Köthe-Bochner Function Spaces

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To my advisors W.J. Davis and D.R. Lewis

Pei-Kee Lin

Kothe-Bochner Function Spaces

Springer-Science+Business Media,

LLC

Pei-Kee Lin Department of Mathematics University of Memphis Memphis, TN 38152 U.S.A.

Library of Congress Cataloging-in-Publication Data Lin, Pei-Kee, 1952-

K(jthe-Bochner function spaces / Pei-Kee Lin p. cm.

Includes bibliographical references and index. ISBN 978-1-4612-6482-8

ISBN 978-0-8176-8188-3 (eBook)

DOl 10.1007/978-0-8176-8188-3 1. Normed linear spaces. 2. Banach spaces. 3. Vector-valued functions. 4. Operator-valued functions. I. Title. QA322.2.L52 2003 515'.732-dc22

2003063007 CIP

AMS Subject Classifications: Primary: 46B20, 46E40; Secondary: 46B22, 46B28, 46842, 46E30, 28B05

ISBN 978-1-4612-6482-8

Printed on acid-free paper.

©2004 Springer Science+Business Media New York Originally published by Bi rkhliuser Boston in 2004

Softcover reprint ofthe hardcover 1st edition

2004

All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher Springer-Science+Business Media, LLC, except for brief excerpts in

connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to property rights.

9 8 7 6 5 4 3 2 1

www.birkhasuer-science.com

SPIN 10953037

Content s

Preface

vii

Notation

xi

1 Classical Theorems

1.1 1.2 1.3 1.4 1 .5 1 .6 1.7 1.8 1.9 1.10 1.11

Preliminaries . . Basic Sequences . . . . . . . . . . . Banach Spaces Containing f1 or Co James's Theorem . . . . . . . Continuous Function Spaces . . The Dunford-Pettis Property . The Pelczynski Property (V*) . Tensor Products of Banach Spaces Conditional Expectation and Martingales Notes and Remarks . References . . . . . . . . .

1

1 8 29 42 48 57 69 74 81 94 96

2 Convexity and Smoothness

101

3 Kothe-Bochner Function Spaces

143

2. 1 2.2 2.3 2.4 2.5

3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8

Strict Convexity and Uniform Convexity . Smoothness . . . . . . Banach-Saks Property Notes and Remarks . References . . . . . . . Kothe Function Spaces . . . . . . . . . . . . . Strongly and Scalarly Measurable Functions . Vector Measure . . . . . . Some Basic Results . . . . . . . Dunford-Pettis Operators . . . The Radon-NikodYm Property Notes and Remarks . References . . . . . . . . . . . .

101 124 130 137 139 143 162 167 177 188 195 21 1 216

vi

CONTENTS

4 Stability Properties I

4.1 4.2 4.3 4.4 4.5

Extreme Points and Smooth Points . Strongly Extreme and Denting Points . . Strongly and w* -Strongly Exposed Points Notes and Remarks . References . . . . . . . . . . . . . . . . .

5 Stability Properties II

5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8

Copies of Co in E(X) . . . . The Diaz-Kalton Theorem . Talagrand's L1 (X)-Theorem . Property (V* ) . . . . . . . . The Talagrand Spaces . . . The Banach-Saks Property Notes and Remarks . References . . . . . . . . . . .

219

219 226 233 242 245

241

248 257 261 278 290 295 307 309

6 Continuous Function Spaces

313

Index

361

6.1 6.2 6.3 6.4 6.5 6.6

Vector-Valued Continuous Functions The Dieudonne Property in C(K, X) The Hereditary Dunford-Pettis Property . Projective Tensor Products Notes and Remarks. References . . . . . . . . . .

313 326 331 348 355 363

Preface This monograph is devoted to a special area of Banach space theory-the Kothe Bochner function space. Two typical questions in this area are: Question 1. Let E be a Kothe function space and X a Banach space. Does

the Kothe-Bochner function space E(X) have the Dunford-Pettis property if both E and X have the same property? If the answer is negative, can we find some extra conditions on E and (or) X such that E(X) has the Dunford-Pettis property? Question 2. Let 1 :::;

p

:::; 00 ,

E a Kothe function space, and X a Banach space. Does either E or X contain an t'p-sequence if the Kothe-Bochner function space E(X) has an t'p-sequence? To solve the above two questions will not only give us a better understanding of the structure of the Kothe-Bochner function spaces but it will also develop some useful techniques that can be applied to other fields, such as harmonic analysis, probability theory, and operator theory. Let us outline the contents of the book. In the first two chapters we provide some some basic results for those students who do not have any background in Banach space theory. We present proofs of Rosenthal's t'l -theorem, James's theorem (when X is separable) , Kolmos's theorem, N. Randrianantoanina's theorem that property (V*) is a separably determined property, and Odell-Schlumprecht's theorem that every separable reflexive Banach space has an equivalent 2R norm. In Chapter 3 we introduce Kothe-Bochner function spaces and prove several basic results regarding those spaces: •

•

•

We prove that Vitali's lemma and Lebesgue's dominated convergence the orem are still true if one replaces by an order-continuous Kothe function space over a finite measure space.

L1

For any order-continuous Kothe function space E and any Banach space X , the Kothe-Bochner function space E(X) is (weakly) uniformly convex if and only if both E and X are (weakly) uniformly convex. For any order-continuous Kothe function space E and any Banach space X , the Kothe-Bochner function space E(X) has the Radon-Nikodym

viii

PREFACE

property if and only if both E and X have the Radon-Nikodym prop erty. Chapter 4 is devoted to the geometric properties of the Kothe-Bochner func tion spaces. One of the fundamental questions regarding the Kothe-Bochner function spaces is the following: Question 3. Let E be a Kothe function space and X a Banach space. Suppose that I is an extreme point of the unit ball of the Kothe-Bochner function space E X ) . Is II f an extreme point of the unit ball of X for almost all t in support of I?

(

{gL

The answer to Question 3 is dependent on whether there is a measurable selection. It is known that there is a nonseparable Banach space X and a unit vector I in X) such that for all t , II is not an extreme point of the unit x ball of X , but I is an extreme point of X) . On the other hand, Zhibao Hu and Bor-Luh Lin showed that if X is separable, then such a measurable selection exists. Using this technique, we prove that for any 1 < p < 00 and any unit vector I in X ) , if for almost all t in support of I, l(t)/ ll/ (t) llx is a strongly exposed point of X , then there is a linear functional F E (X)) * X) at I. which strongly exposes the unit ball of In Chapter 5 we present several deep results on the Kothe-Bochner function spaces:

Lp (/-t,

Am

Lp (/-t,

Lp (

•

Lp(

( Lp

Bourgain's eo-average Theorem. J. Hoffmann-Jorgensen and S. Kwapien first proved that for any 1 � p < 00 , the Lebesgue-Bochner function space X ) contains a eo-sequence if and only if X contains a eo-sequence. Later, J. Bourgain gave another proof. But there is an advantage to Bourgain's proof, i.e. , it does not require In 's to be strongly measurable. It is known that if E is a separable order-continuous Kothe function space, the dual of the Kothe-Bochner function space E(X) is isometrically iso morphic to the space of weak* measurable X* -valued function spaces. By Bourgain's proof and a classical theorem on Banach spaces, for any order-continuous Kothe function space E and any Banach space X , the Kothe-Bochner function space E(X) contains a complemented copy of .e1 if and only if either E or X contains a complemented copy of .e1

Lp (

•

•

The Diaz-Kalton Theorem. Ultraproduct is an important tool for Ba nach space theory. Using the ultraproduct technique, C. Stegall showed that there is a Banach space with the Dunford-Pettis property whose dual does not have the same property. Modifying Stegall's proof, B.W. Johnson showed that there is a separable Banach space X such that for any other separable Banach space Y, y* is isomorphic to a complemented subspace of X*. In Section 5.2, we use this idea again to show that .eoo (X) con tains a complemented .e1-sequence if and only if X contains .er's uniformly complemented.

ix

PREFACE •

{ fn }�= l '

•

1

Talagrand's L1(X)-theorem. B. Maurey, G. Pisier and J. Bourgain proved that for any ::; p < 00 and any uniformly bounded il-sequence in Lp(X) , 1 < p < 00 , there is contains an i1such that {ink subsequence. Later, M. Talagrand proved that for any bounded sequence such there is an essentially normalized iI-block sequence that for almost all is weakly Cauchy or there is either k (depending on w ) such that is an iI -sequence. Using this result, he proved that for any order continuous Kothe function space E, the Kothe-Bochner function space E(X) is weakly sequentially complete if both E and X are weakly sequentially complete.

t E n,

t E n,

{ }� {gn(tgn(t))}�= =lk

(t)} � l

{ ik } r'= l

{gn }�= l

The Bourgain-Cembranos Theorem. The application of the Ramsey The orem in Banach spaces is an important topic in Banach space theory. H.P. Rosenthal proved that every bounded sequence in a Banach space either contains a weakly Cauchy subsequence or an t'l-subsequence. Recently, W. T. Gower discovered a block Ramsey theorem for Banach spaces. U s ing this technique, he also proved that every Banach space X either con tains a subspace Y with an unconditional basis or contains a hereditarily indecomposable subspace Z; i.e. , every infinite dimensional subspace of Z has no nontrivial complemented subspace. In Section 5.6, we use Ram sey's Theory and give a necessary and sufficient condition such that the Lebesgue-Bochner function space Lp (X) , 1 < p < 00, has the Banach Saks property.

Chapter 6 is devoted to C(K, X) and X ®Y. Some interesting results are proved in this chapter. •

The Cembranos-Diestel-Elton-Knaust-Odell Theorem. In Section 6.3, we apply Ramsey's Theorem again and we prove that a Banach space X has the hereditary Dunford-Pettis property if for any weakly null sequence in X , either is a null sequence or contains a eo-subsequence. Using the result, we prove that C(K, X) has the hered itary Dunford-Pettis property if and only if both C(K) and X have the hereditary Dunford-Pettis property.

{Xn}�= 1

•

{xn }�=l

{xn }�= l

The Bu-Diestel Theorem. One question about Banach spaces is: 4. Let X and Y be two Banach spaces with the Radon Nikodym property. Does the projective tensor product X ®Y have the Radon-Nikodym property? If the answer is negative, can we find a non trivial condition on X and Y such that X®Y has the Radon-Nikodym property?

Question

J. Bourgain and G. Pisier showed that there is a Banach space X with the Radon-Nikodym property such that X0X contains a copy of eo (so X®X does not have the Radon-NikodYm property) . In the second part of Section 6.4, we prove that if E is a reflexive Kothe function space and X

x

PREFACE

is a Banach space with the Radon-Nikodym property, then the projective tensor product E@X has the Radon-Nikodym property. To help the reader keep track of what is known, we have adopted the no tation (Problem 1) or (Problem 3.2.5) for solved problems and (Question 1) or (Question 3.2.5) for open questions. The author thanks his friends and colleagues who pointed out and corrected some mistakes and typos in the preliminary versions. These include, in par ticular, H. Hudzik, J. Jamison, N.J. Kalton, A. Kaminska, E. Odell, N. Ran drianantoanina, and Huiying Sun. Particularly, he owes very special thanks to J. Diestel and David Kramer, who read through our manuscript and offered suggestions and corrections (for our mistakes and typos) and M. Talagrand who explained the idea of the extension of his (X)-theorem so that a correct proof could be presented. We also like to thank to the referees and the Executive Editor of Birkhiiuser Boston, Mathematics and Physics, Ann Kostant, for their suggestions and comments. Of course, the author is responsible for any mistakes in this book.

L1

Pei-Kee Lin October 24, 2003

Not at ion We shall stick to standard definitions and terminology. N, Q, IR, C will be the systems of natural, rational, real, and complex numbers. For an infinite subse quence M of N, [M] and [M]< denote the sets of all infinite subsequences and all finite subsequences of M, respectively. Throughout this text, X, Y, Z, . . . will be Banach spaces; typical members will be x, y, z, . . . respectively, perhaps with indices. E, EI will be Banach lattices (or Kothe function spaces); typical members will be g, h, respectively, perhaps with indices. The norm of X will usually be denoted by II . II, but when more precision is desirable, we may use II . Ilx. B(X) will denote the closed unit ball of X, whereas S(X) will be its unit sphere. For any x E X and any positive real E, B (x, E) denotes the closed ball with center at x and radius E. The dual of X is denoted by X*. Its typical member will be denoted by x* , and for any x E X, we shall write (x* , x) (or (x, x* ) ) for the action of x* on x. For a nonempty subset A of a Banach space X, [A] denotes the closed linear span of A. Let E be a Kothe function space over a measure space (0., /-l) and X a Banach space. The Kothe-Bochner function space E(X) is the set of all strongly measurable X-valued functions f such that IlfOllx E E. Its typical members are denoted by f, iI, h. Suppose that E is order-continuous over a measure space (0, /-l) . It is known that the dual E* is also a Kothe function space (we denote its typical members by H, G, HI)' Suppose that p � 1 . We shall denote the set of all p-integrable functions by C , and the Banach space of all p-integrable functions by It is known that if (0, /-l) is a cr-finite measure, then there is linear mapping (lifting) p from Coo (/-l) to oo (/-l) such that for any f E Coo (/-l), p (J) f a.e. and p( af + g) = ap(J) + p(g). Using this result, we shall show that if E is an order-continuous Kothe function space over a cr-finite measure and X is a Banach space, then for any F in (E(X) ) * , there is a weak* measurable X*-valued function FI such that (F,1) (Fl(W), f(w))d/-l(w),

p

Lp.

L

=

=

J

and IIFII(E(x))* IIFI(-)llx*IIE*' In this case, we denote the dual of E( X ) by E* (X, w* ) . A typical member of E* (X, w*) is denoted by F. Let {Xj } jEJ be a family of Banach spaces. =

xii

NOTATION

Let X and Y be any two Banach spaces. We shall denote the set of all operators (respectively compact operators) from X to Y by .c(X, Y) (respec tively K(X, Y)). If X is a dual space, then .cw• (X* , Y) is the set of all weak* to weakly continuous operators form X to Y.

Kothe-Bochner Function Spaces

Chapter

1

C lassical TheoreIns Recall that a Banach space is a complete normed space. In this book, we always assume that X is an infinite-dimensional Banach space. For completeness, in this chapter, we provide some basic results on Banach spaces that we will use in this book. Most of them can be found in the books [8, 12, 17, 18, 21, 28, 44, 73, 79] and the two survey papers [19, 72].

1.1

Preliminaries

For a Banach space, we denote the closed unit ball and the unit sphere of X by B(X) and S(X) , respectively. The dual X* of a normed space X is the set consisting of all bounded linear functionals on X. It is known that the dual of any normed space is a Banach space. Let Y be a subset of X* that separates the points in X ; i.e. , for any X l =1= X2 in X, there is y E Y such that (y , X l ) =1= (y, X2 ) . The topology O" (X, Y) denotes the topology generated by the collection of the following sets:

where Xo is an element in X , and YI , . . . , Yn are finite elements in Y. If Y = X* , then the topology O" (X, Y) is called the weak topology of X . If X is a dual space and if X = Y* , then the topology O" (X, Y) is called the weak* topology of X. X is said to be reflexive if X = X** . The following is a useful observation about weakly null sequences. Fact 1 . 1 . 1 . Let X be a Banach space, and {Xn};::'= l a bounded sequence in X that is not weakly null. Then there are X* E S(X*), f3 > 0, and a subsequence {Yj}�l of {xn}�= l such that for any j E N ,

2

CHAPTER

1.

CLASSICAL THEOREMS

{aj } j = l with :LT= l aj I jt= l aj Yj I

So for any finite nonneg ative sequence

�

=

1,

(3.

Theorem 1 .1.2. (Hahn-Banach Theorem) Suppose that

(1) X is a vector space and Y a subspace of X; (2) p : X

-7

R

satisfies

p(x + y) � p(x) + p(y) and for all

Q

p ( Qx ) = Qp(

x)

� ° and x, y E X;

(3) h is a linear functional defined on Y of X such that -p( - y)'� h(y) � p(y) for all y E Y. Then there is a linear functional 9 defined on X such that -p( - x) � g(x) � p(x) for all x E X, and g(y) = h(y) for all y E Y .

Recall that a subset A of a Banach space X (respectively, a dual space X = Y*) is said to be weakly bounded (respectively, weak* bounded) if sup{ l (x * , x) : x E A and x * E B(X*) (respectively, x* E B(Y))}

Proof: By passing to a subsequence if necessary, we may assume that there is € > 0 such that > 2€ > 0 for all = is Since Let K be the basis constant of and set a Schauder basis, there is an integer such that � 8� ' such that Set = = 0, there exists Since limk�oo f. such that ::; 82f.K' and ::; 82 K· Select set Then select k3 such that < 83f.K· Continuing in this way, we construct a block basis of Note:

k E N. {en} �= 1, Ykt Y l . {en } �= 1 PI 1 2: :'=pt+ 1 an , l en l k2 E N U l 2:��1 an , l en ' an , k P2 1 2:��1 an , k2 en l 1 1 2: :'=P2 an ,k2 en l U2 2:��+ 1 an , k2 en. 1 2:�:' 1 an , k3en l {Ui }� 1 {en }�= I ' 00 00 jL= 1 I Ykj - Uj l < jL= 1 23k� I K < 3� ' I Yk l 1

=

19

1.2. BASIC SEQUENCES

{ Ykj } � 1 is a basic sequence that is equivalent to the {Uj } 1 {en }�= I '

By Proposition 1 .2. 14, block basis � of

0

Proposition 1 .2.18. Let X be either Co or fp, 1 � p
" Thus for any n E N, Yn(c) 1 1 >.. (m - 8(1 + >..) ) 1M>"_ >.. > 1 (m - 28 - M >.. ) -> m - 28. 1 - >"

Note that = and for all n (1 D. By our assumption, there is C such that So for any n we have

We have

sUP c' E C

z (c').

=

�

c' E C

c' E C

-

�

-

_

--

Hence

u(C) limn sup xn(c) =

..... oo

��

n Yn(c) � m - 28.

lim sup

The proof is complete.

..... oo

o

Let C be a closed, convex, and bounded subset of a separable real Banach space X . Then the following are equivalent: ( 1 ) C is weakly compac t. (2) Every functional x* attains it supremum on C. In particular, a separable Banach space X is reflexive if every x* E X* attains its supremum on the unit ball of X . Theorem 1 .4.5. ( James ' s Theorem )

Proof: ( 1 )

=}

(2) is obivious. Conversely, assume that C is not weakly compact. Then there is y** in cow* (C) \ c. By the separation theorem, there are y *** X*** and a > 0 such that I l y*** I I = 1 and

E

( y *** , y ** )

E

> a > sup { ( y *** , y ) : y C} .

CHAPTER 1. CLASSICAL THEOREMS

48

N}

Let {xn : n E be a norm dense subset of C. Since y*** is in the weak* closure of B(X*), for any n E there is x� E B(X*) such that ( y ** , x� ) > Q and for any 1 :::; k :::; n , I (x� y *** , xk) � . So for any k E N, lim (X� , X k) = ( y *** , X k) . n-+oo

N

-

Let

D

1

for all n, L:�= I an = 1 , and M

an l (x * , xn +N) 1 1 < & IL n= 1

for all x * E K.

Theorem 1 .5.3. ( Dieudonne-Grothendieck )

For any compact Hausdorff space K, let B be the set of all finite Borel measure on K. For any bounded subset K of B ( K ) , K is relatively weakly compact if for any sequence {On}�= 1 of mutually disjoint open subsets of K, nlim ->oo sup { I J.l ( On) 1 : J.l E K} = O. Proof: Suppose the theorem is not true. By the Eberlein- S mulian theorem,

K

{v }�

we may assume that is a sequence n = 1 of regular Borel measures that does not contain any weakly convergent subsequence. We claim that for any E > 0 and any compact set � K, there is an open set of K containing for which

C

U

C

for all n E N. Suppose that the claim is not true. Then there are a compact set and an E > 0 such that for any open set containing there is ( depends on in such that > E. We shall construct two sequences of open subsets that satisfy the following conditions:

Co V) K I v l ( V\Co) ( Un)�=o ' (Vn)�=o ( ) Vo = K and Uo = 0. a

(b) For any n E N, there is

Assume that

V

Co,

v

kn such that I Vkn ( Un) 1 � 2� '

{UO , " " Un - I} and {VO , " " Vn - I} have been constructed and

{kl' . . . , kn-I} is selected. By assumption, there is kn such that

51

1.5. CONTINUOUS FUNCTION SPACES

lIkn ' there a Borel set F and a compact set Cn Cn � F � Vn - 1 \ Co, I lIkn (F) I � '5E ' I lIkn I (F \ Cn) < 40E '

By ( 1 .5 ) and the regularity of such that

Therefore,

I lIkn (Cn) 1 � I lIkn (F) I - l lIknl (F \ Cn) � 403E ' Notice that Co and Cn are two disjoint compact sets. There are two disjoint open sets WI and W2 that contain Co and Cn, respectively. Again the regularity of lIkn allows us to assume that I lIkn l ( W2 \ Cn) ::; 4€0 ' Let Vn = WI Vn - 1 and Un = W2 Vn - 1. Then n

n

{Un : E N}

The construction is complete. Note: (c) implies that n are mutually disjoint open subsets of K. From (b) , we get a sequence of mutually disjoint open sets such that for all n

{Un } ;:O= 1

E N,

a contradiction. We have proved our claim. Let 0 be any open subset of K. Then = K \ 0 is a compact subset of K. By the claim, for any E > 0, there is an open set => such that

C

U C l J-Ln l ( U \ C) < E for all n E N .

Let

C1 = K \ U. Then we have

U

We have proved that for any E > 0 and any open set � K there is a closed set of K contained in for which \ ::; E for all n Let II be the measure defined by II = � I

C

U

I lIn l ( U C) lin I ' n � 2 l lln l

E N.

II for all n, by the Radon-Nikodym theorem, there is a sequence { in } ;:O=lin1 of integrable functions (with respect to the measure lI) such that for any Borel set A and any n E N, lIn (A) = i in dll. By Lemma 1 .3.10 and the proof of Theorem 1.3.12, {lin : n E N} is relatively weakly compact if and only if { in : n E N} is uniformly integrable. So there

Since

«

52


{ gn}�=l of { fn }�= l' and a sequence {Bn}�= l of Borel E N,

are E > 0, a subsequence sets in n such that for all n

lI(Bn) ::; 2n+1 1 and is regular. For any n E N, there is an open set Un Bn such that n

Note: 11 lI(Un) ::; 1/2 . Let

E N}

2

00

m=n

Then {Vn : n is a decreasing sequence of open sets that satisfy the following conditions: (d) limn -+oo lI(Vn) = o. (e) Jv,n E. By our claim, for each n there is a compact set � Vn such that for any E ::; n + 1 . 2 JVn \Cn C Vn . Then Let = is a decreasing sequence of � compact sets such that � Vn . So is a compact set of lI-measure Applying our claim to we obtain a closed set of such that for any

I gn l dll �

E N, Cn k E N, r I gk I dll Fn nk= l ck Cn {Fn }�= l n�= l Fn n�= l n�= l Fn O K\n�= l Fn' C K \ n�= l Fn . k E N, JrK\n':'= lFn \C I gk l dll -=-4 . But lI (n�= l Fn) = O. We have JrK\C I gkl dll = JrK\n':'= lFn\C I gkl dll � for all k E N . Note: K \ C is an open set containing n�= l Fn' and {Fn}� l is a decreasing sequence of compact sets. There is an such that Fm � (K \ C). Thus ::;

::;

m

53

1.5. CONTINUO US FUNCTION SPACES

o a contradiction. The proof is complete. For any topological space A, we denote the set of accumulation points of A by A' . A topological space A is said to be perfect if A = A'. A topological space A is said to be dispersed (or scattered) if it contains no nonempty perfect subset. If is an ordinal, the topological derived set of order denoted by A ( a) , is defined by transfinite induction as follows:

Q

Q,

A (O) = A , A ( a+ 1 ) = (A ( a» )' ,

Q

and A ( a) = n A ( {3) {3 < a

Q

if is a limit ordinal. Let be an ordinal number. It is known that any nonempty subset A of has a minimal number 'Y and 'Y � A'. So A(a+ 1 ) = 0. Lemma 1 .5.4. is a compact For any ordinal Corollary

[ 1, Q] Q, [0 , Q] 73, 8. 6 .7] [ dispersed space under the order topology. Lemma 1.5.5. [ 7 3, Theorem 8. 5 . 2] A topological space A is dispersed if and only if there exists an ordinal number Q such that A ( a) = 0 .

Proof: Since the derived set of any set is closed, for any ordinal {3, A ( {3+ 1 ) � A({3) . So there is such that A(a+ 1 ) = A ( a). Clearly, P = A(a) =1= 0 is dense in itself. Hence if A({3) =1= 0 for all ordinals {3, then A is not dispersed.

Q

Conversely, suppose that A is not dispersed. Let C be a nonempty perfect subset of A. Then for any ordinal (3 , 0 =1= C � A({3) . The proof is complete. 0 It is known that for any continuous function f, any compact set A , and any ordinal f ( A(a» ) � f ( A ) (a). We have the following corollary. Corollary 1.5.6. (Pelczynski and Semadeni Let f -is a continuous

Q,

[60])

function from a compact set A onto another compact set. Then for any ordinal Q , f (A(a» ) :2 f (A ) (a) . Hence every continuous image of a dispersed space is dispersed. Let A be a topological space. Recall that A is said to be a-dimensional if it has a base of open-and-closed sets. The topological space A is said to be totally disconnected if for any pair of distinct points {x, y}, there is an open and-closed set 0 such that x E 0 , but y � O. It is known that the Cantor set C is homomorphic to the space {a, l} N under the product topology. So every a-dimensional Hausdorff space is totally disconnected, and the Cantor set is a-dimensional. The following theorem shows that every totally disconnected compact Hausdorff space is a-dimensional.

Theorem 1 .5.7. Let A be a compact Hausdorff space. Then A is totally disconnected if and only if it is a-dimensional. Proof: Clearly, if a topological space A is a-dimensional, then A is totally

disconnected. One direction is plain. Assume that the topological space A is totally disconnected. Let 0 be an open set containing a . We need to show that

54


C A\

there is an open-and-closed 01 such that a E 01 � O . Let = O. Since is totally disconnected, for any x E there is an open-and-closed Ox such that x E Ox but a ¢ Ox ' Note: is a closed subset of a compact set, and so is compact. So there are Xl , . . . , Xn E such that

C, C C

A

C

A\

Let O2 = Ui= 1 0x i and 01 = O2. Then O2 is open-and-closed, and so 01 is 0 also open-and-closed. The proof is complete. Lemma 1 . 5 .8. (Pelczynski and Semadeni [60] )

Let K be a compact Haus

dorff space. The following are equivalent: ( 1 ) K is dispersed. (2) There is no continuous surjection from K onto [0, 1] . (3) dim(K) = 0, and there is no continuous surjection from K onto C . Proof: (1)

=>

(2).

It is known that [0, 1] is not dispersed. By Corollary 1.5.6, if K is dispersed, then there is no continuous surjective mapping from K onto [0, 1] . Suppose that K is not O-dimensional. Then there are two distinct => points x, y E K that cannot be separated by open-and-closed set. Note: K is normal. There is a continuous function 9 from K to [0, 1] such that lg(x) ° and g( y ) = 1 . If there is r E (0, 1) such that r ¢ g(K), then 0 = g - ( [O, r] ) is an open-and-closed subset such that x E 0; but y ¢ O. We get a contradiction. Thus K is O-dimensional. Suppose that there is a continuous surjection from K onto C . Define a function 9 from C to [0, 1] by

(2) (3)

=

i' L i l 2 00

g( 8 ) =

"" S i =

where 8 = ( S i ) E {O, 1 } N .

It is easy to see that 9 is a continuous surjection from C onto [0, 1]. So there is a continuous surjection from K to [0, 1] ' a contradiction. is not dispersed. Then K => (1). Suppose that K is O-dimensional and contains a nonempty perfect set P. Since P is nonempty, there are two distinct points x , Y E P. But K is O-dimensional. There is an open-and-closed Po such that x E Po, but y ¢ Po. Let P1 = K Po . Then Po n P and P1 n P are nonempty perfect subsets of K. Let s be a finite {O, I} sequence. Suppose that Ps is an open-and-closed set such that Ps n P i= 0. Note: Ps n P is a nonempty perfect set. It contains at least two points. Applying the above argument, we obtain two disjoint open-and-closed sets PsU{ O } and PSU{ l } such that

(3)

K

\

Ps U{ O } U PSU{ l } Ps PSU{ O } n P i= 0 i= Ps U{ l } n P. =

55 Then for any x in K, there is ( S i ) E { O , l} N such that for all n E N, x E P( S l ) Define : K C by g(x) = ( Si ) if x E P(S l , . . . , Sn) for all n E N . 1.5. CONTINUOUS FUNCTION SPACES

, . . . ,s ..

·

9

---+

o Then 9 is a continuous surjection from K to C, a contradiction. Since [0, 1] is uncountable, there is no surjection from a compact countable Hausdorff space to [0, 1] . We have the following corollary:

Corollary 1 . 5.9.

Every compact countable Hausdorff space is dispersed.

The following theorem is due to S. Banach and S. Mazur. Theorem 1 . 5 . 10.

If X is a sepamble Banach space, then X is isomorphic

to a subspace of C[O, 1] .

Proof: Let X be a separable Banach space. Then the weak* topology

on B(X*) is metrizable. Hence there is a continuous mapping (B(X* ) , w* ) [41 , p.166] . Let

Q from C onto

Y = { f E C(C) : f(s) = f(t) for any t , S E C with Q(s) = Q(t)}. Then Y is isometrically isomorphic to C(B(X* ) , w* ) , the set of all weak* con

tinuous functions on the unit ball of X*. Hence every separable normed space is linearly isometric to a linear subspace of C(C) . Note: C is homemorphic to a closed subset of [0, 1] . Extending each continuous function on this subset lin early across each open interval of the complement gives a linear isometry of C(C) into C([O, 1] ) . Thus every separable normed linear space is linearly isometric to 0 a linear subspace of C([O, 1] ) . Remark 1 . 5 . 1 1 . A. Pelczynski and Z. Semadeni [60] showed that for any

compact Hausdorff space K, K is a dispersed if and only if for any separable subspace X of C(K) , X* is separable and if and only if C(K) * is isometrically isomorphic to I! 1 (r) for some r. The following theorem is due to Mazurkiewicz and Sierpinski [47] .

Theorem 1 . 5 . 1 2 . (Mazurkiewicz-Sierpinski Theorem) Let K be a first countable compact dispersed space. Then K is homeomorphic to a countable compact ordinal 'Y. Hence if K is a compact countable metric space, then K is homeomorphic to [ l , w . m] for some countable ordinal 'Y and some m < w. '

is a compact dispersed first countable space. Let f3 be A the small ordinal such that A (!3 ) = 0 . Clearly, f3 cannot be a limit ordinal (so f3 is a com pact ordinal) . There is such that f3 = Cl' + 1 . Since A (Q ) is a nonem pty compact set without any accumulation points, it is finite, say it consists of m elements. The pair (Cl', m) is called the chamcteristic system of A . We need to show that A is homeomorphic to [1 , wQ . m] . Proof: Suppose

Cl'


56

We claim that if the theorem is true for a certain pair (a, 1 ) , then it holds for each pair (a, m) for m E N. Suppose A(Q ) consists of m points X l , " " xm . Since X is compact and O-dimensional, there exist disjoint open-and-closed sets 0 1 , . . . , Om such that Xi E Oi and U�1 0i = A. Then for any j � m, we have O)Q ) = {Xj } ' By assumption, OJ is homeomorphic to [wQ (j - 1) + 1 , wQ . j] . Therefore, A is homeomorphic to [1 , wQ . m] . Now assume that the theorem is true for each pair (-y, m) where "( a and m E N. We shall prove that it is true for the pair (a, 1). Let {xo } = A(Q ) . First, we note that if 0 is an open-and-closed subset of A, then for any ordinal f3,

.. : >. E A } be a collection of measur able functions, and for any >. E A, let �>.. be the O"-algebra generated by the set {g� l (O) : 0 is an open subset of lR}. The sequence {gn}�= l is said to be independent if the E>.. are independent. It is known that for any two integrable independent functions g, h, if 9 . h is integrable, then j=l

in 9 . h dlL = in 9 dlL . in h dlL·

Theorem 1.9.2. (Borel-Cantelli lemma) Let ( IL , O) be a probability space and {An}�= l a sequence of measurable subsets.

83

1. 9. CONDITIONAL EXPECTATION AND MARTINGALES

If'L::= l P, (An ) < 00 , then limn-+oo P, (U�=m Am ) = O. (2) If { A n } �=l is independent and if L:�=l P,(An) = 00 , then for any m E N, (1)

n=m Proof: Suppose that L:�=l P,(An) < 00. Then We have proved (1) . Proof of (2) . Suppose that { An } �= l is independent. Then for any n 2: m, n n p, ( n (n \ Ak») = IT p, (n \ Ak ).

k=m

k=m

def

It is known that for any t 2: 0, 1 - t � exp( - t) e -t . Hence n n 00 p, ( n (Q \ Ak ) ) � }�� IT p, (n \ Ak ) � }�11Jo exp ( - L p, (n \ Ak ) )

= O. =k m =k m This is equivalent to p, ( U�= m An ) 1 . The proof is complete. Let { E n } �=l be an increasing sequence of O"-algebras and { fn }�=l a sequence of E n -integrable functions. The sequence {fn } �=l is said to be a martingale (re spectively, submartingale) if fn = £ ( fn l l E n ) (respectively, fn � £ ( fn l I E n » . By Jensen ' s inequality, we have the following theorem:

k=m

o

=

+

+

Proposition 1.9.3. Let cP be a real-valued (increasing) convex function on lR and (fn I En) a (sub) martingale. Suppose that cP 0 f is integrable for all n E N.

Then (cP 0 fn I E n ) is a submartingale. Lemma 1 .9.4.

gale in

Ll(p,).

where hj

=

(Maximal Inequality) Let ( h n I E n ) be a bounded submartin

Then for any c > 0,

{

}

J

p, sup1 hj > c � � sup1 hj dp, j 2: j 2: C

hj V O.

Let Ao = 0 and An = {t : hj (t) > c for some j � n } . Then { An } �=l is an increasing sequence and Proof:

{

lim P, (An) = p, sup1 hj > c n-+oo j 2:

}

.

84

CHAPTER 1 . CLASSICAL THEOREMS

Note: An \ A n - 1 is I; n -measurable and for any t E An \ An - I , hn (t) > c. Hence for any m � n, This implies that

n I1 (nU=l An ) = }�� kL=l I1 (Ak \ Ak - d n1 1 hn dl1 � - nlim -+oo k�=l � ! nlim -+ oo J h� dl1 � ! sup n� l J h� dl1. 00

c

�

Ak \Ak - l

C

c


o

{I;n }�=l be an increasing (J-algebra, and I;oo the (J -algebra generated by U�=l I;n. For any I; -integrable function f , the sequence { fn = £ (f I I; n) : n E N} converges to f a. e., and { fn }�=l converges to f in L1• Theorem 1.9.5. (P.

Levy ) Let

00

Proof:

Let

{ L1(11, I;oo) : £ (f I I;n) converges to f a.e. , and £ (f I I; n) converges to f in L1 } . For any g E L1 (11, I;n), £ (g l I; m ) = g for any n . Hence U L1 (11, I;n) � D, n=l and D is a dense subset of L 1 ( 11, I; oo ). Claim 1 . We claim that for any element f in L1 (11, I; oo ), { £ (f I I; n)}�=l converges to f in L 1 ( 11, I; oo ). For any E 0, there is g E L 1(11, I;n) for some n such that I l f - g i l l < �. Then for any k n, I l f - £ (f I I;k ) l h � I l f - g i l l + I l g - £ (g l I; k ) 1 1 + 1 £ (f - g l I; k ) 1 1 � 2 1 g - f i l l < E. We have proved Claim 1 . Claim 2. We claim that for any element f in L1(11, � oo ), { £ (f I I; n) }�=l converges almost everywhere. Given E and 8 0, there is g E D such that D= fE

m

�

00

>

>

>

°

>

85

1.9. CONDITIONAL EXPECTATION AND MARTINGALES

I I - g i l l � �EJ . Then for all t E n, I £ (f l �n)(t) - £ (f l � k )(t) I � I £ (g l �n)(t) - £ (9 1 � k)(t) 1 + I £ (f - g l �n)(t) 1 + I £ (f - g l � k) (t) 1 � I £ (g l �n)(t) - £ (9 1 � k )(t) 1 + 2 sup £ ( 1 1 - g l l �j )( t). j� l Let hj = £ ( I / -g l l �j) . Since ( hjl �j ) is a nonnegative submartingale and 9 E D, the maximal inequality implies that suPI £ (f I �n)(t) - £ (f l �k )(t) I > E } IL { t : limn , k-+oo � lL { t : 2 sup hj ( t ) > E } � � sup I hj l 1 E '>1 '> 1 = -2 1 I /(t ) - g (t) 1 dlL (t ) = -2 I I - g i l 1 < J. E En J_

J_

E J £ (f l �n) converges almost ev � l ( �n) ( �oo) £ (f l � n) converges to I in I. Theorem 1 .9.6. II (fn l �n) is an L1 -bounded martingale, then { In }�= l converges to a finite limit a. e.

Since both and are arbitrary, it follows that erywhere to a measure function J. , and is dense in L1 1L, Note: U = L1 1L , L1 • This implies j = The proof is complete.

0

{oo} (write for if necessary) if I /i ( t ) 1 < a for i � n - 1 and I /n(t ) 1 � a, (t ) { : if I /i (t ) 1 < a for all i E N. Note that for any k E N, ( = k) E � k . Claim 1 . (fu l\ k l � k) is a martingale. First we note that for any A E �n, A ( > n ) = A \ ( � n ) E �n ' Hence if A E �n' then j Iul\n dlL = j (u n) Iul\n dlL + J (u> n) Iul\n dlL =J (u n) IUI\ (n+ 1 ) dlL + J (u > ) In dlL =J (u n) IUI\(n+ 1 ) dlL + J (u > ) l(n+ 1) dlL = i u l\ n+ 1 dlL· Proof: Fix a > 0 and define

as follows:

CI

CI :

n

-+

CI a

NU

=

CI

n

CI

CI

A

An

�

An

�

An

n

An

�

An

n

I

We have proved claim 1 .

An

CI


86

h : t - sUPk> l I fO'A dt ) I is integrable. From the oo I fO'A k( t) 1 o- (t) < and h (t) a if o- (t) =

Claim 2. The function if definition of = lim k ...... Hence � a . By Fatou's lemma,

h, h 1( 0'=00) h (t) 1(0'< 00) h dJ-L = 1( 0'< 00) klim......oo I fO'Ak (t) 1 dJ-L(t) lim k ......infoo j( 0'< 00) I fO'Akl dJ-L limk......infoo irn I fO'Akl dJ-L. It remains to estimate In I fO'Ak l dJ-L. Notice that ( I fn i I �n ) is a submartingale. 00,

00 .

�

�

�

Thus

and

r

r dJ-L.

hdJ-L � a + j � l in I h l in sup

We have proved claim 2. �n. We claim that for Claim 3. Let be the o--algebra generated by sequence. Cauchy a is any A E { IA Let € > O. Since is integrable, there is a > 0 such that Ie < � such that A 6. < Then for whenever < Choose N and E N any

k�

� oo � oo , fO'Ak dJ-L} � l h J-L(C) & .

For any A E

�oo , let v(A) =

C �N

8

klim ...... oo 1A fO'Ak dJ-L.

U�=l

J-L( C) 8.

h dJ-L

1 . 9. CONDITIONAL EXPECTATION AND MARTINGALES

87

It is clear that 1I is finitely additive. Note that for any A E Eoc , lI(A) � fA h d/1. For any decreasing sequence { Ai }� 1 in E oo with n� I Ai = 0,

This implies that 1I is absolutely continuous with respect to /1 . By the Radon Nikodym theorem, there is an integrable function 9 such that 9 = �� . Since the function fUA n is E n-measurable and ( f A n I En ) is a martingale, it follows that for any A E En , u

r r fUA n d/1 = klim r -+ oo }A fUA k d/1 = lI(A) = }A g d/1.

}A

Consequently, E (g I E n ) = f A n By Theorem 1.9.5, {fUA k}k= 1 converges almost everywhere to g. Now let a vary (a sequence approaches infinity). By the maximal inequality (applied to the sequence { I fk I lk':" I ) , /1({t : O"a (t) : O"a (t) = approaches 1 as a fn (t) when O" approaches infinity. Note that a fU A n (t) = a (t) = 00 . We have proved that {fn}�= 1 converges almost everywhere. The proof is complete. D The proof of the following theorem is similar to the proof of Theorem 1 .9.6. The proof is left to the reader. Theorem 1 .9.7. If (fn I En) is a nonnegative L1-bounded submartigale, then {fn}�= 1 converges to a finite limit a. e. Remark 1 .9.8. For any 9 E L I , let g + = g V O and g - g V O. Let {gn}�= 1 be an L1-bounded martingale. Then {g;t}�= 1 and {g; }�= 1 are nonnegative L l -bounded submartingales. By Theorem 1.9.7 and Fatou's lemma, {g;t}�= 1 and {g; }�= 1 converge almost everywhere to hI and h 2 , respectively, so that u

'

oo} )

=

This implies that {gn}�= 1 converges to h I - h2 a.e., and The following theorem is due to Koml6s: Theorem 1.9.9. (KomI6s's Theorem) For any bounded sequence {gn}�= 1 in L1 (/1), there is a subsequence {gn Jk:: l of {gn }�= 1 such that for any further subsequence {hj } �1 of {gn k }k= I ' ! 2:;: 1 hj converges a. e. Before proving Koml6s's theorem, we need the following three lemmas. Re call that two sequences of measurable functions {fn}�= 1 and {gn}�= 1 are said to be equivalent if 00 L /1(fn =1= gn) < 00 . n= 1

88


{ fn }�= and { }�= are two equivalent se E�= 1 fn - gn converges a. e.1 Hencegn if limn1 --+oo an = then n 1 nlim--+oo -an j� � = 1 (fj - gj ) = 0 a. e.

Lemma 1.9. 10. Suppose that

quences. Then

Proof: Since

00,

{ fn }�= 1 and {gn }�= 1 are equivalent, by the Borel-Cantelli

lemma, for any n E N,

nl�� p (kU=n { fk =1= 9k } ) = O. 00

This means that there exists a null set A such that for any such that for n N . So if ¢:. A, then

fn(w ) = gn(w)

w � nL= 1 (fn(W) - gn(w ))

W ¢:. A, there is N

00

o converges. The proof is complete. Recall that a sequence in L l is said to be a martingale difference sequence ( m.d.s. ) if there is a martingale such that for any n, h + is a martingale difference sequence in L2, then for any n,

{ gn }�= 1 gn = n l - hn. If {gn }�= 1

{ ( hn l �n) }�= 1

L n 9j 9k ) dp I l kt= 1 9k l : = J (kt= 1 I gk l 2 + 2 l �j 0, we have j

Let

::;

J -> OO

j

::;

=

::;

::;

1 . 9. CONDITIONAL EXPECTATION AND MARTINGALES

Thus for any j �

93

nk and i k, �

(1 .9) The construction is complete. Step 7. Let = and E O 1. We claim that the conclusion of the theorem. For any increasing sequence

Then

{gnk }k:. l ' and h satisfies �o {0, O} = {kj }�1 ' let 7]j = S€j_l / 2j (Tj (gnkj ) - hj) , ¢j = 7]j - t' (7]j l �kj - l ) '

{ ¢j }�1 is a martingale difference sequence. Since by Step 2

we have

f== l 1 ¢!21 § f== l 1 7]�dl § f==l I Tj (g�kj ) I §

j

J

� 4

j

J

� 8

j

J

�

96

by (a). By Lemma 1.9.11 and Kronecker's lemma, we have

nlim� oo j� �=l ¢� converges almost everywhere, 1 n nl!..� -;; jL =l ¢j = 0 a.e. J

Therefore,

=h

a.e. (by (1 .9)) .

(1.10)

94


The proof is complete. By Koml6s's theorem, we have the following theorem: Theorem 1.9.13. (Strong Law of Large Numbers) Let (J,L, 0)

be a probabil ity space and { ( hn l � n)}�= 1 a sequence of independent and identically distributed integrable functions . Then {�L:�= 1 hk }�= 1 converges to almost everywhere. Moreover, { � L:�=1 hi } :'= 1 also converges in L1-norm. Proof: By Koml6s's theorem, there is a subsequence {gn}�= 1 of {h n }�=1 N} such that { � L:�= 1 gk } :'= 1 converges almost everywhere. But {gn and {hn E N} have the same distribution, and { � L: �= 1 h k } :'= 1 converges almost everywhere. Let h be the (pointwise) limit of {� L: �=1 h k } :'= 1 ' Then h is integrable. Note: {gn : E N} is uniformly integrable. So { � L: �=1 gk : E N } is also uniformly integrable. By Vitali's lemma, we have that { � L: �= 1 h k } :'= 1 converges to h in ( L d norm. The proof is complete. Remark 1.9. 14. Let (J,L, O) be a probability space, and let { (hn}�= 1 be a sequence of independent and identical distributed random variables. Let = J h I. It is known that the sequence {�L:�=1 h k }�= 1 converges to almost m

:

:

n

n E

n

n

0

m

everywhere.

m

Exercises

Exercise 1.9.1. Prove Theorem 1.9.7. Exercise 1.9.2. Recall that a Banach space X is said to be an

Sp-space if

uniformly complemented, i.e., if there is a A � 1 such that for N, In C(C�, X) and Pn E C(X, C�) such that

X contains c� 's every n E there are operators

E

Suppose that X is an Sp-space. Show that (L: �= o EBCr ) oo contains a comple mented subspace that is isometrically isomorphic to Lp[O, 1] .

1.10

Notes and Remarks

A. Szankowski proved that if X is a separable Banach space such that every closed subspace Y of X has a basis, then X is of type p and cotype q for all p < 2 < q. (For definitions of type and cotype, see Section 3.4.) On the other hand, it is known that every subspace of a Hilbert space is another Hilbert space. So every separable subspace of a Hilbert space has a basis. Johnson [37] showed that every subspace (even every subspace of every quotient space) of the convexified Tsirelson space has a basis. Later, Nielsen and Tomczak Jaegermann [49] showed that any weak Hilbert space (for definition, see [65]) X

T

95

1 . 1 0. NOTES AND REMARKS

that is a Banach lattice has a basis. Pisier [64] proved that every weak Hilbert space has the approximation property. Casazza, Garda, and Johnson showed that there is a Banach space X that is of type p for any p < 2 and cotype q for any q > 2, but X fails the approximation property. In [38, p.277] ' Casazza asked the following question: Question 1 . 1 0. 1 . Let X be a separable Banach space such that every closed subspace Y has a basis. Is X a weak Hilbert space ? The following problem was open for a long time. Problem 1 . 10.2. ( Unconditional Basic Sequence Problem ) Let X be an infinie- dimensional Banach space. Does X contain an unconditional basic se quence ?

In [26 ] , Gowers and Maurey constructed a Banach space X such that no subspace of X has a nontrivial complemented subspace. Such a space is called a hereditarily indecomposable ( H.I. ) space. In the same article, Gowers and Maurey also proved the following facts: ( 1 ) Any H.I. space X is not isomorphic to a proper subspace of X. ( 2 ) Let be a bounded linear operator on an H.I. space. Then there is a scalar >' such that >'1 is strictly singular ( i.e. , the restriction of >'1 to any infinite-dimensional subspace of X is not an into isomorphism ) . In [24] , Gowers showed that there is a Banach space X that does not contain Co , £ 1 , or a reflexive space. In [ 52 ] , Odell asked the following question: Question 1 . 10.3. Recall that a Banach space is said to be B-convex ( or K -convex) if £1 is not finitely representable in X ( see [20, Chapter 1 3] ) . Does every B- convex Banach space contain an infinite-dimensional subspace Y whose

T

T

-

T

-

dual is separable ? Does every B -convex Banach space contain a subspace that is either reflexive or isomorphic to Co ?

Recall the homogeneous Banach space problem: Problem 1 . 10.4. Let X be a separable Banach space. infinite- dimensional subspace Y of X is isomorphic to X .

£2 ?

Suppose that every Is X isomorphic to

Combining the following two theorems of N. Tomczak-Jaegermann and W. T. Gower, we see that the answer to the homogeneous Banach space problem is affirmative. Theorem 1 . 10.5. (N. Tomczak-Jaegermann [ 77] ) Suppose that X is a ho mogeneous space that is not isomorphic to £2 . Then X has a subspace without an unconditional basis.

Theorem 1 . 10.6. (W. T. Gower [ 25] ) Every infinite- dimensional Banach space contains a subspace that either has an unconditional basis or is H. I.

96


1.11

References

[1] S. Banach, Theory of Linear brary, vol. 38 (1987).

Opemtions,

North Holland Mathematical Li

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V*

[3] F. Bombal, On set and Pelczynski 's property 32 (1990), 109-120.

convergence of

( V* ) , Glasgow Math. J.

[4] J. Bourgain, On the Dunford-Pettis property, Proc. Amer. Math. Soc. 81 (1981), 265-272. [5] J. Bourgain, New Classes of CP -spaces, Lecture Notes in Mathematics, Springer-Verlag 889 (1981). [6] J. Bourgain, New Banach Math. 1 5 2 (1984), 1-48.

properties of the disc algebm and Hoo ,

[7] J. Bourgain and F. Delbaen, A (1980), 155-176.

class of special C oo -spaces,

Acta

Acta Math.

145

[8] R.D. Bourgin, Geometric Aspects of Convex Sets with the Radon-Nikodym Property, Lecture Notes in Math. 993, Springer-Verlag, New York (1983) . [9] P.G. Casazza and T.J. Shura, Tsirelson's 1363, Springer-Verlag, Berlin, (1989).

Space,

Lecture Notes in Math.

[10] P.G. Casazza, G.L. Garcia, and W.B. Johnson,

An example of an asymp totic Hibertian space which fails the approximation property, preprint.

[11] J.M.F. Castillo and M. Conzalez, The Dunford-Pettis property three-space property, Israel J. Math. 81 (1993) , 297-299.

is not a

[12] P. Cembranos and J. Mendoza, Banach Spaces of Vector- Valued Functions, Lecture Notes in Math. 1616 Springer-Verlag New York, Berlin, Heidelberg (1997) . [13] KaiLai Chung,

A Course in Probability Theory,

C(K,

Academic Press Inc 1974.

C(K,

Hoo ) have the Dunford [14] M.D. Contreras and S. Dlaz, A) and Pettis property, Proc. Amer. Math. Soc. 1 24 (1996) , 3413-3416.

[15] M.D. Contreras and S. Diaz,

On the Dunford-Pettis property in spaces of vector-valued bounded functions, Bull. Austral. Math. Soc. 53 (1996), 131-

134.

[16] M.M. Day, Normed Linear Spaces ; Springer-Verlag, New-York (1973) .

97

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and Series in Banach Spaces,

Springer-Verlag, New

[19] J. Diestel, A survey of results related to the Dunford-Pettis property, Con temporary Math. vol 2, Proc. of the Conf. on Integration, Topology and Geometry in Linear Spaces, Amer. Math. Soc. , Providence, RI (1980) 1560. [20] J. Diestel, H. Jarchow, and A. Tonge, Absolutely Summing Operators, Cam bridge University Press (1995). [21] J. Diestel and J.J. Uhl, Jr. , Vector Math. Soc., Providence, RI (1977) .

Measures,

Math. Survey 15, Amer.

[22] P. Enfio, A counterexample to the approximation property in Banach spaces, Acta Math. 130 (1970), 309-317. [23] F. Galvin and K. Prikry, Logic 38 (1973) , 193-198.

Borel sets and Ramsey 's theorem,

[24] W.T. Gowers, A space not containing Co, £1 , Amer. Math. Soc. 344 (1994), 407-420.

J. Symbolic

or a reflexive subspace,

Trans.

[25] W.T. Gowers, A new dichotomy for Banach spaces, Geom. Funet. Anal. (1996) , 1083-1093.

6

[26] W.T. Gowers and B. Maurey, The unconditional basic sequence problem, J. Amer. Math. Soc. 6 (1993) , 851-874. [27] A. Grothendieck, d 'espaces du type

[28] S. Guerre,

C(K) ,

Sur les applications lineaires faiblement compactes Canad. J. Math. 5 (1953), 129-173.

Classical Sequences in Banach Spaces,

Marcel Dekker (1992).

[29] P. Harmand, D. Werner, and W. Werner, M -ideals in Banach Spaces Banach Algebras, Springer-Verlag , (1993) .

and

[30] R. Heinrich and P. Mankiewicz, Applications of ultrapowers to the uniform and Lipschitz classification of Banach spaces, Studia Math. 73 (1982) , 225251. [31] R.C. James, Bases (1950) , 518-527. [32] R.C. James, 101-1 19.

and reflexivity of Banach spaces,

Weak compactness and reflexivity,

Ann. of Math.

Israel J. Math.

2

52

(1964) ,

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[33] R.C. James, Uniformly (1964), 542-550. [34] R.C. James, 129-140.

non-square Banach spaces,

Weakly compact sets,

Ann. of Math.

Trans. Amer. Math. Soc.

[35] H. Jarchow, The three space problem and ideals 1 1 9 (1984) , 121-128.

of operators,

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(1964) ,

Math. Nachr.

[36] W.B. Johnson, A

complementary universal conjugate Banach space and its relation to the approximation problem, Israel J. Math. 13 (1972) , 301-310.

[37] W.B. Johnson,

Banach spaces all of whose subspaces have the approxima

tion property, Seminar d' Analyse Fonct. Expose 16, Ecole Polytechnique (1979-1980) ; Special Topics of Applied Mathematics, North-Holland, Am sterdam (1980) , 15-26.

[38] W.B. Johnson and J. Lindenstrauss, Handbook of the Spaces, Volume 1 , Elsevier (2001).

Geometry of Banach

C(l:l. )

[39] W.B. Johnson and M. Zippin, Separable L 1 preduals are quotients of , Israel J. Math. 16 (1973) , 198-202. [40] A. Kaminka and M. Mastylo, The Dunford-Pettis property for symmetric spaces, Can. J. Math. 52 (2000) , 789-813. [41] J.L. Kelley, General Topology, Springer-Verlag (1975). [42] S.V. Kisliakov, On the conditions of Dunford-Pettis, Pelczynski and Grothendieck, Dokl. Akad. Nauk SSSR 225 (1975) , 1252-1255. [43] J. Koml6s, A generalization of a problem of Steinhaus, Acta Math. Acad. Sci. Hungar. 18 (1967), 217-229. [44] J. Lindenstrauss and 1 . Tzafriri, (1977) .

Classical Banach Spaces I,

Springer-Verlag

[45] J. Lindenstrauss and L. Tzafriri, Classical Banach Spaces II, Springer Verlag (1979) . [46] R.H. Lohman, A note on Banach spaces containing .e l l Canad. Math. Bull. 19 (1976) , 365-367. [47] S. Mazurkiewicz and W. Sierpinski, Contribution a La topologie des ensem bles denombrablles, Fund. Math. 1 (1920) , 17-27. [48] V.D. Milman, Infinite-dimensional geometry of the unit sphere in a Banach space, Soviet Math. Dokl. 8 (1967), 1440-1444. [49] N.J. Nelsen and N. Tomczak-Jaegermann, Banach lattices with property (If) and weak Hilbert spaces. Illinois J. Math. 36 (1992) , 345-371 .

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[50] C.P. Nicolescu, Weak compactness in Banach lattices, J. Operator Theory, 6 (1981), 217-231 . [51] E . Odell, Applications of Ramsey theorems to Banach space theory, Univ. of Texas Press, Austin (1981), 379-404. [52] E. Odell, On subspaces, asymptotics structure, and distortion of Banach spaces; connections with logic, in Analysis and Logic, ed C. Finet and C. Michaux 301-376. Cambridge Studies in Advanced Mathematics (2003) . [53] E. Odell and H.P. Rosenthal, A double dual characterization of separable Banach spaces containing '£1 , Israel J. Math. 20 (1975) , 375-384. [54] E. Odell and Th. Schlumprecht, The distortion problem, Acta Math. 1 73 (1994) , 259-281 . [55] M.l. Ostrovskii, Three spaces problem for the weak Banach-Saks property, Math. Notes 38 (1985) , 905-908. [56] A. Pelczynski, A connection between weakly unconditional convergence and weak completeness of Banach spaces, Bull. Acad. Polon. Sci. 6 (1958) , 251253. [57] A. Pelczynski, Banach spaces on which every unconditionally converging operator is weakly compact, Bull. Acad. Polon. Sci. 1 0 (1962) , 641-648. [58] A. Pelczynski, Projection in certain Banach spaces, Studia Math. 1 9 (1960) , 209-228. [59] A. Pelczynski, Banach Spaces of Analytic Functions and Absolutely Sum ming Operators, volume 30 of CBMS conference series in mathematics, Amer. Math. Soc. (1977). [60] A. Pelczynski and Z. Semadeni, Space of continuous functions (III) (Space for 0 without perfect subsets) , Studia Math. 1 8 (1959) , 21 1-222. [61] A. Pelczynski and I. Singer, On non-equivalent basis and conditional basis in Banach spaces, Studia Math. 25 (1964), 5-25. [62] R.S. Phillips, On linear transformations, trans. Amer. Math. Soc. 48 (1940), 516-541 . [63] A. Pietsch, Operator Ideals, North-Holland (1980). [64] G. Pisier, Factorization of Linear Operators and Geometry of Banach Spaces, C.B.M.S. Amer. Math. Soc. 60 (1985) . [65] G. Pisier, The Volume of Convex Bodies and Banach Space Geometry, Cambridge Tracts in Math 94, Cambridge University Press (1990) . [66] N. Randrianantoanina, Complemented copies of .£ l and Pelczynski 's prop erty in Bochner function spaces, Canad. J. Math. 48 (1996), 625-640.

C(O )

(V*)

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[67] H.P. Rosenthal, On relatively disjoint families of measures, Studia Math., 3 7 (1970), 13-36. [68] H.P. Rosenthal, A characterization of Banach spaces containing £1, Proc. Nat. Acad. Sci. U.S.A. 71 (1974), 241 1-2413. [69] Royden, Real Analysis, (3rd ed.) Macmillan Publ. Co. , New York (1988). [70] W. Rudin, Functional Analysis, 2nd edition, McGraw Hill (1991). [71] S.F. Saccone, The Pelczynski property for tight subspaces, J. Funct. Anal. 148 (1997) 86-1 16. [72] E. Saab and P. Saab, On stability problems of some properties in Banach spaces, Function Spaces, Lecture Notes in Pure and Applied Mathematics 136 (ed. K. Jarosz), Marcel Dekker (1992) 367-394. [73] Z. Semademi, Banach Spaces on Continuous Functions, Polish Scientific Publishers (1971). [74] A. Sobczyk, Projection of the space m o n its subspace Co , Bull. Amer. Math. Soc. 47 (1941), 938-947. [75] A. Szankowski, A Banach lattice without the approximation property, Israel J. Math. 24 (1976) , 329-337. [76] A. Szankowski, Subspaces without approximation property, Israel J. Math. 30 (1978) , 123-129. [77] N. Tomczak-Jaegermann, A solution of the homogeneous Banach space problem, Canadian Math. Soc. 1945-1995, Vol. 3, J. Carrell and R. Nurty (editors), CMS, Ottawa (1995), 267-286. [78] A. Ulger, The weak Phillips property, ColI. Math. 87 (2001), 147-158. [79] P. Wojtaszczyk, Banach Spaces for Analysts, Cambridge Studies in Ad vanced Math. 25 (1991). [80] M. Zippin, A remark o n bases and reflexivity i n Banach spaces, Israel J. Math. 6 (1968), 74-79.

Chapter

2

Convexity and S lll o othness

2.1

2.2),

we present some basic In the first part of this chapter (Section and results about various types of convexity and smoothness conditions that the norm of a Banach space may satisfy. For convexity, we consider strictly convex, uniformly convex, locally uniformly convex, fully rotund, and nearly uniformly convex spaces. Particularly, we present a proof of the Schlumprecht-Odell theo rem that shows that every separable reflexive Banach space admits an equivalent norm. On the smoothness, we consider smooth, Frechet smooth, uniformly Gateaux smooth and uniformly smooth spaces. In the second part (Section we introduce the spreading model, and we show that a Banach space X has the weak Banach-Saks property if and only if X does not have an I-spreading model.

2R

£

2 .1

2. 3 ),

Strict Convexity and Uniform Convexity

For a Banach space X , let S(X) and B(X) be the unit sphere and the unit ball of X . Recall that a Banach space X is said to be strictly convex if for any two distinct vectors x , y in the unit ball B(X) of X , Il x + y ll < A Banach space is said to be uniformly convex if for any two sequences {Xn}�=l ' {yn}�= l in the unit ball B(X) of X , limn ---> oo Il xn + Yn I = implies limn---> oo Xn - Yn = O. Clearly, every uniformly convex Banach space is strictly convex.

2.

2

1

< p < 00 , Lp is uniformly convex. the norm of II + = (2) Let denote the norm of and and nor Co is strictly convex. = Neither + + (3) It is easy to see that every finite-dimensional strictly convex Banach space is uniformly convex.

Example 2. 1 . 1 . ( 1 ) Clarkson showed that for any

£1 1 · 1 00 1 · 1 1 e1 e 1 el £ I ( e2 ) ( - 2 ) 1 00 2.

co · e 1 e2 1 1 2

The following definitions generalize uniform convexity and strict convexity.

