Introduction to Banach Spaces and Their Geometry

INTRODUCTION TO BANACH SPACES AND TH.EIR GEOMETRY

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NORTH-HOLLAND MATHEMATICS STUDIES

68

Notas de Matematica (86) Editor: Leopoldo Nachbin Universidade federal do Rio de Janeiro and University of Rochester

Introduction to Banach Spaces and their Geometry BERNARD BEAUZAMY Departement de Mathematiques Universite de L yon I Villeurbanne, France

1982 NORTH-HOLLAND PUBLISHING COMPANY - AMSTERDAM

'

NEW YORK

'

OXFORD

North- Holland Publishing Company. I982

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Lihlar? of I'ongrrss Calalogirg ill Publication Data

Beauzamy, Bernard, 1949I n t r o d u c t i o n t o Banach spaces and their geometry. (North-Holla?d mathematics s t u d i e s ; 68) (Notas de matematica ; 86) Bibliography: p. I n c l u d e s index. .I.Banach spaces. I. T i t l e . 11. S e r i e s . 111. Series: Notas d e matematica (Amsterdam, Netherlands) ; 86. Q.AL.N86 no. 86 [QA322.2] 510s [ 5 1 5 . 7 * 3 2 ] 82-6463 ISBN 0-444.46416-4 ( E l s e v i e r s c i e n c e ) AACR2

PRINTED IN T H E NETHERLANDS

INTRODUCTION

The s t u d y o f Banach Spaces f o r themselves i s a r a t h e r r e c e n t i d e a , s i n c e t h i s branch o f F u n c t i o n a l A n a l y s i s came up, r o u g h l y speaking, a f t e r t h e work o f S. Banach h i m s e l f , i n t h e 1930's. Soon, Banach Spaces Theory, as i t i s now c a l l e d , r e c e i v e d a q u i c k development, c o n s i d e r i n g t h e number o f people w o r k i n g i n t h i s area, as w e l l as t h e importance o f t h e r e s u l t s t h e y o b t a i ned. I n some sense, i t may appear as t h e n a t u r a l c o n t i n u a t i o n o f t h e s t u d y

o f l o c a l l y convex spaces, which c u l m i n a t e d i n t h e y e a r s 1 9 5 0 ' s . But, as w i l l be seen, t h e s p i r i t and t h e methods a r e q u i t e d i f f e r e n t . Meanwhile, courses a r e t a u g h t on Banach spaces i n many U n i v e r s i t i e s , a t elementary o r a t h i g h l e v e l . Several books have a l r e a d y been w r i t t e n on t h i s t o p i c . L e t us mention f i r s t those o f M.M. Day [131 , L i n d e n s t r a u s s - T z a f r i r i [341 ,

E. Lacey [321 , and many o t h e r s which a r e devoted t o some s p e c i a l i z e d

f e a t u r e o f Banach Spaces.

A l l these books a r e i n t e n d e d f o r s p e c i a l i s t s o f t h e f i e l d , and s t u d e n t s who a r e b a r e l y a c q u a i n t e d w i t h t h e b a s i c knowledge i n A n a l y s i s ( t h a t i s : c l a s s i c a l general Topology, c l a s s i c a l Measure Theory, elementary p r o p e r t i e s

o f H i l b e r t spaces) may f i n d them d i f f i c u l t ( o r even almost i m p o s s i b l e ) t o read. The p r e s e n t book i s i n t e n d e d t o f i l l t h i s gap and g i v e an o p p o r t u n i t y t o s t u d e n t s t o become f a m i l i a r w i t h an elementary t h e o r y o f Banach spaces, and t o reach a l e v e l which w i l l a l l o w them t o understand more s p e c i a l i z e d t o p i c s and r e a d r e s e a r c h papers. Rather t h a n r e a c h i n g t h e r e s e a r c h l e v e l on some narrow f i e l d , we p r e f e r t o p r e s e n t t h e b a s i c f a c t s o f t h e t h e o r y and g i v e a general view o f t h e s u b j e c t . Since we s t a r t a t a r a t h e r low l e v e l , we have t o p r e s e n t f i r s t t h e c l a s s i c a l r e s u l t s o f F u n c t i o n a l A n a l y s i s , such as t h e Closed Graph Theorem, t h e Hahn-Banach Theorem, e t c . But, s i n c e we have i n mind t h e s t u d y o f Banach spaces, we have avoided, as f a r as p o s s i b l e , t h e "Bourbaki s t y l e ' ' o f e x p o s i t i o n , which develops a general t h e o r y o f l o c a l l y convex t o p o l o g i c a l V

vi

B. BEAUZAMY

v e c t o r spaces. F o r example, we have c o m p l e t e l y n e g l e c t e d t h e general s t u d y o f t o p o l o g i e s compatible w i t h a d u a l i t y , and we have o n l y presented t h e r e s u l t s which a r e o f f u r t h e r use f o r o u r purpose : t h e y a r e numerous enough! Also, i n t h e c o n t e x t o f normed spaces (endowed w i t h t h e i r norm o r w i t h a weak t o p o l o g y ) many c l a s s i c a l theorems (connected, f o r example, w i t h t h e B a i r e P r o p e r t y ) have a s i m p l e r and more c o n c r e t e p r o o f : we have worked i n t h i s s p i r i t as o f t e n as p o s s i b l e . B u t o f course, when i t m a k e s n o d i f f e r e n c e , we have no reason t o r e s t r i c t o u r s e l v e s . I n c o n t r a s t , we have g i v e n an i m p o r t a n t place, i n t h i s f i r s t p a r t , t o t h e s t u d y o f r e f l e x i v i t y , with, o f course, t h e c l a s s i c a l r e s u l t s about d u a l i t y and weak compactness, b u t a l s o w i t h R.C.

James’ c o n d i t i o n s f o r

r e f l e x i v i t y . These c o n d i t i o n s a r e used s e v e r a l t i m e s i n t h e sequel , and, from t h e s t a r t , t h e y i n t r o d u c e t h e r e a d e r t o t h e s p i r i t o f t h e geometric study o f Banach spaces. The p r o p e r t i e s o f separable Banach spaces a r e a l s o i n v e s t i g a t e d , and we end t h i s f i r s t p a r t w i t h a s t u d y o f t h e d e l i c a t e n o t i o n o f weak s e q u e n t i a l compactness : we prove a weak form (weak, b u t s u f f i c i e n t f o r what f o l l o w s ) o f t h e E b e r l e i n - S m u l i a n Theorem. The second p a r t e n t e r s r e a l l y t h e s u b j e c t . We s t a r t w i t h a s h o r t c h a p t e r about H i l b e r t spaces. The p r o o f s a r e given, b u t we t h i n k t h a t most o f t h e r e s u l t s should be f a m i l i a r t o t h e readers : we i n c l u d e t h e s e r e s u l t s i n o r d e r t o p r e p a r e t h e c o n t r a s t w i t h general Banach spaces. I n a second chapter, we i n v e s t i g a t e t h e n o t i o n o f Schauder b a s i s i n a Banach space : up t o a c e r t a i n e x t e n t , t h i s n o t i o n p l a y s t h e r o l e o f an H i l b e r t i a n b a s i s i n a H i l b e r t space. We a l s o develop a l i t t l e t h e t h e o r y o f u n c o n d i t i o n a l bases , and p r o v e some r e s u l t s , due t o R.C. James, concerning r e f l e x i v i t y i n Banach spaces w i t h Schauder bases. We t h e n t u r n t o t h e s t u d y o f t h e most common Banach spaces :

L p ( lQ p Q t

00

), co , W ( K )

, Lp(l

Q

p

O , ao,...,aN E J K 1 . 0 -. span{ ( xn)n IN 1 a r e t h e i r c l o s u r e s

i s a normed space, and

K a compact i n i t , we s h a l l say t h a t K

xly...,x

forms an € - n e t f o r n and w i t h r a d i u s E .cover K .

i f t h e b a l l s c e n t e r e d on

The space I K ( N ) i s t h e s e t o f Binite sequences o f elements o f IK ( o r , more e x a c t l y , t h e s e t o f sequences which have o n l y f i n i t e l y many nonzero terms). The canonical b a h o f K ( N ) w i l l be c a l l e d ( e n ) n E m ; en i s t h e sequence w i t h

1 a t t h e n t- h rank,

The a!iurne&tr o f a s e t

A

0

otherwise.

i n a normed space

E

is

X,Y

sup

E A

Ilx

-

yll.

Filters. When one deals (as we s h a l l ) w i t h n o n - m e t r i z a b l e t o p o l o g i e s , one cannot use o n l y convergent sequences t o d e f i n e t h e t o p o l o g y : one has t o use f i l t e r s . Therefore, we r e c a l l b r i e f l y t h e necessary d e f i n i t i o n s ( see Bourbaki 1101 f o r d e t a i l s ) .

NOTATIONS AND PRELIMINARIES If

I

i s a s e t , a s e t . 9 o f subsets of

a)

if

X €9 and

b)

if

X,Y € 9 , then

c)

I € 9 .

d)

#&9.

I

3

i s c a l l e d a C;.ietai f :

then Y € 9 .

Y >X,

X n Y E 9.

I f I = IN and 9 i s t h e s e t of X c N 9 i s c a l l e d the Fh&cheA &LLtaon IN.

such t h a t

i s f i n i t e , then

CX

The s e t of f i l t e r s on a s e t I can be ordered by inclusion : 9 i s thinner than 9' i f 9'c 9.An ~1.Lthu6.ietmi s a maximal element f o r this ordering : no f i l t e r i s s t r i c t l y t h i n n e r . If E point xo

i s a topological space, the s e t of a l l neighbourhoods of a given i s a f i l t e r , denoted by 9 V . In any s e t I , t h e s e t of a l l xO

subsets containing a given element i o i s an u l t r a f i l t e r . Such an u l t r a f i l t e r i s c a l l e d a t r i v i a l one. On a topological space E , a f i l t e r 9 i s s a i d t o conve ge t o xo E E i f 9 i s t h i n n e r than 9 V . This means t h a t f o r every V , nei ghbourhood xO

s a cBubteh of xo , t h e r e i s X ~9 with X c V , One says t h a t xo point f o r 9 i f xo i s a c l u s t e r point of a l l the elements x of ,F. I f E i s a metric space ( f o r example, a normed space), 9 i s c a l l e d a Cuuchy &iLtai f , f o r every E > 0 , t h e r e a r e elements X E ,9such t h a t t h e diameter of X i s a t most E . On a compact s e t E , every f i l t e r has a t l e a s t a c l u s t e r p o i n t , a n d every u l t r a f i l t e r i s convergent. I f E i s a complete metric space ( f o r example, a Banach space) , every Cauchy f i l t e r converges.

A l;iLtm bane 3 on a s e t following two p r o p e r t i e s :

3 , B1 n B2

I

contains an element of

1)

If

2)

3 i s not empty, and # $L

Bl,B2

E

i s a s e t of subsets of

I

with the

a.

a.

Then t h e s e t of subsets of I which contain an element of B i s a f i l t e r , c a l l e d l;.ietetr ~ U i t hbane B. In a topological space E , one says t h a t a f i l t e r base 3 converges t o a point xo E E i f the f i l t e r with base 3 converges t o xo .

4

B. BEAUZAMY

Let I be a s e t , w i t h a f i l t e r 8 ; l e t E be a topological space, and f an application from I i n t o E One says t h a t a point y E E i s the l i m i t of f f o r the f i l t e r 9 i f y i s the l i m i t of the f i l t e r base This means a l s o t h a t , f o r any neighbourhood f ( 9 ) . One writes y = lim f

.

of y , t h e r e i s an

V

X E 8 such t h a t

For example, a family the f i l t e r 8 i f y i s t h e f i l t e r 9.This means t h a t X E 9,such t h a t {xi ; i

x. -y 1

9

.

8

.

f(X) C V

.

(xi)i of points of E l i m i t of the a p p l i c a t i o n f o r any neighbourhood V E XI C V . W e then w r i t e

-

converges t o y f o r i x i for t h e of y , t h e r e i s an lim xi = y , o r simply 9

-

In p a r t i c u l a r , y i s the l i m i t o f a sequence ( x ~ i f ) y ~ i s ~the ~ xn f o r the Frechet f i l t e r on IN. One l i m i t of t h e application n checks t h a t t h i s d e f i n i t i o n coi'ncides w i t h the usual one. Let E be a topological space, and F a dense subset in E . For every y E E , l e t 9l be the t r a c e of 9 V y on F ( t h a t i s , the set of a l l X n F, X E 9 V y ) . Then T1 i s a f i l t e r on F , and a f i l t e r base on E Let I be any s e t such t h a t there i s a s u r j e c t i o n from I onto F ; c a l l f : i x i this s u r j e c t i o n . Then f - l [ S , ) on I ; i s a f i l t e r base are also we c a l l 9 the f i l t e r generated by a . Then, the points ( x i ) i y points of E , and x .

.

-

1

9

.

Another way of obtaining the same conclusion i s t h e following : l e t I be the s e t of a l l neighbourhoods of y For every fixed V E I , consider the s e t Bv of the neighbourhoods of y which a r e contained i n V . Then = IBv , V E I } i s a f i l t e r base on I . Let 9 be the corresponding f i l t e r . For each V E I , take xv E F n V . Then xv p y ,

.

PART 1 FUNCTIONAL ANALYSIS

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CHAPTER

I

BAIRE'S PROPERTY AND ITS CONSEQUENCES

- We hag t h a t a tvpaCagicaL hpace

DEFINITION.

evmtry c a u d b L e intemecfitian a6 denbe apen h&

han B a i r e ' s P r o p e r t y id i n E .h a d e a e A & in E

E

.

The i n t e r s e c t i o n needs n o t be open, s i n c e an i n f i n i t e i n t e r s e c t i o n o f open s e t s i s n o t open i n g e n e r a l . F o r example, l e t enumeration o f t h e r a t i o n a l numbers i n

O,, = R

c o u n t a b l e ) , and l e t

n On = n > l dense i n R ) .

IR \ Q

and

-

Iqn)

.

IR

Then

(qn)n>l (one knows t h a t

On

be an

Q is

i s open, f o r each

i s dense, b u t n o t open ( s i n c e

n

>1 ,

Q i t s e l f i s also

I n t h e d e f i n i t i o n , t h e word " c o u n t a b l e " i s e s s e n t i a l , and t h e r e i s no hope t h a t any space c o u l d have t h e same p r o p e r t y , f o r i n t e r s e c t i o n s indexed by

IR

:

x E IR , n Ox

if

IR,

dense i n

but

X€R

put

IR \ {XI , t h e n Ox

Ox =

i s open and

i s empty.

The d e f i n i t i o n i s a l s o e q u i v a l e n t t o t h e f o l l o w i n g ( j u s t t a k i n g complements) : any c o u n t a b l e u n i o n o f c l o s e d s e t s w i t h empty i n t e r i o r has empty i n t e r i o r . The f o l l o w i n g theorem w i l l p r o v i d e two l a r g e c l a s s e s o f t o p o l o g i c a l spaces h a v i n g B a i r e ' s P r o p e r t y . Such a t o p o l o g i c a l space w i l l be c a l l e d a

B a h t hpace. THEOREM 1.

-

Evetly compkeLe meL&Lc hpace and evekg Svca.Ug cvmpact hpace

a m Babe npaceh. PROOF. Let

-

We g i v e i t o n l y f o r complete m e t r i c spaces.

U1,..

. ,U,,..

want t o show t h a t

,

U

we have t o prove t h a t

be dense open s e t s i n

E

i s dense. F o r t h i s , l e t

W

U nW

i s n o t empty.

7

, and U

.

fl Un We n>l be any open s e t i n E ;

=

B. BEAUZAMY

8 d

We c a l l

t h e d i s t a n c e on

t h e open b a l l o f c e n t e r

(go( x , r )

x

, and

E

and r a d i u s

.@o(xyr) = I y E E, d(x,y) < r l r go( x , r ) w i l l be i t s c l o s u r e

.

,

i s c o n t a i n e d i n .@(x,r) = ty E E, d(x,y) G r l

but the inclusion

can be s t r i c t ) .

U1

Since x1 E E

and

U1

i s dense,

>0 ,

rl

have b u i l t two sequences 1 0 < rk < F k = 1, ... ,n

, rk-l)

Uk n330(xk-l

,

, if

, for

. . ,xn

in

k = 2,

Bo ( x k , r k )

n

Then, Un+l

, rn)

go(xn

is

t h u s b u i l t i s a Cauchy sequence : f o r a l l

>

n

,

, x J. )

d(xi

2

Q-

n

go( x n , r n ) , which i s c o n t a i n e d i n

(qn>l converges Bo ( x n , rn),

contained i n

i s contained i n

i s dense, and t h e r e f o r e c o n t a i n s a b a l l 1 rn+l can be chosen s m a l l e r t h a t n+r .

-

sequence

. Assume we

, rly.. . ,rn > 0 , w i t h

E

...,n .

c U1 n W

Un

, where

i,j

i s open and non-empty, and so t h e r e a r e

such t h a t

The sequence n 2 1

W

such t h a t ?Bo (xl,rl)

xl,.

open and non-empty, s i n c e

go( x ~ +, rn+l ~ )

f?

i ~s }

Bo ( x n , rn) , f o r

belongs t o Un's

and

1

W

and t o

. This

n ,

all

proves t h e theorem,

f o r complete m e t r i c spaces. The p r o o f f o r l o c a l l y compact spaces d e a l s w i t h t h e same t y p e of arguments, b u t , i n s t e a d o f b a l l s , one uses compact s e t s , and t h e f a c t t h a t t h e i n t e r s e c t i o n o f a decreasing f a m i l y o f nonempty compact s e t s i s not-empty. The d e t a i l s a r e l e f t t o t h e reader. Theorem 1 has many a p p l i c a t i o n s i n F u n c t i o n a l A n a l y s i s and w i l l be used d i r e c t l y several times throughout t h i s book (see f o r example Second P a r t , But i s g r e a t e s t achievements a r e r e a l i z e d by t h e i n t e r m e d i a t e

Chapter V I ) .

of two v e r y i m p o r t a n t theorems, Banach-Steinhaus Theorem and t h e Closed Graph Theorem, t o which we s h a l l now t u r n . B u t b e f o r e , we s h a l l r e c a l l a few d e f i n i t i o n s , concerning v e c t o r spaces :

- a otrni-nam, on a v e c t o r space E , i s a mapping from E i n t o IR+ , denoted by

*

or

Ic

,

* -

p

,

s a t i s f y i n g t h e f o l l o w i n g two c o n d i t i o n s :

p(xx) = IXlp(x)

,

for all

x

E

E

, and f o r a l l

i s t h e f i e l d o f s c a l a r s o f t h e v e c t o r space p(x

+

y) Q p(x)

a n o m , denoted b y

+

p(y)

for all

x,y E E

E ).

.

II.11 , s a t i s f i e s moreover :

X E

IK

(IK

=

IR

BAIRE'S PROPERTY

*

IIxlI = 0

implies

9

.

x = 0

-

a Banuch n p c e i s a complete normed space.

-

If

E

F a r e normed l i n e a r spaces, and

and

l i n e a r mapping f r o m

llull =

sup

i s a continuous

u

F , we s e t :

into

E

,

IlU(X)llF

II XI1 Q 1

and t h i s f o r m u l a d e f i n e s a norm on t h e v e c t o r space Y ( E , F)

E

l i n e a r mappings from

into

F

and o n l y i f t h e r e i s a c o n s t a n t llu(x)Il,

Q

CllxllE

.

