the order bidual of almost 7^-algebras and 7>algebras · 2018. 11. 16. · the order bidual of...

transactions of theamerican mathematical societyVolume 347, Number 11, November 1995

THE ORDER BIDUAL OF ALMOST 7^-ALGEBRASAND 7>ALGEBRAS

S. J. BERNAU AND C. B. HUIJSMANS

Abstract. It is shown in this paper that the second order dual A" of an

Archimedean (almost) /-algebra A , equipped with the Arens multiplication,

is again an (almost) /-algebra. Also, the order continuous bidual (A')'n of an

Archimedean rf-algebra A is a rf-algebra. Moreover, if the rf-algebra A is

commutative or has positive squares, then A" is again a rf-algebra.

1. Introduction, preliminaries

In this paper we will consider the second order dual A" of an Archimedean

lattice ordered algebra A, equipped with the so-called Arens multiplication,

in great detail. Our main results will be that the order bidual A" of an

Archimedean (almost) /-algebra is again an (almost) /-algebra and that the

order continuous order bidual (A')'n of an Archimedean d-algebra is equally

an Archimedean d-algebra. The question of whether in the latter case the whole

second order dual A" is a af-algebra remains open.

In order to avoid unnecessary repetition we will assume throughout that all

vector lattices and lattice ordered algebras under consideration are Archimedean.

Let us recall some of the relevant notions. For terminology and results on

vector lattices and /-algebras not explained in this paper we refer to [1], [9],

[10], and [14]. Let A be a (real) vector lattice (or Riesz space). The firstorder dual of A is denoted by A' and the second order dual by A" . It is well

known that A' is a Dedekind complete (and hence Archimedean) vector lattice.

Define for each x g A the element x" G A" by x"(f) = f(x) for all / G A',and then define the mapping ct : A -> A" by a(x) = x" for all x e A. We

can easily check that ct is a vector lattice homomorphism of A into A" ([14,

§109]). It is possible that A' is trivial (see for example [14, Examples 85.1 and

85.2]). However, if A' separates the points of A, then ct is injective and we

can identify A with the vector sublattice o(A) of A" . Clearly A" is Dedekind

complete (and, of course, trivial if A' is trivial). We now focus on the algebraic

structure of A" in the case that A is a lattice ordered algebra.

We say that A is a lattice ordered algebra (or /-algebra, Riesz algebra),

whenever A is a vector lattice which is simultaneously an associative (but not

necessarily commutative) algebra such that xy > 0 for all 0 < x, y G A

Received by the editors January 26, 1994.

1991 Mathematics Subject Classification. Primary 46A40, 46A20, 06F25, 13J25.Key words and phrases. Arens multiplication, /-algebra, almost /-algebra, af-algebra, second

order dual, bidual, order continuous bidual.

©1995 American Mathematical Society

4259

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4260 S. J. BERNAU AND C. B. HUIJSMANS

(equivalently, \xy\ < \x\ • \y\ for all x, y g A). In A" a multiplication canbe introduced (the Arens multiplication (see [2, 3]) which is accomplished in

three steps. Given x, y e A, f G A' and F, G e A" , we define /• x G A',

G'feA' and F -G g A" by the equations

(1) (f-x)(y)=f(xy),

(2) (G-f)(x) = G(f-x),

(3) (F-G)(j) = F(G-j).

It is straightforward to show that A" is a lattice ordered algebra with respect to

the Arens multiplication as described in (3) (see [7, Theorem 4.1]). Even if the

original multiplication in A is commutative, the Arens multiplication in A"

need not be. It is easily verified that ct is an algebra homomorphism so when

A' separates points of A we can identify A with the sub lattice algebra a (A)

of A".The band of all order bounded, order continuous linear functionals on A' is

denoted by (A')'n and its disjoint complement in A" by (A')'s, the band of all

order bounded so-called singular functionals on A'. Observe that

A" = (A')'n(B(A')'s,

an order direct sum, as A" is Dedekind complete. Because they are bands in

a Dedekind complete vector lattice, (A')'n and (A')'s are themselves Dedekind

complete vector lattices. It is an easy matter to show that (A')'n is also an

/-algebra with respect to the Arens multiplication ([7, Theorem 4.1]). Further-

more, o(A) c (A')'n which is almost obvious.

If A happens to be commutative (which is the case for Archimedean almost

/-algebras and /-algebras, but not for d-algebras; see [4, Theorem 2.15 and

Example 4.1]), then x"• / = /• x for all x G A, j G A'. Indeed, this followsfrom

(x"-j)(y) = x"(j-y) = (j-y)(x) = j(yx) = j(xy)

= (j-x)(y)

for all y £ A. Moreover, x" • F = F ■ x" for all x G A, F e A" in this

situation. For the proof, note that

(x" ■ F)(j) = x"(F • /) = (F • j)(x) = F(j-x)

= F(x"-j) = (F.x")(j)

for all / G A'. These two properties will be used without further mention.

Let us recall the definitions of an /-algebra, almost /-algebra, and <2f-algebra

(for a survey and most of the elementary properties we refer to [4]). The

/-algebra A is said to be an /-algebra whenever xAy = 0 and 0 < z g A imply

xzNy = zxAy = 0 and an almost /-algebra if x Ay = 0 implies xy = 0. Any

/-algebra is an almost /-algebra, but not conversely. Archimedean (almost)

/-algebras are automatically commutative (and even associativity follows in

case of /-algebras). If A is an almost /-algebra, then xx+ > 0, and x2 > 0

for all x G A . An /-algebra A is a left d-algebra if x Ay = 0 and 0 < z e Aimply zx A zy = 0. If x A y = 0 implies xz Ayz = 0 for all 0 < z G A,

then A is termed a right d-algebra. An /-algebra that is both a left and a

right ^/-algebra is simply called a ^/-algebra. Any /-algebra is a ^/-algebra,License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use

ALMOST F-ALGEBRAS AND D-ALGEBRAS 4261

but not conversely. Almost /-algebras need not have the (/-algebra property

and vice versa. Archimedean (/-algebras need not be commutative nor have

positive squares. Archimedean (/-algebras which are commutative or do have

positive squares are almost /-algebras. Along the same line, that better algebraic

properties force better order properties, note that Archimedean cf-algebras andArchimedean almost /-algebras which are either unitary or semiprime (mean-

ing that 0 is the only nilpotent element) are automatically /-algebras; for details

of this we refer to [4].