102

CHAPTER 2. CONVEXITY AND SMOOTHNESS (WUC) A Banach space X is said to be weakly uniformly convex if for any two sequences { xn }�=l and {yn }�=l in the unit ball B( X ) of X, limn-+oo Il xn + Yn ll 2 implies that { xn - yn }�=l converges to 0 weakly. (LUC) A Banach space is said to be locally uniformly convex if for any x on the unit sphere S( X ) of X and any sequence { xn }�=l in the unit ball of X, limn-+ oo Il x + xn ll 2 implies that { xn }�=l converges to x in B(X) norm. (MLUC) A Banach space X is said to be midpoint locally uniformly con vex if for any x E S( X ) and any two sequences { xn }�=l ' {yn }�=l in B(X), limn-+oo 11 2x - (xn + Yn) ll 0 implies that { xn - Yn }�=l converges to 0 in norm. (WLUC) A Banach space X is said to be weakly locally uniformly convex if for any x in S( X ) and any sequence { Xn }�=l in B(X), limn-+oo II X+xn ll 2 implies that { Xn }�=l converges to x weakly. (URED) A Banach space X is said to be uniformly rotund in every di rection if for any nonzero vector z in X and any two sequences {Xn}�=l ' {yn }�=l in B ( X ) with Xn - Yn E [z] for all n E N (Le., Xn - Yn = An z for some scalar A n ), limn-+oo II Xn + Yn ll 2 implies limn-+oo ( xn - Yn ) O. (kR) For any k � 2, a Banach space X is fully k -rotund if for any sequence { Xn}�=l in X, k 1 1 Xn n k -> n k_ll-��.>- n l -+oo k jL =1 j implies that { Xn }�=l is a convergent sequence. The space X is said to be fully rotund if X is fully k-rotund for some k � 2. (NUC) A Banach space X is said to be nearly uniformly convex if for any f > 0, there is a 8 > 0 such that for any sequence { Xn }�=l in B(X ) with Il xn - Xm I � f for all n =J- m , co{ Xn n E N} n B ( O ; 1 - 8) =J- 0 .

=

=

=

=

=

=

l

1=

:

Recall that a Banach space X has the uniformly Kadec-Klee property if for any f > 0 there is a 8 > 0 such that for any weakly con vergent sequence { xn }�=l in B ( X ) with Il xn - Xm ll � f for all n =J- m, II w- limn-+oo Xn II � 1 - 8 . (KK) Recall that a Banach space X is said to have the Kadec-Klee prop erty (or Radon-Riesz property, or property H) if every weakly convergent sequence in S ( X ) converges in norm. First, we give a easy characterization of the above properties. (UKK)

X

Theorem 2.1.2. Let be a Banach space. Then is uniformly convex if and only if for any two bounded (1) The space

X

{x n }�=l and { Yn }�= l in X, limn-+oo (2 11 xn 1l 2 + 2 11 Yn l1 2 - ll xn + Yn 11 2) 0 implies limn-+oo Il xn - Yn ll = O .

=

sequences

2. 1 . STRICT CONVEXITY AND UNIFORM CONVEXITY

103

(2) The space X is WUC if and only if for any two bounded sequences { xn }�=l and {yn }�=l in X, limn-+ oo (2 1 I xn I 1 2 + 2 11 Yn l1 2 - Il xn + Yn 11 2) = 0 implies { xn - Yn }�=l is weakly null. (3) The space X is ( weakly) LUC if and only if for any Xo in X and any bounded sequence {xn }�=l in X, limn-+ oo (2 1 I xo I 1 2+2 1I xn I1 2- ll xo+xn I 1 2) = 0 implies { xn }�=l converges to Xo in norm ( weakly). (4) The space X is URED if and only if for any nonzero vector z in X and any two bounded sequences { xn }�=l 1 { Yn }�=l in X such that Xn - Yn E [ z ] for all n E N, limn-+ oo (2 1 I xn I 1 2 + 2 11 Ynl1 2 - 1 1xn + Yn 11 2) = 0 implies { xn - Yn }�=l is a null sequence.

All the proofs are similar. We prove only (1). First, we need the following fact. Fact 2 . 1 .3. For any two negative sequences {a n }�=l ' {b n }�=l in IR, Proof:

implies o � n-+ limoo (an - bn) 2 = nlim -+ oo 0 = n-+ limoo (an

- bn ) .

(2a� + 2b� - (an + bn) 2 ) = 0,

Suppose that X is a uniformly convex Banach space that does not satisfy the condition in ( 1 ) . Then there are two bounded sequences { Xn }�=l ' { Yn }�=l in X and an € > 0 such that Il xn - Yn ll > € for all n E N and By Fact 2.1. 3 , we have limn-+ oo ( ll xn ll - II Yn ll ) = O. By passing to subsequences of { xn }�=l and { Yn }�=l ' we may assume that { ll xn ll }�=l is a convergent sequence. If lim n-+ oo Il xn II = 0, then lim ll Yn II = o . n-+oo We get a contradiction. So we may assume that Xn i= 0, Yn i= 0, and lim II xn ll = a i= o . n-+oo Let Z l , n = xn / ll xn ll and Z2 ,n = Yn / Il Yn ll . Then for all n E N, II Z l ,n ll = IIZ2 ,n ll and

=1

This contradicts the hypothesis that X is uniformly convex. Therefore, all uniformly convex Banach spaces satisfy the condition in ( 1 ) . The converse 0 implication is obvious. The following theorem is due to R. Huff [31] .

104

CHAPTER 2. CONVEXITY AND SMOOTHNESS

Theorem 2.1 .4. A Banach space X is nearly uniformly convex if and only if X is a reflexive Banach space with the uniformly Kadec-Klee property. Proof: Suppose that X is a reflexive Banach space with the UKK property. Let E be any positive real number and { xn }�= l a sequence in B(X) such that

Since the closed unit ball of X is weakly compact, by passing to a subsequence, we may assume that { xn }�=l converges to x E co( xn ) weakly. But X has the UKK property,1 and therefore, there is a 6 < 1 such that x E B(O, 6). Therefore, co(xn ) n B (O, 1 6 ) =F 0. This means that X is NUC. Suppose that X is an NUC space. Let { Xn }�=l be a weakly convergent sequence in B(X) such that sep (xn) > E > 0. Since X is NUC, there is a 6 > ° for any n E N, there is Yn E co { Xk : k � n } such that ll Yn I :::; 1 6. Then ' ' Yn w- l1m x w- nl-+1m oo n -+ oo Xn has norm at most 1 6. This implies that X has the UKK property. By James ' s theorem, it is enough to show that for any x* in S(X*), x* attains its norm on B(X). Let { Xn }�= l be a sequence in B(X) such that nlim -+ oo (x * , xn ) = 1 . -

def

=

=

-

If { Xn }�= 1 contains a limit point x , then we have Il x ll

:::;

1 and (x* , x)

=

1.

So we may assume that { xn }�= 1 contains no Cauchy subsequence. By passing to a further subsequence, we may assume that there is an E > ° such that sep (xn ) > E. Since X is NUC, there is a 6 E (0, 1) such that for any N, B (O, 8) n co { xn : n � N} =F 0. On the other hand, for any x E co { xn : n � N}, ) Il x ll � (x * , x ) � ninf ?N (x * , xn . 0 This is impossible. Thus X must be reflexive. The proof is complete. In [39] , J. Lindenstrauss showed that foo does not admit an equivalent weakly locally uniformly convex norm. G.A. Aleksandrov and I.P. Dimitrov [1] proved that foo does not admit an equivalent MLUR norm. Modifying Lindenstrauss ' s proof, Z. Hu, W.B. Moors, and M.A. Smith [30] proved the following theorem: Theorem 2.1.5. The space f oo does not admit an equivalent weakly mid

point locally uniformly convex norm or norm with the Kadec-Klee property.

Proof: Let II · 11 00 deonte the usual sup norm on f oo , and let 1 · 1 denote any equivalent norm on foo . Let Fo {x E foo : Il x ll oo 1 and N \ supp x is an infinite set}, =

=


and

mo Mo

105

inf { l x l : x E Fo}, sup { l x l : x E Fo}. We claim that there are w and two sequences { Yn }�= l and {Zn }�=l in Fo such that { Yn }�=l is not a weakly convergent sequence, {Zn}�=l is weakly null, but it is not norm null, lim n-+oo I w ± Yn l limn-+oo I w ± zn l Iw l · If the claim is true, then (£00' I . I ) is not weakly midpoint locally uniformly convex and does not have the Kadec-Klee property. Let Xl be any element in Fo such that I X I I � i ( 3Mo + mo). Select two distinct integers il, it E N \ SUPPX I . Let =

=

•

•

=

•

=

FI {x E Fo : x and X l agree on SUPP X I U {il, it}}, and m l inf { l x l : x E FIl , MI sup { l x l : x E Fd · Suppose that {Xl, . . . ,xn}, Fn � . . . � Fo, mo � . . . � mn � Mn � . . . � Mo and 2n district integers {il, . . . , in, it , . . . , jn } have been constructed such that for any 1 � k � n, ( ) mk inf { l x l : X E Fk } and Mk sup { l x l : X E Fk }; (b) X k E Fk - l and I X k l � i(3Mk- 1 + mk - l); ( c ) i k , jk rt. SUPPX e for any f � k; (d) Fk consists of all X in Fk - l such that x(£) X k (£) for all £ E SUPPX k- 1 and all £ E {il , . . . , i k , it , · · , jk } ' Let Xn + l be any element in Fn such that I Xn + 1 1 � i( 3 Mn +mn)' Select two distinct integers in + t, jn + 1 E N \ supp Xn \ {i I , . . . , in' it, . . . ,jn} ' Let Fn + l {x E Fn : x and Xn+ l agree on the set supp Xn + l U {i t, . . . ,in + l ' i t , · . . , jn + l}} ' Set mn+ l inf { l x/ : x E Fn+ d, Mn + 1 sup { l x l : x E Fn+ d. For each n E N, let Yn and Zn be the elements in £00 defined by if k i m for some m � n, Yn(k) = { � otherwise; k i m for some 2n � m � n zn(k) { � ifotherwise; =

=

=

a

=

=

=

·

=

=

=

=

=

=

106


and let w be the element in foo defined by �n (k) if k E supp Xn for some n E N, w(k) if k rt. U�=l supp Xn · Then for all n E N, w ± Yn + l, w ± Zn + l E Fn . Note: Both sequences { Yn }�=l and {zn }�=l converge pointwise to O. ( This implies that if { Yn }�=l (respectively, {zn }�=l) converges weakly, then it con verges to 0 weakly ) . Since 0 is not in the closed convex hull of { Yn : n E N}, { Yn }�=l does not converge weakly. On the other hand, for all n E N, Zn E Co . The sequence {zn }�=l is a weakly null sequence. To finish the proof, we need to prove only that lim I w ± Yn l n-+oo lim I w ± zn l I w l · n-+oo =

{

=

=

Note that for any n E N, w, w ± Yn , W ± Yn E Fn . This is equivalent to show that limn-+oo Mn limn-+oo mn. To see this, observe that 2xn - Fn � Fn for all n E N. We have =

1

'2 (3Mn - 1 + mn - l )

So

:::; 2 1 xn l :::; inf { Mn + l y l : Y E Fn } :::; Mn + mn :::; Mn - 1 + mn ·

and n-+oo

n-+oo

o The proof is complete. The following chart summarizes the implications that exist among these various properties:

LUC

UC

MLUC

SC

WLUC

WUC

/

URED

The following two propositions are due to V. Zizler [55] .


107

and (Z, I . l i z) be two Banach spaces such that (Z, II . l i z) is WUC. If T : X --+ Z is a bounded linear operator such that the adjoint operator T* maps Z* onto a norm dense subset of X * , then the norm 1 for every x E X Il x iiT = ( 1 I x ll � + IITxll � ) / 2 . is an equivalent WUC norm on X .

Proposition 2 . 1 .6. Let (X, II

. Il x)

Proof: Since

Il x ll � � Il x ll} = Il x ll � + II Tx ll � � ( 1 + II T I 1 2 ) Il x ll � , we need to show only that I . l iT is a WUC norm. Let { Xn }�= l ' { Yn }�=l be any two bounded sequences in X such that (2.1) nlim --+ oo ( 2 1 I xn ll } + 2 11 Yn ll } - ll xn + Yn ll } ) = O. Since 2a2 + 2b2 � ( a + b) 2 for any two real numbers a , b, the equ;:ttion (2.1) implies that But

Z

is WUC, and so for any z* E Z* ,

By assumption (T* (Z* ) is dense in X * ) , { xn - Yn }�=l converges to 0 weakly. 0 The proof is complete. Proposition 2 . 1 . 7. Let (X, 11

(Z, II ' 1 I z) is URED and if T : X from X into Z, then the norm

and (Z, II · l i z) be two Banach spaces. If Z is a bounded one-to-one linear operator

· ll x)

--+

1 Il x iiT = ( lI x l l � + II Tx ll � ) / 2

is an equivalent

URED

for every

xEX

norm on X .

II . II T is an equivalent norm on X. We need to show only that II . l iT is URED. Let { xn }�= 1 and { Yn }�=l be any two bounded sequences in X , z E X, z =1= 0, and { A n }�= 1 a sequence of lR such that Xn - Yn = An for any n E N, Proof: Clearly,

Z

and

Then { Txn : n E N} and { T Yn }�=l are two bounded sequences in Z such that TXn - TYn = AnT z , and 0 = nlim --+ oo ( 2 I1 Txn ll � + 2 I 1 TYn ll � - II Txn + TYn ll � ) . Since T is one-to-one, T z =1= The proof is complete.

o. By Theorem 2.1.2(4) , we have limn --+oo An

=

O.

0

108

CHAPTER 2. CONVEXITY AND SMOOTHNESS Theorem 2 . 1 . 8 . Every separable Banach space admits an equivalent URED

norm.

Proof: Since every separable space is isomorphic to a subspace of C[O, 1] , we need to show only that C[O, 1] has an equivalent URED norm. Let { fn }�= l be the Schauder system of C[O, 1] , and { en }�= l the unit vector basis of £2 . Define an operator T C[O, 1] £2 by :

-

T (nf=l an fn ) nf= l ;: en . =

Then T is one-to-one. By Proposition 2.1 .7, C[O, 1] admits an equivalent URED 0 norm. The proof is complete. In this section, we need the following two easy propositions. The proofs are left to the reader. Proposition 2 . 1 .9. A Banach space X is MLUC if and only if for any x E S(X) and any sequence { Xn }�= l ' lim n-+oo Il x ± xn ll 1 implies the the sequence { xn }�= l converges to ° in norm. Proposition 2 . 1 . 10. Let 11 · 11 1 and 1 1 · 11 2 be two equivalent norms on X such that 11 · 11 is WUC ( respectively, LUC, MLUC) . Let 1 1 · 1 1 be the norm defined by 2 I I x l 1 2 Il x ll � + Il x ll �· Then 1 1 · 1 1 is an equivalent WUC (respectively, LUC, MLUC) norm on X . The following theorem is due to P. Dowling, Z. Hu, and M.A. Smith [23] . =

=

X be a real Banach space with an unconditional shrinking basis { ej }� o . Then X admits an equivalent WUC and MLUC norm on X that is not LUC. Proof: Let X be a real Banach space with an unconditional shrinking basis { ej }� o . Without loss of generality, we may assume that { ej }� o is a normalized I-unconditional basis. Let I denote the identity mapping on X . For and L:� a , let Theorem 2 . 1 . 1 1 . Let

x

=

n

o jej

For any k E N, let fA and O'k be the functions from X to lR defined by

{

}

x) max IIPo(x) ll , 11 (1 Pk -d ( x ) II , O'k ( X) f3k (X + ) + fA (x - ) ,

fA (

=

=

-

�°

109


and Tk : X � X and T : X � £2 be the operators defined by

Tk (x) Po(x) + ( I Pk )( x ) , =

-

T(x) aoeo + L O'j aj ej. 00

j= 1

=

It is easy to see that {3k , O'k , Tk , and T satisfy the following conditions: (i) For each k 0, (3k and O'k are seminorms on X, and

�

1 2 k (X) � (3k (X) � Il x ll � 2 {31 (X). 0'

�

(ii) max { ll x + ll , Il x - 1I } U¥ . (iii) II T II � 2 , and for any m � 0, II Tm l1 (iv) Since { en }�= 1 is a shrinking basis,

=

1.

{e�}�= 1 is a basis of X * . T* maps

(£ 2 )* onto a dense subset of X * . Let II · IIG l and II . II G be the norms on X defined by

Then

1/2 , 2 2 ({31(x) + nL= 1 Q� O'n(Tn(X)) ) 1 2 2 2 ) ({31 (X) + nL=l Q� O'n(Tn(X)) + II T(x II � ) / 00

Il x llG l

=

Il x l i G

=

00

1 x 2 11 ll � {3 1 (x) � Il x llG l � Il x l i G 00 1 2 2 2 � ( 11 x 1 1 + L Q� (2 1 I x ll )2 + (2 1 I x ll ) ) / � 3 11 x ll ·

n= 1

This implies that 11 · 11 Gl and 11 · 11 G are two equivalent norms on X. By Proposition 2.1 . 6, (X, II · II G ) is WUC. It is easy to see that Il eo llG nlim ---+ oo Il eo + en llG V2, Il ej llG (31 ( ej ) = 1 for all j > 0 . Thus the space (X, II . II G ) is not LUC. It remains to show that (X, II . II G ) is MLUC. Let x 2:;: 0 aj ej be an element in X and { Xk 2:;: 0 a k, j ej }� 1 a se quence in X such that Il x l i G 1 and nlim ---+ oo Il x xk llG 1. =

�

=

=

=

=

By the definition of II . II G, we have limO O'n (Tn (x Xk )) k---+ lim II T(x xk) 11 2 k-+oo

± ±

±

O'n (Tn ( x ) ) , II T(x ) 1 1 2 .

=

�

for all n 0,

(2.2) (2. 3 )

110


Note that Hence

£2 is uniformly convex.

a

By ( 2. 3 ) , we have limk-+oo I T(x k )1 1 2

=

o.

� o.

(2.4) lim k ,) = 0 for all j k-+oo We claim that {X k } k=1 converges to 0 in norm. Suppose it is not true. Then by passing to a subsequence if necessary, we may assume that there exists an € 0 such that I l x kl l € for all On the other hand, {ej }� o is a basis. There are we have 0 such that for any

>

.

k E N. k > K,

ko , K > �

(2.5) € 16 By (2.5) ,

(2.6)

O"ko (Tko + 1 (x) ) ,Bko (Tko+ 1 (x) + ) + ,Bko (Tko+ l(X) - ) :::; l a o l + 11 ( 1 - Pko )(x) + 1 + 1 (1 - Pko )(x) - I :::; l ao l + S€ · =

By (ii) and (2.6) , for any

Fix a

k > K.

k > K,

Without loss of generality, we assume that

(2.7)

ao � 0 and 1 (1 -

Pko )(x;) 11 > i� . Then O"ko (Tko + 1 (x + Xk ) ) ,Bko ((PO + ( 1 - Pko+ 1))((x + Xk ) + ) ) + ,Bko ((Po + (1 -+ Pko+ d)((x + Xk ) - ) ) � ,Bko (Po((x + X k ) ) ) + ,Bko (( 1 - Pko )((X + Xk ) - ) ) =

-> ao - -1€€6 + -17€6 - 1€6 > ao + 4 '

This contradicts (2.2) and (2.7). So (X, I · IIG ) must be MLUC.

o

Remark 2.1. 12. The proof of Theorem 2. 1 . 1 1 shows that every Banach space with an unconditional basis has an equivalent MLUC norm. Proposition 2 . 1 . 13. [55] No equivalent norm on

£1 is WUC.

I . I be an equivalent norm on £1. By Theorem 1.3.20, there exists an iI-sequence { xn}�=1 such that I l xn l 1 and nlim -+oo I l xn + xn + l l 2. Proof: Let

=

=

111


But o

{ Xn - Xn+ l};:O= l does not converge to 0 weakly. So (£1, II . I ) is not WUC.

The following four examples are due to M.A. Smith [50, 51] . Example 2 . 1 . 14. The space £2 admits an equivalent WUC norm that is not MLUC. Let {e k }� l be the unit vector basis of £2' For x = l:;: l aj ej £2 , define

E

Then I . lis is an equivalent norm on £2' Let { aj }� l be a sequence of positive real numbers such that aj ---+ O. Define T : £2 ---+ £2 by

T(I:=l aj ej ) a Iel + I:aj ae j =2 j j. j 00

00

=

Then T is a one-to-one continuous linear mapping whose dual operator T* maps (£2 )* onto a dense subset of (£2 )* ' For any x £2 , define the new norm 1 · llw by

E

II x llw = ( 1 I x l � + I Tx l �) 1 /2 . By Proposition 2.1 .6, I . Ilw is an equivalent WUC norm on £2. We claim that (£2 , 1 · l w ) is not MLUC. Let a = 0 ' x = a el, Xn = a(el + en), and Yn = a (el - en). Then I l x llw = 1 , nlim -+ oo I Yn llw = 1 , -+ oo I xn llw = nlim 2 x = Xn + Yn nlim -+oo I l xn - Yn l w = J2. This implies that the space (£2 , 1 · llw ) is not MLUC. Example 2 . 1 . 1 5. The space £2 admits an equivalent URED and MLUC norm that is not WLUC. Let {ej }� l be the unit basis of £2 , and { aj }� l a sequence of positive real numbers such that aj O. Define an operator T : £2 £2 by T (I:=l aj ej ) = L:= aj aj ej. j j2 Let I . IIF and I . II A be two equivalent norms on £2 defined by (2 .8) al l l + ( L:= l aj l 2 ) 1 /2 , j2 (2.9) ( 1 I x l � + I Tx l � ) 1 /2 for any x E £2 . ---+

---+

00

00

00

112


Since T is one-to one, by Proposition 2 .1.7, (£2 , II . I A) is URED. We claim that ( £2 , I . I A) is MLUC. Let x be a vector in S(X) , and let {Yn = 2:;: 1 bn ,j ej }�= 1 and {zn = 2:;: 1 en ,j ej }�= 1 be two sequences in B(X) such that ( 2.10) nlim ..... oo 1 2x - Yn - zn ll A =

o.

Since £2 is reflexive, without loss of generality, we assume that { Yn }�= 1 converges to Y weakly and {Zn }�=1 converges to Z weakly. Clearly, Y and Z are in B(X) . Notice that I . I F is strictly convex, 2 x Y z, and ( 2.10) implies that

+

=

nlim ..... oo I l zn llF I l x i I F . ..... oo I Yn llF nlim =

=

We have that

Y Z x 2:=1 aj ej . j nlim --+ oo en' 1 a l �oo bn' 1 nlim =

Hence

=

00

=

=

=

and

n ..... oo I Zn llF - l en ' 1 1 1/2 1/2 2 2 m a i . (2: !� oo (2: l en , i I ) i= 2 i=2 l l ) By the uniform convexity of I · 1 2 , we have ( bn i - n i )ei O. nlim ..... oo 1 � Li= 2 , e , l 2 = lim =

00

=

00

=

Thus

nlim ..... oo l Yn - Zn l F = O. We have proved that I . I A is MLUC. On the other hand, I l en l A = 1 for all E N, limn ..... oo I l el + en l A 2, and w- lim n ..... oo en O. We have proved that ( £ 2, I · I A) is not WLUC. Now, we proceed with an elementary observation on infinite series that will be needed in Example 2 . 1 . 17 . Fact 2 . 1 . 16. Let {sn}�= 1 and {tn}�= 1 be two nonnegative decreasing se quences such that limn ..... oo S n 0 and 2: � 1 Sk t k < Then n

=

=

=

00 .

j

2: Sk t k 2: 2: ( Sj - Sj + l)t k 2: 2: ( Sj - Sj + l)t k . 00

k= 1

=

00

00

k=l j=k

=

00

j= l k= 1

113

2. 1 . STRICT CONVEXITY AND UNIFORM CONVEXITY Hence for any perrr dation 7r of natural numbers,

k k ti - L=l t7l" ( i» ) ' Sk tk - L=l Sk t7l" (k) L=l ( Sk - Sk + 1) ( L L =l =l i i k k k 00

00

00

=

Notice that all the terms in the summations are nonnegative. By replacing SHb SH 2 , . . . and tHb tH 2 , . . . with 0 if necessary, we have that l

l

st ' t L =k l Sk k � kL=l k 7l" (k)

where £ is either a finite number or 00 . Furthermore, if t 7l" (k) and t 7l" (m) < t m , then there is j > m such that t 7l" (j) = t m.

=

tk for k
0 such that I l x l oo > € and I l xn �€ for every n N. Let - 1 be the largest natural number satisfying � €/ 6 Let

E

1.

M � no such that for any k �

l ak - bk ,n l < 2 · E

( )E (K) K I l

2, 2,

K,

Hence if k 7rx ({I, . . . , K}) and n > M, then 7r k { I , . . . , K} and I bk,n an i < E. In particular, for n > M, 7r;� X n � and bK,n � aK � < E. SO for any k rt. 7rx ({I, . . . , K}) , we have

+

x, and ( · l i D ) is LUC. operator defined £2 , T : £2 - eo be theI nonlinear co ,

This implies that {Xn}�= 1 converges to For = (ai , a 2 , . . . ) in let by

x

j

I · i l L be a norm on £2 defined by I l x I L I T(x) I D . It is easy to see that I . I l L is an equivalent norm on £2 . We claim that (£ 2 , I . I l L ) is LUC. Let x, Xn E 8(£2 , I . I l L ) such that lim I l x + xn l L 2. n-+oo Then T(x),T(xn ) E B ( (eo, I · l i D) ) and lim I T (x) + T(xn) I D 2 . n-+oo But (eo, I · l i D) is LUC. We have I x l 1 2 2�IIJo I xn l 1 2 2�IIJo I l x +2xn I 1 2 . Therefore, limn -+oo Xn x (because £2 is uniformly convex) . To see that the norm I · I l L is not URED, let Xn e l + V3e n and Yn V3e n for each Then I l xn l L - 1, I Yn l L - 1, and I l xn +Yn I L - 2, but Xn - Yn e l for all

Let

=

=

=

=

=

=

=

n.

=

n.

=

Example 2 . 1 . 18. There is an equivalent URED norm on £1 that is neither

WUC nor MLUC. Let { ,8n }�= 1 be a strictly decreasing null sequence with ,81 be an operator defined by

8 : £1 - £2

8 (nl:= 1 anen ) 00

=

00

l: ,8nan en .

n=1

=

1, and let

116


I .IA

Let be the norm given in Example define the norms and by

I .1M

2 .1.15. For each x = l:� 1 aj ej E £1 ,

I ·IH 00 1 I l x i l M = max {l a 1 , j2:=2 l ajl} , 1 /2 . = I l x l i H ( I x l L + I S(x) l �) By Proposition 2.1.7 and Theorem 2.1.13, I . I H is URED but not WUC. To see that I · I H is not MLUC, let a = � , x = ae1,xn = a ( e1 + e n), and Yn = a (e1 - en) ' Then I l x l i H = 1, I l xn l H -+ 1, I Yn l H -+ 1, and 1 2x-(xn +Yn) I H -+ 0 , but I l x n - Yn l H -+ 2a. K.W. Anderson [2] asked the following question: Problem 2 . 1 . 19 . Let X be a strictly convex Banach space with the Kadec Klee property. Is X midpoint locally uniformly convex ?

.I I 2 . 1 . 18).

of M.A. Smith showed that there is an equivalent strictly convex norm £1 that is not midpoint locally uniformly convex (see Example (Note: £1 has the Schur property, so (£1 , has the Kadec-Klee property.) The following theorem is due to M.1. Kadec which showed that the answer to Problem is affirmative if X does not contain a copy of £1 .

I . I ) [32]

2.1 . 19

£1 .

Theorem 2 . 1 .20. Let X be a Banach space that does not contain a copy of If X is strictly convex and has the Kadec-Klee property, then X is

MLUC.

Xo be a unit vector in X and {xn }�= l a sequence in X such that (2. 16) nlim ---> oo I l xo ± xn l = I l xo l = 1 . Claim 1 . We claim that { xn }�= l has a convergent subsequence. Note that X does not contain any copy of £1 . By Rosenthal ' s £l-theorem and by passing to a subsequence of { Xn }�= l ' we may assume that { Xn }�= l is weakly Cauchy. If Claim 1 is not true, by passing to a further subsequence, there is an € > 0 such that for all E N, I x 2n - X2 n + 1 1 > € > O. So we have 0 = w- nlim ---> oo X2n - X2n l Proof: Let

n

-

and

I l xo l � li�--->s�p l xo ± � (X2n+ 1 - x2n) 1 � lim n--->sup oo �2 ( 1 l xo ± X2n+ l l + I l xo x2n l ) � 1 . =f

But X has the Kadec-Klee property. This is impossible. We have proved Claim

1.

117 Claim 2. The proof of Claim 1 shows that for any subsequence of { xn }�= l ' there is a further convergent subsequence. We claim that every convergent subsequence of {X n }�= l converges to 0 . If the claim is true, then {X n }�= l must be a null sequence. Suppose Claim 2 is not true. By passing to a subsequence of {Xn}�= l ' we can assume that { xn }�= l converges to some y =1= 0. By ( 2 . 1 6 ) , we have I l xo ± y l = I l xo l . But X is strictly convex. This implies that y = 0, a contradiction. Thus 2. 1 . STRICT CONVEXITY AND UNIFORM CONVEXITY

X

0

is MLUC. The proof is complete.

Example 2 . 1 . 2 1 . There is an equivalent norm on £1 which is URED and LUC but not WUC. Let : £1 - £2 be the inclusion mapping. For in £1 , define

I x1 I l x i l E = ( 1 l x l i + I Ix l � ) / 2 . Then I . l i E is an equivalent norm on £1. By Proposition 2.1.7 and Theorem 1 . 3 .20, I · l i E is URED but not WUC. We claim that (£1, I · l i E ) is LUC. Let x = 1:;1 aj ej be a unit vector in (£1, I · l i E ) , and { xn = 1:;1 bj, n ej }�= l a sequence in the unit ball of (£ 1 , I · l i E ) such that limn-+oo I l x + xn l E = 2. Then we must have nlim -+oo I l xn l E 1, nlim -+oo 2( l x l � + I l xn l � ) - l x + xn l � 0, (2 . 17) nlim-+oo 2 ( l x l i + I l xn l i ) - I l x + xn l i - 0, (2.18) nlim-+oo 2( I Ix l � + I Ixn l � ) - I Ix + IXn l � 0. By ( 2 . 17) and (2.18), nlim--+oo bJ"' n aJ" for each j E N. This implies that { xn }�=l converges to x in £1. The properties satisfied by the examples are tabulated in the following table. The symbol + (-) indicates that the space in that row has ( does not have ) the property in that column. =

(£2 , I . 1 2 ) (£2 , I ·. I L ) (£2 , I . I l w ) (£2 , I I A )) (£2 , I ·. I G ) (£l(£1,l II .. lI i EH) (£l l I I I)

UC LUC MLUC WLUC WUC URED SC

+ -

+ ++-

+ + + ++ -

+ + +++-

+ ++-

+ + + + + +-

+ + + + + + + -

118


42

X, [44] x E X, I . I x X [0 , 00 )

V.D. Milman [ ] asked whether for any reflexive Banach space there is an equivalent norm that is E. Odell and Th. Schlumprecht showed that the answer is affirmative if is a separable reflexive Banach space. be the let Let be a real Banach space. For any : symmetric convex function from into jR + defined by

2R.

X

X X

-

for

Y E X.

The following lemma shows that for any x E X, 1 · l x is an equivalent norm on X. The norm I . I l x is called the symmetrized type norm of x. Lemma 2.1.22. For any x E X, I . I l x is an equivalent norm such that 2 1 y I � I Y l x � 2 ( 1 + I x. l ) l y l for all Y E X. Moreover, if 1 · 1 is strictly convex, then for any x E X, I I x is also strictly convex. Proof: We need only to verify the triangle inequality. Since for any fixed

v E X, the function r I ru + v i + I l ru - v i is a convex symmetric u,function, is increasing on [0, 00 ). Hence for any Y b Y2 E X, II Y l + Y2 11 x = 1 I I Y l + Y21 I x + ( Yl + Y2 ) 1 1 + 1 I I Y l + Y21 1 x - ( Y l + Y2) 11 � 11 ( I Y l l + I Y21 1 )x + ( Y l + Y2 ) 11 + 11 ( I Y l l + I Y21 1 ) x - ( Y l + Y2 ) 1 � 1I I Y 1 I x + Y l il + 1 I I Y I i l x - Yl i l + 1 I I Y21 1 x + Y2 11 + 1 I I Y21 1 x - Y2 1 1 = I l y I i l x + I Y21 I x ' 9 :

9

1---+

o


Theorem 2.1 .23. (E. Odell and Th. Schlumprecht) Every separable real reflexive Banach space

X admits an equivalent 2R norm.

Proof: Since X is a separable, there is an equivalent strictly convex norm X. Without loss of generality, we assume that I . I is strictly convex. Let {zn E N} (Zl 0) be a countable dense subset of X that is closed under rational linear combinations. Choose a positive sequence { cn}�= l such that nL= l en(l + I l zn l ) < 00 . Let I . I be the norm on X defined by Il x l = nL= l en l x l zn for all E X.

on

: n

=

00

00

x


Clearly, for every x

119

E X, 00

n= l

00

n= l Thus 1 · 1 1 is an equivalent strictly convex norm on X. We claim that I I · I I is 2R. Let {Xn } �= l be a III . I I -normalized sequence in X such that lim nlim m�oo � oo I xm + xn l = 2 mlim� oo I I xm ll = 1. We need to show only that every subsequence of {Xn}�= l contains a further convergent subsequence. By passing to a subsequence, we may assume that {xn }�= l converges weakly to x. Claim 1. We claim that there is a further subsequence { x�}�= l of {Xn}�=l such that for any y E X and any f31, f32 � 0, we have mlim� oonlim�oo II y + f31x� + f32 x�I I = mlim� oo II y + (f31 + f32 )x� I · Proof of Claim 1 : By passing to subsequence of {xn}�= l' we may assume that for any j, k E N and any nonnegative rationals f31, f32 , the limits exist (first, pass to subsequences for each fixed parameter and then apply a diagonal argument). The hypothesis implies that lim nlim �= l ck l xm + xn l z k = 2 mlim� oo k" m�oo �= l ck/ l xm l zk ' � oo k" and for any k E N, lim / l x m I l z k ' mlim� oo nlim � oo I xm + Xn l zk = 2 m�oo Thus for any rationals 0 ::; f31 ::; f32 and any k E N, 00

00

m� oon� oo 1 I f31xm + f32xn I Zk + mlim� oo II (f32 - f3I)xm / l zk m � oo f3I/ 1 xm l zk + mlim� oo (f32 - f31)/ l xml l z k + nlim � oo f32l 1 xn l zk

::; lim lim ::; lim

{Zk k E N} is a dense subset of X, we have proved that for any nonneg f31, f32 and any y E X,

Since : ative rationals

120


So for any y

E X,

)

(

lim nlim ->oo Il y + (31xm + (32 xn ll + Il y - (31xm - (32 xn ll m->oo = lim lim 11(3 1x m + (32 xn lly = ( (31 + (32 ) lim II X m lly m->oo n ->oo m->oo = J� oo y + ( (31 + (32 )Xm + y - ( (31 + (32 )Xm ·

I)

I lI

(Il

By the triangle inequality, lim nlim ->oo Il y ± ( (31x m + (32 Xn) II m->oo

� J�oo I (3:�y(32 ± (31 Xm I + nl�� I (3::y(32 ± (32 Xn II J�oo ((31 1 (31 : (32 ± Xm l + (32 1 (31 : (32 ± Xm l )

=

=

lim Il y ± ( (31 + (32 )X m II · m->oo

Thus We have proved Claim 1 . Claim 2. We claim that {Xn}�=l converges to x. By Claim 1 , for any z lim nlim ->oo l i z + (xm - x) + ( xn - x) 11 m->oo = lim lim l i z - 2x + xn + x m ll m-> oo n ->oo = lim l i z - 2x + 2x m ll = lim l i z + 2(x m - x) ll · m->oo m->oo By the definition of I I . I I , we have

E X,

lim Il xn - xii · lim n->oo lim 1 1 (x m - x) + (xn - X) I I = 2 m->oo m->oo Replacing Xn by Xn - x, we may assume that {xn}�=l is weakly null. Suppose that Claim 2 is not true. By passing to a subsequence {Xn}�=l ' we may assume that d = limn->oo Il xn II exists. Without loss of generality, we assume that d = 1. By Fact 1 . 1 . 1 , it is enough to show that there is a subsequence {xnJ�l of {xn}�=l such that for any nonnegative finite sequence { (3d � l with �:[:, 1 (3i = 1, 1 (3ixn i I � 2 ' Il L i =l N

Let {8i }�l be a decreasing positive sequence of real numbers such that n: 1 (1 - 8i ) 1 - �. Select Xn 1 such that Il xn l ll 1 - 81 . Since B([Xn l D and [0, 1] are compact and limn ->oo Il xn l l = 1 , by Claim 1 , there are n 2 < N2 such that if m N2 , then for any y B([xn 1 ]), (3 1 , (32 [0, 1],

�

�

E

� E

l I y + (31xn2 + (32 Xm l � (1 - 82 ) ll y + ((31 + (32 )xn2 11 ·

121


{Xn1 , Xn2 , . . . , xn k } and n2 < N2 < n3 < . . . < nk < Nk are NEk -.21 i We have proved Claim 2 . The proof is complete. Suppose that selected. There are

m

:

=

=

-

o

Exercises

{Xn }�= 1 be a sequence of Banach spaces. For any 1 :::; (2:�= 1 \BXn) p is the collection of sequences (xn) such that Xn E Xn for all n E N, and ( 1 l xn I l xn ) E Cpo The norm of (Xn) E (2: �= 1 \BXn) p is defined Exercise 2 . 1 . 1 . Let p :::; 00 ,

by

(2:�= 1 \BXn) p is a Banach space. 1 {Xn }�= 1 is a sequence of strictly convex Banach spaces, then the space (2: �= 1 \BXn) p is strictly convex. (c) Show that there is a sequence {Xn }�= 1 of uniformly convex Banach spaces such that (2::'= 1 EBXn h is not uniformly convex. 1

( a) Show that for any :::; P :::; 00 , (b) Let < p < 00 . Show that if

Exercise 2 . 1 . 2 . ( Milman-Pettis ) Show that every uniformly convex Banach

space is reflexive.

Exercise 2 . 1 .3. ( Clarkson ) Show that for any

uniformly convex.

1
0, ��) (€) � is defined by � � ) (€) inf= 1 inf sup { ll x €y ll - 1 } . = k YEB(Y) I x l Yc x,dim(Y) X is said to be k-uniformly convex (k-UC) if for all € > 0, � � ) (€) > O. Show that a Banach space X is k-UC if and only if X is k- U R. ( ) Suppose that dim(X) k � 2. Show that for each € > 0 there exist k 1 vectors V ! , , Vk 1 such that Exercise 2 . 1 .7. Let X be a Banach space. For any k +

=

a +

•

.

.

+

=

Ilvi ll = € for 1 � i � k + 1 , • €/3 � dist( v i , [VI , . . . , Vi - I]) for 2 � i � k , • Vi O . (b) Show that for any finite-dimensional Banach space X and any 1 > € > •

0,

L 7�11

=

k + t5x( k) ( k (k l)€ k ) 3 (1 + � � ) (€))

�

� � ) (€) . 1 + � � ) (€)

(c) Show that for any finite-dimensional Banach space X and any 1 > € >

0,

Exercise 2 . 1 .8. Let X be a Banach space. For any closed convex subset C � of X, let

A

r (C, x, A) sup{ ll x - y ll : Y E A}, r (C, A) inf {r (C, x , A) : x E C}, diam (C) sup{ ll x - y ll : x, y E C} . If A C, we write r ( C ) for r ( C, A). nonempty closed convex subset C of X is said to have normal structure if for any nonempty closed convex subset K of C, we have either r (K) < diam (C) or r (K) 0 (Le., K is a singleton). A Banach space X is said to have normal structure (respectively, weakly normal =

=

=

A

=

=

if every bounded nonempty closed convex (weakly compact) subset be a bounded sequence in X such that of X has normal structure. Let for any x co :n the limit structure)

E {Xn E N},

{Xn }�= 1

�(X)

=

nlim -+ oo Il x - Xn II

{ n E N},

exists. If � is affine (respectively, constant) on co x : n then we say that is a limit-affine (respectively, limit-constant) sequence. ( a) Show that a Banach space X has (weakly) normal structure if and only if X has no (weakly convergent) limit-constant sequence.

{ Xn }�= l

124


(b) Show that every compact convex subset of a Banach space has normal structure. (c) Show that every UeED space has normal structure. (d) Suppose that X is a Banach space that has the UKK property. Show that X has weakly normal structure. Exercise 2.1.9. A (nonlinear) mapping T : 0 C is said to be nonexpan sive if for any x, Y E O, II T(x) - T( y ) 1 1 � Il x - y l l . A Banach space X is said to have the fixed point property if any nonexpansive mapping on a weakly compact convex subset of X has a fixed point; Le. , there is x E C such that T x = x. Let o be a closed weakly compact subset of a Banach space X. ( a) Show that there is a sequence {Xn}�= 1 of 0 such that limn -+oo Il xn T xn ll = O . The sequence {xn}�= 1 is called an approximate fixed point sequence for T. (b) Let d = inf{lim suPn -+oo Il x - xn ll : x E O} . Show that C1 = {x E O : lim sUPn -+oo Il x - xn I � d } is a closed convex invariant subset of C under T. (c) (Kirk) Show that T has the fixed point property if X has weakly normal structure. ---t

2.2

Smoothness

Let X be a Banach space. (GS) A point x on the unit sphere S(X) of X is said to be a smooth point if for any y E X , the limit

+

Y 11 - Il x ll Il x -t..;:...;.:. .:.:..-. :. -...: ...:.:. 1m t-+ o t 1.

exists. A Banach space X is said to be every x E S(X) is a smooth point. (UGS) A Banach space X is uniformly S(X) and any y in X , the limit

Gateaux smooth

(or smooth) if

Gateaux smooth

if for any x in

+

1. Il x t y ll - Il x ll 1mO t-+ t

.:.:..--�-.:.:.......;.:.

exists and is uniform in x E S (X ) . (FS) A point x E S(X) is said to be Frechet smooth if there is a functional x* such that Il x y ll - ll x ll - (x* , y) O . lim y -+ o I l y ll A Banach space X is said to be Frechet smooth if every x E S(X) is Frechet smooth.

+

=

125 (US) A Banach space X is uniformly smooth if for any x in S(X) and y in X, the limit + ty..:...l -Ilx 1m .:.:.....:..:... - I.:. ---.: l x l ...-:.:. t exists and is uniform in x, y E S(X ) . Since I . I is a convex function, for any x, y E S(X ) , both limits - I.;. + ty----"-l -l x l ...;.;, Il x ' "'-1m I l x + t Y1 1 - I l x l an d l1m --' t t

2. 2. SMOOTHNESS

1.

t->O

1·

tiD

tLO

exist, and lim tLO

I l x + ty l - I l x l - lim I l x + tY1 1 - I l x l - O . t t >

tiD

Hence we have the following lemma:

x a unit vector in X . (1) A point x E S(X) is a smooth point if and only if for any y E X , lim I l x + t Y1 1 + I l x - t y l - 2 1 x l =

Lemma 2 . 2 . 1 . Let X be a Banach space and

t->O

o.

t

x E S(X) is a Frechet smooth point if and only if lim I l x + y l + I l x - Y I - 2 1 x l = O . Il y l Let X be a Banach space and x a vector in S(X) . An element x· E S(X · ) is said to be a supporting functional at x if (x \ x) = 1. By the Hahn-Banach theorem, for any x E S(X ) , the set J ( x ) = { x · E S(X · ) x is a support functional of X at x } is nonempty. The set-valued mapping J x E S(X) J ( x · ) is called a duality mapping of X . (2) A point

y->O

:

*

:

�

Theorem 2.2.2. (Smulian) Let X be a Banach space and of X . Then the following are equivalent:

x a unit vector

x is a smooth point of X ; (2) The set J ( x ) i s a singleton; (3) The mapping J is norm to weak· continuous at x ; i. e . , if {Xn}�= l in S(X) such that lim n ->oo Xn = x , and x � E J ( xn ) for all n E W, then {X�}�= l converges weak· to an element in J ( x ) . (1) The vector

126


Proof: (1) ::} (2) . Let x be a unit vector of X. Suppose that J(x) is not a singleton. Let x* , y * be two distinct elements in J(X). Then there is y E SeX) such that (x* - y* , y) > > Hence

E o.

. Il x + ty ll + Il x - tY 11 - 2 11 x ll 1m III f t -tO t > 1. ( x* , x + ty) + (y* , x - t y) - 2 1I x ll _ 1m - (x * - y * , y » t -tO t

1.

_

_

E,

a contradiction. We have proved the implication (1) ::} (2). (2) ::} Let {yn}�= l be a sequence in S(X) that converges to x. For each n E N, choose y� E J( Yn) . We notice that if y * is a weak* limit point of {y�}�= l ' then y * E J(x). Since the unit ball of X* is weak* compact, the assumption implies that {y�}�= l converges to x* = J(x) weakly. We have proved the implication (2) ::} ::} (2) . Let J(x) = {x* }. Suppose that ::} (1). It is easy to see that x is not a smooth point of X. Then there exist y in SeX), an > and a null sequence {tn}�= l in 1R such that

( 3 ).

(3 ).

(3)

(3)

E 0,

Let x� be any element in J ( I �!::�II ) and y� any element in J ( II�::::::�II ) . Then We have (x� - y�, y) On the other hand,

>

� (2 + dn - (x� , x) - (y�, x) ) � E.

t

± ±

x t ny 1. n -t1moo II X t ny II = x. The assumption implies that both { y�}�= l and {z�}�= l converge weak* to x* , 0 a contradiction. The proof is complete. X.

Lemma 2.2.3. (Smulian) Let X be a Banach space and x a unit vector of Then the following are equivalent:

(1) x E S(X) is a Frechet smooth point of X.

E 0

There is x* E S(X*) for any > there exists 8 > any y * E B(X*) if (y * , x ) > 1 - 8, then we have Il y * - x* I

(2)

(3) J is norm to norm continuous at x, S(X), limn -t oo Xn = x and x� E J(x n ) norm.

Proof: It is easy to see that (2) implications (1) ::} (2) and ::} (1) .

(3)

::}

( 3 ).

0 such that for < E.

i. e. , for any sequence {xn}�= l in imply that {x�}�= l converges in

So we need to prove only the

1 27

2.2. SMOOTHNESS

(1) x. y S(X* )

X, x* 0

x

and a support functional => (2) . Let be a F'rechet smooth point of of Suppose that (2) is not true. Then there is E > for any E N, there is such that �E

x* 11

n

(y� , x) � 1 - -;n . By the definition of the norm of X *, for each n E N, there is Yn E S( X ) such that (y� - x* , Yn) � SO for any n N, Il x + n - 1 . Yn ll + Il x - n - 1 . Yn ll - 2 11 x ll nY- 1 � n ( ( y�, X + : ) + (x * , x - Y: ) - (y� , x) - (x * , x) - :2 ) � (y� - x* , Yn ) - -n � (n -n 1 ) , a contradiction. We have proved the implication ( 1 ) (2) . 3 ) ( 1 ). We notice that the assertion (3 ) implies that J(x) is a singleton. ( Suppose that x is a vector in S( X ) that is not a F'rechet smooth point. Then there are an 0 and a null sequence { Yn }�= l in X such that 1m Il x + Yn ll + Il x - Yn ll - 2 11 xl l n ->oo IIYn l1 For any n N, let y� be any support functional of II:��:II and z� any sup port functional of J ( I :=�:II ) . The assumption implies that both { Y�}�= l and {z�}�= l converge to x*( = J (x)). So . f / * * Yn O = �� \ Yn - Zn ' IIYn l1 ) . f (y� , x + Yn) + ( Z�, x - Yn) - 2 11 x ll 1m m - n ->oo l l Yn II . f II X + Yn ll + Il x - Yn ll - 2 11 x ll m - 1m n ->oo II Yn II a contradiction. We have proved the implication ( 3 ) The proof is (1). complete. IIY� -

> E

E.

and E

E

E

=>

=>

E

>

>

1.

E.

E

1·

> 1.

>

- 1.

=>

E,

o

X be a Banach space. ( 1 ) If X* is strictly convex, then X is smooth. (2) If X* is smooth, then X is strictly convex. Proof: (1) Let X be a Banach space. Suppose that X is not smooth. Then there is a unit vector x of X such that J(x) contains at least two elements x* and y*. Note: For any 0 A � 1, AX* + ( 1 - A )Y* J(x) � S( X*). This implies that X* is not strictly convex. Theorem 2.2.4. Let

�

E

1 28


(2) If X is not strictly convex, then there exist x, Y E SeX) such that x =1= Y and � (x + y ) E SeX). Then for any x* E J( X�Y ) , we have (x* , x) (x * , y) This implies that x, y are two distinct support functionals of x* . So X* is not 0 smooth. =

[16] )

Theorem 2.2.5. (Day A Banach space only if X* is uniformly convex.

X

=

1.

is uniformly smooth if and

Proof: Suppose that X is uniformly smooth. For any f >

0

b > such that

0, there exists a

" I l x + Y l l x + I l x - Y ll x < 2 + fl l � x for all x E SeX) and Y E X with I I Yl l x < b. Suppose that x* , y * E S(X*) with Il x* - Y * llx · > f . Then there is Y E X such that II Y l l x � and (x* - y* , y) > � . Hence =

I l x* + Y* ll x · sup{ (x * + y * , x) : x E SeX)} = sup{ (x * , x + y) + (y * , x - y) - ( (x* - y *), y) : x E SeX)} : x E SeX) � sup I l x + Y l l x + Il x - Y l l x < 2 + f l l y l l x _ fb = 2 _ fb . 8 4 4 This implies that X* is uniformly convex. Suppose that X* is uniformly convex. Then for any f > there exists a 8 > such that I lx* + y * ll x . < 2 - b for all x* , y * E S(X*) with I Ix* - Y * l I x . > f . For any x E SeX) and Y E X with IIY ll x < % ' select x* , y * E S(X*) such that (x* , (x + y ) ) Il x + yl l x and (y*, (x - y ) ) I l x - y l l x . Then Il x * + y * Ilx · (x * + y * , x) (x * , x + y) + (y * , x - y) - (x * - y * , y) = Il x + Y l l x + I l x - Yllx - ( (x * - y * ), y) > 2 - 4 11Yllx 2 - 8. =

-�

{

}

0,

0

=

=

�

=

�

Hence Il x* - y * l l x . < f , and Il x + Yllx + Il x - Yllx = (x * , (x + y ) ) + (y * , (x - y ) ) = ( (x* + y *), x ) + ( (x * - y *), y) � 2 + fll y llx . The proof is complete.

o

([19, p.63 Theorem 6.7] ) A Banach space X is WUC if

Theorem 2 . 2 .6. and only if X* is UGS .

129

2.2. SMOOTHNESS

Proof: Suppose that X* is UGS. Let y * be any unit vector in X* . For any E S(X*) and It I < b,

f > 0, there exists a b > 0 s.u ch that for any x*

(2.20) Il x * + t Y * 11 + Il x * - t Y * 11 � 2 + flt l · Let {Xn }�=l and { Yn }�= l be two sequences in S(X) such that limn -t oo Il xn + Yn ll = 2 . By the Hahn-Banach theorem, there exists a sequence {X�}�=l in S(X*) such that nlim -+ oo (x� , (xn + Yn)) = 2. Since Xn , Yn E S(X), we must have nlim -t oo (x � , Yn) = l . -> oo (x� , xn ) = nlim

By (2.20) , for any It I < b, we have ( (x� + ty * ), Xn ) + ((x� - ty * ), Yn ) � 2 + fltl . Thus there is a number N (independent of y*) such that for any n N and It I � b. (ty * , (xn - Yn) ) � 2 + f b - (x � , xn) - (x� , Yn ) � 2 f b. In particular, for any n > N and any y* E S(X*), we have b(y * , (Xn - Yn)) � 2 f b for every n no . This implies that X is WUC. Conversely, assume that X* is not UGS. Then there exist y * E S(X* ), f > 0, a sequence {x�}�= l in S(X*), and a strictly decreasing null sequence { t n}�= l such that Il x� + tn Y * llx· + Il x� - tn Y * llx· � 2 + ftn . Select {Xn }�= l and { Yn }�= l in S(X) such that ((x� + tny *), xn ) > Il x� + t n Y * Il x - _ tnn and

�

�

1,

Then

nlim -t oo (x� , Xn) = nlim -> oo ((x� + tn Y *), xn) - tn (y * , xn) = nlim -t oo (X� ' Yn) = nlim -+ oo ((x� - tn Y * ), Yn) + tn ( Y * , Yn) = l . We have proved that limn -t oo Il xn + Yn ll = 2. Notice that for any n E

( (x � + tn Y * ), xn ) + ( (x � - tny * ), Yn ) Hence for any n E

N.

(tny * , (xn - Yn))

N,

� 2 + (f - �) tn

� ( f - �) tn + 2 - (x� , xn) - (x� , Yn) � (f - �) tn .

This implies that X is not WUC. The proof is complete.

o

130

CHAPTER 2. CONVEXITY AND SMOOTHNESS Theorem 2 . 2 . 7. For any Banach space X, X is FS if X* is LUC . Proof: Let X be a Banach space such that X* is LUC. By Theorem

Xo

2 . 2 .4

(1), X is smooth. Let any vector in S(X), the unit sphere of X, and let Then J( xn ) converges to {xn}� l be a sequence in S(X) that converges to J(xo) weak* . So we have 1n -+1m· oo ( J(xn ), (xo) ) + (J(xo), xo) , and limn -+oo I j J( xn ) + J(xo) 11 = But X* is LUC. This implies that J(xn) 0 converges to J(xo) in norm. By Theorem X is FS.

Xo.

2.

2

-2

2.2.3,

Example 2.2.8. ( 1 ) For any 1 0 and (3 of {hl ,j }� l such that for all k > 1 ,

>

j hl ,k S; L k =l

9

{gnJ� l of {gn }�=l

for all j.

0, there is a subsequence

{h2, k } k"= 1

Suppose that the claim is not true. Then there are 0 < € < , and a subsequence {h3, d k"= 1 of {hl ,j }� l such that It follows that for any k E N,

k k 1 h = kh 1 (h3,k+ l - (3h3,i) 1 I gl � Il L i=l 3 ' i l l (3 1I 3,k+ - L i=l k k 1 = (3 - 1 I kh 3,k+ l - L ( h 3,k+ 1 - (3 h 3, i ) + + L ( h 3,k+ l - (3 h 3, i ) - II · i =l i=l

k gI l � (3 - 1 1 I kh3,k+ l - L (h3, k+ 1 - (3 h3, i ) + 1 1 � (3 - 1 ( k , - (k(r - E)) = (3 - 1 k € , i=l and this is contradictory for large values of k . We have proved our claim. Now fix 0 < € < � and construct a subsequence {h 2 ,k }k:: l of {hl ,j }� l such that I h 2, k - (3 h 2 , 1 1l � , - € for all k 1, where (3 = 2 I 1 g l / €. Put h 4 , 1 = (3- 1 ((3 h2, 1 - g) + and h4 ,k = ( h2 ,k - (3h2,I) + for all k 1 . It is clear that h4, 1 1\ h4,k = 0 for all k 1. By the choice of the sequence {h2 ,k } k:: l ' we also have that k k h4,j h2, S; g , jL=l S; jL=l j I h4, k l � , - € for k 1 , >

>

>

and

>

Applying this argument to the sequence {h 4,k }k: 2 instead of {hl ,j }� l ' and with €/2 instead of € , we can produce a new subsequence for which the norm of its elements are greater than or equal to , - € - € /2, each partial sum is less

3. 1 . KO THE FUNCTION SPACES

147

than or equal to g, and the first two elements are mutually disjoint and disjoint from the rest of the sequence. Continuing by induction, we obtain a sequence {hs , dk= I ' of mutually disjoint elements of such that I l hs , k l l � , - 2€ and E�= 1 hS ,j � 9 for all k. Hence for any finite sequence { ad � I '

E,

sup{ l aj l : 1 � j � N} . inf { ll hs ,j l l : j E N} N

I l jL= 1 aj hs,j " :c:; I l g ll . sup{ l aj I : 1 � j � N } . This implies that {hs ,j } � 1 is a co-sequence. :c:;

D

It is easy to see that CO is a-complete. Hence the converse of Theorem 3.1.3 is not true. Let r be an uncountable set and X the subspace of £oo (r) spanned by co (r) and the function identically equal to one. Then X is a-order continuous, but not order continuous. On the other hand, £00 is a a-complete Banach lattice that is not a-order continuous. The following proposition shows that a a-complete Banach lattice that is not a-order continuous must contain a copy of £00 ' Proposition 3 . 1 .4. [32, Proposition 1.a.7]) A a-complete Banach lattice that is not a -order continuous contains a subspace isomorphic to £00 '

E

{gn}�=1 {gl -gn}�= 1

Proof: Assume that is a nonconvergent decreasing sequence in with is increasing, order bounded, and = O. The sequence

/\r;:= 1 gn not convergent. It follows from the proof of Theorem 3.1.3 that there exists a sequence {hn}�=1 of mutually disjoint positive elements in E that is equivalent to the unit vector basis and for which 0 < E�= 1 hj :c:; gl, k � 1. For a = { a d� 1 E £00 ' with ak � 0 for every k , we put CO

T

E,

It is easy to see that is an isomorphism from the positive cone of £00 into D and it can be extended uniquely to an isomorphism from £00 into Theorem 3.1.4 shows that a separable a-complete Banach lattice is a-order continuous. It is easily seen from the definition that a separable a-order con tinuous Banach lattice is already order continuous. The following proposition shows that every a-complete and a-order continuous Banach lattice is order continuous. Proposition 3 . 1 . 5 . [32, Proposition 1.a.8] Let be a Banach lattice . Then

E.

E

the following are equivalent: ( 1 ) E is a-complete and a-order continuous . (2) Every order bounded increasing sequence in E converges in the norm.

CHAPTER 3. KO THE-BOCHNER FUNCTION SPACES

148

E is order continuous . (4) E is orner complete and orner continuous.

(3)

Proof: (4) ==} (1) follows directly from the definitions. We need to prove only that (1) (2) , (2) ==} (3) , and (3) ==} (4) . (1) ==} (2) . Let { gn }�=1 be an order bounded increasing sequence. Since is a-complete, Vr:= 1 gn = 9 exists and { g - gn}�= 1 is a decreasing sequence with /\r:= 1 (g - gn) = O. But is a-order continuous. This implies that

E

E

nlim --+oo 9 - 9n = O.

E

==}

(1). Clearly, The statement (2) implies that is a-complete. Let {gn }�= 1 be a decreasing sequence with /\r:= 1 9n = O. Then {91 - 9n}�= 1 is an increasing sequence that is bounded by 91. By (2) , {gl - gn }�= 1 is a convergent sequence, say it converges to h . It is easy to see that 91 - h is the g.l.b. of { gn }�= I' So h = g l, and { 9n }�= 1 converges to O. (2) ==} (3). Let { go Jo EA be a downward directed set such that /\ o EA go = O. If the net { 90 } OEA does not converge to then there are > and a decreasing sequence {90j }� 1 in this net such that 1 90j - gO; + 1 " for every j . But { 901 - 90j }� 1 is an order bounded increasing sequence. This contradicts (2). (3) ==} (4). Let be an order bounded set in By adding to the finite suprema of its elements, we may assume that = { go J o E A is upward directed. Let be the set of all upper bounds of i.e., (2)

0,

U

V

U

& 0 �&

E.

U

U, V = {h : h E : h � 9 for all E U} . We claim that 0 is the g.Lb. of V - U . Let hI � 0 such that hI h - 9 for all h E V and 9 E U . Then h - hI V for all h E V. By induction, h - hI V for all E N. Thus I hI l i = o. By (3), for any > 0, there are an a A and an h E V such that I h - 90" � and therefore, I g,6 - go I for every f3 > a in A. Thus, the net {90} OE A converges to an element which is the Lu.b. of U. Let E be a Banach lattice. Then for any 9, h E, 1 9 1 � I h l implies 1 / 9 1 / � 1 / h 1 / . The lattice E is said to be strictly monotone if 1 9 1 > I h I implies 1 / 9 1 / > 1 / h I / , E is said to be locally uniformly monotone if for any 9 > 0 with 1 / 9 1 / = 1 and > 0, there is a & > 0 such that 9 � h � 0 and 1 / 9 - h l � imply I / h l 1 - &, E is said to be uniformly monotone if for any > 0, there is a & > 0 such that 9 > h � 0 , 1 9 - h l � and 1 / 9 1 / = 1 imply I / h l / � 1 - & . By the definition of locally uniformly monotone, every locally uniformly monotone Banach lattice is 9

E

E

n

€,

�

€

9

o

� €

E

n

E

E

€

€

€

€

�

order continuous.

Example 3.1 .6. (1) It is known that any strictly convex (respectively, lo

cally uniformly convex, uniformly convex) Banach lattice is strictly mono tone (respectively, locally uniformly monotone, uniformly monotone). (2) Ll is not strictly convex; but it is uniformly monotone.