, since C

>0

.-

u

i s continuous i f

such t h a t , f o r e v e r y

x E

E ,

i s complete as soon as

One checks e a s i l y t h a t Y(E, F)

i s complete ( t h e completeness o f

F

we know t h a t

o f continuous

i s n o t needed).

E

be a Ranach ~ p a c c , F a named a @md?y 0 6 C O n t h U V U A fineah mappiMgA &am E i n t v n w e , and (ui)iEI F ( t h e A C L ~ I may have any catldinaLLty). Then, &a c u c a can OCCWL :

THEOREM 2. (Banach-Steinhaus)

ei2he.h t h e m ex,&&

a)

i n t m n e c f i o n ad apen

E

a A ~ AX , which A a countable nay thaX X ,& a G, ) , denne i n E , nuch

in E (We

A&

Let

that sup llUi(X)IlF i E 1 b)

ah

+

PROOF.

-

are i n

E+.

,

dvh&

c a m e , both

06

w

doh a l l

x E

x

.thehe e x h & a p ~ ~ . i t i vnumbetr e M

1 I ui 1 I <M,

and,

=

i E 1 ,

tehmd

We d e f i n e , f o r

un

We p u t

For each

iE I

and t h e r e f o r e

q

nuch t h a t

06 t h e aetehnative excLude each

x E

x , q(x)

= { x E E, q ( x )

, the

function

=

-

>

x

sup llUi ( x ) II i E 1 nI

i s l o w e r semi-continuous

athm.

The values o f

.

llui(x)llF

i s c o n t i n u o u s on

: t h i s i m p l i e s t h a t each

E

,

Un

i s open ( t h i s f a c t can a l s o be checked d i r e c t l y ) .

- I f one o f them, say UN , i s n o t dense i n non-empty i n t e r i o r . T h e r e f o r e t h e r e a r e all

x

in

E

with

llxll

Consequently, f o r a l l all

x

with

IIxll

o i s a v e c t o r space, and i f i t c o n t a i n s FN , as non-empty i n t e r i o r . But FN an open b a l l , i t c o n t a i n s an open b a l l c e n t e r e d a t 0 : FN i s a

basis,

E =

neighbourhood o f

. Therefore

0

E = FN

and

i s f i n i t e dimensional,

E

which c o n t r a d i c t s o u r assumption. We s h a l l now proceed t o t h e second main a p p l i c a t i o n o f Theorem 1 : t h e Closed Graph Theorem and i t s consequences. L e t us r e c a l l t h a t t h e graph o f a mapping

E

the s e t (contained i n

F )

x

u

o f a l l couples

- Let

THEOREM 5. (Closed Graph Theorem).

E, F

, from

E

into

.

x E E

(x,u(x))

F

,

is

be complete meZbLc fineah

bpaceb. Let u be. a Lineah mapping @om E i n t o F . Then u h con-t.imou~ .id and o d y i i j t h e gmph 0 6 u h cloded i n E x F , endowed Lclith t h e phoduct topology. REMARKS.

1)

I f t h e hypotheses on

and

E

F

a r e f u l f i l l e d , t o show t h a t

u

is

0

, and

c o n t i n u o u s , i t i s enough t o e s t a b l i s h t h a t :

.id

h a bequevxe. ad ,point2 in E , convehging t o (Xn)nEIN in F , t o a p o i & y , t h e n y = 0 ( U ( X ~ ) ) , , convehgen, ~~

'6

.

We know t h a t i f F i s a H a u s d o r f f space (which i s t h e case i f i t i s 2) m e t r i c ) t h e graph o f any continuous mapping i s c l o s e d . We o n l y have t o prove t h e converse imp1 i c a t i o n .

- We assume t h a t t h e graph o f u (denoted by G r u ) i s c l o s e d . We s h a l l f i r s t e s t a b l i s h a Lemma. We c a l l g E ( e ) , . g F ( e ) t h e c l o s e d b a l l s , i n E and F , c e n t e r e d a t 0 , and w i t h r a d i u s E : PROOF.

BE(€) = LEMMA 6.

-

gE(6)

Folr

E

{X E

eway u-l(

PROOF o f LEMMA 6. Gn = nu-'( g F ( e ) )

E

gF(E

d(x,O) < e l

.

> 0 , thme

& a numben

1)

- We f i x

.

6

>0

duch t h a t

*

E

>O

The r e u n i o n

, and c o n s i d e r t h e s e t s U

n>l

Gn

covers

E

,

therefore also

12

B. BEAUZAMY

Us .

5

Since E i s a B a i r e space, one o f t h e ' s , say , must n>l have a non-empty i n t e r i o r : i t must c o n t a i n some open b a l l o f r a d i u s q and c e n t e r

, that

xo

d(x,xo)

if

NOW, t a k e y

<TI

with

is :

, then d(y,O)

X

u(

< T I;

a,(€), i f

) E

.

n >N

then

X

9 n -) u( + ) ,

u ( i ) = u( and t h e r e i s a

>O,

XO N 1 such t h a t i f n 2 N1 , b o t h u ( t x o ) and u ( ) n n

%)

belong t o B F (

, and

u(

f

. F o r somme

) E BF(e)

6

> 0 , small

enough,

we have

which proves t h e Lemma. T h i s Lemma uses o n l y t h e l i n e a r i t y o f

u , and n o t

t h e f a c t t h a t i t s graph i s closed. L e t us observe t h a t we can assume t h a t

6 ( )~

<E , for

every

E

>0 .

We come back t o t h e p r o o f o f Theorem 5 . We s h a l l show t h a t , f o r e v e r y E

> O , we have u ( g E ( 6 (

continuous a t So, we f i x

with x

-

x

- x1

-

0 E

, and

u

is

t h e r e f o r e everywhere.

> O , and x

x1 E B E ( 6 (

))

x2 E B E ( 6 ( ) )

( X i ) i>1 , w i t h

% ) ) ) c @IF(€) : t h i s w i l l prove t h a t $ ) ) . We

E aE(6(

, and

then

, and'so

choose

x2 E u - l ( B F (

))

x1 E u - ' ( B F ( ;)), with

on : we b u i l d b y i n d u c t i o n a sequence i and x - z x . E s , ( 6 ( ~ / ~ i t l ) ) ,

xi E u - 1 ( B F ( e / 2 i ) )

j=1 J i = 1,2,.., . Then u(xi) E 9 ? F ( E / 2 i ) , and t h e r e f o r e t h e sequence n i s a Cauchy sequence, which converges t o a p o i n t z , and, zn = Z1 u(xi) for

1

zn E B F ( E )f o r a l l n , we have z E B F ( e ) . n Also, x - z xi E B ' ( 6 ( ~ / ~ n t l ) )C .43E(~/2n++1) , and t h e sequence 1 x n = x 1. t t xn converges t o x But s i n c e u(Xn) = zn , and s i n c e G r u i s closed, we have u ( x ) = z . Therefore, x E u - l ( g F ( e ) ) , which proves t h a t 9 ? E ( 6 ( ~ / 2 ) ) C U - ' ( . % ? ~ ( E ) ), and ends t h e p r o o f o f t h e theorem. since

...

-

.

E is n o t covered by a c o u n t a b l e union o f c l o s e d subsets w i t h empty i n t e r i o r : t h i s i s p r e c i s e l y t h e case f o r any B a i r e Space. T h i s l e a d s t o t h e f o l l o w i n g definition :

REMARK.

We do n o t r e a l l y need t h a t

E

i s complete, b u t o n l y t h a t

BAIRE'S PROPERTY

13

-

76 E h a topalogical? bpace, a b u b b d A ltliee be c&ed meager i n A , oh o f t h e f i r s t category, i6 M h c o n t a i n e d i n a c o u n t a b l e union 06 c l o n e d bub.5e.h ulith empty intehiok. I n t h e c o n v m e c u e , LA w i l l DEFINITION.

be c a e d o f second c a t e g o r y .

-

l e t E be a campteAe meX&c f i n e a n s p a c e , F a f i n e m npace 06 becvnd categvkq i n iLbel6. L e A u be a cantinuoub LLneah b i j e c t i o n dkvm E o n t o F . Then u-' h continuoun (and u h a n ,thmo,tpkinm). COROLLARY 7 .

-

PROOF.

G r u-'

u

If

i s continuous, t h e n

-

Gr u

i s obtained from

v a r i a b l e s : (x,y)

(y,x)

Gru

i s closed i n

by "symmetry",

E

x

F

.

But

t h a t i s by exchanging

, and t h e r e f o r e i t i s a l s o c l o s e d . So u - l

is

continuous by t h e p r e v i o u s theorem. DEFINITION.

-

We bay t h a t a mapping h open id t h e image oQ any open n d by

t k i n appfication

an open b d .

O f course, i f t h e mapping i s i n j e c t i v e , i t i s open i f and o n l y i f t h e i n v e r s e mapping i s continuous. I n t h e sequel, t h e word "opehatoh" w i 11 mean " c o n t i n u o u s 1 i n e a r mapping". THEOREM 8. (Open-mapping Theorem).

bpace,

.

be a c a m p l d e m&c

E

a Haundoh66 t v p o l v g i c d f i n e m bpace,

F

into F

- LeX

Then

u

LLneatl

an apehatoh 6kvm

EMhm u(E) can be covehed by a cvuntabLe u n i o n a6 c l o n e d F , w i t h empty h ~ t & v h (that h , u(E) h meageh i n F 1 -

in

-

Oh

E

:

u(E) = F

,

u .LA open, and

F

be.h,

,

a compLeZe m W c finean

bpace. PROOF.

- We

V = u(E) : t h i s i s a v e c t o r subspace o f

put

-

F

. Assume

that

i t i s n o t meager i n F : then, c e r t a i n l y V has non empty i n t e r i o r . Since V i s a subspace, 0 i s i n i t s i n t e r i o r . T h i s i m p l i e s t h a t = F , and

i s dense i n

F

7

.

S t i l l assuming t h a t n o t meager i n i t s e l f : i f t h e n each and

Fn

V c n Gn n>l

V

i s n o t meager i n

V =

can be w r i t t e n

, and

U Fn

, where

n>l Fn = V n Gn

so one o f t h e

Gn ' s

F

, we s h a l l see t h a t V i s

Fn I s

, where

are closed i n

V

,

Gn i s c l o s e d i n

F

,

, say GN , has non-empty

14

B. BEAUZAMY

-

interior

in

GN

open s e t i n

V

interior i n

V

F

,

V

Since

, contained i n , and t h a t V i s

Now, l e t

El

mapping from

E

-u

: t h i s proves t h a t

FN

El

,

Ker u , 0

by

E

t h e q u o t i e n t map f r o m

;;

i s an open mapping, t h a t

knows t h a t 0

FN has non-empty

n o t meager i n i t s e l f .

be t h e q u o t i e n t o f onto

0

F , V n GN i s a non-empty

i s dense i n

t h e canonical

El

into

F

.

One

i s i n j e c t i v e and continuous,

and t h a t , endowed w i t h t h e q u o t i e n t - d i s t a n c e ,

i

space ( r e c a l l t h a t , i f x E

i

y

i s a complete m e t r i c El x , d ( i , i ) = infId(x,y) ;

i s the class o f

y E ;I).

V

Since

i s n o t meager i n i t s e l f , and s i n c e

space, we can a p p l y C o r o l l a r y 7,and

El

i s a complete m e t r i c

be equipped w i t h t h e m e t r i c brought by

(that is,

d(u(x),u(y)) = d(x,y))

and i s complete f o r t h i s m e t r i c . Since i t was dense i n

-u

F

, we

u i s s u r j e c t i v e and F i s a complete m e t r i c space. Moreover,

, and

u =

REMARK.

- I f , f o r example,

theorem says t h a t , i f still,

u(E)

V

have u

=

F

i s open

b o t h a r e open. T h i s proves t h e theorem.

since

0

can

i s an isomorphism. T h e r e f o r e V

u

E

and

F a r e Banach spaces, t h e p r e v i o u s

i s not surjective,

can be dense i n

F

. This

u(E)

i s meager i n

F

.

But

i s t h e case, f o r example, o f t h e

canonical i n j e c t i o n from L ([0,11 , d t ) i n t o L ([0,11 , d t ) , i f p > q . P q A s e t can v e r y w e l l be meager and s t i l l be dense i n t h e whole space : f o r example, Q i n R

.

To i l l u s t r a t e t h e p r e v i o u s theorem, we s h a l l e s t a b l i s h t h e f o l l o w i n g proposition : PROPOSITION 9. subspace

04

-

LeX

L P ( R ,dt)

, q , Lclith n L (IR , d t )

p

q

.

1 < p,q

L q ( R , d t ) , t h e t o p o l o g i a induced on

E

16 E

l ( f n , f n ) c o n v e r g e s i n Lp x L t o an element ( f , g ) . T h e r e f o r e q fn f in L f g i n L . So we can e x t r a c t from t h e P ' n n3+ q

----

,

BP.IRE'S PROPERTY sequence

(fJn>1

everywhere ;

a subsequencelanJ' , f(

which converges t o

almost

f

.

Therefore f = g , q I i s continuous and s u r j e c t i v e , t h e

t h i s sequence converges t o

I i s c l o s e d . Since

and t h e graph o f

15

g

in

L

converse mapping i s a l s o continuous ( C o r o l l a r y 7 ) a n d I i s an isomorphism :

E

t h e r e f o r e , b o t h t o p o l o g i e s coi'ncide on

, and

t h e p r o p o s i t i o n i s proved.

We s h a l l see i n t h e Second P a r t , Chapter V I , some examples o f subspaces

E

which s a t i s f y t h e assumptions o f t h i s P r o p o s i t i o n .

EXERCISES ON CHAPTER I. EXERCISE 1. (Condensation o f s i n g u l a r i t i e s ) . - L e t m e t r i c v e c t o r space, F a m e t r i c v e c t o r space, and sequence o f l i n e a r mappings from

IN ,

E

into

F

.

E be a complete ) a double (u PYq P E W qEIN We assume t h a t , f o r e v e r y

Lim u ( x ) does q + + m P19 P n o t e x i s t . Show t h a t t h e s e t o f p o i n t s x E E such t h a t Lim u (x) q + t m P ,q e x i s t s f o r no p E IN i s o f second c a t e g o r y i n E p E

there i s a point

x

P

E E

such t h a t t h e l i m i t

.

EXERCISE 2. from E o n t o F

L e t E and F be Banach spaces, u a s u r j e c t i v e o p e r a t o r F Show t h a t , f o r e v e r y sequence o f p o i n t s (Y,),>~ in

.

converging t o a p o i n t E

in

('n)n>l

, there

yo = u ( x o )

-

EXERCISE 3. ( l i f t i n g p r o p e r t y o f

ll ) .

and A a s u r j e c t i v e o p e r a t o r from

E onto

Ll

into

, there

F

i s a sequence o f p o i n t s

, converging t o x 0 , w i t h u ( x n )

z

i s an o p e r a t o r

T

from

Let F

.

L1

E

= yn

and

for all F

n

>1.

be Banach spaces,

For every operator

T

from

E , such t h a t

nto

A?=T. [Show f i r s t t h a t , i f are points for all

E

into

F

x

IN.

n E

E X E R C I S E 4. for all

(

-

(en)n

IN

i s t h e canon c a l b a s i s o f

E~ , such that ~ ~

ll , t h e r e

sup llxn < t m , and Axn = Ten n F o r t h i s , use t h e f a c t t h a t A s an open mapping1

Let

~

i )n

.

E

and

F

be Banach spaces, and

T

an o p e r a t o r f r o m

, w i t h c l o s e d range. Show t h a t t h e r e e x i s t s a K > 0 such t h a t one can f i n d x E E , w i t h Tx = y and I l x l l < KllyII .

y E ImT,

B. BEAUZAMY

16

REFERENCES ON CHAPTER I . Most r e s u l t s which a r e presented i n t h i s f i r s t c h a p t e r a r e due t o S.

BANACH h i m s e l f , o r , a t l e a s t , appear i n h i s book 1 5 1 . The p r o o f o f t h e

Closed Graph Theorem which we g i v e here i s v e r y c l o s e t o t h e one g i v e n by W.

RUDIN [431.

CHAPTER I 1

INFINITE-DIMENSIONAL NORMED SPACES

INTRODUCTION. In this chapter, we s h a l l develop t h e most important r e s u l t s concerning normed l i n e a r spaces. In f a c t , most of them make sense a l s o in t h e more general s e t t i n g of l o c a l l y convex topological vector spaces, and we could have made a more general - and more a b s t r a c t - study. We have chosen t o r e s t r i c t ourselves, both because we have i n mind the study of Banach spaces (endowed with their norm, o r with a weak topology, see below), and because t h i s w i l l allow us, i n many c a s e s , t o present simpler and more concrete proofs. There i s one exception, however : the Hahn-Banach Theorem, i n i t s two forms ( a n a l y t i c and geometric), i s given i n i t s l a r g e s t s e t t i n g ; t o do so does not require more work, and will be useful. Except f o r t h i s r e s u l t , we have avoided as much as we could t o develop a general theory of l o c a l l y convex spaces ; the reader i n t e r e s t e d by such a theory i s r e f e r r e d t o Bourbaki "Espaces Vectoriels Topologiques", Chapitres I 2 IV, [ll]. A vector space E i s .ind.iYLite-d&endond i f one can find i n t h i s space a sequence of points ( e n ) n E M , such t h a t any f i n i t e subset of { ( e n ) n E IN 1 i s made of l i n e a r l y independent vectors. I f t h e space i s equipped w i t h a norm, we may of course assume t h a t , f o r a l l n E IN, llenII

=

1.

In an infinite-dimensional normed space, one can f i n d a sequence of norm-one points which a r e f a r from each o t h e r : PROPOSITION 1. - L e A E be an in&bzLte-d&enniond notmed npace. One can hind i n E a nequence o d poina2 nuch t h a t , 6011 at?.! n 2 1 ,

IlxnII and

=

1

,

d i s t ( x n + l , spanlx l , . . . , x n l ) 2 1

.

T k i n L a 2 condition h p & c ~ o d c o u ~ ~ s ellxn

17

- xmII 21

404

n,m, n # m.

B. BEAUZAMY

18 PROOF.

-

Assume

xl,...

,xn

have been c o n s t r u c t e d . Put

F = spanIxl,. . . ,xn1 , and l e t a E E , a E$! Fn . Put Gn =spanIa,xl ,... ,xnl. n Then Fn i s an hyperplane o f Gn , and so t h e r e i s a l i n e a r f u n c t i o n a l fn

,

d e f i n e d on

fn = 0

on

such t h a t

Fn

( w h i c h i s a f i n i t e - d i m e n s i o n a l space), such t h a t

Gn

. One may o f

course assume

l l ~ ~ += ~1I l and fn(xn+l)

unit ball of

.

scalars :

n

z

1

aixill

2 fn(xn+l

x ~ +be ~

L e t now

= 1 (such a p o i n t e x i s t s , s i n c e t h e

i s compact). Then, f o r e v e r y sequence

Gn

IIX,+~ -

IIfnII = 1

n

- r. a 1. x .1)

= 1

al

,.. . ,a n

of

,

which means t h a t

>1,

, span(x l,...,xn))

dist(xn+l

and t h e P r o p o s i t i o n i s proved.