In the next lines we will recollect some of the history of the order bidual of the

various kinds of lattice ordered algebras. The first result in this direction is due

to B. de Pagter and the second author. They proved in the paper [7, Theorem

4.4] that the order continuous bidual (A')'n of an Archimedean /-algebra A

is a Dedekind complete (hence Archimedean, commutative and Arens regular)

/-algebra with respect to the Arens multiplication. In §3 of the present paper

we will give two alternative proofs of this result which, in our opinion, are

simpler and more straightforward. Some years after [7] was published, the

second author, in [5, Theorem 2.8], proved that the entire second order dual A"

of an Archimedean /-algebra A is an Archimedean /-algebra with respect to

the Arens multiplication. This was also obtained independently by E. Scheffold

in [ 11, Satz 2.2], by means of representations, for the class of Banach /-algebras.

In [5] even more was shown: if F e A", G e (A')'s, then F • G = G ■ F = 0.

This implies in particular that F -G e (A')'n for all F, G e A" and furthermore

that the Arens multiplication in (A')'s is trivial.

In the survey paper [6, Section 8] it was conjectured by the second author

that the order bidual A" of an Archimedean almost /-algebra A (af-algebra

A) is an Archimedean almost /-algebra ((/-algebra) with respect to the Arens

multiplication. Recently, E. Scheffold presented a proof of the fact that the

order continuous bidual (A')'n of a Banach almost /-algebra A is also a Banach

almost /-algebra (equipped with the Arens multiplication), once more by means

of representation theory [12, Theorem 3]. In §2 of the present paper we will

provide the reader with an elementary proof of this result which is valid for the

wider class of Archimedean almost /-algebras that do not carry a lattice norm.

In fact, we will show that the whole order bidual A" of an Archimedean almost

/-algebra A is also an Archimedean (whence commutative) almost /-algebra.

Moreover, the results mentioned above for /-algebras carry over in this case:

F-G = G-F = 0 for all F g A", G G (A')'s and F-G e (A')'„ wheneverF, GG A".

Finally, in [13, Theorem 2.1], E. Scheffold gives a complicated proof that theorder continuous order bidual (A')'n of a Banach (/-algebra A is a left Banach

(7-algebra. Difficulties arise for the class of (/-algebras because Archimedean

(/-algebras need not be commutative. We are able to verify in §4 of this paper

that whenever A is an Archimedean (/-algebra, (A')'„ is so too (i.e., (A')'n is

both a left and a right (/-algebra). So far, we are not able to accomplish a

similar result for the whole order bidual A" of an Archimedean (/-algebra A .

The method of proof we developed in the case of (almost) /-algebras does not

apply.In the final paragraphs of this introduction we list some elementary facts on

vector lattices that play a fundamental role in the sequel. First we quote the

famous Riesz-Kantorovich theorem (see e.g. [1, Theorem 1.13]). If A is a vec-


4262 S. J. BERNAU AND C. B. HUUSMANS

tor lattice, then A' is a Dedekind complete vector lattice. Its lattice operations

satisfy

(g V h)(x) = sup{g(y) + h(z) :x = y + z,0<y,zeA},

(g A h)(x) = inf{g(y) + h(z) : x = y + z ,0 <y, z e A}

for all g,heA',0<xeA.Secondly, if A is a vector lattice and a (A) = {x" : x G A}, then it is a

well-known result of H. Nakano that the order ideal

I (a (A)) = {F G (A')'n : \F\ < x" for appropriate 0 < x G A}

generated by a (A) in (A')'n is order dense in (A')'„ (see e.g. [10, Proposition

1.4.15]). This means that for every 0 < F G (A')'n there exists G G I (a (A))satisfying 0 < G < F. Equivalently, there exist 0 < Fa G I (a (A)) (say 0 <

Fa < x'a for some 0 < xa G A) such that Fa T F .

For F g A" , the absolute kernel or null ideal NF of F is defined by

NF = {feA':\F$\f$ = 0}.It is evident that NF = N\F\ = NF+ n NF- is an order ideal in A' that is (in

general properly) contained in the null space of F . If fe (A')'„ , then NF is

a band in A'. The disjoint complement of Cf = Np is called the carrier of F

and is always a band in A'.For future reference we list some properties of the null ideal and the carrier

in the next theorem.

Theorem 1.1. (i) (Nakano) Ij G, 77 e (A')'„ , then G_H (i.e., \G\ A |77| = 0) ifand only if Cg c Nh (and, by symmetry, if and only if Ch c Ng) ■

(ii) If G G (A')'n , then A' = NG® CG (a band decomposition).

(iii) If 77 G A", then H e (A')'s if and only if Nff = A'.Proof, (i) See [1, Theorem 5.2] or [14, Theorem 90.6].

(ii) [14, Theorem 90.9].

(iii) For the only if part, a possible reference is [8, Theorem 50.4]. In order

to show the converse, take 77 g A" such that Nh is order dense in A'. We

may assume without loss of generality that 77 > 0, as Nh = N\H\ ■ Pick

0 < F g (A')'„ and observe that NH C NHAF implies NHdAF = A'. On the

other hand, 77 A F is order continuous, so Nhaf = NHdAF . Combining these

two equalities we find NHaf = A' ,so 77 A F = 0 for all 0 < F g (A')'n . This

shows that 77 G (A')'s.

Corollary 1.2. If G, 77 g (A')'„ , G_H, and 0 < f G A', then there exist g, h gA' such that f = g + h, g Ah = 0, and G(g) = 0 = H(h).

Proof. By Theorem 1.1 (ii), there exist g G NG and h e CG such that / = g+h ,and (clearly) g A h = 0. By Theorem 1.1 (i), CG c NH , so G(g) = 0 = H(h),as required.

