198914

1003817100381710038171003817

⩽ 31205751+ 120575

2+ 3120575

3⩽ 10120575

3

10038171003817100381710038171198921 minus 11989241003817100381710038171003817 ⩽10038171003817100381710038171198921 minus 1198923

1003817100381710038171003817 + 1198923 minus 1198924

⩽ 1205752+ radic120575

3⩽ 2radic120575

3

(78)

10 Abstract and Applied Analysis

Let 119878 = 120572Ψminus11198794Ψ 119891 = Ψminus1

1198914 and 119892 = Ψlowast

1198924 then we have

119879 minus 119878 =10038171003817100381710038171003817119879 minus 120572Ψ

minus11198794Ψ10038171003817100381710038171003817=10038171003817100381710038171003817120572Ψ119879Ψ

minus1minus 119879

4

10038171003817100381710038171003817

=10038171003817100381710038171198791 minus 1198794

1003817100381710038171003817 ⩽ 31205753

100381710038171003817100381710038171198910minus 11989110038171003817100381710038171003817=100381710038171003817100381710038171198910minus Ψ

minus11198914

10038171003817100381710038171003817=10038171003817100381710038171198911 minus 1198914

1003817100381710038171003817 ⩽ 101205753

100381710038171003817100381710038171198920minus 11989110038171003817100381710038171003817=10038171003817100381710038171198920 minus Ψ

lowast1198924

1003817100381710038171003817 =100381710038171003817100381710038171003817(Ψ

minus1)lowast

1198920minus 119892

4

100381710038171003817100381710038171003817

=10038171003817100381710038171198921 minus 1198924

1003817100381710038171003817 ⩽ 2radic120575

3

⟨119891 119892⟩ = ⟨Ψminus11198914 Ψ

lowast1198924⟩ = ⟨119891

4 119892

4⟩ = 1

10038161003816100381610038161003816⟨119878119891 119892⟩

10038161003816100381610038161003816=10038161003816100381610038161003816⟨120572Ψ

minus11198794ΨΨ

minus11198914 Ψ

lowast1198924⟩10038161003816100381610038161003816= |120572| = 1

(79)

This completes the proof

Finally we may give the proof of the main result in fullgenerality

Proof of Theorem 9 Notice that the Kakutani representationtheorem (see [26] for a reference) says that for every 120590-finitemeasure ] the space 119871

1(]) is isometrically isomorphic to

1198711(119898) for some positive Borel regular measure on a compact

Hausdorff space Then by Proposition 10 there is a universalfunction 120576 997891rarr 120578(120576) gt 0 which gives the BPBp-nu for 119871

1(]) for

every 120590-finite measure ]Fix 120576 gt 0 Suppose that 119879

0isin L(119871

1(120583)) with V(119879

0) = 1

and (1198910 119891

lowast

0) isin Π(119871

1(120583)) satisfy

1003816100381610038161003816⟨119891lowast

0 119879

01198910⟩1003816100381610038161003816 gt 1 minus 120578 (120576) (80)

Choose a sequence 119891119899 in 119871

1(120583) such that sup

119899119879

0119891119899 = 1

and let 119866 be the closed linear span of

119879119899

0119891119898 119899 119898 isin N cup 0 (81)

As 119866 is separable there is a dense subset 119892119899 119899 isin N of 119866

and let 119864 = ⋃infin

119899=1supp119892

119899 where supp119892

119899is the support of 119892

119899

Then the measure 120583|119864is 120590-finite Let

119884 = 119891 isin 1198711(120583) supp (119891) sub 119864 (82)

be a closed subspace of 1198711(120583) It is clear that 119871

1(120583) = 119884oplus

1119885

and 119884 is isometrically isomorphic to 1198711(120583|

119864) So 119884 has the

BPBp-nu with 120578(120576)Now write 119878

0= 119879

0|119884 119884 rarr 119884 consider 119910

0= 119891

0isin

119878119884 119910lowast

0= 119891

lowast

0|119884isin 119878

119884lowast and observe that 119910lowast

0(119910

0) = 1 and

|119910lowast

0(119878

01199100)| = |119891

lowast

0(119879

01198910)| gt 1 minus 120578(120576) Hence there exist 119878 isin

L(119884) and (1199100 119910

lowast

0) isin Π(119884) such that

1003816100381610038161003816119910lowast

0(119878119910

0)1003816100381610038161003816 = 1 = V (119878) 1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

10038171003817100381710038171199100 minus 11991001003817100381710038171003817 lt 120576

1003817100381710038171003817119910lowast

0minus 119910

lowast

0

1003817100381710038171003817 lt 120576

(83)

Finally consider the operator 119879 isinL(1198711(120583)) given by

119879 (119910 119911)

= (119878119910 0) + 1198790(0 119911) ((119910 119911) isin 119871

1(120583) equiv 119884oplus

1119885)

(84)

We have 119879 = 1 (and so V(119879) = 1) Indeed1003817100381710038171003817119879 (119910 119911)

1003817100381710038171003817 =1003817100381710038171003817(119878119910 0)

1003817100381710038171003817 +10038171003817100381710038171198790 (0 119911)

1003817100381710038171003817 ⩽10038171003817100381710038171199101003817100381710038171003817 + 119911 =

1003817100381710038171003817(119910 119911)1003817100381710038171003817

(85)

for all (119910 119911) isin 1198711(120583) and 119879(119910

0 0) = (119878119910

0 0) = 119878119910

0 = 1

Let 119909 = (1199100 0) and 119909lowast = (119910lowast

0 119891

0|119885) Then (119909 119909lowast) isin Π(119871

1(120583))

Moreover we have1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119910

lowast

0119878119910

0

1003816100381610038161003816 = 1 = V (119879) 1003817100381710038171003817119909 minus 1198910

1003817100381710038171003817 =1003817100381710038171003817119910 minus 1199100

1003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowast

0minus 119891

lowast

0

1003817100381710038171003817

= max 1003817100381710038171003817119910 minus 119891lowast

0|119884

1003817100381710038171003817 1003817100381710038171003817119891

lowast

0|119885minus 119891

lowast

0|119885

1003817100381710038171003817

=1003817100381710038171003817119910

lowastminus 119910

lowast

0

1003817100381710038171003817 lt 120576

1003817100381710038171003817119879 minus 11987901003817100381710038171003817 = sup119910+119911⩽1

1003817100381710038171003817119879 (119910 119911) minus 1198790 (119910 119911)1003817100381710038171003817

= sup119910⩽1

1003817100381710038171003817119878119910 minus 11987801199101003817100381710038171003817 =1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

(86)

This completes the proof

5 Examples of Spaces Failing theBishop-Phelps-Bollobaacutes Property forNumerical Radius

Our goal here is to prove that the density of numerical radiusattaining operators does not imply the BPBp-nu Actuallywe will show that among separable spaces there is noisomorphic property implying the BPBp-nu other than finite-dimensionality

We need to relate the BPBp-nu to the Bishop-Phelps-Bollobas property for operators which as mentioned in theintroduction was introduced in [15] A pair (119883 119884) of Banachspaces has the Bishop-Phelps-Bollobas property for operators(in short BPBp) if given 120576 gt 0 there exists 120578(120576) gt 0 suchthat given 119879 isin L(119883 119884) with 119879 = 1 and 119909 isin 119878

119883such that

119879119909 gt 1 minus 120578(120576) then there exist 119911 isin 119878119883and 119878 isin L(119883 119884)

satisfying

119878 = 119878119911 = 1 119909 minus 119911 lt 120576

119879 minus 119878 lt 120576

(87)

We refer the reader to [15 19 25] and references therein formore information and background Among the interestingresults on the BPBp we emphasize that a pair (119883 119884) when119883is finite-dimensional does not necessarily have the BPBp Forinstance if119884 is a strictly convex space which is not uniformlyconvex then the pair (ℓ(2)

1 119884) fails to have the BPBp (this is

contained in [15] see [19 Section 3])

Abstract and Applied Analysis 11

The next result relates the BPBp-nu to the BPBp foroperators in a particular case We will deduce our examplefrom it

Theorem 15 If 1198711(120583)oplus

1119883 has the BPBp-119899119906 then the pair

(1198711(120583) 119883) has the BPBp for operators

Before proving this proposition we will use it to get themain examples of this section The first example shows thatthe density of numerical radius attaining operators does notimply the BPBp-nu

Example 16 There is a reflexive space (and so numerical radiusattaining operators on it are dense) which fails to have theBPBp-nu Indeed let 119884 be a reflexive separable space whichis not superreflexive and we may suppose that 119884 is strictlyconvex Observe that 119884 cannot be uniformly convex since itis not superreflexive Now 119883 = ℓ(2)

1oplus1119884 is reflexive but the

pair (ℓ(2)1 119884) fails the BPBp since 119884 is strictly convex but not

uniformly convex [19 Corollary 33] Therefore Theorem 15gives us that119883 does not have the BPBp-nu

The example above can be extended to get the result thatevery infinite-dimensional separable Banach space can berenormed to fail the BPBp-nu This follows from the factthat every infinite-dimensional separable Banach space canbe renormed to be strictly convex but not uniformly convex(this result can be proved ldquoby handrdquo an alternative categoricalargument for it can be found in [27] and references therein)With a little more of effort we may get the main result of thesection

Theorem 17 Every infinite-dimensional separable Banachspace can be renormed to fail the weak-BPBp-nu (and so inparticular to fail the BPBp-nu)

We need the following result which is surely well knownAs we have not found a reference we include a nice and easyproof kindly given to us by Vladimir Kadets We recall thatgiven a Banach space 119884 the set of all equivalent norms on119884 can be viewed as a metric space using the Banach-Mazurdistance

Lemma 18 Let119884 be an infinite-dimensional separable Banachspace Then the set of equivalent norms on 119884 which are strictlyconvex and are not (locally) uniformly convex is dense in the setof all equivalent norms on119884 (with respect to the Banach-Mazurdistance)

Proof Fix 119890 isin 119878119884and 119890lowast

1isin 119878

119884lowast such that 119890lowast

1(119890) = 1 For a fixed

120576 isin (0 12) denote

119902 (119910) = max (1 minus 120576) 10038171003817100381710038171199101003817100381710038171003817 1003816100381610038161003816119890lowast

1(119910)1003816100381610038161003816 (119910 isin 119884) (88)

Evidently (1 minus 120576)119910 ⩽ 119902(119910) ⩽ 119910 for every 119910 isin 119884 Fix asequence 119890lowast

119896 119896 ⩾ 2 of norm-one functionals separating

the points of 119884 and denote

119901 (119910) = radic

infin

sum

119896=1

1

2119896

1003816100381610038161003816119890lowast

119896(119910)1003816100381610038161003816

2

(119910 isin 119884) (89)

Then 119901 is a strictly convex norm on 119884 119901(119890) ⩾ 1radic2 and119901(119910) ⩽ 119910 for all 119910 isin 119883 Finally write

100381710038171003817100381711991010038171003817100381710038171= (1 minus 120576) 119902 (119910) + 120576

119901 (119910)

119901 (119890)(119910 isin 119884) (90)

Then sdot 1is a strictly convex norm on 119884 and

(1 minus 120576)2 10038171003817100381710038171199101003817100381710038171003817 ⩽100381710038171003817100381711991010038171003817100381710038171⩽ (1 + 120576)

10038171003817100381710038171199101003817100381710038171003817 (119910 isin 119884) (91)

Wewill finish the proof by showing that sdot 1is not uniformly

convex (actually it is not locally uniformly convex) Indeedfor each 119899 isin N we select 119910

119899isin ⋂

119899

119896=1ker 119890lowast

119896with 119910

119899 = 1 and

consider 119890119899= 119890 + (1205764)119910

119899 Then 119902(119890) = 1 119902(119890

119899) = 1 and

119902(119890+119890119899) = 2 At the same time 119901(119910

119899) rarr 0 so 119901(119890

119899) rarr 119901(119890)

and 119901(119890 + 119890119899) rarr 2119901(119890) Consequently

1198901 = 1100381710038171003817100381711989011989910038171003817100381710038171997888rarr 1

1003817100381710038171003817119890 + 11989011989910038171003817100381710038171997888rarr 2 (92)

but 119890 minus 1198901198991= (1205764)119910

1198991⩾ (1 minus 120576)

2(1205764) which means the

absence of local uniform convexity at 119890

Proof of Theorem 17 Let 119883 be an infinite-dimensional sepa-rable Banach space Take a closed subspace 119884 of 119883 of codi-mension two By [28 Proposition 2] the map carrying everyequivalent norm on 119884 to its numerical index is continuousand so the set of values of the numerical index of 119884 upto reforming is a nontrivial interval [28 Theorem 9] ThenLemma 18 allows us to find an equivalent norm | sdot | on 119884 insuch a way that (119884 | sdot |) is strictly convex and is not uniformlyconvex and 119899(119884 | sdot |) gt 0 Now the space 119883 = ℓ(2)

1oplus1(119884 | sdot |)

is an equivalent renorming of 119883 which does not have theBPBp-119899119906 (indeed otherwise the pair (ℓ(2)

1 (119884 | sdot |)) would

have the BPBp for the operator norm and so (119884 | sdot |) wouldbe uniformly convex by [19 Corollary 33] a contradiction)Moreover as

119899 (119883) = min 119899 (ℓ(2)1) 119899 (119884 |sdot|) gt 0 (93)

(see [20 Proposition 2] for instance) 119883 also fails the weak-BPBp-nu by Proposition 6

To finish the section with the promised proof ofTheorem 15 we first see the following stability result

Lemma 19 Let119883 = [⨁infin

119896=1119883

119896]1198880

or [⨁infin

119896=1119883

119896]ℓ1

If119883 has theBishop-Phelps-Bollobas property for numerical radius with afunction 120578 then each Banach space 119883

119894has the Bishop-Phelps-

Bollobas property for numerical radius with 120578nu(119883119894) ⩾ 120578 That

is inf119894120578nu(119883119894

)(120576) ⩾ 120578nu(119883)(120576) for all 0 lt 120576 lt 1

Proof Let 119875119894 119883 rarr 119883

119894and 1198751015840

119894 119883

lowastrarr 119883

lowast

119894be the natural

projections and let 119876119894 119883

119894rarr 119883 and 1198761015840

119894 119883

lowast

119894rarr 119883

lowast be thenatural embeddings

Assume that an operator 119879119894 119883

119894rarr 119883

119894and a pair

(119909119894 119909

lowast

119894) isin Π(119883

119894) satisfy that

V (119879119894) = 1

1003816100381610038161003816119909lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (94)

We define an operator 119879 119883 rarr 119883 and (119909 119909lowast) isin Π(119883) by

119879 = 119876119894∘ 119879

119894∘ 119875

119894 (119909 119909

lowast) = (119876

119894119909119894 119876

1015840

119894119909lowast

119894) (95)

12 Abstract and Applied Analysis

then clearly we see that1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119909

lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (96)

From the assumption there exist 119878 119883 rarr 119883 and a pair(119910 119910

lowast) isin Π(119883) such that

1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 = 1 = V (119878) 119878 minus 119879 lt 120576

1003817100381710038171003817119910lowastminus 119909

lowast1003817100381710038171003817 lt 1205761003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576

(97)

Since this clearly shows that1003817100381710038171003817119875119894 ∘ 119878 ∘ 119876119894

minus 119879119894

1003817100381710038171003817 lt 120576100381710038171003817100381710038171198751015840

119894119910lowastminus 119909

lowast

119894

10038171003817100381710038171003817lt 120576

1003817100381710038171003817119875119894119910 minus 1199091198941003817100381710038171003817 lt 120576

(98)

we only need to show that |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = 1

We first show the case of 1198880sum Since

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817=10038171003817100381710038171003817119875119895119910 minus 119875

11989511990910038171003817100381710038171003817⩽1003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576 (99)

for every 119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽100381710038171003817100381710038171198751015840

119894119910lowast10038171003817100381710038171003817+ 120576 sum

119895isinN119895 = 119894

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817⩽1003817100381710038171003817119910

lowast1003817100381710038171003817 = 1

(100)

This shows that 1198751015840119894119910lowast = 1 and 1198751015840

119895119910lowast= 0 for every 119895 = 119894

So 119910lowast = 1198761015840

1198941198751015840

119894119910lowast and 1198751015840

119894119910lowast(119875

119894119910) = 1 This and the fact that

119910 minus 119876119894119875119894119910 lt 120576 imply that

(119876119894119875119894119910 + (

1

120576) (119910 minus 119876

119894119875119894119910) 119876

1015840

1198941198751015840

119894119910lowast) isin Π (119883) (101)

So we get that (1198761015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910 + (1120576)(119910 minus 119876

119894119875119894119910))V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =10038161003816100381610038161003816(119876

1015840

1198941198751015840

119894119910lowast) 11987811991010038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910 +1

120576(119910 minus 119876

119894119875119894119910))

1003816100381610038161003816100381610038161003816⩽ 1

(102)

and so we get |1198751015840119894119910lowast(119875

119894∘119878 ∘119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

We next show the case of ℓ1sum The proof is almost

the same as that of the 1198880case However for the sake of

completeness we provide it hereSince 1198751015840

119895119910lowast = 119875

1015840

119895119910lowastminus 119875

1015840

119895119909lowast ⩽ 119910

lowastminus 119909

lowast lt 120576 for every

119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽10038171003817100381710038171198751198941199101003817100381710038171003817 + 120576 sum

119895isinN119895 = 119894

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817⩽10038171003817100381710038171199101003817100381710038171003817 = 1

(103)

which shows 119875119894119910 = 1 and 119875

119895119910 = 0 for every 119895 = 119894 Since this

implies (119876119894119875119894119910119876

1015840

1198941198751015840

119894119910lowast+(1120576)(119910

lowastminus119876

1015840

1198941198751015840

119894119910lowast)) isin Π(119883) we get

that |(1198761015840

1198941198751015840

119894119910lowast+ (1120576)(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast))119878(119876

119894119875119894119910)| ⩽ V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =1003816100381610038161003816119910

lowast119878 (119876

119894119875119894119910)1003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast+1

120576(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast)) 119878 (119876

119894119875119894119910)

1003816100381610038161003816100381610038161003816⩽ 1

(104)

and so |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

Proof of Theorem 15 Note that 120578nu(1198711(120583)oplus1119883)(120576) rarr 0 as120576 rarr 0 Fix 0 lt 120576

0lt 1 and choose 0 lt 120576 lt 1 such that

6120576 + 120578nu(1198711(120583)oplus1119883)(120576) lt 1205760 Let 120578(1205760) = 120578nu(1198711(120583)oplus1119883)(120576)Suppose that 119879

0isin L(119871

1(120583) 119883) with 119879

0 = 1 and 119891

0isin

1198781198711(120583)

satisfy

1003817100381710038171003817119879011989101003817100381710038171003817 gt 1 minus 120578 (1205760) (105)

For any measurable subset 119861 let

1198711(120583|

119861) = 119891|

119861 119891 isin 119871

1(120583) (106)

with the norm 119891|119861 = 119891120594

1198611 Then it is easy to see

that 1198711(120583|

119861) is isometrically isomorphic to a complemented

subspace of 1198711(120583) Let 119875

119861 119871

1(120583) rarr 119871

1(120583|

119861) be the

restriction defined by 119875119861(119891) = 119891|

119861for all 119891 isin 119871

1(120583) and

let 119869119861 119871

1(120583|

119861) rarr 119871

1(120583) be the extension defined by

119869119861(119891)(120596) = 119891(120596) if 120596 isin 119861 and 119869

119861(119891)(120596) = 0 otherwise

It is clear that 119875119861119869119861= Id

1198711(120583|119861)and 119869

119861119875119861(119891) = 119891120594

119861for all

119891 isin 1198711(120583) Notice also that 119871

1(120583) is isometrically isomorphic

to 1198711(120583|

119861)oplus

11198711(120583|

119861119888)

Let 119860 = supp1198910and 119892

0= 119875

1198601198910 Then

1003817100381710038171003817119879011986911986011989201003817100381710038171003817 =100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) gt 0 (107)

and define the operator 119879119860 119871

1(120583|

119860) rarr 119883 by 119879

119860119891 =

1198790119869119860119891119879

0119869119860 for every 119891 isin 119871

1(120583|

119860) Then

100381710038171003817100381711987911986011989201003817100381710038171003817 ⩾100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) (108)

Since 120583|119860is 120590-finite 119871

1(120583|

119860)lowast= 119871

infin(120583|

119860) Let 119892lowast

0isin 119878

119871infin(120583|119860)

be a function such that ⟨119892lowast0 119892

0⟩ = 1 and choose 119909lowast

0isin 119878

119883lowast

such that 119909lowast0(119879

1198601198920) = 119879

1198601198920 Define the operator 119878

0isin

L(1198711(120583|

119860)oplus

1119883) by

1198780(119891 119909) = (0 119879

119860119891) ((119891 119909) isin 119871

1(120583|

119860) oplus

1119883) (109)

and observe that 1198780 = V(119878

0) = 1 Indeed

100381710038171003817100381711987801003817100381710038171003817 ⩽ 1 =

10038171003817100381710038171198791198601003817100381710038171003817

= sup 1003816100381610038161003816119909lowast1198791198601198911003816100381610038161003816 119909

lowastisin 119878

119883lowast 119891 isin 119878

1198711(120583|119860)

= sup 1003816100381610038161003816(119891lowast 119909

lowast) 119878

0(119891 119909)

