Locally Solid Riesz Spaces (Pure and Applied Mathematics (Academic Press), Volume 76)

Locally Solid Riesz Spaces

CHARALAMBOS D. ALlPRANTlS OWEN BURKINSHAW Department of Mathematical Sciences Indiana University-Purdue University at lndianapolis Indianapolis, Indiana

ACADEMIC PRESS

New York

San Francisco

A Subsidiary of Harcourt Brace Jovanovich, Publishers

London

1978

COPYRIGHT 0 1978, BY ACADEMIC PRESS, INC. ALL RIGHTS RESERVED. NO PART OF THIS PUBLICATION MAY BE REPRODUCED OR TRANSMITTED IN A N Y FORM OR BY ANY MEANS, ELECTRONIC OR MECHANICAL, INCLUDING PHOTOCOPY, RECORDING, OR ANY INFORMATION STORAGE AND RETRIEVAL SYSTEM, WITHOUT PERMISSION IN WRITINQ FROM THE PUBLISHER.

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United Kingdom Edition published by ACADEMIC PRESS, INC. (LONDON) LTD.

24/28 Oval Road, London NW1 IDX

Library of Congress Cataloging in Publication Data Aliprantis, Charalambos D. Locally solid Riesz spaces. (Pure and applied mathematics; ) Bibliography: p. Includes index. 1. Rieszspaces. I. Burkinshaw, Owen, joint author. 11. Title. 111. Series. QA3.PB [QA322] 510’.8s [515’.73] 77-1 1206 ISBN 0-12-050250-X AMS (MOS)1970 Subject Classifications:46A15,46A40, 06A65,46E05,46E30,47B55 PRINTED IN THE UNITED STATES OF AMERICA

Preface

In 1928, at the International Mathematical Congress in Bologna, F. Riesz in a short address [61] triggered the investigation of what is today called the theory of Riesz spaces. Soon after, in the mid-thirties, H. Freudenthal [27] and L. V. Kantorovich [35, 361 independently set up the axiomatic foundation and derived a number of properties dealing with the lattice structure of Riesz spaces. From then on the growth of the subject was rapid. In the forties and early fifties the Japanese school, led by H. Nakano, T. Ogasawara, and K. Yosida, and the Russian school, led by L. V. Kantorovich, A. I. Judin, and B. Z . Vulikh, made fundamental contributions. At the same time a number of books started to appear on the field. Most of the early work on Riesz spaces, as well as most of the books, was devoted to the so-called “algebraic part” of the theory and little attention was given to the “analytical part.” The bulk of the existing work on the analytical part is in the area of normed Riesz spaces, and in particular in the theory of Banach lattices; we note the work of W. A. J. Luxemburg and A. C. Zaanen appearing in a series of articles [45] and the recent book of H. H. Schaefer [67]. The general theory of topological Riesz spaces seems somehow to have been neglected. The recent book by D. H. Fremlin [25] is partially devoted to this subject; and there are, of course, a number ofbooks on linear ordered spaces (see [32,56,70,71]); but there have been no books completely devoted to topological Riesz spaces. The purpose of this book is to fill this gap and present a unified approach to the theory of locally solid Riesz spaces. It has been our desire to emphasize the relationship between the order structure and the topological structure. The present work is written for graduate students and for persons involved ix

X

PREFACE

in research in functional analysis. The book is self-contained and should be accessible to any student with a standard course in functional analysis. The material has been arranged into seven chapters. As a supplement, at the end of each chapter, we have included a number of exercises of varying degrees of difficulty. We have also listed some open problems to give insight into possible areas of research. Throughout the book the basic concepts are illustrated by numerous examples and counterexamples, and special notes trace the history of the subject. However, it is possible that credit was not given to all contributors to this field. In Chapter 1 a summary of the basic theory of the lattice structure of Riesz spaces is given, with special emphasis on results that are not easily accessible and to proofs that are not well known. The fundamental properties of locally solid topologies are presented in Chapter 2, while Chapter 3 studies the interaction between order convergence and topological convergence (Lebesgue properties) with applications to the representation theorems of abstract L,-spaces. Chapter 4 deals with Fatou topologies and shows the remarkable relationship between order completeness and topological completeness, while Chapter 5 investigates the important properties of metrizable locally solid Riesz spaces. In Chapter 6 locally solid topologies that are also locally convex are considered. Weakly compact sets are studied via the concept of orderequicontinuity. The interplay between the order structure and the topological concept of compactness is demonstrated extensively, and characterizations of semireflexive Riesz spaces are given. Finally, Chapter 7 discusses both the remarkable lattice and topological properties of laterally complete Riesz spaces. This book has been influenced by the discussions the first author had with W. A. J. Luxemburg and the second with S . Kaplan, to whom we express our appreciation for introducing us to the subject. Finally, we would like to thank our wives, Bernadette and Betty, for their help and understanding during the preparation of this book. Indianapolis June 1977

C. D. ALIPRANTIS 0. BURKINSHAW

List of Special Symbols

Numbers that follow each entry are the page numbers on which the symbol first occurs.

Zero element 2 Ideal generated by u 4 Band generated by u 12 Carrier of cp 23 Carrier of the topology t 83 Band generated by the discrete elements 154 Disjoint complement of D 7 Order dual of L 23 Order continuous linear functionals on L 23 a-order continuous linear functionals on L 23 Topological dual of ( L , t ) 28 Dedekind completion of L 10-11 a-Dedekind completion o f L 11 Universal completion of L 175

L equipped with a Riesz seminormp 38 Topological completion of(L,t) 30 Order bounded operators from L into M 19 Order continuous operators of Yb(L,k!) 22 a-order continuous operators of di”,(L,M) 22 Null ideal of cp 23 Quotient seminorm determined by p 39 Seminorm generated by cp 40 Seminorm generated by A 59 Projection onto the band generated by u 13 xi

xii

LIST OF SYMBOLS

lol(L,L') Absolute weak topology on L generated by L' lal(L',L) S(A) TlA

40 Absolute weak topology on L' generated by L 127 Solid hull of A 3 Quotient topology determined by z 30

zi u,

Tu

u,

-,u (0)

u, 1,u

Equivalence class of u in L/A 9 Increasing net with supremum u 5 Order convergence 5 Topological convergence 35

CHAPTER

1

The Lattice Structure of Riesz Spaces

In the first chapter of this book we shall deal with some of the basic properties of Riesz spaces and we hope that the reader is somewhat familiar with the basic theory of Riesz spaces. Most of the proofs of the elementary properties of Riesz spaces can be found in the excellent book “Riesz spaces,” Vol. I, by Luxemburg and Zaanen (cf. [46]), which we shall refer to by L&Z. For this reason a number of the lattice properties of Riesz spaces will be stated without proofs, and the reader will be referred for proofs to the above-mentioned book. We shall provide, however, a number of proofs of results that are either not easily accessible or according to our judgment greatly simplified proofs. 1.

ELEMENTARY PROPERTIES OF RIESZ SPACES

An ordered uector space E is a real vector space E equipped with an order relation 2 (i,e., with a transitive, reflexive, and antisymmetric relation 2 ) compatible with the algebraic structure of E in the sense that it satisfies the following two properties:

(i) If u,u E E satisfy u 2 u, then u + w 2 u + w holds for all w E E . (ii) If u,u E E satisfy u 2 u, then Au 2 h for each A E R with A 2 0. (Throughout this book the symbol R will denote the set of real numbers.) Occasionally the notation u I u will be used to mean u 2 u. The relation u > u (or /I < u) will mean u 2 u and u # u. The zero element of a vector 1

RlESZ SPACES

2

[Chap. 1

space will be denoted by 8. The elements u of E with u 2 8 are called positive elements, and those satisfying u > 8 are called strictly positive elements. The set E + = { u E E : u 2 0) is called the positive cone of E. The positive cone satisfies the following properties:

+ E+ E E + , where E + + E + = ( u + u:u,v E E'}. G E + for 0 5 1 E R,where AE+= { h : uE E ' ) . E + n ( - E+) = { e},where - E + = { - u : u E E + >.

(a) E +

(p) IE' (y)

Any subset C of a real vector space E satisfying properties (a), (p), and (y) is called a cone of E. If C is a cone of E, note that the relation u 2 v whenever u - 21 E C makes E an ordered vector space whose positive cone is precisely C. A nonempty subset A of an ordered vector space E is said to have a supremum (or a least upper bound) if there exists u E E with u 2 a for all u E A and such that if 1' 2 a holds for some u E E and all a E A , then L' 2 u also holds. Clearly a subset A can have at most one supremum, denoted by sup A. A similar definition can be given for the injmum (or the greatest lower bound) of a nonempty subset A of E , denoted by inf A . A Riesz space (or u uector lattice) L is an ordered vector space L with the additional property that the supremum of every finite nonempty subset of L exists. Following the classical lattice notation, we denote the supremum of the set { u p } by u v u, that is, u v u = sup{u,u}. Similarly, u A u denotes the infimum of the set {u,u}. It can easily be seen that in a Riesz space every nonempty finite subset has an infimum. Note. The concept of a Riesz space is due, independently, to Riesz [61,62], Freudenthal [27], and Kantorovich [35-371.

For an element u of a Riesz space L the posiriue part of u is defined by u+ = u v 8, the negutihe purr by u - = ( - u ) v 8, and the absolute value by IuI = u v ( - u). For the following fundamental relations and additional ones see L&Z, pp. 55-66. Theorem 1.1. If u, u, and w are elements of a Riesz space L, then we haue U +U=UVU+UAU.

u = u + - u - and u+ A u- = 8. uVU=(U-u)+ + u = ( u - u ) + +u. IuI = u+ u - (hence IuI = 8 ifund only i f u = 0).

+

+ UI

+

I(uI - lull 5 Iu s IuI IuI (the triangle inequality). u + u v w = (U + U ) V ( U + w)undu + U A W = (u + D ) A ( U n(u v u) = (nu)v ( n u ) and I ( u A t i ) = (nu)A (nu) i f I 2 0. liul = 1A1 lul for all A E R.

+ w).

3

THE BASIC THEORY OF RlESZ SPACES

Sec. 11

(9) u - U A w = ( u - u ) v ( u - w)and u - uv w = ( u - U ) A ( U - w). (10) I U - U l = U V U - U A A . (11) Iu v w - u v wI + Iu A w - u A wI = (u - uI ( B i r k h o f s identity). (Birkhofs (12) I u v w - u v wI IIu - uI and IUA w - U A wI IIu inequalities). (13) l f u , u , w ~L’, then(u + U ) A W IU A W + U A W .

UI

Observe that (3) implies that an ordered vector space L is a Riesz space if and only if u+ exists for every u E L. Every Riesz space L satisfies the injnite distributive law, that is, if sup A exists for a nonempty subset A of L then sup{u A a : u E A } exists for each u E L and satisfies the relation u A sup A = sup{u A a : a E A } . A relation analogous to the last one holds by replacing A by v and sup by inf; see L&Z, p. 62.

+

+

Theorem 1.2. Assume that in a Riesz spuce L the relation IuI IIul ... u,, holds. Then there exist elements u l , . . . ,u, E L satisfying luil I luil for i = 1,. . . ,n, and such thut u = u1 . . . u,. In addition. i f u is positiue, then the u, cun he chosen to he positiue too.

+

+

Proof. By using induction it is enough to establish the result for n = 2. Thus assume IuI s lo1 + u21. Next put u1 = { u v ( - lull)} A lull and note that lull I lull and that 0 Iu l I u holds if u is also positive. Now if u2 = u - u l , then u2

=u

-

However, IuI I lull we get

C~V(-lUI1)}AIulI

+ lu21 implies

-1u2l = ( - l u 2 1 ) A e I (u

Thus lu21 5

={

W u + lu,l)}v(u-

- I u l l - Iu21 Iu I lull

lull).

+ Iu21, from which

+ I u l l ) A \ e Iu2 I0 v ( u - lull) Ilo2[.

1 0 ~ 1 . The proof is now immediate.

4

The Cartesian product L = l l L , of a family of Riesz spaces {L,:a E A } , ordered componentwise, that is, ordered by the relation {u,:a E A } 2 {u,:a E A } whenever u, 2 u, for all CI,becomes a Riesz space. The proof is straightforward; note also that {U,:CI E A } v {o,:a E A ] = (u,v V,:CI E A } holds in L and that a similar formula holds for the infimum. A subset S of a Riesz space L is said to be a solid set if it follows from Iu[ I IuI in L and V E S that ~ E SEvery . subset A of L is contained in a smallest (with respect to the inclusion) solid set called the solid hull of A and denoted by S(A). Note that S ( A ) = { u E L:3u E A with IIuI). Every solid subset S is a balanced set, that is, if u E S then l u E S for each A E R with 121 I1.

11

RIESZ SPACES

4

[Chap. 1

We recall that the conuex hull Co(A) of a subset A of a vector space E is the smallest convex set containing A and consists precisely of all elements Iiuiwhere ui E A , I i 2 0 for i = 1, . . . ,n, and I i = 1. of E of the form Theorem 1.3. T h e conuex hull of a solid set of a Riesz space is a solid set.

Proof. Let S be a solid set of a Riesz space L, u E Co(S)and u E L such that IuI I There exists a finite number of elements of S , say ul, . . . ,u,, and positive constants 1 1 ,. . . ,I, with I i = 1 such that u = liui. By Theorem 1.2 there exist elements u l , . . . ,u, such that luil I IIiuil = Iiluil for each i and with u = ui. Put wi = I1:'ui and note that wi E S for ui = Iiwi E Co(S)and hence Co(S) each i = 1, . . . ,n. But then u = is solid. H

[MI.

cy=

cy=

cy=

Definition 1.4. Let K be a vector subspace of a Riesz space L. We shall say that (i) K is a Riesz subspace of L iffor every pair of elements u,u of K the supremum of u and u taken in L belongs to K ; (ii) K is an ideal of L if K is a solid subset of L. By Theorem 1.1(3) a vector subspace K of a Riesz space L is a Riesz subspace if and only if for every u E K we have u + E K . From the relation 8 I u+ I it follows now that every ideal is a Riesz subspace. A Riesz subspace need not be an ideal, as can be easily demonstrated by examples. However, a Riesz subspace K is an ideal of L if B Iu Iu and u E K imply u E K. Every subset D of a Riesz space L is contained in a smallest ideal A, called the ideal generated by D ;A is the intersection of all ideals containing D.Observe that

(MI

A = { u E L : 3 u l , . . . ,u,

E

D and A,, . . . ,I, E Rf with

n

Iu( I

1Iiluil}.

i= 1

A principal ideal is any ideal generated by a subset containing only one element u ; this ideal will be denoted by A,. An order unit (or a strong unit) is an element 8 < e with A, = L , that is, if for every u E L , there exists I > 0 with Iu( IIe. Note also that if A and B are ideals of L, then their algebraic sum A + B = { u + u : u E A,u E B } is also an ideal of L. We recall that a binary relation 2 directs a nonempty set A if it satisfies the following three conditions:

(a) c( 2 a for all a E A. (b) If a,p,y E A satisfy a 2 p and p 2 y, then a 2 y. (c) For any pair a$ E A, there exists y E A with y 2 a and y 2 p.

Sec. 11

THE BASIC THEORY OF RlESZ SPACES

5

The set A [more precisely the pair ( A , > ) ] is then called a directed set. A net of a set X is a mapping u : A -+ X from a directed set A into X . As usual, u(a) will be denoted by u, for a E A and the net u : A + X by {u,}, omitting the directed set A . The set A will be denoted by { a } (the indexed set of the net {u,}) and its members will be called indices. A net {u,) of a Riesz space L is called increasing (in symbols u, f ) if us 2 u, whenever the indices a$ satisfy 2 a. The symbol u, 1 denotes a decreasing net and its definition is analogous. The symbol u, f u means that the net {u,} is increasing and the supremum of {u,} exists and equals u. The meaning of u, 1 u is defined similarly.

a

Definition 1.5. A net {u,) of a Riesz space L is said to be order convergent to u E L (in symbols u, 2 u) if there exists a net {v,} (with the same indexed set) such that Iu, - uI 5 v, holds for all a and with v, J, 9. (We shall write the last two conditions symbolically as Iu, - uI I v, J 6.)

By using properties (4)and (5) of Theorem 1.1, it can easily be seen that a net in a Riesz space can have at most one order limit. The basic properties of order convergence are summarized in the next theorem; see L&Z, p. 78. Theorem 1.6. Assume that the nets {u,} and { u s } of a Riesz space L (0) satisfy u, 2 u and us -+ u. Then we have (i)

Iu,I

2 [MI;

u,

+

(0)

-+

u + and u;

(0)

-+

-

u .

Note. All nets having a double index (GI$) are considered indexed by the product { a } x where as usual the direction on the product { a } x is the componentwise one.

{a},

{a}

A subset S of a Riesz space L is said to be order closed if it follows from 2 u in L that u E S , that is, S is order closed if it contains its order limits in L. Similarly, a subset S is called o-order closed if S contains its sequential order limits; see L&Z, pp. 78-91, for more about order convergence and order closed sets. A o-order closed ideal is called a a-ideal and an order closed ideal is called a band. Note that an ideal A is a band if and only if 9 I u, t u in L with {u,} _c A implies u E A ; similarly an ideal A is a o-ideal if and only if 9 Iu, f u and {u,} G A implies u E A. Given a Riesz subspace K of a Riesz space L we say that the embedding of K into L preserves arbitrary suprema and injma if for every subset of K {u,} G S and u,

RIESZ SPACES

6

[Chap. 1

whose supremum (or infimum) exists in K , then the supremum (or infimum) of the same subset exists in L and is the same as that in K . This and related properties are included in the next definition. Definition 1.7. A Riesz subspuce K of a Riesz space L is said to be

(i) regular, if the embedding of K into L preserves arbitrary suprema and infirnu; (ii) a-regular, if the embedding of K into L preserves countable suprema and infirnu; (iii) full, iffor every u E L there exists v E K with IuI I v. The proof of the next theorem is straightforward and is left to the reader. Theorem 1.8. For a Riesz subspuce K of a Riesz space L the following stutements are equivalent :

(i) K is a regular Riesz subspace of L. (ii) I f {u,} G K sati.$es u, 18 in K , then u, .1 8 also holds in L. (iii) I f {u,} E K satisjes u, 2 8 in K . then u, 5 6 also holds in L. The reader should be able to formulate by himself the “a-analogue” of Theorem 1.8. Observe also that every ideal is a regular Riesz subspace. A further property dealing with Riesz subspaces is described in the next definition. Definition 1.9. A Riesz subspace K of a Riesz space L is said to be (i) order dense in L, $.for every 0 < u E L there exists v E K with 8
Theorem 1.10. Every order dense Riesz subspace of a Riesz space is a regular Riesz subspace. Proof. Assume that K is order dense in L and ( u a } satisfies u, 16 in K . If u, 18 does not hold in L, then there exists 0 < u E L with 8 < u Iu, for all a. But then since K is order dense in L, there exists 6 < v E K such that 8 < v Iu. So, 6 < v I u, holds in K for all a, contradicting u, 18 in K ; therefore K is a regular Riesz subspace by Theorem 1.8.

THE BASIC THEORY OF RlESZ SPACES

Sec. 11

7

Two elements u and v of a Riesz space are called orthogonal or disjoint (in symbols u I u ) if IuI A = 8. The basic properties of the orthogonality relation are the following (see L&Z, pp. 68-70):

(01

(i) If u I u and u I w, then u I(Au + pw) for all A,p E R. (ii) If u I v, then Iu + v1 = Iu - uI = IuI + IuI = IluI = IuI v 1u1. (iii) If Iu uI = Iu - vJ, then u I u.

IL~II

+

The disjoint complement Dd of a nonempty subset D of a Riesz space L is defined by Dd = { u E L : u I u for all u E 0 ) .It can be seen easily that Dd is an ideal of L, and by using the infinite distributive laws, it can be shown that Dd is actually a band of L. The disjoint complement (Dd)dwill be denoted by Ddd.Observe that D n Dd = { 8 ) and that D G Ddd.Also, if A G B, then Bd E Ad. Theorem 1.11. lf A is an ideal of a Riesz spuce L, then A is order dense in Add.In particular, A is order dense in L if and only if Ad = { O } .

Proof. Assume that A is not order dense in Add.This means that there exists 0 < u E Add with no element u of A satisfying 8 < c I u. In particular (since A is an ideal) we must have A u = 8 for each u E A. Thus u E Ad and hence 11 E Ad n A d d= (e), a contradiction. This contradiction establishes that A is order dense in Add. H

IL~I

The next theorem shows that for every ideal A the ideal A @ Ad is an order dense ideal. Theorem 1.12. For every ideal A of a Riesz space L, the ideal A @ Ad is order dense in L. Proof. If L I E( A @ Ad)d,then u E Ad n Add = {O).So, ( A 0 AdId = [ O ) , and the result follows from Theorem 1.11. H Among the classes of interesting Riesz spaces is the class of the Archimedean Riesz spaces, which is defined next. Definition 1.13. A Riesz space L is said to be Archimedean if u,u E L' and nu Ivfor n = 1,2, . . . imply u = 0. The best known non-Archimedean Riesz space is the lexicographic plane, that is, R2 ordered as follows: ( x l , y l )2 (x2,y2) if either x1 > x2 or else x 1 = x2 and y1 2 y 2 . Observe that the lexicographic plane is linearly ordered, that is, every two elements are comparable. It should be clear that Riesz subspaces of Archimedean Riesz spaces are Archimedean and that Cartesian products of Archimedean Riesz spaces are also Archimedean. In Archimedean Riesz spaces the order denseness has a nice characterization. The details follow.

RIESZ SPACES

0

[Chap. 1

Theorem 1.14. For a Riesz subspace K of an Archimedean Riesz space L the following statements are equivalent:

(i) K is order dense in L. (ii) u = sup { u E K :8 I v Iu} holds in L for each u E Li (or equivalently, for each u E L’ there exists a net {u.} c K with 8 Iu, t u in L). Proof. (i) 3 (ii) Let 8 < u E Land assume that sup { u E K :8 5 u I u} = does not hold. This means that there exists 0 < w E L such that v E K and 8 5 u I u imply u Iu - w. Pick x E K such that 8 < x I wand observe thatx 5 (u - w) + w = u.Thisinturnimplies2x = x + x I (u - w) w = u and so 2x I u - w. By an inductive argument, we get 8 < nx I u for n = 1,2, . . . , contradicting the Archimedeanness of L. (ii) =r (i) Obvious. W II

+

To see that (i) does not imply (ii) in the non-Archimedean case, consider L to be the lexicographic plane and K to be the “y-axis,” that is, K = ((0,y):yE R}. Note that K is an order dense band of L, and that (ii) does not hold. By combining Theorems 1.11 and 1.14, we see at once that every band A in an Archimedean Riesz space satisfies A = Add. Actually, this condition is equivalent to the Archimedeanness. Some characterizations of Archimedeanness that will be used in this book are included in the next theorem; for a proof and more characterizations see L&Z, pp. 112-121. Theorem 1.15. For a Riesz space L the following statements are equivalent:

(i) L is Archimedean. (ii) For ever)?bounded net ( i nof) R with La 4 0 we have A.,u every u E L. (iii) A = Addholdsfor every band A ofL. (iv) For every nonempty subset S of L+ we have S inf(u - u : u ~ und

where T =

( U E L:O

2 0 in L for

U ET

) = 0,

-, M

from a Riesz space L into a

5 u 5 u for all ~ E S ) .

Definition 1.16. A linear mapping n:L Riesz space M is called

(i) a Riesz homomorphism, i f u A v = 8 in L implies n(u)A n(v)= 8 in M ; (ii) a Riesz a-homomorphism, i f n is a Riesz homomorphism and u, 3 8 in L implies n(u,,)2 8 in M ; (iii) a normal Riesz homomorphism, i f 7 1 is a Riesz homomorphism and u, 26 in L implies 7c(ua) 2 8 in M ; (iv) a Riesz isomorphism (into),i f n is a one-to-one Riesz homomorphism.

ORDER COMPLETENESS AND PROJECTIONS

Sec. 21

9

Riesz homomorphisms preserve the lattice structure as the following theorem shows; for a proof see L&Z, p. 98. Theorem 1.17. For a linear mapping n:L + M between two Riesz spaces L and M the following statements are equivalent: (i) (ii) (iii) (iv) (v) (vi)

n is a Riesz homomorphism. n(u A u) = n(u)A n(u)for all u,u E L. n(u v u) = n(u)v n(u)for all u,u E L. n(u v u) = n(u)v n(u),wheneueru A u = 6' holds in L. n(u+)= [ n ( u ) ] + f o rall u E L. n(lu1)= Iz(u)lforall u E L.

It follows from n( lul) = In(u)l that every Riesz homomorphism satisfies n(u)E M + if u E L + , that is, every Riesz homomorphism is a positive operator. An onto Riesz homomorphism n is normal if and only if its kernel K , = { u E L:n(u)= 6') is a band of L and llkewise n is a Riesz a-homomorphism if and only if K , is a a-ideal of L ; see L&Z, p. 103. Two Riesz spaces L and M are called Riesz isomorphic if there exists a Riesz isomorphism from L onto M . Theorem 1.18. Let n : L + M be an onto Riesz homomorphism between two Riesz spaces L and M . Then n carries solid subsets of L onto solid subsets of M .

Proof. Assume that S is a solid subset of L and that In(u)I I In(u)I holds in M for some u E S. Put w = [( - 1 ~ 1 ) v u ] A IuI E L and note that since IwI s IuI holds we have w E S . Also n ( w ) = [( - In(u)l)v n(u)]A 1n(u)I = n(u). Thus n ( S ) is a solid subset of M .

For every ideal A of a Riesz space L the quotient vector space L / A becomes a Riesz space when ordered by ri > d whenever there are u1 E ti and u1 E d with u1 2 u1 (a denotes the equivalence class determined by the element u). The canonical projection n : L + L / A from L onto L / A determined by n(u)= ri = u + A (u E L ) is a Riesz homomorphism and, according to what was mentioned above, the canonical projection becomes a normal Riesz homomorphism (resp. a Riesz a-homomorphism) if and only if A is a band of L (resp. a a-ideal of L ) .

2.

ORDER COMPLETENESS AND PROJECTION PROPERTIES

Let L be a Riesz space. For any two elements u,u E L with u I u, the order interval [u,u] is a subset of L defined by [u,u] = { w E L : u Iw I u } . A subset A of L is said to be (order) bounded from above if there exists u E L

10

[Chap. 1

RlESZ SPACES

with v I u for all u E A. The bounded from below subsets are defined analogously. A subset A of L is called order bounded if it is both bounded from above and from below, or equivalently if A is contained in an order interval. Definition 2.1. A Riesz space L is said to be

(i) Dedekind complete, i f every nonempty subset that is bounded from above has a supremum; (ii) a-Dedekind complete, i f every countable subset that is bounded from above has a supremum; (iii) almost a-Dedekind complete, if L is Riesz isomorphic to a super order dense Riesz subspace of some a-Dedekind complete Riesz space. Clearly Dedekind completeness implies a-Dedekind completeness and a-Dedekind completeness implies almost a-Dedekind completeness. Note also that every o-Dedekind complete Riesz L is Archimedean; indeed, if 0 _< nu I 1’ holds in L for some u,u E L’ and all n, then since w = sup(nu:n = 1,2,. . .) exists in L , the relation nu = ( n + l ) u - u I w-u implies w I w - u, from which follows u I 8 and hence u = 8. Thus all the above classes of Riesz spaces are Archimedean. It can be readily seen that a Riesz space L is Dedekind complete if and only if for every net {ua>of L satisfying 8 I u, t Iu in L we have u, u for some u. Similarly, a Riesz space L is a-Dedekind complete if and only u we have u, t u for if for every sequence {u,,} of L satisfying 8 I u, t I some u.

r

Theorem 2.2. Let L be an order dense Riesz subspace of an Archimedean Riesz space K . If L is Dedekind complete, then L is an ideal of K . Proof. Assume 8 I u I v E L and u E K . Since K is Archimedean and L is order dense in K , there exists a net {u,} of L with 8 5 u, t u in K (Theorem 1.14). Now since L is Dedekind complete, 8 I u, w holds in L for some w E L. But by Theorem 1.10, L is a regular Riesz subspace of K and thus u, 7 w holds also in K . It follows that u = w E L, so that L is an ideal of K .

For the next basic theorem (due to H. Nakano and A. I. Judin) concerning the Dedekind completion of an Archimedean Riesz space see L&Z, p. 191. Theorem 2.3. Let L be an Archimedean Riesz space. Then there exists a Dedekind complete Riesz space Ls (unique up to a Riesz isomorphism) such that Ls contains a full, order dense Riesz subspace Riesz isomorphic to L, which we identify with L. That is, u = sup{u E L:v I u ) = inf{v E L:u 2 u )

for each u E L’.

Sec. 21

11

ORDER COMPLETENESS AND PROJECTIONS

The unique Dedekind complete Riesz space L* is culled the Dedekind completion of L.

The a-Dedekind completion L" of an Archimedean Riesz space L is defined now as follows: L" is the intersection of the collection of all Riesz spaces between L and L* that are a-Dedekind complete in their own right (the collection is nonempty since L* is one of them). Clearly L" is a cr-Dedekind complete Riesz space between L and La. Categorically the Dedekind and a-Dedekind completions are characterized by the following uniqueness property: There exists a Riesz isomorphism TC from L into L* (resp. L") such that given any normal Riesz homomorphism cp from L into any Dedekind (resp. a-Dedekind) complete Riesz space M , there exists a unique normal Riesz homomorphism $ from L* (resp. L") into M satisfying qn = $ n. Almost a-Dedekind complete Riesz spaces L are super order dense in their cr-Dedekind completions L". In particular, note that if L is almost a-Dedekind complete, then for every u E L" there exist two sequences (u,) and ( u , ) of L with u, t u and u, 1 u in L"; see also exercise 19 at the end of this chapter. For more about the a-Dedekind completion see [59]. Definition 2.4. A Riesz space L is said to have the countable sup property

if for euery subset S of L whose supremum exists in L, there exists an at most

countable subset of S having the same supremum us S in L. A Dedekind complete Riesz space with the countable sup propertji is called a super Dedekind complete Riesz space.

Note that a Riesz space L has the countable sup property if and only if for every net i u z ) of L with u, 1 0, there exists a sequence ( i d z , ) G ( u z ) with umn18. Every Archimedean Riesz space with the countable sup property is super order dense in La. Hence every Archimedean Riesz space with the countable sup property is almost a-Dedekind complete and its Dedekind completion is super Dedekind complete. Theorem 2.5. Let L be an Archimedean Riesz space with the countable sup property and let {u,) be a sequence of L and u E L. Then the following statements are equivalent : (i) u, (ii) u,

(0)

u in L. + u in La. -+

(0)

Proof. Since (i) 3 (ii) is obvious we need only to show that (ii) 3 (i). There exists a sequence { u , } of L* satisfying Iu, - Iu, 1 0 in L6. Since L6 is super Dedekind complete and L is full and order dense in Ld,for each k there exists a sequence (uk,,:n= 1,2,. . .) E L with ul,, 1 L'k in La. Put

UI

n

12

[Chap. 1

RlESZ SPACES

w, = inf{vk,,:k,m = 1,. . . ,n}. Then we have v, Iw, for each n and w, in L. Thus Iu, - uI I w, 1 8 holds in L, that is u, 9 u in L.

18

A useful criterion for the countable sup property is included in the next theorem and is due to Luxemburg and Zaanen [45, Note X, Theorem 31.11, p. 4931. Theorem 2.6. Let L and M be two Riesz spaces. Assume that M is Archimedean with the countable sup property und that there exists a strictly positioe operator T : L --* M (that is, T is linear and T ( u ) > 8 whenever u > 8). Then L has the countable sup property. Proof. Observe that since Ma is super Dedekind complete we can assume without loss of generality that M is super Dedekind complete. To see that L has the countable sup property let u, 1 8 in L. Then T(u,) 4 u* 2 8 holds in M for some u*. Pick a sequence {ua,} E {u,} with uan1 in L and with T(u,,) 1 u* in M . To complete the proof it is enough to show that u,, 1 8 holds in L. By way of contradiction assume that there exists u E L with 0 < u I u,, for all n. Since u, 10, there exists j? with u A up < u and so - Up A U U n = (Ma, - Up)'

+

2

(U - Up)'

=U -U

A Up

=W

> 8.

It follows that T(u,") 2 T ( w ) T ( u pA uan)2 T ( w ) + u* for all n, contradicting the fact that T ( w ) > 8. This contradiction completes the proof. We recall that a band is an order closed ideal. Every subset D of a Riesz space L is contained in a smallest band, called the band generated by D and denoted by {D}. A principal band is a band generated by a subset containing exactly one element; the principal band generated by the element u will be denoted by B,. A projection band B of a Riesz space L is a band satisfying L = B 0 Bd. A projection element is any element whose generating principal band is a projection band. An element 8 < e of a Riesz space L is called a weak order unit i f Be = L. Concerning the projection bands, we have the following characterizations, whose proofs can be found in L&Z, p. 133. Theorem 2.7. Let L be a Riesz space. Then we have the following: (i) I n order for a band B of L to be a projection band it is necessary and suflcient that sup{u E B : 8 Iu Iu } exists for all u E L'. (ii) I n order for a principal band B, of L to be a projection band it is necessary and suficient that sup{u A nlul:n = 1,2,. . .} exists for all v E L+.

For every projection band B of a Riesz space L, the projection P , of L onto B [defined by P,(u) = u l , where u = u1 + u 2 , u1 E B, and u2 E Bd] is a normal Riesz homomorphism; moreover, P,(u)

= s u p { u ~ B : 0I uI u)

for all U E L + .

Sec. 21

ORDER COMPLETENESS AND PROJECTIONS

13

The projection determined by a principal projection band B, will be denoted by P,; note in particular that by Theorem 2.7(ii) P , ( v ) = ~ u p { v ~ n ~ u ~ 1,2,. : n = ..} forall U E L + .

Definition 2.8.

A Riesz space L is said to have

(i) the projection property, ifevery band of L is a projection band; (ii) the principal projection property, i f every principal band of L is a projection band; (iii) suficiently many projections, if every nonzero band of L contains a nonzero projection band. Clearly the projection property implies the principal projection property, which in turn implies the property of having sufficiently many projections. We shall see later that no inverse implication holds. All the above classes of Riesz spaces are Archimedean as the following result due to Luxemburg and Zaanen shows; see L&Z, p. 174, or see Exercise 4 at the end of this chapter. Theorem 2.9. Every Riesz space with suficiently many projections is Archimedean.

In view of Theorem 2.7 we can see easily that Dedekind completeness implies the projection property and a-Dedekind completeness the principal projection property. A component of a positive element u of a Riesz space is any element u satisfying v A (U - v) = 8. In terms of the components the projection property was characterized by Kutty and Quinn [41, Theorem 2.6, p. 3091 as follows. Theorem 2.10. The following statements for an Archimedean Riesz space L are equivalent:

(i) L has the projection property. (ii) The components of the elements of L + in L' lie in L, that is, whenever u E L and u E L' satisfy v A (u - v) = 6' we already have v E L. Proof. (i) +-(ii) Assume that v E L' satisfies V A ( U - u ) = 6' for some u E L. Let B, denote the band generated by v in La. Then B = B, n L is a band of L, and hence a projection band of L. Put u = u1 u2 with u1 E B G B, and u2 E Bd. Since L is order dense in L', we get u2 A v = 8, and from this it follows easily that u = u1 E L. (ii) =. (i) Let B b e a bandofLand U E L+. Put u* = s u p { v ~ B : OIv Iu } in L'. If u* A (u - u*) > 8, then there exists w E L with Q < w IU* A (u - u*). In particular, it follows that w E B (since 8 I w Iu* and B is a band) and w u* Iu. But then w u* = sup{w v:u E B ; 8 s u Iu } = u*, a contradiction. This shows that u* A (u - u * ) = 8, and thus from our hypothesis u* E L. Theorem 2.7 now shows that B is a projection band.

+

+

+

+

14

RlESZ SPACES

[Chap. 1

A sequence ( u n ) of a Riesz space L is called u-uniformly Cauchy if there exists a sequence of positive numbers { E , , } with E, -+ 0 satisfying - u,,I s E,U for all n,p = 1,2,. . . . A sequence {u,,} is called u-uniformly convergent to some v E L if there exists a sequence (8,) of positive numbers with c, --t 0 and such that Iu, - < c,u. For Archimedean Riesz spaces the uniform limits are uniquely determined. A Riesz space is called uni$ormll. complete if every u-uniformly Cauchy sequence has a u-uniform limit. For the proof of the next theorem see L&Z, p. 280.

IU,,+~

111

Theorem 2.1 1. For a Riesz space L thefollowing statements are equivalent:

(i) L is Dedekind complete. (ii) L has the projection property and it is uniformly complete. The following theorem, called the main inclusion theorem, shows the various relations between the different classes of Archimedean Riesz spaces. Theorem 2.12. (Main Inclusion Theorem). Consider the following properties that a Riesz space may possess: super Dedekind completeness (SDC), Dedekind completeness (DC), a-Dedekind completeness ((rDC), almost aDedekind completeness (AaDC), the projection property ( P P ) , the principal projection property ( P P P ) ,the property of having suficiently many projections ( S M P ) , and the property of being Archimedean (Arch). Then the following implications (and only these implicafions)hold:

The only implication that has not been already mentioned is PP 3 AaDC. To see this, note first that by Theorem 2.10 the components of the elements of L+ in La lie in L, and then by using Freudenthal’s spectral theorem [L&Z. p. 2571 infer that L is super order dense in La. Next we exhibit examples to show that no other implication holds. The details in the examples are left as exercises to the reader. Example 2.13. (1) ( A Dedekind complete Riesz space that is not super Dedekind complete). Take L to be the Dedekind complete Riesz space consisting of all real-valued functions defined on some uncountable set X . that is, L = RX.Note that the collection S of the characteristic functions of the singletons has as supremum the constant function one. However, note that no countable subset of S has as supremum the function one.

Sec. 21

ORDER COMPLETENESS AND PROJECTIONS

15

(2) ( A o-Dedekind complete Riesz space that is not Dedekind complete and without the projection property). Consider the vector space L consisting of all real-valued Lebesgue measurable functions defined on [O,l]; L becomes a Riesz space if ordered by the pointwise ordering, that is, u > u in L whenever u(x) 2 u(x) for all x E [0,1]. Observe that u, 2 u in L implies u,(x) -, u(x) for all x E [OJ]. Thus, if E is a nonmeasurable subset of [O,l] and {x,} is the net of the characteristic functions of the finite subsets of E, then 8 Ix, T I e [e(x) = 1 for all x E [O,l]] holds but {x,} does not have a supremum; so that L is not Dedekind complete. To see that L does not have the projection property note that the band BE = { u E L : u ( x )= 0 for all x E E } is not a projection band. For the o-Dedekind completeness let 8 I u, T I u in L, and note that u(x) = sup{u,(x):n = 1,2,. . .} < cc for each x E [OJ] satisfies u E L and 0 I u, t u in L. This example also shows that PPP (or SMP) does not imply PP. (3) ( A Riesz space with the projection property that is not Dedekind complete). Consider an infinite set X , and let L be the Riesz space of all real-valued functions defined on X whose range is finite, with the pointwise ordering. Then L is not a Dedekind complete Riesz space, but it has the projection property. (4) (An almost o-Dedekind complete Riesz space with sufJiciently many projections but without the principal projection property). Let L be the Riesz space of all convergent real sequences with the pointwise ordering. The Riesz space L is almost o-Dedekind complete since it is super order dense in I,. Note also that L has sufficiently many projections but L lacks the principal projection property, since if u = (1,0,1/2,0,1/3,0,. . .) then the band generated by u is B, = { {u,} E L:u, = 0 for n even}, which is not a projection band since e = (l,l,l, . . .) has no decomposition with respect to this band. ( 5 ) (An almost o-Dedekind complete Riesz space with the principal projection property that is not o-Dedekind complete and that lacks the projection propertj)). Let L be the Riesz space of all ultimately constant sequences with the pointwise ordering. The space L is super order dense in I , and satisfies the principal projection property, but is not o-Dedekind complete and fails to satisfy the projection property. (6) ( A Riesz space with the principal projection property that is not almost o-Dedekind complete). Let L denote the vector space of all real-valued functionsf of a real variable, with the property that there exists a two-tailed sequence ( x n } [ a n ;of real numbers with * . . < x- < xo < x1 < . . . , such that x, -, co as n -, cc and x-, + - cc as n + cc and such that f is constant on the open intervals (x,,x,+ 1) for all integers k. Note that L is a Riesz space under the pointwise ordering, and that order convergence in L implies pointwise convergence. The space L does not have the projection property,

RlESZ SPACES

16

[Chap. 1

for if B = ( u E L : u ( x )= 0 for all irrational x},

then B is not a projection band. It is clear though that L does have the principal projection property, for if U E L, then the band B, is B, = ( u E L : u ( x )= 0 whenever u ( x ) = 0 ) ; by the restrictions on functions in L, this will always be a projection band. Note, however, that L is not almost o-Dedekind complete. (7) ( A n almost a-Dedekind complete Riesz space without suficiently many projections). Let L = C[O,l], the Riesz space of continuous real-valued functions on [OJ] with the pointwise ordering. Note that L does not have sufficiently many projections, but it does satisfy the countable sup property, since cp: C[0,1] + R, cp(u) = u(.x)dx, is a strictly positive operator (Theorem 2.6). Therefore, L is super order dense in its Dedekind completion, so that L is almost a-Dedekind complete.

s:

For a proof of the next result see L&Z, p. 142. Theorem 2.14. Let A be an ideal of a Riesz space L. If L satisjes any of the properties of the main inclusion theorem, then A considered as a Riesz space in its own right also satisjies the same property. Note. The main inclusion theorem was stated and proven without the class of almost o-Dedekind complete Riesz spaces by Luxemburg and Zaanen [46, pp. 137, 1741. The class of almost o-Dedekind complete Riesz spaces was introduced by Aliprantis and Langford [9] and independently by Quinn [59]. Theorem 2.12 as well as Examples 2.13, (4)-(7) are reprinted from [9] with permission of the publisher American Mathematical Society from the PROCEEDINGS of the American Mathematical Society, Copyright @ 1974, Vol. 44, pp. 421 -426.

A complete disjoint system { e i ) of a Riesz space L is a pairwise disjoint collection of elements of L', that is, ei A e j = 8 if i # j , such that if u A ei = 8 holds for all i, then u = 8. Theorem 2.15. Let { e i :i E 1 ) be a complete disjoint system of a Riesz space L consisting of projection elements. Then for every u E L+ we have u = sup{P,,(u):iE I ) .

Proof. Let u E L+ and 8 I w E L satisfying Pei(u)I u - w for all i E 1. Observe next that 8 Iw Iu - Pei(u)E B:, holds for all i E I , that is, w A ei = 8 for all i E I . Hence w = 8 and so u = sup { P J u ) :i E I ) holds. W

Sec. 21

ORDER COMPLETENESS AND PROJECTIONS

17

A discrete element of a Riesz space L is a nonzero element u of L whose generating ideal A , equals the vector subspace generated by u in L, that is, A , = [3,u:AE R ) . It should be clear from this definition that if u is a discrete element then any other nonzero element v satisfying 1 0 1 I IuI is also a discrete element. The discrete elements of Archimedean Riesz spaces have an additional interesting property. Theorem 2.16. Let u be a discrete element of an Archimedean Riesz space L. Then the ideal A , generated by u is a projection band. Proof. By replacing u by 1% (if necessary) we can assume u E L'. To see that A, is a band let 8 I v, t v in L with { v , } E A,. For each tl pick A, E R with v, = A,u and note that 0 I 1, t holds in R. Since L is Archimedean, A, 3, E R holds and by Theorem 1.15(ii) v, = A,u t 1u also holds. Thus v = Au E A , and so A , is a band of L. To show that A , is a projection band let v E L'. Since U Anu E A,, there exists An E R with u A nu = Anu. Observe next that 0 I An t 3, E R for some 1. It now readily follows that s u p { v A n u : n = 1,2,. . .} = h,so that by Theorem 2.7(ii) A , is a projection band.

A discrete Riesz space is a Riesz space L having a complete disjoint system consisting of discrete elements of L. Note that a Riesz space L is discrete if and only if for every 8 < u E L, there exists a discrete element u of L satisfying 8 < u I u. Indeed, if L is a discrete space, then there exists a complete disjoint system { e i } of discrete elements; thus if 8 < u, then u A e, > 8 must hold for at least one i and so u A e, is a discrete element satisfying 8 < u A e, I u. On the other hand, use Zorn's lemma to choose a maximal family { e , } of pairwise disjoint strictly positive discrete elements and note that if 8 I u satisfies u A e, = 8 for all i, then u = 8, since otherwise by the condition, there should exist a discrete element v satisfying 8 < v I u, and thus v A e, = 8 for all i, contrary to the choice of { e , ) . For a nonempty set X any order dense Riesz subspace of RX is a discrete Archimedean Riesz space. The point is, however, that the only Archimedean Riesz spaces that are discrete are the order dense Riesz subspaces of the Riesz spaces of the form RX.The details follow. Theorem 2.17. For an Archimedean Riesz space L the following statements are equivalent; (i) L is a discrete Riesz space. (ii) L is Riesz isomorphic to an order dense Riesz subspace of some Riesz space of the form RX.

18

RlESZ SPACES

[Chap. 1

Proof. (i) 3 (ii) Choose a complete disjoint system { e i :i E I } consisting of discrete elements. According to Theorems 2.16 and 2.15, we have u = sup (Pet(u):iE I} for each u E L+. Also for each i there exists a unique LiE R with P,,(u) = ,liei. Consider now the mapping n:L+ + (R')' given by n(u)= { A i : i E I } for u E L+ and extend n from L into R' by ~ ( u=) n ( u + )n(u-) for u E L. We leave the details to the reader to verify that n is a well defined Riesz isomorphism (into) and that n(L) is an order dense Riesz subspace of R'. (See also Lemma 3.1.) (ii) 3 (i) Straightforward. W

It should be noted that not every Riesz subspace of an Archimedean discrete Riesz space is discrete. As an example consider R['*'] and its Riesz subspace C[O,l] (the Riesz space of all continuous functions on [0,1]). Note that C[O,l] is not discrete. (Actually C[O,l] has no discrete elements at all.) We shall close this section with a characterization of the Dedekind completion of certain Archimedean Riesz spaces. For every Riesz subspace L of a Riesz space M there exists (at least) one ideal A of M maximal with respect to the property L n A = (Q}, that is, there exists an ideal A of M with L n A = (0). and such that whenever an ideal I of M satisfies A G I and L n I = { 0 } , it follows that A = I . To see this, consider the collection C of all ideals I of M with L n I = { 0 } and note that C is nonempty ((0) E C) and partially ordered by the inclusion relation. Now use Zorn's lemma to obtain a maximal element A E C , and note that A is maximal with respect to the property L n A = {8>. Now assume that L is an Archimedean Riesz subspace of a Riesz space M and that A is a maximal ideal of M with respect to the property L n A = ( O } . Then the canonical projection n:L+ MIA is a Riesz isomorphism, and if L is identified with n(L), the ideal I generated by L in MIA is an Archimedean Riesz space containing L as an order dense Riesz subspace. Indeed, to see first that n : L + M / A is one-to-one, note that if ti = 0 in M I A with u E L, then u E L n A = ( O } , so that u = 8. Now to see that L is order dense in M I A (and hence in I ) let 0 < li E MIA. Since u 4 A the ideal A + A , properly contains A and hence ( A + A , ) n L # I d } ; pick O < w E ( A + A,) n L, write w = w 1 + w 2 with w 1 E A and lw21 I klul for some k, and note that O < ct = w2 I kli. Thus k - ' w E L satisfies O < k - ' w I c i and therefore L is order dense in MIA. Finally, to see that I is Archimedean let 8 I nli I 0 for all n with o E L. If 0 < li holds, pick w E L with 0 < w I li and note that 8 < nw I d holds for all n, contradicting the Archimedeanness of L. If now M is a Dedekind complete Riesz space, then I is the Dedekind completion of L. The details follow. Theorem 2.18. Let L be a Riesz subspace of a Dedekind complete Riesz space M and let A be an ideal of M maximal with respect to the property

Sec. 31

19

ORDER BOUNDED TRANSFORMATIONS

L n A = {0). Then the ideal I generated by L in M I A (where L is ident@ed with its canonical image in M I A ) is the Dedekind completion of L. Proof. According to the preceding discussion, we have to show only that I is Dedekind complete. Since L is order dense and full in the Archimedean Riesz space 1, it is enough to show that for every nonempty set S E L', inf S exists in I. Thus, let S c L + be nonempty. By the Dedekind completeness of M , u = infS exists in M , and we shall show next that Li = infS holds in I. Assume to the contrary that there exists v E M with ti < D I S for all s E S . Now since L is order dense in I there exists 0 < w E L such that 0<w I i. - ti. But then it follows H < nMs I ii for all n (why?), a contradiction. Thus ti = inf S holds in I, so that I is Dedekind complete. Note. Theorem 2.18 is due to Maxey [48, Theorem 1.3, p. 181.

3.

ORDER BOUNDED TRANSFORMATIONS AND THE ORDER DUAL OF A RlESZ SPACE

A linear mapping T :L -+ M between two Riesz spaces L and M is said to be an order bounded linear transformation (or operator) if T ( A )is an order bounded subset of M whenever A is an order bounded subset of L. It should be clear that the collection T b ( L , M )of all order bounded linear transformations from L into M forms a real vector space. A linear mapping T : L -+ M is said to be positive (denoted symbolically by T 2 0 or 8 IT ) if T ( u )2 8 holds in M whenever u 2 0 holds in L. Clearly every positive linear mapping is an order bounded linear transformation. Observe that the relation T I 2 T , whenever T , - T , 2 0 defines an order relation on T b ( L, M ) . We start with a result concerning positive operators. Lemma 3.1. Let L and M be two Riesz spaces with M Archimedean. Assume that T : L +-+ M + is an additive mapping (that is, T ( u v) = T ( u ) T(u)for all u,v E L'). Then T has a unique positive extension to a linear transformation ,from L into M . Moreover, the extension (denoted by T again) is given by

+

T ( u ) = T ( u + )- T ( u - )

for all

+

u E L.

Proof. Observe that if u , - u, = w , - w, holds with uL,wiE L + ( i = 1,2), then it follows from the additivity of T that T ( u , ) - T(u,) = T ( w , ) - T ( w Z ) . Thus, the extension of T from L into M defined by T ( u )= T(u') - T ( u - ) is well defined and clearly this extension is positive. The additivity of this

RIESZ SPACES

20

[Chap. 1

extension can be shown as T(u + w)= T(U+ + w + - ( u - + w-)) =T(u+) T(w+)- T ( K )- T(w-) = T(u)+ T(w).

+

To see that T(Au)= AT(#)holds for each A E R and u E L, note first that for A > 0 rational and u E L+ the relation follows from the additivity of T . Assume now A > 0 and u E L+. Pick two sequences of rational numbers { r , } and {t,} such that r, T ,iand t , L A. Observe next that r,T(u) = T(r,u) IT(Au)IT(t,u) = t,T(u)

implies 1T(u)= T(Au),since M is Archimedean. Linearity of T is now immediate. The uniqueness of the extension is obvious. 1 Observe that the preceding lemma may be false if M is not supposed to be Archimedean. To see this let cp be an additive function from R into R that is not linear, i.e., not ofthe form ~ ( x=) cx, and let M be the lexicographic plane. Consider the mapping T : R + + M + defined by T(x)= (x,cp(x)) for x E R + . Note that T is additive and that if T could be extended to a linear mapping from R into M , then cp should be linear. As a first application of Lemma 3.1 we shall obtain an embedding theorem for Archimedean Riesz spaces. Assume that { e , : a E A } is a complete disjoint system of projection elements of an Archimedean Riesz space L. Let B, denote the band generated by e, in L, and let K be the Cartesian product nB,. It can be readily seen that the mapping n : L + + K + defined by n(u)= { P J u ) : a E A } is additive. K defined by n(u)= n(u+)- n(u-)is the unique Hence the mapping n:L positive linear extension of n. Now by using Theorem 2.15, we conclude that n is a Riesz isomorphism (into). Note, moreover, that n(L) is order dense in K and that { e , : a E A } is a weak order unit of K . By applying now the above technique to La we can see that every Archimedean Riesz space can be embedded as an order dense Riesz subspace in a Dedekind complete Riesz space with a weak order unit. We have thus proven the following theorem. Theorem 3.2. With the preceding notation, n : L + TTB, maps L Riesz isomorphically onto an order dense Riesz subspace of nB,. In particular, every Archimedean Riesz space can be embedded as an order dense Riesz subspace in a Dedekind complete Riesz space with a weak order unit. The following theorem shows that if M is Dedekind complete, then the ordered vector space Y b ( L , M )becomes a Dedekind complete Riesz space. --f

Theorem 3.3. Let L and M be two Riesz spaces with M Dedekind complete. Then the ordered linear space Y , ( L , M ) is a Dedekind complete Riesz space.

Sec. 31

ORDER BOUNDED TRANSFORMATIONS

21

I n particular we have

+

T, v T2(u)= sup(T,(v) T,(w):u,w E L'; u + w = u > T I A T 2 ( u )= inf{T,(u) + T,(w):u,w E L'; u + w = u } for d l T , , T2 E y b ( L , M ) and u E L+. T+(u) = sup { ~ ( u ) u : E L+ ; I9 I v I u } (ii) T-(u) = SUP{ - T ( u ) : u EL'; 8 I uI U} ITI(u) = s~p{IT(u)I:vE L ; i U} for all T E P',(L,M) and u E L'. (iii) T, 2 T in Y,(L,M) implies T,(u) 2 T(u) in M for each u E L. (i)

Iu(

Proof. Let T E yb(L,M).For each u E L' the set {T(u):I9I uI u } is order bounded in M , and hence since M is Dedekind complete S(u) = sup{T(u):O I u I u } exists in M. It is readily seen that S(u) + S(w) I S(u w)for all u,w E L'. On the other hand, if 0 I uI u + w holds in L we can write u = u 1 u2 with 8 I u1 I u, 0 I u2 I w (Theorem 1.2). Thus, S(u) S(w) 2 T(u,) T ( u Z = ) T(u), from which it follows S(u) + S(w) 2 S(u w) for all u,w E L'. Thus S:L' -+ M f is additive, and so by Lemma 3.1 it can be extended uniquely to a positive linear operator from L into M . Clearly 0 I S and T I S hold in Yb(L,M). Now assume that I9 I Q and T I Q hold in Y,(L,M). Then S(u) = sup{T(v):dI u I u } I Q(u) for all u E L', so that S = T v 8 = T' holds in yb(L,M).This shows that Y b ( L , M ) is a Riesz space and establishes the formula for T'. To see that g b ( L , k f )is Dedekind complete let I9 I T, 7 I S in Z b ( L , M ) . For each u E L' put T ( u )= sup (T,(u)>in M , and note that T : L' M f is additive, and hence extendable to all of L, according to Lemma 3.1. But then T, t T holds in T b ( L , M )and T,(u) T(u)in M for each u E L + . This last observation establishes the validity of (iii).The formulas in (i) and (ii) follow T, and T , A T, = T, from the relations T, v T, = (T, - T,)' (TI - T2)'. W

+ + +

+

+

-+

+

Observe that from the formula for the absolute value of T, we get the very useful relation IT(u)l s ITl(lul)for u E L. A mapping p : L -,M between two Riesz spaces is called sublinear if p ( u 2') I p ( u ) p ( r ) , and p(3,u) = 3,p(u) hold for all 1 4 4 E L and all 1- 2 0. The next theorem is a generalization of the classical Hahn-Banach theorem, and its proof is similar.

+

+

Theorem 3.4. (Hahn-Banach). Let p : L -+ M be a sublinear mapping between two Riesz spuces with M Dedekind complete. Assume that T is a linear mapping defined on a linear subspace K with range in M such that T(u) I p(u) for each u E K . Then T can be extended to a linear mapping T * : L -+ M such that T*(u) I p(u)for all u E L.

22

RIESZ SPACES

[Chap. 1

We continue with the dual theorem of Theorem 3.3. Theorem 3.5. Let L and M be two Riesz spaces with M Dedekind complete. Then the following formulus hold: T,,T,; T , + T , = T } (i) T ( u v v ) = sup{T,(u) + T,(v):8 I T ( u A ~=)inf{T,(u) + T,(v):O I T,, T,; T , + T, = T ) for all u,v E L und all t3 I T E Y , ( L , M ) . (ii) T ( u + )= sup{S(u):SE Y b ( L , M ) ;8 s S I T) T(u-) = SUp{-S(u):S E T b ( L , M ) ;0 I S IT } , T(lUl)= SUp{ls(U)l:SE T b ( L , M ) ;IS1 I T} for all u E L und all 8 I T E Yb(L,M). Proof. Note that the formulas of (i) follow from the formulas of (ii). We prove only the first formula of (ii); the proof for the other formulas of (ii) will be similar. Define p ( v ) = T(u+)for v E L and note that p : L + M is a sublinear mapping satisfying p ( v ) I T(o) for all u E L'. Consider next the vector subspace K = { h : L E R ) , and let S be the linear mapping from K into M defined by S ( h ) = Ap(u) = iT(u'). By Theorem 3.4 there exists an extension S* of S to all of L such that S*(v)Ip ( v ) for all v E L. In particular, S* I T . Now since S*(u) = S ( u ) = T ( u ' ) and it follows that 8 I SUp{S(u):SE Y b ( L , M ) ;8 I S IT } IT(tlf) we easily obtain the desired equality. (It should be noted from the course of the proof that the suprema in the above formulas are actually maxima.) H Definition 3.6. A linear mapping T : L + M between two Riesz spaces is said to be order continuous (or a normal integral) if u, 5 8 in L implies T ( ~2 J e in M . Similarly, T is said to he a-order continuous (or an integral) if u, 2 8 in L implies ~ ( u , , 2 ) o in M .

A positive linear mapping T : L + M is order continuous if and only if 1 8 in L implies T(u,) 10 in M . A similar statement holds for positive g-order continuous linear mappings. The collection of all normal integrals of Y b ( L , M )will be denoted by Y n ( L , M ) ,and the collection of integrals of Y b ( L , M )by Y , ( L , M ) . Both Y n ( L , M )and Y c ( L , M )are vector subspaces of Y b ( L , M ) .The next theorem tells us that they are actually bands of Y b ( L , M ) . u,

Theorem 3.7. Let L and M be two Riesz spaces with M Dedekind complete. Then Y,,(L,M) and Y , ( L , M ) are bands of Y , ( L , M ) . Proof. We show the result for Y , , ( L , M ) ; the case Y c ( L , M ) is similar. Note first that 8 I S IT E Y,,(L,M) and u, J 8 implies S(u,) J 8 in M ,

Sec. 31

ORDER BOUNDED TRANSFORMATIONS

23

which means S E Y , ( L , M ) . Assume now T E 2Yn(L,A4)and 0 Iu, T u in L. We shall show T+(u,) 7 T+(u),and this will be enough to establish that Y , ( L , M ) is an ideal. To this end let T f ( u , ) t h I T + ( u )in M . Then by the the formulas (ii) of Theorem 3.3, we have T(v)I h for all u E L satisfying 0 I 11 I u, for some ct. Now if 0 I M! I u, then u, A w 7 w holds in L, and therefore [since TES?,(L,M)] we have T(u,A w ) 3 T(w).But from the above h for all CI and so by Theorem 1.6(iv)we obtain remark, we have T(u,A w) I T(w)I h. Thus by using the formulas (ii) of Theorem 3.3, we get Tf(u) I h, that is, h = T+(u). To see that Y , ( L , M ) is a band of Y,,(L,M), let 0 I T , t T in y h ( L , M ) with IT,} G Y , ( L , M ) and 0 I uI t u in L. Put T(u,) t h IT(u) in M . Note that T,(u,) I h for each a and A. and that for each fixed ct we have T,(u,) 7 T,(u). Now by Theorem 1.6(iv) we obtain T,(u) I h for all ct and I

thus T(u)= sup{ T,(u)} I h. Hence h = T(u), so that T proof is finished. H

E

Y,,(L,M) and the

We continue with a definition. Definition 3.8. For any Riesz space L the Dedekind complete Riesz space Y h ( L , R is ) called the order dual of Land is denoted by L-, that is, L" = Yh(L,R). W e also put L; = S?,(L,R) and L, = Y c ( L , R ) . Note. Theorem 3.3 was originally proven by Riesz [62] for the L- case; the general case was established by Kantorovich [36]. Theorem 3.5 is essentially due to Luxemburg and Zaanen [45, Note VI, Theorem 19.6, p. 6621; see also [4, p. 6691. Theorem 3.7 is due to Ogasawara [54]. The normal integral and integral terminology is due to Luxemburg and Zaanen [45, Notes VI and VIII] ;the order continuous terminology is due to Nakano [SO]. For more about order bounded transformations see [4], [44, Note XV], [45, Note 1x1, [67], and [68].

For a Riesz space L and cp E L- the null ideal of cp is defined by N , = {U E L : Icpl(lul) = 0 ) .