2. The order bidual of almost /-algebras

In this section we will show first that the order continuous order bidual (A')'„

of an almost /-algebra A is again an almost /-algebra (with respect to the

Arens multiplication). Next we will show that H-G = 0 for all 77 G (A')'sand all G G (A')'„ . From this it will follow easily that A" itself is an almost

/-algebra.License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


Proposition 2.1. Let A bean almost f-algebra. IjO<xeA,0<G,H£ (A')'nsatisjy G A 77 = 0 and 0 < G, 77 < x", then G-H = 0.

Prooj. Choose 0 < / G A'. By Corollary 1.2 there exist g, h e A' such that

x".j = j.x = g + h,g Ah = 0 and G(g) = 0 = H(h). By the Riesz-Kantorovich theorem,

0 = (g A h)(x) = inf{g(y) + h(z) :x = y + z,0<y,zeA}.

If e > 0, the formula above gives 0 < y, z G A such that x = y + z, and

g(y) + h(z) < e . (Note that y and z depend on e .)

Now define the functionals 0 < Gx, Hx g (A')'n by

Gx = GA(y-yAz)", Hx = 77 A (z -y A z)" .

Then

0 < G - Gx = (G - (y - y A z)")+ < (x" -(y-y A z)")+

[ ' =y" + z"-(y-yA z)" < 2z" ,

and similarly

(2) 0<H-Hx< 2y".

Observe now that

(y - y A z) A (z - y A z) = 0,

so, because A is an almost /-algebra,

(y - y A z) • (z - y A z) = 0.

Hence,

(3) 0<Gx-Hx <(y-yAz)"-(z-yAz)" = 0.

Now consider ((G - Gx)-H)(f). We have

0<((G-Gi)-H)(j) = (G-Gx)(H.j)<(G-Gx)(x"-j)

(4) =(G-Gi)(j>x) = (G-Gi)(g + h)

< G(g) + (G- Gx)(h) < 0 + 2z"(h) = 2h(z)

(where we use (1) and G(g) = 0). Similarly,

0<(Gx-(H-Hx))(f)<(G-(H-Hx))(f)<(x".(H-Hx))(f)(5) =((H-Hx)-x")(f) = (H-Hx)(f.x) = (H-Hx)(g + h)

< (77 - Hx)(g) + H(h) < 2y"(g) + 0 = 2g(y).

By (3), Gx • Hx = 0, so we easily check that

G • H = (G - Gx)- H + GX-(H - Hx).

Therefore

0<(G-H)(j) = ((G-Gx).H)(j) + (Gx-(H-Hx)(j)

<2h(z) + 2g(y)<2e,

where we use (4) and (5). Since e > 0 is arbitrary, we get (G-H)(j) = 0 for

all 0 < / G A', so G • 77 = 0 and we are done.

The next result, which is an extension of Proposition 2.1 by means of stan-

dard order continuity arguments, was proved by E. Scheffold for Banach almost

/-algebras via representation theory [12, Theorem 3].License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


Theorem 2.2. If A is an almost f-algebra, then the order continuous order bidual

(A')'n is an almost f-algebra with respect to the Arens multiplication.

Prooj. We have to show that if 0 < G, 77 G (A')'n and G A H = 0, then(7-77 = 0. Let 7(ct(^)) be the order ideal in (A')'n generated by o(A); i.e.,

I(o(A)) = {Ke (A')'n : \K\ < x" for appropriate 0 < x G A] .

As observed in the introduction, 7(ct(^4)) is order dense in (A')'n , so there

exist Ga, Hp g I (a (A)) such that 0 < Ga ] G,0 < Hp \ H with, say, 0 <

Ga < x", 0 < Hp < y'L for some 0 < xa, yp £ A. Now G A 77 = 0 implies

Ga A Hp = 0 for all a, fi . Moreover,

0<Ga,Hp<(xa+yp)",

so by Proposition 2.1 we have Ga • Hp = 0 for all a, fi . Choose 0 < / g A'.

Since 0 < Hp(j-x) T H(j-x) for all 0 < x G A, we get 0<Hp-j] H-j.But each Ga is order continuous, so 0 < (Ga • Hp)(j) | (Ga • H)(j). This holds

for all 0 < / G A', so 0 < Ga • Hp T Ga . Now Ga-Hp = 0 for all p , showingthat Ga • H = 0 for all a.

Since 0 < Ga } G, we have 0 < Ga(H ■ j) | G(H ■ j) for all 0 < / G A'which implies 0 < Ga • 77 } G • 77. Combined with Ga • 77 = 0 for all a wefinally may conclude that G • H = 0 and the theorem follows.

In order to prove that the whole order bidual of an almost /-algebra is again

an almost /-algebra we need some preliminaries.

Lemma 2.3 (Cauchy inequality). Ij A is an almost j-algebra and 0 < j g A',

then

\j(xy)\ < sfj^^Jiy2)jor all x, y € A .

Prooj. Fix x, y e A. Since A is commutative and has positive squares, wehave

(x - ay)2 = x2 - 2axy + a2y2 > 0

for all a G R. Hence

/(x2)-2a/(xy) + a2/(y2)>0

for all real a. Therefore, the discriminant of this quadratic polynomial in a

satisfies

4/(xy)2-4/(x2)/(y2)<0.

This gives the desired result.

Noting that (A')'n is an almost /-algebra, and that each element / of A'

defines a linear functional on (A')'n in standard fashion, we have the following

consequence of Lemma 2.3.

Lemma 2.4. Ij A is an almost j-algebra and 0 < j G A', then

\(G.H)(j)\<^G^)-{m(j)

jor all G,He (A')'„.

In particular, if 0 < G G (A')'n and 0 < x G A (take 77 = x"), we get

0<(C7./)(x)<^/W)-v//(x2)

for all 0<f€A'.License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


A corollary of Lemma 2.3 which we include for completeness but which we

do not use subsequently is the following.

Corollary 2.5. Let A be an almost j-algebra, x e A and 0 < j G A'. Then/ • x = 0 ij and only ij j(x2) = 0.

Incidentally, the latter follows also from the fact that x2 = 0 if and only if

xy = 0 for all y G A . In order to show that x2 = 0 implies xy = 0 for ally we may assume that x > 0 and y > 0. Now it follows from (y - nx)2 > 0

thaty2 > 2nxy - n2x2 = 2nxy > 0 (n = l,2, ...).