1003816100381610038161003816

((119891lowast 119909

lowast) (119891 119909)) isin Π (119871

1(120583|

119860) oplus

1119883)

= V (1198780) ⩽100381710038171003817100381711987801003817100381710038171003817

(110)

Abstract and Applied Analysis 13

It is immediate that

(119892lowast

0 119909

lowast

0) 119878

0(119892

0 0) = 119909

lowast

0(119879

1198601198920) =10038171003817100381710038171198791198601198920

1003817100381710038171003817 gt 1 minus 120578 (1205760)

(111)

By Lemma 19 1198711(120583|

119860)oplus

1119883 has the BPBp-nu with the func-

tion 120578 Therefore there exist 1198781isin L(119871

1(120583|

119860)oplus

1119883) (119892

1 119909

1) isin

1198781198711(120583|119860)oplus1119883 and (119892lowast

1 119909

lowast

1) isin 119878

119871infin(120583|119860)oplusinfin119883lowast such that

1003817100381710038171003817(1198921 1199091) minus (1198920 0)1003817100381710038171003817 lt 120576

1003817100381710038171003817(119892lowast

1 119909

lowast

1) minus (119891

lowast

0 119909

lowast

0)1003817100381710038171003817 lt 120576

10038171003817100381710038171198781 minus 11987801003817100381710038171003817 lt 120576 ⟨(119892

lowast

1 119909

lowast

1) (119892

1 119909

1)⟩ = 1

1003816100381610038161003816(119892lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 = V (119878

1) = 1

(112)

Claim 1 We claim that 1199091= 0

Otherwise

1 = Re ⟨(119892lowast1 119909

lowast

1) (119892

1 119909

1)⟩

=100381710038171003817100381711989211003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

+100381710038171003817100381711990911003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

⟩ ⩽ 1

(113)

We deduce that

((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

(119892lowast

1 119909

lowast

1)) (

(0 1199091)

100381710038171003817100381711990911003817100381710038171003817

(119892lowast

1 119909

lowast

1))

isin Π (1198711(120583|

119860) oplus

1119883)

(114)

Since 1198781((0 119909

1)119909

1) = (119878

1minus 119878

0)((0 119909

1)119909

1) lt 120576 we

get that

1 =1003816100381610038161003816⟨(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)⟩1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

100381710038171003817100381711989211003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

)⟩

+100381710038171003817100381711990911003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

)⟩

100381610038161003816100381610038161003816100381610038161003816

⩽100381710038171003817100381711989211003817100381710038171003817 V (1198781) + 120576

100381710038171003817100381711990911003817100381710038171003817 lt100381710038171003817100381711989211003817100381710038171003817 +100381710038171003817100381711990911003817100381710038171003817 = 1

(115)

a contradiction This proves the claimWe define the operator 119878

2 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883

by 1198782(119891 119909) = 119878

1(119891 0) for every 119891 isin 119871

1(120583|

119860) and for every

119909 isin 119883 Then we have

V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 = 1

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ 120576 (116)

Indeed from Claim 1 we have

V (1198781) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 ⩽ V (119878

2)

(117)

On the other hand we see that1003816100381610038161003816(119891

lowast 119909

lowast) 119878

2(119891 119909)

1003816100381610038161003816

=1003816100381610038161003816(119891

lowast 119909

lowast) 119878

1(119891 0)

1003816100381610038161003816 ⩽10038171003817100381710038171198911003817100381710038171003817 V (1198781) ⩽ V (119878

1)

(118)

for every ((119891lowast 119909lowast) (119891 119909)) isin Π(1198711(120583|

119860)oplus

1119883) Therefore

V(1198782) ⩽ V(119878

1) Also

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ sup

119909isin119878119883

10038171003817100381710038171198781 (0 119909)1003817100381710038171003817

= sup119909isin119878119883

10038171003817100381710038171198781 (0 119909) minus 1198780 (0 119909)1003817100381710038171003817 ⩽ 120576

(119)

Claim 2 There exists 1198783 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883 such

that 1198783(119892

1 0) = 119878

3 = 1 119878

3(0 119909) = 0 119878

3(119891 119909) isin 0oplus

1119883

for every (119891 119909) isin 1198711(120583|

119860)oplus

1119883 and 119878

3minus 119878

2 lt 4120576

Indeed write 1198781= (119863

1 119863

2) where 119863

1 119871

1(120583|

119860)oplus

1119883 rarr

1198711(120583|

119860) and119863

2 119871

1(120583|

119860)oplus

1119883 rarr 119883 We have that

sup 1003816100381610038161003816119892lowast119863

1(119892

1 0) + 119909

lowast119863

2(119892

1 0)1003816100381610038161003816

119909lowastisin 119878

119883lowast ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860) +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⩽ V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119892

lowast

1119863

1(119892

1 0) + 119909

lowast

1119863

2(119892

1 0)1003816100381610038161003816

(120)

This implies that

1003816100381610038161003816119909lowast

1119863

2(119892

1 0)1003816100381610038161003816 =10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

1003816100381610038161003816119892lowast

1119863

1(119892

1 0)1003816100381610038161003816

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119871

infin(120583|

119860)

(121)

Therefore |119892lowast1| equals 1 on the support of 119863

1(119892

1 0) As

|⟨119892lowast

1 119892

1⟩| = 1 we also have that |119892lowast

1| equals 1 on the support

of 1198921 Changing the values of 119892lowast

1by the ones of 119891lowast

0on 119860

(supp(1198631(119892

1 0)) cup supp(g

1)) we may and do suppose that

|119892lowast

1| = 1 on the whole 119860We also have 119863

2(119892

1 0) gt 0 Indeed

10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)1003817100381710038171003817

⩽10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198921 0)

1003817100381710038171003817 +10038171003817100381710038171198780 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817

lt 2120576 + 120576 = 3120576

(122)

So we have

10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

⩽10038171003817100381710038171198631(119892

1 0)1003817100381710038171003817 +10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

14 Abstract and Applied Analysis

=1003817100381710038171003817(1198631

(1198921 0) 119863

2(119892

1 0)) minus (0 119879

1198601198920)1003817100381710038171003817

=10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817 lt 3120576

(123)

and 1198632(119892

1 0) gt 119879

1198601198920 minus 3120576 ⩾ 1 minus 120578(120576

0) minus 3120576 gt 0

Finally define the operator 1198783by

1198783(119891 119909) = (0119863

2(119891 0) + 119892

lowast

1(119863

1(119891 0))

1198632(119892

1 0)

119909lowast

1119863

2(119892

1 0))

(124)

for (119891 119909) isin 1198711(120583|

119860)oplus

1119883 It is clear that

100381710038171003817100381711987831003817100381710038171003817 ⩽ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) (125)

Notice also that

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817

⩽10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 +10038171003817100381710038171198632(119891 0) minus 119879

1198601198911003817100381710038171003817

=1003817100381710038171003817(1198631

(119891 0) 1198632(119891 0)) minus (0 119879

119860119891)1003817100381710038171003817

=10038171003817100381710038171198782 (119891 119909) minus 1198780 (119891 119909)

1003817100381710038171003817

(126)

for all (119891 119909) isin 1198711(120583|

119860)oplus

1119883 Hence we have

10038171003817100381710038171198783 minus 11987821003817100381710038171003817

= 2 sup119891isin1198781198711(120583|119860)

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 ⩽ 210038171003817100381710038171198782 minus 1198780

1003817100381710038171003817 lt 4120576(127)

On the other hand let 119866 1198711(120583|

119860) rarr 119871

1(120583|

119860) be defined by

119866(119891) = 119892lowast

1119891 for every 119891 isin 119871

1(120583|

119860) Then we have

V (1198782)

= sup 1003816100381610038161003816119911lowast11987821199111003816100381610038161003816 (119911 119911

lowast) isin Π (119871

1(120583|

119860) oplus

1119883)

⩾ sup10038161003816100381610038161003816100381610038161003816

119909lowast119863

2(119866(

1

120583 (119862)120594119862) 0)

+119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119909lowastisin 119878

119883lowast 119862 isin Σ

119860 120583 (119862) gt 0

= sup10038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119862)120594119862) 0)

10038171003817100381710038171003817100381710038171003817

+

10038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119862 isin Σ119860 120583 (119862) gt 0

(128)

where Σ119860is the family of measurable subsets of 119860

Hence for any simple function 119904 = sum119899

119894=1((120572

119894120583(119860

119894))120594

119860119894

) isin

1198781198711(120583|119860) where 119860

119894119894is a family of disjoint measurable subsets

with strictly positive measure we have

V (1198782)

119899

sum

119894=1

10038161003816100381610038161205721198941003816100381610038161003816 (

100381710038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381710038171003817100381710038171003817100381710038171003817

+

100381610038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381610038161003816100381610038161003816100381610038161003816

)

⩾10038171003817100381710038171198632(119866 (119904) 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119866 (119904) 0)

1003816100381610038161003816

(129)

Since |119892lowast1| = 1119866 is an isometric isomorphism so for each119891 isin

1198781198711(120583|119860)there exists a sequence of norm-one simple functions

(119904119896) such that 119866(119904

119896) converges to 119891 Therefore

V (1198782) ⩾ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) ⩾100381710038171003817100381711987831003817100381710038171003817

(130)

On the other hand we see that100381710038171003817100381711987831003817100381710038171003817 ⩾1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

3(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119909

lowast

1119863

2(119892

1 0) + 119892

lowast

1119863

1(119892

1 0)1003816100381610038161003816 = V (119878

2) = 1

(131)

Therefore 1 = 1198783 = 119878

3(119892

1 0) which proves Claim 2

Finally set 1198783= (0 ) for a suitable 119871

1(120583|

119860)oplus

1119883 rarr

119883 and define the operator 1198791 119871

1(120583) rarr 119883 by

1198791(119891) = 119879

0(119891120594

119860119888) + (119875

119860119891 0) (132)

for every 119891 isin 1198711(120583) Then we have

10038171003817100381710038171198791 (119891)1003817100381710038171003817 ⩽100381710038171003817100381711987901003817100381710038171003817

10038171003817100381710038171198911205941198601198881003817100381710038171003817 +1003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171198911205941198601003817100381710038171003817 =10038171003817100381710038171198911003817100381710038171003817 (133)

for every 119891 isin 1198711(120583) so 119879

1 ⩽ 1 Also

10038171003817100381710038171198791 (1198691198601198921)1003817100381710038171003817 =10038171003817100381710038171198783 (1198921 0)

1003817100381710038171003817 =100381710038171003817100381711987831003817100381710038171003817 = 1 (134)

so 1198791attains its norm at 119869

1198601198921isin 119871

1(120583) and

10038171003817100381710038171198691198601198921 minus 11989101003817100381710038171003817 =10038171003817100381710038171198921 minus 1198920

1003817100381710038171003817 lt 120576 (135)

We also have that for any 119891 isin 1198781198711(120583)

10038171003817100381710038171198790 (119891) minus 1198791 (119891)1003817100381710038171003817 =100381710038171003817100381710038171198790(119891120594

119860) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790 (119869119860119875119860119891) minus 119879119860 (119875119860119891)

1003817100381710038171003817

+10038171003817100381710038171003817119879119860(119875

119860119891) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790119869119860 minus 119879119860

1003817100381710038171003817 +10038171003817100381710038171198780 minus 1198783

1003817100381710038171003817

lt 120578 (1205760) + 6120576

(136)

Hence 1198790minus 119879

1 ⩽ 120578(120576

0) + 6120576 lt 120576

0

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Abstract and Applied Analysis 15

Acknowledgments

The authors thank Gilles Godefroy and Rafael Paya forfruitful conversations about the content of this paper andVladimir Kadets for providing Lemma 18 They also appre-ciate anonymous referees for careful reading and fruitfulsuggestions about revision The second author was sup-ported by Basic Science Research Program through theNational Research Foundation of Korea (NRF) fundedby the Ministry of Education Science and Technology(NRF-2012R1A1A1006869) The third author was partiallysupported by Spanish MICINN and FEDER project noMTM2012-31755 and by Junta de Andalucıa and FEDERGrants FQM-185 and P09-FQM-4911

References

[1] F F Bonsall and J Duncan Numerical Ranges of Operators onNormed Spaces and of Elements of Normed Algebras vol 2 ofLondon Mathematical Society Lecture Note Series 2 CambridgeUniversity Press London UK 1971

[2] F F Bonsall and J Duncan Numerical Ranges II CambridgeUniversity Press London UK 1973 London MathematicalSociety Lecture Notes Series 10

[3] B Sims On numerical range and its applications to Banachalgebras [PhD thesis] University of Newcastle New SouthWales Australia 1972

[4] I D Berg and B Sims ldquoDenseness of operators which attaintheir numerical radiusrdquoAustralian Mathematical Society A vol36 no 1 pp 130ndash133 1984

[5] C S Cardassi ldquoDensity of numerical radius attaining operatorson some reflexive spacesrdquo Bulletin of the Australian Mathemat-ical Society vol 31 no 1 pp 1ndash3 1985

[6] C S Cardassi ldquoNumerical radius attaining operatorsrdquo inBanach Spaces (Columbia Mo 1984) vol 1166 of Lecture Notesin Mathematics pp 11ndash14 Springer Berlin Germany 1985

[7] C S Cardassi ldquoNumerical radius-attaining operators on119862(119870)rdquoProceedings of the AmericanMathematical Society vol 95 no 4pp 537ndash543 1985

[8] MDAcostaOperadores que alcanzan su radio numerico [PhDthesis] Universidad de Granada Granada Spain 1990

[9] M D Acosta and R Paya ldquoNumerical radius attaining oper-ators and the Radon-Nikodym propertyrdquo The Bulletin of theLondon Mathematical Society vol 25 no 1 pp 67ndash73 1993

[10] R Paya ldquoA counterexample on numerical radius attainingoperatorsrdquo Israel Journal of Mathematics vol 79 no 1 pp 83ndash101 1992

[11] Y S Choi and S G Kim ldquoNorm or numerical radius attainingmultilinear mappings and polynomialsrdquo Journal of the LondonMathematical Society vol 54 no 1 pp 135ndash147 1996

[12] M D Acosta J Becerra Guerrero and M Ruiz GalanldquoNumerical-radius-attaining polynomialsrdquo The Quarterly Jour-nal of Mathematics vol 54 no 1 pp 1ndash10 2003

[13] M D Acosta and S G Kim ldquoDenseness of holomorphicfunctions attaining their numerical radiirdquo Israel Journal ofMathematics vol 161 pp 373ndash386 2007

[14] S G Kim and H J Lee ldquoNumerical peak holomorphicfunctions on Banach spacesrdquo Journal of Mathematical Analysisand Applications vol 364 no 2 pp 437ndash452 2010

[15] M D Acosta R M Aron D Garcıa and M MaestreldquoThe Bishop-Phelps-Bollobas theorem for operatorsrdquo Journal ofFunctional Analysis vol 254 no 11 pp 2780ndash2799 2008

[16] A J Guirao and O Kozhushkina ldquoThe Bishop-Phelps-Bollobasproperty for numerical radius in ℓ

1(119862)rdquo Studia Mathematica

vol 218 no 1 pp 41ndash54 2013[17] J Falco ldquoThe Bishop-Phelps-Bollobas property for numerical

radius on 1198711rdquo Journal of Mathematical Analysis and Applica-

tions vol 414 no 1 pp 125ndash133 2014[18] A Aviles A J Guirao and J Rodriguez ldquoOn the Bishop-

Phelps-Bollobas property for numerical radius in119862(119870)-spacesrdquoPreprint

[19] RM Aron Y S Choi S K Kim H J Lee andMMartin ldquoTheBishop-Phelps-Bollobas version of Lindenstrauss properties Aand Brdquo Preprint

[20] V Kadets M Martın and R Paya ldquoRecent progress and openquestions on the numerical index of Banach spacesrdquo Revista dela Real Academia de Ciencias Exactas Fısicas y Naturales A vol100 no 1-2 pp 155ndash182 2006

[21] M Martın J Merı and M Popov ldquoOn the numerical index ofreal 119871

119901(120583)-spacesrdquo Israel Journal of Mathematics vol 184 pp

183ndash192 2011[22] A Iwanik ldquoNorm attaining operators on Lebesgue spacesrdquo

Pacific Journal of Mathematics vol 83 no 2 pp 381ndash386 1979[23] HH SchaeferBanach Lattices and Positive Operators Springer

New York NY USA 1974[24] J Duncan C M McGregor J D Pryce and A J White ldquoThe

numerical index of a normed spacerdquo Journal of the LondonMathematical Society Second Series vol 2 pp 481ndash488 1970

[25] Y S Choi S K Kim H J Lee and M Martin ldquoThe Bishop-Phelps-Bollobas theorem for operators on 119871

1(120583)rdquo Preprint

[26] H E Lacey The Isometric Theory of Classical Banach SpacesSpringer New York NY USA 1974

[27] G Godefroy ldquoUniversal spaces for strictly convex Banachspacesrdquo Revista de la Real Academia de Ciencias Exactas Fısicasy Naturales A vol 100 no 1-2 pp 137ndash146 2006

[28] C Finet M Martın and R Paya ldquoNumerical index andrenormingrdquo Proceedings of the American Mathematical Societyvol 131 no 3 pp 871ndash877 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Mathematical PhysicsAdvances in

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Stochastic AnalysisInternational Journal of

Page 10: Research Article On the Bishop-Phelps-Bollobás Property ...downloads.hindawi.com/journals/aaa/2014/479208.pdf · Research Article On the Bishop-Phelps-Bollobás Property for Numerical

10 Abstract and Applied Analysis

Let 119878 = 120572Ψminus11198794Ψ 119891 = Ψminus1

1198914 and 119892 = Ψlowast

1198924 then we have

119879 minus 119878 =10038171003817100381710038171003817119879 minus 120572Ψ

minus11198794Ψ10038171003817100381710038171003817=10038171003817100381710038171003817120572Ψ119879Ψ

minus1minus 119879

4

10038171003817100381710038171003817

=10038171003817100381710038171198791 minus 1198794

1003817100381710038171003817 ⩽ 31205753

100381710038171003817100381710038171198910minus 11989110038171003817100381710038171003817=100381710038171003817100381710038171198910minus Ψ

minus11198914

10038171003817100381710038171003817=10038171003817100381710038171198911 minus 1198914

1003817100381710038171003817 ⩽ 101205753

100381710038171003817100381710038171198920minus 11989110038171003817100381710038171003817=10038171003817100381710038171198920 minus Ψ

lowast1198924

1003817100381710038171003817 =100381710038171003817100381710038171003817(Ψ

minus1)lowast

1198920minus 119892

4

100381710038171003817100381710038171003817

=10038171003817100381710038171198921 minus 1198924

1003817100381710038171003817 ⩽ 2radic120575

3

⟨119891 119892⟩ = ⟨Ψminus11198914 Ψ

lowast1198924⟩ = ⟨119891

4 119892

4⟩ = 1

10038161003816100381610038161003816⟨119878119891 119892⟩

10038161003816100381610038161003816=10038161003816100381610038161003816⟨120572Ψ

minus11198794ΨΨ

minus11198914 Ψ

lowast1198924⟩10038161003816100381610038161003816= |120572| = 1

(79)

This completes the proof

Finally we may give the proof of the main result in fullgenerality

Proof of Theorem 9 Notice that the Kakutani representationtheorem (see [26] for a reference) says that for every 120590-finitemeasure ] the space 119871

1(]) is isometrically isomorphic to

1198711(119898) for some positive Borel regular measure on a compact

Hausdorff space Then by Proposition 10 there is a universalfunction 120576 997891rarr 120578(120576) gt 0 which gives the BPBp-nu for 119871

1(]) for

every 120590-finite measure ]Fix 120576 gt 0 Suppose that 119879

0isin L(119871

1(120583)) with V(119879

0) = 1

and (1198910 119891

lowast

0) isin Π(119871

1(120583)) satisfy

1003816100381610038161003816⟨119891lowast

0 119879

01198910⟩1003816100381610038161003816 gt 1 minus 120578 (120576) (80)

Choose a sequence 119891119899 in 119871

1(120583) such that sup

119899119879

0119891119899 = 1

and let 119866 be the closed linear span of

119879119899

0119891119898 119899 119898 isin N cup 0 (81)

As 119866 is separable there is a dense subset 119892119899 119899 isin N of 119866

and let 119864 = ⋃infin

119899=1supp119892

119899 where supp119892

119899is the support of 119892

119899

Then the measure 120583|119864is 120590-finite Let

119884 = 119891 isin 1198711(120583) supp (119891) sub 119864 (82)

be a closed subspace of 1198711(120583) It is clear that 119871

1(120583) = 119884oplus

1119885

and 119884 is isometrically isomorphic to 1198711(120583|

119864) So 119884 has the

BPBp-nu with 120578(120576)Now write 119878

0= 119879

0|119884 119884 rarr 119884 consider 119910

0= 119891

0isin

119878119884 119910lowast

0= 119891

lowast

0|119884isin 119878

119884lowast and observe that 119910lowast

0(119910

0) = 1 and

|119910lowast

0(119878

01199100)| = |119891

lowast

0(119879

01198910)| gt 1 minus 120578(120576) Hence there exist 119878 isin

L(119884) and (1199100 119910

lowast

0) isin Π(119884) such that

1003816100381610038161003816119910lowast

0(119878119910

0)1003816100381610038161003816 = 1 = V (119878) 1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