149

3. 1 . KO THE FUNCTION SPACES (3)

The mapping

T

T Roo - R2 is defined by T(aj ) = (�; ) . :

Then is a one-to-one positive operator. By Proposition the norm 00 1 = +L

1 1 (aj ) 1 1 ( 1 (aj ) l � j = 1

I�;j1 2 ) / 2

2.1.7, Roo with I .I E

is a lattice is uniformly rotund in every direction. Clearly, this norm norm. So is strictly monotone. Let be a complete measure space. A real Banach space consisting of equivalence classes modulo equality almost everywhere of locally integrable real-valued functions on is called a Kothe function space if the following con ditions hold. (i) If a.e. on 0, with measurable and E then E and < 00 , the characteristic function (ii) For every A E E with of A belongs to Clearly, every Kothe function space is a Banach lattice in the obvious order 0 if � 0 a.e.). By (i) , every Kothe function space is O"-order complete. The assumption that every E is locally integrable implies that for every is finite measurable set A E the positive functional In well defined and thus bounded; i.e. , it is an element of In general, every measurable function on 0 such that for every E defines an E L1 element in by ( , = In Any functional of this form on is called is called an integral, and we denote the set of all integrals by the Kothe dual of It is an immediate consequence of the Radon-Nikodym theorem that a functional E is integral if and only if for every sequence {gn} in with 1 0 a.e., we have ( , n - O. In particular, if is order continuous, then = E' . We claim that the converse is also true. Suppose that = Let {gn}�= 1 be an order bounded increasing sequence of positive elements in that converges pointwise a.e. to some in By Lebesgue ' s dominated convergence theorem, for any , = nlim

(Roo , I . I ) (O, �, J.L )

� Igg (w ) 1 h:::; I h (w) 1 I l l :::; I l l · E. (g � g(w )

J.L(A)

g E �,

G Gg (J.L ) E* G g) G(w )g (w) dJ.L. E .) G E* E gn(w ) GI g ) I E*

g E 1A

h E,

9

9

(W ) 1 A (W ) dJ.L g E *. g E E E'. (E' 1--+

E* E'. E

E

E. 9 G E E' oo j Ggn dJ.L j G9dJ.L' This implies that gn - 9 weakly. We claim that I l gn - gi l - O. Suppose the claim is not true. Then there exists a sequence {Hn}�= 1 of positive linear functionals such that (Hn' 9 - gn) > {3 > 0 for some {3. Hence for any � we have -+

n

m,

Let H be any w* limit point of {Hn }�= I' Then for all n E N, (H, {3.

g - gn) �


150

E

This is a contradiction. So is order continuous. Theorem 3.1 .7. ( Lebesgue ' s Dominated Convergence Theorem [9] )

Sup pose that E is an order continuous Kothe function space over a measure space (n, IL) . Let be an element in E, and {h n }�= l a sequence in E such that I hn l � and for almost all w E n, limn-+ oo hn (w) = h ew) . Then {hn }�= l converges to h in E . 9

9

E

Proof: Let be an order continuous Koth function space over a measure space (n, IL), 9 a positive element in and {hn }�= l ' Suppose that for all n E N, I hn l � 9 and that the sequence {hn }�= l converges to h almost everywhere. For any n E N, let

E

00 k=n

Then { ¢n }�= l is a decreasing sequence such that continuous. We have lim O. n-+ oo ¢n =

I\�= l ¢n = O. But E is order

o This implies that {hn }�= l converges to h . Let be a Banach lattice. Recall that an element g E E is said to be a weak unit if for h E 9 1\ h = 0 implies h = O. Clearly, every separable Banach lattice has a weak unit. Indeed, if {Xn }�= l is a dense set of then l:�= l I �� I is a weak unit. The following theorem shows that any order continuous Banach lattice with a weak unit is lattice-isomorphic to a Kothe function space over a probability space. For a proof, see [32 , Theorem l .b.14] .

E

E,

B(E),

Theorem 3.1 .B. Let E be a separable order continuous Banach lattice . Then there exist a probability space (n' �' IL) ' an ideal E of Ll (n, � , IL), and a lattice norm I . l i E on E such that (1) E is order isometric to (E, I . l i E) ' (2) E is dense in Ll (n, � ' IL) , and Loo (n, � , IL) is dense in E. (3) I l g l l � I l g i l E � 2 11 g l 00 whene ver E Loo(n , � , IL) · (4) The dual oj the isometry given in ( 1 ) maps the space E* onto the Banach lattice E* of all measurable functions G for which I G I E • = sup { 1n Gg dlL : I l g i l E � 1 } < 00. The value taken by the functional corresponding to G at g E E is In Gg dlL · 9

E be a Banach lattice. Recall that a sublattice El is said to be an ideal if y E El whenever I Y I � I x l for some x E El . The following theorem is due to Let

H. Nakano

[32, Theorem l .b.16] .

3. 1 . KO THE FUNCTION SPACES Theorem 3 . 1 .9. A

is an ideal of E** .

151

Banach lattice E is order continuous if and only if E

Proof: Suppose that E is an ideal of E** . Let {xn}�= l be an increasing

sequence of positive elements of E that is bounded in order by an element x E E. Then for any positive x* E E, { (x* , xn) }�= l is a bounded increasing sequence. So {xn}�= l is necessarily a weak Cauchy sequence in E. Let x** E E** be the weak* limit of {xn } �= l ' Note: For any positive x* E E* , (x** , x * ) = nlim -+ oo (x* , xn) :::; (x * , x) (i.e., x** < x ).

By the assumption (E is an ideal of E** ), x** E E. We have proved that {xn}�= l converges to x** weakly (in E). By Mazur's theorem, for any there is a finite sequence {ad:= l of nonnegative real numbers such that 2::7= 1 ai = 1 and

E>

0,

aixi II < E. I l x** - L i= l Note: x** � Xn + l � Xn for all n E N . Thus for any n � k, k

I I

k

Il x ** - Xn :::; l x ** - L aixi " < i= l

E.

We have proved that E is order continuous. Conversely, suppose that E is an order continuous Banach lattice. Let x E E and x** E E** be two vectors such that :::; x** :::; x. Then Px (x**) = x** . Let El = Px (E) and E2 = Px (E** ). By Theorem 3. 1 .8, there are probability space (0, Jl) and two a-algebras El � E2 such that El and E2 are Kothe function spaces over the probability spaces (0, El l JlI ) and (0, E2 , Jll) ' By the proof of Theorem 1.b.14 and Proposition of 1 .b.15 of [32] , we may assume that El is a a-subalgebra of E2 and /-l2 1 E l = Jll . Note that El is order continuous and any element x* in Ei is El-measurable. Hence for any x* E Ei ,

0

in x** . x* d/-l in £(x** l EI) . x * d/-l. =

This implies that x** = £ (x** I El ) E El . The proof is complete.

o

Let E be an order continuous Kothe function space over a finite measure space (0, Jl) such that for any g E E, I l g ll l � I l g i I E . For a bounded subset A of E, the following are equivalent: ( 1 ) A is weakly precompact. (2) For each H E E* , the set HA = {H · g : g E A} of Ll (/-l) is uniformly integrable. Lemma 3.1.10.


152

(2) . Let H be any element in E* . Since E is order continous, H is an integral, and the mapping 9 � H . 9 from into Ll (f-L) is continuous (has norm I H IIE .) . Hence if A is a weakly precompact subset of E and if H is an element in then H A = {H . 9 : 9 E A} is weakly precompact in L 1 ( f-L). By Corollary 1 .3.10, HA is uniformly integrable. (2) :::} (1). Suppose that A is not weakly precompact. By Rosenthal ' s i1theorem, A contains an iI-sequence {gn}�=I' Since the natural embedding from E to L l is bounded, by Koml6s' s theorem, there is a subsequence {gnk }k.: l of {gn}�= 1 such that { � 2:� 1 gn k } := 1 converges almost everywhere to a function h. Fix an H E E' = E . Since {H . gn : n E N} is uniformly integrable, . { ! H 2: � l gnk }�= l is also uniformly integrable. By Vitali 's lemma, we have Proof: ( l )

:::}

E

E*,

*

H t gn k df-L = irn H · h df-L . k=1 This implies that { ! 2: � 1 gn k } := 1 is weakly Cauchy, a contradiction (since {gn}�= 1 is an iI-sequence) . The proof is complete. lim � r

m-+ oo m

in

0

Let A be a subset of an order continuous Kothe function space over a finite measure space that is.not weakly precompact. By the above proof, there is an operator T : E - Ll such that T(A) is not uniformly integrable. By Proposition 1 .3.8 and Lemma 1 .3. 10, we have the following theorem.

Let E be an order continuous Banach lattice. Any i I -sequence of X contains a complemented iI -subsequence. Theorem 3 . 1 . 1 1 . (Tzafriri)

By Theorem 1 .7.7 and Theorem 3.1 . 1 1 , we have the following theorem. Theorem 3.1. 12. Let E be a Banach lattice. An order continuous Banach lattice E has property (V*) if ( and only if) E is weakly sequentially complete. Let E be an order continuous Kothe function space over a probability space (0, E , f-L ) such that for any g E E, II g l h ::; I l g i I E . A bounded sequence {gn }�=l in E is said to be equi-integrable (p-uniformly integrable if E = Lp for some 1 ::; < ) if for any > 0, there is M such that for any n E N, p

00

€

Vitali showed that any uniformly integrable sequence in L1 ( f-L) that converges pointwise converges in norm. The following lemma is an extension of Vitali ' s lemma.

Let E be an order continuous Kothe function space over a probability space (0, f-L) such that for any f E E, I f i l l ::;; I l f i I E. For any bounded equi-integrable sequence {gn}�=I ' if {gn}�= 1 converges to almost everywhere, then g E E, and {gn }�= 1 converges to in E. Lemma 3.1. 13. (Vitali ' s Lemma)

9

9


153

Proof: Let E be an order continuous Banach lattice, and { gn }�=1 a bounded equi-integrable sequence that converges to 9 a.e. Fix > O. Then there is M > 0 such that for any n E N,

E

I l gn . l[1 gn l�Ml l iE < E. Note: { (gn -M) 1\ M}�= 1 converges to (g - M ) 1\ M almost everywhere. By Lebesgue ' s dominated convergence theorem, we have that {(gn - M ) 1\ M}�=1 converges to (g - M ) 1\ M in E. Since E is an arbitrary positive real number, this implies that { (g - n ) 1\ n} �=1 is a Cauchy sequence. Clearly, it converges to g. It is easy to see that { gn}�= 1 also converges to g. The proof is complete. Let E is an order continuous Kothe function space over a finite measure space (n, f.1) It is easy to see that a bounded subset K of E is equi-integrable if for any E > 0, there is a 8 > 0 such that for any measurable subset A of n with f.1(A) < 8 and any E K , we have I I g . 1 A IIE < E. Lemma 3 . 1 . 14. Let E be an order continuous Kothe function space over a probability space (n, f.1) such that for any g E E, I l g l i � I g i l E . Then for any bounded subset A of E, if A is equi-integrable, then A is relatively weakly compact. Proof: Let E be an order continuous Kothe function space over a proba bility, and A a bounded equi-integrable subset of E. Then for any H E E*, {gH : E A } is uniformly integrable. By Lemma 3.1 . 10, A is weakly precom pact. Let {gn}�=1 be any weakly Cauchy sequence in A. The proof of Lemma 3.1.10 shows that there are a measurable function h and a subsequence { gn k }k:: l such that (a) { � 2: :=1 gn k } :=1 converges to h a.e. (b) For any H E E *, limn ->oo f H . gn df.1 = f H . h df.1. Note: { gn : n E N} is equi-integrable. {� 2::=1 gnk } :=1 is also equi integrable. By Vatali ' s Lemma, h E E. The proof is complete. Recall that a Banach space X is said to have property ( ) if for every weakly Cauchy sequence { Xn }�=1 in X, there exists a wuc series 2: �= 1 Yn such that the sequence {Xn - 2:7=1 Yj }�=1 converges weakly to zero. Theorem 3 . 1 . 1 5 . ( Tzafriri [ 32, Proposition 1 .c.2 ] ) Any order continuous Banach lattice E has property ( ) V

V

V

V

V

D

9

9

D

u

u .

Proof: We first observe that it suffices to prove the assertion for separable

lattices. Thus, by Theorem 3.1.8, we may assume, without loss of generality, that E is a Kothe function space on some probability measure space (n, �, f.1) and that every element of E* is an integral ( Le., E* = E'). Let {gn}�=1 be a weak Cauchy sequence of functions in E. Note: L oo � E* and Ll is weakly sequentially complete. The sequence { gn}�=1 converges weakly


154

Ll .

{gn � l Ll . G = ( { Gg dJL . ( ) { }�= (g, gn) � l ) gn l {ad� l ::; ::; PI ql P2aixdk:q2 ai = l {G · gn : n JL G JL nlim -+ oo j G . gn dJL = klim � -+ oo j G · � = k aigi d = j g d .

Fix a in E* E' the Kothe in Let 9 be the weak limit of } = in dual of E . We claim that } = converges to J Note the the limit exists since is weakly Cauchy. Proof of claim: By Mazur ' s theorem, there are a sequence of non . . . negative real numbers and a sequence 0 < such that < < for any k E N, ,,£i:: k converges to 9 1, and the sequence { ,,£ ;!pk P almost everywhere. Note: E N} is uniformly integrable. By Vitali ' s lemma,

iP

We have proved our claim. Note that the claim implies that 9 is an element in E** and } = converges to 9 in the weak* topology. Now set

{gn � l

Bn = {w E n : n - 1 � I g (w) 1 < n } , Cn = Uf=n+ 1 Bj , hn = 1 Bn , for n E N. 9 .

Then, for any

G E E* , we have

and

nlim -+ oo inr (gn - jt= l hj ) G dJL r = nlim-+oo j (gn - g) . G dJL + in\cn gG dJL = O . 0


3.1.16. [32, Theorem l .cA] Let E be a Banach lattice. The following are equivalent: (1) The Banach lattice E is weakly sequentially complete. (2) No subspace of E is isomorphic to (3) Every norm-bounded increasing sequence in E converges in norm. (4) The Banach lattice E is a ( projection) band of E** . Theorem

co .

()

is not weakly sequentially complete, the implication 1 =? (2) is true for any Banach space. (2) =? (3). Suppose the assertion (3) is not true. Let 0 � � � be an increasing nonconvergent sequence in E with I I � 1 for all E N. Then there are 8 > 0 and an increasing sequence such that for any k E N, Proof: Since

CO

{nk }k:l xnl

X l nX2

•

•

•


155

Yk = Xn k +l - Xn k has norm at least 8. Note: For any finite sequence { ak } k=1 of [ - 1, 1] ,

ak k � xn k 1 . Il t i=1 Y l 1 I l l � By Lemma 1.3.6, the sequence { Yn}�= 1 is wuc. By the proof of Theorem 1 .3.14, E contains a Co-sequence, a contradiction. We have proved the implication (2) ( 3 ) . (3 ) (4). The assumption implies that E is order continuous. Let { ga : a E f} be a maximal family of positive elements in E with disjoint support. By Theorem 3.1.9 and the assumption, E is an ideal of E**, and for any a E f and any positive element X** E E**, Pg", (x ** ) = V�=1 (x ** 1\ nga ) E E. We claim that for any positive element x**, the set f1 = { a : Pg", (x **) i= O} is countable and L Pg", (x ** ) E E. aEr l Suppose not. Then there are > 0, x ** E ( X** ) + , and a subsequence { gan }�=1 of { ga : a E A} such that for any n E N. I Pg"' n (x **) 1 > Then the sequence {L�=1 Pg"' k (x** )} :'= 1 is increasing and bounded (bounded by I l x** I ), but it does not converge in norm. This contradicts the fact E is order continuous. Define a projection P E** E by =}

=}

E

:

E.

---+

aEr aEr where x +* , x �* E E+* and x** = x +* -x �*. The P is a band projection from E** to E. (4) ( 1 ) . By Theorem 3.1.9, E is order continuous. Let { gn }�=1 be a weakly Cauchy sequence of E . Then there is a separable Banach sublattice that contains { gn n E N } . Without loss of generality, we assume that E is separable. By Theorem 3.1.8, E can be considered as a Kothe function space over a probability space (0, /-L) and for any G E E*, there is a measurable function H such that for any h E E, (G, h ) = In Hh d/-L. Since the embedding from E to L1 (/-L) is continuous, { gn}�= 1 is a weakly Cauchy sequence of L 1 ( /-L ) . Note: L1 (/-L) is weakly sequentially complete. The sequence {gn}�= 1 converges weakly in L 1 ( /-L), say it converges to weakly in L 1 ( /-L) . By the argument as in the proof of Lemma 3.1. 10, for any H E E*, nlim -+ oo inr Hgn d/-L = inr Hg d/-L. =}

:

9


156

This implies that 9 let

E

E** and 9 is a weak* limit of {gn};::'= 1 ' For each n E N,

An = { w : I g (w) 1 � n }. Then for any n E N, g . I A n E E and I g l . I A n � I g l · But v;::'= 1 I g l . I A n = I g l · The statement (4) implies that I g l and 9 belong to E. We have proved that E 0 is weakly sequentially complete. The proof is complete.

Theorem 3.1.17. [32, Theorem 1.c.5] Let E be a Banach lattice. The following are equivalent: (1) E is reflexive . (2) No subspace of E is isomorphic to either .e1 or CO .

Proof: The implication ( 1 )

=}

(2) holds trivially in every Banach space. Assume that E is a Banach lattice that does not have any subspace iso morphic to co . By Theorem 3. 1 . 16, E is weakly sequentially complete. By Rosenthal 's .e1-theorem, if E is a Banach lattice that does have have any sub 0 space isomorphic to CO or .e1 , then E is reflexive. Recall that an element g E E is said to be order continuous if limn -+oo h n = 0 whenever {hn };::'= 1 is a decreasing sequence such that h n � I g l for all n E N and I'm hn = O. By Proposition 3.1.5, a a-complete Banach lattice E is order continuous if and only if every g E E is order continuous. The following lemma gives a characterization of order continuous elements in a a-complete Banach lattice. Lemma 3.1. 1S. [29] Let be an element in a a-complete Banach lattice E. The following are equivalent: (1) is order continuous . (2) Let {Pn};::'=1 be a decreasing sequence of band projections and let P = 1\;::'= 1 Pn ' Then limn-+oo (Pn - P) (g) = O. (3) For any sequence Qn of mutually disjoint band projections and g E E, lim Qn ( g) = O . n-+oo 9

9

Proof: Clearly, (1)

(2) � tions, let

�

(2) . (3). For any sequence {Qn};::'= 1 of mutually disjoint band projec 00

Then {Pn};::'= 1 is a decreasing sequence of band projections such that for all O. So (2) � (3) .

n E N, Qn = Pn - Pn+ 1 and 1\;::'= 1 Pn

.


(2 )

157

===> (1). Let 9 be a unit vector in E, and { gn }�= l any decreasing sequence such that I\�=lgn = and gn :::; 9 for all n E N. For a fixed let

0

E > 0,

Pn = P(9n-E I 91)+ , 00

{ Pn }�= l is decreasing. By (2 ), nlim -+oo (Pn - P)(g) = o. lf P(g) =j:. 0, then for any n E N, gn � EPn(g) � EP(g) > O. This contradicts I\�= l gn O. Thus P(g) = 0 and P(gn) = 0 for all n E N. Select n such that I Pn(g) 1 = I (Pn - P)(g) I :::; E. Notice that { gn }�= l is a decreasing sequence and gn :::; for all n E N . For any ml , m2 � n, I l gm l - gm2 1 = 1 (1 - P)(gm i - gm2 ) 1 :::; I (I - Pn)(gm i - gm2 ) 1 + I (Pn - P)(gm i - gm2 ) 1 :::; I Eg l 1 + II (Pn - P)(g) 1 :::; 2 E. Since E is arbitrary, { gn }�= l is a convergent sequence. (3) 2 ). Let { Pn } be a sequence of decreasing band projections. Replac ( ing Pn by Pn - I\ r= l Pk , we may assume that Then

=

9

===>

00

(2 )

Suppose that the statement is not true. By passing to a subsequence of { Pn }�= l ' we may assume that there is a such that I Pn(g) 1 � We claim that there is an such that for any n E N,

E>0

8>0

8.

sup { I (Pn - Pm )(g) 1 : m � n} > E. lf the claim is not true, then there exists an increasing sequence {n k : k E N} such that II (Pn i - Pn HI )(g) 1 :::; -dr. Hence if -ir < � and k > ni , then 2 8 )(g) (Pn Pn j j 1 I I Pk (g) 1 :::; I Pni (g) 1 :::; L + :::; 2i < 2 ' l j=i 00

We have obtained a contradiction, so the claim must be true. The claim implies that there are and a sequence {Qj = Pnj - Pnj + 1 }� l of mutually disjoint band projections such that I Qj(g) 1 which is a contradiction. The proof is 0 complete. It is known that 1 (0, 00 ) is a strongly extreme point in L oo 00 ) . So there is a strongly extreme point in L oo that is not order continuous. The following lemma shows that every denting point of the ball (E) of a Kothe function space is order continuous.

E>0

> E,

B

(0,

158


Lemma 3 . 1 . 19 . [29] Let E be a Kothe function space and let h be a unit vector in E. If h is a denting point of the unit ball of E, then h is order continuous. Proof: Suppose that the element h E S(E) is not order continuous. By

Lemma 3.1.18, there exist € > a and a sequence {Qn}�= l of pairwise disjoint band projections such that

o This implies that h is not a denting point of the unit ball of E. Since every LUC (locally uniformly convex) point is a denting point, we have the following corollary: Corollary 3 . 1 . 20. (J. Cerda, H. Hudzik, and M. Mastylo) Every LUC

Kothe function space is order continuous. Theorem 3 . 1 . 2 1 . Let E be a WUC (weakly uniformly convex) Kothe func tion space. Then both E and E* are order continuous. Proof: It is known that every WUC Banach space does not contain a copy

of fl . Let E be a wuc Kothe function space. By Theorem 3.1.5, if E (E* ) is not order continuous, then E (E* ) contains a copy of foo . It is known that if E* contain foo , then E contains f1 (Corollary 1.3 . 9). In either case, E contains a copy of fl . This is impossible if E is wuc. So both E and E* must be order continuous. 0 Recall that a Kothe function space E is said to have the Fatou property if {9n}�= 1 is an increasing nonnegative sequence in E and sup{ 11 9n ll : n E N} < 00 implies 9 E E and I l g l = nlim I l gn I , ..... where g (w) = lim n ..... oo gn (w ). A Banach lattice E is said to be a Kantorovich Banach space (KB-space) if for any bounded nonnegative increasing {gn}�=l in E, 9 = n gn E E and 9 = nlim gn' ..... oo It is clear that every KB-space is order continuous and has the Fatou property. Theorem 3.1.16 shows that a Banach lattice E is a KB-space if and only if E does not contain a copy of Co . Lemma 3 . 1 .22. (N. Randrianantoanina) Let E be a KB-space ( weakly se quentially complete Banach lattice) and J the embedding from E to L 1. Then J(B(E)) is a closed subset of L 1. oo

V


159

E

E

Proof: Suppose that is a KB-space. By Theorem 3.1.16, is weakly sequentially complete. Let {gn }�=1 be a sequence in such that {gn }�=1 converges to 9 in L1 . We need to show that I l g i l E ::; 1 . By a change of sign, we may assume that 9 is nonnegative. B y passing to a further subsequence, we may also assume that {gn}�=1 converges to 9 almost everywhere. Let 00 hn = (g 1\ /\ gn ) V 0.

B ( E)

j =n

Then {hn } �= 1 is an increasing sequence that converges to 9 a.e. By the defini 0 tion of KB-space, g E and I l g ilE ::; 1. The proof is complete.

E

Let (0, (J, f..l ) be one of the measure spaces { 1 , 2, . . the usual measure ) . For a measurable function g, the defined by

.

}, [0,1] ' or (0, 00) (with distribution function IS

dg(t ) = f..l ( {w E O : I g (w) 1 > t}). Two measurable functions and h are said to be f..l - equimeasurable if dg = dh. A Kothe function space E on (O, �, f..l ) is said to be a rearrangement invariant ( r.i. ) space if the following conditions hold: (i) If g E E and if h is a measurable function such that g and h are f..l equimeasurable, then h E E and I l g l = I l h l . (ii) Either E is order continuous or E has the Fatou property. The Rademacher functions ri , i = 1 , 2, . . . on [0, 1] are defined by rn(w) = 1 n sign sin 2 7l"w. Since f0 ri (w)rj (w) df..l = c5i , j , the Rademacher functions form an orthonormal basis sequence in L 2 (f..l ) . Recall that a set { fa } a EA of random variables is said to be independent if for any finite subset { Qi } i=1 � A and any choice of open sets {Ddi=1 in X , we have n f..l {w E O : fai (W) E Di , 1 ::; i ::; n}) = II f..l ( {w E O : fa (w) E Dd) · i=1 9

;

Example 3.1.23. (1) Every Banach space with a symmetric basis is an

r.i. space over N . (2) (Lorentz Function Spaces ) Let w be 1a positive nonincreasing function on (0, 00) such that lim t�oo w(t) = 0, f0 w(t)dt = 1, and fooo w(t) = 00. For any 1 ::; p < 00, the Lorentz function space Lw,p(O, 00) (respectively, Lw,p[O, 1]) is the space of all measurable functions 9 on (0, 00) ( respectively, [0, 1]) for which (g* (t))PW(t) dt < 00.

j

For any 9 E Lw ,p, the norm of 9 is defined by It is easy to see that the Lorenta spaces are r.i. spaces.

160

CHAPTER 3 . KO THE-BOCHNER FUNCTION SPACES

(3) (Orlicz Function Spaces) Recall that an Orlicz function is a continuous convex function M from [0, 00) (or [0 , 1] ) to [0, 00) such that M (O) = 0, and limt-+oo M (t) = 00 . The Orlicz function space L M (O, 00) (respectively, L M (O, 1)) is the space of all measurable functions 9 on [0 , 00 ) (respectively, [0 , 1]) such that

J M ( l g (t) 1 1p)dt < 00 for some p < 00. The norm of E L M(O, 00 ) (respectively, E L M(O, 1)) is defined by

9

I g il

=

{

inf p > 0 :

J M( l g (t) l lp)dt � I } .

9

It easy to see that the Orlicz spaces are r.i. spaces. It is known that the Rademacher functions ri form a sequence of independent J.L-equimeasurable random variables on [0 , 1] So {ri }� 1 generates a symmetric space, Le. , any permutation 7r of { I , . . . , n} and for any finite real numbers { k } k=1 ,

.

a

l i kt=1 l ak l rk l = I kt=1 ak r� (k) l I ·

Recall that a Kothe function space E is said to be characteristically order con tinuous if lim I IA liE = 0. Il ( A ) -+ O Lemma 3.1.24. Let E be a characteristically order continuous Kothe func

tion space on [0 , 1]. Then for any {3 > 0, there is an integer k such that for any finite sequence {ni }7=1 of N, k

rn < k. Il L i =1 i l i E (3

1 1 [0, 1] liE = 1.

By as sumption, there is , > ° such that I IA liE � � whenever J.L ( A ) < , Note: The Rademacher functions ri in L 2 (0, 1) form a sequence that is equivalent to the unit vector basis of £2. We have that 1 L: �=1 rk 2 = 1 / 2 for all For any let Proof: Without loss of generality, we assume that

I

k E N,

k

l k

Then limk-+oo J.L(Bk , ,B/2 ) = 0. Select such that J.L (B k,.B/2 ) < , . For any finite sequence {ni }7=1 of let

N,

. k E N.

161


J-L (C) = J-L(Bk, /3/2 ). This implies that 1I ( k rn ) � /3 · 1 1 [0, IJ liE + 1 1 el i E < /3 . h� t; l i E 2

Since {rn}�=1 is Li.d.,

!

o The proof is complete. Let us recall the following fact from Chapter 1: Fact 1 . 1 . 1 . Suppose that {Xn}�=1 is a bounded sequence in X that is not weakly null. Then there are x* E S ( X* ) , /3 > 0, and a subsequence { Y }� 1 of j {xn }�=1 such that for any j E N,

So for any finite nonnegative sequence

{aj }�1 with L;: 1 aj = 1,

a I If: j =1 j Yj l l � /3 . We have the following corollary: Corollary 3 . 1 . 2 5 . Let E be a

characteristically order continuous Kothe function space on [0 , 1] . Then the system {ri }� 1 of Rademacher functions con verges to zero weakly. It is easy to see that if E is an r.i. function space on [0, 1] that is not equal to L oo [O , 1] up to equal norm, then E is characteristically order continuous. By Corollary 3.1 .25, the Rademacher functions form a weakly null sequence. On the other hand, the system of Rademacher functions in L oo [0 , 1] is equivalent to the unit basis of .e 1 . So we have the following theorem. Theorem 3 . 1 .26. [ 32, Proposition 2.c. 10] For any r. i . function space E on [0 , 1], the Rademacher functions form a weakly null sequence in E if and only if E is not equal to L oo (O, 1 ) up to an equivalent norm. Exercises

Exercise 3 . 1 . 1 . Let X be a Banach space. Recall that a subspace Y of X* is said to be a norming subspace of X* if for any x E X, I l x l = sup { l (x* , x ) 1 : x* E B( Y ) } . (a) Show that Y is a norming subspace of X* if and only if B( Y) is weak* dense in B ( X* ) . (b) Let E be a Kothe function space over a O"-finite complete measure. Show that the Kothe dual E' of E is a norming subspace of E* if and only if for any nonnegative sequence {gn}�=1 and 9 in E, gn(w ) i g(w) a.e. implies I l gn li -+ I l g l · Hence if E has the Fatou property, the Kothe dual E' is a norming subspace of E* .

162


( )

Exercise 3.1.2. Diaz Let E be a Banach lattice such that Co is not a quotient space of E . This implies (a) Show that E does not contain a complemented copy of

£1.

that E* has no co-sequence ( Le., E* is a KB space ) . (b) Show that E is a Grothendieck space. Thus a Banach lattice E is Grothendieck if and only if CO is not a quotient space of E.

(

E be an order con Show that either E is reflexive or E contains a )

Exercise 3.1.3. S. Dlaz and G. Schliichtermann Let

tinuous Kothe function space. quotient isomorphic to CO .

3.2

Strongly and Scalarly Measurable Functions

Let (0, E, tt) be a complete measure space and X a Banach space. A function f : 0 - X is said to be simple if there exist {Xi }i= 1 in X and {Ei } i:: 1 in E such that f = L: �1 a i l Ei ' A function f is said to be (strongly) measurable if there exists a sequence of simple functions {fn }�= l with limn-+oo I l fn - f l = ° a.e. A function f is said to be weakly measurable if for every x* E X, ( x* , f( ·)) is a measurable function. A function f : 0 - X* is said to be weak* measurable if for any X E X, (f ( . ) , x ) is measurable. The following examples show that in general, weak* scalar measurability is much weaker than scalar measurability.

Some examples of w* scalarly measurable functions. (1) (Birkhof/) Let X = £2 ([0 , 1]), and let ( t ) ; E [0, 1] be the canonical basis of X*. Define F : [0 , 1] - X* by F(t) = ; Then for x E X we have (F(t), ) = ° except on a countable set. Hence F is w* scalarly measurable. However, for s =1= t, I F(s) - F (t) 1 = y'2, so F is not strongly measurable . (2) ( Birkhof/) Let X L1(0, 1). Then X* = L oo ( O, 1). Define F : [0 , 1] L oo ( O, l ) by F(t) = l [o,tj ' For any h = h+ - h_, the functions t f-+ (F(t), h + ) and t _ (F(t), h_ ) are increasing, and so are measurable. For s =1= t, I F(t) - F(s) 1 = 1, so F is not Bochner measurable. However, a function ¢ : 0 - X is scalarly measurable if and only if it is

Example 3.2.1.

e

e .

x

=

weak* scalarly measurable when considered as valued in X** . Theorem 3.2.2. ( Pettis ' s Measurability Theorem ) A function ¢ : 0 - X

is Bochner measurable if and only if it is scalarly measurable and essentially valued in a separable subspace of X . Proof: The necessity is obvious. For the converse, one can as well assume

that X is separable. Let { xn : n E N} be a countable dense subset of B ( X ) and

163

3.2. STRONGLY AND SCALARLY MEASURABLE FUNCTIONS

{X� : n E N} of S(X*) such that for any n E N, (x� , xn) I l xn l . Let B(x, ) denote the closed ball of X with center x and radius Then f - l (B(x, » ) n {w : I (x�, f(w) x) 1 � } is measurable. n= l Now we define a sequence {fn } �= 1 of simple functions such that limn-> fn f · Let Al , l = f - l (B(X l' 1 » ) and if w E Al , l, JI(w) = { � l otherwise. Let A2, 1 = f - l (B(X l, �)) and A2 , 2 f - l (B(X 2 , �) ) \ A 2, 1. Define i if w E A2 : i , f2 (W ) { XJI(w) if w � A2 l A 2, 2 . Suppose that {JI, . . . , fk } are defined. Use induction to define j- l l Ak+ l ,j = f - (B(Xk+ l, : 1 ) ) \ U=l Ak+ l , i i and Ak+ 1 , i for some i � 1, fk+ l (w ) -- { Xfki (W) ifif ww �E U7!l Ak+ 1 , i ' Clearly, {fn}�=l converges pointwise to f . The proof is complete. Let (0, J-L ) be a measurable space and let L oo ( J-L ) denote the set of all bounded measurable functions. A mapping p from L oo ( O, J-L ) into itself is called a lifting if p satisfies the following conditions: (I) For any E L oo ( J-L ) , p (g) a.e. (II) If gl = g 2 a.e. , then P (gl) p (g2 ) ' (III) p (l) = 1 , where 1 is the constant 1 function. (IV) p (g) � a for all � O. (V) For any scalars a , b and any gl, g 2 E L oo ( J-L ) , p (agl bg2 ) ap(gl) bp (g2 ) ' (VI) For any gl, g 2 E L oo ( J-L ) , p (gl . g 2 ) p (gl) . p (g2 ) ' Lemma 3.2.3. Let p be a lifting of L oo ( J-L ) . (1) For any continuous function h : IR IR, (3.1) p (h g ) h p(g) for all E Loo (J-L) . (2) Suppose that gl, g 2 be two uniformly bounded measurable functions such that supp (gI) supp (g2 ) = 0 . Then supp (p (gI) supp (p (g2 » 0 . E.

E

=

00

E

=

E

-

00

=

=

=

u

k

k

+

0

9

= 9

=

9

+

+

=

=

--+

0

n

=

0

9

n

=

164


Proof: (2) follows from (VI) . We note that by (II) , (V), and (VI) , I p (f)l l oo = lI flloo, and (3.1) is true for any polynomial function h. Now (3. 1) follows from

the Stone-Weierstrass theorem. By (IV), we have p( l f l ) = I p(f) 1

o

It is known [20, Chapter 4] that for a complete u-finite measure space (0, J-t) , liftings always exist. Let L O (J-t) denote the set of all measurable functions f such that for any c > 0, J-t ( {t : I f (t ) 1 > c} ) < 00. Lemma 3.2.3 shows that there is a mapping p from Lo(J-t) into itself that satisfies (I)-(V). Let us assume that E is an order continuous Kothe function space. So E* is also a Kothe function space. Let E* (X* , w* ) denote the set of all weak* scalarly measurable function F such that I F ( ' )l l x * E E* . For any F E E* (X* , w*), the mapping f f-+ (F (t ) , f(t»)J-t(t)

J

is a bounded linear functional. We denote it by ( F, 1 ) . It is easy to see that

In this book, the norm of F E E* (X*, w* ) , is defined by

II F IIE * (X * ,w * ) = sUP { J (F (t ), f(t ») dt I l f I E(X) � 1 } . Two weak* scalarly measurable functions FI , F2 E E* (X* , w*) are said to be weak* scalarly equivalent if for each E X, we have (Ft (w) , ) = (F2(w), ) a.e. Note: In general, I F IIE * (X * , w * ) =J II II F O l x * II E * ' But the following theorem shows that if E is an order continuous Kothe function space over a finite measure :

x

x

x

and X is a Banach space, then the dual of E(X) is isometrically isomorphic to E* (X* , w* ). Moreover, for any F E E* (X* , w* ) , there is aweak* scalarly measurable function FI such that FI is weak* scalarly equivalent to F and

I F I (E(X))* = II l Fd ' )l l x * I I E * ' Theorem 3.2.4. Let E be an order continuous Kothe function space over a probability space (O, �, J-t), and X a Banach space. Then there is a natural mapping T from E(X)* to E* (X* , w*) that satisfies the following conditions: (1) For any F E (E(X) * and f E E(X), (F, 1 ) J ((T(F)(t ), f(t))dJ-t(t). (2) For F and G in (E(X» * and real numbers a and b, T (aF bG) = aT (F) + bT ( G) . =

+

3.2. STRONGLY AND SCALARLY MEASURABLE FUNCTIONS (3)

165

II I T (F) l I x * I I E * = I F I (E(X)) * for all F E (E(X » * .

F

Proof: Let be any element on the unit sphere S((E(X» * ) of E(X), and A the set of all strongly measurable X-valued functions f such that for all t E O, I f (t ) lIx = 1 . Note that for any h E E and f E A, f · h E E(X). Hence for each f E A, we can define a linear functional Ff on E by

(Ff' h) = (F, f . h ).

Clearly, I Ff I ::; 1. Since E is order continuous, there is an integral functional Gf E B(E*) such that

(Ff' h) = j Gf ' h dJ-i.

{fd k=l in A, let n Ck = n { t : I Gfj (t ) 1 (4), suppose that (4) fails. Then there exist 8 > 0 and a sequence {An}�=l of mutually disjoint members of L. for which s UP r E r II mr II (An) ;::: 48 > 0 holds for all n E N. By Proposition 3.3.2 (2), for each n E N, there is Cn � An such that Proof: (1)

=>

sup II mr (Cn) II ;::: 8 > rEr

o.

Thus (3) fails to hold. This shows that (3) implies (4). (4) => (5). Suppose that the set { I x* 0 mr l : 7 E r, X* E B(X* )} is not uniformly strongly additive. Then there exist a sequence {An}�=l of mutually disjoint members in L. and a 8 > 0 such that for all k E N, one has

{ n=k 00

sup l: I x * 0 mr l (An )

:

7

Thus there is an increasing sequence all j E N,

kj+l sup { l: I x * n= kj + l

0

mr l (An )

: 7

E

r, X * E B(X * )

}

;:::

28 > O.

{ kj }� 1 of positive integers such that for E r , x* E

B(X * )

}

kj + 1 = sup { l x * mr l ( U An ) : E r , X * E B(X* ) } ;::: 8 > o. n=kj + l Let Hj = U��j + 1 An. Then {Hj }� l is a sequence of mutually disjoint members of L. such that 0

7

{

sup{ II mr l l (Hj ) : 7 E r} = sup l x * 0 mr l (Hj )

: 7

E

}

r , x * E B(X * ) ;::: 8 > o.

(4). We have proved that (4) implies ( 5). (5) (1) is obvious. Proposition 3.3.4. [14, p.8, Corollary 18] Let m be an X -valued vector measure. The following statements are equivalent: (1) The vector measure m is strongly additive. (2) The set {x* m : x* E B(X*)} of measures is uniformly strongly additive .

This denies

0

=>

0

171

3.3. VECTOR MEASURE

The vector measure m is strongly bounded i. e . if {An}�:::::: l is a sequence of mutually disjoint members of E, then limn-+oo I l m(An) 1 O. (4) The measure I m il is strongly bounded, i.e . if {An}�:::::: l is a sequence of mutually disjoint members of E, then limn-+oo I m i l (An) O. (5) The set { x* m : x* E B(X*)} of measures is uniformly additive . (6) For any nondecreasing monotone sequence {An}�:::::: l of members of E, the limit limn-+oo m(An) exists. (7) For every nonincreasing monotone sequence {An}�:::::: l of members of E, the limit li mn -+oo m(An) exists . (3)

=

=

0

Proof: The equivalence of statements (1) through (5) is clear from Propo

sition 3.3.3. The equivalence of (6) and (7) follows from the identity

m(A) + m(n \ A) m(n). To see that (1) implies (6), let {An }�:::::: l be a nondecreasing sequence of members of E. Then the sequence {Aj + l \ Aj }� 1 consists of disjoint members of E, and so the limit n nlim -+oo m(An) m(Ad + nlim -+oo j�= 2 m(Aj + 1 \ Aj ) =

'"

=

exists. We have proved the implication (1) implies (6). (6) =} (1). Let {An}�:::::: l be a sequence of mutually disjoint members of E. Then limn -+oo m(U�=l A k ) exists by (6). Thus

o This completes the proof. Let (0, E, /-L) be a measure space, and (0, E, m) be an X-valued vector mea sure. Recall that the vector measure m is said to be /-L- continuous if

lim

J.t(A}-+O

m(A)

=

O.

Let E be a a-algebra, m : a countably additive vector measure and /-L a finite nonnegative real valued measure on E. Then m is /-L -continuous if and only if m vanishes on sets of /-L-measure zero . Proof: One direction is clear. Suppose that m is an X-valued countable E

Theorem 3.3.5. (Pettis [14, p.lO, Theorem 1])

---+

X

additive vector measure that vanishes on sets of /-L-measure zero, but lim sup II m(A) II > O. J.t(A} -+ O


172

Then there exist € > 0 and a sequence

E E N,

{ An}�=1 in � such that for all n E N,

For each n , select x� B(X*) such that (x�, m(An)) � €/2. By Proposition 3.3.4 and Proposition 3.3.3, the family { I x� 0 m l }�=1 is uniformly strongly ad ditive. For each n let Cn = U� n Aj . Then /-L( n�=1 Cn) = O. Consequently, m vanishes on every D � that is contained in C = n�=1 Cn. Particularly, we have (3.2) I x� 0 m l (C) = 0 for all n Let D1 = n \ C1 and Dn + 1 = Cn \ Cn + ! for all n � 1. Then {Dn}�=1 is a sequence of mutually disjoint members of � for which

E

E N.

00

Ck- 1 \ C U Dj . j =k =

By the uniformly strong additivity of the family { I x * 0 m l : x* 00

00

E B(X*)},

nlim �oo I Xk m l (jU=n Dj ) nlim � oo I Xk m l (Cn- r) nlim � oo j�=n I Xk m l (Dj) 0

=

uniformly in n. Select n

0

=

0

'"

=

0

E N such that

Then a

o

contradiction. The proof is complete.

Theorem 3.3.6. [14, Section1.2, Theorem 4] Let {mr r E A} be a uni formly bounded family of countably additive X -valued vector measures on a algebra �. The family {mr : r E A} is uniformly countably additive ( uni formly strongly additive) if and only if there exists a nonnegative real-valued countably additive measure /-L on � such that {mr : r E A} is uniformly /-L continuous; i. e., /L (lim A)� O I l mr(A) 1 0 uniformly in r E A. The measure /-L is called the control measure of m . Proof: One direction is clear. By Proposition 3.3.3, {mr r E r} is uniformly count ably additive if and only if { x* mr : r E r , x* E B(X* )} is uniformly countably additive. Therefore, we may assume that {mr r E r} is :

=

=

0

:

:

a bounded family of uniformly countably additive scalar-valued measures.

(1-

173

3.3. VECTOR MEASURE

We claim that for any E > 0, there exists a finite family {7"1 , " " 7"n } � r, such that sUPl< i < n I m Ti I (A) = 0 implies sUP TE r I 7"T (A) I < €. Suppose the claim is false. Fix 7"1 r. There are A l � and 7"2 r such that I m T1 1 (A r ) = 0 but I m T2 (Ad l 2:: E . Continuing this process produces a sequence {An }�= 1 of members of � and a sequence {7"n }�= 1 of members of r such that for any n m i (An ) = 0 but I m Tn+ 1 (An ) 1 2:: E . l �sup i� n I T I

E

E

E

E N,

E N,

For any n let Cn = U�n Aj . Then {Cn }�= 1 is a nonincreasing sequence in � such that for any i ::; n, I m Ti I (Cn ) = Thus limn-+oo m Tk (Cn) = O. By the proof of Proposition 3.3.3, { l m Tk (Cn ) I }�= 1 converges uniformly to 0; i.e.,

o.

On the other hand, m Tn+ 1 (Cn ) = /-LTn+ 1 (A n) + /-LTn+ 1 (Cn An ) = /-LTn+l (An ) 2:: E . This implies that lim sup sup I m Tk+l (Cn ) I 2:: E , n-+oo k a contradiction. We have proved our claim. By induction, there are a sequence {mn }�= 1 of {m T : 7" r} and a sequence {Mn }�= 1 of finite subsets of such that for all n

\

E

E N,

N

sup m (A) 0 implies sup I m T I (A) j E Mn I j l = TE r Let /-L be the measure defined by

. be a finitely additive scalar measure and J-L is a finite measure on �. The following are equivalent: (1) The finitely additive scalar measure >. is J-L-continuous. (2) The finitely additive scalar measure >. is J-L-continuous at some A E �.

The finitely additive scalar measure >. is J-L-continuous at 0 . Moreover, (1) -(3) imply that >. is countably additive. (3)

(2) � (3) . Suppose that the >. is a finitely additive scalar measure that is J-L-continuous at 0 . Let {An}�= 1 be a decreasing sequence of measurable sets such that n�= 1 An = 0 . then limn -.cxl >'(An ) = O. So >. is a (count ably additive) measure. (3) � (1). Since d),. is J-L-continuous at 0, for any E > 0, there is 8 > 0 such that J-L(C) < 8 implies >'(C) < E. Hence if J-L(ALlB) < 8, then >.(ALlB) < E. SO 0 >. is J-L-continuous. Let X be a Banach space, and let /C be a family of finitely additive X -valued vector measures defined on �. We say that /C is equi-J-L-continuous at A E � if for each E > 0 there is a 8 > 0 such that if C E � and d� (A, C) ::; 8, then d),. (A, C) ::; E for all >. E /C. A family /C of finitely additive X-valued vector measures is said to be uniformly equi-J-L-continuous if for each E > 0 there is a 8 > 0 such that if A, C E � with d� (A, C) ::; 8, then d),. (A, C) ::; E for all >. E /C. It is easy to see that the family /C of X-valued vector measures is uniformly equi-J-L-continuous if and only if /C is equi-J-L-continuous at some A E �. Let {>'n}�=1 be a sequence of J-L-continuous X-valued measures. Suppose that for any A E �, lim >' n (A) exists. n�oo By Baire ' s classification theorem, the set { A : {>'n}�= 1 is equicontiuous at A } is second category in (�, d�). We have the following theorem. (For a proof, see [12, pp.87-89] or [14, p.23, Theorem 8] .) Theorem 3.3.9. (Vitali-Hahn-Saks Theorem) Let (O, �, J-L) be a finite mea sure space, and {mn }�= 1 a sequence of X -valued measures on � each absolutely continuous with respect to J-L. If limn�oo mn(A) = m(A) exists in X -norm for each A E �, then {mn : n E N} is uniformly strongly additive, and m is a countably additive X -valued measure on �. Proof: Clearly, (1)

�

177

3.4. SOME BASIC RESULTS

Recall that a Y-valued vector measure m is said to be weakly countably additive if for any y * E Y* , y * m is a count ably additive measure. The following theorem shows that every weakly count ably additive vector measure m on a a-algebra is countably additive. Theorem 3.3.10. (Pettis) Any weakly countably additive vector measure m defined on a a-algebra � is countably additive. Proof: Suppose that m is a weakly count ably additive vector measure that is not count ably additive. Let I l m i l be semivariation of m. Then (by Theorem 3.3.4) there are > 0 and a decreasing sequence {An}�=l such that n�=l An = 0 and for any n E N, I m ( A n) I ;::: Let �o be the algebra generated by {An : n E N}, and �1 the a-algebra generated by �o , Z the subspace of Y generated by {m ( A) : A E �o } . Since �o is an algebra generated by countable elements, �o is countable [14, Lemma III.8.4] . Thus Z is a separable Banach space. We claim that for any A E �1 ' m ( A) E Z. If the claim is not true, then there are y * E y* and A E � 1 such that y* E ZJ.. and (y*, m (A)) > O. But (y*, An) = 0 for all n E N. This contradicts the unique extension of countably additive measure. Let z� be a unit vector in Z * such that (z� , m ( An)) ;::: Since Z is sep arable, there is a subsequence {z�Jk:: l of {Z�}�=l such that for any z E Z, the limit lim k-+oo (Z� k ' z) exists. By the Vitali-Hahn-Saks-Nikodym theorem, {Z� k m : k E N} is uniformly strongly additive. This contradicts the assump tion (Z� k ' m ( An k)) � The proof is complete. 0

E

E.

E.

0

E.

0

Exercises

Exercise 3.3.1. Show that a finitely additive vector measure of bounded

variation is countably additive if and only if its variation is also countably additive. Exercise 3.3.2. It is easy to see that any countably additive vector measure over a a-algebra is strongly additive. Show that m is strongly additive if m is a vector measure of bounded variation. Exercise 3.3.3. Suppose that X is a Banach space that does not contain a copy of Co . Let � be a a-algebra. Show that if m : � � X is a vector measure of bounded semivariation, then m is strongly additive.

3.4

Basic Results in Kothe-Bochner Function Spaces

Let X be a Banach space, and E a Kothe function space over a complete measure space (0, /-L). The Kothe-Bochner function space E(X) is the set of all

178


strongly measurable functions f : 0 � X such that f E E(X) is defined by

I l f l = II l f Ol l xll E '

I l f Ol i x E E. The norm of

One may ask the following question: Question 3.4.1. Let P be a geometric properly. Suppose that both E and X have properly P. Does E(X) have property P ? Conversely, do both E and X have properly P if E(X) has the same properly? Recall that a property P is said to be a hereditary property if every closed subspace of a Banach space X has property P. It is easy to see that strict convexity, uniform convexity, and smoothness are hereditary properties. Let x (respectively, g) be an element in S (X) (respectively, S (E )) . Then {hx : h E E} (respectively, {gy : y E X}) is a complemented subspace of E(X) that is isometrically isomorphic to X ( respectively E). Hence if P is a hereditary property and if E(X) has property P , then both E and X have property P. Theorem 3.4.2. For any Kothe function space E and any Banach space X, E(X) is strictly convex if and only if both E and X are strictly convex. Proof: Since strict convexity is a hereditary property, we need to show only

that if E and X are strictly convex, then E(X) is strictly convex. Let !1 and h be two unit vectors in E( X) such that ! 1 !1 + h I E(X) = 1. Then 11 11!1 0 11x 11 E I l lIhOl l x l E = 1 1 1 !1 ; h o ll ll E = 1 .

=

Since E is strictly convex, for almost all w E 0,

x

1 !1(w) lIx = I l h (w) llx = 11 !1 +2 h (w) I I x · By the strict convexity of X, for almost all w E 0, !1 (w) = h (w). Recall that a Banach space X is weakly uniformly convex if for any E > 0, there is 8 > 0 such that for any x * E B(X*) and x , y E B(X), I l x - y l � E implies I (x*, x + y ) I � 2(1 - 8) . A Banach space X is said to be uniformly rotund in every direction if for any z E X \ {O} and E > 0, there is 8 > 0 such that I l x + y llx � 1 - 8 whenever for nonzero X,y E B(X), x - y = az and I l x - y l � E . Theorem 3.4.3. For any Kothe function space E and any Banach space X, E(X) is weakly uniformly convex if and only if both E and X are weakly uniformly convex. 0

Proof: Weakly uniform convexity is a hereditary property. Hence one di

rection is obvious. Let X be a Banach space and E a Kothe function space such that both E and X are weakly uniformly convex. By Theorem 3.1.21 and Theorem 3.2.4, E and E* are order continuous ( thus we may assume that E is separable ) , and the

179 dual of E(X) is isometrically isomorphic to E*(X*, w*). For any X* E S ( X*) ( respectively, G E S(E*)) and 0 < E < 1, let 8(X*, E) sup { 1 - � l x + Y I : X, Y E B(X) and I (x*, x - y) 1 � E } , 8( G , E) sup { 1 - � l g + h l : g, h E B(E) and I (G,g - h ) 1 � E } . Since both X and E are WUC, for any x * E S(X*) and G E S ( E* ) , we have (3.4) 8 ( G, . ). lim 8 *, ) 0 lim €lO (x · tlO Fix E > 0, and F E S(E *(X*, w*)). Then G I F (· ) l l x is a unit vector in E. By the upper semicontinuity of 8(G, . ), the set An { t E supp (F) : 8 ( II FF(t)(t)Ilx . ' 16E ) � ;;1 } is measurable for all n E N. Note that U�=l An supp F, and E* is order continuous. There is no E N such that


=

=

=

=

=

def =

=

By the definition of weakly uniform convexity, there is 1 > 8 > 0 such that for any g, h E E, I (F, g - h ) 1 6 4�o . max { l g IIE , I l h lIE } implies that

�

I l g + h ilE ::; 2 ( 1 - 8) max{ l g I E , I h I E } . We claim that for any fI , 12 E S ( E ( X )) , I (F, fI - 12 ) 1 � E implies that I l fI + 12I I E(x) ::; 2 ( 1 - 8). It is easy to see that the claim implies that E ( X ) is WUC. Proof of the claim: Suppose that il , 12 are two unit vectors in E ( X ) such that (F, fI - h ) � E. Let A { t E Ana : I (F (t), ( fI - 12 ) (t)) 1 � � max{ l fI (t) llx , 1 I 12 (t)l l x } , C { t E Ana : min { l I fI (t ) I , 1I12 (t) l I } � 2n;n� 1 . max{ l I fI (t ) I , 1I12 (t) I } } . =

=

Then

I (F, ( fI - h) . 1 A) 1 � I (F, ( fI - 12 )) I - I (F, ( fl - h) . 1 n\Ana ) I - I (F, ( fI - h) ) . 1 Ano\A ) 1 � E - 2 1 F · 1 n\Ana IIE·(X· , w O) - � ( l I fII I E(X) + 1 I 12IlE(x) E 2E E > E - 2 8 4 -

Note:

-

-

-

=

- .

180

CHAPTER 3. KO THE-BOCHNER FUNCTION SPACES (i) By the definition of the set

Ano' for any t E Ano and any x, Y E X ,

1 \ I !cm , x - Y) I ;:::: If6 max( lI x llx , I Y llx ) implies that x

I l x Y l x � 2nono- 1 max{ l x l x , I Y llx }. (ii) By the definition of the set C again, for any t E C, we have I l h (t) l x 1 I 12 (t) l x - l h (t) 12 (t) llx ;:::: 2nono- 1 . max{ ll h (t) llx , 1 I 12 (t) Ilx } - 2nono- 1 . max{ l h (t) llx , 1I12 (t) llx } 1 max{ l h (t) l x , 1I12 (t) llx } . ;:::: 2 no (iii) By the definition of the set C, for any t E A \ C, I ll h (t) l x - I 12 (t) llx l ;:::: 2�o max{ l h (t) l I x , 1 I 12 (t) llx } . Case 1 . I (F, ( h - h ) . I e ) I ;:::: � . Let gl and g2 be the functions defined by gl(t) = 2"1 ( l h (t) llx Ilh (t) l x ) , g2 (t) - { !:i ll( 1hI h(t)(t ) l x12 (t)1 I II12x(t) Il x ) ifif tt EE nC.\ C, +

.

+

+

+

+

+

Then

(G, gl - g2 ) = � \ G, ( 1 l h Ol l x I l 120 l x - l h O 12 0 l x ) . 1 e ) 1 (G, max{ l hOl l x , I 12 0 l x } . 1 e ) (by (ii) ) ;:::: 4no 1 (C, ( 1 I fl(' ) l x I l hOl l x ) · l e ) ;:::: 8no € . ;:::: 8no1 I(p, ( h - 12 )l e ) l ;:::: 4no Note: I l g d lE � 1 and I g2112 � 1. By the choice of we have 1 1 2" l h h i l E � 2" l g l g211 E � 1 Case 2. I (F, ( h - h ) . 1 e ) 1 < � . In this case, we have I (p, ( /1 - h ) . 1 A \ dl ;:::: �. +

+

+

6

1m' 00 h0'nk - h e ,nk

{nk } k=l such that for

exists, Since E is URED, we must have he = 0 for () = �, 1; i.e. , if () = � , 1, and 'Y E r, then Let 'Y be an element in r such that f ( r ) =J. O. Note: X-y is URED . The equation ( 3.6 ) implies that lim k ..... oo Cn k = O. We shall notice that the above proof shows that every subsequence of {Cn}�=l contains a further subsequence that converges 0 to O. Then { cn }�=l must be a null sequence. The proof is complete.

[ ] There exists an equivalent URED norm 1 · 1 1 on Roo and a sequence {Xn }�=l ofURED Banach spaces such that E (Xn) is not URED . Example 3.4.5. (M. Smith 3 9 )

183


Let {aj }� 2 be a sequence of positive real numbers such that For any x = (Xj ) � 1 E frx» define

2::;2 a; = 1.

I I . I I is an equivalent full function norm on foo such that II . 1 00 � I . I I � v's l . 1 00 ' We claim that (foo , I I . I I ) is URED. Suppose that the claim is not true. Then there are a nonzero vector Z = ( Zj ) and two sequences { x k = (x � )}� l {y k = (Y� )} � l in the unit ball of (foo , 1 1 · 1 1 ) such that x k - y k = Z and lim k->oo Il x k Yk l = 2. This implies that lim 2 1 1 x k l 1 2 2 1 1 y k l 1 2 - l l xk y k l 1 2 0 = k->oo = kl!..� [2 1 I xk l � 2 1 y k l � - ( 1 l xk y k ll� 00 L= aJ (2( l xt l I xjl ) 2 2 ( l y � 1 l yjl ) 2 - I xt y � 1 2 - I xj yjI 2 ) ] j2 00 � kl!.� . [ ( l x k l oo - l y k I (0 ) 2 L a; (( I x t l - I Y � I ) 2 ( l xjl - l yjl ) 2 ) ] . j =2 Clearly,

'

+

+

+

+

+

+

+

+

+

+

+

So for any j E N,

+

+

lim sup I xjl - I yjl = 0,

k->oo lim sup I xj yj l - I yjl - I xjl = 0, k->oo Zj = k->oo 11m· Xjk - Yjk = . +

and

°

We thereby obtain a contradiction. Let E = ( foo , 1 1 · 1 1 ). For any i E N, let be the two-dimensional fi + 1-space, bi = ( 1 ( 4 ' and f, fl,n, h,n the three functions defined by -

Xi

i+ l l / Ci+ l) ) )

{ { (- I,

f ( i ) = ( 1 , 0 ) for all i E N, if i = 1 , (0, 0 ) ft,n( i ) = ( 4 , bn ) if i = n , ( 1 , 0 ) if i # 1 , n, if i = 1 , 0) ( - 4 , bn ) if i = n , h,n( i ) if i # 1 , n . (0, 0 ) =

184


Then for all n � 2 ,

11!l , n IIE(X; ) = V2, Ilh , n IIE( X;) = ( 2 + 3a�) 1 /2 , 1 !l , n + h , n IIE(X; ) = [4b� + 4 + (4b� + 4bn - 3 )a�] 1 /2 , !l , n - h ,n = f. Clearly, limn-> oo bn 1 and lim n -> oo an = O. We have nlim ->oo Ilh, n IIE( X -) = V2, nlim -> oo 11!l , n + h , n I E ( Xi) = 2 V2. =

•

We have proved that E(Xi ) is not URED. Remark 3.4.6. Let Xi , 1 � i < 00 , and E be as defined in Example 3.4.5. By Theorem 3 .4.4, X = (2: : 1 EBXi ) 2 is URED. But Example 3.4.5 shows that E(X) is not URED. The following question is still open: Question 3.4.7. Suppose that E is an order continuous Kothe function space. Is the Kothe-Bochner function space E(X) URED if both E and X are URED ? In [28] , I.E. Leonard posed the following problem. Problem 3.4.8. Let 1 < P < 00. Does Lp ({L, X) have the Kadec-Klee property if X has the Kadec-Klee property? The following theorem is due to M. Smith and B. Turett [40]). It shows that if Lp ([O , 1], X) has the Kadec-Klee property, then X must be strictly convex. Theorem 3.4.9. Let E be a characteristically order continuous function space over [0 , 1]. If X is not a strictly convex Banach space, then E(X) does

not have the Kadec-Klee property.

x, y E X such that I l x l x I l x ± y l x 1 and y =I O. Let {rn }�=l be the system of the Rademacher functions. By Theorem 3.1.24 , {rn }�=l is weakly null. Let f, fn E E(X) be the functions defined by f X1 [O , 1] , fn X1 [O , 1 ] + yrn · Note: fn - f = yrn · So {fn}�=l converges to f weakly. But I l fn - f IIE( X ) = I l y l x ' I l rt i l E' This implies that E(X) does not have the Kadec-Klee property. Proof: Suppose X is not strictly convex. Then there are =

=

=

=

o

By the definition of unconditional basis, we have the following theorem.

185


Let E be a Kothe function space on (0, 1) and X a Banach space with an unconditional basis {ed� l ' Then {ei rd� l is an unconditional basic sequence in E (X ) that is equivalent to {ek }� l' Theorem 3.4. 10.

Remark 3.4. 1 1 . ( 1 ) In [ 7] , Clarkson introduced the notion of uniform convexity in Banach spaces. He proved that for any 1 < p < 00 , and are uniformly convex. In 1940, Boas [2] showed that for any < p, < 00 ,

1

I!p Lp q

Lp(l!q ), I!p (Lq ), and Lp(Lq ) are uniformly convex. He conjectured that Lp(j-L, X ), 1 < < is uniformly convex if X is uniformly convex. M.M. Day [10] proved the Boas conjecture. H. Hudzik and T.R. Landes [19] showed it is still true if one replaces Lp, 1 < < by any uniformly convex Kothe function space. M.M. Day also proved that for any 1 < < Lp(X ) is strictly convex if and only if X is strictly convex. (2 ) Let 1 < < M.A. Smith and B. 'Thrett proved that Lp(j-L, X ) is weakly uniformly convex if X is weakly uniformly convex and X* has the Radon-Nikodym property (Le., X is an Asplund space) . G. Emmanuele and A. Villani [16] showed that the assumption "X has the Radon-Nikodym property" can be removed. Later, A. Kaminska and B. 'Thrett [25] showed that it is still true if one replaces Lp by a weakly uniformly convex KB p

00 ,

p

00 ,

p

p

00 ,

00 .

space. (3 ) G. Emmanuele and A. Villani [16] proved that for any 1 < p < 00 , » * is W*UC if and only if is W*UC. ( 4 ) M.A. Smith and B. 'Thrett [40] proved that for any 1 < p < 00 , is URED implies that is URED. A. Kaminska and B. 'Thrett [25] showed that it is still true if one replaces by a uniformly convex Kothe function space. (5) Recall that a Banach space is said to have the ( weak) sum-property if does not contain a bounded strictly increasing limit-affine sequence. Let E be a finite-dimensional uniformly convex space with a 1-unconditional basis. Landes show that E(X) has normal structure if has normal struc ture. He also proved has normal structure if and only if has the sum-property [26, 27] . (6) M.A. Smith and B. 'Thrett [40] proved that if has normal structure, then has normal structure. The same argument shows that if E is a uniformly convex Kothe function space such that for any g E E, I l g ilE � I l g l h , and if has normal structure, then E has normal structure.