By the Archimedean property, xy = 0.Another way to obtain this result is by observing that (y - nx)(y - nx)+ > 0.

0 < xy - xy A nx2 < x(y - y A nx) < -y2

(n=l,2, ...). Hence if x2 = 0, then 0 < xy < n~xy2 (n = l,2, ...). TheArchimedean property gives xy = 0.

A more subtle result than Corollary 2.5 which is interesting in its own right

and which we do need in the sequel reads as follows.

Proposition 2.6. Let A be an almost j-algebra and 0 < / G A'. Then

(j.x)+ = j-x+

jor all x G A. In other words, j'• A = {j-x : x G A} is a vector sublattice

oj A' (and (j-x)~ = j-x~ , |/-x| = /• |x| for all x G A). Furthermore, the

mapping Tf : A —» A' defined by Tf(x) = j - x is a lattice homomorphism.

Prooj. Since x+y >xy for all 0 < y G A , we have

(j-x+)(y) = j(x+y) > /(xy) = (f-x)(y)

for all positive y, i.e., f-x+ > f-x. Also, f-x+ > 0 showing that f-x+ >

[f-x)+-For the reverse inequality, take 0 < y £ A and put g = (/-x)+ . Now the

almost /-algebra property together with x+ A (x~ Ay) = 0 yields x+(x~ Ay) =

0. Similarly, (x--x~Ay)A(y-x~Ay) = 0 results in (x~-x~Ay) • (y-x~Ay) =0, so

x~(y-x~ Ay) = (x~ Ay)-(y-x~ Ay).

Hence,

g(y) > g(y -x~Ay)>(f- x)(y - x~ A y)

= f(xy-x(x~ Ay))

= f(x+y - x~y -x+(x~ Ay) + x~(x~ Ay))

= f(x+y-x~(y-x~ Ay))

= /(x+y - (x~ A y)(y - x~ A y))

>{f-x+)(y)-f(y2).

Take (l/«)y in place of y to get

g(y)>(f-x+)(y)-^f(y2) (n = l,2,...).



Consequently, g(y) > (f -x+)(y). This holds for all 0 < y g A , so (j -x)+ >j • x+ and the proof is complete.

We now come to a key result which seems to need the full force of Propo-

sition 2.6. First notice that the adjoint T'f : A" -> A' of the above lattice

homomorphism Tf : A —► A' is defined by

(T'fG)(x) = G(Tfx) = G(j-x) = (G-f)(x)

for all x G A , G G A" , so T'fG = G ■ j for all G £ A" .

Proposition 2.7. Let A be an almost j-algebra, 0 < j £ A' and 0 < 77 G A" . Ijg £ A' satisfies 0 < g < H • j, then there exists K £ A" such that 0 < K < 77and g = K • j. Hence, ijH£ (A')'n, then K £ (A')'n as well. In other words,

T', is interval preserving (has the Maharam property).

Prooj. By Proposition 2.6, if x G A and j-x = 0, then j-x+ = 0. Hence

0 < g(x+) < (H-j)(x+) = H(f-x+) = 0 gives g(x+) = 0. Similarly g(x~) =0 and hence g(x) = 0. Accordingly we may define a linear functional K on

the vector sublattice f • A of A' by

K(f-x) = g(x)

for all x G A (this definition makes sense by the above remarks). Clearly, K

is a positive linear functional on / • A . Indeed, if / • x > 0, then / • x =

(f-x)+ = f-x+ gives K(f-x) = K(f-x+) = g(x+) > 0. Moreover, in this

case g < 77 • / implies

K(f-x) = g(x+) < (H-f)(x+) = H(f-x+) = H(f-x),

so K is a positive linear functional on the vector sublattice f-A of A' satisfy-

ing 0 < K < 77 on / • A . By the Hahn-Banach extension theorem for positivelinear functionals (see e.g. [1, Theorem 2.2]) there exists a positive extension

of K to the whole of A' which is still dominated by 77. If we denote thisextension again by K , we have 0 < K < 77 on A' and for all x G A we have

(K.j)(x) = K(j-x) = g(x),so g = K-j.A slightly different approach to this result is obtained by using the result

that the adjoint T'f of the lattice homomorphism Tf is interval preserving

(see e.g. [1, Theorems 7.4 and 7.8]). Indeed, our proof of Proposition 2.7 is

identical in concept with the proof of [1, Theorem 7.4] and both proofs use the

Hahn-Banach theorem. Hence

T}[0, 77] = [0,7)77] = [0,77./],

so if 0 < g < 77 • /, then g = T'f(K) = K ■ f for some 0<K<H.

Lemma 2.8. Let A be an l-algebra and Ga } 0 in A". Then G2 1 0 as well.

Proof. Suppose that 0 < / G A'. For fi fixed and all a with Ga < Gp we

have

0 < Gl(f) < (Ga ■ Gp)(f) = Gn(Gp • /) I 0,

from which the claim follows.

The next proposition plays a key role in the sequel. The proof is in the same

spirit as the method used in the verification of [5, Proposition 2.2].License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


Proposition 2.9. Let A be an almost f-algebra and Ga J, 0 in (A')'n. Then

F-Ga[0 jor all 0<F g A" .

Prooj. Choose 0 < j £ A'. By Lemma 2.8 there exists a subsequence {Gk}kxL1

of the net {Ga} such that G\(j) < k~4 (k = 1,2, ...). For 0 < x G A,Lemma 2.4 gives

OO OO - - - oo .

o< £(cv/)(*) <Y,\/Gk(f)-y/nx2) $ V^x2)- Ef < °°-fc=l fc=l fc=l

The functional g : A+ —> R defined by

oo

g(x) = £(GV/)(x)fc=l

is additive on ^+ and has therefore (via standard arguments) a positive linear

extension to the whole of A ([14, Lemma 83.1]) which we shall denote by g

again.

We have to show that (F-Ga)(j) j 0. To this end we note that, for all

positive integers m,

m m

F(g)_Y,F(Gk-f) = YJ(F'Gk)(f).k=\ k=x

It follows that Y^LiiF • Gk)(f) is convergent and hence that (F -Gk)(j) -* 0as k —> oo. It is now immediate that (F • Ga)(f) | 0, as required.