10038171003817100381710038171199100 minus 11991001003817100381710038171003817 lt 120576

1003817100381710038171003817119910lowast

0minus 119910

lowast

0

1003817100381710038171003817 lt 120576

(83)

Finally consider the operator 119879 isinL(1198711(120583)) given by

119879 (119910 119911)

= (119878119910 0) + 1198790(0 119911) ((119910 119911) isin 119871

1(120583) equiv 119884oplus

1119885)

(84)

We have 119879 = 1 (and so V(119879) = 1) Indeed1003817100381710038171003817119879 (119910 119911)

1003817100381710038171003817 =1003817100381710038171003817(119878119910 0)

1003817100381710038171003817 +10038171003817100381710038171198790 (0 119911)

1003817100381710038171003817 ⩽10038171003817100381710038171199101003817100381710038171003817 + 119911 =

1003817100381710038171003817(119910 119911)1003817100381710038171003817

(85)

for all (119910 119911) isin 1198711(120583) and 119879(119910

0 0) = (119878119910

0 0) = 119878119910

0 = 1

Let 119909 = (1199100 0) and 119909lowast = (119910lowast

0 119891

0|119885) Then (119909 119909lowast) isin Π(119871

1(120583))

Moreover we have1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119910

lowast

0119878119910

0

1003816100381610038161003816 = 1 = V (119879) 1003817100381710038171003817119909 minus 1198910

1003817100381710038171003817 =1003817100381710038171003817119910 minus 1199100

1003817100381710038171003817 lt 120576

1003817100381710038171003817119909lowast

0minus 119891

lowast

0

1003817100381710038171003817

= max 1003817100381710038171003817119910 minus 119891lowast

0|119884

1003817100381710038171003817 1003817100381710038171003817119891

lowast

0|119885minus 119891

lowast

0|119885

1003817100381710038171003817

=1003817100381710038171003817119910

lowastminus 119910

lowast

0

1003817100381710038171003817 lt 120576

1003817100381710038171003817119879 minus 11987901003817100381710038171003817 = sup119910+119911⩽1

1003817100381710038171003817119879 (119910 119911) minus 1198790 (119910 119911)1003817100381710038171003817

= sup119910⩽1

1003817100381710038171003817119878119910 minus 11987801199101003817100381710038171003817 =1003817100381710038171003817119878 minus 1198780

1003817100381710038171003817 lt 120576

(86)

This completes the proof

5 Examples of Spaces Failing theBishop-Phelps-Bollobaacutes Property forNumerical Radius

Our goal here is to prove that the density of numerical radiusattaining operators does not imply the BPBp-nu Actuallywe will show that among separable spaces there is noisomorphic property implying the BPBp-nu other than finite-dimensionality

We need to relate the BPBp-nu to the Bishop-Phelps-Bollobas property for operators which as mentioned in theintroduction was introduced in [15] A pair (119883 119884) of Banachspaces has the Bishop-Phelps-Bollobas property for operators(in short BPBp) if given 120576 gt 0 there exists 120578(120576) gt 0 suchthat given 119879 isin L(119883 119884) with 119879 = 1 and 119909 isin 119878

119883such that

119879119909 gt 1 minus 120578(120576) then there exist 119911 isin 119878119883and 119878 isin L(119883 119884)

satisfying

119878 = 119878119911 = 1 119909 minus 119911 lt 120576

119879 minus 119878 lt 120576

(87)

We refer the reader to [15 19 25] and references therein formore information and background Among the interestingresults on the BPBp we emphasize that a pair (119883 119884) when119883is finite-dimensional does not necessarily have the BPBp Forinstance if119884 is a strictly convex space which is not uniformlyconvex then the pair (ℓ(2)

1 119884) fails to have the BPBp (this is

contained in [15] see [19 Section 3])

Abstract and Applied Analysis 11

The next result relates the BPBp-nu to the BPBp foroperators in a particular case We will deduce our examplefrom it

Theorem 15 If 1198711(120583)oplus

1119883 has the BPBp-119899119906 then the pair

(1198711(120583) 119883) has the BPBp for operators

Before proving this proposition we will use it to get themain examples of this section The first example shows thatthe density of numerical radius attaining operators does notimply the BPBp-nu

Example 16 There is a reflexive space (and so numerical radiusattaining operators on it are dense) which fails to have theBPBp-nu Indeed let 119884 be a reflexive separable space whichis not superreflexive and we may suppose that 119884 is strictlyconvex Observe that 119884 cannot be uniformly convex since itis not superreflexive Now 119883 = ℓ(2)

1oplus1119884 is reflexive but the

pair (ℓ(2)1 119884) fails the BPBp since 119884 is strictly convex but not

uniformly convex [19 Corollary 33] Therefore Theorem 15gives us that119883 does not have the BPBp-nu

The example above can be extended to get the result thatevery infinite-dimensional separable Banach space can berenormed to fail the BPBp-nu This follows from the factthat every infinite-dimensional separable Banach space canbe renormed to be strictly convex but not uniformly convex(this result can be proved ldquoby handrdquo an alternative categoricalargument for it can be found in [27] and references therein)With a little more of effort we may get the main result of thesection

Theorem 17 Every infinite-dimensional separable Banachspace can be renormed to fail the weak-BPBp-nu (and so inparticular to fail the BPBp-nu)

We need the following result which is surely well knownAs we have not found a reference we include a nice and easyproof kindly given to us by Vladimir Kadets We recall thatgiven a Banach space 119884 the set of all equivalent norms on119884 can be viewed as a metric space using the Banach-Mazurdistance

Lemma 18 Let119884 be an infinite-dimensional separable Banachspace Then the set of equivalent norms on 119884 which are strictlyconvex and are not (locally) uniformly convex is dense in the setof all equivalent norms on119884 (with respect to the Banach-Mazurdistance)

Proof Fix 119890 isin 119878119884and 119890lowast

1isin 119878

119884lowast such that 119890lowast

1(119890) = 1 For a fixed

120576 isin (0 12) denote

119902 (119910) = max (1 minus 120576) 10038171003817100381710038171199101003817100381710038171003817 1003816100381610038161003816119890lowast

1(119910)1003816100381610038161003816 (119910 isin 119884) (88)

Evidently (1 minus 120576)119910 ⩽ 119902(119910) ⩽ 119910 for every 119910 isin 119884 Fix asequence 119890lowast

119896 119896 ⩾ 2 of norm-one functionals separating

the points of 119884 and denote

119901 (119910) = radic

infin

sum

119896=1

1

2119896

1003816100381610038161003816119890lowast

119896(119910)1003816100381610038161003816

2

(119910 isin 119884) (89)

Then 119901 is a strictly convex norm on 119884 119901(119890) ⩾ 1radic2 and119901(119910) ⩽ 119910 for all 119910 isin 119883 Finally write

100381710038171003817100381711991010038171003817100381710038171= (1 minus 120576) 119902 (119910) + 120576

119901 (119910)

119901 (119890)(119910 isin 119884) (90)

Then sdot 1is a strictly convex norm on 119884 and

(1 minus 120576)2 10038171003817100381710038171199101003817100381710038171003817 ⩽100381710038171003817100381711991010038171003817100381710038171⩽ (1 + 120576)

10038171003817100381710038171199101003817100381710038171003817 (119910 isin 119884) (91)

Wewill finish the proof by showing that sdot 1is not uniformly

convex (actually it is not locally uniformly convex) Indeedfor each 119899 isin N we select 119910

119899isin ⋂

119899

119896=1ker 119890lowast

119896with 119910

119899 = 1 and

consider 119890119899= 119890 + (1205764)119910

119899 Then 119902(119890) = 1 119902(119890

119899) = 1 and

119902(119890+119890119899) = 2 At the same time 119901(119910

119899) rarr 0 so 119901(119890

119899) rarr 119901(119890)

and 119901(119890 + 119890119899) rarr 2119901(119890) Consequently

1198901 = 1100381710038171003817100381711989011989910038171003817100381710038171997888rarr 1

1003817100381710038171003817119890 + 11989011989910038171003817100381710038171997888rarr 2 (92)

but 119890 minus 1198901198991= (1205764)119910

1198991⩾ (1 minus 120576)

2(1205764) which means the

absence of local uniform convexity at 119890

Proof of Theorem 17 Let 119883 be an infinite-dimensional sepa-rable Banach space Take a closed subspace 119884 of 119883 of codi-mension two By [28 Proposition 2] the map carrying everyequivalent norm on 119884 to its numerical index is continuousand so the set of values of the numerical index of 119884 upto reforming is a nontrivial interval [28 Theorem 9] ThenLemma 18 allows us to find an equivalent norm | sdot | on 119884 insuch a way that (119884 | sdot |) is strictly convex and is not uniformlyconvex and 119899(119884 | sdot |) gt 0 Now the space 119883 = ℓ(2)

1oplus1(119884 | sdot |)

is an equivalent renorming of 119883 which does not have theBPBp-119899119906 (indeed otherwise the pair (ℓ(2)

1 (119884 | sdot |)) would

have the BPBp for the operator norm and so (119884 | sdot |) wouldbe uniformly convex by [19 Corollary 33] a contradiction)Moreover as

119899 (119883) = min 119899 (ℓ(2)1) 119899 (119884 |sdot|) gt 0 (93)

(see [20 Proposition 2] for instance) 119883 also fails the weak-BPBp-nu by Proposition 6

To finish the section with the promised proof ofTheorem 15 we first see the following stability result

Lemma 19 Let119883 = [⨁infin

119896=1119883

119896]1198880

or [⨁infin

119896=1119883

119896]ℓ1

If119883 has theBishop-Phelps-Bollobas property for numerical radius with afunction 120578 then each Banach space 119883

119894has the Bishop-Phelps-

Bollobas property for numerical radius with 120578nu(119883119894) ⩾ 120578 That

is inf119894120578nu(119883119894

)(120576) ⩾ 120578nu(119883)(120576) for all 0 lt 120576 lt 1

Proof Let 119875119894 119883 rarr 119883

119894and 1198751015840

119894 119883

lowastrarr 119883

lowast

119894be the natural

projections and let 119876119894 119883

119894rarr 119883 and 1198761015840

119894 119883

lowast

119894rarr 119883

lowast be thenatural embeddings

Assume that an operator 119879119894 119883

119894rarr 119883

119894and a pair

(119909119894 119909

lowast

119894) isin Π(119883

119894) satisfy that

V (119879119894) = 1

1003816100381610038161003816119909lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (94)

We define an operator 119879 119883 rarr 119883 and (119909 119909lowast) isin Π(119883) by

119879 = 119876119894∘ 119879

119894∘ 119875

119894 (119909 119909

lowast) = (119876

119894119909119894 119876

1015840

119894119909lowast

119894) (95)

12 Abstract and Applied Analysis

then clearly we see that1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119909

lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (96)

From the assumption there exist 119878 119883 rarr 119883 and a pair(119910 119910

lowast) isin Π(119883) such that

1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 = 1 = V (119878) 119878 minus 119879 lt 120576

1003817100381710038171003817119910lowastminus 119909

lowast1003817100381710038171003817 lt 1205761003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576

(97)

Since this clearly shows that1003817100381710038171003817119875119894 ∘ 119878 ∘ 119876119894

minus 119879119894

1003817100381710038171003817 lt 120576100381710038171003817100381710038171198751015840

119894119910lowastminus 119909

lowast

119894

10038171003817100381710038171003817lt 120576

1003817100381710038171003817119875119894119910 minus 1199091198941003817100381710038171003817 lt 120576

(98)

we only need to show that |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = 1

We first show the case of 1198880sum Since

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817=10038171003817100381710038171003817119875119895119910 minus 119875

11989511990910038171003817100381710038171003817⩽1003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576 (99)

for every 119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽100381710038171003817100381710038171198751015840

119894119910lowast10038171003817100381710038171003817+ 120576 sum

119895isinN119895 = 119894

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817⩽1003817100381710038171003817119910

lowast1003817100381710038171003817 = 1

(100)

This shows that 1198751015840119894119910lowast = 1 and 1198751015840

119895119910lowast= 0 for every 119895 = 119894

So 119910lowast = 1198761015840

1198941198751015840

119894119910lowast and 1198751015840

119894119910lowast(119875

119894119910) = 1 This and the fact that

119910 minus 119876119894119875119894119910 lt 120576 imply that

(119876119894119875119894119910 + (

1

120576) (119910 minus 119876

119894119875119894119910) 119876

1015840

1198941198751015840

119894119910lowast) isin Π (119883) (101)

So we get that (1198761015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910 + (1120576)(119910 minus 119876

119894119875119894119910))V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =10038161003816100381610038161003816(119876

1015840

1198941198751015840

119894119910lowast) 11987811991010038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910 +1

120576(119910 minus 119876

119894119875119894119910))

1003816100381610038161003816100381610038161003816⩽ 1

(102)

and so we get |1198751015840119894119910lowast(119875

119894∘119878 ∘119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

We next show the case of ℓ1sum The proof is almost

the same as that of the 1198880case However for the sake of

completeness we provide it hereSince 1198751015840

119895119910lowast = 119875

1015840

119895119910lowastminus 119875

1015840

119895119909lowast ⩽ 119910

lowastminus 119909

lowast lt 120576 for every

119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽10038171003817100381710038171198751198941199101003817100381710038171003817 + 120576 sum

119895isinN119895 = 119894

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817⩽10038171003817100381710038171199101003817100381710038171003817 = 1

(103)

which shows 119875119894119910 = 1 and 119875

119895119910 = 0 for every 119895 = 119894 Since this

implies (119876119894119875119894119910119876

1015840

1198941198751015840

119894119910lowast+(1120576)(119910

lowastminus119876

1015840

1198941198751015840

119894119910lowast)) isin Π(119883) we get

that |(1198761015840

1198941198751015840

119894119910lowast+ (1120576)(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast))119878(119876

119894119875119894119910)| ⩽ V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =1003816100381610038161003816119910

lowast119878 (119876

119894119875119894119910)1003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast+1

120576(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast)) 119878 (119876

119894119875119894119910)

1003816100381610038161003816100381610038161003816⩽ 1

(104)

and so |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

Proof of Theorem 15 Note that 120578nu(1198711(120583)oplus1119883)(120576) rarr 0 as120576 rarr 0 Fix 0 lt 120576

0lt 1 and choose 0 lt 120576 lt 1 such that

6120576 + 120578nu(1198711(120583)oplus1119883)(120576) lt 1205760 Let 120578(1205760) = 120578nu(1198711(120583)oplus1119883)(120576)Suppose that 119879

0isin L(119871

1(120583) 119883) with 119879

0 = 1 and 119891

0isin

1198781198711(120583)

satisfy

1003817100381710038171003817119879011989101003817100381710038171003817 gt 1 minus 120578 (1205760) (105)

For any measurable subset 119861 let

1198711(120583|

119861) = 119891|

119861 119891 isin 119871

1(120583) (106)

with the norm 119891|119861 = 119891120594

1198611 Then it is easy to see

that 1198711(120583|

119861) is isometrically isomorphic to a complemented

subspace of 1198711(120583) Let 119875

119861 119871

1(120583) rarr 119871

1(120583|

119861) be the

restriction defined by 119875119861(119891) = 119891|

119861for all 119891 isin 119871

1(120583) and

let 119869119861 119871

1(120583|

119861) rarr 119871

1(120583) be the extension defined by

119869119861(119891)(120596) = 119891(120596) if 120596 isin 119861 and 119869

119861(119891)(120596) = 0 otherwise

It is clear that 119875119861119869119861= Id

1198711(120583|119861)and 119869

119861119875119861(119891) = 119891120594

119861for all

119891 isin 1198711(120583) Notice also that 119871

1(120583) is isometrically isomorphic

to 1198711(120583|

119861)oplus

11198711(120583|

119861119888)

Let 119860 = supp1198910and 119892

0= 119875

1198601198910 Then

1003817100381710038171003817119879011986911986011989201003817100381710038171003817 =100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) gt 0 (107)

and define the operator 119879119860 119871

1(120583|

119860) rarr 119883 by 119879

119860119891 =

1198790119869119860119891119879

0119869119860 for every 119891 isin 119871

1(120583|

119860) Then

100381710038171003817100381711987911986011989201003817100381710038171003817 ⩾100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) (108)

Since 120583|119860is 120590-finite 119871

1(120583|

119860)lowast= 119871

infin(120583|

119860) Let 119892lowast

0isin 119878

119871infin(120583|119860)

be a function such that ⟨119892lowast0 119892

0⟩ = 1 and choose 119909lowast

0isin 119878

119883lowast

such that 119909lowast0(119879

1198601198920) = 119879

1198601198920 Define the operator 119878

0isin

L(1198711(120583|

119860)oplus

1119883) by

1198780(119891 119909) = (0 119879

119860119891) ((119891 119909) isin 119871

1(120583|

119860) oplus

1119883) (109)

and observe that 1198780 = V(119878

0) = 1 Indeed

100381710038171003817100381711987801003817100381710038171003817 ⩽ 1 =

10038171003817100381710038171198791198601003817100381710038171003817

= sup 1003816100381610038161003816119909lowast1198791198601198911003816100381610038161003816 119909

lowastisin 119878

119883lowast 119891 isin 119878

1198711(120583|119860)

= sup 1003816100381610038161003816(119891lowast 119909

lowast) 119878

0(119891 119909)

1003816100381610038161003816

((119891lowast 119909

lowast) (119891 119909)) isin Π (119871

1(120583|

119860) oplus

1119883)

= V (1198780) ⩽100381710038171003817100381711987801003817100381710038171003817

(110)

Abstract and Applied Analysis 13

It is immediate that

(119892lowast

0 119909

lowast

0) 119878

0(119892

0 0) = 119909

lowast

0(119879

1198601198920) =10038171003817100381710038171198791198601198920

1003817100381710038171003817 gt 1 minus 120578 (1205760)

(111)

By Lemma 19 1198711(120583|

119860)oplus

1119883 has the BPBp-nu with the func-

tion 120578 Therefore there exist 1198781isin L(119871

1(120583|

119860)oplus

1119883) (119892

1 119909

1) isin

1198781198711(120583|119860)oplus1119883 and (119892lowast

1 119909

lowast

1) isin 119878

119871infin(120583|119860)oplusinfin119883lowast such that

1003817100381710038171003817(1198921 1199091) minus (1198920 0)1003817100381710038171003817 lt 120576

1003817100381710038171003817(119892lowast

1 119909

lowast

1) minus (119891

lowast

0 119909

lowast

0)1003817100381710038171003817 lt 120576

10038171003817100381710038171198781 minus 11987801003817100381710038171003817 lt 120576 ⟨(119892

lowast

1 119909

lowast

1) (119892

1 119909

1)⟩ = 1

1003816100381610038161003816(119892lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 = V (119878

1) = 1

(112)

Claim 1 We claim that 1199091= 0

Otherwise

1 = Re ⟨(119892lowast1 119909

lowast

1) (119892

1 119909

1)⟩

=100381710038171003817100381711989211003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

+100381710038171003817100381711990911003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

⟩ ⩽ 1

(113)

We deduce that

((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

(119892lowast

1 119909

lowast

1)) (

(0 1199091)

100381710038171003817100381711990911003817100381710038171003817

(119892lowast

1 119909

lowast

1))

isin Π (1198711(120583|

119860) oplus

1119883)

(114)

Since 1198781((0 119909

1)119909

1) = (119878

1minus 119878

0)((0 119909

1)119909

1) lt 120576 we

get that

1 =1003816100381610038161003816⟨(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)⟩1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

100381710038171003817100381711989211003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

)⟩

+100381710038171003817100381711990911003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

)⟩

100381610038161003816100381610038161003816100381610038161003816

⩽100381710038171003817100381711989211003817100381710038171003817 V (1198781) + 120576

100381710038171003817100381711990911003817100381710038171003817 lt100381710038171003817100381711989211003817100381710038171003817 +100381710038171003817100381711990911003817100381710038171003817 = 1

(115)

a contradiction This proves the claimWe define the operator 119878

2 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883

by 1198782(119891 119909) = 119878

1(119891 0) for every 119891 isin 119871

1(120583|

119860) and for every

119909 isin 119883 Then we have

V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 = 1

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ 120576 (116)

Indeed from Claim 1 we have

V (1198781) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 ⩽ V (119878

2)

(117)

On the other hand we see that1003816100381610038161003816(119891

lowast 119909

lowast) 119878

2(119891 119909)

1003816100381610038161003816

=1003816100381610038161003816(119891

lowast 119909

lowast) 119878

1(119891 0)

1003816100381610038161003816 ⩽10038171003817100381710038171198911003817100381710038171003817 V (1198781) ⩽ V (119878

1)

(118)

for every ((119891lowast 119909lowast) (119891 119909)) isin Π(1198711(120583|

119860)oplus

1119883) Therefore

V(1198782) ⩽ V(119878

1) Also

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ sup

119909isin119878119883

10038171003817100381710038171198781 (0 119909)1003817100381710038171003817

= sup119909isin119878119883

10038171003817100381710038171198781 (0 119909) minus 1198780 (0 119909)1003817100381710038171003817 ⩽ 120576

(119)