A l i n e a r f u n c t i o n a l on an i n f i n i t e - d i m e n s i o n a l space a p p l i c a t i o n from of s c a l a r s

, as

i n t o the f i e l d o f scalars o f

E

.

E

IK,

we a l r e a d y mentioned, i s denoted by

E

i s a linear

For us, t h e f i e l d and i s R , o r

(i:

.

A t o p o l o g i c a l v e c t o r space i s a v e c t o r space endowed w i t h a topology, f o r which t h e a d d i t i o n o f v e c t o r s , and t h e m u l t i p l i c a t i o n o f a v e c t o r by a s c a l a r a r e continuous o p e r a t i o n s . I f t h i s t o p o l o g y i s g i v e n by a m e t r i c , we have a m e t r i c v e c t o r space. A s p e c i a l case i s g i v e n by t h e normed spaces. The m e t r i c v e c t o r spaces were a l r e a d y

considered i n t h e f i r s t

c h a p t e r . A l l o u r t o p o l o g i c a l spaces w i l l always be H a u s d o r f f .

If E : E*

E

i s a t o p o l o g i c a l v e c t o r space, we c a l l

E*

i s t h e s e t o f continuous l i n e a r f u n c t i o n a l s on

finite-dimensional,

E

(if

E

is

any l i n e a r f u n c t i o n a l i s a u t o m a t i c a l l y continuous ;

E

t h i s i s n o t so i f we define a norm on (1)

t h e dual o f

i s infinite-dimensional).

*

E

If

E

i s a normed space,

by t h e formula :

Ilr; II = sup Ir;(x)( , E* llxll < l E

and we know t h a t

E*

i s always complete, even i f

r e c a l l e d i n Chapter I , p. 9 here, F =

, for

Y(E,F)

E

i s n o t ( t h i s was

, where F i s a Banach space :

IK ) .

The f a c t t h a t t h e r e e x i s t , on an i n f i n i t e - d i m e n s i o n a l normed space, n o n - t r i v i a l continuous l i n e a r f u n c t i o n a l s ( t h a t i s , d i f f e r e n t from t h e z e r o

19

INFINITE-DIMENSIONAL NORMED SPACES

l i n e a r f u n c t i o n a l ) i s n o t obvious : i t i s a consequence o f t h e f o l l o w i n g theorem, which p l a y s an e s s e n t i a l r o l e i n A n a l y s i s .

Q 1. HAHN-BANACH THEOREM

: ANALYTIC FORM.

-

LeZ

be a 4 e d uecto4 clpace, and a d d i t i v e and p o c l i t i v e l y homogeneous &~ncLLonon E , thaL & :

THEOREM 1. (Hahn-Banach).

E

be a vectoh subclpace 0 6 o n F and dominated b y p : LeA

E

7

Then thehe & an extevuion

-

PROOF.

Let

e E E

and

where

06

, f a fineah &ncLLond, dedined

t o E , b.tieR dominated by p

f

f

We f i r s t show t h a t we can extend

which c o n t a i n s

F

E

e

.

F

x E F

,e

,X

E

.

F

G

Let

+ xc

G p(x

. We

IR

and we s h a l l choose f(x)

t o a v e c t o r subspace o f

and has "one dimension more" t h a n

define

c = F(e)

+

xe)

,

F

.

be t h e v e c t o r subspace of

y E G

Then e v e r y p o i n t

-f ,< p

?(y) = f ( x ) t x c

i n order t h a t for all

x E F for

, for

, all

A = 0

.

every

G

on

y = x t Xe, y E G

< p(

X

t e

and t h e r e f o r e , we must have

X >O

c G i n f { p ( x l t e) For

X

1

all

g o E E*

E

.

for

:

xl,...,xn

E E

.

t h i s topology, which i s complete y d e f i n e d by t h e

p r e v i o u s fundamental system o f neighbourhoods. I f

sequence o f elements o f for a l l

x E E

neighbourhoods

* E , it

En(x) V

E YX

(Eo) ,

converges t o

Eo(x)

(En n E N i s a t o E E* i f and o n l y i f ,

( t o see t h i s ,

u s i n g o n l y one p o i n t

j u s t take the

x ) . T h i s i s t h e reason

why t h i s t o p o l o g y i s c a l l e d " p o i n t - w i s e convergence on t h e elements o f E

'I.

But i t s h o u l d be n o t e d t h a t t h i s d e f i n i t i o n w i t h sequences does n o t s u f f i c e t o characterize the topology, since i t i s not metrizable i f E i s i n f i n i t e dimensional.

8. BEAUZAMY

24

*

The t o p o l o g y o ( E ,E)

i s weaker t h a n t h e norm-topology :

a neighbourhood o f t h e f o r m (1) c o n t a i n s t h e b a l l centered a t E o

I

radius

.

IIE, - toll

norm, t h e n

-

IEn(x)

t n , if In n3+m. E o

max IlxiII (for a sequence i=l,. . ,n

e

Eo(x)l

, and

0

< llEn -

tll

. llxll

thus, f o r a l l

x E

E ,

, with

for the

*

is in

0 ) . The t o p o l o g y o ( E ,E)

n++m

general s t r i c t l y weaker, as shows t h e example o f t h e v e c t o r s o f t h e canonical b a s i s o f o(L2

on

, L2)

. The

L2 , which a r e o f norm 1, b u t t e n d t o zero, f o r

*

t o p o l o g y o ( E ,E)

i s o f t e n c a l l e d t h e weak

*

topology

.

E*

The main i n t e r e s t o f t h i s t o p o l o g y i s t h a t i t w i l l a l l o w us t o o b t a i n compactness r e s u l t s . One knows t h a t , i f a normed space

E

i s infinite

dimensional, i t s u n i t b a l l ( f o r us, " u n i t b a l l " w i l l always mean " c l o s e d u n i t b a l l " ) i s n o t compact f o r t h e norm topology. T h i s f a c t i s t h e R i e s z ' s Theorem : t h e u n i t b a l l o f a normed space i s norm-compact i f and o n l y i f t h e space i s f i n i t e - d i m e n s i o n a l . Up t o some extend, t h e weak t o p o l o g y w i l l a l l o w us t o a v o i d t h i s drawback. THEOREM 1. ( A l a o g l u ) .

-

PROOF. unit

-

d ball! od

cumpuct doh. u(E*,E)

E*

*

E* , which i s d e f i n e d w i t h r e s p e c t t o t h e norm o f E , by 1 ( l ) , and t h e weak t o p o l o g y o(E*,E) . The u n i t b a l l gE* i s

n o t a neighbourhood o f

,in

0

of

o(E*,E)

, so

f a c t , f o r any neighbourhood o f

does n o t a p p l y t o i t , and BE*can

0

t h e R i e s z ' s theorem (which i s

,in

compact when

*

E

L e t us f i r s t observe t h a t we can i d e n t i f y I K E , by a s s o c i a t i n g t o each

{E(x), x E E l

.

E

* E E

E*

is

w i t h a subset o f t h e

t h e s e t o f i t s values :

One checks immediately on t h e d e f i n i t i o n o f neighbourhoods

t h a t t h e t o p o l o g y u(E*,E) E*

a l o c a l l y convex space)

be u(E*,E)

infinite-dimensional. product

.

L e t us observe f i r s t t h a t t h e statement i n v o l v e s two t h i n g s : t h e

ball o f

formula true

The u

i d e n t i f i e s i t s e l f w i t h t h e t o p o l o g y induced on

by t h e p r o d u c t t o p o l o g y on

IKE

. Moreover,

the u n i t b a l l o f

E*

becomes a subset o f t h e p r o d u c t :

I =

l-l- {A XEE

E

IK ;

1x1 < Ilxlll

which i s a p r o d u c t o f compacts, and t h e r e f o r e i s compact by T y c h o n o f f ' s

INFINITE-DIMENSIONAL NORMED SPACES

25

theorem. A l l we have t o show i s t h e r e f o r e t h a t BE* i s c l o s e d i n x,y E E

F o r each P

x YY

(5)

= 5(x) + 5(y)

E

These f u n c t i o n s a r e continuous on

, and

E*

the u n i t b a l l o f

is

I : o u r theorem i s proved.

which i s a c l o s e d subset o f

3. THE BIDUAL

ME

I with

the i n t e r s e c t i o n o f

5

.

IK , l e t us c o n s i d e r t h e f u n c t i o n s E ( x + Y ) and $x,X(C;) = # ( A X ) - X t ( x ) .

, each X

-

I

AND THE WEAK TOPOLOGY ON

E**

We have seen t h a t

E*

, endowed

E

.

w i t h t h e norm d e f i n e d by t h e f o r m u l a 1,

(l), was a Banach space. So we can, again, c o n s i d e r i t s dual : i t i s a Banach space

which i s c a l l e d t h e b i d u a l o f

There a r e , i n

** , some E

-

E

,

and i s denoted by

.

E**

s p e c i a l elements, which can be immediately

-

i d e n t i f i e d . These a r e t h e elements o f t h e f o r m :

(1)

x0 E E

t5 E E*

( t h a t i s : t o each element o f point

5(xo)1

, we a s s o c i a t e i t s v a l u e a t a f i x e d

E*

xo E E ) . T h i s a p p l i c a t i o n w i l l be denoted by

,t >

<xo

. By

f o r m u l a 1, ( 2 ) , we have :

I
z ( 5 ) t o i n d i c a t e t h e a c t i o n o f z E E** on C; E E* , we can 0

g i v e t h e form o f a fundamental system o f neighbourhoods o f E*)

o(E**,

for all

E

,all

>O

, all tl,...,tn~E

n > l

A nequence o f elements E*)

u(E**,

Here again,

for

: these a r e t h e s e t s o f t h e form :

* . zo

( z ~ ) ~ ~ , converges , , t o an element

,

i f and o n l y i f , f o r a l l C; E E*

, C; >-< n + + m

.

t h i s d e f i n i t i o n u s i n g sequences does n o t s u f f i c e t o characte-

r i z e t h e topology. L e t us now l o o k f o r t h e t r a c e on E o f t h e t o p o l o g y has a meaning, s i n c e

** . A E

i s a subspace o f

E

E*)

u(E**,

: this

fundamental system o f

neighbourhoods o f t h e o r i g i n i s , by d e f i n i t i o n , c o n s t i t u t e d b y t h e intersection with

E

of

t h e s e t s ( 2 ) . Therefore, t h i s fundamental system

i s g i v e n by t h e s e t s :

for all

E

>O

,

We s h a l l c a l l

all

n

>1,

u(E, E )

all

tl,...,5n

* .

E E

t h i s t o p o l o g y : i t i s t h e weak t o p o l o g y on

E

From t h e f o r m o f t h e neighbourhoods ( 3 ) , one deduces t h a t i t i s t h e a nequence o(E,

E*)

( x ~E )N ~ o f elements o f

i f and o n l y i f , f o r every

E

converges t o a p o i n t

C; E E*

< xn , 5 >

w i t h t h e new n o t a t i o n s ,

. I n particular,

E*

t o p o l o g y o f p o i n t - w i s e convergence on t h e elements o f

, C;(x,) <xo

x0 f o r

t(xo) Y

E

(or,

>).

i s equipped w i t h a norm, by a formula analogous t o 1, (1). By E** A l a o g l u ' s theorem, i t s u n i t b a l l BE** i s compact f o r o(E**, E*) . But the u n i t b a l l WE o f

E

*

has no reason t o be o ( E , E )

compact : we

s h a l l come back on t h i s p o i n t i n t h e n e x t c h a p t e r . We s h a l l now t u r n o u r i n t e r e s t t o t h e spaces w i t h t h e i r weak t o p o l o g i e s : o(E, E*) PROPOSITION 1.

-

and

E

E)

.

E* , equipped

t h e w e a k a t t o p o l o g y on E* all t h e fineah &~ncLiona& t E E* -+ < x , t > ,

The t o p o l o g j o(E*,

wkich h e n d m ConLinUOUn doh x E E .

and o(E*, E)

.

27

INFINITE-DIMENSIONAL NORMED SPACES

-

PROOF.

E

F i r s t , i t i s obvious t h a t i f

*

i s endowed w i t h

o(E

, E) ,

these l i n e a r f u n c t i o n a l s a r e continuous : t h e i n v e r s e image o f a 0 , i n M y {[XI < e l , i s the s e t , I < x , E > ( < e l , which i s o f t h e t y p e 2, ( 1 ) .

neighbourhood o f

It

E E*

Conversely, l e t Z be a t o p o l o g y on functionals

r;

points i n

,

o(E*,

E)

E

--f

E

< x, t > , x E E , a r e > 0 , and VEiX 1 . . xn I .

. Since,

,

continuous f o r Z

. The

we c a l l i t

Oi

open f o r Z

, and

than

i = l,...,n

for

It

the set

intersection

V, ;x

it i s

continuous. L e t

xlY...,xn

a neighbourhood o f

Y

0

be for

, t h e mapping 5 - + < x i ,E > is E* ; I< xi , E >(<E l i s open f o r z ; n Oi o f these open s e t s i s s t i l l i=l,.. . ,n . t h i s proves t h a t Z i s s t r o n g e r

1,.**,Xn

.

u(E*, E)

E

such t h a t a l l t h e l i n e a r

E*

*

One proves t h e same way :

-

PROPOSITION 2.

The t o p o l o g y u ( E , E*)

LA t h e weahent t o p o L o g y on

which kendekb coM.tinuoub all t h e f i n e a h d u n c f i o n a h 604

x E E

* .

--t

< x,

5

E

>,

t E E

-

PROPOSITION 3.

76

endowed w a h o(E*, E )

E*

be i d e n t i d i e d Mlith E

: i n

fineatr d u n c i i o n u b on

E*

othck

,

tQtrm6,

,

the d u d

06

can

E*

the,tc me no o t h m continuoub

than t h e appficaLLons

t E E*

-+

< x,

t

>,

X E E .

PROOF.

-

We have seen t h a t t h e l i n e a r f u n c t i o n a l s

t

-+

< x, E >

continuous, we s h a l l see t h a t t h e y a r e t h e o n l y ones. L e t

* E

f u n c t i o n a l on E

> 0 , there

-5 q ; x

, continuous f o r > 0 , and xl,..

is q

,

l,...’Xn

then

lf(t)l

It(X1)l 0 nuch t h a t

.

PROOF. - We do t h e same as i n t h e p r e v i o u s lemma, and a p p l y c o r o l l a r y 4. We s h a l l now a p p l y these t o o l s t o t h e convex s e t s , c l o s e d under t h e weak o r under t h e s t r o n g t o p o l o g i e s , i n a Banach space.

:

34

5

B. BEAUZAMY

IN

5. CLOSED CONVEX SETS AND BOUNDED SETS TOPOLOGIES.

5

We now come back t o t h e s e t t i n g o f form o f t h e neighbourhoods g i v e n f o r

THE WEAK AND I N THE STRONG

2 and 3. It i s obvious on t h e

o(E*,

, u(E**,

E)

E)

, u(E, E*) ,

t h a t these neighbourhoods a r e convex. Also, i f two p o i n t s a r e d i s t i n c t , one can f i n d a continuous l i n e a r f u n c t i o n a l which does n o t t a k e t h e same v a l u e on b o t h : t h e r e f o r e , these t o p o l o g i e s a r e H a u s d o r f f . We have t h u s

5

which we s h a l l be a b l e t o a p p l y t h e t o o l s o f

,

E

i n t r o d u c e d H a u s d o r f f l o c a l l y convex t o p o l o g i e s on

E* o r

, to

E**

4.

The n e x t r e s u l t i s a s p e c i a l case o f a theorem o f G. Mackey :

-

PROPOSITION 1.

Let

be u nomed npuce. Ewetry conwex oubhet od &bed doh o(E, E*) .

E

E

w k i c h i n cLo4ed do& t h e r m h m .h &o

(The converse i s obvious, s i n c e t h e norm-topology i s s t r o n g e r t h a n o(E, E * ) ) . PROOF.

-

5

4, t h e r e is a r e a l l i n e a r f u n c t i o n a l

IR , w i t h form t y E E ;

cy

be a convex s e t , c l o s e d f o r t h e norm. L e t

C

Let

corollary 5, E

f(xo) < a

C

u(E,

E

C

E*)

.

xo

6

C

.

By

and a number But a s e t o f t h e

(since

i s open, by d e f i n i t i o n o f t h e neighbourhoods) ; t h i s

, and

be i n t h e c l o s u r e o f

y

>a for all

f ( y ) 2011 i s c l o s e d f o r

Iy E E ; f ( y ) , x E E . We say t h a t a Banach space i s trel;lexive i f a l l t h e continuous l i n e a r f u n c t i o n a l s on E* a r e o b t a i n e d

**

t h i s way ; i n o t h e r terms, i f

E

. Since E**

= E

t h e norm, t h i s e q u a l i t y i s p o s s i b l e o n l y i f

i s always complete f o r

i s complete, t h a t i s , i f

E

E

i s a Banach space. B u t n o t a l l t h e Banach spaces a r e r e f l e x i v e : we have seen t h a t i f

lp (1 < p

, then

E = co

< +m )

or

L

E**

. On

lm

=

(1 < p

< +m

)

t h e c o n t r a r y , t h e spaces are r e f l e x i v e .

P I n t h i s paragraph, we s h a l l s t u d y t h e p r o p e r t i e s o f r e f l e x i v e Banach

spaces, and g i v e "geometric" c o n d i t i o n s f o r r e f l e x i v i t y . i s r e f l e x i v e , t h e u n i t b a l l BE i s

E

First, i f

s i n c e we know t h a t gE** i s

o(E, E )-compact,

*

.

E )-compact, and LBE = gE**

o(E**,

Conversely, t h i s p r o p e r t y c h a r a c t e r i z e s r e f l e x i v i t y : PROPOSITION 1.

*

-

E

iA hedlexive i d and avLey i d i,t~u n i t b a l l BE iA

o(E, E )-campact. PROOF.

-

We assume BE t o be

embedding o(E**,

E**

of

j

E*)

. Since

E

into

PROOF. is

-

o(E*,

E**

(by d e f i n i t i o n o f

i s continuous from o(E, E*))

i t i s dense i n .!BE**

we have gE**= BE , j COROLLARY 2.

*

o(E, E )-compact. Since t h e c a n o n i c a l

-

16

E

,*fE

o(E

*is

,

i s s u r j e c t i v e , and

* , E)-compact.

49

in

E ) E i s reflexive.

o(E

compact, and by p r o p o s i t i o n 1,

o(E**,

into

* E )-compact

( c h a p t e r 11, § 5 , p r o p . 4 ) ,

iA t l e ~ l e x i v e ,.iA d u d E*

We know t h a t BE* i s

** E )

for

o(E, E*)

E*

iA &a If

E

tre@?exive. i s r e f l e x i v e , gE*

i s reflexive.

6. BEAUZAMY

50

-

PROOF.

Let

F

be a c l o s e d subspace. The u n i t b a l l BF i s

*

compact, s i n c e o(F, F ) =

BF i s

-

i s r e f l e x i v e , so i s 16 E

**

o(F, F*)-

7, prop. l ) , and s i n c e

( c o r . 2 ) , and so i s

E

( c o r . 3).