Our next result will lead directly to the final proof that A" is an almost

/-algebra whenever A is so.

Theorem 2.10. Let A be an almost j-algebra, G £ (A')'n and H £ (A')'s. ThenH-G = 0.

Prooj. By splitting into positive and negative parts, if necessary, we may assume

that 0 < G £ (A')'„ and 0 < 77 g (A')'s . We have to show that H(G-j) = 0for all 0 < / G A'. Fix such an / and put

J = {L £ (A')'n :L>0, (H-L)(j) = H(L-j) = 0}.

Clearly, / is closed under addition and multiplication by positive scalars,

and is a lattice ideal in the positive cone of (A')'n (i.e., if Lx £ J, 0 < L2 < Lx ,

then L2£ J and if Lx, L2 £ J , then Lxv L2£ J). The latter follows from

0 < (H-(LxVL2))(f) < (H-Lx)(f) + (H-L2)(j) = 0.

We claim that / is order dense in the positive cone of (A')'n . For this purpose

we have to show that for each 0 < M £ (A')'n there exists an element L £ J

satisfying 0 < L < M. If (H • M)(f) = 0, we may choose L = M, so

suppose (H-M)(f) = H(M-f) > 0. Since Nff = A' there exists g £ NH (so77(g) = 0), such that 0 < g < M•/. We note in passing that, by Theorem

1.1 (iii), this makes full use of the fact that 77 is singular. By Proposition 2.7,

g can be written in the form g = L- f with L £ (A')'n, and 0 < L < M.Hence, H(g) = H(L • /) = 0, showing that L £ J. Because g > 0 we have

L > 0 as well.License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


Now consider the set

JG = {L £ J : L < G} .

Since J is a lattice ideal, JG is upwards directed. Moreover, JG is bounded

above by G in the Dedekind complete vector lattice (A')'„ , so Go = supJG

exists in (A')'n and Gq < G. Since, by Proposition 2.9,

(H ■ (Go-L))(f) 10 (L£JG),

we have (77 • Go)(f) = 0, so Go £ JG ■ We assert that G = Go . Suppose on the

contrary that G- G0 > 0. Since J is order dense in the positive cone of (A')'„,

there exists L £ J , with 0 < L < G-Go . From this we obtain 0 < L+Go < G.Together with L + G0 £ J we get L + G0 G JG, so L + Go < Go which isat variance with L > 0. We have obtained a contradiction, so G = Gq £ J.

Hence, (77 • G)(f) = 0. This holds for all 0 < / e A! , and therefore 77 • G = 0.This finishes the proof of the theorem.

Corollary 2.11. Ij A is an almost j-algebra and 77 g (A')'s, then 77 • / = 0 jor

all J£A'. Hence, F • 77 = 0 jor all F £ A", 77 e (A')'s.

Prooj. By Theorem 2.10, 77-x" = 0 for all x £ A. Consequently 0 =(H ■ x")(j) = (x" ■ H)(j) = x"(H ■ j) = (H ■ j)(x) for all x£A,andf£A'.Therefore, 77 • / = 0 for all / G A'. It follows that 0 = F(H ■ j) = (F • H)(f)for all J£A', and all F £ A" ; in other words, F • 77 = 0 for all F £ A" .

We are now in a position to prove the main result of this section.

Theorem 2.12. Ij A is an almost j-algebra, then A" is a (Dedekind complete

and hence Archimedean) almost j-algebra with respect to the Arens multiplica-

tion.

Prooj. Take 0 < F, G G A" with F A G = 0 and decompose F and G

according to the order direct sum A" = (A')'n © (A')'s, so

F = Fn + Fs (0<Fn£(A')'n,0<Fs£(A')'s),

G = Gn + Gs (0<Gn£ (A')'n ,0<GS£ (A')'s).

Since FnAGn = 0 and, by Theorem 2.2, (A')'n is an almost /-algebra, we have

Fn-G„ = 0. By Theorem 2.10, Fs • G„ = 0. By Corollary 2.11, F • Gs = 0. Itfollows that F ■ G = 0 and we are done.

The proof of Theorem 2.12 shows that, for any F , G £ A" , we have F -G =

F„-G„ . This fact, coupled with the observations that (A')'n is commutative and

closed under multiplication, makes the following two corollaries immediate.

Corollary 2.13. Let A be an almost j-algebra. Then F • 77 = 77 • F = 0 jor allF £A" ,77 G (A')'s. In particular, H2 = 0 jor all 77 G (A')'s.

Corollary 2.14. Let A be an almost j-algebra. Then F • G £ (A')'n jor all

F ,G£A".

3. The order bidual of /-algebras

The second author and B. de Pagter [7, Theorem 4.4] proved that the or-

der continuous bidual (A')'n of an Archimedean /-algebra is an Archimedean

/-algebra with respect to the Arens multiplication and the second author [5,License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


Theorem 2.8] extended this to A" itself (for the lattice normed case this last

result was obtained independently by E. Scheffold in [11, Satz 2.2]). In thissection we will show that these results can also be obtained by the methods of

Proposition 2.1 and Theorem 2.10.

Theorem 3.1. Let A be an f-algebra. Then (A')'n is an j-algebra with respect

to the Arens multiplication.

Prooj. We will exploit the ideas used in the proof of Proposition 2.1. First,

take 0 < G, 77 G (A')'„ , and suppose that G A 77 = 0 and that 0 < G, 77 < x"for some 0 < x G A. Assume also that F £ (A')'n satisfies 0 < F < x". We

will show that (F-G) AH = 0. To this end, fix 0 < / G A' and use Corollary1.2 to obtain g, h £ A' such that

j + j.x = g + h, with g Ah = 0 and G(g) = 0 = H(h).

Suppose e > 0; because g A h = 0 there exist elements 0 < y, z £ A

satisfying x = y + z, and g(y) + h(z) < e . As before, put

Gx = G A (y - y A z)" and Hx=HA(z-yA z)".

Then again,

0 < G - Gx < 2z" and 0 < 77 - 77, < 2y".