Note that N , is actually an ideal of L, and that if cp is a normal integral, N , is a band of L. The carrier C, of cp E L* is defined by C, = NPd. Theorem 3.9. Let L be an Archimedean Riesz space, 8 I cp E L , and 0I (I E L-, Then the following statements are equivalent: (i) cp I t,b in L-, i.e., cp A t,b = 0 in L-. (ii) C, E N,. (iii) C, I C,, i.e., IuI A = e f o r all u E C, u E C,

101

I

24

[Chap. 1

RlESZ SPACES

Proof. (i) (ii) Let 8 .c u E N,d = C , and let E > 0. Pick a sequence {u,} G L'; 8 I u, I u for all n and such that +(u,) + cp(u - u,) I 2 - " ~for all n (Theorem 3.3). Put w, = inf{u,:l I i I n } , n = 1,2, . . . and note that w, .1 0 in L. Indeed, if 8 I w I w, holds for all n, then 0 I +(w) I 2 - " ~ for cp E Lz, cp(u - w,) cp(u), all n, so that + ( w ) = 0 and hence w = 0. Since 0 I and since 0 I cp(u - w,) = cp(sup{u - u , : l I i I n}) I

--

n

C cp(u - u,) s E,

i= 1

we get 0 I cp(u) I E for all E > 0. Thus q ( u ) = 0 and so C , E N , . (ii) (iii) Assume 8 I UEC,, 8 I U E C,. Then 8 I U A u E C, c N , and thus u A u E C , n N , = (8). Hence u A u = 8, so that C, I C,. (iii) (i) Note that C , I C , implies C, c CVd= NVdd= N , , where the last equality holds by virtue of the Archimedeanness of L. Now if 8I u = u1 u2 E N , 0 N,d, then

+

0 I (cp A +Nu) = (cp A M U l )

+

+ (cp A + ) ( U 2 )

5 +(uJ

+ cp(u2) = 0,

and thus cp A = 8 on the order dense ideal N , 0 C,. Now use Theorem 1.14 and the fact cp A $ E L: (Theorem 3.7) to obtain cp A = 0 on L. W

+

As a corollary of the previous result we have the following theorem.

Theorem 3.10. Let L he un Archimedean Riesz space and q,+ E Lz. Then the following statements are equiualent : (i) (ii) (iii) (iv)

cp

1 +.

C, c N,. C, E N,. C , I C,.

A generalization of the above theorem to 9 , ( L , M ) is not possible as the following example (due to Luxemburg) shows. Consider L = M = L , ( [ O , l ] ) (=the Riesz space of the equivalence classess of all Lebesgue integrable functions on [0,1]), and note that both L and M are super Dedekind complete Riesz spaces. Now let S and T be the positive linear operators from L into M defined by s(u) = u,

T ( u )=

(J:

u ( x )ds) e

for u E L,

where e(.u)= 1 for .Y E [0,1].Note that T,S E Y n ( L , M )and N , = N , = (0;. Thus C, = C, = L. It can be seen, however, that S A T = 0 in Y , ( L , M ) .

Sec. 31

25

ORDER BOUNDED TRANSFORMATIONS

For any Riesz space L and any ideal I of L', there exists a natural embedding u H ii of L into 1,; defined by

i i ( f )= f ( u )

for

JE

I.

The properties of this embedding are included in the next theorem. Theorem 3.1 1. Let L he a Riesz space, I an ideal of L', and u H ii the en.lherlding of' L into 1, rlqfined h!, E ( , f ) = f ( u ) .for .f' E I . Theri the jijllowing statements hold:

(i) u H E i s a Riesz homomorphism. (ii) u H ii is a Riesz isomorphism (into)if and only if l o = ( u E L : ( p ( u ) = O , f o r a l l c p E I )= ( 0 ) .

(iii) I f L is Archimedecrn rrnd I c L;, then i i H ii is homomorphism whose image L# i s order dense in 1,;.

LI

iiormal Riesz

Proof. (i) Let 6' I cp E I and u E L. By using Theorems 3.5 and 3.3 and and the fact that I is an ideal of L', we have Y

u'(cp) = cp(u+)= sup{$(u):0 I $ I cp) = sup{ii($):$ E I; 8 I $ I cp). so that

z

= (6)' in 1,

= (S)+(cp),

and thus I I H ii is a Riesz homomorphism. (ii) The proof is straightforward and is left to the reader. (iii) To see that u H ii is a normal Riesz homomorphism let u, 5 8 in L and note that since E,(cp) = cp(u,) 1 0 for all 8 I cp E I , it follows from Theorem 3.3 that ijl 10 in I;. So, it remains to be shown that L# is order dense i n I,:. To this end let 0 < cp E 1;. Without loss of generality we can assume that there exists u E L with cp I ii. Indeed, pick 8 < f E C,, and then choose 0 < u E C;, $ S A cp = 8, then by Theorem 3.10 ii(C,) = (0) so that i i ( f )= f ( u ) = 0 contrary to the choices of u and f . Replacing now cp by cp A 6, we establish our claim. Now pick 0 < M < 1 with $ = (cp - ME)' > 0. Next choose 8 < ,f E C, and then 0 < u E C,. Put w = u A M U and we claim 8<EI cp. To see that E > 0 assume E = 8. Then i j E ~= I3 and so (since 0 I $ I ii) T A $ = 0. Thus by Theorem 3.10 F(C,) = {O) and so i j ( f )= f (u ) = 0, a contradiction. To see that E I cp assume E $ cp. So, cpl = (E - cp)' > 8. Choose 0 < h E C, and note that [since 8 I cpl I ( M E - cp)' = (cp - M E ) - ] cpl I $ and so by Theorem 3.10 (again!) C, I C,. Hence h I f and so by using Theorem 3.10 once more, we get h(C,) = (0). In particular, we have h(u)= 0. But 0 < cp,(h) = (E - cp)+(h)I @(h)I E(h)= h(u) = 0, II

+

a contradiction. This contradiction completes the proof.

26

RIESZ SPACES

[Chap. 1

Note. Theorem 3.10 was proven first by Nakano under the condition that L is 8-Dedekind complete [50, Theorem 20.1, p. 741; the general case is due to Luxemburg and Zaanen [45, Note IX, Theorem 31.2(ii), p. 3731. Theorem 3.1 1 is due to Nakano [SO, Theorem 22.6, p. 831. If L is an Archimedean Riesz space, M a Dedekind complete Riesz space, and 0 I T E 9,,(L,M)define T'(u*) = sup{T(v):v E L and u I u * } for every u* E L' (the Dedekind completion of L). It should be clear that T': L' + M is a positive linear operator satisfying T'(u) = T(u)for all u E L. Also, if u,* 8 in L', then there exists a net { u J of L with v A 1 8 in L and such that for each A, there exists tl with v A 2 u,*; so by the order continuity of T, we get T'(u,*) 1 8 in M , that is, T' E 9,,(L6,M). Therefore, we have defined a mapping T H T' from ( 9 , , ( L , M ) ) +into (9,,(L',M))+, which is in addition onto and one-to-one. Furthermore, T H T' is additive on (9,,(L,M))+, and hence by Lemma 3.1, it extends to a linear mapping from 9,(L,M) onto 9,,(L',M), by the formula T' = (T')' - (T-)'. Now if T A S = 8 in 9,,(L,M),then for every 6' I u E L we have

+

6' I T' A S'(u) = inf(T'(v*) Ss(w*):8I u*,w* E La;V* + W* = U } I inf{T(v) T(w):v,wE L + ;v w = u } = TAS(U)= 6'.

+

+

So, T' A Sd(u)= 6' for all u E L and hence [since L is order dense in L' and T' A S' E 9,,(L',M)], we have T' A S'(u*) = 6' for all u* E L', that is, T' A S' = 0 in Y,,(L',M). We have therefore established the following theorem. Theorem 3.12. I f L is an Archimedean Riesz space and M a Dedekind complete Riesz space then the mapping T H T' (defined above) is a Riesz isomorphism between Y,,( L,M) and 9,,( L',M). In particular .for every Archimedean Riesz space L the Riesz spaces L,' and (L'));;are Riesz isomorphic via q H q'. The final result of this section shows the relation between the Riesz homomorphisms of a Riesz space and the discrete elements of its order dual. Theorem 3.13. For a Riesz space L and 0 < cp E L' the following statements are equivalent: (i) cp is a Riesz homomorphism (from L into R). (ii) cp is a discrete element of L', i.e., A , = (Acp:A E R } . Proof. (i) (ii) Let $ E L' satisfy 6' I$ Icp. Then u E Kercp implies It,b(u)l I $(lul) 5 cp(ly1) = Icp(u)l = 0, and hence $(u) = 0, so that Kercp c Ker $. The last relation is, however, equivalent to the property that $ = Acp

for some A E R and we are done.

Sec. 41

LINEAR TOPOLOGIES

27

(ii) =- (i) Note first that since 1 $ 1 I cp implies $ = Acp for some 1J.1 I 1, we have I$(u)l = (Acp(u)I= 121 Iq(u)l IIcp(u)l for all u E L. It now follows from Theorem 3.5 that qO(lu()= sup [ l $ ( u ) l : $

E

L-;

1$1

I q ~ I j Iq~(u)l

for all u E L,

and hence since Icp(u)l I ( ~ ( 1 ~ 1 ) is always true, we get ( ~ ( 1 ~ 1 = ) Icp(u)l for all u E L. Thus q is a Riesz homomorphism, and the proof is finished.

4.

LINEAR TOPOLOGIES ON VECTOR SPACES

In this section we shall discuss briefly (without proofs) the basic properties of the linear topologies needed for this book. For a complete study of the theory of topological vector spaces, we refer the reader to [23], [31], [40], [63], and [66]. For simplicity all vector spaces will be assumed'to be real vector spaces. By a linear topology z on a vector space E , we mean a topology z on E that makes the addition and the scalar multiplication continuous, that is, makes the two functions (x,y) H x

+y

(A,x)c,Ax

from E x E into E from R x E into E

continuous. A topological vector space (E,z)is a vector space E equipped with a linear topology z. Every linear topology z on a vector space E has a basis { V } for the neighborhoods of zero satisfying the properties : (a) Each I/ E { V } is a balanced set, that is, lu E V holds for every A E R with 111 I 1 and u E V . (fi) Each V E { V ) is an absorbing set, that is, for every u E E, there exists l > 0 such that l u E V. (7) For each V E { V } there exists W E { V } with W + W G V .

Conversely, if a collection { V } of subsets of a vector space E forms a filter basis (i.e., every V E { V } is nonempty and for every pair V,W E { V } there exists U E { V } with U s V n W )satisfying properties (a),(p), and (y), then there exists a unique linear topology z on E having as a basis for zero the collection { V } . A topological vector space (E,z) is called metrizable if there exists a metric generating the topology 7. The metrizable topological vector spaces can be characterized by the following countability theorem.

28

RlESZ SPACES

[Chap. 1

Theorem 4.1. A Hausdorf topological vector space (E,z) is metrizable if and only i f 5 has a countable neighborhood basis of zero.

A subset A of a topological vector space (E,z) is called topologically bounded (or briefly z-bounded) if for every r-neighborhood V of zero there exists I > 0 such that LA E V . A locally convex topology z on E is a linear topology having a basis for zero consisting of convex sets. [A subset A of E is called convex if I u + (1 - A)u E A whenever u,u E A and 0 5 I I 1.1 A seminorm p on a vector space E is a function p : E + R satisfying the following properties : (1) p(u) 2 0 for all u E E . (2) p(u v) I p ( u ) p(u) for all u p E E . (3) ~ ( I u=) lIlp(u) for all I E R and u E E.

+

+

For any E > 0 note that the set { u E E : p ( u ) I E} (the E-closed ball of p ) is a convex set of E. Every family { p a } of seminorms on a vector space E defines a locally convex topology on E by taking as a basis for the neighborhoods of zero the finite intersections of the closed balls of the pa. In other words a subset I/ of E is a neighborhood of zero for this topology if there exists E > 0 and E for a finite number of indices, say M , , . . . ,a,, such that { u E E:p,,Ju) I m = 1,. . . ,n} E V. This topology is called the locally convex topology generated by the family of seminorms { p a > . Theorem 4.2. For a linear topology z on a vector space E the following statements are equivalent : (i) z is a locully convex topology. (ii) There exists u fumily { p a } of seminorms on E that generates the topology t .

Note that a subset A of a locally convex space (E,z) whose topology is generated by the family of seminorms ( p , } is z-bounded if and only if p,(A) is a bounded subset of R for each M . The topological dual E’ of a topological vector space (E,z) is the collection of all the continuous linear functionals of E . With the usual operations E‘ is a vector space; actually E‘ is a vector subspace of the algebraic dual E* of E , which consists of all linear functionals on E . The weak topology o(E,E’) of a locally convex space (E,t) is the locally convex topology on E generated by the family of seminorms { p s : f E E ‘ } , where ps(u) = If(u)l for u E E. The weak topology a(E,E’)is the smallest locally convex topology on E for which every f E E’ is continuous. The topological dual of (E,a(E,E’)) is precisely

Sec. 41

LINEAR TOPOLOGIES

29

E’. Similarly, the weak topology o(E‘,E) is the locally convex topology on E‘ generated by the family of seminorms {p,:u E E } ; where p,( f )= I f ( u ) J for f E E’. The strong topology P(E’,E) on E’ is defined to be the locally convex topology generated by the family of seminorms { p A: A E 9i?}, where PA(f)

= s u p { I f ( u ) l : uE A3

and 9i? is the collection of all o(E,E’)-bounded subsets of E. The strong topology P(E’,E) is the topology of uniform convergence on the o(E,E’)bounded subsets of E. The topology P(E,E‘) is defined similarly. The Mackey topology z(E,E’) on E is the locally convex topology of uniform convergence on the balanced, convex and o(E’,E)-compact subsets of E . If (E,z) is Hausdorff, then we have z G t(E,E‘). For a locally convex topology z* on E, in order for the topological dual of (E,z*) to be E‘, it is necessary and sufficient for T * to satisfy a(E,E’) G T * G z(E,E’). The topology z(E‘,E) is defined similarly. Theorem 4.3. Let (E,T)be a Hausdorf locally convex space and V Then the following statements hold:

c E.

(i) V i s z-bounded if and only if V is o(E,E’)-bounded. (ii) I f V is convex, then I/ is z-closed if and only if I/ is o(E,E’)-closed. A net { u n ) of a topological vector space (E,z) is called a z-Cauchy net if for every neighborhood V of zero, there exists u1 such that u, - up E I/ for all a,P 2 a ] . A subset A of (E,z) is called z-complete if every z-Cauchy net of A z-converges to some element of A . In particular, (E,T)is called topologically complete (or briefly z-complete) if every z-Cauchy net of E z-converges in E. A function f : A E (E,z) -+ (El,zl), from a subset A of one topological vector space (E,z) to another (El,zl), is called uniformly continuous if for every z,-neighborhood V of zero of E l , there exists a z-neighborhood W of zero of E such that f ( u ) - f ( v ) E V holds whenever u,v E A satisfy u - u E W . Concerning the uniformly continuous functions, the following important extension theorem holds.

Theorem 4.4. Let f : A E ( E J ) + (E,,z,) he a uniformly continuous function. If (E,,z,) is a HausdorfSz,-complete topological vector space, then there exists a unique uniformly continuous extension 7off to the z-closure A of A. Two topological vector spaces (E,T) and (El,tl) are called isomorphic if there exists a one-to-one, onto linear mapping cp: E + El that is at the same time a homeomorphism, in other words, if there exists a one-to-one mapping from E onto El preserving both structures, the algebraic and the topological.

30

RIESZ SPACES

[Chap. 1

For the next theorem dealing with the topological completion of a Hausdorff topological vector space see in particular [40, p. 331. Theorem 4.5. For every Hausdor8 topological vector space (E,z) there exists u unique (up to an isomorphism) Hausdor- topological vector space (I!?,?) having the following properties: (i) (I!?,?) is ?-complete. (ii) There exists a vector subspace E of I!? that is algebraically isomorphic to E (thus identifying E with E , , we consider E as a vector subspace of I!?). (iii) T h e topology z^ induces 7 on E. (iv) E is ?-dense in B.

In particular, if { V )Ais a basis of zero for T, then the collection { V ) of the ?-closures of the V in E is a basis for zero for z’. The unique Hausdorfltopological vector space (I!?,?) is called the topologicul completion of (E,z). We recall that for every mapping f from a topological space ( X , z ) into some set Y the finest topology zf on Y for which f is continuous is called the quotient topology determined by f . Obviously zf = { A G Y : f - ’ ( A ) E 73. If now (E,z) is a topological vector space, F a vector space, and f :E -+ F is an onto linear mapping, then the quotient topology zf on F is a linear one. The details are included in the next theorem. Theorem 4.6. Let 1’:( E , z )---t F be u linear mapping fiom u topologicul rector space ( E , t ) onto a vector space F. Then the following statements hold:

(i) T h e quotient topologji Tf is a linear topology and the linear mapping f :(E,z) -+ (F , z f ) is U M open mupping. (ii) For every basis { V }for the 7-neighborhoods of zero in E , { f ( V ) ) is a basis for the Tf-neighborhoods of zero in F. As a particular case of the preceding theorem consider a topological vector space (E,z) and a vector subspace A of E. Then the canonical projection n :E + E / A defined by n(u)= ir = u + A for u E E, is a linear mapping onto the quotient space, and hence its quotient topology T~ (which we shall denote by z / A ) is a linear topology on the quotient space EIA. We summarize the above in the next theorem. Theorem 4.7. For every uector subspace A of a topological vector space (E,z) the quotient topology T / A on E / A satisfies the following properties:

( i ) The canonicul projection n:(E,T) -> ( E / A , r / A )is continuous uncl open. (ii) For ewry basis [ V ) .for the 7-neighborhoods of zero qj E, [n(V ) ) is a basis.for the z/A-neighborhoods of zero in EIA. (iii) (E/A,.r/A)is a Hausdorff space if and onlji ifA is t-closed.

EXERCISES

31

EXERCISES

1. Give an example of a convex subset of an Archimedean Riesz space whose solid hull is not convex. 2. Prove Theorem 1.6. 3. Let B be the band generated by an ideal A of a Riesz space L and let U = { u E L+ :3{u,} c A with 0 I u, t u in L } . Show that B = { u - u:u,uE U } and that Bf = U . State and prove a similar statement for the a-ideal generated by A in L. 4. Prove Theorem 2.9. (Hint: If I3 Inu I u holds for all n and u # 19,choose a nonzero projection band B G B, and show that P,(o) = sup{w E B : e I I L l } = 8.1 5. Prove Theorem 2.14. 6. Consider an uncountable set X and let L be the Riesz space consisting of all real-valued functions f defined on X , such that there exists a real number f(00)(depending upon f ) for which the set {x E X : f ( x )# f(a)} is finite. Determine L" and L'. Is L almost a-Dedekind complete? 7. Let L be a full Riesz subspace of a Dedekind complete Riesz space M . Show that if an ideal A of M is maximal with respect to the property L n A = ( O } , then MIA is the Dedekind completion of L. 8. Show that the Riesz space of Example 2.13(6) is not almost aDedekind complete. 9. A nonzero element u of a Riesz space L is called an atom whenever U A w = 8 and 8 I u g (uI, 8 I wI IuI, imply u = 8 or w = 8. (i) Show that every discrete element of a Riesz space is an atom and give an example of an atom that is not a discrete element. (ii) If L is Archimedean, show that an element is an atom if and only if it is a discrete element. (iii) Show that C[O,l] does not contain any discrete elements. 10. (Judin). If every disjoint subset of an Archimedean Riesz space L is finite, show that L is Riesz isomorphic to some R". (Hint: Use the previous exercise to show that L is a discrete space and then use Theorem 2.17.) 11. Show that an Archimedean Riesz space that is totally ordered (i.e., every two elements are comparable) is either trivial or else it is Riesz isomorphic to R. 12. (Aliprantis [4]). Let L be a Riesz space with L- # ( 0 ) and let M be any Riesz space. Consider -4Vh(L,M)partially ordered by S I T whenever S(u) I T(u)for all u E L'. Pick 8 < cp E L - and define u H T , from M into Y h ( L , M )by T,(o) = cp(u)u for all u E L and u E M . (i) Show that T :M -, Zb(L,M)is a positive, one-to-one linear mapping. (ii) If M is Dedekind complete show that T :M + Zh(L,M) is a normal Riesz isomorphism (into); and thus if L- # ( O } we can consider M as a regular Riesz subspace of .IU,(L,M).

32

RIESZ SPACES

[Chap. 1

(iii) By using the embedding T :M -+ yh(L,M) show that if Th(L,M)is a Dedekind complete Riesz space, then M is also Dedekind complete (the converse of Theorem 3.3). 13. Let L be a Riesz space, and let M be a super Dedekind complete Riesz space. (i) Show that if L has an order unit, then Tb(L,M)is super Dedekind complete. (ii) Show that if L is Archimedean with a weak order unit, then Tn(L,M) is super Dedekind complete. 14. Let L be a a-Dedekind complete Riesz space, M a super Dedekind complete Riesz space, and T E 5Yc(L,M). (i) Show that C, = N T d is super Dedekind complete and that T restricted to C, is a normal integral. (ii) Show that C, is a projection band of L. (iii) Show that T E Y , ( L , M ) if and only if N T is a band of L. 15. Let L = C(X,), where X , is the one point compactification of an uncountable set X , considered with the discrete topology. If {xl,xz,.. .) is a countable subset of X and cp:L -+ R is defined by cp(u) = u(00) + 2-"u(x,) for all u E L, show that cp E L,", that N , is a band of L, and that cp 6 Lz. Why does the last claim not contradict part (iii) of the previous exercise? 16. Let L = C[O,l]. Show that L i = Lz = to}. Determine also L'. (Hint: For L" use the Riesz representation theorem.) 17. (Luxemburg-Zaanen [45,Note VI]; Kaplan [38, Note I]). Let L = C ( X ) ,where X is a topological space, and let {u,} G L satisfy u,(x) 1 0 for all x E X . Show that cp(u,) 10 holds for every 0 I cp E L". (Hint: If cp(u,) 2 E > 0 holds for all n, then w = C:= (u, - ~ u ~ / 2 c p ( u ~E) C ) +( X ) satisfies

cpw = a.1

18. Give an example of a Riesz homomorphism 71 from a Dedekind complete Riesz space L onto itself, which is not a discrete element of Th(L,L) and deduce from this that Theorem 3.13 cannot be generalized. (Hint: Consider 7 1 : R2 + R2 defined by x(u,v) = (u,2v).) 19. For a RiesL space L show that the following statements are equivalent : (i) L is almost o-Dedekind complete. (ii) Whenever 0 I u, 1 holds in L, there exists a sequence (v,} with 8 I v, T, uk I u, for all n,k and such that u, - v, 1 8.

CHAPTER

2 Locally Solid Topologies

This chapter presents the basic theory of locally solid Riesz spaces. As we shall see, this concept is a natural topological condition relating to the ordering. This chapter will illustrate the interplay between the order and the topological structures of such spaces. We begin by characterizing a locally solid topology in terms of the uniform continuity of the lattice operations, and then study topological convergence, bands, ideals, and solid sets. We then focus our attention on locally convex-solid Riesz spaces and we draw a number of interesting results about such spaces and their duals. Finally, we investigate which order properties of a Hausdorff locally solid Riesz space are inherited by its topological completion.

5.

CHARACTERIZATIONS AND BASIC PROPERTIES

We start with the definition of locally solid Riesz spaces. Recall that a subset S of a Riesz space L is solid if IuI I and u E S imply u E S.

101

Definition 5.1. A linear topology z (not necessarily Hausdorff) on a Riesz space L is said to be locally solid if z has a basis for zero consisting of solid sets. A locally solid Riesz space (L,z)is a Riesz space L equipped with a locally solid topology z. The next result characterizes the linear topologies that are locally solid. 33

34

LOCALLY SOLID TOPOLOGIES

[Chap. 2

Theorem5.2. Let z be a linear topology on a Riesz space L. Then the following statements are equivalent : (i) (L,z)is a locally solid Riesz space. (ii) The mapping (u,u) H u v u from (L,z) x (L,z) into (L,z) is uniformly continuous. (iii) The mapping (u,u) H u A u from (L,z) x (L,z) into ( L J )is uniformly continuous. (iv) The mapping u H IuI from (L,z) into (L,z)is uniformly continuous. (v) The mapping u +-+ 1 4 - from (L,z)into (L,z)is uniformly continuous. (vi) The mapping u H u + from (L,z)into (L,z)is uniformly continuous. Proof. (i) * (ii) Use the inequality Iu v u - f v g1 I Iu - f l + Iu - 91. (ii) * (iii) Use the equality u A u = - ( - u) v ( - u). (iii) * (iv) Use the equality IuI = -(u A ( - u ) ) . (iv) * (v) Use the equality u- = $( IuI - u). (v) (vi) Use the equality u+ = ( - u ) - . (vi) (i) Let U be a z-neighborhood of zero. Choose a balanced zneighborhood V of zero with V V E U . By the uniform continuity of u H u + , there exists a balanced z-neighborhood W of zero such that u u E W implies u+ - u+ E V. In particular, note that if u E W ,then u+, u - E V . Next choose a balanced z-neighborhood W, of zero with W, + W, E W , and then pick a balanced z-neighborhood W, of zero such that u - u E W, implies u+ - u+ E W,. We claim that the solid hull of W, satisfies S( W,) E U . Indeed, if IuI I IuI with u E W,, then u+,u- E W, and so IuI E W. Thus IuI = u+ - (u+ - Iul) E W implies u+ = ( u + ) + - (u+ - lo[)+ E V ; similarly, u - E v and hence u = u+ - u - E V + V E U . Therefore t is a locally solid topology, and the proof is finished. R

+

It may be asked at this point whether the continuity of u H u+ at zero is enough to ensure that the linear topology z is locally solid. The following example shows that the answer is negative. Example 5.3. Let L be the Riesz space consisting of the continuous piecewise linear functions on [0,1], ordered with the pointwise ordering. Define a norm p on L by

where u’ is the derivative of u (note that u’ exists for all but a finite number of points and that u’ is a step function). It is left to the reader to verify that p is a norm on L satisfying p(u) = p(lu1) for all u E L. Note also that 0 I u I u in L does not imply p(u) I p(u), that is, p is not a Riesz norm.

Sec. 51

THE BASIC PROPERTIES

35

Let z be the Hausdorff linear topology generated by p. Since for each u E L, lull = I(u+)’l + I(u-)’l holds for all but a finite number of points of [0,1], it follows that limp(u,) = 0 implies limp(u,’) = 0, i.e., that u H u + from (L,z) into (L,z) is continuous at zero. However, it is easy to see that e is not a locally solid topology. We continue with a number of properties of locally solid Riesz spaces. Theorem 5.4. Let (L,z)be a locally solid Riesz space. Then we have

(i) The order bounded subsets of L are z-bounded. (ii) The z-closure of a solid subset of L is a solid subset of L. (iii) The z-closure of a Riesz subspuce of L is a Riesz subspace (and hence the z-closure of an ideal is also an ideal). . )

Proof. (i) Let [u,u] be an order interval of L and let I/ be a solid zneighborhood of zero. Pick 1> 0 such that 1(lul + E V . Now since u I w 5 u implies IwI I IuI + and V is a solid set, we easily get A[u,u] E V , so that [ u , ~ ]is t-bounded. (ii) Assume that S is a solid subset of L and IuI IIuI with u E S. Pick a net ( v , ) G S with t‘, D and define u, = (U A ~ ( - 1 ~ ~ 1 for ) all CI. Note leal holds for all a.Thus {u,; E S and by Theorem 5.2 u, A u. that Iu,I I Hence u E S and so S is solid. (iii) Let M be a Riesz subspace of L. Clearly M is a vector subspace of L. Now if LI E R,then there exists a net [u,) of M with u, 1,u. But then by I Theorem 5.2 the net [u,’; G M satisfies I/,’ -+ u t , and so u + E M , that is, M is a Riesz subspace of L. The parenthetical part follows now easily from part (ii).

101)

11

111~1)

Assume now that {(L,,z,)} is a collection of locally solid Riesz spaces. Let L = IIL, be the Cartesian product of the collection of Riesz spaces {L,). Then L when ordered componentwise is a Riesz space, and if it is equipped with the product topology e = nz,,it is easy to see that it becomes a locally solid Riesz space. Hence we have the following theorem. Theorem 5.5. Let {(L,,za))be a collection of locally solid Riesz spaces und let L = l l L , and z = n ~ ,Then . (L,r) is a locally solid Riesz space.

More properties for the Hausdorff locally solid Riesz spaces are included in the next theorem. Theorem 5.6. For HuusdorfS locally solid Riesz spaces (L,z) we have (i) L is Archimedean. (ii) The cone Li of L is z-closed.

36

LOCALLY SOLID TOPOLOGIES

[Chap. 2

(iii) If u, 1,u and u, t in L, then u, t u holds in L . Similarly, u, 1,u and 1 implies u, 1 u. (iv) Every band of L i s z-closed. (v) If two ners {u,} and {ti,} of L satisfjj 8 I u, 1 I u, and v, - u, 1,8, then u, 1 u in L i f and only i f u, 1 u in L.

u,

u

Proof. (i) Assume 8 I nu I u for all n and some U,L' E L+. Then 8 I I n- u 1,8, and hence u = 8, since r is a Hausdorff locally solid topology.

Hence L is Archimedean. (ii) By Theorem 5.2 u H u- from (L,r) into (L,T) is continuous. The result now follows from L+ = { u E L:u- = 0). (iii) Fix an index a.Then for p 2 a we have

8I u v u ,

-u

I u v u , - u I124,

-

UI

i,0 /J> z

and hence u v u, - 21 = 8. Thus u, I I I holds for all a.Now let u, I u for all cx and some 1' E L. Then 0 I c - u, 1,I' - u, and so (since L+ is r-chsed) we get c - u E L', i.e., u I u. This shows u, t u in L. For the other case apply the above arguments to the net { -u,). (iv) Observe that for every nonempty subset D of L the set Dd = { u E L:IuI A IvI = 8 for all v E D ) is r-closed. Indeed, if u E then u, u holds for some net {u,} c Dd, and hence 8 = IuI A Iu,I 1, A IuI holds for each u E D. Since r is a Hausdorff topology IuI A IuI = 8 for all u E D, that is, u E Dd and therefore Dd is r-closed. To obtain the result now, observe that since L is Archimedean every band A satisfies A = Add and thus is t-closed. (v) If u, 1 u and v, 2 u 2 u for all a,then u, - u, 2 (c - u,)' 2 8 for all a and hence ( t i - u,)' 1,8. But ( v - u,)+ t (a - u)' holds and hence by part (iii) ( r - u)' = 0, i.e., u = 1'. Thus 11, 2.1. Now if u, 1 u holds in L, then 8 5 u - u A u, I u, - u, implies u u A u, -& 0. Now since u - u A u, t, we get by part (iii) that 0 I u - u A u, t 0, and thus u - u A u, = 0 for all N, or u I u, for all a.It is easy to see now that u, 1 u holds in L.

101

We shall show next that the topological dual of a locally solid Riesz space is an ideal of its order dual. [The topological dual E' of a topological vector space (E,r) is the vector space consisting of all t-continuous linear functionals on E.] Theorem 5.7. Let (L,T)be a locally solid Riesi space. Then the topological dual L' of (L,z) is an ideal of the order dual L' (and hence it is a Dedekind complete Riesz space in its own right).

THE BASIC PROPERTIES

Sec. 51

37

Proof. Observe first that L' is a vector subspace of L'. Indeed, by way of contradiction, assume that for some cp E L' we have cp q! L'. Then there exists u E L f and a sequence { u n } G [8,u] satisfying Icp(u,)l 2 n for all n. It now follows from 0 I n-'u, I n - ' u & 0 that n - l u , 1,8 and so (since cp E L') limcp(n-'u,) = 0, contrary to lcp(n-lu,)l 2 1 for all n. To see that L' is an ideal of L-, assume IcpI _< I$I with $ E L' and cp E L". Let u, 1,6 and E > 0. By Theorem 3.3 there exists a net {v,) satisfying Iu,I I Iu,I and I$I(Iu,I) 5 I$(u,)l e for all a. Clearly u, 8, and since

+

Idua)l

I~

l~I(luu1)

~

5

~

I($ ( u u~) l

+G ~

a

~

)

we get 0 I lim suplcp(u,)l 5 c for all E > 0. Thus lim suplcp(u,)( = 0, and so lim cp(u,) = 0. Hence cp E L', and the proof is finished. Note. Theorem 5.7 was proven by Nakano [50, Theorem 31.1, p. 1341 for the normed case; see also [53, Theorem 31.1, p. 1341.

The following example shows that L' in general is not a band of L' and need not even be a a-ideal of L'. Example 5.8. Let L denote the Riesz space consisting of all real sequences that are eventually zero, with the pointwise ordering. Now consider the linear functional cp-defined by m

d u )=

1 nub)

for u

E

L.

n= 1

Clearly cp 2 0 and so cp E L'. Now let r be the Hausdorff locally solid topology generated by the (Riesz) norm p ( ~=) sup(lu(n)I:n= 1,2,. . .).

Next consider the sequence (u,) c L defined by u,(k) = 0 if k # n, u,(n) = n-lI2 and note that p(u,) = n-'I2 + 0. But cp(u,) = n1I2 + cc and hence cp $ L'. On the other hand, define the sequence {cp,) E L' by cp,(u) = xi= ku(k) ( u E L), n = 1,2,. . . . Now since Icp,(u)l I (xi= k ) p ( u ) we have (cp,} E L' and clearly 8 I cp, t cp in L'. This shows that L' is not a a-ideal of L'. The next theorem tells us that quotients of locally solid Riesz spaces equipped with their quotient topologies become locally solid Riesz spaces. Theorem 5.9. Let (L,s)be a locally solid Riesz space and let 71 he a Riesz homomorphism from L onto a Riesz space M . Then M equipped with the quotient topology z, is a locally solid Riesz space. In particular., it ,fOllow.s that if A is an ideal of L then (L/A,s/A)is a locally solid Riesz space.

30

LOCALLY SOLID TOPOLOGIES

[Chap. 2

Proof. Let { V }be a basis of zero for T consisting of solid sets. By Theorem 4.6 { " ( V ) }forms a basis of zero for T ~ But . by Theorem 1.18 each n ( V )is a solid set, and hence the topology T , is locally solid. The last part follows by noting that the canonical projection of L onto L / A is a Riesz homomorphism. (Note that in this case we denote the topology by T/A.) H 6.

LOCALLY CONVEX-SOLID TOPOLOGIES

A linear topology T on a Riesz space L that is at the same time locally solid and locally convex will be called a locally convex-solid topology. A locally convex-solid Riesz space (L,T)is a Riesz space L equipped with a locally convex-solid topology T . By Theorems 1.3 and 5.4(ii) we see that for a locally convex-solid Riesz space (L,T)the collection of T-closed, convex, and solid z-neighborhoods of zero forms a basis for the z-neighborhoods of zero. A seminorm p on a Riesz space L is said to be a Riesz seminorm if IuI I IvI in L implies p(u) I p(u), or equivalently, if p(u) = p(lu1) holds for all u E L and 6 5 u I u in L implies p ( u ) I p(u). A Riesz norm is a Riesz seminorm that is also a norm. A seminormed (resp. normed) Riesz space L, is a Riesz space L equipped with a Riesz seminorm (resp. norm) p. Observe that a seminorm p on a Riesz space L is a Riesz seminorm if and only if its open unit ball B, = {u E L : p ( u )< 1) is a solid subset of L. Note first that if p is a Riesz seminorm then clearly B, is a solid subset of L. On the other hand, ifB, is a solid set and 1111 I IcI holds in L, then I(p(u) + E ) - luI I J(p(c)+ E ) - 'vI holds for all E > 0 and since ( p ( u )+ E ) - ~ LE B, we get (p(o) + E ) - ~ Eu B,. Thus p ( u ) I p(u) + E holds for all E > 0, and therefore p ( u ) I p(v), i.e., p is a Riesz seminorm. Similarly, a seminorm p is a Riesz seminorm if and only if its closed unit ball U , = { u E L : p ( u )I 1) is a solid set. A complete normed Riesz space is called a Banach lattice. It should be clear that any collection of Riesz seminorms generates (in the usual manner) a locally convex-solid topology. The point is, that every locally convex-solid topology on a Riesz space L is generated by a family of Riesz seminorms.

Theorem 6.1. Let L be a Riesz space and T a linear topology on L. Then the following statements are equivalent: (i) T is a locally convex-solid topology. (ii) There exists a family { p a } of Riesz seminorms that generates the topology T .

Proof. (i) * (ii) Let (l/b} be a basis of zero for T consisting of z-closed, solid, and convex sets. Let p. denote the Minkowski seminorm generated

Sec. 61

LOCALLY CONVEX-SOLID TOPOLOGIES

39

by V,, that is, p,(u) = inf{A > 0 : u E A V , )

for u E L. Then pa is a t-continuous seminorm whose closed unit ball equals the solid set V,. By the above discussion pa is a Riesz seminorm. The result now follows by noting that the family { p a }generates the topology t. (ii) =. (i) Obvious. Assume now that L is a Riesz space and p a Riesz seminorm on L. If A is an ideal of L, we denote by 2i the element of L/A whose representative is u, i.e., ti = n(u),where n is the canonical projection of L onto L/A (we also write V = n ( ~for) v G L). It is well known that the function p defined on L / A by

p(2i) = inf(p(v):u E L a n d o E ti},

ti E L/A,

is a seminorm, which is actually a Riesz seminorm. This follows immediately from Theorem 1.18 by noting that the open unit ball of 0 is solid since it is the image of B, under the canonical projection of L onto L/A. The following result should now be clear. Theorem 6.2. Let (L,z) be a locally convex-solid Riesz space whose topology z is generated by the family of Riesz seminorms ( p , } , and let A be an ideal of L. Then the quotient topology T ~ Aon LIA is a localljj convex-solid topology generated by the family of Riesz seminorms { p a } .

The discussion next will give to the reader an idea as to how far the locally solid topologies are from the locally convex-solid ones. A Riesz pseudonorm p is a function from a Riesz space L into R ( p :L + R ) satisfying the following properties : p(u) 2 0 for each u E L. (2) p(u + u ) I p(u) + p(u) for all u,v E L. ( 3 ) p ( h ) -, 0 as A + 0, for each u E L. (4) p(u) 5 p(u) whenever IuI IIvI holds in L.

(1)

Clearly every Riesz seminorm is a Riesz pseudonorm. Theorem 6.3. Let L be a Riesz space and the following statements are equivalent:

t

a linear topology on L. Then

(i) t is a locally solid topology. (ii) There exists a family { p a } of Riesz pseudonorms that generates the topology T. Note. The introduction of the notion of the Riesz pseudonorm as well as Theorem 6.3 is due to Fremlin. For a proof of Theorem 6.3 see [25, 22C Proposition, p. 391.

40

LOCALLY SOLID TOPOLOGIES

[Chap. 2

We recall that a locally convex topological vector space (E,z) is said to be a barrelled space whenever every absorbing, balanced, convex, and zclosed subset of E is already a z-neighborhood of zero. Barrelled Riesz spaces have the property that their topological duals are bands in their order duals, as the next theorem shows. Theorem 6.4. Let (L,z) be a barrelled locally convex-solid Riesz space. Then L’ i s a band of L‘. Proof. Assume 0 I cpa t cp holds in L’ with (cp,} E L’. Then the set V = ( u E L: Icp,(u)I 5 1 for all a } is an absorbing, balanced, convex, and zclosed subset of L, as can be easily verified. Thus Vis a z-neighborhood of zero. But Icp,(u)l I 1 for each u E V and all a implies that Icp(u)l 5 1 for each u E V , since cpz t cp (Theorem 3.3). Therefore, cp E L’, since cp is bounded on the z-neighborhood T/ of zero, and thus it follows that L’ is a band of L“.

For any Riesz space L observe that for each cp E L‘ the function p,(u) = Icpl(lul) for u E L defines a Riesz seminorm on L, and thus any nonempty

subset A of L- defines a locally convex-solid topology (not necessarily Hausdorff) on L via the collection of Riesz seminorms {pB:cpE A ) . This topology is called the absolute weak topology generated by A on L. Definition 6.5. Let L be a Riesz space and let A be a nonempty subset of L‘. The absolute weak topology lol(L,A)generated by A on L is the locally convex-solid topology generated on L by the collection of Riesz seminorms {pq:cp E A ) , where p,(u) = Icpl(lul)for all u E Land cp E A.

Observe that lol(L,A) = lol(L,I(A)), where I(A) is the ideal generated by A in L-. Indeed, note first that lol(L,A) E lol(L,I(A)) and that for every $ E Z(A), there exist (cpl,. . . ,cp,) G A and positive constants I , , . . . ,I, with I Ailcpil;hence p + I Aip,, holds and so p+ is a lal(L,A)continuous Riesz seminorm. This implies lol(L,I(A)) G lol(L,A) and thus lal(L,A) = Iol(LJ(A)). I 2 1 p v z ( ~implies ~) that every Note also the relation I$(u)l I $ E I ( A ) is a lol(L,A)-continuous linear functional on L. The next theorem tells us that Z(A) contains precisely all the lol(L,A)-continuous linear functionals of L.

c;=l

1$1(1~11)

c:=l

Theorem 6.6. Let L be a Riesz space, A a nonempty subset of L’, and

I ( A ) the ideal generated by A in L-. Then

(L,lol(L,A)1’ = I(A). Proof. As noted above every cp E I(A) is lol(L,A)-continuous and so we need only to show that every cp E (L,Io/(L,A))’belongs to I(A). To this end suppose that cp is lol(L,I(A))-continuous. Then there exist I,$ E I ( A ) and

Sec. 61

41

LOCALLY CONVEX-SOLID TOPOLOGIES

II > 0 such that Icp(u)l 4 Ap&u) for all u E L. But then 0 s )cpl(u)I ;ip&u) = i l $ ( ( u ) holds by Theorem 3.3(ii) for all u E L f , and hence IcpI I Al$I. Therefore, cp

E

I(A) and the proof is finished.

Now let (L,z) be a locally convex-solid Riesz space. We define the absolute weak topology on L to be the topology lol(L,L’),that is, lol(L,L’) is the topology generated by the family of Riesz seminorms {prp:cpE L’}. Theorem 6.7.

have

Let (L,z) be a locally convex-solid Riesz space. Then we o(L,L’)E lal(L,L’)G 7.

In particular, it follows that \ol(L,L) is a Hausdorff topology if and only

ifz is Hausdorfl

- -

lal(L,L‘)

Proof. Assume u, 0. Then for each cp EL‘ the relation Icp(u,)l I o(L,L’) \cp\((u,\) implies limcp(u,) = 0. Thus u, 0 and so a(L,L) c lol(L,L‘). On the other hand if u, Q holds, then Iu,( 5 Q also holds since z is a locally solid topology. But then lim p,(u,) = lim Icpl( lu,l) = 0 for each cp E L‘ a I(L L’) and so u, 0. Thus lal(L,L’)E z, and the proof is finished. H

-’

Note. Theorem 6.6 has appeared in various forms in the work of many authors, the result as it is stated here is due to Kaplan [38, Note 11, Proposition 10.1, p. 3491.

Some conditions for the absolute weak topology Ja\(L,L’)to coincide with the weak topology o(L,L’)are given in the next theorem. Theorem 6.8. Let (L,z)be a Hausdorff locally convex-solid Riesz space. Then the following statements are equivalent:

(i) (ii) (iii) (iv) (v)

o(L,L‘) = Jol(L,L’). T h e weak topology o(L,L’) is locally solid. T h e mapping u w IuI is weakly continuous at zero. Every order interval of L’ lies in a finite dimensional vector space. Every principal ideal of L’ is of Jinite dimension.

Proof. (i) * (ii) * (iii) Obvious. (iii) (iv) It is enough to show that [O,cp] spans a finite dimensional vecI1) tor space for each 0 5 cp E L‘.So, let 8 Icp E L‘.Then V = ( u E L :cp( is a o(L,L’)-neighborhood of zero, since the weak continuity of u H IuI at zero implies the continuity at zero of u H cp(lul) from (L,o(L,L’))into R. Thus there exist E > 0 and c p l , . . . ,cpn E L’ such that

\MI)

{ u E L:Jcp,(u)J IE

for i

=

I , , . . ,n} G V.

42

LOCALLY SOLID TOPOLOGIES

[Chap. 2

Now assume $ E L' satisfies 8 I$ Icp. If u E Kercp,, i = 1,. . . ,n [that is, cpi(u) = 0 for i = 1 , . . . ,n], then cpi(ku)= 0 for i = 1,. . . ,n and all k, and so ku E V for all k. Thus kcp( Iul) I1 for all k and so cp( Iul) = 0. It follows that 0 II$(u)l I$ ( l u l ) = 0 and so u E Ker $. Thus we have

n Ker n

i= 1

'pi

c Ker $.

Note that the last condition is equivalent to the property that IC/ is in the linear span of {cpl, . . . ,cp,} (see, for example, [31, p. 1863). Hence [8,cp] spans a finite dimensional vector space in L', and we are done. (iv) + (v) Note that given cp E L' the ideal generated by cp in L' equals the vector subspace generated by the order interval [8,lcpl] in L'. The result now follows. (v) + (i) Observe first that if 8 < cp E L' is a discrete element of L' then ( ~ ( 1 ~ 1 )= Icp(u)l for all u E L (since cp is also a discrete element of L-, and hence a Riesz homomorphism by Theorem 3.13). Now let 8 < cp E L' and let A , be the ideal generated by cp in L'. By hypothesis A , is of finite dimension and hence A , is Riesz isomorphic to some R" [46, p. 1521. Thus there exists a basis cpl, . . . ,cpn of A , consisting of positive discrete elements of L'. In particular, we have cp = A i q i for some n(L L ' ) appropriate real numbers I , , . . . ,A,,.Thus if u a - - - 4 8 holds, then cp(Iual) = IaKL L') Aicpi( lu,l) = Ailcpi(ua)Iimplies that IuKl-8. It now follows that a(L,L') = lal(L,L'),and the proof is finished. H

c:=

Remark. A glance at the above proof shows that if a(L,L') is locally solid, then L' is a discrete space.

We are now in a position to characterize the finite dimensional normed Riesz spaces in terms of the weak topology. Theorem 6.9. For a normed Riesz space L, the following statements are equivalent :

(i) L is of jinite dimension, that is, L is Riesz isomorphic to some R" with its natural ordering. (ii) The weak topology a(L,,L,') is a locally solid topology. Proof. (i) * (ii) Obvious. (ii) (i) It is enough to show that L,' is of finite dimension. To this end assume that L,' contains a sequence {cp,} of linearly independent functionals. Without loss of generality we can assume that each cpn has norm one. Next observe that since L,' is norm complete the series 2-"lcp,l norm converges in L,'. Put cp = xF=,2-"lq,l and note that [O,cp] spans an infinite dimensional vector space, in contradiction with Theorem 6.8. This contradiction establishes that L,' is of finite dimension and completes the proof. H

Sec. 71

TOPOLOGICAL COMPLETION

43

The following example shows that the implication (ii) 3 (i) of Theorem 6.9 is no longer true if the hypothesis of L , being a normed Riesz space is replaced by local convexity. Example 6.10. Let X be an infinite set, and let L = RX and z be the Hausdorff locally convex-solid topology of pointwise convergence. [In other words, z is generated by the family of Riesz seminorms { p x : xE X } , where p,(u) = lu(x)l for u E L.] If x l , . . . ,x, is a finite number of points of X and , I l , . . .,,I,are , real constants, then the linear functional cp defined by n

q ( f )=

1 Aif(xi)

i= 1

for

f EL,

is a z-continuous linear functional on L. The point is that every cp E L' is of the above form. To see this, note that if cp E L', then there exist E > 0 and a finite number of points x l , . . . ,x,, of X such that Icp( f ) l I 1 holds whenever f E L satisfies 1f ( x i ) )s E for i = 1, . . . ,n. Put Ai = cp(x~,,,)( i = 1, . . . ,n) and note that for each positive k the function gk = k( f , f ( x i ) x c x , lsatisfies } gk(xi)= 0 for i = 1,. . . ,n. Thus klq(f )l if(xi)I I 1 holds for k = 1,2, . , . , and so cp( f)= lif ( x i )holds for all f E L. It follows now easily that o(L,L')= lol(L,L')= z. Note, however, that L has infinite dimension.

cy=

x;=

Note. Theorem 6.9 is due to Peressini [55, Theorem 2.4, p. 2081; see also [56, 2.14 Proposition, p. 1341. 7.

TOPOLOGICAL COMPLETION

As we saw in Theorem 4.5 every Hausdorff topological vector space (E,z) has a unique topological completion (I?,?) in the sense that (I?,?) is ?-complete, containing (E,z) as a vector and topological ?-dense subspace. In the next theorem we shall show that if in addition E is a Riesz space and z is a locally solid topology, then (I?,?) has a natural Riesz space structure that induces the original on E. The details of this important result are included in the next theorem. Theorem 7.1. Let (L,z) be a Hausdor- locally solid Riesz space and let its topological completion. Then the ?-closure of L f in t is a cone of and ($?) equipped with this cone becomes a Hausdor- locally solid Riesz space containing L as a Riesz subspace. I n particular, the ?-closure of every solid subset of L is a solid subset oft.

(L,?) be

e,

Proof. We show first that f,' (the ?-closure of L + in t)is a cone o f t . To this end note that the relations t' + L' c t' and A t f G f,+ for l 2 0

LOCALLY SOLID TOPOLOGIES

44

[Chap. 2

follow from the corresponding relations of L + , and the continuity of the addition and scalar multiplication in To see that L' n ( - E') = (8) let ii E n and then choose two nets {u,} and (wp} of L+ with u, ii and wp A -6. But then 0 I u, I u, + wp I ,8 implies u, 1,6, and thus (a.B) 6 = 6. Hence L' is a cone of 2. We show next that t is a Riesz space with this cone. The mapping u H u+ from ( L 7 t into ) (L7?)is uniformly continuous, and because (L,?)is ?-complete, it can be extended uniquely to a uniformly continuous mapping p from L into L (Theorem 4.4).We show next that p(6) = 6 v 8 holds in t for each ii EL. Indeed, if ii E f,, pick a net {u,} E L with u, f ii and note that p(u,) - u, = u,' - u, E L+ implies p ( 6 ) - ii E L', i.e., p(ii) 2 ii holds in L. Clearly p(ii) 2 8 also holds in L. On the other hand, if ii I B and 8 5 v^ hold in L7 then pick ( w l } and { u , ) in L+ (with the same directed set) such that i i F w, -+ v^ - ii and v, + 6. Note that f, = v, - w, -,ii and u, - p(f,) = v, -fa' 2 6. Hence by taking the ?-limits7 we get v^ - p(ii) E 2' and so p ( 6 ) = ii+ holds in 2. This shows that L is a Riesz space containing L as a Riesz subspace. The uniform continuity of ii H'ii now implies that (L,?)is a locally solid Riesz space (Theorem 5.2). For the last assertion, let S denote the ?-closure of a solid set S of L. Assume (iil I 161 with 27 E S. Pick a net (i,) of S with c, f 6 and a net ( u p ! of L with up f t. Put w , , ~= [( - )v,1) v up] A Iu,), and observe that s Ic,1 holds for all a$. Thus { w , , ~ }E S . But we also have w , , ~ t , so that ii E S, and hence S is a solid subset of L. (The last property shows that if ( V } is a basis of zero for T consisting of solid sets, then { V } is a basis of zero for ? consisting of solid sets.)

e+ (-e+),

e.

)W,,~I

Note. The preceding proof can be found in [2, Theorem 2.1, p. 1091.

The next two theorems deal with the lattice relation of L with respect to L. Theorem 7.2. For a Hausdorfllocally solid Riesz space (L,r) the following statements are equivalent:

(i) L is a regular Riesz subspace of L. (ii) Every r-Cauchy net {u,} of L f with u,

48 in L satisfies u,

0.

Proof. (i) * (ii) Let (u,) c L+ be a r-Cauchy net such that u, 2 0 in L. Pick a net (L;)c L+ with 8 I 11, s v, 1 0 in L. By hypothesis L', .1 8 holds also in L. Since [u,) is a r-Cauchy net of L, there exists ii E L such that u, f ii. But for each fixed 3!, we have 6 s 11, I 0, I up for all a 2 p, and thus 19I ii I up holds for all p. Hence ii = 8, and so u, 1,8.

Sec. 71

45

TOPOLOGICAL COMPLETION

e

(ii) * (i) Let u, 1 0 in L. Assume 0 ID I u, holds in for all a. Pick a net (c2j of L + with cA 4i; and note that JL., A 11, - 61 = lvA A u, - F A id,( I Ic2 - i;l for all crJ. This shows that c2 A ii, f F. In particular, we have that (2.a) {oAA uai is a 7-Cauchy net of L. But we also have 8 I o2 A u, I u, 1 8 ( 2 4

in L, and hence from our hypothesis we get i‘, A I I , 0.~ Thus 1‘ = 0 and hence u, -1 6, in f.. We recall that a subset A of a topological vector space ( E , t ) is said to be z-complete whenever every 7-Cauchy net of A z-converges to some element (necessarily unique if z is Hausdorff) of A . (A.1)

Theorem 7.3. Let (L,z) be a Hausdorff locally solid Riesz space. Then the following statements are equivalent:

(i) L is an ideal of f. (and hence an order dense ideal). (ii) Ererj, order interval of L i s z-complete. I n purticular. if L is an ideal of L then for everis 0 < ii E t there exists a net (u,) c L + with u, T ii and u, f 11. Proof. (i) 3 (ii) If 8 I ii E L, pick a net [up,)c L+ with u, f ii and note that since L is an ideal o f t we have {u, A ii) G L’. Thus by taking the finite suprema of the net (u, A ii), we get a net ( L . ~ ]c L f such that 8 I T ii and r , f ii, and so in particular L is order dense i n Now since [ U J ’ ] = u + [8,v - L I ] it is enough to establish that every interval [8,u] ( u E L + ) is 7-complete. To this end let [ u , ) be a 7-Cauchy net of [fl,~].Then 11, 5 ii holds in I?,. But then 0 I L? I u also holds, and since L IS an ideal of L, we have ir E L and so [O,ii] is 7-complete. ii I u E L. Pick a net (u,; g L+ with u, f ii. We (ii) 3 (i) Assume fl I may suppose that 8 I u, I u holds for each cr; otherwise we replace (if necessary) the net (u,) by {u, A u ) . It follows now that ii E L, and from this (since L is a Riesz subspace of L), we get that L is an ideal of W

e.

e.

Given a topological vector space (E,z) it is natural to ask what conditions will imply that every z-Cauchy net of E converges with respect to z to some element of E . That is, when is (E,z) topologically complete? Clearly for a Hausdorff topological vector space (E,z) the above question can be phrased as follows: When does E = I? hold? The following discussion will answer this question for the Hausdorff locally solid Riesz spaces. But first we give a definition. Definition 1.4. A locally solid Riesz space ( L J )is said to satisfy the monotone completeness property (denoted hji MCP) if ei’erji monotone z-Cauchy net of L z-converges in L.

46

LOCALLY SOLID TOPOLOGIES

[Chap. 2

Note that a locally solid Riesz space (L,z) satisfies MCP if and only if every increasing z-Cauchy net of L+ is z-convergent. It should be also clear that every topologically complete locally solid Riesz space satisfies MCP. However, there exist Hausdorff locally solid Riesz spaces satisfying MCP or having their order intervals topologically complete without being by themselves topologically complete. Some examples follow. Example 7.5. Let X be an infinite set and let L = l a ( X ) , i.e., let L be the Riesz space consisting of all bounded real-valued functions defined on X with the pointwise ordering. Let z be the Hausdorff locally solid topology of pointwise convergence. Note that the order intervals of L are z-complete, but (L,z)is not z-complete. Observe also that = R[’.”. Example 7.6. Consider L = C[O,l] and let C, denote the collection of all countable and compact subsets of [0,1]. Then for each K E C, define a Riesz seminorm pK(u)= sup{ lu(x)l:x E K }

for u E L.

Define also p ( u ) = A! lu(x)l dx for u E L. Now let z be the (nonmetrizable) Hausdorff locally solid topology generated by the collection of Riesz seminorms { p K : KE C,} u { p ) . In particular, observe that u, 1,u implies u&x) -,u ( x ) for each x E [0,1]. The locally solid Riesz space (L,z)has the following properties : (1) (L,r)is not z-complete. To see this proceed as follows. For each E > 0 and each K E C,, pick u , , ~E C[O,1] such that 8 Iu , , ~I1, JA u,,,(x)~x 2 1 - E and u , , ~ ( x= ) 0 for all x E K . (Since the Lebesgue measure of K is zero and K is compact, functions with the above properties exist.) Next consider the net ( u , , ~ : ( E , K )E ( 0 , ~x) C,], directed by ( E ~ , K 2 ~( c)2 , K Z )whenever E , IE , and K , 2 K , . It can easily be seen that { u ~ ,is~ a} r-Cauchy net. However, if u , , ~1,u holds in L then since u , , ~ ( x + ) u(x) must hold for each x E [0,1], we must have u = 8, contradicting p ( ~ , ,G) ~ ) 1. This shows that (L,z)is not z-complete. We shall next identify the topological completion of (L,z).Let z, be the Hausdorff locally solid topology generated by the family of Riesz seminorms { p K : KE C,} defined above. Note that (C[O,l],z,) is z,-complete. Let z2 be the Hausdorff locally solid topology generated by the integral norm p on Ll([0,1]). Obviously (Ll([O,l]),z,) is r,-complete. Let M = C[O,l] x L,( [0,1]) and r 3 = z, x z2 (the product topology). We shall next show that ( M , z 3 )is isomorphic to To this end consider the mapping f ++q ( f ) = (f,[f]) from L into M ([f]= the equivalence class determined by f in L,([O,l])). Note that f H q ( f ) is a Riesz isomorphism from L onto q(L), which is in

(e,?).

Sec. 71

47

TOPOLOGICAL COMPLETION

addition a homeomorphism from (L,T)onto (q(L),z3). Thus, it remains to be shown that q ( L ) is 7,-dense in M . To this end let (g,[ f ] ) E M , E > 0, and K E C,. Select a function u E C[O,l] with u(x) = g(x) for each x E K and A j If ( x ) - u(x)l dx < E , and note that (u,[u]) E q ( L ) satisfies p K ( u - g) = 0 and p([f] - [u]) < E . This shows that q ( L ) is r,-dense in M (since [ p K : K E C,) is closed under finite maxima). (2) L is not a a-regular Riesz subspace o f t .

To see this pick a sequence {u,,} of L with u, 16' in L and with the set {x E [O,l]:inf{u,(x)} > 0} having positive Lebesgue measure. Then u, 1 6' does not hold in t. (3) (L,T)satisfies the property: I f u , u, 1,6'.

16' and

{u,} is

a z-Cauchy net then

Indeed, if u, 1 0 and {u,} is a z-Cauchy net, then {u,) converges uniformly on every countable compact subset of [OJ], say to some function u. But then u,(x) J, u(x) holds for every x E [O,l], and since u is continuous on every countable compact subset of [0,1], u E C[O,l]. Hence u = 6' (since u, 1 u also holds in L). Now by Dini's theorem { u,} converges uniformly to zero on [0,1], and from this it follows that u, 1,6'.

(4) (L,T)satisJies MCP. Indeed, if 6' I u, t and {u,) is a z-Cauchy net, then u,(x) u ( x ) holds for each x E [OJ]. Since {u,} converges uniformly on every countable compact subset of [0,1], u E C[O,l] and hence u, 1,u. R We give next a general characterization of the topological completeness of the Hausdorff locally solid Riesz spaces. Theorem 7.7. For a Hausdorff locally solid Riesz space (L,T)the following statements are equivalent

(i) (L,T)is z-complete, i.e., L = t. (ii) (L,T)satisjies MCP and every order interval of L is z-complete. Proof. (i) 3 (ii) Obvious. (ii) 3 (i) Since the order intervals of L are z-complete, L is an ideal of t by Theorem 7.3. Let 6' I ii E Then there exists a net {u,} G L+ such that 6' I u, t ii and u, f ii, by Theorem 7.3 again. But then { u,} is a z-Cauchy u holds in L for some u. It follows that net of L, and so by hypothesis u, ii = u E L, so that L = and therefore (L,T)is t-complete.

e.

The next theorem gives a condition for the Dedekind and a-Dedekind completeness to be inherited by t.

40

LOCALLY SOLID TOPOLOGIES

[Chap. 2

Theorem 7.8. Let (L,z) be a HausdorfS locally solid Riesz space having its order intervals z-complete. Then we have

(i) I f L is a-Dedekind complete, then so is t. (ii) I f L is Dedekind complete, then so is t. Proof. Note first that according to Theorem 7.3 L is an ideal of t. (i) Assume 6' I ii, I ii holds in Pick a net { wa> c L+ with 6' I w, ii and w, ii. For each a the sequence {w, A ii,} is a sequence of L+ satisfying 0I w, A ii, t I w,. Put v, = sup,{w, A ii,} in L. Now since lw, A ii, - wDA ii,l

T

r

e.

n

I 1 w, - wsl for each a$ and n, we get Iu, - vsl I [w, - wsl, which shows that {v,} is a z-Cauchy net of L. Let t3 I v, DE and therefore by Theorem 5.6(iii),we have u, T D in Thus w, A ii,* i v, I v^ holds for each a and n, and so ii, I v^ for all n. Now assume that ii, I i3 holds in f, for n = 1,2, . . . . Then we have w, A ii, I 6 for all a and all n and so v, I G for all a. Thus v^ I G and hence ii, t D holds in t,so that is a-Dedekind complete. (ii) Repeat the arguments of the previous proof.

e,

e.

e

Note. Theorem 7.2 was proven first by Kawai [39, Theorem 4.1, p. 2961 for the locally convex case; the general case was established in [2, Theorem 2.3, p. 1101. Theorem 7.3 was proven by Duhoux [22] and independently in [2]. Example 7.6 is due to D. H. Fremlin (personal communication). Theorem 7.7 is due to the authors [ 5 , Theorem 5.3, p. 321; results similar to Theorem 7.6 were obtained by Amemiya in [lo] under the assumption that L is Dedekind complete. Theorem 7.8 is essentially due to Luxemburg [44, Note XVI, Theorem 66.5, p. 6651. Theorem 7.9. Let (L,z) be a Hausdorff locally solid Riesz space and let A he a projection band of L. Then the ?-closure of A is a projection band Oft.