E

h ) ~ e 6 L ~ x i u e&oh , any covLtinuaun 6unctLonaR an

at C e u t one p o i n t xo in ,the

them

0

.

E* h ) ~ e 6 L e x i u e no , h E

16

- I f E*

COROLLARY 5 .

(chap. 11,

E*)lF

E

o(E,

COROLLARY 4. PROOF.

* )-closed.

a(€,

u n i t b a l l a6

E

buch t h a t :

E

*

L e t t E E . Then E i s continuous on BE endowed w i t h (chap. 11, 5 3, prop. 4) ; t h i s s e t i s compact i f E i s r e f l e o ( E , E*) x i v e . Therefore, t h e r e i s a p o i n t xo E BE such t h a t sup I t ( x ) l Replacing, if necessary, x by f x ( i f the 0 0 Ik(xo)l = llxllrL 91 PROOF.

-

.

space i s r e a l ) o r by

This p o i n t

E

if3

xo

IIxoll = 1

and, o b v i o u s l y ,

which

e

~

.

( i f t h e space i s complex), we o b t a i n ( l ) ,

(which i s n o t unique i n g e n e r a l ) i s c a l l e d a p o i n t a t

xo

a t t a i n s i t s norm.

Conversely, l e t us g i v e an example o f a Banach space and a l i n e a r

-

f u n c t i o n a l on i t which does n o t a t t a i n i t s norm. We c o n s i d e r t h e Banach space

co =

linear functional Yo=O

Y

Y n = p

y

xn

n

0

IR.

for all

,

n

g i v e n by t h e element o f

, for

< x , y > = L: xnyn . that

; xn E

{(XnInED\I

n >, 1

We have and

.

llyll

suplxnl n

n o t a t t a i n i t s norm on Bc

0

.

If

5

x E co

,x

-t

+

1, : y =

,

, and t h e (Y,),~,, , with 01

x = (xn)nEN

= 1, but, f o r every

< 1 , then I

L: xnynI n

, then

x = ( x ~ ) , , ~such ~

k

Fva e w a y 0 , 0 i n E , uLith IIxkII

c)

poi&

I

n G k

id id

= 0

pCibh%e

< 1 , t h e m iA a bequcnce dvh all k , buch t h a t :

> 1 , all

k { Z

k G

aixi

;k

K , id k

.Z

‘

01.

i=1

=

(

x

~

v6)

. ,ak ,ak+l,.. . ,aK

al,..

K

X a.

i=k+l

> 1 , a.1 > 0 ,

1 (which we c a l l

Ixl,...,xk}

I i

= 1

nurnbeho ba-tindying

Since t h e s e t hull of

0 ,

( f o r James). One i n t e r e s t i n g t h i n g

about t h i s c o n d i t i o n i s t h a t i t a l l o w s t o check t h a t a space i s nonr e f l e x i v e w i t h o u t knowing i t s d u a l .

~

~

~

B. BEAUZAMY

52 PROOF. j

-

a)

=+

b ) . We assume t h a t

t h e canonical embedding o f

c l o s e d subspace o f E**

E**

i s n o t r e f l e x i v e . As b e f o r e , we c a l l

E

E

into

E**

. We

know t h a t

j(E)

( f o r t h e norm), which i s n o t t h e whole space

**

: t h e r e f o r e , we can f i n d a continuous l i n e a r f u n c t i o n a l on

o f norm 1, which i s equal t o z e r o on choose

, with

F E E**

is a

IlFll

< 1 , such

j(E)

that

. Let

B

,

a(F) > B

0 0 , and cl,.. . ,c n h c d m . Then, 604 e v a y E > 0 , one can hind a p o i n t x E E , w i t h llxll = M t E and fk(x) ck 604 k = 1, ..., n , id and onLy id, 604 evetry bequence

LEMMA 7 . ( H e l l y ' s C o n d i t i o n ) .

al

,.. . ,an 06

n c d m , one h a :

PROOF OF LEMMA 7.

-

Assume t h a t t h e r e i s a p o i n t

and f k ( x ) = Ck , f o r al,...,an , we have :

.

k = l,...,n

Then,

x E E with

llxll = M t

E ,

f o r e v e r y sequence o f s c a l a r s

from which (H) f o l l o w s . Conversely, assume (H) t o be s a t i s f i e d . We may assume t h a t t h e f i l s

i , and we t a k e any x w i t h llxll = M + E ) Among t h e n f i l s , we can f i n d a t most r ( r 9 n ) which a r e l i n e a r l y independent. We assume, f o r t h e sake o f s i m p l i c i t y , t h a t i t i s fl,...,fr . We c a l l Hi = I x E E ; f i ( x ) = x .11 ,

a r e n o t a l l equal t o z e r o ( o t h e r w i s e , we have

.

ci = 0

for a l l

53

REFLEXIVE BANACH SPACES

i = 1,

for

...,n

n Hi i s an a f f i n e subspace o f ,r i = l, From (H) f o l l o w s a l s o t h a t F = n Hi i = l ,...,n

.

r

codimension

,

and

0 , o f radius

centered a t

5

Theorem (chap. 11,

, which

.. .

a = inf{llxll ; x E F}

We p u t

F

F =

. Then

. By t h e

a

F

does n o t meet t h e open b a l l ,

g e o m e t r i c f o r m o f Hahn-Banach

4, t h . l ) , t h e r e i s an a f f i n e hyperplane H c o n t a i n i n g

does n o t meet t h i s open b a l l . T h i s means t h a t t h e r e i s a

continuous l i n e a r f u n c t i o n a l F c H =tx E

E

; f(x) = c}

f

, and

and

But

F c H , and t h e r e f o r e t h e

II f II

.

scalar

c

such t h a t

.

llxll = - I >C aI

imply f ( x ) = c such t h a t : n f = C aifi i=l

a

we have :

inf

x €H

.

n

conditions

fi(x)

It follows that there e x i s t scalars

and

n

c =

C a.c.

i=l

1 1

= ci

ai

,i

,i=

= 1,

...,n

1, ...,n

,

.

By (H) , we have :

n

and t h e r e f o r e xo E F

with = 0

fi(x')

a

x

choose

< 14 .

This i n d i c a t e s t h a t f o r every

IlxoII G M t

E

.

NOW, t a k e

f i r s t that, since

>

fl(xl) that

IF(fl)l = 0

xk

.

= 0

can f i n d

, we

choose

xl€

, with

llxlII

. Therefore,

= 0

.

= 1

Thus

, and

T h i s i s t h e f i r s t s t e p o f t h e i n d u c t i o n . L e t us now assume

and

-

fk have been chosen, f o l l o w i n g t h e p r e s c r i b e d c o n d i t i o n s

.

k = p 1 We now choose f , w i t h Ilf II = 1 , F ( f ) = 0 and P P P f (x ) = 0 i f k < p T h i s c h o i c e i s p o s s i b l e , because t h e f o l l o w i n g P k H e l l y ' s Condition i s s a t i s f i e d : up t o

may

i n t h e theorem. We observe

, we have IlFll > 0 , such t h a t F(fl) E

X EIR

some

Ilflll = 1

dist(F, j(E)) > 0

* , with . We t h e n

fl E E

t h e r e i s an

a ) =+. b)

> 0 , we

such t h a t

, i = 1, ...,n , and IIx'II = 1 . Then, f o r = Axo + (1 - X ) x ' . T h i s proves t h e lemma.

L e t us come back t o t h e p r o o f o f

Ilflll

x' E E

E

.

54

B. BEAUZAMY

We must now choose

1

if

P

sequences

x

. We

P

t

(

b) * c ) .

x

~

Let

K

FH~**

,

for a l l

x w i t h Ilx II 1 , and P P by i n d u c t i o n t h e c o n s t r u c t i o n o f t h e t o choose

and ) ~( Y ,~) , ~~~ , and shows

>1, k

1< k

z ai

p o s i t i v e numbers, w i t h

sly...,a p - l

have :

< 1 , i t i s possible i < p : t h i s gives

IlFll

Since f.(x ) = 8

P- 1 i=lL: ai j ( x i )

1)

1 ,

,

such t h a t

(hm(Yn))nEIN

. We

n'"'

'

(hl(xA1)))nEIN ~

( h 2 ( x n( 2 )

(x,!,~)),,~~ converges i n

m

>1 ,a

converges i n

IN -- (xn( n ) )nEIN

K

o( E , E*)

s h a l l prove t h a t

If(Yn ) k

- f(Yo)l > f

.

IK

. We

consider

. By c o n s t r u c t i o n ,

converges.

. Let

yo

(Y,),~,,

A

has an

be an accumulation p o i n t o f i s weakly convergent t o yo

Assume t h a t t h i s i s n o t t h e case. Then, we can f i n d E and a subsequence ( Y , ~ ) ~ ~ , , , such t h a t : Of (Yn)n E IN (1)

,

subsequence

i s weakly r e l a t i v e l y compact, e v e r y sequence i n

accumulation p o i n t f o r (ynInEN

into a linear

same way, t h e r e i s a subsequence

(m) , such t h a t (h,(xp)))n ('n ) n E IN t h e "diagonal subsequence", (yn)n

A

By

(since i t i s contained

, such ) ~ t h a~t

By r e p e a t i n g t h i s process, we e x t r a c t , f o r a l l

Since

IK

be a c o u n t a b l e

* , Eo) .

i s compact) ; t h e r e f o r e t h e r e i s a subsequence

('n ( 1 ) ) n E IN , c o n t a i n e d i n

f o r each m

be t h e

o(Eo

H

H : hl,h 2,...,h

i s bounded i n

H

for

Hahn-Banach Theorem, we can extend each element o f E

Eo

v e c t o r space i s separable, and

i s weakly separable. L e t

s e t o f l i n e a r f u n c t i o n a l s on f u n c t i o n a l on

. Let

a) sequence ~ ~ o f~ p o i n t s i n A

for a l l

k

>1.

>0 ,

f E E*

.

61

REFLEXIVE BANACH SPACES

) has i t s e l f an accumulation p o i n t , which 'k k E N The p o i n t y; belongs t o Eo , and h(ynk) m + VY;) 3

But t h e sequence

.

we c a l l

y;

for all

h E H

each

E

y

is

(by d e f i n i t i o n o f an a c c u m u l a t i o n p o i n t , f o r each

o(E*,

nk

-

l i m h(ynk) Ik ++m H

f

> 0 , the

many of t h e

(y

E)

ty E E ;

set

Is

, therefore

h(y;)/

-

< E ) . Therefore,

dense i n

f ( y o ) = f(y;)

Ih(y)

the l i m i t

E*

for all

h(y;)I

<el

we o b t a i n

f E E

, that

is

,

contains i n f i n i t e l y

l i m h(yn ) k+tm k h ( y o ) = h(y;)

, and

hEH

satisfies

, for

yo = y;

all

H

. But

,

which c o n t r a d i c t s ( 1 ) , and proves o u r p r o p o s i t i o n . COROLLARY 2.

-

I n a t e d l e x i v e Apace, e v m y bounded nequence hub a weakey

conv mg ent bu bh equence.

Conversely, we have seen as a c o r o l l a r y o f James' Theorem

(5

1) t h a t ,

iA t h e u n i t b u l l ad t h e Apace, i f A i s n o t weakly compact ( t h a t

when A

i s , i f t h e space i s n o t r e f l e x i v e ) , t h e r e i s a sequence o f p o i n t s i n A which has no weakly convergent subsequence. T h i s i s t r u e more g:nerally,

b u t i s harder t o prove : t h i s f a c t i s

known as t h e E b e r l e i n - Smulian Theorem : v THEOREM (EBERLEIN-SMULIAN).

E

, A

Lt-6 ck!oduAe

Then

don

-

o(E,

LeA E be a k n a c h Apace, E*)

.

A iA weakly compact id and o&y id,

A

a bubo&

06

dhom e v m y A~cjUenCc?i n

one can exfA.act a bubbequence which h weakey c o n v m g e d ( t o a poi& 06

A

A).

We s h a l l n o t have t h e o p p o r t u n i t y , here, t o use t h i s r e s u l t i n i t s f u l l s t r e n g t h : P r o p o s i t i o n 1 and t h e C o r o l l a r y o f James' Theorem which we j u s t r e c a l l e d w i l l be s u f f i c i e n t f o r us. T h e r e f o r e , we s h a l l n o t p r o v e t h i s r e s u l t , and we r e f e r t h e r e a d e r t o ( f o r example) Dunford-Schwartz

1161 (p.430).

See a l s o e x e r c i s e 10 below.

B. BEAUZAMY

62

EXERCISES ON CHAPTER 111.

-

EXERCISE 1.

Let

E

be a r e f l e x i v e space and

Prove t h a t t h e q u o t i e n t

-

EXERCISE 2. "Condition J

EXERCISE 3.

I'

a c l o s e d subspace o f E

F

.

i s reflexive.

ElF

Find i n the u n i t b a l l o f

.

- L e t (En)nGm

a sequence s a t i s f y i n g

co

be a sequence o f Banach spaces. We s e t , f o r

l < p < t m ,

(

nE:IN

; xn E En

En)p = {(x,,),,~~

for a l l

n

,

and

Z IIx nEm

!Ip < + 1 00

En

endowed w i t h t h e norm

identified with

(

c)

m

1 P

En)p

17En)p

EXERCISE 4.

-

of

E

dual o f

1 q

.

En's

i s r e f l e x i v e and

1 Ilx -

= Ilx-yll

2

+

2tRe <x-y, and f i n a l l y Re

y-z

<x-y,

> + t2 I l y - z l l 2 y-z

>

+

20

>O

.

tz...

z-Y

> + Ily - zII 2

z

C

. Then,

E

.

for all

Then :

x-y+t(y-2)

2bie < x - y ,

and t h e r e f o r e

for a l l VzE C

y1I2

z ' = (1 - t ) y

= <x-y+t(y-z),

2 Re

.

t(y-z)

> > + t2 I l y - z l l 2 ,

HILBERT SPACES

L e t us now c o n s i d e r t h e case when

-

C

71

i s a c l o s e d subspace.

.

be a cloned vectoh bubbpace 06 H The phojection P on F h a fineah appfication 06 nohm one (aw i l l be calXed "ohthogond pho jection" I PROPOSITION 3.

F

1eX

.

PROOF.

-

has : die

Let

<x-y, z' E F

for a l l and i n t o

y

, one

(c

so

P(Xx)

has

z

z = Px

Take L

IIPxII

=

< x,

We denote by v x E FI

P(xl

.

, one

>=

,

.

-

= <xl-Pxl,

z

Px2

.

<x -

Px, z

>=

0

> < llxll

IIPxII

= Pxl

,

and

I t i s a c l o s e d subspace, and

t o be closed,

F l l =

PROPOSITION 4.

-

F

y, z

0

> +<x2-Px2,

z

>=

,

0

. Then IIPxII

t h e orthogonal o f

G = F1

> =X < x -

:

t

+ x2)

in Px

F

. Finally

APx

<x1tx2-(Px1+Px2), and t h e r e f o r e

z E F

> GO . T h e r e f o r e , s i n c e F i s a v e c t o r space, < x - y , z ' > < 0 , and, changing z ' i n t o - 2 ' , o b t a i n s < x - y , z ' > = 0 , f o r a l l z ' E F . So

X E

, and

. For a l l

die

i s orthogonal t o

For a l l

F

i t s p r o j e c t i o n onto

z-y

,

, one

iz'

x - y = x-Px yz€ F

,

x E E

F

< llxll , f o r F1 =

Fly = F

Iz

(if

all

E H,

.

= 0

F was n o t assumed

) . We o b t a i n :

The ohthogond

06 t h e vhthogond phojecaon g o n d phojecfion on F1 , and

P

06 a doned nubnpace P the Kmnd on ttkin nubnpace ; I - P h t h e ohtho1 H h t h e dirrect b U m 0 6 F and F F1

.

F o r subspaces and q u o t i e n t s , we have : PROPOSITION 5.

- E v t v ~ ydoned bubnpace

06

a HLLbeJLt b p c e h a HLLbat

npace. T h i s i s obvious, s i n c e i t i s equipped w i t h t h e induced s c a l a r p r o d u c t , and i s complete f o r t h e norm. PROPOSITION 6.

HiF

-

LeX

F

h a HLLbtvLt npace.

be a dosed bubnpace 06

H

; the

cjuo,tient bpace

72

6. BEAUZAMY

T h i s i s c l e a r , s i n c e i t i s i s o m e t r i c t o t h e Kernel o f t h e orthogonal p r o j e c t i o n onto

F

.

From now on, we denote by

PF

the p r o j e c t i o n onto

F

.

PROPOSITION 7. a)

16

F1

and

F2

m e ebbed b u b b p a c u , u l i t h

F1

C

F2

, then

PROOF.

Is obvious. For b ) , i t i s enough t o c o n s i d e r t h e case when t h e

a) Fn‘s

are increasing : the other w i l l f o l l o w by t a k i n g orthogonals. I f

x E H , x = P x t ( I - PF)x , and ( I - PF)x i s orthogonal t o F and so F 0 , and we j u s t have t o prove t h e t o a l l Fn’s T h e r e f o r e PF ( I - PF)x n r e s u l t f o r P F x , t h a t i s , t o assume x E F . Put

.

-

L = { x E F, PF x n + t m n i f x E L and x + x P P

But

x E Fm

L

. SO

c o n t a i n s each

-

L 3 U Fn = F

x 1

. This

i s a subspace, which i s c l o s e d since,

:

Fm

.

, because,

- A @ m d y (ei)iEI

if

n

>m ,

P

x = x Fn

if

dement6 oh a H d b & opace H A c a l l e d orthogonal id, d o t all i, j , i # j , < ei , e . > 0 . 7 R b J orthonormal id i t b o h t h v g o n d and lleiII = 1 do4 all i 7 t b an orthonormal b a s i s id, motLeourn, H A t h e d o b e d f i n e m bpace g e n r n d e d DEFINITIONS.

06

.

by

(ei)i €1

*

PROPOSITION 8.

-

Let

(en (0 G n 9 N )

be a

oh i n ~ i n i t e (N =

.

t

00

)

ohthonahmd bequence, and F = -{(en) ; 0 < n < N} Then t h e appficaN ( a n l o e n oQn=

0

, and

so

x

en

> I2

Q

(

< x,

H

.

-

Let y = C < x , n

y

en

>)

1 1 ~ 1 1 ,~ f o r

p E lN

all

, and

. Therefore

E L2 n€H

en > e n

i s orthogonal t o

F

. For a l l

, and

k

,

.

y = PF x

Also :

Z : I < X Y e n > I 2 = IIYII n

2

F i n a l l y , the application onto

,

,

-+ C

1

x E H

For a l l

<x -

, and

a n e n i s d e f i n e d ; i t i s an i s o m e t r y , n 2 2 " 2 anen(/ = l i m anen(\ = l i m c lan12 = Cia,, n p - f t m n

I . n

t2

happen^ id and

74

B. BEAUZAMY I t can be proved t h a t e v e r y H i l b e r t space has an orthonormal b a s i s .

We s h a l l prove i t o n l y f o r separable H i l b e r t spaces, b u t then t h e f a m i l y

w i l l be countable.

( e i ) i €1

-

PROPOSITION 10.

bmh

.

-

PROOF.