Now (y-yAz)A(z-yAz) = 0 and A is an /-algebra so x(y-yAz)A(z-yAz) =

0. Hence,

0 < (F ■ Gx) A Hx < x"(y - y A z)" A (z - y A z)" = 0

gives (F • Gx) A Hx = 0. We use the commutativity of A to get

0<(F.(G-Gx))(f)<(x"-(G-Gx))(f) = ((G-Gx)-x")(f)

= (G-Gi)(f-x)<(G-Gi)(f + f-x)= (G- Gx)(g + h)< G(g) + (G- Gx)(h) < 0 + 2z"(h) = 2h(z).

Analogously,

0 < (77 - Hx)(f) < (77 - 77,)(/ + f-x) = (77 - Hx)(g + h)

<(H-Hx)(g) + H(h)<2g(y).

It follows from

0<(F-G)AH<(F-(G-Gx)AH+(F'Gx)A(H-Hx) + (F-Gx)AHx

<F-(G-Gx) + (H-Hx) + 0

that

0 < ((F ■ G) A H)(f) <(F-(G- Gx))if) + (77 - 77,)(/)< 2h(z) + 2g(y) <2e.

The inequality 0 < (F-G AH)(j) < 2e holds for all e > 0, so((F • G)AH)(j) = 0. This being true for all 0 < / G A' we deduce (F • G)AH =0. Similarly it is shown that (G • F) A H = 0 (again we use the commutativity

of A). Summarizing, we have shown so far that, if 0 < F, G, 77 < x" (for

some 0 <x £ A) and G A 77 = 0, then (F ■ G) A 77 = 0 = (t7 • F) A 77. Inthe general case when F, G, H are arbitrary positive elements of (A')'n , with

G A 77 = 0 we obtain (F • G) A 77 = 0 = (G • F) A 77 from the fact that 7(ct(^))


4270 S. J. BERNAU AND C. B. HUUSMANS

is order dense in (A')'n and the same standard arguments that were used in the

proof of Theorem 2.2.

Now, every /-algebra is an almost /-algebra so all the arguments used in

§2 (Lemma 2.3 up to Corollary 2.11) and particularly Corollary 2.14 hold true.

Suppose now that G A 77 = 0 in A" and 0 < F £ A" . Decompose all three of

F, G and 77 into their order continuous and singular parts as follows.

F = Fn + Fs, G = Gn + Gs, 77 = 77„ + 77$.

As in Corollary 2.14, F-G = F„-G„ £(A')'„,so (F-G)AHS = (Fn-G„) AHS =0. Since G„ A Hn = 0 and (A')'n is an /-algebra, by Theorem 3.1, we have

Fn • Gn A 77„ = 0. Taking all these observations into account we find

(F-G)AH = (Fn-Gn) A (Hn + Hs) = (Fn-Gn)AHn + (Fn-Gn)AHs = 0.

Similarly, (G • F) A 77 = 0, so A" is an /-algebra. We summarize these results

in the next theorem.

Theorem 3.2. (cf. [15, Theorem 2.8]). If A is an f-algebra, then A" is a(Dedekind complete and hence Archimedean) j-algebra with respect to the Arens

multiplication.

In the remainder of this section we present yet another (even simpler and

essentially different) proof of the fact that the order continuous bidual (A')'n of

an Archimedean /-algebra is again an /-algebra. We first need a lemma.

Lemma 3.3. Let A be an j-algebra, 0 < a £ A and 0 < j £ A'. Then(i) ax < x + a2x jor all 0 < x £ A ;

(ii) j-a<j + j-a2.

Prooj. (i) Multiplying the first term in

(x - x A ax) A (ax — x A ax) = 0

by a we get

(ax - a(x A ax)) A (ax - x A ax) = 0,

which leads to

ax = a(x A ax) V (x A ax) < a(x A ax) + x A ax < a2x + x.

(ii) By (i) and the positivity of / we have

(j-a)(x) = j(ax) < j(x) + f(a2x) = f(x) + (f-a2)(x)

for all 0 < x G A . This is the desired result.

Now we are in a position to present an alternative proof of Theorem 3.1.

Theorem 3.4. The order continuous order bidual (A')'n of an f-algebra A is

again an j-algebra (with respect to the Arens multiplication).

Prooj. Observe first that Lemma 3.3(H) implies that for each 0 < x G A and

0 < G £ A" we have

ix" • G)(j) = G(j-x) < G(j) + G(j-x2) = G(j) + «x")2 • G)(j)

for all 0 < / G A', so

x"-G<G + (x")2-G.License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


In this inequality replace x" by (l/n)x" to obtain

n(x"-G)<n2G + (x")2-G (n = 1,2,...).

Now take 0 < G, 77 g A" such that G AH = 0. We claim that (F-G)AH =0 for all 0 < F £ (A')'n. First consider the case that 0 < F < x" for some

0 < x G A . By the above,

0 < n((x" - G) A 77) = n(x" -G)AnH

< n2G AnH+ ((x")2 - G) A nH < (x")2 -G (n = 1, 2, ...).

Since A" is Dedekind complete, and hence Archimedean, we find (x" • G)AH =

0. Consequently, (F • G) A 77 = 0. If F is an arbitrary positive element of

(A')'n , there exist Fa £ I(o(A)) (say 0 < Fa < x'l) such that 0 < Fa ] F. Bythe first part of this proof (Fa - G) A 77 = 0 for all a. Since 0 < Fa - G T F - Gwe find (F ■ (?) A 77 = 0.

Next consider 0 < F, G £ (A')'n , and 0 < 77 G A" such that G A 77 = 0.Since x" • G = G • x" for all x G A , we deduce from the earlier results that

(G • x") A77 = 0 for all 0 < x G A . With the same notation as above, (G • Fa) A

77 = 0 for all a. The order continuity of G combined with 0 < Fa * Fimmediately gives 0 < G • Fa } G • F and (G - F) A 77 = 0 follows.

The previous results make it clear that, if G A 77 = 0 in (A')'n, and 0 < F £

(A')'n , then

(F -G) A 77 = 0 = (G-F) A 77 = 0,

so (A')'n is an /-algebra.

From this point we can use the ideas of the first half of this section or follow

the proof used in [5] to show that the whole order bidual A" is an /-algebra

whenever A is. Our next theorem will describe yet another approach.

