mathematics without boundaries ||

Mathematics Without Boundaries

Themistocles M. Rassias Panos M. PardalosEditors

Mathematics WithoutBoundaries

Surveys in Pure Mathematics

2123

EditorsThemistocles M. Rassias Panos M. PardalosDepartment of Mathematics Dept. Industrial & Systems EngineeringNational Technical University of Athens University of FloridaAthens, Greece Gainesville, FL, USA

ISBN 978-1-4939-1105-9 ISBN 978-1-4939-1106-6 (eBook)DOI 10.1007/978-1-4939-1106-6Springer New York Heidelberg Dordrecht London

Library of Congress Control Number: 2014940540

Mathematics Subject Classification (2010): 05A16, 05A, 15, 05A18, 11B75,30C45, 26D15, 31A10, 45P05, 47G10, 47A07, 46B04, 46B20, 46A16, 46E40

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Preface

Mathematics Without Boundaries: Surveys in Pure Mathematics consists of chap-ters written by eminent scientists from the international mathematical community,who present significant advances in several theories, methods and problems ofmathematical analysis, discrete mathematics, geometry and their applications.

These contributions focus on both old and recent developments in functionalanalysis, harmonic analysis, complex analysis, operator theory, combinatorics,functional equations, differential equations as well as a variety of applications.

Furthermore some review works are published in this book which could prove tobe particularly useful for a broader audience of readers in mathematical sciences andespecially to graduate students who search for the latest information.

It is our pleasure to express our thanks to all the contributors of chapters inthis book.

We would like to thank Dr. Michael Batsyn for his invaluable help during thepreparation of this publication. Last but not least, we would like to acknowledgethe superb assistance that the staff of Springer has provided for the publication ofthis book.

Athens, Greece, Themistocles M. RassiasGainesville, FL, USA Panos M. Pardalos

v

Contents

Some Unexpected Connections Between Analysisand Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Dorin Andrica and Eugen J. Ionascu

The Hilali Conjecture for Hyperelliptic Spaces . . . . . . . . . . . . . . . . . . . . . . . . 21Javier Fernndez de Bobadilla, Javier Fresn, Vicente Muozand Aniceto Murillo

Aveiro Discretization Method in Mathematics:A New Discretization Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37L.P. Castro, H. Fujiwara, M.M. Rodrigues, S. Saitoh and V.K. Tuan

Variational Inequality Models Arising in the Studyof Viscoelastic Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93O. Chau, D. Goeleven and R. Oujja

Lucjan Emil Bttcher and his Mathematical Legacy . . . . . . . . . . . . . . . . . . . 127Stanisaw Domoradzki and Magorzata Stawiska

Spectral Properties of Toeplitz Operators Acting on Gabor TypeReproducing Kernel Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163H. G. Feichtinger, K. Nowak and M. Pap

Which Numbers Simplify Your Problem? . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181Paolo Giordano

Approximation Properties of Phillips Operators . . . . . . . . . . . . . . . . . . . . . . 221N. K. Govil and Vijay Gupta

Geometric Patterns in Uniform Distributionof Zeros of the Riemann Zeta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245Mehdi Hassani

vii

viii Contents

Fractional Cauchy Problem in Sense of the Complex HadamardOperators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259Rabha W. Ibrahim

Studies on Generalized Fractional Operatorsin Complex Domain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273Rabha W. Ibrahim

On Completely Monotonic and Related Functions . . . . . . . . . . . . . . . . . . . . . 285Stamatis Koumandos

HyersUlam Stability of Set-Valued Mappings . . . . . . . . . . . . . . . . . . . . . . . . 323Jung Rye Lee, Choonkil Park and Themistocles M. Rassias

On the Generalized HyersUlam Stability in Multi-Banach SpacesAssociated to a Jensen-type Additive Mapping . . . . . . . . . . . . . . . . . . . . . . . . 337Fridoun Moradlou and Themistocles M. Rassias

Elliptic Problems with Nonhomogeneous Differential Operatorsand Multiple Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357Dumitru Motreanu and Patrick Winkert

Isotone Retractions onto the Positive Cone of the OrderedEuclidean Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381A. B. Nmeth and S. Z. Nmeth

On the Circle Preserving Property of Mbius Transformations . . . . . . . . . 397Nihal Ylmaz zgr

Extended Crystal PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415Agostino Prstaro

Multiplicative Arithmetic Functions of Several Variables: A Survey . . . . . 483Lszl Tth

Sequential Maximality Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515Mihai Turinici

Univalence Conditions and Properties for Some New IntegralOperators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549Nicoleta Ularu and Daniel Breaz

On the Study of Hyperbolic Triangles and Circles by HyperbolicBarycentric Coordinates in Relativistic Hyperbolic Geometry . . . . . . . . . . 569Abraham Albert Ungar

Contents ix

Multidimensional Half-Discrete Hilbert-Type Inequalitiesand Operator Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 651Bicheng Yang

On the Extension Problems of Isometricand Nonexpansive Mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 725Xiuzhong Yang and Xiaopeng Zhao

Advances in Opials Type Integral Inequalities . . . . . . . . . . . . . . . . . . . . . . . . 749Chang-Jian Zhao and Wing-Sum Cheung

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 779

Contributors

Dorin Andrica Faculty of Mathematics and Computer Science, Babes-BolyaiUniversity, Cluj-Napoca, Romania

Javier Fernndez de Bobadilla Instituto de Ciencias Matemticas CSIC-UAM-UC3M-UCM, Consejo Superior de Investigaciones Cientficas, Madrid, Spain

Daniel Breaz University of Alba Iulia, Alba Iulia, Romania

L.P. Castro CIDMA-Center for Research and Development in Mathematics andApplications, Department of Mathematics, University of Aveiro, Aveiro, Portugal

O. Chau University of La Runion, PIMENT EA4518, La Runion, France

Wing-Sum Cheung Department of Mathematics, The University of Hong Kong,Hong Kong, China

Stanisaw Domoradzki Faculty of Mathematics and Natural Sciences, Universityof Rzeszw, Rzeszw, Poland

H. G. Feichtinger Faculty of Mathematics, University Vienna, Wien, Austria

Javier Fresn LAGA, UMR 7539, Institut Galile, Universit Paris 13,Villetaneuse, France

H. Fujiwara Graduate School of Informatics, Kyoto University, Kyoto, Japan

Paolo Giordano Fakultt fr Mathematik, University of Vienna, Wien, Austria

D. Goeleven University of La Runion, PIMENT EA4518, La Runion, France

N. K. Govil Department of Mathematics and Statistics, Auburn University,Auburn, USA

Vijay Gupta Department of Mathematics, Netaji Subhas Institute of Technology,Dwarka, New Delhi, India

Mehdi Hassani University of Zanjan, University Blvd., Zanjan, Iran

Rabha W. Ibrahim Institute of Mathematical Sciences, University Malaya, KualaLumpur, Malaysia

xi

xii Contributors

Eugen J. Ionascu Math Department, Columbus State University, Columbus,GA, USA

Stamatis Koumandos Department of Mathematics and Statistics, Universityof Cyprus, Nicosia, Cyprus

Jung Rye Lee Department of Mathematics, Daejin University, Pocheon, SouthKorea

Fridoun Moradlou Department of Mathematics, Sahand University of Technology,Tabriz, Iran

Dumitru Motreanu Dpartement de Mathmatiques, Universit de Perpignan,Perpignan, France

Vicente Muoz Facultad de Ciencias Matemticas, Universidad Complutense deMadrid, Madrid, Spain

Aniceto Murillo Departamento de lgebra, Geometra y Topologa, Universidadde Mlaga, Mlaga, Spain

A. B. Nmeth Faculty of Mathematics and Computer Science, Babes BolyaiUniversity, Cluj-Napoca, Romania

S. Z. Nmeth School of Mathematics, University of Birmingham, Birmingham,UK

K. Nowak Wayne, PA, USA

Department of Computer Science, Drexel University, Philadelphia, PA, USA

R. Oujja University of La Runion, PIMENT EA4518, La Runion, France

Nihal Ylmaz zgr Faculty of Science and Literature, Department of Mathemat-ics, Balkesir University, Balkesir, Turkey

M. Pap Faculty of Sciences, University of Pcs, Pcs, Hungary

Choonkil Park Research Institute for Natural Sciences, Hanyang University, Seoul,South Korea

Agostino Prstaro Department SBAI-Mathematics, University of Roma LaSapienza, Roma, Italy

Themistocles M. Rassias Department of Mathematics, National TechnicalUniversity of Athens, Athens, Greece

M.M. Rodrigues CIDMA-Center for Research and Development in Mathemat-ics and Applications, Department of Mathematics, University of Aveiro, Aveiro,Portugal

S. Saitoh Department of Mathematics, Institute of Reproducing Kernels, Kiryu,Japan

Contributors xiii

Magorzata Stawiska Mathematical Reviews, Ann Arbor, MI, USA

Lszl Tth Institute of Mathematics, Universitt fr Bodenkultur, Vienna, Austria

Department of Mathematics, University of Pcs, Pcs, Hungary

V.K. Tuan Department of Mathematics, University of West Georgia, Carrollton,GA, USA

Mihai Turinici A. Myller Mathematical Seminar, A. I. Cuza University, Iasi,Romania

Nicoleta Ularu University of Pitesti, Pitesti, Romania

Abraham Albert Ungar Department of Mathematics, North Dakota StateUniversity, Fargo, ND, USA

Patrick Winkert Technische Universitt Berlin, Institut fr Mathematik, Berlin,Germany

Bicheng Yang Department of Mathematics, Guangdong University of Education,Guangzhou, P. R. China

Xiuzhong Yang College of Mathematics and Information Science, Hebei NormalUniversity, and Hebei Key Laboratory of Computational Mathematics andApplications, Shijiazhuang, Hebei, P. R. China

Chang-Jian Zhao Department of Mathematics, China Jiliang University,Hangzhou, Peoples Republic of China

Xiaopeng Zhao Department of Mathematics, Zhejiang University, Hangzhou,Zhejiang, P. R. China

Some Unexpected Connections Between Analysisand Combinatorics

Dorin Andrica and Eugen J. Ionascu

2000 AMS Subject Classification: 05A16 (05A15; 05A18; 11B75)

Abstract We go through a series of results related to the k-signum equation 1k 2k nk = 0. We are investigating the number Sk(n) of possible writingsand the asymptotic behavior of these numbers, as k is fixed and n . Theresults are presented in connections with the ErdsSurnyi sequences. Analyticmethods and algebraic ones are employed in order to predict the asymptotic behaviorin general and to study in detail various situations for small values of k. Somesimplifications and further ramifications are discussed in the end about the recentproof of AndricaTomescu conjecture.