Moreover = (A)d(where Ad is the disjoint complement (A)dis the disjoint complement of A in

e).

of A in L, and

Proof. Combining the last part of Theorem 7.1 with Theorem 5.4(ii), we get immediately that both A and Ad are ideals of L. Now assume ii E t. t Then there exists a net {u,} ofL with u, -ii. iWrite u, = 0, + w, with (u,) SE A and { w , } E Ad (this is possible since L = A @ Ad). Since now Iu, - u I Iu, - up[ + Iw, - wpI both {u,} and {w,) are r-Cauchy nets of L. Let v, GEA and w, f 6 ~ 2Thus . ii = D + G E A + 2,that is, = A + 2.To complete the proof we have to show only that A n 2 = id}. Therefore, let 6 5 ii E Z n 2 and then pick two nets {u,} G A and ( v , } E Ad with u, A ii

Sec. 7)

TOPOLOGICAL COMPLETION

and c, f 6. But then 0 = Iu,I A f ii A ii = ti, and so fi = (A)dshould now be immediate. 1

~

7

~

1

49

= 0.

The relation

Theorem 7.10. Let (L,t) be a Hausdorf locally solid Riesz space. If L is order dense in t,then we have

(i) I f L has suficiently many prqjectinns, then has suficiently manjf project ions. (ii) I f L has the projection property, then is Dedekind complete.

e.

Proof. (i) Let A be a nonzero band of Then the order denseness of L in t implies that A n L is a nonzero band of L. Pick a nonzero projection band B of L with B s A n L G A . By Theorem 7.9 B is a projection band of t and by Theorem 5.6(iv) B E A . (ii) Let B be a band o f t . Then B n L is a projection band of L and so by Theorem 7.9 B n L E B is a projection band of L. Now if 8 I ii E B, then 8 5 u, t ii holds in L for some net {u,) G B n L. Hence ii E Bn. Thus B = B X , and hence has the projection property. Now since t is uniformly complete, t must be Dedekind complete according to Theorem 2.11.

e,

If L is order dense in then it is not always true that the principal projection property is inherited by L. An example follows. Example 7.11. Let X be an infinite set and let L be the Riesz space of all real functions f defined on X for which there exists a constant f ( w ) such that {x E X : f ( x )# f ( c o ) )is a finite subset of X , with the pointwise ordering. Let z be the Hausdorff locally solid topology generated by the sup norm. The topological completion 2 of (L,z) is the Riesz space consisting of all real-valued functions defined on X for which there exists a constant f ( w ) such that for every E > 0 the set { x E X : ( f ( x )- f(w)l > E ) is finite. [Alternatively = C(X,), where X , is the one-point compactification of the set X considered with the discrete topology.] Clearly z^ is the topology generated by the sup norm on t. Note that L has the principal projection property, that L is order dense in I!., but that does not have the principal projection property. However, it does have sufficiently many projections; compare this with Theorem 7.10.

e

e

Note. Theorem 7.9 was proven first by Luxemburg for normed Riesz spaces [44, Note XVI, Theorem 66.3, p. 6641; the extension to the general case can be found in [2, Lemma 7.1, p. 1221.Theorem 7.10 is due to Aliprantis [2, Theorems 7.4 and 7.5, p. 1231. Example 7.11 is due to Luxemburg [44, Note XVI, Example 65.6, p. 6631.

50

[Chap. 2

LOCALLY SOLID TOPOLOGIES

EXERCISES

1. Let (L,z)be a locally solid Riesz space. Denote by { V } the collection of all z-neighborhoods of zero and define the kernel of z by K , = n { V :VE { V } }. Show that K , is a z-closed ideal of L. Show also that T is a Hausdorff topology if and only if K , = {0}. 2. Let (L,z) be a Hausdorff locally solid Riesz space, and let A E L. If u E A, show that u = inf{uv v : v E A } = s u p { u ~ v:v E A}. 3. Let {u,} be a net of a Hausdorff locally solid Riesz space (L,z)such that u, T. Show that if {u,} has an accumulation point u, then u, t u and u, 1,u hold in L. 4. Let (L,z) be a Hausdorff locally solid Riesz space, and let L be order dense in a Riesz space M . If z can be extended to a locally solid topology z* on M , show that z* is necessarily a Hausdorff topology. 5. If W denotes the collection of all weak order units of a locally solid Riesz space (L,z),show that either W = @ or else that W is z-dense in L'. 6. Let (L,z)be a Hausdorff locally solid Riesz space. Show that every z-dense ideal of L is order dense. 7. If (L,z) is an Archimedean locally solid Riesz space, show that there exists a locally solid topology z* on L' that induces z on L. 8. Let (L,r)be a Hausdorff locally solid Riesz space, and let L' and denote the topological duals of (L,z) ahd respectively. For every cp E L' let $ be the unique continuous extension of cp to (see Theorem 4.4). Show that cp H $3 is a Riesz isomorphism from L' onto 9. Show that a Hausdorff locally solid Riesz space ( L J )is sequentially z-complete if and only if every order interval of L is sequentially z-complete and each increasing z-Cauchy sequence of L+ is r-convergent. 10. Let L, be a seminormed Riesz space and { u n } E L + a disjoint sequence. Show that lim p ( u - u A u,) = p ( u ) for every u E L'. 11. (Luxemburg). Show that every seminorm p on a finite dimensional Archimedean Riesz space L that satisfies p(u) = p(1ul) for all u E L is a Riesz seminorm. (Compare this with Example 5.3.) 12. Let L be the Riesz space consisting of all real sequences with only a finite number of nonzero terms, equipped with the sup norm. Show that L' = 11, and give an example of some cp E L: with cp $ L . 13. Show that if the topological dual L' of a Hausdorff locally convexsolid Riesz space (L,z) admits an order unit, then lal(L,L') is generated by a Riesz norm and z = lol(L,L'). 14. Consider L = C[O,l] equipped with the Hausdorff locally solid topology z generated by the family of Riesz seminorms { p x : xE [OJ]} u { p } , where p,(u) = lu(x)( and p ( u ) = JA lu(x)l dx for all u E C[O,l] and x E [OJ]. Determine [Hint: Show that = R[O,'] x L,([O,l]).]

(e,?),

(e,?).

e

(e)'

e (e)'.

EXERCISES

51

15. Let (L,t) be a locally solid Riesz space. If L is separable (i.e., if it contains an at most countable t-dense subset) show that L' is super Dedekind complete. 16. Let L be an Archimedean Riesz space. Consider L , and (La), identified as in Theorem 3.12. If icp,) c ( L ; ) + show that (cp,,] converges to cp for o(L;.L) if and only if (cp,] converges to cp for CJ(L,,L'). ( H i n t : Approximate an element of L' from above and below in the order sense.) 17. Let (L,t) be a locally solid Riesz space. If L+ has a nonempty interior, show that L' = L'.

CHAPTER

3 Lebesgue and Pre-Lebesgue Topologies

In this chapter we shall consider locally solid topologies satisfying an additional continuity property. A Lebesgue topology T will be defined to be a locally solid topology satisfying u, 1,0 whenever u, J, 8; and a preLebesgue topology will be a locally solid topology T such that {u,,] is a u, t I u holds. z-Cauchy sequence, whenever (I I These additional properties on the topology will impose a number of interesting conditions on the structure of the Riesz r-sce. We shall see, for instance, that the Lebesgue property is equivalent to the property that every topologically closed ideal is a band and that the pre-Lebesgue property becomes equivalent to the property that every disjoint order bounded sequence of the Riesz space converges topologically to zero. On the other hand, a number of results will be obtained relating the above properties to the Dedekind and topological completeness. 8.

EXAMPLES AND PROPERTIES OF LEBESGUE TOPOLOGIES

The relation so far between order and topology has been through the locally solid topologies. In this section we shall describe a number of properties connecting order and topological continuity. These properties are described in the following definition. Definition 8.1. Let (L,T)be a locally solid Riesz space. Then we say that:

(i) (L,z) satisjies the o-Lebesgue property (or that topology) if u, I e in L implies u, 1,8. 52

T

is a a-Lebesgue

Sec. 81

LEBESGUE PROPERTIES

53

(ii) (L,z)satisfies the Lebesgue property (or that z is a Lebesgue topology) if u, 18 in L implies u, 1,O. (iii) (L,z) satisfies the pre-Lebesgue property (or that z is a pre-Lebesgue topology) $ 8 I u, 7 I u in L implies that { u,} is a z-Cauchy sequence of L. It should be clear from the above definition that a locally solid Riesz space (L,z)satisfies the a-Lebesgue property if and only if u, 1,8 whenever u, 5 8, (L,z) satisfies the Lebesgue property if and only if u, 8 whenever u, 8, and (L,z)satisfies the pre-Lebesgue property if and only if {u,] is a z-Cauchy sequence of L whenever 8 I u, 4 holds. Note. The a-Lebesgue property seems to have first appeared in the works of Kantorovich [37, p. 1531. For normed Riesz spaces Nakano was the first to make a systematic study of these properties. In his terminology a Riesz norm with the a-Lebesgue property is referred to as a continuous norm, a Riesz norm with the Lebesgue property as a universally continuous norm, and a Riesz norm with the pre-Lebesgue property as a strictly monotone norm; see [50, pp. 126-1341. In the Luxemburg-Zaanen terminology the a-Lebesgue, Lebesgue, and pre-Lebesgue properties are referred to as the (A,$ (A,$, and (A,iii) properties, respectively [45, Note X, p. 5011. The above terminology is due to Fremlin [25]. The a-Lebesgue and Lebesgue names are justified by the well-known Lebesgue dominated convergence theorem from the theory of integration. For a justification of the preLebesgue name see Theorem 10.5.

We next illustratk ‘ h e above properties with examples. Example 8.2. Take L to be the Riesz space consisting of all continuous functions on [O,l] with the pointwise ordering, i.e., L = C[O,l]. Pick 1 I p < 00 and let z be the Hausdorff locally solid topology generated by the Riesz norm

Then z is a pre-Lebpsgue topology on L. Indeed, if 6’ I u, 7 I u holds in L then u,(x) u(x) < co holds for each x E [O,l]. Clearly u E L,([O,l]) and so P(un+k - U n ) I

P(Un+k

- 0)

+ P(un - 0 )

+

0

as n,k + m. This shows that {u,) is a z-Cauchy sequence. Note that the of (L,z) is Lp([0,1]) with z^ being the topology topological completion generated by the L,-norm. It is easy to see that z is not a a-Lebesgue topology. Actually, a stronger result holds: There is no HausdorfSlocally solid topology on C[O,l] with the a-Lebesgue property.

(e,?)

54

[Chap. 3

LEBESGUE TOPOLOGIES

To see this assume that there exists a Hausdorff locally solid topology

z on C[O,l] with the o-Lebesgue property. Let {r1,r2, . . .} denote all rational

numbers of [0,1]. Now for every pair m,n of positive integers, there exists urn,,E C[O,l] such that (1) 0 I u,,,(x) I 1 for all x E [0,1]. (2) = 1 for all m,n. (3) u,,,(x) = 0 for x E [0,1] with Ix - r,, > 2-"m-'. Set urn,, = ~ u p { u , + ~=: i1 , . . . ,n). Then

t

L',,,

for each fixed m and since

urnJrn)= 1, umSnt e holds in L for each fixed m, where e(x) = 1 for all

x E [0,1].

n

Now let V be a solid z-closed neighborhood of zero. Next choose a sequence {K} of solid z-neighborhoods of zero such that V,,, + V,,, G V, holds for all n and with V, = V . Now since urn,, t e and z has the cr-Lebesgue n

property, urn,, 2e for each fixed rn. Thus for each positive integer m, (n-a) there exists n, with e E V,, Observe now that by condition (3) the set {x E [0,13:um,,(x)> 0} is at most of Lebesgue measure 2m-' for every n, and in particular for n = n,. Now put w, = inf{u,,nrn:m= 1, . . . ,n} for n = 1,2,. . . . Then w, and the set {x E [O,l]:w,(x) > 0} is at most of Lebesgue measure 2n- '. Hence w, 1 8 holds in L and so w, 8. But

,.

8se-wn=sup{e-v,,nm:l < m < n } < G

n

1 (e-urn,,,)EV2+...+

m= 1

K,,

V, = V .

Therefore, e E V for every solid z-closed neighborhood I/ of zero, that is, e = 8, which is a contradiction. This establishes the claim that C[O,l] does not admit any Hausdorff a-Lebesgue locally solid topology. Example 8.3. Again take L to be C[O,l] and let z be the Hausdorff locally solid topology generated by the Riesz norm p ( u ) = sup{lu(x)I:xE [0,1]},

2.4 E

L.

It can easily be seen that (L,z)is z-complete without having the pre-Lebesgue property. Example 8.4. Once more let L = C[O,l], and let z be the Hausdorff locally solid topology of pointwise convergence, ie., the topology generated by the family of Riesz seminorms

P A 4 = lu(x)(,

u E L,

x E [OJ].

Sec. 81

55

LEBESGUE PROPERTIES

However (L,T) does satisfy the pre-Lebesgue property, and ( L , T ) is not 7-complete. Note that L = R[o~’land that z^ is the topology of pointwise convergence in R[o*’l. Example 8.5. Let L = 1,(X) where X is an infinite set, i.e., let L be the Riesz space of all bounded real-valued functions defined on X , with the pointwise ordering. Again let T be the Hausdorff locally solid topology of pointwise convergence. Then z is a Lebesgue topology. Note also that L = RX with z^ being again the topology of pointwise convergence. Example 8.6. Let 0 < p < co,and let L be the Riesz space consisting of all the real-valued, Lebesgue measurable functions f defined on [O,l] such that JA If(x)lPdx < co,with the pointwise ordering. Note that the elements of L are functions and not equivalence classes, that is, functions differing at one point are already considered to be distinct. Now given a positive integer n, E > 0, and a finite subset F of [0,1], define the set

W,,,,,

= {f E L :

J:

If(x)lPdx < n - ’ and If(x)l < E for all x

E

I

F .

As F runs over the finite subsets of [0,l], n over the natural numbers, and E over (O,co), we obtain a family of sets { W,,,,,} that is a filter basis for a linear topology z on L . Clearly each W,,,,, is a solid set. It should also be clear that z is a Hausdorff topology. Thus, (L,z) is a Hausdorff locally solid Riesz space. Note that T is actually the “supremum topology” of the topology of pointwise convergence and the one generated by the “L,-norm.” In other j Iua(x)IPdx 0. words, u, 1,8if and only if u,(x) -,0 for each x E [0,1] and A Observe also that for p 2 1 the topology T is a locally convex-solid topology; (L,T)has the following interesting properties. (1) L is a-Dedekind complete but not Dedekind complete. --f

Indeed, if 8 I u, t I u holds in L, then u,(x) t u(x)holds for each x E [0,1]. It follows that u E L and that u, 1u holds in L . To see that L is not Dedekind complete notice first that u, t u in L implies u,(x) T u(x) for all x . Now consider a nonmeasurable subset E of [O,l], and define the net u, = x,, for all finite subsets c1 of E . Then {u,} c L, 0 5 u, I e, where e ( x ) = 1 for all x , and u,(x) t xE(x)for all x . It follows easily that sup{u,} does not exist in L .

r

(2) (L,z) satisfies the a-Lebesgue and pre-Lebesgue properties but not the Lebesgue property. Indeed, if u, 1 f3 in L, then u,(x) 1 0 for each x E [0,1], and the a-Lebesgue property follows from the Lebesgue dominated convergence theorem. For the pre-Lebesgue property note that if f3 s u, t 5 u, then u, u must hold

r

56

LEBESGUE TOPOLOGIES

[Chap. 3

in L for some u and so u, 1,u, from which it follows that { u,} is a z-Cauchy sequence. To see that (L,t) does not satisfy the Lebesgue property consider where M runs over the set of all finite subsets of [O,l] directed the net by M 2 p if M 2 p. Note that 8 I x, t e holds in L but xa e does not hold, since lt le(x) - ~ , ( x ) I P d x= 1 for all a.

{x,},

(3) L contains a z-closed a-ideal that is not a band.

Let A = { u E L : u = 0 a.e.}. Obviously A is a a-ideal of L. To see that A is z-closed let u, 5 u in L with {u,} E A. Then we have, in particular, Jollu(x)lPdx= Jollu(x)- u,(x)IPdx -, 0,

from which follows A j lu(x)lPdx= 0, so that u E A. To see that A is not a band of L use the net {x,} of (2). (Compare this with Theorems 8.7 and 8.9.) (4) (L,z)is sequentially z-complete but not z-complete. To see that (L,z) is not z-complete use the net {x,} of (2). The sequential z-completeness follows from the Lebesgue dominated convergence theorem. Next we determine the topological completion (L,?) of (L,z).Consider the Rieszspace K = R[O*'] x Lp([0,1]), where R[O.']is equipped with the topology of pointwise convergence, and Lp([0,1]) is equipped with the topology generated by the "L,-norm." Note that K , if equipped with the product topology, is a topologically complete Hausdorff locally solid Riesz space. Define now the mapping n:L -P K by n(u)= (u,[u]),where [u] denotes the equivalence class determined by u in L,([O,l]). It is not difficult to see that n is a Riesz isomorphism (into) whose image n ( L ) is topologically dense in K . Hence = K . Observe also that since, for 0 < p < 1, (L,?) is not locally convex-solid, neither is (L,z). [For the computation of the topological dual L' of (L,T)see Application 16.11.1 Note. The result that C[O,1] does not admit any Hausdorff a-Lebesgue locally solid topology is essentially due to Luxemburg and Zaanen [45, Note VII, Example 24.5(ii), p. 6741. Example 8.6 is due to Aliprantis [l, Example 2.5, p. 4451; the proof about the topological completion of this example is due to L. C. Moore, Jr. (personal communication).

We continue by characterizing the Lebesgue and 0-Lebesgue properties. Theorem 8.7. For a locally solid Riesz space (L,r) the following statements are equivalent:

(i) (L,z)satisjes rhe Lebesgue property. (ii) Every z-closed ideal of L is a band.

Sec. 81

57

LEBESGUE PROPERTIES

Proof. (i) * (ii) Let A be a z-closed ideal of L, and let f3 I u, t u in L with (u,) G A . Then u, 1,u and so u E A , that is, A is a band of L . (ii) * (i) Assume that 8 i u, u holds in L and that U is a solid Tneighborhood of zero. Pick another solid z-neighborhood V of zero with V + V c U . Now pick 0 < 6 < 1 such that (1 - 6)u E V . For each a let v, = (u, - 6u)+, and let A be the ideal generated by {u,} in L. Since 8 I v, t (1 - 6)u holds in L, it follows that u E ( A } (the band generated by A in L). But by hypothesis the closure A of A in L [which is an ideal by Theorem 5.4(iii)] is a band of L and hence { A ) E A. In particular, it follows that u E A and thus (u + V) n A # Pick 0 I w E A such that u - w E V . Now with 8 I w i nvp. since w E A, there exists some positive integer nand some /l But then (us - 6u)- A vs = 0 implies (us - 6u)- A w = 6, so that (up - 6u)W A U = ( u s - 6u)- V ( W A U ) I u and hence 0 I (us - 6u)- i u - U A W I Iu - wI E V . Thus (us - 6u)- E V. Now for a 2 p we have

a.

+

e I - u, I u - up = (1 I ( 1 - 6)u

Thus u - u, finished.

E

+ (us

-

U holds for all a 2

+ (6u - up) 6u)- E v + vG u. -

61u

p, that is,

u,

5u

and the proof is

For Archimedean Riesz spaces we can add another characterization of the Lebesgue property. The details follow. Theorem 8.8. Lei (L,z) be an Archimedean locally solid Riesz space (in particular a HausdorfSlocally solid Riesz space). Then the following statements are equivalent: (i) (L,T)satisfies the Lebesgue property. (ii) Every order dense ideal of L is z-dense. (iii) Every z-closed ideal of L is a band.

Proof. (i) * (ii) follows immediately from Theorem 1.14 and (iii) * (i) was proven in the previous theorem. Thus, to complete the cycle we have to show that (ii) * (iii). To this end let A be a z-closed ideal of L . By Theorem 1.12, A @ Ad is order dense in L, and hence by hypothesis A 0 Ad is z-dense in L. Thus if u E Add, then there exists a net {u, + v,) of A @ Ad with u, + u, A u. Now since Iu, + u, - uI = Iu, - uI + u, 1,u and so u E A = A . Thus A = Add, and the proof is finished. 1

Iv,~,

As a sequential analog of Theorem 8.7, we have the following result, whose proof is similar and is omitted.

58

LEBESGUE TOPOLOGIES

[Chap. 3

Theorem 8.9. Let (L,T)be a locally solid Riesz space. Then the following statements are equivalent :

(i) (L,T)satisfies the o-Lebesgue property. (ii) Every r-closed ideal of L is a o-ideal. The next result shows that the quotient topologies always inherit the pre-Lebesgue property. Theorem8.10. Let (L,T)be a locally solid Riesz space with the preLebesgue property and let A be an ideal of L. Then ( L / A , z / A )satisfies the pre- Lebesgue property. Proof. Assume that 0 I ti, J holds in L / A . Put w, = inf{u,+ : 1 I k I n } for all n and note that k,= li, for n = 1,2,. . . and 8 I w, 1 holds in L. It now follows from the pre-Lebesgue property of (L,z)that { w,} is a z-Cauchy sequence. From this it follows that {Li,] is a z/A-Cauchy sequence of L / A and the proof is finished. Quotient Riesz spaces of Archimedean Riesz spaces always inherit the Lebesgue propety. The details follow. Theorem 8.11. Let (L,T)be an Archimedean locally solid Riesz space with the Lebesgue property. Then for any ideal A of L the quotient locally solid Riesz space (L/A,z/A)satisfies the Lebesgue property. Proof. Assume that tia 4 8 holds in L/A. Consider the net { v,]consisting of the finite infima of the net (m,+). Then it should be clear that 8 I u, J holds in L and ii, 4 8 in L/A. In particular, note that if u E L satisfies 0 Iu I u, for all 1,then b = 0 and so u E A. On the other hand, note that since L is I u, Archimedean, there exists an increasing net {wo)of L f with 8 I wo holding for each 1 and with ui. - wp J 8 in L [Theorem 1.15 (iv)]. By the (,.a) Lebesgue property of (L,T), we now infer that u, - wp 8 and since li, = ir, - w p , we get 6,

that li,

d.4 +

-, 8. Now it follows from the construction of

r/A

8 and the proof is finished.

{u,}

H

Note. Theorems 8.7 and 8.9 are essentially due to T. And6, the proof of Theorem 8.7 can be found in [44, Note XIV, Theorem 47.3, p. 2441. Theorems 8.10 and 8.11 are due to Langford and Aliprantis [42]. If {(L,,z,)} is a family of locally solid Riesz spaces such that each topology

z, is a Lebesgue topology, then it is easy to see that the product topology

on L = IIL, is likewise a Lebesgue topology. A similar result holds if each (La,za) satisfies the o-Lebesgue or the pre-Lebesgue property. Thus we have the following theorem. T = IIT,

Sec. 91

59

LOCALLY CONVEX LEBESGUE TOPOLOGIES

Theorem 8.12. I f every member of a family of locally solid Riesz spaces {(Lu,zu)}satisjies the Lebesgue (resp. the o-Lebesgue or the pre-Lebesgue) property, then (llL,,llz,) satisjies also the Lebesgue (resp. the o-Lebesgue or the pre-Lebesgue) property.

9.

LOCALLY CONVEX-SOLID TOPOLOGIES RND THE LEBESGUE PROPERTY

Recall that given a Hausdorff locally convex space (E,z),the strong topology /?(E',E)on its topological dual E' is the topology of uniform convergence on the z-bounded subsets of E (a subset A of E is said to be z-bounded if for every z-neighborhood V of zero, there exists 13 > 0 such that ;LA G V ) . The strong topology P(E',E) is the Hausdorff locally convex topology on E' generated by the family of seminorms { p e : B E g}; where is the collection of all z-bounded subsets of E and ps('p) = sup( l'p(u)l:u E B ) for each B E B. The bidual E" of E is the topological dual of (E',P(E',E)). If now (L,z) is a Hausdorff locally convex-solid Riesz space then to of Iall s generate /?(L',L)it is enough to restrict ourselves to the collection @ r-bounded solid subsets of L. In this case each p A becomes a Riesz seminorm. I I'pI in L'. Then by Theorem 3.3 and the To see this, assume A E BSand solidness of A , we have for every u E A I$(c)l I I$l(lul) I l'pl(lul) = suPcI'p(u)l:IuI I Iul}

I PAP)

and hence p A ( $ ) I pA('p). so that p A is a Riesz seminorm. Thus /?(L',L)is a Hausdorff locally convex-solid topology on L', and hence L" is an ideal of (Ly. There is a natural embedding u H u' of L into the bidual L" [that is, the topological dual of (L',P(L',L))] defined by ii('p) = ' p ( u )

for cp E L'.

By the Hahn-Banach theorem it is easy to see that u H ii is an algebraic isomorphism (into)that is also a Riesz isomorphism. To see this let 0 I cp E L , u E L, and then use Theorems 3.5 and 3.3 to obtain rv

Ll+('p)

N

= ' p ( U + ) = sup($(u):$ E L',

0I $I 'p} = sup(u'(l)):$ E L', 8 I $ 'pj = I (u')+(cp),

that is, u+ = (E)' ; see also Theorem 3.11. In the sequel we shall consider L embedded in L" according to the above Riesz isomorphism u H ii and will make no distinction between an element of L and its image in L". Note also that under this identification we have

60

LEBESGUE TOPOLOGIES

[Chap. 3

L G (L');.Indeed, if cp, J 0 holds in L', then u(cp,) = cp,(u) 1 0 holds for all E L + by Theorem 3.3 and the result follows. Regarding the Lebesgue property in Hausdorff locally convex-solid Riesz spaces we have the following characterizations. II

Theorem 9.1. For a Hausdorfl locally conuex-solid Riesz space (L,z) the following statements are equiualent : (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)

(L,z)satisjes the Lehesgue property. L' E L;. Euery band of L' is o(L',L)-closed. L (embedded as above) is a regular Riesz suhspace of L". L is an order dense Riesz suhspace of (L');. T h e null ideal N , is a hand of L for each cp E L'. Euery order dense ideal of L is 5-dense. Euery 5-closed ideal of L is a band.

-

Proof. (i) =. (ii) Obvious. a(L',L) cp (ii) + (iii) Let B be a band of L', and let {q,} G B be such that cp, holds in L'. Since L' is Dedekind complete (Theorem 5.7), we can write cp = cpl + cp2 with cpl E B and cp2 E Bd. It follows that the net {$,), where a(L' L ) $, = cp, - cpl E B, satisfies $,cp2. Now if cpz # 0 then either cp2- # 0 or q 2 +# 0; we shall assume the latter. Since cp2' E Lz, we have CB2+# ( 0 ) . So, choose 0 < u E Co2+. But then by Theorem 3.10 we get cp2-(u) = 0 and $,(ti) = 0 for all a, and consequently it follows from lim$,(u) = cp2(u)= cp2+(u)- cp2-(u) that cp2'(u) = 0, in contradiction with the choice of u. Hence cp2 = 8 and this shows that the band B is o(L',L)-closed. (iii) + (iv) Let u, 1 0 in L and assume that 0 < cp Iu, holds in L" for all a and some cp. Fix an index a and let u = u,; thus 0 < cp I u holds in L". Next choose 0 < 6 < 1 with $ = (cp - 6u)' > 0 and then select 0 < f E C,. Observe that f # Bfd ( B f = the band generated by f in L') and that Bfd is o(L',L)-closed by hypothesis. It now follows from the Hahn-Banach theorem that there exists u E L such that f ( u ) = 1 and g ( u ) = 0 for all g E Bfd; we can assume that u > 0. Put w = u A 6u and note that w # 8. [Indeed, if w = 0 then U A U = 8 and so U A$ = 0, since 8 < $ I u ; but then Theorem 3.10 implies f ( u ) = u ( f ) = 0, a contradiction.] Now set = ( w - cp)' and note that since 8 I (w - cp)' I (6u - cp)' = (cp - 6u)-, (w - cp)' A $ = 8. Now for every h E C,, we have h If and so (by the choice of u above) h(u) = 0. Consequently, if 8 I h E C , , , then we have 0 I $,(h) I w(h) I u(h) = h(u) = 0 and so $,(h) = 0. But then C,, = { O } , and therefore = 8. Hence 0 < w I cp holds and thus 0 < w I u, holds in L for all a, a contradiction. Thus u, -1 0 holds in L", so that L is a regular Riesz subspace of L".

Sec. 91

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61

(iv) => (v) Assume u, 1 8 holds in L. Then by hypothesis we have u, 1 8 in L" and so u,(cp) = cp(u,) J 0 holds for every 8 Icp E L ' . It now follows that L' G L; and the conclusion follows from Theorem 3.11. (v) * (vi) Let 8 I cp E L ' , and let 8 Iu, t u in L with ( u a }E N,. Then 8 I u, u also holds in L", and therefore 0 = cp(u,) = u,(cp) t u(cp) = cp(u) also holds. It follows that cp(u) = 0 and so u E N,. Hence N , is a band of L. (vi) * (vii) Let A be an order dense ideal of L, and let A be its z-closure. If A # L, then there exists 8 < u E L with u A. Now by the Hahn-Banach theorem there exists cp E L ' with cp(u) # 0 and q ( u ) = 0 for all u E A . But since A is an ideal A c N , holds and since A is order dense in L and by hypothesis N , is a band of L , N , = L ; so that cp = 8, a contradiction. This shows that A = L, and we are done. (vii) * (viii) and (viii) * (i) follow by Theorem 8.8 and hence the proof of the theorem is now complete. The next result follows immediately from the above theorem and Theorem 2.2. Theorem 9.2. Let &,T) be a Hausdorff Dedekind complete locally convexsolid Riesz space. Then the following statements are equivalent;

(i) (L,z)satisjes the Lebesgue property. (ii) L is an order dense ideal of (L');. Note. Most of the equivalent statements of Theorem 9.1 have been studied by a number of authors. They have appeared (with a different terminology) in a paper by Kawai [39, Theorem 5.1, p. 2991 and at the same time in the works of Schaefer [64]. Statement (iii) of Theorem 9.1 is due to Luxemburg and Zaanen [45, Note XI, Theorem 36.2, p. 5171. Other places where similar results can be found include [56, pp. 90-1001, [25], and [71, Chapter 131.

As an application of the last result we shall characterize the perfect Riesz spaces. To do this we need their definition and some preliminary discussion. For a Riesz space L let us denote the Dedekind complete Riesz space (L;): by L,,. By Theorem 3.11 there exists a natural Riesz homomorphism u H C from L into L,, defined by i i ( f ) = f ( u ) for ,f E L;. The Riesz space L is said to be a perfect Riesz space if u H ii is an onto Riesz isomorphism (from L onto L,,). Obviously every perfect Riesz space must be Dedekind complete. Definition 9.3. A locally solid Riesz space (L,T)is said to satisfy the Levi property (or that T is a Levi topology) if every increasing z-bounded net qf L f has a supremum in L.

62

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[Chap. 3

By replacing the word “net” with “sequence” in the preceding definition, we obtain the a-Levi property. Clearly every locally solid Riesz space with the Levi (resp. a-Levi) property must be Dedekind (resp. a-Dedekind) complete. Also a a-Levi topology on a Riesz space must be Hausdorff (see Exercise 1 at the end of this chapter). We are now ready to characterize the perfect Riesz spaces. Theorem 9.4. For a Riesz space L the following statements are equivalent:

-

(i) L is a perfect Riesz space. (ii) The absolute weak topology lal(L,Lz)i s a Levi topology. Proof. (i) (ii) Assume that 0 I u, t holds in L with {u,} lal(L,Lz)bounded. Then F ( q ) = sup{cp(u,)J < 00 holds for all 0 I cp E L i , and as is easily checked, F : (L;)’ -+ R + is additive. Therefore, by Lemma 3.1 F is extendable as an order bounded linear function to all of L i , that is, F E (Lz)-. Now observe that if each u, is identified with its image in L,,, then 0 5 u, 7 F holds in (LJ“, and so F E L,, [since L,, is a band of (LL)-].Thus F = u holds for some u E L, since L is perfect. But then u, t u holds in L, and we are done. (ii)*(i) Observe that (L,[al(L,L;))is a Hausdorff locally convexsolid Riesz space with the Lebesgue property, whose topological dual is L i (Theorem 6.6).It is easy to see now that u H u‘ is a Riesz isomorphism from L into L,, and so L (identified with its image) is an order dense ideal of L,, by Theorem 9.2. We need only to show that L = L,”. To this end let 0 I F E L,,. Pick a net {u,} of L+ with 0 I u, 7 F in L,,. Since IcpI(u,) = uz(IcpI) I F(lcp1) holds for all c( and all ~ I E L ; (u,) , is Ial(L,L;)-bounded and hence by hypothesis, 8 I u, u holds in L for some u. But then since L is order dense in L,,, u, u also holds in L,, and hence F = u, that is, L = L,,. Thus, L is a perfect Riesz space, and we are done.

r

In the following results two classes of perfect Riesz spaces are described. Theorem 9.5. For every Riesz space L its order dual L“ is a perfect Riesz space. Proof. According to Theorem 9.4 we have to show that IcTI(L-,(L-);) is a Levi topology. To this end note that if 8 I (pa t holds in L’ with (qI,)IaI(L-,(L-);;)-bounded, then q ( u ) = sup{q,(u)} < co holds for all u E L+ and cp:L+-+ R + is additive. Thus by Lemma 3.1 cp can be extended to an order bounded linear functional on L, that is, cp E L‘. It follows now easily that cp, cp holds in L- and so lal(L-,(L”);) is a Levi topology.

r

Theorem 9.6. Ever)’ band of a perfect Riesz space i s a perfect Riesz space. Proof. Let A be a band of a perfect Riesz space L. We have to show that lal(A,Az) is a Levi topology. To this end note first that every cp E L,

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63

restricted to A is a normal integral of A. Now if 0 I u, t holds in A with (u,) lal(A,Ai)-bounded, then it follows from the above observation that (u,} is lal(L,LJ-bounded, and hence since L is a perfect Riesz space u, t u holds in L for some u. But then since A is a band, we have u E A and u, t u in A . The proof is now complete. H Note. Levi topologies were first considered by Nakano in [51]. The term “Levi topology” is due to Fremlin [25,231 Definition, p. 491, although his definition is slightly different than Definition 9.3; the terminology is justified by Levi’s theorem from the theory of integration. Theorems 9.4-9.6 are due (with a different terminology) to Nakano [50, $241; see also [45, Note VIII].

The sequential analogue of Theorem 9.1 is presented next. Theorem 9.7. For a Hausdorff locallj~convex-solid Riesz space (L,z) the following statements are equivalent : (i) (ii) (iii) (iv) (v)

(LJ) satisfies the a-Lebesgue property. L‘ c L,. L is LI a-regular Riesz subspace of’ (I.’);. T h e null ideal N, is a a-ideal of L for each cp E L‘. Every 7-closed ideal of L is a a-ideal.

Proof. (i) * (ii) Obvious. (ii) =-(iii) Assume u, 1 0 in L, then cp(u,) 1 0 holds for each 0 I cp E L’ and so u,, 1 0 holds also in (L’);, that is, L is a a-regular Riesz subspace of (L’);. (iii) (iv) Assume 9 5 cp E L’ and {u,) G N , with 8 5 u, t u in L. Since u, u holds in (L’); by hypothesis, it follows that 0 = cp(u,,) t cp(u). and so cp(u) = 0. Hence LI E N , and therefore N , is a a-ideal. (iv) e.(v) Let A be a 7-closed ideal of L, and let {u,} s A with 0 Iu, t u in L. If u $ A , then by the Hahn-Banach theorem there exists cp E L’ such that cp(v) = 0 for all u E A and with cp(u) # 0. But since A is an ideal of L, it follows from Theorem 3.3 that A s N , and so (since N , is a a-ideal) u E N , . Thus cp(u) = 0, a contradiction. Therefore, A is a a-ideal. (v) => (i) Use Theorem 8.9. H

The final result of this section deals with a useful continuity property of the locally convex-solid Riesz spaces. Theorem 9.8. Let (L,z) be u Huusdorff locally conuex-solid Riesz space and let u, 0 in L. Then the following statements are equivalent:

(i) u,&o. O(L,L’) (ii) u,0.

64

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Proof. (i) + (ii) Obvious. (ii) (i) Note first that if (ii) holds, then zero belongs to the a(L,L’)closure ofthe convex hull Co(u,} of {ua}.But by Theorem 4.3(ii) the o(L,L’)closure and z-closure of Co{u,} coincide, and hence zero is in the z-closure of CO{U,}. Now let V be a solid z-neighborhood of zero. By the previous remark, V n Co{u,) # 0, and so there exists a finite number of indices, say a l , . . . ,a,, and positive constants Al, . . . ,I, with Am = 1 and such that Amu,, E V. Now pick an index /3 such that B 2 aifor i = 1, . . . ,n and note that since

ym=

ym=

0 I up = we get u,

E

V for all a 2

10.

n

n

m= 1

m= 1

1 A m U p I 1 AmU,_E v,

fi, so that u, 1,0, and the proof is finished. W

LEBESGUE PROPERTIES AND Lp-SPACES

We recall that by a disjoint sequence of a Riesz space L we mean a sequence {u,} of L satisfying A IumI = 0 if n # m. In general, we shall call a sequence {u,,} of a Riesz space a k-disjoint sequence (where k is a positive integer) if we have inf{ luil:i E I } = 0, whenever I is a set of positive integers with at least k elements. The following important result characterizes the pre-Lebesgue property in terms of disjoint and k-disjoint sequences.

Iu,~

Theorem 10.1. For a locally solid Riesz space (L,z) the following statements are equivalent :

(i) (ii) (iii) (iv)

(L,z)satisjes the pre-Lebesgue property. I f 0 I u, I u holds in L. then (u,} is a z-Cauchy net of’ L. Every order bounded disjoint sequence of L is 7-convergent to zero. Every order bounded k-disjoint sequence of L is z-convergent to zero.

Proof. (i) + (ii) Let 0 I u, I u in L. If {u,} is not a z-Cauchy net, then there exists a r-neighborhood V of zero-such that given any a there exists a pair of indices /3,y 2 a with up - uy # V . Choose (inductively) a sequence {a,} of indices with a, I a,+ for all n and such that ua,+, - umn# V. But then 0 5 uant I u and (u,,} is not a z-Cauchy sequence, in contradiction with our hypothesis. (ii) e-(iii) Let { u n )be an order bounded disjoint sequence of L. Then w, = luil = sup{(uil:i= 1,. . . ,n} ( n = 1,2,. . .) is an increasing order bounded sequence of L+ and hence ( w , } is a z-Cauchy sequence of L. It

I;=,

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65

follows now easily that Iu,, 1,8 and from this (since z is a locally solid topology) that u, 1,0. (iii) * (iv) We shall use induction on k. Since the only l-disjoint sequence is the zero sequence, the statement is true for k = 1. Assume that every order bounded k-disjoint sequence of L is z-convergent to zero. Let {u,) be a (k 1)-disjoint sequence of [Q]. Put u1 = 8 and u, = (u, - n 111: ui- n - 'u)+ for n 2 2. Clearly 8 I u, I u for all n and we claim that {u,} is a disjoint sequence of L. For if m > n > 1, then the chain of inequalities

+ /

(

Im - ' u -

m- 1

\t

n- 1

1 u i - u, -m-'u

i= 1

n-lu

n- 1

+ n 1 ui- u,)+

)+-(

< n-'u-

= (u,, - n

n- 1

1 ui-

i=l n- 1

1 ui- n-1u

11,

>'

i= 1

i= 1

implies u, A u, = 8.Hence by hypothesis u, 1,8. Now put w 1 = 8 and w, = u, A ( H ui)for n 2 2 and note that {w,} is a k-disjoint sequence of [8,u]. Indeed, for the k integers 1 I n , < . . . < nk we have trivially inf{wni:i= 1,. . . ,k} = 8 if n , = 1, and if 1 < n , then

1;:;

0 Iinf{wni:i= 1,. . . ,k) Iu,,

A . . . A U,,A i= 1

nl-1

I n1

1 (un1

A.

. . A u,,

A

ui)= 8,

i= 1

where every term of the last sum is zero by the ( k + 1)-disjointness of {u,}. Thus by the induction hypothesis w, 1,0. Now for n 2 2 the inequalities 0
i= 1

= u,

+ w, + n-'u

imply u, 1,0. The induction is now complete and (iv) follows. (iv) (i) Assume that 8 < u, t i u holds in L. If {u,} is not a z-Cauchy sequence, then (by passing to a subsequence if necessary) we can assume that there exists a solid z-neighborhood V of zero such that u,+ - u, I/ for all n. Next choose a solid z-neighborhood W of zero with W + W G V and then pick a positive integer k such that k-'u E W.

[Chap. 3

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66

Define now the sequence w, = (u,+ - u, - k-'u)+ ( n = 1,2,. . .)and note that (w,> is a k-disjoint order bounded sequence of L + . Indeed, observe first that 0 I w, I u holds for all n and if I is any set consisting of k positive integers then since w = inf { wi: i E I } satisfies

0=wA =wA

is1

,

(ui+ - ui - k-'u)- = w A

1(k- ' u + ui - ui+

1)+

is1

C ( k - ' u + ui - ui+, + wi) = W A

(ui+ - ui) +

iel

C wi)

iel

we get w = 0, so that ( w , } is a k-disjoint order bounded sequence. Now from our hypothesis we get w, 1,0 and thus w, E W for all n 2 n,. But we then have U , + ~ - U , = W , + ( U , + ~ - U , ) A ~ - ~ ~ E W +forall W C I /n > n , ,

a contradiction. This contradiction establishes (i) and completes the proof of the theorem. Note. A version of Theorem 10.1 was proven for Banach lattices in [49], see also [67, 5.13 Lemma, p. 921. The general result was established in [25], 24H Proposition, p. 561. The above simple proof is due to the authors [8].

The next two results deal with the different relations among the Lebesgue properties. Theorem 10.2. For Archimedean locally solid Riesz spaces the Lehesgue property implies the pre- Lehesgue property. Proof. Let (L,?)be an Archimedean locally solid Riesz space with the Lebesgue property and let 0 I u, T 2 u in L. Since L is Archimedean there exists a net {u,) of L such that u, - u, 1 0 holds in L (Theorem 1.15). But (a,,)

then since z is a Lebesgue topology u, - u, easily that {u,) is a 7-Cauchy sequence. H

-&0, and from this it follows

Observe that Example 8.2 shows that the pre-Lebesgue property does not imply the Lebesgue property even in the Archimedean case.

Theorem 10.3. For a Hausdorff ?-complete locallj~solid Riesz space (L,?) the following statements are equivalent : (i) ( L , t ) satisfies the Lebesgue property. (ii) (L,?) satisfies the pre-Lebesgue property. In this case it follows that L is Dedekind complete.

-

Sec. 101

L,-SPACES

67

Proof. (i) (ii) It follows immediately from the previous theorem by observing that L is Archimedean. (ii) 3 (i) Let u, 1 0 in L. Then {u,) is a z-Cauchy net of L, according to Theorem 10.1, and hence (since L is z-complete) u, 1,u holds in L for some u. Now by Theorem 5.6(iii) we have ti, 1 u and so u = 19.Thus u, 1,8, and we are done. To see that in this case L is Dedekind complete, let 6, 5 u, 5 u in L. Then {u,) is a z-Cauchy net of L and hence r-convergent in L, say u, A L’. It now follows that 11, t 11 holds in L, and thus L is Dedekind complete.

r

The next example shows that the a-Lebesgue property does not imply the pre-Lebesgue property.

Example 10.4. Let X be an uncountable set and let L be the Riesz space consisting of all real-valued functions .f defined on X with the property that there exists a real number .f(a)such that the set {x E X : If(.x) - .f(co)(> E ) is finite for every E > 0. In other words, L = C(X,,), where X , is the onepoint compactification of the set X topologized with the discrete topology; see also Example 7.1 1. Let T be the Hausdorff locally solid topology generated by the Riesz norm p ( u ) = sup{lu(.x)l:x E X ) (if E L). Then T is a a-Lebesgue topology. To see this let u, 1 I9 in L. Observe that u,(x) 1 0 for each x E X and since X is uncountable, there exists y e X with u,(y) = u,(o3) for all n. Thus u,(x) 1 0 holds for every x E X,. It follows now from Dini’s classical theorem that u, 1,0. To see that (L,r) does not satisfy the pre-Lebesgue property note that 0 I X, e holds in L ; where ct runs over the finite subsets of X and e(x) = 1 for all .Y E X . However {x,) is not a z-Cauchy net of L. H

r

The next theorem justifies the name of the pre-Lebesgue property.

Theorem 10.5. For a Hausdorfl locally solid Riesz space (L,T)the following statements ure equivalent : (i) ( L,T)sutisjes the pre-Lebesgue property. (ii) (L,?) satisjes the pre-Lebesgue propert!!. (iii) (L,?) satisfies the Lebesgue property. Proof. (i) (ii) Assume 0 I 6, 1 in L and that U is a solid ?-neighborhood of zero. Pick a solid ?-neighborhood I/ of zero such that V V V G U . Next choose a sequence {K) of solid ?-neighborhoods of zero with V,, + V,, G V , for all n and with V, = I/. Now for each positive integer n choose v, E L+ such that 6, - v, E V,+ and then put u, = inf{v,: i = 1, . . . ,n} for all n. It should be clear that 8 I u, 1

+ +

,

,

[Chap. 3

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60

holds in L and so since (L,r) satisfies the pre-Lebesgue property, (14,) is a r-Cauchy sequence of L. In particular, we have u, - urnE V for all n,m 2 n,. But since

ii,

- u, = 6,

- inf{ui:l I iI n} = sup{ii,

I sup{iii - u , : l I i 5 n} I

and

c pi

- ui:l I iI

n}

2 liii - uil n

i= 1

fl

u, - 6, I

i= 1

- Uil,

cy=

+

+

it follows that lii, - u,I I liii - uil E V, . . V,+ c V , so that 16, - u,I E V holds for n = 1,2, . . . . Finally for n,m 2 n, we have lii, - ii,l I 16, - u,I Iu, - urn[ Iu, - ii,l E V I/ V c U and hence ii, - 6, E U for all n,m 2 n,. Thus {ii,,} is a 5^-Cauchysequence and we are done. (ii) + (iii) It follows from Theorem 10.3. (iii) e. (i) It follows from Theorem 10.2. H

+

+

3

+ +

Combining Theorems 10.5 and 10.3, we also obtain the following result. Theorem 10.6. If a Hausdorff locally solid Riesz space (L,t) satisjes either the pre-Lehesyue or the Lebesyue property, then sati.$es the Lebesyue property and L is Dedekind complete.

(e,?)

Note. The equivalence of statements (i) and (ii) of Theorem 10.5 as well as Theorem 10.6 were proven first by Luxemburg for normed Riesz spaces [44, Note XVI, Theorems 64.1 and 64.3, p. 6581. and by Kawai in [39]; they were generalized in [2, p. 1121.

Let us call two locally solid Riesz spaces (L,r) and (L1,?,) Riesz homeomorphic if there exists a Riesz isomorphism from L onto L, that is at the same time a homeomorphism; in other words if there exists a mapping from L onto L, that preserves all the structures, the algebraic, the lattice, and the topological. Let us also say that a locally solid Riesz space (L,r) is embeddable in the locally solid Riesz space ( L l , t l ) if there exists a Riesz subspace M of L, with the property that (L,r) is Riesz homeomorphic to (M,r,). The best-known Banach lattice (complete normed Riesz space) without the pre-Lebesgue property is the space I,, that is, the Riesz space consisting of all bounded real sequences with the pointwise ordering and equipped with the sup norm. To see this, consider the sequence (u,} of 1, defined by u,(k) = 0 if k I n and u,(k) = 1 otherwise ( n = 1,2, . . .); note that u, 1 8 holds in I , and that p,(u,+,, - u,) = sup { lu,,+p(m) - u,(rn)I:m = 1,2,. . .>= 1 holds for all n,p = 1,2, . . . .

Sec. 101

L,-SPACES

69

Surprisingly enough, the pre-Lebesgue property in a-Dedekind complete locally solid Riesz spaces depends upon the embeddability of I , into those spaces. The details follow. Theorem 10.7. For a a-Dedekind complete locally solid Riesz space (L,z) the following statements are equivalent: (i) ( L,z) satisfies the pre-Lehesgue property. (ii) I , is not embeddable in (L,z).

Proof. (i) => (ii) Assume that I , can be embedded in (L,z) in the sense defined above. Consider a sequence {u,} of 1 , with 0 I u, 1in 1, and which is not a Cauchy sequence. But then { u,} is not a z-Cauchy sequence of L (considered as a sequence of L under the embedding). This shows that (L,z) does not satisfy the pre-Lebesgue property, a contradiction. This contradiction establishes that 1, is not embeddable in (L,r). (ii) 3 (i) Assume that z is not a pre-Lebesgue topology. Then by Theorem 10.1 there exists a disjoint sequence {u,} of L + with 0 I u, I u for all n and some u E L, and with {u,} not z-convergent to zero. By passing to a subsequence if necessary; we can assume that there exists a solid z-neighborhood V of zero with u, $ V for all n. Next observe that aiui,order for every a = {a,}E 1, the sequence { v,} of L defined by v, = converges in L (here we use the fact that L is o-Dedekind complete) and as . have thus defined a usual we denote this order limit by ~ ~ = l a , u ,We a,u,, for a = {a,}E I,. It can be readily mapping n : l , -+ L by n(a)= seen that n is an into Riesz isomorphism, which is also continuous, because

c:=

c,"=

1n(a)J=

ILl I 1 a,u,

02

=

1Ia,Iu, n=l

= sup{lanlun:n= 1,2,. . .} I p,(a)u.

On the other hand, to show that 7 c - l is continuous consider a net { u A }of I, for which .(a,,) A8 and assume that { a ' ) does not converge to zero in I,. By passing to a subnet if necessary, we can assume that there exists 6 > 0 such that p,(al) 2 26 for all A. Now choose I,, such that n(aA)E 6V for all A 2 A1 and then fix an index A 2 A l . Thus, if a, = {a,'} choose k such that lak'l > 6 and then note that since I.(al)l

=

an'u,

I

00

=

1 lan'lun

n=l

2

Iak'luk

2

6Uk

2 0,

we have 6u, E 6 V . Thus uk E V , in contradiction to the choice of the neighborhood V . Hence { a n ) converges to zero in I,, and therefore ?I embeds 1, in (L,z).The last conclusion is, however, in contradiction with our hypothesis and hence the result follows.

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Since by Theorem 10.3 the Lebesgue and pre-Lebesgue properties are equivalent in Hausdorff topologically complete locally solid Riesz spaces the following result should be clear. Theorem 10.8. For a topologically complete HausdorfS a-Dedekind complete locally solid Riesz space (L,7)the following statements are equivalent:

(i) (L,z)satisjes the Lebesgue property. (ii) 1, is not embeddable in (L,7). As an application of Theorem 10.7 we also have the following interesting result. Theorem 10.9. Let (L,z) be a a-Dedekind complete locally solid Riesz space that is also separable (in the sense that it contains an at most countable 7-dense subset). Then (L,7)satisjies the pre-Lehesgue property. Proof. According to Theorem 10.7 we need only to show that 1, cannot be embedded in L. So, by way of contradiction assume 1, embedded in L. Let A denote the points of I , whose coordinates are either zero or one; it should be clear that A is uncountable and that p,(x - y ) = 1 for every pair x,y of A with x # y. Now choose a solid 7-neighborhood I/ of L such that I/ n I , G B,where B = ( u E 1, :p,(u) < 11, and then select another solid 7-neighborhood W of zero with W + Ll( z I/. Clearly x - y $ V for every pair x,y of A with x # y. On the other hand, if C is a countable 7-dense subset of L, then for each x E A choose u, E C with x - u, E W and note that if x # y then u, # uy (since otherwise, u, = uy implies

1.x - yI I Ix - u,( +lux

- uyl

+ 14’ - uyl = Ix - u,( + l y - uy( w + w E v, E

contradicting x - y $ V ) . But then it readily follows that C must be uncountable, which is a contradiction. Hence 1, is not embeddable in (L,7), and the proof is finished. Note. Theorem 10.8 for the Banach lattice case can be found in [43, Proposition 2.1, p. 1211, [49], and [67, Theorem 5.14, p. 941. Theorem 10.9 for normed Riesz spaces has appeared in [44, Note XIV, Theorem 43.1, p. 2291; see also the corollary on p. 95 of [67].

The rest of this section is devoted to the study of the properties and the characterizations of abstract Lp-spaces. A Riesz seminorm p on a Riesz space L is called p-additive (1 I p < co) whenever p(u + u)” = p(u)” + p(v)P holds for every u p E L + with u A u = 0. A Banach lattice L, is called an abstract L,-space if the Riesz norm p is

Sec. 101

L,-SPACES

71

p-additive for some 1 I p < co. For the rest of this chapter instead of the letter p, we shall use the conventional symbol 11. 1 1 to indicate the norm. The next theorem tells us that normed Riesz spaces with p-additive norms always satisfy the pre-Lebesgue property.

Theorem 10.10. Every normed Riesz space with p-additive norm satisjies the pre-Lebesgue property (and hence every abstract L,-space is Dedekind complete and satisfies the Lebesgue prGperty). Proof. Let L be a normed Riesz space with p-additive norm, and let {u,} be a disjoint sequence satisfying 0 I u, I u for all n. Now the p-additive property of the norm implies

5

i= 1

lluillp =

llil

ui/l

=

Ilsup{ui:l 5 i

5

llullp

for all n and hence llunllP converges. It follows now that lirnllu,ll = 0, so that by Theorem 10.1 L satisfies the pre-Lebesgue property. The parenthetical part follows from Theorem 10.3. W

Theorem 10.11. The norm completion of a normed Riesz space with a padditive norm is a Dedekind complete abstract L,-space with the Lebesgue property. Proof. Let L be a normed Riesz space with norm completion f,, and let 6,s E ?, satisfy ii A v^ = 0. Pick two sequences {u,} and {u,} of L + with limllu, - iill = 0 and lim)lu, - 011 = 0 and observe that (u, - u, A u,) A (u, - u, A u,) = 0 holds for all n and that limIIu, A u,11 = 0. Hence I ( ( u n - un A 0,)

+ (On - u n A un)IIp = IIun - un A UnllP +

IIun

- un A

unIIp

holds for all n and so by taking the limits in both sides we get

116

+ v^p=

I(iillP

+

IIDIIP.

The rest of the theorem follows from the previous theorem. Examples of normed Riesz spaces with p-additive norms are provided by the classical Banach lattices L,(X,C,p), where (X,Z,p) is an arbitrary measure space and the norm on L,(X,C,p) is defined as usual by llfll, = (Slfl”dpL)l’p (1 I p < a). Actually those are the only abstract L,-spaces; more precisely we shall show next that every abstract L,-space is norm and Riesz isomorphic to some L,(X,Z,p). We recall that for a nonempty subset X of Rf we put E x e X x= s u p { ~ , , , x : F c X and F finite}.

Theorem 10.12. For every abstract L,-space L there exists a measure space (X,&p) and a norm-preserving Riesz isomorphismfrom L onto L,(X,C,p).

LEBESGUE TOPOLOGIES

72

[Chap. 3

Proof. Let L be an abstract L,-space. Then by Theorem 10.10 L is Dedekind complete and satisfies the Lebesgue property. Assume first that L contains a weak order unit e > 0. Let denote the Boolean algebra of all components of e, that is, B = ( u E L : u A(e - u) = O), and observe that B is also Dedekind complete. Now let Lo be the Riesz subspace of L consisting aiui with u l , . . . , u, E B pairwise of all elements f of the form f = disjoint and note that Lo is order dense (and hence norm dense) in L. Indeed, if 0 < u E L and 0 < 6e I u does not hold for every 6 > 0 (otherwise we are done) then (6e - u)+ > 0 holds for all . 6 > 0. Choose 6 > 0 such that (6e - u)- > 0 and note that if P denotes the projection on (he - u)+ we have (by virtue of w I P,+(w) = w+)6e - u I P(6e - u) and hence 6(e - P ( e ) )I u - P ( u ) I u ; so that 0 < 6(e - P ( e ) )E Lo holds and thus Lo is order dense in L. By the well-known Stone's representation theorem (see [46, Theorem 6.6, p. 271) there exists a set X and a Boolean algebra d of subsets of X such that 98 is isomorphic to d,say via u HS,. Now by defining p ( S , ) = l l u l l p we can see easily that p defines a o-additive measure on d,in the sense that if S, = Sunand the Sunare disjoint then p ( S , ) = p(Su,) holds (this follows easily from the Lebesgue property). Next, by the standard arguments generate the exterior measure p* of p and then let C be the o-algebra of all measurable subsets of X. Now, consider the Banach lattice L,(X,C,p) and let n:Lo -, L,(X,C,p) be defined by n ( f )= aiuiE Lo. aixS,,,for all f = Note that n is a norm-preserving Riesz isomorphism (into) whose image is order dense (and hence norm dense) in L,(X,C,p). It now follows easily from the last observation that n extends (in the obvious manner) to a normpreserving Riesz isomorphism between L and L,(X,C,p) and so the result is established for this case. Now if L does not contain an order unit consider a complete disjoint system {ei:iE I > of strictly positive elements and note that by what was shown before Bi = B,,, is norm and Riesz isomorphic to some Lp(Xi,Ci,pi), say via ni:Bi -, Lp(Xi,Ci,pi).We can consider Xi n X j = 0 if i # j . Put X = U i E I X i , C = ( A g X : A n X i E C i f o r a l l i E I ) , p ( A ) = C i , l p i ( An x i ) for A E C and note that (X,C,p) is a measure space. Now by Theorem 2.15 for each 0 4 u E L we have u = sup(u,:i E I } , where ui= P J u ) E Bi and so by the p-additivity of 11. 1 1 we have llullp = C i s ,IIuilIp. On the other hand, n : L + -,Lp+(X,C,p)defined by u H n(u)= C i eIni(ui)is one-to-one, onto, lattice and norm preserving. We leave the details for the reader to complete the proof by verifying that n extends to a norm-preserving Riesz isomorphism between L and L,(X,C,p).

cy=

u:=

By Theorem 10.11 and the above result the following result should be clear now.

Sec. 101

L,-SPACES

73

Theorem 10.13. For euery normed Riesz space L whose norm is p-additive there exists a norm-preserving Riesz isomorphism.from L onto a Riesz subspace of some Lp(X,Z,p). Note. The representation Theorem 10.12 for the abstract L,-spaces is due to Kakutani [33, Theorem 7, p. 5321, while Bohnenblust in [14] and Nakano [52, p. 1211 established the general result under some additional conditions.

If L is a normed Riesz space, then the strong topology p(L’,L) on L‘ is generated by the Riesz norm

llcpll = sup(lcp(u)l:IIuIII 1) = sup(lcp(u)l:u E L +

and

llull = 1).

Observe that L‘ is norm complete and the natural embedding of L into L” is norm preserving. A normed Riesz space L is called an M-space (or that the norm is an M-norm) if 111.4 + 011 = max{ IIuII,IIuII} whenever u A u = 8. A Banach lattice that is at the same time an M-space is called an abstract M-space. By arguments similar to those of the proof of Theorem 10.11 it is easily seen that the norm completion of an M-space is an abstract M-space. A typical abstract M-space with an order unit is the Riesz space C(X), where X is a Hausdorff compact topological space, equipped with the norm = sup{If(x)I:x E X } ; note that the closed unit ball is simply [-e,e] [e(x) = 1 for all x E XI and that every Riesz subspace of C(X) (with the induced norm) is also an M-space. Another class of M-spaces is provided by the Riesz subspaces of the abstract M-spaces of the form c,(A) = {f E R A :{x E A : If(x)l > E } is finite for every E > 0},

with the sup norm. Observe that each c,(A) satisfies the Lebesgue property and that it admits a strong unit if and only if A is finite (see also Exercise 12 at the end of this chapter). Let us call as usual a normed Riesz space with 1-additive norm an L-space. Lemma 10.14. Let L be a normed Riesz space, cp,$ E L’ with cp A $ = 0 and E > 0. Then there exist u,u E L + with K A u = 8,IIuII,IIuII I 1, and such that

llcpll 5 c p h ) + E

and

11$11

5

$(4 + E .

Proof. Let cp,$ E L’ satisfy cp A $ = 8, and let E > 0. Choose f,g E Lf with Ilfll,llgll I 1 and such that llcpll I cp(S) 43, 11$11 I $ ( g ) 43. Now since (cp A $)(f)= 0, there exist f i f 2 E L + with fi + f2 = f and such that cp(fl) + $(f2) 4 3 ; similarly choose gl,g2 E L + with g1 + g 2 = g and such , that cp(gl) + $ ( g 2 ) < 43. Put u = f 2 - g , ~ fu =~g1 -, g1 ~ fand~note

+

-=

+

[Chap. 3

LEBESGUE TOPOLOGIES

74

that u,u, E L', IIuII,IIuII I 1, u A u = 8,

cp(4= c p ( f 2 )

-d

s , AfZ)

and similarly $ ( u ) 2

ll$ll

2

- E.

cp(f2)

-4

3 = df) -dfl)

- E/3

Hence the lemma is established.

llcpll - 8,

2

W

The next theorem describes an important duality property between Lspaces and M-spaces. Theorem 10.15. For a normed Riesz space L we have:

(i) L is an L-space i f and only i f L' is an M-space. (ii) L is an M-space ifand only i f L is an L-space.