Evehg n e p a m b l e H i L b m t Apace h a c o u n t a b l e ohthono4rna.t

It w i l l be o b t a i n e d f r o m t h e Gram-Schmidt orthogonal i z a t i o n

process : L e t (Un)nEm such t h a t n 2 1

.

u

"1 Then :

uhl) = 0

a dense sequence i n

.

# 0

n

if

Put

e

< nl ,

-

. Let

H

"" and < u;'),

be t h e f i r s t index

u h l ) = un -

1 - 1Iun.H 1 and

nl

>=

el

0

< un ,el > el , f o r

for all

n

.

Now choose n2 t o be t h e f i r s t i n d e x n such t h a t )' ,!u, # 0 , put (1) and ~ ( = ~u ( l 1) - < u:'), e2 > e2 , Then uh2) = 0 i f n n e2 =--JTJ- , n ,, n < n2 ,Land < u i 2 ) , e2 > = 0 f o r a l l n And so on : one chooses as

I

'"'

and so,

I

.

< ek ,

un( k )

>=

0

for all

a r e l i n e a r combinations o f t h e combination o f t h e

n

. For

(u:~-'))~

(u,!,~))~ , and so

,

If

< ek ,

k

all

el

>1 ,

.t > k

,

>=

. The

0

el

ek

and

(un( k )

)n

i s a linear sequence

i s t h e r e f o r e orthonormal. But t h e ui's a r e themselves l i n e a r ( e n ) n E IN ( t h i s i s checked by i n d u c t i o n ) , and so t h e span combinations o f t h e e n ' s o f the

en's

contains a l l the ui's

sequence

, span{ (en)n E M 1

-

Therefore,

= H

, and,

since

( u ~ ) ~ i s a dense

.

two separable H i 1 b e r t spaces a r e i s o m e t r i c : choose

orthonormal bases (en)n o f t h e f i r s t , (fn)nEIN o f t h e second. Then t h e a p p l i c a t i o n en 3 fn d e f i n e s an i s o m e t r y between these spaces. T h i s and L2 : i n l2 , the i s t h e case, f o r example, o f t h e spaces l2

HILBERT SPACES sequences

(en)n E M

75

k f n , form an k = n , L2 , t h i s i s t h e case f o r t h e complex e x p o n e n t i a l s

d e f i n e d by

{

en(k) = 0

if

= 1 if

orthonormal b a s i s ; i n (ein6' (which can be arranged i n a sequence indexed by )n E I L

IN ).

We t u r n now t o t h e dual o f a H i l b e r t space. I t can be i d e n t i f i e d w i t h t h e space i t s e l f : PROPOSITION 11.

- Fv4 dl a

EH

, t h e mapping x

+

< x,

a

>

h a

*

cvntinuvun fineah dvmn on H , called a* . The a p p f i c d o n a -+ a b i j e c a w e , cvntinuouh, h v m u k i c , and a w n e m ( t h a t h (xa)* = ha" ) * 64om H o n t o H

.

PROOF.

-

The c o n t i n u i t y o f

1 < x,

> I < llxll

y

. llyll

a*

,

f o l l o w s from Cauchy-Schwarz i n e q u a l i t y for a l l

A l l t h e announced p r o p e r t i e s o f

.

x,y E H a

+

*

a

a r e immediately checked. L e t

us j u s t show t h a t i t i s s u r j e c t i v e . L e t 5 E H*, t h i s i s a c l o s e d hyperplane. Choose a ' x

< x,

-+

Write

x =

and E

> < x,

a'

i s a l i n e a r form on a'

> a'

comes from a p o i n t o f

H

orthogonal t o

H

+ y , with y

.

. Let

E # 0 F

.

F = Ker E

Then

, which has t h e same k e r n e l as E

= P F x . Then

E(x) =

:

< x,

a' >E(a')

,

,

Therefore, a H i l b e r t space i s r e f l e x i v e , and i t s u n i t b a l l i s compact for

a(H,

* .

H )

To end t h i s chapter, l e t us see how f i n i t e - d i m e n s i o n a l H i l b e r t spaces are r e a l i z e d : PROPOSITION 12.

nom

-

N llxll = C ailxi/ i=l

Le,t

N 2 1

2 , i d

.

16

x = (x,,

ahe pvhLLive

al,...,aN

...,xN)

E

MLunbem,

the

KN

ai xi yi, and t h e ~ ~ e t ( o 4 eKN 1 equipped with t k i n nvhm h a H d b & Apace. The u n i t b a l l h t h e ne,t N I ( x l ,... ,xN) ; al/x1I2 + + a N / x N I 2 < 11 , which h aneh5phoXcfiiMM

coma 6 4 v m t h e n c d a h ptvducz

< x,

y

>=

...

So we have seen t h a t t h e geometric s t r u c t u r e o f a H i l b e r t space appears

r a t h e r s i m p l e : a l l subspaces and q u o t i e n t s a r e H i l b e r t spaces, a l l subspaces a r e complemented ( i . e . t h e r e i s a p r o j e c t i o n on them), and

76

B. BEAUZAMY

separable spaces have bases. U n f o r t u n a t e l y none o f these f a c t s h o l d s i n general Banach spaces : t h e f o l l o w i n g chapters a r e devoted t o t h e study o f these q u e s t i o n s .

EXERCISES ON CHAPTER I.

-

EXERCISE 1.

E

a E H

E

Let

be a c l o s e d convex subset, i n a H i l b e r t space

F

H

b E F which minimizes

H*. Show t h a t t h e r e i s a unique p o i n t

the function 2

XEF-Ilx-all

b

and t h a t t h i s p o i n t Forall

y

-

EXERCISE 2.

x

-+

ex

k

-

EXERCISE 4.

-

elements w i t h n

i s characterized by : b + y € F ,

bie(Z+E(y))>O.

Find the best approximation p r o j e c t i o n o f the f u n c t i o n

, consider

I n L2

1

>1

qn(z) =

t h e sequences

1,

compact i n L2

for

x>,

on t h e subspace o f p o l y n o m i a l s o f degree

EXERCISE 3.

for all

such

t < 5 ,

I n a H i l b e r t space Ilxnll

1.

( x ~ ) ~ , - ~be* a sequence o f For a l l

z E H

we d e f i n e ,

:

l n I: IIZ 1

xjn

.

Show t h a t vn

i s a convex f u n c t i o n , which a t t a i n s i t s minimum a t a unique n th 1 Z xj Show t h a t i f H = L2 , t h e kp o i n t sn , and t h a t sn = n l c o o r d i n a t e o f sn sn(k) minimizes n 2 . t - + - C It xj(k)l j=1

.

-

Study t h e same q u e s t i o n s , when

H = L2

, i f vAp)

1 9 p < + =

y

i s d e f i n e d by p # 2 .

77

HILBERT SPACES

-

EXERCISE 5. Let

The n o t a t i o n s a r e t h e same as i n t h e p r e v i o u s e x e r c i s e .

(Zn)n>l

then

-

zn

be a sequence i n

n-.+- 0

sn

~ ~ ( >2q n)( s n )

71

t

-

llz

.

H

. For

H

in

snll

2

- qn(sn)

Show t h a t i f q n ( z n )

0,

t h i s , e s t a b l i s h the formula :

vzEH .

,

- Same n o t a t i o n s . Assume t h a t a subsequence ( s ) nL L E N converges weakly t o some p o i n t a E L2 , and t h a t a n o t h e r subsequence converges weakly t o b E L2 . P u t a = Ila - b1I2 . Prove t h a t EXERCISE 6.

(S"B)LEIN t h e r e e x i s t s Lo nL 1 - Z Ilb

-

'L j=1

>1 x.ll J

For t h i s , take

.

Z lb(k)I2 < e 2

nL

nL 1

Ib(k)

EXERCISE 7.

-

A

a

Let

be

>-'L1

>0

E

k >K

1Z

2

-

If

nL

Z: Ila

j=1

Let

< Aa -

- sn n-t+m 0

.

K

that

> -1

. Show

Asn

t

t

2

t

1

-

la(k)

( n o n - n e c e s s a r i l y l i n e a r ) c o n t r a c t i o n from

Assume t h a t

Lo

such t h a t , i f

G O .

i s a fixed point f o r

A

,

that i s

Af = f

.

B. BEAUZAMY

78

- L e t A be as i n t h e p r e v i o u s e x e r c i s e . F i x e

EXERCISE 8 . x

j

.

= AJe

as i n e x e r c i s e 4.

D e f i n e sn

Use e x e r c i s e 5 t o show t h a t

Asn

show t h a t any weak l i m i t o f t o show t h a t

(sn)

-

sn

0

E C

, and

. Use e x e r c i s e

i s afixed point o f

A

take

7 to

. Use e x e r c i s e

(sn) i s weakly convergent. So one o b t a i n s t h e f o l l o w i n g

6

theorem :

-

THEOREM. (J.B. BAILLON) Hdb&

dpace, and

doh any e E C , ,the p a i d od A

.

A

Let

C

be a cLoned conuex bounded o e t i n a

a (nun-fineat] cavttnaotian C e h a t o aumagu

1

n

Z: A j e n l

&om

C

into C

. Then,

conumge weakly ,to a hixed

REFERENCES ON CHAPTER I . Almost every book devoted t o F u n c t i o n a l A n a l y s i s c o n t a i n s a more o r l e s s e x t e n s i v e s t u d y o f H i l b e r t space. F o r a more d e t a i l e d s t u d y t h a n o u r s , we r e f e r t h e r e a d e r t o t h e book by L . SCHWARTZ 1451. The theorem o b t a i n e d i n e x e r c i s e 8 was proved by J.B. B a i l l o n . The p r o o f g i v e n here ( v i a t h e f u n c t i o n s

vn ) i s simpler than the o r i g i n a l

p r o o f , and comes from a paper by P. E n f l o and t h e a u t h o r ( "Th&or@mes de p o i n t s f i x e s e t d ' a p p r o x i m a t i o n " , t o appear), which a p p l i e s a l s o t o

1< p

0

x

~

nuch t h a t ,

604

...

, and & s c d m a,

REMARK.

-

( a

q '

h) basic ~ i~ d and ~ o n l y id t h e m all

iVUkJehA

p

and

q

, wdh

one han :

P , from qYP have norms u n i f o r m l y bounded by K .

T h i s c o n d i t i o n means t h a t t h e p r o j e c t i o n s

spanCx o,...,x

q

1 o n t o spantx o,...,x

1

P

PROOF. a)

L e t us assume f i r s t t h a t

( x ~ ) , , ~ i ~s b a s i c . We s h a l l use t h e

f o l l o w i n g lemma, t h e p r o o f o f which i s elementary and i s l e f t t o t h e r e a d e r : LEMMA 2. xn

# 0

604

conumgen}

L&

F

.

, equipped

Il(an)nEm Then

( x ~ ) , , ~a ~sequence o d p o i &

all n

L&

i n a Banach space E , w i t h

( n c d e m l , nuch that

F =

C anxn

with the now : N

I I =~ ,,,~hI

anxnll

.

h a Banach space.

NOW, i f x

E El

?$%t(xn)nE,,,l

, x has, b y d e f i n i t i o n , a unique

x = C a x , and so t h e a p p l i c a t i o n (a,) E F 3 x = 2 a x n n n n nEIN n n i s a l i n e a r b i j e c t i v e mapping from F o n t o El B u t t h i s mapping i s

decomposition

.

continuous, s i n c e :

~

~

~

B. BEAUZAMY

82

st By t h e open mapping theorem (Ip a r t , c h a p t e r I , theorem 8 ) , t h e i n verse mapping i s continuous : t h e r e i s a c o n s t a n t

for a l l

(a,)

E F nElN

. Therefore,

for a l l

p, q

K

>0

with

such t h a t

p ~q

,

and (1) i s proved.

-

Conversely, assume (1) t o be s a t i s f i e d . Take x E s p a n I ( x n ) n E N 1 b) Then t h e r e e x i s t s a sequence (yk)kEIN , each yk belonging t o s p a n { ( x n ) n ~ m} Y with

Yk k + t r n t x

From (1) we deduce, l e t t i n g ges, one has, f o r a l l

L e t us c a l l

q

p E

-+

t m

1

Z anxn n

conver-

P t h e p r o j e c t i o n d e f i n e d by P

IN , we

.

obtain

X

: l e t us c a l l i t

deduce t h a t t h e c o e f f i c i e n t s a(k)

can be w r i t t e n :

that i f a series

t m

and i t f o l l o w s t h a t , f o r any f i x e d k

yk

p

P ( C anxn) = C anxn P n n Qp For a l l

-+

. Each

.

have a l i m i t

al,..

.

P

p

, Pp

. Taking

0

when

yk

successively

have a l i m i t

k

3

has a l i m i t i n

t m

.

a.

E

p = 0,1,2y...,

when we

, that the coefficients

Therefore

Xp =

Z aex~

L QP

.

83

SCHAUDER BASES I N BANACH SPACES From ( 3 ) , we o b t a i n , f o r a l l

Let

E

>0

. We

can f i n d

k

and

p

:

such t h a t , i f

ko

k

> ko ,

llyk

-

XI] < f

.

which proves o u r a s s e r t i o n . The uniqueness o f t h e decomposition i s c l e a r :

if

x = X aLxL

and

e

for all

p

,

n

, t h e n 2 (aa. - fl,)x,

x = C flexL

a.

X (ae

e n j j y

0

"shrinking").

PROOF

1)

Assume f i r s t n

E

,

for a l l

m

>0 ,a

o f i n t e g e r s , and a sequence and

3,

has a decompo-

a r e t h e c o o r d i n a t e f u n c t i o n a l s , we

,

and

,

and so

fm(un) = am i f

m

max Bk

t h e sequence

j

Therefore, we have proved t h e e q u i v a l e n c e o f a l l i t e m s , e x c e p t d ) . f) *d).

We s h a l l i n t r o d u c e on

E

a new norm, e q u i v a l e n t t o t h e

o r i g i n a l one, and h a v i n g i n t e r e s t i n g p r o p e r t i e s : LEMMA 2.

x =

-

16

x = Z anen n

Z anen

n EA

t h e ~ u n ~ . t i o nt

-+

06

,

y =

Ix

+

, put

:

Z bnen n€B

,

an even, conuex t;unc.tivn

tyl

tEJR.

PROOF OF LEMMA 2.

Since we have a l s o

2)

If

B1

, B2

('n)n €B1 UB2

of

1x1 2 llxll

, the

two norms a r e e q u i v a l e n t .

a r e d i s j o i n t subsets o f

+1

,

if

x =

N , we

Z: anen : n EB1 UB2

have, f o r e v e r y sequence

B. BEAUZAMY

92

and a l s o

and consequently

from which f o l l o w s 1x1

'I

Zanenl

Y

i s monotone i n t h e norm

and

1 1 .

Is obvious.

3)

L e t us now prove d). L e t (bn)nEIN , w i t h , f o r a l l n , l b n l Q l a n [ Put b, = tnan , w i t h I t n /Q 1 , and tn = an t iP, , w i t h an,Pn E IR , lan/ Q

, IPnI

1

Q 1

have t o prove t h a t

. So,

i n o r d e r t o prove t h a t

Z ananen

and 2 Onanen

ZEnanen) n

,

n f o r t h e f i r s t , we have :

1

Z a a e

n

n n n

1

Q

I

with

E

n

Z b,en

n converge.

= sgn an

converges, we But, f o r example,

Vn,

by 3 ) o f t h e p r e v i o u s lemma. T h i s l a s t s e r i e s converges when C anen i n t h e norm

II*lI , s i n c e t h e norms a r e e q u i v a l e n t . F i n a l l y , d )

=*

c)

does is

obvious, and o u r p r o p o s i t i o n i s proved. The s i m p l e s t example o f u n c o n d i t i o n a l b a s i s i s t h e canonical b a s i s o f

.

One can a l s o show ( b u t i t i s much h a r d e r ) lp (1 < p < t m ) t h a t t h e Haar system i s an u n c o n d i t i o n a l b a s i s o f Lp( [0,11 , d t ) , i f cO

or

1nk series, there i s a integer

n;

> nk

such t h a t

I

t: i>nl;

ai( k ) ei

I 0 ,

E

o f i n t e g e r s , a sequence o f p o i n t s :

IN

and

(tk)kEA are scalars :

But s i n c e c o n v e r s e l y

we o b t a i n :

s o , if ( e n I n E p j from = I ( X ~ ) ~ ~ , , from

-

i s the canonical basis o f

L1

, the

1 i n t o C 1 , d e f i n e d by T xk = ek

spanE(xk)kEIN 1

onto

operator

,is

T

an isomorphism

L 1 , which proves t h e p r o p o s i t i o n .

EXERCISES ON CHAPTER 11.

EXERCISE 1.

-

Let

(en)nEN

be a Schauder b a s i s i n

t h e c o o r d i n a t e f u n c t i o n a l s . Show t h a t f o r e v e r y n l i m C E(ek)fk = t n++= 1

, for

~(E*,E)

.

E

, and

,

(fn)nEIN

E E E* , one has :

€3. BEAUZAMY

96

-

EXERCISE 2.

Let

(en)nEM

be a Schauder b a s i s i n

, and

E

(fn)nEM

t h e c o o r d i n a t e f u n c t i o n a l s . Show t h e equivalence o f t h e f o l l o w i n g two properties : a)

(fn)nEN

b,

(en)nEm

-

E X E R C I S E 3. (fn)n EN sequence i n

i s a Schauder b a s i s o f

E

i s a shrinking basis o f

Let

(en)nEN

such t h a t , f o r every

(yk)k

IN

m E

Show t h a t t h e r e i s a subsequence b a s i c sequence made o f b l o c k s on t h e EXERCISE 4.

-

E

Let

.

be a normalized Schauder b a s i s i n

the coordinate functionals. L e t E

.

E

.

en's

IN

and

E

a normalized

fm(yk)

f

w b

+

0.

which i s e q u i v a l e n t t o a

Em

be a Banach space w i t h u n c o n d i t i o n a l b a s i s

(en)nEIN.

be t h e c o o r d i n a t e f u n c t i o n a l s . Assume t h a t t h e r e i s a

Let ( f k k E I N sequence ( Zn) E IN that

IN ,

k E

for all

( q n E m

such t h a t

but

llznll

0 , such

that

f(un)

go(x) = 1

-

,

Let

of

>6

h,(t)

gn(x) =

1

X

EXERCISE 6.

E

( f k ) k >1

1)

-

for a l l

-

un

, and

'n)n E IN

hn(t)dt

0

. Let

n

0

.

n-t+m 0 Show

L1-basis.

-

z' 0 , for a n n++w

that there i s

f E E*

and

1

n

>1

. Show

[0,11

(n

> 1) .

t h a t t h e sequence

Put (gn)n>,o

( c a l l e d t h e Schauder b a s i s o f 'if [0,11)). ( be a monotone s h r i n k i n g b a s i s o f a Banach

Let (Pn)n>l

be t h e p r o j e c t i o n s a s s o c i a t e d t o t h e b a s i s , and

the coordinate functionals.