Claim 2 There exists 1198783 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883 such

that 1198783(119892

1 0) = 119878

3 = 1 119878

3(0 119909) = 0 119878

3(119891 119909) isin 0oplus

1119883

for every (119891 119909) isin 1198711(120583|

119860)oplus

1119883 and 119878

3minus 119878

2 lt 4120576

Indeed write 1198781= (119863

1 119863

2) where 119863

1 119871

1(120583|

119860)oplus

1119883 rarr

1198711(120583|

119860) and119863

2 119871

1(120583|

119860)oplus

1119883 rarr 119883 We have that

sup 1003816100381610038161003816119892lowast119863

1(119892

1 0) + 119909

lowast119863

2(119892

1 0)1003816100381610038161003816

119909lowastisin 119878

119883lowast ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860) +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⩽ V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119892

lowast

1119863

1(119892

1 0) + 119909

lowast

1119863

2(119892

1 0)1003816100381610038161003816

(120)

This implies that

1003816100381610038161003816119909lowast

1119863

2(119892

1 0)1003816100381610038161003816 =10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

1003816100381610038161003816119892lowast

1119863

1(119892

1 0)1003816100381610038161003816

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119871

infin(120583|

119860)

(121)

Therefore |119892lowast1| equals 1 on the support of 119863

1(119892

1 0) As

|⟨119892lowast

1 119892

1⟩| = 1 we also have that |119892lowast

1| equals 1 on the support

of 1198921 Changing the values of 119892lowast

1by the ones of 119891lowast

0on 119860

(supp(1198631(119892

1 0)) cup supp(g

1)) we may and do suppose that

|119892lowast

1| = 1 on the whole 119860We also have 119863

2(119892

1 0) gt 0 Indeed

10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)1003817100381710038171003817

⩽10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198921 0)

1003817100381710038171003817 +10038171003817100381710038171198780 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817

lt 2120576 + 120576 = 3120576

(122)

So we have

10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

⩽10038171003817100381710038171198631(119892

1 0)1003817100381710038171003817 +10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

14 Abstract and Applied Analysis

=1003817100381710038171003817(1198631

(1198921 0) 119863

2(119892

1 0)) minus (0 119879

1198601198920)1003817100381710038171003817

=10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817 lt 3120576

(123)

and 1198632(119892

1 0) gt 119879

1198601198920 minus 3120576 ⩾ 1 minus 120578(120576

0) minus 3120576 gt 0

Finally define the operator 1198783by

1198783(119891 119909) = (0119863

2(119891 0) + 119892

lowast

1(119863

1(119891 0))

1198632(119892

1 0)

119909lowast

1119863

2(119892

1 0))

(124)

for (119891 119909) isin 1198711(120583|

119860)oplus

1119883 It is clear that

100381710038171003817100381711987831003817100381710038171003817 ⩽ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) (125)

Notice also that

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817

⩽10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 +10038171003817100381710038171198632(119891 0) minus 119879

1198601198911003817100381710038171003817

=1003817100381710038171003817(1198631

(119891 0) 1198632(119891 0)) minus (0 119879

119860119891)1003817100381710038171003817

=10038171003817100381710038171198782 (119891 119909) minus 1198780 (119891 119909)

1003817100381710038171003817

(126)

for all (119891 119909) isin 1198711(120583|

119860)oplus

1119883 Hence we have

10038171003817100381710038171198783 minus 11987821003817100381710038171003817

= 2 sup119891isin1198781198711(120583|119860)

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 ⩽ 210038171003817100381710038171198782 minus 1198780

1003817100381710038171003817 lt 4120576(127)

On the other hand let 119866 1198711(120583|

119860) rarr 119871

1(120583|

119860) be defined by

119866(119891) = 119892lowast

1119891 for every 119891 isin 119871

1(120583|

119860) Then we have

V (1198782)

= sup 1003816100381610038161003816119911lowast11987821199111003816100381610038161003816 (119911 119911

lowast) isin Π (119871

1(120583|

119860) oplus

1119883)

⩾ sup10038161003816100381610038161003816100381610038161003816

119909lowast119863

2(119866(

1

120583 (119862)120594119862) 0)

+119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119909lowastisin 119878

119883lowast 119862 isin Σ

119860 120583 (119862) gt 0

= sup10038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119862)120594119862) 0)

10038171003817100381710038171003817100381710038171003817

+

10038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119862 isin Σ119860 120583 (119862) gt 0

(128)

where Σ119860is the family of measurable subsets of 119860

Hence for any simple function 119904 = sum119899

119894=1((120572

119894120583(119860

119894))120594

119860119894

) isin

1198781198711(120583|119860) where 119860

119894119894is a family of disjoint measurable subsets

with strictly positive measure we have

V (1198782)

119899

sum

119894=1

10038161003816100381610038161205721198941003816100381610038161003816 (

100381710038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381710038171003817100381710038171003817100381710038171003817

+

100381610038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381610038161003816100381610038161003816100381610038161003816

)

⩾10038171003817100381710038171198632(119866 (119904) 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119866 (119904) 0)

1003816100381610038161003816

(129)

Since |119892lowast1| = 1119866 is an isometric isomorphism so for each119891 isin

1198781198711(120583|119860)there exists a sequence of norm-one simple functions

(119904119896) such that 119866(119904

119896) converges to 119891 Therefore

V (1198782) ⩾ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) ⩾100381710038171003817100381711987831003817100381710038171003817

(130)

On the other hand we see that100381710038171003817100381711987831003817100381710038171003817 ⩾1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

3(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119909

lowast

1119863

2(119892

1 0) + 119892

lowast

1119863

1(119892

1 0)1003816100381610038161003816 = V (119878

2) = 1

(131)

Therefore 1 = 1198783 = 119878

3(119892

1 0) which proves Claim 2

Finally set 1198783= (0 ) for a suitable 119871

1(120583|

119860)oplus

1119883 rarr

119883 and define the operator 1198791 119871

1(120583) rarr 119883 by

1198791(119891) = 119879

0(119891120594

119860119888) + (119875

119860119891 0) (132)

for every 119891 isin 1198711(120583) Then we have

10038171003817100381710038171198791 (119891)1003817100381710038171003817 ⩽100381710038171003817100381711987901003817100381710038171003817

10038171003817100381710038171198911205941198601198881003817100381710038171003817 +1003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171198911205941198601003817100381710038171003817 =10038171003817100381710038171198911003817100381710038171003817 (133)

for every 119891 isin 1198711(120583) so 119879

1 ⩽ 1 Also

10038171003817100381710038171198791 (1198691198601198921)1003817100381710038171003817 =10038171003817100381710038171198783 (1198921 0)

1003817100381710038171003817 =100381710038171003817100381711987831003817100381710038171003817 = 1 (134)

so 1198791attains its norm at 119869

1198601198921isin 119871

1(120583) and

10038171003817100381710038171198691198601198921 minus 11989101003817100381710038171003817 =10038171003817100381710038171198921 minus 1198920

1003817100381710038171003817 lt 120576 (135)

We also have that for any 119891 isin 1198781198711(120583)

10038171003817100381710038171198790 (119891) minus 1198791 (119891)1003817100381710038171003817 =100381710038171003817100381710038171198790(119891120594

119860) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790 (119869119860119875119860119891) minus 119879119860 (119875119860119891)

1003817100381710038171003817

+10038171003817100381710038171003817119879119860(119875

119860119891) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790119869119860 minus 119879119860

1003817100381710038171003817 +10038171003817100381710038171198780 minus 1198783

1003817100381710038171003817

lt 120578 (1205760) + 6120576

(136)

Hence 1198790minus 119879

1 ⩽ 120578(120576

0) + 6120576 lt 120576

0

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Abstract and Applied Analysis 15

Acknowledgments

The authors thank Gilles Godefroy and Rafael Paya forfruitful conversations about the content of this paper andVladimir Kadets for providing Lemma 18 They also appre-ciate anonymous referees for careful reading and fruitfulsuggestions about revision The second author was sup-ported by Basic Science Research Program through theNational Research Foundation of Korea (NRF) fundedby the Ministry of Education Science and Technology(NRF-2012R1A1A1006869) The third author was partiallysupported by Spanish MICINN and FEDER project noMTM2012-31755 and by Junta de Andalucıa and FEDERGrants FQM-185 and P09-FQM-4911

References

[1] F F Bonsall and J Duncan Numerical Ranges of Operators onNormed Spaces and of Elements of Normed Algebras vol 2 ofLondon Mathematical Society Lecture Note Series 2 CambridgeUniversity Press London UK 1971

[2] F F Bonsall and J Duncan Numerical Ranges II CambridgeUniversity Press London UK 1973 London MathematicalSociety Lecture Notes Series 10

[3] B Sims On numerical range and its applications to Banachalgebras [PhD thesis] University of Newcastle New SouthWales Australia 1972

[4] I D Berg and B Sims ldquoDenseness of operators which attaintheir numerical radiusrdquoAustralian Mathematical Society A vol36 no 1 pp 130ndash133 1984

[5] C S Cardassi ldquoDensity of numerical radius attaining operatorson some reflexive spacesrdquo Bulletin of the Australian Mathemat-ical Society vol 31 no 1 pp 1ndash3 1985

[6] C S Cardassi ldquoNumerical radius attaining operatorsrdquo inBanach Spaces (Columbia Mo 1984) vol 1166 of Lecture Notesin Mathematics pp 11ndash14 Springer Berlin Germany 1985

[7] C S Cardassi ldquoNumerical radius-attaining operators on119862(119870)rdquoProceedings of the AmericanMathematical Society vol 95 no 4pp 537ndash543 1985

[8] MDAcostaOperadores que alcanzan su radio numerico [PhDthesis] Universidad de Granada Granada Spain 1990

[9] M D Acosta and R Paya ldquoNumerical radius attaining oper-ators and the Radon-Nikodym propertyrdquo The Bulletin of theLondon Mathematical Society vol 25 no 1 pp 67ndash73 1993

[10] R Paya ldquoA counterexample on numerical radius attainingoperatorsrdquo Israel Journal of Mathematics vol 79 no 1 pp 83ndash101 1992

[11] Y S Choi and S G Kim ldquoNorm or numerical radius attainingmultilinear mappings and polynomialsrdquo Journal of the LondonMathematical Society vol 54 no 1 pp 135ndash147 1996

[12] M D Acosta J Becerra Guerrero and M Ruiz GalanldquoNumerical-radius-attaining polynomialsrdquo The Quarterly Jour-nal of Mathematics vol 54 no 1 pp 1ndash10 2003

[13] M D Acosta and S G Kim ldquoDenseness of holomorphicfunctions attaining their numerical radiirdquo Israel Journal ofMathematics vol 161 pp 373ndash386 2007

[14] S G Kim and H J Lee ldquoNumerical peak holomorphicfunctions on Banach spacesrdquo Journal of Mathematical Analysisand Applications vol 364 no 2 pp 437ndash452 2010

[15] M D Acosta R M Aron D Garcıa and M MaestreldquoThe Bishop-Phelps-Bollobas theorem for operatorsrdquo Journal ofFunctional Analysis vol 254 no 11 pp 2780ndash2799 2008

[16] A J Guirao and O Kozhushkina ldquoThe Bishop-Phelps-Bollobasproperty for numerical radius in ℓ

1(119862)rdquo Studia Mathematica

vol 218 no 1 pp 41ndash54 2013[17] J Falco ldquoThe Bishop-Phelps-Bollobas property for numerical

radius on 1198711rdquo Journal of Mathematical Analysis and Applica-

tions vol 414 no 1 pp 125ndash133 2014[18] A Aviles A J Guirao and J Rodriguez ldquoOn the Bishop-

Phelps-Bollobas property for numerical radius in119862(119870)-spacesrdquoPreprint

[19] RM Aron Y S Choi S K Kim H J Lee andMMartin ldquoTheBishop-Phelps-Bollobas version of Lindenstrauss properties Aand Brdquo Preprint

[20] V Kadets M Martın and R Paya ldquoRecent progress and openquestions on the numerical index of Banach spacesrdquo Revista dela Real Academia de Ciencias Exactas Fısicas y Naturales A vol100 no 1-2 pp 155ndash182 2006

[21] M Martın J Merı and M Popov ldquoOn the numerical index ofreal 119871

119901(120583)-spacesrdquo Israel Journal of Mathematics vol 184 pp

183ndash192 2011[22] A Iwanik ldquoNorm attaining operators on Lebesgue spacesrdquo

Pacific Journal of Mathematics vol 83 no 2 pp 381ndash386 1979[23] HH SchaeferBanach Lattices and Positive Operators Springer

New York NY USA 1974[24] J Duncan C M McGregor J D Pryce and A J White ldquoThe

numerical index of a normed spacerdquo Journal of the LondonMathematical Society Second Series vol 2 pp 481ndash488 1970

[25] Y S Choi S K Kim H J Lee and M Martin ldquoThe Bishop-Phelps-Bollobas theorem for operators on 119871

1(120583)rdquo Preprint

[26] H E Lacey The Isometric Theory of Classical Banach SpacesSpringer New York NY USA 1974

[27] G Godefroy ldquoUniversal spaces for strictly convex Banachspacesrdquo Revista de la Real Academia de Ciencias Exactas Fısicasy Naturales A vol 100 no 1-2 pp 137ndash146 2006

[28] C Finet M Martın and R Paya ldquoNumerical index andrenormingrdquo Proceedings of the American Mathematical Societyvol 131 no 3 pp 871ndash877 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 11: Research Article On the Bishop-Phelps-Bollobás Property ...downloads.hindawi.com/journals/aaa/2014/479208.pdf · Research Article On the Bishop-Phelps-Bollobás Property for Numerical

Abstract and Applied Analysis 11

The next result relates the BPBp-nu to the BPBp foroperators in a particular case We will deduce our examplefrom it

Theorem 15 If 1198711(120583)oplus

1119883 has the BPBp-119899119906 then the pair

(1198711(120583) 119883) has the BPBp for operators

Before proving this proposition we will use it to get themain examples of this section The first example shows thatthe density of numerical radius attaining operators does notimply the BPBp-nu

Example 16 There is a reflexive space (and so numerical radiusattaining operators on it are dense) which fails to have theBPBp-nu Indeed let 119884 be a reflexive separable space whichis not superreflexive and we may suppose that 119884 is strictlyconvex Observe that 119884 cannot be uniformly convex since itis not superreflexive Now 119883 = ℓ(2)

1oplus1119884 is reflexive but the

pair (ℓ(2)1 119884) fails the BPBp since 119884 is strictly convex but not

uniformly convex [19 Corollary 33] Therefore Theorem 15gives us that119883 does not have the BPBp-nu

The example above can be extended to get the result thatevery infinite-dimensional separable Banach space can berenormed to fail the BPBp-nu This follows from the factthat every infinite-dimensional separable Banach space canbe renormed to be strictly convex but not uniformly convex(this result can be proved ldquoby handrdquo an alternative categoricalargument for it can be found in [27] and references therein)With a little more of effort we may get the main result of thesection

Theorem 17 Every infinite-dimensional separable Banachspace can be renormed to fail the weak-BPBp-nu (and so inparticular to fail the BPBp-nu)

We need the following result which is surely well knownAs we have not found a reference we include a nice and easyproof kindly given to us by Vladimir Kadets We recall thatgiven a Banach space 119884 the set of all equivalent norms on119884 can be viewed as a metric space using the Banach-Mazurdistance

Lemma 18 Let119884 be an infinite-dimensional separable Banachspace Then the set of equivalent norms on 119884 which are strictlyconvex and are not (locally) uniformly convex is dense in the setof all equivalent norms on119884 (with respect to the Banach-Mazurdistance)

Proof Fix 119890 isin 119878119884and 119890lowast

1isin 119878

119884lowast such that 119890lowast

1(119890) = 1 For a fixed

120576 isin (0 12) denote

119902 (119910) = max (1 minus 120576) 10038171003817100381710038171199101003817100381710038171003817 1003816100381610038161003816119890lowast

1(119910)1003816100381610038161003816 (119910 isin 119884) (88)

Evidently (1 minus 120576)119910 ⩽ 119902(119910) ⩽ 119910 for every 119910 isin 119884 Fix asequence 119890lowast

119896 119896 ⩾ 2 of norm-one functionals separating

the points of 119884 and denote

119901 (119910) = radic

infin

sum

119896=1

1

2119896

1003816100381610038161003816119890lowast

119896(119910)1003816100381610038161003816

2

(119910 isin 119884) (89)

Then 119901 is a strictly convex norm on 119884 119901(119890) ⩾ 1radic2 and119901(119910) ⩽ 119910 for all 119910 isin 119883 Finally write

100381710038171003817100381711991010038171003817100381710038171= (1 minus 120576) 119902 (119910) + 120576

119901 (119910)

119901 (119890)(119910 isin 119884) (90)

Then sdot 1is a strictly convex norm on 119884 and

(1 minus 120576)2 10038171003817100381710038171199101003817100381710038171003817 ⩽100381710038171003817100381711991010038171003817100381710038171⩽ (1 + 120576)

10038171003817100381710038171199101003817100381710038171003817 (119910 isin 119884) (91)

Wewill finish the proof by showing that sdot 1is not uniformly

convex (actually it is not locally uniformly convex) Indeedfor each 119899 isin N we select 119910

119899isin ⋂

119899

119896=1ker 119890lowast

119896with 119910

119899 = 1 and

consider 119890119899= 119890 + (1205764)119910

119899 Then 119902(119890) = 1 119902(119890

119899) = 1 and

119902(119890+119890119899) = 2 At the same time 119901(119910

119899) rarr 0 so 119901(119890

119899) rarr 119901(119890)

and 119901(119890 + 119890119899) rarr 2119901(119890) Consequently

1198901 = 1100381710038171003817100381711989011989910038171003817100381710038171997888rarr 1

1003817100381710038171003817119890 + 11989011989910038171003817100381710038171997888rarr 2 (92)

but 119890 minus 1198901198991= (1205764)119910

1198991⩾ (1 minus 120576)

2(1205764) which means the

absence of local uniform convexity at 119890

Proof of Theorem 17 Let 119883 be an infinite-dimensional sepa-rable Banach space Take a closed subspace 119884 of 119883 of codi-mension two By [28 Proposition 2] the map carrying everyequivalent norm on 119884 to its numerical index is continuousand so the set of values of the numerical index of 119884 upto reforming is a nontrivial interval [28 Theorem 9] ThenLemma 18 allows us to find an equivalent norm | sdot | on 119884 insuch a way that (119884 | sdot |) is strictly convex and is not uniformlyconvex and 119899(119884 | sdot |) gt 0 Now the space 119883 = ℓ(2)

1oplus1(119884 | sdot |)

is an equivalent renorming of 119883 which does not have theBPBp-119899119906 (indeed otherwise the pair (ℓ(2)

1 (119884 | sdot |)) would

have the BPBp for the operator norm and so (119884 | sdot |) wouldbe uniformly convex by [19 Corollary 33] a contradiction)Moreover as

119899 (119883) = min 119899 (ℓ(2)1) 119899 (119884 |sdot|) gt 0 (93)

(see [20 Proposition 2] for instance) 119883 also fails the weak-BPBp-nu by Proposition 6

To finish the section with the promised proof ofTheorem 15 we first see the following stability result

Lemma 19 Let119883 = [⨁infin

119896=1119883

119896]1198880

or [⨁infin

119896=1119883

119896]ℓ1

If119883 has theBishop-Phelps-Bollobas property for numerical radius with afunction 120578 then each Banach space 119883

119894has the Bishop-Phelps-

Bollobas property for numerical radius with 120578nu(119883119894) ⩾ 120578 That

is inf119894120578nu(119883119894

)(120576) ⩾ 120578nu(119883)(120576) for all 0 lt 120576 lt 1

Proof Let 119875119894 119883 rarr 119883

119894and 1198751015840

119894 119883

lowastrarr 119883

lowast

119894be the natural

projections and let 119876119894 119883

119894rarr 119883 and 1198761015840

119894 119883

lowast

119894rarr 119883

lowast be thenatural embeddings

Assume that an operator 119879119894 119883

119894rarr 119883

119894and a pair

(119909119894 119909

lowast

119894) isin Π(119883

119894) satisfy that

V (119879119894) = 1

1003816100381610038161003816119909lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (94)

We define an operator 119879 119883 rarr 119883 and (119909 119909lowast) isin Π(119883) by

119879 = 119876119894∘ 119879

119894∘ 119875

119894 (119909 119909

lowast) = (119876

119894119909119894 119876

1015840

119894119909lowast

119894) (95)

12 Abstract and Applied Analysis

then clearly we see that1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119909

lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (96)

From the assumption there exist 119878 119883 rarr 119883 and a pair(119910 119910

lowast) isin Π(119883) such that

1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 = 1 = V (119878) 119878 minus 119879 lt 120576

1003817100381710038171003817119910lowastminus 119909

lowast1003817100381710038171003817 lt 1205761003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576

(97)

Since this clearly shows that1003817100381710038171003817119875119894 ∘ 119878 ∘ 119876119894

minus 119879119894

1003817100381710038171003817 lt 120576100381710038171003817100381710038171198751015840

119894119910lowastminus 119909

lowast

119894

10038171003817100381710038171003817lt 120576

1003817100381710038171003817119875119894119910 minus 1199091198941003817100381710038171003817 lt 120576

(98)

we only need to show that |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = 1

We first show the case of 1198880sum Since

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817=10038171003817100381710038171003817119875119895119910 minus 119875

11989511990910038171003817100381710038171003817⩽1003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576 (99)

for every 119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽100381710038171003817100381710038171198751015840