Keywords Derivative Partition of integers Asymptotic formula Integralrepresentation Random variable ErdsSurnyi sequence

1 Introduction

One of the simplest differential equations, f = g, for a given function g, leadsto a classic theory in analysis, part of which, the usual calculus is just the firstapproximation. If the derivative f is understood in the usual sense, it turns out thattwo necessary conditions on g are well known: it must be of first class Baire and itneeds to have the Darboux property. This is almost a characterization since, by G.Choquets result, if a function g is of the first Baire class and it has the Darbouxproperty, there exists a differentiable function f and a homeomorphism h such thatg = f h.

D. Andrica ()Faculty of Mathematics and Computer Science, Babes-Bolyai University,Str. Kogalniceanu 1, 400084, Cluj-Napoca, Romaniae-mail: [email protected]

E. J. IonascuMath Department, Columbus State University, 4225 University Avenue,31907, Columbus, GA, USAe-mail: [email protected]; [email protected]

T. M. Rassias, P. M. Pardalos (eds.), Mathematics Without Boundaries, 1DOI 10.1007/978-1-4939-1106-6_1, Springer Science+Business Media, LLC 2014

2 D. Andrica and E. J. Ionascu

We begin Sect. 2 with some results in this area about some elementary functionshaving discontinuities of the second kind and characterize when these are derivatives,using mostly elementary methods. In the end, we arrive at an analytical interpretationof the number of vectors [ 1,1, . . .,1]

n

whose dot product with a fixed vector

[a1, a2, , an] Nn gives a desired output. Of course, there is room for askingquestions like, what is the range of outputs? or what is the most favorable output?.Before reading further, we challenge the reader to find the derivative at zero for thefunction

f (x) = x

0cos

(

1

t

)

cos

(

3

t

)

cos

(

5

t

)

cos

(

7

t

)

dt , x R.

In Sect. 3, we look at classical sequences that have been studied in connectionwith questions mentioned earlier. An integral formula of the main combinatorialnumbers at interest is derived here. An asymptotic conjecture is formulated as ageneralization of AndricaTomescu [6] using an argument due to S. Finch [18].Certainly, probabilistic methods can be employed here and we give in detail thisargument based on a variant of central limit theorem due to A. C. Berry [9].

There are famous problems that appeared over the years, like in the Putnamcompetition or some other mathematics olympiads, which relate intimately to thissubject, in Sect. 4.

Finally, in the last section, we go over Sullivans proof [31] of AndricaTomescuconjecture and point out what can be shown in general and what are the difficultiesin the general case.

2 Derivatives

In this section, we introduce the class D consisting all derivatives defined on thereal axis, and in Theorem 1, we derive a result about the product of some specialderivatives, which opens the door to some difficult problems involving partitions ofintegers.

Definition 1 A function f : R R is called a derivative if there exists a differ-entiable function g : R R such that g = f . We denote by D the class of allderivatives.

From examples and properties, we noticed that any continuous function is in D.It is known that every function in D has the Darboux property, that is, it transformsany open interval into an interval (see [27]). It is then clear that D is a linear spaceunder usual addition of functions and multiplication with real scalars. Because ofthe Darboux property any function of the form

a,b(x) ={

a if x = 0b if x = 0, (1)

Some Unexpected Connections Between Analysis and Combinatorics 3

is in D if and only if a = b.Lemma 1 For a R \ {0} and b R, the function

fa,b(x) =

cos ax

if x = 0

b if x = 0,is in D if and only if b = 0.Proof We consider the function

h(x) =

x2 sin ax

if x = 0

0 if x = 0,which is easy to see that it is differentiable with

h(x) =

2x sin ax a cos a

xif x = 0

0 if x = 0,.

Hence, we see that we can write

fa,b(x) = 2a

x sin ax

if x = 0

0 if x = 0 1

ah(x) +

0 if x = 0

b if x = 0

= 2ak(x) 1

ah(x) + 0,b(x).

Since k is continuous, 2ak 1

ah D, which implies that fa,b D if and only

if 0,b D, where 0,b is a function defined in (1). The previous observation thengives us the claim.

Lemma 1 allows us to show that D is not closed under usual multiplication offunctions. Indeed, if we take f 2a,0 with fa,0 as in Lemma 1, we observe that

f 2a,0 =1

2

(

f2a,0 + 1,0)

and so, if f 2a,0 D that would lead to the contradiction 1,0 D. For other remarksinvolving such examples we refer to the paper of D. Andrica and S. Buzeteanu (85,[4]).

M. Iosifescu (58, [21]) showed that if f1 and f2 are in D, then the productf1f2 may not even have the Darboux property. We can recapture this fact by taking


= f1,0 D and

(x) =

cos2 1x

if x = 0

1/2 if x = 0,= 1/2,1/2(x) + 1

2f2,0(x)

which shows that D. Hence, the functions f1 = + and f2 = are inD. Then, we calculate f1f2 = 2 2 that can be simplified to

(f1f2) (x) = 14

sin2 2x

if x = 0

1 if x = 0.It is clear that this function transforms every interval containing 0 into [0, 1]{1},

hence, it does not have the Darboux property.In comparison to the previous example, a totally surprising result of V. Hruska

(46, [19]) says that for two functions f , g D, f/g does have the Darboux property(g = 0). Other properties of functions in D are given in the paper of A. M. Brucknerand J. L. Leonard (66, [12]).

The following lemma is an easy exercise of mathematical induction but it is veryuseful when we want to get the linear form of a product of cosines.

Lemma 2 For x1, x2, . . . , xm arbitrary real numbers, we have

cos x1 cos x2 cos xm = 12m

cos ( x1 x2 xm),

where the sum is over all possible choices of signs + and .Using this identity one can prove the following generalization of Lemma 1.

Theorem 1 (D. Andrica, 00, [3]) Let n1, , nk be positive integers such thatn1 + nk 1, 1, . . .,k mutually distinct nonzero real numbers, and the functiong1,...,kn1,...,nk : R R be defined by

g1,...,kn1,...,nk (x) ={

cosn1 1x cosnk k

xif x = 0

if x = 0.

Then, g1,...,kn1,...,nk D if and only if = 12n1++nk S(n1, . . ., nk;1, . . .,k), where theinteger S(n1, . . ., nk;1, . . .,k) is given by the number of choices of signs + and such that

1 1

n1 times

k k

nk times

= 0.

Proof Using Lemma 2, we can write

g1,...,kn1,...,nk (x) =


12n1++nk

+, cos (1

x 1

x

n1 times

kx

kx

nk times

) if x = 0

if x = 0,=

1

2n1++nk

+,=0f(1 1

n1 times

k k

nk times

,0)+

{

12n1++nk S(n1, . . ., nk;1, . . .,k) if x = 0 if x = 0 ,

where the sum

+,=0 f(1 1

n1 times

k k

nk times

,0) is taken over all com-

binations 1 1

n1 times

k k

nk times

different from 0. According to

Lemma 1, each function f(1 1

n1 times

k k

nk times

,0), corresponding to

such a combination, belongs to D. Finally, it follows that g1, ,kn1, ,nk D if and only if 1

2n1++nk S(n1, ,nk ;1, ,k ), D,

and this is equivalent to the relation = 12n1++nk S(n1, , nk;1, ,k).

As an immediate application to Theorem 1, we obtained that the constant forwhich the function g1n : R R, defined by

g1n(x) ={

cosn 1x

if x = 0 if x = 0,

is a derivative, is given by 12n S(n; 1), where S(n; 1) is the number of choices of signs+ and such that 1 1

n times

= 0. If n is odd, clearly we have S(n; 1) = 0. If n is

even, we have S(n; 1) = ( nn/2

)

, hence

=

0 if n is odd

12n(

n

n/2

)

if n is even.

This result was obtained with a different method by the first author in [2].Another application may be obtained when the distinct nonzero real numbers

1, ,k are linear independent over Q. In this case, we can write the relation1 1

n1 times

k k

nk times

= 0 as ( 1 1)

n1 times

1 ( 1 1)

nk times

k = 0,


and from the linear independence, we get 1 1

n1 times

= = 1 1

nk times

= 0.

Therefore, the function g1,...,kn1,...,nk in Theorem 1, is a derivative if and only if

=

12n1++nk

(n1n1/2

) ( nknk/2

)

if n1, , nk are even

0 otherwise.

3 A Problem About the Number of Partitions

This section is a part of the work done by the first author and I. Tomescu in [6] aboutthe interpretation of the number S(n1, . . ., nk;1, . . .,k). Let n1, . . . , nk be naturalnumbers and let M be the multiset

M = {1, ,1

n1 times

, ,k , ,k

nk times

}.

In [6], it is shown that S(n1, . . ., nk;1, . . .,k) equals the number of orderedpartitions of M having equal sums, i.e., the number of pairs (C1,C2) such that

(i) C1 C2 = M and C1 C2 = (ii)

xC1x =

xC2x = 12

ki=1 nii .

If we assume that M is a set of k positive integers then S(n1, . . ., nk;1, . . .,k) isthe term not depending on z in the expansion

F (z) =(

z1 + 1z1

)n1(

z2 + 1z2

)n2

(

zk + 1zk

)n1

.

Let us observe that we can write F (z) = S(n1, . . ., nk;1, . . .,k) + jZ\{0}

cj zj ,

for some cj Z. If we set z = cos t + i sin t , then we get the equivalent form

2n1++nkk

j=1

(

cosj t)nj = S(n1, . . ., nk;1, . . .,k) +

jZ\{0}cj ( cos j t + i sin j t).