(Lp(j-L, X

X

Lp(j-L, X )

X

X

X

Lp

X

l!i (X )

Lp(X )

X

X

X

(X )

The following theorem is due to Kahane. For a proof, see [ 2 3] . Theorem 3.4. 12. (Kahane ' s Inequality) For any ° < p, q < 00 ,

there exists a positive constant Kp, q such that for any Banach space X and any finite subset (Xj YJ=l X, the following inequality holds: c

186


Recall that a Banach space X is said to have type p, 0 < p if there is C < 00 such that for any n vectors Xl, . . . , Xn in X, we have

The smallest constant Tp(X) satisfying the above inequality is called the type p constant of X . A Banach space X is said to have cotype q, 0 < q :5 00 , if there is C < 00 such that for any n vectors Xl, . . . , Xn in X , we have

We denote the smallest constant that satisfies the above inequality by Cq (X) . Cq (X) is called the cotype q constant of X. Remark 3.4.13. (1) By Khinchine ' s inequality, X = {O} is the only Ba nach space that can have type > 2 or cotype < 2. (2) Since for any Xl , X 2, , Xn in X, •

•

•

every Banach space X has type 1. On the other hand, for any Xl , . . . , Xn E X, there are j :5 n and X* E B (X* ) such that ( x * , Xj) = maxk :$n I l x kl l . Note: For any X* E X*,

We have proved that every Banach space has cotype 00 . (3) If a Banach space X has type p > 1 , then it also has type p for any 1 :5 P :5 p and Tp(X) :5 Tp(X) . If a Banach space X has cotype q < 00 , then it also has cotype ij for any q � ij � 00 and Cq(X) � Cq (X) . (4) Suppose that X is finitely representable in Y. If Y has type p (respec tively, cotype q ), then X also has type p (respectively, cotype q ). Hence by the principle of local reflexivity, for any Banach space X, X and X** have the same type and cotype. (5) Let {e n}�= l be the unit vector basis of fr o Then

Hence fr (respectively, Lr) does not have type p (respectively, cotype q ) for r < p (respectively, q < r ) .

187


Recall that a Kothe function space E over [0, 1] is said to be p-convex (re spectively p-concave ), 1 � p < 00, if there is a constant M such that

1 /P I E � M (t I l gil l � r iP g il I ) P I (t i= 1n i= 1n (respectively, (2: I l gil l � ) 1 � � M i l (2: 1 9i1 P ) 1 � l i E ) i= 1 i= 1 for any g b . . . , gn E E. The smallest possible value of M is denoted by M ( p) (E) (respectively, M (p) (E)).

Remark 3.4.14. (1) It is easy to see that every Banach lattice is I-convex

and oo-concave, and every p-convex (respectively, q-concave) Banach lattice is r-convex (respectively, s-concave) for all r � p (respectively, s � q) . (2) (Figiel and Johnson [32, Theorem 1 .d.8]) Let 1 < p � 2 ' � q < 00 . Figiel and Johnson Showed that any Banach lattice (E, that is p convex and q-concave has an equivalent p-convex and q-concave norm such that both the p-convexity and q-concavity constants are 1 . It is known [32, Theorem 1 .f.l] that if E is a p-convex arid q-concave Banach lattice with M (p) (E) = M(q) (E) = 1, then E is uniformly convex and uniformly smooth. Hence if E is p-convex and q-concave for some 1 < p < q < 00 , then E has an equivalent uniformly convex and uniformly smooth norm. (3) [32, Proposition 1.f.3] Any q-concave Banach lattice is of cotype q V 2. Any r-convex Banach lattice with 1 < p � 2 that is also q-concave for some q < 00 is of type p. Theorem 3.4. 15. (A. Kaminska) Let q � 2, and E a Kothe function space over a measure space (0, J-t) . Suppose that E is q-concave, and X is co type q. Then E(X) is co type q.

I . I)

of

of

. . . , fn } be any finite sequence in E(X). Then n n l/q 1 Iq f x f ( ) il l I l l l i (2: ) ( �(x») 2:l i= 1 i= 1 l - n l � II � M(q) (E) I (2: I l fi ( ' ) I 1- ) q i l E (E is q-concave) i= 1 1 � M(q) (E)Cq (X) I l 1 I t h ( ' )ri (t) ll x dt I E o i= 1 1 � M(q) (E)Cq (X) l 1 1 I t fi ( ' ) ri (t) ll x II E dt o i= 1

Proof: Let {iI,

=

188

CHAPTER 3. KO THE-BOCHNER FUNCTION SPACES o

The proof is complete. Exercises

Exercise 3.4.1. If E be a Kothe function space and X a Banach lattice. Show that the Kothe-Bochner function space E(X ) is also a Banach lattice with the trivial order II 2: h if and only if II (w ) 2: h (w ) a.e. Let X and E be two Kothe function spaces. Show that the Kothe-Bochner function space E(X) is order continuous if and only if both E and X are order continuous. Exercise 3.4.2. Show that for any Kothe function space E and any Banach space X, E(X) is uniformly convex if and only if both E and X are uniformly convex. Exercise 3.4.3. Let E be a weakly uniformly convex Kothe function space. Show that the Kothe-Bochner function space E(X) is uniformly rotund in every direction if X is a Banach space that is uniformly rotund in every direction. Exercise 3.4.4. (J. Cerda, H. Hudzik, and M. Mastylo [8] ) Let E and X be two Kothe function spaces. Show that the Kothe-Bochner function space E(X) is strictly monotone (respectively, uniformly monotone, locally uniformly monotone) if and only if both E and X are strictly monotone (respectively, uniformly monotone, locally uniformly monotone). Exercise 3.4.5. (Partington) Let X be a Banach space that is not uniformly convex. Show the Lebesgue-Bochner function space Lp([O, 1], X) does not have the uniform Kadec-Klee property. Exercise 3.4.6. (Kaminska) Let 1 ::; p ::; 2 . It is known that if a Banach lattice E is of type p, then E is q-concave for some q < 00 [32, Corollary 1.f.13]. Show that the Kothe-Bochner function space E(X) is of type p if E is p-convex and X is of type p. Exercise 3.4.7. (Kaminska) Let E be a Kothe function space over a mea sure space (0" J.L ) that is not q-concave for q 2: 2. Show that E( L q ) is not of cotype q.

3.5

Dunford-Pettis Operators

Recall that an operator T : X Y is said to be a Dunford-Pettis operator if T maps weakly convergent sequences to norm convergent sequences. Let ([0, 1]' J.L) be the Lebesgue measure on [0 , 1], In ,k [(k - 1)2 -n , k 2 - n ], and hn ,k = lIn,k ' For any n En N u { a}, let �n denote the a-algebra generated by the intervals In , b 1 � k ::; 2 , and £' ( . , �n) the conditional expectation with respect to � n ' It is known that if f E L oo ( J.L , � n' X) and 9 E L1 (J.L), then -?

=

189

3.5. DUNFORD-PETTIS OPERATORS

{fn}�=1 in L1 (1-£, X ) is said be a martingale if for any n � 1 , £(fn+ 1 1 �n) fn . A martingale {fn}�=1 in L1 (1-£, X) is said to be uniformly bounded if su { l fn l oo : n E N} < We shall ask the reader to proof that there is a one-to-one correspondence between the bounded operator T : L1 X and the uniformly bounded martingale (fn l �n) of L1 ( 1-£, X). For any f E L1 ( 1-£, X) , the Pettis-norm of f is defined by Il f l l sup {J I (x* , f(t »)l dt : x* E B(X * ) } . A sequence {fn }� 1 in L1 ( 1-£, X) is said to be Pettis-Cauchy if {fn}� 1 is a Cauchy sequence in the Pettis norm. In this section, we present a result of Bourgain [4] that shows that an oper ator T : L1 X is Dunford-Pettis if and only if the corresponding martingale A sequence

=

00 .

P

�

=

�

is Pettis-Cauchy. First, we need the following theorem. the proof is left to the reader.

3 .5.1. Let X be a Banach space and {fn l �n }� 1 a uniformly bounded martingale in L1 ( 1-£, X ). Then there exists a bounded operator T from L1 into X such that (3.7) T(h) nl� J fn(t)h(t ) dt h E L1 . Conversely, if T is a bounded operator from L1 to X and if 2n n fn(t) 2 L hn , k (t )T(hn ,k ), k=1 then {fn l �n }� 1 and T satisfy the equation (3. "I). For 1 ::; ::; let i p denote the canonical injection from L p to L1 . It is known that for any 1 < ::; i p is weakly compact. The following theorem is due to J. Bourgain [4]. Theorem 3.5.2. For an operator T : L1 X, the following are equivalent: (1) T is a Dunford-Pettis operator. (2) For any 1 < ::; T o ip is a compact operator. (3) T o i oo is a compact operator. Proof: Note: For any 1 < ::; ip is a weakly compact operator. We have (1) (2). (2) (3) is obvious. (3) (1) . The assumption implies that T maps B ( L oo ) to a relatively compact set. It is known that every weakly compact subset of L1 is uniformly integrable. By Exercise 1.1 .6, if A is a weakly compact subset of L1, then the closure of the convex hull of T (A) is compact. The proof is complete. Theorem

=

=

p

00 ,

p

00,

�

p

=?

=?

=?

00 ,

p

00,

0

The following lemma gives a characterization of martingales that is Pettis Cauchy.

190


Lemma 3.5.3. (Bourgain [4]) Let X be a Banach space. A martingale {fn I En}�=1 of L1(/-L, X) is Pettis-Cauchy if and only if limn -+oo I J fnhn llx 0 whenever {hn}�=1 is a uniformly bounded (i.e., an Loo -bounded) weakly null sequence in L1. =

Proof: Suppose { fn }�= 1 is not Pettis-Cauchy. Then there are a 6 > 0 and an increasing sequence {n k }k'= 1 such that Il fnk - fn k _1 1 1 > 6 for every k E N. Let be an unit vector in such that J I (x k , fn k (t) - fn k-l (t)) I dt > 6 . For each k E N, let

xk

X*

Then gk is a E n k -measurable function such that

I l gkl l oo � 1 and

This implies that I JUn k - fnk -I ) · gk l x > 6 . Let h k = 9k - £ (9k I Enk _ I ). It is known that • Every bounded martingale difference sequence in L 2 is weakly null. • For any 1 < p < 00 , the weak topology of L1 and the weak topology of L p are equivalent on B(L oo ). We have

I l hkl l oo � 2 , w- k-+oo lim h k = 0,

(in

L1)

and

j(x*, fnk . gk - fn k . £(9kI En k _ 1)) 1 I x O E B (X o ) sup I j(x*, fnk · 9k - £Un k I E n k _l ) • gk ) I xO E B(X O ) sup I j(x* , fnk · 9k - fn k-l . 9k ) I x O E B (Xo ) sup

Conversely, let {hn }�= 1 be a uniformly bounded weakly null sequence in Ll and 6 > 0 such that I l hn l oo � 1 and lim suPn -+ oo I J fnhn l x > 6. By passing to a subsequence, we may assume that I J fnhn l x > 6 for all n E N . Since for any n E N, £ (· I E n) is a compact operator (from L oo to L oo ), there is an increasing sequence {n k } � 1 such that

191


Notice that for any nk ,

II I fnk hn k I l x > 8. Thus for any k E N,

Il fnk - fnk - l il sup

1 I (xk' (fnk (t ) - fnk- l (t )) ) 1 dt � sup 1 I (X k' ( fnk (t) - fn k- l (t )) hn k (t )) I dt x * E B (X* ) � 1 1 ( fn k - fn k _ l )hn k I l x � I l fnk hn k I l x - l fnk - 1 hn k I l x > �. x * E B (X* )

So the martingale { fnk I �n k }k: l is not Pettis-Cauchy. The proof is complete. D

[4] ) A uniformly bounded X -valued martingale is Pettis-Cauchy if and only if the corresponding operator T : L1 X is Dunford-Pettis . Proof: Let X be a Banach space, and T an operator from Ll to X . Let ( fn l �n) be the corresponding martingale of T. Suppose that T is not a Dunford Pettis operator from Ll to X. By Theorem 3.5.2, there is an L oo -bounded Theorem 3.5.4. (Bourgain

{ fn }�= 1

�

weakly null sequence {9n}�= 1 in Ll for which lim su Pn --+ oo II T(gn) 11 > O. Note: U�= 1 L1 (J.L, � n ) is dense in L1 . Without loss of generality, we may assume that gn is �kn -measurable. Then T(gn) J An ' 9n. By Lemma 3.5.3, { fn }�= 1 is not Pettis-Cauchy. Conversely, suppose { fn l �n}�= 1 is a uniformly bounded martingale that is not Pettis-Cauchy. By Theorem 3.5.3, there is an L oo-bounded weakly null sequence {gn}�= 1 in Ll such that limn--+ oo l fn ' 9n x > O. Let T denote the corresponding operator of ( fn l �n ) . Then for any m > n, =

il

l

1 fn . 9n 1 fn . £ (9n l �n ) 1 £ ( fm l �n) . £ (gn l �n) 1 fm . £ (9n l �n) T(£ (gn l �n)). =

=

=

=

We claim that £ (9n l �n) is a weakly null sequence. If the claim is true, then T is not Dunford-Pettis. Proof of the claim: Fix m > O. Let h be any element in L oo (J.L, �m ) ' Then

1

1

1

. lim-+ oo h · 9n O. nlim --+ oo h · £ (gn l �n) dt nlim --+ oo £ (h l �n) gn dt nNote that the set U�= 1 L oo (J.L, � m ) is a dense subset of Loo (J.L) in the Ll norm, and { £ (gn l � n ) : n E N } is uniformly bounded. The sequence { £ (9n l �n)}�= 1 D converges to 0 weakly. The proof is complete. =

=

=

192


Let Lt denote the set of all nonnegative functions in L1 • An operator

T : Ll ---+ Ll is positive if T(Lt ) � Lt . For any T E £(L1 I L1) and E Lt , let T+ (g) sup{T(h) : ° � h � g}, I T I (g) sup{T(h) : I h l � g }. 9

=

=

It is easy to see that £(Ll ' L1) is a Banach lattice under this order. Let 2 denote the Borel O"-algebra of [0 , 1]. For any f E Ll (�, J1" Ll (2, v)), ¢f : ( [0 , 1] x [0 , 1], � x 2) ---+ lR is defined by

¢f (t , u ) f(t ) (u ) . By Fubini ' s theorem, the mapping f ---+ ¢f is an isometry from Ll (� ' J1" L1 (2, v)) onto Ll (� x 2, J1, x v). For any f E Ll (�, J1" Ll (2, v)), let f + be the function =

defined by

It is easy to see that ¢f + = ( ¢f ) + . For any n E N U {O}, let Qn = �n ® 2 on [0 , 1] x [0 , 1]. One can easily verify that (fn l � n) is a uniformly bounded L1-valued martingale if and only if ( ¢fn ' Qn ) is a martingale such that sup

{ I l f I ¢fn (t, u) l du l i oo : n E N }

< 00 .

For a bounded operator T : Ll ---+ L1 I let (fn l � n) be the corresponding L1-valued martingale. For any A E Qm , let gn E Ll ([O, 1], L1 (J1,)) be defined by

gn(t )(U) fn(t )(u) . lA(t, u) . =

Then (gn l � n )n>m is still a uniformly bounded L1-valued martingale. This allows us to introduce the operator TA from Ll to L1 : Lemma 3.5.5. (Bourgain) Let T E £(Ll L1) and A E U � 1 Qn. Then TA � T+ . Hence if T is Dunford-Pettis, then T' A is also Dunford-Pettis. Proof: Let T be an operator from Ll to L1 I and let (fn l � n) be the corre sponding martingale. Let A U ;:I Im, i x A i be an element of � m ® 2, and let gn(t )(u) fn(t ) (u) . lA(t, u). Then there are a measurable partition OJ, 1 � j � k, of [0 , 1] and � m measurable sets Dj , 1 � j � k, such that A U7 1 Dj x OJ. Fix n � =

=

=

=

=

m.

193


Let h be a nonnegative �n-measurable function. Then

k TA( h) = 1 gn(t)h(t) dt 'L ( 1 fn(t)h(t)lVj (t) dt) 1 ej ) == 1 k k + = 'L T(h . 1 Vj ) . 1 ej ::; 'L T ( h) . 1 ej = T + (h). j =1 j =1 We have proved that TA � T+ . Now assume that T is a Dunford-Pettis oper ator . Let {hn }�== 1 be a weakly null L oo -bounded L 1 ([0, 1] , Ll (V)) sequence. Then for any k ::; 2 m , {hm, k . hn }�= m is still a weakly null Loo -bounded L 1 ([0, 1] , Ll (V)) sequence . For any k ::; 2m , let A k be the measurable subset of ([0, 1] ' 3) such that A U �: I Im , k x Ak • Then 1 1 gn(t )hn(t) dt I 1 1 2m 1 11 gn(t)(u) hn(t) dt l du = / 1 1 k'L=1 1 Im , kXAk fn(t)(U)hn(t) dt l du 2m ::; 'L 1 1 1 1 Im , kXAk fn(t)(u) · hn(t) dt l du k=1m 2 2m = 'L J II, fn(t)(u)hn(t) dt l dU ::; 'L 1 1 1, fn(t)(u)hn(t) dt l du k=1m Ak Im , k k=1 Im, k m 2 2 = 'L ll fn( t)h m,k (t)hn(t)dt l l l 'L II T ( h m ,k hn) 1 1 1 � ° as n � 00 . k== 1 l k=1 By Lemma 3.5.3 and Theorem 3.5.4, TA is a Dunford-Pettis operator . =

=

=

=

0

Lemma 3.5.6. Let (fn l � n) be a uniformly bounded L1-valued martingale, and let ( ¢fn ' Qn ) be the correspondin g martingale in Ll ( � x 3, J.L x v) . Then for any > 0, there are an m E N and an A E Qm such that lI ¢in - ¢fn 1 A I I < for all n � m . E

Proof:

l

Select an m such that II ¢jJ l l � sup{ ll ¢jJ I : n E N } - E. Let

A {t : (PJ,,, ( t) � O}. =

Then A is a Qm-measurable set. Hence for any n � m, we have

E

1 94


Consequently,

I l cPt - cPfn 1 A I 1 1

1An [4>fn fn �O] cPfn dJl x + 1Ac cPt dJl x � - L cPfn dJl x + I cPt dJl x 1I

1I

=

1I

1I

1I

=

II cPjJ I1 - llcPjJ l 1 < e .

1I

1I

o


Let T L1 L1 be a bounded operator, and (fn l �n) the corresponding martingale of T. Then for any h E L�(�n ' Jl), T+ (h) = nl�� I ft(t)h(t) dt. Moreover, if T is Dunford-Pettis, then T+ is Dunford-Pettis. Proof: Let ( cPn, (}n) be the corresponding martingale of T in L1(Jl x ) By Lemma 3.5.6, for any e > 0, there are an and a (}m -measurable set A such that I l cPt - cPfn 1 A II 1 < e for all n � Let {gn}�=l be the corresponding martingale of TA. Then for any n � and any h E L oo , II I ft(t)h(t ) dt - I gn(t)h(t)dt I 1 1 = I I I [cPt (t, u) - cPfn (t, u)lA( t, u)] h(t ) dt l du � Il cPt - cPfn 1 A II 1 ' 1 l h l oo � e l h l oo . Since {J gn(t)h(t) dt}�=1 is a convergent sequence, there is an N such that I I gn(t)h(t) dt - I gm (t)h(t) dt l 1 1 � e for all n, � N. This implies that for any h E L oo , {J cPt (t)h(t)}�= l is a Cauchy sequence. Let S(h) = nlim - oo I cPt (t)h(t) dt. Theorem 3.5.7.

:

�

1I .

m

m.

m

m

3.6. THE RADON-NIKOD YM PROPERTY

195

Then

1 8 (h) l l � li��p J I l f;t (t) l l l h(t) 1 dt � li��p J I l f(t) l l l h(t) 1 dt � I T I · l h l I Since L oo is dense in Ll, 8 can be extended to Ll with 1 8 11 � I I T I I . The above computation shows that for any E > 0, there is A E U�=l Qn such that I TA 0 i oo - 8 0 i oo l t < E. By Theorem 3.5.2, Lemma 3.5.5, and Exercise 1.1.6, if then 8 0 i oo is compact, and hence 8 is Dunford-Pettis. T is Dunford-Pettis, It remains to show that 8 T + . It is clear that 8 � T + . For the reverse inequality, fix h E L�. Because TA( h ) � T + ( h ) for all A E U n Qn, the above computation shows that 8(h) � T + (h). By a density argument, we have 8(h) � T+ ( h ) for all h E Lt . The proof is complete. =

0

We have the following theorem: Theorem 3.5.8. (J. Bourgain [4] )

is a sublattice of £( L1, L I) .

The ideal of all Dunford-Pettis operators

Exercises

/-l be the Lebesgue measure on [0 , 1] and X a Banach (a) Let {fn}�=l be a uniformly bounded martingale in Ll (/-l, X). Show that for any h E Lb the limit T(h) limn---> oo J fn(t)h(t) dt exists and I T(h) 1 � I l h l l . supn Il fn l oo . (b) Let T be a bounded operator from L l ( /-l ) to X, and let h n , k denote the characteristic function of Ik, n [ ( k - 1)2 - n , k2 - n ]. For each n, let 1:n be the a-algebra generated by {In , k 1 � k � 2 n }, and define the function fn E Ll (/-l, X) by 2n n fn(t) 2 L= hn , k (t)T( hn , k ) ' kl Show that Un l En) is a uniformly bounded martingale and for any h E Ll (/-l), T(h) n--->limoo J fn . h d/-l.

Exercise 3.5.1. Let

space.

=

=

:

=

=

3.6

The Radon-Nikodym Property

Let (0, 1:, /-l) be a finite measure space and El a a-subalgebra of 1:. By the Radon-Nikodym theorem, for any simple function f = L �=l X i 1 Ai E Ll(/-l, X ), there is a El-measurable function

n 9 = L xit' (1Ai lE I) i=l

196


such that for any C E �1 '

1 I dJ.L 1 g dJ.L . =

It is easy to see that for any simple function I E Ll (J.L, X), 11 £ ( fI �I) II L 1 ( X ) � 1I / II L 1 ( X ) ' Furthermore, the operator £ ( ' I �I) can be extended as a contraction on Ll (J.L, X). For any I E Ll (J.L, X) , the function £ ( fI �I) is called the conditional expectation of I relative to �1 ' Let {�n}�= 1 be an increasing sequence of cr algebras. A sequence {In }�=l of integrable functions is said to be a martingale if for each n E N, In is � n -measurable and £ ( fn + 1 l � n ) In . It is easy to see that Theorem 1 .9.5 is still true for Ll (J.L, X). =

Theorem 3.6. 1. (P. Levy) Let X be a Banach space. Suppose that {� n }�= 1 is an increasing cr-algebra. Let �oo denote the cr-algebra generated by U�= 1 �n . Then lor any I E Ll (� oo , X), In £ ( fl �n ) converges to I a.e., and {In }�= 1 converges to I in Ll (X) , =

Let K be a closed, bounded, convex subset of a Banach space X. Then K is said to have the Radon-Nikodym property (RNP) if for any finite measure (n, �, J.L) and any X-valued measure m on � that is absolutely continuous with respect to J.L, m(A)/ J.L(A) E K for all A E � with J.L(A) > ° implies that there is an I E Ll (J.L, X) such that for any A E �, A closed subset K of a Banach space X is said to have the martingale conver gence property (Mep) if every uniformly bounded K-valued martingale ( fn , �n ) converges in L1 (J.L, X). A Banach space X is said to have the Radon-Nikodym property (respectively, martingale convergence property) if every closed bounded subset of X has the Radon-Nikodym property (respectively, martingale conver gence property) . Example 3.6.2. [1 , p.l02] We give three examples of vector measures that

do not have a Radon-Nikodym derivative. (1) Let X = L1 (0, 1), and let J.L be the Lebesgue measure on [0, 1] . Let m be the L1-valued vector measure defined by m(A)

=

l A for any measurable subset A of [0, 1] .

It is easy to see that m is absolutely continuous with respect to J.L. We claim that there does not exist an L1-valued function I E Ll (J.L, L1) such that for any measurable subset A of [0, 1] , m(A)

'=

L l(t)dJ.L.

( 3.8 )

197

3. 6. THE RADON-NIKOD YM PROPERTY

Suppose the claim is not true. Then there is an L1-function J on the unit square [0 , 1] x [0 , 1] such that for any A E �, we have

meA) i J ( s, t ) dt 1 A. This implies that for any two measurable subsets A, B � [0 , 1], Jl(A B ) L i J(s, t)dtds. Consequently, J vanishes outside the diagonal, which is impossible. (2) Let X Co, and let Jl be the Lebesgue measure on [0 , 7r] . Let m be the Co-valued vector measure on [0 , 1] defined by Jl(A) ( JrA eikt dt) 00k=O for any measurable subset of [0 , 27r] . If m has a Radon-Nikodym derivative J, then we must have J(t) (eikt )k::o . But J(t) tj. Co for any t E [0, 27r] . This implies that m does not have a Radon-Nikodym derivative. (3) Let X be the space of all continuous functions on [0 , 1] ' and let Jl be the Lebesuge measure on [0 , 1] . Let m be the C(O, 1) vector measure defined by m(A)(t) meA [0 , t] ) for any measurable subset A of [0 , 1] . If J is a Radon-Nikodym derivative of m with respect to Jl, then for any measurable subset A of [0 , 1] ' meA [0 , t]) i J(s, t)ds. =

=

n

=

=

=

=

n

=

n

=

This implies that

almost all s � t, J(s, t) { � for for almost all s > t . This is a contradiction, since J ( s, · ) tj. C(O, 1) . Remark 3.6.3. The above examples show that iJ X contains a copy oj L1 or Co, then X does not have the RNP. Let (0, Jl) be a probability space and let P(Jl) denote the set =

If A is a measurable subset of n and Jl(A) > 0, then P (Jl, A) denotes the set

P (Jl, A)

=

{g

E P(Jl)

:

supp (g) � A}.


198

Recall that a bounded operator T : L 1 (p,) X is a uniformly bounded measurable function f : [0, 1] 9 E L 1 [0, 1] , ---t

representable if there is X such that for any

---t

Theorem 3.6.4. [1 , Proposition 5.5] A closed, bounded, convex set C in X has the RNP if and only if every bounded linear operator T : L 1 (p,) X such that T E C for all E P(p,) is representable. Proof: (Sufficient condition) Let m be an X -valued vector measure such that for any measurable set A with p,(A) > 0, we have :(�} E C. Define a linear operator T : L 1 (p,) X by T(lA) :(�} E C. For any simple function l:7= 1 ,\ lAi in P (p,), 9

---t

9

9 =

---t

This implies that for any 9 E P ( p,), T(g) f E L oo ( p" X ) such that

=

E C.

By the assumption, there is an

Thus for any measurable set A with p,( A ) > 0, We have proved that C has the RNP. (Necessary condition) Conversely, assume that C has the RNP and T is an operator from L 1 ( p, ) to X such that T(g) E C for all 9 E P ( p, ). Let m be the vector measure defined by m(A) = T( lA ) Then m « p, and :(�} E C for all measurable sets with p,( A ) > 0. By assumption, there is an f E L oo ( p" X) such that for any measurable set A, '

It is easy to see that


o


199

Lemma 3.6.5. [1 , Lemma 5.6] Let (0, f-L ) be a probability space and T a bounded linear operator from L 1(f-L) to X. The operator T is representable if and only if for any measurable set A with f-L(A) > 0, and any E > 0, there is a measurable subset B of A with f-L(B) > ° such that the set fB T(P(f-L, B)) (3.9) has diameter at most E . de f

Proof: The necessary condition is obvious. We need to prove only the sufficient condition. A simple exhaustion argument shows that for each k E fiJ, we can decompose n into a countable disjoint union UA k, j such that for each j, A k, j is contained in A k- 1 , i for some i and diam (f A n , ]· ) < � . n Let 00 Xn ,j = T(IAn , )f-L(An ,j )) and fn = L xn ,j lAn , j '

Then for any n, m > N , we have

j =l

I l fn fm l oo � N1 ' I I T (g) J fngdf-L I I � 1 � 1 for all g E L 1(f-L). Let f lim k->oo /k . Then for any g E L1 ( f-L ), we have T(g) -

-

=

=

In gf df-L. The

proof is complete. Recall that a closed, bounded, convex subset D of X is said to be if for any E > 0, there are x* E X* and a > ° such that the slice S(D, x* , a) de f { y E C : (x * , y) � sup{ (x * , x) : x E

0

dentable

C} a } has diameter less than E. A point x E D is said to be a denting point of D if x -

is contained in slices of arbitrarily small diameter. Lemma 3.6.6. Let T : L 1( f-L ) � X and define the sets fB as in (3.9) . Fix a measurable set A c n with f-L(A) > 0, an x* E X* , and an a > 0. Then there is a measurable subset C of A such that f-L(C) > ° and fc � S(fA , x* , a) . Proof: Let

"Y sup { (x * , x) : x E fA} a, and let L1 ( A) be the set of all functions in L1 that are supported on A. Since a > 0, there is g E P ( f-L, A) such that (x* , T g) > "Y. So the set K { h : ° � h E L1(A) and (x * , Th ) � "Y l h l l I} =

=

-

200


is a closed convex cone in L1 (A) that does not contain g. By the separation theorem, there is G E LCX) (A) such that

! gG dJ-t

>

s UP

{ ! hG dJ-t : h E K } .

Notice that K is a cone. We have s UP

{ ! hG dJ-t : h E K }

=

O.

Let C be the set { w : G(w) > O}. Then C has positive measure (since f gG dJ-t fc gG dJ-t > 0). It is easy to see that for any 0 < h E L1 (C), f h G dJ-t > O. This 0 implies that r c � s (r A , X* , a). The proof is complete. =

Let K be a closed, bounded, convex subset of a Banach space. Then the following statements are equivalent: (1) K has the RNP . ( 2) K has the MCP. (3) Every closed convex subset of K is a dentable subset of X . Theorem 3.6.7. [1, Theorem 5.8]

Since every bounded operator T : L1 (J-t) ---+ X corresponds to a uniformly bounded X-valued martingale, ( 2 ) '* (1). The implication (1) '* ( 2 ) follows from the following two theorems.

If X has the RNP, then every L1(X ) -bounded martingale (fn l�n) converges a.e . Theorem 3.6.8. (Chatterji)

Proof: Let X be a Banach space and let (fn l �n) be an L1 (X)-bounded

martingale that converges almost everywhere. Fix a > 0 and define a function a : n ---+ N U { oo} (we shall wrire a for a if necessary) as follow:

a(t ) = { :

a

if /fi ( t ) / < a for i $ n - 1 and /fn ( t ) / � a, if /fi ( t ) / < a for all i E N.

In the proof of Theorem 1 .9.5, we have proved the following facts: (i) The sequence (fu/\ k / � k ) is a martingale. (ii) For each k E N, the function h : t f-+ s UP k / I fu/\ k ( t ) / 1 is integrable. (iii) Let �CX) be the a-algebra generated by U�= l �n . Then for any A E �CX) , {JA fu/\ k dJ-t} k'= l is a Cauchy sequence in X. Since the proof of the claim is similar to the proof of Theorem 1.9.5, we leave it to the reader. For any k E N and A E �CX) , let


201

Then for any k E N, m k is absolutely continuous with respect to p,. By (iii) and the Vitali-Hahn-Saks theorem, m k (A) m(A) def klim -+ oo is a count ably additive X -valued vector measure that is absolutely continuous with respect to p,. Hence there exists a Bochner integrable function 9 such that for any A E �oo , m(A) = g dp,.

L

By (i), for each n E N, the function fuAn is �n measurable and the sequence (fuAn l �n) is a martingale. So for any A E �n ,

JA fuAn dp,

=

lim

J fuA k dp,

k-+oo A

=

m( A) =

JA dp" 9

and for any n E N" £ (g l �n) = fUAn ' By Theorem 3.6.1 , {fUA k }k:: l converges almost everywhere to g. Now let a approach infinity. By the maximal inequality, we have that p,( { t : a (t) : a (t) = oo}) approaches 1 as a approaches infinity. Note: If a ( t ) = 00 , then fUaA n ( t ) = fn ( t ). We have proved that {fn}�= 1 0 converges almost everywhere. The proof is complete. Let 9 be a nonnegative integrable function in L1 (f!, p,), and {fn}�= 1 a se quence in L1 (n, p" X). We say that the sequence {fn : n E N} is dominated by 9 if for all n E N, Il fn (w) llx ::; g(w) a.e. By Theorem 3.6.8, we have the following theorem. Theorem 3.6.9. Suppose that X has the RNP. If {(fn l �n)}�= 1 is a mar tingale that is dominated by an integrable junction, then {fn}�= 1 converges in Ll (X) , (3) => (1). Let p, be a probability space, and T : Ll (p,) --+ X a bounded linear operator with T(g) E K for all 9 E P(p,) . Fix N E N. Let A be a set with p,(A) > O. By (3) , Lemma 3.6.6, and a simple exhaustion argument, we can decompose 0. into a countable disjoint union UjCN,j such that for all j E N, CN,j � CN - l, i for some i and diam (r eN) < ]; . Let fN be the X-valued Loo-function defined by a

a

a

Let �N be the a-algebra generated by {CN,j : j E N }. Then (fn l �n) is a uni formly bounded X-valued martingale such that Il fn - fm ll oo ::; max { l / n , 1 1 m} . Let f be the limit of {fn}�= I ' It is easy to see that for any 9 E L l (p,), T (g) =

J fg dp,.


202

(2) =? (3) . Let B(x, E) denote the ball with center x and radius E. Assume that (3) does not hold. We may assume that K itself is not dentable, and hence there is an E > ° such that for all x E K, x E co(K\B(x, E) . We claim that there are an increasing sequence of finite subalgebras En of [0, 1] and En-measurable functions gn : [0, 1] K such that ( a) I gn(t) - gn + !(t) 1 � E for all n and all t E [0, 1] . (b) 2: :'= l ll gn - £ (gn + 1 I E n) 1 1 � i · Suppose that the claim has been proved. Let in = limk ..... £ (gkI En). Then Un l En) is a martingale such that limn-+ I l in - gn l 1 1 � E/4. This implies that K does not have the MCP (since I l in - in + I i l l � � ) . Proof of claim: Let E1 = { 0 , [0, I] } and gl = X 1 [0 ,1 ] for some x E K. Assume that En and gn have already been constructed. Let [0, 1] = U� l A i be the decomposition of [0, 1] into atoms of En, and write gn = 2: :: 1 Xi 1 A i ' By the choice of E, for each i � m, we can find vectors {X i,j } ;:: 1 in K and A i,j � ° such that for all i � m, ki ki E E, and x i - L Ai,jXi,j I I � 2 n +3 ' Ai,j = 1, I l xi - Xi,j I � L j=l j= l ----+

co

co

•

Il

Partition each A i into disjoint sets {A i,j : j � ki } with m(A i,j ) = A i,j m(Ai) for all i and j. Let E n + 1 be the algebra generated by {A i,j : i � m, j � ki } and gn+ ! = 2::: 1 2:;!: 1 xi,j 1 Ai ,j " Then (a) and (b) hold with this construction. The 0 proof is complete. Remark

3.6.10. It is easy to see that a bounded closed convex set C has

the MCP if and only if every closed convex separable subset of C has the MCP. Thus, C has the RNP if and only if every closed convex separable subset of C has the RNP. Theorem 3.6.11. [1, Theorem 5.11] (1) Every weakly compact convex subset of a Banach space has the RNP . (2) A sepamble w* -compact convex set in a dual space has the RNP . Proof: (1) By the above remark, we may assume that C is a weakly compact separable convex set. Let D be the weak closure of extreme points of C. By the Krein-Milman Ttheorem, D is nonempty and C = co(D) . Fix E > ° and let {Xn}�= l be a dense sequence in C. Since D is weakly compact and D � U�= l B(Xn ' E ) , by Baire ' s category theorem, there are a weakly open set V and an integer no such that 0 =I D n V � B(Xno ' E ) . Let C1 = co(V n D) and C2 = co(D \ V) . Then diam (Cd � 2E and C = co(Cl U C2). Since D is the weak closure of the extreme points of C, there is a Zo E D n that is an extreme point of C. Also, since D \ V is weakly compact, all the extreme points of C2 belong to D \ V. In particular, Zo 1. C2. For any 0 < 8 < 1, let Ks be the set

V


203

The set Ks is a closed convex subset of C that does not contain zoo If u , v E C \ Ks , then u = A l Z l + (1 - Ady! ' v = A2 Z2 + (1 - A2) Y2 for some Zi E C1 , Yi E C2 , and A i > S . Thus Il u - vii ::; I AI - A2 1 · Il z l ll + diam (Cl ) + (1 - s ) ( ll y l i I + IIY2 11 ) · Note: C is bounded, I AI - A2 1 ::; 1 - s , and diam (C1 ) < 2€. It follows that if s is close enough to 1 , then diam (C \ Ks) < 3€. Thus separating Zo from Ks yields a slice of C of diameter less than 3€. We have proved that C is dentable. (2) Assume that C is a w*-compact and norm separable convex set in Y = X*. Then there is a separable subspace Xo of X such that the canonical map X* -+ XO' is an isometry on C; hence there is no loss of generality to assume that X is separable. Let (fn I E n)�=l be a uniformly bounded martingale from (n, E, 11-) to C, and {xn : n E N} a dense subset of B(X) . By the martingale convergence theorem, there is a null set A such that limn --+ oo (fn (w ) , xj ) exists for all j E N and w ¢ A. Then for any w E it \ A and any (weak*) limit point y of {fn (t ) : n E N}, we have ( Y , xn) = limn--+ oo (fn (w) , xn) . This shows that for any w E it \ A, {fn(w)}�=l converges weak*; say it converges to f(t ) weak* . Since C is norm separable, it follows that f is separably valued. Moreover, since every norm open ball in C is a countable union of w* -compact sets (balls of smaller radius) , the separability of C implies that the w*-Borel structure of C and its norm Borel structure coincide. Hence they also coincide with the weak Borel structure of C. Since (f( ) , x ) is measurable for every x E X, it follows that ( x** , f(·)) is measurable for every x** E y* = X** . By Pettis ' s measurability theorem, f is measurable. It is easy to see that £ (f I En) = fn . By Theorem 0 3.6.1, {fn}�= l converges to f. The proof is complete. We have the following corollary: Corollary 3.6.12. (1) Every separable dual spaces has the RNP. (2) Every reflexive space has the RNP. Let C be a closed convex subset of X. A point x in C is called a strongly exposed point of C if there is an x* E X* such that (x* , x - y) � 0 for all y E C and lim diam ( S(C, x* , a) ) = O. '

aLO

A functional x* for which this holds is said to be a strongly exposing functional (at the point x) . Let f be a real-valued function on C. For any a > 0, let S(C, f, a) = { y E C : fe y ) � f(x) - a}. A real-valued function f is said to have a strong maximum at x E C if f(x) = sup{f(y ) : y E C} and lim diam ( S(C, f, a» ) = O. aLO

We need the following lemma whose proof is left to the reader. Lemma 3.6.13. Let K be a set, and g, h two real-valued functions on K that are bounded from above. If a > 0 and if 8 = sup{ l g(x) - h(x) 1 : x E K} < a/2,

then

S(K, h, a - 2 8 ) c S(K, g, a) .

204


Theorem 3.6. 14. Let K be a closed, bound, convex set with the RNP in a Banach space X and J a real-valued upper semicontinuous function on K that is bounded Jrom above . Then Jor any E > 0, there is an elemnet x * E X* with Il x* 11 � E such that J + x* attains a strong maximum on K. Proof: We claim that for any 1 > E > 0, there are a positive real number a > 0 and an element x* E EB(X*) such that diam ( S(K, J + x* , a) ) � 2E.

Suppose the claim is not true. For any n > - log2 E, let 00

An = U {S(K, J + x * , 4 - n ) : Il x * 11 � E - 2 - n }. n =l We shall show that for any n � - log2 E, and for any y E X, (3. 10) where B(y, E) is the ball with center y and radius E. Assume that (3.10) is false for some n > - log2 E and for some y E X. Since the set on right-hand side of (3. 10) is convex (and nonempty because diam ( S(K, J + x * , 4- n ) ) > 2E) , there is a vector x* E X* with Il x* 11 � E - 2- n and an element x E S(K, J + x* , 4- n ) that can be separated from the set onl the right-hand side of (3. 10) by a y* E X* with Il y* 11 = 1. Set z* = x* + 2 - ( n + )y* . Then I l z* I I � E - 2-( n + 1 ) . By the choice of y* , we have

(y * , x) � (y* , z ) + 4 . 2 - n for all z E S(K, J + z* , 4-4(n + l ) ) \ B(y, E) . Thus for any z E S(K, J + z* , 4-4( n + 1 )) \ B(y, E) , we have J(x) + (z * , x) = J(x ) + (x * , x) + 2 -( n + l ) (y * , x) � J (z) + ( x* , z ) - 4 -n + 2 -(n + 1 ) ( y* , x ) l � J(z) + (x * , z ) - 4 - n + 2 -(n + 1 ) (y * , z ) + 4 . 2 - n · 2 - ( n + ) = J(z) + (z *, z ) + 4 - n , a contradiction. We have established (3. 10). We now verify that the set

n = no

m=n

is a nondentable subset of K. This contradicts the assumption that K has the RNP. By (3. 10), co(An ) C co ( An + 1 ) + 4 . 2- n B(X). A simple iteration argument shows that A is nonempty. In fact, for any n � - log2 E, we have

coAn C A + 8 · 2 - n B(X) .

(3. 11)


205

The same argument implies that for any m � n � - log 2 E and any y E X, m (3.12) co U Ak C CO(Am + 1 \ B( y , E)) + 16 · 2 - n B(X). = n k Let x E A, and fix n > 4 - log 2 E. By the definition of A, there are m > n and u E co U;= n (Ak \ B(X, E)) such that Il x - u ll ::s; 2 - n . By (3. 12) (with y = x), there is v E co(An+ ! \ B(x, -E))n such that Il u - vii ::s; 16 · 2 -n . Hence Il x - vii ::s; Il x - u ll + Il u - vii ::s; 17 · 2 . Represent v as a convex combination v = 2: >"jVj , where Vj E Am + ! and II Vj - xii � E. By (3. 11) (with n replaced by 1 n m + 1), there are Wj E A with Il wj - Vj I ::s; 8 · 2 - m - ::s; 8 · 2 - < E/2 for all j E N. Thus II Wj - xii > E/2 for all j E N, while Il x - 2: >"j Wj I ::s; 25 · 2 - n . Since n is as large as we please, x E co (A \ B(X, E/2)) . Thus A is not dentable. We have proved our claim. First, we notice that f upper semicontinuous implies that for all x* E X* , the sets S(K, f + x * , 0:: ) are closed. Fix E > O. By our claim, there are xi E E/2B(X*) and 0:: 1 > 0 such that diam (S(K, f + xi , o:: I )) ::s; � . Let E2 = min{E/4, 0:: I /2} and use the claim and Lemma 3.6.13 to obtain x 2 E E2B(X*) such that

and

S(K, f + x i + x; , 0:: 2 ) c S(K, f + x i , 0:: 1 )' Continuing inductively, we obtain xk and a decreasing sequence of sets of the form S(K, f + 2:�= 1 xk ' O::n ) = Sk whose diameters tend to O. Moreover, the sequence {xk}k� l can be chosen such that 2:� n Il xk ll ::s; Q n2- 1 for every n > 1 and 2:� 1 Il x k ll ::s; E. By Lemma 3.6.13 again, f + 2:� 1 xk attains a strong maximum at the unique point of intersection of the sequence {Sk }� l ' The D proof is complete.

Let K X be a closed, bounded, and convex set with the RNP. Then K is the closed convex hull of its strongly exposed points. The functionals that strongly expose points of K form a dense Go set in Theorem 3.6.15. [1, Theorem 5.17]

C

E* .

Proof: Let Gn

= {x* : diam (S(K, x* , 0:: ) )

l/n for some 0:: > o } . By Lemma 3.6.13 and Theorem 3.6. 14, Gn is a dense open subset of X*. Hence G = n�= l Gn , the set of functionals that strongly expose a point of K, is a dense Go subset of X*. In particular, K has at least one strongly exposed point. Let K 1 be the closed convex hull of the strongly exposed points of K. We claim that K = K1 • Suppose the claim is not true. Then there are x E K and x* E G such that (x * , x) > sup{ (x* , y ) : y E Kd. This implies that the point exposed by x* does D not belong to Kl , a contradiction. The proof is complete.
.)) with the RNP and X a Banach space with the complete conti nuity property. Then E(X) has the complete continuity property. Proof: For any

n

2n ,

{O} and k � let In,k - [(k - 1)2 - n k2 - n ]

E NU

"

and let En be the a-algebra generated by {In,k : k � 2 n }. Let T be any bounded operator from LdO, 1] into E(X). Since LdO, 1] is separable and E has the RNP, we may assume that E is a Kothe function space over a finite measure space (0, >.) such that L oo (>') is dense in E and for any 9 E L oo (>'), 1 Il g ll oo � Il g il E � 2" ll g ll l ' Hence L oo (>.) is dense in E* , and for any G E E* n L oo (>.) , we have (3.13) Let (fn I En) be a uniformly bounded E(X)-valued martingale. By Theorem 3.5.4, we need to show that {fn}�=1 is Pettis-Cauchy. Let 9n = En X 3. Then ( fn I 9n ) can be considered as an L1-bounded martingale on L1 ([0, 1] x 0, J-t x >.) . Fix a > 0. We need to define a stopping time a such that for almost all w E 0, (fcrl\n(· , w) I En) is a uniformly bounded martingale. Unfortunately, the stopping time defined in the proof of Theorem 3.6.8 does not have this property. Thus we need a modification. For t E [0, 1] ' let In,t denote the atom in E n such that t E In,t , and let tn,t In - 1 , t \ In,t . Note that be any element in (i) For any In,k , the restriction of fn to In,k is constant.


209

= 2� ' Hence for any w E 0, 2fn - l (t, W) = fn (t, w) + fn(tn ' w) .

(ii) For any k � 2 n , m(In,k)

Thus if Il fn - l (t, W) llx < a and I l fn (t, w) llx � 5a, then Define a stopping time u : °

-?

N U {oo} as follows:

if Il h (t, w) llx � a, n - 1 if Il fi (t, w) llx < a, Il fi (ti ,t , w) llx � 5a for i � n - 1 , and max { ll fn (tn ,t , w) llx , Il fn(t, w) llx } � 5a, n if Ilfi (t) (w) llx < a for i � n - 1 , 5a � Il fn(t, W) llx � a, and II fi (ti, t , w) lI x � 5a for i � n , 00 if II fi (t) (w ) I x < a for all i E N. 1

u (t, w) =

Then for any w we have ( 3.14 )

II fcr (t, w) lIx � max { 5a , I Ih (t, w) lI x } .

By Fubini ' s theorem and ( 3.14 ) , for almost all w E 0, (fcrl\ k(·, w) l � k) is a uni formly bounded martingale. Since X has the complete continuity property, by Theorem 3.5.4, for any f > 0 and N (selected later) , there is M such that for any m, k � M, I l fcrl\ m ( " w) - fcrl\ k (', w) 1I1 1 sup I (x * , fcrl\ m (t, w) - fcrl\ k (t, w)) I dt : x * E B(X*) =

{1

< fiN.

}

( 3.15 )

By the proof of Theorem 3.6.17, there are ¢ E E + and a subsequence {fnk }k=l of { fn}�=l such that that for all k E N and t E [0, 1 ] , By ( 3.13 ) , for any N E N and G E B(E*), Jl{w : I G(w) 1 � N} � N2 '

Fix f > O. Since ¢ is an element in Ll ( E ) and the space Ll (E) is order contin uous, there is N such that for any G E B(E* ) and n E N,


210

Hence for any j, k E N with nj , nk > M, I I fO"Ani - fO"Ank I I - 2 € f 1 ( F, fO"A ni (t) - fO"A nk (t)) dt - 2 € sup 1 =

�

F E B (( E ( XW ) Jo

1

l

t f

sup

F E B (( E (XW ) Jo J( II F (w)llx* '.5 N )

�f

(by (3. 15)).

(F(w) , fO"A nj (t, w) - fO"A nk (t, w)) dl-L l dt

This implies that (fO"An I E n) is a uniformly bounded Pettis-Cauchy martingale. Thus the corresponding operator Ta of (fO" A n I E n ) is Dunford-Pettis. Note: fn(t)(w) = fO"An (t) (W) if 1 4>(t, w) 1 � a. So lima-+oo Ta T, and T is Dunford 0 Pettis. The proof is complete. =

Remark 3.6.20. In [36] , N. Randrianantoanina and E. Saab used a theorem

of N. Kalton to show that if X has the complete continuous property, then Lp (l-L, X), 1 < p < 00 , has the complete continuous property. But the following problem is still open. Question 3.6.21. Let E be an order continuous Kothe function space and X a Banach space. Does the space E(X) have the complete continuous property if both E and X have the complete continuous property'? Let X be a Banach space, and ( 0 , I-L ) a probability space. A weak measurable function f : ° --+ X is said to be Pettis integrable if for any measurable subset A of 0, there is X A (p)- fA f(t) dt in X such that for any X * E X * , =

(X* , XA )

=

i (x* , f( ·) ) dt.

A mapping T : LI fO, 1] --+ X is said to be Pettis representable if there exists a weakly measurable, uniformly bounded Pettis integrable function f : [0, 1] --+ X such that for any 9 E LI fO, 1] , 1 T ( g ) ( p ) - fg dt . =

1

A Banach space X is said to have the weak Radon-Nikodym property if every operator T from L1 ( 0, I-L ) to X is Pettis representable. It is known that on a Banach lattice, the weak Radon-Nikodym property is equivalent to the Radon Nikodym property. By the proof of Theorem 3.6.19, we have the following theorem: Theorem 3.6.22. Let E be a Kothe function space, and X a Banach space. E(X) has the weak Radon-Nikodym property if (and only if) both E and X have the weak Radon-Nikodym property.

211

3. 7. NOTES AND REMARKS

Remark 3.6.23. E. Saab [37, 38] proved that if X is a dual space or a com plemented subspace of a Banach lattice, and if every Dunford-Pettis operator from Ll to X is Pettis-representable, then X has the weak RNP.

3.7

Notes and Remarks

Let E be a Kothe function space over a finite measure space (0, p ) . A subset K of E is called equi-integrable if for any f > 0, there is a 8 > 0 such that for any 9 E K and measurable subset A with p(A) � 8, we have I l g · lA l E � f . By the definition of order continuous, a subset K of an order continuous Kothe function space over a finite measure space (0, p ) is equi-integrable if for any f > 0, there is an M > 0 such that 11( l g l - M . In) V O II E � f for all 9 E K. Let E be a Kothe function space over a finite measure space (0, p ) . The Kothe function space E is said to have the subsequence splitting properly if for any bounded sequence {gn};::'= 1 in E, there are a sequence {Ak }k:: l of mutually disjoint measurable sets and a subsequence {gn k } �= 1 of {gn};::'= 1 such that { gn k . I n\ Ak } �= 1 is equi-integrable. Let E be a Kothe function space (over a finite measure space) that is order isometric to Co . Then the unit vector basis {en};::'= 1 is not uniformly bounded (if it is not true, then E contains a copy of loo ). We ask the reader to verify that E does not have the subsequence splitting property. In [22] , Kadec and Pelczynski proved that Ll [0 , 1] has the subsequence splitting property. We shall show that if E is a Kothe function space with a nontrivial cotype, then E has the subsequence splitting property. Let U be any free ultrafilter of N. For any Banach space X, the ultraproduct X of X (with respect to U) is defined as the quotient X loo (X)jN, where N ((gn) E loo (X) : U- lim I gn l O}. n =

=

=

It is known that if X is a Banach space (respectively, Banach lattice, Lp-space), then the ultraproduct E of E is also a Banach space (respectively, Banach lattice, Lp-space) . Let E be an order continuous Kothe function space over a probability space (0, p ) , and let I n be the element (I n) in E. Eo denotes the band Eo {g : g E E and Pin 9 g } of E. The following lemma and theorem are due to L.W. Weis [47] : Lemma 3.7. 1 . Let E be an order continuous Kothe function space over a probability space (0, p ) , and let 9 (gn) be an element in E. Then (a) 9 E E� if and only if U - limn gn 0 with respect to the topology of =

=

=

=

convergence in measure. (b) 9 E Eo is order continuous if and only if there is a representation 9 (gn) with an equi-integrable sequence (gn) in E. Proof: (a) A sequence (gn) in Xf if and only if U- lim n l l l gn 1 mIn I 0 for all m E N. =

A

=

212


(a) is obvious. (b) Suppose that the sequence {gn : n E N} is equi-integrable. Then for any E > 0, there is a 8 > 0 such that for any n E N and any measurable set A with p (A) < 8, we have I l gn . lA 1 < E . Hence for any measurable set (An) E n with p (An) < 8 for n E N, we have SUPn I l gn . I An I � E . This implies that 9 = (gn) is order continuous. Now suppose that 9 = (gn) is order continuous. Select a sequence {a m }�=l of real numbers such that By induction, there is an increasing sequence {n m }�=l such that for any k � m, we have II ( l gn k l - am ln) + 11 � 2 m . We claim that { gn k }� l is equi-integrable. Let E be any positive real number. Then there is m such that 2 m < E . Since {gnt ' . . . , gn",, } is equi-integrable, there is bm � a m such that for any k � m, we have -

-


o

Let E be an order continuous Kothe function space over a probability space. Then E has the subsequence splitting property if and only if the sublattice Eo of E is order continuous. Theorem 3.7.2.

Proof: Suppose that Eo is not order continuous, but E has the subsequence splitting property. Then there are 8 > 0, (gn) E Eo, and a sequence {A n = (A�) �=l }�=l of measurable sets that satisfy the following conditions: (i) For all n E N, {A�}�= l is mutually disjoint. (ii) For all m E N, U- limn p(A�) = 8m > 0 exists, and lim m -> oo 8m = (iii) For any m E N, U- limn I l gn . 1 A ;:" I > 8. By (iii) and the definition of the band Eo , there is a sequence (.em ) of real numbers such that

o.

By induction, there is an increasing sequence {nm }�=l of natural numbers such that for any k � m , we have Since E has the subsequence splitting property, there are a subsequence {hj = gn k)� l of {gn k } k.: l and two sequences {hl ,j }� l' {h2,j }� 1 in E that satisfy the following conditions: (iv) The h 2, j have mutually disjoint support. (v) {hl , j : j E N} is equi-integrable.

3. 7. NOTES AND REMARKS (vi) For each j E N, hj

213

o.

hI,j + h2,j and I hI ,j I A I h2,j l = By (v), there is f > 0 such that if IL(A) < f, then for all j E N, we have =

Om

Select an m such that < �. Note: limj --> oo lL (supp h2, j ) 0, and E is order continuous. There is j 2 m such that Il l h2,j l A . 1n ll � % . Fix j 2 m that satisfies the above inequality. Set C A� i . Then =

fm

=

a contradiction. We have proved that if Eo is not order continuous, then E does not have the subsequence splitting property. Suppose that Eo is order continuous. Let (gn) be a bounded sequence in E. Then there are two sequences (hI,n) E Eo and (h2,n) E (Eo)l.. such that (gn) (hl,n) + (h2,n) (in E). By Lemma 3.7. 1 , (vii) U- limn h2,n 0 with respect to the topology of convergence in mea sure. (viii) {hl,n : n E N} is equi-integrable. (ix) U- lim n (gn - hl,n - h2,n) This implies there are a sequence (n k ) and a sequence (h3,n k ) with mutually disjoint support such that lim k --> oo gn k - hl,n k - h3,n k O. The proof is complete. =

=

=

D

o.

=

Remark 3.7.3. (1) In [47] , L.W. Weis showed that if E is a rearrangement

invariant function space over a probability space, E has the subsequence splitting property if and only if E contains no copy of Co . (2) It is known that every uniformly monotone Banach lattice is order continuous and any ultraproduct of a uniformly monotone Banach lattice is still uniformly monotone. By Theorem 3.7.2, every uniformly monotone Banach lattice has the subsequence splitting property. It is also known that if E is a q-concave Kothe function space with q < 00 , then E has an equivalent q-concave norm whose q-concave constant is 1 (Proposition l .d.8 [32] ) . Note: If is E is a Banach lattice that is lattice isomorphic to a Banach lattice with the subsequence splitting property, then E has the subsequence splitting property. Thus a Banach lattice E with a nontrivial cotpye (this implies that E is q-concave for some q < 00 ) has the subsequence splitting property. It is natural to ask whether E has a copy of CO if E does not have the subsequence splitting property. The following example gives a negative answer to this question.

214


Example 3.7.4. (Figiel, Ghoussoub, and Johnson [ 17] ) Let decreasing sequence of positive real numbers such that

a=

{ fd� l be a

00

I= l >k < 41 ' k

For any k E N, let (� k , J.t k ) be the Cantor set {O, l } N equipped with the prod uct measure /-L k = f� 0�= 1 /-L k, n , where /-Ln ,k is a measure on {O, 1 } such that /-Lk, n {O} = 1 - ft and /-Lk, n {1} = ft · Let E = (E � l EBL 2 ( �k ) ) l ' and for each k, n E N, let h k, n be the function in L 2 ( � k ) defined by where (tn) �= l

if k ::; n, otherwise,

E �k . It is easy to see that E � l I /-Lkl ::; � and for any k ::; n E N, I l hk,n IIL2(�k) = 1. For each n E N, let hn = E�= l h k, n and let I · I h be the greatest lattice norm on E such that I g l 1 1 ::; I l g ilE for all f E E, I l hn llE ::; 1 for all n E N. (The unit ball of (E, I . liE ) is the convex hull of the unit ball of (E, I . 1 1) and { E�= l ± hk, n : n E N} . ) Let (F, I · IIF ) be the completion of (E, 1 · 1 1), and for each k E N, let e k = (h k, n ) �= l E F. It is easy to see that {ek }� l is a bounded sequence in Fo with disjoint support, and IIE7= 1 e k l p = 1 . Note: For any n, m E N, f�3 1 n � E�� h k, n ' This implies that en ::; f;; 3 1 n and en E Fo For all n E N. By Theorem 3.7.2, F does not have the subsequence splitting

property. We claim that F contains no copy of eo . First, we note that the natural embedding from L 2 ( � i ) into F is an isometry. Suppose that the claim is not true. Let {gk }� l be a eo-sequence in F, and for each i E N, let Pi be the band projection from F to L 2 (� i ) ' Since eo has the Dunford-Pettis property and {gk } k= l is a weakly null sequence, we have lim k -> oo Pi gk = 0 for all i E N. By passing to a subsequence of { gd� l if necessary, we may assume that there are a sequence {gD k= l and an increasing sequence {n k }� l such that I l gk - gk llF ::; 2 - k for all k E N, Pi gk = 0 if i � [nk ' nk+ 1 - 1] . By a theorem of James (cf. Theorem 1 .3.20) , there is a normalized block basis {ik} k=O of [gk : k E N] such that ik � 0 for all k E N and for all n E N, Choose 1 = mo < m 1 < . . . such that Pi fk = 0 for all i � [m k- 1, mk - 1] . Now for fixed k E N with l i fo + fkllF < 1 + a such that there is a finite sequence

215

3. 7. NOTES AND REMARKS

( S i )��;) of nonnegative real numbers, and a positive element 91 ,k in E+ for which n(k) n (k) fo + !k :::; 91 , k + I:=l sj hj and 1 91 ,k 1 E + I:=l Sj < 1 + a. j j Then for any k E N , n (k) mk - l !k :::; I: Pj 91 , k + I: s j hj , j =mk j =mk - l l mk Pj 91 , k I l E · 1 91 , k 1 E = " ?==l Pj 91 , k 11 E + II .I: = mk J J 00

We have

Let

m l - l n (k) mk- 1 mk - l Pi ( = Sj hj ) . !1 ,k = jL=l Pj 91 ,k + jL=l s j hj , and h ,k = L mk i=l j L Then for any k E N , (3.16) 11!1 ,k IIF :::; a and fo :::; !1 , k + h ,k . Choose 1 = k(l) < k(2) < ' " such that n(k(i)) < m k(i+ l ) for all i E N. By averaging over k(1 ) , k(2) , . . . , k( N ) , we obtain

N N fo :::; N1 I:=l fl ,k(i) + N1 I:=l f2,k(i) j i

and by (3. 16) ,

N (3. 17) I: l i l.= 1 h, k (i) II F > 1 - a Note: For each k E N , the h k, n (k :::; n) are uniformly bounded independently identically distributive integrable functions and L-;�'% k Sj :::; 1 + a. By Koml6s ' s theorem and by passing to a subsequence, we may assume that 1 N


216

Thus we have

This contradicts (3.17) , since a < �. The proof is complete. It is known that every weakly compact operator from a Banach space to another Banach space factors through a reflexive Banach space. Talagrand [44] showed that there is a positive weakly compact operator from £1 to C( [O, l]) that cannot be factored through any reflexive Banach lattice. But the following problem is still open: Question 3.7.5. ( The Factorization Problem ) Let T be a positive compact operator acting between two Banach lattices. Does T factor (with positive com pact factors if possible) through a reflexive Banach lattice ?

3.8

References

[1] Y. Benyamini and J. Lindenstrauss, Geometric Nonlinear Functional Anal ysis, vol 1 , Amer. Math. Soc. Colloquium Publications Vol 48. (2000) . [2] R.P. Boas, Jr. , Some uniformly convex spaces, Bull. Amer. Math. Soc. 46 (1940) , 304-3 1 l . [3] R. D. Bourgin, Geometric Aspects of Convex Sets with the Radon-Nikodym Property, Lecture Notes in Math. 993, Springer-Verlag, New York (1983) . [4] J. Bourgain, Dunford-Pettis operator on L l and Radon-Nikodym property, Israel J. Math. 37 (1980) , 34-47. [5] J. Bourgain and H.P. Rosenthal, Application of the theory of semiembed dings to a Banach space theory, J. Funct. Anal. 52 (1983) , 149-188. [6] A.V. Buhvalov, Radon-Nikodym property in Banach spaces of measurable vector functions, Math. Notes, 26 (1979) , 939-944. [7] J.A. Clarkson, Uniformly convex spaces, Trans. Amer. Math. Soc., 40 (1936) , 396-414. [8] J. Cerda, H. Hudzik, and M. Mastylo, Geometric properties in Kothe Bochner spaces, Math. Proc. Camb. Phil. Soc. 120 (1996) , 521-533. [9] S. Chen and B.-L. Lin, Strongly extreme point in Kothe-Bochner spaces, Rocky Mountain J. Math. 27 (1997) , 1055-1063 . [10] M.M. Day, Some more uniformly convex spaces, Bull. Amer. Math. Soc. 47 (1941) , 504-507. [11] J. Diestel, Geometry of Banach Spaces: Selected Topics, Lecture Notes in Math. 485, Springer-Verlag, New York (1975) . [12] J. Diestel, Sequences and Series in Banach Spaces, Springer-Verlag, New York (1984 ) . [13J J. Diestel, H. Jarchow, and A. Tonge, Absolutely Summing Operators, Cam bridge University Press (1995) . .