Proof. We show first that the dual of an L-space is an M-space. To this end let L be an L-space and cp,$ E L' with cp A $ = 8. Since obviously Ilcp++ll 2 maxCllcpll,ll$ll>. it is enough to show IIcp + $11 I ~ ~ ~ { ~ ~ c p ~ ~ So, given E > 0 choose w E L' with llwll I 1 and such that IIcp + $11 I (cp $)(w) E. Now since cp A $(w) = 0, there exist u,u E L' with u + u = w and such that cp(u) $(v) < E. But then we have

+

IIq +

+

$11

+

+

+ cp(4+ $(u) + E I cp(4 + $(u) + 28 + 3E

5 cp(w) + $ ( I 4 + & = d u ) $(u) 5 ( p ( L ' - U A U ) $(U - U A U )

+

I max{llcpll,ll$ll>(IIu -

5 maxCllcpll,ll@ll)+ 38

A

u(I + IIu - u A u l ( ) + 38

uI I

where the last inequality holds by virtue of IIu - u A + IIu - u A uJI = IIu + u - 2 u uII ~ I llwll I 1, due to the fact that L is an L-space and (o - u A u) A (u - u A u ) = 8. The result now follows. Next we show that the dual of an M-space is an L-space. Thus, let L be an M-space and cp,$ E L' with cp A $ = 8. By virtue of the triangle inequality it is enough to show Ilcpll ll$ll I IIq $11. To this end let c > 0. By Lemma 10.14 there exist u,u E L + with u A u = 8, llull,llull I 1, and such that llcpll I q ( u ) E and Il$ll 5 $(u) E. Now since L is an M-space we have IIu + = ~~x{~[uII,IIuII> I 1 and so llcpll+ 11$11 I cp(u u ) + $(u v) 28 5 IIcp $11 2~. The result now follows. To complete the proof note that if the topological dual L' of a normed Riesz space L is an L-space, then its second dual L" is an M-space and hence L G L" is also an M-space. A similar observation holds if L' is an M-space, and the proof of the theorem is finished. W

+

+ +

+

+

+ +

uI I + +

If L is an abstract L-space, then by Theorem 10.12 L is norm and Riesz isomorphic to some L1(X,C,p) and hence Ilu + = llull + llull holds for all u,u E L'. This observation will be used in the proof of the next theorem, which characterizes the abstract M-spaces.

uI I

Sec. 101

L,-SPACES

75

Theorem 10.16. For an abstract M-space the following statements are equivalent:

(i) There exists a norm-preserving Riesz isomorphism from L onto a C ( X ) , for some Hausdorff compact topological space X . (ii) L admits an order unit e > 0 such that [ -e,e] is the closed unit ball of L, i.e., [ -e,e] = { u E L: llull II}. In particular, a normed Riesz space L is an M-space if and only if there exists a norm-preserving Riesz isomorphism from L onto a Riesz subspace of some C ( X ) .

Proof. (i) =. (ii) Obvious. (ii) ==(i) Put Y = (cp E (L')+:[lcpll= 1) and note that Y = ('p E (L')+: cp(e) = llcpll = l}. It is clear that Y is nonempty, convex, and a(L',L)compact. We shall show next that.cp E Y is an extreme point of Y if and only if cp is a Riesz homomorphism (we recall that cp E Y is called an extreme point o f Y i f c p = k p , + ( 1 -A)cp,withO
uI I

76

[Chap. 3

LEBESGUE TOPOLOGIES

\lull if u A v = 0. We shall show next that those are the only Riesz norms which have this characteristic property. To do this we need a lemma.

Lemma 10.17. Let f :R + x R + + R+ be a continuous function. Then f + qP)lipforsome 0 < p < 00 o r f ( 5 , q ) = max{t,q} if and only if the function f satisjies the following five conditions: is oftheform f ( 4 , q ) = ( < p

( 1 ) f ( z 4 , z d = $(4,'1)for all G 4 9 ' 1 E R + . (2) f(t,'1)I f ( 5 1 , V l ) i f 0 I 5I 4 1 and 0 I '1 I '11. (3) f(t,r)= f(q,t)for all 4,Y E R + . (4) f(0,l) = 1 . ( 5 ) f(tf(W) ) = f(f(4,'1),.)for all 49'1,z E R + . Proof. It should be clear that iff is of the prescribed form thenf satisfies the listed properties. On the other hand, i f f satisfies the five properties for n = 2,3, . . . and define the sequence {a,} by a , = 1 and a, = f(l,a,note that:

(i) a,,+, = f(a,,a,)

for all n,m.

To see this use induction on m. For m = 1 we have a,+ = f(l,a,) and if a, + ,= f(a, ,a,) then a n + m + 1 = f(1,an + m) = f ( l f ( a m9

4

=f ( a , , l )

) =f(f(l,am),an)

= f ( a m + 1 ,an) = f(an r a m + 1 ) .

(ii) a,, = anamfor all n,m. Again use induction on m. For m = 1 the result is trivial and if a,, holds, then anam+ 1 = a n f ( l * a m ) =f(an,anam) = f(an,anm) = an+,, = an(,+

= anam

1)'

Now since {a,) is nondecreasing two cases arise: Case 1. a2 = 1, that is,f(l,l) = 1 . Then for t 2 '1 > 0 we have and hence f ( 4 , q ) = max { t,~}. Case 2. a2 > 1 and hence a, > 1 for all n. Let m,n > 1 . Then for every positive integer k determine h such that mh I nk I m h + l , and note that since (a,)h = a , h and similarly for h + 1, we get hloga, I kloga, I (h + l)loga,. Now hlogm I klogn I (h + 1)logm implies loga, logu, loga, loga, loga, ~ - _< _ -I-+logm klogn logn logm klogn

77

EXERCISES

for all k, and so log a,/log m = log a,/log n, that is, the value of log a,/log n is independent of n. Put p = log2/loga2 and derive that a, = nil’’. But then it follows easily that f ( l , r l i P )= ( 1 + r)’/P holds for all positive rationals r and from this by using the continuity off we get f ( t , q ) = ((P + q P ) l / p .

Theorem 10.18. For a normed Riesz space L of dimension at least three, the following statements are equivalent:

(i) The norm of L is either an M-norm or a p-additive norm. (ii) There exists a function f : R t x Rf 4R’ satisfying IIu + f(lluII,IIvll)for every pair u,v, E L+ with u A u = 8.

uII

=

Proof. (i) * (ii) If the norm is an M-norm f ( t , q ) = max{<,q}, and if the norm is p-additive f ( t , q ) = (tp+ (ii) 3 (i) Assume that the norm satisfies the stated property. Since L has dimension at least three there exist three pairwise disjoint elements u p , and w of L+ each of norm one. But then,f(c,q)= ))
+

Note. The representation Theorem 10.16 is due to Kakutani [34] and Bohnenblust and Kakutani [15]. Lemma 10.17 as well as Theorem 10.18 and their proofs are due to Bohnenblust [141.

EXERCISES

1. Show that every locally solid Riesz space (L,z) with the a-Levi property is a Hausdorff space. (Hint: Consider the intersection K , of all z-neighborhoods of zero and note that if 0 I u E K,, then {nu} is an increasing z-bounded sequence.) 2. Show that a Hausdorff locally convex-solid topology z on a Riesz space L is Lebesgue if and only if lol(L,L’)is Lebesgue. (Hint: Use Theorem 9.8; for another proof see Theorem 11.7.) 3. Show that a Hausdorff locally convex-solid Riesz space (L,z) satisfies the Lebesgue property if and only if C, # (0) for each 8 < cp E L‘. (Hint: Show that every order dense ideal is z-dense.) 4. Let (L,z) and (M,z*) be two Hausdorff locally convex-solid Riesz spaces with z a o-Lebesgue topology. Assume that T:(L,o(L,L’)) -, (M,o(M,M’))is positive, linear, and continuous. Show that T is an integral.

70

LEBESGUE TOPOLOGIES

[Chap. 3

5. Let (L,z) be a Hausdorff locally solid Riesz space with the preLebesgue property. Show that if L is infinite dimensional, then the positive cone L+ does not have interior points. (Hint: If 8 < u is an interior point of L', then [ - u,u] is a z-neighborhood of zero and u is an order unit.) 6. By u topological completion of a locally solid Riesz space (L,z) we mean any 7"-complete locally solid Riesz space (L*,z*) (not necessarily Hausdorff) containing (L,t) as a topological z*-dense Riesz subspace (that is, L is a Riesz subspace of L* and z* induces z on L). (i) Give an example of a non-Hausdorff topological completion of some Hausdorff locally solid Riesz space. (ii) If (L*,z*) is a topological completion of a locally solid Riesz space (L,z), show that (L,z) satisfies the pre-Lebesgue property if and only if (L*,z*)does. 7. Show by a counterexample that the hypothesis of a-Dedekind completeness in Theorem 10.9 cannot be dropped. 8. Let L = I , and let A be the ideal of L consisting of all null sequences, that is, A = co. Show that A; = I , and A,, = I,, and derive from this example that an ideal of a perfect Riesz space need not be perfect. 9. Show that a Riesz space L is perfect if and only if it admits a Lebesgue locally convex-solid topology with the Levi property. 10. Generalize Theorem 9.5, that is, show that if L is any Riesz space and M any perfect Riesz space, then Y b ( L , M )is a perfect Riesz space. 11. Let L be a Riesz space, A an ideal of L, and p a p-additive Riesz seminorm on L. Show that the quotient Riesz seminorm p is p-additive on L/A. Prove a similar statement if p is an M-seminorm. 12. (Bohnenblust). Show that an abstract M-space L, satisfies the Lebesgue property if and only if L, is norm and Riesz isomorphic to some c o ( X ) , with the sup norm. ( H i n t : Show first that L is a discrete space and then use Theorem 2.17.) 13. Show that every abstract L,-space is a perfect Riesz space. 14. Let L be an almost a-Dedekind complete Riesz space. Show that every cr-Lebesgue topology on L is necessarily a pre-Lebesgue topology. 15. Let M be a regular Riesz subspace of a Riesz space L. If z is a Hausdorff locally convex-solid Lebesgue topology on L, show that M is order dense in L if and only if M is t-dense in L.

Open Problem Is the a-Lebesgue property preserved under the process of topological completion? (The answer is affirmative if the topology is metrizable; see Theorem 17.11.)

CHAPTER

4

Fatou Topologies

As we have seen in the previous chapter the full richness of locally solid Riesz spaces may be realized when there is a further interaction between the order structure and the topological structure. In this chapter we shall see that when the topology has a basis at zero consisting of solid and order closed neighborhoods we are led to some of the deepest results of the theory of locally solid Riesz spaces. Two of the main results of this chapter are the striking theorem that all Hausdorff Lebesgue topologies on a Riesz space induce the same topology on the order bounded sets, and Nakano's theorem relating Dedekind completeness and topological completeness. 11.

BASIC PROPERTIES OF THE FATOU TOPOLOGIES

Recall that a subset S of a Riesz space is called order closed if it follows from u, -, u and { u,) E S that u E S . Note that a solid subset S is order closed if and only if it satisfies the following weaker property: If (u,} c S satisfies 8 5 u, t u, then u E S . Indeed, if the solid set S has the last property and a net {u,} E S satisfies u, + u, then there exists a net {u,} with Iu, - Iu, 18; so from the relations (lul - u,)' I Iu,I and 8 I (lu( - 0,)' IuI we easily obtain u E S . Similarly, a solid subset of a Riesz space is 0-order closed if and only if u belongs to S whenever 8 I u, t u holds for some sequence {u,} G S. (0)

MI

(0)

Definition 11.1. A locally solid Riesz space (L,T)is said to satisfy the Fatou property (or that T is a Fatou fopology) ifft has a basisf o r zero consisting of solid and order closed sets. 79

FATOU TOPOLOGIES

80

[Chap. 4

Similarly, (L,T)satisjes the a-Fatou property i f z has a basis for zero consisting of solid and a-order closed sets. A solid and order closed neighborhood will be called a Fatou neighborhood. The a-Fatou neighborhoods are defined analogously. Clearly the Fatou property implies the a-Fatou property. Also every locally solid Riesz space (L,z) with the Lebesgue property satisfies the Fatou property. To see this, observe that if (Lz)satisfies the Lebesgue property, then every solid z-closed subset of L is order closed. A Riesz pseudonorm p on a Riesz space L is said to be a Fatou pseudonorm (or that p satisfies the Fatou property) if 8 Iu, t u in L implies p(u,) T p(u). Similarly, a Riesz pseudonorm p is said to be a a-Fatou pseudonorm (or that p satisfies the a-Fatou property) if 8 Iu, t u implies p(u,) t p(u). Example 11.2. Clearly any topology generated by a Fatou (resp. aFatou) pseudonorm is a Fatou (resp. a-Fatou) topology. The sup norm on C[O,l] satisfies the Fatou property, and so the Hausdorff locally solid topology t generated by the sup norm on C[O,l] is a Fatou topology. However, note that z is not a Lebesgue topology. Example 11.3. Take L = C[O,l] x R[','] and z the product of the Fatou topology generated by the sup norm on C[O,l], and the topology of pointwise convergence on R[ov'l. Observe that (L,z)satisfies the Fatou property, but it does not satisfy the Lebesgue property. The next two results will explain why the Fatou properties are so named. Theorem 11.4. Let p be a Riesz seminorm on a Riesz space L. Then the following statements are equivalent:

(i) p satisfies the a-Fatou property, i.e., 8 Iu, t u in L implies p(u,) (0) (ii) u, -+ u implies p(u) Iliminfp(u,). (iii) The closed unit ball U p= { u E L : p ( u )I1) is a-order closed. Proof. (i)

t p(u).

(ii) Assume u, + u. Pick a sequence {u,) E L with Iu, - uI Iu, 8 in L. Now use the relations (lul - u,)' IIu,I and 8 I (IuI - u,)' t IuI to obtain that p(u) Iliminfp(u,). (ii) (iii) Assume that {u,} G U p satisfies u, 2 u. Note that p(u,) I 1 for all n and p(u) Iliminfp(u,) imply p(u) I 1 and so u E U p .Hence U p is a-order closed. (iii) 3 (i) Let 8 Iu, t u in L. Put a = limp(u,). We need only to show p(u) Ia. Observe that given E > 0 the relations (a + E)-'u, 2 (a + E ) - % and (a E ) - ' u , E U pfor all n imply (a + E ) - ' u E U p ,and hence p(u) Ia + E holds for all E > 0. Thus p ( u ) Ia, and the proof is finished.

+

(0)

Sec. 111

BASIC FATOU PROPERTIES

81

For Fatou seminorms the above result holds for nets. A proof of the next result can be found in [25, 23B Theorem, p. 441. Theorem 11.5. Let (L,t)be a locally solid Riesz space. Then the following statements are equivalent:

(i) t is a Fatou topology. (ii) There exists a family { p a } of Fatou pseudonorms that generates the topology z. Note. The term Fatou seminorm is due to Luxemburg and Zaanen [45, Notes I1 and 111] and it was obviously inspired by the well-known Fatou lemma from the theory of integration, where L = L , ( X , p ) and p(u) = [ IuI dp. The term Fatou topology was subsequently introduced by Fremlin in [24]. Fatou and a-Fatou norms were first studied by Nakano [53,§30].

We have seen so far that the Lebesgue property implies the Fatou property. The next theorem characterizes the Lebesgue property in terms of the Fatou property. Theorem 11.6. Let (L,z) be a Hausdorff locally solid Riesz space. Then the following statements are equivalent:

(i) ( L , t )satisfies the Lebesgue property. (ii) (L,z) satisfies both the Fatou property and the pre-Lebesgue property. Proof. (i) =-(ii) It follows immediately from Theorem 10.2. (ii) 3 (i) Assume u, LO. Let I/ be an order closed z-neighborhood of zero. By Theorem 10.1 {u,} is a T-Cauchy net and so for some ctl, u, - up E V (0) for all a$ 2 a l . But for each fixed a 2 aI we have u, - up u,, and so

-

B

since I/ is order closed, we get u, E I/ for all a 2 al. Hence u, (L,z)satisfies the Lebesgue property.

L 0 and

so

As an application of the last theorem we have the following result. Theorem 11.7. Let (L,T)be a Hausdor-locally convex-solid Riesz space. Then the following statements are equivalent:

(i) lal(L,L’)is a Fatou topology. (ii) lol(L,L‘) is a Lebesgue topology. (iii) L’ c LE. (iv) t is a Lebesgue topology. Proof. To see that (i) 3 (ii) observe that (L,(ol(L,L’))satisfies the preLebesgue property, Indeed, if 6 < u, t < u holds in L, then {cp(u,)} is a Cauchy sequence for every Q Icp E L’, and so {u,} is a lol(L,L’)-Cauchy

FATOU TOPOLOGIES

02

[Chap. 4

sequence. The result now follows from Theorem 11.6. The other implications now follow from Theorem 9.1 and the relations o(L,L') G lol(L,L')E z. Theorem 11.8. Let (L,z) be a locally solid Riesz space with the Fatou property, and let M be a regular Riesz subspace of L (in particular an order dense Riesz subspace or an ideal of L). Then t induces a Fatou topology on M . Proof. Note that if V is a Fatou z-neighborhood of zero then V n M is a solid order closed subset of M , since M is a regular Riesz subspace of L. w The next example shows that the hypothesis of M being a regular Riesz subspace in the above theorem cannot in general be dropped. Example 11.9. Let L = R[op'l and z be the Hausdorff Lebesgue topology of pointwise convergence. Take M = C[O,l], and note that M is not a regular Riesz subspace of L. Observe also that z restricted to M is not a Fatou topology. We continue with an important extension theorem. Theorem 11.10. Let L be an Archimedean Riesz space. If z is a Fatou topology on L, then there exists a unique Fatou topology t6 on the Dedekind completion Ls of L that induces z on L. I f in addition t is HausdorfS, then so is t6 and if(L,t) satisfies the Lebesgue property, then so does (Ls,z'). Proof. Let { V } be a basis of zero for t consisting of solid and order closed sets. Given V E { V >define the subset of La V s = {u* E L'::3{ua}E V with 6' Iu,

t

(u*l in La}.

Observe that since I/ is solid and order closed in L , we have V 6 = {u* E L':u

E

L and 6'

suI

Iu*l implies u E V } .

Note that each I/' is a solid, absorbing, and order closed subset of La. To see that V 6 is order closed let 6' 5 u,* t u* in La, with {ua*>c V'. For each u,* there exists a net of V n L + that increases to u,*. The net consisting of the finite suprema of these nets is a net of V n L + that increases to u*, so that u* E V'. Observe also that ( V n W)' E V s n W' and that if W + W c V in L , then W s W s c V 6 holds in La. Note also that V6 n L = V. Thus the collection { V 6 }is a basis for zero for a Fatou topology zs on Lg that induces z on L. Now let t* be a Fatou topology on L6 that induces t on L. If W is a Fatou t*-neighborhood of zero, then V = W n L is a Fatou t-neighborhood of zero for L, and hence V s G W , i.e., W is a z6-neighborhood of zero. If V E { V } , then U n L G V for some Fatou t*-neighborhood of zero U . Note that if

+

THE STRUCTURE OF FATOU TOPOLOGIES

Sec. 121

83

u* E U , then 0 I u, Iu*l for some net (u,) c L. So, {u.} E U n L and hence u* E Vd, i.e., U c V', which shows that V bis a 7'-neighborhood of zero. Thus t* = t6 and hence z6 is uniquely determined. If z is Hausdorff, note that the order density of L in La implies that z6 is

also a Hausdorff topology. If now (L,T)satisfies the Lebesgue property, then it also satisfies the Fatou property, and hence t can be extended to t6 as above on Lb. It now follows easily that t6is likewise a Lebesgue topology. Following the arguments of the preceding proof we obtain the sequential analog of the previous theorem.

Theorem 11.11. Let L be an almost o-Dedekind complete Riesz space. If t is a a-Fatou topologjl on L, then there exists a unique a-Fatou topology t"on L" (the a-Dedekind comp!etion of L ) that induces t on L. I f in addition z is a Hausdorftopology, then so is z" and if ( L J ) satisfies the a-Lebesgue propert)', then so does ( L",P). Note. Theorem 11.10 was first proven by Kawai for Hausdorff locally convex-solid topologies with the Lebesgue property [39, Theorem 1.5, p. 2881; the theorem as it is stated here was proven by the authors [5, Theorem 2.1, p. 261. 12.

THE STRUCTURE OF THE FATOU TOPOLOGIES

The carrier of a locally solid topology will be introduced next. It will be a useful device for studying the structure of the Fatou topologies. We start with the following definition. Definition 12.1. A sequence {S,; of subsets of a vector space is said to be S,, E S, holds for all n. normal if&+

+

Now let L be a Riesz space and {S,} a normal sequence of solid subsets of L. We define the null ideal N of {S,} by

n s,. cc

N

=

n= 1

Note that N is actually an ideal of L, and that N is a band of L if the normal sequence {S,) consists of solid order closed sets. Definition 12.2. Let (L,t) be a locally solid Riesz space, and let JV denote the collection of all normal sequences consisting of solid z-neighborhoods of zero. We define the carrier C , of the topology z by

c, =

u ( P : N n v,;cv,} m

=

n= 1

E

~ 1 .

[Chap. 4

FATOU TOPOLOGIES

04

The disjoint complements Nd appearing in the definition of the carrier will be called the components of C,. Observe that C, is an ideal of L. The basic properties of C, are included in the next theorem. Theorem 12.3. Let (L,z) be a Hausdorff locally solid Riesz space. Then we have

(i) T h e carrier C, is a o-ideal of L. (ii) I f z is also a Fatou topology, then C , is order dense in L (and so $ L has the countable sup property then C, = L). (iii) I f z satisjies also the Lebesgue property, then C, has the countable sup property. Proof. (i) Assume that {u,} E C,. For each k pick a normal sequence { K k : n = 1,2, . . .} of solid r-neighborhoods of zero such that uk

E

Nkd;

n Kk m

where

Nk =

n= 1

for all k.

Set W, = V,i for all n, and note that { W,) is a normal sequence of solid r-neighborhoods of zero satisfying m

N = (7 W, E n= 1

m n=

1

Kk= N k

for all k.

Thus NkdE Nd for k = 1,2,. . . , and so {u,} E Nd. Now if 0 I u, t u holds, then u E Nd E C,, so that C, is a o-ideal. (ii) Let 8 < u E L. Pick a Fatou z-neighborhood I/ of zero of L with u 4 V . Next choose a normal sequence (K)of Fatou t-neighborhoods of V, is a band of L and so N = Ndd, zero with V, = V . Note that N = since L is Archimedean. But then since u 4 N = Ndd,there exists 0 < u E Nd such that w = u A v > 8. Therefore, 8 < w I u and w E C, hold in L and so C, is order dense in L. The parenthetical part now follows immediately from part (i). (iii) Assume u, 1 0 in C,. We can assume I9 I u, I u holds in C, for each CI and some u E C,. Pick a normal sequence { of Fatou z-neighborhoods of zero with u E Nd. Thus, {u,} E Nd.Since z is a Lebesgue topology u, 0. Now for each n pick u , E~ V,. We can assume uUn1 in C,. If 0 I w I uan holds for all n, then w E N n Nd = { e } , which shows that uan18.Hence C, has the countable sup property.

v,)

The next theorem describes a metrizability property of the Fatou topologies. Theorem 12.4. Let L be a Riesz space with a HausdorfS Fatou (resp. Lebesgue) topology z. If L has the countable sup property and admits an at

Sec. 121

THE STRUCTURE OF FATOU TOPOLOGIES

85

most countable complete disjoint system (in particular a weak order unit), then L also admits a metrizable Fatou (resp. Lebesgue) topology z’ coarser than z, that is, z’ E z.

Proof. By Theorem 12.3(ii) we have C, = L. Let {u,} be a complete disjoint sequence of L. As in the proof of Theorem 12.3(i)there exists a normal sequence { V,} of Fatou z-neighborhoods of zero such that {u,} G Nd,where N = [ V,:n = 1,2, . . .). Since {u,) is a complete disjoint system, it follows that L = Nd.Clearly { V,} defines a metrizable Fatou topology z’ on L (which must necessarily be Lebesgue, if z is Lebesgue) satisfying z‘ c z.

0

One of the key tools for the discussion in this chapter will be the following lemma. Lemma 12.5. Let L be a a-Dedekind complete Riesz space, {S,} a normal sequence of solid and a-order closed subsets of L , and { u,} an order bounded sequence of L satisfying u,+1 -U,ES.+2

for all n. Then w = lirn inf u, and u = lim sup u, satisfj w - u, E S, and u - u, E S , for all n.

Proof. We shall show that w - u, E S, for n = 1,2,. . . . The proof for the lim sup is similar. Start by observing that

0I u, - inf{u,:n I rn I n + p } = sup{u, < sup{lu, - uml:nI mI n +p} n+p-l

I

1

m=n

Ium + 1 - urn1 E S , + 2

- um:nIm In

+p}

+ . . + S , +p + 1 c S , + 1

implies u, - inf{u,:n Im In + p } E S n + lfor n,p = 1,2,. . . . Thus, if w, = inf { urn:m 2 n } ,then u, - inf { urn: n I mI n + p } t u, - w, and consequently (since S,+ is o-order closed) u, - w, On the other hand if w = lim infu,

P

S,+ for all n. = sup{w,:n 2 1) the relations

E

Iw,+p - u,I I Iw,+p - %+PI + I%+P - 4 5

Iwn+p

- un+pl +

n+p- 1

1

m=n

/urn+1 - urn1 E S , + p + 1

+ S , + , c S,,

and and the o-order closedness of S, imply Iw - u,I E S, for all n. Since now each S, is also solid, we get w - u, E S, for all n, and the proof is finished.

86

FATOU TOPOLOGIES

[Chap. 4

As a first application of the preceding result we have the following. Theorem 12.6. Let (L,z) be a Hausdorff locally solid Riesz space with the Fatou property. Assume that L has the countable sup property and that {u,} is an order bounded net of L. If u, 1,u holds in L, then there exists an increasing sequence of indices (0) {a,} E { a } such that uan+ u in L.

Iu,~

I u E L for all a. Use Proof. Assume { u ~c} L satisfies u, 4 u and Theorem 11.10 to extend z to the Fatou topology z* on the Dedekind completion L* of L ( L is Archimedean by Theorem 5.6(i)). Obviously L6 has the countable sup property and so by Theorem 12.3(ii) C+ = La. Choose next a normal sequence { V,} of Fatou z6-neighborhoods of zero such that u E Nd ( N = V,). Since { u,} is a z*-Cauchy net of La, there exists a a,+ ( n = 1,2,. . .) with uan+,- uanE sequence of indices (a,} c ( a ) , a, I V,, and uan- u E V , for all n. Put u* = lim sup u,, and w* = lim inf uanin La. By Lemma 12.5 we get uan- u* E V , for all n and so u* - u E V , for all n. Thus u* - u E N n Nd, and therefore, u* = u. Similarly w* = u. But then 5 u holds in La,and consequently u,, 5 u holds also in L (Theorem 2.5), and the proof is finished.

The next result tells us that the solid order closed sets of a Hausdorff locally solid Riesz space with the Fatou property are topologically closed. Theorem 12.7. Let (L,z)be a Hausdorff locally solid Riesz space with the Fatou property. Then euery solid and order closed subset of L is z-closed. Proof. Assume first that L is also Dedekind complete. Let S 2 L be a solid and order closed subset and u E S (the z-closure of S). Assume also that 8 5 u E C, satisfies 8 I u I (u(.Since S is a solid subset (Theorem 5.4) u E S. Pick a normal sequence { V,} of Fatou z-neighborhoods of zero with u E Nd,where N is the null ideal of { V,}.Then for each n there is u, E S with u, - u E V , + 3 . Without loss of generality we can assume 8 I u, 5 u for all n (and thus {on} c Nd).Note next that u,+ - u, E V,, 2 , and so by Lemma 12.5 we get w = lim inf u, satisfies w - u, E V,. Consequently w - u E V , for n = 1,2,. . . . But then w - u E N n Nd, and therefore u = w E S . Now by Theorem 12.3 there exists a net {u,} E C, with 8 I u, (u(.By the above we have {u,} E S and so by the order closedness of S, IuI E S. But then since S is also solid, we get u E S , i.e., S is z-closed. Now if L is not Dedekind complete use Theorem 11.10 to extend z to z* on Ld and then put S* = {u* E Ld:3{u,}c S with 8 Iu,

T Iu*l in La}.

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a7

THE STRUCTURE OF FATOU TOPOLOGIES

Then S6 is solid and order closed in L6, and so S6 is ?-closed. Thus S = S' n L is z-closed in L , and we are done. The following theorem shows a striking relation between Fatou and Lebesgue topologies. Theorem 12.8. Assume that a Riesz space L is equipped with a HausdorFatou topology z1 and a Lebesgue topology z 2 . Then the topology induced by z1 on any order bounded subset of L isjner than the topology induced by t 2 on the same set.

Proof. According to Theorem 11.10, we can assume without loss of generality that L is Dedekind complete. Let 8 < u E L. It is enough to be shown that the identity mapping from [8,u] to itself is z1 - z2 continuous at zero. So, let (u,} E [8,u] with u, 2 8 and assume that u, ';4 0. Then there exist two solid z,-neighborhoods ofzero Vand U , with V + I/ c U and such that given any a there exists fi 2 a with u, 4 U. Now since the carrier C,, of 71 is order dense in L and since z2 is a Lebesgue topology, there exists w E C,, with 8 I w I u and u - w E V. Pick a normal sequence { V,} of Fatou z,-neighborhoods of zero with w E Nd, where N is the null ideal of {K}. Since w A u, 8, there exists a sequence {a,,} of indices such that uan4 U and w A U , E~ V,, for all n. But

sup{w A uam:n _< m I n

+ p ) I 1 w A u , E~ V,, + . . + V,,,, ,E V , n+p

*

m=n

for n,p = 1,2,. . . , and so sup{w A uQm:m2 n} E V , holds for all n. Thus w 1 = limsup w A ua, E V , for n = 1,2, . . . , and so w1 E N n Nd, i.e., w1 = 8. This shows that w A u,, 2 8 holds in L and so w A u," % 0 must also hold since z2 is a Lebesgue topology. But we then have 8
As another corollary of Theorem 12.8 we have the following. Theorem 12.10. Let T:(L,z)+ (M,z') be a normal Riesz homomorphism. Assume that z is a HausdorfS Fatou topology and z' a Lebesgue topology. Then T restricted to any order bounded subset of L is continuous.

FATOU TOPOLOGIES

00

[Chap. 4

Proof. Let { V ) be a basis for zero for 7’.Then { T I ( V ) } is a basis for a locally solid topology z* on L. Since T is order continuous and z’ is a Lebesgue topology, z* is a Lebesgue topology. Thus T:(L,z*)--t (M,z’) is continuous and by Theorem 12.8 z* G 7 holds on every order bounded subset of L. The result is now immediate.

The fact that all Hausdorff Lebesgue topologies agree on the order bounded subsets is by no means enough to ensure that a Riesz space can admit at most one Hausdorff Lebesgue topology. The following examples illustrate this possibility as well as Theorem 12.8. Example 12.11. Let L = L,([O,l]) be the super Dedekind complete Riesz space consisting of the equivalence classes of the bounded (almost everywhere) Lebesgue measurable functions on [0,1]. Let z1 and 72 be the Hausdorff Lebesgue topologies on L generated respectively by the Riesz norms pl(u) = Jol lu(x)ldx

and

p2(u) =

(J:

lu(x)12dx)1’2

for u E L.

Note that z1 # r 2 . Actually the Hausdorff Lebesgue topologies generated by the Riesz norms

are all distinct for 1 I p < co. Example 12.12. Let X be an infinite set and let L = l m ( X ) .Let z1 be the Hausdorff Fatou topology (but not Lebesgue) generated by the sup norm and z2 the Hausdorff Lebesgue topology of the pointwise convergence. Then z2 G tl,but z1 # z 2 . Compare this with Theorem 12.8.

The next theorem tells us about the relation between the countable sup property and the Hausdorff Lebesgue topologies. Theorem 12.13. Let (L,z) be a Hausdorf locally solid Riesz space with the Lebesgue property. Then the following statements are equivalent:

(i) L has the countable sup property. (ii) C , = L (where C , is the carrier of t). (iii) z induces a metrizable topology on each order interval of L. Proof. (i) 3 (ii) It follows immediately from Theorem 12.3. (ii) * (iii) Let u E L’. Then G Nd must hold for some normal sequence ( K } of Fatou t-neighborhoods of zero, where N is the null ideal of (K}. Now z restricted to Nd satisfies the Lebesgue property. Also { Vn}

[@,MI

Sec. 121

89

THE STRUCTURE OF FATOU TOPOLOGIES

defines a metrizable Lebesgue topology on Nd. But then according to Theorem 12.9, both topologies agree on [6,u] and so z induces a metrizable topology on [O,u]. (iii) + (i) Assume u, J. 8 in L. Without loss of generality we can suppose that 0 5 u, I u holds in L for all a and some u E L. Let { V,} be a normal sequence of Fatou z-neighborhoods of zero such that { V , n [8,u]} is a basis of zero for the topology induced on [O,u] by z. For each n pick an index a, with u,, E V , n [6,u] (this is possible since the topology z is Lebesgue). uanholds for all n in L, then u E V , n [6,u] We can assume u,* 1in L. If 6 I u I for all n. Hence u = 6 and so uan 1 6 holds in L. This shows that L has the countable sup property. We close this section with a comparison property of the Fatou topologies. Theorem 12.14. Assume that un Archimedean Riesz space L is equipped with a Fatou topology z and a locally solid topology z’. Then the following statements are equivalent:

(i) (ii) (iii) carrier

is coarser than z’, that is, z c z’. z induces a coarser topology on its carrier C, than z’. z induces a coarser topology than z‘ on each component Nd of its z

C,.

Proof. The implications (i) G- (ii) and (ii) 3 (iii) are obvious. (iii) + (i) Let V be a Fatou z-neighborhood of zero. Pick a normal sequence { V , ) of Fatou z-neighborhoods of zero with V, + V, c V and note V,. By hypothesis there that N @ Nd is order dense in L, where N = exists a solid z’-neighborhood U of zero such that U n Nd E Vl n Nd. Now let u E U . Then since L is Archimedean and N @ Nd is order dense in L, there exists a net {u, + u,} c N 0 Nd such that 6 5 u, + u, t IuI in L. But then { u,} c N c V, and since { u, ) G Nd,8 I u, I IuI for all a, and U is solid, we have { u,} c U n Nd c V, n Nd E V, .Thus {u, u,} G Vl Vl C V and hence lul E V, since V is order closed. Therefore, u E V and so U c V; which shows that z E z’.

n;=l

+

+

Note. The carrier C, of a locally solid topology z appeared implicitly in the work of Kawai [39] and later in [25, p. 501. Theorem 12.7 was proven by Amemiya for the Dedekind complete case [lo, Theorem (B,6), p. 221; the general case was established by Fremlin [25, 23L Proposition, p. 501. Theorem 12.9 (for the Dedekind complete case) appeared in a paper by Amemiya and Mori [ l l , p. 1351 and the same theorem as well as Theorem 12.8 were proven for the Dedekind complete case by Amemiya [10,Theorems (B,4) and (B,5), pp. 21-22] by a Baire-type category argument; the general proof given here is due to the authors [7].

FATOU TOPOLOGIES

90

13.

[Chap. 4

TOPOLOGICAL COMPLETENESS AND FATOU TOPOLOGIES

We start this section with one of the deepest results of the theory of locally solid Riesz spaces.

Theorem 13.1. If (L,r) is a Dedekind complete locally solid Riesz space with the Fatou property, then the order intervals of L are r-complete. Proof. Assume first that z is also a Hausdorff topology. By Theorem 7.3 we have only to show that L is an ideal of L. We shall establish this by showing that L is order dense in A(L),the ideal generated in t by L ; it will then follow from Theorem 2.2 that L = A(L). To this end let 8 < ii I u E L. Pick a normal sequence of Fatou z-neighPick a net borhoods { V,) of zero of L with ii q! Vl (the ?-closure of V, in { v a } of L', with 8 I u, I u for all CI and v , 4 ii. Next, choose a sequence {a,} E { a } such that u, = van(n = 1,2,. . .) satisfies

e).

ii-U,E

and

u,+,,-

for all n and p .

u,E~/,+~

It follows from Lemma 12.5 that w = lim sup u, satisfies w - u, n and thus ii - w E for all n. Now let

E

V , for all

n v,}. m

S + ~ L + : ~ - ~ ~ N = n= 1

Obviously S is nonempty. Put s = infS in L and note that since N is a band of L, s - W E N and thus s > 8 (s= 8 implies W E N and consequently ii E V, ; a contradiction). Now let W, be a Fatou z-neighborhood of zero for L with W, c V,. Pick a normal sequence { W,} of Fatou z-neighborhoods of zero of L with W, c V , for n = 1,2,. . . . Apply the same arguments as above to obtain an element a E L' with ii - a E W,for all n. Next use Theorem 12.7 to get a - w E N. Thus s I a and so 8 I (s -')i I ( a -')i E W,. Hence (s -')i E Wl for all the Fatou z-neighborhoods of zero W, with W , c V,. But then (s - ')i = 8 and so 8 < s Iii, that is, L is order dense in A(L). Now if z is not a Hausdorff topology, consider the kernel of the topology K , = { V :V is a z-neighborhood of zero) and note that K , is a band of L. Thus L = K , 0 K,d and by Theorem 11.8 z induces a Hausdorff Fatou topology on Kid, and since K,d is Dedekind complete by the above proof, the order intervals of K,d are z-complete. Finally, since every net of K , is z-convergent to zero and the order intervals of K,d are z-complete, we infer that the order intervals of L are z-complete. A sequential analog of the preceding theorem is presented next.

Sec. 131

TOPOLOGICAL COMPLETENESS

91

Theorem 13.2. Let (L,?)be a Hausdorff Dedekind complete locally solid Riesz space with the a-Fatou property. Then the order intervals of L are sequentially z-complete. Proof. Put La = {ii E L : 3 { u n }c L with u, A ti]. Note that La is a Riesz subspace of L containing L as a Riesz subspace and that L is sequentially t-complete if and only if L = La. Now assume 0 < ii I u E L with ii E La. Pick a sequence {u,} c L f with 8 Iu, Iu for all n and with u, ii. Let V be a a-Fatou t-neighborhood of zero of L with ii f$ V . Choose a normal sequence of a-Fatou z-neighborhoods {K} of zero of L with V, = V . By passing to a subsequence (if necessary) we may suppose u,+ - u, E V,,, for all n. By Lemma 12.5 w = lim infu, in L satisfies w - u, E V, c V for all n. Note that 8 < w, for otherwise w = 8 implies ii E V , which is a contradiction. On the other hand, if W is any other a-Fatou ?-neighborhood of zero for L, choose a normal sequence { W,] of a-Fatou z-neighborhoods of zero of L with W, = W.Next pick a subsequence ( v , } of (u,} with u,+ - v, E W,, for all n. Set v = lim inf v, in Land note that 8 < w I u and v - u, E W, holds for all n (by Lemma 12.5 again!). But then 8 I ( w - C)' I ( v - ')i E W implies (w - ')i = 8, i.e., 0 w I ii. Hence L is order dense in La and so by Theorem 2.2 L is an ideal in La.Therefore, the order intervals of L are sequentially t-complete. Remark. Following closely the preceding proof, we readily see that a more general result holds. Namely, if (L,?)is a Hausdorff a-Dedekind complete Riesz space with the a-Fatou property, then for every 0 Iii E La there exists a net {u,) c L satisfying 8 I u, t ii and u, 1,ii. The next example shows that the Dedekind completeness in Theorem 13.1 cannot be weakened to the projection property.

,

-=

Example 13.3. Let L be the Riesz space of all real-valued functions defined on an infinite set X whose range is finite, with the pointwise ordering. It can be seen easily that L has the projection property but it is not Dedekind complete. Let t be the Hausdorff Lebesgue topology of pointwise convergence. Note that the order intervals of L are not z-complete. Observe also that L= RX and L' = lm(X). Additional important properties of the Fatou topologies are included in the next theorem. Theorem 13.4. Let (L,?)be a Hausdorff locally solid Riesz with the Fatou property. Then we have (i) L is order dense in its topological completion L. (ii) T h e topological completion (E,?) of (L,?) also satisJies the Fatou property.

FATOU TOPOLOGIES

92

[Chap. 4

Proof. Assume first that L is also Dedekind complete. Then, according to Theorem 13.1 L is an ideal of t and so by Theorem 7.3 L is order dense in L. Now let { V } be a basis of Fatou 5-neighborhoods of zero and V be the ?-closure of V in L. We know V is a solid subset o f t . To see that V is also order closed let H I ii, T ii in t with {ii,} c V . According to Theorem 7.3 there exists a net { vA} c L with 6 I uA 7 ii and vA 1,ii. We shall show that {uA}G I/ and this will imply that ii E V . For each CI there exists a net of L + that increases to 6,. Since ii, E V this net must lie in V . Taking the finite suprema of all those nets, we get a net {w,} E V with H I w, 7 ii. But for each fixed A, 6 I w, A t’A7 uA holds in L and since I/ is order closed we get U

uA E I/ for all A. Now since { Vj forms a basis of zero for ? consisting of solid and order closed sets, z^ is a Fatou topology. If L is not Dedekind complete, extend z to 5’ on Ld according to Theorem 11.10. T h F as above (Ld,,zd)satisfi? the Fatou propegy and Ld is order dense in L*. Now observe that the r’-closure of L in L* is the topological completion of (L,z). Hence L is order dense in Land by Theorem 11.8 (t,?) satisfies the Fatou property. Observe also that if ( V } is a basis of Fatou t-neighborhoods for zero, then { where

v},

P = {ii E t : u E L and 8 I L; I liil implies u E V } forms a basis of Fatou neighborhoods of zero for f., As an application of the last theorem we have the following.

Theorem 13.5. Let (L,.r) be a Hausdorf locally convex-solid Riesz space whose topology is generated by a family of Fatou seminorms { p a } .For each CI let p^, denote the unique continuous extension of p, to L. Then we have (i) Each ba sutisjies p^,(ii) = sup(p,(u):u E L and H s u I liil} for 6 E (ii) Each p^, is a Fatou seminorm of (iii) T h e family {p,} generates the Fatou topology f.

e.

t.

Proof. (i) If p is a z-continuous Fatou seminorm on L, define p*(ii) = sup{p(u):u E L-and 8 I u I liil} for ii E t,and note that since p^ is a Riesz We shall show that seminorm of L, we have p*(ii) I p^(ii) for all ii E p* = p^ on t,and this will be enough to establish (i). To do this we need to establish the triangle inequality for p*. So let ii,6 E L. By Theorem 13.4 uy T 1 61 in L. there exist two nets { u p } and {u,) of L with 6 I up T liil and 0 I Now if w E L satisfies H I w I liil + 161,then 8 I (up + u y )A w t w holds in

e.

(B.Y)

L and so since p is a Fatou seminorm of L, we get p( (up + vy) A w ) 7 p(w). But then p( (up + v,) A w) I p(up u y ) I p ( u p ) p(o,) I p*(ii) p*(6) implies p ( ~I) p*(ii) + p*(G), so that p*(ii + 6) I p*(ii) + p*(v^).

+

+

+

Sec. 131

TOPOLOGICAL COMPLETENESS

Now let ii E inequalities

93

e. Choose a net {u,} s L with u, f ii, and note that the

p*(ii) 2 p*(u,) - p*(u, - ii) 2 p(u,) - p^(u, - ii) = p^(u,) -,p^(u, - a)

imply p*(ii) 2 p^(ii). Hence p* = p^ on L. (ii) Let p be a t-continuous Fatou seminorm on L, and let 8 I ii, t ii in L. Put a = lim /j(ii,). We have to show p^(@ I a. For each ii, there exists a net of L+ that increases to 6,. By taking the finite suprema of these nets we get a net ( u s ) of L f satisfying 0 I us ii; in particular, it follows that lim p(us) = a. Now if w E L satisfies 0 5 w I ii, then the relations

8I

A

w w in L,

p(up A w) I p ( u p ) I a,

and the fact that p is a Fatou seminorm of L imply p(w) I a. Thus p^@) and we are done. (iii) Obvious.

I a,

Theorem 13.6. Let L he a Riesz space with the projection property. Then we hare

(i) l f t is a Hausdorf Farou topology on L, then is Dedekind complete. (ii) If L admits a Hausdorff locally solid topology t for which the order intervals qf L are t-complete, then both L and Lare Dedekind complete. Proof. (i) Apply Theorem 7.10. (ii) By Theorem 7.10 L is Dedekind complete and hence so is L, since L is an ideal off. (Theorem 7.3). Remarks. (1) If a Hausdorff locally solid Riesz space (L,t) has the Fatou property and L has the projection property, then the Dedekind completion Ld of L can be taken to be A(L),the ideal generated in by L. Hence by Theorem 11.10, t6must be the restriction of ? to A(L). So, in this case we have the following Riesz subspace inclusions:

e

(L,r) E (La,?)

c (L,?).

(2) If (L,t) is a Dedekind complete Hausdorff locally solid Riesz space with the Fatou property, then is also Dedekind complete.

e

Note. Theorem 13.1 is due to Nakano [51, Theorem 3.3, p. 931. Nakano's proof, however, contained a number of gaps, filled later by Schaefer [65, Theorem 1, p. 3041. The proof presented here is due to the authors [5, Theorem 3.2, p. 281 ; see also [ 6 ] . For some interesting applications of Theorem 13.1 see [30]. The first part of Theorem 13.4 appears in a paper of Kawai [39, p. 2981 for the locally convex case; the theorem as it is stated here was proven first by Fremlin [24, Theorem 1, p. 3431. The proofpresented here is due to the authors [ S , Theorem 4.11.

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We recall that a locally solid Riesz space (L,z)satisfies the Levi property if every increasing z-bounded net of L+ has a supremum in L (Definition 9.3). Note that if (L,z) satisfies the Levi property, then L is Dedekind complete and t is a Hausdorff topology (see Exercise 1, Chapter 3).

Definition 13.7. A locally solid Riesz space (L,z) is said to be a Nakano space if(L,t)satisfies both the Levi property and the Fatou property. The next three theorems give conditions for the topological completeness of locally solid Riesz spaces.

Theorem 13.8. Let (L,z) be a locally solid Riesz space with the Levi property. I f the order intervals of L are z-complete, then (L,z) is z-complete. Proof. Let 8 IQ E t.By Theorem 7.3 there exists a net {u,) G L f with 8 Iu, t 6. Since {u,} is z-bounded, 8 Iu, f u must hold in L for some u E L, and so since L is an ideal off,, u, f u also holds in t.Hence Q = u E L

and so L = t, i.e., (L,z)is r-complete.

Theorem 13.9. Every Nakano space (L,z) is z-complete. Proof. Apply Theorems 13.1 and 13.8. We rec$l that a locally solid Riesz space (L,z) satisfies the monotone completeness property (MCP) if every monotone z-Cauchy net of L is z-convergent (Definition 7.4).

Theorem 13.10. Let (L,t)be a Hausdorff locally solid Dedekind complete Riesz space with the Fatou property. If (L,z) satisfies MCP, then (L,z) is z-complete. Proof. By Theorem 13.1, L is an ideal of Theorem 7.7.

f, and

the result follows from

Corollary 13.11. Let (L,z) be a Hausdorff locally solid Riesz space with the Lebesgue property. I f (L,z) satisfies MCP, then (L,z) is z-complete (and hence L must be also Dedekind complete). Proof. We need only to show that L is Dedekind complete. To this end let 8 Iu, T I u in L. Then {u,} is a z-Cauchy net of L (Theorem 10.1) and so by hypothesis u, 1,v holds in L. It follows now from Theorem 5.6 that 8I u, T u holds in L and we are done. We continue next with a theorem concerning sequential t-completeness.

Theorem 13.12. Let (L,t) be a Hausdorf a-Dedekind complete locally solid Riesz space with the a-Fatou property. I f (L,z) satisjes MCP, then (L,z) is sequentially t-complete.

Sec. 131

95

TOPOLOGICAL COMPLETENESS

Proof. By the remark following the proof of Theorem 13.2 we know that for every 8 I ii E L, = {ii E L : ~ . ( Kc! }L with u, f ii} there exists a net {u,} of L with 8 I u, f ii and u, 1,6. In particular, note that {u,} is a zCauchy net of L. Now since {u,} is monotone and L satisfies MCP, we get that u, 1,u holds for some u E L. Thus ii = u E L and hence L = L,, so that L is sequentially t-complete.

Note. Theorem 13.8 is due to the authors [ 5 , Theorem 5.4, p. 321. Theorem 13.9 is due to Nakano [51, Theorem 4.2, p. 961; the simple proof presented above is due to the authors [5, Corollary 5.5, p. 321. We close this section with an example due to Fremlin, which shows that a Hausdorff locally solid Riesz space with the Fatou property and with MCP need not be topologically complete. Example 13.13. Let K , be the set of all real-valued functions f defined on [0,1] such that lim,l,f(t) exists in R for every x E [0,1) (alternatively f belongs to K , whenever limf(t,) exists in R for every strictly decreasing sequence (t,,) (i.e., t,+ < f,, for all n) of [0,1]). With the usual operations of addition and multiplication and with the pointwise ordering K , is an order dense Riesz subspace of Rt0*'](note that x{,)E K , for every t E [OJ]). Note that f o r f e K , and E > 0 the set

,

Ob = { t E [O,l]:the oscillation off at t is greater than

EJ

does not contain strictly decreasing sequences and thus OEmust be countable; therefore, the set of discontinuities off is at most countable. Now let S , be the collection of all sequences s = {t0,t,,t2, . . .} of [0,1] with t, 1t o(including the constant sequences). For every s = {t,,f,,t,, . . .} E S , define the Fatou seminorm ps on K by p , ( f ) = sup{lf(tk)l :k = 0,1, . . .} forf E K , [the supremum is finite since limf(tn) exists in R ] . Clearly the family of Fatou seminorms { p s : sE S , } defines a Hausdorff Fatou topology t1on K , , and as can be easily seen, (K,,z,) is 7,-complete. Similarly, let K , be the order dense Riesz subspace of R[O*']consisting of all functions f for which lim,t,f(t) exists for every x E (0,1]. Let S , be the set of all sequences s = {to,rlrt,, . . .} of [OJ] with t , t t o and define p,(f) = sup{ If(tk)l:k = 0,1,. . .} forf E K,. Let z2 be the Hausdorff Fatou topology generated by { p s : sE S , } on K,. Note that (K,,z,) is z,-complete. Now put K = K , x K , and z3 = z1 x z,. Then ( K , z 3 )is a Hausdorff locally solid Riesz space, with the Fatou property, that is also z,-complete. Next define L = { ( u p )E K :u(t) = u ( t ) if u and u are both continuous at t } .

FATOU TOPOLOGIES

96

[Chap. 4

Then L is a Riesz subspace of K (this requires the fact that members of both K , and K , are continuous except on countable sets). Also, since L contains (x,,),O) and (O,xI,,) for every t E [0,1], it follows that L is order dense in K and hence the topology t induced by t3 on L is a Hausdorff Fatou topology. We leave it to the reader as an exercise to verify the following: (1) (L,t) satisfies the Monotone Completeness Property. (2) (L ,t)is not t-complete. ( 3 ) L is t,-dense in K , that is, (L,?) = (K,tg). Note also that the set S = {(u,O)E L : u ( l )= 0 and (u(r)l I 1 for t E [OJ]} is a solid and order closed subset of L, which is also t,-closed in K = t. However, S is not order closed in L. Thus, the closure in of a solid and order closed subset of a Hausdorff locally solid Riesz space (L,t) with the Fatou property need not be an order closed subset of L.

e

14.

QUOTIENT RlESZ SPACES AND FATOU PROPERTIES

In this section we shall study the conditions under which the quotient Riesz spaces of a locally solid Riesz space (L,t) with the Fatou property (or the o-Fatou property) inherit the same property. To indicate that general results cannot be obtained, we shall start with two examples. In the first example we shall present a normed Riesz space L, with the Fatou property that is almost o-Dedekind complete with the principal projection property and a band A of L such that the quotient space L / A (with the quotient topology) fails to satisfy the a-Fatou property. Example 14.1. Let L be the Riesz space of all real sequences that are ultimately constant with the pointwise ordering. Note that L is almost oDedekind complete with the principal projection property but without the projection property and without being o-Dedekind complete. Let t be the Hausdorff Fatou topology on L generated by the Fatou norm p(u) = sup(lu(n)l:n = 2,4,6,.

. .} + sup{2-"lu(n)l:n

= 1,3,5,.

. .}

for every u = { u ( n ) }E L. Now let A be the band A = { u E L : u ( n )= 0 for n = 1,3,5, . . .}.

Observe that the quotient Riesz norm p that generates the Hausdorff locally solid topology t/A on L / A is given by

p(a) = lu(co)l + sup{2-"lu(n)l:n = 1,3,5,. . .},

where u(co) denotes the ultimate constant value of u.

Sec. 141

97

QUOTIENT RlESZ SPACES

Now by defining u, E L by u,(k) = 1 for k = 1 , . . . ,n and u,(k) = 0 for k > 11, we see that 0 I u, t e = ( l , l , l , . . .) in L and since A is a band of L it follows that I9 I 12, P in L/A. Now note that p(Li,,+,, - Li,) I 2-" implies that (Li,) is a P-Cauchy sequence and so for a a-order closed t/A-neighborhood I/ of zero the relation I9 I ti,+,, - Li, t P - Li, in L/A implies P - Li, E V for all sufficiently large n. P

Thus if L/A satisfies the a-Fatou property, then Li, + P, i.e., p(Li, - P) -+ 0, and in particular, p(Li,) -+ p(P). However, the last conclusion is impossible since P(Li,) = 2- for all n and P ( P ) = 3/2, so that (L/A,t/A) does not satisfy the a-Fatou property. rl.4

As a second example we present a a-Dedekind complete locally solid Riesz space with the Fatou property and a band A of L such that the quotient space L/A fails to satisfy the Fatou property. Example 14.2. Let L be the Riesz space of all bounded real-valued functions u on the unit interval with the property that u ( x ) # u ( 0 ) holds for at most countably many x E [O,l]. Note that L is a a-Dedekind complete Riesz space. Let z be the locally solid topology generated by the Riesz seminorm p(u) = sup(lu(x)l:$ I x I 1).

Now consider the band A = { u E L:u(.y) = 0 for s E [O,))). Even though I9 I u, 7 u in L does not necessarily imply u,(x) t u ( x ) for all x E [0,1], it can be seen that p is a Fatou seminorm and so that z is a Fatou topology. But (L/A,t/A) fails to have the Fatou property as we now show. Let a range over the family of finite subsets of (0,1], and let u, = xb be the characteristic function of a. Clearly 6' I u, t e, where e(x) = 1 for all x, holds in L and so 0 I Li, P holds in L/A, since A is a band. Observe now that p($) = 0 and p(P) = 1. It follows now that (L/A,z/A) does not satisfy the Fatou property. We continue next with some affirmative results.

Theorem 14.3. Let (L,T) be u a-Dedekind complete Riesz space with the a-Futou property, and let A he u a-ideul of L. Then the quotient Riesz space (L/A,z/A) satisjies the a-Futou propert!'. Proof. It is enough to show that the canonical image of a a-Fatou tneighborhood of L is a a-Fatou z/A-neighborhood of L/A. To this end, let I/ be a a-Fatou z-neighborhood of L. Note first that = n ( V ) is a solid subset of L/A (Theorem 1.18). On the other hand, observe that if 0 I Li, t Li holds in L/A with {Li,,) E P, then we can assume without loss of generality that f3 I u, I u and (u,} G I/ both hold in L, and so by the a-order closedness

98

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[Chap. 4

of V , we get w = lim infu, E V . But since A is a a-ideal the canonical projection of L onto L/A is a Riesz a-homomorphism and thus ti = sup (ti,}

that is,

= lim inf ti, = \ii E

V,

p is a-order closed and the proof is finished.

Now assume that A is a projection band of a Riesz space L, i.e., assume L = A @ Ad. If 4 denotes the restriction to Ad of the canonical projection n : L + L/A, then q :Ad + L / A is an onto Riesz isomorphism. If in addition z is a locally solid topology on L, then q is also continuous if Ad is equipped with the induced topology z and L / A with z/A. On the other hand, the diagram below ( P denotes the projection of L onto A d ) (L,z)

'(L/A,T/A)

k i[

(Ad!; '

and the basic properties of the quotient topologies imply that 4 - is also continuous, and hence (L/A,z/A)and (Ad,z)are Riesz homeomorphic. In particular, it follows that if z is a Fatou (or a-Fatou) topology, then so is its restriction on Ad. Therefore, we have proven the following theorem. Theorem 14.4. Let (L,z) be a locally solid Riesz space with the Fatou property (resp. a-Fatou property), and let A be a projection band of L. Then (L/A,z/A)satisfies the Fatou (resp. the a-Fatou) property. There is one more positive result, concerning the a-Fatou property, that we will establish. Theorem 14.5. Let (L,z) be an almost a-Dedekind complete Riesz space with the a-Fatou property, and let A be a a-ideal of L. Suppose that the ideal A* generated by A in the a-Dedekind completion L" of L is likewise a a-ideal of La. Then (L/A,z/A)satisfies the a-Fatou property. Proof. By Theorem 11.11 we can extend z to a a-Fatou topology on L"; where if V is a solid a-Fatou neighborhood of zero for z, then V* = ( u E L":3{u,) E V with 8 I u, t is a a-Fatou za-neighborhood of zero for L" and V = V* n L. Now let V be a a-Fatou z-neighborhood of zero of L ; we need to show that n ( V )= is a a-Fatou z/A-neighborhood of L/A. To this end let 0 I a, t ti in L/A with {a,} G I'; without loss of generality we can assume 8 < u, < u for all n and {u,} c V . Then u = lim inf u, E V* and we claim u - u E A*. Indeed, since L is super order dense in L", there exists a sequence {xk)C L+ with xk t u - u. We shall show that {xk) c A and this will

1.)

Sec. 141

QUOTIENT RIESZ SPACES

99

imply that u - u E A* since A* is a a-ideal. For each fixed n set y , inf{ui:n I i I n m>and note that for each k,

+

8 Ixk I u-uI u - inf{u,:i 2 n ) = SUP,{U

r

=

- y,}

implies ( u - ym) A xk f xk in L and so (ti - j m )A kk kk in L / A . But then m

m

A kk = ( d - j,) A kk for each m and so ( d - ti,) A ik= kk for all n. It follows now from ti, t li that kk = 8, so that xk E A holds for all k . Since A* is the ideal generated by A in L" and u - u E A*, there exists w E A with u - u I w. Thus (u - w)' Iu E I/* and so (u - w)' E V . But we Hence 3 is a-order closed, and therefore then have ti = (d - w)' E ( L / A , r / A )has the a-Fatou property.

(d - ti,)

v.

The next theorem gives a condition for A* to be a a-ideal in L". Theorem 14.6. Let L be an almost a-Dedekind complete Riesz space, and let A be a a-ideal of L with the property that every band B contained in Add is a projection band. (This is equivalent to the assumption that Add is a projection band of L, and that Add,considered as a Riesz space in its own right, has the projection property.) Then the ideal A* generated by A in L" is a a-ideal.

r

Proof. Suppose that 8 I u,* u* with { u n * } G A*. Since each u,* is the upward limit of a sequence of positive elements of L, using a diagonal u, t u*. There process, we can find a sequence {u,) of elements of A with 0 I exists v E L with u* 5 v ; let u be the projection of u onto the band Add.Since {u,} E A c Add, certainly u, Iu for all n and thus u* I u E Add. Let B,, be the band generated in L" by u* and let B = B,, n Add. Then B is a band of L and since B c Add,B is a projection band. Let x = P,(u). For every n, clearly u, E B so that P,(u,) = u,, and therefore, u, = P,(u,) I P,(u) = x E B, implying that u* I x. But x E B E B,, implies that x = sup{x A nu* : n = 1,2, . . .} in L". Let t, = x A nu, E A . Since u, 7 u*, it follows that 8 5 t , t x in L. But {r,} c A and A is a a-ideal so that x E A . Since u* I x, necessarily u* E A*, and we are done. As an application of the last two results we have the following theorem. Theorem 14.7. Let (L,z) be a locally solid Riesz space with the a-Fatou property and let A be a a-ideal of L. I f L has the projection property then ( L / A , t / A )satisfies the a-Fatou property.

Proof, Apply Theorems 14.5 and 14.6 by noting that L is almost aDedekind complete (Theorem 2.12). W Note. All the results of this section were first proven by Langford and Aliprantis for seminormed Riesz spaces in [42, Section 5, pp. 205-2101; see also [3].