Integrating this last identity over [0, 2 ], we obtain the following integral formulafor the number S(n1, . . ., nk;1, . . .,k) :

S(n1, . . ., nk;1, . . .,k) = 2n1++nk

2

2

0

k

j=1

(

cosj t)nj

dt. (2)


Fig. 1 Graph of f5 and f12 on [0, ]

One particular case here, which brings into our attention an old combinatorialproblem (see [17, 18, 23] 1), is if n1 = n2 = = nk = 1 and j = j , j = 1,2, , k.

Let us shorten the notation by setting S(n) := S(1, 1, , 1; 1, 2, 3, , n), whichis the number of ways of choosing + and such that 123 n = 0. Theabove equation is called by S. Finch [18] the signum equation. This is the sequenceA063865 in the The On-Line Encyclopedia of Integer Sequences. As an example,we have S(40) = 5830034720.

In the paper [6], it is shown that:(i) S(n) = 0 if and only if n 1 or 2 (mod 4)(ii) for n 8 we have S(n) 6S(n 4)(iii) limn S(4n)

14n = limn S(4n+ 3) 14n+3 = 2,

and it was conjectured the following asymptotic formula

limn

n0 or 3(mod4)

S(n)2n

nn

=

6

,

which was recently proved using analytic methods by B.D. Sullivan (2013) (see[31]).

From (2), we easily get

S(n) = 2n

0cos t cos 2t cos ntdt. (3)

1 The problem A5 on the William Putnam Competition (1985) asked to find all all positive integersm 10 such that 20

mj=1 cos jxdx is not zero.


Using this integral representation one can check for instance, that S(n) = 0 ifn 1 or 2 (mod 4), by showing that the integrant in (3), fn(t) := nj=1 cos jt isa function symmetric with respect to the point (/2, 0). For clarity, we include thegraphs of f5 and f12 in Fig. 1.

For small n, with no difficulty, integrals as in (3) can be computed using, forinstance, Mathematica or Maple. It takes only a few seconds to get S(24) = 187692.

Another particular case is if n1 = n2 = = nk = 1 and j = j 2, j = 1, 2, ,k. In this case, S2(n) := S

(

1, 1, , 1; 12, 22, 32, , n2) is the sequence A158092or the number of ways to choose + and such that 1 22 32 n2 = 0.In this case, it seems that a similar asymptotic expression yields

S2(n) = 2n

10

n5(1 + o(1)),

and Pietro Majer has one more on OEIS (see A158118)

S3(n) := S(1, 1, , 1; 13, 23, 33, , n3) = 2n

14

n7(1 + o(1))

as n , n 0 or 3 (mod 4).S. Finch [18] has a very interesting probabilistic but heuristic approach to arrive

at above two formulas, which allows one to find similar such asymptotic results. Theidea of proof suggested by H.-K. Hwang [20] does not seem to generalize to higherpowers.

For the sake of completeness, let us include next, Finchs heuristic argument aboutthe asymptotic behavior in general. For k > 0, we consider the random variable Ejtaking two values: jk or jk with probability 1/2 each. For i = j , each Ei isindependent of Ej . We then let the random variable

Sn := E1 + E2 + + En. (4)Since each of the Ej has expectation zero, we have E(Sn) = nj=1 E(Ej ) = 0.

Because each of the Ej has variation 2j = V ar(Ej ) = E(|Ej E(Ej )|2

) = j 2k ,and the random variables are independent, we get

s2n = V ar(Sn) =n

j=1V ar

(

Ej) =

n

j=1j 2k.

We would like to use the so called BerryEsseen theorem (see for example [9]),that is a better version of central limit theorem in this case.

Theorem 2 Let X1, X2, . . . , be independent random variables with E(Xj ) = 0,E(

X2j

)

= 2j > 0 and j = E(|Xj |3

)

< . If Gn = (X1 + X2 + +Xn)/

n

j=1 2j , then

supxR

|P (Gn x) 12

x

e

t22 dt | C1(

n

j=1 2j )

1/2 max1jn

j

2j, (5)


for some universal constant C1.In our case, if we take Xj = Ej , we get j = j 3k and so max

1jnj

2j= nk . The

inequality (5) then becomes

supxR

|P (Sn xsn) 12

x

e

t22 dt | C1 n

k

sn.

In particular for y < x, we have

|P (ysn < Sn xsn) 12

x

y

et22 dt | 2C1 n

k

sn.

Let us denote by Sk(n) the sequence S(1, 1, , 1; 1k , 2k , 3k , , nk). For an ar-bitrary (0, 1) and x = 1

sn, y = 1

sn, the inequality above can be written

as

1

2nSk(n) 1

2

1sn

1sn

et22 dt

2C1 nk

sn.

It is easy to show that for > 0 we have

(n) :=n

j=1j = n

+1

+ 1 +O(n), (6)

which implies sn = nk+1/22k+1 (1 + o(1)). Putting these together we see that there existan universal constant C such that

1

2nSk(n)

2(2k + 1)n2k+1

Cn.

This inequality suggests that we may conjecture in general that for every k N

limn

n0 or 3(mod4)

Sk(n)2n

nkn

=

2(2k + 1)

, (7)

but at this moment we do not have a proof for k 2.

4 Connections to ErdsSurnyi Type Problems

In this section, we present a special class of sequences of distinct positive integers,which give special representations of the integers. Some general results are presentedin Theorems 3 and 4, and then we concentrate on specific examples.


Definition 2 We say that a sequence of distinct positive integers {am}m1 is a ErdsSurnyi sequence if every integer may be written in the form

a1 a2 anin infinitely many ways.

As an example of a ErdsSurnyi sequence, for every k N, an = nk (seeJ.Mitek [25]). For instance, for k = 1 we have

m = ( 1 + 2) + ( 3 + 4) + + ( (2 m 1) + 2 m) + +

[(n+ 1) (n+ 2) (n+ 3) + (n+ 4)]

=0For k = 2, we have the original result of Erds and Surnyi mentioned as a problemin the book [17], which is based on the identity 4 = (m + 1)2 (m + 2)2 (m +3)2 + (m+ 4)2 and on the basis cases

1 = 12, 2 = 12 22 32 + 423 = 12 + 22, 4 = 12 22 32 + 42.

For k = 3, one may use the identity(m+ 1)3 + (m+ 2)3 + (m+ 3)3 (m+ 4)3

+(m+ 5)3 (m+ 6)3 (m+ 7)3 + (m+ 8)3 = 48and induction with a basis step for the first 48 positive integers.

An important result concerning the ErdsSurnyi sequences is the following:

Theorem 3 (M.O. Drimbe, 83, [15]) Let {am}m1 be a sequence of distinct positiveintegers such that a1 = 1 and for every n 1, an+1 a1 + + an + 1. If thesequence contains infinitely many odd integers, then it is a ErdsSurnyi sequence.

Proof Recall that a sequence of distinct positive integers is complete if every positiveinteger can be written as a sum of some distinct of its terms. From the first hypothesis,it follows that the sequence {am}m1 is complete. For details, we refer to the papersof J. L. Brown Jr. ([10, 11]) and that of M. O. Drimbe ([15]).

Following the recent papers of M. Tetiva ([32, 33]), let us denote by un the partialsum a1 + + an, n 1. Because the sequence {am}m1 has infinitely many oddterms, it follows that up is even and odd for infinitely many ps. Let m be an arbitrarypositive integer, and let p be sufficiently large such that up > m, and up and m areof the same parity. The integer s = (up m)/2 is less than up. Hence, it is the sumof some distinct terms of the sequence {am}m1 with indices p, that is

1

2

(

a1 + + ap m) = 1a1 + + pap,

for some 1, , p {0, 1}. The last relation is equivalent to m = (1 21)a1 + + (1 2p

)

ap, with 1 21, , 1 2p {1, 1}, and the proof is finishedsince p can be selected in infinitely many ways.


Unfortunately, the sequences an = nk do not satisfy the condition an+1 a1 + +an+1, n 1, in Theorem 3, we cannot use this result to prove they are ErdsSurnyi sequences.These examples of ErdsSurnyi sequences are particular casesof the following result.

Theorem 4 (M.O. Drimbe, 88, [16]) Let f Q[X] be a polynomial such thatfor any n Z, f (n) is an integer. If the greatest common factor of the terms of thesequence {f (n)}n1 is equal to 1, then {f (n)}n1 is an ErdsSurnyi sequence.

Using this result, we can obtain other examples of ErdsSurnyi sequences:an = (an 1)k for any a 2, and an =

(n+ss

)

for any s 2. Note that it is difficultto obtain a proof by induction for these sequences, similar to those given for theprevious examples. Other discussions are given in the recent paper of D. Andricaand E. J. Ionascu (13, [5]).

In the same spirit with formula (2), we obtain the following.

Theorem 5 (D. Andrica and D.Vacaretu,06, [7]) Given a ErdsSurnyi sequence{am}m1, then the number of representations of k [ un, un], where un = a1 + + an, in the form a1 a2 an, denoted here by An(k), is given by

An(k) = 2n

0cos (kt)

k

j=1cos (aj t)dt. (8)

Proof Consider the function Fn is associated to the first n terms of the sequence{am}m1

Fn(z) = (za1 + 1za1

) (zan + 1zan

).

We may writeFn(z) = unj=unAn(j )zj ,

and observe thatAn(k) is the term not containing z in zkFn(z). For z = cos t+i sin t ,we have

zkFn(z) = 2n( cos kt i sin kt) cos a1t cos ant =

An(k) +j =kAn(j )( cos (j k)t + i sin (j k)t).Integrating the last identity over the interval [0, 2 ], we get the result in formula

(8).

Let us look at the range of the variable Sn defined earlier in (4). For a fixed k, weare going to denote the range of Sn by Rk(n). In other words, the set Rk(n) consistsof all numbers

1k 2k nk. (9)To determine R1(n) was a 2011 Romanian Olympiad problem, let us include

an answer to this problem. The greatest element of the set R1(n) is the triangular


number Tn := 1+ 2+ +n = n(n+1)2 , and the smallest element of R1(n) is clearlyTn. Also, the difference of any two elements of R1(n) is an even number. Hence,all elements of R1(n) are of the same parity

We claim that

R1(n) = {Tn,Tn + 2, , Tn 2, Tn}. (10)Let us define a map on the elements of R1(n)\ {Tn} having values in R1(n). First,

if x R1(n) \ {Tn} is an element for which the writing begins with 1, then bychanging 1 by +1, we get x + 2 R1(n). If the writing of x begins with +1, thenconsider the first term in the sum with sign . Such a term exists unless x = Tn. Inthis case, we have

x = 1 + 2 + + (j 1) j n.By changing the signs of terms j 1 and j , it follows that x + 2 R1(n). This

shows the claim in (10).One can wonder what happens if we work within classes modulo m with repre-

sentations of the form (9). Of course, taking all the numbers in (10) modulo a smallm, the chances are that all classes are covered. There is an interesting and a moreprecise result related to this question which appeared as a problem in the monthly[24].