119894119910lowast10038171003817100381710038171003817+ 120576 sum

119895isinN119895 = 119894

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817⩽1003817100381710038171003817119910

lowast1003817100381710038171003817 = 1

(100)

This shows that 1198751015840119894119910lowast = 1 and 1198751015840

119895119910lowast= 0 for every 119895 = 119894

So 119910lowast = 1198761015840

1198941198751015840

119894119910lowast and 1198751015840

119894119910lowast(119875

119894119910) = 1 This and the fact that

119910 minus 119876119894119875119894119910 lt 120576 imply that

(119876119894119875119894119910 + (

1

120576) (119910 minus 119876

119894119875119894119910) 119876

1015840

1198941198751015840

119894119910lowast) isin Π (119883) (101)

So we get that (1198761015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910 + (1120576)(119910 minus 119876

119894119875119894119910))V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =10038161003816100381610038161003816(119876

1015840

1198941198751015840

119894119910lowast) 11987811991010038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910 +1

120576(119910 minus 119876

119894119875119894119910))

1003816100381610038161003816100381610038161003816⩽ 1

(102)

and so we get |1198751015840119894119910lowast(119875

119894∘119878 ∘119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

We next show the case of ℓ1sum The proof is almost

the same as that of the 1198880case However for the sake of

completeness we provide it hereSince 1198751015840

119895119910lowast = 119875

1015840

119895119910lowastminus 119875

1015840

119895119909lowast ⩽ 119910

lowastminus 119909

lowast lt 120576 for every

119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽10038171003817100381710038171198751198941199101003817100381710038171003817 + 120576 sum

119895isinN119895 = 119894

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817⩽10038171003817100381710038171199101003817100381710038171003817 = 1

(103)

which shows 119875119894119910 = 1 and 119875

119895119910 = 0 for every 119895 = 119894 Since this

implies (119876119894119875119894119910119876

1015840

1198941198751015840

119894119910lowast+(1120576)(119910

lowastminus119876

1015840

1198941198751015840

119894119910lowast)) isin Π(119883) we get

that |(1198761015840

1198941198751015840

119894119910lowast+ (1120576)(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast))119878(119876

119894119875119894119910)| ⩽ V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =1003816100381610038161003816119910

lowast119878 (119876

119894119875119894119910)1003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast+1

120576(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast)) 119878 (119876

119894119875119894119910)

1003816100381610038161003816100381610038161003816⩽ 1

(104)

and so |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

Proof of Theorem 15 Note that 120578nu(1198711(120583)oplus1119883)(120576) rarr 0 as120576 rarr 0 Fix 0 lt 120576

0lt 1 and choose 0 lt 120576 lt 1 such that

6120576 + 120578nu(1198711(120583)oplus1119883)(120576) lt 1205760 Let 120578(1205760) = 120578nu(1198711(120583)oplus1119883)(120576)Suppose that 119879

0isin L(119871

1(120583) 119883) with 119879

0 = 1 and 119891

0isin

1198781198711(120583)

satisfy

1003817100381710038171003817119879011989101003817100381710038171003817 gt 1 minus 120578 (1205760) (105)

For any measurable subset 119861 let

1198711(120583|

119861) = 119891|

119861 119891 isin 119871

1(120583) (106)

with the norm 119891|119861 = 119891120594

1198611 Then it is easy to see

that 1198711(120583|

119861) is isometrically isomorphic to a complemented

subspace of 1198711(120583) Let 119875

119861 119871

1(120583) rarr 119871

1(120583|

119861) be the

restriction defined by 119875119861(119891) = 119891|

119861for all 119891 isin 119871

1(120583) and

let 119869119861 119871

1(120583|

119861) rarr 119871

1(120583) be the extension defined by

119869119861(119891)(120596) = 119891(120596) if 120596 isin 119861 and 119869

119861(119891)(120596) = 0 otherwise

It is clear that 119875119861119869119861= Id

1198711(120583|119861)and 119869

119861119875119861(119891) = 119891120594

119861for all

119891 isin 1198711(120583) Notice also that 119871

1(120583) is isometrically isomorphic

to 1198711(120583|

119861)oplus

11198711(120583|

119861119888)

Let 119860 = supp1198910and 119892

0= 119875

1198601198910 Then

1003817100381710038171003817119879011986911986011989201003817100381710038171003817 =100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) gt 0 (107)

and define the operator 119879119860 119871

1(120583|

119860) rarr 119883 by 119879

119860119891 =

1198790119869119860119891119879

0119869119860 for every 119891 isin 119871

1(120583|

119860) Then

100381710038171003817100381711987911986011989201003817100381710038171003817 ⩾100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) (108)

Since 120583|119860is 120590-finite 119871

1(120583|

119860)lowast= 119871

infin(120583|

119860) Let 119892lowast

0isin 119878

119871infin(120583|119860)

be a function such that ⟨119892lowast0 119892

0⟩ = 1 and choose 119909lowast

0isin 119878

119883lowast

such that 119909lowast0(119879

1198601198920) = 119879

1198601198920 Define the operator 119878

0isin

L(1198711(120583|

119860)oplus

1119883) by

1198780(119891 119909) = (0 119879

119860119891) ((119891 119909) isin 119871

1(120583|

119860) oplus

1119883) (109)

and observe that 1198780 = V(119878

0) = 1 Indeed

100381710038171003817100381711987801003817100381710038171003817 ⩽ 1 =

10038171003817100381710038171198791198601003817100381710038171003817

= sup 1003816100381610038161003816119909lowast1198791198601198911003816100381610038161003816 119909

lowastisin 119878

119883lowast 119891 isin 119878

1198711(120583|119860)

= sup 1003816100381610038161003816(119891lowast 119909

lowast) 119878

0(119891 119909)

1003816100381610038161003816

((119891lowast 119909

lowast) (119891 119909)) isin Π (119871

1(120583|

119860) oplus

1119883)

= V (1198780) ⩽100381710038171003817100381711987801003817100381710038171003817

(110)

Abstract and Applied Analysis 13

It is immediate that

(119892lowast

0 119909

lowast

0) 119878

0(119892

0 0) = 119909

lowast

0(119879

1198601198920) =10038171003817100381710038171198791198601198920

1003817100381710038171003817 gt 1 minus 120578 (1205760)

(111)

By Lemma 19 1198711(120583|

119860)oplus

1119883 has the BPBp-nu with the func-

tion 120578 Therefore there exist 1198781isin L(119871

1(120583|

119860)oplus

1119883) (119892

1 119909

1) isin

1198781198711(120583|119860)oplus1119883 and (119892lowast

1 119909

lowast

1) isin 119878

119871infin(120583|119860)oplusinfin119883lowast such that

1003817100381710038171003817(1198921 1199091) minus (1198920 0)1003817100381710038171003817 lt 120576

1003817100381710038171003817(119892lowast

1 119909

lowast

1) minus (119891

lowast

0 119909

lowast

0)1003817100381710038171003817 lt 120576

10038171003817100381710038171198781 minus 11987801003817100381710038171003817 lt 120576 ⟨(119892

lowast

1 119909

lowast

1) (119892

1 119909

1)⟩ = 1

1003816100381610038161003816(119892lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 = V (119878

1) = 1

(112)

Claim 1 We claim that 1199091= 0

Otherwise

1 = Re ⟨(119892lowast1 119909

lowast

1) (119892

1 119909

1)⟩

=100381710038171003817100381711989211003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

+100381710038171003817100381711990911003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

⟩ ⩽ 1

(113)

We deduce that

((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

(119892lowast

1 119909

lowast

1)) (

(0 1199091)

100381710038171003817100381711990911003817100381710038171003817

(119892lowast

1 119909

lowast

1))

isin Π (1198711(120583|

119860) oplus

1119883)

(114)

Since 1198781((0 119909

1)119909

1) = (119878

1minus 119878

0)((0 119909

1)119909

1) lt 120576 we

get that

1 =1003816100381610038161003816⟨(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)⟩1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

100381710038171003817100381711989211003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

)⟩

+100381710038171003817100381711990911003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

)⟩

100381610038161003816100381610038161003816100381610038161003816

⩽100381710038171003817100381711989211003817100381710038171003817 V (1198781) + 120576

100381710038171003817100381711990911003817100381710038171003817 lt100381710038171003817100381711989211003817100381710038171003817 +100381710038171003817100381711990911003817100381710038171003817 = 1

(115)

a contradiction This proves the claimWe define the operator 119878

2 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883

by 1198782(119891 119909) = 119878

1(119891 0) for every 119891 isin 119871

1(120583|

119860) and for every

119909 isin 119883 Then we have

V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 = 1

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ 120576 (116)

Indeed from Claim 1 we have

V (1198781) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 ⩽ V (119878

2)

(117)

On the other hand we see that1003816100381610038161003816(119891

lowast 119909

lowast) 119878

2(119891 119909)

1003816100381610038161003816

=1003816100381610038161003816(119891

lowast 119909

lowast) 119878

1(119891 0)

1003816100381610038161003816 ⩽10038171003817100381710038171198911003817100381710038171003817 V (1198781) ⩽ V (119878

1)

(118)

for every ((119891lowast 119909lowast) (119891 119909)) isin Π(1198711(120583|

119860)oplus

1119883) Therefore

V(1198782) ⩽ V(119878

1) Also

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ sup

119909isin119878119883

10038171003817100381710038171198781 (0 119909)1003817100381710038171003817

= sup119909isin119878119883

10038171003817100381710038171198781 (0 119909) minus 1198780 (0 119909)1003817100381710038171003817 ⩽ 120576

(119)

Claim 2 There exists 1198783 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883 such

that 1198783(119892

1 0) = 119878

3 = 1 119878

3(0 119909) = 0 119878

3(119891 119909) isin 0oplus

1119883

for every (119891 119909) isin 1198711(120583|

119860)oplus

1119883 and 119878

3minus 119878

2 lt 4120576

Indeed write 1198781= (119863

1 119863

2) where 119863

1 119871

1(120583|

119860)oplus

1119883 rarr

1198711(120583|

119860) and119863

2 119871

1(120583|

119860)oplus

1119883 rarr 119883 We have that

sup 1003816100381610038161003816119892lowast119863

1(119892

1 0) + 119909

lowast119863

2(119892

1 0)1003816100381610038161003816

119909lowastisin 119878

119883lowast ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860) +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⩽ V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119892

lowast

1119863

1(119892

1 0) + 119909

lowast

1119863

2(119892

1 0)1003816100381610038161003816

(120)

This implies that

1003816100381610038161003816119909lowast

1119863

2(119892

1 0)1003816100381610038161003816 =10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

1003816100381610038161003816119892lowast

1119863

1(119892

1 0)1003816100381610038161003816

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119871

infin(120583|

119860)

(121)

Therefore |119892lowast1| equals 1 on the support of 119863

1(119892

1 0) As

|⟨119892lowast

1 119892

1⟩| = 1 we also have that |119892lowast

1| equals 1 on the support

of 1198921 Changing the values of 119892lowast

1by the ones of 119891lowast

0on 119860

(supp(1198631(119892

1 0)) cup supp(g

1)) we may and do suppose that

|119892lowast

1| = 1 on the whole 119860We also have 119863

2(119892

1 0) gt 0 Indeed

10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)1003817100381710038171003817

⩽10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198921 0)

1003817100381710038171003817 +10038171003817100381710038171198780 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817

lt 2120576 + 120576 = 3120576

(122)

So we have

10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

⩽10038171003817100381710038171198631(119892

1 0)1003817100381710038171003817 +10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

14 Abstract and Applied Analysis

=1003817100381710038171003817(1198631

(1198921 0) 119863

2(119892

1 0)) minus (0 119879

1198601198920)1003817100381710038171003817

=10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817 lt 3120576

(123)

and 1198632(119892

1 0) gt 119879

1198601198920 minus 3120576 ⩾ 1 minus 120578(120576

0) minus 3120576 gt 0

Finally define the operator 1198783by

1198783(119891 119909) = (0119863

2(119891 0) + 119892

lowast

1(119863

1(119891 0))

1198632(119892

1 0)

119909lowast

1119863

2(119892

1 0))

(124)

for (119891 119909) isin 1198711(120583|

119860)oplus

1119883 It is clear that

100381710038171003817100381711987831003817100381710038171003817 ⩽ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) (125)

Notice also that

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817

⩽10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 +10038171003817100381710038171198632(119891 0) minus 119879

1198601198911003817100381710038171003817

=1003817100381710038171003817(1198631

(119891 0) 1198632(119891 0)) minus (0 119879

119860119891)1003817100381710038171003817

=10038171003817100381710038171198782 (119891 119909) minus 1198780 (119891 119909)

1003817100381710038171003817

(126)

for all (119891 119909) isin 1198711(120583|

119860)oplus

1119883 Hence we have

10038171003817100381710038171198783 minus 11987821003817100381710038171003817

= 2 sup119891isin1198781198711(120583|119860)

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 ⩽ 210038171003817100381710038171198782 minus 1198780

1003817100381710038171003817 lt 4120576(127)

On the other hand let 119866 1198711(120583|

119860) rarr 119871

1(120583|

119860) be defined by

119866(119891) = 119892lowast

1119891 for every 119891 isin 119871

1(120583|

119860) Then we have

V (1198782)

= sup 1003816100381610038161003816119911lowast11987821199111003816100381610038161003816 (119911 119911

lowast) isin Π (119871

1(120583|

119860) oplus

1119883)

⩾ sup10038161003816100381610038161003816100381610038161003816

119909lowast119863

2(119866(

1

120583 (119862)120594119862) 0)

+119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119909lowastisin 119878

119883lowast 119862 isin Σ

119860 120583 (119862) gt 0

= sup10038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119862)120594119862) 0)

10038171003817100381710038171003817100381710038171003817

+

10038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119862 isin Σ119860 120583 (119862) gt 0

(128)

where Σ119860is the family of measurable subsets of 119860

Hence for any simple function 119904 = sum119899

119894=1((120572

119894120583(119860

119894))120594

119860119894

) isin

1198781198711(120583|119860) where 119860

119894119894is a family of disjoint measurable subsets

with strictly positive measure we have

V (1198782)

119899

sum

119894=1

10038161003816100381610038161205721198941003816100381610038161003816 (

100381710038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381710038171003817100381710038171003817100381710038171003817

+

100381610038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381610038161003816100381610038161003816100381610038161003816

)

⩾10038171003817100381710038171198632(119866 (119904) 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119866 (119904) 0)

1003816100381610038161003816

(129)

Since |119892lowast1| = 1119866 is an isometric isomorphism so for each119891 isin

1198781198711(120583|119860)there exists a sequence of norm-one simple functions

(119904119896) such that 119866(119904

119896) converges to 119891 Therefore

V (1198782) ⩾ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) ⩾100381710038171003817100381711987831003817100381710038171003817

(130)

On the other hand we see that100381710038171003817100381711987831003817100381710038171003817 ⩾1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

3(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119909

lowast

1119863

2(119892

1 0) + 119892

lowast

1119863

1(119892

1 0)1003816100381610038161003816 = V (119878

2) = 1

(131)

Therefore 1 = 1198783 = 119878

3(119892

1 0) which proves Claim 2

Finally set 1198783= (0 ) for a suitable 119871

1(120583|

119860)oplus

1119883 rarr

119883 and define the operator 1198791 119871

1(120583) rarr 119883 by

1198791(119891) = 119879

0(119891120594

119860119888) + (119875

119860119891 0) (132)

for every 119891 isin 1198711(120583) Then we have

10038171003817100381710038171198791 (119891)1003817100381710038171003817 ⩽100381710038171003817100381711987901003817100381710038171003817

10038171003817100381710038171198911205941198601198881003817100381710038171003817 +1003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171198911205941198601003817100381710038171003817 =10038171003817100381710038171198911003817100381710038171003817 (133)

for every 119891 isin 1198711(120583) so 119879

1 ⩽ 1 Also

10038171003817100381710038171198791 (1198691198601198921)1003817100381710038171003817 =10038171003817100381710038171198783 (1198921 0)

1003817100381710038171003817 =100381710038171003817100381711987831003817100381710038171003817 = 1 (134)

so 1198791attains its norm at 119869

1198601198921isin 119871

1(120583) and

10038171003817100381710038171198691198601198921 minus 11989101003817100381710038171003817 =10038171003817100381710038171198921 minus 1198920

1003817100381710038171003817 lt 120576 (135)

We also have that for any 119891 isin 1198781198711(120583)

10038171003817100381710038171198790 (119891) minus 1198791 (119891)1003817100381710038171003817 =100381710038171003817100381710038171198790(119891120594

119860) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790 (119869119860119875119860119891) minus 119879119860 (119875119860119891)

1003817100381710038171003817

+10038171003817100381710038171003817119879119860(119875

119860119891) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790119869119860 minus 119879119860

1003817100381710038171003817 +10038171003817100381710038171198780 minus 1198783

1003817100381710038171003817

lt 120578 (1205760) + 6120576

(136)

Hence 1198790minus 119879

1 ⩽ 120578(120576

0) + 6120576 lt 120576

0

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Abstract and Applied Analysis 15

Acknowledgments

The authors thank Gilles Godefroy and Rafael Paya forfruitful conversations about the content of this paper andVladimir Kadets for providing Lemma 18 They also appre-ciate anonymous referees for careful reading and fruitfulsuggestions about revision The second author was sup-ported by Basic Science Research Program through theNational Research Foundation of Korea (NRF) fundedby the Ministry of Education Science and Technology(NRF-2012R1A1A1006869) The third author was partiallysupported by Spanish MICINN and FEDER project noMTM2012-31755 and by Junta de Andalucıa and FEDERGrants FQM-185 and P09-FQM-4911

References

[1] F F Bonsall and J Duncan Numerical Ranges of Operators onNormed Spaces and of Elements of Normed Algebras vol 2 ofLondon Mathematical Society Lecture Note Series 2 CambridgeUniversity Press London UK 1971

[2] F F Bonsall and J Duncan Numerical Ranges II CambridgeUniversity Press London UK 1973 London MathematicalSociety Lecture Notes Series 10

[3] B Sims On numerical range and its applications to Banachalgebras [PhD thesis] University of Newcastle New SouthWales Australia 1972

[4] I D Berg and B Sims ldquoDenseness of operators which attaintheir numerical radiusrdquoAustralian Mathematical Society A vol36 no 1 pp 130ndash133 1984

[5] C S Cardassi ldquoDensity of numerical radius attaining operatorson some reflexive spacesrdquo Bulletin of the Australian Mathemat-ical Society vol 31 no 1 pp 1ndash3 1985

[6] C S Cardassi ldquoNumerical radius attaining operatorsrdquo inBanach Spaces (Columbia Mo 1984) vol 1166 of Lecture Notesin Mathematics pp 11ndash14 Springer Berlin Germany 1985

[7] C S Cardassi ldquoNumerical radius-attaining operators on119862(119870)rdquoProceedings of the AmericanMathematical Society vol 95 no 4pp 537ndash543 1985

[8] MDAcostaOperadores que alcanzan su radio numerico [PhDthesis] Universidad de Granada Granada Spain 1990

[9] M D Acosta and R Paya ldquoNumerical radius attaining oper-ators and the Radon-Nikodym propertyrdquo The Bulletin of theLondon Mathematical Society vol 25 no 1 pp 67ndash73 1993

[10] R Paya ldquoA counterexample on numerical radius attainingoperatorsrdquo Israel Journal of Mathematics vol 79 no 1 pp 83ndash101 1992

[11] Y S Choi and S G Kim ldquoNorm or numerical radius attainingmultilinear mappings and polynomialsrdquo Journal of the LondonMathematical Society vol 54 no 1 pp 135ndash147 1996

[12] M D Acosta J Becerra Guerrero and M Ruiz GalanldquoNumerical-radius-attaining polynomialsrdquo The Quarterly Jour-nal of Mathematics vol 54 no 1 pp 1ndash10 2003

[13] M D Acosta and S G Kim ldquoDenseness of holomorphicfunctions attaining their numerical radiirdquo Israel Journal ofMathematics vol 161 pp 373ndash386 2007

[14] S G Kim and H J Lee ldquoNumerical peak holomorphicfunctions on Banach spacesrdquo Journal of Mathematical Analysisand Applications vol 364 no 2 pp 437ndash452 2010

[15] M D Acosta R M Aron D Garcıa and M MaestreldquoThe Bishop-Phelps-Bollobas theorem for operatorsrdquo Journal ofFunctional Analysis vol 254 no 11 pp 2780ndash2799 2008

[16] A J Guirao and O Kozhushkina ldquoThe Bishop-Phelps-Bollobasproperty for numerical radius in ℓ

1(119862)rdquo Studia Mathematica

vol 218 no 1 pp 41ndash54 2013[17] J Falco ldquoThe Bishop-Phelps-Bollobas property for numerical

radius on 1198711rdquo Journal of Mathematical Analysis and Applica-

tions vol 414 no 1 pp 125ndash133 2014[18] A Aviles A J Guirao and J Rodriguez ldquoOn the Bishop-

Phelps-Bollobas property for numerical radius in119862(119870)-spacesrdquoPreprint

[19] RM Aron Y S Choi S K Kim H J Lee andMMartin ldquoTheBishop-Phelps-Bollobas version of Lindenstrauss properties Aand Brdquo Preprint