100

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[Chap. 4

EXERCISES

1. Prove Theorem 11.11. 2. Let (L,T)be a Hausdorff almost a-Dedekind complete Riesz space with the a-Fatou property. Show that T can be extended to a o-Fatou Hausdorff topology on Ld. 3. Show that the following stronger version of Theorem 12.10 holds: If T :(L ,z)-+ (M,T') is a positive order continuous linear operator and if T is a Hausdorff Fatou topology and T' a Lebesgue topology, then T restricted to the order bounded subsets of L is continuous. (Hint: If V is a solid 7'-neighborhood of zero let V* = { u E L :T(lul) E V } . ) 4. Let (L,z)be a locally solid Riesz space and let {N,:a E A } denote the family of all null ideals of the normal sequences of solid z-neighborhoods of zero. Show that C, is order dense in L if and only if A Nzd = {Bj. 5. Let (L,z) be a Hausdorff locally solid Riesz space with the Fatou property and consider the Fatou extension 2' o f t to Ld (see Theorem 11.10). Show that C,a n L = C,. 6. If T is a Fatou topology on a Riesz space L with the projection property, show that C, is .t-dense in L. 7. Consider the Hausdorff locally solid Riesz space (L,z)of Example 8.6. Determine the carrier C, and verify that C, is order dense but not z-dense in L. Is z a Fatou topology? 8. For a Hausdorff locally solid Riesz space (L,t)with the Fatou property show that C, c C, holds for every cp EL-. 9. Let L be a Riesz space with the projection property. If T is a locally solid topology on L with the a-Fatou property, show that: (i) The kernel K , of T (i.e., the intersection of all z-neighborhoods of zero) is a z-closed a-ideal of L. (ii) (L/Kr,z/K,) is a Hausdorff locally solid Riesz space with the a-Fatou property. 10. If T is a Fatou topology on a Riesz space L with the projection property, show that (L/K,,z/K,) is a Hausdorff locally solid Riesz space with the Fatou property. 11. Consider the Hausdorff locally solid Riesz space (L,z) of Example 13.13. Verify the three claims about (L,z). 12. If p is a Fatou norm on a Riesz space L [i.e., 0 < u, u implies p_(u,) t p(u)] show that the norm extension p^ of p to the norm completion L, is a Fatou norm and that p^(ii) = sup{p(u):u E L and B I ti I 161) for all

nuE

ii E L,,.

13. Let L be an Archimedean Riesz space and p a Fatou seminorm on L. Define the ideal I,, = ( u E L : p ( u )= O} and show that the quotient Riesz norm p on L/I, is a Fatou norm.

EXERCISES

101

14. (Aliprantis [3]). Let L be an almost a-Dedekind complete Riesz space and let p be a a-Fatou seminorm of L [i.e., 8 I 11, u implies p(u,) T p ( u ) ] . If I, = ( u E L : p ( u )= 0) show that the quotient Riesz norm 0 on L/I, is a o-Fatou norm. 15. (Aliprantis [3]). Let X be an uncountable set, X , the one-point compactification of X considered with the discrete topology, and {.Y, ,x2,. . .) a countable subset of X . Define the Riesz seminorm p on L = C ( X , ) by p ( u ) = Iu(rx;)I sup{Iu(xi)I:i= 1,2,. . .$and show that: (i) p satisfies the a-Lebesgue property but p is not a Fatou seminorm. (ii) I , = j u E L : p ( u )= 0) = ( u E L:u(x,)= 0 for i = 1,2,. . and infer from this that I , is a band of L. (iii) The quotient norm 0 on L/I, is not a o-Fatou seminorm. Why does this not contradict the previous exercise? 16. Show that a Fatou topology z on a Riesz space L is a Hausdorff topology if and only if C, is order dense in L.

+

.I,

OPEN PROBLEMS

1. If (L,z) is a Dedekind complete Hausdorff locally solid Riesz space with the monotone completeness property, does it follow then that (L,T)is z-complete? 2. Does the conclusion of Theorem 13.2 remain true if the hypothesis of Dedekind completeness is weakened to a-Dedekind completeness'? 3. Is the o-Fatou property preserved under the process of topological completion?

CHAPTER

5

Metrizable Locally Solid Riesz Spaces

In this chapter we shall deal with metrizable locally solid topologies, that is, with topologies that in addition to being locally solid are generated by a metric. This is equivalent, of course, for Hausdorff linear topologies to the existence of a countable basis of zero consisting of solid sets (Theorem 4.1). In particular, the class of metrizable locally solid Riesz spaces includes the class of normed Riesz spaces. The metrizability of the topology has a number of interesting consequences. Among the remarkable results concerning metrizable locally solid Riesz spaces (L,z),we shall find that the upper and lower elements of L are enough to describe the structure of the order denseness of L in becomes equivalent to a simple continuity property (u, 1 8 and { u,} z-Cauchy implies u,, 5 8) and the Lebesgue property becomes equivalent to the pre-Lebesgue and 0-Lebesgue properties taken together. On the other hand, we shall see that a Riesz space can admit at most one metrizable locally solid topology that is topologically complete.

e;

15.

e

UPPER AND LOWER ELEMENTS

We start with the definition of the upper and lower elements of a Hausdorff locally solid Riesz space. Definition 15.1. Let (L,r) be a Hausdorfl locally solid Riesz space. We shall call an element 8 I u^ E an upper element of L (resp. a lower element

e

102

Sec. 151

UPPER AND LOWER ELEMENTS

of L) if there exists a net {u,} G L+ with u, i u, -+ 2).

103

t ii and u, 5 ii (resp. u, 1ii and

Note that by Theorem 5.6, u, t and u, A ii are actually enough to ensure t ii in L. It should be also clear that finite sums, finite suprema, or finite infima of upper elements (resp. lower elements) are upper elements (resp. lower elements). The next theorem tells that for metrizable locally solid Riesz spaces the upper and lower elements can be described in terms of sequences. u,

Theorem 15.2. For a metrizable locally solid Riesz space (L,z) an element is an upper element (resp. a lower element) of L if and only if there exists a sequence {u,) c L + with 0 I u, t ii (resp. u, 2) and u, f ii.

8I GE

Proof. Let { V,} be a normal countable basis of zero for (L,z) consisting of solid sets and 0 I ii E be an upper element of L. Pick a net {aa}c L+ with 0 I u, ii and u, ii. For each positive integer n pick an index CI, with ii - uUnE We can choose the sequence (cI,] so that CI, I a,+ holds for all n. It now follows easily that u," f ii and 0 I uUnt ii holds in L, and the proof follows.

E.

The next theorem gives some information about the structure of the topological completion of a metrizable locally solid Riesz space. Theorem 15.3. Let ( L J )be a metrizable locally solid Riesz space. Then euery element of 2 is the difference of two upper elements of L. Proof. It is enough to establish the result for the positive elements of L. So, let 0 I ii E I?,. Pick a normal countable basis { I/n) of zero for z consisting of solid sets. Choose a sequence {u,} c L+ with u, f ii. By passing to a subsequence (if necessary) we can assume without loss of generality that u,+ - u, E V, holds in L for all n. Now put u1 = w 1 = 0 and n- 1

n- 1

un

= u1+ f k= 1

(uk+ 1

-

uk)+,

wt,

= u1-

+ k1 (uk+ = 1

1

- uk)-

for n = 2,3, . . . and observe that both {u,} and {w,}are increasing z-Cauchy sequences of L'. Hence there exist 6,iG E such that u, f 6 and w, f iG. In particular, note that both 6 and iG are upper elements of L. Now observe that u, = u, - w, holds for n 2 2 and so by taking the ?-limits we get ii = v^ - 6,and the proof is finished.

e

Next we continue with two important approximation theorems concerning metrizable locally solid Riesz spaces.

METRIZABLE TOPOLOGIES

104

[Chap. 5

Theorem 15.4. Let (L,z) be a metrizable locally solid Riesz space. Assume that 8 I 3, L and V is a z-neighborhood of zero for (L,z). Then there exists an upper element ii of L with f I ii and ii - f E In particular, it follows that there exists a sequence {ii,} of upper elements of L with ii, J f and 6, f f in

v.

e.

Proof. Pick a normal countable basis {K} of zero for z consisting of solid sets. Choose a neighborhood V, from { V,}, and pick a sequence { f,} c L + with f, f . Now choose a subsequence {g,) of { f,} such that g n + p - gn E h+,+Zfor n,p = 1,2,. . . . Set u, = gn+ 1 + C Y = l 1 g i + 1 - gil for n=1,2, . . . andnotethatsinceu,,,, - ~ , = g , + ~ - g , + , + 1 g , , + 2 - g n + l 1 2 0 for all n, we have 0 I u, t in L. On the other hand, the relation n+p

8 I u n + p - uii = Y n + p +

1

- gn+

1

+ C Igi+ 1 - Yil E i=n+ 1

V,+n+3

+

V,+n+2

c V,

for all n and p implies that {u,} is a z-Cauchy sequence. Hence u, ii holds Thus ii is an upper in for some ii, and therefore u, t ii also holds in element of L. lgi+ - gil E V, ( n = 1,2, . . .), we get Now since 8 I u, - gn+ = 0 I ii - ,f; E & and we are done. The last part follows from the first part by noting that the infimum of any finite collection of upper elements of L is again an upper element of L.

e

z.

,

Theorem 15.5. Let (L,t) be a metrizable locally solid Riesz space, 8 I and V a z-neighborhood of zero for (L,z). Then there exists a lower element ii of L with ii ~f a n d f - ii E In particular, it follows that there exists a sequence {ii,,} of lower elements of L with ii, 7 f and ii, f in L.

f

E

e,

v.

Proof. Assume first that f I u also holds for some u E L. Pick an upper element G of L with u - f I G and G - (u - f ) = f - (u - G) in V . Put ii = (u - G)+ and note that ii is a lower element of L such that ii I f and f - i i E V .

+

Now pick a solid z-neighborhood U of zero with U U G I/ and a sequence {f,} c L f withf, A f .Fix a positive integer n such thatf, - f E r f , and note that f, A ? - f E r f . But then by what was shown above, there exists a lower element ii of L with ii I f, A f; and such that f,A ? - ii E r f . Thus f - ii = ( 7 -.f A j,,) + C f A .f, - ii) E 0 0 c Y , and so f - G E V , and we are done.

+

Sec. 151

UPPER AND LOWER ELEMENTS

105

Note. Theorems 15.4 and 15.5 are essentially due to Luxemburg [44, Note XVI, Theorems 60.3 and 60.6, p. 6481.

We recall that a Riesz subspace L of a Riesz space K is called full in K if for every u E K , there exists u E L with IuI I u. Let us also call a subset A of K full in K if the ideal generated by A is a full Riesz subspace. The next theorem characterizes the metrizability of the absolute weak topology lal(L,L’). Theorem 15.6. For a Hausdorf locally convex-solid Riesz space the following statements are equivalent : (i) T h e absolute weak topology lol(L,L’)is metrizable. (ii) L’ contains a countable full subset. Proof. (i) * (ii) There exists a sequence {q,} G L‘ such that the family of Riesz seminorms {p,,} generates lol(L,L’) [p,,(u) = Iq,l( Iul)]. Put II/, = sup{ Iqkl:lI k I n>for all n. Then 8 I t+b, f holds in L’ and {p,”) generates lol(L,L’).Now since q E L’ is bounded on the closed unit ball of some p*n, there exists a positive number A such that lql I A$,,. This shows that {$,} is full in L‘. (ii) * (i) Observe that the relation a(L,L’)G lal(L,L‘)E z shows that lol(L,L’)is a Hausdorff topology. On the other hand if (q,} is a countable full subset of L‘, then the sequence {p,J generates lal(L,L‘) and hence lal(L,L‘)is metrizable.

For a Riesz space L let S, denote the collection of all Riesz seminorms on L. Clearly the relation p 1 I p z if pl(u) I p2(u) for all u E L+ partially orders S,. It should also be clear that for every pair p1,p2 E S, the function p’(u) = max{p,(u),p,(u)} for u E L defines a Riesz seminorm on L that, as easily seen, is the least upper bound of p1 and p2 in S,. Also if pI,p2E S,, then the function p” defined on L by p”(u) = inf{pl(ul)

+ p2(u2):u1,u2 L + ;u1 + u2 = 1.3 E

is a Riesz seminorm that is the greatest lower bound of p 1 and p2 in S,. To see this note first that p”(u) = p”(lu1) for all u E L. Now if 0 I u I u= u 1 u2 with u1,u2E L + ,write u = u1 + uq with 0 I v1 I u l , 0 Iv2 I u2 and pl(ul) p 2 ( u 2 )I p l ( u l ) p 2 ( u 2 )so that p”(u) I p”(u). For note that p”(v) I the triangle inequality assume u = u1 + u 2 , u = u1 + v2 with the u,’s and vi’s in L+ and then note that

+

+

+

P ” b + 4 I PI(”1 + u1) + P 2 ( 4 + u 2 ) I [Pl(Ul) + PZ(U2)l + [Pl(Ul) + pz(u2)I from which follows p”(u + u) I p”(u) + p”(u); thus for u,u E L p”(u

+

21)

= p”(Iu

+ vl) I p”(Iu1 + Iul) Ip”(u) + p”(v)

[Chap. 5

METRIZABLE TOPOLOGIES

106

holds. It should be also clear that p" is the infimum of p1 and p z in S,. We have thus shown the following theorem. Theorem 15.7. I f S , is the collection of all Riesz seminorms on a Riesz space L ordered by p1 Ip z whenever pl(u) I p z ( u ) for all u E L + , then S , is a lattice. Moreover, the lattice operations are given by the formulas Pl v P Z ( 4 = max{Pl(uXPz(u)l p1 A P , ( U ) = inf(pl(ul)+ pZ(u2):ul,u2E L + ; u1

+ ~2 = )1.

for all p1,p2 E S , and u E L. It should be noted that S , is in general a nondistributive lattice; see Exercise 10 at the end of this chapter. We recall that for a Riesz space L the ideal consisting of all integrals of L- is denoted by L, and the ideal consisting of the normal integrals of Lby L;. For a locally solid Riesz space (L,T)we put L,' = L' n Lzand L,' = L' n Lr. Theorem 15.8. I f (L,z) is a metrizable locally convex-solid Riesz space, then L,' is an order dense ideal of L,, and L,' is an order dense ideal of L;. Proof. We shall prove that L,' is an order dense ideal of L,, the other case will then follow immediately. We need to show that for every B < cp E L," there exists t,b E L,' with 8 < t,b I cp. To this end let { p , } be a sequence of Riesz seminorms generating T such that p n I pn+ for all n and 0 < cp E L;. Choose u E L + such that 0 < cp(u) < 1 and then select 6 > 0 so that cp(u) > 6 > 0. Observe that p(v) = cp(lv1) is a Riesz seminorm of L. For each n, let pn* = pn A p, and we claim that p,*(u) > 0 for at least one n. For suppose if possible that p,*(u) = 0 for all n. Put c1 = cp(u) - 6 > 0, and then choose 6 I w1 5 u such that p l ( w I )+ cp(u - w l ) < E ~ . Then cp(wl)= cp(u) - cp(u - w l ) > cp(u) - E~ = 6 and p l ( w l )< 1. Assume 8 I w, I u has been constructed such that w, I w , - ~ (wo = u), cp(w,) > 6, and p,(w,) < n - ' . Put E , + ~ = cp(w,) - 6 > 0 and note that since p,*+,(w,) = 0, there exists 8 I w , + ~5 w, such that Pn+

Thus cp(W,)

A+l(W,+

-

&I+ 1

1)

= 6.

1(w,+ 1) + d w , - w,+ 1) < ( n + 1)-

< ( n + 1)-

and

d W , + 1)

l&,+ 1.

= d w , ) - d w , - w,+

1)

>

Thus we can construct inductively a sequence 0 I w, 1 such that cp(w,) > 6 > 0 and p,(w,) < n - for all n. Then w, 4 0, for if 0 Iw I w, holds for all n, then for each fixed k, we have P k ( W ) I Pk+,(Wk+,) I( n + k)-' for all n, so that P k ( W ) = 0 for all k and hence w = 6. But now cp(w,) > 6 > 0 for all n contradicts cp E L,, and thus our claim is established.

UPPER AND LOWER ELEMENTS

Sec. 151

107

Now pick n so that p,*(u) > 0 and let E to be the vector subspace generated = by u, i.e., E = { h : AE R } . Define the linear mapping $: E + R by $(h) Ap,*(u), and note that by Theorem 3.4 $ has an extension to L satisfying $ I pn*. In particular, note that II/ E L' and $ I cp. But then II/+ I cp also holds and hence t j + E Lc'.The result now follows by observing that $+ > 8 due to the fact that $+(u) 2 $(u) = p,*(u) > 0. The next results describe some properties of the metrizable o-Fatou topologies. Theorem 15.9. Let (L,z)be a metrizable locally solid Riesz space with the o-Fatou property, and let {u,} be a sequence of L satisfying 8 Iu, I u for all n and u, A u. Then there exists a subsequence { u,) of { u,} and a net { w,} of L+ such that:

(i) For each ct there exists n, (depending upon a ) satisfying w, I u, for all n 2 n,. (ii) 8 I w, u holds in L.

r

Proof. Let { V,] be a normal basis of zero consisting of o-Fatou z-neighborhoods. Pick a subsequence {u,} of {u,} satisfying u - u, E V, for all n. Now let A = { a E L+ : 3n such that a I u k for k 2 n}. Observe that a,b E A imply a v b E A. Thus by defining w, = ct for ct E A , we get the increasing net { w , } of L+ satisfying (i) by construction and bounded from above by u. So, it remains to be shown that 8 I w, t u holds in L. To establish this let w be an upper bound for {w,}. By taking w A u (if necessary) we can assume that w, I w Iu holds for all ct. For each fixed positive integer m put S = {s E L+ :s I uk for all k 2 m } , and note that since L is Archimedean by Theorem 1.15, inf{uk - s:k 2 m, s E S } = 8. But then the relation 8 I

(uk

- w)'

inf(u,vw:k 2 m } - w

I uk - s for k 2 m and s E S implies = inf{(uk -

w)+:k 2 m } = 8,

and so w = inf{ukv w:k 2 m}. Now observe that 8Iu-W=SUp{u-UkVW:k2m}=SUp{Xk:k2

I},

where xk = sup{u - uiv w:m I i I m + k } , and on the other hand note that Iu - ui v wI I u - ui E V;. for all i implies

8 I xk I

m+k

C

i=m

Iu - uil

E

Vm

+ . . . + V m + k c V,-

for all m.

But then it follows from 8 I xk 7 u - w and the a-order closedness of Vm- that u - w E V,- for all m. Thus u - w = 8 and hence u = w, that is, 8 I w, t u holds in L, and the proof is finished.

METRIZABLE TOPOLOGIES

1D8

[Chap. 5

For metrizable locally solid Riesz spaces Theorem 12.7 can be strengthened as follows. Theorem 15.10. Let (L,t) he a metrizahle locallj~solid Riesz space with the a-Fatou property. Then every solid and order closed subset of L is r-closed.

s.

Proof. Let S be a solid and order closed subset of L and let u E Pick a sequence [u,) E L+ n S with u, A IuI. By replacing {u,) by { U , A )1. (if necessary) we can assume (9 I u, I IuI for all n, and so according to Theorem 15.9 there exists a net {w,} of S with w, T 1111 in L. But then lul E S and so u E S; this shows that S is 5-closed and the proof is finished. Theorem 15.11. Let (L,t) he a metrizable locally solid Riesz space with the a-Fatou property. Then the topology induced by T on any order hounded subset of L isfiner than the topologj) induced by any Lebesgue topology of L on the same set. Proof. Let t* be a Lebesgue topology on L, and let {u,,) E [H,u] be such that u, 1 ,0. It is enough to show that u, I :0. Assume to the contrary that u, 5 I9 does not hold. Then there exists a z*-neighborhood I/ of zero and a subsequence {v,) of {u,} with L’,$ Vfor each n. Put x, = u - r, ( n = 1,2, . . .), and note that 0 5 x, I u holds for all n and s,, 5 u. By Theorem 15.9 there exists a subsequence {xk,} of {x,) and a net (w,} of L with 19I w, T u and such that for any LY there exists k with w, I xk, for all n 2 k. Now since t* is a r* Lebesgue topology, w, I :u holds and therefore x k , = u - v k , -, u also holds. But then L’k, 2 I9 holds, in contradiction to t i k , $ I/ for all n. This contradiction completes the proof. We recall that a Riesz seminorm p on a Riesz space L satisfies the a-Fatou property if I9 I u, t u implies p(u,) p(u).

r

Theorem 15.12. I f a Riesz norm p on a Riesz space L satisfies the a-Fatou property, then the continuous extension of p to its norm completion E, also satisfies the a-Fatou property. Proof. We denote the continuous extension of p to L, by p again. Assume 0 I 6, t ii in t,.Since limp(6,) I p(6) we need only to show p(ii) I limp(li,,), For each n pick an upper element 6, of L with 0 I ii - ii, I 6, and p ( 6 - ii, - 6,) < n-’. We can assume F, 1 and so by Theorem 5.6(v) 6, 1 0 must also hold. Now let E > 0. For each n pick u, E L + and u E L+ with p ( 6 , - u,) < E and p(G - u) < E ; in particular, note that p(u,) I p(ii,) + E I lim p(ii,) E and p(6) - E < p(u).

+

Sec. 161

109

FRECHET TOPOLOGIES

+

Next put 6, = Iu - iil Iu, - ii,l and note that p(0,) < 2~ and u - u,, I Iu - u,I I Iu - iil Iu, - ii,l lii - fi,l 4 G,, P,. Now for each n pick y, E L f with 8 I y, I 6, and p(F, - y,,) < 2 - k Let x, = inf{ yl,. . . ,y,} for all n. Then 8 I x, I B,, p(F, - x,) < E , and x, 1 8 in L. Hence u - u, I 6, G, = 6, D, - x,, x, implies ( u - x,)' 5 6, t;, - .Y, u, and so

+

+

+

+

+

+

p( ( u - x,)+) I ~(6,) p(D, - x,)

+

+

+

+ p(u,) < 4.5 + lim p(ii,).

Since ( u - x,)' T u in L and p is a o-Fatou norm, it follows from the last relation that p(u) I ~ Iimp(ii,), E and hence p(ii) - E < 4~ + Iimp(ii,). So, p(ii) < limp(ii,) 58 for each c > 0. Therefore, p(ii) I limp(ii,), and the proof is finished.

+

+

Note. Theorem 15.6 is due to Peressini [55, Proposition 1.3, p. 2041 and Theorem 15.9 is hidden in [25, p. 511. For normed Riesz spaces Theorem 15.8 is due to Luxemburg and Zaanen [45, Note XII, Theorem 37.1, p. 5201. Theorem 15.12 was proven in [ 3 ] ; the above proof is due to Luxemburg (personal communication). 16.

FRECHET TOPOLOGIES

In this section we shall deal with metrizable locally solid topologies that are in addition topologically complete. We start by giving some characterizations of this property. Theorem 16.1. For a metrizable locally solid Riesz space (L,t)thefollowing statements are equioalent :

(i) (L,t)is t-complete, that is, L = L. (ii) Ecery increasing t-CauchJi net of' L+ is t-convergent, that is, (L,t) satisjes MCP. (iii) Every increasing t-Cauchj' sequence of L is t-convergent. +

Proof. Clearly (i) (ii) and (ii) * (iii). To see that (iii) * (i) note first that the upper elements of L must lie in L, and then use Theorem 15.3 to obtain that L = L. Observe that Example 7.6 shows that statements (i) and (ii) of Theorem 16.1 are no longer equivalent if the metrizability hypothesis is dropped [although (i) (ii) is obvious].

In the next theorem we shall characterize the norm completeness of normed Riesz spaces. Recall that for a positive sequence (u,) we denote by u, the supremum (if it exists) of the sequence of partial sums uk), that is, we put u, = sup{C;=, u k : n= 1,2,. . .}.

c:ll

(I;=

[Chap. 5

METRIZABLE TOPOLOGIES

110

Theorem 16.2. For a normed Riesz space L, the following statements are equivalent:

(i) L, is a Banach lattice, that is, L, i s norm complete. (ii) I f {u,,] E L+ and I, p(u,) " < 00, = then C;= u, exists. (iii) I f {u,) G L+ and p(u,) < w ,then u, exists and P(C,"= 1

ufl)

En^=d u n ) . 1

Proof. (i) * (ii) Assume that (u,) c L+ satisfies p(u,) < co. Put w',,= uk ( n = 1,2,. . .), and note that (w,) is a p-Cauchy sequence. Hence since L, is p-complete, there exists u E L, with limp(w, - ZI) = 0, and therefore u = u, holds according to Theorem 5.6. (ii) * (iii) Assume that (u,: 5 L+ satisfies p(u,) < x,and p(C,", 11,) > I, p(u,). " Then = there exists some E > 0 such that p(C,", u,,) > E p(u,). Observe that by the triangle inequality we have p(C,",, u,) > E c,"=,p(u,) for all m. Next choose a strictly increasing sequence of positive integers { m k ) such that C;==,,p(u,,) < kK3 for all k. Put wk = k ~ ~ E ; ; u ,and , note that p(w'k) s k - 2 < GO; so by hypothesis w= wk exists. But then w 2 kC;=,,,, 11, for each k implies p(w) 2 kp(C,"=,, u,) 2 kE for all k, which is a contradiction. Statement (iii) now follows immediately. (iii) => (i) Assume (u,) is a p-Cauchy sequence of Lf satisfying B I u, in L. Pick a subsequence ( o n ) of (u,) with p ( ~ ' , +-~ on) < 2-" for all n and note that C,"= p(o,+ - u,) < co. Hence c:=k(Li,+ - u,) exists in L for all k . Put u = CF= (u,+ - il,,), and note that since

C,"=

C,"=

,

I,"= +

c:=

C=:

n=k

+

fl=

1

n= 1

we get

+ U1 - Uk) = P

c oc

(n=k

) c m

(on+ 1

-

ufl)

5

n=k

1

-

Un),

from which it follows that (0,) is norm convergent to v + til, and therefore (ufl> norm converges in L. Now use Theorem 16.1 to complete the proof of the theorem. Note. The equivalence (ii) o (iii) of Theorem 16.2 is due to Luxemburg and Zaanen [45, Note I, Theorem 4.2, p. 141; and Note VIII, Theorem 26.3, p. 1051 while the equivalence (i) o (iii) is due to Halperin and Luxemburg; see [45, Note VIII, Theorem 26.21.

Sec. 161

FRECHET TOPOLOGIES

111

Remark. Statement (ii) of Theorem 16.2 is referred to as the RieszFischer property.

A metrizable locally solid topology z on a Riesz space L is said to be a FrPchet topology if (L,r) is 7-complete. Definition 16.3. A Frkchet space (L,T)is a Riesz space L equipped with a Frkchet topology z. A FrPchet lattice (L,z)is a Frkchet space that is in addition locally convex. The Hausdorff quotient spaces of Frkchet spaces are also Frechet spaces. The details follow. Theorem 16.4. I f (L,z) is a Frkchet space and A i s a z-closed ideal, then (L/A,s/A)is a FrPchet space. Proof. Observe first that z / A is a Hausdorff metrizable locally solid topology. In particular, note that if {K)is a normal basis for zero of (L,t) consisting of solid sets then { kj forms a normal basis for zero of (L/A,z/A) consisting of solid sets. We need only to show that (L/A,z/A)is z/A-complete. To this end let {Li,} be a z/A-Cauchy sequence of L/A. Choose a subsequence [ f i n ) of (Li,,) with f i n + , - zj, E V, for n = 1,2,. . . . For each n choose x, E V , with i , = fin+ - 6, and then define wl = u1 and w,,= zil + .y1 + . . . + x,,for n 2 2. Observe that ( w , ) is a z-Cauchy sequence of L. Hence there exists v E L with w, A u in L, and so kn= i., 2 i also holds in LIA. This shows that (L/A,z/A) is s/A-complete, and the proof is finished.

,

The following result should be obvious. Theorem 16.5. For an at most countahlefamilj~{(Lfl,zn);of Fre'chet spaces the Cartesian product (IIL,,IIz,) is a FrPchet space. Concerning Frechet spaces the following remarkable theorem holds. Theorem 16.6. Let (L,z,) be a FrPchet space and ( M , z 2 ) a locally solid Riesz space. Then every positive linear mapping T : ( L , z l )-+(M,z,) is continuous. Proof. Let ( V , ) be a normal basis of zero for z1 consisting of solid neighborhoods. Assume that T : ( L , z , )+ (M,z,) is a positive linear mapping that is not continuous. Then there exists a sequence (u,) E L+ with u, 2 8 and a solid 7,-neighborhood W of zero such that T(u,) 6 W for n = 1,2, . . . . By passing to a subsequence (if necessary) we can assume that nu, E V , for all n. Put w, = ku, ( n = 1,2,. . .), and note that {w,) is a 7,-Cauchy sequence satisfying 8 I w, 7 . Now since (L,z,) is a Frechet space, there exists

112

METRIZABLE TOPOLOGIES

[Chap. 5

w E L such that w , L w and so by Theorem 5.6 we get 8 I w, w in L. In particular, it follows that 8 5 nu,, I w for all n, and therefore (since T is positive) 8 5 T(u,) 5 n - ' T ( w ) holds for all n. But then it follows that T(u,) 3 8, contradicting the choice of the sequence (u,}. This contradiction completes the proof. R Theorem 16.7. Let (L,z)he a FrPchet space. Then any other locally solid topology z' on L is coarser than z, i.e., z' E z. Proof. Consider the identity mapping I:(L,t)-+ (L,z') and apply the previous theorem. R Theorem 16.8. A Riesz space L can admit at most one FrPchet topology. In particular, it follows that all the Riesz norms on a Riesz space L that make L a Banach lattice must be equivalent.

We already know that L' c L" and if in addition (L,t)is a Frechet space then by Theorem 16.6 every cp E L" is continuous and hence L' = L" holds. We thus have the following theorem. Theorem 16.9. For every Frdchet space (L,z)we have L' = L". I n particular, it follows that every Banach lattice L, satisjes L,' = Lp.

As an application of the preceding theorem we shall exhibit an example of a Riesz space L without order bounded linear functionals. Recall that Lp([O,l])(0 < p < a)denotes the Riesz space consisting of the equivalence classes of the real-valued Lebesgue measurable functions f on [OJ] with j:, If(x)lPdx< a. For each n let

V , = { 8} and for each n there exists m such Then each V , is a solid set,=:)( that V, + V, 5 V,. Thus { V,} defines a metrizable locally solid topology z on L = L,([O,l]). By standard arguments from the theory of integration it can be shown that (L,z) is z-complete (or by using Theorem 16.1). So, according to Theorem 16.8 the above topology is the only Frkchet topology that Lp([0,1])can carry. If p 2 1 then (L,z) is a Frechet lattice (actually a Banach lattice). For 0 < p < 1, (L,z)is a Frkchet space but not a Frechet lattice as the next theorem will show. Theorem 16.10. Let L = L,([O,l]) with 0 < p < 1. Then L" = {8}. In particular, it follows that L does not admit any Hausdorff locally convex-solid topology.

Sec. 161

FRECHET TOPOLOGIES

113

+

Proof. Since t P I t 1 holds for all t 2 0, it follows easily that Ll([0,1]) is an ideal of Lp([0,1]). Also since the constant function one is a weak order unit, it follows from the Lebesgue dominated convergence theorem that L1([O,l]) is r-dense in Lp([O,l]). Now by Theorem 16.9 we have L‘ = L ; and so every positive linear functional of L is r-continuous. Thus in order to show that L’= { O ) it is enough to establish that every 0 I cp E L‘ vanishes on Ll([OJI). To this end let 6 I cp E L’. Choose a positive integer n such that i f f E L satisfies A j I f ( x ) l p d xI n - ’ then Iq(f)l I 1, and so for every f E L we have Iq(f)lI n1lP(SAI f ( ~ ) l ~ d x ) In ” ~ particular, . if f E L1([O,l]) satisfies If(x)l d x I1, then the last inequality implies

Hence cp restricted to L,([O,l]) is norm continuous and thus by the Riesz representation theorem there exists 0 5 g E L,([O,l]) such that q ( f ) = j A g ( x ) f ( x ) d xfor all f E L,([0,1]). We claim that g = 6. If g # 6, then there exists a Lebesgue measurable subset E of [0,1] with A(E) > 0 (A is the Lebesgue measure on [0,1]) and such that g ( x ) 2 E for some E > 0 and all x E E . Now for each positive integer k pick a measurable subset Ek of E with 0 < A(Ek)5 k - and note that

’,

E m )

5 Jol 9 ( X ) X E k ( X ) d x = c p ( X E k )

Hence k l i p - ’ I [A&)]’I & - ‘ n 1 l p for all k, a contradiction. Thus g = 8 and so cp = 6; so that L“= (0). The last assertion follows immediately from the Hahn-Banach theorem. Remark. A glance at the arguments of the last proof shows that we can actually establish the following stronger result: For every nonatomic Jinite measure space ( X , Z , p ) and 0 < p < 1 the Riesz space L = Lp(X,C,p)satisjies L ‘ = (0).

Note. The Banach lattice version of Theorem 16.8 is due to Goffman [29, Theorem 2, p. 5371; for more general results on this subject see [25, pp. 61-68]. The fact that Lp([O,l]) (0 < p < 1) does not admit any nonzero continuous linear functionals was proven by Day [20].

As an application of the above results we shall determine the topological dual of the locally solid Riesz space studied in Example 8.6.

114

METRIZABLE TOPOLOGIES

[Chap. 5

Application 16.11. Let (L,t) be the Hausdorff locally solid Riesz space described in Example 8.6. We shall show that: (1) If 0 < p < 1, then for every cp E L' there exists a finite number of points x l , . . . ,xn of [0,1] and real constants a l , . . . ,an(all depending upon cp) such that ~ ( u= )

C aiu(xi)

i= 1

holds for all u E L. (2) If 1 I p < co, then for each cp E L' there exist a finite number of points x l r. . . ,xn of [0,1], real constants a,, . . . ,an, and u E L,([O,l]); l/p l/q = 1 (all depending upon cp) such that

+

holds for all u E L. To establish the above claims let cp E L', and note that by the continuity of cp there exist a finite number of points x l , . . . ,xnof [0,1] and E > 0 such that (q(u)l I 1 holds whenever u E L satisfies lu(xi)l I E for i = 1, . . . ,n and j i Iu(x)Ipdx I E. Put cti =cp(x,,,J, i = 1, . . . ,n. Now if u,u E L satisfy u(x) = o(x) for almost all x, put

and note that gk = kw satisfies &(Xi) = 0 for i = 1, . . . ,n and j i Igk(x)IPdx= 0 for k = 1,2,. . . . It follows easily that l q ( g k ) )= kIcp(w)l I 1 holds for k = 1,2,. . . and so n

CP(U) -

1 aiu(xi) =

i= 1

n

~ ( 0)

1 ai~(xJ.

i= 1

Hence the linear functional $ on Lp([OJ]) defined by n

$ ( [ u ] ) = q(u) -

1 aiu(xi)

i= 1

for u E L

is well defined, where [u] is the equivalence class determined by u E L in Lp([OJl). Next observe that $ is a continuous linear functional on Lp([OJ]). Indeed, assume that [u] E L,([O,l]) ( u E L ) satisfies lu(x)lPdx s E and consider u E L defined by u(xJ = 0 ( i = 1, . . . ,n) and u(x) = u(x) otherwise. Then [v] = [ u ] and I$([u])I = I$([u])I = Iq(u)I I 1, and hence is bounded on a neighborhood of zero; therefore $ is continuous.

PSEUDO LEBESGUE PROPERTIES

Sec. 171

115

If now 0 < p < 1 holds, then by Theorem 16.10we get I) = 0 and (1) follows. If 1 I p < co holds, then by the Riesz representation theorem $([u]) = ~ ~ u ( x ) L ' (holds x ) ~ xfor some [u] E L,,([O,l]), where l/p + l/q = 1, and so (2) also follows. W 17.

THE PSEUDO LEBESGUE PROPERTIES

We start this section by introducing another continuity property, which is weaker than the a-Lebesgue property. Definition 17.1. Let (L,T)he a locally solid Riesz space. Then we say that:

(i) (L,.r)satisJies the pseudo a-Lebesgue propert!' if u, 1 6' and (u,) TCuuchji implies un 1,8. (ii) (L,T)saliSJies the pseudo Lehesgire property if u, 1 6' and {u,) r-Cauchy implies u, 1,U. It should be clear that the pseudo Lebesgue property implies the pseudo a-Lebesgue property. The converse is, however, false as Example 8.6 shows. Every topologically complete Hausdorff locally solid Riesz space (L,T)satisfies the pseudo Lebesgue property. Indeed, if L I , J 0 and { u,) is 7-Cauchy, then u, A u holds for some u E L and so by Theorem 5.6 u, 1 u ; thus u = 6' and hence u, A 0. Similarly, every sequentially z-complete Hausdorff locally solid Riesz space satisfies the pseudo a-Lebesgue property. Also if (L,r)is a Hausdorff locally solid Riesz space such that L is a regular (resp. a-regular) Riesz subspace off,, then ( L,r) satisfies the pseudo Lebesgue (resp. pseudo a-Lebesgue) property. Indeed, if L is a regular Riesz subspace of f, and {u,) is a r-Cauchy net of L with ii, 1 0 in L, then u, f ii holds in f, for some ii. But then u, 1 ii holds in f,. So, ii = 0 and the claim is established. More properties about the pseudo Lebesgue properties are included in the next theorem.

Theorem 17.2. For a metrizahle locally solid Riesz space (L,T)the following staternenis are equicalent : (i) ( L J )satisfjes the pseudo Lehesgue propertj'. (ii) (L,T).sati.sfieies the pseiido a-Lehesgue propert!).

Proof. (i) (ii) Obvious. (ii) =. (i) Assume that a r-Cauchy net (u,: satisfies u, 18 in L. Then u, f ii holds in f, for some 8 _< ii; in particular, we have u, 1 ii in f,. If 0 < ii holds in f,, then by Theorem 15.5 there exists a strictly positive lower element F B of L such that 0 < B I ii. Let (w,) E L f be such that w,, 1 8 and w, +

METRIZABLE TOPOLOGIES

116

[Chap. 5

e.

in By virtue of u, J 8 in L we get w, 1 8 in L. Now since {w,} is z-Cauchy and (L,z) satisfies the pseudo a-Lebesgue property, we get w, 8. Hence B = 8, contradicting ii, > 8. Therefore, ii = 8 and so u, 1,8. Theorem 17.3. statements hold:

Let (L,z)be a locally solid Riesz space. Then the following

(i) I f (L,z)satisjies the Fatou property, then (L,z)also satisfies the pseudo Lebesgue property. (ii) I f (L,r) satisfies the a-Fatou property, then (L,z) also satisjies the pseudo a-Lebesgue property. Proof. (i) Assume that the z-Cauchy net {un} satisfies u,

1 f3 in L. Let

I/ be a Fatou 7-neighborhood of zero. Then there exists an index ctl such that

u, - uBE V for all ct,P 2 al. But for each fixed ct 2 a1 we have u, - u g x u , (0)

and hence u, E I/ for all ct2 ctl ;thus u, 1,8, and we are done. (ii) Repeat the arguments of the previous proof. The following theorem gives some characterizations of the pseudo Lebesgue property in metrizable locally solid Riesz spaces.

6-

Theorem 17.4. For a metrizable locally solid Riesz space (L,e) the following statements are equivalent: (i) (ii) (iii) (iv)

(L,r)satisjes the pseudo a-Lebesgue property. L is order dense in t. L is a regular Riesz subspace of L. L is a a-regular Riesz subspace of L.

Proof. (i) * (ii) Let 8 < ii E.,!I By Theorem 15.5 there exists a lower element ii, of L such that 6 < B 5 2. Next pick a sequence {w,} c L+ with w, J B and w, f B;in particular, note that {w,} is a z-Cauchy sequence of L. We now claim that there exists an element 8 < w E L such that f3 < w 5 6. If not, then w, 1 8 holds in L and so by the pseudo a-Lebesgue property of (L,z) we get w, A 8. But then ii, = 0, a contradiction. Thus L is order dense in L. (ii) * (iii) It follows immediately from Theorem 1.10. (iii) * (iv) Obvious. (iv) * (i) Assume u, 1 8 and {u,} is z-Cauchy. Then u, f ii holds in L and so u, 1 ii also holds in L. But then since L is a a-regular Riesz subspace of L, we get ii = f3 and so u, 1 ,8 ; the proof of the theorem is now complete. It may be asked at this point whether statements (ii) and (iii) of the preceding theorem are equivalent without assuming metrizability. Clearly (ii) (iii) but (iii) no longer implies (ii), as the following example shows.

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PSEUDO LEBESGUE PROPERTIES

117

Example 17.5. Let L = C[O,l] and let 7 be the Hausdorff locally solid topology generated by the Riesz seminorms PK,,KI(u)

= SUP(IU(X)I:X E K , uK 2 )

for u E L ,

where K , is a compact subset of rationals of [OJ] and K , is a compact subset of irrationals of [O,l]. Observe that 7 is not a metrizable locally solid topology. By the classical Tietze extension theorem it is easy to see that t c a n be identified with the Riesz space of all real-valued functions on [0,1] whose restrictions to the rationals and irrationals of [O,l] are both continuous. Note that 8 < ii = xE E %, where E = Q n [OJ] and Q denotes the rational numbers of R, but there is no nonzero function of C[O,l] that is below ii. Thus L is not order dense in 2. On the other hand if u, 4 8 holds in L and 0 < ii Iu, holds in f, for all a, then there exists a real number xo such that ii(xo) > 0. If xo is rational, then there exists a neighborhood of xo for which ii(x) 2 E for all rationals of that neighborhood and some E > 0. But then u,(x) 2 E > 0 holds for all x in some neighborhood of xo, and all a. It now follows easily that u, 1 8 does not hold in L, a contradiction. The same argument can be applied if xo is irrational. Hence u, 4 8 holds in t. We have thus proven that L is a regular Riesz subspace of but L is not order dense in 2.

e,

Theorem 17.6. Let (L,z)be a metrizable locally solid Riesz space with the pseudo o-Lehesgue property. If L is a-Dedekind complete, then L is an ideal of (and hence the order intervals of L are 7-complete). I n particular, it follows in this case that L is super order dense in f,. Proof. Note first that if ii is a lower element of L then u, 1 ii and u, f ii hold in t for some sequence {u,} c L'. Now since L is a-Dedekind complete, u, 1 u must hold in L for some u E L + .But by Theorem 17.4 u, u must also hold in and hence ii = u E L, that is, every lower element of L lies in L. Now let 8 Iii E t.By Theorem 15.5 there exists a sequence {u,} of L+ with u, ii and u, 4 ii. Now if ii I u holds for some u E L + , then u, w holds in L for some w E L + and so by Theorem 17.4 we get ii = w E L, that is, L is an ideal of %. The last part should be now obvious from the above arguments.

Theorem 17.7. For a metrizable locally solid Riesz space (L,7) with the pseudo a-Lebesgue property the following statements hold: (i) If L is a-Dedekind complete, then so is t. (ii) If L is Dedekind complete, then so is t. (iii) If L is super Dedekind complete, then so is I?.

118

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Proof. (i) and (ii) follow immediately from Theorems 17.6 and 7.8. To show that (iii) holds, note first that is Dedekind complete and that L is a super order dense ideal of by Theorem 17.6. Now assume 8 I ii, ii in t. Pick a sequence {u,} c L' with 6 Iu, 7 ii and note that since, for each k, 0 Iuk A ii, itk holds in the super Dedekind complete Riesz space L , there a existsasequenceofindices (C(k,n:n=1,2,... } w i t h s u p ( u , ~ i i , , , ~ : n 1,2 = ,...): uL in L. It now follows easily that sup(ii,,,,,:n,k = 1,2,. . .} = ii holds in L, so that is super Dedekind complete, and the proof is finished. W

e

e

r

The next theorem characterizes the Lebesgue property in terms of the pseudo a-Lebesgue property in metrizable locally solid Riesz spaces. Theorem 17.8. Let (L,z) be a metrizable locally solid Riesz space. Then the following statements are equivalent:

(i) (L,z)satisfies the Lebesgue properly. (ii) ( L J )satisfies the pseudo o-Lebesgue property and the pre-Lebesgue property. (iii) (L,z) satisjies the o-Lebesgue property and pre-Lebesgue property. Proof. (i) (ii) Obvious. (ii) 3 (iii) Assume u, 18 in L . Since (L,z)has the pre-Lebesgue property { u n } is a z-Cauchy sequence and so u, A 6, that is, (L,z) satisfies the oLebesgue property. (iii) + (i) Assume u, I 0 in L. Then since (L,z) satisfies the pre-Lebesgue property, {u,} is a z-Cauchy net (Theorem lO.l), and so u, 1 ii and u, f ii holds in for some ii. But then by Theorem 17.4 u, 1 6 must also hold in 2 and hence ii = 6, so that u, 1,8.

.!,

Remark. Note that if a metrizable locally solid Riesz space (L,z)satisfies the equivalent conditions of Theorem 17.8 then L has the countable sup property. Theorem 17.9. For a metrizable locally solid Riesz space (L,z) the following statements are equivalent: (i) (L,z) satisfies the Lebesgue property and L is super Dedekind complete. (ii) (L,z) satisfies the o-Lebesgue property and L is o-Dedekind complete.

Proof. (i) * (ii) Obvious. (ii) * (i) Note that (L,z) must satisfy the pre-Lebesgue property and so by the previous theorem (L,T)satisfies the Lebesgue property. To see that L is super Dedekind complete assume 6 I u, 7 I u in L. Then by Theorem 10.1 {u,) is a z-Cauchy net and hence u, f ii must hold in f,, and clearly u, T ii. Since (L,r) is metrizable, we can select a sequence {u,,} c {u,} with

Sec. 171

PSEUDO LEBESGUE PROPERTIES

119

r

0 I u , ~ fi and u,,, f li in f . But then since L is o-Dedekind complete, there exists u E L+ with 6' I uan t LI I u in L, and so by Theorem 17.4 fi = v E L. This shows that L is super Dedekind complete, and the proof is finished. H Note that Example 8.6 shows that Theorems 17.8 and 17.9 are false for nonmetrizable Hausdorff locally solid Riesz spaces. By comparing Theorems 10.6 and 17.9 the following result should be clear now. Theorem 17.10. I f a metrizable locally solid Riesz space (L,z) satisfies the pre-Lebesgue property, then satisjies the Lebesgue property and f is super Dedekind complete.

(e,?)

As an illustration of the above result and Theorem 10.6 consider L = C[O,l]. If z is the metrizable locally solid topology generated by the Riesz norm p ( u ) = 1; lu(x)l dx, then L,([O,l]) is a super Dedekind complete Riesz space. O n the other hand if z* is the Hausdorff locally solid topology of pointwise convergence on L, then f = RIO,ll is a Dedekind complete Riesz space that is not super Dedekind complete. The next theorem shows that for metrizable locally solid Riesz spaces the o-Lebesgue property is preserved under the process of topological completion.

e=

Theorem 17.11. If a metrizable locally solid Riesz space (L,z) satisfies the a-Lebesgue property, then so does its topological completion

(e,?).

Proof. Let { V,} be a normal basis of zero for z consisting of solid sets. Assume that f, 4 0 holds in We have to show that ,L f 0 holds in For each n let ii, be an upper element of L such that I ii, and ii, - A E (Theorem 15.4). Put 9, = inf{fi,, . . . ,fiJ for all n and note that each G, is an upper element of L, 9,L and 9, - f, E Since 1 8, then 9, J 6' also holds in f by Theorem 5.6(v). This shows that we can assume without loss in generality that { f,} is a sequence of upper elements of L. Now let V, be a neighborhood from { V , ) . For each n choose an element gn E L+ such that - g, E E+n+2 and 0 I g, I A. Put u, = inf{g,, . . . ,g,} for all n and note that u, J 0 holds in L. It follows from the a-Lebesgue property of L that u, E V,, for all n 2 no. Now since

e.

e.

c.

e

- c, = sup{X

c (J

- g,:1 I i In) I sup [ J - gi:l I i I n )

n

I

i= 1

we get f, - u, E that is,

-

gJ

E

&+

3

+ . . . + G+,+2 G K +

E+ for n = 1,2, . . . . Thus

f 8, and the proof is finished. H

1

[Chap. 5

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120

The last theorems of this section deal with the pseudo o-Lebesgue property in quotient locally solid Riesz spaces. The first theorem follows immediately from the discussion preceding Theorem 14.4 and its proof is omitted. Theorem 17.12. Let (L,t) be a locally solid Riesz space with the pseudo o-Lebesgue property, and let A be a projection band of L. Then the quotient locally solid Riesz space (L/A,t/A) satisfies the pseudo a-Lebesgue property.

In Theorem 14.3 we saw that the a-Dedekind completeness of L was sufficient to guarantee that the o-Fatou property was inherited by (L/A,z/A). This works also in the pseudo o-Lebesgue case; interestingly enough, we had to assume that A was a a-ideal in Theorem 14.3, but it suffices to have a simple ideal here, provided that z is metrizable. Theorem 17.13. Suppose that (L,z) is a o-Dedekind complete metrizable locally solid Riesz space with the pseudo o-Lebesgue property, and that A is an ideal of L. Then (L/A,z/A) satisfies the pseudo o-Lebesgue property.

Proof. Suppose ti, 16' in L/A and that {ti,} is a z/A-Cauchy sequence. Pick a normal basis { V,} of zero for z and note that { is a normal basis for zero in L/A. By passing to a subsequence (if necessary) we can assume ti, - ti,,+ E V,+2 for all n. Now for each n choose 8 I x, E V,+, with k, = ti,-ti,+,andputu, = ~ ~ + a n d u , = u , + - ( x +...+ , x,-,)forn=2,3 , . . . . Observe that u, 1 holds in L and d, = ti, for all n. Taking positive parts we have 8 I u,' 1 and so by the a-Dedekind completeness of L, 0', -1 u 2 8 holds in L for some u E L'. Now 8 I u I u,' for all n, so that 8 I d I d,' = ti, holds for all n. It follows that b = 6 and thus if w, = u,+ - u, we have w, 1 0 in L with w, = ti, for all n. But {w,} is a t-Cauchy sequence, since if n,k are arbitrary

c}

8 I w,

- w n + k = u',

- u,'+,

< 10,

=X,+"'+X,+k-lE~+2+"'+

- u,+kl V,+k+1

&T/n.

By the pseudo o-Lebesgue property of (L,z) we have w, rlA ti, = w, -,e.

8 and hence

Note. The pseudo o-Lebesgue property was introduced by Luxemburg as the (A,O)property [44, Note XVI, p. 6501. Theorem 17.4 was also proven by Luxemburg [44, Note XVI, Theorem 61.5, p. 6521 for the normed case; the above generalization can be found in [2, Theorem 6.1, p. 1191. Example 17.5 is due to Fremlin (personal communication). For the normed case, Theorem 17.8 is due to Luxemburg and Zaanen [45, Note X, Theorem 33.8, p. 5051 and Theorem 17.9 is due to Nakano [52, p. 3211;the abovegeneralizations have appeared in [2, pp. 121-1221. Theorems 17.12 and 17.13 were proven for normed Riesz spaces in [42, pp. 210-2111.

Sec. 181

THE 6-PROPERTY

18.

121

THE 6-PROPERTY

In this section we shall introduce a property that is stronger than the pre-Lebesgue property and independent from the Lebesgue property. The definition of this property follows. Definition 18.1. A locally solid Riesz space (L,z) is said to satisfy the Bproperty if it follows from 0 Iun T in L and {un} z-bounded that {un} is a z-Cauchy sequence.

It should be clear that the B-property implies the pre-Lebesgue property. The Lebesgue property is, however, independent from the B-property, as the following examples show. Example 18.2. Let X be an infinite set, and let L be the Riesz space consisting of all real-valued functions defined on X that vanish outside finite sets, with the pointwise ordering. Let z be the Hausdorff locally solid topology on L generated by the sup norm. Note that (L,t)satisfies the Lebesgue property, but it does not satisfy the B-property. For, if { x l , x 2 , .. .} is a countable subset of X , then the sequence u, = qx,,,,,,~ “ n ~= ,1,2,. . . satisfies 0 I u, T and it is z-bounded. However, {un} is not a z-Cauchy sequence. W Example 18.3. C[O,l] with the Hausdorff locally solid topology generated by the Riesz norm p(u) = lu(x)ld x satisfies the B-property, but it does not satisfy the a-Lebesgue property; see Example 8.2. W

1;

We continue with a simple characterization of the B-property, the easy proof of which is left to the reader. Theorem 18.4. For a locally solid Riesz space (L,z) the following statements are equivalent:

(i) (L,z)salisjes the B-property. (ii) 0 I u, t in L and (u,) z-bounded implies that {u,} is a 7-Cauchy net. It should be noted that if a Hausdorff locally solid Riesz space satisfies the B-property then it also satisfies the following “weak Lebesgue property”: If {u,} G L+ satisfies 0 Iu, T ii in t,then u, 1,u^. Theorem 18.5. For a metrizable locally solid Riesz space (L,z)the following statements are equivalent: (i) (L,?) satisfies the B-property. (ii) ( L , t )satisjies the B-property.

In particular, it follows in this case that with the Lebesgue property.

(t,?) is super

Dedekind complete

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Proof. (i) = (ii) Obvious. (ii) =. (i) Pick a normal basis { V,} of zero for z consisting of solid sets. Assume that 8 t is a ?-bounded sequence of f,. Pick a V, from {V,). For each n choose g, E L f such t h a t x - g, E E+ Define u, = 0 if n < k + 3 and u, = ~ u p { g ~. + . . ~,gn) , if n 2 k + 3, and note that 8 I u, t in L. For n 2 k + 3 we have

IX

u,

-L=sup{gi -x:k + 3 I i s n } I sup{gi-i: k + 3 Ii In } I 1 Igi -XI, i=k+3

and so Iun - j;,l

that is, u,

-A

I

E &+2

1

i=k+3

1gi - j;l

for n 2 k

E K+4

+

-

*

*

+ v,+

1

2

K+2,

+ 3.

Now put Gn= I i - gil ( n = 1,2, . . .) and note that (9,) is a 6Cauchy increasing sequence of t ;so K>,t5 R holds in t,and therefore G,, t 9 also - u,I I R for all n and thus - u,} is holds. In particular, we get ?-bounded; consequently u, = f, - (f,- u,)(n = 1,2, . . .) is z-bounded and hence by our hypothesis { u,} is a r-Cauchy sequence. In particular, we have Iu, - urn(E V,+, for all n,m 2 no 2 k 3. But then, for n,m 2 no, we have

IX

{A

1

IL -.Ern1

I

lX - unI + Iun - urn1 +

+

ITrn -

Hence & - &, E K for n,m 2 no, that is, required.

urn1 E

K+2

{ A ) is a

+ & + 2 + K + 2 c E. ?-Cauchy sequence, as

The last part follows from Theorem 17.10. Note. The B-property was introduced as property (By$ by Luxemburg and Zaanen [45, Note XI, p. 5071; Theorem 18.5 is due to Aliprantis [2, Theorem 5.3, p. 1181.

EXERCISES

1. Show that if (L,z)is a metrizable locally solid Riesz space with the pre-Lebesgue property, then every order bounded disjoint subset of L is at most countable. (Hint: If { V,} is a basis of zero for T and A is an order bounded disjoint subset of L, then { u E A : u $ V,,} is finite for each n.) 2. Show that if (L,z)satisfies the hypotheses of the previous exercise, then L has the countable sup property.

EXERCISES

123

3. Show that a Hausdorff locally solid Riesz space (L,T)satisfies the Lebesgue property if and only if it satisfies the pre-Lebesgue and pseudo Lebesgue properties together. 4. Let L be a Riesz space and z a metrizable linear topology on L. Show that if the solid hull of every z-bounded subset of L is z-bounded, then z is a locally solid topology. (Hint: Pick a basis {K} of zero for z, and then show that given a z-neighborhood I/ of zero there exists n such that S ( ( 1 h ) K )E v.1 5. Let (L,z)be a metrizable locally solid Riesz space with the 0-Lebesgue property, and let A be an ideal of L. Show that A is z-closed if and only if A is a a-ideal. 6. Let (L,z) be a metrizable locally solid Riesz space and (M,z*) a Hausdorff locally solid Riesz space with the a-Lebesgue property. Assume that T:(L,z)+ (M,z*) is a positive, continuous linear operator that is also an integral. Show that the unique continuous extension ?:(L,?) + (A,?*) is also an integral. (Hint: Use upper elements.) 7. Show that every separable Frechet space contains a weak order unit. u holds in L, then 8. Let (L,z) be a Frechet space. Show that if u, there exists a subsequence {u,) of {u,) and u E L' such that Iu, - uI 4 (l/n)u for all n. 9. A proper ideal A of a Riesz space L is called maximal if it follows from A G B, B # A , and B an ideal of L that B = L. Show that every maximal ideal of a Frechet space (L,r)is z-closed. 10. Consider the super Dedekind complete Riesz space L = L,([O,l]) and the three Riesz norms defined on L by

p 3 ( u ) = ess suplu(.

(i) Show that p 1A p2(u) = lk lu(x)ldx for all u E L. (ii) Put e ( x )= 1 for all x E [0,1] and show that ( p l A p 2 )v p3(e) = 1. (iii) Show that ( p l v p 3 ) ~ ( p v2 p 3 ) ( e )2 2(1 + $)/(3 + $) > 1, and conclude from this that S , is not distributive. 11. Let X be a nonempty set, and let L be the Riesz space consisting of all real-valued functions defined on X whose range is finite. (i) Show that L satisfies the projection property. (ii) If p denotes the sup norm on L, show that L, satisfies the Fatou property.

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124

(iii) Determine the norm completion of L,. (Compare your answer with Theorems 17.4 and 7.10.) 12. Show that the topological completion of the Hausdorff locally solid Riesz space of Example 17.5 consists of all real-valued functions defined on [0,1] whose restrictions to the rationals and irrationals of [0,1] are both continuous. (Hint: Use Tietze's extension theorem.) 13. Consider L = C[O,l] equipped with the Riesz norm p(u) = jh lu(x)l dx. Show that L, satisfies the pre-Lebesgue property and that it fails to satisfy the pseudo a-Lebesgue property. 14. Let L = L,,([O,l]) with 0 < p < 1 and M = C[O,l]. Show that Y,,(L,M) = {O), a super Dedekind complete Riesz space. Why does this not contradict Exercise 12(iii) of Chapter l ? 15. Let (L,z) be a metrizable locally convex-solid Riesz space. Show that the bidual L" is a a-ideal of (L')-. (Hint: Use the fact that L" is sequentially a(L",L')-complete; see [40, p. 3961.) 16. Let p be a a-Fatou norm on a Riesz space L. Show that the conof p to t,satisfies tinuous extension ;

;(ii)

= inf(1im p(u,): {u,) G

for all ii E t,.

L', u,

r and u,

A

liil

7 liil in L,)

OPEN PROBLEMS

1. Is the B-property preserved under the process of topological completion without assuming metrizability ? 2. Let (L,z) be a metrizable locally convex-solid Riesz space. Is L" a band of (L')"? (See Exercise 15 of this chapter.) 3. Let (L,r) be a Hausdorff locally convex-solid Riesz space and set as usual L,' = L' n LLand L,' = L' n L;. Then under what conditions:

(i) Is L,' order dense in L,? (ii) Is L,' order dense in L,?

Remark. Observe that if (i) is true then (ii) must also be true, and that statement (i) is true for metrizable spaces by Theorem 15.8. However, there are Hausdorff locally convex-solid Riesz spaces (L,z) for which L,' is not order dense in L;.

CHAPTER

6

Locally Convex-Solid Riesz Spaces

A number of properties of locally convex-solid Riesz spaces have been presented in the previous chapters as they were needed. In this chapter a systematic study of these spaces will be made. Some of the basic properties of locally convex-solid Riesz spaces are derived first, and then the compact subsets of L, and Lz are characterized (in the various topologies) in terms of the concept of order-equicontinuity. Results concerning the weak convergence of sequences in the topological dual will be obtained and conditions for weak sequential compactness will be given. The solid z-compact subsets as well as the weakly compact solid subsets will be characterized, and some of the most fascinating results will deal with solid compact sets. For instance, we shall see that if a subset A of a z-complete Hausdorff locally convex-solid Riesz space (L,z) has the property that every disjoint sequence of its solid hull is z-convergent to zero, then A must be a(L,l')-relatively compact. Finally, these results will be used to give necessary and sufficient conditions, in terms of disjoint sequences, for a locally convex-solid Riesz space to be semireflexive. 19.

LOCALLY CONVEX-SOLID RlESZ SPACES AND THEIR DUALS

Briefly, we shall recall below the basic theorems of locally convex spaces needed for our discussion. We have already mentioned a number of results in Section 4. 125

126

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[Chap. 6

For a nonempty set X the real vector space RX equipped with its product topology becomes a locally convex vector space (actually a locally convexsolid Riesz space). The product topology on RX is called the topology of poinrwise convergence, and it is generated by the family of Riesz seminorms (ps:x E X ) , where p , ( f ) = If(x)l for each f E RX.For any nonempty set A of real functions defined on X (i.e., A G Rx),the weak topology o ( A , X ) on A (also called the topology of pointwise convergence on A ) is the Hausdorff topology induced on A by the product topology of RX.Note that a net [,Li c A satisfies ji f' in A if and only i f f ' E A and fi(.x) + f(x) for all x E X. If A is a vector subspace of RX,then a ( A , X ) is a linear topology on A. For a linear space E the algebraic dual E* of E is the vector subspace of RE consisting of all linear functionals on E; note that E* is closed in RE. For every linear topology t on E we have the vector subspace inclusions: E' c E* G RE. For a subset A of a topological vector space (E,z) the polar of A is defined 1 for all u E A } ; clearly A' is o(E',E)-closed. If V is by A' = {cp E E':Icp(u)I I a t-neighborhood of zero, then V o is pointwise closed in RE. Indeed, if (cp,} G V o satisfies (pa + cp in RE,then cp E E* and Icp(u)l 5 1 holds for every u E V and hence cp E E', since cp is bounded on V. On the other hand as it is readily seen, V o is topologically bounded in RE and thus V o is a compact subset of RE [and hence of E' for o(E',E)] by Tychonoff's classical theorem. We have, therefore, proven the Alaoglu-Bourbaki theorem. O % )

Theorem 19.1. (Alaoglu-Bourbaki). For every t-iiei~hbOrhOot/V qf zero of' a topological rector space (E,s), the polar V o is a o(E',E)-c.ompuct subset o f E'.