Let p be an odd prime. The 2p1

2 numbers 1 2 p12 represent eachnonzero residue class mod p the same number of times.

Indeed, let z = e 2ip be the standard primitive root of order p of the unity, anddenote by

S =

i{1,1}z1+22++ p12 p1

2 = a0 + a1z + + ap1zp1,

for some integers aj .It is clear that aj is exactly the number of ways the residue j is represented by

the numbers 1 2 p12 . We need to show that a1 = a2 = = ap1.Our analysis is going to also give the common value. As we have seen several times

before, we have S = p1

2j=1

(

zj + zj ). We observe thatp1

j=1

(

zj + zj ) = S p1

j= p+12

(

zj + zj ) = S p1

2

j=1

(

zpj + zjp) = S2.

On the other hand, we can easily compute

p1

j=1

(

zj + zj ) = 1zp(p1)

2

p1

j=1

(

1 + z2j ) =p1

j=1

(

1 + zj ) = 1,

where we used the fact that j 2j is a bijection of the nonzero remainders modulop, and that

p1j=1 (1 + zj ) = 1. This last relation immediately follows from the

polynomial identityp1

j=1 (X zj ) = Xp1

X1 by taking X = 1.


Putting together these facts, we conclude that S2 = 1 and so S = 1. Then, therelation a0 1+a1z+ +ap1zp1 = 0, implies a0 S = a1 = a2 = = ap1([1], Proposition 4, p. 47). Since we must have a0 + a1 + + ap1 = 2 p12 , we getS + pa1 = 2 p12 , which implies a1 = 2

p12 Sp

and also

S 2 p12 (

2

p

)

( 1) p218 (modp).

Here, we used Eulers Criterion ([26], p. 101) and a standard result on quadratic

residues. Hence, S = ( 1) p218 and

a1 = a2 = = ap1 = 2p1

2 ( 1) p218p

.

For k = 2, the situation withR2(n) is almost the same as in case k = 1 but there isan interesting new phenomenon, although expected since {m2}m1 is a ErdsSurnyisequence as we have seen. Let us define the set

R2(n) = {

2(n),

2(n) + 2, ,

2(n) 2,

2(n)},

where

2(n) = 12 + 22 + + n2 = n(n+1)(2n+1)6 .Theorem 6 For n N, R2(n) = R2(n) \ E2(n), where

E2(n) = {(

2 (n) 2j)

: j E} and

E := {2, 3, 6, 7, 8, 11, 12, 15, 18, 19, 22, 23, 24, 27, 28, 31, 32, 33,

43, 44, 47, 48, 60, 67, 72, 76, 92, 96, 108, 112, 128}.

(11)

Proof The set E is known (see A001422) as being the set of all positive integerswhich cannot be written as a sum of distinct squares. This is connected to an oldresult of E. M. Wright ([35]) and to the papers of R. Sprague, P. T. Bateman, A. J.Hildebrand, and G. B. Purdy (see [29] and [8]). Also, one can find a simple prooffor this fact that was given by P. Jain ([22]). Let us observe that the equation

12 22 32 n2 =

2(n) 2j

can be written equivalently as

2j = (1 1)12 + (1 1)22 + (1 1)32 + (1 1)n2 orj = c112 + c222 + c332 + + cnn2, with ci {0, 1}.

So, a representation is possible if and only if j can be written as a sum of distinctperfect squares. It is known that the only positive integers that cannot be written asa sum of distinct squares are the ones in E.


For k = 3, a similar result can be stated. The list of the numbers (A001476) whichcannot be represented as a sum of distinct cubes has 2788 terms. This was obtainedby R. E. Dressler and T. Parker in [14]. For k = 4, the exceptional set of numbers(A046039) has 889576 elements.

In [30], Pn(k) denotes the number of partitions of k into distinct parts from1, 2n, 3n, . . . , and it is proved that for each n, there are only a finite number ofintegers which are not the sums of distinct nth powers. That is, there is a positiveinteger Nn depending only on n such that Pn(k) > 0 for all k > Nn. This result wasextended by H. E. Richert ([28]) to a more general class of sequences.

For a sequence of distinct positive integers a = {am}m1, define the exceptionalset of a to be the set E(a) constituting all positive integers that cannot be representedas a sum of distinct terms of a. Also, define the range Ra(n) of the random variableSn = E1 +E2 + +En, where Ej takes two values: aj or aj with the probability1/2 each. If a is a ErdsSurnyi sequence, then using a similar argument as inTheorem 6, we get Ra(n) = Ra(n) \ E(a), where E(a) = {(un 2j ) : j E(a)},un = a1 + + an, and Ra(n) = {un,un + 2, . . ., un 2, un}.

As we have seen for the sequence of squares and of cubes, it is challengingto determine the exceptional set for a given ErdsSurnyi sequence. For a betterunderstanding of the difficulty of this problem, we mention here three more examples.

The sequence t of triangular numbers Tn = n(n+1)2 , (n 1), satisfies for everym 1 the relation Tm+3 Tm+3 Tm+3 + Tm = 2. Since we may write 1 = T1 and2 = T1 + T2, it follows by induction that t is an ErdsSurnyi sequence. On theother hand, according to the result of H. E. Richert [28], its exceptional set is E(t) ={2, 5, 8, 12, 23, 33}, hence the range Rt(n) = Rt(n) \ {(un 2j ) : with j E(t)},where un = T1 + + Tn = n(n+1)(n+2)6 .

It is also known that the sequence of primes is an ErdsSurnyi sequence. Anice proof based on Theorem 3 combined with Bertrands postulate is given by M.O.Drimbe ([15], Proposition 4). According to the result of R. E. Dressler [13], everypositive integer, except 1, 2, and 6, can be written as the sum of distinct primes, thatis the exceptional set of the sequence p of primes is E(p) = {1, 4, 6}. Therefore, fora given n, the range Rp(n) = Rp(n) \ {un + 2,un + 8,un + 12, un 12, un 8, un 2}, where un = p1 + + pn.

Finally, it is not difficult to check that the hypotheses in Theorem 3 are satisfied forthe Fibonacci sequence F0 = 0, F1 = 1, Fn+2 = Fn+1 +Fn, n 0. Thus, {Fn} is anErdsSurnyi sequence. On the other hand, it is more or less known Zeckendorfstheorem in [34], which states that every positive integer can be represented uniquelyas the sum of one or more distinct Fibonacci numbers in such a way that the sum doesnot include two consecutive Fibonacci numbers. Such a sum is called Zeckendorfrepresentation and it is related to the Fibonacci coding of a positive integer. In thiscase, the exceptional setE(f) of the Fibonacci sequence, say f, is the empty set. Then,for a given positive integer n, the rangeRf (n) consists of all positive integers between(Fn+2 1) and Fn+2 1 of the same parity as Fn+2 1 = F1 + F2 + + Fn.


5 A Proof of AndricaTomescu Conjecture

Let us observe that, in general for k N, as we have seen in the previous sections

Sk(n) = 2n1

2

0cos t cos 2kt cos nktdt.

We let fk,n(x) = cos t cos 2kt cos nkt , t R, which is a 2 -periodic and aneven function. Hence, we can write the above as

Sk(n) = 2n

0cos t cos 2kt cos nktdt.

It is easy to see that each x cos (jkx) is symmetric with respect to the point(

2 , 0

)

if j is an odd integer and symmetric with respect to the vertical line x = /2if j is even. This means Sk(n) = 0 if n 1 or 2 (mod 4). So, we may assume thatn 0 or 3 (mod 4) in what follows, but we will see that after the next reduction thismakes no difference. The reduction is to shrink the domain of fk,n even further andwrite

Sk(n) = 2n+1

/2

0cos t cos 2kt cos nktdt. (12)

Let us emphasize that, the asymptotic behavior is the same regardless of what nis, modulo 4, as long as Sk(n) is newly defined as in (12) (the new values of Sk(n) forn 1 or 2 (mod 4) are not integers anymore, in general, but the asymptotic behaviorfor these values of n is the same, but probably no combinatorial interpretation existsfor these values).

Sullivans idea [31] is to split the integral into two parts, let us say, one over[0, 4nk ] and the other over the interval

[

4nk ,2

]

. Then show that the first one is

(1+ o(1)))

2(2k+1)2nk+

12

and the second is o

(

1

nk+ 12

)

. This is clearly enough to conclude

that (7) is true. We will follow this argument and see what parts can be obtained ingeneral.

First, let us observe that each function x cos (jkx) (j = 1. . .n) on the interval[0, 4nk ] is positive and in fact bounded below by

12. Hence, we can write fk,n(x) as

an exponential:

fk,n(x) = exp [n

j=1ln ( cos (jkx))] = exp [

n

j=1gj (x)],

with gj ,k(x) = ln ( cos (jkx)), j = 1. . .n. We need to compute some derivativesof gj ,k and use the Taylor formula to estimate fk,n(x) for x [0, 4nk ]. First,we have gj ,k(x) = jk tan jkx and so gj ,k(0) = 0. Then, gj ,k(x) =


j 2k sec2 jkx = j 2k(1 + tan2 jkx) which gives gj ,k(0) = j 2k . Finally gj ,k(x) =2j 3k tan jkx sec2 jkx. Then, we have

fk,n(x) = exp [12

(n

j=1j 2k)x2x

3

3

n

j=1j 3k tan (jkcx) sec

2 (jkcx)], x, cx [0, 4nk

].

Since 0 tan t sec2 t 2 for t [0,/4], this implies that

exp ( x2

22k(n) 2x

3

33k(n)) fk,n(x) exp ( x

2

22k(n)), x [0,

4nk],

(13)

where as before (n) = nj=1 j . We need then the following result.