3.B.

REFERENCES

217

[14] J. Diestel and J. J. Uhl, Jr., Vector Measures, Math. Survey 15, Amer. Math. Soc. , Providence, RI (1977) . [15] N. Dinculeanu, Vector Measures, Pergamon Press, New York (1967) . [16] G. Emmanuele and A. Villami, Lifting of rotundity properties from E to LP(/-L, E) , Rocky Mountain J. Math. 17 (1987) , 617-627. [17] T. Figiel, N. Ghoussoub, and W.B. Johnson, On the structure of nonweakly compact operator on Banach lattices, Math. Ann 257 (1981), 317-314. [18] H. Hudzik, A. Kaminka, and M. Mastylo, Montonicty and rotundity prop erties in Banach lattices, Rocky Mountain J. Math. 30 (2000), 933-950. [19] H. Hudzik and R. Landes, Characteristic of convexity of Kothe function spaces, Math. Ann. 294 (1992) , 1 1 7-124. [20] A. Ionescu-Tulcea and C. Ionescu-Tulcea, Topics in the Theory of Liftings, Springer-Verlag, Berlin (1969) . [21] W.J. Johnson, B. Maurey, G. Schechtman, and L. Tzafriri, Symmetric Structures in Banach Spaces, Memoirs of Amer. Math. Soc. vol. 217, (1979) . [22] M.1. Kadec and A. Pelczynski, Bases, lacunary sequences and comple mented subspaces in the spaces Lp, Studia Math. 21 (1962) , 161-176. [23] J.P. Kahane, Some Random series of Functions, 2nd ed. Cambridge Studies in Advanced Mathematics 5, Cambridge University Press (1985) . [24] A. Kaminska, Type and cotype in vector-valued function spaces, Math. Japonica 37 (1992) , 743-749. [25] A. Kaminska and B.Turett, Rotundity in Kothe spaces of vector-valued functions, Can. J. Math. 41 (1989) , 659-675. [26] T. Landes, Permanence properties of normal structure, Pacific J. Math, 110 (1984) , 125-143. [27] T.R. Landes, Normal structure and the sum-property, Pacific J. Math, 123 (1986) , 127-147. [28] I.E. Leonard, Banach sequence spaces, J. Math. Anal. Appl. 54 (1976) , 245-265. [29] P.K. Lin and H. Sun, Extremal structure in Kothe function spaces, J. Math. Analysis and Applications 218 (1998) , 136-154. [30] J. Lindenstrauss and H.P. Rosenthal, The Cp -spaces, Israel Math. J. 7 (1969) , 325-349. [31] J. Lindenstrauss and 1. Tzafriri, Classical Banach Spaces, Lecture Notes in Math. 338, Springer-Verlag (1973) . [32] J. Lindenstrauss and L. Tzafriri, Classical Banach Spaces, II, Springer Verlag, (1979) . [33] H.P. Lotz, N.T. Peck, and H. Porta, Semiembeddings of Banach spaces, Proc. Edinburgh Math. Soc. 22 (1979), 233-240. [34] B. Maurey and G. Pisier, Series de variables aleatoires independantes et properietes geometriques des espaces de Banach, Studia Math. 58 (1976) , 45-90.

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[35] R. Pluciennik and P. Kolwicz, P -convexity of Orlicz-Bochner spaces, Proc. Amer. Math. Soc. 126 (1998) , 2315-2322. [36] N. Randrianantoanina and E. Saab, The complete continuity property in Bochner function spaces, Proc. Amer. Math. Soc. 117 (1993) , 1109-1 114. [37] E. Saab, On the Dunford-Pettis operators, Canad. Math. Bull. 25 (1982), 207-209. [38] E. Saab, On the Dunford-Pettis operators that are Pettis-representable, Proc. Amer. Math. Soc. 85 (1982) , 363-365. [39] M.A. Smith, Product of Banach spaces that are uniformly rotund in every direction, Pacific J. Math. 73 (1977) , 215-219. [40] M. Smith and B. Turett, Rotundity in Lebesgue-Bochner function spaces, Trans. Amer. Math. Soc. 257 (1980) , 105-1 18. [41] K. Sundaresan, The Radon-Nikodym Theorem for Lebesgue-Bochner func tion spaces, J. Funct. Anal. 24 (1977) , 276-279 . [42] M. Talagrand, Weak Cauchy sequences in L l (E) , Amer. J. Math. 106 (1984), 703-724. [43] M. Talagrand, Pettis Integral and Measure Theory, Memoirs AMS vol 307 (1984) . [44] M. Talagrand, Some weakly compact operators between Banach lattice do not factor through reflexive Banach lattices, Proc. Amer. Math. Soc. 96 (1986) , 95-102. [45] B. Turett and J.J. Uhl, Jr., Lp(j.t, X) has the Radon-Nikodym property if X does by martingales, Proc. Amer. Math. Soc. 61 (1976) 347-350. [46] L. Tzafriri, Reflexivity in Banach lattices and their subspaces, J. Funct. Anal. 10 (1972) , 1-18. [47] L.W. Weis, Banach lattices with subsequence splitting property, Proc. Amer. Math. Soc. 105, (1989), 87-96. [48] P. Wojtaszczyk, Banach Spaces for Analysts, Cambridge Studies in Ad vanced Math. 25 (1991) .

Chapter

4

S t ability P rop ert ies I Let X be a Banach space. Recall that a unit vector x in X is said to be an extreme point if for any y, Z E B(X), x � (y + z) implies x = y z. A unit vector x in X is said to be a smooth point of X if there is a unique x* E S(X*) such that ( x * , x ) 1. A natural question in the study of Kothe Bochner function spaces is the following: =

=

=

Problem. Let E be an order continuous Kothe function space, and X a

Banach space. Suppose that f E E(X) is an extreme point (respectively, a smooth point) of E(X). Is it true that for almost all t E supp (1) , f(t)/ lI f(t) llx is an extreme point (respectively, a smooth point) of X? Conversely, suppose that Il f O ll x is a smooth point of E and for almost all t E supp (1) , f(t )/ ll f(t) llx is a smooth point of X. Is f a smooth point of E(X)? In this chapter, we consider the above questions and provide some positive and negative answers.

4.1

Extreme Points and Smooth Points

Recall that for any closed convex set C of a Banach space X, a point x in C is said to be an extreme point of C if for any y, z E C, y�Z x implies that x y = z. If C is the unit ball B(X) of X, then we say that x is an extreme point of X. A unit vector x in X is called a smooth point of X if the set J(x) = { y * E S(X*) : ( y * , x ) 1 } consists of only one point. A Banach space X is said to be smooth if every unit vector x in X is a smooth point. A unit vector x in X is said to be an exposed point of X if there exists a unit vector x* in X * such that for any y E B (X) \ { x }, 1 ( x * , x ) > Re ( x * , y ) . In this case, we say that x* exposes B(X) at x . If X y* for some Y and if y E Y exposes B(X) at x, then we say that x is a weak* exposed point and that =

=

=

=

=

CHAPTER 4. STABILITY PROPERTIES I

220

y weak* exposes B ( X* ) at x. It is known ( Theorem 2.2.2) that x is a smooth point of X if and only if there is x* E S ( X* ) such that x exposes B ( X* ) at x* . It is natural to ask the following question: Problem 4. 1 . 1 . Let f be an extreme point (respectively, a smooth point, an exposed point) of B(E(X ) ) . Is f(t)/ ll f(t) ll x an extreme point (respectively, a smooth point, an exposed point) of B(X) for almost all t E supp f ? In this section, we give some positive and negative solutions to the above problem. First, we need the following lemma. Let (0, E, /-l ) be a measure space. The outer measure /-l* of /-l is a function from 211 (the power set of 0) to lR defined by /-l* (A) inf { /-l ( C ) : A � C E E } . Suppose that jl* (A) < 00 . Then there is a sequence {Cn : n E N} of E measurable sets such that for all n E N, A � Cn and /-l(Cn) � /-l* (A) + �. Let C = U�= l Cn . Then /-l(A* ) = /-l ( C ) . A measure /-l is said to be complete if A E E for all A � 0 with /-l* (A) = O. Lemma 4. 1.2. (Z. Hu and B.-L. Lin [12]) Let (0, /-l ) be a finite complete measure space, and A a subset of 0 such that /-l* (A) = /-l(0) < 00 . Let (M, d) be a metric space. If 9 is a mapping from 0 to M with separable range, then for any € > 0, there exist a measurable partition {En : n E N} of 0 and a sequence {An � A}�= l of sets such that for all n E N, =

Proof: Let (M, d) be a metric space, and (0, /-l ) a finite complete measure

space. For any € > 0 and x E M, let B(x, €) denote the set B(x, €) = {y E M : d(x, y) < €} .

Suppose that 9 is a mapping from 0 to M with separable range. Fix € > O. There exists a sequence {x n E M : n E N} of g(O) such that 00

g(O) � U B(xn , €/2) . n= l For each n E N , let Cn = g - l (A n B(x n , €/2)). Since /-l* (Cn) � /-l(0) < 00 , there is a measurable set Dn such that Cn � Dn and /-l* (Cn) = /-l(Dn). Let n- l n- l An = Cn \ U Dk and En = Dn \ U Dk. k= l k= l

Then {An}�= l and {En}�= l are two sequences of subsets of 0 such that for all n E N, En is measurable,

4. 1 . EXTREME POINTS AND SMOOTH POINTS

221

o The proof is complete. Note: In Theorem 3.2.4, we proved that if E is an order continuous Kothe function space over a finite measure space (0, /-l), then for any F E (E(X)) * , there is a weak* measurable function F : 0 -4 X* such that (i) for any f E E(X) , ( F, 1) = fn ( F (t) , f(t) ) dt;

(ii) II F II ( E (X » ' = 1I II F O ll x · II E' ·

The following theorem is due to Z. Hu and B.-L. Lin [12] : Theorem 4.1.3. Let X be a separable Banach space, and E an order con tinuous Kothe function space over a finite complete measure space (0, /-l) . Then F E B((E(X))*) = B (E* (X* , w *)) is an extreme point of B((E(X))*) if and only if II F( · ) llx . is an extreme point of B(E* ) and for almost all t E supp F, F(t)/ II F(t) llx . is an extreme point of B (X * ) . Proof: One direction is trivial. Hence we need to show only that if F is an extreme point of B((E(X))*), then for almost all t E supp f, F(t)/ II F(t) llx . is an extreme of B(X*) . Let F be any vector on the unit sphere S (E* (X* , w *)) of E* (X, w * ) . Suppose that the set { t .. F(t) IS. not an extreme pomt . of X * A II F(t) 11 has outer measure greater than O. Then there is a function H : 0 -4 X* such that for all t E O, II F(t) ± H (t) llx = II F(t) llx H (t) =I 0 for all t E A. Let {xn : n E N} be a countable dense subset of S(X), and let d be the metric on X* defined by

}

_ -

00

d(x * , y * ) = L 2 - n l ( (x * y * ) , xn ) l · n= 1 It is known that the metric d and the weak* topology coincide on every bounded subset of X* . Since A has outer measure greater than 0, there is N E N, such that C = t : d (H(t) , 0) ;::: and II F(t) I ::; N } has outer measure greater than O. Note: /-l(0) < 00 . There is a measurable set Eo such that C � Eo and /-l(Eo) = /-l* (C). Let M = N B(X * ) . Applying Lemma 4.1.2 to the functions H , F, and the set A = C with € = 2 - 1 , we obtain the sequences {Dn , d �= 1 and {En , 1 } �= 1 of subsets of 0 which satisfy the following conditions: • {En , d �= 1 is a sequence of mutually disjoint measurable sets such that U�= 1 En , 1 = Eo . -

{

�

222


For any n E N , Dn,l � En, l and 1-£* (Dn , I ) = I-£ (En, d . 1 • For any n E N , d-diam (H(Dn,I)) < 2 - and d-diam (F(Dn, l)) � 2 - 1 . Since the function F is w*-measurable, we may assume that for each n E N , d-diam (F(En, l)) � 2 - 1 • Replace € 2 - 1 by € = 2 - 2 . Applying Lemma 4.1.2 to the functions H, F and each set Dn, l , n E N , and then reordering the subsets, we obtain two sequences {Dn, 2 }�= 1 ' {En, 2 }�= 1 of subsets of it that satisfy the following conditions: • {En, 2 }�= 1 is a sequence of mutually disjoint measurable sets such that U�= l En, 2 Eo . • For any n E N , Dn,2 � En, 2 n Dj , l for some j E N and 1-£* (Dn, 2 ) = I-£ (En, 2 ). 2 2 • For each n E N , d-diam (H(Dn, 2 )) < 2 - and d-diam (F(En, 2 )) � 2 - . Continuing by induction, we obtain two double sequences { {Dn,k}�=d: 1 ' { {En,k}�=d: 1 that satisfy the following conditions: • For any k E N , {En,k}�= l is a sequence of mutually disjoint measurable sets such that U�= l En,k = Eo . • For any k , n E N , Dn k � En,k n Dj,k 1 for some j E N , and I-£ * (Dn,k) = I-£ (En,k). k and d-diam (F(En,k)) � 2 - k . • For any n, k E N , d-diam (H(Dn,k)) < 2For each n , k E N , select an element t n,k E Dn,k , and for any k E N , let Hk and Fk be the functions defined by •

=

=

,

It is easy to see that for all t E it and any k , m E N , II Fk (t) ± Hk (t) ll x · = II Fk (t) l l x · , d(Fk (t) , F(t ) ) � 2 - k , d(Hk (t) , Hk +m (t)) � 2 - k . Note that for any k E N and t E it , Fk is weak* measurable and II Fk (t) lI x = II F(t ) l l x . So for any k E N , Fk belongs to the unit ball B (E(X))*) of (E(X)). We claim that {Fk}k: 1 converges to F weak* . Let f be any element in E(X ) . Then for any t E it , we have I ( F'; (t ) , f(t )) 1 � II F(t ) lI x · ll f(t ) ll x , klim -+ oo (F'; (t) , f(t)) = ( F(t) , f(t)) .

223


By Lebesgue's dominated convergence theorem, the sequence { ( Fn , J) }�= 1 con verges to ( F, I ) . We have proved our claim. Note that (B(X*), d) is a compact metric space. A similar argument shows that {Hd� l converges in the weak* topology. Let Ho be the weak* limit of {Hdk=l ' Then for any t E C, d ( O, Ho (t)) 2:: l iN (this implies that Ho =1= 0) and II F(t) Ho (t) llx . � I I F(t) llx We have proved that F is not an extreme 0 point of B ( (E(X))* ) . The proof is complete. Modifying the proof of the above theorem (replacing d by the norm I . I l x ), we have the following theorem.

±

• .

Theorem 4.1.4. Let X be a separable Banach space, and E a Kothe func

tion space over a finite complete measure space (0, J-L) . A unit vector f in E(X) is an extreme point of B(E(X)) if and only if I l fO ll x is an extreme point of B (E), and for almost all t E supp f, f(t)/ ll f(t) llx is an extreme point of B(X) .

The following example is due to P. Greim [5] : Example 4.1.5. There is a nonseparable Banach space W such that for any

1 < p < 00, there is an f E S(Lp([O, 1] , W)) such that f is an extreme point of B(Lp([O, 1] , W) , but for all t E [0, 1] , f(t)/ ll f (t) ll w is not an extreme point of B(W) . Let B be the convex hull of Then there is a three-dimensional normed space Wo such that B(Wo) = B. Let h : [0, 1] - Wo be the function defined by

(

(� ) ( � )) .

h(t) = 0, cos t , sin t

It is easy to see that h(t) is an extreme point of B if and only if t =1= o. Let

{

W = IT Wo = g : [0, 1] - Wo : sup Il g(t) ll wo < t E [O, l ] [0, 1 ]

oo },

be an foo-product of uncountably many copies of Wo , and let f be a function from : [0, 1] to W defined by f(s) (t) = h( l s + t - 1 1 ).

Then f( s) (t) is an extreme point of B(Wo) if and only if t =1= (1 - s). This implies that for any s E [0, 1] , f(s) is not an extreme point of B(W) . We claim that f is an extreme point of B(Lp([O, 1] , W)). Let gl , g2 be any two vectors in the unit ball of W such that 2f = g l + g2. Then for almost all t E [0, 1] , if 1 - s =1= t, then gl (t) (S) = f(t) (s) = g2 (t) (S). Without loss of generality, we assume that for any t E [0, 1] and any s E [0, 1] \ { 1 - t },


224

gl (t) (S) = f(t) (s) = g2 (t) (S) . We claim that for almost all t E [ 0 , 1] gl (t) (1 - t) = f(t) (1 - t) = g2 (t) (1 - t) . If the claim is true, then we have gi = g2 = f · Fix € > 0 . By Lusin ' s theorem, there is a measurable subset C � [ 0 , 1] such that m ( [O, 1] \ C) < € , and the functions gi l e (the restriction of gi to C), g2 1 e and f i e are continuous. Recall that a point t E (0 , 1) is called a Lebesgue point of C if 1. p, (( t - € ,t + €) n C) = 1. 1m 2€ dO Clearly, every Lebesgue point of C is a limit point of C. Let t E C be a Lebesgue point of C. Then there is a sequence {tn }�= 1 � C \ {t} such that limn ..... oo t n = 1 - t. This implies '

f(t) (1 - t) = nlim t) ..... oo f(tn)(1 t) = g2 (t) (1 - t) . = nlim ..... oo g2 (tn)(1 -

It is known that for any measurable subset C of (0 , 1 ) ,

m ( {t E C :

t is a Lebesgue point of C} ) = m ( C).

We have proved that for almost all t E C, we have gl (t) = g2 (t) = f (t). Since € is an arbitrary positive real number, f = gl = g2 a.e. The proof is complete. Using the technique in the proof of Theorem 4 . 1 . 3 , we have the following theorems. We leave all proofs to the reader. Theorem 4.1.6. Let E be an order continuous Kothe function space over

a finite complete measure space, and X a Banach space . A unit vector f of the Kothe -Bochner function space E(X) is a smooth point of E(X) if and only if Il fO llE is a smooth point of E, and for almost all t E supp (I) , f(t)/ ll f(t) ll x is a smooth point of X . Theorem 4.1.7. Let X be a separable Banach space, and E an order con

tinuous Kothe function space over a finite complete measure space (0, p,). If f E S(E(X)) is an exposed point of B(E(X)), then Il fO llx is an exposed point of B(E), and f(t)/ ll f (t) llx is an exposed point of B(X) for almost all t E supp f. Theorem 4.1.8. (Z. Hu and B.-L. Lin [12] ) Let X be a separable Banach

space and E an order continuous Kothe function space over a finite complete measure space. If F is a weak* exposed point of B ((E(X))*) , then II F( · ) ll x. is a weak* exposed point of B (E*), and F(t)/I I F(t) ll x. is a weak* exposed point of B (X*) for almost all t E supp F.


225

p < 00 , let q = pS . In [12] , Z. Hu and B.-L. Lin modified Example 4. 1 .5 and showed that there are a Banach space X and a weak* extreme point of ( Lp(X))* such that for all w E supp F, F(w)/ II F(w) 11 is not a weak* extreme of X. They asked whether every extreme point of Lq(/-l, X*) is an extreme point of (Lp(/-l, X))* for any Banach space X. (2) The extreme points of Lebesgue-Boncher function spaces were first con sidered by J.A. Johnson [14] and K. Sundaresan [28] . J.A. Johnson proved Theorem 4. 1 .4 when E is an Lp-space for some 1 < p < 00 and n is a Polish space. (3) The smooth points of Kothe-Bochner function spaces were first consid ered by W. Deeb and R. Khalil [4] . They proved Theorem 4. 1 .6 if E is an Lp-space for some 1 < p < 00 . Then J. Cerda, H. Hudzik, and M. Mastylo [2] showed that this result still holds if E is a strictly monotone Kothe func tion space and if the dual X* of X is separable. In [8] , W. Hensgen also considered the exposed point in Hp(X). He proved the following results: (a) Let X be a separable reflexive Banach space. The a unit vector f in the Hardy-Bochner space Hp(X), 1 < p < 00 , is an exposed point of B(Hp(X)) if there is a subset C of [0, 1] with positive measure such that for any t E C, f(t)/ ll f(t) llx is an exposed point of X. (b) There is an exposed point f in B(Hp(co)) such that for all t E [0, 1] ' f(t)/ ll f(t) lloo is not an extreme point. (4) In [2] , J. Cerda, H. Hudzik, and M. Mastylo proved Theorem 4. 1 .7 with a stronger assumption. We do not known whether the converse of Theorem 4.1.7 is true. We do not know whether the assumption "X is separable" in Theorem 4.1.7 is necessary. (5) Let X be a separable Banach space, and E an order continuous Kothe function space over a finite complete measure space. Suppose that II F( · ) llx . is a weak* exposed point of B(E* ) and for almost all t E n, F(t)/ II F (t) llx . is a weak* exposed point of B(X*). In [12] , Z. Hu and B.-L. Lin proved that F is a weak* exposed point of B ( (E(X))* ) .

Remark 4.1.9. (1) For any 1
0, x r:J. co ( M ( x , € X) ) . By the Hahn-Banach theorem, x is a denting point of B(X) if and only if for any € > 0, there are a unit vector x* in X* and ° > ° such that x belongs to the slice S( x * , 0, X) and the diameter of the slice S( x * , 0, X) is less than €. A unit vector x in X is said to be a locally uniformly convex point of X if for any sequence {xn} �= 1 in the unit ball of X, limn-+oo IIxn + x II = 2 implies { Xn}�=l converges to x . A unit vector x in X is said to be a weakly locally uniformly convex point of X if for any sequence {x n }�= l in the unit ball of X, lim n II x n + x II = 2 implies { Xn}�= l converges to x weakly. A Banach space X is said to be midpoint locally uniformly convex (re spectively, locally uniformly convex, weakly locally uniformly convex) if every x E S(X) is a strongly extreme (respectively, locally uniformly convex, weakly locally uniformly convex) point of B(X). A Banach space X is said to have property G if every x E S(X) is a denting point of B(X). Lemma 4.2.1. Let E be a Kothe function space, and X a Banach space. If f E E(X) is a strongly extreme point (respectively, denting point, locally uniformly convex point, weakly locally uniformly convex point) of E(X), then I f ( . ) II X is a strongly extreme point (respectively, denting point, locally uniformly convex point, weakly locally uniformly convex point) of E. -

,

.... oo

Proof: Let x be a unit vector in X, and g

(t ) =

{ �f<mx

g

the function defined by if f (t) =1= 0, otherwise.

4.2. STRONGLY EXTREME AND DENTING POINTS

227

The mapping T : E -. E(X) defined by T(h) (t) = h(t) . g( t )

is an isometry such that T ( II 1 ( . ) II x ) = I . SO if 1 is a strongly extreme point (respectively, denting point, locally uniformly convex poipt, weakly locally uni formly convex point) of E(X) , then 11/ (') llx is a strongly extreme point (respec tively, denting point, locally uniformly convex point, weakly locally uniformly 0 convex point) of E . Lemma 4.2.2. Let 1 be a unit vector in E(X) such that 11/ 0 I lx is a denting

point 01 the unit ball 01 E. For any € > 0, Il /O llx does not belong to the set

{ ll h O llx : h E co { h E E(X) : Il h O ll x E M ( II /Ollx, € , E) } } .

Proof: Assume that the lemma is not true. Then there are € > 0 and sequences {Nn E N}�= l ' {{ a� E IR}�=d�: l ' {{ I:: E E(X)}�=d�: l that satisfy the following conditions: (i) a � > 0 and E �: l a� = 1 for all n E N and m � Nn. (ii) For all n E N and m � Nn , II /::- O l i x E M ( II / O llx , € , E) . (iii) limn-+oo I E �: l a � I::- O ll x

I

Let

=

II /O llx .

Nn

gn = L Ila� I:O llx, m= l 9

=

I I /Ollx, hn gn - 2 ( (gn - g) V 0) . =

Then Il gn ll E � 1 and I h n l � gn (because 9 ;::: 0) . Note that

( ( 11/0llx

) V0 Nn I I L a� I:O ll x ) V O. m= l Nn

(g - gn ) V 0 = 11/0 llx - L a� II /:O llx m= l �

-

This implies that limn-+oo ((g - gn) V 0) = 0, and limn-+oo (gn 1\ g) = g. So we have hn + gn ' gn 1\ 9 = g. 1· 1m l1m n-+oo 2 = n-+oo Note that 9 is a strongly extreme point of B (E) . This implies that limn-+oo gn = g,. But this is impossible (since 9 is a denting point of B(E) and for any € > 0, 0 9 E M(g, € , E)) . The proof is complete.


228

Theorem 4.2.3. Let E be a Kothe function space and X a Banach space.

A unit vector f in E(X) is a denting point of B (E(X)) if and only if Il f O ll x is a denting point of B(E) and for almost all t E supp f, IIf1m x is a denting point of B (X) . Proof: Sufficient conditions. Let f be an element in B(E(X)) such that

Il f O l l x is a denting point of B(E) , and for almost all t E supp f, I l<m x is a denting point of B(X) , but f is not a denting point of B (X) . Then there are € > 0, a sequences {Nn}�= l of natural numbers, a sequence { a� : m � Nn}�= l of positive real numbers, a sequence {f: : m � Nn}�= l of E(X) that satisfy

the following conditions: (i) a� � 0 and

l:�:1 a� = 1 for all n E N and m � Nn .

(ii) f: E M(j, € , E(X)) for all

n

E N and m < Nn.

. . . ) l ' mn ( III I � oo "Nn L..J m = l anm fnm = f .

We claim that for any 0 > 0, nlim � oo

�

�

I l f;:' ( .) II x EM( llf ( ·) Il x 8 E) ,

a�

=

O.

,

Assume that the claim is not true. Let A = M( ll f O ll x , 0, E) .

By passing to a subsequence, we may assume that there is a 01 > 0 such that bn = l: lIf;:' ( - ) lI xE A a� > 01 . By Lemma 4.2.2, there is a 02 > 0 such that

L a�f: /bn E { iI E E(X) : Il g ( ' ) l l x E M( ll f( · ) l l x , 02 , E) } . IIf;:' ( - ) ll x E A Note: The sequence

converges to Il f O ll x , and Il f( ' ) l l x is a strongly extreme point of E. This implies that either lim n � oo bn = l or { l: lIf;:' ( -)llx EA a� f: /bn }�= l converges to Il f O ll x (in E) , a contradiction. We have proved our claim. By the above claim, we may assume that for all n E N and m � Nm , f Il � O l l x = Il f O ll x . By passing to a further subsequence if necessary, we may also assume that the sequence l: �':1 a� f� converges to f almost everywhere.

229

4.2. STRONGLY EXTREME AND DENTING POINTS

For any n E N, m � Nn, and t E n, let

{

i }

f t) Bn, m = t E n : Il f: (t) - f(t ) 11 > € ll II ,

{

S (n) = t E n : S=

00

00

i},

L a� >

tEBn,m

n U S(k).

n= l k=n Note: S � supp f. If t E S, then t E S(n) infinitely often, and l I !cm x is not a denting point of B ( X ) . By assumption and Lemma 3. 1. 19, we have jj(S) = 0, and Il / O ll x is order continuous. Hence if n is large enough, we have € II I . 1 u%"=n s( n) II E (X) � "4

and

Nn € � L a� ll /: - f IIE (X) m= l Nn � L a: II /: o - f O llx . 1 Bn,m + ll f:o - I O llx . 1 0\Bn,1n E E =l m Nn � L a� 2 11 /o llx . 1 Bn,1n \ S ( n) E + 2 11 / 0 ll x . 1 S ( n ) E + m= l € < 2 · -8 + 2 · -4€ + -8€ < € '

(I I

-

(1

li

Il 1l

li ) i

1

Il )

a contradiction. Thus I must be a denting point of E(X). Necessary conditions. By Lemma 4.2. 1, we need to show only that if f is a denting point of E(X) , then for almost all t E supp I, l(t)/ ll f(t) llx is a denting point of X. Let

{

An, m , l = x E X : Il x ll 1 there exist X l , X 2 , " " x 2 n E M(x , 1 /i , X) n such that 11 2In 2i = l X i - x ii < mI . =

,

}

L

If x is not a denting point of B ( X ) then :3Nm:3n such that x E An, m , l ' It is easy to see that An, m+ 1, l is relatively open in S(X ) , and An, m+ 1, l � An, m , l � An + 1, m , l � An + 1, m , l+ 1 ' Suppose that I be an element in S(E(X)) such that

{ t E supp I : 1I !�;�l x is not a denting point of B (X ) } I(t) E An, , l } = { t E supp ! : 11 / (t) llx lld Dl n�1 m 00

00

00

230


has positive measure. Then there exists L such that

{

L An � l l ild mCl n , m ,£ } Dl n�l An , m , L } = C 00

f(t) t E supp f : I f(t) x E / lI f(t) E = t E supp f : f(t) Il ll x

{

00

00

00

has positive measure. Let

Then

f(t) Cn , m = { t E supp : Il f(t) E An , m , L } . ll x 00

C U�=l Cn ,m

00

c = n U Cn , m '

m=l n=l

Note: � � supp f for all m E N and Il fO llx is order continuous. For any mo > 0 there exists no such that

Fix M > 0 and let mo = 4 max{M, L}. Note: f is strongly measurable. We may assume that the range of f is separable. Let {Yl , Y2 , . . . } be a count able dense subset of the range of f. Fix i E N. Suppose that the intersec tion of B(yd II Yi ll x , l imo ) and is nonempty. Select an element w in B (yd I lYi ll x , l imo ) By the definition of the set there exist vectors Zi, l , Zi, 2 , . . . , Zi,2 n o E M(w , 11 L, X) such that 1 Il w - Zi,j Il x > L Vj = 1 , 2, . . . , 2 , no 1 2 w - - � t , ) X - mo

2no

o Ano m , L , nAno , mo , L .

Ano , mo , L , no

I l 2no j�z" =l 1 < _1 .

Suppose that w' is another element in B (Yi / II Yi ll x , l imo ) . Then 2 no 2 no w' Z" j � w Z" j + IIw ' - wllx x x 3 2 1 1 = -n n--1 I I F(t) ll x o ' 1I f (t ) ll x and Il x ll < Il f (t ) ll x } ,

inf{ E > 0 : S(t, n) � B(f(t) , E n , inf{d(t", n) : n E N} .

Let { X k : k E N} be a countable dense subset of X. Fix 8 > k, n E N, let An , k be the set An, k

=

n-1 { t : (F(t) , Xk ) > nIl x k ll x

.

O.

For any

Il f(t) llx . II F(t) ll x o ,

}

< Il f(t) l l x , and Il f(t) - x k ll � 8 .

Then for each n, k E N, An, k is measurable. Note: (i) The set { t : d(t, n) � 8} is equal to U%"= l An, k ' So e is measurable.

(ii) II FftWx strongly exposes at II f{m x if and only if e(t) O. Hence we need to show only that e = 0 a.e. Suppose that e =1= 0 a.e. Then there are a 8 > 0 and a measurable subset A of { t : e ( t) > 8} with J-L (A) > O. For each n E N, let =

0

if t E A n An, k \ U ::l1 An, i if t t/: A.

Then

n-1 n

and Il f - gn II E(X) � 8 1 1 1 A I I · o We get a contradiction. The proof is complete. Let E be an order continuous Kothe function space over a finite measure space. Since B(E* (X*)) is weak* dense in B((E(X))*), every weak* strongly exposed point of (E(X » )* is in E* (X* ). We have the following corollary: Corollary 4.3.2. Suppose that X is a Banach space such that X* is sep arable, and E is an order continuous Kothe function space over a finite mea (F, g ) � n

-

sure space. If F E (E(X))* is a weak* strongly exposed point of (E(X) * , then II F( ' ) l l x o is a weak* strongly exposed point of E* , and for almost all t E supp F, x is a weak* strongly exposed point of X* . lI

.:(ml

4. 3.

23 5

STRONGLY AND W* -STRONGLY EXPOSED POINTS

Theorem 4.3.3. (Z. Hu, B.-L. Lin, P.K. Lin and H. Sun [1 1 , 22]) Let E be a order continuous Kothe function space over a complete measure space (O, �, J.L ) , and let X be a Banach space. A unit vector f in E(X) is a strongly exposed point of B(E(X» if Il f O ll x is a strongly exposed point of B(E) and for almost all t E supp f, II x is a strongly exposed point of B(X).

tm

For the proof of Theorem 4.3.3, we need the following two lemmas. First, we need the following notation: Let C be a closed, convex, bounded subset of X, and x* a nonzero element in X* . Let b = sup { (x* , x) : x E C } . For any f > 0, the slice S (x* , f) is the set S(X * , f) = {x E

C : (x* , x) � b - f}.

Lemma 4.3.4. (Z. Hu and B.-L. Lin ) Let X be a Banach space. A unit

vector x in X is a strongly exposed point of X if and only if there exist two de creasing null sequences {fn}�=l and {On}�=l and a sequence {X�}�= l in B(X* ) such that for any m � n, and any x, y E B(X), the inequalities (x� , y) > 1 - Om and (x� , x) > 1 - Om implies that Il x - y ll � f m . In particular, every weak* limit point x* of {x� : n E N} strongly exposes B(X) at x. Proof: The necessary conditions are clear. We need to prove only the

sufficient conditions. Let x* be a weak* limit point of {x� } . Then (x* , x) = 1. We claim that for any y , z E B(X) , min { (x* , y) , (x* , z ) } � 1 - � implies that Il y - z ll � 2 fn . Proof of the claim. Suppose that (x* , y) � 1 - � and (x* , z) � 1 - � . Then there is m > n such that (x� , y ) > 1 - on and (x� , z) > 1 - on . Notice that the assumption implies that the diameter of S(x� , on ) is less than 2 f n . So 0 Il y - z ll � 2 fn• The proof is complete. Lemma 4.3.5. (Z. Hu and B.-L. Lin ) Let E be an order continuous Kothe function space, and X a Banach space. Let f be any positive real number, F an element in S« E(X» * ) , and II an element in S(E(X» . Let G = II F( · ) l l x * E S(E * )

and g = Il lI O ll x E S(E) .

Suppose that (G, g) = 1 and that there are o(f, G) > 0, M > 1, and a measurable subset A of supp G that satisfy the following conditions:

(1) For any h E E, (G, h ) > 1 - 0 implies I l h - g i l E < � . ( 2) (G, g . 1 n\A ) < � . (3) For any t E A, diam ( S ( I FftWx * it)) < � . (4) ror any t E A ' D

(

'

h (t) F(t) 1 I h (t ) lI x E S II F(t) lI x *

'

1 M

)

.

Then for any 12 E B(E(X» , the inequality (F, h ) > 1 II II - h I I E( x ) < €.

3it implies that

3


2 6

�' 3 unit vector in X and 12 the measurable function defined by Proof: Suppose that 12 E B(E(X)) and (F, h ) > 1 _

h (t)

=

{

Xo

-

Let Xo be any

if t E supp (I), otherwise.

h (t) IIh (t) ll x

Since (G, IIhO ll x) � ( F, h ) > 1

-

3� � 1

-

8,

Il g · 12 - h II E ( x ) = I l g - ll h O ll x ll E
1 - �. o


Proof of Theorem 4.3.3: Let f be a unit vector in E(X) such that Il f O ll x

is a strongly exposed point of B (E) and for almost all t E supp f , f(t)/ ll f(t) l l x is a strongly exposed point of B(X) . Let 9 = Il f O ll x E E, and let G be an element in S (E* ) such that G strongly exposes B(E) at g . Let a : supp f � S (X*) be any function from supp f to S (X*) such that a (t) strongly exposes B(X) at f(t)/ ll f(t) I l x for each t E supp f. For each n, m E N, let D(m, k) be the set

{

D(m, k) = t E supp f : diam

Note: D(m, k) may not be measurable, but D(m + 1, k) 00

U D (m, k)

:>

(S (a (t) , � )) < 41k } '

D(m, k) for all

for all k E N.

supp f

m= l

k , n E N,

( 4 . 3) (4 . 4)

Let J-L* be the outer measure of J-L . Since f is a denting point of B(E(X) ) , supp f is a-finite. Hence for any subset A of supp f , there is a measurable subset M (A) of supp f such that for any measurable set B,

n

n

J-L * (B A) = J-L (B M (A)) .

Note: G strongly exposes B(E) at g. There is 8n > 0 such that for any h E B(E) , the inequality ( G, h ) > 1 - 8n implies that Il g - h i l E < 4� ' Step 1 . We claim that for each n E N, there are an increasing sequence {mn : n E N} and sequences {A(n, j) : j :::; n} and {C(n, j) : j :::; n} of subsets

of n that satisfy the following conditions: (i) {A(n, j) : j :::; n} is a finite sequence of disjoint subsets of D(mn , n) such that for any k :::; n, A(n, k) � (ii) For each n E N and

k :::; n,

n

n D(mj , j) .

j =k

C(n, k) = M (A(n, k)) , J-L (C(n, j) C(n, k) ) = 0 if j -I k :::; n.

n

238

CHAPTER

4.

STABILITY PROPERTIES I

(iii) For any k � n, -

( G, g . ( 2:= 1C(n ,j) ) ) > 1 ; j 1 k

_

0 .

We shall prove the claim by induction. By (4 . 4), for any subset A of f!,

(G, 9 . 1 M(A) )

=

J{M (A)

( G , g) dJ-L

=

lim

1

m-> oo M (AnD(m, 1 ))

(G, g) dJ-L.

So there is m1 E N such that

( G , g . l M (D(ml , 1 )) ) > 1 - � .

Let A(l, 1 ) D(m l , 1) and C(l, 1) M (A(l, 1)) . Assume that {mj : j � n} , {C(i, j) : j � i � n} and {A(i, j) : j have been constructed. By (ii) and (iii), for any k � n, =

=

�

i � n}

-

(G, g . ( �= 1 1 M (A(n ,j)) ) ) > 1 - ; k

0

3

.

By ( 4 . 3 ) and (4 .4), there is mn + 1 > mn such that for all j < n, and For j � n, set A ( n + 1, k) C(n + 1 , k)

=

=

A(n, k) n D(mn+ l , n + 1)

n+ 1

� n D(m i ' i) ; i= k

M (A(n + 1 , k)) ,

and for j = n + 1, set n

A(n + 1 , n + 1) C(n + 1 , n + 1)

=

=

D(mn + l , n + 1 ) \

U C(n + 1 , i)

i= 1 M (A(n + 1, n + 1)) .

Clearly, the sets {A(n + 1 , 1), . . . , A(n + 1 , n + 1)} and {C(n + 1 , 1), . . . , C(n + 1 , n + 1)} satisfy (i)-(iii) . We have proved the claim.

4.3.

239


Step 2. Let {mn : n E N}, {An , j : j � n E N} and {C(n,j ) : j � n E N} be the sets that we obtain in the above claim. Note: f is a measurable function. Without loss of generality, we assume that {f(t ) : t E O} is separable. We claim that for any n E N, there are a finite increasing sequence {n ( n, 0) 0 � n (n, 1) � . . . < n ( n, n - 1 ) � n (n, n) < oo}, finite collections {E(n,j ) : j � n ( n, n) } of measurable sets and a finite set {t n,k E O : k � n ( n, n) } that satisfy the following conditions: =

E A(n, i ) for n ( n, i - 1) < j � n(n, i ) . (v) For any j � n E N, E(n, j ) { t E supp f : 11 !;��l x E S (a(tn,j ) , �n ) } n C(n,j ) , and for any f � k :S n, (iv) t n,j

=

We shall use induction to prove this claim. Let { Sj } � 1 be a sequence in A(l, 1) such that { f (sj ) : j E N} is a dense subset of f(A (l, 1)) . For each j E N, set E(l, 1 , J ) t E supp f : Il ff(t(t) )l l x E S a(sj ) , m11 ) } n C(l, 1 ) . Since

-

. {

(

=

A(l, 1) � U { sE A ( 1 , 1 )

f (t) t E supp f : ( ) Il f t 11 x

and

00

E S (a(s), ;;;-1 ) } U= 1 E(l, 1 , J ) j 1

=

-

.

1C( 1 , 1 ) (G, g) dJ-L > 1 - 381 '

there exists n(l, 1) such that

For all j � n(l, 1 ) , set

Sj and E(l, j ) E (l, 1 , j ) . Then {tl,l, . . . , t1,a ( 1 , 1 ) } and {E(l,j ) : j � n(l, 1)} satisfy (iv) and (v) . Assume that for all k < n, we obtain {n(k, 0) 0 � . . . � n(k, k) } , {E ( k , j ) : j :S n ( k , k) }, and { tk,j j � n(k, k)} that satisfy (iv) and (v) . We shall apply the above argument to each A ( n, k) , 1 � k � n . t 1 ,j :

=

=

=

240


Let {Sj : j E N} be a countable subset of An , l such that {f( sj ) : j E N} is dense in f(A(n, l)). (Note: { Sj : j E N} is a new countable subset of A.) For each j E N, set

( 1 )} f( sj ) t E supp f : E S a( sj ) ' { ' mn Il f ( sj ) llx

-

E(n, 1 , ) )

=

n

e ( n, 1 ) .

Note: A(n, 1) � U;:l E (n, 1 , j) and

1 j n,j (G, g) , du > 1 - 3"8k u =l C(

for all k :S n. There is 0' ( n,

)

1 ) E N such that

For all j :S O'(n, 1), set

t n ,j

=

Sj and E(n, j ) = E ( n, 1 , j) .

Then {E(n, l) , . . . , E(n, O'(n, l))} and {tn , l, . . . , tn , a ( n , l ) } satisfy (iv) and (v) . Assume that for some k :S n, we obtain finite sequences {tn , !, . . . , tn , a ( n ,k - l ) } and {E(n, 1 ) , . . . , E(n, O'(n, k - 1)) } that satisfy (iv) and (v) . Let {Sj : j E N } be a countable subset of A(n, k) such that {f( sj ) : j E N} is dense in f(A(n, k)) . For each j E N, set

-(

f( sj ) (a( sj ) , 1 ) } t E supp f : E S { ' mn Il f( sj ) llx

E n, k, ) )

=

n

C ( n, k).

Note: A(n, k) � U;:l E (n, k, j) , and for any k :S f :S n,

1

", (n, k - l ) E ( u j=l

n ,]

' ) uu jl = C( k

�. 1-du ( > g) G, 3 n ,] ')

There is a natural number O'(n, k) � O'(n, k - 1) such that for any k :S f :S n,

1

u j",=(nl , k - l ) E (

n,]

' ) U u ",(n , k ) - ", ( n , k - l ) E -( j=l

n , k ,]

' ) uulj= + 1 C ( k

�. 1-dj-t (G, > g) 3 n , ]'

)

For each O'(n, k - 1) < j :S O'(n, k) , set

tn ,j

=

Sj

-

a n ,k l (

-

)

and E(n, j)

=

E (n, k, j - O'(n, k - 1)).

Then {E(n, 1 ) , . . . , E(n, O'(n, k)) } and {tn , l , " " We have proved our claim.

tn , a (n , k) } satisfy (iv) and (v) .

4.3.


241

For each k, n, k ::; n E N, let B (n, k) be the set a:(n , k ) B (n, k)

and let Fn : n

--+

=

U E(n, j),

j =l

X* be the functions defined by a:(n , n )

Fn

=

(a(tn ,t ) . l supp f\B(n ,n) + jL=l a(tn ,j ) · 1 E(n ,j)\U1:; E(n ,i) ) . G .

Then Fn is measurable for all n there is k l ::; k such that a: (n, k l

-

1)

E

N. Let t be an element in B(n, k). Then

< j ::; a: (n, k1 ) ::; a: (n, k) and t E E(n, j).

((

:k )) f(t) 1 1 E S ( a(t n , j ) , S ( a(tn ,j ) , ) . ) mk mn Il f(t) l l x 1

< 4k '

diam S a(tn , j ) ,

�

We claim that for any k, n, k ::; n, and 12 E B (E(X)), (Fn , h ) > 1 implies that 1 11 11 - h II E( x ) < "k .

-

3� k

Suppose the claim has been proved. By Lemma 4.3.4, f is a strongly exposed point of B(E(X)) (where € k i and 8k 3� k ) · Proof of claim.: For any k ::; n, let 11 f, F Fn , M m k , € i , A B(n, k) , and 8 8k • Then F, 11 satisfy the assumptions of Lemma 4.3.5. By Lemma 4.3.5, we have =

=

=

=

=

=

11 11 - h Il E(x ) < €

=

=

=

1

k·

0 The proof is complete. The proof of the following lemma is similar to that of Lemma 4.3.4. We leave it to the reader.

Lemma 4.3.6. Let X be a Banach space. A unit vector x* in X* is a

strongly exposed point of X if and only if there exist two decreasing null se quences {€n}�=l and {8n }�= 1 and a Cauchy sequence {Xn}�=l in B (X) such that for any m ::; n, and any y*, z* E B(X* ), the inequalities (y* , xn) > 1 - 8m and (y* , xn ) > 1 - 8m imply Il x* - y* I I ::; Em . In particular, every limit point x of {xn : n E N} weak* strongly exposes B(X*) at x* .

242

CHAPTER

4.

STABILITY PROPERTIES I

Using Lemma 4.1.7 and the technique employed in the proof of Lemma 4.3.5 and Theorem 4.3.3 (add the condition "diam a( A ( n , i)) � t"), we have the following theorem. Theorem 4.3.7. (Z. Hu and B.-L. Lin) Let X be a separable Banach space and E a Kothe function space over a finite measure space (0, p,) . A unit vector F in (E(X))* is a strongly exposed point if II F O ll x * is a weak* strongly exposed point of E* and for almost all t E supp F, F(t)/ II F(t) llx * is a weak* strongly exposed point of X* . Remark 4.3.8. J.A. Johnson [15] first considered the strongly exposed

point of the unit ball of Lp(X) . He proved that if f is a unit vector in Lp(X) , 1 < p < 00 , and if there is a measurable function F : 0 � B ( X* ) such that II F O ll x * strongly exposes at Il f O ll x and for almost all t E supp (I) , F(t) is strongly exposes f(t)/ ll f (t) llx , then f is a strongly exposed point -of E(X) . Later, Z. Hu and B.-L. Lin [11] showed that one needs to assume only that for almost all t E supp (I) , f(t)/ ll f(t) llx is a strongly exposed point of X. P.K. Lin and H. Sun showed that the same result is still true if one replaces Lp by any order continuous Kothe function space. In [6, 7] , P. Greim proved that for any separable space X and 1 < p < 00 , if f is a strongly exposed point of Lp(p" X) , then for almost all t E supp f, II tm x is a strongly exposed point. Moreover, he also proved that the assumption can be removed if there is a measurable function F E L q (p" X*) such that F(t) is strongly exposed at II f�m x for almost all t E supp f. Therefore, the assumption "X is separable" in Corollary 4.3.2 is redundant. Note: Lusin ' s theorem is not true for w* -measurable functions. We do not know whether the assumption "X is separable" in Theorem 4.3. 1 is necessary.

4.4

Notes and Remarks

In [27] , M.A. Smith and B. Turett showed that if X is a strictly convex reflexive Banach space with the Kadec-Klee property, then for any 1 < p < 00 , Lp(X) has the Kadec-Klee property. We will show that this result is still true if X does not contains a copy of f l . First, we need the following characterization of denting points. Let X be a Banach space, and Y a subspace of X* that separates the points in X. Let T be the weakest topology such that for any y* E Y, the mapping x (y* , x) is T-continuous. If Y X* , then we call T the weak topology of X. Let K be a closed, bounded, convex subset of X. A point x E K is a T-point of continuity if for any sequence {xn}�=l in K, {xn}�= l converges to x in T implies that {xn }�= l converges to x. A point x E K is said to be a T- denting point of K if for any € > 0, there are y* E Y and 8 > 0 such that the set I---t

=

243

4.4. NOTES AND REMARKS

contains x and has diameter less then E. Let k be the closure of K in (X, T ) ** . It is known that (k, T ) is compact. A point x E K is said to be a T-extreme point if x is an extreme point of k. Clearly, every T-denting point is a T-point of continuity, and every T-denting point is a T-extreme point. The following theorem shows that if x is a T-point of continuity and x is a T-extreme point, then x is a T-denting point. Theorem 4.4.1. Let K be a closed, bounded, convex subset of X . x E K is a T-denting point of K if and only if x is a T-extreme point and x is a T-point of continuity.

Proof: One direction is clear.

Let K be a closed, bounded, convex subset of X . Suppose that x is a T extreme point of K and a T-point of continuity of K. For any E > 0, there exist 8 > ° and yi , . . . , y� E Y such that the set

Y2,

A=

{z E K

has diameter at most For i 1 , 2 , let =

:

(yj ,

z) � (yj , x) - 8 for all j ::; n }

E.

Mi =

{ z E K : (y; , z) � (y; , x) - 8}.

We claim that x ¢. T-co ( MI U M2 ) ' Suppose the claim has been proved. By the Hahn-Banach theorem, there are y * E Y and 8 > 8' > ° such that

)

(y * , x �

sup

z E CO(M1 u M2 )

(y * , z) - 8' .

This implies that

{ Z E K : (y * , z) � (y * , x) - 8' and (yj , z ) � (yj , x) - 8' for all 3 ::; j ::; n }

is a subset of A. Repeating this argument, we conclude that there is a T-open slice that contains x and is contained in A . So x is a T-denting point. Proof of claim: Suppose the claim is not true. Since x is a T-point of continuity, there exist E MI , E M2, and E [0, 1] such that + (1 converges to x in norm. Notice that x E MI U M2 , and k is T-compact. There are E MI , Z E M2 and E (0, 1) such that

- an)zn }� 1 y

Yn y

Zn

an

a x = ay + (1 - a ) z .

{anYn

o Clearly, =1= x. We get a contradiction. The proof is complete. It is known that every closed convex subset of a Banach space is weakly closed. We have the following corollary: Corollary 4.4.2. Let K be a closed, bounded, convex subset of a Banach space. x E K is a denting point if and only if x is a w* -extreme point of K and x is a point of continuity (with respect to weak topology) for K.


244

Suppose that X is a strictly convex Banach space with the Kadec-Klee property and X does not contain a copy of il . (i) Theorem 2.1.20 shows that X is midpoint locally uniformly convex. Hence every point on the unit sphere is a w*-extreme point in B(X). (ii) Rosenthal ' s iI-theorem ensures that every bounded sequence in a Ba nach space has either a weakly Cauchy subsequence or an iI-subsequence. Let X be a Banach space with the Kadec-Klee property. If X contains no iI-sequence, then the norm topology on the unit sphere of X is equiva lent to the weak topology (Le., every point on the unit sphere is a point of continuity of the unit ball) . We have following theorem: Theorem 4.4.3. Let X be a strictly convex Banach space with the Kadec Klee property. If X does not contain a copy of iI , then Lp(/-l, X) has the Kadec Klee property.

The following theorem is due to Z. Hu and B.-L. Lin [9]: Theorem 4.4.4. Every point of continuity of Lp([O , 1], X), 1 4 such that 1 l: � =1 rkllE ::; ;l!t . ( ii ) implies that {l:�!� + 1 (Xk - Xp+d )}� l is weakly null. By Corollary 1 .5.2, there exists a nonnegative sequence { A q }� 1 such that l:� 1 A q 1 and =

(q+ 1 )d (Xk - X (q+ 1 )d ) l l x ::; 8. q ±A 2:: 1 2:: =l = 1 q k qd+ 00

Thus

(q+ 1 ) d 8d ::; 2:: 2:: A q rk x k ll x 1 q=l k =qd+ 1 E( ) (q+ 1 )d (q+ 1 )d q 1 rk X (q+ 1 )d II E( X) ::; 1 q2::=l Aq k=2::qd+ 1 rk (xk - X (q+ 1 )d ) li E(x ) + 1 2:: q=l A k=2:: qd+ (q+ 1 )d ::; 8 · l l ( o , l ) IIE + 11 2:: A q 2:: rk x (q+ 1 )d II x E( ) q =l k=qd+ 1 d d8 M ::; 8 + 2:: A q l1 2:: r k ll ::; 8 + 2 ' q=l k=l E 00

00

00

00

00

o We get a contradiction. The proof is complete. Let E be a Kothe function space such that I g l 1 1 ::; I l g ilE for every g E E, and X a Banach space. A bounded subset A of E ( X) is said to be uniformly integrable if for any E > 0, there is an M > 0 such that for any f E A,

I k · l [ 1 I J ( ' ) l I x �Ml I 1 1 < E.

The proof of the following lemma is similar to that of Corollary 1 .3.10, we leave it to the reader. Lemma 5.1.4. Let X be Banach space and E a Kothe function space over a probability space (0, J-t) such that for any g E E, II g l 1 ::; I g I E . If {fn}�=l

is a bounded sequence in E ( X ) that is not uniformly integrable . Then {fn }�=l contains a complemented £1 -subsequence. The following theorem is due to J. Bourgain [7] . Theorem 5.1.5. Let X be a Banach space and {fn }�=l

a co-sequence in L dJ-t, X). Then there is w E O such that {fn(w) }�=l contains a co-subsequence . Proof: Let {fn }�=l be a co-sequence in L1 ( J-t, X). There are 0 < 8 < M such that for any finite sequence {ad k= 1 ,

251 E(X) Note that the set U� l supp in is a-finite. We may assume that (n, J.t) is a probability space. By Lemma 5.1.4, {in}�=l is uniformly integrable. So there is an r > 0 such that for any measurable set D with J.t(D) < r, we have in I l in(w) IIx dJ.t < � for all n E N. 5. 1 .

COPIES OF Co IN

Let

(5.1) We claim that (3 def J.t ( B ) is positive. Suppose the claim is not true. Then there exists an no such that the set

Bl = { w : I fn (W) IIx � � for some n � no } has measure less than r. Hence if n > no, then a contradiction. We have proved our claim. Let {rn }�=l be the system of Rademacher functions on [0 , 1] . For each n E N , define �n : n � lR by

By Lemma 5.1.2, such that

{�n }�=l is an increasing sequence of measurable functions

Let (5.2) Then {A n }�=l is a decreasing sequence and J.t ( An) � 1 � for all n E N. This implies that B n n�=l An =1= 0. Fix an w ' E B n n�=l An ' By (5.1 ) , (5.2) , and Lemma 5.1.2, for any subset C of {I, . . . , n } , -

and there is an increasing sequence {i k }k:: l such that for any k E N ,

252

CHAPTER 5 . STABILITY PROPERTIES II

and

ik ( 5.4 ) i= l 2(1 + ;- ' ) M . = (1 + 4 - ' ) l' l i t, T, (t) f; ( W') l dt < x We claim that { fi k (w ' ) }k:: l is wuc. If the claim is true, then by Theorem 1 . 3 .14, X contains a Co-sequence. Let io = 0, and for any k E N and €j = ± 1, 1 � j � k , let � (1 + 4 - k ) I L ri(·) fi(W ' ) 11 L 1 ( m , X )

Then

k I nL=l €n fin (W' ) llx < p [€ l, . . . , € k], ik l i t; rn ( - )fi(w ' ) I Ll(m , X ) J p [r l (t), . . . , rk (t)] dt. By ( 5 .3), (5.4), and ( 5 .6), we have p [€ l, . . . , €k ] � P [€ b . . . , €k , €k+ 1] -

and

(5.5 ) ( 5 .6)

J p [rl (t), . . . , rdt), rk+ l (t)] - (1 + 4 - k ) p [rl (t), . . . , rk (t)] dt � 0.

Fix €l, . . . , €k+ l = ± 1, and let C=

Then

{ t E [0 , 1] : rj (t) = €j for all j � k + 1}.

2 - k - 1 (p [€ l, . . . , €k , €k+ l] - (1 + 4 - k ) p [€ 1 (t), . . . , €k (t)] ) = l ( p [rl (t), . . . , rk (t), rk + l(t)] - (1 + 4 - k ) p [rl(t), . . . , rk (t) ] ) dt � J{ l ( 1 + 4 - k )p [r t{t), . . . , rk (t)] - p [rl ( t ), . . . , rk (t), rk + l (t)] dt [O, )\C � (1 + 4 - k - 1 ) J p [rl(t), . . . , rk (t)] dt < - 42Mk(3 · -

253

5. 1 . COPIES OF Co IN E(X)

Ei ± 1, 1 � i � k + 1, p [E !, . . . , Ek, Ek+ 1 l � (1 + 4 - k ) p [E !, . . . , Ekl + 2 - k +3 M,B- 1 . Iterating the inequality (5.7), we have obtained n . �M kfik ( ) l l T j =l k=l for all n E No The proof is complete. This implies for any

w

=

'

(5.7)

00

x

o

The main theorem of this section is the following: Theorem 5.1.6. Let X be a Banach space, and let E be a Kothe function space over a finite measure space (0, J-L). Then the Kothe-Bochner space E(X) contains a copy of Co if and only if either E or X contains a copy of co . Before proving the above theorem, we need the following two lemmas: Lemma 5.1.7. Let E be an order continuous Kothe function space over a finite measure space (0, J-L). Suppose that I 1 n llE � 2, and I g l 1 1 � 2 1 g l i E for all g E . Then for any bounded sequence {gn }�=l in E that converges to 0

E

a.e., there are a subsequence {gn k }� l of {gn}�=l and a sequence {Ak }� l of disjoint measurable sets such that gn k . In \ Ak l i E � 2. L: I l k=l 00

Proof: Let E be an order continuous Kothe function space over a finite measure (0, J-L ) , and let {gn}�=l be a bounded sequence in E that converges to 0 almost everywhere. By induction, there is a sequence {nl 1 < n 2 < n3 < . . } such that if for any 1 < j N we set

E ,

=

.

Bj { E O : I gnj I < 2j�3 } ' then we have I l gni . In\ Bj l i E � 2H1 2 for all i < j. Set B1 0, and for each j E N, set Aj 0 \ (Bj UU:j + l ( O\Bi) ) . Then for any j < k , (0 \ Aj ) U (0 \ A k ) � (0 \ Bk ) U B k 0 =

w

=

=

and

=

' gn kL:=l ll k In\ Ak l i E � kL:=l (1 l gn k IBk l i E + j =L:k+ 1 I l gnk In\Bj l i E) 1 1 � L: ( 2 k+2 ' 1 l ln llE + L: 2j + 2 ) k=l j= k+ 1 1 1 � L: ( 2 k + 1 + 2 k+ 1 ) 1 k=l 00

.

00

00

.

00

00

=

254

CHAPTER 5. STABILITY PROPERTIES II D

The proof is complete

Lemma 5.1.S. [11, Theorem 2.1.4] Let X be a Banach space and E an order continuous Kothe function space over [0 , 1] such that II g l1 1 � Il g ilE for all g E E . Let I E([0 , 1], X) L1([0 , 1],X) be the canonical embedding and Y a subspace of E. Either the restriction of I to Y is an isomorphic embedding or there are a sequence {An }�=l of disjoint measurable sets and a bounded sequence { fn }�=l in S (Y) such that L Il fn . 1 [ l \ J E � 2 . n =l :

�

00

O J A ,

i to Y is not an isomorphic embedding. Theil there is a sequence { fn }�=l in Y such that Il fn ll = 1 for all E N , and nlim -+oo I Un ) = 0. Proof: Suppose that the restriction I l y of

n

By passing to a subsequence, we may assume that Il fn O llx converges to ° a.e. The lemma follows from Lemma 5.1.7 ( by passing to a further subsequence of D { fn}�=l ) ' By Proposition 1.2.14 and Lemma 5.1.8, we have the following corollary.

5.1.9. ( Cembranos and Mendoza [11, Corollary 2.1 .1]) Let E be an order continuous Kothe function space over [0 , 1] such that II g l1 1 � Il g ilE for all g E E. (1) Suppose that E has no copy of Co. Then every Co-sequence in E([O , 1], X) is also a Co-sequence in L1 ([0 , 1], X). (2) Suppose that E has no copy of £1. Then every £l-sequence in E([O , 1], X) is also an £l-sequence in L I( [O , 1], X). (3) For any 1 < r,p < =1= p, every £r -sequence in Lp(J-L, X) is also an £r -sequence in L dJ-L, X). Proof of Theorem 5.1.6 : Clearly, if either E or X contains a copy of Co , then E(X) contains a copy of Co. One direction is trivial. Suppose that E ( X) contains a copy of Co , but E contains no copy of Co . Then E is order continuous ( Proposition 3.1.4), and we may assume that E is separable. By Theorem 3.1.8, E can be considered a Kothe function space over a probability space (o, � , J-L) Corollary

00 ,

r

such that • E is dense in L I( O , � , J-L) and L oo ( O , � , J-L) is dense in E.

Suppose that E has no Co-sequence, and {fn}�=l is a Co-sequence in E ( X). By Corollary 5 .1.9, { fn }�=l is also a Co-sequence of the Lebesgue-Bochner space

L1 (0 , J-L, X).

255

5. 1 . COPIES OF Co IN E ( X )

Note that for any measure space (n, E, J.l) , there is an atomless measure space (nb Eb J.ld such that L1 (n, E, J.l) is isometrically isomorphic to a sublattice of L1 (n1 , E1 , J.l1) ' So we may assume that (n, E, J.l) is atomless. By Theorem

5.1.5,

D X has a eo-sequence. Note: In all proofs in this section, we do not use the fact that fn is strongly measurable for all n E N. We have the following theorem:

Let E be an order continuous Kothe function space, and X a Banach space. Then the Kothe-Bochner function space ( E ( X ) ) * contains a copy of eo if and only if either E* or X* contains a copy of eo. Hence the Kothe-Bochner function space E (X ) contains a complemented copy of f1 if and only if either X or E contains a complemented copy of fl . Theorem 5.1.10.