[20] V Kadets M Martın and R Paya ldquoRecent progress and openquestions on the numerical index of Banach spacesrdquo Revista dela Real Academia de Ciencias Exactas Fısicas y Naturales A vol100 no 1-2 pp 155ndash182 2006

[21] M Martın J Merı and M Popov ldquoOn the numerical index ofreal 119871

119901(120583)-spacesrdquo Israel Journal of Mathematics vol 184 pp

183ndash192 2011[22] A Iwanik ldquoNorm attaining operators on Lebesgue spacesrdquo

Pacific Journal of Mathematics vol 83 no 2 pp 381ndash386 1979[23] HH SchaeferBanach Lattices and Positive Operators Springer

New York NY USA 1974[24] J Duncan C M McGregor J D Pryce and A J White ldquoThe

numerical index of a normed spacerdquo Journal of the LondonMathematical Society Second Series vol 2 pp 481ndash488 1970

[25] Y S Choi S K Kim H J Lee and M Martin ldquoThe Bishop-Phelps-Bollobas theorem for operators on 119871

1(120583)rdquo Preprint

[26] H E Lacey The Isometric Theory of Classical Banach SpacesSpringer New York NY USA 1974

[27] G Godefroy ldquoUniversal spaces for strictly convex Banachspacesrdquo Revista de la Real Academia de Ciencias Exactas Fısicasy Naturales A vol 100 no 1-2 pp 137ndash146 2006

[28] C Finet M Martın and R Paya ldquoNumerical index andrenormingrdquo Proceedings of the American Mathematical Societyvol 131 no 3 pp 871ndash877 2003

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Stochastic AnalysisInternational Journal of

Page 12: Research Article On the Bishop-Phelps-Bollobás Property ...downloads.hindawi.com/journals/aaa/2014/479208.pdf · Research Article On the Bishop-Phelps-Bollobás Property for Numerical

12 Abstract and Applied Analysis

then clearly we see that1003816100381610038161003816119909

lowast1198791199091003816100381610038161003816 =1003816100381610038161003816119909

lowast

119894119879119894119909119894

1003816100381610038161003816 gt 1 minus 120578 (120576) (96)

From the assumption there exist 119878 119883 rarr 119883 and a pair(119910 119910

lowast) isin Π(119883) such that

1003816100381610038161003816119910lowast1198781199101003816100381610038161003816 = 1 = V (119878) 119878 minus 119879 lt 120576

1003817100381710038171003817119910lowastminus 119909

lowast1003817100381710038171003817 lt 1205761003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576

(97)

Since this clearly shows that1003817100381710038171003817119875119894 ∘ 119878 ∘ 119876119894

minus 119879119894

1003817100381710038171003817 lt 120576100381710038171003817100381710038171198751015840

119894119910lowastminus 119909

lowast

119894

10038171003817100381710038171003817lt 120576

1003817100381710038171003817119875119894119910 minus 1199091198941003817100381710038171003817 lt 120576

(98)

we only need to show that |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = 1

We first show the case of 1198880sum Since

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817=10038171003817100381710038171003817119875119895119910 minus 119875

11989511990910038171003817100381710038171003817⩽1003817100381710038171003817119910 minus 119909

1003817100381710038171003817 lt 120576 (99)

for every 119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽100381710038171003817100381710038171198751015840

119894119910lowast10038171003817100381710038171003817+ 120576 sum

119895isinN119895 = 119894

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817⩽1003817100381710038171003817119910

lowast1003817100381710038171003817 = 1

(100)

This shows that 1198751015840119894119910lowast = 1 and 1198751015840

119895119910lowast= 0 for every 119895 = 119894

So 119910lowast = 1198761015840

1198941198751015840

119894119910lowast and 1198751015840

119894119910lowast(119875

119894119910) = 1 This and the fact that

119910 minus 119876119894119875119894119910 lt 120576 imply that

(119876119894119875119894119910 + (

1

120576) (119910 minus 119876

119894119875119894119910) 119876

1015840

1198941198751015840

119894119910lowast) isin Π (119883) (101)

So we get that (1198761015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910 + (1120576)(119910 minus 119876

119894119875119894119910))V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =10038161003816100381610038161003816(119876

1015840

1198941198751015840

119894119910lowast) 11987811991010038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910 +1

120576(119910 minus 119876

119894119875119894119910))

1003816100381610038161003816100381610038161003816⩽ 1

(102)

and so we get |1198751015840119894119910lowast(119875

119894∘119878 ∘119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

We next show the case of ℓ1sum The proof is almost

the same as that of the 1198880case However for the sake of

completeness we provide it hereSince 1198751015840

119895119910lowast = 119875

1015840

119895119910lowastminus 119875

1015840

119895119909lowast ⩽ 119910

lowastminus 119909

lowast lt 120576 for every

119895 = 119894 we have

1 = 119910lowast(119910) = sum

119895isinN

1198751015840

119895119910lowast(119875

119895119910) ⩽ sum

119895isinN

100381710038171003817100381710038171198751015840

119895119910lowast10038171003817100381710038171003817

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817

⩽10038171003817100381710038171198751198941199101003817100381710038171003817 + 120576 sum

119895isinN119895 = 119894

1003817100381710038171003817100381711987511989511991010038171003817100381710038171003817⩽10038171003817100381710038171199101003817100381710038171003817 = 1

(103)

which shows 119875119894119910 = 1 and 119875

119895119910 = 0 for every 119895 = 119894 Since this

implies (119876119894119875119894119910119876

1015840

1198941198751015840

119894119910lowast+(1120576)(119910

lowastminus119876

1015840

1198941198751015840

119894119910lowast)) isin Π(119883) we get

that |(1198761015840

1198941198751015840

119894119910lowast+ (1120576)(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast))119878(119876

119894119875119894119910)| ⩽ V(119878) = 1

Hence we have

1 =1003816100381610038161003816119910

lowast1198781199101003816100381610038161003816 =1003816100381610038161003816119910

lowast119878 (119876

119894119875119894119910)1003816100381610038161003816

=

1003816100381610038161003816100381610038161003816(1 minus 120576) (119876

1015840

1198941198751015840

119894119910lowast) 119878 (119876

119894119875119894119910)

+120576 (1198761015840

1198941198751015840

119894119910lowast+1

120576(119910

lowastminus 119876

1015840

1198941198751015840

119894119910lowast)) 119878 (119876

119894119875119894119910)

1003816100381610038161003816100381610038161003816⩽ 1

(104)

and so |1198751015840119894119910lowast(119875

119894∘ 119878 ∘ 119876

119894)119875

119894119910| = |(119876

1015840

1198941198751015840

119894119910lowast)119878(119876

119894119875119894119910)| = 1

Proof of Theorem 15 Note that 120578nu(1198711(120583)oplus1119883)(120576) rarr 0 as120576 rarr 0 Fix 0 lt 120576

0lt 1 and choose 0 lt 120576 lt 1 such that

6120576 + 120578nu(1198711(120583)oplus1119883)(120576) lt 1205760 Let 120578(1205760) = 120578nu(1198711(120583)oplus1119883)(120576)Suppose that 119879

0isin L(119871

1(120583) 119883) with 119879

0 = 1 and 119891

0isin

1198781198711(120583)

satisfy

1003817100381710038171003817119879011989101003817100381710038171003817 gt 1 minus 120578 (1205760) (105)

For any measurable subset 119861 let

1198711(120583|

119861) = 119891|

119861 119891 isin 119871

1(120583) (106)

with the norm 119891|119861 = 119891120594

1198611 Then it is easy to see

that 1198711(120583|

119861) is isometrically isomorphic to a complemented

subspace of 1198711(120583) Let 119875

119861 119871

1(120583) rarr 119871

1(120583|

119861) be the

restriction defined by 119875119861(119891) = 119891|

119861for all 119891 isin 119871

1(120583) and

let 119869119861 119871

1(120583|

119861) rarr 119871

1(120583) be the extension defined by

119869119861(119891)(120596) = 119891(120596) if 120596 isin 119861 and 119869

119861(119891)(120596) = 0 otherwise

It is clear that 119875119861119869119861= Id

1198711(120583|119861)and 119869

119861119875119861(119891) = 119891120594

119861for all

119891 isin 1198711(120583) Notice also that 119871

1(120583) is isometrically isomorphic

to 1198711(120583|

119861)oplus

11198711(120583|

119861119888)

Let 119860 = supp1198910and 119892

0= 119875

1198601198910 Then

1003817100381710038171003817119879011986911986011989201003817100381710038171003817 =100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) gt 0 (107)

and define the operator 119879119860 119871

1(120583|

119860) rarr 119883 by 119879

119860119891 =

1198790119869119860119891119879

0119869119860 for every 119891 isin 119871

1(120583|

119860) Then

100381710038171003817100381711987911986011989201003817100381710038171003817 ⩾100381710038171003817100381711987901198910

1003817100381710038171003817 gt 1 minus 120578 (1205760) (108)

Since 120583|119860is 120590-finite 119871

1(120583|

119860)lowast= 119871

infin(120583|

119860) Let 119892lowast

0isin 119878

119871infin(120583|119860)

be a function such that ⟨119892lowast0 119892

0⟩ = 1 and choose 119909lowast

0isin 119878

119883lowast

such that 119909lowast0(119879

1198601198920) = 119879

1198601198920 Define the operator 119878

0isin

L(1198711(120583|

119860)oplus

1119883) by

1198780(119891 119909) = (0 119879

119860119891) ((119891 119909) isin 119871

1(120583|

119860) oplus

1119883) (109)

and observe that 1198780 = V(119878

0) = 1 Indeed

100381710038171003817100381711987801003817100381710038171003817 ⩽ 1 =

10038171003817100381710038171198791198601003817100381710038171003817

= sup 1003816100381610038161003816119909lowast1198791198601198911003816100381610038161003816 119909

lowastisin 119878

119883lowast 119891 isin 119878

1198711(120583|119860)

= sup 1003816100381610038161003816(119891lowast 119909

lowast) 119878

0(119891 119909)

1003816100381610038161003816

((119891lowast 119909

lowast) (119891 119909)) isin Π (119871

1(120583|

119860) oplus

1119883)

= V (1198780) ⩽100381710038171003817100381711987801003817100381710038171003817

(110)

Abstract and Applied Analysis 13

It is immediate that

(119892lowast

0 119909

lowast

0) 119878

0(119892

0 0) = 119909

lowast

0(119879

1198601198920) =10038171003817100381710038171198791198601198920

1003817100381710038171003817 gt 1 minus 120578 (1205760)

(111)

By Lemma 19 1198711(120583|

119860)oplus

1119883 has the BPBp-nu with the func-

tion 120578 Therefore there exist 1198781isin L(119871

1(120583|

119860)oplus

1119883) (119892

1 119909

1) isin

1198781198711(120583|119860)oplus1119883 and (119892lowast

1 119909

lowast

1) isin 119878

119871infin(120583|119860)oplusinfin119883lowast such that

1003817100381710038171003817(1198921 1199091) minus (1198920 0)1003817100381710038171003817 lt 120576

1003817100381710038171003817(119892lowast

1 119909

lowast

1) minus (119891

lowast

0 119909

lowast

0)1003817100381710038171003817 lt 120576

10038171003817100381710038171198781 minus 11987801003817100381710038171003817 lt 120576 ⟨(119892

lowast

1 119909

lowast

1) (119892

1 119909

1)⟩ = 1

1003816100381610038161003816(119892lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 = V (119878

1) = 1

(112)

Claim 1 We claim that 1199091= 0

Otherwise

1 = Re ⟨(119892lowast1 119909

lowast

1) (119892

1 119909

1)⟩

=100381710038171003817100381711989211003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

+100381710038171003817100381711990911003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

⟩ ⩽ 1

(113)

We deduce that

((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

(119892lowast

1 119909

lowast

1)) (

(0 1199091)

100381710038171003817100381711990911003817100381710038171003817

(119892lowast

1 119909

lowast

1))

isin Π (1198711(120583|

119860) oplus

1119883)

(114)

Since 1198781((0 119909

1)119909

1) = (119878

1minus 119878

0)((0 119909

1)119909

1) lt 120576 we

get that

1 =1003816100381610038161003816⟨(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)⟩1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

100381710038171003817100381711989211003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

)⟩

+100381710038171003817100381711990911003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

)⟩

100381610038161003816100381610038161003816100381610038161003816

⩽100381710038171003817100381711989211003817100381710038171003817 V (1198781) + 120576

100381710038171003817100381711990911003817100381710038171003817 lt100381710038171003817100381711989211003817100381710038171003817 +100381710038171003817100381711990911003817100381710038171003817 = 1

(115)

a contradiction This proves the claimWe define the operator 119878

2 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883

by 1198782(119891 119909) = 119878

1(119891 0) for every 119891 isin 119871

1(120583|

119860) and for every

119909 isin 119883 Then we have

V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 = 1

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ 120576 (116)

Indeed from Claim 1 we have

V (1198781) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 ⩽ V (119878

2)

(117)

On the other hand we see that1003816100381610038161003816(119891

lowast 119909

lowast) 119878

2(119891 119909)

1003816100381610038161003816

=1003816100381610038161003816(119891

lowast 119909

lowast) 119878

1(119891 0)

1003816100381610038161003816 ⩽10038171003817100381710038171198911003817100381710038171003817 V (1198781) ⩽ V (119878

1)

(118)

for every ((119891lowast 119909lowast) (119891 119909)) isin Π(1198711(120583|

119860)oplus

1119883) Therefore

V(1198782) ⩽ V(119878

1) Also

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ sup

119909isin119878119883

10038171003817100381710038171198781 (0 119909)1003817100381710038171003817

= sup119909isin119878119883

10038171003817100381710038171198781 (0 119909) minus 1198780 (0 119909)1003817100381710038171003817 ⩽ 120576

(119)

Claim 2 There exists 1198783 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883 such

that 1198783(119892

1 0) = 119878

3 = 1 119878

3(0 119909) = 0 119878

3(119891 119909) isin 0oplus

1119883

for every (119891 119909) isin 1198711(120583|

119860)oplus

1119883 and 119878

3minus 119878

2 lt 4120576

Indeed write 1198781= (119863

1 119863

2) where 119863

1 119871

1(120583|

119860)oplus

1119883 rarr

1198711(120583|

119860) and119863

2 119871

1(120583|

119860)oplus

1119883 rarr 119883 We have that

sup 1003816100381610038161003816119892lowast119863

1(119892

1 0) + 119909

lowast119863

2(119892

1 0)1003816100381610038161003816

119909lowastisin 119878

119883lowast ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860) +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⩽ V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119892

lowast

1119863

1(119892

1 0) + 119909

lowast

1119863

2(119892

1 0)1003816100381610038161003816

(120)

This implies that

1003816100381610038161003816119909lowast

1119863

2(119892

1 0)1003816100381610038161003816 =10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

1003816100381610038161003816119892lowast

1119863

1(119892

1 0)1003816100381610038161003816

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119871

infin(120583|

119860)

(121)

Therefore |119892lowast1| equals 1 on the support of 119863

1(119892

1 0) As

|⟨119892lowast

1 119892

1⟩| = 1 we also have that |119892lowast

1| equals 1 on the support

of 1198921 Changing the values of 119892lowast

1by the ones of 119891lowast

0on 119860

(supp(1198631(119892

1 0)) cup supp(g

1)) we may and do suppose that

|119892lowast

1| = 1 on the whole 119860We also have 119863

2(119892

1 0) gt 0 Indeed

10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)1003817100381710038171003817

⩽10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198921 0)

1003817100381710038171003817 +10038171003817100381710038171198780 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817

lt 2120576 + 120576 = 3120576

(122)

So we have

10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

⩽10038171003817100381710038171198631(119892

1 0)1003817100381710038171003817 +10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

14 Abstract and Applied Analysis

=1003817100381710038171003817(1198631

(1198921 0) 119863

2(119892

1 0)) minus (0 119879

1198601198920)1003817100381710038171003817

=10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817 lt 3120576

(123)

and 1198632(119892

1 0) gt 119879

1198601198920 minus 3120576 ⩾ 1 minus 120578(120576

0) minus 3120576 gt 0

Finally define the operator 1198783by

1198783(119891 119909) = (0119863

2(119891 0) + 119892

lowast

1(119863

1(119891 0))

1198632(119892

1 0)

119909lowast

1119863

2(119892

1 0))

(124)

for (119891 119909) isin 1198711(120583|

119860)oplus

1119883 It is clear that

100381710038171003817100381711987831003817100381710038171003817 ⩽ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) (125)

Notice also that

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817

⩽10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 +10038171003817100381710038171198632(119891 0) minus 119879

1198601198911003817100381710038171003817

=1003817100381710038171003817(1198631

(119891 0) 1198632(119891 0)) minus (0 119879

119860119891)1003817100381710038171003817

=10038171003817100381710038171198782 (119891 119909) minus 1198780 (119891 119909)

1003817100381710038171003817

(126)

for all (119891 119909) isin 1198711(120583|

119860)oplus

1119883 Hence we have

10038171003817100381710038171198783 minus 11987821003817100381710038171003817

= 2 sup119891isin1198781198711(120583|119860)

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 ⩽ 210038171003817100381710038171198782 minus 1198780

1003817100381710038171003817 lt 4120576(127)

On the other hand let 119866 1198711(120583|

119860) rarr 119871

1(120583|

119860) be defined by

119866(119891) = 119892lowast

1119891 for every 119891 isin 119871

1(120583|

119860) Then we have

V (1198782)

= sup 1003816100381610038161003816119911lowast11987821199111003816100381610038161003816 (119911 119911

lowast) isin Π (119871

1(120583|

119860) oplus

1119883)

⩾ sup10038161003816100381610038161003816100381610038161003816

119909lowast119863

2(119866(

1

120583 (119862)120594119862) 0)

+119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119909lowastisin 119878

119883lowast 119862 isin Σ

119860 120583 (119862) gt 0

= sup10038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119862)120594119862) 0)

10038171003817100381710038171003817100381710038171003817

+

10038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119862 isin Σ119860 120583 (119862) gt 0

(128)

where Σ119860is the family of measurable subsets of 119860

Hence for any simple function 119904 = sum119899

119894=1((120572

119894120583(119860

119894))120594

119860119894

) isin

1198781198711(120583|119860) where 119860

119894119894is a family of disjoint measurable subsets

with strictly positive measure we have

V (1198782)

119899

sum

119894=1

10038161003816100381610038161205721198941003816100381610038161003816 (

100381710038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381710038171003817100381710038171003817100381710038171003817

+

100381610038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381610038161003816100381610038161003816100381610038161003816

)

⩾10038171003817100381710038171198632(119866 (119904) 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119866 (119904) 0)

1003816100381610038161003816

(129)

Since |119892lowast1| = 1119866 is an isometric isomorphism so for each119891 isin

1198781198711(120583|119860)there exists a sequence of norm-one simple functions

(119904119896) such that 119866(119904

119896) converges to 119891 Therefore

V (1198782) ⩾ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) ⩾100381710038171003817100381711987831003817100381710038171003817

(130)

On the other hand we see that100381710038171003817100381711987831003817100381710038171003817 ⩾1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

3(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119909

lowast

1119863

2(119892

1 0) + 119892

lowast

1119863

1(119892

1 0)1003816100381610038161003816 = V (119878

2) = 1

(131)

Therefore 1 = 1198783 = 119878

3(119892

1 0) which proves Claim 2

Finally set 1198783= (0 ) for a suitable 119871

1(120583|

119860)oplus

1119883 rarr

119883 and define the operator 1198791 119871

1(120583) rarr 119883 by

1198791(119891) = 119879

0(119891120594

119860119888) + (119875

119860119891 0) (132)

for every 119891 isin 1198711(120583) Then we have

10038171003817100381710038171198791 (119891)1003817100381710038171003817 ⩽100381710038171003817100381711987901003817100381710038171003817

10038171003817100381710038171198911205941198601198881003817100381710038171003817 +1003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171198911205941198601003817100381710038171003817 =10038171003817100381710038171198911003817100381710038171003817 (133)

for every 119891 isin 1198711(120583) so 119879

1 ⩽ 1 Also

10038171003817100381710038171198791 (1198691198601198921)1003817100381710038171003817 =10038171003817100381710038171198783 (1198921 0)

1003817100381710038171003817 =100381710038171003817100381711987831003817100381710038171003817 = 1 (134)

so 1198791attains its norm at 119869

1198601198921isin 119871

1(120583) and

10038171003817100381710038171198691198601198921 minus 11989101003817100381710038171003817 =10038171003817100381710038171198921 minus 1198920

1003817100381710038171003817 lt 120576 (135)

We also have that for any 119891 isin 1198781198711(120583)

10038171003817100381710038171198790 (119891) minus 1198791 (119891)1003817100381710038171003817 =100381710038171003817100381710038171198790(119891120594

119860) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790 (119869119860119875119860119891) minus 119879119860 (119875119860119891)

1003817100381710038171003817

+10038171003817100381710038171003817119879119860(119875

119860119891) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790119869119860 minus 119879119860

1003817100381710038171003817 +10038171003817100381710038171198780 minus 1198783

1003817100381710038171003817

lt 120578 (1205760) + 6120576

(136)

Hence 1198790minus 119879

1 ⩽ 120578(120576

0) + 6120576 lt 120576

0

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Abstract and Applied Analysis 15