Lemma 3 For k 1, we have L1 = L2 =

(2k+1)2 , where

L1 := limn n

k+1/2

4nk

0exp

(

x2

22k(n)

)

dx =

(2k + 1)2

, and

L2 := limn n

k+1/2

4nk

0exp

(

x2

22k(n) 2x

3

33k(n)

)

dx.

Proof For the first limit, let us just change the variable: x2k(n) = t . Then, the

limit becomes

L1 = limn

nk+1/22k(n)

2k (n)

4nk

0et

2/2dt =

(2k + 1)2

,

since we have seen that limn2k (n)nk+1/2 = 12k+1 and

0 e

t2/2dt = 2 . For thesecond limit, we do the same change of variable and obtain

L2 = limn

nk+1/22k(n)

2k (n)

4nk

0et2/2 2t33k (n)

32k (n)3/2

dt ,

which is clearly satisfying L2 L1. Since ex 1+ x for all x, and a

0 et2/2t3dt =

2 2exp(

a22)

a2exp(

a22)

2 as a , we have

L2 L1 limn

nk+1/22k(n)

23k(n)

32k(n)3/2

2k (n)

4nk

0et

2/2t3dt = L1.

This shows that L2 = L1.


From (13) and Lemma 3, we can conclude that

limn n

k+1/2

4nk

0fk,n(x)dx =

(2k + 1)2

.

For k = 1, in order to finish the proof, we simply need to show that

limn n

3/2 /2

4n

f1,n(x)dx = 0.

Sullivans approach here is to use the ArithmeticGeometric Mean inequality andobserve that

[f1,n(x)2]1/n 1

n

n

j=1cos2 jx

= 12n

n+n

j=1cos 2jx

= 12+cos ((n+ 1)x) sin nx

2n sin x.

Hence, for x [ 2n , 2 ] we get[

f1,n(x)2]1/n 1

2+ 1sin /2n

/2n

1

c := + 2

2< 1.

This is enough to conclude that

0 limn n

3/2 /2

2n

|f1,n(x)|dx limn n

3/2

(

+ 22

)n/2

= 0.

We observe that there is still a missing interval left unaccounted for, but bothlimits can be extended a little more so that the two new intervals will cover [0, 2 ].Indeed, let us consider an (0, 1) that is going to be determined later. The first limitcan be shown to be true for the interval

[

0, 2n (1 )]

since we only need to havethe functions x cos jx bounded from below by a positive constant. The constantc above becomes (1)+22 (1) which is still less than 1 if is small enough ( = 1/2 isjust not good enough).

Let us close by formulating the equivalence of the conjecture (7) for k 2:

limn n

k+1/2 /2

(1)2nk

fk,n(x)dx = 0 for some (0, 1).

This seems to require a very fine analysis especially in the absence a good upperbound for the sums

nj=1 cos (2jkx) for x

[ (1)

2nk ,2

]

.

Acknowledgement The present material was elaborated in the period of the Fall Semester 2013when the first author was a Mildred Miller Fort Foundation Visiting Scholar at Columbus StateUniversity, Georgia, USA. He takes this opportunity to express his gratitude for the nice friendshipshowed and for the excellent facilities offered during his staying.


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The Hilali Conjecture for Hyperelliptic Spaces

Javier Fernndez de Bobadilla, Javier Fresn, Vicente Muozand Aniceto Murillo

Abstract The Hilali conjecture predicts that for a simply connected elliptic space,the total dimension of the rational homotopy does not exceed that of the rationalhomology. Here, we give a proof of this conjecture for a class of elliptic spacesknown as hyperelliptic.

Keywords Rational homotopy Sullivan models Elliptic spaces Tor functors

1 Introduction

Let X be a simply connected CW-complex. Then, X is said to be of elliptic typeif both dim H (X,Q) < and dim (X) Q < . For these spaces, Hilaliconjectured in [5] the following:

Conjecture 1 If X is a simply connected CW-complex of elliptic type, then

dim (X) Q dim H (X,Q) .

J. Fernndez de Bobadilla ()Instituto de Ciencias Matemticas CSIC-UAM-UC3M-UCM, Consejo Superior deInvestigaciones Cientficas, C/Nicols Cabrera, no 1315, Campus Cantoblanco UAM,28049 Madrid, Spaine-mail: [email protected]

J. FresnLAGA, UMR 7539, Institut Galile, Universit Paris 13, 99, Avenue Jean-Baptiste Clment,93430 Villetaneuse, Francee-mail: [email protected]

V. MuozFacultad de Ciencias Matemticas, Universidad Complutense de Madrid, Plaza de Ciencias 3,28040 Madrid, Spaine-mail: [email protected]

A. MurilloDepartamento de lgebra, Geometra y Topologa, Universidad de Mlaga, Ap. 59,29080 Mlaga, Spaine-mail: [email protected]


22 J. Fernndez de Bobadilla et al.

By the theory of minimal models of Sullivan [3], the rational homotopy type ofX is encoded in a differential algebra (A, d) called the minimal model of X. This isa free graded algebra A = V , generated by a graded vector space V = k2 V k ,and with decomposable differential, i.e., d : V k (2V )k+1. It satisfies that:

V k = (k(X)Q) ,Hk(V , d) = Hk(X,Q) .

Therefore, the Hilali conjecture can be rewritten as follows: for a finite-dimensional graded vector space V (in degrees bigger or equal than two), wehave

dim V dim H (V , d)for any decomposable differential d on V .

An elliptic space X is called of pure type if its minimal model (V , d) satisfiesthat V = V even V odd , d(V even) = 0 and d(V odd) V even. Also X is calledhyperelliptic if d(V even) = 0 and d(V odd) +V even V odd .

In his thesis [5] in 1990, Hilali proved Conjecture 1 for elliptic spaces of pure type.The conjecture is known to hold [6, 7] also in several cases: H-spaces, nilmanifolds,symplectic, and cosymplectic manifolds, coformal spaces with only odd-degree gen-erators, and formal spaces. Hilali and Mamouni [6, 7] have also proved Conjecture 1for hyperelliptic spaces under various conditions in the homotopical and homologicalEuler characteristics.

The main result of this paper is the following:

Theorem 1 Conjecture 1 holds for hyperelliptic spaces.We shall start by proving it for elliptic spaces of pure type in Sect. 3. This requires

reducing the question to a problem about Tor functors of certain modules of finitelength over a polynomial ring. We solve it by using a semicontinuity result for theTor functor. Then, in Sect. 4 we prove Theorem 1 for hyperelliptic spaces. For thiswe have to prove a semicontinuity result for the homology of elliptic spaces, andapply it to reduce the general case to the case in which the minimal model only hasgenerators of odd degree and zero differential. We give two different proofs of aninequality from which the result follows.

2 Minimal Models

We recall some definitions and results about minimal models [2]. Let (A, d) be adifferential algebra, that is, A is a (positively) graded commutative algebra overthe rational numbers, with a differential d which is a derivation, i.e., d(a b) =(da) b + ( 1)deg (a)a (db), where deg (a) is the degree of a. We say that A isconnected if A0 = Q, and simply connected if moreover A1 = 0.

The Hilali Conjecture for Hyperelliptic Spaces 23

A simply connected differential algebra (A, d) is said to be minimal if:

1. A is free as an algebra, that is, A is the free algebra V over a graded vectorspace V = k2V k , and

2. Forx V k , dx (V )k+1 has no linear term, i.e., it lives inV>0V>0 V .Let (A, d) be a simply connected differential algebra.A minimal model for (A, d) is

a minimal algebra (V , d) together with a quasi-isomorphism : (V , d) (A, d)(that is, a map of differential algebras such that : H (V , d) H (A, d) is anisomorphism). A minimal model for (A, d) exists and it is unique up to isomorphism.

Now consider a simply connected CW-complex X. There is an algebra of piece-wise polynomial rational differential forms

(

PL(X), d)

defined in [3, Chap. VIII].A minimal model of X is a minimal model (V , d) for

(

PL(X), d)

. We have that

V k = (k(X) Q) ,Hk(V , d) = Hk(X,Q) .

A space X is elliptic [1] if both

dim k(X) Q < and dimHk(X,Q) < . Equivalently, if (V , d) is the minimal model, we require that bothV and H (V , d) are finite dimensional. For elliptic spaces, the EulerPoincar andthe homotopic characteristics are well defined:

=

i0( 1)i dim Hi(V ,Q),

=

i0( 1)i dim i(X) Q = dim V even dim V odd .

We refer the reader to [2, Theorem 32.10] for the proof of the following:

Proposition 1 Let (V , d) be an elliptic minimal model. Then, 0 and 0.Moreover, < 0 if and only if = 0.

In his thesis [5], M. Hilali conjectured that for elliptic spaces:

dim (X) Q dim H (X,Q) .In algebraic terms, this is equivalent to

dim V dim H (V , d) ,whenever (V , d) is a minimal model with dim V < . Note that finiteness of bothdim H (X,Q) and dim (X)Q is necessary. Otherwise, one can easily constructcounterexamples such as X = S3 S3.


3 Proof of the Hilali Conjecture for Elliptic Spaces of Pure Type

A minimal model (V , d) is of pure type if V = V even V odd , withd (V even) = 0, d (V odd) V even.

An elliptic space is of pure type if its minimal model is so. These spaces are widelystudied in [2, 32]. By Proposition 1, we have that dim V even dim V odd 0. Letn = dim V even and n + r = dim V odd , where r 0. Write x1, . . . , xn for thegenerators of even degree, and y1, . . . , yn+r for the generators of odd degree. Then,dxi = 0, and dyj = Pj (x1, . . . , xn), where Pj are polynomials without linear terms.

In this section, we prove the following:

Theorem 2 The Hilali conjecture holds for elliptic spaces of pure type.