The polar of a subset B of E' of a topological vector space (E,t) is defined 1 for all cp E B),and clearly Bo is o(E,E')-closed. by Bo = { u E E:Icp(u)I I The bipolar of a subset A of E is the subset (Ao)' of E, denoted as usual by A''. The following basic theorems can be found in most of the books on topological vector spaces. See [23], [31], [40], [63], and [66]. Theorem 19.2. (Bipolar Theorem). For ecery subset A of a topological i'ector space (EJ) the bipolar Aoo is the balanced, conz>e.x,o(E,E')-closed lzull of A. In particular, if (E,t) is a Hausdorff locally convex space and A is a balanced, convex, and t-closed subset of E, then A = Aoo. Theorem 19.3. (Banach-Steinhaus). Lrt E be N Banuch space. Then a subset A of E' is norm hounded ifand onlji f A is a(E',E)-bounded.

Recall that a subset A of a topological vector space (E,T) is called srelatively compact if the t-closure of A is t-compact.

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Theorem 19.4. (Eberlein-Smulian). For a subset A E the following statements are equivalent:

of a Bunach spuce

(i) A is a(E,E')-relatiuely compact. (ii) Euerji sequence of A has a subsequence that is o(E,E')-convergent to an element of E. For a locally solid Riesz space (LJ), the polar of any solid subset of L is a solid subset of L'. Indeed, by Theorem 3.3 we have for u E L and cp E L', and so if S is a solid subset of L, then So = {cp

E

L':Icp(u)I I1 for u E S )

= {cp E

~':lcpl(Jul) I 1 for u E s),

so that So is a solid subset of L'. Likewise, if S is a solid subset of L', then So (use here Theorem 3.5) is a solid subset of L. Finally, if A is an ideal of L, then A' is a band of L' and moreover A' = [cp

E

L':cp(u)= 0 for all u E A ) = (cp

E

~':lcpl(lul)= 0 for all ti E A } .

Theorem 19.5. Let (L,T)be a loc.allj~solid Riesz space, and let the bands A and B sarisfy L = A @ B. Then L' = A' @ Bo [and so fi P is the projection of L' onto A', then cp(u) = Pcp(u)for ull u E B and cp E L'].

Proof. By the above discussion A' and Bo are bands of L' satisfying A' n B' = i d } , as is easily observed. Let Q be the projection of L onto B. For cp E L', let cpl = cp 0 Q and cp2 = cp - cpl. Then q l ( u ) = 0 for all u E A and cp2(u)= 0 for all u E B and thus cpl E A' and cp2 E B'. Hence cp = cpl cpz E A' 0 Bo, and therefore L' = A' @ B'; the parenthetical part follows by observing that cpl = Pcp and that cp2(u) = 0 for all u E B. 1

+

Now let L be a Riesz space, B a subset of L, and A an ideal of L". The absolute weak topology (o((A,B)on A is the locally convex-solid topology generated by the family of Riesz seminorms {p,,:u E B } , where p,(cp) = lq((lu1) for all cp E A . Note that lal(A,B)= /ol(A,Z(B)), where Z(B) is the ideal generated by B in L.

Theorem 19.6. For a Riesz space L and an ideal A of L" the absolute weak topology lo((^,^) is a Hausdor-Lebesgue topology satisjjling ~ ( A , LG) loJ(A,L).

Proof. To see that lol(A,L)is Hausdorff and finer than o(A,L)use the relation Icp(u)[ I Icpl(lul) and the fact that o(A,L) is always a Hausdorff topology. Now if cpa 1 0 in A , then cpm J 0 in L" and so cp,(u) J 0 for every u E L'. Thus { q U )is lol(A,L)-convergent to zero and so lol(A,L)is a Lebesgue topology. 1

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Recall that a locally solid topology z on a Riesz space L is called a Levi topology if every increasing z-bounded net of L + has a supremum (Definition 9.3). The basic properties of the absolute weak topology lol(L",L) are stated in the next theorem. Theorem 19.7. For every Riesz space L the absolute weak topology (o((L",L)is a HausdorfS, Lebesgue, Levi topology. Moreover, L" is lol(L",L)complete.

Proof. To see that (o((L",L)is a Hausdorff Lebesgue topology use Theorem 19.6. For the Levi property consider 8 I cpa t in L- with {(pa} lol(L",L)-bounded. Put cp(u) = sup{cp,(u)} for every u E L + , and note that cp:L+ + R+ is an additive mapping. By Lemma 3.1 cp can be extended to all of L by cp(u) = cp(u') - cp(u-). It is now readily seen that cp, t cp holds in L' and so lo((L',L) is a Levi topology. The topological completeness of lol(L',L) now follows from Theorem 13.9. H Theorem 19.8. Let (L,z) be a locally convex-solid Riesz space. Then the lal(L',L)-closure of an ideal A of L is the band generated by A in L'.

Proof. By Theorem 1.11 A is order dense in Addand hence, since lo((L',L) is a Lebesgue topology, A is lal(L',L)-dense in Add. The result now follows by observing that Addis lal(L',L)-closed [Theorem 5.6(iv)]. H Theorem 19.9. Let (L,r) be a locally convex-solid Riesz space. Then the topological completion of (L',lol(L',L)) is the band generated by L' in L". In particular, L' is (ol(L',L)-completei f and only i f L' is a band of L".

Proof. Note that lol(L',L) induces the topology lol(L',L) on L' and so by Theorem 19.7 the topological completion of (L',[ol(L',L)) is the lol(L-,L)closure of L'. Now since lol(L-,L) is a Lebesgue topology this closure is a band of L-, which is precisely the band generated by L' in L". Remark. If A is an ideal of L", then by Theorem 6.6 there always exists a locally convex-solid topology on L, namely, z = lo((L,A),such that the topological dual of (L,z)is exactly A. Hence by Theorem 19.9 the topological completion of (A,Iol(A,L))is the band generated by A in L".

Recall that for a Riesz space L and any ideal A of L-, there exists a natural embedding U H ii of L into A; defined by u"(cp) = cp(u) for all cp E A . By Theorem 3.1 1 this embedding is a Riesz homomorphism that is one-to-one if and only if A' = { O } . So, if A' = { O } we can identify L with a Riesz subspace of A,, and as usual we shall make no distinction between an element u of L and its image ii in A,. Now if (L,z) is a Hausdorff locally convex-solid Riesz space, then by the Hahn-Banach theorem (L')' = { 0 } , and thus as above, L is a Riesz subspace

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of (L’);, which is pol necessarily an order dense Riesz subspace (however, this is the case if and only if T is a Lebesgue topology, by Theorem 9.1). The next result shows that L is lal((L‘);,L’)-dense in (I,’):. Theorem 19.10. If (L,r) is a HausdorfS locally convex-solid Riesz space, then L is lol( (L’);,L’)-dense in (I,’);. Proof. Put M = (I,’);. Then L‘ (embedded in Mr according to Theorem 3.1 1) is an order dense Riesz subspace of M ; and so by Theorem 2.2 L‘ is an ideal of M , (and hence of MJ. Now by Theorem 6.6 the topological dual of (M,lal(M,L’))is the ideal generated by L’ in M“, which is exactly L’. Now let L be the lol(M,L’)-closure of L in M . If E # M , then there exists 8 < v E M with v .$ L, and so by the Hahn-Banach theorem there exists a lal(M,L’)-continuous linear functional of M , that is, some cp E L’, with ~ ( u=) 0 for all u E L and v(cp) > 0. But cp(u) = 0 for all u E L implies cp = 8, a contradiction, and hence, L is lal( (L’);,L’)-dense in (I,’):. W

Now if (L,z) is a Hausdorff locally convex-solid Riesz space, then lol( (L’)i,L’)induces the topology (al(L,L’)on L ; this follows immediately from the fact that u,+O holds in L for (al((L’);,L‘) if and only if limlcpI(Iu,l) = 0 for every cp E L‘, which is equivalent to u, + 8 in L for lal(L,L’). Note next that by the remark following Theorem 19.9 (I,’): is lol( (L’);,L’)-complete and hence by combining this with the last theorem, we obtain the following result. Theorem 19.11. For every Hausdorff locally convex-solid Riesz space (L,z) the topological completion of (L,lal(L,L’)) is (I,’); equipped with the absolute weak topology (I.[ (,!,’);,I,’).

We have seen before (Section 9) that for a Hausdorff locally convex-solid Riesz space (L,z) the strong topology p(L’,L) is a locally convex-solid topology on L’ generated by the family of Riesz seminorms { p a : A E Bs}, where p,(cp) = sup{ Icp(u)I:uE A } , and Bs is the collection of all z-bounded solid subsets of L. Since for every u E L the set A = [ - IuI,IuI] belongs to Bs and ~ c p ~ (=~ yA(cp) u ~ ) for all cp E L‘, it follows easily that p(L’,L) is finer than lal(L’,L).Thus the following inclusions hold: o(L’,L)G lap&) E P(L’,L). The strong topology p(L’,L) is always a Fatou topology as the next theorem shows. Theorem 19.12. For every Hausdorfl locally convex-solid Riesz space (L,z)the strong topology p(L‘,L)is a Fatou topology. Proof. We shall show that every Riesz seminorm p A ( A E as) is a Fatou seminorm on L’. So, let A E gSand 0 Icp, cp in L’. If p,(cp,) p,(cp) does not hold, then there exists E > 0 such that p,(cp,) Ip,(cp) - E for all CI.Now

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[Chap. 6

pick 8 I u E A with p,(cp) - E < cp(u), and note that {cp,(u)) does not converge to cp(u), contradicting 6’ I cp, ? cp in L‘. Thus p,(cp,) t p,(cp) holds and wearedone. W Theorem 19.13. Let (L,z) be a Hausdor-locally convex-solid Riesz space. Then the order intervals of L’ are P(L’,L)-complete. l f in addition L’ is a hand of L-, then L‘ is P( L’,L)-complete. Proof. By the previous theorem P(L’,L) is a Fatou topology. The first part follows now immediately from Theorem 13.1 by observing that L‘ is Dedekind complete. Now if L‘ is a band of L-, then P(L’,L) is a Levi topology. To see this let 8 I cpa t in L’ with { ( p a ) p(L’,L)-bounded. Then there exists cp E L‘ such that cp, t cp in L- and hence cp E L’, since L’ is a band of L’, so that P(L’,L)is Levi. Thus by Theorem 13.9 it follows that L‘ is P(L’,L)-complete. W

The next theorem tells us that the order intervals of L‘ are always o(L’,L)compact. Theorem 19.14. For every locally convex-solid Riesz space (L,z) the order intervals of L’ are a(L,L)-compact. Proof. Let [O,cp] be an order interval of L‘ G RL. We have to show that

[6’,cp] is bounded and closed in RL. It should be evident that [8,cp] is bounded in RL. Now if cpa -+ rl/ holds in RL with {qZ)c [O,cp], then from the pointwise t+b I cp and so rl/ E L’, that is, [O,cp] convergence it should be clear that 0 I is closed in RL and hence [O,cp] is cr(L’,L)-compact. W

Recall that a Riesz space L is called uniformly complete if for every E,V for all n,p, some L‘ E L + , sequence (u,) of L that satisfies Iu,+,, - u,I I E, -, 0, there exists LI E L + with Iu, I E,U and some real sequence 0 I for all n.

UI

Theorem 19.15. Let (L,z) be an Archimedean locally convex-solid Riesz space and A c L’. l f L is uniformly complete, then the following statements are equivalent:

(i) A is lol(l’,L)-bounded. (ii) A is o(L’,L)-bounded. Proof. (i) =+. (ii) It follows from the relation a(L’,L) E lol(L’,L). (ii) =-(i) Let A be o(l’,L)-bounded and let u E L f . Define the Riesz norm p on the ideal A, generated by u in L by p(v) = inf{;i > 0:lvl I

h)

for v E A , ,

and note that [I-u,u] is the closed unit ball of p in A,. Now since L is uniformly complete it follows easily that A, is p-complete.

Sec. 191

DUALSPACES

131

On the other hand if ( P E A , then sup{Icp(v)(:u~ [ - u , u ] } < co and so cp restricted to A , is bounded on the closed unit ball [-u,u] of A , , and therefore is p-continuous. Next note that the restrictions of the elements of A to A , are a(A,’,A,) bounded (where A,’ is the norm dual of A”). Hence by the Banach-Steinhaus Theorem, A (restricted to A , ) is norm bounded (Theorem 19.3), that is, there exists a number c > 0 such that lcp(u)I I c for all u E [-u,u] and all cp E A . But then by Theorem 3.3(ii) we have Icpl(u) = sup{ lcp(u)( : u E [ - u,u]) I c for all cp E A , that is, A is (al(L’,L)-bounded,and the proof is finished. Theorem 19.16. Let (L,T)be a locally convex-solid Riesz space. If L’ is a 0-ideal of L‘, then for a subset A of L’ the following statements are equivalent:

(i) A is (ol(L’,L)-bounded. (ii) A is P(L‘,L)-bounded. Proof. (i) * (ii) If A is not B(L‘,L)-bounded then neither is its solid hull S. This means that there exists a solid a(L,L’)-bounded subset B of L u, E B and such that sup{ lcp(u)J:cp E S ; u E B ) = co. For each n, pick 8 I 8 I cp, E S with cp,,(u,) 2 n3. Now define the sequence $, = k-’cp, ( n = 1,2,. . .) and note that since A (and hence S ) is lo)(L’,L)-bounded, {$,,} is (a((L’,L)-bounded.Thus 8 I $, t $ holds in L- for some $ E L-, and therefore since L‘ is a 0-ideal of L” we have $ E L‘. Now observe that since {u,) is (ol(L,L‘)-bounded the sequence {n- ‘u,J must be Ial(L,L’)-convergent to zero in L, which is in contradiction with

$(n-’u,,) 2 n-’cpn(n-lun) 2 1 for all n. Hence A is P(L’,L)-bounded. (ii) 3 (i) Use the relation laJ(L’,L)C p(L‘,L). Recall that the bidual L” of a Hausdorff locally convex-solid Riesz space (L,z) is the topological dual of (L’,/j(L’,L)).Since p(L’,L) is a locally solid topology, it follows that L” is an ideal of (L’)-. Moreover, L is a Riesz subspace of L“ under its natural embedding (see Theorem 3.11). Theorem 19.17. Let (L,T)he a HausdorfS localljl convex-solid Riesz space, and let 19 I u, be a z-bounded net of L. Then 0 I u, t f holds in L” for some f E L”. Proof. For every 8 Icp E L‘ definef(cp) = sup{cp(uJ} and note that f ( c p ) is finite since {urn)is z-bounded. Obviously f:(L’)+ + R + is additive, and hence by Lemma 3.1 f extends to an order bounded linear functional from L’ into R. Note that 0 I 11, f holds in (L‘)-. To complete the proof we show that f E L“. Let A be the solid hull of {urn)in L; A is z-bounded. Now cp E L‘ we have 0 I cp(m,) I sup{cp(u):u E A } = p,(cp) for all CI and so for 8 I f ( q )I p,(cp). But then 1f(cp)l I f(1cpJ) I p,(cp) holds for all cp E L’ and

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so f is p(L',L)-continuous, because the Riesz seminorm p A is p(L',L)continuous. The next theorem describes an important property of Hausdorff locally convex-solid Riesz spaces with the pre-Lebesgue property. Theorem 19.18. I f a Hausdorff locally convex-solid Riesz space (L,t) satisfies the pre-Lebesgue property, then t and lol(L,L) induce the same topology on the order bounded subsets of L. Proof. Let A be an order bounded subset of L. By Theorem 10.6 the of (L,t) satisfies the Lebesgue property and topological completion hence by Theorem 11.7 the Hausdorff locally solid topology lal(L,L')likewise does. But A is also an order bounded subset of 2and hence by Theorem 12.9 ? and I o l ( ~ , ~both ' ) induce the same topology on A . Now since each q E L' has a unique continuous extension @ on t and cp H @ is a Riesz isomorphism between L' and 2'(see Exercise 8 of Chapter 2), we see that lol(L,L') and lol(L,L') both induce the same topology on A and the result follows.

(z,?)

If a Hausdorff locally convex-solid Riesz space (L,z)satisfies the Lebesgue property, then L is order dense in (L'); (Theorem 9.1) and hence the ideal generated by L in (L'); (which is the same as the ideal generated by L in L") is the Dedekind completion of L. However, if t is not a Lebesgue topology then L is no longer order dense in (L');. In this case the Dedekind completion of L can be obtained from L" by using Theorem 2.18; the details follow. Theorem 19.19. Let (L,z)be a Hausdorff locally convex-solid Riesz space and let A be an ideal of L" that is maximal with respect to the property L n A = ( O } . Then the Dedekind completion of L is the ideal generated by L in L"/A (where L is identijied with its canonical image in L ' / A ) . Note. Most of the results of this section have appeared in many forms in the works ofvarious authors, see [45], [52], [55], and [71]. Theorem 19.18 is due to Kawai [39, Theorem 5.3, p. 3011. 20.

WEAK COMPACTNESS IN THE ORDER DUAL

In this section the weakly compact subsets of L,, L,, and L" will be characterized in terms of the order structure of L. We start with some basic properties of L-. Theorem 20.1. Let L be a Riesz space. Then for a subset A of L - the following statements are equivalent:

(i) A is lal(L",L)-bounded. (ii) The solid hull of A is o(L-,L)-relatively compact.

WEAK COMPACTNESS

Sec. 201

133

Proof. (i) (ii) Since lal(L-,L) is a locally solid topology, the solid hull S of A is also lol(L",L)-bounded and hence a(L",L)-bounded. Thus in order to establish that S is a(L-,L)-relatively compact, it is enough to show that the o(L*,L)-closure of S in L* lies in L'[since a o(L*,L)-closed and bounded subset of L* is a(L*,L)-compact]. To this end let cp be in the a(L*,L)closure ofSand u E L'. Now for u E [O,u] choose $ E S with Icp(u) - $(v)l < 1, put 6 = sup{l~)l(u):$E S} < a,and note that (cp(o)l 5 Icp(u) - $(u)l I$(u)l I 1 6. Thus cp carries order bounded subsets of L onto order bounded subsets of R, and hence cp E L". (ii) (i) Let S be the solid hull of A. Then S is a(L',L)-relatively compact and so for U E L' there exists 6 > 0 such that I S for all $ E S. But if cp E A, then IcpI E S and thus 0 5 Icpl(u) I 6 for all cp E A . So, A is lol(L',L)bounded, and the proof is finished.

-

+

+

I$(u)~

Theorem 20.2. Let L be an Archimedean uniformly complete Riesz space. Then L - is a(L",L)-sequentiallp complete. Proof. Let {cp,} be a a(L",L)-Cauchy sequence of L'. Then {cp,,} is a(L',L)-bounded and hence by Theorem 19.15 Ial(L-,L)-bounded. But then by Theorem 20.1 {cp,} is a(L',L)-relatively compact and hence has a o(L",L)accumulation point cp. It now follows easily that {cp,} is a(L',L)-convergent to cp, and we are done. Let (L,z) be a locally convex-solid Riesz space. Recall that for a subset A of L', we define p,(u) = sup{lq(u)l:cp E A } I co for each u E L. Note that if A is a(L',L)-bounded, then p A is a seminorm on L and if in addition A is solid, then pA is a Riesz seminorm (see the discussion at the beginning of Section 9). Clearly if A c B, then pA(u)I p e ( u ) for all u E L. A sequence {u,} of L is called p,-Cauchy if for each E > 0 there exists k such that pA(un - u,) < E for all n,m 2 k. The a(L,,L)-compact subsets of L, will be characterized first for Archimedean Riesz spaces. The next theorem is, in fact, a very weak result, as Theorem 20.7 and Example 20.8 will later show. As we shall see, if in addition L is assumed to be almost a-Dedekind complete, then much stronger results hold. Theorem 20.3. Let L be an Archimedean Riesz space. For a subset A of L,' the following statements are equivalent: (i) The solid hull S of A is o(L,,L)-relatively compact. (ii) limp,(u,) = 0 whenever u, J 0 in L. (iii) limp,(u,) = 0 whenever u, 1 8 in L. Proof. (i) = (ii) Suppose that (ii) does not hold. Then there exist E > 0 and u, 1 8 in L with pA(u,) > 26 for all n. For each n choose cp, E A such that

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lcpn(u,l)l> 2 ~ Let . 0I cp E L, be a a(L,,L)-accumulation point of {lcpn(j. Given n, choose m 2 n such that ( Icprnl - cp)(uJ < E and note that since 2~ < l~rn(urn)l5

l~rnl(urn)I lVrn((hln) =

(IVrnl - cp)(un)

+

cp(un)

I E + cp(un),

we have cp(u,) > E for all n. But this contradicts u, 1 0 in L and cp E L, . (ii) =-(iii) If (iii) is false, there exist E > 0, u, 1 0 in L, and {cp,} E A such that Icp,((u,) > 28. By Theorem 3.3(ii) for each n there exists Iw,I 5 u, with lcpn(w,)l > 2~ and so either Icp,(w,+)I > E or ~cp,(w,,-)~ > E. Set L', equal to w,+ or w,- so that ~ q , , ( v n> ) ~ E . Now since v I v I[,, 1 u 1 and cpl E L,, there exists k l such that Icpl(ul v u k l ) l > E. Similarly, since L'kl v hi, 1 vkl thereexists k , > k , such that ((okl(vklv uk2)1> E . Proceeding inductively, we obtain a sequence k1 < k , < k , < ' ' ' Satisfying I4?k,(L1k, V Uk,,+ ,)I > E for all 12. Put X , = C k , V uk, 5 x, I u, -1 0). But then P,(x,) 2 uk,,+,,and note that x, 1 0 (for .Y,+ I Icpk,(x,)l > E holds for all n, contradicting (ii). (iii) + (i) Note first that S is lol(L,,L)-bounded. Indeed, for u E L + we have n- ' u 1 0 since L is Archimedean; so there exists k such that ps(n- 'u) < 1 for all n 2 k, and thus Icpl(u) I k for all cp E S. By Theorem 20.1 S is a(L",L)-relatively compact and to complete the proof we shall show that the a(L-,L)-closure of S in L- is contained in Lz. To this end let $ be in the a(L-,L)-closure of S and E > 0. For u E L pick cp E S such that I(cp - $)(u)l < E , and note that since I$(u)(I [(cp - $)(u)I + Icp(u)l I E ps(u), it follows that ($(u)(I ps(u) holds for each u E L. So, if u, 1 0 holds in L, then lim $(u,) = lim ps(u,) = 0, so that ij E L,. +

+

The concept of order-equicontinuity is introduced next in order to study the deeper properties of the compact subsets of L; and L,. Definition 20.4. Let (L,z) be a locally convex-solid Riesz space. A subset A of L is called order-equicontinuous on L if (u,) is u p,-Cauchy sequence whenever 0 I u, I u holds in L.

r

It should be clear from the above definition that subsets of order-equicontinuous sets are also order-equicontinuous. It should also be clear that A L L' is order-equicontinuous if and only if 0 I u, I u in L implies that (ua) is a p,-Cauchy net (see Theorem 10.1).

r

Theorem 20.5. Let (L,z) be u locally convex-solid Riesz space. If A E L' i s order-equicontinuous on L, then the following statements hold:

(i) A is la((L',L)-bounded. (ii) T h e convex solid hull of A is order-equicontinuous (and hence the solid hull of A is also order-equicontinuous). Proof. (i) If A is not lol(L',L)-bounded, then there exist u E L + and a sequence {cp,} c A such that Icp,l(u) > 2"+l for all n. Now for each n by

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WEAK COMPACTNESS

Theorem 3.3(ii) there exists 0 I u, I u such that Icp,(v,)l > 2,. But then we have 0 I w, = 2-'vi I u and pA(w, - w,- ,) 2 Icpn(2-"u,)I > 1 for all n; so that { w , ) is not a p,-Cauchy sequence, in contradiction with our hypothesis. (ii) Let S and C be the solid and convex solid hulls of A , respectively. pc(u) for all u E L. On the other hand if cp E C, Since S s C, we have p,(u) I then there exist positive constants R,, . . . ,Afl with lii= 1 and \cp,, I . . . ,cp,,> E S such that cp = i+cpi(see Theorem 1.3) and so

,

xr=,

I;=

holds for all u E L, from which follows pc(u) I ps(u). Hence p s = pc and so it is enough to show that S is order-equicontinuous. Assume that S is not order-equicontinuous. Then there exist E > 0 and 0I 11, t I u in L such that ps(un+, - u,) > 2.5. Now for each n pick cp, E A with Icp,,I(u,,+, - u,) > 2 ~ and , then use Theorem 3.3(ii) to select t' I z', I w, = vi t I 11 and u,+ - u, such that [cp,,(~,,)I > c. But then 0 I pA(w, - w,2 Icp,(~,, I > E holds for all n, contradicting the fact that (w,) is a p,-Cauchy sequence. H Remark. If A is a solid set, then the order-equicontinuity of A expresses the fact that pA generates a pre-Lebesgue topology on L.

The following characterizations of the order-equicontinuity show how powerful the property is. It will later become evident that the concept of order-equicontinuity is the single thread that ties together the various results on weak compactness.

Theorem 20.6. Let (L,z)be a locally convex-solid Riesz space. Thenfbr a subset A of L' the following statements are equivalent: (i) A is order-equicontinuous on L. (ii) For each L I EL+ and E > 0 there exist 6 > 0 and a jinite subset \f c p l , . . . ,y,) of A such that if v E [fl,u] and Icpil(a) < 6 for i = 1, . . . ,n then lql(~1) < Efor all cp E A . (iii) For ecrch u E L + and E > 0 there r.uists 0 i,b in the ideal generutecl by A S U C ~that (/PI - $ ) + ( u<) for all ~p E A . (iv) A is lol(L',L)-bounded and every disjoint sequence in the solid hull of A is 101 (L',L)-convergent to zero. (v) A is lol(L',L)-bounded and limp,(u,) = 0 holds whenezier {u,} is an order hounded disjoint sequence of L. Proof. (i) * (ii) By Theorem 20.5 we can assume that A is solid. If (ii) does not hold, then there exist u E L + , E > 0, and sequences {u,) G [R,u],

136

{cp,,)

[Chap. 6

LOCALLY CONVEX-SOLID TOPOLOGIES

E A n L + such that for each

cPn(Un)

> 36.

n we have cp,(u,) < 2-" for 1 I k < n and

+

For each fixed n the sequence uk,, = sup{ui:n I i I n k ) ( k = 1,2,. . .) is an order bounded increasing sequence. Hence, by the order-equicontinuity ~ all 6 I cp E A of A there exists an integer r, such that q ( u k , , - urn,,) < 2 - " for and k 2 r,. Set w, = urn,, and note that (urn- w,)+ I (urn,, - w,)+ for m 2 n implies ~ ( ( L I , - w,)') I 2 - " ~for all 6 I cp E A . Now put z, = inf{wi:l I iI n } and observe that since (u, - 2,)' I (u, - wi)+ we get

xr=

n

1

Cp,((u, - Z,)+) I c p , ( ( L f , I=

so that cp,(z,)

1

WJ+)

c2-iE I n

s

E,

i= 1

2 cp,(u,) - cp,( (u, - z,,)') 2 3.5 - E = 2.5, for all n. But since c1

( ~ n ( z n +1)

I (PAW,+

1)

I

1

i=n+

x (xi

1

Yn(ui) I

i=n+ 1

2-I

= 2-",

we obtain cpn(z, - z,+ 1 ) 2 2.5 - 2-" for all n, and in particular, p,(z, - z,+ 1) 2 q,,(z, - z,, 1 ) > E for all sufficiently large n. But then 0 I z1 - z, t I z1 and {zl - z,} is not p,-Cauchy, contradicting the order-equicontinuity of A . (ii) (iii) Let it E L + and 0 < E < 1. Choose 6 > 0 and { q l ,. . . ,cp,} E A such that (ii) holds. Then pick 0 < 2. < 6 such that Icp,I(Lu) < 6 for i = 1, . . . ,a, and note that Icpl(Au)< E for all cp E A . Put $ = L-2(Icp11 + . . . + Icp,,l). Now if cp E A and t9 I uI u satisfy (lql - $ ) ( u ) 2 0, then lcpl(u) 2 $ ( t i ) 2 2.-21cpil(u) and so lcpil(u) I A21cpl(u) I LE < 6 for i = 1,. . . ,n. Therefore, lcpl(u) < E and hence (191- $ ) ( u ) I Icpl(u) < E. But then by Theorem 3.3(ii) we infer that (lcpl - $)+(u)I c, and we are done. (iii) * (iv) Let {q,} be a disjoint sequence of the solid hull of A . For I$,[. Now let u E L + and E > 0. Choose each n pick I),E A such that lcp,I I 8I $ E L' such that (IqI - $)'(u) < E for all cp E A , and note that ( Iqnl $)+(u)I (I$, / - $)'(ti) < E for all n. Since { Icpnl A $) is an order bounded lcpkl A $(u) I $(u) for all n and u E L + , disjoint sequence of L', we have and so lim Icpnl A $(u) = 0. Pick k such that Iq,l A $(u) < E for n 2 k , and note that

+

0I JYnl(U) = (lcpnl - $)+(u) Icpnl A $(u) < 2~

for all n 2 k,

that is, {cp,} is lo((L',L)-convergentto zero. The lol(L',L)-boundedness of A follows from the relation lcpl(ti) = (lql - $ ) + ( u ) + IcpI A $b). (iv) * (v) If (v) is false, then there exist E > 0, an order bounded disjoint sequence {u,] E [e,u], and a sequence {cp,} c A such that Jcp,l(u,) > E. Consider L embedded in (L'): and for each n denote by P , the projection of L' onto the carrier band Cum (recall that L' = Nun @ C,,, where Cum= N t n and

WEAK COMPACTNESS

Sec. 201

137

Nun= {q E L': lq((u,,) = 0)). For each n put also $,#= P,( Iq,l). S'ince 11, A u, = i3 for n # m, it follows from Theorem 3.9 that C,,,?I C,,,?for n # m and so {$,} is a disjoint sequence in the solid hull of A. But Theorem 19.5 now implies Il/n(u) 2

Ic/n(un)

= pn(lpnI)(LLn)= l q n l ( U n )

>E

for all n, contradicting our hypothesis that (+hn) is lol(L',L)-convergent to zero and the desired implication follows. (v) * (i) Let S be the solid hull of A. Since A is lol(L',L)-bounded, ps is a Riesz seminorm on L. Now let {u,} be an order bounded disjoint sequence and E > 0. Since P d l l n ) z=

s ~ P ( l ~ ) ( ( ~ En A) )} :=q S U P { ( ~ O ( E~ 'A) ;~( 0:1 CI ~ Iunl},

for each n there exist q,,E A and Iu,I I (u,I with ps(u,) - E 5 Iq,(u,)l 5 pA(un). Now limpA(r,)= 0 (since {u,} is an order bounded disjoint sequence) and so limsupps(u,) I E for all E > 0, so that limp,(u,) = 0. Thus ps generates a pre-Lebesgue topology by Theorem 10.1; so S is order-equicontinuous and hence A is also order-equicontinuous on L. Note. Theorem 20.6 is a combination of a result of Burkinshaw [16, Theorem 3.2, p. 1921 and a theorem of Fremlin [25, Theorem SlH, p. 2231.

Theorem 20.7. Let L be an almost o-Dedekind complete Riesz space. Then for a subset A of Lz the ,following statements are equivalent: (i) (ii) (iii) (iv)

The solid hull S cf A is a(L,~,L)-relutioelycompact. limp,(u,) = 0 whenever u, J H in L. limp,(u,) = 0 wheneuer 11, J 0 in L. A is order-equicontinuous on L.

Proof. The equivalence of (i), (ii), and (iii) has already been established in Theorem 20.3. (iii) (iv) If 8 I u, I u holds in L, then since L is almost o-Dedekind u, for all n and complete, there exists a positive sequence u, 1in L with urnI m, and such that u, - 11, J 8 [see Definition 2.l(iii) and related discussion]. u , + ~- u, I u, - u, holds, we Now since ps is a Riesz seminorm and 0 I get =j

L l A ( u n + p - un)

I P s ( U n + p - u,) I pdvn -

11,)

10

from which it follows that {id,) is a p,-Cauchy sequence. (iv) * (ii) Let u, J 6 in L and E > 0. For each n choose qnE A with pA(u,) i ~q,,(u,)~ E. Now pick k so that pA(un- u,) < E for n,m 2 k, and then for fixed n > k choose m > n such that (p,(u,)( < E (this is possible

+

130

since cp,,

LOCALLY CONVEX-SOLID TOPOLOGIES E

[Chap. 6

LL). But

I 14Dfl(un)l+ C 5 Icpn(% - urn)[ + Icp,(u,)[ + E 53.5 Therefore, lim pA(u,) = 0 and the proof is complete. PA(%)

for n 2 k.

The following example shows that the hypothesis of almost a-Dedekind completeness in the previous theorem cannot be weakened to Archimedeanness. (Compare this example with Theorem 20.3.) Example 20.8. Consider a normed Riesz space L, with the a-Lebesgue property but without the pre-Lebesgue property (the normed Riesz space of Example 10.4 is such a space). Note that under these circumstances L is not almost a-Dedekind complete and that L' c L, (see Exercise 14 of Chapter 3). Now if U , is the closed unit ball of L,, then since U , is solid it follows that A = U p o is a solid subset of L' and by Theorem 19.1 a a(L',L)-compact subset of L'. Thus A is a solid o(L,,L)-compact subset of LL. O n the other hand since p = pA holds, A cannot be order-equicontinuous on L, since otherwise L, would satisfy the pre-Lebesgue property. Thus there are solid a(l,,L)-compact sets that are not order-equicontinuous. W The next important theorem shows that the concepts of order-equicontinuity and o(L,,L)-relative compactness in L; coincide for a-Dedekind complete Riesz spaces.

Theorem 20.9. Let L be u a-Dedekind complete Riesz spuce. Then for a subset A of L, the jdlowing statements ure equivalent: (i) A is a(L,,L)-relutiuely compuct. (ii) A is order-equic,ontinuous on L.

Proof. (i) * (ii) If A is not order-equicontinuous, then there exist E > 0, 0 I u, T I u in L, and (cp,) E A such that lcpn(un+, - u,)l > 46 for all n. Put (0) w,= u,+ - 11, and note that w, -+ 6; indeed, since L is a-Dedekind complete 0 I u, t v holds in L and so 6 5 w, s v - u, 4 0. Let cp be a o(L,,L)accumulation point of {cp,). Next we shall construct by induction a strictly increasing sequence (k,} of positive integers satisfying for every n 2 2, I(cpi - (p)(wk,)I < 2 - " ~and l(Vkn - cp)(wi)I < 2-"& for i = 1,. . . , k n P l .To see this start with k , = 1 and (0) assume that k, has been chosen. By virtue of w, + 6 and cpI - cp E L, there exists m > k, with I(cpi - q)(wk)l < 2 - ( " + ' ) ~for all i = 1,. . . ,k, and all k 2 m. Now since cp is a o(Lz,L)-accumulation point of {cp,,), there exists k,, > m > k, so that I((Pk,+l - cp)(wi)I < 2 - ( " + ' ) ~for i = 1,. . . ,k,, and thus the positive integer k,, I can be chosen to satisfy the above properties.

,

Sec. 201

WEAK COMPACTNESS

139

zii, and note that since L is a-Dedekind Set $, = qk,,vn = wk,,,zn = complete 8 I z, z holds in L. Now for m > n the relations n- 1 l($n

1

- q)(zn-l)l

i= 1

n- 1 l($n

- q)(ui)l

<

1 2-ie <

i= 1

i=n+ 1

i=n+ 1

imply I($n

- q)(zm)l 2

I$n(vn)I

> 4E

-

I ~ ( u n ) l- l ( $ n

- lc'p(Un)l

-

~ ) ( z n - - l($n

- E - E = 2E - l q ( U n ) I .

-

q)(zm - z n ) l

Thus by letting m -+ m we get (since I),, - cp E L,) I($, - q)(z)l2 2c lq(url)lfor all n. Now by taking into consideration that tirl 2 0 and cp E L,", we infer that there exists k such that - q)(z )I > c for all n 2 k . Let $ be a o(L,,L)-accumulation point of {t,hrlj. Then since

I($

- q)(vi)I 5

I($

- $n)(ui)J

+

l($n

- q)(vi)( I

-

t,hn)(Ui)I

+2 - n ~

for all n sufficiently large, we get $(ui) = q(q) for all i and so $ ( z ) = q ( z ) , since $,q E L,. But the last conclusion contradicts I($, - q ) ( z ) l > E for n 2 k, and hence our result follows. (ii) * (i) It follows from Theorem 20.7. 4 Since a set is order-equicontinuous if and only if its solid convex hull is order-equicontinuous the next result follows immediately from the previous theorem.

Corollary 20.10. Let L be a o-Dedekind complete Riesz space and A c L,. Then A is a(L,,L)-relatively compact i f und only if its solid convex hull is a(L;,L)-reluticely compact. We now turn our attention to the solid o(Li,L)-compact subsets of L,. The next theorem tells us that the solid o(Li,L)-compact subsets of L i are always order-equicontinuous on L and hence satisfy the equivalent conditions of Theorem 20.6. Theorem 20.11. Let L he an Archimedean Riesz space. Then for a subset A of L,' the jdlowing statements are equivalent (i) The solid hull S of A is o(L,,L)-relatively compact. (ii) limp,(u,) = 0 whenever u, 1 0 in L. (iii) A is order-equicontinuous on L.

LOCALLY CONVEX-SOLID TOPOLOGIES

140

[Chap. 6

Proof. (i) * (ii) Note that ps is a Riesz seminorm on L. Now let u, 1 8 in L and E > 0. For each a choose 8 I cp, E S such that ps(u,) I cp,(u,) + E. Next pick a a(LL,L)-accumulation point 8 I cp E L, of (cp,), and then select an index a l so that cp(u,) < E for all M 2 cxl. Now choose a2 2 a , such that [(cp,, - cp)(u,,)l < c, and note that for M 2 a2 we have PA(^ 5 Ps(ua)

+ E (PaZ(Ua1) + E +E <3 ~ .

Pdua,) I ya2(ua,)

+

= ( V a , - c~)(um,) ~ ( u m , )

Thus lim pA(u,) = 0. (ii) * (iii) Apply the same arguments used in the proof of Theorem 10.2. (iii) * (i) Note first that since L; is the topological dual of (L,lal(L,LZ)), it follows from Theorem 20.5 that the solid hull S of A is Ial(L; ,L)-bounded and thus lal(l",L)-bounded. Consequently S is a(L',L)-relatively compact by Theorem 20.1. To complete the proof we shall show that the a(L-,L)-closure of S lies in L,. To this end let cp be in the o(L',L)-closure of S and u, 1 8 in L. Since S is also order-equicontinuous we can see that {u,) is a ps-Cauchy net, and so given E > 0 there exists a1 such that ps(u, - up) < E for a,p 2 al.But for fixed a 2 a1 there exists $ E S such that I($ - cp)(u,)I < E, and so if p 2 a, we have

I($

+ J$(ua)l I E + l$(ua - up)( + /$(up)/ I E + Ps(ua - up) + J$(up)( 2~ + l$(up)l*

IV(ua)JI - cp)(ua)l

But since lim $(up) = 0, it follows that Icp(u,)l I 28 for x 2 al. Therefore, E L i and the proof is complete. 4

cp

It is important to observe that if L is a-Dedekind complete, then it follows from Theorem 20.9 and L, G L, that every a(L;,L)-compact subset of Lx must be order-equicontinuous on L. Thus the following result is immediate from Theorems 20.5 and 20.11. Corollary 20.12. Let L be a a-Dedekind complete Riesz space. Then a subset of L; i s a(Lt,L)-relatively compact if and only if its convex solid hull is a(L~,L)-relativelycompact.

By Theorem 3.11 for every locally convex-solid Riesz space (L,z) there exists a Riesz homomorphism from L into (L');. If in addition T is a Hausdorff topology, then this embedding is a Riesz isomorphism (into) and in this case as usual we consider L as contained in (L');. Observe that since L" is an ideal of (L')-, the ideal I generated by the image of L in (L'), is the same as the ideal generated in L". Now by once more applying Theorem 3.11 to the pair L' and I, we obtain that the em-

Sec. 201

WEAK COMPACTNESS

141

bedding of L' into 1; is a Riesz isomorphism, and moreover by using Theorem 2.2 we have that L' is an order dense ideal of I,'. The next theorem gives a condition for equality of L' and I;.

Theorem 20.13. Let (L,z) be a locally convex-solid Riesz space and let I be the ideal generated by the imuge of L in L". If L' is u band of L', then L' = 1;. Proof. As observed above, L' is an order dense ideal of 1;. Let 8 I cp E 1;. Pick a net { q Z )G L' with 8 I cpa t cp in I,' and note that since [cpcl) is lol(l;,I)-bounded {cp,} is also lol(L',L)-bounded. Therefore, by Theorem 19.7 cpn t $ holds in L' and since L' is order dense in In-, we must have $ = cp, so that L' = I;. The solid compact subsets of L,' were characterized in terms of the oLebesgue property, while those in L; were characterized in terms of the Lebesgue property. Order-equicontinuity has been the central property relating compactness in the spaces L, and LI. The next theorems will show that there is actually a deeper relationship.

Theorem 20.14. Let (L,z) be an Archimedean locally convex-solid Riesz space such that L' is a band of L'. Let I be the ideal generated by the image of L in L". Then for a subset A of L' the following statements are equivalent: '(i) A is order-equicontinuous on L. (ii) A i s o(l',l)-relatioely compact.

Proof. (i) =-(ii) Let 0 I v E I and E > 0. Pick u E L with u I u, and then use Theorem 20.6 to select $ E L' such that (lcpl - +)+(u) < E for all cp E A. Hence (lqI- $ ) + ( u ) < E for all cp E A and thus A is order-equicontinuous on I (by Theorem 20.6 again). Now by Theorem 20.13 L' = I,', and therefore A is o(li,l)-relatively compact (Theorem 20.11), that is, A is o(L',l)-relatively compact. (ii) ==(i) Since L E I and L' = I;, by Theorem 20.11 A is order-equicontinuous on I and hence order-equicontinuous on L. Consider a compact Hausdorff topological space X , and let L = C ( X ) be the Riesz space of all real-valued continuous functions on X equipped with the sup norm. If e denotes the constant function one on X, then e is an order unit of L and in addition e is an order unit of the norm bidual L". Indeed, since llcpll = cp(e) holds for all 8 I cp E L', we have for 8 If E L" 0 I f(cp) I llfll llcpll = Ilfllcp(e) for all 8 I cp E L', so that 8 s f I llflle holds in L''Land therefore e is an order unit of L". In particular, we have that the ideal I generated by L in L" is exactly L" and thus by virtue of L' = L', Theorem 20.14 gives that a subset of L' is o(L',L")-relatively compact if and only if it is order-equicontinuous on L. Finally, by noting that a sequence

142

LOCALLY CONVEX-SOLID TOPOLOGIES

[Chap. 6

of C ( X )is order bounded if and only if it is norm bounded, we get the following classical result of A. Grothendieck on weak compactness in the norm dual of C ( X )(see [23, p. 3891 or [67, Theorem 9.8, p. 1261). Theorem 20.15. Consider L = C ( X ) with the sup norm for a HausdorfS compact topological space X . Then for a norm bounded subset A of L' the following statements are equivalent; (i) A is o(L',L")-relatively compact. (ii) For every positive norm bounded disjoint sequence { f.} of C ( X ) it follows that lim cp( f,)= 0 uniformly for cp in A. Now by observing that the topological dual of (L,lol(L,L')) is L", we have the following important special case of Theorem 20.14. Theorem 20.16. Let L be an Archimedean Riesz space, and let I be the ideal generated by the image of L in (L-);. Then for a subset A of L - the following statements are equivalent; (i) A is order-equicontinuous on L. (ii) A is o(L-,l)-relatiuely compact. If L is a-Dedekind complete, then the order-equicontinuous subsets of L; and Lz are actually relatively compact in a number of topologies. By combining Theorems 20.9 and 20.16 we have the following. Theorem 20.17. Let L be a o-Dedekind complete Riesz space and let I be the ideal generated by the image of L in (L-):. For A E L, the following statements are equivalent: (i) A is order-equicontinuous on L. (ii) A is o(L,,l)-relatively compact. (iii) A is o(L,,l)-relatively compact. Now if we identify Lz and (La), according to Theorem 3.12 and take into consideration Theorems 20.11 and 20.16, we obtain the following result concerning Archimedean Riesz spaces. Theorem 20.18. Let L be an Archimedean Riesz space, and let I be the ideal generated by the image of L in (L-);. For a subset A of L: the following statements are equivalent;

(i) (ii) (iii) (iv)

A is order-equicontinuous on L. The solid hull S of A is o(L,,L)-relatively compact. A is o(L:,L')-reIatively compact. A is o(L:,l)-relatiuely compact.

Sec. 201

WEAK COMPACTNESS

143

The next Theorem relates weak sequential convergence with the concept of order-equicontinuity. It is a result of basic importance and will be used to characterize weak sequential compactness. To prove the Theorem we need ui = sup{cr= u,}, if the the following simple lemma; remember that supremum exists.

,

Lemma 20.19. Let L be a a-Dedekind complete Riesz space and a sequence c L + with u, I u. Then for every 8 I cp E L' and E > 0, there exists a subsequence {u,} of {u,} (depending on cp and E ) suCh that u = u, satisfies q ( u ) < E. {u,}

,

Proof. Let 8 I cp E L" and E > 0. Choose a countable partition { N , } of the natural numbers, with each N , infinite. Put w, = CieN,ui for all n. Since N in N j = @ for i # j we obtain 0 I wi I uiI u for all n. Thus cp(wi) I cp(u) < GO, and hence cp(wi) < E for some i. Write N i = {kl ,k2 ,. . .} with { k , } increasing, put u, = u k , for all n, and we are done.

C:=

2%

Theorem 20.20. Let (L,z) he a a-Dedekind complete locally convex-solid Riesz space. Then every a(L,L)-convergent sequence of L' is order-equicontinuous on L. Proof. Without loss of generality we can only consider sequences which are convergent to zero. If the result is false, then there exist E > 0,8 I u, t u in L, and a a(L',L)-convergent to zero sequence {cp,) c L' with (q,(u,+ - u,)l > 3.2 for all n. Put w, = u , + ~- u, 2 8, note that w, = u - u1 5 u and Jq,(w,)l> 3~ for all n. We shall construct subsequences {u,,") (k = 1,2,. . .) of {w,) and positive integers n, < n2 < . ' . such that for each k the following statements hold:

(i) {u,,"} is a subsequence of {$I}, (ii) w,,is a member of {ui- '}, (iii) C!:; IVnk(Wn,)I< E, (iv) IP,,l(uk) < E, where uk = uflk.

C,"=

We use induction on k. For k = 1, apply Lemma 20.19 to {w,,};extract a subsequence {u,'} with u1 = unl satisfying J(pll(ul) < E, and put n, = 1. Assume that {unk} and nk have been selected satisfying properties (i) through (iv). Since limcp,(wi) = 0 for 1 s i 5 nk, pick n k f l > nk so large that Icpnr+ l(w,,)l < E and w , , + ~is a member of {unk}. Now apply Lemma 20.19 to {u,'}. Choose a subsequence (u,k+'} with ((P,,,+,~(U~+ ,) < E ; the induction is complete. w,,. Then Next put w =

If=,

8Iw =

k

W

k

1wn, + i = k + 1 wn,5 i1 wnx+ u k i= = 1 1

144

[Chap. 6

LOCALLY CONVEX-SOLID TOPOLOGIES

holds for all k, and hence,

for all k. Combining the last relation with the triangle inequality we get Iqf2k(w)l

’

lqnk(wfIk)l

k-1

-

i= 1

Iqnk(Wn,)l

-

’

for all k, which contradicts lim, qnk(w)= 0. The proof is now complete.

W

It is interesting to observe that the classical Vitali-Hahn-Saks Theorem on convergence of measures [23, p. 1581 can be obtained by combining Theorem 20.20 with Theorem 20.6(ii). Note. Theorems 20.16 and 20.20 are due to Burkinshaw [16]. The proof of Theorem 20.20 presented here is based on the techniques of Dodds [21]. As an application of Theorem 20.20 we have the following.

Theorem 20.21. Let (L,t) be a a-Dedekind complete locally convex-solid Riesz space and let I be the ideal generated by the image of L in L”. Then every a(L’,L)-convergent to zero sequence of L’ is also a(L‘,I)-convergent to zero. Proof. Assume that {q,}G L‘ is a(L’,L)-convergent to zero. By Theorem 20.20 { q n }is order-equicontinuous on L. Now if J is the ideal generated by the image of L in (L-):, then by Theorem 20.16 (q,)is o(L‘,J)-relatively compact and hence {cp,} is a(L-,J)-convergent to zero. I I$I(u) Now if 8 5 f E I , then there exists u E L with 8 If Iu. So, holds for all $ E L’ and thus by Theorem 3.4, there exists F E (L‘)- such that F ( q ) = f ( c p ) for all q E L’ and with IF(q)l I Icpl(u) for all cp EL-; thus IF1 Iu and so F E J . Finally, since { q n }is a(L‘,J)-convergent to zero, we have limf(q,) = lim F(q,) = 0 and thus { q n )is a(L’,I)-convergent to zero, and the proof is complete. W

If($)

-

Corollary 20.22. Let (L,t) be a a-Dedekind complete locall)) convex-solid dL’ L) Riesz space and let P be the projection of L’ onto a band B. I f cpn L0, u(L’L ) then P(q,) 8. u(L’ L )

Proof. Assume cpn8. Let I be the ideal generated by the image of L in L”. Since L’ = B 0 Bd by applying Theorem 19.5 to (L’,l~l(L’,l)), we obtain that I = Bo 0 (Bd)O.Now for u E L write u = F + G with F E Bo and G E (Bd)’, and note Pqn(u)= ‘(Pqn)

+ G ( f ‘ ~ n=) G(Pqn)= G(qn) - G(qn - Pqn) = G(c~n).

Sec. 201

WEAK COMPACTNESS

But lirn C(cp,) = 0 (by Theorem 20.21) and thus lirn Pcp,(u) Hence Pcp, p(L',L!O and the proof is finished.

145

= 0 for each u E

L.

By using the preceding result we get the following version of Theorem 20.2. Theorem 20.23. I f L is a a-Dedekind complete Riesz space, then every band of L- is a(L",L)-sequentially complete. I n particular, L, and L: are both a(L-,L)-sequentially complete.

The next result shows the close relationship between order-equicontinuity and sequential compactness. Theorem 20.24. I f L is a-Dedekind complete, then for a subset A of Lthe following statements are equivalent:

(i) A is order-equicontinuous on L. (ii) For each u E L+ and sequence {q,>of A there exist cp E L- and a subsequence {(Pk,}of {cp,) satisfJJing limcpkn(u)= cp(u) for all u E [O,u]. Proof. (i) (ii) Let I be the ideal generated by the image of L in (L-);. For u E L + let I , be the ideal generated by u in I . Then L' = N u @ C,, where N u = {cp E L-: Icpl(u) = 0) and C,, = Nud.Now let {cp,} c A . For each n set (P, = f , + g, with f , E N u and gn E C, and note that cp,(z~) = g,(u) for u E [o,~]. Since the solid hull of A is order-equicontinuous on L, (9,) is also orderequicontinuous on L, and thus by Theorem 20.16 { g , ) is a(L-,I)-relatively compact. Now C, is lal(L-,Z)-closed and hence by Theorem 4.3(ii) a(L-,I)closed [since the topological dual of (L-,IcJI(L-,I))is exactly I ] . It now follows that { g , } is a(C,,I)-relatively compact and hence a(Cu,Zu)relatively compact. But C, equipped with the Riesz norm p,(cp) = lcpl(u)(cpE C,) has norm dual I , (Theorem 6.6), and thus by the Eberlein-Smulian theorem (Theorem 19.4), there exists a subsequence { g k , } of { g , } and cp E C, such that { g k , } is a(C,,I,)-convergent to cp. Therefore, we have

lirn ( P k , ( u )

= lirn g k , ( v ) = cp(v)

for all u E [O,u].

(ii) (i) If A is not order-equicontinuous, then there exists E > 0, cp E L', O I u, I u in L, and {q,} E A such that Icp,(~,+~ - u,)l > E and with lim cp,(u) = cp(u) for all v E [O,u]. Let A , be the ideal generated by u in L. For each n let t+b, denote the restriction of cp, to A,. Then {$"; c A; and ($,) is a(A;,A,)-convergent to cp. Thus by Theorem 20.20 {I),*} is orderequicontinuous on A,, which is in contradiction to /t+b,(u,+, - un)l = ~cpn(u,+ - u,)l > cfor all n. The proof of the theorem is now complete. W

Note. Theorem 20.21 and Corollary 20.22 are due to Schaefer [67, Theorem 10.5, p. 1331. Theorem 20.24 is due to Burkinshaw [16, Proposition 4.6, p. 1991. For additional results on order-equicontinuity see [16].

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We continue with a useful variation of Theorem 12.4. Recall that for cp E L' the Riesz seminorm pa on L is defined by p,+,(u) = Iql(lul) for all u E L. Lemma 20.25. Let (L,z) be a Hausdor- locally convex-solid Riesz space with the Lebesgue property. I f L has the countable sup property, then for every sequence {u,} of L there exists a sequence {q,,}of L' such that {pa,,} dejnes LI rnetrizahle Lehesgue topology on the hand generated hjf tun) in L. Proof. Let B be the band generated by a sequence {u,,} E L. Since lal(L,L') is a Lebesgue topology, by Theorem 12.13 the carrier CICIequals L. Thus as in the proof of Theorem 12.3(i) there exists a normal sequence { V,} of Ic((L,L')-neighborhoods of zero such that {u,} c Nd,where N ==:)-( V,. For each n choose (P, E L' such that the open unit ball of plpnis contained in V,. It now follows easily from B E Nd that {p,,} defines a metrizable Lebesgue topology on B. H We close this section by characterizing the o(L,L')-sequential completeness. Theorem 20.26. Let (L,z) be a Hausdorf locally convex-solid Riesz space with the Lebesgue property. If L has the countable sup property, then the following statements are equivalent:

(i) L is a(L,L')-sequentially complete. (ii) Every monotone a(L,L')-Cauchy sequence of L+ is o(L,L')-convergent. (iii) (L,.r) satisfies the a-Levi property, that is, 8 s u, t in L and { u n } zbounded implies that sup{u,,} exists in L. Proof. The implications (i) * (ii) (iii) are obvious. (iii) * (i) Let {u,,} be a o(L,L')-Cauchy sequence of L. Embed L in (L'): and note that L is order dense in (L'); (Theorem 9.1). Then (u,} is a a( (L')i,L')-Cauchy sequence and thus by Theorem 20.23 there exists f E (L'): such that {u,} is a( (L')l,L')-convergent to f .To complete the proof we show next that f belongs to L. To this end let A be the ideal generated by {u,} in L and let B be the band generated by A in (L');. By Theorem 9.1 B is a( (L'):,L')-closed [remember that (L'); is the topological dual of (L',lal(L',(L')z))] and thus .f E B. Pick a net {w,] c L with 0 I w, f in (L');. Now by Lemma 20.25 there exists a sequence {cp,} E L' such that {p,.} generates a metrizable topology on A . But {pvp,}generates (always) a Lebesgue topology on B [here pqn(g)= IgI(Icp,,I) for all g E (L');] that is also metrizable due to the order denseness of L in (L'):, and thus there exists a sequence {wan}c { w,} with 8 I wan t f in (L');. But then since clearly {w,,} is cr(L,L')-bounded, {w,,} is z-bounded and hence by hypothesis w , ~t w holds in L. Now by the order denseness of L in (L'); we get f = w E L. Similarly, f - E L; thereforef = f - f - E L, and the proof is finished. H +

+

+

+

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147

Note. Nakano apparently was the first to study the compact subsets of Lz in terms of order-equicontinuity (called in his terminology universalequicontinuity). Theorems 20.1 1 and 20.12 were proven by Nakano for super Dedekind complete Riesz spaces [53, $281; the compact subsets of L: have appeared also in the works of Kaplan and Fremlin. An extensive study of compact sets can be found in Chapter 8 of [25]. A Banach lattice version of Theorem 20.26 was proven by Meyer-Nieberg [49, Theorem 1.7, p. 3081; the result as stated here is due to Burkinshaw and Dodds [18, Corollary 4.21. Finally, the hypothesis of a-Dedekind completeness can be weakened in many of the results of this section; for such generalizations see the paper by Dodds [21]. 21.

COMPACT SOLID SETS

The weakly compact (respectively z-compact) solid subsets of a locally convex-solid Riesz space (L,z) will be characterized in terms of disjoint sequences. We start with some basic results on weakly compact intervals. Theorem 21.1. Let (L,z) be a HausdorfS locally convex-solid Riesz space. Then f o r u E L f the following statements are equivalent:

(i) [&u] is o(L,L‘)-compact. (ii) [ -u,u] is a solid subset of (L’);. (iii) [ - u,u] is a solid subset of L“. (iv) T h e principal ideal A , generated by u in L is Dedekind complete and z induces a Lebesgue topology on A , . (v) Every disjoint sequence of [d,u] is z-convergent to zero and [@u] is r-complete. (vi) Each disjoint sequence of [e,u] is r-convergent to zero and every increasing z-Cauchy net of [e,u] is z-convergent. Proof. (i) 3 (ii) It follows from Theorem 19.10 that the interval [O,u] c L is lo[((L’);,L‘)-dense in ( 0 , ~ = ) ( v E (L’)::d I u I u ) , since if v E (Q,u),and {urn}c L is lal( (L’);,L’)-convergent to u, then { Iv,I A u } c [O,u] is lo[((L’)z,L’)-convergent to u. Clearly [6,u] is a((L’),,L’)-dense in ( 0 , ~ ) . However, since [O,u] is a(L,L’)-compact [and hence o((L’); ,L’)-compact] it follows that [d,u] = (d,u), and thus [ - u,u] is a solid subset of (L’):. (ii) 3 (iii) It follows from the fact that L” is an ideal of (L’)-. (iii) (iv) By assumption the principal ideal A , generated by u in L coincides with the principal ideal generated by u in L“ and hence A , is Dedekind complete. Now if { u,> E A , satisfies u, 4 0 in L, then it also satisfies u,JO in L” and consequently lim q(u,) = 0 for each cp E L‘ and thus by

I 48

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-

Theorem 9.8 it follows that u, 1,8. Therefore, T induces a Lebesgue topology on A,. (iv) (v) Since A , is Dedekind complete and t induces a Lebesgue topology on A,, it follows from Theorem 13.1 that each order interval of A , is z-complete. Finally, since the topology induced by T on A, is a Lebesgue topology (and hence a pre-Lebesgue topology by Theorem 10.2), it follows easily that each disjoint sequence of [B,u] is z-convergent to zero. (v) =-(vi) Obvious. (vi) + (i) By Theorem 10.1 the topology induced by T on A , is a preLebesgue topology. Hence each increasing net of [8,u] is z-Cauchy and thus by hypothesis z-convergent. Therefore, it follows from Theorem 5.6 that A , is Dedekind complete and that z induces a Lebesgue topology on A , . Now since lal(L,L’) c z, it follows that lal(L,L‘) also induces a Lebesgue topology on A, and thus by Theorem 13.1 the order interval [B,u] is lal(L,L’)complete. But by Theorem 19.10 [8,u] is lol((L’):,L’)-dense in (0,u) and thus [ B , ~ ]= (e,u>. Finally, by applying Theorem 19.14 to L‘ with the topology lal(L‘,(L’)L), we obtain that each order interval of (L’); is a((L’);,L‘)-compact, and therefore [B,u] is a( (L’)z,L’)-compact.Hence [B,u] is o(L,L’)-compact and the proof is finished. H The previous characterizations of the weakly compact order intervals give an interesting insight into the convex solid and weakly compact sets. Theorem 21.2. Let (L,z) he a Hausdorff locally convex-solid Riesz space. For a convex solid subset A of L the following statements are equivalent: (i) A is a(L,L’)-compact. (ii) Every disjoint sequence of A is a(L,L‘)-convergentto zero and for each net {u,} c A with B I u, t in L there exists u E A such that u, & u.

Proof. (i) * (ii) Note that A considered as a subset of (L’); is a( (L‘): ,L’)compact. Hence by Theorems 20.11 and 20.6 each disjoint sequence of A is a( (L’),,L’)-convergent to zero and thus a(L,L’)-convergent to zero. u, t in Finally, A is a((L’)x,L’)bounded and hence if (u,) E A satisfies 0 I L, then by Theorem 19.17 u = sup{u,) exists in (I,’):. Also cp(u,) t cp(u) for each 8 I cp E L‘, and so u E A, since A is a( (L’):,L’)-closed. But then by u. Theorem 9.8 we get ti, (ii) (i) Observe that A is lol(L,L’)-bounded, for suppose sup{cp(lul): u E A } = co for some 8 I cp E L’. Choose B I u, E A such that cp(u,) > n2” for each n. But then since A is solid and convex, we have 2-’ui} c A and hence Cy= 2-’ui t u for some u E A . So, cp(u) 2 cp(2-”un)2 n for all n, a contradiction.