Show t h a t i f

fk(zn)

o f c o n s e c u t i v e b l o c k s on t h e

be t h e Haar system on

i s a b a s i s o f g( [0,11)

space

IN ,

( z ~ ) ~ , -does ~ n o t converge weakly t o

has a subsequence e q u i v a l e n t t o t h e

[Show t h a t t h e r e i s a sequence (u,) m n e n ' s ( t h a t i s un = I: aiei ) , w i t h mn-l+ 1

EXERCISE 5.

n E

t E E**

, ttP,(t)

=

n

Z E(fi)ei

i=l

, and

SCHAUDER BASES I N BANACH SPACES

97

Conversely, i f i s a sequence o f s c a l a r s such t h a t n ** , E*) supl aieiII < t m , show t h a t any l i m i t C i n E** , f o r o ( E n. n J n o f a subsequence Z aiei s a t i s f i e s E ( f . ) = ai Deduce t h a t C aiei 1 1 1 converges t o C i n t h i s t o p o l o g y . 2)

,

.

3)

Show t h a t

E**

can be i s o m e t r i c a l l y i d e n t i f i e d w i t h t h e space o f n sequences ( a n ) n >1 o f s c a l a r s such t h a t sup/l aieill < + m , t h e n correspondence b e i n g C (C ( fj ) ) 5. 2 1

-

EXERCISE 7. (The James' space J ) . - L e t J be t h e v e c t o r space o f a l l o f r e a l numbers, such t h a t x ( k ) k + -+ 0 and sequences x = (x(k))k>l

,

n > l

1)

Show t h a t

llxllJ

... < p 2 , , I < t m .

pl
l

.

3)

, and

J

en

,

n

of

J

i s a Banach space.

*

IK('

> 1 , show

)

that

i s a monotone

J

i s not

reflexive.

4)

Show t h a t

(en)n>l

i s a shrinking basis o f

J

.

( I f n o t , t h e r e i s a sequence o f c o n s e c u t i v e n o r m a l i z e d b l o c k s the

en's 'k

C -[?-

k= 1

,a

> 0 , and

S

belongs t o

a

f E J*

with

f(uk) > S

for all

6)

J ).

; lJ

Show t h a t i f

on

k ; but

Using e x e r c i s e 6, show t h a t J** i s t h e space o f sequences n aiei < t m , t h a t i s such t h a t Il(an)llJ < t such t h a t supll n

5)

uk

Il(an)llJ

1 Show t h a t J by C o and

J

O

, and

J

that

.

.

J**

i s isomorphic t o

J

.

B. BEAUZAMY

98

REFERENCES ON CHAPTER I 1 For t h e v a r i o u s n o t i o n s o f convergence and t h e l i n k s reader may c o n s u l t M.M.

between them, t h e

DAY's book [131. I n t h i s book, beside u n c o n d i t i o n a l

and a b s o l u t e convergence, several o t h e r types o f convergence a r e s t u d i e d . P r o p o s i t i o n 1,

5

1, occurs i n e v e r y book d e a l i n g w i t h t h i s t o p i c , f o r

example LINDENSTRAUSS-TZAFRIRI [34

,vol .

11 , o r I . SINGER 147, v o l . 11.

P r o p o s i t i o n 3 i s i n BANACH's book E l . Theorems 5

(5

1) and 2

(5

2 ) a r e due t o R.C.

JAMES [241 ; t h e v e r s i o n

DAY's book

which we g i v e i s n o t complete. The r e a d e r i s r e f e r r e d t o M.M.

[131 f o r t h e complete statements and p r o o f s , and f o r some e x t e n s i o n s . The v a r i o u s c h a r a c t e r i z a t i o n s o f u n c o n d i t i o n a l convergence appear i n

LINDENSTRAUSS-TZAFRIRI [341 .

JAMES' r e s u l t . See [131. E x e r c i s e 4

E x e r c i s e s 1 and 2 a r e p a r t o f R.C.

i s i n SINGER [471 , e x e r c i s e 6 i n LINDENSTRAUSS-TZAFRIRI [34

, vol.

11 , and

James'space i s g i v e n i n R.C. JAMES [241, and a l s o i n LINDENSTRAUSS-TZAFRIRI [34

, v o l . 11 .

COMPLEMENTS ON CHAPTER 11.

A Banach space

-

The Approximation P r o p e r t y .

i s s a i d t o have t h e Approximation P r o p e r t y (A.P. i n

E

s h o r t ) i f , f o r e v e r y compact s e t f i n i t e rank o p e r a t o r

T

, with

K C E

IlTx

-

xII

l e n t f o r m u l a t i o n i s : f o r e v e r y Banach from

Y

into

E

If

into

, E

the f i r s t

E

, there

converging t o

< E

Y

E

> 0 , there

for a l l

, every

x E K

exists a

. An

compact o p e r a t o r

equivaT ,

e x i s t s a sequence o f f i n i t e rank o p e r a t o r s , from Y T

i n norm.

has a Schauder b a s i s ,

n

, every

E

has

A.P.

( c o n s i d e r t h e p r o j e c t i o n s on

c o o r d i n a t e s ) . ENFLO's example 1191 i s a separable space

w i t h o u t A.P. See I341 , v o l . I , f o r a d e t a i l e d s t u d y o f these n o t i o n s .

CHAPTER I 1 1 COMPLEMENTED SUBSPACES I N BANACH SPACES

I n the previous chapter

we have i n t r o d u c e d and s t u d i e d a n o t i o n o f

b a s i s f o r Banach spaces, wh ch may be c o n s i d e r e d as an e x t e n s i o n o f t h e notion o f H i l b e r t i a n basis

n a H i l b e r t space. Another r e s u l t , which we

have mentioned i n c h a p t e r I

and which i s t y p i c a l o f H i l b e r t s p a c e s , i s t h e

f a c t t h a t e v e r y c l o s e d l i n e a r subspace i s t h e range o f a l i n e a r p r o j e c t i o n ( o f norm 1). T h i s i s n o t t r u e i n e v e r y Banach space, as we s h a l l see on an example. We s h a l l t h e n i n v e s t i g a t e t h i s q u e s t i o n i n some common Banach spaces : d'(1 Q p

l

basis o f N

P

co

t h e canonical the r e s t r i c t i o n o f

.

fn

L e t d be t h e d i s t a n c e on E* , t h e r e s t r i c t i o n o f which t o aE* a) d e f i n e s t h e topology a(E*, E) ( s e e f i r s t p a r t , c h a p . I I I y § 2 , p r o p . 4 , p . 5 7 ) . Check t h a t t h i s d i s t a n c e i s t r a n s l a t i o n i n v a r i a n t : d ( y , z ) = d ( y - z , Vy,z€E

*

.

-

.

0)

Show t h a t any accumulation p o i n t o f t h e P u t M = BE*n F1 b) sequence ( fn 1 > 1 belongs t o M

-

c)

Deduce t h a t

d)

Let

, 9,)

d(fn

-+

, M)

-

0

.

be a sequence o f elements i n

(g,)

0

d(in

.

,

that i s

x E E :

Show t h a t

Px E co

REFERENCES

ON CHAPTER 111.

fn

and t h a t

-

P

gn

0

M such t h a t

o(E*,

. Put,

f o r every

i s a p r o j e c t i o n o f norm a t most 2.

The example o f non-complemented subspace i n L [311.

E)

P

i s t a k e n from Kothe

The r e s u l t g i v e n i n e x e r c i s e 1 i s due t o Sobczyk ; t h e p r o o f suggested i s t h a t o f Veech, and i s taken from LINDENSTRAUSS-TZAFRIRI

I34 , v o l . 11.

,

CHAPTER I V

L

THE BANACH SPACES

I-

L

SUBSPACES OF

P

The space (x(k)) k

+

L

+

P

03

)

(1

0

LeR

E

. Evehy

a cLvned nubnpace

be a Banach Apace, uLith Schaudeh b a n b

(en)nElN ;

cloned i n , 5 i n L t e - d i m e ~ n i o n dnubnpace F ad

E

G , which ha^ a no/rma&zed Schaudeh b m b

equivdent t o a n v m f i z e d b m h

(u

~

)

~made o d blvctiu on

covLta,h

(Y,),~~ (en)nElN

,

and buch that

REMARK.

-

Consequently, i f

(en)n E N

i s unconditional,

G

has an uncon-

d i t i o n a l b a s i s , because e v e r y sequence made o f b l o c k s on an u n c o n d i t i o n a l b a s i s i s i t s e l f u n c o n d i t i o n a l , and e v e r y sequence e q u i v a l e n t t o an uncondit i o n a l b a s i c sequence i s a l s o u n c o n d i t i o n a l .

6. BEAUZAMY

112

-

PROOF OF LEMMA 4. in

F

an element

(1)

L e t us f i r s t show t h a t , f o r e v e r y

can f i n d

.

n>P

Assume t h i s i s f a l s e . Then, f o r some

P ( Z anen) = Z anen P n nGp

d e f i n e d by

, we

, which i s n o t 0 , and which i s o f t h e f o r m :

y

Z anen

y =

p E IN

(since f o r every y E F

,y #

closed. But we know t h a t

P

, the

p E IN

, restricted

to

projection

, would

F

P

P be i n j e c t i v e

P y # 0 ) . I t s image i s a f i n i t e P , and t h e r e f o r e i s dimensional l i n e a r space ( c o n t a i n e d i n s p a n { ( e j ) j 0

implies

i s continuous. T h e r e f o r e i t would be an

P isomorphism on i t s image, and

F would be f i n i t e - d i m e n s i o n a l ,

which c o n t r a -

d i c t s o u r assumption. NOW, choose

yo E F

Since t h e p a r t i a l sums

(1

, with

. Then

yo

can be w r i t t e n :

converge t o

yo

, we

IJyoII = 1

Z an(0) en npn .i n order t o obtain a,!,’)en// < ; By i n d u c t i o n , we b u i l d t h i s way a Z 2 n >PI

such t h a t

.

.

(1

sequence

y

j

F

in

, with ,

integers

finally

I

C llyj j 20

-

ujII

IIy.II = 1 f o r a l l J such t h a t

< E

j

, and

a sequence o f

.

The u . ’ s a r e b l o c k s on t h e e n ‘ s , and consequently, i f M i s t h e J b a s i s c o n s t a n t o f t h e sequence (en)nEIN , we o b t a i n , i f p < q , f o r any f i n i t e sequence o f s c a l a r s

(an)nEN

:

lp (1 E > 0 . n + + m

0

. Then,

f o r some

E

L e t us now choose

rl

Now, choose

nl

with

such t h a t

k>rl

Ixn ( k ) 1

Assun

nl,.

((x (1 > E nl

, rl,...,rp-l

.,npp-l

, that

is

C

k€IN

Ixn ( k ) I 1

> E

have been chosen. We choose

such t h a t :

n >n P P-1

z

keIN and

r

P- 1

k=l

< z.

Ixn ( k ) I p

We choose a l s o

=

k>r

Ixn ( k

p

2p

rP

>

I

k } f

p o i n t s a t which one i n b(K)

.

f

i s upper-semi-continuous

i s closed. Since

K

i s compact, t h e e x i s t e n c e o f

reaches i t s maximum f o l l o w s . The problem i s t o f i n d

L e t us c o n s i d e r t h e f a m i l l y 9 o f t h e c l o s e d subsets have t h e f o l l o w i n g p r o p e r t y : i f an open segment Ia,b meets

F

least

K

,

i t i s completely contained i n

.

F

[

K

, which in

K

,

This family 9 contains a t F

need n o t be

i s , i n t h e plane, t h e c l o s e d u n i t d i s k , any

c l o s e d subset o f t h e u n i t c e r c l e belongs t o 9

K

K

F of

, contained

i t s e l f . I t s h o u l d b e o b s e r v e d t h a t these subsets

convex. F o r example, i f contained i n

i f , f o r every

can meet such an

F

.

, since

no open i n t e r v a l

The f a m i l y 9 i s s t a b l e under i n t e r s e c t i o n : i f

n Fi

(Fi)iEI E 9,t h e n 9.L e t 9* be t h e s e t o f non-empty elements o f 9.On 9*,

E

i€ 1 we p u t t h e o r d e r g i v e n by i n c l u s i o n : F1

> Fz

order, t h i s s e t i s i n d u c t i v e : i f

i €n1 Fi

'lorn's

if

F1 3 Fz

. With

this

i s t o t a l l y ordered, t h e n

(Fi)iEI i s a minorant (non-empty s i n c e a l l a r e compact). Therefore, by

9* c o n t a i n s minimal elements : we s h a l l show t h a t such an

axiom,

element i s reduced t o one p o i n t , and t h a t t h i s p o i n t i s an extreme p o i n t . So, l e t function

be a minimal element o f

X

,

f

from

K

IR , p u t

to

9*. F o r any upper-semi-continuous

Xf = I x E X ; f ( x ) = s u p ( f l X ) l

i s a non-empty s e t , which belongs t o 9 : i f an open i n t e r v a l intersects

Xf

, it

X

intersects

convex and upper-semi-continuous

, it

of l a , b [ i s minimal and

, and

: if

must be c o n s t a n t on

X 3 Xf

, then

X = Xf

. This

la,b[

.

so i s c o n t a i n e d i n X But f i s f reaches i t s maximum a t a p o i n t

, and

la,b[

, and

X

so

] a , b [ c Xf

. Since

X

must be reduced t o a s i n g l e

p o i n t : i f i t c o n t a i n e d two d i s t i n c t p o i n t s , one c o u l d f i n d a continuous l i n e a r f u n c t i o n a l which separates them ( f i r s t p a r t , chap. 11,

5

4, c o r . Z ) ,

and so i t would be a convex upper-semi-continuous f u n c t i o n which would n o t be c o n s t a n t on

X

.

So we have seen t h a t i f t h e s e t o f p o i n t s where Fo

of

i s convex and upper-semi-continuous on

,

K

reaches i t s maximum c o n t a i n s a minimal element

9* , and t h a t Fo i s a s i n g l e p o i n t { x o }

see t h a t b # xo

f

f

.

I t i s now easy t o

i s an extreme p o i n t : we cannot have xo = a t b , a # xo , - atx, btx, would meet Fo , b u t would since the i n t e r v a l , xo

[

'not be c o n t a i n e d i n Fo . T h i s f i n i s h e s t h e p r o o f o f Bauer's Theorem.

,

EXTREME POINTS OF COMPACT CONVEX SETS

- Euehq conuex compact h u b b d 0 4 a HLCTVS iA

THEOREM 2. (KREIN-MILMAN).

t h e c h e d conuex ~LLU 06

-

PROOF.

K

Let

Lth

ex,OLme point^.

be a compact convex s e t . Put

K' C K

.

a >O

and a r e a l l i n e a r f u n c t i o n a l

I f t h e r e was a p o i n t

f o r every b(K)

. This

y E K'

, which

125

x

in

K

f

K' =

but not i n with

cOnv

f(x) > a

b(K)

.

Obviously

, we c o u l d f i n d a

K'

, and

f(y) < a

l i n e a r f u n c t i o n a l would n o t reach i t s maximum on

c o n t r a d i c t s t h e p r e v i o u s theorem.

.

be a compact conuex 4ubheX 06 a 4 e d HLCTVS E Euehq exahwe p o i n t ad K huh a bacse 0 4 n&ighbou&hoadh ( i n K 1 made o d open h f i C C h , t h a t A, b e h oh t h e &am :

-

PROPOSITION 3.

-

PROOF.

Since

c o f n c i d e on point E

>0

.

K

x E K

1eA K

K

V

Let

*)lK , of

E

u(E,

a

: by d e f i n i t i o n (see f i r s t p a r t , c h a p t e r 11), one can f i n d

V 3 I y E K ; Ifi(x)

fly... ,fn such t h a t

-

fi(y)I

2 fi(x)

+ €1

. Therefore

< E

,i

or

I y E K ; fi(y)

= l,...,nl

a r e compact, convex, and do n o t c o n t a i n

If x

u(E, E )

and

be an open neighbourhood, f o r

and r e a l l i n e a r f u n c t i o n a l s

I y E K ; fi(y)

E

i s compact, t h e t o p o l o g i e s o f

.

x

d fi(x)

the sets

- €1

( i = 1,

i s an extreme p o i n t , t h e i r convex h u l l does n o t c o n t a i n

...,n )

x

x = C a x , f i n i t e decomposition w i t h i i i and t h i s c o n t r a d i c t s t h e e x t r e m a l i t y o f x

e i t h e r : o t h e r w i s e , we c o u l d w r i t e xi E K

, ai

2 0

,X

ai

= 1

,

.

But t h i s convex h u l l i s a l s o compact ( i f previous sets,

convIKil

Ki

i s t h e image o f

( i = 1,...,2n)

A

2n

x

2n

application C(al,...,a2,,), separating

,in

i=l

K

, by

the

) ) -+ C aixi , where A c lR2n i s ((al,...,a2n),(xl,..., '2n 1 2n ai > 0, 2 ai = 11) So we can f i n d an hyperplane s t r i c t l y 1 x and convIKil ( f i r s t p a r t , c h a p t e r 11, 5 4, c o r . 5 ) :

.

f

there i s a linear functional I y E K, f ( y ) > a 1

proposition.

17Ki

are the

contains

x

and a r e a l

a

such t h a t

and i s c o n t a i n e d i n

V

; t h i s proves t h e

126

B. BEAUZAMY We s h a l l say t h a t a hyperplane H = { z ; f ( z ) =

(f real) i s a

CY}

nuppohting hyperplane f o r a convex K i f K i s c o n t a i n e d i n one o f t h e half-spaces { z ; f ( z ) < a 1 o r { z ; f ( z ) >,a1 , and i f K n H i s nonempty

.

-

LeA K be a compact convex AubneA i n a k e d HLCTVS E , and U a compact nubhe2 06 K . The doUoL1Ling conditionn me eqLLivdevtt : PROPOSITION 4. a)

K

LA t h e c l o n e d convex h u l l od

b)

U

me&

the intehnecfion 0 6

U

,

06

and any

K

nuppohting

hqpehpLanen, c)

U

.

contai~n b(K)

PROOF.

*

Iz ; f ( z ) = a 1 such t h a t K n H n U = Q . Assume f o r example t h a t f ( x ) > a , f o r a l l x E K . Then f ( x ) > a f o r a l l x E U , and, s i n c e U i s compact, p = i n f f ( x ) > a . So U i s c o n t a i n e d i n t h e h a l f - s p a c e Iz ; f ( z ) > , P I , a)

b ) . Assume t h a t t h e r e i s a s u p p o r t i n g hyperplane H =

XEU

and

cOnv b)

U

a l s o , and

U

* c).

. Then

t h e r e i s a neighbourhood

V

and a r e a l

and

Iz

x

. Assume

K

x

U

of

K

x

. does n o t belong t o

V nU

such

i s empty. By

f(x)

> cr

= a)

s t r i c t l y separates

. Then,

f o r some y

i s a s u p p o r t i n g hyperplane, which cannot meet

> CY , U

U

.

I s a consequence o f K r e i n - M i l m a n ' s Theorem.

.

i s a compact s e t , we c a l l

valued continuous f u n c t i o n s on

The dual o f

of

Iz ; f ( z )

H

f o r example t h a t

2. THE BANACH SPACES %(K)

If

V

cOnv

c o n t a i n s an open s l i c e : t h e r e i s a l i n e a r f u n c t i o n a l such t h a t

CY

; f ( z ) = a1

c) *a).

5

cannot be equal t o

Assume t h a t some extreme p o i n t

p r o p o s i t i o n 3, f

K

V(K)

,

M(K)

,is

K

%(K)

, with

t h e Banach space o f r e a l -

t h e norm o f u n i f o r m convergence :

t h e space o f Radon measures on ' K

.