3.1 Expressing the Homology as a Tor Functor

To work over nice modules, we would like to reorder the generators y1, . . . , yn+r ,so that P1, . . . ,Pn form a regular sequence in (x1, . . . , xn). Recall that this meansthat the image of Pi in (x1, . . . , xn)/(P1, . . . ,Pi1) is not a zero divisor, for anyi = 1, . . . , n. But this is not possible in general, as shown by the following example.Example 1 Let V = Qx1, x2, y1, y2, y3, where deg (x1) = 2 and deg (x2) = 6.Define a differential d on V by

dy1 = x61 + x22 , dy2 = x91 + x32 , dy3 = x41x2 + x1x22 .Then, (V , d) is a pure minimal model. It can be proved that it is elliptic if andonly if there exist exact powers of x1 and x2. This is the case, since 2x101 =d(

x41y1 + x1y2 x2y3)

and 2x42 = d(

x22y1 + x2y2 x51y3)

. But for the same rea-son, models

(

(x1, x2, yi , yj ), d)

are not elliptic for any choice of indices i, j . Thisamounts to say that dyi , dyj are not a regular sequence in (x1, x2).

However, Halperin showed in [4, Lemma 8] that pure models always admit a basisz1, . . . , zn+r of V odd such that dz1, . . . , dzn is a regular sequence in (x1, . . . , xn).This basis is not necessarily homogeneous but it is possible to preserve the lowergrading induced by the number of odd elements, that is

(V )pq =(

V even qV odd)p .This grading passes to cohomology and by taking into account the quasi-

isomorphisms

((x1, . . . , xn, y1, . . . , yn+r ), d) ((x1, . . . , xn, z1, . . . , zn+r ), d)

((x1, . . . , xn, z1, . . . , zn), d) ((x1, . . . , xn)/(dz1, . . . , dzn), d)


with respect to the lower grading, one deduces that:

H(V , d) = H((x1, . . . , xn)/(dz1, . . . , dzn) (zn+1, . . . , zn+r ), d).So let z1, . . . , zn+r be a basis such that dz1, . . . , dzn form a regular sequence. Put

Pj = dzj for j = 1, . . . , n+ r and consider the moduleM = Q[x1, . . . , xn]/(P1, . . . ,Pn)

over the ringR = Q[x1, . . . , xn] .

Consider the ringS = Q[1, . . . , r ]

and the map f : S R, i Pn+i . Then, M becomes an S-module.Consider also the S-module

Q0 = S/(1, . . . , r ).Then, we have the following:

Proposition 2 H(V , d) = TorS(M ,Q0).Proof Let U = z1, . . . , zn, W = zn+1, . . . , zn+r so that V odd = U W . Then,the map (V even U , d) (M , 0) is a quasi-isomorphism. Actually, the Koszulcomplex

R nU R n1U . . . R 1U R Mis exact, which means that (R U, d) (M, 0).

Therefore,

(V , d) = (R U W , d) (M W , d ) , (1)is a quasi-isomorphism, where the differential d is defined as zero on M , andd zn+i = Pn+i M . This can be seen as follows: the map (1) is a map of dif-ferential algebras. Grading both algebras in such a way that kW has degree k, weget two spectral sequences. The map between their E1-terms is

H (R U, d)W M W .As this is a quasi-isomorphism, it follows that the map in the E-terms is also an

isomorphism. The E-terms are the homology of both algebras in (1). So the map(1) is a quasi-isomorphism.


Finally, we have to identify H (M W , d ) = TorS(M ,Q0). Note that thehomology of (M W , d ) is computed as follows: Take the Koszul complex

S rW S r1W . . . S 1 W S Q0 ,and tensor it with M over S (with the S-module structure given above), to get

(

M S (S W ), d ) = (M W , d ) .

The homology of this computes TorS(M ,Q0).

Lemma 1 Under our assumptions,

dim Tor0S(M ,Q0) n+ 1 and dim TorrS(M ,Q0) n+ 1.

Proof Clearly,

Tor0S(M ,Q0) = M S Q0 = M/(

Pn+1, . . . , Pn+r) = R/ (P1, . . . ,Pn+r ) .

As all the polynomials P1, . . . ,Pn+r have no linear part, this module contains theconstant and linear monomials at least, so dim Tor0S(M ,Q0) n+ 1.

For the other inequality, note that TorrS(M ,Q0) is the kernel of M rW M r1W , i.e., the kernel of

(Pn+1, . . . ,Pn+r ) : M M (r). . . M . (2)Now we use the following fact: As M is a complete intersection R-module (it is

the quotient of R by a regular sequence), it has Poincar duality in the sense that

there is a map M Q such that : M M mult M Q is a perfect pairing.Take elements ,j M , j = 1, . . . , n, such that

(

, xj) = 0, j = 1, . . . , n, (, 1) = 1,

(

j , xk) = jk , j , k = 1, . . . , n,

(

j , 1) = 0,

(,Q) = (j ,Q) = 0, for any quadratic Q R.

Since the elements ,j are in the kernel of (2) and they are linearly independent,we get dim TorrS(M ,Q0) n+ 1.

3.2 Semicontinuity Theorem

We are going to prove a semicontinuity theorem for the Tor functors TorkS(M ,Q0)for flat families of modules M of finite length (i.e., finite dimensional as Q-vectorspaces).


Consider a variable t . A family of S-modules is a module M over S[t] such thatfor each t0, the S-module

Mt0 = M/(t t0)is of finite length. We say that M is flat over Q[t] if it is a flat Q[t]-module, underthe inclusion Q[t] S[t]. Consider M as a Q[t]-module. Then

M = Q[t]N Q[t](t t1)b1

. . . Q[t](t tl)bl ,

for some N 0, l 0, 1 b1 . . . bl . The module is flat if and only if thereis no torsion part, i.e., l = 0 (to see this, tensor the exact sequence 0 Q[t] ttiQ[t] Q[t]/(t ti) 0 with M). Note that for generic , length(M ) = N .Therefore, the flatness is equivalent to M/(t ti) being of length N , i.e.,

M is flat length(Mt ) = N , t .

Lemma 2 For any flat family M,

dim TorkS(M0,Q0) dim TorkS(M ,Q0),for generic Q.Proof Let us resolve M as an S[t]-module:

0 S[t]ar . . . S[t]a0 M 0 . (3)

As M is flat as Q[t]-module, if we tensor the inclusion Q[t] t Q[t] by M overQ[t], we have that M t M is an inclusion. Hence, the sequence

0 M t M M/(t) 0is exact. But this sequence is the sequence 0 S[t] S[t] S[t]/(t) 0tensored by M over S[t]. Hence, Tor1S[t](M, S[t]/(t)) = 0. ObviouslyTorjS[t](M, S[t]/(t)) = 0 for j 2 (since the resolution S[t]/(t) has two terms).

Using the above, we can tensor (3)S[t]S[t]/(t) to get an exact sequence:0 Sar . . . Sa0 M0 0 . (4)

Now we tensor (4) by SQ0 and take homology to obtain TorS(M0,Q0). But(4) S Q0 = (3) S[t] Q0 = ((3) S[t] Q[t]) Q[t] Q[t]/(t) = (5) Q[t] Q[t]/(t) ,

where Q0 = S[t]/(1, . . . , r , t), and0 Q[t]ar . . . Q[t]a0 F = M/(1, . . . , r ) 0. (5)


(This is just a complex, maybe not exact.) Analogously,

TorS(M0,Q ) = H ((5) Q[t] Q[t]/(t )) .So it remains to see that for a complex L of free Q[t]-modules like (5), it holds

thatdim Hk(L Q[t]/(t )) dim Hk(L Q[t]/(t)),

for generic . (Tensor products are over Q[t], which we omit in the notationhenceforth.) For proving this, just split (5) as short exact sequences

0 Zi Li Bi1 0, (6)and note that Zi ,Bi are free Q[t]-modules, being submodules of free modules. SoZi = Q[t]zi and Bi = Q[t]bi . Now 0 Bi Zi Hi(L) 0 gives that

Hi(L) = Q[t]zibi torsion.For generic , we have dim Hi(L Q[t]/(t )) = zi bi . Hence,

0 Zi Q[t]/(t) Li Q[t]/(t) Bi1 Q[t]/(t) 0 ||

0 Zi(L Q[t]/(t)) Li Q[t]/(t) Bi1(L Q[t]/(t)) 0 .The first sequence is (6) tensored by Q[t]/(t). Thus, the last vertical map is

surjective, and the first vertical map is injective.Therefore, we get

dim Hi(L Q[t]/(t)) = dim Zi(L Q[t]/(t)) dim Bi(L Q[t]/(t)) dim Zi Q[t]/(t) dim Bi Q[t]/(t)= dim Hi(L) Q[t]/(t) dim TorQ[t]1 (Hi(L),Q[t]/(t))= zi bi ,

where we have used in the third line that there is an exact sequence

0 TorQ[t]1 (Hi(L),Q[t]/(t)) Bi Q[t]/(t) Zi Q[t]/(t) Hi(L) Q[t]/(t) 0,

and in the fourth line that dim (N Q[t]/(t)) = dim TorQ[t]1 (N ,Q[t]/(t)) for atorsion Q[t]-module N .


3.3 Proof of Theorem 2

We proceed to the proof of the Hilali conjecture for elliptic spaces of pure type. Wehave to prove that

dim H (V , d) 2n+ r.By Proposition 2, we need to prove that dim TorS(M ,Q0) 2n+ r . Consider the

family

M = Q[t , x1, . . . , xn](P1 + tx1, . . . ,Pn + txn) .

For small t , the hypersurfaces P1 + tx1, . . . ,Pn + txn intersect in N points nearthe origin accounted with multiplicity, where N = length(M). Therefore, M is aflat family. By Lemma 2, it is enough to bound below dim TorS

(

M ,Q0)

. But forgeneric t , the hypersurfaces P1 + tx1, . . . ,Pn + txn intersect in N distinct points(at least, it is clear that they intersect in several points and the origin is isolated ofmultiplicity one). Therefore,

TorkS(

M ,Q0) = TorkS(Q0,Q0) .

This is easily computed to have dimension(r

k

)

(using the Koszul complex).Therefore , using Lemma 1,

dim TorS(M ,Q0) (n+ 1) +r1

k=1dim TorkS(M ,Q0) + (n+ 1)

2n+ 2 +r1

k=1dim TorkS

(

M ,Q0)

= 2n+ 2 +r1

k=1

(

r

k

)

= 2n+ 2r 2n+ r .

Remark 1 The above computation works for r 1. If r = 0 then we have to provethat length(M) 2n. But then computing the degree 2 nonzero elements in M , wehave that they are at least

(n+1

2

) n. So for any n,

length(M) 1 + n+(

n+ 12

)

n = 12

(n+ 1)n+ 1 2n.