Proof: One direction is clear. Suppose that ( E ( X )) * contains a eo-sequence,

but E* contains no eo-sequence. Then E* is an order Kothe function space. By the proof of Theorem 5.1.6, X has a eo-sequence. The proof is complete. D Exercises

Exercise 5.1.1. ( Batt and Hiermeyer ) Let (n, J.l) be a probability space and X a Banach space. (a) Let {xn}�= l be a weakly Cauchy sequence in X, and F any element in L oo (J.l, X* , w* ) . Show that { ( F ( . ), xn ) }�=1 converges a.e. (b) Let { gn }�= l be a weakly null sequence of B (L1(J.l)) , and {Xn }�=l a

bounded sequence of X . Show that the sequence {xn }�=l is weakly precom pact if and only if {xn gn }�=l is a weakly null sequence in L1 (J.l, X) . (c) Let { en }�= l be the unit vector basis of fl . Show that there is a weakly null sequence in S(Lp(J.l)) such that {gnen }�=l is weakly null in Lp(J.l, X) . Exercise 5.1.2. For any

i,j E N, let

where the {ri } � l is the system of the Radamacher functions on {nd� l be a sequence of natural numbers, and for each k E N , let nk

fk = L rk, =l m

m

e

m

[0 , 1].

Let

E L2 ( eo ).

is a weakly null sequence in L 2 ( eo ) . Exercise 5.1.3. Let E be a Kothe function space over a probability space (n, J.l) , and X a Banach space.

Show that

{ !k }� 1

256

CHAPTER 5. STABILITY PROPERTIES II

( a) Let { Gn } �= l be a weak*-null sequence in the Kothe dual of E. Show that • for any f E E ( X ) , limn oo In Gn . f dJ-L 0; • for any bounded sequence { X � } �= l in X* , { x � . Gn } � l is a weak* null sequence in ( E ( X )) * . (b) Let E be a Kothe function space over [0, 1] such that for any g E E, I g l 1 1 � I l g i I E . Show that the system of the Radamecher functions on [0 , 1] is a weak* null sequence in the Kothe dual E' . Show that if X contains a copy of Co , then E ( X ) contains a complemented copy of co . (c) For any 1 � p < 00 , Hp( X ) denotes the set =

-+

=

Hp(X) { f E Lp([O , 1], X ) : J f(t) . ei2mrt dt ° } . Show that Hp(X) contains a complemented copy of if X contains a copy of Co . (d) Let E be a Kothe function space over [0, 1] such that for any g E E, I g i r 1 � I l g i I E . Show that Co is a quotient space of E ( X ) if X* contains an =

=

Co

ll-sequence. Exercise 5.1.4. Prove the following theorem: Theorem. ( Diaz ) Let X be a Banach space, and (0, J-L ) a measure space. Then X contains a copy of Co if L oo ( X ) contains a complemented copy of Co. Let (0, J-L) be a measure space and X a Banach space. Suppose that X contains no Co-sequence, but L oo ( X ) has a complemented Co-sequence. ( a) Show that for almost all w E 0, the sequence {fn(w)}�=l is wuc. ( Thus we may assume that for all w E 0, {fn } � l is wuc. ) (b) Show that for any increasing sequence {nk}�"I and any w E 0, the series L: �=1 fnk (w) converges. (c) For any subset M of N, define =

f( M)(w) nlim -+oo k� n and ke M Show that for all M � N, f(M) is measurable and {f(M ) : M � N} is a bounded subset of L oo ( X ) . (d) For any (ak) E B(l oo ) , define =

Show that T is well-defined and that it can be extended as a linear operator from loo to L oo ( X ) . We still denote this operator by T.

257

5.2. THE DtAZ-KALTON THEOREM

(e) Let S denote the projection from L oo (X) onto [ in : n E N] . Show that S o T is not weakly compact from foo to (We get a contradiction.) Co .

Exercise 5.1.5. Let E be a Banach space with a I-unconditional basis.

Show that for any sequence {Xn}�= l of Banach spaces, (€BXn )E contains a complemented copy of Co if either E contains a complemented copy of Co , or Xn contains a complemented copy of Co for some n E N. Exercise 5.1.6. Let E be an order continuous Kothe function space over [0, 1] such that I lg ll l � I l g I I E for any g E E. Suppose that E(X) is Grothendieck. By Exercise 3.1.2 and 3.1.3, E is reflexive. Show that X is reflexive if E(X) is Grothendieck. (Note: Since X is Grothendieck, X* is weakly sequentially complete.)

5.2

The Diaz-Kalton Theorem

In Section 5 . 1 , we proved that for any Kothe function space E and any Ba nach space X, E(X) contains a complemented fl-sequence if and only if either E or X has a complemented fl-sequence (Theorem 5.1.10) . In this section, we prove the Diaz-Kalton Theorem, which shows that f oo (X) contains a comple mented fl-sequence if and only if X contains f?'s uniformly complemented. Lemma 5.2.1. [U, Lemma 5.2.2] Let f be any nonempty set. For any E > 0, there exists a continuous linear operator J : foo (f, X**) ( foo (f, X) r * ---+

such that II J II � 1 + and Jl loo (r,x) = idloo (r,x) , Proof: Let G be any finite-dimensional subspace of foo (f, X** ) . Then there E

exists a sequence (F, ),y er of finite-dimensional subspaces of X** such that

G ( 2: €BF, ) c

, er

00

c

foo (f, X **).

By the local reflexivity principle (see [19, p.178] ) , for each 'Y E f, there exists an injective operator u, : F, ---+ X such that U, l xnF"Y = idxnF"Y and l I u, 1 I � 1 + E .

Define a nonlinear operator Ic : foo (f, X**)

---+

foo (f, X) by

if (x, ) E G , otherwise.

Let U be the directed set of all finite-dimensional subspaces of foo (f, X**) and define Ic (x) E f oo (X) ** . J ( x) = w* - clim -.u

258


l(xI ) E (ioo (f, X)) ** ,

l l t oo (r , X)

= idtoo (r , x ) ,

l(Xl + X2 ) = l(Xl ) + 1(x2 ) ,

� 1 + E.

11 1 11

D The proof is complete. Before proving the main theorem in this section, we need the following the orems. For proofs, see [30, Theorem II.5. 1] and [19, Chapter 14] . Theorem 5.2.2. For 1 � p � 00, let q = p/(p 1). Then for any Ban ach

space X, the following are equivalent: (1) X contains i� 's uniformly complemented. (2) X* contains i� 's uniformly complemented. (3) X** contains i� 's uniformly complemented. Theorem 5.2.3. (Maurey-Pisier) A Banach space X is of cotype q for some q < if and only if X does not contain t� 's uniformly. Hence, X does not have a finite cotype if and only if X contains t� 's uniformly complemented. -

00

Let f be a nonempty set. Let ioo (f, X) and eo (f, X) be the sets { (X-Y)-Y Er : x-y E X and sup Il x-y 11 < oo}, -y Er eo(f, X) = ( (x -y ) E ioo (f, X) : for any positive real number E, the set {, : Il x -y II > E} is finite} .

ioo (f, X) =

The following theorem is the main result in this section. Theorem 5.2.4. [1 1, Theorem 5.2.3] Let (0, E, /-L) be any measure space,

X

a Banach space, and f the ordinal of E . Then the following are equivalent: (1) The space Loo (/-L, X) contains a complemented subspace isomorphic to LI lo , 1] . (2) There is a nonempty set f, ioo (f, X) contains t'1 's uniformly comple mented. (3) The space eo (X) contains i1 :s uniformly complemented. (4) The space X is an Sl-space; i.e., X contains i1 's uniformly comple mented. Proof: (4) =} (1). Suppose that X contains i1 's uniformly complemented.

Then (2::= 2 EBii) 00 is isomorphic to a complemented subspace of ioo (X ) . It is known that ioo (X) is isomorphic to a complemented subspace of Loo(/-L, X). It is enough to show that i1 is isomorphic to a complemented subspace of (2::= 2 EBii )oo . Let { e : n E N} be the unit vector basis of ib and ii the subspace spanned by { el , . . . , e m }. For each n, let Pm be the natural projection n

259

5.2. THE DtAZ-KALTON THEOREM

from £ 1 to £i. Then for any n < m E N, Pn 0 Pm Pn Pn 0 Pm . Let U be a free ultrafilter on N . For each n E N , define an operator 8n from (2: :'= 2 E9£i) 00 to £1 by =

=

Since £1 is finite-dimensional for any n E N , the limit limm-+u Pn(xm ) exists for all (x m ) E (2::'= 2 E9£i) 00 ' It is easy to see that 1 8n l 1 for all n E N, and Pm 0 8n 8m for all m < n E N. Fix an element (xm ) in (2::'=2 E9£i) 00 ' The above argument implies that there is a sequence {a k } k: 1 of real numbers such that n =

=

8n((xm ) ) L a k e k , k=l =

n sup I l x ml l x I (x m ) l oo � 1 8n(x m ) l el' L l a kl · m k=l We have proved that complete, n nlim -+oo 8n ((xm ) ) nlim -+oo k=l ak ek k=l ak e k E £1 ' Let Y be the set m { (I:=l a k ek ) :=2 : L=l a k e k E £1 }, k k and let 8 be the mapping from (2::'=2 E9£i) to Y defined by and

=

=

=

'""" L-t

=

00

'""" L-t

00

00

Then • Y is isometrically isomorphic to £1; • 8 is well-defined, i.e. , 8((x m ) ) E Y for all (x m ) E (2::'= 2 E9£i) 00 ; 1, and 8(y) y for any y E Y. • 1 81 The verification of the above assertions is left to the reader. We have proved the implication ( 4 ) =} (1). (1) =} (2). By (1), Loo (/-L, X) is an Sl-space. So for any n E N, there is a A-complemented n-dimensional subspace Wn of L oo (/-L, X) that is A-isomorphic to £1' Fix n E N. By a small perturbation of Xn , we may assume that • Wn has a normalized basis { !n ,i : i � n } ; • there are € > 0 (depending on Wn) , disjoint measurable sets {An ,i, "Y : I E f n , i } ( I fn,i l � I f ! ), and a subset {Xn , i ,"Y : I E rn,i } of the unit ball B(X) of X such that =

=

!n,i(W) = Xn ,i,"Y

260


for all i ::; n and all w E An, i , "Y ' and for all i ::; n and /1 Let An be the set

=1 /2 .

I l xn , l· , "Yl - xn , l. ,"Y2 I - € >

An {An , l m n . . . n An , n ,"Yn : /i E fn ,i } {An ,"Y : / E fn } =

=

Then • for any A, B E An, either A n B 0 or A B; • the cardinality of An (Le. , the cardinality of f n) is less than or equal to the cardinality of E. Let k be a natural number such that I6€ > 2 - k , and set =

=

n jk f L=l 2 - · fn ,j . j Note: For any w E 0 and i ::; n, I l fn ,i (W) l x ::; 1 . Then for any w, w' E 0, either f(w) f(w') or I l f(w) - f(w' ) 1 � 2 - nk - 1 . This implies that for any subset A of f n, the set U>' E AAn , >. belongs to E. Let En be the power set of f n and define a measure Vn on (fn , ) by Vn (A) J1. (U>' E AAn , >. ) . Let On , E� be the sets =

=

()"

E�

=

{ >'E AUA A>. : A is a subset of fn }. n, A

Then the space Loo (On , E � , J1., X) is isomorphic to Loo (fn , En , Vn , X). So there is a A-complemented n-dimensional subspace W� of Loo (f n, E n , Vn , X) that is A-isomorphic to l']: . Let Yn be the set

{f E foo(fn, X) : vn (supp f) O}. It is known that Loo (fn, E n , Vn , X) is just the quotient space foo(f n, X)/Y. This implies there is a quotient mapping from foo (fn , X) to an n-dimensional space that is A-isomorphic to £";' By the proof of Theorem 1.3.8, for any € 0, foo (f n, X) contains a A €-complemented subspace that is A €-isomorphic to =

fl .

+

+

>

It is known that for any two sets f and A, if the cardinality of f is less than or equal to the cardinality of A, then foo (A, X) has a I-complemented subspace that is isometrically isomorphic to foo(f, X). Let f be the cardinality of E. Since

26 1

5.3. TALAGRAND'S L1 (X)-THEOREM

for any n E N, the cardinality of r n is less than or equal to r, we have proved that for any n E N and any E > 0, f oo (r, X) contains a A + E-complemented subspace that is A + E-isomorphic to f l . (2) ::} (3) . Suppose that foo (r, X) contains f1 ' s uniformly complemented. There are A > 0, a sequence {Fn} of finite-dimensional subspaces of f oo (r, X), and a sequence of projections Pn : f oo (r, X) Fn such that -

for all n E N. Let J : foo (r, X**) (f oo (r, X)) ** be the operator that we obtained in Lemma 5.2.1 . Then Pn oJ is a projection from ( foo (r, X)) ** onto Fn. Therefore, foo (r, XU) Co (r, X)** contains f1's uniformly complemented. By Theorem 5.2.2, Co (r, X) contains f1 's uniformly complemented, too. For each n E N, select a A-complemented subspace Wn of Co(r, X) that is A-isomorphic to [I ' and let { fn,k : k � n} be a basis of Wn . Let ro be the set -

=

ro =

U

k $ n EN

supp fn,k '

Then Co (ro, X) Co(X) contains f1 's uniformly complemented. We have proved the implication (2) ::} (3) . ( 3 ) ::} (4). Suppose that Co (X) contains f1 ' s uniformly complemented. By Theorem 5.2.2, ( Co ( X)) * f1 (X*) contains [� 's uniformly complemented. By Theorem 5.2.3 and Theorem 3.4.15, neither f1 (X*) nor X* has finite cotype. By Theorem 5.2.3 and Theorem 5.2.2 again, X contains f1 's uniformly comple 0 mented. The proof is complete. =

=

Remark 5.2.5. Dlaz

[14] proved Theorem 5.2.4 when J.L is a a-finite measure

and X is a Banach lattice. He also proved all directions of Theorem 5.2.4 except ( 2 ) ::} ( 3 ) . P. Cembranos and J. Mendoza used Kalton 's result (Lemma 5.2.1 ) to prove Theorem 5.2.4 when (n, J.L ) is a-finite.

5.3

Talagrand 's L 1 ( X ) -Theorem

Let X be a Banach space, and (n, �, J.L ) a probability. For convenience, we denote the sequence {hn }�= l of L1 (/1-, X) by (hn). A sequence (9n) is said to be essentially normalized f1-block of (hn) , denoted by (9n) « (hn) , if there exist k E N and sequences {Pn E N : n � k } , {qn E N : n � k } , and {ak E jR + : k � Pk} such that k � Pk � qk < Pk + 1 � qk + 1 < . . . , aqn =I 0, �i;;;'Pn a i 1, and 9n � i;;;' Pn aihi. Rosenthal's f1-theorem asserts that for any bounded sequence {xn }�= l in X, {xn }�=l contains either a weakly Cauchy subsequence or an f1-subsequence. Let { fn}�= l be a uniformly bounded sequence in L1 (/1-, X) . It is natural to ask whether there is an increasing subsequence {ndk:: 1 of the positive integers =

=

262


such that for almost all w E 0, { fnk (W)}� l is weakly Cauchy or { fnk (W)}� l is an .e1-sequence. Unfortunately, the Rademacher functions provides a negative solution to the above question. Indeed, Talagrand showed that there exists a Banach space X and a weakly null sequence { fn }�= l in L1 ( [0 , 1 ] , X) such that for any W E [0 , 1] ' any subsequence { fnk (W)}� l of {fn(W) }�= l contains a further i1-subsequence (see Remark 5.5.5 and [47] ). In this section, we present a result of Talagrand [46] that shows that if { fn }�= l is a uniformly bounded sequence in L1 (/1, X), then there exists a sequence {gn}�=l ' h « g, such that for almost all W E 0, either {gn(W)}�= l is weakly Cauchy or for some k > 0, {gn(w)}�= k is an i1-sequence. Theorem 5.3.1. (Talagrand's L1 (X)-Theorem) Let Un) be a sequence of measumble functions 0 B(X) . Then there exists a sequence (gn) « Un) and two disjoint measumble sets C and L of 0, /1( C U L) 1 such that (1) for all W E C , the sequence {gn(w)}�=l is weakly Cauchy; (2) for all W E L, there is a k E N such that the sequence {gn(w)}�= k is an i1 -sequence. -

=

Proof: Step 1 . First, we need the following elementary lemma. The proof

is left to the reader. Lemma 5.3.2. (1) If (h n) « (gn ) and (gn) « Un) , then (h n) « Un). (2) Suppose that for each k E N, U� + 1 ) « U�) . Then there exists an increasing sequence {kn }�= l such that (gn) clef U�n ) « U�) for all k E N. In the sequel, letters f, g, h, with or without superscript, will denote se quences of measurable functions from 0 into X. Since the fi are strongly mea surable, by Pettis's measurability theorem, the Ii are essentially valued in a separable subspace of X. Without loss of generality, we assume that X is sep arable. Thus the unit ball B (X *) of X* with the weak* topology is compact and metrizable. (In this section, for any subset C of B(X* ) , C denotes the w*-closure of C in B(X*) .) Let d denote a distance on B(X*) that was induced by the weak* topology on B(X*), and {On }�= l a countable basis for this topol ogy, with 01 B(X*) and the set of all weak* compact subsets of B(X*). A set-valued map W � K(w) from 0 to is said to be measumble if for any n E N, the set {W E O : K(w) n On :f: 0} is measurable. This is equivalent to saying that for any w*-Borel set A of B(X*), the set {w E O : K(w) n A :f: 0} is measurable. Step 2. Let Kex : w � Kex (w) be a measurable set-valued map from 0 to Then for any k E N, the set =

K

K

K.

5.3. TALAGRAND 'S

Ll(X)-THEOREM

is measurable. For a measurable function [ - 1, 1] be the function defined by

263

9n : n B(X), let 9n , k ,a : n -+

-+

if W E Ck,a , otherwise.

From the following fact, it is easy to see that for any n, k E N , 9n , k ,a is measur able.

Let {y; q E N} be a countable weak* dense subset of B(X* ) . For any W E Ck,a , 9n , k ,a (W) � a if and only if for any p E N, there is qp such that :

Fact 5.3.3.

qp such that l l (y;p , 9n(W)) 2: a - p- and max{d(y;p ' Ok ), d(y;p ' Ka (w )) } < p- . Let y* be a weak* limit point of {y;p }� l ' Then y* E O k n K(w) and Proof: (=}) is obvious. Suppose that for any p E N, there is

o

We have proved the fact.

The orginal 9n ,k,a is defined by 9n,k,a (W) = sup { (y * , 9n {W)) : y * E Ok n Ka (W) } . (So we assume that 9n ,k, a (W) = - 1 if Ok nKa (w) = 0.) We do not know whether these 9n , k , a (which are defined above) are measurable. For any sequence (9n) of measurable functions, let (h , a (9) be the function defined by Ok, a (g)(W ) limsuP9n n-+oo , k ,a (W ). Remark 5.3.4.

=

We have the following lemma:

Let (gn) be a sequence of measurable functions from n to B(X). If (hn) (9n) , then Ok, a ( h) � Ok,a (g). Proof: Since (hn) (gn), there exist a sequence { Ai } � l of nonnegative real numbers, a strictly increasing sequence {mj } � l ' and an N such that Lemma 5.3.5. «

«

mj +l

L

i = mj + l

Ai = l

and

mj+l

hj = =L Ai9i i l mj+

for all j 2: N.

2 64


So for any W E {w : K(w) n O k =1= 0},

\ y;, i=mjL+1 Ai9i (W) ) mj+l sup (y;, Aigi (W)) :::; lim L P-+OO i=mj +1 max{d(y� ,KQ (w» ,d(y� ,Ok ) } < ; mj+l = L lim sup (y;, Aigi (W)) � < } ,ok i=mj + 1 P-+OO max{d(y� ,KQ (w» ,d(y� )

hj,k, ex (W) = lim

mj+l

sup

P-+OO max{ d(y� ,KQ (w» ,d(y� ,o k ) } < ;

mj+l

=

Ai 9i,k, ex (W) :::; s.up 9i,k, ex (W) , L mj < ��mi +l i= +1 mj

o

We have proved that Ok,ex (h) (w) :::; Ok,ex (g) (w) for all w E n.

Let Kex : W Kex (w) be a measurable set-valued map from n to and (un) a sequence of measurable functions from n to B(X). For any k E N, there exists (gn) « (un) such that for any h = (hn) « (gn), h Ok , ex (h) = Ok ,ex (g) and nlim -+oo I Ok , ex (g) - n,k,ex 11 1 = O. �

Lemma 5.3.6. IC ,

Proof: Let g l =

U.

such that for each p > 1 ,

By induction, we can construct sequences gP = (9h)

J Ok,ex (gP ) dJ-t :::; 2 -P + inf { J Ok,ex (V) dJ-t

: v

}

« g P- 1 .

By Lemmal 5.3 .2, there is 9 = (gn) such that 9 « gP for all p E N . (In particular, g « u = g .) Claim 1 . Let h be a sequence such that h « g. We claim that Ok,ex (h) = Ok,ex (g) , a.e. In Lemma 5.3.5, we have proved that for any k E N, Ok,ex (h) :::; Ok , ex (g) a.e. Thus for any p E N, h « 9 « gP and inf

{ J Ok, ex (V) dJ-t

:

V « gP

}

:::; J Ok,ex (h) dJ-t :::; J Ok,ex (g) dJ-l :::; J Ok , ex (gP+ 1 ) dJ-l :::; inf { J Ok,ex (V) dJ-l « gP } + 2 - p - 1 . : v

But P is an arbitrary positive integer. We have J Ok,ex (h) dJ-l = J Ok , ex (g)dJ-l. Combining the fact that Ok,ex (h) :::; Ok,ex (g), we have proved Claim 1 . Claim 2 . Let h be a sequence such that h « g. We claim that { h i,k, ex}� l converges wea�* to Ok,ex (g) in L oo ( n J-t) i.e. Ok , ex (g) is the uniq�e weak* clus ter point of (hi,k,ex) ' Let v be any weak* cluster point of (h i,k,ex) ' Since lim su Pi -+oo hi , k , ex :::; Ok,ex (g) ,

, ,

2 65

5.3. TALAGRAND'S L1 (X)-THEOREM

We notice that the set {f E L1 (0., 1-£) : I f(w) 1 � 1 for every w E n} is weakly compact. The measure is also a weak cluster point of {h i, k ,a }� l in L 1( 1-£ ) . By Mazur's theorem, there are an increasing sequence {ki }� l of N and a sequence { a i } � 1 of nonnegative real numbers such that for all n E N,

1I

kn+l kn+l L a i 1 and 1 1I L a i iii, k ,a II � 2 - n . i= kn+ 1 i= kn+ 1 kn+l kn+l lin = =L 1 ai h i , k, a , and h'n =L 1 a i hi. i kn+ i kn+ =

Let

-

=

We have

This implies that

lI (w) nlim -+oo lIn(w) a.e. By the proof of Lemma 5.3.5 and Claim 1, Ok, a (g) Ok,a (h' ) limn-+oosup h� , k ,a � nlim -+oo lin = 1I � Ok,a (g). We have proved Claim 2 . The definition of Ok,a (g) implies that for almost all w E n, nlim -+oo (hn " k a (W) - Ok , a (g)(W)) V 0 O. =

=

=

=

By the bounded convergence theorem, we have Claim 2 implies that

nlim -+ oo I ( hn ,k,a - Ok,a(g)) V 0 1 1 O. =

/ (Ok,a (g) - iin,k,a ) V O dl-£ � J�� [/ Ok,a (g) - hn , k ,a dl-£ + / (hn , k ,a - Ok, a (g)) V 0 dl-£] 0

o � nl��

=

and

The proof is complete. For a sequence (gn) of measurable functions from 0. to functions 9n,k, a and cPk,a from 0. to [ - 1, 1] by

9n ,k,a (W) inf { (y * , gn ,k,a (W)) cPk,a (g)(w) limn-+ooinf 9n " k a (W), =

=

:

y*

o

[ - 1, 1] ' define the

E O k n Ka (w) } ,

266


where inf 0 = 1 . So if (h n) = ( - gn) , then for all n E N,

9n , k , a (W) = - ( hn , k ,a )(W),

¢k,a(g)(w) = - ( Ok,a (h))(w) . Applying Lemma 5.3.6 to the sequence ( - gn) , we have the following lemma. Lemma 5.3.7. Let (un) be a sequence of measurable functions from n to B(X), Ka n K a set-valued measurable functions and k E N . Then there exists = (gn) (Un) such that for any h = (hn) (gn), we have ¢k, a (g) = ¢k,a (h) and 9

:

and

-

«

«

Step 3. Construction. Let us fix - 1 < a < b < 1, and let T denote the first uncountable ordinal. Let h� = in and Ko(w) = B(X * ) . For all a < T, we shall construct a sequence h a = (h�) and a measurable set-valued map Ka : n that satisfy the following conditions. -

K

f3 < a < h a h{3 . (b) a = f3 1 < for some f3 � O. In this case, for any h h a , we have Ok, {3 (ha ) = Ok, {3 ( h ) , nlim oo I Ok ' (3 (ha ) - hn k {311 1 = 0, --fo ¢k, {3 (ha ) = ¢k,{3 (h), nlim -+oo I ¢k ' (3 (ha ) - hn, k , (31 1 1 = O. Set Ka (w) = { t E K{3 (w) Vk, t ¢ Ok or (Ok, {3 (ha )(w) > b, and ¢k, {3 (h a )(w) < a) } . ( ) If a is a limit ordinal, then we set Ka (w) = n {3 < a K{3 (w). Assume that the construction has been done for all f3 < a . Suppose that a T,

(a) For all

+

«

T

«

1

,

:

c

is a limit ordinal. Let

Since a is countable, there exists an increasing sequence {f3n}�=l of ordinals with a = limn --> oo f3n. By Lemma 5.3.2, there is h a such that h a « h {3n for each n E N. So if f3 < a, then for some n, f3 < f3n and h a « h {3n « h {3 . This shows that (a) is satisfied. We have completed the construction when a is a limit ordinal. Suppose a is not a limit ordinal. Then there is f3 � 0 such that a = f3 + 1 . Applying Lemmas 5.3.6 and 5.3.7 repeatedly, we can construct a sequence g k with g l = h{3 and g k+ 1 « g k such that for each h « g k , we have

Ok, {3 (l ) = Ok, {3 (h), J!"'� I Ok, {3 (gk ) - hn- ,k,{3 11 1 = 0, ¢k, {3 (gk ) = ¢k, {3 (h), nl!...� II ¢k, {3 (gk ) - hn- ,k, {311 1 = O.

2 67

5.3. TALAGRAND'S L 1 (X)- THEOREM

By Lemma 5.3.2, there is a sequence h O with h O « g k for all k � O. Thus k fh,l3 (g )

=

Ok,l3 (h O ) a.e. and cPk ,l3 (l )

=

cPk ,l3 (h O ) a.e.

We have proved that h O satisfies the first part of ( b) . Define Ko (w) by Ko(w) { y * E KI3(w) : Vk, y * � O k or ( Ok ,l3 (h O ) (w) > b, cPk ,l3 (h O ) (w) < a) } KI3(w) \ Ok ) . U 9k,P (hQ)(w )� b or ¢k,p (hQ)( w )�a =

(

=

Then for each w E 0, Ko(w) is closed. It remains to show that the mapping w � Ko (w) is measurable, i.e., for any (w* ) compact set L, {w : Ko (w) nL 0} is a measurable set. Let [N] <w denote the collection of all finite subsets of N, and w an element of { w : Ko(w) n L 0} . We have =

=

U But the set Ko(w) n L is compact. There is a finite subset F of or cPk ,l3 (h O ) (w) � a } such that Ko(w) n L � U kE F Ok . This implies that {k : Ok ,l3 (h O ) (w) � b {w :

where

ZF

=

Ko (w) n L 0} =

=

ZF, U F E [Nj<w

{ w : KI3(w) n L �, kEUF Ok } n n (( Ok ,l3 (h O ) � b) U (cPk ,l3 (hO ) � a)) . ke F

But [N] < w is a countable set, and for any F E [N] < w , ZF is a measurable set. This implies that the set { w : Ko (w) n L 0} is measurable. The construction is complete. =

Remark 5.3.8. We do not know of any proof to show that Ko is measurable without using the fact B(X*) is (weak* ) compact.

Step 4. When to stop the construction. We claim that there is 0 < T such that Ko(w) KO +1 (w) a.e. Fix a k E N, and set A I: { w : Ko(w) n Ok 0} . =

=

=

Then the sequence ( J.L (AI:))o (x * , g� (w) ) ;::: an or lim sUPn-> oo (x * , g� (w) ) � b n. (k) For all 2 � f � k and w E Le \ Le - I , (g� (w)) is 8e 2/(be - ae) equivalent to the unit vector basis of fl . ( I ) g k+1 « g k . The result of steps 1 to 5 shows that there are g l « g, CI , and LI satisfying condition (i)-(k) . Assume that Ck, Lk, (g�) have been constructed and satisfy (i)-(l). We can apply the result of steps 1 to 5 with a = ak+I and b bk + I , in g� . There exist a sequence (gn ) « (g�) and measurable sets e' , l' such that J.-L( e' U 1') = 1 and • for any W E C' and y* E B(X * ), either lim inf n->oo (y* , gn (w) ) � ak+ I or lim sUPn->oo (y * , gn (w)) � bk +1 ; • for all W E L', {gn(W)}� j is an frsequence for some j > O. Set Ck+ ! = C' n Ck and Lk + I L' U L k. It is clear that conditions (k) and (1) are satisfied. We finish the construction. Finally, set e n� I Ck, L U� I Lk, and let hn g::. For any W E C, y* E B(X*), and k E N, we have inf (y * , hn(w)) ;::: a(k) or lim sup(y * , hn (w)) � b (k). either lim n-+oo n -> oo Thus for any w E C, lim(y*, hn (w) ) exists. We have proved that for any w E C, the sequence {hn(W)}� I is weakly Cauchy. Let w be an element in L . Then w E L m for some m E N . By Step 5 , there is k such that {g� (w)}�= k is an fI-sequence. It is known that any normalized fI-block (for a definition, see Exercise 5 . 3 . 3) of an fI-sequence is still an fI-sequence. Thus for any w E L, there is k E N such that the sequence {hn (w)}�= k is equivalent to the unit vector basis of fl . The proof is complete. 00

=

=

=

=

=

D

=

=


272

Lemma 5.3.9. Suppose that E is an order continuous Kothe function space and X is a Banach space. For any bounded sequence { fn }�= 1 in E(X), if there is a subset of positive measure of n such that for any w E there is k such that {fn(w)}�= k is an iI -sequence, then there exists k ' such that { fn }� k l is an i I -sequence.

L

by

L

a

Proof: For w E n and k E N, let (k , w ) and ,B (k , w ) be the numbers defined

a (k , w )

=

{ ll nt=k anfn(w) l x / n�k L l an l : finite sequence of rationals } ,

inf

{an}� 1 is a

,B (k , w) inf{ I fn(w) llx : n ;::: k}. It follows that a (k , w) is a measurable function of w. The assumption implies that limk-> oo a(k, w),B(k, w) > 0 for all w E L. Hence there exist k , a > 0, and ,B > 0 such that the set L' = { w E L : a (k, w) > a and ,B (k , w ) > ,B } =

has positive measure. Thus for any sequence {an}�= k '

ti nL= k anfn ll E(X) ;::: I i a nL= k l an l ' ll fn O ll x ' 1 L' I E ;::: a,B 1I 1 L' liE nL=k l an l · 00

00

00

o


Theorem 5.3.10. Let E be an order continuous Kothe function space over

X

a probability space, and a Banach space. For a bounded convex subset A of E(X), A is weakly precompact if and only if the following two conditions hold:

:

( 1) The set V(A) = { ll f O ll x f E A} is weakly precompact in E. ( 2) For each sequence {fn }�= 1 in A, the set

{ w E n : there is a k such that { fn }�= k is an iI -sequence} has measure O .

(2) follows from Lemma 5.3 . 9. Suppose that V(A) is not weakly precompact. Then there is G E E* such that { ( G O , II f O ll x ) : f E A } is not uniformly integrable. So there are € > 0, Proof: Suppose that A is weakly precompact. Then

a subsequence { fn } � l in A, and a sequence {Ck � supp fn } � l of mutually disjoint measurable sets such that k

k

5.3.

273

TALAGRAND'S Ll(X ) -THEOREM Let f be the function defined by Ink (W) if w E Ck n supp fnk , f(w) = �fnk (W ) l Ix if w 1:- U� l (Ck n supp fnk ) '

{

Then f E L l (X ) , By the Hahn-Banach theorem, there is Fl E L oo ( X* , w * ) such that II FI il L oo (X , w - ) = 1 and ( Fb i) = 1. Then II Fl (w) llx - :5 1 and ( Fl O , f(·) ) = I l f(w) llx a.e. Set F = Fl ' G. Then F E E* (X*,w*) and the set { ( F( ' ) , fnk ( ')) : k E N} is not uniformly integrable. Note that the function f 1---+ ( F, 1) is a continuous linear operator from E(X) to Ll . This implies that if V(A) is not weakly precompact, then A is not weakly precompact. We have proved (1). To prove the converse, we need the following two lemmas: Lemma 5.3. 1 1 . Let E be an orner continuous Kothe function space. Sup

(1)

{fn}�=l �=

(2)

pose that in and of Theorem 5. 3. 1 0 hold. Then for any sequence « A, there is a sequence (gn) (fn) such that for almost all w E n, {gn (w )} 1 is weakly Cauchy. Proof: Since E is order continuous, there is a separable sublattice El of E

that contains { ll fn O ll x : n E N} . By Theorem 3 . 1.8, we may assume that E is a Kothe function space over a probability space such that for any g E E, we have Il g ll i :5 2 11 g 11E . Let ¢n = Il fn O ll x . Note: The natural embedding from E to Ll is bounded. The sequence { ¢n : n E N} is uniformly integrable ( in L l) , and it has a weak cluster point ¢ in Ll. By Mazur 's theorem, there is « (¢�) (¢n) such that II ¢ - ¢� ll l :5 2 2n . Then {¢� } �= l converges to ¢ a.e., and sup{¢� (w) : n E N} < 00 a.e. If ¢� 2::p � n Q�¢p , then we set -

=

f� = L Q� fp , p� n

Let Zp be the set Zp

Then

=

{w E n : for all E N , ¢� (w)

:5 pl .

Il f� O llx :5 ¢� for all n E N; s up{ ll f� (w) 1I : n E N} < sup{¢� (w) : n E N} < 00 a.e.; J.l ( U� l Zp) = 1 . We shall use induction over p tol construct p sequences g P = (g�)�l of E(X) such that g� f�, gP « gP- , and for almost all w E Zp, the sequence {g� (w)}� l is weakly Cauchy. The first step is similar to the general step in the proof of Theorem 5.3.1. Assume that the construction has been done up to p. There is a sequence l p 1 + + P P g « g such that the restriction of g to Zp+ l satisfies the conclusion of l + ' Theorem 5.3.1 ( Talagrand s Ll(X) theorem ) . Since g� E A, hypothesis (2) =


274

forces { g�+ 1 (w) }� l to be weakly Cauchy for almost all w E Zp. This concludes the construction. By Lemma 5 . 3 . 2, there is a sequence h such that h « gP for 0 all p E N. Then h satisfies the conclusion of Lemma 5.3.11. Lemma 5.3.12. Let E be an order continuous Kothe function space over a probability space, and { gn }� l a bounded sequence of E(X) that is weakly precompact. (a) If for almost all w E n , {gn (w) }� l is weakly Cauchy, then { gn }� l is

weakly Cauchy. (b) If there is 9 E E(X) such that for almost all w E n , {gn (w)}�l converges weakly to g (w) , then { gn} �= l converges weakly to g . Proof: Let ¢ E ( E ( X » * . By Theorem

3.2 . 4, there is a weak* measurable

function F : w � F(w) from n to X* such that 1I II F O llx* II E* for any h E E(X), ¢, h ) F(w), h (w» ) d� (w) .

\

=

=

II ¢ IIE (X) * and

J\

By Lemma 3 . 1.10(1) and Theorem 5.3.10, { II F( · ) llx* . Il gn O l iX : n E N} is uniformly integrable. Since for almost all w E n , the sequence (F(w) , gn (w») converges, by Vitali ' s lemma, the limit

exists. Thus { gn } �= l is weakly Cauchy. Now suppose that for almost all w E n , {gn(W ) }� l converges to g (w) weakly. Note: { I IF(') ll x* ' 11 gn O IIX : n E N } is uniformly integrable. By Vitali ' s lemma again, we have

J!"'� J (F(w) , gn(w») d� (w) = J (F (w ),g (w »)d� (w ). o This implies that { gn } �= l converges to 9 weakly. Proof of Theorem 5.3.10: Suppose that ( 1 ) , (2) hold and A is not weakly precompact. By Rosenthal ' s £l-theorem, there is a sequence f (In) � A equivalent to the unit vector basis of £1 . On the other hand, by Lemma 5.3.11, there is 9 « f such that {gn(W)}�l is weakly Cauchy for almost all w E n. Then by Lemma 5.3.12, { gn}� l is weakly Cauchy. But { gn } �= l is an £10 sequence. This is impossible. The proof is complete. =

Note: the closed convex hull of any weakly precompact set is weakly pre compact. We have the following theorem: Theorem 5.3.13. Suppose that X is a Banach space that contains no copy of £1 , and E is an order continuous Kothe function space over a finite measure space (n, � ) . A bounded subset A of E(X) is weakly precompact if and only

TALAGRAND'S Ll(X)-THEOREM

5.3.

275

X

if the set { ll f O llx : f E A} is weakly precompact. Hence if neither E nor contains a copy of then the Kothe-Bochner function space E(X) has no copy of

.el l

.e l ·

X

Theorem 5.3. 14. (Talagrand) Let be a Banach space, and E an order continuous Kothe function space over a finite measure space (0, J.L ) . For any weakly Cauchy sequence {fn}�= l in E(X), fn can be written as fn = gn + hn, where {hn}�l converges weakly to zero and for almost all E 0, {gn (W)}�l is weakly Cauchy. Hence if E and are weakly sequentially complete, then the Kothe-Bochner function space E(X) is weakly sequentially complete.

w

X

Proof: Let {fn}�l be a weakly Cauchy sequence in E(X) and

sequence given by Lemma 5.3. 1 1 . Then nlim --+ oo ( F, gn)

=

9

«

f the

nlim --+ oo ( F, fn)

for all F E (E(X))* , and the sequence {gn(w)}� l is weakly Cauchy for almost all W E 0. Let hn = fn - gn . Then {hn}�l is a weakly null sequence. We have proved the first assertion. Suppose that both E and X are weakly sequentially complete. To prove the second assertion, it is enough to prove that {gn}�=l converges weakly. Let 9 be given by 9 (w) w- limn gn (w ). Then 9 is scalarly measurable and essentially valued in a separable subspace of X . By Pettis ' s measurability theorem, 9 is strongly measurable. By Lemma 5.3.12, we need to show only that 9 E E(X ) , i.e., Il g( · ) llx E E . Let h lim infn--+ oo Il gn O ll x . Then Il g( ' ) llx ::; h. Fix k E N. By Lemma 3.1 .10 and Fatou's lemma, for any positive H E B (E* ) , =

=

(H, II (g ( ' ) lI x /\ k)

- ! H ( ll g( · ) llx /\ k) dJ.L ::; ! H ll g( ' ) llx dJ.L < r H h dJ.L ::; lim inf ! H ll gn O ll x d $ lim n--+inf n --+oo oo ll gn I I E (X) < 00. J /-l

We proved that { ll g( ' ) llx /\ k } k:: l is a bounded increasing sequence that converges to 9 a.e. But E is a KB space (weakly sequentially complete Banach lattice). This implies Ilg( ' ) llx E E. By the definition of the Kothe-Bochner function 0 space E(X ) , we have proved that 9 E E(X ) . The proof is complete. Now we can give a characterization of weakly compact subsets of Ll (X) that was proved by Dlaz, Diestel, Ruess, Schachemayer, and Ulger [12, 18, 48] . Theorem 5.3.15. Let E be a KB space over [0, 1] such that for any g E E,

Il g ll l � Il g i I E . For any subset A of E(X), the following are equivalent: (1) A is relatively weakly compact. (2) The set { ll f O ll x : f E A} is relatively weakly compact in E and for any sequence {fn}� l in A, there exists a sequence (g) « (I) such that for almost all W E [0, 1] , {gn (W )}� converges in norm.

l


276

(3) As in (2), except that for almost all w E [0, 1], {gn (W)}�=l converges

weakly. (4) The set { II f (-) II x : f E A} is relatively weakly compact in E and, A is relatively weakly compact in L1 (X) , Proof: The implication (2)

::::}

(3) is obvious. (3) ::::} (1). Clearly, (3) implies that A is bounded. By the Eberlein-Smulian theorem, it is enough to show that A is weakly sequentially compact. Let {fn}�l be a sequence in A. B'y Rosenthal ' s i1-theorem and by passing to a further subsequence, we may assume that either {fn}�=l is weakly Cauchy or that it is an i1-sequence. Suppose that {fn}�l is an i1-sequence. By Lemma 5.3.12, there are (g) « (I) and a positive measurable set M such that for any w E M, there is Nw such that {gn (w) }�= N..., is equivalent to the unit vector basis of i1 . This contradicts (3). Thus we may assume that {fn}�= l is weakly Cauchy. By (3) , there is (g) « (I) such that for almost all w E [0, 1] ' {gn (w) }�l converges weakly, say it converges to h (w) weakly. By the proof of Theorem 5.3.14, {gn}�= l converges to h weakly. It is easy to see that {fn}�l also converges to h weakly. (1) ::::} (2). Suppose that A is a relatively weakly compact subset of E(X). By Lemma 5.3.10, V (A) def { ll f(-) llx : f E A} is a weakly precompact subset of E. But E is a KB space (weakly sequentially complete Banach lattice). So V(A) is relatively weakly compact. Let {fn}�=l be a sequence in A. Without loss of generality, we may as sume that {fn}� l converges weakly; say it converges to f weakly. By Mazur ' s theorem, there is (gn) « (In) such that limn oo gn = f . Hence there is a subsequence {gnk }k:: 1 of {gn}�=l that converges to f almost everywhere. (4) {=9 (3). Notice that the natural embedding from E(X) to L 1 ( X ) is bounded. The implication (4) {=9 (3) follows from the fact that (1), (2) , and 0 (3) are equivalent (when E = LI ) . From the proof of the above theorem, we have the following theorems: Theorem 5.3.16. ( U lger) Let X be a Banach space and {fn}�=l a sequence -+

in L 1 (X) such that for almost all w E S1, {fn (w)}� l is weakly Cauchy. If {fn}�=l converges weakly to f, then for almost all w, {fn (w)}�=l converges to f(w) weakly. Theorem 5.3. 17. Let E be a KB space over [0, 1] such that for any f E E, f II l1 1 li l li E . Then (a) {g E E I } is a weakly compact subset of E. (b) Let X be a Banach space and {fn}� l a sequence in B (Loo (X) ) . Then {fn}�=l converges weakly in L1 (X) if and only if it converges weakly in E(X ) . Example 5.3.18. Let { en}� l be the unit vector basis of i2, and fn a

�

: I g l i oo �

sequence in eo(i2) defined by

fn (k)

=

{ �n .

if k � n, otherwise.

2 77

5.3. TALAGRAND'S Ll (X)-THEOREM

Then {fn}� 1 is a uniformly bounded sequence such that for any k E N, en = O. w- nlim fn (k) = w- nlim -+ oo -+ oo This implies that { fn }� 1 is a weakly null sequence. On the other hand, { ll fn O ll x }� 1 is a weakly Cauchy sequence in Co that does not converge weakly. Thus, in Theorem 5.3.15, we cannot replace the KB space E by Co (note: Co is order continuous). Remark 5.3.19. Let (O, }:; , J-L) be a finite measure space and X a Banach space. C. Abbott, E. Bator, R. Bilyeu, and P. Lewis [1] showed that a bounded subset C of L1 (X) is 0'(L1 (X), L oo (X » -compact if and only if the set { ll f O ll x : f E C} is precompact in L1 (J-L) and for all A E }:;, the set {fA f(w) dJ-L : f E C} is pre compact in X. Let E be a Kothe function space over (0, J-L) , and E' the Kothe dual of E. In [34] M. Nowak proved that for any Banach space X, C c E(X) is O'(E(X), E' (X* » -compact if and only if the following two conditions hold. (a) The set { li f O Ilx : f E C} of E is O'(E, E')-compact. (b) For any measurable set A, the set {fA f(w) dJ-L : f E A} is weakly pre compact in X. Exercises

Exercise 5.3. 1. Prove Lemma 5.3.2. . Exercise 5.3.2. Let E be an order continuous Kothe function space over a finite measurable space, and X a Banach space. Show that for any subset A of E(X), if the set { ll f O ll x : f E A } is not weakly precompact, then A is not a (V*)-set. Exercise 5.3.3. Suppose that X* contains an iI-sequence, but has no weak* null il sequence. (a) Let { X � }�= 1 be an iI-sequence in X*. A block basis { z � } 00= 1 is said to be a normalized iI-block of { x� }�= 1 if there exist a sequence {PI � ql < P2 � q2 < . . . of natural number and a sequence {an}� 1 such that z� = 'E- i::'Pn a i x i and 'E- i::'Pn l a i l = 1 for all n E N. For any iI-sequence { Y�}� 1

in X* , let 0, € be the two functions defined by O(Y�) = sup lim sup l (y� , x) l , xES ( X) n -+oo

€(y�) = inf { o(z�) : { z� } is a normalized iI-block of {Y�}�=d . Show that there is a normalized iI-block {y�} of {x� } for which holds for all normalized iI-blocks { z�}� 1 of { Y�}� I

'


278

(b) Let Y = {xl [y;;. : n E NJ : x E X}. Show that Y contains an .e1-sequence (cf.

the proof of Rosenthal ' s .e1-theorem). Since Y is a quotient space of X, this implies that X contains a copy of .e 1 . Theorem. (Hagler and Johnson) Suppose that X * contains an .e1-sequence,

but has no weak* null .e1-sequence. Then X contains a copy of .e1 . Exercise 5.3.4. Suppose that X contains a copy of .e1 • Show that .e2 is a quotient space of X. Hence if X contains a copy of .e1 , then X* does not have the Schur property. (Hint: Every operator from .e1 to .e2 is absolutely 2-summing, and L oo (/1-) is an injective space.) By Exercise 5.3.3, we have the following theorem. The Josefson-Nissenzweig Theorem, For any infinite-dimensional Ba nach space X, there is a weak* null sequence in X* that is not a null sequence. Exercise 5.3.5. (S. Diaz and A Fernandez [15]) Suppose that X has no .e1sequence. Show that X has a complemented Co-sequence if X has a Co-sequence.

5.4

Property (V* )

Recall thatoo a subset A of a Banach space X is called a (V*)-se t if for any . . wuc senes L.. �m =l xn* m X* , nlim -+oo

(suP{ I (x� , x) 1 : x E A} ) = o.

X is said to have the (weak) property (V*) if every (V*)-set of X is relatively weakly compact (respectively, weakly precompact) . In [35] , Pelczynski intro duced the notion of property (V* ) . It is easy to see that if X has property (V*), then every subspace Y of X has property (V*). F. Bombal showed that if A is a (V*)-set of X and {xn }�= l is a sequence in A that is equivalent to the unit vector basis of .e 1 , then there is a subsequence {xnk } k:: 1 that spans a complemented subspace of X (see Theorem 1.7.6). Using these notions and the Heinrich-Mankiewicz theorem, N. Randrianantoanina proved that a Banach space X has property (V* ) if (and only if) every separable subspace of X has property (V*) (see Theorem 1 .7.10). In this section, we prove that for any Kothe function space E and any Banach space, if E and X have property (V*) , then E(X) has property (V*). Since the techniques of the proofs are similar to those in the proof of Talagrand ' s L1 (X)-theorem, we give only a sketch of the proof. Since in this section we use both unit vector bases of Co and .e1 , to avoid confusion, we will denote the unit vector basis of Co by {en}�= l ' while that of .e1 by { e �}�= l . Lemma 5.4. 1. (N. Randrianantoanina) A subset A of a Banach space X is a (V* ) -set if and only if for every sequence { X n}�= l in A, the sequence {xn 0 en}�= l is weakly null in X0 1r Co .

279

5.4. PROPERTY (V* )

Proof: By Theorem 1 .8.2, C(X, fI ) is isomorphic to (X®7rCo)* . Suppose that A contains a sequence { Xn }�=1 such that {xn 0 en}�=1 is not weakly null. Then there is T E C(X, fI) such that { (T(xn)' en)}� 1 does not converge to zero. This implies that T is not compact. By Corollary 1 .3.8, { xn }� 1 contains a complemented f1-subsequence. We have proved that A is not a (V*)-set. Conversely, suppose that A is not a (V*)-set. Then there is a complemented f1-sequence {Xn}� 1 in A. Let P be the projection from X onto [xn : n E N] , and S an isomorphism from [xn : n E N] onto fl such that S(xn) = e�. Then S o P E C(X, f1 ) and S o P(xn), en ) = ( e�, en) = 1. This implies that 0 { xn 0 en}� 1 is not weakly null.

\

The following theorem is due to N. Randrianantoanina. Theorem 5.4.2. Let X be a Banach space, and (n, E, /1-) a probability space. For any bounded sequence {f of bounded functions in L l (/1-, X), there exists a sequence 9 « f and two measurable subsets C and L of n, /1-(C U L) = 1, such that

n}� 1

{gn(W ) 0 en}: 1 is a weakly Cauchy sequence in X@7rCo; E L, there is a k E N such that {gn(W) 0 en} n ? k is an £1 -

(1) for any w E C, (2) for any

W

sequence in X @7rCo.

Proof: We claim that it is enough to prove the theorem when X is separable.

Suppose that the theorem is true when X is separable. Let {fn}� 1 be any uniformly bounded sequence in L1 (/1-, X). Since each fn has (essentially) separable range, there exists a separable subspace Xo of X such that for almost all w E n, fn(w) E Xo. By the Heinrich-Mankiewicz theorem (Theorem 1 .7.9), there are a separable subspace Z and an isometric embedding J : Z* -+ X* such that Xo � Z and for all z E Z, z* E Z*, ( Jz * , z)

=

(z * , z) .

By the assumption, there exist (gn) « (In) and measurable sets C and L of n with /1-(C U L) = 1 such that for all W E C, the sequence {gn(w) 0 en}� 1 is weakly Cauchy in Z@7rCo and for W E L, the sequence {gn(w)0en}� 1 has an f1subsequence in Z@7rCo. By the proof of Proposition 1 .8.7, the same conclusion holds if we replace Z@7rCo by X@7rCo. We have proved our claim. Suppose that X is a separable Banach space. Let {fn}�= 1 be any bounded sequence in L1 (/1-, X) such that { l fnO llx }� 1 is uniformly integrable. First, we assume that {fn : n E N} is uniformly bounded. Since X is separable, X@7rCo is separable, and the unit ball B ( C(X, fI ) ) = B ( (X@7rCo)* ) endowed with the weak* topology is compact metrizable. Let /C denote the collection of all weak* compact sets of B (C(X, fI) ) , and {On : n E N} a basis of the weak* topology on B ( C(X, fI ) ) with 01 B ( C(X, £I ) ) . Let Kex be a set-valued measurable function from n to /C. As in the proof of Talagrand ' s L l (X)-theorem (Theorem =

CHAPTER 5. STABILITY PROPERTIES II 5.3.1), for any (gn) « Un) with a representation gk = L.;� P k adi ( aqk =f. 0), set 280

£? qk {

}

9k,j , 0: (W) = sup sup / T (gk (w) ) , e £ ) : T E OJ n Ko: (w) , \

OJ,o: (g) (w) = lim sUP 9k,j, 0: (w) , k -+ oo

¢j,o: (g) (w) = likm-+00 inf gk,j,o: (w). We note that inf 0 = 1 , sup 0 = - 1 , and the definitions of 9k and gk are heavily dependent on the representation of gk as a block convex combination of the fn . As in step 2 in the proof of Talagrand's Ll(X)-theorem, one can show that the mapping w sup { (T, gn (w) ® e k ) : T E O J n K0: (w) } is measurable. For any j, k E N, 9k,j, 0: is the supremum of countable measurable functions and so it is measurable. Thus for any j E N, OJ ,o: (g) is me�urable. Fix a sequence 9 and set h = - g . Then for any j, k E N, gk,j, o: = - hk,j,o: and ¢j, o: (g) = - (OJ,o: (h)) . The above argument implies that for any k, j E N , gk,j, o: and ¢j,o: are also measurable. 1---+

Lemma 5.4.3. Let Ko: be a measurable set-valued mapping from n to

K

and j a natural number. For any sequence (un) of measurable functions from n to B(X), there exists (gn) « (un) such that for any h = (hn) « g, we have lin 11 1 OJ,o: (g) = OJ ,o: (h) , nlim -+ oo II OJ'' 0: (g) - ,J' ,o: ¢j ,o: (g) = ¢j, o: (h), Proof: Note: ¢j ,o: (g) = - (OJ,o: ( - g)) for any sequence

to show only that there exists (gn) « (un) such that for have

=

0,

9

= (gn). We any h = (hn) «

need 9 we

By the proof of Lemma 5.3.6, there exists (gn) « (un) such that for any h = (hn) « (gn), O( h ) = O( g ) . Since the proof is similar to the proof of Lemma 5.3.6, we show only that OJ,o: (g) � OJ,o: (h) . Let

qn hn = L Q: i fi' i=Pn

hn =

bn

{3 gj , j=Lan j

dn

gn =

L 1£ 1£

£=Cn

be the representations of gn and hn (Q:qn , (3bn , and Idn are nonzero for all n E N.) Note: If Cj ::; i ::; dj , then Q:i = {3j li (so Q: i = 0 if and only if either {3j = 0 or

281

5.4. PROPERTY (V* )

Ii = 0) . Since d i ::; qn for all an ::; i ::; bn , hn,j, Q (w) = sup k ?, q.. TEOjnsupKoc(w) (T(hn(w)), e k ) b.. sup sup !3i ( T(gi (W)) , e k ) k?, q.. TEOjn Ka (w) iL = a .. ( T(gi (W)), e k ) ::; sup k?, q.. TEOjnsupKa(w) a.. sup $ i $b.. sup ( T(gi (W)), e k ) = sup sup a n $ i $b.. k ?, q.. TEOj nKoc(w) sup ( T(gi(W)), e k ) = sup 9i,j, Q (W) , sup < a.. sup a.. $ i$b.. $ i $b.. k ? di TEOjnKa (w) Note that limn -+oo an = 00 . This implies that Bj , Q (h) ::; Bj, Q ( g ). The proof is 0 complete. Let T denote the first uncountable ordinal, and set h� = in , Kl(w) B(C(X, f d ). Fix - 1 < a < b < 1. As in step 3 in the proof of Talagrand ' s Ll(X)-theorem, for a < T , we can construct a sequence hOI = (h�), and mea surable maps KQ (KQ(j, a, b, . )) : n -+ IC , that satisfy the following conditions: ( a ) For !3 < a < T , hOI « h {3 . (b) If a = !3 + 1 for some !3 � 0, and h « hOI , then we have Q h Bk , {3 (hQ ) = Bk , {3 (h), nlim -+<X) I Bk ' (3 (h ) - n , k , (3 ll l = 0, Q h ¢k,{3 (hQ ) = ¢k , {3 (h), nlim -+oo II ¢k ' (3 (h ) - n , k , (3 ll l = O. =

=

In this case, we set

KQ(w) = { T E K{3 (w) : Vk, either T � Ok or (Bk, {3 (hQ )(w) > b and ¢k, {3 (h Q )(w) < a) } . ( c ) If a is a limit ordinal, let KQ (w) = n {3 < QK{3 (w), Step 4. As in the proof (step 4) of Talagrand ' s Ll (X)-theorem, there is a < such that KQ(w) = KQ + 1(w) for almost all w E n. Let C = {w : KQ(w) 0}, M = {w : KQ(w) = KQ + 1(w) # 0}, h = hQ + 1 . Clearly, J.t( C U M) = 1. We claim that C has the following property: Lemma 5.4.4. For any w E e, T E B (.C( X, f d) , and « h, we have either lim sup ( T(u n (w)) , en) ::; b or lim n -+inf oo ( T(un(w)) , e n } � a. n -+ oo T

=

U


282

and T E B (.c ( X , f l ) ) ' Fix (un) « h « j, with Un = E �;' a n adi (abn i= 0) for all n E N. Let S be the bounded linear operator from Co into Co defined by Proof: Let W E

0

if j = bn for some n E N, otherwise. Clearly, II S II = 1. Since S* 0 T E K1 (w) = B (.c ( X , f l ) ) and S* 0 T � Ko; (w), there is a least ordinal f3 for which S* o T � K{3 (w), By (c) , f3 cannot be a limit ordinal; so there is a 'Y such that f3 'Y + 1 and S* 0 T � K",( (w). By (b) , there exists k E N such that =

Since U « h{3 , we have either n-+oo or

n-+oo

lim inf ( T(un (w)), en ) n-+oo

lim inf ( S* 0 T(un(w)) , ebJ n -+oo � b and _ ( m (2)) (w ) < a } , { w E M : g_n(2) gn(2) , 1 ,o , I ,o

2- 2 . By the definition of 9 and g, for any w E B� , we have

{ : T E Ko {w) } < a, sup sup { ( T (gn(2) { W )) , ek ) : T E Ko (w) } > b . qn(2) $ k$ m (2) inf

q n(2) $ k$ m (2)

in f ( T (gn(2) {w)) , ek )

Note that for any x E X , the maps T

1---+

T

1---+

sup

( T x , ek) ,

inf

( T x , ek)

qn(2) $k$ m (2) qn(2) $k$ m (2)

are continuous. The sets

sup ( T x , ek » b } , { T E B (C {X f ) ) : qn(2) $k$ m (2) { T E B (C {X, fd ) : qn(2) $infk$m (2) (T x , ek ) < a } ,

1

are open. As in step 5 in the proof of Talagrand ' s L 1 {X)-theorem, there are measurable maps Q o O and Q1 0 from M to N satisfy the following conditions . • 0 Q o ( w) n Ko {w) i= 0 , and inf { ( T{ gn (2 ) ( w )), ek ) : qn(2) ::; k ::; m ( 2 ) } < a for all T E 0 Qo (w) ' •

0Q l (W ) n Ko {w) i= 0 , and sup { ( T{gn(2) (w)) , ek ) : qn(2) ::; k ::; m ( 2 ) } > b for all T E 0 Q l (W ) ' Since p is a probability space, there exists an integer f such that if we set B1

{ w E B i : max{ Q o {w) , Q 1 ( W ) } < e } , then p ( M \ Bd ::; 2- 1 . Let o (w) if w E B b Q ( (O) , w) if w E M \ B 1 ; and �1 (w) if w E Bl l Q({l ) , w ) if w E M \ RI . =

=

=

{� {

5.4. PROPERTY (V* )

285

Then Q ( (O) , w) , Q ((l), w) , and B l satisfy (h)-(k). Suppose that the result has been proved for some i � 1 . Let £,

= sup{Q( s, w) : l si

Since Ka + 1 ( W ) Ko (w) for all w k E N and each w E M, =

E

=

i , w E Bi } .

M, condition (d) implies that for each

either Ko (w) n Ok = 0 or (¢k ,a ( g) (W ) < a and (}k , a ( g ) (w) From (b) , we deduce as in the case i the set B�� l

=

{w E M

=

> b) .

1 that there is n(i + 1 ) > n(i) such that

: Vk � £, either Ok n Ka (w) = 0 or

(9n(i+ 1), k,a (W ) < a and gn(Hl ), k,a (W) > b) }

satisfies J-l(M \ B��l ) � 2-i-3 • Using a similar argument to that employed as in the case i = 1 , one can pick an integer m(i + 1) > qn( i+ 1 ) such that the set

{

i

B� + 1 = w E M : Vk � either Ok n Ka (w) 0 or -(m( H l» ( - ( m(H 1 » gn(Hl ), k ,a ) b and gn(H l) ,k,a ( w ) < a

(

=

w>

satisfies J-l(M \ B� + 1 ) � 2-i- 2 . Fix w E B�+ 1 and s E S with l si Q ( s , w) � i and Ko (w) n Oq ( s ,w) =1= 0. We have 9n (p+1) ,Q ( s , w) ,a (w) < a and gn (p+l),Q ( s ,w) ,o (w)

)}

=

i. Then

> b.

Note that the set B(X*) is normal, and for any Ok � OJ , gn(p + l) , k ,a (W ) � gn(p+ l),j,o (w) , 9n(p+l), k ,a (W ) � 9n (p+1), k ' , o ( W ) .

For each k � i, apply the above method (p 1 ) to { w : Q ( s , w) k } and all OJ such that OJ � Ok . We obtain two measurable maps Qo((s, 0), . ) and Q1((S , 1 ) , ·) from M to N satisfy the following conditions: =

•

•

=

{

For all T E OQo « s ,O)w) , inf (T( gn(i + 1) (w) ) , e k) : qn(H l) � k � m ( i + I)} < a, O Qo « s ,O) , w) n Ka(w) =1= 0 , and O Qo« s ,O), w) � O Q« s ,O), w) '

For all T E OQ l « s , l) w) , sup { (T( gn( Hl ) ( W )) , e k) : qn( i + l ) � k � m(i + I)} > b , OQl «s ,l ) , w) n Ko (w) =1= 0 , and O Q l « s ,l),w) � O Q« s, l) , w) ' Then there exists an integer £' such that if BH 1 = { w E B�+ 1 : Vs E S, l s i

=

i, max(Q o ( (s , 0) , W) , Q1 ( ( S , l ), w ) ) � e } ,


286

Q((s, l ) , w) Q((s, 0), w)

=

=

{ � l ( ( S , l ) , w)

if w E BH l , otherwise;

{ �O (( S , 0), w)

if w E BHb otherwise.

Let M' U� l n i > k Bi . From (i), we have J.L(M \ M') O. To conclude the proof, we show that if w E n i � k Bi , then { gn(c(i» (W) ® ei h � k is a sequence in X@1rCo that is 2/(b a )-equivalent to the unit vector basis of .f l . Let p � k , and { a k ' . . . , ap} be a finite sequence. Let =

=

-

P

=

{ c (j ) : aj � 0 and p � j � k},

k defined by , { 01 ii rt.E P,P. Si

and let s , s' E S be the finite sequences with l s I S.