Acknowledgments

The authors thank Gilles Godefroy and Rafael Paya forfruitful conversations about the content of this paper andVladimir Kadets for providing Lemma 18 They also appre-ciate anonymous referees for careful reading and fruitfulsuggestions about revision The second author was sup-ported by Basic Science Research Program through theNational Research Foundation of Korea (NRF) fundedby the Ministry of Education Science and Technology(NRF-2012R1A1A1006869) The third author was partiallysupported by Spanish MICINN and FEDER project noMTM2012-31755 and by Junta de Andalucıa and FEDERGrants FQM-185 and P09-FQM-4911

References

[1] F F Bonsall and J Duncan Numerical Ranges of Operators onNormed Spaces and of Elements of Normed Algebras vol 2 ofLondon Mathematical Society Lecture Note Series 2 CambridgeUniversity Press London UK 1971

[2] F F Bonsall and J Duncan Numerical Ranges II CambridgeUniversity Press London UK 1973 London MathematicalSociety Lecture Notes Series 10

[3] B Sims On numerical range and its applications to Banachalgebras [PhD thesis] University of Newcastle New SouthWales Australia 1972

[4] I D Berg and B Sims ldquoDenseness of operators which attaintheir numerical radiusrdquoAustralian Mathematical Society A vol36 no 1 pp 130ndash133 1984

[5] C S Cardassi ldquoDensity of numerical radius attaining operatorson some reflexive spacesrdquo Bulletin of the Australian Mathemat-ical Society vol 31 no 1 pp 1ndash3 1985

[6] C S Cardassi ldquoNumerical radius attaining operatorsrdquo inBanach Spaces (Columbia Mo 1984) vol 1166 of Lecture Notesin Mathematics pp 11ndash14 Springer Berlin Germany 1985

[7] C S Cardassi ldquoNumerical radius-attaining operators on119862(119870)rdquoProceedings of the AmericanMathematical Society vol 95 no 4pp 537ndash543 1985

[8] MDAcostaOperadores que alcanzan su radio numerico [PhDthesis] Universidad de Granada Granada Spain 1990

[9] M D Acosta and R Paya ldquoNumerical radius attaining oper-ators and the Radon-Nikodym propertyrdquo The Bulletin of theLondon Mathematical Society vol 25 no 1 pp 67ndash73 1993

[10] R Paya ldquoA counterexample on numerical radius attainingoperatorsrdquo Israel Journal of Mathematics vol 79 no 1 pp 83ndash101 1992

[11] Y S Choi and S G Kim ldquoNorm or numerical radius attainingmultilinear mappings and polynomialsrdquo Journal of the LondonMathematical Society vol 54 no 1 pp 135ndash147 1996

[12] M D Acosta J Becerra Guerrero and M Ruiz GalanldquoNumerical-radius-attaining polynomialsrdquo The Quarterly Jour-nal of Mathematics vol 54 no 1 pp 1ndash10 2003

[13] M D Acosta and S G Kim ldquoDenseness of holomorphicfunctions attaining their numerical radiirdquo Israel Journal ofMathematics vol 161 pp 373ndash386 2007

[14] S G Kim and H J Lee ldquoNumerical peak holomorphicfunctions on Banach spacesrdquo Journal of Mathematical Analysisand Applications vol 364 no 2 pp 437ndash452 2010

[15] M D Acosta R M Aron D Garcıa and M MaestreldquoThe Bishop-Phelps-Bollobas theorem for operatorsrdquo Journal ofFunctional Analysis vol 254 no 11 pp 2780ndash2799 2008

[16] A J Guirao and O Kozhushkina ldquoThe Bishop-Phelps-Bollobasproperty for numerical radius in ℓ

1(119862)rdquo Studia Mathematica

vol 218 no 1 pp 41ndash54 2013[17] J Falco ldquoThe Bishop-Phelps-Bollobas property for numerical

radius on 1198711rdquo Journal of Mathematical Analysis and Applica-

tions vol 414 no 1 pp 125ndash133 2014[18] A Aviles A J Guirao and J Rodriguez ldquoOn the Bishop-

Phelps-Bollobas property for numerical radius in119862(119870)-spacesrdquoPreprint

[19] RM Aron Y S Choi S K Kim H J Lee andMMartin ldquoTheBishop-Phelps-Bollobas version of Lindenstrauss properties Aand Brdquo Preprint

[20] V Kadets M Martın and R Paya ldquoRecent progress and openquestions on the numerical index of Banach spacesrdquo Revista dela Real Academia de Ciencias Exactas Fısicas y Naturales A vol100 no 1-2 pp 155ndash182 2006

[21] M Martın J Merı and M Popov ldquoOn the numerical index ofreal 119871

119901(120583)-spacesrdquo Israel Journal of Mathematics vol 184 pp

183ndash192 2011[22] A Iwanik ldquoNorm attaining operators on Lebesgue spacesrdquo

Pacific Journal of Mathematics vol 83 no 2 pp 381ndash386 1979[23] HH SchaeferBanach Lattices and Positive Operators Springer

New York NY USA 1974[24] J Duncan C M McGregor J D Pryce and A J White ldquoThe

numerical index of a normed spacerdquo Journal of the LondonMathematical Society Second Series vol 2 pp 481ndash488 1970

[25] Y S Choi S K Kim H J Lee and M Martin ldquoThe Bishop-Phelps-Bollobas theorem for operators on 119871

1(120583)rdquo Preprint

[26] H E Lacey The Isometric Theory of Classical Banach SpacesSpringer New York NY USA 1974

[27] G Godefroy ldquoUniversal spaces for strictly convex Banachspacesrdquo Revista de la Real Academia de Ciencias Exactas Fısicasy Naturales A vol 100 no 1-2 pp 137ndash146 2006

[28] C Finet M Martın and R Paya ldquoNumerical index andrenormingrdquo Proceedings of the American Mathematical Societyvol 131 no 3 pp 871ndash877 2003

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Stochastic AnalysisInternational Journal of

Page 13: Research Article On the Bishop-Phelps-Bollobás Property ...downloads.hindawi.com/journals/aaa/2014/479208.pdf · Research Article On the Bishop-Phelps-Bollobás Property for Numerical

Abstract and Applied Analysis 13

It is immediate that

(119892lowast

0 119909

lowast

0) 119878

0(119892

0 0) = 119909

lowast

0(119879

1198601198920) =10038171003817100381710038171198791198601198920

1003817100381710038171003817 gt 1 minus 120578 (1205760)

(111)

By Lemma 19 1198711(120583|

119860)oplus

1119883 has the BPBp-nu with the func-

tion 120578 Therefore there exist 1198781isin L(119871

1(120583|

119860)oplus

1119883) (119892

1 119909

1) isin

1198781198711(120583|119860)oplus1119883 and (119892lowast

1 119909

lowast

1) isin 119878

119871infin(120583|119860)oplusinfin119883lowast such that

1003817100381710038171003817(1198921 1199091) minus (1198920 0)1003817100381710038171003817 lt 120576

1003817100381710038171003817(119892lowast

1 119909

lowast

1) minus (119891

lowast

0 119909

lowast

0)1003817100381710038171003817 lt 120576

10038171003817100381710038171198781 minus 11987801003817100381710038171003817 lt 120576 ⟨(119892

lowast

1 119909

lowast

1) (119892

1 119909

1)⟩ = 1

1003816100381610038161003816(119892lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 = V (119878

1) = 1

(112)

Claim 1 We claim that 1199091= 0

Otherwise

1 = Re ⟨(119892lowast1 119909

lowast

1) (119892

1 119909

1)⟩

=100381710038171003817100381711989211003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

+100381710038171003817100381711990911003817100381710038171003817Re⟨(119892

lowast

1 119909

lowast

1) (0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

⟩ ⩽ 1

(113)

We deduce that

((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

(119892lowast

1 119909

lowast

1)) (

(0 1199091)

100381710038171003817100381711990911003817100381710038171003817

(119892lowast

1 119909

lowast

1))

isin Π (1198711(120583|

119860) oplus

1119883)

(114)

Since 1198781((0 119909

1)119909

1) = (119878

1minus 119878

0)((0 119909

1)119909

1) lt 120576 we

get that

1 =1003816100381610038161003816⟨(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)⟩1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

100381710038171003817100381711989211003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((119892

1 0)

100381710038171003817100381711989211003817100381710038171003817

)⟩

+100381710038171003817100381711990911003817100381710038171003817⟨(119892

lowast

1 119909

lowast

1) 119878

1((0 119909

1)

100381710038171003817100381711990911003817100381710038171003817

)⟩

100381610038161003816100381610038161003816100381610038161003816

⩽100381710038171003817100381711989211003817100381710038171003817 V (1198781) + 120576

100381710038171003817100381711990911003817100381710038171003817 lt100381710038171003817100381711989211003817100381710038171003817 +100381710038171003817100381711990911003817100381710038171003817 = 1

(115)

a contradiction This proves the claimWe define the operator 119878

2 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883

by 1198782(119891 119909) = 119878

1(119891 0) for every 119891 isin 119871

1(120583|

119860) and for every

119909 isin 119883 Then we have

V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 = 1

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ 120576 (116)

Indeed from Claim 1 we have

V (1198781) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 119909

1)1003816100381610038161003816 =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

1(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816 ⩽ V (119878

2)

(117)

On the other hand we see that1003816100381610038161003816(119891

lowast 119909

lowast) 119878

2(119891 119909)

1003816100381610038161003816

=1003816100381610038161003816(119891

lowast 119909

lowast) 119878

1(119891 0)

1003816100381610038161003816 ⩽10038171003817100381710038171198911003817100381710038171003817 V (1198781) ⩽ V (119878

1)

(118)

for every ((119891lowast 119909lowast) (119891 119909)) isin Π(1198711(120583|

119860)oplus

1119883) Therefore

V(1198782) ⩽ V(119878

1) Also

10038171003817100381710038171198781 minus 11987821003817100381710038171003817 ⩽ sup

119909isin119878119883

10038171003817100381710038171198781 (0 119909)1003817100381710038171003817

= sup119909isin119878119883

10038171003817100381710038171198781 (0 119909) minus 1198780 (0 119909)1003817100381710038171003817 ⩽ 120576

(119)

Claim 2 There exists 1198783 119871

1(120583|

119860)oplus

1119883 rarr 119871

1(120583|

119860)oplus

1119883 such

that 1198783(119892

1 0) = 119878

3 = 1 119878

3(0 119909) = 0 119878

3(119891 119909) isin 0oplus

1119883

for every (119891 119909) isin 1198711(120583|

119860)oplus

1119883 and 119878

3minus 119878

2 lt 4120576

Indeed write 1198781= (119863

1 119863

2) where 119863

1 119871

1(120583|

119860)oplus

1119883 rarr

1198711(120583|

119860) and119863

2 119871

1(120583|

119860)oplus

1119883 rarr 119883 We have that

sup 1003816100381610038161003816119892lowast119863

1(119892

1 0) + 119909

lowast119863

2(119892

1 0)1003816100381610038161003816

119909lowastisin 119878

119883lowast ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860)

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816

⟨119892lowast 119892

1⟩ = 1 119892

lowastisin 119878

119871infin(120583|119860) +10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

⩽ V (1198782) =1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

2(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119892

lowast

1119863

1(119892

1 0) + 119909

lowast

1119863

2(119892

1 0)1003816100381610038161003816

(120)

This implies that

1003816100381610038161003816119909lowast

1119863

2(119892

1 0)1003816100381610038161003816 =10038171003817100381710038171198632(119892

1 0)1003817100381710038171003817

1003816100381610038161003816119892lowast

1119863

1(119892

1 0)1003816100381610038161003816

= sup 1003816100381610038161003816119892lowast119863

1(119892

1 0)1003816100381610038161003816 ⟨119892

lowast 119892

1⟩ = 1 119892

lowastisin 119871

infin(120583|

119860)

(121)

Therefore |119892lowast1| equals 1 on the support of 119863

1(119892

1 0) As

|⟨119892lowast

1 119892

1⟩| = 1 we also have that |119892lowast

1| equals 1 on the support

of 1198921 Changing the values of 119892lowast

1by the ones of 119891lowast

0on 119860

(supp(1198631(119892

1 0)) cup supp(g

1)) we may and do suppose that

|119892lowast

1| = 1 on the whole 119860We also have 119863

2(119892

1 0) gt 0 Indeed

10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)1003817100381710038171003817

⩽10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198921 0)

1003817100381710038171003817 +10038171003817100381710038171198780 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817

lt 2120576 + 120576 = 3120576

(122)

So we have

10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

⩽10038171003817100381710038171198631(119892

1 0)1003817100381710038171003817 +10038171003817100381710038171198632(119892

1 0) minus 119879

1198601198920

1003817100381710038171003817

14 Abstract and Applied Analysis

=1003817100381710038171003817(1198631

(1198921 0) 119863

2(119892

1 0)) minus (0 119879

1198601198920)1003817100381710038171003817

=10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817 lt 3120576

(123)

and 1198632(119892

1 0) gt 119879

1198601198920 minus 3120576 ⩾ 1 minus 120578(120576

0) minus 3120576 gt 0

Finally define the operator 1198783by

1198783(119891 119909) = (0119863

2(119891 0) + 119892

lowast

1(119863

1(119891 0))

1198632(119892

1 0)

119909lowast

1119863

2(119892

1 0))

(124)

for (119891 119909) isin 1198711(120583|

119860)oplus

1119883 It is clear that

100381710038171003817100381711987831003817100381710038171003817 ⩽ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) (125)

Notice also that

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817

⩽10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 +10038171003817100381710038171198632(119891 0) minus 119879

1198601198911003817100381710038171003817

=1003817100381710038171003817(1198631

(119891 0) 1198632(119891 0)) minus (0 119879

119860119891)1003817100381710038171003817

=10038171003817100381710038171198782 (119891 119909) minus 1198780 (119891 119909)

1003817100381710038171003817

(126)

for all (119891 119909) isin 1198711(120583|

119860)oplus

1119883 Hence we have

10038171003817100381710038171198783 minus 11987821003817100381710038171003817

= 2 sup119891isin1198781198711(120583|119860)

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 ⩽ 210038171003817100381710038171198782 minus 1198780

1003817100381710038171003817 lt 4120576(127)

On the other hand let 119866 1198711(120583|

119860) rarr 119871

1(120583|

119860) be defined by

119866(119891) = 119892lowast

1119891 for every 119891 isin 119871

1(120583|

119860) Then we have

V (1198782)

= sup 1003816100381610038161003816119911lowast11987821199111003816100381610038161003816 (119911 119911

lowast) isin Π (119871

1(120583|

119860) oplus

1119883)

⩾ sup10038161003816100381610038161003816100381610038161003816

119909lowast119863

2(119866(

1

120583 (119862)120594119862) 0)

+119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119909lowastisin 119878

119883lowast 119862 isin Σ

119860 120583 (119862) gt 0

= sup10038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119862)120594119862) 0)

10038171003817100381710038171003817100381710038171003817

+

10038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119862 isin Σ119860 120583 (119862) gt 0

(128)

where Σ119860is the family of measurable subsets of 119860

Hence for any simple function 119904 = sum119899

119894=1((120572

119894120583(119860

119894))120594

119860119894

) isin

1198781198711(120583|119860) where 119860

119894119894is a family of disjoint measurable subsets

with strictly positive measure we have

V (1198782)

119899

sum

119894=1

10038161003816100381610038161205721198941003816100381610038161003816 (

100381710038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381710038171003817100381710038171003817100381710038171003817

+

100381610038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381610038161003816100381610038161003816100381610038161003816

)

⩾10038171003817100381710038171198632(119866 (119904) 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119866 (119904) 0)

1003816100381610038161003816

(129)

Since |119892lowast1| = 1119866 is an isometric isomorphism so for each119891 isin

1198781198711(120583|119860)there exists a sequence of norm-one simple functions

(119904119896) such that 119866(119904

119896) converges to 119891 Therefore

V (1198782) ⩾ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) ⩾100381710038171003817100381711987831003817100381710038171003817

(130)

On the other hand we see that100381710038171003817100381711987831003817100381710038171003817 ⩾1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

3(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119909

lowast

1119863

2(119892

1 0) + 119892

lowast

1119863

1(119892

1 0)1003816100381610038161003816 = V (119878

2) = 1

(131)

Therefore 1 = 1198783 = 119878

3(119892

1 0) which proves Claim 2

Finally set 1198783= (0 ) for a suitable 119871

1(120583|

119860)oplus

1119883 rarr

119883 and define the operator 1198791 119871

1(120583) rarr 119883 by

1198791(119891) = 119879

0(119891120594

119860119888) + (119875

119860119891 0) (132)

for every 119891 isin 1198711(120583) Then we have

10038171003817100381710038171198791 (119891)1003817100381710038171003817 ⩽100381710038171003817100381711987901003817100381710038171003817

10038171003817100381710038171198911205941198601198881003817100381710038171003817 +1003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171198911205941198601003817100381710038171003817 =10038171003817100381710038171198911003817100381710038171003817 (133)

for every 119891 isin 1198711(120583) so 119879

1 ⩽ 1 Also

10038171003817100381710038171198791 (1198691198601198921)1003817100381710038171003817 =10038171003817100381710038171198783 (1198921 0)

1003817100381710038171003817 =100381710038171003817100381711987831003817100381710038171003817 = 1 (134)

so 1198791attains its norm at 119869

1198601198921isin 119871

1(120583) and

10038171003817100381710038171198691198601198921 minus 11989101003817100381710038171003817 =10038171003817100381710038171198921 minus 1198920

1003817100381710038171003817 lt 120576 (135)

We also have that for any 119891 isin 1198781198711(120583)

10038171003817100381710038171198790 (119891) minus 1198791 (119891)1003817100381710038171003817 =100381710038171003817100381710038171198790(119891120594

119860) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790 (119869119860119875119860119891) minus 119879119860 (119875119860119891)

1003817100381710038171003817

+10038171003817100381710038171003817119879119860(119875

119860119891) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790119869119860 minus 119879119860

1003817100381710038171003817 +10038171003817100381710038171198780 minus 1198783

1003817100381710038171003817

lt 120578 (1205760) + 6120576

(136)

Hence 1198790minus 119879

1 ⩽ 120578(120576

0) + 6120576 lt 120576

0

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Abstract and Applied Analysis 15

Acknowledgments

The authors thank Gilles Godefroy and Rafael Paya forfruitful conversations about the content of this paper andVladimir Kadets for providing Lemma 18 They also appre-ciate anonymous referees for careful reading and fruitfulsuggestions about revision The second author was sup-ported by Basic Science Research Program through theNational Research Foundation of Korea (NRF) fundedby the Ministry of Education Science and Technology(NRF-2012R1A1A1006869) The third author was partiallysupported by Spanish MICINN and FEDER project noMTM2012-31755 and by Junta de Andalucıa and FEDERGrants FQM-185 and P09-FQM-4911

References

[1] F F Bonsall and J Duncan Numerical Ranges of Operators onNormed Spaces and of Elements of Normed Algebras vol 2 ofLondon Mathematical Society Lecture Note Series 2 CambridgeUniversity Press London UK 1971

[2] F F Bonsall and J Duncan Numerical Ranges II CambridgeUniversity Press London UK 1973 London MathematicalSociety Lecture Notes Series 10

[3] B Sims On numerical range and its applications to Banachalgebras [PhD thesis] University of Newcastle New SouthWales Australia 1972

[4] I D Berg and B Sims ldquoDenseness of operators which attaintheir numerical radiusrdquoAustralian Mathematical Society A vol36 no 1 pp 130ndash133 1984

[5] C S Cardassi ldquoDensity of numerical radius attaining operatorson some reflexive spacesrdquo Bulletin of the Australian Mathemat-ical Society vol 31 no 1 pp 1ndash3 1985

[6] C S Cardassi ldquoNumerical radius attaining operatorsrdquo inBanach Spaces (Columbia Mo 1984) vol 1166 of Lecture Notesin Mathematics pp 11ndash14 Springer Berlin Germany 1985

[7] C S Cardassi ldquoNumerical radius-attaining operators on119862(119870)rdquoProceedings of the AmericanMathematical Society vol 95 no 4pp 537ndash543 1985

[8] MDAcostaOperadores que alcanzan su radio numerico [PhDthesis] Universidad de Granada Granada Spain 1990

[9] M D Acosta and R Paya ldquoNumerical radius attaining oper-ators and the Radon-Nikodym propertyrdquo The Bulletin of theLondon Mathematical Society vol 25 no 1 pp 67ndash73 1993

[10] R Paya ldquoA counterexample on numerical radius attainingoperatorsrdquo Israel Journal of Mathematics vol 79 no 1 pp 83ndash101 1992

[11] Y S Choi and S G Kim ldquoNorm or numerical radius attainingmultilinear mappings and polynomialsrdquo Journal of the LondonMathematical Society vol 54 no 1 pp 135ndash147 1996

[12] M D Acosta J Becerra Guerrero and M Ruiz GalanldquoNumerical-radius-attaining polynomialsrdquo The Quarterly Jour-nal of Mathematics vol 54 no 1 pp 1ndash10 2003

[13] M D Acosta and S G Kim ldquoDenseness of holomorphicfunctions attaining their numerical radiirdquo Israel Journal ofMathematics vol 161 pp 373ndash386 2007

[14] S G Kim and H J Lee ldquoNumerical peak holomorphicfunctions on Banach spacesrdquo Journal of Mathematical Analysisand Applications vol 364 no 2 pp 437ndash452 2010