4 The Hyperelliptic Case

A minimal model (V , d) of elliptic type is hyperelliptic if V = V even V odd , andd (V even) = 0, d (V odd) +V even V odd . (7)


An elliptic space is hyperelliptic if its minimal model is so. Note that ellipticspaces of pure type are in particular hyperelliptic.

By Proposition 1 we have that dim V even dim V odd 0. Let n = dim V evenand n + r = dim V odd , where r 0. Write x1, . . . , xn for the generators of evendegree, and y1, . . . , yn+r for the generators of odd degree. Then, dxi = 0, anddyj = Pj

(

x1, . . . , xn, y1, . . . , yj1)

, where Pj do not have linear terms.In this section, we prove the following:

Theorem 3 The Hilali conjecture holds for hyperelliptic spaces.

4.1 Semicontinuity for Elliptic Minimal Models

Lemma 3 Let V be a graded rational finite-dimensional vector space, and let dbe a differential for V Q[t] such that dt = 0, where t has degree 0. Take anon-countable field k Q, Vk = V k. We denote by d the differential inducedon Vk = V k[t]/(t ), for k. Then

dim H(

Vk, d) dim H (V , d0) ,

for generic k.Proof Write

0 K V k[t] I 0 ,where K and I are the kernel and image of d , respectively. Note that both K and Iare free k[t]-modules, being submodules of V k[t].

Denote by k = k[t]/(t ). Then, we have a diagram0 K k (V k[t]) k I k 0

|| 0 K Vk I 0.

(8)

(Here, the tensor products of all k[t]-modules are over k[t], and the tensor productV k[t] is over the rationals.) Therefore, the last vertical map is a surjection, andthe first map is an injection.

We have0 I K H (V k[t], d) 0 ,

which is an exact sequence of k[t]-modules. Then, H (V k[t], d) contains a freepart and a torsion part. The torsion is supported at some points, which are at mostcountably many. Therefore, for generic k,

0 I k K k H (V k[t], d) k 0is exact. As I k I K and I k K k K , we have that the lastmap in (8) is an injection, therefore an isomorphism, and thus the first map is alsoan isomorphism by the snake lemma.


Note that also, when tensoring with k(t), we have an exact sequence

0 I k(t) K k(t) H (V k[t], d) k(t) 0 .AlsoH (V k[t], d)k(t) = H (V k(t), d), since k(t) is a flat k[t]-module.

Hence,

dim H (Vk, d ) = dim K dim I= dim K k dim I k= dim H (V k(t), d) .

In the first line, we mean dim K dim I = d0(

dim Kd dim I d).Take now = 0. The map K K K/I factors as K/I K/I . Tensor this

map by k0 to get (K/I ) k0 K/I . Note that there is an exact sequenceI k0 K k0 (K/I ) k0 0,

but the first map may not be injective. Then, there is a map

K k0Im(I k0)

= (K/I ) k0 K/I .

By (8), this is an inclusion. Now we have:

dim H (V , d ) = dim H (V k(t), d)= dim (K/I ) k(t) dim (K/I ) k0

= dim K k0Im(I k0)

dim K/I= dim H (Vk, d0)= dimQ H (V , d0) .

4.2 Perturbing the Minimal Model

Let x1, . . . , xn denote generators for V even, and y1, . . . , yn+r generators for V odd .Here, dxi = 0 and dyj = Pj

(

x1, . . . , xn, y1, . . . , yj1)

.


We consider the algebra

(W , d) = (V , d) (y1, 0) ,where deg (y1) = deg (x1) 1. Then

dim H (W , d) = 2 dim H (V , d) .Consider now the differential on W such that xj = 0, yj = 0 and y1 = x1.

Hence, 2 = 0 and d = d = 0. So,dt = d + t

is a differential on W k[t].For generic k, (Wk, d ) verifies that d y1 = x1. So, for nonzero , there

is a KS-extension [8, 1.4](

(x1, y1), d) (Wk, d

) ((x2, . . . , xn, y1, . . . yn+r ), d) .As H ((x1, y1), d ) = k, we have that

H (Wk, d ) = H ((x2, . . . , xn, y1, . . . yn+r ), d) .Now we apply Lemma 3 to this to obtain that

dim H ((x2, . . . , xn, y1, . . . yn+r ), d) dim H (W , d) = 2 dim H (V , d) .Repeating the argument n times, we get that

dim H ((y1, . . . yn+r ), d) 2n dim H (V , d) .But the hyperelliptic condition says that d = 0 for the first space, so

2n dim H (V , d) dim H ((y1, . . . yn+r ), d) = 2n+r .This gives

dim H (V , d) 2r . (9)

4.3 Another Proof of Inequality (9)

In this section, we present a different proof of the inequality dim H (V , d) 2r forhyperelliptic spaces. Recall that if A is a commutative graded differential algebra,and if M ,N are differential graded A-modules, the differential Tor is defined as

Tor(M ,N ) = H (P A N ),where P

M is a semifree resolution, i.e., a quasi-isomorphism from a semifreeA-module P to M (see [2, 6]).


Lemma 4 Let C A B be morphisms of commutative differential graded

algebras. There exists a convergent spectral sequence

Ep,q2 = Hp(B) TorqA(Q,C) Torp+qA (B,C).

Proof Decompose and as

A

A W

A

A U

B C

Then, : AW B is a semifree resolution of B regarded as A-module, soTorA(B,C) = H ((AW ) A C).

Moreover, Id : (A W ) A A U (A W ) A C is a quasi-isomorphism and (AW )A (AU ) = AW U. Therefore, one getsa rational fibration

AW AW U U ,whose associated Serre spectral sequence has the form

Ep,q2 = Hp(AW ) Hq(U ) Hp+q(AW U ).

On the one hand, H (AW ) = H (B).On the other hand, since is a semifreeresolution of C, we have that

H (U ) = H ((AU ) A Q) = TorA(Q,C).Putting all pieces together, we get

Ep,q2 = Hp(B) TorqA(Q,C) Torp+qA (B,C).

Theorem 4 Let (V , d) be a hyperelliptic minimal model. Then

dim H (V , d) 2r .Proof Write as usual x1, . . . , xn for generators of X = V even and y1, . . . , yn+rfor generators of Y = V odd . When we apply the previous lemma to morphismsQXV we get a spectral sequence

E2 = H (V , d) TorX(Q,Q) TorX(V ,Q).On the one hand,

TorX(Q,Q) = H ((x1, . . ., xn), 0) = (x1, . . . , xn),where (x1, . . ., xn, x1, . . ., xn)

Q is a semifree resolution of Q regarded asX-module. Hence, xi are all of odd degree.


On the other hand, V is already X-semifree, so

TorX(V ,Q) = H (V X Q) = H ((y1, . . ., yn+k), 0) = (y1, . . ., yn+k).Then, the inequality

dim H (V , d) dim TorX(Q,Q) dim TorX(V ,Q)coming from the spectral sequence translates into

2n dim H (V , d) 2n+r ,so the result follows.

4.4 Proof of Theorem 3

Now we prove the inequality dim H (V , d) 2n+ r , for the hyperelliptic minimalmodel.

If r = 0, then = 0. So [2, Prop. 32.10] says that the model is pure, and thiscase is already covered by Remark 1.

If r > 0, then < 0. So by Proposition 1, = 0, and hence, it is enough toprove that

dim H even(V , d) n+ r2 .Suppose that r = 1, 2. As the degree 0 and degree 1 elements give always nontrivialhomology classes, then dim H even(V , d) n+ 1, and we are done.

Therefore, we can assume r 3. We use the following fact: if P (x) is a quadraticpolynomial on the x, and P (x) = d, V , then must be linear, V odd anddenoting by do the composition

V odd +V even V odd +V even ,we have P (x) = do. So there are at least

(n+1

2

) (n + r) quadratic terms in thehomology. Conjecture 1 is proved if

{

either 1 + n+ (n+12) (n+ r) n+ r2 ,

or 2r 2n+ r . (10)

So now assume that (10) does not hold. Then

2r r 2n 1 , (11)

and 1 + (n+12) n < 32 r , i.e.,

(2n 1)2 12r 11. (12)


Putting together (11) and (12), we get 2r r 12r 11, i.e., 2r r +12r 11. This is easily seen to imply that r 3. So, r = 3 and n = 3.It remains to deal with the case n = 3, r = 3, and do is an isomorphism of the

odd degree elements onto 2 V even. Let x1, x2, x3 be the even-degree generators, ofdegrees d1 d2 d3, respectively. The degrees of x21 , x1x2, x22 , x1x3, x2x3, x23 arethe six numbers

2d1 d1 + d2 2d2, d1 + d3 d2 + d3 2d3.We have two cases:

Case 2d2 d1+d3. We can arrange the odd generators y1, . . . , y6 with increasingdegree and so that doy1 = x21 , doy2 = x1x2, doy3 = x22 , doy4 = x1x3, doy5 =x2x3, doy6 = x23 . Clearly, dy1 = x21 . Then, dy2 = x1x2 + P (x1), where P (x1) isa polynomial on x1, i.e., of the form c xn1 , n 2. But this can be absorbed bya change of variables y2 y2 c xn21 y1. So, we can write dy2 = x1x2. Nowthe even-degree closed elements in (x1, x2, x3, y1, y2) are again polynomials onx1, x2, x3. So, we can assume dy3 = x22 as before. Continuing the computation,the even-degree closed elements in (x1, x2, x3, y1, y2, y3) are either polynomialson the xis or a multiple of the element x22y1y2 x1x2y1y3 +x21y2y3 = d(y1y2y3),which is exact. Therefore, we can again manage to arrange that dy4 = x1x3.

Case 2d2 > d1 + d3. Then, we have that doy3 = x1x3 and doy4 = x22 . Asbefore, we can arrange dy3 = x1x3. Now the even-degree closed elements in(x1, x2, x3, y1, y2, y3) are polynomials on the xis or a multiple of x3y1y2 x2y1y3 + x1y2y3. But this element has degree 3d1 + d2 + d3 2 > 2d2, so it mustbe dy4 = x22 .

In either case, dy1, dy2, dy3, dy4 are x21 , x1x2, x

22 and x1x3. Let us assume that we are

in the first case to carry over the notation.Now we compute the even-degree closed elements in (x1, x2, x3, y1, y2, y3, y4).