{I;=

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149

Consider A L (L’);; then by Theorem 20.6 [condition (iv)] A is orderequicontinuous on L‘ and thus by Theorem 20.1 1 A is a( (L’)L,L‘)-relatively compact. To complete the proof we show that A is a( (L‘),,L’)-closed. Since for every u E A , the interval [8,(u1]satisfies statement (vi) ofTheorem 21.1 it follows that A is a solid subset of (L’):. Also it follows from our hypothesis that A is an order closed subset of (L‘);, and hence since the topology lol((L’);,L’) is a Hausdorff Fatou topology, A is lal( (L’);,L’)closed by Theorem 12.7. Now by Theorem 6.6 the topological dual of (L’); with the topology la[((L’);,L’) is L’, and since A is convex, it follows by Theorem 4.3(ii) that A is a( (L’);,L’)-closed, and we are done. Lemma 21.3. Let (L,7) be a Hausdorf locally convex-solid Riesz space and B be the band generated by a sequence {q,)in L’. If A is a o(L,L’)-relatively compact subset OfL, thenfor every sequence {u,) G A there exists a subsequence o(L,B) {v,} of {u,} and u E L such that u, u.

-

Proof. Let A be a(L,L‘)-relatively compact and consider A E (L‘);. Then A is a( (L’);,L’)-relatively compact and so by Theorems 20.11 and 20.12 A is order-equicontinuous on L‘ (since L’ is Dedekind complete). Now if {u,} E A , then by Theorem 20.24 there exists a subsequence {u,,’; of {u,} such that lirncp(o,’) exists for all cp E [O,ly,,l].Next if the sequence {u,’} has been defined, use Theorem 20.24 again to obtain a subsequence {u: I } of {u,’} such that lim,cp(uhf’) exists for all cp E [6,1cpi+ll]. Thus by induction we can select the sequences {u,‘) ( i = 42, . . .) with the above properties. Put v, =,:v and note that { u,} is a subsequence of { u,} such that lim cp(v,) exists for all cp in the ideal I generated by {cp,) in L’. Now let u be a a(L,L’)-accumulation point of {u,,}; then lim cp(u,,) = cp(u) for all cp E I . Also since I is order dense in the band B generated by { q n }in $a $ in L’. Now since L‘, for given 8 5 $ E B, there exists ($,} G I with 8 I { u,} is order-equicontinuous on L’, it follows from Theorem 20.1 1 [applied to L’ and (L’);] that {$,j converges to $ uniformly on {v,,), and therefore lim $(u,) = +(u), and we are done.

Recall that a subset A of a topological vector space (E,7) is called 7relatively sequentially compact whenever each sequence of A has a subsequence that is z-convergent to some point of E. Theorem 21.4. Let (L,7) be a Hausdorff locally convex-solid Riesz space.

If L has the countable sup property, then each solid o(L,L’)-relatively compact subset of L is o(L,L’)-relatively sequentially compact.

Proof. Let S be a solid o(L,L’)-relativelycompact set and (u,,} E S. Since each order interval of S is o(L,L’)-compact, it follows from Theorem 21.1 that S is a solid subset of (L’);. Hence lol(L,L‘)induces a Lebesgue topology

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[Chap. 6

on the ideal A generated by (u,} in L. Also by Lemma 20.25 there exists a positive sequence {cp,} c L’ such that {p,,) defines a metrizable Lebesgue topology on A . Let B be the band generated by {cp,} in L‘. Then L‘ = B 0 Bd and by Lemma 21.3 there exists a subsequence {u,} of {u,,} such that limcp(u,) exists for each cp E B. Also (L‘): = B” @ (Bd)’ by Theorem 19.5, and we claim {on} E (Bd)O. Indeed, for n fixed write Iu,I = w1 w2 with w1 E Bo and w2 E (€Id)’. Since S is a solid subset of (L‘);, it follows that w1 E L, and since w1 E B”, we have p,p,(wI)= cp,(wl) = 0 for all m. Hence w1 = tl and so { u n } c (Bd)’. Now let u E L be a o(L,L’)-accumulation point of {u,,}. Then limcp(u,,) = cp(u) holds for all cp E B and also I+!I(u)= I+!I(u,) = 0 for all I+!I E Bd. Therefore, {u,} is a(L,L’)-convergent to u, and the proof is complete.

+

For a proof of the following result see [40, Theorem 24.7, p. 3131.

-

Lemma 21.5. Let (E,T) be a metrizable locally convex topological uector space and M a a(E,E)-relatively compact subset of E. I f u belongs to the a(E E ’ ) u. a(E,E‘)-closure of M , then there exists a sequence {u,} c M such that u, Theorem 21.6. Let (L,T)be a Hausdorff locally convex-solid Riesz space and suppose L’ has a countable complete disjoint system. Then for a subset A of L the following statements are equiualent: (i) A is a(L,L’)-relatively compact. (ii) A is o(L,L‘)-relatively sequentially compact. Proof. (i) + (ii) Let {cp,} be a complete disjoint sequence of L’ and note that the band generated by {cp,} in L‘ is all of L‘. Now if {u,,} c A, then by Lemma 21.3 there exists a subsequence {u,,} of {u,} that is a(L,L‘)-convergent to some u E L. (ii)+(i) Consider A c (L’):. Since each sequence of A contains a o(L,L’)-convergent subsequence, it follows from Theorem 20.24 that A is order-equicontinuous on L’. Thus A is a( (L’)x,L’)-relatively compact by Theorem 20.11. Let {q,) c L’ be a complete disjoint sequence of L‘ and let I be the ideal generated in L’ by {cp,). Note that since a( (L’);,I) is a coarser topology than a( (L’);,L’), A is a( (L’),I)-relatively compact. But (L’); is metrizable under the topology lal((L’);,I) by Theorem 15.6. Now by apit follows that the topological plying Theorem 6.6 to (L’); with la[((L‘)i,f) dual of (L’): with the topology lal((L’)~,I)is exactly 1. To complete the proof we show that the a( (L’),,I)-closure of A is contained in L. To this end let u E (L’): be in the a( (L’),,I)-closure of A . By Lemma 21.5 [applied to the metrizable topology lo[((L’);,I)] there exists a sequence tun} c A such that limcp(u,,)= cp(u) for all cp E I . But since A is o(L,L’)relatively sequentially compact, there exists u E L and a subsequence {u,}

Sec. 211

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151

of {u,} such that lim+(u,) = +(u) for all t,b E L'. Clearly cp(u) = q ( u ) for all cp E I and since I is order dense in L', it follows that cp(u) = cp(u) for all cp E L', i.e., u = u E L, and we are done. In order to investigate additional properties of solid compact sets we need to extend Theorem 20.6. Theorem 21.7. Let (L,z) be a HausdorfS locally convex-solid Riesz space with the monotone completeness property. Let A he a nonempty solid o(L,L')bounded subset of L and B a nonempty solid o(L',L)-bounded subset of L' such that sup{lq\(lu\):'pE B, u E A } < m. Then the following statements are equivalent :

(i) lim pA(cp,) = 0for each disjoint sequence {cp,} E B. (ii) lim pB(u,) = 0 for each disjoint sequence {u,} E A. (iii) For each E > 0 there exist 8 I in the ideal generated by B in L' and 6' I u in the ideal generated by A in L such that pa( ((401 - +)+) < E for all cp E B and ps( (Ju(- u ) + ) < E for all u E A.

+

Proof. (i) * (ii) If (ii) does not hold, then there exist E > 0, a positive sequence {cp,} E B, and a positive disjoint sequence {u,} G A such that cp,(u,) > E for all n. Denote by P, the projection of L' onto the carrier band CUn(recall that L' = Nun0 C, where Nun= {'p E L': Icpl(u,) = 0 } and CUn= N:,,). Put t,b, = P,('p,) for each n and observe that {+,} E B, since B is solid. But by Theorem 3.10 {$,} is a disjoint sequence and pA(t,bn)2 P,(cp,)(u,) = cp,(u,) > E for all n contradicts our hypothesis. (ii) 3 (iii) Put 6 = sup{ \cp\( \ u \ ) :cp E B, u E A } . By way of contradiction, assume that there exists E > 0 such that for every 0 I u in the ideal generated by A there exists u E A (depending on u) with pB( ((ul - u)') > 3 ~ . Then there exists a positive sequence (u,} s A satisfying

for all n. Since z is a Hausdorff locally convex topology and {u,,} c A is z-bounded, it follows that 2-iui t is z-Cauchy and hence u = 2-'ui = sup(Cy,, 2-'ui:n 2 l } exists in L. For each n set u, = (u,+ - 4"17= ui - 2-"u)+. Clearly { t i , , ) E A and we claim (u,} is a disjoint sequence. Indeed, for m > n 2 1 the inequalities m

4-mU,+1

-

C Ui - 2-3mU

i= 1

< (4-mUm+1

- U,+l)'

LOCALLY CONVEX-SOLID TOPOLOGIES

152

+

[Chap. 6

4"1;=,

imply U , A V , = 8. Now since u, 2-"u 2 ( u , + ~ui)+, we have ui)+)and SO pB(u,) 2 3~ - 2-"6 which pB(u,) p,(2-"u) 2 pB((u,+ - 4" implies p , ( ~ , ) > E for n sufficiently large, in contradiction to (ii). Thus we have proven that given E > 0 there exists 8 I z; in the ideal generated by A in L such that pB( (lul - u ) + ) < E for all u E A. It is now an easy consequence of Theorem 1.2 and the above proven statement that lim pB(u,) = 0 for each order bounded disjoint sequence of the ideal generated by A in L. But then by Theorem 20.6 there exists 8 I $ in the ideal generated by B in L' such that - $)+(v) < E for all cp E B and so

+

([PI

(l4? - $)+(lul) = (lcpl - $)+((lul + (I(PJ - $)+(lul A I pB((lul - l ' ) + ) f (lql - $ ) + ( u ) < 2&

4

for each u E A. Thus pa( (IqI - $)+)< 2~ for all (P E B. (iii) 3 (i) Let E > 0 and {(P,} c B be a positive disjoint sequence. By assumption there exist u E L+ and 0 I $ E L' such that pA( (lcpl - $)+) < E andp,((lul - v ) + ) < ~ f o r e v e r y c p ~ B a n d u ~ A . Now since {cp,, A $} is an order bounded disjoint sequence, lim cp, A $ ( v ) = 0, and so from the relations (Pn A

$(Iu[)

I (Pn A

$((Iul

-

we infer that pA(cpnA $) I p,(Cp,)

I pa( (40,

-

$1')

E

v)'

+ (P,

+ ( P n A $(u) I P B ( (lul - u ) + ) + A

(Pn A $(l'),

$(u). But since pa is a Riesz seminorm, A $(4, from which (4 fol-

+ pa(cpnA $1 I 2~ + cp,

lows, and the proof is finished.

Remark. It should be clear that if a solid set A G L satisfies any one of the equivalent conditions of Theorem 21.7 then the convex hull of A also satisfies the equivalent statements of Theorem 21.7. Note. Theorem 21.7 is due to Burkinshaw and Dodds [17, Proposition

2.21.

The next result gives a sufficient condition for a solid set to be weakly compact. Theorem 21.8. Let ( L J ) be a Hausdorf locally convex-solid Riesz space with the monotone completeness property. Let S c L be solid and z-bounded. If every disjoint sequence of S is z-convergent to zero, then S is o(L,L')-relatioely compact. Proof. Let S be the z-closure of S. Then every disjoint sequence of S is z-convergent to zero. To see this let {u,,} E S be a positive disjoint sequence and Vbe a solid z-neighborhood of zero. For each n choose 0 Iu, E S such

COMPACT SOLID SETS

Sec. 211

153

that u, - u, E V . Then u, A u, 5 0, since { u , A P,}is a disjoint sequence of S. Choose m so that u, A u, E V for all n > m and note that 0 I ~ i ,= (u, - u,)’ + u, A u, E I/ V for n > m; therefore u, 1,0. Thus without loss of generality we can assume that S is z-closed and by the remark following Theorem 21.7 we can also assume that S is convex. By c S satisfies 8 s u, 7 , then Theorem 21.2 it is enough to show that if (zI,} u, 1,u for some u E S. To this end let E > 0 and V a closed convex solid z-neighborhood of zero. By Theorem 21.7 there exists a positive element u in the ideal generated by S in L such that lcpl(lul - lul A u) < E for all u E S and all cp E V o . By Theorem 21.1 we know that S is a solid subset of (L’): and thus the interval [0,u] E L is a(L,L’)-compact. Hence ( U ~ A U is) TCauchy by Theorem 21.1 (again). Thus there exists an index a1 such that (cpl( Iv, A u - up A < r, for all q~ E V o and a$ 2 al, and so

+

01)

I’PI(Ivn

-

IVI(h -

A 0)

+ lVl(luaA L: - up A L’I) + lVl(up

-

up A 0)

38

for all a$ 2 a l and cp E Vo.Therefore, { u , ) is z-Cauchy and so LI, A u holds S is r-closed we get u E S, and the proof is finished. in L. But since Next we give a sufficient condition in terms of (L,T)for a solid subset of L‘ to be o(L’,L”)-compact.

Theorem 21.9. Let the HausdorfS locally convex-solid Riesz space (L,T) be s-complete and L’ be p(L’,L)-complete. Let S E L‘ be solid and p(L‘,L)bounded. I f limp,(u,) = 0, whenever (u,) is a z-bounded disjoint sequence of L, then S is a(L’,L”)-relatively compact. Proof. By Theorem 21.7 for every z-bounded subset B of L lim p,(cp,) = 0 holds whenever {cp,} is a disjoint sequence of S. But this merely says that each disjoint sequence of S is p(L‘,L)-convergent to zero and the result follows from Theorem 21.8 by observing that the topological dual of (L’,p(L’,L))is L“. H Recall that an abstract L-space is a Banach lattice L, whose norm satisfies + u) = p(u) p(u) for all u,u E L’. In the next theorem we shall show that for abstract L-spaces the converse of Theorem 21.8 holds and that for a weakly compact set its closed solid hull is also weakly compact.

p(u

+

Theorem21.10. Let L, be a n abstract L-space. Then for a subset A of L, the following statements are equivalent: (i) A is a(L,,L,‘)-relatiuely compact. (ii) A is norm bounded and each disjoint sequence of the solid hull of A is norm convergent to zero.

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[Chap. 6

LOCALLY CONVEX-SOLID TOPOLOGIES

Proof. First define cp(u) = p(u’) - p ( u - ) for U E L, and note that 8 I cp E L,‘. From this it follows easily that the norm topology coincides with IoI(L,,L,’). Thus lal(L,,Lp’) is a complete topology and hence L, = (L,’); by Theorem 19.11. The equivalence of (i) and (ii) now follows from Theorem 20.11 and Corollary 20.12 applied to the pair of Riesz spaces L,’ and (L,’):. Recall that a discrete element of a Riesz space L is an element u of L such that A, = { A u : E ~ R } . For Archimedean Riesz spaces an element u > 8 is discrete if and only if it follows from w Iu and w,u E [8,u] that u = 8 or w = 8; see Exercise 9 of Chapter 1. The band generated by the discrete elements of a Riesz space L will be denoted by 9.Note that an Archimedean Riesz space L is discrete if and only if 9 = L. Observe also that if u E L is a discrete element then Icp(u)I = Icpl(luI) holds by Theorem 3.3 for all cp E L‘. A subset A of a topological vector space (E,T)is called totally bounded if for each r-neighborhood I/ of zero there exists a finite subset { x l ,. . . ,xn} of A such that A G xi + I/. It should be clear that subsets of totally bounded sets are also totally bounded. The following result dealing with totally bounded subsets can be found in [63, p. 601 and will be needed.

u;=l

Theorem 21.11. Let (E,T)he a Hausdorff topological vector space and A a subset of E. Then A is z-compact fi and only if A is totally bounded and z-complete. Theorem 21.12. Let (L,T) be a Hausdorff locally conuex-solid Riesz space. Then for u E L f the following statements are equivalent: (i) [O,u] is a(L,L‘)-compact and u E 9. (ii) [O,u] is lal(L,~’)-compact. (iii) [O,u] is z-compact. (iv) Each a(L,L’)-convergent sequence of [O,u] is z-convergent and [8,u] is z-complete. (v) Each o(L,L‘)-Cauchy net qf [Q,u]is z-convergent.

-

Proof. (i) (v) If (v) is false, then there exists a net {u,} G [O,u] and a a(L L‘) closed convex solid z-neighborhhod I/ of zero such that u, u and with u, - u 4 I/ for all a. For each a choose cp, E V’such that Icp,(u, - u)( > 1. Let cp be a a(L’,L)-accumulation point of {cp,}. Pick an index a1 such that Icp(u, - u)I <$for a 2 a,. Then [(cp, - cp)(u, - v)l > 1 - L - L- 2 , and so Iu - u,I I u implies lcpa - cpl(u) 2 Icp, - cp1 (Iv, - u [ ) 2 i, for a 2 a,. Since by Theorem 1.11 the ideal generated by the discrete there exists a net { un} such that O I un t u, elements of L is order dense in 9, where each ui. is a linear combination of discrete elements. But by Theorem 21.1 we have that un 5 u and so there exists an index ;1such that Icp, - cpl(u- un)

Sec. 211

COMPACT SOLID SETS

155

<$foralla.ThusIcp, --cp1(11~,)2*-= $for all c1 2 a l . But u1 where each xi is a discrete element and thus

=

cixi,

n

for all c1 2 a,. However, this contradicts that cp is a o[L',L)-accumulation point of the net {cp,) and the result follows. (v) * (iv) Obvious. (iv) * (iii) Note that each disjoint sequence of [8,u] is z-convergent to zero and so by Theorem 21.1, [O,u] is a(L,L')-compact. We will show that [ 4 u ] is totally bounded for z and the result will follow from Theorem 21.11. If [0,u] is not totally bounded, there exist a closed convex solid z-neighborhood V of zero and a sequence {u,} c [O,u] such that u, - urn$ I/ for n # rn. Now if E > 0, then by Theorem 21.7 there exists 8 I $ E L' such that (Icpl$)+(u) < E for all cp E V o . Let B be the band generated by $ in L' and write L' = B 0 Bd. By Theorem 21.1 the ideal A , generated by u in L is Dedekind complete and z induces a Lebesgue topology on it. Therefore the ideal F = {ti E A,:IqI(IuI) = 0 for all cp E Bd} is a band of A, (for if {u,} E F and 8I u, t u in A , then u, A u and so u E F). Let P denote the projection of A, onto F, so [8,Pu] c [O,u]. By Lemma 21.3 there exists a subsequence (Pu,} of {Pun>such that cp(Pu,) converges for all cp E B. But cp(Pu,) = 0 for all cp E Bd, so that {Pun} is a(&!,')-convergent and by hypothesis z-convergent. Then IVI(Iun - urn[)=

(Iv(

I 2E

- $)+(ll~n -

Litri1)

+ 191* $ ( l [ f n

-

~ ~ m l )

+ $((u, - urn[)= 2E + $(IPU" - Purnl)I3 E

for all cp E V oand some rn,n sufficiently large. Therefore, I/,, - LI,, E V for some n # rn, a contradiction. Thus [8,u] is totally bounded and the result follows. (iii) => (ii) It follows immediately from lol(L,L')G z. (ii) (i) Suppose that 8 < u 4 9.Then by Theorem 1.12 there exists 8 < u I u with u E gd. Thus the interval [8,u] contains no discrete elements and is IaI(L,L')-compact; so by replacing u by u if necessary we can assume that [8,u] contains no discrete elements. Clearly [&u] is o(L,l')-compact, and so by Theorem 21.1 the ideal A , generated by u is Dedekind complete, cp E L' such that and each cp E L' is a normal integral on A,. Choose 8 I c p ( 4 0. Next we claim that given 0 I 6I q ( u ) there exists a component u of u [i.e., u A (u - u) = 81 with cp(u) = 6. To see this let C be the collection of all components w of u with cp(w) I 6. Obviously C # 12/ (0 E C) and it is partially ordered by IAlso . if K is a chain of C, then w = sup K is a component of u and since cp is a normal integral on A , , p ( w ) = sup{cp(u):u E K } 5 6, i.e.,

'

156

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[Chap. 6

w E C. It now follows from Zorn’s lemma that C has a maximal element, say If q ( u ) < 6, choose E > 0 such that q ( v ) + E < 6, and note that there exists a strictly positive component w of u - u (and hence of u) with q ( w )I E. Indeed, since u - v is not a discrete element, there exists a component 8 < u , Iu - v with q ( v l )5 +q(u - v), and so by repeating the process the existence of w can be established. But then the component v + w of u satisfies u < u + w and q ( v + w) = q ( u ) + q ( w ) I q(o)+ E < 6, contradicting the maximality property of u. Hence q ( u ) = 6. Now by using induction on k, for each k we obtain elements { z t : i = 1,. . . ,2k}satisfying: u.

(a) zik A zjk = d if i z j (b)

2k

1

Z:

=u

i= 1

(c) q ( z ; ) = 2-kq(u)for i = 1,. . . ,2k k i l k+l (d) zi k -- z2i-l z2i .

Cj”:

+

Set xk = ( - 1)jzjk ( k = 1,2, . . .) and note that q ( l x , - x,l) = b ( u ) for n # m, in contradiction with the lol(L,L‘)-compactnessof [O,u]. Thus u E 9 and the proof is finished. Corollary 21.13. Let (L,z)be a Hausdorff locally convex-solid Riesz space. Then the following statements are equivalent:

(i) L is discrete, Dedekind complete and z is Lebesgue. (ii) Each order interval of L is Iol(L,L‘)-compact. (iii) Each order interval of L is z-compact. (iv) Each order bounded o(L,L’)-conuergent sequence is 7-convergent and each order bounded increasing z-Cauchy net is z-convergent. (v) Each order bounded o(L,L’)-Cauchy net of L is z-convergent. Remark. If a locally convex-solid Riesz space (L,s) satisfies statement (i) of Corollary 21.13 then by Theorems 2.2 and 2.17, it follows that L is an order dense ideal of a Riesz space of the form RX for some X. Theorem 21.14. Let (L,z)be a HausdorfSlocally convex-solid Riesz space with the Lebesgue property. For a sequence {un>of 9 the following statements are equivalent:

(i) {u,} is o(L,L’)-conoergent to zero. (ii) {un>is IoI(L,L’)-convergent to zero. Proof. (i) 3 (ii) Let A be the set of all positive discrete elements of L, and let A be the set of all finite subsets of A . For each A E A put P , = Pa,

xael

Sec. 211

157

COMPACT SOLID SETS

where P, is the projection of L onto the ideal generated by a (see Theorem 2.16). Observe that t i = sup{P,(u):i, E A} for each 0 I u E 9. Let QA be the and let Q be the projection of L' projection of L' onto the band ( onto the band It follows from Theorem 20.20 [applied to the spaces L' and (L'):] that the sequence {u,; is order-equicontinuous on L'. Since Qncp Q q for each H I cp E L', it follows from Theorem 20.11 that for E > 0 there exists jLIsuch that (Qcp - Qj,lcp)(l~inl) < c for all n. But then since (cp(u)I = [q[( lul) for each discrete element u and each cp E L', we have ( Q a c p ) ( ( U n l ) = cp(P,lu,l) =

Icp(P,u,)l

for all n, positive discrete elements a, and 0 I q q ( l U n l ) = (Qq)(lunJ) = (QqE

+at,1 (Q,cp)(IU.l) 1

= E

I(Q,cp)(u,)(

L'. Thus

QAI(P)(lunl) + (Q~,q)(Iunl)

=E

+

c I(Q,cp)(u,)I

ad.,

for all 0 I cp E L'. Now since {u,] is a(L,L')-convergent to zero and A1 is a finite set, there exists m so that for a fixed H I cp E L' we have cp(Iu,I) < 2r: for n > m, that is, {u,f is Ial(L,L')-convergent to zero. (ii) * (i) Obvious, from the relation o(L,L') E (o((L,L'). The characterizations of the t-compact order intervals given in Theorem

21.12 will now be used to study the z-compact solid subsets of a locally

convex-solid Riesz space (L,t).A surprisingly simple relationship between a solid z-compact subset and its disjoint sequences is included in the next result.

Theorem 21.15. Let the Hausdorfl locally convex-solid Riesz space (L,t) be t-complete. For a t-bounded solid subset S of L thefollowing statements are equivalent:

(i) S is z-relatively compact. (ii) Each disjoint sequence of S is z-convergent to zero and S c 9. (iii) Each a(L,L')-convergent sequence of S is z-convergent and S is o(L,L')-relatively compact. Proof. (i) (ii) Let {u,} G S be a disjoint sequence, and let V be a solid t-neighborhood of zero. Since { u n } is a totally bounded sequence (since S is), it follows that there exists m such that {u,] G ui V . If n > m, then u,, - ui E V for some 1 I i I m. Since the sequence {u,} is (ui( = Iu, - uil E V for n > m, so that disjoint, it follows that Iu,I I u, E V for n > m, and therefore u, 0. Finally, since each order interval of S is z-compact it follows from Theorem 21.12 that S c 9.

ur=,+

Iu, ~ +

LOCALLY CONVEX-SOLID TOPOLOGIES

158

[Chap. 6

(ii) 3 (iii) Note first that by Theorem 21.8 S is o(L,l')-relatively compact. Now as in the proof of Theorem 21.8 we can show that every disjoint sequence in the closed convex hull of S is also z-convergent to zero. Thus we can assume without loss of generality that S is convex and z-closed. a(L L') Let (u,} c S such that u, --% u, V a closed convex solid z-neighborhood $ E L' such that (lcpl of zero and E > 0. By Theorem 21.7 there exists 8 I $)+(lul) < E for all cp E V o and all u E S . Since by Theorem 21.1 each interval of S is o(l,L')-compact, it follows (again from Theorem 21.1) that T induces a Lebesgue topology on the ideal A generated by S in L. But then (ol(A,L')is also Lebesgue and hence by Theorem 21.14, applied to (A,lol(A,L')),{ Iu, is lol(A,L')-convergent to zero. Thus there exists m such that $(lu, - u l ) < E for n 2 m. Then lql(lu,- u l ) I (Iq(- $)+(Iu, - u l ) $(lu, - u ( )I3~ for all cp E V o and n 2 m. Now from the bipolar theorem we get u, - u E V for n 2 m and hence u, 1,u. (iii) (i) We will show that S is totally bounded for T. If not, there exists a closed convex solid z-neighborhood V of zero and a sequence {u,} E S such that u, - u, 4 V I/ V for n # m. Then by Theorem 21.7 u in the ideal generated by S such that Iql((IuI - .v)+)< 1 there exists 0 I for all cp E V o and all u E S ; therefore, (lul - u)+ E V for all u E S. Now write v = Aiui with 8 I vi E S and ;li > 0 for each i. It follows from Theorems 21.12 and 21.1 that each interval [8,;livi] is z-compact and so since a finite sum of totally bounded sets is also totally bounded [O,v] = [8,1ivi] is totally bounded. Hence {u, A v } E [8,v] is totally bounded uiA u + V . But and so there exists an integer m such that {u,, A v } c then for any fixed k > m there exists 1 I iI m with uk A u - uiA u E V, and so

MI)

+

+ +

x;=

(uk-

uy=

5 ( l u k l - 0)'

+ Iuk A 0 - ui A V I + ( [ M i l - 0)

+

E

1/

+ I/ + v,

which contradicts the choice of {u,}, and the proof is finished.

By combining Theorems 21.15 and 21.4 the following result is now immediate. Corollary 21.16. Let (L,z)be a Hausdorff .r-complete locally convex-solid Riesz space. If L has the countable sup property, then for a solid subset of L the following statements are equivalent: (i) S is relatively z-compact. (ii) S is relatively sequentially z-compact. Note. Theorem 21.8 was proven for Banach lattices by Meyer-Neiberg [49, Theorem 2.6, p. 3131 and the general case by Burkinshaw and Dodds [17, Proposition 3.23. Theorem 21.9 can be found also in [17, Proposition 3.31. Corollary 21.13 is due to Kawai [39, Theorem 5.4, p. 3021. Theorem 21.14 can be found in [21, Theorem 4.7, p. 4001.

Sec. 221

SEMIREFLEXIVE RIESZ SPACES

22.

159

SEMIREFLEXIVE RlESZ SPACES

A HausdorfT locally convex topological vector space (E,T)is called semirejexive if E coincides with its bidual E". Clearly if a locally convex-solid Riesz space (L,z) is semireflexive, then L must be an ideal (in fact a band) of its bidual L". We begin by giving necessary and sufficient conditions for L to be an ideal (respectively a band) of L". Theorem 22.1. For a Hausdorff locally convex-solid Riesz space (L,z) the following statements are equivalent: (i) L is an ideal of L". (ii) L is Dedekind complete and z is a Lebesgue topology. (iii) Each order bounded disjoint sequence of L is z-convergent to zero and each order interval of L is z-complete. (iv) Each order interval of L is a(L,L')-compact.

-

Proof. (i) (ii) Since L" is Dedekind complete and L is an ideal of L", it follows that L must be also Dedekind complete. That z is a Lebesgue topology now follows from Theorem 9.1(iv). (ii) (iii) Since z is a Lebesgue topology, each order bounded disjoint sequence is z-convergent to zero (Theorem 10.1). By Theorem 13.1 each order interval of L is z-complete. (iii) (iv) It follows from Theorem 21.1. (iv) (i) Again by Theorem 21.1 each order interval of L is a solid subset of L". Thus L is a solid subset of L" and so L is an ideal of L".

-

A sequence {un}of a topological vector space (E,z) is said to have zbounded partial sums if the sequence u k } is z-bounded.

(IF=

Theorem 22.2. For a Hausdorff locally convex-solid Riesz space (L,z) the following statements are equivalent: (i) (ii) (iii) partial (iv) (v) (vi)

L is a band of L". z is a Levi, Lebesgue topology. L is z-complete and each disjoint sequence of L' with z-bounded sums is z-convergent to zero. Each increasing z-bounded net of L+ is z-convergent. L is (a((L,L')-complete. L = (I,');.

Proof. (i) * (ii) That z is a Lebesgue topology follows from Theorem 9.1. For the Levi property let {u,} c L be a 7-bounded net with 8 Iu, t in L. Then by Theorem 19.17 we have u, t u in L" and so u E L (since L is a band of L") and the Levi property follows.

160

LOCALLY CONVEX-SOLID TOPOLOGIES

[Chap. 6

(ii) + (iii) The z-completeness of (L,z) follows from Theorem 13.9. Now uk} are let {u,} c L+ be a disjoint sequence whose partial sums uk t u z-bounded. Since z is a Levi topology, there exists u E Lf with and so u, 6' since t is also a Lebesgue topology. (iii) * (iv) Let {u,} E L be z-bounded with 6' 5 u, in L, V a solid closed and convex z-neighborhood of zero and E > 0. By Theorem 21.7 there exists u E L + such that lqI( (ma - u ) + ) < E for all cp E V o and all a. Now since 6' I u, A u t I u and [6',u] is o(L,L')-compact by Theorem 21.1, it follows that {u, A u } is a z-Cauchy net and so there exists a, such that IcpI(IuaA v - up A < E for all cp E V o and all a$ 2 a l .

,

uI)

Consequently, for all cp E V o and all a$ 2 a, IVl(lua

- upl) I IqI(ua

- ua A

0)

+ IqI(Iua A

- up A 01) + IqI(up - up A u) I3 ~ .

By the Bipolar Theorem, (u,} is z-Cauchy and hence z-convergent. (iv) * (v) By Theorem 4.3(i) every lol(L,L')-bounded set is z-bounded, so if {u,} c L is lol(L,L')-bounded and 6' I u, 7 holds in L, then there exists u E L with u, 1,u. But then by Theorem 5.6(iii) u, t u and hence lol(L,L') is a Levi topology. Also since z is a Lebesgue topology Iol(L,L') is likewise so, and thus by Theorem 13.9 L is lol(L,L')-complete. (v) (vi) It follows from Theorem 19.11. (vi)=>(i) Since L" is an ideal of (L')- and (I,'): is a band of (I,')", the equality L = (L'): shows that L is a band of L", and the proof is finished. Note. Theorems 22.1 and 22.2 are due to Burkinshaw and Dodds [17, Propositions 5.1 and 5.21. The next theorem gives a number of conditions in order for the band generated by L in L" to be all of L". Theorem 22.3. For a Hausdor- locally convex-solid Riesz space (L,z), the following statements are equivalent:

(i) Ld = {6'} holds in L". (ii) L" is the band generated by L in L". (iii) L" c (L');. (iv) p(L',L) is a Lebesgue topology. (v) Each order bounded disjoint sequence of L' is P(L',L)-convergent to zero. Proof. (i) * (ii) If Ld = {Q},then the band generated by L in L" equals Ldd = L". (ii) (iii) Since L c (I,'): and L" is an ideal of (L')-, it follows that the band generated by L in L" is contained in the band generated by L in (L'); and hence L" c (L'):.

Sec. 221

SEMIREFLEXIVE RlESZ SPACES

161

(iii) * (iv) Since L” is the topological dual of (L’,p(L’,L)),the result follows from Theorem 9.1. (iv) +-(v) Since the Lebesgue property implies the pre-Lebesgue property, the result follows from Theorem 10.1. (v) * (i) By Theorem 19.12 p(L’,L)is a Fatou topology and by hypothesis p(L‘,L) is a pre-Lebesgue topology; therefore p(L‘,L) is a Lebesgue topology by Theorem 11.6. Hence by Theorem 9.1 L“ c (L’): and so by Theorem 19.10 L is Jol(L”,L’)-densein L“. Now let 0 I v E L” and v disjoint from L. Since L is lol(L”,L’)-densein L”, there exists a net {va} E L + such that { v a } is lo)(L”,l’)-convergentto v ; in particular {v, A u } is IoJ(L”,LI)-convergentto u. But v, A v = 8 for each a, and so v = 8, and the proof is finished. By combining the conditions given in Theorems 22.1, 22.2, and 22.3, we can now characterize the semireflexive Riesz spaces in terms of disjoint sequences. Theorem 22.4. For a Hausdorff locally convex-solid Riesz space (L,z) the following statements are equivalent;

(i) L is semirejexive. (ii) t is a Lebesgue, Levi topology and p(L’,L) is a Lebesgue topology. (iii) L is z-complete, each z-bounded disjoint sequence of L+ is o(L,L’)convergent to zero, and each disjoint sequence of L+ whose sequence of partial sums is z-bounded is z-convergent to zero. (iv) L is JoJ(L,L)-complete and each z-bounded disjoint sequence of L + is o(L,L’)-convergent to zero. Proof. (i) (ii) From Theorem 22.2 it follows that z is a Levi, Lebesgue topology. Also p(L’,L) is a Lebesgue topology by Theorem 9.1, since L“ G (L’);. (ii) (iii) That (L,z)is z-complete and each disjoint sequence of L + with z-bounded partial sums is z-convergent to zero follows from Theorem 22.2. Now let {u,,} E L + be a z-bounded disjoint sequence and let S be the solid hull of { u,}. Let 8 I cp E L‘ and denote by P, the projection of L‘ onto the band CUn= N:“. For each n put $,, = P n q and observe that by Theorem 3.10 {I),,}is a disjoint sequence of L’ that is also bounded by cp. Since p(L’,L) is a Lebesgue topology we have limp,($,,) = 0; so 0 I q(u,,) = $“(u,,) I pS(I),,) implies that {un>is o(L,L’)-convergentto zero. (iii) (iv) By Theorem 22.2 we have that L is Iol(t,L’) complete, and by hypothesis each z-bounded disjoint sequence of L+ is o(L,L’)-convergent to zero; hence the condition holds. (iv) 3 (i) Let S be a solid z-bounded subset of L. Since each disjoint sequence of S is o(L,L’)-convergentto zero, it follows from Theorem 21.7 that lim ps(cp,) = 0 for each order bounded disjoint sequence (q,,}of L’.

-

LOCALLY CONVEX-SOLID TOPOLOGIES

162

[Chap. 6

Thus each order bounded disjoint sequence of L' is b(L',L)-convergent to zero and so L" E (L'):by Theorem 22.3. But by Theorem 22.2 we have (L'); = L ; therefore L = L" and the proof is complete. Since for Banach spaces the concepts of reflexivity and semireflexivity coincide, Theorem 22.4 applied to Banach lattices yields the classical result of Ogasawara [54]; see also [45, Note XIII, Theorem 40.1, p. 5301. Theorem 22.5. For a Banach lattice L, the following statements are equivalent: (i) L, is reflexive. (ii) L, and L,' both have the Lebesgue property and 8 I u, sup{p(u,)} < 00 implies that sup{u,} exists in L,.

t

in L, with

Lemma 22.6. Let (L,z) be a Hausdorff locally convex-solid Riesz space. If

L' is lol(L',L)-separable, then each principal ideal A, of L is z-separable.

Proof. Let {q,}be IcI(L',L)-dense in L' and let 8 < u E L. For each n use Theorem 3.3 to choose u, with Iu,I 4 u and such that Iq,(u,)l 2 ilq,,l(u). Let M be the set consisting of all (finite) rational linear combinations of {u,}. We claim that the z-closure A of M contains A,. Indeed, if A , $ A, then there exists u E A , with u $ A and by the HahnBanach theorem there exists q E L' such that q ( u ) = 1 and q ( w ) = 0 for all w E M . But for every E > 0 there exists an integer k (depending upon E ) satisfying Iq - (pkI(u) < E and consequently I

21qk(uk)l

= 21((Pk - (P)(uk)l I 219, - ql(u)< 2E*

So, Iq((u)I I q k l ( u ) + Iq - (Pkl(U) < 26 + E = 3.5 for all E > 0 and hence Iql(u) = 0. Thus Iq((x)= 0 for all x E A,, and therefore, q ( x ) = 0 for all x E A,, contradicting q ( v ) = 1 and u E A , . Hence A , G iGI and the result follows by observing that M is a countable subset of A,. Theorem 22.7. Let ( L , t ) be a z-complete Hausdorff locally convex-solid Riesz space. If L" is P(L",L')-separable then L is semirejlexiue. Proof. By Theorem 10.9 P(L",L') is a pre-Lebesgue topology and hence a Lebesgue topology by Theorems 11.6 and 19.12. Now let {u,} G L+ be a disjoint sequence with z-bounded partial sums. By Theorem 19.17, uk t u holds in L" for some u E L" and hence {u,} is P(L",L')-convergent to zero. From this it follows that u, & 8, for if I/ is a z-neighborhood of zero then I/' is P(L',L)-bounded and so {u,} converges to zero uniformly on .'/I Therefore, by Theorem 22.2(iii) we conclude that L = (L');. Now since each principal ideal A , of L' is separable for fl(L',L) (Lemma 22.6), it follows from Theorem 10.9 that P(L',L) induces a pre-Lebesgue

EXERCISES

163

topology on A , that is in fact a Lebesgue topology on A , (Theorems 11.6 and 19.12 again). So each . f L" ~ is a normal integral on A,, from which it follows that L" G (L');.Therefore L = L" and the proof is complete. w Note. Theorems 22.4 and 22.7 are due to Burkinshaw and Dodds [17, Propcsitions 5.4 and 5.101. For additional characterizations of reflexive Banach lattices see [43].

EXERCISES

1. Let A be an ideal of a locally solid Riesz space (L,z)and let P be the projection of L' onto Show that

Pcp(u) = sup{cp(u):vE A and 8 I u I u} holds for all 0 I cp E L' and u E L+. 2. Let (L,z)be a Hausdorff locally convex-solid Riesz space. Assume that a convex solid subset S of L has the property that every positive increasing sequence of S has a supremum in L. Show that S is z-bounded. 3. Let (L,z)be a Hausdorff locally convex-solid Riesz space. Show that the topological dual of (L',la/(L',L))equals L if and only if L is Dedekind complete and z is a Lebesgue topology. 4. Let L be a a-Dedekind complete Riesz space with L, separating the points of L. Show that the Mackey topology t ( L , L z )is a locally solid topology on L. 5. Let (L,r) be a Hausdorff locally convex-solid Riesz space with an order unit. Show that a convex, solid, and a(L,L')-compact subset of L must be order bounded. 6. Let L,, be an abstract L-space. If a subset A of L is o(L,,L,,')-relatively compact, show that the solid hull of A is also a(L,,L,')-relatively compact and that A is contained in a principal band of L. 7. Let (L,z)be a Hausdorff locally convex-solid Riesz space and let I be the ideal generated by L in L". If S is a convex solid subset of L', show that S is a(L',l)-compact if and only if S is order-equicontinuous on L and the supremum of every increasing positive net of S exists in L' and belongs to s. 8. Let (L,z)be a r-complete Hausdorff locally convex-solid Riesz space and let L' be p(L',L)-separable. Show that (L,z) is semireflexive if and only if (L,z)has the a-Levi property. 9. Let L be an Archimedean Riesz space with an order unit c'. Define the Riesz norm p(u) = inf(i. > 0:IuI 1.e;. If L is an ideal of (L-):, show that L, is reflexive.

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[Chap. 6

10. Let (L,T)be a Hausdorff locally convex-solid Riesz space. An element

u E L+ is called a quasi-interior point if the ideal A, generated by u in L is

z-dense in L. Show that if u E L+ is a quasi-interior point and [8,u] is o(L,L’)-compact, then each order interval of i is o(f+i’)-compact. 11. Let (L,t) be a Hausdorff locally convex-solid Riesz space and for u E L+ put (8,u) = { B E L:8 I B I K}. Show that (0,u) is weakly compact if and only if for each disjoint sequence {u,} G (8,u) n L it follows that u,

: 8.

Let (L,T)be a Hausdorff locally convex-solid Riesz space such that o(L’,L)is a locally solid topology (that is?a(L‘,L) = lol(L’,L)).Show that L is Dedekind complete, 7 is a Lebesgue topology and that both L and L‘ are discrete spaces. ( H i n t : Use Exercise 3 and the fact that each order interval of L’ is loI(L’,L)-compact.) 13. Show that every band of a semireflexive space with the induced topology is semireflexive in its own right. 14. Let (L,T)be a semireflexive Riesz space and A a band of L. Show that ( L / A , z / A ) is semireflexive. 15. Let L be a Riesz space with Lz separating the points of L. Show that (L,lal(L,L,)) is semireflexive if and only if L is a perfect Riesz space and fi(L;,L) is a Lebesgue topology. 12.

OPEN PROBLEMS

1. Are the conditions of Theorem 21.7 equivalent without assuming any kind of topological completeness? 2. Consider an Archimedean Riesz space L and S c L“. It is easy to see (from Theorem 20.1 1) that if S is o(L“,(L’);l)-relatively compact then each increasing lol(L,L”)-bounded sequence of L+ is ps-Cauchy. Is the converse of this statement true? In other words, if every increasing lol(l,L‘)-bounded sequence of L+ is ps-Cauchy, is then S a o(L-,(L-)L)relatively compact subset of L”?

CHAPTER

7

Laterally Complete Riesz Spaces

The final chapter of this book deals with laterally and a-laterally complete Riesz spaces, that is, with Riesz spaces having the property that the supremum of every disjoint subset or disjoint sequence of positive elements exists. The first part discusses the remarkable lattice properties of these spaces. For instance, it will be shown that Archimedean a-laterally complete Riesz spaces satisfy the principal projection property, that Archimedean laterally complete Riesz spaces satisfy the projection property, and that in Archimedean Riesz spaces lateral and Dedekind completeness are independent properties. In the second part we shall deal with locally solid topologies on laterally or a-laterally complete Riesz spaces. It will be seen that on a a-laterally complete Riesz space every Hausdorff locally solid topology must satisfy the preLebesgue and a-Lebesgue properties ; that a a-laterally complete Riesz space can carry at most one Hausdorff Fatou topology (necessarily Lebesgue) and that if it can carry this topology then every other Hausdorff locally solid topology is finer and agrees with this Fatou topology on an order dense a-ideal. 23. THE LATTICE STRUCTURE OF LATERALLY COMPLETE RlESZ SPACES

We start this section with the definition of laterally and a-laterally complete Riesz spaces. 165

166

LATERALLY COMPLETE SPACES

[Chap. 7

Definition 23.1. A Riesz space L is said to be

(i) a-laterally complete, if the supremum of every disjoint sequence of L+ exists; (ii) laterally complete, fi the .supremum of every disjoint subset of L+ exists. Note. In the terminology of Luxemburg and Zaanen a laterally complete Riesz space is referred to as a universally complete Riesz space [46, Definition 47.3, p. 3231; here a universally complete Riesz space will be defined to be a laterally and Dedekind complete Riesz space (see Definition 23.17).The term “laterally complete” is taken from the lattice group theory; see [12] and [19]. Observe that the lexicographic plane provides an example of a laterally complete Riesz space that is not Archimedean. Theorem 23.2. For a laterally complete Riesz space L the following statements hold;

(i) Every band of L is a laterally complete Riesz space in its own right. (ii) I f L is Archimedean, then L contains weak order units. (iii) I f L is Archimedean, then every band of L is a principal band. Proof. (i) Let B be a band of L and let S be a disjoint subset of positive elements of B. Then u = s u p s exists in L and u belongs to B, since B is a band of L. But then u = sup S holds in B, and this shows that B is laterally complete. (ii) Let {e,} be a complete disjoint system of positive elements of L+. Since L is laterally complete, e = sup(e& exists in L and since L is Archimedean, e must be a weak order unit of L. (iii) Let B be a band of L. By (i) B is laterally complete and so by (ii) B contains a weak order unit e. But then, since L is Archimedean, B = Be (the band generated by e in L ) holds, and the proof is finished. H The following result can be proven as in (i) above. Theorem 23.3. Every a-ideal of a a-laterally complete Riesz space is a a-laterally complete Riesz space in its own right. One of the most remarkable properties of the Archimedean laterally complete Riesz spaces is included in the next theorem. Theorem 23.4. For an Archimedean Riesz space L the following statements hold:

(i) I f L is a-laterally complete, then L satisjes the principal projection property.

Sec. 231

THE LATTICE STRUCTURE

167

(ii) If L is laterally complete, then L satisfies the projection property. Proof. (i) Let u p E L+. Set w,

=((n

+ 2)u - v)+ nu -

( n = O,1,2,. . .)

0)-

and observe that w, A w , + ~= I9 holds for all n and all p 2 2. Observe also that since I9 I w, I ( n + 2)u holds for all n, we have {w,}c B,. Now the relations [(n

+ l)((n + 2)u + (n + 2 ) ~ , + ~ + I)(( n + 2)u - u)+ + ( n + 2)w,+ v U)+]AU

2 [(n

A

+ I)(( n + 2)u - u)+ + ( n + 2){(( n + 3)u - u)+ A ( ( n + 1)u - u ) - } ] A u 2 [(n + 2)(( n + 3)u r u ) + ] A [ ( n + 1)(( n + 2)u - u) + ( n + 2)(u - ( n + I)u)] A u

= [(n =

[(n + 2)(( n + 3)u - u ) + ]

A

u

imply that n

C ( r + I)wr 2 [(n+ l)((n + 2)u

r=O

-

u)']

But since L is a-laterally complete and w, suprema

A

A

u 2 (nu - u)'

A

u.

w, = 6' if Irn - n( 2 2, the

+ 2 ) ~ * , + ~2: n0) both exist in L and obviously f and g belong to B,, and thus u (f + g) E B,. f = sup((2n + l)w,,:n 2 0)

and

g = sup((2n

A

We shall show next that u - u A (f+ 9 ) E Bud.To this end observe that for every nonnegative integer k we have /

k

< UA(U -u ~ ( ku u)+) I U A ( U- u r \ ( k u - u ) ) = UA(2U-2UAku)=(U+2u)A(U+ku)A20-2UAku

= 2UA(k -k 1)U - 2UAkU

and thus n [ u ~ ( i i - u ~ ( ( f + g ) )I] ~ ~ ~ ~ [ 2 l )uu -~2 u(n kku ]+ s 2u holds for n = 1,2, . . . . Now, since L is Archimedean, u A ( u - u A (f+ 9 ) ) = 8 and so u - u A (f+ g) E B,d. Thus L = B, 0 Budand this shows that L has the principal projection property. (ii) The claim follows immediately from Theorem 23.2(iii) and part (i) above. The next four results are applications of the preceding theorem. The first theorem characterizes the Riesz spaces that are Dedekind and laterally complete.

168

[Chap. 7

LATERALLY COMPLETE SPACES

Theorem 23.5 are equivalent:

For an Archimedean Riesz space L thefollowing statements

(i) L is Dedekind complete and laterally complete. (ii) L is a-Dedekind complete and laterally complete. (iii) L is uniformly complete and laterally complete.

-

Proof. (i) = (ii) and (ii) = (iii) are obvious. To see that (iii) (i) note first that by Theorem 23.4(ii) L has the projection property, and then that the uniform completeness and the projection property are equivalent to the Dedekind completeness, by Theorem 2.1 1. W Similarly, it can be shown that a Riesz space is a-Dedekind and a-laterally complete if and only if it is uniformly and a-laterally complete. We continue with a useful lemma. Lemma 23.6. Let L be an Archimedean a-laterally complete Riesz space and let u, 1 8 in L. Thenfor every E > 0 there exists u E L' (depending upon E ) such that 8 I u, I 2-"u + eul holds for all n. Proof. Let u, -1 8, E > 0, and let P, denote the projection on the band generated by (u, - E U ~ ) ' in L (the projection P, exists by Theorem 23.4). Put u, = P,(ul), w, = u1 - u, for all n, and observe that w, A (u, - E U ~ ) += 8. Since (u, - wl)+ = P,(u, - E U ~ = ) P,(u,) - EU,, we get 8 I E U , 5 u , ~and hence v, 8. Note next that {2"+1(v,- u , + ~ ) }is a disjoint sequence of L', and hence 8 I 2"' ' ( v , - vn+1) I u must hold in L for all n and some u. Also, 8 I (u, A w, - &u1)+I (u, - m1)+A w, = 8 implies u, A w, I &ul for all n, and thus 8 <

U, = U , A U 1 = U, A ( U ,

+ W,)

u,

EM1

+

= (0, - v k ) f

+

holds in L for all k > n. Since u, for n = 1,2,. . . . H

U, A U,

+ U, A W ,

uk

1 8, we obtain easily that 8 I

u, I 2 - k

+

E U ~ ,

Theorem 23.1. Let T :L -+ M he a positive linear mapping between two Archimedean Riesz spaces. If L is a-laterally complete, then T is a-order continuous. Proof. Let u, 1 8 and let E > 0. According to the previous lemma, there exists u E L + such that 8 I u, I 2-"u + m1holds in L for all n. Now if 8 I w I T(u,) holds in M for all n, then it follows from the positivity of T that 8 5 w I T(u,) I 2 - " T ( u ) + ET(ul) for all n. Now because

Sec. 231

169

THE LATTICE STRUCTURE

M is Archimedean, 0 I w I &T(ul) holds for all E > 0 and thus by using the Archimedeanness of M once more we get w = 8. Hence T(u,) 1 8 holds in M , so that T is a-order continuous. Note that a a-order continuous positive operator between two Archimedean Riesz spaces need not be order continuous even if its domain is laterally complete. To see this let L = RX for some uncountable set X , and let A be the a-ideal of L : A = {f

E

RX:{x E X :f(x) # 0} is at most countable).

So, LIA is Archimedean. However, since A is not a band of L, the canonical projection of L onto LIA is not order continuous. Theorem 23.8. For an Archimedean Riesz space L thefollowing statements hold: (i) l f L is a-laterally complete and A is a o-ideal of L, then L / A is an Archimedean a-laterally complete Riesz space. (ii) I f L is laterallj~complete and A is a band of L, then LIA is an Arckimedean laterallj~complete Riesz space.

Proof. (i) Let {ti,) be a disjoint sequence of positive elements of L / A . We can assume {u,,) E L+. Set u1 = u l , and inductively if u l , . . . p, have been defined, put w, = u l + ' . . + u,, and define

u,+

1

= Un+

1

- suP{41+1 A mw,:m

2 1>= u,+

1

- Pw,(un+1).

Observe that {u,) is a disjoint sequence of L+. Indeed, for m < n we have and hence urn I w,- 1, so P,_(u,) IPwn~ ~ ~ , A ~ , ~ = ~ ~ A ( ~ , - P ~ , , _P,,,,(~~,))EB,,,, , ( ~ , ) ) ~nBt,,,= ~ , A ( {O], ~ , -

that is, urnA u, = 8. Now put t i = sup(u,} in L, and note that 6, = ti, for all n and that the canonical projection of L onto L/A is a a-Riesz homomorphism (since A is a a-ideal). Hence D = sup (ti,,) holds in LIA, so that LIA is a-laterally complete. The Archimedeanness of L / A follows from the fact that A is a a-ideal. (ii) By Theorem 23.4(ii) A is a projection band of L and hence L / A is Riesz isomorphic to Ad. The result should now be immediate. Note. Theorem 23.4 was proven first by Veksler and Geiler [69, Theorem 8, p. 321 via the representation theorems of the Archimedean Riesz spaces; the elegant proof presented here is due to Bernau [13, Theorem 1, p. 3201. Theorem 23.7 is a generalization of a result of Fremlin [26, Corollary 1.13, p. 781. Theorem 23.8(i) as well as the example preceding this theorem can be found in [26, pp. 78-79].

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[Chap. 7

An important tool for the study of laterally complete Riesz spaces is the notion of the dominable set, which is introduced next.

Definition 23.9. Let L be a Riesz space, and let A be a nonempty subset of L'. Then A is said to be a dominable set i f for every 8 < u E L there exists 8 < v E L and a positive integer k satisfying (ku - a)' 2 u for all a E A. Note. The notion of the dominable set is due to Fremlin [26, Definition l.l(c), p. 721.

We shall see later (Theorem 23.22) that the dominable sets of an Archimedean Riesz space are precisely the subsets of L' that are order bounded in its universal completion. We continue with a simple but very useful lemma. Lemma 23.10. Let L be an order dense Riesz subspace of a Riesz space M , and let A be a nonempty subset of L'. Then A is a dominable subset of L ifand only i f A is a dominable subset of M . Proof. Assume first that A is dominable in M . Let 8 < u E L. Then there exist 0 < u* E M and a positive integer k such that (ku - a)' 2 u* for all a E A . Now pick 6 < v E L with 8 < u I u*, and note that (ku - a)' 2 u for all a E A , that is, A is a dominable subset of L. On the other hand if A is a dominable subset of L and 8 < u* E M , pick u E L with 8 < u I u*. Now choose 0 < v E L and a positive integer k satisfying (ku* - a)' 2 (ku - a)' 2 v for all a E' A. Hence A is a dominable subset of M and we are done. H

It should also be clear that subsets of dominable sets are also dominable. More important properties of the dominable sets are included in the next results. Theorem23.11. Let L be a Riesz space with the principal projection property, and let A be a dominable subset of L'. Then for each 8 < LI there exists v, 0 < v I u, and a positive integer n satisfying P,(a) nv for all a E A . Proof. Let 8 c u. Since A is a dominable subset, there exists a positive integer n and 8 < w satisfying

(*I

a A nu I nu - w

for all a E A.

Observe now that 8 < w I nu implies w A u > 8 and so t' = Pw(u) satisfies 8 < v. It now follows from (*) that f,(a) A no I nu - w for all a E A . Now for every a E A the element f = (P,(a) - nu)' satisfies f E B, and f E Bwd [since f I(P,(a) - nu)- 2 w > 81, and so f = 8. Hence P,(a) I nu

Sec. 231

171

THE LATTICE STRUCTURE

for all a E A. Now since u E B, implies P,(a) E B, for all a E A , we get P,(a) = P,(P,(a)) = P,(P,(a)) 5 nP,(u) = nu and we are done.

for all a E A,

I

Theorem23.12. Let L be a Riesz space with the principal projection property and let A be a dominable subset of L +. Then there exists a complete disjoint system {e,:i E Z} of strictly positive elements of L such that for each i E Z there exists a positive integer ni (depending upon i ) satisfying P,,(a) 5 niei for all a E A . Proof. Consider the collection of sets of the form {ej:i E Z} where ei > 8, ei A ej = 8 if i # j , and such that for each i E Z there exists a positive integer n, (depending upon i ) satisfying Pei(a)5 niei for all a E A. By Theorem 23.1 1 this collection is nonempty and by Zorn's lemma there exists a maximal disjoint set with respect to the above property, say {ei:i E I } . We claim now that {e,:i E I } is a complete disjoint system of strictly positive elements, i.e., {e,:i E = L. Indeed, if 8 < u E L satisfies u A e, = 8 for all i E I, then by Theorem 23.1 1 there exists 8 < u I u and a positive integer n satisfying P,(a) Inu for all a E A . But then by adding v to the collection {e,:i E I } we contradict the maximality property of the collection. This completes the proof.

We are now in the position to characterize the dominable sets of the Archimedean laterally complete Riesz spaces. Theorem 23.13. Let L be an Archimedean laterally complete Riesz space, and let A be a nonempty subset of L'. Then the following statements are equivalent: (i) A is order bounded from above in L. (ii) A is a dominable set of L. Proof. (i) * (ii) Assume that 8 I a I u holds for all a E A. Let w > 8. Since L is Archimedean, there exists a positive integer k such that v = (kw - u)' > 8. But then (kw - a)+ 2 (kw - u)+ = u holds for all a E A , and hence A is dominable. (Note that the lateral completeness is not needed here.) (ii) (i) According to Theorems 23.4 and 23.12, there exists a complete disjoint system {e,:iE I) of strictly positive elements with the property that for each i E I there exists a positive integer n, satisfying P,,(a) I niei for all a E A. Now since L is laterally complete, u = sup{n,e,:i E Z} exists and so by

[Chap. 7

LATERALLY COMPLETE SPACES

172

Theorem 2.15 we have a = sup{P,,(a):iE I } I sup{n,e,:iE I } = u

for all a E A.

Thus A is order bounded, and the proof is finished. Among the most remarkable properties of the laterally complete Riesz spaces are the ones included in the next theorem. Theorem 23.14. Let L be an order dense Riesz subspuce of an Archimedean Riesz space M . Then we have the following:

(i) If L is laterally complete, then L is full in M . (ii) If L is laterally and Dedekind complete, then L = M . Proof. (i) Let u E M . Since L is order dense in M and M is Archimedean, there exists a net {u,} c L+ with u, t Iu( in M (Theorem 1.14). But then {u,} is a dominable subset of M, and hence by Lemma 23.10 a dominable subset of L. Thus according to Theorem 23.13 there exists w E L satisfying u, I w for all CI. It now follows that IuI I w holds, so that L is full in M. (ii) Use part (i) and Theorem 2.2. Lemma 23.15. Let L be an order dense Riesz subspace of an Archimedeun Riesz space M . I f L is Dedekind complete, then every element of M + is the supremum of a disjoint system of elements of L + . Proof. Note that L is an ideal of M (Theorem 2.2). Now let 6 < u E M . Since M is Archimedean and L is order dense in M , there exists a net {u,} of L with 0 I u, t u in M. It follows that {u,} is dominable in L + and so by Theorem 23.12 there exists a complete disjoint system {ei:iE I } of L + such that for each i E I we have P,,(u,) I niei for all CI and some fixed positive integer n,. Now for each i E I define wi = supa{Pei(ua)}I niei E L + ,and note that {wi:i E I } is a disjoint set of L'. To complete the proof we shall show that u = sup{wi: i E I } holds in M. Observe first that Pei(u,) I u, I u for all s( and i implies wi I u for all i E I. Now assume that wi I u E M holds for all i E I . Then Pei(ua)I u holds u for all i and CI.But then by Theorem 2.15 we get u, = sup(P,,(u,):i E I} I for all CI,and hence u Iu. The proof is now complete. 0 We continue with an important extension theorem. Theorem 23.16. Let cp: L -, K he a normal Riesz homomorphism from a Dedekind complete Riesz space L into an Archimedean laterally complete Riesz space K . I f L is an order dense Riesz subspace of an Archimedean Riesz space M , then cp has a unique normal Riesz homomorphic extension from M into K .

Sec. 231

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THE LATTICE STRUCTURE

In particular, it follows that ifcp is a normal Riesz isomorphism (into), then so is its extension. Proof. Let 0 < u E M . By Lemma 23.15 there exists a disjoint system {u,} of L+ with u = sup{u,} in M . Then {cp(u,)} is a disjoint system of K + (since cp is a Riesz homomorphism), and hence since K is laterally complete, u* = sup{cp(u,)) exists in K . Now, if (u,} G L+ satisfies sup{u,} = u in M , then for each fixed A we have t', = sup, {v, A u,) in L and hence from the order continuity of cp we get cp(v,) = sup,{cp(zi, A u,)} I u*. On the other hand if cp(u,) 5 v* holds in K for all 1,then from u, = sup,{u, A v,} and the order continuity of cp it follows that cp(u,) I u* holds for all a and hence u* = sup{cp(v,)} holds in K . Thus the element u* is independent of the system chosen from L+. So, we extend cp to M + by q ( u ) = u*. Observe that this extension of cp from M + into K + satisfies cp(u v) = cp(u) cp(v) for all u,u E M + . Thus by Lemma 3.1 cp can be extended linearly

+

+

to all of M . The rest of the proof is left as an exercise for the reader.

We continue with the definitions of universally and a-universally complete Riesz spaces. Definition 23.17. A Riesz space that is both laterally and Dedekind complete is called a universally complete Riesz space. Similarly, a a-Dedekind and a-laterally complete Riesz space will be called a a-universally complete Riesz space.