[15] M D Acosta R M Aron D Garcıa and M MaestreldquoThe Bishop-Phelps-Bollobas theorem for operatorsrdquo Journal ofFunctional Analysis vol 254 no 11 pp 2780ndash2799 2008

[16] A J Guirao and O Kozhushkina ldquoThe Bishop-Phelps-Bollobasproperty for numerical radius in ℓ

1(119862)rdquo Studia Mathematica

vol 218 no 1 pp 41ndash54 2013[17] J Falco ldquoThe Bishop-Phelps-Bollobas property for numerical

radius on 1198711rdquo Journal of Mathematical Analysis and Applica-

tions vol 414 no 1 pp 125ndash133 2014[18] A Aviles A J Guirao and J Rodriguez ldquoOn the Bishop-

Phelps-Bollobas property for numerical radius in119862(119870)-spacesrdquoPreprint

[19] RM Aron Y S Choi S K Kim H J Lee andMMartin ldquoTheBishop-Phelps-Bollobas version of Lindenstrauss properties Aand Brdquo Preprint

[20] V Kadets M Martın and R Paya ldquoRecent progress and openquestions on the numerical index of Banach spacesrdquo Revista dela Real Academia de Ciencias Exactas Fısicas y Naturales A vol100 no 1-2 pp 155ndash182 2006

[21] M Martın J Merı and M Popov ldquoOn the numerical index ofreal 119871

119901(120583)-spacesrdquo Israel Journal of Mathematics vol 184 pp

183ndash192 2011[22] A Iwanik ldquoNorm attaining operators on Lebesgue spacesrdquo

Pacific Journal of Mathematics vol 83 no 2 pp 381ndash386 1979[23] HH SchaeferBanach Lattices and Positive Operators Springer

New York NY USA 1974[24] J Duncan C M McGregor J D Pryce and A J White ldquoThe

numerical index of a normed spacerdquo Journal of the LondonMathematical Society Second Series vol 2 pp 481ndash488 1970

[25] Y S Choi S K Kim H J Lee and M Martin ldquoThe Bishop-Phelps-Bollobas theorem for operators on 119871

1(120583)rdquo Preprint

[26] H E Lacey The Isometric Theory of Classical Banach SpacesSpringer New York NY USA 1974

[27] G Godefroy ldquoUniversal spaces for strictly convex Banachspacesrdquo Revista de la Real Academia de Ciencias Exactas Fısicasy Naturales A vol 100 no 1-2 pp 137ndash146 2006

[28] C Finet M Martın and R Paya ldquoNumerical index andrenormingrdquo Proceedings of the American Mathematical Societyvol 131 no 3 pp 871ndash877 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 14: Research Article On the Bishop-Phelps-Bollobás Property ...downloads.hindawi.com/journals/aaa/2014/479208.pdf · Research Article On the Bishop-Phelps-Bollobás Property for Numerical

14 Abstract and Applied Analysis

=1003817100381710038171003817(1198631

(1198921 0) 119863

2(119892

1 0)) minus (0 119879

1198601198920)1003817100381710038171003817

=10038171003817100381710038171198782 (1198921 0) minus 1198780 (1198920 0)

1003817100381710038171003817 lt 3120576

(123)

and 1198632(119892

1 0) gt 119879

1198601198920 minus 3120576 ⩾ 1 minus 120578(120576

0) minus 3120576 gt 0

Finally define the operator 1198783by

1198783(119891 119909) = (0119863

2(119891 0) + 119892

lowast

1(119863

1(119891 0))

1198632(119892

1 0)

119909lowast

1119863

2(119892

1 0))

(124)

for (119891 119909) isin 1198711(120583|

119860)oplus

1119883 It is clear that

100381710038171003817100381711987831003817100381710038171003817 ⩽ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) (125)

Notice also that

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817

⩽10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 +10038171003817100381710038171198632(119891 0) minus 119879

1198601198911003817100381710038171003817

=1003817100381710038171003817(1198631

(119891 0) 1198632(119891 0)) minus (0 119879

119860119891)1003817100381710038171003817

=10038171003817100381710038171198782 (119891 119909) minus 1198780 (119891 119909)

1003817100381710038171003817

(126)

for all (119891 119909) isin 1198711(120583|

119860)oplus

1119883 Hence we have

10038171003817100381710038171198783 minus 11987821003817100381710038171003817

= 2 sup119891isin1198781198711(120583|119860)

10038171003817100381710038171198631(119891 0)

1003817100381710038171003817 ⩽ 210038171003817100381710038171198782 minus 1198780

1003817100381710038171003817 lt 4120576(127)

On the other hand let 119866 1198711(120583|

119860) rarr 119871

1(120583|

119860) be defined by

119866(119891) = 119892lowast

1119891 for every 119891 isin 119871

1(120583|

119860) Then we have

V (1198782)

= sup 1003816100381610038161003816119911lowast11987821199111003816100381610038161003816 (119911 119911

lowast) isin Π (119871

1(120583|

119860) oplus

1119883)

⩾ sup10038161003816100381610038161003816100381610038161003816

119909lowast119863

2(119866(

1

120583 (119862)120594119862) 0)

+119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119909lowastisin 119878

119883lowast 119862 isin Σ

119860 120583 (119862) gt 0

= sup10038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119862)120594119862) 0)

10038171003817100381710038171003817100381710038171003817

+

10038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119862)120594119862) 0)

10038161003816100381610038161003816100381610038161003816

119862 isin Σ119860 120583 (119862) gt 0

(128)

where Σ119860is the family of measurable subsets of 119860

Hence for any simple function 119904 = sum119899

119894=1((120572

119894120583(119860

119894))120594

119860119894

) isin

1198781198711(120583|119860) where 119860

119894119894is a family of disjoint measurable subsets

with strictly positive measure we have

V (1198782)

119899

sum

119894=1

10038161003816100381610038161205721198941003816100381610038161003816 (

100381710038171003817100381710038171003817100381710038171003817

1198632(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381710038171003817100381710038171003817100381710038171003817

+

100381610038161003816100381610038161003816100381610038161003816

119892lowast

1119863

1(119866(

1

120583 (119860119894)120594119860119894

) 0)

100381610038161003816100381610038161003816100381610038161003816

)

⩾10038171003817100381710038171198632(119866 (119904) 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119866 (119904) 0)

1003816100381610038161003816

(129)

Since |119892lowast1| = 1119866 is an isometric isomorphism so for each119891 isin

1198781198711(120583|119860)there exists a sequence of norm-one simple functions

(119904119896) such that 119866(119904

119896) converges to 119891 Therefore

V (1198782) ⩾ sup

119891isin1198781198711(120583|119860)

(10038171003817100381710038171198632(119891 0)

1003817100381710038171003817 +1003816100381610038161003816119892

lowast

1119863

1(119891 0)

1003816100381610038161003816) ⩾100381710038171003817100381711987831003817100381710038171003817

(130)

On the other hand we see that100381710038171003817100381711987831003817100381710038171003817 ⩾1003816100381610038161003816(119892

lowast

1 119909

lowast

1) 119878

3(119892

1 0)1003816100381610038161003816

=1003816100381610038161003816119909

lowast

1119863

2(119892

1 0) + 119892

lowast

1119863

1(119892

1 0)1003816100381610038161003816 = V (119878

2) = 1

(131)

Therefore 1 = 1198783 = 119878

3(119892

1 0) which proves Claim 2

Finally set 1198783= (0 ) for a suitable 119871

1(120583|

119860)oplus

1119883 rarr

119883 and define the operator 1198791 119871

1(120583) rarr 119883 by

1198791(119891) = 119879

0(119891120594

119860119888) + (119875

119860119891 0) (132)

for every 119891 isin 1198711(120583) Then we have

10038171003817100381710038171198791 (119891)1003817100381710038171003817 ⩽100381710038171003817100381711987901003817100381710038171003817

10038171003817100381710038171198911205941198601198881003817100381710038171003817 +1003817100381710038171003817100381710038171003817100381710038171003817

10038171003817100381710038171198911205941198601003817100381710038171003817 =10038171003817100381710038171198911003817100381710038171003817 (133)

for every 119891 isin 1198711(120583) so 119879

1 ⩽ 1 Also

10038171003817100381710038171198791 (1198691198601198921)1003817100381710038171003817 =10038171003817100381710038171198783 (1198921 0)

1003817100381710038171003817 =100381710038171003817100381711987831003817100381710038171003817 = 1 (134)

so 1198791attains its norm at 119869

1198601198921isin 119871

1(120583) and

10038171003817100381710038171198691198601198921 minus 11989101003817100381710038171003817 =10038171003817100381710038171198921 minus 1198920

1003817100381710038171003817 lt 120576 (135)

We also have that for any 119891 isin 1198781198711(120583)

10038171003817100381710038171198790 (119891) minus 1198791 (119891)1003817100381710038171003817 =100381710038171003817100381710038171198790(119891120594

119860) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790 (119869119860119875119860119891) minus 119879119860 (119875119860119891)

1003817100381710038171003817

+10038171003817100381710038171003817119879119860(119875

119860119891) minus (119875

119860119891 0)10038171003817100381710038171003817

⩽10038171003817100381710038171198790119869119860 minus 119879119860

1003817100381710038171003817 +10038171003817100381710038171198780 minus 1198783

1003817100381710038171003817

lt 120578 (1205760) + 6120576

(136)

Hence 1198790minus 119879

1 ⩽ 120578(120576

0) + 6120576 lt 120576

0

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Abstract and Applied Analysis 15

Acknowledgments

The authors thank Gilles Godefroy and Rafael Paya forfruitful conversations about the content of this paper andVladimir Kadets for providing Lemma 18 They also appre-ciate anonymous referees for careful reading and fruitfulsuggestions about revision The second author was sup-ported by Basic Science Research Program through theNational Research Foundation of Korea (NRF) fundedby the Ministry of Education Science and Technology(NRF-2012R1A1A1006869) The third author was partiallysupported by Spanish MICINN and FEDER project noMTM2012-31755 and by Junta de Andalucıa and FEDERGrants FQM-185 and P09-FQM-4911

References

[1] F F Bonsall and J Duncan Numerical Ranges of Operators onNormed Spaces and of Elements of Normed Algebras vol 2 ofLondon Mathematical Society Lecture Note Series 2 CambridgeUniversity Press London UK 1971

[2] F F Bonsall and J Duncan Numerical Ranges II CambridgeUniversity Press London UK 1973 London MathematicalSociety Lecture Notes Series 10

[3] B Sims On numerical range and its applications to Banachalgebras [PhD thesis] University of Newcastle New SouthWales Australia 1972

[4] I D Berg and B Sims ldquoDenseness of operators which attaintheir numerical radiusrdquoAustralian Mathematical Society A vol36 no 1 pp 130ndash133 1984

[5] C S Cardassi ldquoDensity of numerical radius attaining operatorson some reflexive spacesrdquo Bulletin of the Australian Mathemat-ical Society vol 31 no 1 pp 1ndash3 1985

[6] C S Cardassi ldquoNumerical radius attaining operatorsrdquo inBanach Spaces (Columbia Mo 1984) vol 1166 of Lecture Notesin Mathematics pp 11ndash14 Springer Berlin Germany 1985

[7] C S Cardassi ldquoNumerical radius-attaining operators on119862(119870)rdquoProceedings of the AmericanMathematical Society vol 95 no 4pp 537ndash543 1985

[8] MDAcostaOperadores que alcanzan su radio numerico [PhDthesis] Universidad de Granada Granada Spain 1990

[9] M D Acosta and R Paya ldquoNumerical radius attaining oper-ators and the Radon-Nikodym propertyrdquo The Bulletin of theLondon Mathematical Society vol 25 no 1 pp 67ndash73 1993

[10] R Paya ldquoA counterexample on numerical radius attainingoperatorsrdquo Israel Journal of Mathematics vol 79 no 1 pp 83ndash101 1992

[11] Y S Choi and S G Kim ldquoNorm or numerical radius attainingmultilinear mappings and polynomialsrdquo Journal of the LondonMathematical Society vol 54 no 1 pp 135ndash147 1996

[12] M D Acosta J Becerra Guerrero and M Ruiz GalanldquoNumerical-radius-attaining polynomialsrdquo The Quarterly Jour-nal of Mathematics vol 54 no 1 pp 1ndash10 2003

[13] M D Acosta and S G Kim ldquoDenseness of holomorphicfunctions attaining their numerical radiirdquo Israel Journal ofMathematics vol 161 pp 373ndash386 2007

[14] S G Kim and H J Lee ldquoNumerical peak holomorphicfunctions on Banach spacesrdquo Journal of Mathematical Analysisand Applications vol 364 no 2 pp 437ndash452 2010

[15] M D Acosta R M Aron D Garcıa and M MaestreldquoThe Bishop-Phelps-Bollobas theorem for operatorsrdquo Journal ofFunctional Analysis vol 254 no 11 pp 2780ndash2799 2008

[16] A J Guirao and O Kozhushkina ldquoThe Bishop-Phelps-Bollobasproperty for numerical radius in ℓ

1(119862)rdquo Studia Mathematica

vol 218 no 1 pp 41ndash54 2013[17] J Falco ldquoThe Bishop-Phelps-Bollobas property for numerical

radius on 1198711rdquo Journal of Mathematical Analysis and Applica-

tions vol 414 no 1 pp 125ndash133 2014[18] A Aviles A J Guirao and J Rodriguez ldquoOn the Bishop-

Phelps-Bollobas property for numerical radius in119862(119870)-spacesrdquoPreprint

[19] RM Aron Y S Choi S K Kim H J Lee andMMartin ldquoTheBishop-Phelps-Bollobas version of Lindenstrauss properties Aand Brdquo Preprint

[20] V Kadets M Martın and R Paya ldquoRecent progress and openquestions on the numerical index of Banach spacesrdquo Revista dela Real Academia de Ciencias Exactas Fısicas y Naturales A vol100 no 1-2 pp 155ndash182 2006

[21] M Martın J Merı and M Popov ldquoOn the numerical index ofreal 119871

119901(120583)-spacesrdquo Israel Journal of Mathematics vol 184 pp

183ndash192 2011[22] A Iwanik ldquoNorm attaining operators on Lebesgue spacesrdquo

Pacific Journal of Mathematics vol 83 no 2 pp 381ndash386 1979[23] HH SchaeferBanach Lattices and Positive Operators Springer

New York NY USA 1974[24] J Duncan C M McGregor J D Pryce and A J White ldquoThe

numerical index of a normed spacerdquo Journal of the LondonMathematical Society Second Series vol 2 pp 481ndash488 1970

[25] Y S Choi S K Kim H J Lee and M Martin ldquoThe Bishop-Phelps-Bollobas theorem for operators on 119871

1(120583)rdquo Preprint

[26] H E Lacey The Isometric Theory of Classical Banach SpacesSpringer New York NY USA 1974

[27] G Godefroy ldquoUniversal spaces for strictly convex Banachspacesrdquo Revista de la Real Academia de Ciencias Exactas Fısicasy Naturales A vol 100 no 1-2 pp 137ndash146 2006

[28] C Finet M Martın and R Paya ldquoNumerical index andrenormingrdquo Proceedings of the American Mathematical Societyvol 131 no 3 pp 871ndash877 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 15: Research Article On the Bishop-Phelps-Bollobás Property ...downloads.hindawi.com/journals/aaa/2014/479208.pdf · Research Article On the Bishop-Phelps-Bollobás Property for Numerical

Abstract and Applied Analysis 15

Acknowledgments

The authors thank Gilles Godefroy and Rafael Paya forfruitful conversations about the content of this paper andVladimir Kadets for providing Lemma 18 They also appre-ciate anonymous referees for careful reading and fruitfulsuggestions about revision The second author was sup-ported by Basic Science Research Program through theNational Research Foundation of Korea (NRF) fundedby the Ministry of Education Science and Technology(NRF-2012R1A1A1006869) The third author was partiallysupported by Spanish MICINN and FEDER project noMTM2012-31755 and by Junta de Andalucıa and FEDERGrants FQM-185 and P09-FQM-4911

References

[1] F F Bonsall and J Duncan Numerical Ranges of Operators onNormed Spaces and of Elements of Normed Algebras vol 2 ofLondon Mathematical Society Lecture Note Series 2 CambridgeUniversity Press London UK 1971

[2] F F Bonsall and J Duncan Numerical Ranges II CambridgeUniversity Press London UK 1973 London MathematicalSociety Lecture Notes Series 10

[3] B Sims On numerical range and its applications to Banachalgebras [PhD thesis] University of Newcastle New SouthWales Australia 1972

[4] I D Berg and B Sims ldquoDenseness of operators which attaintheir numerical radiusrdquoAustralian Mathematical Society A vol36 no 1 pp 130ndash133 1984

[5] C S Cardassi ldquoDensity of numerical radius attaining operatorson some reflexive spacesrdquo Bulletin of the Australian Mathemat-ical Society vol 31 no 1 pp 1ndash3 1985

[6] C S Cardassi ldquoNumerical radius attaining operatorsrdquo inBanach Spaces (Columbia Mo 1984) vol 1166 of Lecture Notesin Mathematics pp 11ndash14 Springer Berlin Germany 1985

[7] C S Cardassi ldquoNumerical radius-attaining operators on119862(119870)rdquoProceedings of the AmericanMathematical Society vol 95 no 4pp 537ndash543 1985

[8] MDAcostaOperadores que alcanzan su radio numerico [PhDthesis] Universidad de Granada Granada Spain 1990

[9] M D Acosta and R Paya ldquoNumerical radius attaining oper-ators and the Radon-Nikodym propertyrdquo The Bulletin of theLondon Mathematical Society vol 25 no 1 pp 67ndash73 1993

[10] R Paya ldquoA counterexample on numerical radius attainingoperatorsrdquo Israel Journal of Mathematics vol 79 no 1 pp 83ndash101 1992

[11] Y S Choi and S G Kim ldquoNorm or numerical radius attainingmultilinear mappings and polynomialsrdquo Journal of the LondonMathematical Society vol 54 no 1 pp 135ndash147 1996

[12] M D Acosta J Becerra Guerrero and M Ruiz GalanldquoNumerical-radius-attaining polynomialsrdquo The Quarterly Jour-nal of Mathematics vol 54 no 1 pp 1ndash10 2003

[13] M D Acosta and S G Kim ldquoDenseness of holomorphicfunctions attaining their numerical radiirdquo Israel Journal ofMathematics vol 161 pp 373ndash386 2007

[14] S G Kim and H J Lee ldquoNumerical peak holomorphicfunctions on Banach spacesrdquo Journal of Mathematical Analysisand Applications vol 364 no 2 pp 437ndash452 2010

[15] M D Acosta R M Aron D Garcıa and M MaestreldquoThe Bishop-Phelps-Bollobas theorem for operatorsrdquo Journal ofFunctional Analysis vol 254 no 11 pp 2780ndash2799 2008

[16] A J Guirao and O Kozhushkina ldquoThe Bishop-Phelps-Bollobasproperty for numerical radius in ℓ

1(119862)rdquo Studia Mathematica

vol 218 no 1 pp 41ndash54 2013[17] J Falco ldquoThe Bishop-Phelps-Bollobas property for numerical

radius on 1198711rdquo Journal of Mathematical Analysis and Applica-

tions vol 414 no 1 pp 125ndash133 2014[18] A Aviles A J Guirao and J Rodriguez ldquoOn the Bishop-

Phelps-Bollobas property for numerical radius in119862(119870)-spacesrdquoPreprint

[19] RM Aron Y S Choi S K Kim H J Lee andMMartin ldquoTheBishop-Phelps-Bollobas version of Lindenstrauss properties Aand Brdquo Preprint

[20] V Kadets M Martın and R Paya ldquoRecent progress and openquestions on the numerical index of Banach spacesrdquo Revista dela Real Academia de Ciencias Exactas Fısicas y Naturales A vol100 no 1-2 pp 155ndash182 2006

[21] M Martın J Merı and M Popov ldquoOn the numerical index ofreal 119871

119901(120583)-spacesrdquo Israel Journal of Mathematics vol 184 pp

183ndash192 2011[22] A Iwanik ldquoNorm attaining operators on Lebesgue spacesrdquo

Pacific Journal of Mathematics vol 83 no 2 pp 381ndash386 1979[23] HH SchaeferBanach Lattices and Positive Operators Springer

New York NY USA 1974[24] J Duncan C M McGregor J D Pryce and A J White ldquoThe

numerical index of a normed spacerdquo Journal of the LondonMathematical Society Second Series vol 2 pp 481ndash488 1970

[25] Y S Choi S K Kim H J Lee and M Martin ldquoThe Bishop-Phelps-Bollobas theorem for operators on 119871

1(120583)rdquo Preprint

[26] H E Lacey The Isometric Theory of Classical Banach SpacesSpringer New York NY USA 1974

[27] G Godefroy ldquoUniversal spaces for strictly convex Banachspacesrdquo Revista de la Real Academia de Ciencias Exactas Fısicasy Naturales A vol 100 no 1-2 pp 137ndash146 2006

[28] C Finet M Martın and R Paya ldquoNumerical index andrenormingrdquo Proceedings of the American Mathematical Societyvol 131 no 3 pp 871ndash877 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 16: Research Article On the Bishop-Phelps-Bollobás Property ...downloads.hindawi.com/journals/aaa/2014/479208.pdf · Research Article On the Bishop-Phelps-Bollobás Property for Numerical

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of