1 _

{ 01 ii rt.E P,P

and

=

I s' l

=

=

From (k) , it follows that there are Tb T2 E B(.c ( X , .f l )) satisfy the following conditions: •

If k ::; i ::; p and i E P, then

•

if k ::; i ::; p and i rt. P, then

sup { ( Tl ( gn(c(i» (W)), em ) : qn(c(i» ::; m ::; m( c (i))} > b, inf { (T2 ( gn (c(i» (w)) , em ) : qn(c(i» ::; m ::; m( c ( i ))} < a . inf { (Tl ( gn(c(i» (w)) , em ) : qn(c(i» ::; m � m( c ( i ))} < a sup { (T2 ( gn(c(i» ( W)), em ) : qn(c(i» S; m S; m( c (i)) } > b.

Now for p ::; i ::; k, choose k (i) and k' ( i ) such that

•

•

if k ::; i ::; p and i E P, then

sup { ( Tl (gn(c(i» (w)) , em ) : qn(c(i» ::; m ::; m( c (i))} inf{ (T2 ( gn(c(i» (w)) , em ) : qn(c(i» S; m ::; m( c ( i ))}

if k ::; i ::; p and i rt. P, then inf { ( Tl ( gn(c(i» (w)) , em ) : qn(c(i » sup { ( T2 ( gn(c(i» (w) ), em ) : qn (c(i»

m .$ m(c( i)) } .$ m ::; m( c ( i ))}

::;

= =

= =

( Tl ( gn(c(i» (w)) , ek (i» ) ' ( T2 ( gn(c(i» (w) ) , ek' (i» ) ;

( Tl (gn(c(i» (w)), ek( i» ) ' ( T2 ( gn(c(i» (w)) , ek' (i» ) '

5. 4 .

PROPERTY (V* )

287

Let 81 and 82 be the linear operators from Co into Co such that This implies that

o We have proved our claim. F. Bombal proved (Theorem 1.7.7) that a Banach space X has property (V*) if and only if X is weakly sequentially complete and any iI-sequence of X has a complemented iI-subsequence. It is known (Theorem 3 . 1 . 1 1 ) that if E is an order continuous Kothe function space, then every iI-sequence {gn}�= 1 of E has a complemented iI-subsequence. Thus an order continuous Kothe function space has property (V* ) if and only if E is weakly sequentially complete. The following theorem is due to N. Randrianantoanina: Theorem 5.4.6. (N. Randrianantoanina) For any Banach space X and any Kothe function space E, the Kothe-Bochner function space E(X) has property (V*) if and only if both E and X have property (V*) .

Proof: It is known that every subspace of a B"anach space with property

(V*) has property (V*). Suppose that E(X) has property (V*). Since E and X are isomorphic to subspaces of E(X), X and E have property (V* ) . Hence one direction is clear. Assume that both X and E have property (V*). By Theorem 1 .7.2 and The orem 5.3. 14, E, X, and E(X) are weakly sequentially complete. Particularly, E is order continuous. Note: Every separable subspace of an order continuous Banach lattice is contained in some separable sublattice. By Theorem 1.7.10, we may assume that both E and X are separable. By Theorem 3.1 .8, we may also assume that E is a Koth function space defined over a probability space (O, �, /-£) such that L oo ( /-£ ) c E c LI( /-£ ) and for any 9 E L oo ( /-£ ), 1 2 11g 1 1 I :S II g ll E

:s

Il g ll oo .

By Theorem 1.7.7, we need to show only that for any iI-sequence {fn}�= 1 in E(X) , {fn }�= 1 contains a complemented iI-subsequence. If {fn : n E N } is not uniformly integrable, then {fn : n E N } contains a complemented iI-subsequence. This implies that the set {fn : n E N } cannot be a (V* )-set. Hence we may assume that {fn : n E N} is uniformly integrable. By Talagrand ' s L I (X)-theorem, there exist a sequence (gn) « (In) and a measurable subset 0' of 0 with /-£(0') > 0 and such that for each w E O', there exists k E N such that {gn (W)}n� k is equivalent to the unit vector basis of i l in X. (Note: Since {fn}�=l is equivalent to the unit vector basis of iI , /-£(0') > 0.) Applying Theorem 5 . 4.2 to the sequence g, we obtain two disjoint measurable sets C, L of 0 and a sequence ( O. By Lemma 1 .8.6, we can identify L1 {Il, X)0Co with Ll (Il, X 0Co). By Lemma 5.3.9, there is k > 0 such that the sequence {CPn 0 en}�= k is equivalent to the i1 basis in L1 (Il, X®Co) = L1 {Il, X)0Co. So it cannot be a weakly null sequence. By Lemma 5.4.1, the sequence {CPn }�= k contains a complemented iI-subsequence in L1 (Il, X). Notice that the inclusion map from E{X) into L1 (Il, X) is continuous. So the sequence {CPn}�=l contains a complemented iI-subsequence in E{X). As a consequence, the set { 0 W E Of ,. (3) The map w F(w)x is norm-measurable for each x E X. Let 0 be a Hausdorff space. Then 0 is said to be a Polish space if 0 is homeomorphic to a complete separable metric space. A subset A of a Polish space 0 is said to be an analytic space if A is the continuous image of a Polish �

space. The following measurable selection theorem is due to N.J. Kalton, E. Saab, and P. Saab. For a proof, see [27] . Theorem 5.4.8. (N. J. Kalton, E. Saab and P. Saab) Let A and B be two complete metric spaces, and H a nonempty Borel subset of A x B. Let PI denote

289

5.4. PROPERTY (V* )

the projection from A x B onto A. For any analytic subset M of PI (H) , there is a universally measurable map S : M B such that (a, S ( a)) E H for all a E M. ----+

Proof of Proposition 5.4.7: We need a few steps to prove the proposition.

Notice that since X is separable, so is the space X011'Co, and therefore, the unit ball of its dual B (C(X, £I ) ) is compact metrizable in the weak* topology (particularly, it is a Polish space). The space B ( C(X, £I ) ) x (X011'Co) N with the product topology is a Polish space. Let A be the subset of B (C( X, £ I)) X (X011'Co) N defined by

It is easy to see that the set A is a Borel subset of (C(X, £I ) ) x (X011'Co) N . Let II be the projection from B (£(X, £I ) ) x (X011'Co) N onto (X011'Co) N . Then the operator II is, of course, continuous, and therefore, II(A) is analytic. By Theorem 5.4.8, there is a universally measurable map e : II(A) B ( C(X, £I) ) such that the graph of e is a subset of A. Fix w E nf. Since T(w) satisfies inequality (5.11), ----+

So for any w following:

E nf , (T(w) , (1Pn (w) 0 en)n) E A. Define F : n for w E nf,

----+

B (C(X, £I )) as

otherwise. The map F is the composition of a universally measurable map e and the /-L measurable map w (1Pn (w) 0 en) n ' So F is measurable for the weak*-topology. Note that for any x E X, the range of F(w)(x) E C(X, £I ) is contains in £l l and £ 1 is a separable dual. By the proof of the Pettis ' s measurability theorem, F(·)(x) is measurable. By the definition of F and e , for any w E nf, f---+

So (by the definition of A) we have that for any w E nf , lim ( F(w) (1Pn (W)), en } > O . n-+oo 0 The proof is complete. By Proposition 5.4.7. (3), for any f E E(X) � L 1 (X) , the function w F(w) ( J(w) ) is measurable. Let ')'(w) limn-+oo ( F(w) (1Pn(w)) , en } . Then the map ')' is measurable, and for each w E nf , ')'(w) > O. Now define S : E(X) £ 1 by S(f) r F(w) (f(w)) d/-L(w) for each f E E(X) � Ll (A, X). =

=

Jnt

f---+

----+

CHAPTER 5. STABILITY PROPERTIES

290

II

Since I F(w)1 1 � 1 for all w E 1"2, the operator S is linear and II S I I � 1 . One can easily verify that lim (S( 1/;n ) , en) r ,(w) d >.. ( w) II > O. n-+oo ln' We proved that there exists N E N such that for n � N , (S( 1/;n ) , en) > � . By the proof of Lemma 5.4. 1 , {1/;n }�= l contains a complemented t'l-subsequence. 0 Thus the set {In ; n E N} is not a (V*)-set. The proof is complete. =

5.5

=

The Talagrand Spaces

In [6] , Bourgain showed that for any compact Hausdorff space K and any measurable space (1"2, /l) , both C(K, Ll (/l)) and Ll ( C(K)) have the Dunford Pettis property. It is natural to ask the following question: Problem 5 . 5 . 1 . Suppose that X has the Dunford-Pettis property. Does Ll (/l, X ) (respectively, C([O, 1] , X)) have the Dunford-Pettis property? In this section, we show that the answers to the above question is negative. Let Ll U�= l {O, l} n be the usual dyadic tree. An element ¢ E Ll is said to be a node of order n if ¢ E {O, l} n . In this case, we write I¢I n. For any two nodes ¢ and 1/;, 1/; � ¢ if 1/; extends ¢ . The Talagrand spaces Tp , 1 � p < 00 , are the completion of the spaces of all finitely nonzero functions x : Ll ---.. lR under the norm l Il x lI Tp supl ( L sup{ l x(1/;) I : 1/; � ¢ } p ) ip . n � 1¢I = n Enumerate Ll as {1/;n }�= l ' Let e1/Jn : Ll ---.. lR be the function that is 1 at 1/;n and o elsewhere. It is easy to see that {e 1/Jn }�= l is a I-unconditional basis for Tp . Lemma 5.5.2. For any 1 � p < 00 , let {x n}�= l be a sequence in T such that 0 < 0 � inf{ ll xn l l Tp : n E N} � sup{ ll xn l l Tp : n E N} � a. Supposep that limn-+oo xn ( 1/; ) 0 for all 1/; E Ll. Then {Xn }�= l has a eo-subsequence. =

=

=

=

Proof: Without loss of generality, we assume that a 1 . By perturbing { Xn }�= l and by passing to a subsequence of {Xn}�= l ' we may assume that there is an increasing sequence {kn : n E N} such that for any n E N, =

Xn

=

For any ¢ E Ll, let

t( ¢ , m) 1/Jsup� ¢ I Xm (1/;) I . By the definition of the norm of Tp , for any n and m in N, L= tP( ¢ , m) � IIxm ll �p � 1 . =

1¢I n

291

5.5. THE TALAGRAND SPACES

For any infinite subset M of N, [M] denotes the collection of all infinite subsets of M. We claim that there are an infinite subset {ml 1, m 2 , . . . } of N and an infinite subset {Nl N, N2 , • • • } of [N] that satisfy the following conditions: =

=

(ii) For any n < kmi +l ' l: 1 4>I = n s UPm E Ni +l

tP( c/J, m) � 2P+ I.

First, let NI N and m l 1. Suppose that m l , . . . , mi and No, Nl , . . . , Ni are selected. There is an infinite subset Ni H � Ni such that for any c/J E � with order at most km i H , the set {tP ( c/J, m) : m E Ni + d has diameter less than 2 - k""i +1 • Select an mi + l > mi in Ni + l . Then for any n < kmi + l , =

=

sup tP ( c/J, m) L m E 1 4>I = n Ni +l � 1 L= (t(c/J, mi+ l ) 2 - k""i +1 ) P 4>I n � 2P 1 4>IL= sup {t (c/J, miH ) , 2 - krni +1 }P � 2P+ I . +

n

We have proved (ii). The construction is complete. We claim that for any sequence c ( Ci) E Co , =

i

sup

S

By (ii) , we have


By Lemma 5.5.2 and Theorem 1.2.30, we have the following theorem.

o

Theorem 5.5.3. (Talagrand) For any 1 < the dual of the Talagrand space Tp is separable and has Schur property.� Thus both T; and Tp have the Dunford-Pettis property. p

00 ,


292

II

Proof: By Lemma 5.5.2, Tp has a shrinking unconditional basis { e ..p : ¢ E Ll}. So the sequence of coefficient functionals { e ; : ¢ E Ll} forms a 1unconditional basis for T;. By Theorem 1 . 2.30, T; has the Schur property. 0 This implies that both T; and Tp have the Dunford-Pettis property. The following theorem is due to M. Talagrand. Theorem 5.5.4. For any 1 < p < 00 , have the Dunford-Pettis property.

C([O , 1 ] , Tp) and L1 ( [0, 1 ] , T;) do not

{O, I} N is isomorphic to ° = IT:'= 1 {O, l} n , it is enough to show that C(O, Tp) and L 1 ( 0 , A, T;) do not have the Dunford-Pettis property, where A is the product Haar measure on 0. Let Pn be the canonical projection from n ° onto {O, l} . For w E 0, set fn(w) = epn (w) , Fn(w) = epn (w) ' Then fn E C(O, Tp), Fn E L 1 ( A , T;) � C(O, Tp)*, I l fn IIL oc ( '\ , Tp ) � 1 , and I Fn IIL 1 (,\, T;) � 1 . We claim that {fn}�=1 and { Fn }�=1 satisfy the following conditions: (i) {fn}�=1 converges weakly to ° in C(O, Tp). (ii) {Fn}�=1 converges weakly to ° in L 1 ( A , T;). N9tice that (Fn, fn ) = 1 for all n E N. If the claim is true, then neither C(O, Tp) nor L 1 (A , T;) has the Dunford-Pettis property. Proof of (i) . Recall the following fact: Fact 1 . 1 . 1 . Suppose that { xn }�=1 is a bounded sequence in X that is not weakly null. Then there are X* E S ( X*), 13 > 0, and a subsequence { Yj }� 1 of { Xn }�=1 such that for any j E N, Proof: Since

So for any finite nonnegative sequence

{aj }}= 1 with L:;: 1 aj = 1 ,

1 jf:=1 ajYj I � 13·

It is enough to show that for any k > 0, and any m 1 < m2 < ' " < mk ,

l i t; fmi I I L oc (A ,Tp ) � k 1/p • k

ml (w) , . . . , fmk (w ) have disjoint support, for any w E O , k (w ( � 1 . i =1 fmi ) 1/J) 1 1/JSUp? ..p I L

Since f

5 . 5 . THE TALAGRAND SPACES

293

If 1 ¢ 1 1 1 ¢2 1 and ¢ 1 =J ¢2, then the sets {"p E � : "p � ¢t} and {"p E � : "p � ¢2} are disjoint. So for any n E N, we have =

k

P � k. ¢ 2: ,pSUpl2: fm;( ) )( W 1 1 4> I =n ? 4> i= 1 This implies that I 2: i < k fm i ( W ) I Tp � k 1 /p for all w E 0, and 11 2: 7= 1 fm J � k 1 /p • We have proved (i). Proof of (ii). Let �. By Fact 1 . 1 . 1 again, it is enough to show that for any 0 1 =£ We claim that 11 4>2:E A e; 11 Tp. � d · 2£/ q . Let {Ai : i � d} be a partition of A such that for any ¢ E � with I ¢ I i, Ai contains at most one element "p � ¢. Then for each i � d, Ai contains at =

=

most 2£ elements. So

£/q T sup ( 2: I x ( ¢) I P f /P x E p , I x l 9 4>E A ; sup ( 2: ( sup I x ("p) I ) P f / P � 2 £/ q . � 2 £/ q x E Tp , I x l 9 1 4> 1 =£ ,p ? 4>

�2

This implies that

We have proved our claim. Notice that

k

k

11 ( �?m . ) t (" T; ) D I �?m. (W ) II T/>' (W ) . =

By the above claim, we have

k

I (t; Fmi ) II Ll( >' ,T; ) � 2£/q J dk (W)dA (w),


294

where dk (W) sUP l cl> l = i card {i :=:; Let hT be a function defined by =

k : Pm; (W ) ;::: ¢}. Fix ¢ E {O, l }i and i E N.

hT (w) { �

if ¢ :=:; Pm ; (w), otherwise.

=

Then

dk (W)

II

k

sup L hT(w). IcI> l = i i = 1 But {h f } � 1 is a sequence of LLd. random variables with J hf 2 -i . By the weak (or strong) law of large numbers, f 2: 7= 1 hT converges to 2 - i in measure. This implies that =

=

k 1 1 lim - j dk (w)d>"(w) k-+oo lim j sup L hT(w)d>"(w) k-+oo k IcI> l =i i= 1 k =

-

=

2-i . o


w E 0 and any increasing sequence { n k } k:: l of natural numbers, the sequence {Fn k (w)} k:: l that is defined Remark 5.5.5. It is easy to see that for any

in the proof of Theorem 5 . 5 . 4 co ntains an iI-subsequence. The following example shows that there is a strictly convex Banach lattice X with the KK (Kadec-Klee) property such that L 2 ( [0, 1] , X) does not have the KK property. This gives a negative answer to a question of Smith and Thrett [43]

.

T;;) oo , eo (1, 0), and en (0, e�J for n ;::: 1 . Then {en} �= 1 is a normalized unconditional basis for X. Let { Qn }�= o be a strictly decreasing null sequence with aD 1 , and I . l i e the Example 5.5.6. (P.K. Lin) Let X

=

(JR

EB

=

=

=

norm on X defined by

I ( b, (an)) I1& I ( b , ( an)) IIi=

00

n= 1 Qna� .

+ b2 + L

Then (X, 11 · 11 ) is a strictly convex Banach space with the Schur property. So X has the KK property. It is easy to see that

nlim -+oo l i eD + enl l e l I eol l e v'2. We claim that for any 1 < < L ( , (X, I . l i e )) does not have the KK n property. Let 0 11 �= 1 { O, l} , E the Borel (T-algebra of subsets of 0, and the normalized Haar measure on O. For each n E N, let Pn be the canonical projection from 0 onto {O, I} n . Define 9n : 0 ---+ X by 9n(W) � ( O, e pn(W» ) for all w E O. The proof of Theorem 5 . 5.4 shows that { 9n} �= 1 converges weakly =

p

=

00 ,

p

=

/-l

/-l

=

295

5. 6. THE BANACH-SAKS PROPERTY

to zero in Lp(J-L, X) for each 1 < p < 00. Let h : 0 -7 X be the function defined by h(w) � eo for all w E 0. Then h + 9n h weakly in Lp(J-L, X) for all 1 < p < 00 . Notice that for each w E 0, -7

=

nlim -+oo I I h(w) + 9n(w) ll e

=

II h(w) li e

=

1

and II h (w) + 9n(w) lI e ::; 2. By the bounded convergence theorem, for alI I < p < 00 . Thus Lp(J-L, (X, II . li e )) fails to have the KK property. The proof is complete. Remark 5.5.7. Schachermayer [42] showed that there is an interpolation

S between T1 and £1 such that S has the Banach-Saks property, but L 2 ([0, 1] , S) does not have the Banach-Saks property (also see Example 5.6.9) .

space

Exercises

{Xn} be a sequence of Banach spaces, and E a Banach space with a I-unconditional basis {en}�=l ' Show that Exercise 5.5.1. (Castillo-Gonzalez) Let

E(Xi ) has the Dunford-Pettis property if and only if E and the Xi have the Dunford-Pettis property.

5.6

The Banach-Saks Property

Recall that a Banach space X is said to have the (weak) Koml6s property if for any bounded (weakly null) sequence {fn }�=l in L 1 (J-L, X) there exist a subsequence {fn k }k:: 1 of {fn}�=l and f E L1 (J-L, X) such that for any further subsequence {h k } k=l of {fn k } k=l' m

1 L h k converges to f almost everwhere. m

k=l

Koml6s (Theorem 1 9 . 9 ) showed that the scalar field has the Koml6s property. Let (0, J-L) be an atomless probability space. J. Bourgain [10] proved that for any 1 < p < 00 , Lp(O, X) has the Banach-Saks property if and only if X has the Koml6s property. Modifying Bourgain ' s proof, P. Cembranos [10] showed that for any 1 ::; p < 00 , Lp(O, X) has the weak Banach-Saks property if and only if X has the weak Koml6s property. In this section, we show that the Bourgain Cembranos result is still true if one replaces Lp by any order continuous Kothe function space E with the subsequence splitting property. We also show that there is a Banach space X with the Banach-Saks property such that L 2 ( X ) does not have the Banach-Saks property. .


296

Let {fn}�=l be a sequence in the space L 1 (J-l, X ) such that for any subsequence {f�}�= l of {fn}�=l and € > 0, there is a further subsequence {f::}�= l of {f�}�=l satisfying Lemma 5.6. 1 . (Bourgain-Cembranos [10])

Then there is a subsequence {f�}�=l of {fn}�=l such that for any subsequence {f::}�=l of {f�}�=l ' 1 k a e. klim -+oo -k nL =l f:: 0 . =

Proof: Let [N] denote the set of all infinite subsets of N with the pointwise topology (identify [N] as a subset of { O , 1 } N ). Fix € > O. For each N, m, n E N,

is an open subset of [N] . Thus

is a Borel set. By Ramsey's theorem, there is M E [N] such that either [M] � A or [M] n A 0 . Our hypotheses preclude the latter possibility, and so we can find M E [N] such that [M] � A. By induction and the diagonal method, we can find (mj) E [N] such that for any £ E N and (Pj ) E [mj : j E N] , =

This implies that for any £ E N,

297

5 . 6. THE BANACH-SAKS PROPERTY

Thus

and

k =0 j =l

lim -k1 "" � fv' k-+oo J

a.e. 0


[10]) Assume that L1 ({l, X) has the weak Banach-Saks properly. For any weakly null sequence {fn }�=l in L1 ({l, X) and > 0, there is a subsequence {f�}�=l such that Lemma 5.6.2. ( Bourgain-Cembranos f.

{fn }�=l be a weakly null sequence in L1 ({l, X ) and > O. By Lemma 2.3.8 , Theorem 2 .3.9, and our hypothesis, there exist a subsequence {f�}�=l of {fn }�=l and f. E N such that �

Proof: Let

f.

< � whenever n1 < n2 < . . . < nf. . (5.12) f � i I �L 1 1 1 i=l Notice that {f� : n E N} is uniformly integrable. There are M > 0 and a sequence of measurable subsets {An}�=l such that the sequence {f:: =f� . 1 A,J�=1 satisfies the following conditions: ( 5 .13) 1 f:: - f� I L1(IL ' X ) < 4 for all n E N, ( 5.14 ) I l f� l oo < M for all n E N. Applying Koml6s ' s theorem to the sequence { gn = 1 ( 1:: - f�) Ol l x }�=l' we obtain a subsequence {9n = 1 ( /:: - 1� )( ' ) l x }�=l of {gn}�= l such that for any further subsequence {9�}�=1 of {gn}�=l' { k L�=l 9� } : 1 converges almost everywhere. By Fatou ' s lemma, we have f.

For r E N, let


298

Then by (5.12) and (5.13) , we have

(r+ l )f (r+ l )f 1 I l hr l h £ 1 n=L l l� II L ( + £ n=L l 11 1� - l� II L l (lL.x ) < � + � � . rf+ rf+ Applying Koml6s ' s theorem and Fatou ' s lemma to {h r }� l again, we obtain a subsequence {h rJ� l of {hr }� l such that k k l h l (5.16) hr � f � I l k� � L i =l ri l l \�� l l � L i=l i ll l � . 1 �

=

1 IL. X )

Let N be the set

E N : £ri + 1 � � (ri + 1)£, i E N} , and denote the sequences (f�)n EN and (f�)n EN by { ¢n} �=l and { en}�= l' re spectively. We claim that the sequence {l�} n EN satisfies the required condition. For any k � £, there exists j E N such that j£ � k < (j + 1 )£ . By (5.14 ) , for N

=

n

{n

any t E O,

1 k

I h� nL= l en(t) ll x

l

By (5. 16) , f lim suPk-+oo l i L:: � =l en o l x dJ.L � � . On the other hand, using ( 5.15 ) , we have

Thus

299

5. 6. THE BANACH-SAKS PROPERTY

o

The proof is complete. The following lemma is due to J. Bourgain [8] .

Let A be a weakly precompact subset of L l ( J-L, X), and C a bounded subset of Loo . Then the set A . C {f . 9 : f E A and 9 E C} is weakly precompact. Lemma 5.6.3.

=

Proof: Without loss of generality, we may assume that C is a subset of the unit ball of L oo . Suppose the theorem is not true. Then by Rosenthal ' s iI-theorem, there exist c5 > 0, {fn : n E N} in A, and {9n : n E N} in C such that for any finite sequence {a k }k=l of scalars,

By Lemma 5 . 1 .2,

l i kt=l an ' fn ' rn l l L1 CnX [O, l] ,X) l an ' fn(w) · rn l L1 ( [O , lj , X) dJ-L(w) Jo{ i l kt =l =

n

1

n

� Jo( I l kL=l an ' fn(w) . 9n(W ) . rn l L1 C[O, l] , X) dJ-L(w ) � c5 kL=l l an l ·

{fn } �= l contains an iI-subsequence, a contradiction.

By Lemma 5 . 1 .3, proof is complete.

The 0

) Let {fn } �=l be a weakly null sequence in L l (J-L , X). For any 0, there exist a subsequence {f� } �=l of {fn } �=l and a uniformly bounded weakly null sequence {jn } �=l in Ll(J-L, X) such that for all n E N, Lemma 5 .6.4. ( Cembranos [ 10] €

>

{fn }�=l be a weakly null sequence in L l (J-L , X), and a positive real number. Then there are a sequence { An } �=l of measurable sets and M 0 such that the sequence {In IA n . fn } �= l satisfies I l fn - fn l LdlL , X ) < 4 and I l fn l oo < M for all n E N . By Lemma 5 . 6 . 3, { !n }�=l is a weakly precompact subset of Ll (J-L, X). By passing to a subsequence of {fn }�=l we may assume that { !n }�=l is weakly €

Proof: Let

-

=

-

€

'

>

300

CHAPTER

5.

STABILITY PROPERTIES II

Cauchy. By Mazur ' s theorem, there are natural numbers PI < q l < and nonnegative scalar numbers { a d� 1 such that qn

I:: ai 1

i =Pn

=

P2

0, there is a 8 > 0 such that

sup { l g · 1 A IIE J-L (A) < 8, g E K} < :

E.

An order continuous continuous Kothe function space E is said to have the subsequence splitting property if for any bounded sequence {gn}�=l in E, there are a sequence {A k }� l of disjoint measurable sets and a subsequence {gn k }� 1 of {gn}�= l such that {gn k . 1 n \ Ak }� 1 is equi-integrable. It is known that if E is a q-concave Kothe function space for some q < 00 (Le., E is a Kothe function space with nontrivial cotype), then there is an equivalent norm of E whose q concave constant is 1 . Thus if a Kothe function space E has a nontrivial cotype, then E has the subsequence splitting property. Theorem 5.6.5. (Bourgain and Cembranos [10]) For any Banach space X,

the following are equivalent:

5.6.

30

THE BANACH-SAKS PROPERTY

1

The space L1 ( [0, 1] ' X) has the weak Banach-Saks property; Let E be an order continuous Kothe function space over [0, 1] such that for any f E E I f 11 1 � I f i l E . A ssume that E has the subsequence split ting property and the weak Banach-Saks property. Then the Kothe-Bochner function space E(X) has the weak Banach-Saks property. (3) The space X has the weak Koml6s property. (1) (2)

,

(1)

Proof: =} (3) follows from Lemma 5.6.1 and Lemma 5.6.2. (3) =} (2) . Suppose that X is a Banach space with the weak Komlos prop

erty, and E an order continuous Kothe function space with the subsequence splitting property. Let {fn }�= l be a weakly null sequence in E(X) . By the def inition of the subsequence splitting property, there are a subsequence {f�}�= l of {fn}�=l and a sequence of mutually disjoint measurable sets {An}�= l such that { I f�(l n - lA J Ol i x : n E N} is equi-integrable. Set

fn f� · I \ n and in f� · l n • We claim that { in E N} is weakly null. If the claim �s true, then {fn}�=l is n A

=

A

=

: n

also weakly null. Suppose that the claim is not true. Then there is F E E* (X* , w* ) such that ( F, n ) does not converge to zero. But the functions n have disjoint support. This implies that { ( F( . ), n ( ) ) : n E N} is not uniformly integrable. By Proposition 1 .3.8, {in}�= l contains an f1-subsequence, which contradicts Lemma 5.6.3. Notice that the embedding from E(X) to L1 (J-l, X) is bounded. By the hypothesis, there is a subsequence {f�}�= l of {fn}�= l such that for any further subsequence {f:: }�=l of {f�}�= l' we have

i

i

· 11m

1- Lk ,-Ifn

k--+CXl k n=l

i

-

=

0

a . e.

{ II fnO lix E N} is equi-integrable. So the set {l l L:�=1 f::O llx : k E N } is also equi-integrable. By Vitali ' s lemma (Lemma 3.1.13), the sequence {l L: �=1 f::r:= l converges to zero in norm. On the other hand, the sequence {i�}�=l is a weak null sequence with disjoint support. { l i h Ol i x }�= 1 is weakly null in E. But E has the weak Banach-Saks property. There is a subsequence {i:: }�=l of {i�}�= l such that : n

But

1 \ t i::\ \ k--+CXl k k= l E(X ) lim

=

o.

Thus {f:: f:: + i:: }�= l has norm convergent Cesaro mean and we conclude that E(X) has the weak Banach-Saks property. (2) =} Suppose that the Kothe-Bochner function space E(X) has the weak Banach-Saks property. Let {fn}�= l be a weakly null sequence in L1 (J-l, X), =

(1 ).


302

and £ a natural number. By Lemma 5 . 6 . 4, there exist a subsequence {f�}�=l and a uniformly bounded weakly null sequence { gn }�=l in L1(/-L, X) such that

II gn - f� l l < ;£ for all n E N. By Lemma 5 .3.1 7 (2) , {gn}�=l is a weakly null sequence in E (X). By Lemma 2.3.8 and Theorem 2·. 3.9, there exist E N and {nl < n 2 < . . . } such that for any kl < k2 < . . . < kml l mi

Then

ml 1 ml 1 ml f f gn 1 gn II � L i= l ± ki l l E i=l ± �k i ll l $ � L i= l l �k i - ki l + II � L 1

i

i

=

1 1 1 2£ + 2f "'i ' By the diagonal method, there is a subsequence for any £ E N and any £ < nl < n 2 < . . . < nml ,
O. Note that the sequence {Q� (F)}�= 1 converges to 0 a.e., and I Q� (F ) Ol l l l � II F( ' ) l l l for all n E N . By Lebesgue ' s dominated convergence theorem, there is N E N such that I Q:N (F ) I � E. This Let

=

=

implies that

lim sup l ( F, fn) I
I F I ,

k =l n k =l

a contradiction. Since € is an arbitrary positive number, the above argument 0 also shows that I m l � I F I . The proof is complete. Now we can give a characterization of weakly convergent and weakly Cauchy sequences of C(K, X). Since the proof is similar to that of Theorem 1.5.1, we leave it to the reader.

For a compact Hausdorff space K and a Banach space X, let be a sequence of continuous X -valued functions defined on K . (1) The sequence {fn}�=l converges to f weakly if and only if {fn}�=l is uniformly bounded, and {fn (w)}�=l converges weakly to few) for all w E K. (2) {fn}�=l is a weakly Cauchy sequence if and only if {fn}�=l is uniformly bounded, and {fn (w)}�=l is weakly Cauchy for all w E K. Theorem 6 . 1 .6. {fn}�=l

318

CHAPTER 6. CONTINUO US FUNCTION SPACES

Let C be a nonempty closed convex subset of a Banach space X . For any is said to be a best approximant of x from C if

x E X, Y E C

I l y - x i d (x , C) =

The

def

inf

zE C

li z - x i .

metric projection Pc from X to 2 c is the set-valued function Pc ( x ) = {y : y is a best approximant of x from C} .

Note that Pc ( x ) may be the empty set. It is known that if C is a closed convex subset of a strictly convex reflexive Banach space X , then for any x E X , Pc ( x ) contains exactly one point. We need the following lemma. The proof is left to the reader.

Let C be a nonempty closed convex subset of a Hilbert space Then the metric projection Pc is nonexpansive. Lemma 6.1 .S. Let Ko be a separable closed subset of a compact Hausdorff space K, and F a continuous mapping from Ko into a bounded subset of a locally convex space X . Then F can be continuously extended to K with all values in

H.

Lemma 6.1 .7.

co(F(A)) .

Proof: Let Ko be a separable closed subset of a compact Hausdorff space

K, and F a continuous mapping from Ko into a bounded subset of a locally convex space X . Then F(Ko) is a separable compact subset of X . By the Hahn-Banach theorem, there exists a countable set { x i , x ; , . . . } � X* such that for any n E N, SUPX E F(Ko ) I (x� , x ) 1 ::; 1 and for any x, y E co(F(Ko)), ( x; , x ) = ( x; , y ) for all j E N implies that x = y. Define the mapping f from co(F(Ko)) to £ 2 by Since f is a one-to-one continuous affine mapping, f(co(F(Ko)) is homeomor phic to C = co(F(Ko)). (Note: co(F(Ko)) is compact.) Hence we can consider F(Ko) as a subset of 1 e a : � £2 . a k k kl ::; l = 1 k Note: K is normal. By the Tietze extension theorem, there is a continuous extension gk : K [ - 11k, 11k] of t f-+ (x icl k F(t ) ) . It is easy to see that the mapping j : t E�= 1 9k (t)e k is continuous. Let r be the metric projection 0 from £2 onto C. Then f - 1 0 r 0 j has the desired property. Now we can prove a version of the Borsuk-Dugundji theorem that is due to Bombal and Cembranos [3, Theorem 1] .

{ l: 00

-t

-t

k} ,

319

6. 1 . VECTOR- VALUED CONTINUOUS FUNCTIONS

Theorem 6.1.9. (Borsuk-Dugundji Theorem) Let Ko be a closed subset of a compact Hausdorff space K, X a Banach space and Y a separable subspace of C(Ko, X) . Then there is an operator from Y to C(K, X) such that 1 1 8 11 � 1 and 8( f ) is a continuous extension of f such that for all t E K, 8 f (t) E co(J(Ko)) . Proof: First, let us recall the definition of the strong operator topology.

Let W and Z be two locally convex topological vector space, and let C(W, Z) be the set of all continuous linear operator from W to Z . For any for any neighborhood V of Oz (the zero vector in Z) and any w E W, let Ov,w denote the set Ov,w {T E C(W, Z) : T ( w ) E V} Recall that the strong operator topology of C(W, Z) is the topology generated by the sets =

{T + Ov, w : T E

C(W, Z), V is a neighborhood of Oz ,

W

E

W}.

(A net T", E C(W, Z) converges to T in the strong operator topology if and only if T", (w ) converges to T(w ) for all w E W.) Let Y be a separable subspace of C(Ko , X ) , and let T be the strong operator topology of C(C(Ko, X ) , X ) . It is known that (C(C(Ko , X) , X ) , T) is a locally convex topological vector space. For each t E Ko, let 1St be the mapping from C(Ko, X ) defined by ISt ( f )

=

f (t) .

First, we assume that Ko is separable. Define a mapping F : Ko ---+ by F (t) 1St . Let t", be any net in Ko that converges to t E Ko. Then for any f E Y � C(Ko, X ) , (C ( Y, X ) , T )

=

lim F(t", ) ( f ) '"

=

lim f (t", ) '"

=

f (t) = F (t) ( f ) .

By the definition of strong operator topology, F is a continuous function. By Lemma 6.1 .8, there is a continuous extension F : K ---+ C(X, Y ) of F with values in M co {1St : t E Ko}. If we set 8 : Y ---+ C(K, X ) by 8( f ) ( t ) F (t) ( f ) , then for any f E C(K, X ) ( a) the function 8 ( f ) is a continuous function (by the definition of strong operator topology) ; (b) for any t o E Ko , =

=

8( f ) ( t o )

=

F (to ) ( f )

( c ) for any t E K and f E Y, 8 f (t) E

This implies that

11 811 � 1 ,

=

f (to ) ( f )

=

ISto ( f )

co { f ( s ) : s E Ko}.

and for any f E Y,

=

f (to) ;

CHAPTER 6 . CONTINUOUS FUNCTION SPACES

320 •

8(f) is a continuous extension of f; • for all t E K, 8 f( t ) E co (f(Ko)) . General Case. Let { iI , 12 , . . . } be a dense subset of Y . Define an equivalence relation ", on K by w '" w' whenever w w', or w, w' E Ko and fn(w) fn (w') for all n E N. Note that for each n E N, fn (Ko) is separable (a compact subset of a metric space) and Ko/ '" is homeomorphic to a subspace rr �= l fn (Ko) (with the product topology) . So the quotient space Ko/ '" is separable. Let q be the quotient mapping and Z the subspace of C(Ko/ "') defined by Z {¢ : there is f E Y such that ¢( [t J ) f( t )}. It is known that q induces an isometry from Z onto Y. Let R denote the inverse mapping. By Case 1 , there is an operator 8 from Z to C(K/ "', X) such that 11811 � 1, and 8(¢) is a continuous extension of ¢ for all ¢ E Z. Then 0 R 0 8 0 q : Y C(K, X) has the desired property. For a sequence {fn : n E N} in C(K, X), let K/ '" be the quotient space defined in the proof of the above theorem. Note: The countable product of compact metric spaces is compact and metrizable. So K/ '" is compact and metrizable. We have the following lemma. Lemma 6.1. 10. Let K be a compact Hausdorff space and {fn}�=l a se quence in C(K, X) . Then there are a compact metric space K', an isometric isomorphism from C(K', X) to C(K, X), and a sequence {f� }�= l such that 8(f� ) fn for all n E N. Let B be the Borel algebra of a compact Hausdorff space K, and m an X valued vector measure on B. Recall that the semivariation of m the extended nonnegative function function II m ll whose value on a set A E B is given by =

=

=

=

--+

=

II m ll (A)

=

{

}

sup l (x * , F(' )) I (A) : x * E B(X * ) ,

where l (x* , F(')) 1 is the variation of the real measure (x* , F(·)) . Let m be a vector measure. It is known that for any A E B, II m ll (A)

=

sup

a i m(Ci) I } , { lt i= l

where the supremum is taken over all finite collections of mutually disjoint Borel subsets Ci � A an scalars a i with l a i l � 1 . A vector measure m is said to be of bounded semivariation if II m l l (K) < 00. It is easy to see that every vector measure of bounded variation is of bounded semivariation. But there is a vector measure of bounded semivariation that is not of bounded variation (see Example 3.3. 1 ) .

Let T be a bounded operator from C(K, X) into Y. Then T * : Y* (C(K, X))* M(K, X) , T**' : (C(K, X))** :::) L oo (K, X) Y** --+

=

--+

321

6 . 1 . VECTOR- VALUED CONTINUOUS FUNCTIONS

are w*-w* continuous linear operators. For any Borel subset A of K, the map ping X t---+ T ** ( x . 1 A ) is a bounded linear operator from X into Y**. We denote this operator by meA). Then m is a finitely additive y* -regular C ( X, Y** )-valued vector measure. For any y* E Y*, let my. = T*(y*). Since T* is w*-w* continuous, the mapping y* t---+ my. is w* -w* continuous. The vector measure m is called the representing measure of T. Using the techniques in the proof of Theorem 6.1 .5, one can show that I T I = I l m l ( K ). Conversely, let m be a Y*-regular C(X, Y**)-valued vector measure of bounded semivariation. Define an operator T : C ( K, X) � y** by

T (f ) = J fd m nl�� t= m(Ck )(f(tk )), k 1 where {Cl, . . . , Cn} is a partition of K such that for any j � n and E Cj , I l f(sl) - f( s ) l � lin and tj E Cj . Then T is a bounded operator from C(K, X) into y** with the norm I T I = I l mi l ( K ). For any y* E Y*, let my. be the measure in M ( K, X*) such that for any Borel set A and x E X, (my. (A), x) = (m (A ) (x) , y*). If the mapping y * my. is w*-continuous from y* to M (K, X*), then T f is a w*-w continuous function on Y* , and T f E Y. Hence we have the following theorem: Theorem 6.1.11. (Dinculeanu-Singer [16, p.182] ) Let Y be a Banach space and let T be a bounded linear transformation from C(K, X) into Y. Then there is a finitely additive C (X, Y**)-valued set function m on B with the following properties: (1) m is C (X, Y**)-valued vector measure that is Y * -regular and of bounded semivariation . (2) The mapping y* my. from y* into M (K, X*) = C ( K, X)* is contin uous with respect to the Y -topology and the C (K, X)-topology of C(K� X)*. (3) T f = iK f(t)d m for all f E C ( K, X). (4) I T I = I l m l (K). (5) T*y* = my., y* E Y*, in the sense of the isometric isomorphism between the spaces C ( K, X)* and M(K, X*). Conversely, a set function m on B with the properties ( 1 ) and (2) defines a linear bounded transformation T from C(K, X) into Y by (3) . Then we have (4) and (5). The condition (2) ensures that the integral (3) that defines T takes values in Y, though m has values in C (X, Y**). Theorem 6.1.12. (Bartle-Dunford-Schwartz [16, p. 105, Theorem 5]) Let Y be a Banach space, and let T be a bounded operator from C ( K) to Y. Let m be the representing measure of T. Then the following are equivalent. de f

2

S I , S2

x

t---+

t---+

CHAPTER 6. CONTINUOUS FUNCTION SPACES

322

The operator T is weakly compact. The representing measure m of T is a Y -valued vector measure, and the set {y* m : y* E B(Y*)} of measures is uniformly strongly additive. (3) The representing measure m of T is a strongly additive Y -valued vector measure . ( 4) Let B denote- the Borel algebra on K. Then the representing measure m of T is a Y -valued vector measure, and the set m(B) is relatively weakly compact. (1) (2)

0

Proof: By Proposition 3.3.4, (2) => (3) .

Claim 1. We claim that if m is not strongly additive, then a sequence {Bk H':" l of mutually disjoint measurable sets such that {m(Bk )}� l is equiva lent to the unit vector basis of eo. Suppose Claim 1 has been proved. Then we would have the implication (4) => (3) . Suppose that the operator T C(K) Y is weakly compact. Then T** is weakly compact. This implies that the repre senting measure m is Y-valued, and the closure of m( B ) � T**(B(C( K )) ) is weakly compact. We have proved the implication (1)=>(3). Proof of Claim 1 : Let M sup { ll m(A) 1 1 : A E B } II T II . Suppose that m is not strongly additive. Then there is a sequence {A n}� l of mutually disjoint members of 2: such that L: �=l m(Ak ) does not converge in norm; i.e., there are f > 0 and an increasing sequence {n k }� l such that for any j E N, :

=

M�

=

m(Ai) 1 \ � f. II i=� n2j _l n 2j Bj = U Ai . i n2j- l

for each j E N, set

--+

=

Then L: � l m(Bk ) is a wuc series that does not converge. By passing to a subsequence, we may assume that {m(B k ) } k=l is a basic sequence. It is easy to see that {m(Bk )}� l has a eo-subsequence. The verification is left to the reader. Claim 2. We claim that {T(J) : 11 /11 00 � 1 and 0 � I} is contained in the closure of the convex hull of T** (B) . Suppose that Claim 2 has been proved . Let C be the closure of the convex hull of T** (B) m( B ) . Then { T ( J ) : 11 / 11 00 � 1 } is a subset of C - C { e' - e : e' , e E C}. So if m( B ) is relatively weakly compact, then T is weakly compact. We have the implication ( 4 ) => (1). Proof of Claim 2. For any 1 E B(C(K)) + and any f > 0, there exist a sequence {An }�=l of mutually disjoint measurable sets and a finite decreasing sequence {a k } k=l of nonnegative real numbers such that al � 1 and =

=

6. 1 . VECTOR- VAL UED CONTINUOUS FUNCTIONS

323

Then

a d A k ) II II T** (f - t=1 a k 1 A k ) I � € II T II , II T( f ) - T** (t =1 k k =

and

•

n T** ( I: a d A k ) k=1 n- 1 k (1 - a1)m( 0 ) + I: (a k - a k+ 1 ) m ( U Aj ) =1 =1 j k n . + an m ( U Aj ) E co (m (B) ) . j=1 =

We have proved Claim 2. (3) ::::} (4) . Suppose that m is strongly additive and bounded. By Proposition 3.3.4, the family {x* 0 m : x* E B (X*) } is a uniformly strongly additive family of countably additive (finite) real-valued measures on B. By virtue of Theorem 3.3.6, there exists a nonnegative real-valued countably additive measure J1, on B such that {mr : T E A } is uniformly J1,-continuous, i.e., lim JL ( A } ->O I l mr(A) 11

=

0 uniformly in T E A .

Let (go:) be a weak* converging net in L oo (J1,) , say (go:) converges to go in the weak* topology. Then for any y* E Y* ,

where hy * = dy�:m is the Radon-Nikodjrn derivative of y * om with respect to J1,. This implies that T** is a weak* to weak continuous linear operator. Thus T** maps the weak* compact set { g E L oo (J1,) : I l g l oo � 1 } onto a weakly compact set C of X. But

{m(A) : A E B}

=

{T** (1 A ) : A E B} � {T** (g) :

I l g l oo � 1 } � C.

o The proof is complete. Let X, Y be two Banach spaces. Recall that an operator T : X Y is said to be unconditionally converging if T maps all wuc series to unconditionally convergent series. -

If T is an unconditionally converging operator from C(K, X) into Y, then the representing measure m of T is an C(X, Y)-valued countably additive vector measure . Theorem 6.1. 13. (Swartz [41, Theorem 2.7] )

324

CHAPTER 6 . CONTINUO US FUNCTION SPACES

Proof: Let T be an unconditionally converging operator from C(K, X) into Y, and m the representing measure of T. We claim that m is an C(X, Y}valued vector measure. For any x E X, define Sx : C(K) ---+ Y by

Sx(g) T(g . x). =

Then Sx is unconditionally converging. By Proposition 1.7.1 and Theorem 1.7.5, Sx is weakly compact. Let m� be the representing measure of Sx . By Theorem 6 .1.12, m� is a Y-valued vector measure. But for any Borel set A,

m�(A) m(A)(x). =

This implies that m is an C(X, Y)-valued vector measure. If m is not count ably additive, then by the proof (claim 1 ) of Theorem 6.1.12, there are 8 > 0 and a sequence {An}�=l of disjoint measurable subsets such that I l m(An) 1 > 8 for all n E N. Since m is Y*-regular, we may assume that An is closed for each n E N. Then by the proof of Theorem 6.1.5, there is a sequence {fn}�=l of functions with mutually disjoint support such that I l fn lloo S 1, and ilf fdmll � 8/2. Note that {fn }�= l is a uniformly bounded sequence such that for all w E K, {fn(W)}�=l converges to o. By Theorem 1.5.1, {fn}�= l converges to 0 weakly. This contradicts the fact that T is unconditionally converging. The 0 proof is complete. Exercises

Exercise 6.1.1. Let C be a nonempty closed convex subset C of a Banach

space X. For any x in X, Pc ( x ) denotes the set of all best approximants of x from C i.e. Pc ( x ) E C and I l x - Pc ( x ) 1 1 infc Ec I l x - ell . (a) Show that if X is strictly convex, then for any x E X, Pc ( x) contains at most one point. (b) Show that if X is reflexive, then for any x E X, Pc ( x ) is nonempty. Exercise 6.1.2. A mapping T : X ---+ X (not necessarily linear) is said to be nonexpansive if for any X, y E X, I T x - TY I s I l x - Y I . (a) Suppose that C is a subspace of a Hilbert space X. Show that Pc is a (linear) nonexpansive mapping. (b) Suppose that C is a nonempty closed convex subset of a Hilbert space X. Show that Pc is nonexpansive. (Hint: Let X , Y be two elements in X. By translation, we may assume that Pc(x) o . Let M be the subspace spanned by { Pc (y)}. Show that I PM ( X ) - PM (y ) 11 � II Pc(x) - Pc ( y ) I .) Exercise 6.1 .3. Let K be an infinite compact Hausdorff space, and X an infinite-dimensional Banach space. Show that C(K, X) contains a comple mented subspace isomorphic to CQ . (Hint: The Josefson-Nissenzweig theorem) =

=

325

6 . 1 . VECTOR-VALUED CONTINUOUS FUNCTIONS Exercise 6.1 .4. Let {

1/ = 2: %: 1 � I m kl ·

mk } � l be a bounded sequence in M(K, X*) and let

[Xk : k E N] , the closed linear span of {mk : k E N} , is isomorphic to a subspace of Ll (K, 1/, X* , w* ) . (b) Show that if { m d� l is a eo-sequence, then X* contains a eo-sequence. (a) Show that

(Hint: Theorem 5.1 .10.) Thus C(K, X) contains a complemented copy of fl if and only if X contains a complemented copy of fl .

Exercise 6.1.5. Let 1/ be a Borel measure on K, and X a separable Banach space. Let { Fn }�= l be a uniformly bounded sequence in Ll (1/, X*, w*).

(a) Show that there is a sequence (Gn) « (Fn) such that for almost all w E K, either { Gn(w) }�=l converges weak* or for some k E N, { Gn(w) }�= k

is an fl-sequence. (Let X = Y* . Use the technique in the proof of Talagrand ' s Ll(X)-theorem, but replace the weak topology of X by the weak* topology.) (b) Talagrand proved if for almost all w E K, { Gn(w) }�=l is weakly Cauchy, then { Gn }�= l is weakly Cauchy. Suppose that eo is not a quotient space of X. Show that if { Gn }�= l is an fl-sequence, then X* contains a copy of fl . Exercise 6.1 .6. Let Wo be the first limit ordinal. For an infinite set A, let

M

I}

[A] denote the set of all infinite subsets of A. A subset = { Ua : Q E of [A] is said to be if for any Q =j:. /3 , Ua n Uf3 is finite. Show that there is an almost disjoint family = { Ua : Q E of N such that = 2WO • (Hint:

almost disjoint

U

I}

lU I

Consider all convergent sequences of rational numbers whose limit is irrational.) Exercise 6.1 .7. Let

T be a weakly compact operator from fOC) to X. Show

that for any a = ( an ) E fOC) , 2: :=1 an T(e n ) converges.

Exercise 6.1 .8. (Drewnowski) Let T be a weakly compact operator from

fOC) to X and let S : fOC) - X be the operator defined by OC) S(a) = 2: an T(en ) where a = ( a n ) E fOC).

n=l

Let m and me be the representing measures of T and S.

(a) Show that S is weakly compact. Hence both m and me are strongly additive. (b) Let m o = m - me, and {Nt lt E [ o , l ) an almost disjoint family of N. Show that for any € > 0, the set { t : I m l (Nt ) � € } is countable. (c) By (b) , there is t E [0, 1 ] such that I mo l (Nt) = O. Let M be the subset of N such that II mo ll ( M) = O. Show that for any a = ( a n ) E fOC)(M) , T(a) =

anT(en). nL E M

326

CHAPTER 6. CONTINUO US FUNCTION SPACES Exercise 6.1.9. (J. Mendoza ) Show that for any 1 �

p

0; for some 1 � q � p(n), L =l i then there is a v* E A l that satisfies the following conditions: (i) E f=l max{O, (v*, Xl k; ) } > 0; ; (ii) (V*, Xl k ' ) (U*, Xl k ' ) for 1 � i < n; (iii) I ( V*, XlJ I � €n for all i < k1. Let A A l and € € 1. We shall apply Lemma 6.3 .3 p(1) times. First, set 0 ot and [N] [N] and apply Lemma 6.3.3 to obtain a sequence L1 . Replacing ot by Oi, [N] by [LU , and then applying Lemma 6.3.3 , we obtain L�. Continue this way until we obtain L 1 L; ( l ) by Lemma 6.3.3 with 0 O;(1) , N L;( l ) l Let .e1 min Ll and F1 Hl1 · and L m Suppose that we obtain .e 1, , .em , Fi Hl; for 1 � i � { .e 1, .e2 , , .em , sn+ 1, . . . } that satisfy (i)-(iii) for all n � Let L � L m \ { .e 1, , .em }.

Let

=

•

=

=

.

•

=

=

_

'

=

=

=

=

•

•

•

•

=

=

•

•

•

=

m.

•

=

•

.

.

m,

=

CHAPTER 6. CONTINUO US FUNCTION SPACES

336

...

(t i ) mi= l E n mi = l Fi , set Ar +1 { x * E A l : (x *, xeJ t i for all 1 � i � m}. This partitions1 A l into finitely many sets. We apply Lemma 6.3.3 repeatedly with A A r + , E Em + 1, C C;: + 1 , and L beginning with L L'm, rand q range over all possibilities. Let L� + l be the subsequence ultimately obtained and .em + ! min L� + l ' Let Lm + ! L� + l U { .e 1, . . . , .em } and Fm +1 Hetn + 1•

For t

=

=

=

=

=

=

as

=

=

=

=

By induction, we get a sequence {.e i } i= l and a sequence {Fi }� l of finite subset of [ - 1 , 1] that satisfy (i)-(iii) for all n E No Let Yk xe k and Yk Ylk for all =

=

k E N.

Claim 1. We claim that for any z* E A l with E�= I max {O, (z*, Yk.) } > � . M for some finite sequence {kdf= l ' there exists w* E A l that satisfies the following conditions:

Ef= l max{O, (W*,Yki ) } > � . ( v ) I (w*, Yi ) 1 ° if i > kp, ( v i) I (w*, Yi) I < E i if i ¢ {k I , . . . , kp } or (w*, Yi ) < 0 . (iv )

=

Suppose that Claim 1 has been proved. Let y* be any element in the unit ball B ( X*) of X* such that E�= I (y*, YkJ > M. By Claim 1 , there is w* E A l that satisfies (iv)-(vi). Let D { i : i =I kj for any j � p} U { i : (w *, Yi ) < O} and set '\""" ( w * , Yi ) Yi* ' x * w * - L...J =

=

i ED

p I l x* 1 � Il w * 1 + L= Ei � 3, i 1

By (vi), we have and

if i E {k1, . . . , kp} and otherwise.

(W*' Yi) � 0 ,

This implies that

p P P M *, ; (X *, xe > (X*, , X {(w Y O} ) m k L k L= ax k.) 2 ' ) L i= l i= l i l Note: w* E A I . For any i E N, (w*, Xi ) E Hi . Thus (w*, Yi ) (x *, xe. ) E Hei Fi . We have proved the lemma (if Claim 1 is true) . Claim 2. Let k1 < k2 < . . . < kp be a finite sequence of natural number. Suppose that z* is an vector in A l such that E�= l m ax { (z *, YkJ, O} > � . M. We claim that there exist a finite sequence {Co, C1, • • • , Ckp} in [0, 1], a finite sequence { ,Bo, ,B1, " " ,Bkp } in N, and a finite sequence {w(j, wi, . . . , Wkp } of A l that satisfy the following conditions: =

=

=

=

6.3. THE HEREDITARY DUNFORD-PETTIS PROPERTY

337

Wo = z*, Co = �M, and {30 = o. ( v iii) For any 1 ::; r ::; kp, 0 ::; (Cr - l - {3r - d - Cr < fr. (ix) For any 1 ::; r ::; kp, L: { i : k i � r } max {O, (W ; ' Yki ) } > Cr . (x) For any 1 ::; r < kp, if r = ki for some 1 ::; i ::; p, {3r { 0maxi (w;, Yr), O} otherwise. (xi) l (w;, Yr)1 ::; fr for 1 ::; r ::; 8 , provided that either r rf:. {kl, . . . , kp} or (W;, Yr) < 0. Proof of Claim 1 : Assume that Claim 2 has been proved. Let z* be an l element in A such that L: f=l max {O, ( Z* ' Yki ) ) > �M. Applying Claim 2 to z* and the sequence { Yk1 , , Ykp } ' we obtain finite sequences {Co, . . . , Ckp} C JR. , {{30 , . . . , {3kp} � [0, 1] ' and {wo, . . . , wkp } C A l that satisfy conditions (vii)-(xi). Let W* = P:p (wkp ) (where Pn denotes the projection from [Xj : j E N] onto [Xb . . . , xn D . Then the vector W* satisfies conditions (v) and (vi). p max{O, (W * ' Yk J} L =l i kp - l p = L {3ki (= L {3i + {3kp) (by (x)) i =O i=l kp (by (viii)) � L (Ci - 1 - Ci - fr ) + {3k p =l i (observing that {3kp � Ckp by (ix)) i=l 3 1 1M > M = -M -4 4 2 Thus the vector w* satisfies condition (iv). We have proved Claim 1 (if Claim ( v ii)

-

•

•

•

2 is true). Proof of Claim 2: Again the construction is done by induction. Suppose that we obtain finite sequences {Co, . . . , Cs } of [0, 1] and {{30 , . . . , {3s } of N and {wo, wi, . . . , w;} of A l for some 8 < kp. Choose Cs + 1 = C: + I for some 1 ::; q ::; p(8 + 1 ) such that

(Cs - {3s) - Cs + 1 < fs + !. Case 1 . 8 + 1 = kj for some 1 ::; j ::; p and (w;, Ys + I) � o . Set W;+ l = w; and {3s + ! = (W;,Ys + I ). Then the vector w;+ l and the constant Cs + 1 satisfy conditions(viii) - (x). Case 2. (8 + 1 = kj and ( w;, Ys + !) < 0) or kj < 8 + 1 < kj + l for some j ::; p. By our construction (apply it to kj ::; 8 + 1 < kj + ! < kj + 2 < . . . < kp, w;, and Cs + d , there is w;+ l E A l that satisfies the following conditions: o ::;

338


Ef=H 1 max{O, (W;+ 1 , Yk. ) } > C8+ 1. (xiii) (W; + 1 ' Yk. ) = (w;, Yk.) for any j + 1 ::; i ::; p. (xiv ) I ( w;+ 1 ' Y8+ 1 ) I � 1: 8+ 1 ' (xii)

Let

if s + 1 = ki , otherwise. satisfy conditions (viii)-(xi).

Then vector w;+ 1 and the constants {38+ 1, C8+ 1 The proof is complete. The following three theorems are due to J. Elton Theorem 6.3.5. Let

X.

{ xn }�=1

[18].

0

be a normalized weakly null sequence of a satisfying the of There is a subsequence

{ Yn }�=1 {Xn}�=1

Banach space following condition: Let N be a fixed natural number. For any two natural numbers j, m ::; N and a finite sequence the inequality ::; implies I ::; 2 m +i + 1 for any F � i : l a l �

II E�= N aiYi ll 2m -i /6 { i 1/2i }. EiE F aiYi l 1 Proof: We shall use induction to construct the sequence { Yk }k:: 1 ' By passing to a subsequence { xn k }k:: 1 of { xn }�=1 and then renorming the space [xn l l . . . ], we may assume that {Xn }�=1 is a bimonotone normalized basic se quence. Applying the construction in the proof of Lemma 6.3.4 and using the diagonal method, we get a subsequence {yd �=1 of {Xn}�=1 such that if there are y * E B(X*), N ::; k1 < . . . < kp, with Ef=1 (y* , Yk. ) > 2 m for some m ::; N, then there exists x* E 3B(X) satisfying the following conditions: (i) E f=1 (x*, Yki ) � 2; . (ii) (X*' Yi) = 0 if i ¢ {kl, . . . , kp}, or i = ki and (X*, Yi) < O. (iii) (x*, Yi) � 0 for all i E N. Suppose the theorem is not true. Then there are j, m � N, p E N, a finite sequence {kl" ' " kp } of N, and Y = Ef= N a iYki such that 2m --i ' (6.1) 6 but for some subset F � { I a i l � 1/2i and N ::; i ::; p }, (6.2) I I iELF aiYki I > 2m +i + 1 . Inequality (6. 2) implies that there is y * E B(X*) such that iYki ) > 2m+H 1 . a ( Y*, L iE F { ai }f= N '


Let

D+ D_

Then

= =

339

{i E F : (y* , Yk. ) > 0 and a i � O} {i E F : (Y*' Yk. ) < 0 and a i < O}.

y * , aiYk. ) � 2m+j ( L iE D+

y * , aiYk. ) � 2m+j . ( L iE D_ Without loss of generality, we assume that E iED+ ( y * , Y k. ) � 2 m+j . Then *, a iYki ) ( y L iED+ (since 1 � 2j a i for all i E F) � L 1/2 j ( y *, YkJ iED+ either

or

� 2m .

So there is x * E 3B(X*) that satisfies conditions (i)-(iii) , i.e., p 2m � 2'

=N iL

(X*, Yk. ) (x* , Yk. ) � 0 (x*, Yk. ) = 0

if i E D+ , if i ¢ D+ .

This implies that 3

p

p

L (x*, a iYki ) 1 iL=N aiYki II � iL=N (x* , aiYki ) iED+ 1 ( x* , Yki ) � 2m -j- 1 . � 2j iED+ =

"""" L-

This contradicts inequality ( 6 . 1 ) . The proof is complete

o

Let { xn }�=l be a normalized weakly null sequence that does not contains a eo-subsequence. Then there is a subsequence { Yn }�=l of { xn }�=l such that for any subsequence {zn}�=l of { Yn }�=l and any sequence {an}�=l E foo \ eo, we have Theorem 6.3.6.

{ xn } �= 1 and then renorming the sub space generated by the subsequence, we may assume that {Xn}�=l is a bimono Proof: By passing to a subsequence of

tone basic sequence that satisfies the conclusion of Theorem 6.3.5. Suppose the


340

theorem is not true. Then there is (an) E B(loo) \ CO such that .L:�=l anxn E X. So there exist j E N, b with I b l � � and a sequence {n k }k:: l of N such that for any k E N, l a nk - b l � W and l a nk I � 1 /2i . Let M = SUPn .L: �=l akXk and let N be a fixed number such that 6 . �:+i � M. By passing to a further subsequence of {n k } r=l ' we may assume that nl > N. By Theorem 6.3.5, for any finite subsequence {mi g=l of {nk }k:: l ' we have

1

I

for all l E N. By Lemma 6.3.2, there is a subsequence of { xnk }k:: l that is 0 equivalent to the unit vector basis of CO . We get a contradiction. Theorem 6.3.7. (J. Elton) If X has the hereditary Dunford-Pettis property,

then X has property (8) .

Proof: Suppose the theorem is not true. Then there is a Banach space

X that has the hereditary Dunford-Pettis property, but does not have prop

erty (8) . Let { x n}�=l be a normalized weakly null sequence that has no co subsequence. By Theorem 6.3.6 and by passing to a subsequence, we may assume that { xn }�=l is a basic sequence, and we have

n (6.3) sup n l k2:=l akxk l 00 for all scalar sequence {an}�=l rt. Let {X�}�=l be the coefficient functionals, and for each n E N, let Pn be the CO .