Theorem 3.5. The order bidual A" ojan j-algebra A is also an j-algebra.

Prooj. We have shown in the proof of Theorem 3.4 that (F • G) A 77 = 0 when-

ever 0 < F £ (A')'„ and 0 < G, 77 G A" satisfy G A 77 = 0. We will showthat the previous hypotheses also lead to (G ■ F) A H = 0 . Indeed apply Lemma

3.3(h) to the positive elements F of (A')'n and / of A', considered as a linear

functional on (A')'„ , to obtain

F-j<j + F2-j

for all 0 < / G A'. Since 0 < G £ A" we have

(G-F)if) = G(F-j) < G(j) + G(F2-j) = G(j) + (G-F2)(j)

for all 0 < / G A', so

G-F <G + G-F2.

Since G A77 = 0, we have (G-F) AH < G AH + (G-F2) AH = (G-F2) AH.Replacing F with (l/n)F leads to

(G-F)AH<(l/n)(G-F2)AH (n = 1,2,...).

Because A is Archimedean, (G ■ F) A 77 = 0.

It was shown in Proposition 2.2 of [5] that F £ (A')'n and G £ A" imply

G-F £ (A')'„ . (The proof is technical, but elementary, and can be simplified

to look very much like the proof of Proposition 2.9.) Take 0 < G £ (A')'s, 0 <License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


F G (A')'n . By the result just cited G-F £ (A')'n . Also G A 77 = 0 for all 0 <

77 G (A')'n , so by the earlier observations G-F AH = 0 for all 0 < 77 g (A')'n ,

i.e., G-F £ (A')'s. Hence, G-F = 0 for all 0 < G £ (A')'s, and 0 < F £ (A')'n .This is precisely the outcome of Theorem 2.10. For the rest of the proof, follow

the arguments from Corollary 2.11, Corollary 2.14 and Theorem 3.2.

4. The order continuous order bidual of (/-algebras

Recently, E. Scheffold showed [13, Theorem 2.1] that the order continuous

order bidual (A')'n of a Banach (/-algebra is a left Banach (/-algebra with respect

to the Arens multiplication. He was not able to show that (A')'n is also a right

Banach (/-algebra, the main reason for this being that Archimedean (/-algebras

need not be commutative (see e.g. [4, Example 4.1]). In the present section we

will generalize Scheffold's results considerably by showing that the order contin-

uous order bidual (A')'n of an Archimedean (/-algebra is an Archimedean (and

even Dedekind complete) (/-algebra with respect to the Arens multiplication.

The (/-algebra A we will consider need not carry a lattice norm and we will

show that (A')'n is not only a left but also a right (/-algebra. Our proofs will be

completely different from Scheffold's proofs and, we believe, more elementary.

So far we are unable to prove or disprove the corresponding result for the

whole order bidual A" of an Archimedean (/-algebra A , the main reasons being

that (/-algebras need not be commutative nor need they have positive squares.

(For an example of the latter, see [4, Example 4.1 ].) This means that the method

used in §2 for almost /-algebras (and in particular Lemma 2.3) does not work

in this case. In the proof of the next theorem we are able to use the same basic

methods that were developed in the proof of Proposition 2.1 and Theorem 3.1.

Theorem 4.1. Let A be a d-algebra. Then (A')'n is a d-algebra with respect to

the Arens multiplication.

Prooj. We have to show that if 0 < F ,G,H £ (A')'„ and G A 77 = 0, then(F-G) A(F-H) = 0 (so (A')'n is a left (/-algebra) and (G-F) A (H-F) = 0(so (A')'n is a right (/-algebra). We may assume that 0 < F, G, 77 < x" for

some 0 < x £ A . The standard order continuity arguments (use that I(o(A))

is order dense in (A')'n) then yield the general result. Take 0 < / G A' and

define the positive linear functional / * x by

(f * x)(y) = f(yx)

for all y G A (the dual functional of f-x). Of course f-x = f*x whenever

A is commutative, but Archimedean (7-algebras need not be! By Corollary 1.2

again, there exist 0 < g, h £ A' with g A h = 0, and G(g) = 0 = H(h) such

that

f-x + f*x = g + h.

Suppose e > 0; since g A h = 0 there exist elements 0 < y, z £ A with

x = y + z , and g(y) + h(z) < e . As before we define Gx = GA(y-yAz)" , and

77, = 77 A (z - y A z)" . Again we have 0 < G - Gx < 2z" , 0 < 77 - 77, < 2y" .It follows from

0<(Gx-F)A (Hi ■ F) < (Gx - x") A (Hi - x")

< ((y - y A z)" ■ x") A ((z - y A z)" • x") = 0License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use

ALMOST F-ALGEBRAS AND B-ALGEBRAS 4273

(here we use the right (/-algebra property in A) that

(Gx-F)A(Hx-F) = 0.

Observe that

(x"-j)(y) = x"(j-y) = (j-y)(x) = j(yx) = (/* x)(y)

for all y £ A , so x" • f = f * x. Hence,

0 < ((G - Gi) • F)(f) < ((G - Gx)-x")(f) = (G- Gx)(x" - f)

= (G-Gi)(f*x)<(G-Gi)(f-x + f*x)= (G- Gx)(g + h)< Gig) + (G- Gx)(h) < 0 + 2z"(h) = 2h(z).

Analogously,

0<((H-Hx)-F)(j)<2g(y).

Now,

(G-F)A(H-F) = (G-GX+GX)-F A(H-H{+Hi)-F

<(G-Gi)-F + (H-Hi)-F + (Gi-F)A(Hx-F)= (G-GX)-F + (H-HX)-F

shows that

0<((G-F)A (H-F))(j) < ((G-Gx)-F)(f) + ((77- 77.) - F)(f)<2h(z) + 2g(y) <2e.

Hence, since e > 0 is arbitrary, we get ((G-F) A (77• F))(f) = 0 for all 0 <

/ G A', so (G • F) A (77 • F) = 0. Similarly, (F - G) A (F - 77) = 0 and the proofis complete.