These are polynomials on xis or combinations of

x22y1y2 x1x2y1y3 + x21y2y3 = d(y1y2y3),x3y1y2 x2y1y4 + x1y2y4,x1x3y2y3 x22y2y4 + x1x2y3y4 = d(y2y3y4), andx1x3y1y3 + x21y3y4 x22y1y4 = d(y1y3y4).

Only the second one is nonexact, but its degree is strictly bigger than d2 + d3. Soagain we can arrange that dy5 = x2x3.

Finally, the minimal model is:

dy1 = x21 ,dy2 = x1x2,dy3 = x22 ,dy4 = x1x3,dy5 = x2x3,dy6 = x23 + P (xi , yj ).


The even-degree closed elements in (x1, x2, x3, y1, y2, y3, y4, y5) contain at least

1 = x3y2y3 + x1y3y5 x2y2y5 ,2 = x3y1y2 x2y1y4 + x1y2y4 .

At most, one of them does not survive in H (V , d), so proving the existence ofat least another even-degree cohomology class. Hence, dim H (V , d) 10 9, asrequired.

References

1. Flix, Y.: La Dichotomie Elliptique-Hyperbollique en Homotopie Rationnelle, Astrisque 176.Socit mathmatique de France Paris (1989)

2. Flix, Y., Halperin, S., Thomas, J.-C.: Rational Homotopy Theory. Graduate Texts inMathematics, vol. 205. Springer, Berlin (2000)

3. Griffiths, P.A., Morgan, J.W.: Rational Homotopy Theory and Differential Forms. Progress inMathematics, vol. 16. Birkhuser, Basel (1981)

4. Halperin, S.: Finiteness in the minimal models of Sullivan. Trans. Am. Math. Soc. 230, 173199(1977)

5. Hilali, M.R.: Action du tore Tn sur les espaces simplement connexes. Ph.D. thesis, Universitcatholique de Louvain, Belgique (1990)

6. Hilali, M.R., Mamouni, M.I.: A conjectured lower bound for the cohomological dimension ofelliptic spaces. J. Homotopy Relat. Struct. 3, 379384 (2008)

7. Hilali, M.R., Mamouni, M.I.: A lower bound of cohomologic dimension for an elliptic space.Topol. Appl. 156, 274283 (2008)

8. Oprea, J., Tralle, A.: Symplectic Manifolds with no Khler Structure. Lecture Notes inMathematics, vol. 1661. Springer, Berlin (1997)

Aveiro Discretization Method in Mathematics:A New Discretization Principle

L.P. Castro, H. Fujiwara, M.M. Rodrigues, S. Saitoh and V.K. Tuan

Abstract We found a very general discretization method for solving wide classes ofmathematical problems by applying the theory of reproducing kernels. An illustrationof the generality of the method is here performed by considering several distinctclasses of problems to which the method is applied. In fact, one of the advantagesof the present methodin comparison to other well-known and well-establishedmethodsis its global nature and no need of special or very particular data conditions.Numerical experiments have been made, and consequent results are here exhibited.Due to the powerful results which arise from the application of the present method,we consider that this method has everything to become one of the next-generationmethods of solving general analytical problems by using computers. In particular,we would like to point out that we will be able to solve very global linear partialdifferential equations satisfying very general boundary conditions or initial values(and in a somehow independent way of the boundary and domain). Furthermore, wewill be able to give an ultimate sampling theory and an ultimate realization of theconsequent general reproducing kernel Hilbert spaces. The general theory is herepresented in a constructive way, and contains some related historical and concreteexamples.

S. Saitoh ()Department of Mathematics, Institute of Reproducing Kernels,Kawauchi-cho, 5-1648-16, Kiryu 376-0041, Japane-mail: [email protected]

L.P. Castro M.M. RodriguesCIDMACenter for Research and Development in Mathematics and Applications,Department of Mathematics, University of Aveiro, 3810193 Aveiro, Portugale-mail: [email protected]

M.M. Rodriguese-mail: [email protected]

H. FujiwaraGraduate School of Informatics, Kyoto University, Kyoto, 606-8501 Japane-mail: [email protected]

V.K. TuanDepartment of Mathematics, University of West Georgia,1601 Maple Street, Carrollton, GA 30118, USAe-mail: [email protected]


38 L.P. Castro et al.

Keywords Reproducing kernel Discretization Computer Numerical PDE ODE Integral equation Numerical experiment Generalized inverse Tikhonovregularization Real inversion of the Laplace transform Matrix Convolution Singular integral equation Sampling theory Analyticity Smoothness of function

1 Introduction

Nowadays, the importance and influence of reproduction kernel Hilbert spaces isidentified in several different subareas of mathematics and their applications. Insome cases, such importance is increased when convenient modifications and ap-plications of different techniques are additionally implemented. A major ingredientwhich gave rise to important and global results was simply the consideration of linearmappings within the framework of Hilbert spaces in a reproduction kernels identifi-cation process. This gave rise to a very powerful and also natural use of reproducingkernels in the framework of Hilbert spaces (as it may be seen in [35]) which, in turn,had several very significant consequences during the last decades.

As a somehow natural consequence of the research project proposed by S. Saitohin the University of Aveiro some years ago, and being developed until now, it wasintroduced as a new line of research within the Functional Analysis and ApplicationsGroup of this university (involving not only researchers from this university but alsocollaborators from abroad). Among some other fruitful consequences of that projectwithin reproduction kernel Hilbert spaces, in here we would like to present the so-called Aveiro Discretization Method in Mathematics. Interestingly enough, this namewas identified by S. Saitoh in the night just before one of his talks presented in the9th International Conference on Mathematical Problems in Engineering, Aerospaceand Science (Vienna, Austria, 1014 July 2012). In such a talk, it was presented forthe first time (in an oral and nondetailed way) the main ideas of the method that weare now presenting in here in a complete form.

As main ingredients for the starting points of the method, we may identify thegeneral theory of reproducing kernels, the Tikhonov regularization procedure, conse-quent approximate solutions for bounded linear operator equations on Hilbert spaces,and optimal numerical solutions.

In particular, as examples of possible applications, we will be able to solve verygeneral linear partial differential equations (PDEs) satisfying very general boundaryconditions and initial valuesindependently of the boundary and domains. Fur-thermore, we will be able to give clearly an ultimate sampling theory and ultimaterealizations of general reproducing kernel Hilbert spaces. We would like to developthe general theory in a self-contained manner with some related history and manyconcrete examples.

The remaining part of this work is organized in the following way. In Sect. 2,we establish a general formula giving an optimal inverse by using a finite numberof output data in the framework of Hilbert spaces. In Sect. 3, we shall establishthe convergence property of our approximate inverses in Sect. 2. In order to show

Aveiro Discretization Method in Mathematics: A New Discretization Principle 39

our history for some fundamental inverses with the typical example of the Laplacetransform, we shall refer to several typical inverses in Sects. 39 and in Sect. 10,we shall give our final and new approximate inversion of the Laplace transform. InSect. 11, we shall introduce the PaleyWiener spaces and Sobolev spaces (due to thenecessity of these spaces in our new discretization principle). In Sect. 12, we shallderive results for general linear ordinary differential equations, and in Sect. 13, wederive concrete results for various typical partial differential equations. In Sects. 14and 15, we will introduce results for convolution and singular integral equations,respectively. In Sect. 16, we shall give several numerical experiments, and we endup with a general discussion in Sect. 17 (mostly from the viewpoint of numericalanalysis).

2 Principal Formulation of the Inverse by Using a FiniteNumber of Data

We shall first establish our fundamental inversion formula by using a finite numberof data in the framework of Hilbert spaces. In view of this, following [35, 38], weshall start by introducing a general theory for linear mappings in the framework ofHilbert spaces.

Let H be a Hilbert (possibly finite-dimensional) space, and consider E to be anabstract set and h a Hilbert H-valued function on E. Then, we are able to considerthe linear transform

f (p) = (f , h(p))H, f H , (1)fromH into the linear spaceF(E) comprising all the complex valued functions onE.In order to investigate the linear mapping (1), we form a positive definite quadraticform function K(p, q) on E E defined by

K(p, q) = (h(q), h(p))H on EE. (2)Then, we obtain the following fundamental result.

Proposition 1

(I) The range of the linear mapping (1) by H is characterized as the reproducingkernel Hilbert space HK (E) admitting the reproducing kernel K(p, q) whosecharacterization is given by the two properties: (i)K(, q) HK (E) for anyq Eand, (ii) for any f HK (E) and for any p E, (f ( ),K( .p))HK (E) = f (p).

(II) In general, we have the inequality

f HK (E) fH.Here, for any member f of HK (E), there exists a uniquely determined f Hsatisfying

f (p) = (f, h(p))H on E

40 L.P. Castro et al.

and

f HK (E) = fH. (3)(III) In general, we have the inversion formula in (1) in the form

f f (4)in (II) by using the reproducing kernel Hilbert space HK (E).

The typical ill-posed problem (1) will become a well-posed problem, because theimage space of (1) is characterized as the reproducing kernel Hilbert space HK (E)with the isometric identity (3), which may be considered as a generalization of thePythagorean theorem.

However, this viewpoint is a mathematical one and is not a numerical one, as weshall refer to it later.

We shall consider the inversion in (4) formally. However, this idea will be veryimportant for the general inversions and for our new discretization method.

Following the above general situation, we shall recall the general theory in thefollowing general situation (cf. also [34, 35]): Let {vj } be a complete orthonormalbasis for H. Then, for

vj (p) =(

vj , h(p))

H ,

we have

h(p) =

j

(

h(p), vj)

H vj =

j

vj (p)vj .

Hence, by setting

h(p) =

j

vj (p)vj ,

we may write

h( ) =

j

vj ( )vj .

Thus, we shall define(

f , h(p))

HK=

j

(

f , vj)

HKvj .

For simplicity, we will write as follows:

HK = HK (E),here and in the sequel. Then, we have the following result.

Aveiro Discretization Method in Mathematics: A New Discretization Principle 41

Proposition 2 We assume that for f HK , it holds(

f , h)

HK H

and for all p

mathematics without boundaries ||

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