127

EXTREME POINTS OF COMPACT CONVEX SETS

,

Please n o t e t h a t i n t h i s paragraph

V(K)

The spaces

,

K

needs n o t be convex.

U( [0,11)

and s p e c i a l l y

, play

i n t h e c l a s s i f i c a t i o n o f Banach spaces, as a l r e a d y

an i m p o r t a n t r o l e

shows t h e f o l l o w i n g

p r o p o s i t i o n , due t o Banach and Mazur : PROPOSITION 1.

od

V([ O Y l I )

PROOF.

-

If

.

- Evmy nepahable Banach

npace ,LA ,LAom&c

t o a nubnpace

i s separable, t h e u n i t b a l l BE* o f i t s dual i s compact

E

* , E)

and m e t r i z a b l e f o r

( f i r s t p a r t , chap. 111,

o(E

5

11, p r o p o s i t i o n 3 ) .

We need a lemma :

- Evehy m&zable

LEMMA 2.

nex

c .

[0,1]

.

04

t h e Canton

-

We f i r s t r e c a l l b r i e f l y t h e c o n s t r u c t i o n o f t h e Cantor 1 , 32 [ f r o m [0,11 At A t t h e f i r s t stage, withdraw I

PROOF OF LEMMA 2. set i n

compact next i~ a coVLtinuoun h u g e .

t h e second, withdraw

.

]

1

th remain. A t t h e n - stage

,2

J 7 , 8 [ : so 2'

[ and

one has

2n

closed i n t e r v a l s

i n t e r v a l s o f same l e n g t h ; each i s th (n+l)-

'+'

d i v i d e d i n t o t h r e e p a r t s , and t h e m i d d l e one i s removed a t t h e stage, so

2

closed i n t e r v a l s remain.

let ('kk)k>l

be any sequence o f s t r i c t l y p o s i t i v e n a t u r a l

numbers ( n o t n e c e s s a r i l y i n c r e a s i n g ) . Then any sequence

m, G 2nk

f o r every

3

(mk)k21

with

k 2 1 d e t e r m i n a t e s a p o i n t o f t h e Cantor s e t : a t

'! .

Znl

i n t e r v a l s , we choose t h e ml This n, i n t e r v a l s ( a t the stage i n t e r v a l w i l l i n i t s t u r n be devided i n t o 2

tie

nl

stage, t h e r e a r e

'

2n1+n2 ) , we s h a l l choose t h e

k

, the

th m2 -

,

and so on. Since

nk 2 1 f o r a l l

diameters o f these i n t e r v a l s t e n d t o zero, and t h e r e i s o n l y one

point i n t h e i r intersection. Let

K

be a m e t r i z a b l e compact s e t , and

d e f i n e s t h e t o p o l o g y . F i r s t we cover o f closed b a l l s o f radius

1

.

K

d

be t h e d i s t a n c e which

by a f i n i t e number (say

Then t h e i n t e r s e c t i o n w i t h

t h e s e c l o s e d b a l l s w i l l be covered by a f i n i t e number (say

K

2n1 )

o f each o f

Zn'

) o f balls

B. BEAUZAMY

128 o f radius

1 , and so on : t h e sequence

i s constructed i n d u c t i -

(nk)n21

vely. We s h a l l now c o n s t r u c t a s u r j e c t i v e c o n t i n u o u s mapping from C

K

. Take

(mk)k21

any p o i n t

x EC

, with

< Znk

: i t determinates b i u n i v o c a l l y a sequence

f o r a l l k 2 1 ( ml i s t h e number o f t h e . nl ( ) t o which x belongs, m2 i s t h e number,

mk

i n t e r v a l of l e n g t h among t h e

i n t e r v a l s o b t a i n e d i n t h e p r e v i o u s one a t t h e 1 nl+n which c o n t a i n s stage, o f t h e i n t e r v a l o f l e n g t h ( )

pn2

the

pn2

pnl

2

nltn

x

2 t-h

, and

so

i n i t s t u r n determinates a p o i n t o f

on). Then t h i s sequence among t h e

onto

1

b a l l s o f radius

1

balls o f radius

, take

t h e b a l l o f number

ml

K :

; among

,

m2

c o v i r i n g i t , t a k e t h e b a l l o f number

and so on : these b a l l s form a sequence o f decreasing compacts, so t h e i n t e r s e c t i o n i s non-empty, and c o n s i s t s i n a s i n g l e p o i n t , s i n c e t h e r a d i i t e n d t o zero. Since f o r each point o f

K

k , the b a l l s o f radius

3

cover

K , each

belongs t o such a chain, and t h e mapping i s s u r j e c t i v e . I t i s

a l s o continuous : t a k e any

yo E K

, and

o f this ball is,

any open neighbourhood o f i t : i t

, for

c o n t a i n s some c l o s e d b a l l o f r a d i u s

3 by d e f i n i t i o n o f t h e mapping, ,of

which i s a neighbourhood, i n C

the point

k 2 1

. The

converse image

an i n t e r v a l o f

, converse

xo

,

[0,11

image of

yo.

T h i s ends t h e p r o o f o f o u r lemma. If

t E C

,let

ft

t h e element o f BE* which corresponds t o

t h e a p p l i c a t i o n g i v e n by t h e lemma. L e t f u n c t i o n on

[0,11

if

extended l i n e a r l y on

, by

y(t)

.

i s continuous : i f o(E*,

E)

, and

tn E C

and

tn -+to , t h e n

so

[O,ll \

c,

i s a f f i n e , t h e r e f o r e continuous. NOW,

with

t

d e f i n e a continuous

t EC

[0,11 \ C

for y(t)

. We

by :

y(t) = ft(x)

This function

x E E

l e t us compute

f ( x ) = llxll

( J Y ( t ) I I$g( [0,1])

. By c o n s t r u c t i o n ,

=

sup

OGtGl

there i s a

Iy(t)l

to E C

. Choose such t h a t

f E BE*, f = ft 0

.

EXTREME POINTS

So we have

Iy(to)I =

Ift

OF

( x ) I = llxll

COMPACT CONVEX SETS

, and

129

since

0

l y ( t ) I = I f t ( x ) I G IIftll

llxll

< llxll ,

for all

t E

c ,

we o b t a i n

Let E

U be t h e a p p l i c a t i o n x -+y ( t ) : t h i s i s a l i n e a r i s o m e t r y from , and o u r p r o p o s i t i o n i s proved.

i n t o Gf( [0,11)

T h i s r e s u l t shows t h a t U( [0,11)

is a

" u n i v e r s a l space" f o r separable

Banach spaces : i t i s separable ( s i n c e p o l y n o m i a l s w i t h r a t i o n a l c o e f f i c i e n t s a r e dense i n i t ) , and c o n t a i n s a l l separable spaces. So t h e r e i s no

, b u t one can wonder

need t o wonder what a r e t h e subspaces o f U ( [ O , l I )

what a r e t h e complemented subspaces. The answer t o t h i s q u e s t i o n i s n o t p e r f e c t l y known i n general (see a l s o e x e r c i s e 7, below), b u t a d e s c r i p t i o n can be g i v e n i n some s p e c i a l cases. Among them, t h e r e i s t h e case o f 1-complemented subspaces o f U(K)

. We

s h a l l mention t h i s r e s u l t w i t h o u t

g i v i n g i t s p r o o f ( t h e r e a d e r i s r e f e r r e d t o LINDENSTRAUSS-TZAFRIRI 1341). Let

uL = I d ) of

be an i n v o l u t i v e homeomorphism ( i . e .

u

i t s e l f . Call

t h e s e t o f f u n c t i o n s o f V(K)

gu(K)

f(ux) = -f(x) f o r a l l

x E K

,

0

t h e symmetry o f c e n t e r

( f o r example i f

% '

K)

K

K

into

which s a t i s f y

i s t h e u n i t c e r c l e and

u

i s t h e s e t o f f u n c t i o n s such t h a t

f(-x) = -f(x)).

PROPOSITION. ( J . LINDENSTRAUSS

Aom&c h hum&c

D. WULBERT).

-

A Banach opace

E

A

t o a 1-complemented xbopace 06 a Apace W ( K ) id and o n l y id 2 t o a opace gU(H) , doh home compact H and A U m C invuluLLve

homeomotphhme

u

06

H

.

, t h e f o l l o w i n g two

Concerning t h e complemented subspaces o f V(K)

r e s u l t s a r e known, b u t do n o t p r o v i d e a complete d e s c r i p t i o n :

-

If

t o W(K)

-

X @ Y

.

i s isomorphic t o a V(K)

,

either

Every complemented subspace o f U( [0,11)

isomorphic t o U( [0,11)

X

or

Y

i s isomorphic

w i t h non-separable dual i s

(H.P. ROSENTHAL 1421).

0 . BEAUZAMY

130

. The

I t i s i n t e r e s t i n g t o compare two W(K)-spaces

f i r s t result i n

t h i s d i r e c t i o n i s due t o Banach and Stone : THEOREM 3. (BANACH

-

-

STONE).

h p a c u W(K) and V(H) hvmevmvtrpkic.

LeA

K and

be AWO cvmpct he&. The

H

m e LwmeLLLc id and o d y id

and

K

H me

PROOF. a)

If 9

GR(K)

f E

i s an homeomorphism between

K

and

H

, the

application

f 0 9 E V(H)

--+

.

and V(H)

i s a s u r j e c t i v e i s o m e t r y between W ( K )

To show t h e converse i m p l i c a t i o n , t h e b a s i c t o o l w i l l be t h e b) We s h a l l f i r s t extreme p o i n t s o f t h e u n i t b a l l s o f W(H) and g ( K )

.

d e s c r i b e them :

- The CLXLUYWp ~ i n t n ~ u n d o n n f nuch X t h a t I f ( x ) I

t h e ul.Zit b a l l 0 4 V(K) 1 6vh all x E K

LEMMA 4.

PROOF OF LEMMA 4.

- Obviously

Conversely, i f f o r some f

fl+ f 2

fl(xo)

with

#'f2(xo)

REMARK.

(fl =

,

-

xo

Ifl(x)I

, and

f

,

such an If(xo)I

Q 1

f

i s an extreme p o i n t o f

< 1 , one

, (f2(x)I

Q

W(K) * may f i n d a decomposition

1 for all

, but

x

i s n o t extreme.

Since t h e f u n c t i o n s a r e r e a l - v a l u e d y t h e f u n c t i o n s

1 a r e constant, w i t h values t1 o r

component o f points o f

the CVVILL~~COU~

.

=

K

93

.

WK)

If K

.

-1

, on

f

with

each connected

i s connected, t h e n t h e r e a r e o n l y two extreme

: t h e . f u n c t i o n c o n s t a n t l y equal t o

+1

and t h e func-

-1 So 93 i s c e r t a i n l y n o t t h e c l o s e d convex h u l l o f W( K) i t s extreme p o i n t s ( c l o s e d f o r t h e norm, o r f o r u ( W ( K ) , M ( K ) ) ) . But t h i s

t i o n equal t o

does n o t c o n t r a d i c t Krein-Milman's theorem, s i n c e none o f these t o p o l o g i e s .

93

w K)

LEMMA 5. - The exLteme pvintn 0 6 t h e u n i t b a l l 9?* 0 6 V i t a e meahmu + SX , x E K

.

i s compact f o r

M(K)

ahe t h e

131

EXTREME POINTS OF COMPACT CONVEX SETS

-

PROOF.

,x

A = { kdx

Put

E K1.

E x t 3?* c A

L e t us show f i r s t t h a t

1)

O f course,

E

there i s a

E

> a , and

v x E K . Since If(x)

3?* , E $ TGG A

f E %(K)

function

I 0 such t h a t

q E

IIE II

cOnv

. llfll > a

A

.

This implies

, then

If(x)l < a so

< 1 , and

llfll

*

.

Conversely, l e t us show t h a t any element o f

A

is

i s an extreme p o i n t .

p 1 + P p 2 , p l , p 2 E &(K) , 6x = 1 I = 1 , a, P 2 0 , a + P = 1 ( t h e argument a p p l i e s a l s o t o 2 - S x ) . We c a l l l K t h e c o n s t a n t f u n c t i o n , equals t o 1 on K Then :

Assume f o r example t h a t 11cl

II = 1

llcl

.

Take now Then

, w i t h Ilfll < 1 , f 2 0 on K and f ( x )

f E V(K)

IllK - fll W K )

= 1

, and,

= 0

again :

and t h e r e f o r e

"(lK

-

P1(f)

= P2(f) = 0

-

f) = 1

IIfll

n(E) ,

Deduce t h a t

Ifn(t)I

(use ( Q )

[0,11 \ A

on

< E

fndpn

0

.

o(A([0,11),A

supl(pnll < t 00 . Show t h a t , f o r e v e r y n , w i t h p ( A ) < 6 ( ~ ) , and a number n(e)

E

in

Show t h a t

t E [0,11

Show t h a t a set

.

p,

(f0,ll)).

> 0 , there

is

such t h a t , f o r

and E g o r o f f ' s Theorem).

.

Observe t h a t , by p r o p o s i t i o n 1,

5

2, and e x e r c i s e 5, e v e r y separable,

r e f l e x i v e , i n f i n i t e - d i m e n s i o n a l Banach space i s i s o m e t r i c t o an uncomplemented subspace o f U ( [0,11)

.

REFERENCES ON CHAPTER V. The p r o o f o f B a u e r ' s maximum p r i n c i p l e comes from G. CHOQUET's book [121. P r o p o s i t i o n 2 ,

5

I, f o l l o w s BOURBAKI [Ill.

The p r o o f o f Banach-Mazur's Theorem

(5

11) i s t h e o r i g i n a l p r o o f

(S. BANACH [51) ; t h a t o f Banach-Stone's Theorem, u s i n g extreme p o i n t s , can

be found, f o r example, i n LINDENSTRAUSS-TZAFRIRI [341. E x e r c i s e 2 comes from BOURBAKI [ I l l , E x e r c i s e s 3 t o 7 f r o m LINDENSTRAUSS-TZAFRIRI [34]

.

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CHAPTER V I THE BANACH SPACES

Lp ( 5 2 , d , p )

We have a l r e a d y s t u d i e d t h e case 52 =

, if

d e f i n e d by P ( A ) = I A l

A c

IN

( IAl

,

1< p < t

IN , d = 9(IN) , and

P

A

i s the cardinality o f

A ) : we o b t a i n e d t h e

i . e . t h e number o f p o i n t s i n

.

1 -spaces P

,

We s h a l l now c o n s i d e r t h e case o f a p u r e l y non-atomic measure P : f o r all

w E 51

, ~ ( I w l )=

separable and i f P to

Lp ([0,11,

dt)

. One

0

can show t h a t i f t h e space

i s p u r e l y non-atomic, t h e n

. Therefore,

measurable f u n c t i o n s

L f

P

L (51,d,~) is P

(S2,d,~) i s isometric

P we s h a l l r e s t r i c t o u r s e l v e s t o t h e s t u d y

o f t h i s l a s t space, which we c a l l L e t us r e c a l l t h a t

L

Lp

i n short.

i s t h e space o f ( r e a l o r complex) c l a s s e s o f

such t h a t

( ,:,f(t)

lpdt)l/P

0 , one

B. BEAUZAMY

140

can approximate any f u n c t i o n

f

in

L

c o n s t a n t on dyadic i n t e r v a l s and s a t i s f i e s

, and

s p a n I ( h m ) m > O l = Lp PROPOSITION.

-

P

by a f u n c t i o n

IJf-

gll

P

g

which i s

; therefore

< E

we have o b t a i n e d :

The Hum S g ~ t mh u Schudm 6 u h 5 in

Lp (1 G p

2 :

ri(t)12dt)1/2

= ( Z ai) 2 112

i=l

By H i i l d e r ’ s

i=1 nequal i t y ,

.

THE BANACH SPACES

Lp

, 1

0

llfll

P

The s e t s

evefiy

r

p

,

1d r

U O<E

IIfJlp -

dt

> (1 -

EpI/fIIF

Ep)IIfIIF

,

and t h e lemma i s proved. LEMMA 7.

-

LeR

, 1< p < t

p

nom-one 6uncLLon~i n

A(E,~) , 0

n&

LP 0

A(E ,p

A(E1,p),..

fn's

LP

.

which do n o t belong t o

A(e,p)

o n l y a f i n i t e number o f f u n c t i o n s ,

.

Since. L

,ek

>0

such t h a t then

fn E

E

>0 ,

.

there are

Indeed, assume t h a t

fnly..

A ( E , ~ ) , there i s e l

U O < E < l

,

min(e,el,...,ck)

=

06 the

iA iAomohphic t o Lp 1,

L e t us f i r s t observe t h a t , f o r e v e r y g i v e n

i n f i n i t e l y many

06

which iA eqLLivdent t o t h e cano-

(f;l)n>l

(and thehe6ohe span{(f;l)n>ll

P

be a nequence

which iA n o t cvmpLukly contained i n any

Then thehe iA a dubbequence n i c d bunin

(fn)n>l

and LeR

. ,f

'k

>0

do n o t such t h a t

E A(ek,p) , and so, i f 'k A ( E ' , ~ ) f o r a l l n > 1 , which c o n t r a f

d i c t s t h e assumption. From t h i s remark f o l l o w s t h a t , f o r every there i s

n 2 no

such t h a t

fn g A(E ,p)

.

E

>0

and e v e r y

no

>1,

THE BANACH SPACES

>o ,

L e t now q

q


e k , w i t h ,

, integers for

:

l -

9p

Al; = Ak \

L e t us d e f i n e now

m u t u a l l y d i s j o i n t . We have :

9p

21-qkp

( -1t -

1

qP

qP-1

f u n c t i o n s : t h e sequence (fl;)k>l o f f? , and spant(fl;)k21} is

P

j >k

,

for

- 'illp

'

Ilfnk

-

fnk

and

= ( 1

Therefore

and

-

sets are

4kp'

is

1 - e q u i v a l e n t t o t h e canonical b a s i s

1-complemented i n

F i n a l l y , we have : IIfnk

... These

k = 1,2,

'~l;ll~

IIfni

' ~ l ;-

fill,

L

P

by p r o p o s i t i o n 4.

THE BANACH SPACES I f we choose TI

L

P

(n,&,c()

8

s m a l l enough t o have

lemma f o l l o w s now f r o m c h a p t e r I V ,

5 I,

TI

, 1< p < t

15 1

00

1 , that i s 0 , there

f E X

.

c A(E ,p)

, and

X

M

>0

such t h a t

i s isomorphic t o

L2

, and

.

f, E S , n o t b e l o n g i n g t o A(€ ,p) 1 T a k i n g s u c c e s s i v e l y E = ?i , n 2 1 , we o b t a i n a sequence (f,) , with 2 1 )(fnII = 1 , and fn F A( , p ) , f o r a l l n . By lemma 7, t h e sequence P 2 c o n t a i n s a subsequence (fr;)n>l which i s e q u i v a l e n t t o t h e (fn)n > I

or, f o r a l l

E

canonical basis o f

l

P

, and

is a

spanC(f;l)n211

i s complemented i n

LP

.