=

projection from Y = [x k : k E N] (the closed linear span of {x k : k E N}) onto [X k : 1 � k � n] defined by

n Pn{x) 2:= l (Xk ' X)Xk. k =

Note: For any X** E [X k : k E N] * , X** is just a weak* limit of .L:�=l (x** , Xk)Xk . So we have

n sup l 2: (x ** , xk)x k l � sup II P� * x ** II � sup II Pn l1 ' 1I x ** II < 00 . n k=l n n

Equation ( 6.3 ) implies that limn-+oo (x** , x�) = O. Since x** is an arbitrary element in Y** , we have proved that {X�}�=l is a weakly null sequence. On the other hand, (x�, xn ) 1 . This contradict the fact that every subspace of X has 0 the Dunford-Pettis property. The proof is complete. First, we notice that if { Xn}�=l is an M-bad sequence, then {xn}�=l cannot be a norm null sequence. Hence there is a subsequence of {Xn}�=l that is a =


341

basic sequence. Recall that a collection (X?kn e N � B (X) is called an army in X if { x?}� l E C(X) for all n E N. An array (y ? ) is a subarray of the array (x? ) if there exists (mn) �=l E [N] such that for all n E N, (y f )� l is a subsequence of (x � n )� 1 An array (x?) is a bad army if there exists an increasing sequence {Mn }�=l such that limn-+oo Mn 00 , and (x? )� l is an Mn-bad sequence for all n E N. A bad array (x?) satisfies the army procedure (ARP) if there exist a subarray (yf ) of (x?) and a sequence {a n }�=l of positive real numbers with :E�=1 an � 1 such that if Yi :E�= 1 anyf, then for any subsequence {Z i } � 1 of '

=

{Yi }� l ' {Zi }� l ¢ C(X).

=

We say that X satisfies the ARP if every bad array in X satisfies the ARP. Suppose that X satisfies the ARP. Assume that X does not have property ( U S ) . Then X contains a bad array (x? ). By passing to a further subarray, we may assume that { x? : i, n E N} is a basic sequence in the lexicographical order. (Thus the order is x �, x ?, x�, XI, x �, x§, . . ) By ARP, we get a seminormalized weakly null sequence { Yi }� l such that any subsequence {zn}�=l of {yn }�=l does not belong in C(X) (i.e. any subsequence of {zn}�=l cannot be equivalent to the unit vector basis of eo). Hence if X satisfies the ARP, then X has property (U S) if and only if X has property (S). .

.

Lemma 6.3.8. Let X be a Banach space and ( Xn) a sequence of Banach spaces such that Xn satisfies the ARP for all n E N. Let (X?) i,n be a bad army in X, Y [x? : i , n E N] , and {mn}�=l an increasing sequence such that limn -+oo mn = Suppose that for any m E N, there is a norm one opemtor Tm Y Xm such that {TmXi}� l is an m -bad sequence in Xm . Then (X?) i,n satisfies the ARP. Proof: Let (X ? ) i , n be a bad array in X. Fix m. Suppose that there is a bad subarray of (Tm (X?))n , i in Xm . Then there are a further subarray (Tm (yf)kn of (Tm (xf ))n ,i and real numbers an > 0 with :E�1 an � 1 such that if =

:

--+

00 .

00

Wi L= l anTm (yf ), n then no subsequence of {wd � l belongs to C ( Xm ) =

null sequence.) Let

00

'

(Clearly, {Wd� l is a weakly

Zi = L=l any? n Then any subsequence of {Z i }� l does not belong to C(X). Hence we may assume that for any m E N, (Tm (x? )) does not contain any bad subarray in Xm . By passing to a further subarray of (X? )n , i , we may assume that for any m E N, there is Mm such that for any finite subset F of N and n E N, we have I ieLF Tm (x? ) 1 I � Mm .

342


Note that for all n E N, that

{ xi}� 1 E C(X). foe each n E N, there is Nn E N such i xi l i � Nn IL =1 i for any f E N and f E N. By induction, we shall select sequences {mn }�=1 of N and a sequence {an }�=1 of positive real numbers such that a1 = �, an + ! � � for all n E N, and n- 1 anmn > max { n, 4 L= 1 aj Nmi , 4Mmn =L 1 aj } , j j n+ for all n E N. Suppose that the finite sequences {at, . . . , a n } and {mI, . . . , mn - d are con structed. Select mn > mn - 1 such that n- 1 anmn > max { n, 4 L=1 aj Nmj } . j Then choose a bn such that 0 4Mm n b n. Finally, let an + 1 = bn/2 . We claim that if Yk = 2:}: 1 ajxr;:i , then the sequence { Yk }k:: 1 has no sub sequence in C(X) . Note: for all n E N, {Tm n (x ;n n )}� 1 is an mn -bad sequence. For each n E N, there is in E N such that I 2: �: 1 Tm n (x;;; n ) I > mn. Then for any n E N, we have 00

in in n - 1 j � I L L ajx� l - 1 L L aj x �i l i=1 j =1 i=1 j =n n- 1 in j in j � I Tm n ( L L ajx� ) l - L l I aj L x;: l ( 1 Tml l � 1) i= 1 j =n j =1 i= 1 n- 1 in in j n � an I L Tm n (x;;; ) l I - L aj l L Tm n (x � ) l I - L aj Nm; i=1 j =n+ 1 i=1 j =1 anmn >- an mn - L a ·Mmn - -4 j =n+ 1 00

00

00

00

J


n an n S Up L Yki II � sup : = l nEN n E N i= l

This implies

343

00 .

0 The proof is complete. Let Wo denote the first limit ordinal. For any ordinal a, let eo ([I , a] , X) denote the Banach space eo ([I , aj , X) = ((x-y ) E C([l , a] , X) : Xo = O} with the norm I (X-y ) 11 00 = sup I I x-yl l ·

-y< o

{Xn}�= l be a sequence of Banach spaces satisfying the ARP . Then (EBXn)co also satisfies the ARP. In particular, if K is a compact countable metric space, then C ( K ) satisfies the ARP . Proof: Let {Xn}�=l be a sequence of Banach space that satisfy the ARP and let Rm be the natural projection from X (2:: EBXn) Co onto Xm . We claim that for any (M + 3)-bad sequence { Yn }�=l in the unit ball B( EBXn)co) of (EBXn)co, there is m E N such that {Rm ( Yn)}�=l contains an M-bad subse quence. Suppose the claim has been proved. Let (xf) be a bad eo-array in X. By passing to a subarray of (xf) , we may assume that there is an increasing sequence {N(n)}�=l of natural numbers such that {RN (n) (xf)}� l is an n-bad sequence for all n E N. By Lemma 6.3.8, (xf ) satisfies the ARP. This proves the first Lemma 6.3.9. Let

=

assertion of the lemma (if the claim is true) . Proof of claim: Suppose the claim is false. By a gliding hump argument, there are a subsequence (Zi ) � l of (Yi) � l and an increasing sequence {md� l of natural numbers which satisfy the following conditions: (i) For any j E N, I l xj l l oo � 1 and sUPm > mj II Rmzj llx � 2 - j • (ii) For any two natural numbers p � j, s UP m � mi I L: f=l Rm zi l lxm � M . Fix p E N. Select m E (mi - l, m i l (depending on p) for some i E N (mo = 0) such that ...

Then

�

i -j L =j l 2

+ 1

+ M < M + 3.

344


This implies that (Zi)� 1 is not an (M + 3)-bad sequence, a contradiction. We have proved our claim. It is known that for any countable limit ordinal a, and any increasing se quence { ,8n }�=o with ,8n i a (,80 = 0) , C([ I , a] ) '" eo ([ I , aD '"

<Xl

C([ , n I ]) (I:: n=1 EB I ,8 - ,8n ) -

�

.

By induction and the argument in the above proof, one can show that C([ I , a] ) satisfies the ARP for all countable ordinals a. It is known that if K is a compact countable metric space, then there is countable ordinal a such that K is home omorphic to [0 , a] for some a < WI (the Mazurkiewicz-Sierpinski theorem). We have proved that if K is a compact countable metric space, then C(K) has the 0 ARP. The proof is complete.

{Xi }� 1 be an M-bad sequence that is a monotone basic sequence. Then there exist a subsequence { Yn }�=1 of {Xn}�=1 and a w* compact countable subset K such that { Yn I K }�=1 is an Ml6-bad sequence in . Proposition 6.3.10. Let

C(K)

{Xi }� 1 be an M-bad monotone basic sequence. By Lemma 6.3.4. there are a sequence {fi }� 1 of natural numbers and a sequence {Fi }� 1 of finite subsets of [ - 1, 1] such that 0 E Fi for all i E N, and for any y * E X* and any finite increasing sequence {ki } f=1 of natural numbers, Ef=1 (y * , Xl i ) > M implies there is x* E 3B(X*) which satisfies the following conditions: (i) E f=1 (x* , Xl i ) � Pf · (ii) (x* , Xl"., ) E Fk i and (x * , Xl ". ) � 0 for all i E N. (iii) (x* , Xl.) = 0 if i rt {kl, . . . , kp} or if i = kj and (x* , XlJ < O. Replacing Xi by Xl i if necessary, we may assume that fi = i for all i E N. Let Y = [xn : n E N] be the subspace spanned by { xn : n E N} , and for each m E N, let Q m be the natural projection from Y onto [X i : 1 � i � m] . Then I Q mll � 1. Let p /C = { (kl, . . . , kp) : II I:: X ki II � M for all r M } , i =1 . i=1 Proof: Let

"

"

.

r

and Then K is a weak*-closed countable subset of B(Y* ) . Fix kn = ( kl, . . . , kp) E /C . Since II E� I X ki II > M, there is a unit vector y * E X* such that Ef=1 (y* , X k. ) > M. So there is x* E 3B(X) such that

345


(iv) l: f= 1 (X * , Xk J � !If ; (v) (X* , XkJ E Fki and (x * , XkJ � 0 for all i E N; (vi) (X* , Xi ) = 0 if i rI: {kb . . . , kp} or if i = kj and (X* , Xi < O. Let z* = x * / 3. Then z* E K. By (iv), the sequence {xi I K } : 1 is an � -bad 0 sequence. The proof is complete. Theorem 6.3. 1 1 . (H. Knaust and E. Odell) Every Banach space satisfies the ARP. Hence every Banach space with property (S) X has property (US) . Proof: Let (xi) be a bad eo-array in X. By passing to a subarray if necessary, we may assume that for n E N, (xi) � 1 is an Mn -bad eo-sequence with Mn 6n, and { xi : i, n E N } is a basic sequence (in the lexicographical order) . Let Y = [xi : i, n E N] . Renorming Y if necessary, we may assume that { xi : i, n E N} is a bimonotone basic sequence. By Lemma 6.3.10, for each n E N, there are a subsequence (Yi )� 1 of (xi)� 1 and a sequence {Kn }�= 1 of w*-compact countable subsets of B(Y* ) such that for each n E N, (Yi I Kn )� 1 is an n-bad eo-sequence. By Lemma 6.3.9, C(Kn) satisfies the ARP. Define Tn Y � C ( Kn) by Tn(x) X I Kn' By Lemma 6.3.8, (Yi) satisfies the ARP; thus (xi) satisfies the ARP. The proof >

:

=

0

is complete.

Lemma 6.3.12. (Cembranos [9]) Suppose that X is a Banach space that has the hereditary Dunford-Pettis property. Then eo(X) has the hereditary Dunford Pettis property. Proof: Let { (xi)� d�= 1 be a normalized weakly null sequence in eo(X). Hence for any j E N, { xj}�= 1 is a weakly null sequence. Since X has property (U S ) , by passing to further subsequence and applying the diagonal method, there exist a strictly increasing sequence { mn } �= 1 ( mo = 0) and M 0 satisfy the following conditions: (i) For any j E N and k � mj , I l xL l � 21] . (ii) For any mj - l < k � mj and (an) E eo, 00 II �= alxf ll x � M I (an) l oo. t. 1 Hence for any (an) E eo and mj - l < k � mj , j-l 00 a a x x + � II L: t. f ll x II L: alxf ll x � (M + 2) I (an) 1 00 . ll IIL: f t. x 1 � �1 �j >

00

346


We have proved that 00

at (xf} I co(X) � (M + 2) II (an ) 1 1 00 ' ilL = 1 t

This implies that every normalized weakly null sequence of Co ( X ) contains a D eo-subsequence. The proof is complete. Before proving the main theorem in this section, we need the following lemma. Theorem 6.3.13. (Pelczynski and Szlenk [14 , Theorem 5] ) Let K be a com pact Hausdorff space. Then C(K) has the hereditary Dunford-Pettis property if and only if K is dispersed and the w-derived set of K is empty. Proof: Necessary conditions.

(a) Suppose that K is not dispersed. Then C( K ) contains a copy of C([O, 1]), and C(K) does not have the hereditary Dunford-Pettis property. (b) First, we notice that CO has the weakly Banach-Saks property. Thus if X has the hereditary Dunford-Pettis property, then X has the weak Banach Saks property. Suppose that the w-derived set of K is not empty. By Theorem 1 .5.13, there is a quotient mapping from K onto [l , wWl , where w is the first limit ordinal. Since C([l , W W] ) does not have the weak Banach-Saks property (Theorem 2.4.2), C(K) does not have the hereditary Dunford-Pettis property. Sufficient conditions: Let K be a dispersed set such that the w-derived set of K is empty. Suppose that C(K) does not have the hereditary Dunford-Pettis property. Then there is a subspace Y of C(K) that does not have the Dunford Pettis property. So there are a weak null sequence {gn}�= 1 in Y and a weak null sequence {On}�= 1 in Y'" such that (On' gn) 1 . Without loss of generality, we may assume that {gn : n E N} spans Y. Define a relation on K by t t' if gn(t) gn(t') for all n E N . Then there exist a metrizable quotient space f< of K and a sequence (9n) � C (K ) such that 9n(7r(t)) gn(t) for all t E K and n E N, where 7r : K K is the quotient map. Let Q denote the mapping from Y to C (K) defined by Q g(7r(t)) g (t). It is known that Q is an isometry from Y into C(K). Since K is a dispersed compact metric space and the w-derived set of K is empty, by the Mazurkiewicz-Sierpinski theorem, K, is homeomorphic to [1 , w"" . ml for some m, ,,( < w. By Exercise 1.5.1 (b) , C(K) is isomorphic to CO or .e� for some n E N . This implies that Q(Y) and Y have the Dunford-Pettis property, D a contradiction. The proof is complete. =

=

f'J

=

---..

=

Let K be a compact Hausdorff space, and X a Banach. Then C(K, X) has the hereditary Dunford-Pettis property if and only if both C(K) and X have the hereditary Dunford-Pettis property. Theorem 6.3. 14.


347

Proof: Since X (respectively, C(K)) is isomorphic to a subspace of C(K, X), one direction is clear. Assume that both C(K) and X have the hereditary Dunford-Pettis property. By Theorem 6.3.13, K is dispersed and and the w-derived set of K is empty. Let {fn}�=l be a normalized weakly null sequence in C(K, X). Define a relation on K by t t' if fn ( t ) = fn ( t' ) for all n E N. Then there exist a metrizable quotient space K of K and a sequence ( in) � C(K, X) such that rv

in(1r(t )) = fn(t)

for all t E K and n E N,

where 1r : K K is the quotient map. Since K is a dispersed compact metric space and the w-derived set of K is empty, by the Mazurkiewicz-Sierpinski theorem, K is homeomorphic to [ 1 , w' . m] for some m, ,,( < w. By Exercise 1 .5.1 (b), C ( K , X ) is isomorphic to co (X) or t'�(X ) for some n E N. By Lemma 6.3. 12, C(K , X) has the hereditary Dunford-Pettis property. Since the embedding is an isomorphism, {in}�= l is also a normalized weakly null sequence. By the claim, {in }�=l contains a co-subsequence. Thus {fn}�=l 0 contains a co-subsequence. The proof is complete. ---+

Exercises

X be a Banach space. Show that X have property ( ) if X has property (8) . (Hint: Let {Xn}�=l be a weakly Cauchy sequence. Use Ramsey ' s theorem to show that there is N E [N] such that for any M = (m k ) E [N] , there is K > a such that 11 2:�=1 €i ( Xm2i - Xm2i+1 ) 11 � K for all €i = ± 1 and all n E N.) Exercise 6.3.2. Show that if X has the hereditary Dunford-Pettis property, then t'1 ( X ) has the hereditary Dunford-Pettis property . Exercise 6.3.3. (Bourgain) Let X be a Banach space that is the closure of the union of an increasing sequence {Xn }�=l of subspaces of X. Assume that (2: �=1 ffiXn) oo has the Dunford-Pettis property. Show that X has the Dunford-Pettis property. Assume that X does not have the Dunford-Pettis property. Let {Xn}�= l and {X�}�=l be weakly null sequences in X and X* such that { (x�, Xn) }�=l does not converge to O. By passing to a further subsequence, we may assume that {Xn}�=l is a basic sequence. (a) There exists {n k }k':: l such that Xn/c contains Xj for all j � k. Note: (2:%: 1 ffiXn/c) is complemented in (E�=l ffiXn)oo ' We may assume that Xn/c = Xk• (b) For each n E N, let in be the injection from Xn to X, and Pn the projection from ( 2: �=1 ffiXn) oo to Xn. Fix a free ultrafilter U on N. Let Exercise 6.3. 1. Let

00

u


348

if n < i, if n � ij

' m Pn�n * . * ( X*i ) . 1}i = * - nl1--+U Show that the sequence {(1}i , e i ) } � 1 does not converge to O. (c) Show that both { e i }� I ' and {1}d� 1 are weakly null. ( Hint: { xn }�= 1 is a weakly null sequence. By Mazur ' s theorem, for any K > 0 and € > 0, there is a finite nonnegative sequence {a i H=1 such that 2: : 1 a i = 1 and 1 2:�=l aixK+ i ll < €. Show that 1 2: �= l ai 1}K+i ll < 2K€, where K the basis constant of { xn : n E N} .) W

6.4

Projective Tensor Products

It is known that for any Banach space X, C(K, X) = C ( K ) ® X and L1(X) = L1 ®X (Exercise 1 .8.2 and Proposition 1.8.6) . M. Talagrand showed that there is a Banach space X such that X* has the Schur property, but neither C(K, X) nor L1 (X*) has the Dunford-Pettis property. In the first part of this section, we present some results of F. Bombal and 1. Villanueva about the Dunford-Pettis property of the tensor product spaces. First, we need the following lemma. Lemma 6.4. 1 . [15, Corollary 4.16] Suppose that X contains a copy of £1 . Then £2 is a quotient space of X . Proof: Let q denote the quotient mapping from £1 to £2 and let J denote the natural canonical mapping from Loo to L2. It is known that every operator from £ 1 to £2 is 2-integral ( see Exercise 1.8.6). There are operators S : £1 ---+ L oo and T : L2 ---+ £2 such that q = T o J 0 S . Note: L oo is an injective space ( see Exercise 1.6.2) . There is an extension S : X ---+ Loo of S, ( Le. , Sit l = S) . This implies that T o J 0 S is a ( 2-integral) quotient mapping from X to £2 ' The 0 proof is complete. By the above lemma, if X contains a copy of £1 , then X * contains a copy of £2. We have the following corollary:

Let X be a Banach space . If X * has the Schur property, then X does not contain a copy of £1. Lemma 6.4.3. [35] Let X and Y be two Banach spaces such that both X* and Y* have the Schur property. Then (X®Y)* has the Schur property. Proof: Let {Tn }�=l be a weakly null sequence in C(X, Y*) = (X®Y)*. For any X E X and y ** E Y**, the mapping T f-7 (y**, T(x )) Corollary 6.4.2.

6.4. PROJETIVE TENSOR PRODUCTS

349

is a bounded linear function on C(X, Y* ) . Thus for any x E X and any y ** E Y** , we have limn�oo ( Y ** , Tn(x)) = O. But y* has the Schur property. This implies that limn�oo Tn(x) = 0 for all x E X . For any x** E X** and y ** E Y** , the mapping

T (x **, T*(y**)) is also a bounded linear function on C(X, Y* ) . Note: X* has the Schur property. We have limn�oo T:: (y**) = 0 for all y** E Y** . We claim that {Tn }�= l converges to 0 in norm. Suppose not. Then there are > 0 and a sequence {xn}�= l in B(X) such that for all n E N, I Tn(xn) 1 I > lf {Tn(xn) E Y}�= l converges weakly to 0, then {Tn( xn) }�= 1 converges to 0 in norm, a contradiction. Hence by Fact 1 . 1 . 1 and by passing to a subspace of {xn }�= l we may assume that there is 6 > 0 and y ** E y** such that for all �

E

E.

n E N,

'

This implies that

(T�(y ** ), xn) > 6,

a contradiction (since {T:: ( Y ** )}�= l converges to 0 in norm) . The proof is Ll complete. By Theorem 1.6.6 and Corollary 6.4.2, we have the following corollary: Corollary 6.4.4. [3 5] Suppose that both X and Y have the Dunford-Pettis property and contain no copy of £1 . Then X ®Y has the Dunford-Pettis property and contains no copy of £1 . Let X and Y be two Banach spaces and Cw* (X* , Y) the set of all weak* to weakly continuous operators. By the argument in the proof of Lemma 6.4.3, Cw* (X* , Y) has the Schur property if both X and Y have the Schur property. Note: X®Y is isometrically isomorphic to a subspace of Cw* (X* , Y) . We have the following theorem: Theorem 6.4.5. [28] Let X and Y be two Banach spaces . If both X and Y have the Schur property, then X ®Y has the Schur property. Recall that an operator T : X Y is completely continuous if T maps weakly Cauchy sequences to Cauchy sequences. The following lemma and theorem are due to F. Bombal and 1. Villanueva [5] . Lemma 6.4.6. Let X, Y be two Banach spaces such that every opemtor from X to y* is completely continuous. For any weakly null sequence {xn }�= l in X and any bounded sequence {yn }�= l in Y, { xn 0 yn }�= l is a weakly null sequence in X ®Y . ---+

be a linear functional on X ®Y. By Theorem 1 .8.2, there is 1 . The following facts are known: -t

6. 4.

351

PROJETIVE TENSOR PRODUCTS

(1) Let q = p': l ' The dual of any .cp-space is an .cq-space. (2) Any operator from an .coo-space to an .c1-space is absolutely 2-summing

[27, Theorem 4.3] .

(3) (Lewis and Stagall [26]) Let X be an infinite-dimensional separable .coo-

space. Then X* is isomorphic to either £1 or ( e[O, 1] ) * . Moreover, X* is isomorphic to £1 if and only if X does not contain a copy of £1 .

Since every .c oo -space has the Dunford-Pettis property, every operator from an .c oo -space to an .c1 space is compact. We have the following corollary.

Suppose X and Y are two .coo -spaces. If X does not have the Schur property and Y contains a copy of £1 , then X0Y does not have the Dunford-Pettis property. Corollary 6.4.8.

In the second part of this section, we consider the following question.

Let X and Y be two Banach spaces . Does the projective tensor product X 0Y have the RNP if both X and Y have the RNP ? Problem 6.4.9.

Bourgain and Pisier [7] showed that there is a Banach space X with the RNP such that the projective tensor product X0X contains a copy of Co. Thus the answer to the above problem is negative. In the second part of this section, we provide some positive solutions to the above problem.

Let E be a separable reflexive Kothe function space over a finite measurable space, and X a Banach space with the RNP. Then the projective tensor product E0X has the RNP . Theorem 6.4.10. (Q . Bu and J. Diestel)

Let X be a Banach space, and E a separable Kothe function space with the RNP. Define the mapping u : E0X � E(X) by U

00

00

9i ® Xi ) = 2: 9i ' X i · (2: =l n =l i

Then u is a one-to-one operator such that Il u ll = 1. Let us review some facts about the RNP. ( a) The RNP is a separably determined property (Le. , a Banach space X has the RNP if every separable subspace of X has the RNP). (b) (Proposition 1 .8.7) For any separable subspace Z � E0X, there is a

separable subspace Xl of X such that Z � E0X1 and for any u E E0X,

( c ) (Theorem 3.6.17) The Kothe-Bochner function space E(X) has the RNP if both E and X have the RNP.

352


(d) ( Theorem 3.6.16) Let X be a separable Banach space. Suppose that there exists a semiembedding T from X to Y and Y has the RNP. Then the space X has the RNP. Let (n, J.L) be a finite measure space and X a Banach space. Recall that a function f from n to X is said to be weakly measurable if for any x* E X* , the function w 1--+ ( x* , f (w ) ) is measurable. Suppose that E is an order continuous Kothe function space over (n, J.L). Then the dual of E is also a Kothe function space over (n, J.L). A weakly measurable X*-valued function F is said to be weakly E*-integrable if for any x** E X** , (x** , F ( · ) ) E E* . E* ( X* , weak) is the set of all weakly E*-integrable functions with the seminorm II F II E* CX o ,weak)

=

sup

x O O E B (X o o )

I (x** , F( · ) ) 1 EO .

It is easy to see that the seminorm of E* ( X* , weak) is weaker than the semi norm of E* ( X* , w* ) defined in Section 3.2. It is easy to see that for any F E E* ( �* , w* ) , we have (6.4)

A strongly measurable X-valued function E -integrable if sup

{ in l (F (W ) , f (

w

) ) 1 dJ.L(w)

:F

f :

n

�

X is said to be

strongly

}

E B ( E * ( X * , weak)) < 00 .

The Banach space E ( X, strong) is the set of all strongly E-integrable X-valued functions with the norm I l f II ECX,strong)

=

sUP

{ in l ( F (W ) , f (w)) 1 dJ.L(w) : F E B ( E* ( X * , weak)) } .

By (6.4), for any f E B ( E ( X, strong )) , II f I l E ( x ) � 1 . The proof of the following two propositions are due to Q. Bu and J. Diestel. Proposition 6.4. 1 1 . Let E be an order continuous reflexive Kothe function space and X a Banach space . Let h be the mapping from E®X to E ( X, strong)

defined by

00

11 (L: gn ® xn )

The h is an isometry.

n =l

=

00

L: xn gn ·

n =l

Proof: It is known that for any two Banach spaces X, Y, the natural in clusion map I of X ®Y to X ® y** is an isometry [16, p.258, Corollary 14] . Hence we need to show only that I h I � 1 and that there is an operator 12 : E ( X, strong) � E ® X** such that 11 12 11 � 1 and 12 0 h (u) = l(u) for all u E E® X.

353

6.4. PROJETIVE TENSOR PRODUCTS

Let u be an element in E®X, and E any positive real number. By Lemma 1 .8.5, there are sequences {gn }�=l in E and { Xn }�=l in X such that u L�=l gn 0 Xn , I lxn ll 1 for all n E N, and L�=l Il gn llE ::; l I u liA + E . Let F be any unit vector in E* (X* , weak). Then =

=

::; 1n nf:=l I (F(w), gn(w)xn) 1 d/l(w) � in I (F(W) , 9n (W)xn) 1 dl'(w) ::; nL=l I (F( . ), Xn ) l i E II gn il E

=

00

.

•

L 00

II gn llE ::; l I u ll E®X + E . n =l Since E is an arbitrary positive real number, this implies that II U Il E (X , strong) ::; lI u Il E®x ' We have proved that 11 11 11 ::; 1. Let K be the set (B(E) , weak) x ( B ( X **), w*), and let J be the operator from E* (X* , weak) to C ( K ) be defined by :s;

J(F)((g, x* * )) In (x **, F(w))g(w) d/l(w) for all (g, x**) E K. Then for any F E E* (X* , weak) , sup sup ( g(w)(x ** , F(w)) d/l (w) I J(F) I I x··EB(X··) gEB(E) in sup I (x ** , F(·)) 1 E· II F IIE· (X· , weak) · X" EB(X ) =

00

=

=

=

••

Fix an element f in E(X, strong) , and then define a linear functional J(E* (X* , weak) ) by

( J(F) ) In (F(w), f(w)) d/l(w). ef ,

el

on

=

Then el has norm II f I l E ( X , strong) ' By the Hahn-Banach theorem and the Riesz representation theorem, there is a regular Borel measure A I on K such that I A/ I = II f Il E (X , strong) and for any F E E* (X* , weak), we have

In (F(w), f(w)) d/l (w ) i J(F ) dAf(9, x** ) =

=

i In (x**, F(w))g(w) d/l(w) dAf(9, x** ).

354


Let G : K ---+ B(E) and H : K ---+ B(X** ) be the functions defined by G((g, x ** )) = 9 and H((g, x ** )) = x ** . Then both G and H are continuous, and II G lloo � 1 , II H lloo � 1. Hence, for any n E N, there are a finite sequence { In,k }�l C B (E) and a measurable partition {An , d �1 of K such that

Let and

m"

Un = L In,k ® X��k ' k =l

Then

mn

mn

lI un ll E®X � L Il /n,k ll ' ll x��k ll � L II X��k ll � I A/ I = 1I / IIE (x , strong) ' (6.5) k =l k =l Note: If An,j n Al I i is nonempty, then II X��j - Xl,i I � l i n + 11f . The argument in the above proof also shows that II Un - ul ll E®x � ( l i n + 1/f) I A I · So {un} �= 1 is a convergent sequence. Let UI = limn oo Un and define an operator 12 from E(X, strong) to E® X** by 12 (f) = ul ' By (6.5), we have 11 12 11 � 1 . Note: For any 1 = g ® x E E ® X , lIIl ( f ) is the point charge at (g , x ). So I2 (f ) = 9 ® x. This implies that for any U E E®X, 12 0 I1 (u) = I (u ) . The 0 proof is complete. -+

Proposition 6.4. 12. Let E be an order continuous Kothe function space over a finite measure space (0, J.t) , and X a Banach space. Then the natural embedding from E(X, strong) to E(X) is a semiembedding.

Proof: Let I be the natural embedding from E(X , strong) to E(X). Clearly, I is one-to-one and 11 1 11 � 1 . So we need to show only that T(B(E(X, strong)))

is closed in E(X), i.e., T is a semiembedding. Let { In }�=l be a sequence in the unit ball B(E(X , strong)) of E(X , strong) such that { In }�=l converges to I in E(X) . We claim that I E B(E(X, strong)). By passing to a subsequence { In }�=l ' we may assume that {In}�=l con verges to I a.e. Let F be any unit vector in E* (X* , weak). Since In E E(X, strong) for all n E N,

In I (F(w ) , In(w ) } 1 dJ.t(w) � 1 .

6.5.

355

NOTES AND REMARKS

Note: For almost all w E 0, (F(w), f(w)) = nlimoo (F(w), fn(w)) . By Fatou's lemma, we have -+

10 I (F(w), f(w)) 1 d/-t (w)

:5

1.

This implies that T(!) E B(E(X)). The proof is complete.

o

Remark 6.4. 13. Let X and Y be two Banach spaces such that every weakly compact operator from X into y* is compact. G. Emmanuele and W. Hensgen [20] proved that the completed projective tensor product X®Y of two Banach spaces has property (V) if both X and Y have property (V). They also showed if X* or Y has the MAP, then this condition is also necessary.

Exercises

Exercise 6.4. 1. Let X and Y be two infinite-dimensional C oo -spaces. Show

that if X®Y has the Dunford-Pettis property, then either X and Y have the Schur property or X* and y* have the Schur property. Exercise 6.4.2. Let X and Y be two Banach spaces such that X has no complemented i1-sequence, but Y has a eo-sequence {Yn}�= l ' ( a) Show that for any bounded sequence { Xn }�=l in X, {Xn ® Yn }�=l is a weakly null sequence in X®Y. (b) Show X®Y does not have the Dunford-Pettis property.

6.5

Notes and Remarks

Let 0 be a compact Hausdorff space. First, we would like to extend Ta lagrand's L1 (X)-theorem to the space M(O, X*) = ( C ( O, X))* . The author thank to Professor Talagrand for the explanation of his proof.

Theorem 6.5. 1 . [42, Theorem 13 and Theorem 14] Let X be a Banach space and 0 a compact Hausdorff space . Let l/ be a measure on K and {Fk}k:: 1 a uniformly bounded sequence in L(O, l/, X* , w*) � M(O, X*). There exist a sequence (Gn) « (Fn) and two disjoint measurable subsets C and L of 0 such that l/ ( C U L) 1 and (1) for w E C, the sequence { Gn(W)}�= l is weakly Cauchy in X* ; (2) for any w E L , there is k E N such that {Gn(w ) }n�k is an i1-sequence. =

356


Sketch Proof: Without loss of generality, we assume that II Fk(w) l l x* ::; 1 for all k E N and w E n. For each w E n, let Yw be a Banach space that is isometrically isomorphic to f 1 , and let {ew ,n }�= 1 be the unit vector basis of Yw ' (If there is no confusion, we will write en for ew ,n.) Fix w E n and define a linear operator Tw from Yw to Zw = [Fn (w) : n E N] by

Since T;; is w*-w* continuous (with II T II ::; 1), T;; (B(Zw)) is a weak* com pact subset of B(Yw) = B(foo) , the unit ball of foo . We shall denote the set T;; (B(Z�)) by Ko(w). We claim that the set-valued mapping Ko : w 1--+ 2 B ( l oo ) is measurable, where 2 A is the power set of A. Let D be the collection of all weak* open subset 0 of B(foo ) of the form

where O k is an open subset of [ - 1 , 1] and there is n E N such that O k = [ - 1 , 1] for k ;::: n. It is known that Ko is measurable if and only if for any 0 E D , the set {w : Ko (w) n 0 =1= 0} is measurable. Note: X* is weak* dense in X*** , and for any k E N, the Fk is weak* measurable. The set {w : Ko(w) n O =l= 0} = {w : there is x *** E B ( X *** ) , (x *** , Fk (W) ) E Ok for all k ::; n } = {w : there is x * E B ( X * ) , (Fk (w), x * ) E O k for all k ::; n}

is measurable. We have proved our claim. Let be the set of all weak* compact subsets of B(foo ) , and let {On }�= 1 be a countable basis of B (foo ) with the weak* topology such that 0 1 = B(foo) , and Bn =1= 0 for all n E N . Fix a sequence g , 9 = (gn = 2:;;:1 ak,nik) « f, and a measurable set-valued map Ka. from n to Let {e: ,n } be the biorthogonal functions associated with the the basis {ew ,n : n E N}. For any t = (tn) E B(f oo ) , define a function t from n to Uw e oY; by

K

K.

00

t(w) = t w = L tne: ,n E Y; . n= 1

Fix an element t = (tn) in B (foo) . Let Ct be the set defined by 00

1 Ct = {w : tw E Ka. (w)} = n { w : B (t - w, -) n Ka. (w) =1= 0 } P p= 2

6.5.

357

NOTES AND REMARKS

is a measurable set. Let (t, gn (-)) be the function from Ct to [ - 1 , 1] defined by (t, gn (w ») = (tw , gn (w »)

=

mn

L tnak,n '

k= 1

Then (t, gn( ' » ) is a measurable function. For any k E N, let Ck, Ot. be set Since KOt. is a measurable set-valued function, the set Ck ,Ot. is measurable. Define the functions gn,k, Ot. and 9n,k, Ot. from 0 to [-1 , 1] by { (t, gn (w » ) : t E O k { sup 1 - k, Ot. (W) - { inf { (t, gn(W » ) : t E O k gn, 1

gn, k, Ot. (w) _

_

_

n n

KOt.(w) } KOt.(w) }

if w E Ck, Ot. , otherwise; if w E Ck , 01. , otherwise.

Applying the six steps in the proof of Talagrand ' s theorem to h� = In and Ko(w) = Tw ( (B(Z; » ) , we obtain a sequence 9 « I of functions, and two disjoint measurable subsets C, L of 0 that satisfy the following conditions: ( a) C u L = O . (b) For almost all w E C and t E Ko(w) , { (t, gn (w » ) }�= 1 is Cauchy.

( c ) For almost all W E L, there is k such that {gn (w)}�= k is an iI-sequence

in the space Xw with the seminorm 1 · 1 defined by Ixl

= sup{(t, x)

: t E Ko(w)}.

Suppose that gn is the form l:�:::'t ak,n ik (from the six steps in the proof of Talagrand ' s LI (X)-theorem). Then we set mn

Gn = L ak,n Fk . k= 1 Then G = (Gn ) « F. Fix an element x** in B(X** ) and for each w E O, let S k = s(X ** , W ) k = (x ** , Fk (w » ) . Then s (x** , w) = (S k ) is an element in Ko(w) and II gn(w) 1 I = sup{ (X ** , gn (w » ) : x ** E B(X** )} = sup{ (s(x ** , w) , Gn (w » ) : x ** E B(X** )} = sup{ (tw , Gn (w » ) : tw E Ko(w)}. By (b) and (c) , we obtain the following conclusion:

358


•

For almost all w E C, the sequence {Gn (w)} is weakly Cauchy.

•

for almost all w E L, there is k such that {Gn (w)}�= k is an f1-sequence. The proof is complete.

o

Let X be a Banach space and n a compact Hausdorff space. We shall show that if the dual X* of X is weakly sequentially complete, then M(n, X*), the dual of C(n, X), is weakly sequentially. Before proving it, we need the following theorem which is due to M. Talagrand [42, Theorem 15] : Theorem 6.5.2. Let n be a compact Hausdorff space, (0" A ) be a probability space, X a Banach space, and {mn }�= 1 a sequence of M(n, X*) such that for all n E N, I mn I < A. Then for each n E N, mn can be considered as an element in L1 (A, X* , w*). ( 1 ) If for almost all w E n, {mn (w)}�= 1 is weakly Cauchy, then {m n }�= 1 is weakly Cauchy. (2) If for almost all W E K, {m n (w)}�= 1 converges weakly, then {m n }�= 1 converges weakly. Proof: Let m' be a uniformly bounded function in L1 (A, X* , w* ) and C the

set of all functionals ¢ in M(K, X*) such that n ¢(m) = L (x � * , m(Ai )) , i= 1

where {Ai }i:: 1 is a finite measurable partition of 0, and {xi* } i= 1 a bounded sequence in X** . It is easy to see that ¢ has norm at most maxi $ n II xi* II . Let C1 be the set Note that C1 is convex, and II m ll

=

sup{ (¢, m ) : ¢ E Cd

for all m E M(K, X*). The set C1 is a weak* dense subset of B ( (M(K, X*) ) * ) . For any ¢ E (M(K, X*))* and m' E M(K, X*), define a measure Z(¢, m') on K by Z(¢, m') (A) = (¢, m' · l A ) . Then Z (¢, m') « A. By the Radon-NikodYm theorem, there is a measurable function 9 such that for any measurable set A, we have Z(¢, m') (A) =

i 9 dA.

We still denote the function 9 by Z(¢, m') . We have the following facts:

6.5.

359

NOTES AND REMARKS

(a) (¢, m') = In Z(¢, m') d)" for any m' E M(!1, X·). (b) It is easy to see that for any h E L oo ( ).. ) ,

in Z(¢, m') . h d)" = in Z(¢, m' . h ) d)". So the mapping ¢ f-+ Z (¢, m') is a weak· to weak continuous operator from ( M(!1, X·) ) * to L 1 ().. ) .

( ) For any finite measurable partition {A i } f= l of !1 and any bounded se quence {xi· } f= l in X .. , let 'IjJ be the element in ( M(!1, X·) ) * defined by c

n

'IjJ (m) = L (x t , m(Ai) } , i= l

Then for almost all w E A i ,

Z('IjJ, m' ) (w) = ( xi · , m'(w) ) . Let {mn }�= l be a uniformly bounded sequence in L1 ( ).. , X* , w· ) such that for almost all w E !1, {mn (w)}�= l is weakly Cauchy. Note that we have proved the following facts . •

For each q E N, the mapping ¢ f-+ Z(¢, mq) is a weak· to weak continuous function from ( M(!1, X) ) * to L 1 ().. ) . Thus for any n E N , The mapping ¢ f-+ (Z(¢, mI ) , . . . , Z(¢, mn)) is weak· to weak continuous .

•

The set C1 is weak· dense, convex subset of B ( (M(!1, X·))· ) .

Fix an element ¢ in B ( ( M(!1, X·) ) * ) and an n in N. By Mazur's theorem, there is ¢n E C1 such that ¢n (m) =

(xi , m(Ai,n)) , i�Lkn ,�

and for all q :s; n , we have One can easily verify that for any q E N, the sequence {Z ( ¢n, mq) } �= 1 converges to Z ( ¢, mq) almost everywhere. Let U be any free ultrafilter of N. For each w E !1, let F(w) E B(X· .. ·) be the weak· limit along U of the sequence {x�t (·n,w ) ,n }�= 1 ' where i(w, n ) is the

360


unique i � kn such that W E A i ( w ,n) ,n ' Then for almost all W E 0, =

=

nlim -to oo ( x:(: " n ) n' m q (w) ) ( F(w), m q (w) ) .

But for almost all W E 0, {mq (w)}�=l is weakly Cauchy. By the bounded convergence theorem, nlim - oo ( , mn)

=

nlim -oo }rn Z(, mn ) (w ) d)"

=

nlim - oo }rn ( F(w), mn (w ) ) d)"

exists. We have proved (1). (2) follows from the same argument. (Note: in this 0 case, we have F(w) E B(X** ).) Combing the results of Theorem 6.5.1 and Theorem 6.5.2, we have the fol lowing theorem. Theorem 6.5.3. Let ° be a compact Hausdorff space and X a Banach

space. Then the space M(O, X*) (0(0, X) r is weakly sequentially complete if X* is weakly sequentially complete. =

Let X be a Banach space and (O, �) a a-algebra. Let cabv(O, X) denote the set of all countably additive X -valued vector measures that are of bounded variation. It is natural to ask whether cabv(O, X) is weakly sequentially com plete if X is is weakly sequentially complete. The following example is due to Talagrand [43] which shows that the answer to the above question is negative. Example 6.5 .4. Let I-" be the Lebesgue measure on [0, 1] and 1-" 2 the product measure on [0, 1] 2 . Let p : [0, 1] 2 -+ [0, 1] be the mapping define by p(x, y) x, and let E be the set of all measurable functions f E E on [0, 1] 2 such that =

IfI � h o p + g

It is easy to see that E is a Kothe function space. We claim that E satisfies the following conditions: (1) E is order continuous. (2) E is a KB space, (Le., every bounded increasing sequence in E converges) . (3) cabv([O, 1] , E) is not weakly sequentially complete.

36 1

6.5. NOTES AND REMARKS

Proof of (1): Let { In }�=l be a decreasing sequence in E that converges to

° a.e. and h E L 1 (/-£), 9 E L 2 (/-£ 2 ) such that II � h o p + g. For any E > 0, there is K > ° such that I l h - (h 1\ K) II 1 � E and II g - (g 1\ K) II 2 � E. Since

In � In 1\ 2K + (h - h 1\ K) + (g - 9 1\ K)

and limn -too II /n 1\ 2K II 2 = 0, we have lim sup II In liE � 2E. n -t oo This implies that E is order continuous. Proof of (2) : Let {In}�=l be a bounded increasing sequence in E. Then there are M > 0, {hn : n E N } � Li (/-£) , and {gn : n E N } � Lt (/-£ 2 ) such that for all n E N. Let U be a free ultrafilter of N and k a fixed natural number. Since { hn 1\ k : n E N } is relatively weakly compact in L1 (/-£) and {gn : n E N } ) is relatively weakly compact L 2 (/-£2 ), the weak limits h k = w- nlim -tU (h n 1\ k ) and 9 = w- nlim -tU gn exist. Notice that {h k }k:: 1 is a bounded increasing sequence in L 1 (/-£). By the monotone convergence theorem, it converges to some h E Ll (/-£). Thus for any k, n E N, h o p + 9 2:: (h 0 p 1\ k ) + 9 2:: In 1\ k. This implies that sup In = l E E. n EN By (1) , {In}�=l converges to f in norm. We have proved (2) . Proof of (3) : Let A be a measurable subset of [0, 1 ] . Define an E-valued vector measure m on [0, 1] by Since for any measurable subset G of [0 , 1 ] lc 0 p 2:: lc xA , it follows that '

Thus for any measurable subset A of [0, 1 ] , we have II m Ail � 1 . We claim that if /-£(A) > /-£( G ) , then II I cx A i l E = /-£(G) . Suppose that the claim has been proved. Let A be any measurable subset of [0, 1 ] with positive measure, and { G1 , G2, . . . , Gn} a partition of [0 , 1 ] such that /-£(Gj ) � /-£(A) for all j � n. Then n n n II m A il 2:: L II m A ( Gj ) l i E = L I I 1 Cj x A il E = L /-£( Gj ) = 1 . j =l j =l j =l

362


This implies that for any measurable subset A of [0, 1] with J.t(A) > 0, Il mA 11 1 . So if the claim is true, then cabv ( [O, 1] , E) contains a copy of L oo (J.t) and it cannot be weakly sequentially complete. Proof of the claim: Since II I c xAllE J.t(C) , we need to prove only that II I cx A IlE � J.t(C) for any measurable subset C such that J.t(A) � J.t( C) > 0. Suppose that J.t(A) � J.t(C) > 0. For any € > 0, there are two nonnegative functions h E L 1 (J.t) and 9 E L 2 (J.t2 ) such that =

�

II h l1 1 + II g l12

Note:

� II I c x A l l E + €

r 2 iCXA h o p dJ.t By Holder's inequality, we have

� � �

and h o p + 9 � lc xA '

=

Il h ll i . J.t(A) .

J.t(A) . J.t(C) - J.t(A) · ll h ll 1 2 r iCXA (lc xA - h o p) V O dJ.t r 9 dJ.t 2 II g l1 2 . II 1 Ax c I1 2 iCXA II g l1 2 . J.t(A) .

� � �

�

So J.t(C) II h il I + Il g 11 2 . Since € is an arbitrary positive real number, we have 0 J.t(C) II I c x A IIE . The proof is complete . Let E be a Kothe function space over ([0, 1] ' J.t) such that for any g E E, II g l1 1 Il g iIE . S. Diaz ( Exercise 5 . 1 .3) proved that if X* contains a copy of f 1 , then E(X) contains a quotient isomorphic to Co . It is natural to ask the following question. Problem 6.5.5. Let E be a Kothe function space over ( [0, 1] , J.t) such that 1 o lI [ , l ) II E 2 and for any g E E II g l1 1 Il g iIE . Let X be a Banach space. Suppose that E(X) has a quotient isomorphic to Co but neither E nor X has a quotient isomorphic to Co . Does X* contain a copy of f 1 ? The following facts are known: ( 1 ) Co is a quotient space of fl . (2) If E is an order continuous Banach lattice and E contains a copy of f1 ' then E contains a complemented copy of f 1 (Theorem 3 . 1. 11). (3) Let X be a Banach space which does not have any copy of fl . S. Diaz and A Fernandez proved that X has a copy of Co if and only if X has a complemented copy of Co . ( See Exercise 5 . 3 . 5 and [13, Lemma 3.4] . ) Thus for any order continuous Banach lattice E, either E is reflexive or E has a quotient isomorphic to Co . Diaz and G. Schliichtermann [13 , Theorem

�

,

�

6. 6. REFERENCES

363

3.2] showed the answer to the above question when E is order continuous. On the other hand, if E = l oo and if X contains £''1 ' s uniformly complemented, then E(X) contains a complemented l l -space. In this case, we have a negative solution to the above question. A. Pelczynski asked the following question [30, p.645, Remark 1] . Question 6.5.6. Let 0 be a compact Hausdorff space and X a Banach space. Does C(O, X) have property (V) if X has property (V) ?

In [1 1] , P. Cembranos, N. J . Kalton, E. Saab and P. Saab proved that if X has property ( u ) and contains no copy of lI , then C([2, X) has property (V). N. Randrianatoanina [33] tried to use Talagrand ' s technique to show the answer to question 6.5.6 is affirmative. In his proof, he used the strong operator topology instead of the weak* topology. It is known that the unit ball of C(X, Y) in the strong operator topology is not necessary to be compact. But in the proof of Talagrand ' s L1-Theorem, we did use the fact the unit ball of the dual space is w*-compact (see Step 2 and 3 of the proof Talagrand ' s L1 (X)-Theorem) . We believe that Question 6.5.6 is still open.

6.6

References

[1] R.G. Bilyeu and P.W. Lewis, Vector measures and weakly compact operators on continuous function spaces: a survey, in Geometry of Normed Linear Spaces, Conference on Measure Theory and its Applications, Proceedings of the 1980 conference at Northern Illinois University. Contemp. Math. , vol. 52, 96-101 [2] F. Bombal and P. Cembranos, The Dieudonne property for C(K, E), Trans. Amer. Math. Soc. 285 (1984) , 649-656. [3] F. Bombal and P. Cembranos, Characterization of some classes of operators on spaces of vector-valued continuous functions, Math. Proc. Camb. Phil. Soc. 97 (1985), 137-146. [4] F. Bombal and P. Cembranos, Dieudonne operators on C(K, E) , Bulletin Polish Acad. Sci. 34 (1986) , 301-305. [5] F. Bombal and 1. Villanueva, On the Dunford-Pettis property of the tensor product of C(K) spaces, Proc. Amer. Math. Soc. 129 (2001), 1359-1363. [6] J. Bourgain, Hoo is a Grothendieck space, Studia Math. 75 (1983) , 193-216. [7] J . Bourgain and G. Pisier, a construction of C oo -spaces and related Banach spaces, Bol. Soc. Brasil. Mat. 14 (1983) , 109-123. [8] Qingying Bu and J. Diestel, Observations about the projective tensor prod uct of Banach spaces, I-lP®X, 1 < p < 00, Quaestiones Math. 24 (2001), 1 - 15.

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[9] P. Cembranos, The hereditary Dunford-Pettis property for £ l ( E) , Proc. Amer. Soc. Math. 108 (1990), 947-950. [10] P. Cembranos, The hereditary Dunford-Pettis property for C(K, E ) , Illinois J. Math. 31 (1987) , 365-373. [1 1] P. Cembranos, N.J. Kalton, E. Saab, and P. Saab, Pelcynski 's property (V) on C(O, E) spaces, Math. Ann. 271 (1985), 91-97. [12] F. Delbaen, Weakly compact operators on the disc algebra, L. Algebra 45 284-294. [13] S. Dfaz and G. Schliichtermann, Quotients of vector-valued function spaces, Math. Proc. Camb. Phil. Soc. 126 (1999), 109-1 16. [14] J. Diestel, A survey of results related to the Dunford-Pettis property, in Topology and Geometry in Linear Spaces, Proc. of the Conf. on Integration, Contemporary Math. vol 2, Amer. Math. Soc., Providence, RI (1980) , 1560. [15] J. Diestel, H. Jarchow, and A. Tonge, Absolutely Summing Operators, Cam bridge University Press (1995) . [16] J. Diestel and J.J. Uhl, Jr., Vector Measures, Math. Survey 15, Amer. Math. Soc., Providence, RI (1977). [17] N. Dunford and J. Schwartz, Linear Operators, I, Interscience Publishers, New York 1958. [18] J. Elton, dissertation, Yale University, New Haven, CT. [19] G. Emmanuele, Another proof of a result of N. J. Kalton, E. Saab and P. Saab on the Dieudonne property in C{K, E), Glasgow Math. J. 31 (1989), 137-140. [20] G. Emmanuele and W. Hensgen, Property (V) of Pelczyriski in projective tensor products, Proc. Roy. Irish Acad. Sect. A 95 (1995), 227-231. [21] M. Gonzalez and J.N. Gutierrez, The Dunford-Pettis property on tensor products, Math. Proc. Camb. Phil. Soc. 131 (2001), 185-192. [22] A. Grothendieck, Sur les applications lineaires faiblement compactes d 'espaces du type C(K), Canad. J. Math. 5 (1953) , 129-173. [23] W. Hensgen, A simple proof of Singer's representation theorem, Proc. Amer. Math. Soc. 124 (1996) , 321 1-3212. [24] N.J. Kalton, E. Saab, and P. Saab, On the Dieudonne property for C(K, E ) , Proc. Amer. Math. Soc. 96 (1986), 50-52. [25] H. Knaust and E. Odell, On Co sequence in Banach spaces, Israel J. Math. 67 (1989), 153-169.

6. 6 . REFERENCES

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[26] D.R. Lewis and S. Stegall, Banach spaces whose dual are isomorphic to £ l ( r ) , J. Funet. Anal. 12 (1971) , 167-177. [27] J. Lindenstrauss and A. Pelczynski, Absolutely summing operators and their application, Studia Math. 26 (1968) , 275-326. [28] F. Lust, Produits tens oriels injectifs d 'espaces de Sidon, Colloq. Math. 32 (1975) , 286-289. [29] T.V. Panchapagesan, A simple proof of the Grothendieck theorem on the Dieudonne property of Co (T) , Proc. Amer. Math. Soc. 129 (2000) , 823831. [30] A. Pelczynski, On Banach spaces on which every unconditionally converging operator is weakly compact, Bull. Acad. Polon. Sci. 10 (1962) , 641 -648. [31] A. Pelczynski and W. Szlenk, An example of a non-shrinking basis, Rev. Roumaine Math. Pures Appl. 10 (1965) , 961-966. [32] N. Randrianantoanina. Pelczynski 's property (V) on spaces of vector-valued junctions, Colloquium Math. 71 (1996) , 63-78. [33] N. Randrianantoanina. Complemented copies of £1 in spaces of vector mea sures and applications, Math. Nachr. 202 (1999) , 109-123. [34] H.L. Royden, Real Analysis, ( 3rd ed. ) Macmillan Publ. Co. , New York (1988). [35] R.A. Ryan, The Dunford-Pettis property and projective tensor products, Bull. Polish Acad. Sci. Math. 35 (1987) , 785-792. [36] E. Saab and P. Saab, A stability property of a class of Banach spaces not containing a complemented copy of £1 , Proc. Amer. Math. Soc. 84 (1982), 44-46. [37] E. Saab and P. Saab, On stability problems of some properties in Banach spaces, in Function Spaces, Lecture Notes in Pure and Applied Mathematics 136 ( ed. K. Jarosz ) , Marcel Dekker (1992) , 367-394. [38]

Singer, Linear functions on the space of continuous mappings of a com pact Hausdorff space into a Banach space (in Russian ) , Rev. Roum. Math. Pures Appl. 2 (1957) , 301-315.

[39]

1.

[40]

1.

1.

Singer, Sur les applications lineaires integrales des espaces de fonctions continues, I Rev. Roum. Math. Pures Appl. 4 (1959) , 391-401.

Singer, Best Approximation in Normed Linear Spaces b y Elements of Linear Subspaces, vol. 171 Springer-Verlag, Berlin, Heidelberg, New York (1970) .

366

CHAPTER 6 . CONTINUO US FUNCTION SPACES

[41] C. Swartz, Unconditionally converging operators on the space of continuous functions, Rev. Roumaine Math. Pures Appl. 17 (1972) , 1695-1702. [42] M. Talagrand, Weak Cauchy sequences in L l (E) , Amer. J. Math. 106 (1984) , 703-724. [43] M. Talagrand, Quand l 'espace des mesures a variation bomee est-l faible ment sequentiellent complet? Proc. Amer. Math. Soc. 90 (1984) , 285-288.

Index admissible, 25 I-admissible, 26 totally, 304 Alaoglu ' s theorem, 3 approximation property, 24 ARP, 341-345 Baire ' s theorem, 3 Banach lattice, 143 a-complete, 147 a-order continuous, 144-147 a complete, 148 band, 144 band projection, 144 complete, 144 ideal, 144, 150 order continuous, 144-156 order isometric, 144 order preserving, 144 positive operator, 143 strictly monotone, 148 sublattice, 144 uniformly monotone, 148 weak unit, 150 Banach-Mazur distance, 8 Banach-Saks property, 130-139, 295307 weak, 130-138, 295-307, 346 basis bimonotone, 8 block, 16-20, 27 boundedly complete, 10-12, 24 constant, 8 equivalent, 15 Haar, 13 monotone, 8 Schauder, 8-12

shrinking, 10-12, 22 summing, 13 symmetric, 25 unconditional, 20-27, 257 Bishop-Phelps theorem, 44 Bochner integrable, 166 Borsuk-Dugundji theorem, 319 Co-sequence, 13, 32, 36, 250-255, 334 Cauchy sequence, 279 closed graph theorem, 3 complemented Lp, 94 .e l l 19, 33, 71, 152, 250, 255, 257-261 .e� 's uniformly, 258 i2 , 62 ip, 19 .e� ' s uniformly, 258 Co, 19, 37, 256, 324 complete cont�nuity property, 208211 conditional expectation, 13, 82, 196 convex B-convex, 95 K-convex, 95 fully, 118 locally uniformly, 101-1 17 midpoint locally uniformly, 1021 17, 232 nearly uniformly, 102 strictly, 101 uniformly, 101-106, 122, 128 weakly locally uniformly, 102 weakly uniformly, 1 17, 158, 178 cotype, 94, 186-258 constant, 186

368 Diaz-Kalton Theorem, 257-261 Day ' s norm, 113 dentable, 199 derived set, 53 Dieudonne property, 326-331 dispersed, 53 distortable, 38 Dunford-Pettis property, 57-67, 214, 291-294, 349-351 hereditary, 331-347 Eberlein-Smulian theorem, 4 Ekeland ' s variational principle, 43 equi-integrable, 152, 300 Fatou property, 158 Gantmacher ' s theorem, 6 Gowurin integral, 314 Grothendieck space, 41, 257 Holder ' s inequality, 5 Hahn-Banach theorem, 2 Hardy space, 65 Heinrich-Mankiewicz theorem, 72, 78, 279 independent, 82 injective space, 67 injective tensor product, 75-78 James ' s theorem, 42-48 Jensen ' s inequalty, 5 Josefson-Nissenzweig theorem, 278 Kothe dual, 149 Kothe function space, 149-151 p-concave, 187 p-convex, 187 characteristically order continuous, 160, 249 KB space, 158 Kadec-Klee property, 104, 184, 294 uniformly, 102, 188 Kahane ' s inequality, 185 KK property, see Kadec-Klee prop erty

INDEX

Koml6s property, 295-307 Koml6s ' s theorem, 87-94, 295 Krein-Milman property, 244 Krein-Milman theorem, 7 Kronecker ' s lemma, 89 .cp-space, 68, 350 fI-sequence, 13, 29, 70, 249, 254, 262, 272, 279 fp-sequence, 13, 254 Lebesgue ' s dominated convergence theorem, 150 lifting, 163 lifting property, 29 limited, 36-38 LUC, see convex, locally uniform martingale, 83-88, 189, 196 martingale difference sequence, 88 maximal inequality, 83 Mazur ' s theorem, 2 Mazurkiewicz-Sierpinski theorem, 55, 344, 346, 347 metric projection, 318 Minkowski ' s inequality, 5 NikodYm boundedness theorem, 174 normal structure, 123 weakly, 123 norming space, 161 open mapping theorem, 3 operator p-integral, 76 absolutely p-summing, 79 adjoint, 6 Banach-Saks property, 139 compact, 6 completely continuous, 58, 349 Dunford-Pettis, 188-195 nuclear, 79 unconditionally converging, 73, 323 weakly compact, 6, 322 weakly completely continuous, 326

INDEX

partition of unity, 313 perfect set, 53 Pettis integrable, 210 Pettis representable, 210 Pettis's measurability theorem, 162 Pettis-Cauchy, 189 Pettis-norm, 189 Phillips's lemma, 38 point denting, 158, 226-231 exposed, 219, 224 extreme, 219-224 locally uniformly convex, 226, 232 smooth, 124, 219, 224 strongly exposed, 203, 233-241 strongly extreme, 226 weak* exposed, 219, 224 weak* extreme, 221 weak* strongly exposed, 241, 242 projective tensor product, 74-79, 348355 property (8) , 331 property (u ) , 73, 153, 347 property (U 8 ) , 331 property (V), 69, 74 property (V*), 69-74, 152, 278-290 quotient, 41 r.i. function space, 159-160, 249 Rademacher functions, 35, 249-253 Radon-NikodYm property, 196-208, 351-355 weak, 210 Radon-NikodYm theorem, 5, 81, 149 Radon-Riesz property, see KadecKlee property Ramsey's Theorem, 29, 296, 332 reasonable crossnorm, 74 reflexive, 1 regular, 48 Riesz representation theorem, 5 Rosenthal's iI-theorem, 29, 64, 249, 274 Rosenthal's lemma, 38

369 rotund k- uniformly, 122 fully, 102, 118 fully k, 102 uniformly rotund in every di rection, 102-1 17 scattered, see dispersed Schauder system, 14 Schauder's theorem, 6 Schur property, 27, 60, 307, 348 semiembedding, 206 separation theorem, 2 slice, 199 smooth, 124 Fh�chet, 124 uniformly, 125 uniformly Gateaux, 124 Sobczyk's theorem, 41 space lI ' 28, 41, 156, 249, 254, 308 l oo , 38, 104, 147 lp, 254 Co, 24, 145, 154, 156, 248-255, 322 Baernstein, 25 Lorentz function, 159 Lorentz sequence, 27 Orlicz function, 160 Orlicz sequence, 26 Schlumprecht, 26 Schreier, 25 Talagrand, 290-295 Tsirelson, 26 spreading model, 130, 132-136 ll , 134-136 lp, 139 strong law of large numbers, 94 strong maximum, 203 strong minimum, 42 strongly E-integrable, 352 strongly exposing functional, 203 strongly measurable, 162 strongly operator topology, 319 submartingale, 83

370 subsequence splitting property, 211216, 300 symmetrized type, 118 Talagrand ' s L1 (X ) -theorem, 261-277, 279 three-space property, 66 totally disconnected, 53 type, 94, 186 constant, 186 ultraproduct, 68, 211 UMD, 309 Uniform boundedness principle, 3 uniformly bounded, 189 uniformly integrable, 34-36, 151, 250253 URED, s e e rotund, uniformly rotund in every direction ( V* ) -set, 70, 278

variation, 314 vector measure bounded semivariation, 167, 320 bounded variation, 314 contral measure, 172 count ably additive, 167 finitely additive, 167 representing measure, 321 strongly additive, 169 uniformly strongly additive, 322 w* regular, 314 w*-count ably additive, 314 weakly countably additive, 177 Vitali ' s lemma, 40, 152 Vitali-Hahn-Saks theorem, 176, 201 w.u.c., 69-72 weak Phillips property, 73 weak topology, 1 weak* measurable, 162 weak* scalarly equivalent, 164 weak* sequentially compact, 4 weak* topology, 1 weakly E* -integrable, 352 weakly bounded, 2

INDEX weakly Cauchy sequence, 22, 29, 262 weakly compact, 42 weakly compactly generated, 4 weakly conditionally compact, see weakly pre compact weakly measurable, 162, 352 weakly precompact, 34, 151, 272277, 299-300 weakly sequentially complete, 22-36, 152, 154, 275, 358-362 weakly unconditionally convergent, see wuc WLUC, see convex, weakly locally uniform WU C, see convex, weakly uniform wuc, 32-36

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