Another way of obtaining the result of Theorem 4.1 is by means of the theory

of lattice homomorphisms. Indeed, if A is an /-algebra, then A is a (/-algebra

if and only if for each 0 < x G A the right multiplication, Rx , by x (defined by

F-x(y) = yx for all x G A) and the left multiplication, Lx , by x (Lx(y) = xy

for all y £ A) are lattice homomorphisms of A . Hence, the first adjoints R'xand L'x are interval preserving and consequently the second adjoints R'x and

L'x are lattice homomorphisms of A" (for details, see [1, Theorems 7.4, 7.8

and 7.9]).We claim that R'X(G) = G-x" for all G £ A" (so 7^' = Rx„ in A"). Indeed,

if / G A', then

(Kf)(y) = f(R*y) = f(yx) = (f-y)(x) = x"(f-y) = (x"-f)(y)

for all y £ A implies R'xf = x" • f. Consequently, if G £ A" , then

(RxG)(f) = G(R'xj) = G(x"-j) = (G-x")(j)

for all f£A' yields R'X(G) = G-x" (x£A,G£ A"). Similarly, L'xf = f-x(x£A,f£A') and L'^(G) = x"-G(x£A,G£ A"), so L'x = Lx„ in A" .

Now take 0 < x G A and 0 < F, G £ A" such that F A G = 0. Since L"xis a lattice homomorphism of A" , we get

(x" • F) A (x" ■ G) = L"X(F) A L"X(G) = 0.

By the standard order continuity arguments we find that for all 0 < 77 G (A')'n

we have (H • F) A (H • G) = 0. In particular, (A')'„ is a left (/-algebra. (In factLicense or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


we proved a slightly stronger result.) If F A G = 0 in (A')'n, then as above

(F-x") A (G-x") = R'X(F) A R'X(G) = 0. This implies easily that (F ■ H) A

(G-H) = 0, so (A')'„ is also a right (/-algebra. (But this time we need theorder continuity of F and G!)

As observed before, we have not been able to prove that the whole second

order dual A" of a d-algebra A is a (/-algebra as well, mainly because we have

not found a way to handle the singular part of A" . We have observed already

that A need not be commutative, neither need A have positive squares. If we

impose one of these additional conditions on A the situation becomes simple.

Theorem 4.2. Let A bea d-algebra which is commutative or has positive squares.

Then A" is a Dedekind complete, hence Archimedean, d-algebra with respect

to the Arens multiplication. Moreover, A" is commutative and its squares are

positive.

Prooj. By [4, Theorem 4.5] each of the conditions on A implies that A is an

almost /-algebra. By Corollary 2.14, F-G = Fn-G„ for all F, G £ A" , whereFn and G„ denote the order continuous parts of F and G, respectively. Now

assume that G A 77 = 0 in A" and 0 < F g A". Decompose all three of

F, G, H into their order continuous and singular components as follows.

F = Fn + Fs, G = Gn + Gs, H = Hn + Hs,

according to the band decomposition A" = (A')'n © (A')'s. Since, by Theorem

4.1, (A')'n is a (/-algebra and Gn A Hn = 0, we have

(Fn - G„) A (Fn-Hn) = 0 = (Gn-F„) A (Hn-F„).

By Corollary 2.14, F - G = F„ - G„ , F -H = Fn-Hn, so (F-G)A(F-H) = 0.Similarly, (G-F) A (H-F) = 0, so A" is a (/-algebra. By Theorem 2.12, A"is an Archimedean almost /-algebra; in particular A" is commutative and its

squares are positive.

Acknowledgments

We thank J. J. Grobler for pointing out to us that we could use the facts that

the adjoint of a lattice homomorphism is interval preserving (in the different

approach in the proof of Proposition 2.7) and that the bi-adjoint of a lattice

homomorphism is again a lattice homomorphism (in the alternative proof of

Theorem 4.1). We also thank the referee for providing the bibliographical in-

formation for [13].

References

1. CD. Aliprantis and O. Burkinshaw, Positive operators, Academic Press, Orlando, FL, 1985.

2. R. Arens, Operations induced in function classes, Monatsh. Math. 55 (1951), 1-19.

3. _, The adjoint of a bilinear operation, Proc. Amer. Math. Soc. 2 (1951), 839-848.

4. S. J. Bernau and C. B. Huijsmans, Almost f-algebras and d-algebras, Math. Proc. Cam-

bridge Philos. Soc. 107 (1990), 287-308.

5. C. B. Huijsmans, The order bidual of lattice ordered algebras. II, J. Operator Theory 22

(1989), 277-290.6. _, Lattice ordered algebras and f-algebras: A survey, Studies in Economic Theory 2,

Positive Operators, Riesz Spaces and Economics (C. D. Aliprantis, K. C. Border and W. A.

J. Luxemburg, eds.), Springer, Berlin, 1991, pp. 151-169.



7. C. B. Huijsmans and B. de Pagter, The order bidual of lattice ordered algebras, J. Funct.

Anal. 59(1984), 41-64.

8. W. A. J. Luxemburg, Notes on Banach function spaces. XVa, Indag. Math. 27 (1965),

415-429.

9. W. A. J. Luxemburg and A. C. Zaanen, Riesz spaces. I, North-Holland, Amsterdam, 1971.

10. P. Meyer-Nieberg, Banach lattices, Universitext, Springer, Berlin, 1991.

11. E. Scheffold, Der Bidual von F-Banachverbandsalgebren, Acta Sci. Math. 55 (1991), 167—179.

12. _, Ueber den ordnungsstetigen Bidual von FF-Banachverbandsalgebren, Arch. Math.

60(1993), 473-477.

13. _, Ueber Bimorphismen und das Arens-Produkt bei kommutativen D-Banachverbands-

algebren, Rev. Roumaine Math. Pures Appl. 39 (1994), 259-270.

14. A. C. Zaanen, Riesz spaces. II, North-Holland, Amsterdam, 1983.

Department of Mathematical Sciences, University of Texas at El Paso, El Paso, Texas

79968-0514E-mail address: simonQmath. utep. edu

Department of Mathematics, Rijksuniversiteit Leiden, P.O. Box 9512, 2300 RA Leiden,

The Netherlands

E-mail address: chuij smansQrulcri. leidenuniv. nl


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