Note. Again we remind the reader that in the terminology of Luxemburg and Zaanen a laterally complete Riesz space is referred to as a universally complete Riesz space [46, Definition 47.3, p. 3231. Fremlin, inspired by Theorem 23.14(ii), calls a universally complete Riesz space an inextensible Riesz space [26, Definition l.l(a), p. 721, while Vulikh calls the universally complete Riesz spaces extended Riesz spaces [70, pp. 140-1441, As an application of Theorem 23.16 we have the following result. Theorem 23.18. Assume that two universally complete Riesz spaces L and M contain two order dense Riesz subspaces L , and M respectively, that are Riesz isomorphic, via n say. Then n extends uniquely to a Riesz isomorphism between L and M .

,,

Proof. Note first that since M , is order dense in M , n:L1 -, M is a normal Riesz isomorphism, and then that the ideal generated by L , in L, say A, is the Dedekind completion of L,. Next extend 71 to A + by n(f) = sup{n(u):u E L + ; 8 5 u ~ f if }f~ A' and then to all of A by n ( f )= n(f +) - n( f - ) for f E A ; note that this extension is a normal Riesz isomorphism (into). Now by Theorem 23.16, 7~ can be extended to a normal Riesz

174

LATERALLY COMPLETE SPACES

[Chap. 7

isomorphism from L into M. To complete the proof it suffices to show that the above extension of n is onto. To this end let 0 < u* E M. Then there exists a net {u,} of M I + with 8 I u, t u* in M. Pick a net { u Z ) of L: with u, = n(u,) for each a. Now observe that (u,} is a dominable subset of M and hence of M I . Thus {u,} is a dominable subset of L , and consequently of L. Theorem 23.13 implies now that t ) I u, f u* holds in L for some u* E L, and therefore u, = n(u,) t n(u*)= u*, that is, 71 is onto. Next we continue with the definitions of the universal and a-universal completions. Definition 23.19. A universal (resp. a-universal) completion of a Riesz space L is a universally (resp. o-universally) complete Riesz space K having an order dense (resp. super order dense) Riesz subspace M that is Riesz isomorphic to L. (We shall think of L and M as identical.) It should be clear that only Archimedean Riesz spaces can have universal completions and by Theorem 23.18 any two universal completions of a Riesz space must be Riesz isomorphic. Next we discuss the relation between universally complete Riesz spaces and extremally disconnected topological spaces. It is well known that for every topological space X the set Cm(X)consisting of all extended realvalued continuous functions f on X for which the open set {x E X : If(x)l < cc}is dense in X is in general not a Riesz space under the ordinary operations [46, p. 2951; [70, p. 1171. If, however, the topological space X is extremally disconnected (that is, if the closure of every open set of X is also open), then C m ( X )forms a Riesz space under the ordinary operations, which is universally complete [46, Theorem 47.4, p. 3231; see also [50, Chapter 11, p. 291 and [70, Theorem V.2.2, p. 1261. Thus Cm(X)forms a universally complete Riesz space for every extremally disconnected topological space X . (It should be noted that the property that makes Cm(X)a Riesz space is the is a continuous extended real-valued function following: I f f : 0 + [ - CO,OO] defined on an open subset 0 of an extremally disconnected topological space X , then j has a unique continuous extension to the closure 6 of 0 ;see [46, Theorem 47.1, p. 3221.) The Hausdorff compact extremally disconnected topological spaces appear naturally in connection with Dedekind complete Boolean algebras. We shall recall briefly this relationship and we shall refer the reader for details to [46, Chapter 11; see also Exercise 1 at the end of this chapter. An ideal of a Boolean algebra 93 is a nonempty subset I of 93 such that u v u E I , if u,u E I and if it follows from w I u with u E I, that w E I. An ideal I is called a prime ideal if u A u E I implies u E I or u E 1. Let R denote the

Sec. 231

THE LATTICE STRUCTURE

175

collection of all proper prime ideals of LB. Then the collection of subsets of R of the form R, = {aE Q : u $ w ) ( u E 98)forms a basis for a topology on R, called the hull-kernel topology. Now if in addition 3 is a Dedekind complete Boolean algebra, that is, if every nonempty subset of LB has a supremum, then R with its hull-kernel topology is a compact, Hausdorff, and extremally disconnected topological space, called the Stone space of the Dedekind complete Boolean algebra 249. The following result holds: For a given HausdorfS, compact, and extremally disconnected topological space X , there exists a unique (up to a Boolean isomorphism) Dedekind complete Boolean algebra LB whose Stone space is precisely X . Now if L is an Archimedean Riesz space then the collection of all bands (ordered by inclusion) forms a Dedekind complete Boolean algebra 91L. The following basic theorem is due to Maeda and Ogasawara [47] and deals with the existence of universal completions. A proof can be found in [46, Theorem 50.8, p. 3401. Theorem 23.20. Let L be an Archimedean Riesz space and let R be the Stone space of .BL.Then K = Ca(R) is a universal completion of L. Since by Theorem 23.18 any two universal completions of a Riesz space are Riesz isomorphic, it makes sense to talk about “the universal completion.” The universal completion of an Archimedean Riesz space L will be denoted by L”. Note also that if L is an order dense Riesz subspace of an Archimedean Riesz space M , then the universal completion M” of M serves equally well as the universal completion of L, since obviously L “seats” order densely in M”. Thus the universal completion L“ is Riesz isomorphic to MUby Theorem 23.18. So, if L is order dense in M , then the universal completion L”“contains” M as an order dense Riesz subspace. More precisely, we have the following theorem. Theorem 23.21. I f an Archimedean Riesz space L is Riesz isomorphic to some order dense Riesz subspace of un Archimedean Riesz space K , then K is Riesz isomorphic to an order dense Riesz subspace of L” (and hence L and K have the “same” universal completion). Proof. Assume n is a Riesz isomorphism between L and an order dense Riesz subspace, say K of some Archimedean Riesz space K . Note that K” is the universal completion of K , , and then use Theorem 23.18 to extend n to a Riesz isomorphism between L” and K ” . The result now follows easily.

In terms of the universal completior, the following characterization of the dominable sets can be given.

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Theorem 23.22. Let L be an Archimedean Riesz space, and let A be a nonempty subset of L+. Then the following statements are equivalent: (i) A is a dominable subset of L+. (ii) A is order bounded in L" (and hence sup A exists in L"). Proof. Apply Lemma 23.10 and Theorem 23.13. H

A sequential analog of the preceding theorem follows. Theorem 23.23. For a o-Dedekind complete Riesz space L the following statements are equivalent : (i) L is o-laterally complete (and hence o-universally complete). (ii) Every countable dominable subset of L+ is order bounded in L (and hence it has a supremum).

Proof. (i) (ii) Assume first that L contains a weak order unit e. Let {u,} be a countable dominable subset of L+. Then by Theorem 23.22 u* = sup{u,} exists in L". Observe that for each u E L + , w = sup{u, A v} exists in L (since L is o-Dedekind complete) and so since L is order dense in L", U * A V = W A U E L. Now if w, = ( ( n 2)e - u*)+~ ( n e u*)- ( n = 0,1,. . .), then by what was just noted {w,} c L+. Now proceeding as in the proof of Theorem 23.4(i) (and working in L"), we see that the projection of u* on e (which equals u*) can be written as u* A (f g ) where

+

f

= sup((2n

+ l)w,,:n

+

2 0}

and

g

= sup

+

{(2n 2)w,,,+ :n 2 0 } ,

which by the o-lateral completeness of L both belong to L. Thus u* = u* A ( f + g ) I f + g E L , so that 0 I u, I f + g holds in L for all n, and the proof in this case is finished. We shall show next that {u,} is included in a principal band Be of L. We can then apply the above arguments to the o-universally complete Riesz space Be to show that {u,} is order bounded in Be, and hence in L. To this end, define v1 = u1 and then inductively if v l , . . . ,v, have been defined, put w, = v1 + + v,, and define u,+ = u,+ - Pw,(un+l). As in the first part of the proof of Theorem 23.8, it can be shown that {v,} is a disjoint sequence of L'. Put e = sup{ v,} in L, and note that since 0 I v, I w, = sup{v,, . . . ,v,} I e, we have Pw,(un+1), v, E Be for all n and thus u,+ = PW,(un+1 ) + v,+ E Be for all n. Hence { u,} E Be, and we are done. (ii) 3 (i) Let {u,} be a disjoint sequence of L'. Then {u,} is order bounded in L", and therefore {u,} is a dominable subset of L". It follows that {u,} is a dominable subset of L and thus order bounded by hypothesis. Therefore, sup { u,} exists in L and the proof of the theorem is complete. H The previous theorem has a number of important consequences, some of which are included in the next results.

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177

Theorem23.24. For a Dedekind complete Riesz space L the following statements are equivalent:

(i) L is universally complete, that is, L = L". (ii) L is a-universally complete and has a weak order unit. Proof. (i) 3 (ii) Obvious. (ii) =-(i) Since L is Dedekind complete, L is an ideal of L" (Theorem 2.2). Now let 6' < e E L be a weak order unit of L and let 6' < u* E L". Since e is also a weak order unit of L", we have 6' I u* A ne t u*. Now { u* A ne} c L and so {u* A ne} is a dominable sequence of L. By Theorem 23.23 {u* A ne} is order bounded in L and thus u* E L. Hence L = L" and the proof is finished. H The next example shows that the Dedekind completeness cannot be dropped from Theorem 23.24. Example 23.25. Consider the Riesz space L consisting of all real-valued Lebesgue measurable functions defined on [OJ] (with the pointwise ordering). Then L is a-universally complete, has a weak order unit ( e ( x )= 1 for all x E [0,1] is a weak order unit of L),and it is neither laterally complete nor Dedekind complete. H Another example also shows that the existence of the weak order unit in statement (ii) of Theorem 23.24 cannot be dropped. Example 23.26. Consider an uncountable set X and let L = f f E

RX:( x E X :f ( x ) # 0 } is at most countable}, with the pointwise ordering. It can

be seen easily that L is Dedekind complete and a-universally complete without weak order units. Note, however, that L is not universally complete. H

We are now in the position to characterize the Riesz spaces that have a a-universal completion. Theorem23.27. For a Riesz space L the following statements are equivalent: (i) L is almost a-Dedekind complete. (ii) L has a a-universal completion (determined uniquely up to a Riesz isomorphism). Proof. (i) +- (ii) Assume first that L is a-Dedekind complete. Put U ( L + )= {u* E L":3{u,} E L+ with 6' s u, t u* in L"}, and define M = {u* - v*:u*,v* E U ( L + ) } .We have clearly the following Riesz subspace inclusions L E M E L". We shall show next that M + = U(L+).To this end let 6' I u* - v* with u*,v* E U(L+).Pick two sequences {u,} and {v,} of L+ with 6' I u, t u* and 6' I v, 7 v* in L". Note that since L is a-Dedekind

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complete u*

A

u is in L for each u E L + . Now 6

and u, - u* defined by

A

U,

- U*

A U,,

= U* A U, - U*

A U,

5

U* - U*

u, .+ u* - u* in L" imply that the sequence {w,} of L+ (0)

w, = sup{ui - u * A u i : i = 1,. . . ,n}

satisfies 0 Iw, t u* - u* in L". Thus u" - u" E U ( L + )and so M + = U ( L + ) . It can be proven now easily that M is a o-universally complete Riesz space that contains L as a super order dense Riesz subspace. Now if L is almost o-Dedekind complete, embed L super order densely in a 0-Dedekind complete Riesz space K and then embed K as above super order densely to a o-universally complete Riesz space. Thus L has a 0universal completion. The uniqueness of the o-universal completion can be derived easily by applying Theorem 23.23; the details are left to the reader as an exercise. (ii) (i) It follows immediately from the definition of the almost o-Dedekind completeness. Note. The preceding three theorems are due to Fremlin [26, pp. 73-76].

By a stepfunction s on a topological space X, we shall mean any function of Cm(X)for which there exists a collection { &: i E I ) of mutually disjoint closed-open subsets of X whose union is dense in X and such that s is constant on each F. In other words, s is a step function on X if it has a cixv, with each closed and open, representation of the form s = & n F = @ i f i # j , a n d w i t h t h e o p e n s e t L o = U ( K : i E I } denseinX. If the step function s has an at most countable representation, that is, if s = c,xv,, then it will be called a o-step fincrion. We shall denote by S"(X) (resp. by Sum(X))the collection of all step functions (resp. the collection of all o-step functions) on a topological space X.

CieI

c."=

Theorem 23.28. Let X be an extremally disconnected topological space. Then the following statements hold: (i) Sm(X)is an Archimedean laterally complete Riesz space. (ii) Sum(X)is an Archimedean o-laterally complete Riesz space. (iii) The following Riesz subspace inclusions hold: Sum(X)E S"(X) E Cm(X),with Sum(X)order dense in Cm(X). Proof. (i) Note first that Sm(X)is a vector subspace of Cm(X).Indeed, ifs and tare in Sm(X)with representations s = cixv,and t = bjxu,, then As = ilcixv, E Sm(X) for each ilE R and

Xiel

ciEI

cjeJ

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179

since U { v n U j : i E I , j E J }= ( l J { v : i ~ n ~(}U){ U j : j E J } ) is dense in X . For the Riesz space structure of Sm(X)note that sv t =C

2 max{ci,bj}Xvinu,E Sm(X)

id j d

and sA t =

C 1min{ci,bj}XvinujE Sm(X). is1 j c J

Since clearly Sm(X)is Archimedean, it remains to be shown that Sm(X)is a laterally complete Riesz space. To this end let {s,} be a disjoint system of positive elements of S m ( X ) .For each LX pick a representation for s,. Then for each LX collect all the closed-open sets I/ appearing in the representation of s, for which s,(x) > 0 for x E I/. Now the collection of all those closed-open sets forms a mutually disjoint family that we shall denote by { E: i E I } . Note that for each V;. there exists ci > 0 and exactly one c( such that cixb appears in the representation of s,. Put 0 = U{V;.:iE I ) and U = X - 0(closedcixvi Oxv. on the open dense set 0 u U . Note that s open) and let s = can be extended to a function of Cm(X)[46, Theorem 47.1, p. 3221 and hence s E Sm(X).Now it should be clear that s = sup{s,} holds in S m ( X )and this completes the proof of part (i). (ii) Proceed as in (i). (iii) The proof is straightforward and is left to the reader.

xitl

+

The next theorem shows the relation between Archimedean laterally complete Riesz spaces and the spaces of the form Sm(X),with X extremally disconnected. Theorem 23.29. Let L be an Archimedean laterally complete Riesz space. Then there exists a compact, HausdorJY; extremally disconnected topological space R satisfying thefollowing Riesz subspace inclusions: Sm(R)G L L Cm(R). Proof. Let K = Cm(R)be the universal completion of L, where R is Hausdorff, compact, and extremally disconnected. Since L has a weak order unit e, the embedding of L into Cm(R)can be taken to carry e to the constant function 1 on R. Now since L has the projection property [Theorem 23.4(ii)], xv E L for each closed-open subset I/ of R (Theorem 2.10). It now follows easily from the lateral completeness of L that Sm(R)G L E Cm(R). Examples of Archimedean laterally and a-laterally complete Riesz spaces will be presented next. Example 23.30. ( A n Archimedean laterally complete Riesz space that is not Dedekind complete.) By a theorem of Gleason [28, Theorem 3.2, p. 4861

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[Chap. 7

there exists a Hausdorff, compact, and extremally disconnected topological space X and an onto continuous function cp:X + [OJ] that does not carry any proper closed subset of X onto [O,l]. Note that cp E C(X) c Cm(X),and we shall show next that cp 4 Sm(X).Indeed, if cp E Sm(X),then cp must be constant on some nonempty proper closed-open subset V of X. But then rp maps the closed (actually compact) subset X-V onto [0,1], in contradiction with the properties of rp. Now if Sm(X)is Dedekind complete then Sm(X)= Cm(X)by Theorem 23.14(ii), in contradiction to cp 4 Sm(X).Thus Sm(X)is the desired example. (Observe also that Cm(X)is the universal completion of C[O,l]; the mapping T :C[O,l] + Cm(X),defined by T ( f )= f 0 cp for f E C[O,l], embeds C[O,l] order densely into Cm(X).) Example 23.31. (An Archimedean a-laterally complete Riesz that is neither a-Dedekind complete nor laterally complete.) If K is the Riesz space of Example 23.25 and Sm(X)is as determined in Example 23.30, then the Riesz space L = K x Sm(X)has the stated properties. Note. The proof that cp 4 Sm(X)in Example 23.30 is due to J. Quinn (personal communication). We devote the rest of this section characterizing the laterally complete Riesz spaces of the form RX.Note that two such Riesz spaces RX and RY are Riesz isomorphic if and only if X and Y are of the same cardinality (i.e., if there exists a one-to-one mapping from X onto Y).The characterizations of the finite dimensional RX spaces are given first. Theorem 23.32. For an Archimedean a-laterally complete Riesz space L the following statements are equivalent: (i) L is Riesz isonzorphic to some R". (ii) L admits an order unit. (iii) L admits a Riesz norm. Proof. (i) * (ii) Obvious. (ii) =. (iii) If e is an order unit of L , then p(u) = inf{A > 0:lul I Ae} for u E L is a Riesz norm on L. (iii) * (i) Let p be a Riesz norm on L. If {u,,} is a disjoint sequence of positive elements of norm one, then {nu,,} is also a disjoint sequence of L+ and so u = sup{nu,,} exists in L . But then we have the contradiction n I p(u) < 00 for all n. Thus every set of disjoint elements of L is finite and hence L is Riesz isomorphic to some R" (see Exercise 10 of Chapter 1). Theorem 23.33. For an Archimedean laterally complete Riesz space L the following statements are equivalent: (i) L is Riesz isomorphic to some RX. (ii) L admits a HausdorfS locally concex-solid Lebesgue topology.

Sec. 241

TOPOLOGIES ON LATERALLY COMPLETE SPACES

(iii) (L:)' = { u E L : 1p,1(lul) = 0 for all p E (iv) L is a discrete Riesz space.

=

181

{O).

Proof. (i) (ii) The topology of pointwise convergence on RX is a Hausdorff locally convex -solid Lebesgue topology. (ii) => (iii) Let z be a Hausdorff Lebesgue topology on L that is also locally convex. By Theorem 9.1 the topological dual L' of (L,r) is an ideal of L; and the result follows from the Hahn-Banach theorem. (iii) => (iv) Let 0 < u E L. Pick 0 < cp E L; with cp(u) > 0, and note that N , = { v E L :q(Ivl) = 0} is a band of L and so by Theorem 23.4(ii)L = N , 0 C,; where C , = Nqd.But C , admits a Riesz norm [namely p(v) = cp(lvl)] and so by the previous theorem C , is Riesz isomorphic to some R". Now let v be the projection of u on C,. Then there exists a discrete element w of C,, and hence of L, such that 6 < w I ZI I u. (iv) * (i) By Theorem 2.17 L is Riesz isomorphic to an order dense Riesz subspace of some RX.But then since L is laterally complete, the Riesz isomorphism must be onto and the proof of the theorem is complete. Note. The equivalence (ii) o (iv) of Theorem 23.33 was established in [60, Theorem 3.1, p. 2391. 24.

LOCALLY SOLID TOPOLOGIES ON LATERALLY COMPLETE RIESZ SPACES

In this section we shall study the locally solid topologies that the alaterally complete or laterally complete Riesz spaces can carry. We start with the following theorem. Theorem24.1. For a a-laterally complete Riesz space L the following statements hold: (i) Every disjoint sequence of L' converges to zero with respect to any locally solid topology on L. In particular, every locally solid topology on L satisjes the pre-Lebesgue property. (ii) If in addition L is Archimedean, then every locally solid topology on L satisjes the a-Lebesgue property. Proof. (i) For the first part note that if {u,,} is a disjoint sequence of L+, then so is {nu,,} and hence u = sup{nu,,} exists in L. The result now follows immediately from the relation 0 Iu, I n-'u for all n. For the last part apply Theorem 10.1. (ii) Let z be a locally solid topology on L and let u, 4 6' in L. If I/ is a solid r-neighborhood of zero, choose another r-neighborhhod U of zero with U + U 5 V and then E > 0 such that E U . Now, by Lemma 23.6

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[Chap. 7

there exists u E L+ such that 0 I u, I 2 - " u + w1holds for all n. Pick a u, I positive integer k with 2-"u E U for all n 2 k. But then we have 0 I 2-94 E U ~E U + U E V for all n 2 k, so that u, 5 0, and the proof is finished. H

+

We continue with more properties of the locally solid topologies on o-laterally complete Riesz spaces. Theorem 24.2. Let (L,T)be a Hausdorff locally solid Riesz space with the Fatou property. Then the following statements hold: (i) Every increasing z-bounded net of L+ is dominable. (ii) If L is o-laterally complete, then z is a Lebesgue topology. (iii) If L is universally complete, then (L,z) is z-complete.

u, t is z-bounded and that u > 0. Pick a Fatou Proof. (i) Assume 0 I z-neighborhood V of zero with u # r/: and then choose a positive integer n such that n-'u, E V for all a. Since V is solid and order closed, there exists 0 < w E L such that n- '11, A u I u - w for all ct (otherwise, if 0 I n- ' u , A u t u then u E V , a contradiction). But then (nu - u,)' = n(u - n-'u, A u ) 2 nw > 0 for all a, so that (u,) is dominable. (ii) Note that by Theorem 24.l(i) z is a pre-Lebesgue topology. The result now follows immediately from Theorem 11.6. (iii) By Theorem 13.4L is order dense in its topological completion i and by Theorem 23.14(ii)we get L = t,i.e., that (L,T)is z-complete. H It is natural to ask when a Hausdorff Fatou topology on a Riesz space can be extended to a Hausdorff Fatou topology on its universal completion. Clearly such a topology must be necessarily Lebesgue. The following theorem tells us what Hausdorff Lebesgue topologies on a Riesz space have such an extension; this theorem will be the key result for the discussion in this section. Theorem 24.3. For a H a u s d o r - locally solid Riesz space (L,T)with the Lebesgue property the following statements are equivalent:

(i) z extends to a Lebesgue topology on L". (ii) T extends to a locally solid topology on L". (iii) z is coarser than any Hausdor-a-Fatou topology on L. (iv) Every dominable set of L is z-bounded. (v) Every disjoint sequence of L+ is z-convergent to zero. (vi) Every disjoint sequence of L + is z-bounded. (vii) T h e topological completion t of (L,T)is Riesz isomorphic to L", that is, t is the universal completion of L.

TOPOLOGIES ON LATERALLY COMPLETE SPACES

Sec. 241

103

Proof. (i) =. (ii) Obvious. (ii) => (iii) Let z* be a locally solid extension of z to L" and let z' be a Hausdorff 0-Fatou topology on L. By Theorem 12.14 it is enough to show that z c z' holds on every component Nd of the carrier C , . So, let N = V,, where [ V , } is a normal sequence of solid z-neighborhoods of zero. Note that every disjoint subset of strictly positive elements of Nd is at most countable. Indeed, if A is a disjoint subset of strictly positive elements of Nd, then A = A,, where A , = ( u E A : u # V,!, and so if some A , is infinite it must contain a sequence {u,} that must be z*-convergent to zero in L" [Theorem 24.l(i)] and hence t-convergent to zero in L, contradicting u, $ V, for all n ; thus each A , is finite and hence A is at most countable. By Zorn's lemma there exists an at most countable complete disjoint system. Now since Nd admits a metrizable Lebesgue topology (its basis is { V , n Nd))it has the countable sup property and hence the restriction of z' to Nd, say zl, has the Fatou property. By Theorem 12.4 there exists a metrizable Fatou topology, say ,z, on Nd that is coarser than zl. Next consider (Ad,?,), and note that since Nd is order dense in fid(use Theorem 13.4 or 17.4),and the universal completion of Nd is the band generated by Nd in L", we have the Riesz subspace inclusions Nd E Ad E L" (see Theorem 23.21). Now since F,,, is a Frechet topology, it follows from Theorem 16.7 that z* restricted to Ad is coarser than F,. Hence z is coarser than z' on Nd and we are done. (iii) * (iv) First use Theorem 11.10 to extend z to the Hausdorff Lebesgue topology zb on La. Let A be a dominable subset of L. Then A is a dominable subset of Lb. By Theorem 23.12 there exists a complete disjoint system (ei:i E I } of strictly positive elements of Lb such that for each i E I , P,,(a) Iniei holds for all a E A and some positive integer ni. Let Bi be the band generated by ei in Lb and then use Theorem 3.2 to embed L order densely in nBi. But if zi denotes the Hausdorff Lebesgue topology of the restriction of z6 to Bi, then nziis a Hausdorff Lebesgue topology on nBiand hence its restriction to L is also a Hausdorff Lebesgue topology. The result now follows by observing that A (as embedded in n B i ) is nIt,-bounded and that n z i restricted to L is finer than z by our hypothesis. (iv) + (v) If {u,] is a disjoint sequence of L+, then so is {nu,).. Now since (nu,). is order bounded in L" it is dominable and hence by hypothesis z-bounded. Now let I/ be a z-neighborhood of zero that is also solid. Pick a positive integer k such that nu, E kV for all n. But then u, E I/ for n 2 k and thus u, 1,0. (v) + (vi) Obvious. (vi) * (vii) Let (L,F) be the topological completion of (L,z). Then L is order dense in L (Theorem 13.4),(t,?) satisfies the Lebesgue property, and L is Dedekind complete (Theorem 10.6).So, we have only to show that is

o:=

Ul=

e

[Chap. 7

LATERALLY COMPLETE SPACES

184

laterally complete. Let {ii,} be a disjoint sequence of ,'f and let V and W be solid z-neighborhoods of zero such that W + W G V. Since L is order dense in f, and ? is a Lebesgue topology, for each n there exists u, E L+ with f3 Iu, 5 ii, and 6, - u, E I T'. Observe that {u,} is a disjoint sequence of L+ and hence z-bounded. Pick a positive integer k with {u,} E kW and note that we have

ii,

= (6, - u,)

+ u, E kW + kW G kP

for all n.

Thus {fin} is ?-bounded and so every disjoint sequence of ' i is ?-bounded. Now assume that {ii,} is a disjoint system of f,'. Let A be the set of all finite subsets of the indexed set {a). For each A E A put C, = i,. Then 0 I5, 7 holds in f,, and we claim that { C,} is in fact a ?-Cauchy net. Indeed, if {C,) is not a ?-Cauchy net, then there is a solid z-neighborhood I/ of zero such that for any A E A there exist &,A2 E A with &,A, 2 A and with 6, CL2 $ P. Thus there exists a sequence {A,} E A with A,-E An+ and G, = CLn+ ,- CLn $ P for all n. But {G,} is a disjoint sequence of Lf and hence so is {nG,}. Thus by what was shown above {nGn} is ?-bounded and so there exists k with {nG,,} E k P . In particular, it follows that G, E V for sufficiently large n, a contradiction. Hence (6,) is a 5-Cauchy net and so 5, f ii holds in f, for some ii. But then 6, ii holds also in i.It now follows that 6 = sup{iiu} holds in t,so that f, is laterally complete. (vii) * (i) Observe that z^ is a Lebesgue extension of z to f, (Theorem 10.6).

xaE,

Remark. If a Hausdorff Lebesgue topology z on a Riesz space L extends to a Lebesgue topology z" on its universal completion L", then any basis { V } of zero for z consisting of Fatou z-neighborhoods of zero extends to a basis { V " }of zero for z" consisting of Fatou z"-neighborhoods of zero, where V" = {u*

E Lu:3{uu} E

I/ with f3 Iu,

t (u*l in L"}.

Note, in particular, that z" is determined uniquely. We are now in the position to state the main topological properties of laterally complete Riesz spaces.

Theorem 24.4. If a a-laterally complete Riesz space L admits a HausdorfS Fatou topology z, then any other Hausdorff locally solid topology on' L is finer than z. Proof. Assume that z is a Hausdorff Fatou topology and z* a Hausdorff locally solid topology on a a-laterally complete Riesz space L. Then by Theorem 24.l(i) z satisfies the pre-Lebesgue property and hence the Lebesgue property (Theorem 11.6). Likewise, by Theorem 24.l(ii) z* satisfies

Sec. 241

TOPOLOGIES ON LATERALLY COMPLETE SPACES

185

the a-Lebesgue property and hence the a-Fatou property. But by Theorem 24.1 z satisfies condition (v) of Theorem 24.3 and hence condition (iii) of the same theorem holds. Thus z E z* holds and we are done. w As an obvious consequence of the last result we have that the Hausdorff Fatou topologies on o-laterally complete Riesz spaces are uniquely determined. Theorem 24.5. A o-laterally complete Riesz space admits at most one Hausdorff Fatou topology, which must be necessarily Lebesgue.

The next theorem follows immediately from the preceding result. Theorem24.6. An Archimedean Riesz space can admit at most one Hausdorff Lebesgue topology that extends to its universal completion as a locally solid topology.

The following two theorems deal with metrizable locally solid topologies on a-laterally complete Riesz spaces. Theorem 24.7. If a a-laterally complete Riesz space L admits a metrizable locally solid topology z, then z is the only Hausdorff locally solid topology L can carry, and in addition z is a Lebesgue topology. Proof. Let z be a metrizable locally solid topology on a a-laterally complete Riesz space L. Then by combining Theorems 24.1 and 17.8, we infer that z is a Lebesgue topology. In particular, it follows that L must satisfy the countable sup property. Thus every other Hausdorff locally solid topology on L must be a Lebesgue topology, and the result follows from Theorem 24.5. Theorem 24.8. For a laterally complete HausdorfS locally solid Riesz space (L,z) the following statements are equivalent:

(i) z is metrizable (and hence uniquely determined). (ii) L has the countable sup property. Proof. (i) 3 (ii) By Theorem 24.7 z is a Lebesgue topology and so L must have the countable sup property. (ii) (i) Note first that z is a Fatou topology. Hence by Theorem 12.4 (since L has a weak order unit), there exists a metrizable locally solid topology z' on L, and by Theorem 24.7 we have z = z', that is, z is metrizable, and we are done.

The next result gives some conditions for the normal Riesz homomorphisms to be continuous.

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[Chap. 7

Theorem 24.9. Assume that z is a HausdorfSo-Fatou topology on a Riesz space L, T‘ a Lebesgue topology on a o-laterally complete Riesz space M , and that T : ( L , z )+ (M,z‘)is a normal Riesz homomorphism. Then T is continuous. Proof. If { V } is a basis for zero for T‘ consisting of Fatou neighborhoods, note that since T is a normal Riesz homomorphism, { T - ’( V ) }is a basis of zero for a Lebesgue topology, say z l , on L. Clearly T : ( L , z , )+ (M,z’) is continuous. Observe also that z1 (which is not necessarily Hausdorff) satisfies condition (v) of Theorem 24.3. Indeed, if ( u n j is a disjoint sequence of L’, then since T is a Riesz homomorphism { T(u,)} is a disjoint sequence of M + and hence by Theorem 24.l(i) T(u,) 1;6, so that u, 2 6. Now put N = n { T - ’ ( V ) :V E { V }}, and note that N is a band of L and that z1 induces a Hausdorff Lebesgue topology on Nd. Thus by Theorem 24.3 z1 must be coarser than z if both are restricted to Nd,that is, T restricted to Nd is continuous. Obviously T restricted to N is continuous and hence T restricted to the order dense ideal N 0 Nd is continuous. Now let {u,} c L+ with u, 6 and V,U E { V ) such that V + V E U . Since N 0 Nd is order dense in L, for each o! there exists u, E N 0 Nd with 6 5 u, 5 u, and T(u,) - T(u,) E V [we use here that z1 is Lebesgue and that T:(L,z,) --, (M,z’) is continuous]. But then u, It B holds in L and since {u,} c N 0 Nd, it follows T(u,) 2 0 and thus T(u,) E V for all o! 2 u l . Hence T(u,) = [T(u,) - T(u,)] + T(u,) E V

+VGU

for u 2 a l ,

so that T:(L,z)+ (M,z’) is continuous, and the proof is finished. The next result describes a basic property of the unique Hausdorff Fatou topology on a a-laterally complete Riesz space. Theorem 24.10. l f a a-laterally complete Riesz space L admits a HausdorfS Fatou topology z (necessarily unique and Lebesgue) then T is coarser than any HausdorfS locally solid topology on L and finer than any Lebesgue topology on L. Proof. The “coarser” part is just Theorem 24.4. To see the other part let z* be a Lebesgue topology on L , and consider the identity mapping l:(L,z)3 (L,z*). Now apply Theorem 24.9 to get T* E z. We now state a theorem giving some conditions under which an order continuous operator is continuous. Theorem 24.11. Let T be a positive, order continuous linear operator from a o-laterally complete Hausdor- locally solid Riesz space (L,T)into a locally solid Riesz space (M,z‘)with the Fatou property. Then T:(L,z)+ (M,Ti) is continuous.

Sec. 241

TOPOLOGIES ON LATERALLY COMPLETE SPACES

187

Proof. Let { V } be a basis of zero for z’ consisting of Fatou t’-neighborhoods. For each I/ E { V ]define V* = {u E L : T(lu1)E V } ,and note that { V * } defines a Fatou topology z* on L (and hence t* is Lebesgue) such that T : ( L , z * )-+ (M,z’) is continuous. Now since z is a-Lebesgue, it follows from Theorem 24.9 that the identity mapping Z:(L,z)-P (L,z*) is continuous and hence T:(L,z) (M,z’) must be continuous. -+

We can ask at this point whether there exist Archimedean o-laterally complete Riesz spaces admitting no Hausdorff locally solid topologies. The following example gives an affirmative answer to this question. Example24.12. Let L be the universal completion of C[O,l]. Assume that L admits a Hausdorff locally solid topology z. Then by Theorem 24.l(ii) the topology z is a o-Lebesgue topology and hence z restricted to C[O,l] is also a Hausdorff o-Lebesgue topology, contradicting the fact that C[O,l] does not admit Hausdorff o-Lebesgue topologies; see Example 8.2. This contradiction establishes that L does not admit any Hausdorff locally solid topology. 4

Regarding the quotient Riesz spaces of o-laterally complete Riesz spaces we have the following result. Theorem 24.13. Let L be a a-laterally complete Riesz space carrying a (necessarily unique) Hausdorff Lebesgue topology z, and let A be a band of L. Then z l A is the unique Hausdorff Lebesgue topologji on the o-laterally complete Riesz space LIA. Proof. Note that L / A is a-laterally complete (Theorem 23.8), z / A is Hausdorff (since every band is z-closed) and z / A is a Lebesgue topology by Theorem 8.11, and hence unique by Theorem 24.5.

For the a-laterally complete Riesz spaces admitting the unique Hausdorff Fatou topology the following interesting result holds. Theorem 24.14. If a o-laterally complete Riesz space L admits a Hausdorf Fatou topology z (necessarilyunique and Lebesgue), then there exists an order dense a-ideal of L such that any other Hausdorff locally solid topology on L induces the same topology on this a-ideal as z. Proof. By Theorem 11.6, z is a Lebesgue topology, and by Theorem 12.3, the carrier C , of z is an order dense a-ideal of L with the countable sup property. In particular, note that C , is a a-laterally complete Riesz space in its own right. Now if z* is any other Hausdorff locally solid topology on L , then z* satisfies the o-Lebesgue property (Theorem 24.1) and so, since C , has the

Iaa

LATERALLY COMPLETE SPACES

[Chap. 7

countable sup property, T* restricted to C, must be a Lebesgue topology. The result now follows immediately from Theorem 24.5. 1 As an illustration of the preceding theorem we give the following example.

Example 24.15. Consider the universally complete Riesz space L = RX, for some uncountable set X . Let T be the Hausdorff Lebesgue topology of pointwise convergence on L. An easy verification shows that the carrier C, of z consists of all functions of L whose support is at most countable, that is, C, = {f E RX:{x E X : f ( x )# 0 ) is at most countable}.

According to Theorem 24.14, every Hausdorff locally solid topology on L induces the topology of pointwise convergence on C,. 1 Note. Theorems 24.1, 24.2, 24.5, 24.8, 24.9, and 24.11 were proven by Fremlin for universally complete Riesz spaces in [26]. The generalizations and the approach of the results in this section are due to the authors [7]. EXERCISES

1. Let 93 be a Boolean algebra. (i) Show that a proper ideal I of B is prime if and only if I is a maximal ideal. (ii) If an ideal I of B does not contain an element u, show that there exists a prime ideal J of 93 such that I c J and u # J . (iii) Let 51 denote the collection of all proper prime ideals of 93. For each ~ € 9 put 3 R , = { w ~ R : u q i c o } Show . that { 5 1 , : u ~ 9 3 }forms a basis for a Hausdorff, compact topology T on 51, called the hull-kernel topology. (iv) Show that A G R is closed-open with respect to z if and only if there exists u E B with A = 51,. (v) Show that (0,~) is extremally disconnected if and only if 93 is a Dedekind complete Boolean algebra. (vi) Show that if X is a compact, Hausdorff, and extremally disconnected topological space, then there exists a unique (up to a Boolean isomorphism) ) homeoDedekind complete Boolean algebra B whose Stone space ( 5 1 , ~ is morphic to X . 2. Show that if L is an Archimedean Riesz space, then the collection BLof all bands of L ordered by inclusion forms a Dedekind complete Boolean algebra whose Boolean operations are given by A A B = A n B, A v B = ( A + B)dd,and A' = Ad for all A,B E B L . 3. Let L be the Riesz space consisting of all real-valued, bounded everywhere, Lebesgue measurable functions on [0,1] with the pointwise

EXERCISES

189

ordering (everywhere). Determine the a-universal and universal completions of L. 4. Let L be an Archimedean a-laterally complete Riesz space. Show that if cp E L", then C , is a projection band of L. [Compare this with Exercise 14(ii) of Chapter I.] 5. Let L be an Archimedean a-laterally complete Riesz space. Show that L has no discrete elements if and only if L; = lo). 6. Complete the details of the proof of Theorem 23.27. 7. (Bernau [13]). Let L be an Archimedean Riesz space. The lateral completion LA of L is defined as the intersection of all laterally complete Riesz subspaces between L and L". Show that (LA)'= (L')A= L". 8. Let L = RX for a nonempty set X. Show that 15: = L-. (Hint: Use Exercise 17 of Chapter 1 or use Theorem 23.7.) 9. A set X is said to have a nonmeasurable cardinal if there is no measure p defined on all subsets of X that vanishes on the finite subsets of X and with p(X) = 1. Show that if a set X has a nonmeasurable cardinal, then for every cp E (R')" there exist xl, . . . ,x, E X and real numbers A,, . . . ,A, (all depending upon cp) such that q ( f ) = l , f ( x i )for all f E RX and derive that in this case ( R x ) r = (R')'. [For more about this, see W. A. J. Luxemburg, Is every integral normal? Bull. Amer. Math. Soc. 73 (1967), 685-688.1 10. Let L = RX,where X is a nonempty set and let T be the topology of pointwise convergence on L. For a sequence {u,} of L show the following: (i) u, 5 6 if and only if u, 3 6 in L. (ii) If z1 and z2 are two Hausdorff locally solid topologies on L, then u, 2 0 if and only if u, 3 8. 11. Let (L,t)be a a-laterally complete locally solid Riesz space, { V,} a normal sequence of solid z-neighbwhoods of zero and N = (?{ V,:n = 1,2,. . .}. Show that Nd is an Archimedean laterally complete Riesz space with the countable sup property. 12. Give an alternate proof of Theorem 24.9 by using Theorem 12.14. 13. Verify that the carrier of Example 24.15 is given by C, = {f E R X :{x E X:f(x) # 0} is at most countable}. 14. Let ( X , C , p ) be a finite measure space whose finite subsets have measure zero and let L be the Riesz space of all real-valued p-measurable functions defined on X (with the pointwise ordering everywhere). (i) Show that L is a a-universally complete Riesz space whose unique Hausdorff Fatou topology z is the topology of pointwise convergence. (ii) Let A = { f L ~ : f ( x )= 0 a.e.1. Show that A is a a-ideal of L and that L / A is a universally complete Riesz space.

190

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[Chap. 7

(iii) For each n put

Show that { V,} is a basis of solid order closed neighborhoods of zero for the unique Hausdorff Fatou topology on LIA. (iv) What is zlA? Compare your last answer with Theorems 24.13 and 14.3. 15. If z is a locally solid topology on a o-laterally complete Riesz space, show that C, is a o-laterally complete Riesz space with the countable sup property and that z restricted to C, is a Hausdorff Lebesgue topology. 16. Show that a Hausdorff locally solid topology z on a o-laterally complete Riesz space is Lebesgue if and only if C, is z-dense. 17. Let (L ,t) be a Hausdorff 0-laterally complete locally solid Riesz space. Show that L admits a Hausdorff Fatou topology (necessarily unique) if and only if C, is order dense in L. 18. Give an example of a o-universally complete Riesz space with a Hausdorff locally solid topology that is not Lebesgue. 19. Let X be a nonempty set. If every Hausdorff locally solid topology on RX is Lebesgue, show that X has a nonmeasurable cardinal. (Hint: If p is a finite measure on all subsets of X and vanishing on the finite sets, then the Hausdorff locally solid topology on RX generated by the Riesz pseudonorms

is not a Lebesgue topology.) OPEN PROBLEMS

1. Is every Hausdorff locally solid topology on a universally complete Riesz space, necessarily Lebesgue? (See [26, Section 3, pp. 84-88], where this question is studied in connection with the nonmeasurable cardinal problem.) 2. If a universally complete Riesz space admits a Hausdorff locally solid topology does it follow that it must admit a Hausdorff Lebesgue topology also?

References

I . C. D. Aliprantis, Some order and topological properties of locally solid linear topological Riesz spaces. Proc. Amer. Math. Soc. 40 (1973). 443-447. 2. C. D. Aliprantis, On the completion of Hausdorff locally solid Riesz spaces. Trans. Amer. Math. Soc. 196 (1974), 105-125. 3. C . D. Aliprantis, Riesz seminorms with Fatou properties. Proc. Amer. Math. Soc. 45 (1974), 383-388. 4. C. D. Aliprantis, On order properties of order bounded transformations. Canudian J . Math. 27 (1975), 666-678. 5 . C. D. Aliprantis and 0. Burkinshaw, A new proof of Nakano’s theorem in locally solid Riesz spaces. Math. 2. 144 (1975), 25-33. 6. C. D. Aliprantis and 0. Burkinshaw, Nakano’s theorem revisited. Michigan Math. J . 23 (1 976), I73 - 176. 7. C. D. Aliprantis and 0. Burkinshaw, On universally complete Riesz spaces. PuciJic J . Math. 71 (1977). 1-12. 8. C . D. Aliprantis and 0. Burkinshaw, A simple proof of the order continuity of the norm in L,-spaces. Preprint. 9. C. D. Aliprantis and E. Langford, Almost u-Dedekind complete Riesz spaces and the main inclusion theorem. Proc. Amer. Math. Soc. 44 (1974), 421 -426. 10. I . Amemiya, On ordered topological linear spaces. In Proc. Internat. Symp. Linear Spaces, Jerusalem. 1960 pp. 14-23. Israel Acad. Sci. Humanities, Jerusalem, 1961. I I . I. Amemiya and T. Mori, Topological structures in ordered linear spaces. J . Math. Soc. Japan9(1957), 131-142. 12. S . J. Bernau, Orthocompletion of lattice groups. Proc. London Math. SOC.16 (1966), 107- 130. 13. S. J. Bernau, Lateral and Dedekind completion of Archimedean lattice groups. J. London Math. Soc. 12 (1976), 320-322. 14. H. F. Bohnenblust, O n axiomatic characterization of I,,-spaces. Duke Math. J . 6 (1940), 627-640.

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15. H. F. Bohnenblust and S. Kakutani, Concrete representations of (M)-spaces. Ann. Math. 42 (1941), 1025-1028. 16. 0. Burkinshaw, Weak compactness in the order dual of a vector lattice. Trans. Amer. Math. SOC.187 (1974), 183-201. 17. 0 . Burkinshaw and P. Dodds, Disjoint sequences, compactness and semireflexivity in locally convex Riesz spaces. INinois J . Math. 21 (1977), 759-775. 18. 0. Burkinshaw and P. Dodds, Weak sequential compactness and completeness in Riesz spaces. Canadian J. Math. 28 (1976), 1332-1339. 19. P. Conrad, The lateral completion of a lattice-ordered group. Proc. London Math. Soc. 19 (1969), 444-480. 20. M. M. Day, The spaces LPwith 0 < p < I . BUN. Amer. Math. Soc. 46 (1940), 816-823. 21. P. Dodds, Sequential convergence in the order dual of certain classes of Riesz spaces. Trans. Amer. Math. Soc. 203 (l975), 391 -403. 22. M. Duhoux, Le completb d’un treillis localement solide. Bull. Soc. Math. Belg. 24(1972), 133-151. 23. N. Dunford and J. T. Schwartz, “Linear Operators,” Vol. I. Wiley, New York, 1958. 24. D. H. Fremlin, On the completion of locally solid vector lattices. PaciJic J. Math. 43 (1972), 341 -347. 25. D. H. Fremlin, “Topological Riesz Spaces and Measure Theory.” Cambridge Univ. Press, London and New York, 1974. 26. D. H. Fremlin, Inextensible Riesz spaces. Math. Proc. Cambridge Phil. Soc. 77 (1975), 71 -89. 27. H. Freudenthal, Teilweise geordnete Moduln. Nederl. Akad. Wetensch. Proc. Ser A 39 (1936), 641-651. 28. A. M. Gleason, Projective topological spaces. INinois J . Math. 2 (1958), 482-489. 29. C. Goffman, Compatible seminorms in a vector lattice. Proc. Nut. Acad. Sci. U.S.A. 42 (1956), 536-538. 30. C . Goffman, Completeness in topological vector lattices. Amer. Math. Monthly 66 (1959). 87-92. 31. J. Horvath, “Topological Vector Spaces and Distributions,” Vol. I. Addison-Wesley, Reading Massachusetts, 1966. 32. G. Jameson, “Ordered Linear Spaces” (Lect. Notes in Math. 141). Springer-Verlag, Heidelberg, 1970. 33. S. Kakutani. Concrete representation of abstract (L)-spaces and the mean ergodic theorem. Ann. Math. 42 (1941), 523-537. 34. S. Kakutani, Concrete representations of abstract (M)-spaces. Ann. Mafh. 42 (1941). 994- 1024. 35. L. V. Kantorovich, Sur les proprietes des espaces semi-ordonnts linbaires. CRL Acad. Sci. Paris 202 (1936), 813-816. 36. L. V. Kantorovich, Concerning the general theory of operations in partially ordered spaces. DAN SSSR I , 271-274, 1936. (Russian.) 37. L. V. Kantorovich, Lineare halbyeordnete Raume. Math. Sbornik 44 (1937). 121-168. 38. S. Kaplan, On the second dual of the space of continuous functions. Trans. Amer. Math. Soc., Note I 8 6 (1957), 70-90; Note I1 93 (1959), 329-350; Note 111 101 (1961), 34-51; Note IV 113 (1964), 512-546. 39. I. Kawai, Locally convex lattices. J . Math. SOC. Japan 9 (1957), 281-314. 40. G. Kothe, “Topological vector spaces I . ” Springer-Verlag, New York, 1969. (English transl. from the German.) 41. K. K. Kutty and J. Quinn, Some characterizations of the projection property in Archimedean Riesz spaces. Canadian J . Math. 24 (1972). 306-31 1.

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42. E. Langford and C. D. Aliprantis, Regularity properties of quotient Riesz seininorms. Nederl. Akad. Wetensch. Proc. Ser. A 78 (1975), 199-212. 43. H. P. Lotz, Minimal and reflexive Banach lattices. Math. Ann. 209 (1974). 117-126. 44. W. A. J. Luxemburg, Notes on Banach function spaces. Nederl. Akad. Wetensch. Proc. Ser. A , Note XIV 68 (1965). 229-248; Note XV 68 (1965). 415-446; Note XVI 68 (1965), 646 -667. 45. W. A. J. Luxemburg and A. C. Zaanen, Notes on Banach function spaces. Nederl. Akad. Wetensch. Proc. Ser. A , Note I 6 6 (1963) 135-147; Note 11 66 (1963), 148-153; Note 111 66 (1963). 239-250; Note IV 66 (1963), 251-263; Note V 66 (1963). 496-504; Note VI 66 (1963), 655-668; Note VII 66 (1963). 669-681; Note VIII 67 (1964). 104-119; Note IX 67 (1964). 360-376; Note X 67 (1964). 493-506; Note XI 67 (1964). 507-518; Note XI1 67 (1964), 519-529; Note XI11 67 (1964), 530-543. 46. W. A. J. Luxemburg and A. C. Zaanen, “Rieszspaces I.” North-Holland Publ.. Amsterdam, 1971. 47. F. Maeda and T. Ogasawara, Representation of vector lattices. J . Sci. Hiroshima Uniu. A 12 (1942). 17-35. 48. W. Maxey, The Dedekind completion of C(X)and its second dual. Ph.D. Thesis, Purdue Univ., 1973. 49. P. Meyer-Nieberg, Zur schwachen kompaktheit in Banachverbanden. Math. Z . 134 (l973), 303-3 15. 50. H. Nakano, “Modulared Semi-ordered Linear Spaces.” Maruzen Co., Tokyo, 1950. 51. H. Nakano, Linear topologies on semi-ordered linear spaces. J . Far. Sci.Hokkaido Uniu. Ser. I 1 2 (1953). 87-104. 52. H. Nakano, “Semi-ordered Linear Spaces.” Japan Society for the Promotion of Science, Tokyo, 1955. 53. H. Nakano, “Linear Lattices.” Wayne State Univ. Press, Detroit, Michigan, 1966. 54. T. Ogasawara. “Vector Lattices I and 11.” Tokyo, 1948. (In Japanese.) 55. A. L. Peressini, On topologies in ordered vector spaces. Math. Ann. 144 (1961), 199-223. 56. A. L. Peressini, “Ordered Topological Vector Spaces.” Harper & Row, New York, 1967. 57. A . G. Pinsker, On the extension of partially ordered spaces. DAN S S S R 21 (1938). 6-10. (Russian.) 58. A. G. Pinsker, Extension of partially ordered groups and spaces. Leningrad Gos. Ped. Insr. Uchen. Zap. 86 (1949), 285-315. (Russian.) 59. J. Quinn, Intermediate Riesz spaces. Pacific J . Math. 56 (1975), 225-263. 60. J. @inn and R. Reichard, Characterizing discrete vector lattices. Colloq. Math. 31 (1974). 235-242. 61. F. Riesz, Sur la decomposition des operations lineaires. A f f i . Congr. Bologna 3 (1928), I43 148. 62. F. Riesz, Sur quelques notions fondamentales dam la theorie generale des operations linkaires. Ann. Math. 41 (1940), 174-206. (This work was published first in 1937 in Hungarian.) 63. A. Robertson and W. Robertson, “Topological Vector Spaces,” 2nd ed. Cambridge Univ. Press, London, 1973. 64. H. H. Schaefer, Halbgeordnete IokalkonvexeVektorrHume. Math. Ann., Note I 135 (1958), 115-141; Note I1 138 (1959), 259-286; Note 111 141 (1960), 113-142. 65. H. H. Schaefer, On the completeness of topological vector lattices. Michigan Math. J . 7 (1960). 303-309. 66. H. H. Schaefer, ‘‘Topological Vector Spaces.” Springer-Verlag, Heidelberg, 1971. 67. H. H. Schaefer, “Banach Lattices and Positive Operators.” Springer-Verlag, Heidelberg, 1974. -

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68. A. R. Schep, Kernel operators. Ph.D. Thesis, Univ. of Leiden, 1977. 69. A. I. Veksler and V. A. Geiler, Order and disjoint completeness of linear partially ordered spaces. Siberian Math. J . 13 (1972), 30-35. 70. B. Z. Vulikh, “Introduction to the Theory of Partially Ordered Spaces.” Wolter-Hoordhoff, Groningen, Netherlands, 1967. (English transl. from the Russian.) 71. Y. C. Wong and K. F. Ng, “Partially Ordered Topological Vector Spaces.” Clarendon Press, Oxford, 1973.

Index

Birkhoff’s inequalities, 3 Boolean algebra, 174, I88 Bounded set from above, 9 from below, 10 order, 10, 87, 132, 163 topologically, 28, 35, 130

A Absolute value, 2, 21, 22 Absolute weak topology, 40, 62, 81, 105, 127 Absorbing set, 27 Abstract L-space, 73, 153, 163 Abstract L,,-space, 70, 71, 78 Abstract M-space, 73, 78 Algebraic dual, 28, 126 Algebraic sum of two ideals, 4, 127 Almost a-Dedekind completeness, 10, 83, 98, 177 Archimedean Riesz space, 7, 14, 35 Atom, 3 1, see ulso Discrete element and Discrete Riesz space

B B-property, 121 Balanced set. 3, 27 Banach lattice. 38. 71, 110, 112 Band, 5,56, 127 principal. 12 projection, 12,48, 144 Barrelled space, 40 Bidual, 59, 131, 153, 159 Bipolar of a set, I26 Birkhoff’s identity, 3

C Canonical projection, 9 Carrier of locally solid topology, 83, 100 of order bounded linear functional, 23, 24. 77,100 Cartesian product, 3, 20, 35, 11 1 Cauchy net, 29 Compactness, 130, see also Relatively compact set and Relatively sequentially compact set Complete disjoint system, 16, 20, 171 Complete topologically, 29,45, 90 Component of carrier, 84, 89 of element, 13 Cone, 2, see also Positive cone Convex hull, 4, 126, 139 Countable sup property, I I , 12, 84, 86, 146

195

196

INDEX

I

D

Decreasing net, 5, 36, 63 Dedekind complete Boolean algebra, 175 Riesz space, 10, 48,90 Dedekind completion, 1 I, 18, 82, 132 Direct sum, 7, 127 Directed set, 5 Discrete element, 17, 26, 154 Discrete Riesz space, 17, 156, 181 Disjoint complete system, 16, 171 complement, 7 elements, 7, 23, 154 sequence, 64, 15I , 166, I8 1 Distributive law, 3, 123 Dominable set, 170 Dual space order-l', 23, 62, 128 order continuous-L,, 23, 26, 60, 106, 139 u-order continuous-lc-, 23,63, 106, 137 topological-l', 36,40, 128, 153

E Embeddable locally solid Riesz space, 68 Extremally disconnected space, 174 F

Fatou neighborhood, 80 Fatou pseudonorm, 80 Fatou seminorm, 80,92 Fatou topology, 79, 129, 184 Frechet lattice, 111 Frechet space, 111, 123 Frechet topology, 11 1- 115 Full Riesz subspace, 6, 172 Full subset, 105 G

Greatest lower bound, 2

Ideal, 4, 36, 45 order closed, 5 order dense, 6, 7, 45, 61 principal, 4, 17 u-order closed, 5 Increasing net, 5 , 36, 131 Infimum, 2 Infinite distributive law, 3 Integral linear functional, 22,63, 133 normal, 22, 26, 60, 139 transformation, 22 Isomorphic topological vector spaces, 29

K k-disjoint sequence, 64 Kernel of Riesz homomorphism, 9 of topology, 50, 100

L L-space, 73, 153 L,-space, 7 1, 1 12 Lateral completion, 189 Laterally complete Riesz space, 166 Least upper bound, 2 Lebesgue topology (property), 53, 81, 88, 118, 128, 146, 159, 182 Levi topology (property), 61,94, 128, 159 Lexicographic plane, 7 Linear mapping integral, 22 order bounded, 19 order continuous, 22 normal integral, 22 positive, 19 u-order continuous, 22, 168 strictly positive, 12 Linear topology, 27 Locally convex-solid topology, 38, 59, 106, 125, 180 Locally convex topology, 28 Locally solid topology, 33 Locally solid Riesz space, 33 Lower element, 102

H Hahn-Banach Theorem, 21 Hull-kernel topology, 175, 188

M M-norm, 73 M-space, 73,75, 142

INDEX

Mackey topology, 29, 163 Main Inclusion Theorem, 14, 16 Maximal ideal, 123 Metrizable topology, 27, 84, 102, 146, 185 Minkowski seminorm, 38 Monotone Completeness Property (MCP), 45, 94, 109

N Nakano space, 94 Negative part of an element, 2.22 Neighborhood Fatou, 80 normal sequence of, 83 a-Fatou, 80 Net, 5 Nonmeasurable cardinal, 189, 190 Normal integral, 22, 26, 60, 132 Normal Riesz homomorphism, 8. 25, 172. 186 Normal sequence of subsets, 83 Normed Riesz space, 38,42, 71, 110 Null ideal of linear functional, 23, 60, 63 of normal sequence of subsets, 83 0 Order bounded set, 10, 87, 132, 163 Order bounded transformation, 19, 22 Order closed set, 5, 79, 86, 108 Order continuous linear functional, 22, 60, 106, 139, 186 Order convergence, 5. 36, 63 Order dense subspace. 6, 10,25,45,9I . 1 72 Order dual, 23, 36, 62, 112, 128 Order-equicontinuous set, 134-146 Order ideal, 4 Order interval, 9, 45, 90, 130. 147, 154 Order unit, 4, 50, 75 Ordered vector space, 1 Orthogonal elements, 7, sei2 also Disjoint

P p-additive norm, 70 Perfect Riesz space, 61, 78 Pointwise convergence, I26

197

Polar of a set, 126, 127 Positive cone, 2, 35, 43, 78 Positive part of element, 2, 22 of linear transformation, 21 Positive transformation, 19 Pre-Lebesgue topology (property), 53, 64, 69, 132, 181 Prime ideal of a Boolean algebra, 174 Principal band, 12 Principal ideal, 4 Principal projection band, 12, 163, 166 Principal projection property, 13, 166 Product space, see Cartesian product Product topology, 35, 59, 1 1 1 Projection band, 12,48 element, 12, 16, 17 property, 13,49, 167 Pseudo Lebesgue property, 1 15 Pseudonorm, 39,80 Pseudo a-Lebesgue property, 115

Q Quasi-interior point, 164 Quotient seminorm, 39 space, 9, 30, 37, 58, 96-99 topology, 30, 37. 58, 96-99

R Reflexive Banach lattice, 162 Regular Riesz subspace, 6,44, 60, 78, 82 Relative uniform convergence, see Uniformly convergent sequence Relatively compact set, 126, 132, 152, 157 Relatively sequentially compact set, 149, 158 Riesz-Fischer property, 11 1 Riesz homomorphism, 8, 25, 26, 186 a-homomorphism, 8 Riesz homeomorphic spaces, 68 Riesz isomorphism, 9, 25, 173 Riesz isomorphic spaces, 9, 173 Riesz norm, 38, 71 Riesz pseudonorm, 39 Riesz seminorm, 38, 59, 80, 92, 106 Riesz space, 2

198

INDEX

Riesz subspace, 4,35 order dense, 6, 10, 25,45,60, 91, 175 regular, 6,44, 60, 78, 82 super order dense, 6, 117, 174

S a-Dedekind completeness, 10, 14, 48 a-Dedekind completion, 11, 83 a-Fatou neighborhood, 80 a-Fatou pseudonorm, 80 a-Fatou seminorm, 80, 108, 124 a-Fatou topology, 80, 9 I , 97, 107 a-homomorphism, see Riesz homomorphism a-ideal, 5, 97, 131, 169 a-laterally complete Riesz space, 166- 190 a-Lebesgue topology, 52,63, 83, 118, 181 u-Levi topology, 62, 77, 146 a-order closed set, 5, 79, 85 a-order continuous, 22,63, 168 a-regular Riesz subspace, 6,63 a-step function, 178 a-universal completion, 174, I77 a-universally complete Riesz space, 173 Seminorm, 28, 38, 80,92, 133 Semireflexive Riesz space, I59 Separable space, 51.70, 162 Sequentially compact set, 149, 158 Sequential completeness, 50.91, 133. 145 Solid hull, 3, 139 Solid set, 3, 9, 35 Stone space, 175 Step function, 178 Strictly positive linear mapping, I2 Strong topology, 29, 129, 131 Strong unit, 4, see also Order unit Sublinear mapping, 21 Sufficiently many projections, 13, 49 Super Dedekind complete space, 11, 51, 117 Super order dense subspace. 6 Supremum, 2

T Topological completion, 30, 43, 68, 129 Topological dual, 28, 36, 40, 50, 60, 112, 125

Topological vector space, 27 Topologically bounded set, 28, 130 Topologically complete set, 29,45,47,90 Topology absolute weak, 40,62, 105, 127 Fatou, 79, 129, 184 Lebesgue, 53, 81, 88, 118, 128, 146, 159 locally convex-solid, 38, 59, 106, 125, 180 locally solid, 33 Mackey, 29, 163 of pointwise convergence, 126 product, 35,59, I 1 I quotient, 30, 37, 58, 97 a-Fatou, 80, 91, 97, 107 u-Lebesgue, 52,63, 118, 181 strong, 29, 129 weak, 28,41, 126 Totally bounded set, 154 Triangle inequality, 2

U Uniformly Cauchy sequence, 14 Uniformly complete Riesz space, 14, 130 Uniformly convergent sequence, 14, 123 Uniformly continuous function, 29, 34 Unit ball, 38, 80 Universal completion, 174, 175 Universally complete space, I73 Unit, see Order unit Upper element, 102 V Vector lattice. 2 W

Weak order unit, 12, 50, 85, 166 Weak topology, 28.41, 126 Weakly compact set. 130, 132. 148, 152

A

B C D E F 6 H 1 1

8 9 O 1 2 3 4 5