This

proves t h e f i r s t p a r t o f t h e theorem. F o r t h e second p a r t , we assume t h a t show t h a t LEMMA 9.

i s contained i n a s e t

X

-

LeR

X

Then, 6oh 60me

E

PROOF. find i n

X

A(e,p)

i s isomorphic t o

.

be a bubbpace 06 L 2

a } < 6 , and so

proves t h e e q u i - i n t e g r a b i l i t y o f

F

.

We sha 1 now i n v e s t i g a t e t h e l i n k s , f o r a sequence o f f u n c t i o n s , between e q u i - i n t e g r a b i l i t y and convergence f o r t h e t o p o l o g y PROPOSITION 3.

-

A

[0,11

06

, t h e Lm ia

e x h h and h @~Lte, h egui-integkabbe. M o ~ u w e h , t h e beguence

integhabbe ~unc2ion f

PROOF. on

3

, L),

.

E w a q beguence (fn)n21 06 i n t e g k a b b e dunc-tivnb buch

that, doh all m e a m a b b e bubbeL5

dt

o(L1

IAf(t)dt

( f n ) n >1 c v n w a g u , doh u(L1

, and,

604

all

A

meamabbe i n

l i m jAfn(t)dt n+tm

, L,)

[0,11

, tv an ,

.

- We s h a l l f i r s t f i x some n o t a t i o n s .

We c a l l d t h e Bore1 a - f i e l d o f [0,11 ( o r , more g e n e r a l l y a o - f i e l d , and c a l l 2 t h e q u o t i e n t o - f i e l d by t h e subsets o f measure

156

B. BEAUZAMY

zero : t h a t i s , we i d e n t i f y two measurable s e t s which d i f f e r o n l y by a s e t o f measure zero. I f A is

i s t h e symmetric d i f f e r e n c e between two s e t s ( t h a t

A A A' = A U A' \ A n A'), d(A, A ' ) = P ( A A A ' )

t h e n we can d e f i n e a d i s t a n c e on

(An)nEBV

2

I

( lAn)n i s Cauchy i n

f o l l o w s t h a t t h e sequence converges t o a f u n c t i o n almost everywhere t o

-

IIA(t) n

If

fo

. Since

,

fo t a k e s o n l y t h e values

LEMMA 4.

-

lA(t)Idt

a subsequence o f

9

i s a function i n

f

IK

PROOF.

fo

, and

L1

therefore converges

(lAn)nEN and

0

,

l(a.e.)

fo i s a c h a r a c t e r i s t i c f u n c t i o n : fo = lA , and s i n c e

and so,

into

i s a complete m e t r i c space. To see

be a Cauchy sequence. From t h e f o r m u l a :

,

I

by

,

and, equipped w i t h t h i s d i s t a n c e , this, l e t

2

L1

A)

*

we c o n s i d e r t h e a p p l i c a t i o n

9

, from

[ f(t)dt .

d e f i n e d by 9 ( A ) =

-

A

d

The app&cation

We w r i t e

[Af(t)dt

-

jAlf(t)dt =

and thus

and t h i s l a s t t e r m can be made s m a l l e r t h a n a g i v e n

E

>O

if

P(A A A ' )

i s small enough. T h i s proves t h e lemma. From t h i s lemma f o l l o w s t h a t , f o r any {A

€ 3,

IjAf(t)dtl

<E

1

'

E

> 0 , any

i s closed i n

2

f E L1

, the

set

f o r t h i s topology.

L

THE BANACH SPACES

P

, 1< p l

sequence o f i n t e g r a b l e f u n c t i o n s , such t h a t , f o r a l l

l i m jAfn(t)dt n-tt m F o r any FN = { A

€2

Each

FN

, by

every

A

€ 2,

, we

€2,

V

;

>N ,

m,n

..,

N = l,Z,.

define, f o r

/ I A ( f m ( t )- f n ( t ) ) d t l

t h e p r e v i o u s remark, i s c l o s e d i n

>N ,

if m,n

A

be a

e x i s t s and i s f i n i t e . 0

E>

157

00

fn(t)dt

-

/IA(fm(t)

2.

1

< E

By assumption, f o r

e x i s t s , and we can f i n d a

fn(t))dtl

< E

. This

.

N

such t h a t ,

-.

u FN = d

says t h a t

N>1

By B a i r e ' s P r o p e r t y ( f i r s t p a r t , c h a p t e r I ) , we can f i n d an i n t e g e r such t h a t and

r

F

>0

has non-empty i n t e r i o r : t h i s means t h a t t h e r e i s

NO

A

such t h a t , f o r a l l

€2 w i t h

P(A A Ao)

< r , then

A,

€ 2,

A E FN 0

that i s

Now, t a k e any

B

fn(t))dtl

€ 2. We

if

<E

m,n 2 No

No

,

.

have :

and (Ao U 8 ) A A.

C

B

(Ao \ B) A A.

C

B

Therefore, i f

P(B)

.

l

to

sup Ilf 1 < + 00 . F o r t h i s , we n l n > l a p a r t i t i o n o f [0,11 i n K s e t s o f measure a t most

p r o p o s i t i o n 2, i t remains t o show t h a t

( A j ) j = 1,.. . ,K

take

. For

min(r, r ' )

j =

, we have, f o r a l l m > 1 ,

l,...,K

J

and so

which proves t h e f i r s t p a r t o f t h e p r o p o s i t i o n . NOW, i f

A

that the qn's

€ 2 , we

put

Q(A) =

a r e equi-continuous,

l i m lAfn(t)dt n+tm

.

From t h e f a c t

one deduces e a s i l y t h a t

Q

is

u - a d d i t i v e , and, o b v i o u s l y , i t i s a b s o l u t e l y continuous w i t h r e s p e c t t o t h e

[0,11

Lebesgue measure on f E L1

Q(A) =

such t h a t

I

.

A

f(t)dt

We s h a l l now show t h a t We know t h a t , f o r a l l

By Radon-Nikodym Theorem, t h e r e i s a f u n c t i o n

,

for all

fn

A

f

€ 2,

A

for

IAfn(t)dt

€ 2. u(L1

--f

.

, L),

I A f ( t ) d t . Therefore, i f

g

i s a s t e p - f u n c t i o n ( t h a t i s , a l i n e a r combination o f c h a r a c t e r i s t i c f u n c t i o n s ) , then

I

fn(t)g(t)dt

f u n c t i o n , t h e r e i s a sequence to

h

in

Lm

. We

can w r i t e :

---f

f(t)g(t)dt

(g ) q q>l

.

I f now

h

i s a bounded

o f s t e p - f u n c t i o n s which converges

We know t h a t

supllf (1 = M n '1

00

l a r g e enough so t h a t i f

Now, f o r a g i v e n

,

- gqllm < %,

( M t Ilfll,)Ilh

l a r g e enough t o have no

0 , we q

and then,

h

choose

q

being fixed,

. This

l

fn E L 1( a , d y P 0 )

n+tm

(a,&).

i~ t h e phvbabLLLtq dedined by

, t h e bequence

(Ai)

SUP

Pn(Ai) i * +

-

Pn(A) = 0

If

mt

1 f E L (R,dyPo)

, whehe

, buch

bea2

that Q = f

buch t h a t Po(Ai)

. Thehedvhe,

Po

~+

m b

0 , then

0 .

Po(A) = 0 for a l l

= fn Po

h equL-integnabLe, and cvnuehgeb

(f,)

0 a d e a e c u i n g ~ a m i l q0 6

id

PROOF.

Po

, each Pn can be luttitten P,

weukey t v a duncfivn

n

16

n

.

, for

A ~

dt h e n,

Z 2-n Pn(A) = 0

n>l

, and

Lp (n,d,p), 1 < p

THE BANACH SPACES

Pn

Therefore, each

fn i n

proposition 3 f o r the

, on

,

in

fn's

L1

fn

(n,d,Po))

. We

u ( L 1 (.Q,d,Po),Lm

f

l i m Pn(A) = n ++m

F i n a l l y , if (Aili

f

for

,

.

Pn ' s

Q

are probabilities.

i s a probability.

i s a decreasing family, w i t h

EN

,

By assumption, f o r a l l

fn dPo = ],fdPo

l i m Pn(n) = 1 , s i n c e t h e n++m Q ( A ) 2 0 f o r a l l A E d , and

Moreover,

Po

L e t us use

have

Q(n)=

Also,

.

e x i s t s . So t h e r e i s a f u n c t i o n

I

Q(A) =

.

(n,d,P0)

such t h a t

Pn = fn - P o

such t h a t

lim f,, dPo = l i m Pn(A) n++m i A n++m

L1 (n,d,Po)

16 1

i s absolutely continuous w i t h respect t o

and so t h e r e i s a f u n c t i o n

A E&

0 such that (1)

lim a++-

sup

f € S

(because the function a

jlf(t)ldt =

f

-

6

>O

If I >a} sup f E S

Ilf(t)Idt

is a decreasing function

{ I f I >a)

of a ) . There is an increasing sequence tending to infinity, such that

of positive real numbers,

Therefore, there is a sequence (fn)n21 of elements of S such that, for all n 2 1 :

6. BEAUZAMY

164

We p u t

gn = fn 1

have :

1

n l

1 0 , we a n+tm n

, and hn

= fn

< PIlgnl > O l < P I l f n l

-

P1:(gnl > e ( ( g Since

(1

l f n l >an}

c o n t a i n e d i n any o f t h e s e t s

-

gn

. F o r every 1

>a,}

Gan

see t h a t t h e sequence

A(e,1)

, introduced

shows t h a t we can e x t r a c t from t h e sequence

in

5

$

1. Lemma 7,

ll

.

hn = fn

-

i s not

gn

5

1,

a subsequence

which i s e q u i v a l e n t t o t h e canonical b a s i s o f (gA)n 21 3s observe t h a t i t f o l l o w s from ( 3 ) t h a t < llg’II GT ) n l spant(g,’,)n>l} i s complemented i n L1 We s h a l l now t u r n t o t h e sequence

> 0 , we

.

(gn)n>l

(gn)n>l

e

, and

( l e t us

and such t h a t

show t h a t i t i s

equi-integrable. For any

n

> 1 ,if

p G n

,

the sets

IIh

therefore :

P

I >

an}

a r e empty, and

But

So we o b t a i n :

which proves t h e e q u i - i n t e g r a b i l i t y o f t h e sequence from t h i s sequence o n l y t h e subsequence

(h;l)n>l

. We keep (hn)n>l w i t h i n d i c e s correspon-

: we s t i l l have, o f course, an e q u i - i n t e g r a b l e sequence.

d i n g t o (g,’,)n>l From p r o p o s i t i o n 8 (and f i r s t p a r t , c h a p t e r 111, f o l l o w s t h a t we can e x t r a c t from t h e sequence

(h)n,’,>l

5

3, prop. 1, p. 60 ) a subsequence

THE BANACH SPACES

>1

, 1< p < + m

Lp ( C Z , . d , p )

which converges weakly t o a l i m i t

ho

. We

165

keep t h e subsequence

h i k - hik+l i s i n the closure ( f o r the

w i t h t h e same i n d i c e s . The c o n s e c u t i v e d i f f e r e n c e s

( g i ) n >1 converge weakly t o zero, and, consequently, norm) o f t h e s e t

conv{hik

i n c r e a s i n g sequence with, f o r a l l

,

j

-

hik+l

; k

0

> 1) .

Therefore, we can f i n d an

(k-) o f i n t e g e r s , and p o s i t i v e numbers J j>l kj+l Z: a . = 1 , such t h a t , i f we s e t : 1 i = k .+l J

kj+l

-

u = Z: a i ( f i i j i=k;+l

(ai)i

fii+l)

J

v

=

j

w = j

then

u

j

kj+l

:ai i = kZ.+l J kj+l

:ai i = kZ.+1 J = v.

+

w

~

, and

:

j

We have seen t h a t t h e sequence equivalent t o the canonical basis o f complemented i n

L1

.

(g;l)n>l

( nd

Is0

(!qn>l ) was

Ll , and t h a t span{(g;l)n>ll

We s h a l l see t h a t t h e sequence

(vj)j>l

was

,

which i s

made o f b l o c k s on t h e p r e v i o u s one, has t h e same p r o p e r t i e s :

PROOF. g; ' s

-

The f i r s t f a c t i s obvious, s i n c e t h e

, with

k.i+l

-Z: ai = 1

k,+l J

f o r e v e r y f i n i t e sequence

v.'s J

: t h e r e a r e two c o n s t a n t s (t.) J

o f scalars :

a r e b l o c k s one t h e C1

, C2 , such

that,

B. BEAUZAMY

166

and, i n p a r t i c u l a r ,

C1

< llv.II < C2 , 3 1

for a l l

.

j

(except f o r t h e n o r m a l i s a t i o n , t h e p r o o f i s t h e same as i n c h a p t e r I V , prop. 1, p. 108).

j> 1

F o r t h e second claim, l e t us choose, f o r v

j

, with

E Lm = (L1)* m

we p u t

Pf =

, and

v.(v.) = 1 J J

, an

.

C2

element If

kjtl

f = 2 t .Jg !J j

,

C tig;)vj . We d e f i n e t h i s way a p r o j e c t i o n from i = k .+l J , which i s continuous, s i n c e : onto spanI(vj)jB1l

c v*j (

j=l

spanI(g;)i>ll

where

* Ilvjllm
ll. spanI(g;)j>ll ( s i n c e -I(v 1emma.

I f w e compose w i t h t h e p r o j e c t i o n from

, we

o b t a i n a p r o j e c t i o n from

L1

onto

onto F I ( v j ) j 2 1 1

i s a subspace o f spanI(g;)j211

)

j j

L1

onto

) . T h i s proves t h e

L e t us now come back t o t h e p r o o f o f t h e p r o p o s i t i o n . We know t h a t IIuj

- v j I I 1 j ++.

0

.

I f , once more, we e x t r a c t a subsequence, and

renumber, we may assume t h a t , for a l l

j2 1

.

E

It follows t h a t

chosen small enough,

(uj)j>l

>0

being given, we have

Z JIuj

-

v.11

< E

, and,

IIuj

-

if

E

vjII1l J 1 i s e q u i v a l e n t t o t h e canonical b a s i s o f

i s complemented i n L1 and span{(uj)jrl} T h i s ends t h e p r o o f o f our p r o p o s i t i o n .

.L1 ( c h a p t e r I V , lemma 3, p. 1 1 0 ) .

THE BANACH SPACES

L

P

(a,d,p), 1 < p

1) , p r o p o s i t i o n 8 can be P ROSENTHAL [411 as shown t h a t i f a subspace F o f L1

As we mentioned f o r improved : H.P.

reflexive, there i s a

Lr

r

>1

such t h a t

F

is

i s isomorphic t o a subspace o f

*

§ 3. BANACH VALUED FUNCTIONS.

I n t h e p r e v i o u s paragraphs, we considered s c a l a r - v a l u e d f u n c t i o n s ; we t u r n now t o Banach-valued ones. Let

a Banach space, and

E

A function

f

, from

R

(52,d,~) a measurable space.

into

E

,will

be c a l l e d a n h p l e 6uncLLon i f

i t can be w r i t t e n

f = X x . 1

i

1

Ai

,

where t h e sum i s f i n i t e , t h e

xi's

are points i n

m u t u a l l y d i s j o i n t measurable subsets o f

52

.A

E

, and

function

f

the

,

Ails

are

from R

into

E , w i l l be c a l l e d measurable if t h e r e i s a sequence o f s i m p l e f u n c t i o n s (fn)n

which converges t o

>I

f o r almost a l l w

,

f

Ilfn(w)

a l m o s t everywhere, t h a t i s :

- f(w)11- n - t t

o

.

PROPOSITION 1.

- 7 6 f : ( 5 2 , d , p ) -+ E b memutable, .then, d o t evehy d o d e d ( o h open n d ) i n E , f - l ( C ) E nt . -

16

in E ,

h n e p m a b l e and 44, doh evehy cloned n e t C f - l ( C ) ~d , ,then f h mematable. E

det

C

(04open b e t )

PROOF.

1')

I t i s enough, o b v i o u s l y , t o prove o u r a s s e r t i o n when

b a l l B . Then, l e t f be measurable, (fn)n>l f u n c t i o n s c o n v e r g i n g t o f a.e. . F o r each n FnYk =

{ a € a, d i s t ( f n ( w ) ,

1 B) G K 1

.

C

i s a closed

a sequence o f s i m p l e k > 1 , we p u t

>1 ,

B. BEAUZAMY

168

Then, each

F

n,k

i s o b v i o u s l y measurable, and

n l i m inf F t m n,k k>l n

f-l(B)

-f

and so,

2”)

f-l(B) E d

Now, we assume

sequence i n

E

. E

. For each

t o be separable. L e t n

> 1 , we

be a dense

consider t h e f o l l o w i n g l i s t o f closed

balls :

If f ( w )

if

f(w)

1< k

< n)

fn(W) = x j

does n o t belong t o any o f these b a l l s , we p u t f n ( w ) = 0 ; 1 belongs t o a t l e a s t one o f them, l e t B ( x j , F ) (1 < j < n ,

.

be t h e f i r s t i n t h e l i s t t o which

The f u n c t i o n

b e l o n g s . Then we p u t

fn t h u s d e f i n e d t a k e s o n l y f i n i t e l y many values

), and f o r each

f a c t : O,xl,...,xn

f(w)

(in

,

f i l ( r x . 1 ) i s measurable ( s i n c e J f - l ( B ) i s measurable f o r e v e r y b a l l B ) , so fn i s a s i m p l e f u n c t i o n . We s h a l l see t h a t t h e sequence (fn)n>l converges t o f everywhere. For t h i s , l e t t h e sequence I(x

mO

B(xj

, and

let e

( x ~ >1 ) ~ i s dense i n

- f(w)II 0 . E

,

f o r . n 2 max(no

Take

t h e r e i s an index

, mo) , t h e r e

t h i s corresponds t o t h e f a c t t h a t

, %1 ) , and

k >mo

. Therefore,

we g e t

We denote by

Lp (52,d,r~,; E) ( 1

f

f(w)

Ilfn(w)

mO

no

is a

.

Since

such t h a t j

such t h a t

belongs t o a s e t

- f(w)()

0 , there

E

D c E

We say t h a t a subset i s an

xe E D

i s convex, t h i s means t h a t one can withdraw from

t o t a l diameter

< E

, and

is

such t h a t

a slice of

D

what remains i s s t i l l convex).

Then we have (see 1141 , 1151 ) : PROPOSITION. ( R i e f f e l

.id and

only

-

Maynard

- Huff -

Davis

-

Phelps).

-

E

had R.N.P.

id evehy bounded n u b n d 06 E iA devLtabCe.. can be c h a r a c t e r i z e d

For t h e dual o f a separable space, t h e R.N.P. simply : PROPOSITION. ( S t e g a l l ) .

.id

-

76

E

iA nepuhabLe,

han R.N.P.

E*

id and o n l y

0 bepanab&.

E*

Another geometric p r o p e r t y i s t h e f o l l o w i n g . We say t h a t

-

E

has

Milman P r o p e r t y (K.M.P. i n s h o r t ) i f e v e r y c l o s e d bounded convex s e t i n E i s t h e c l o s e d convex h u l l o f i t s extreme p o i n t s . Then R.N.P. Krein

(LINDENSTRAUSS, see [141) ; t h e converse i s n o t known.

i m p l i e s K.M.P.

Lp(n,d,cr

The d u a l i t y between t o t h e R.N.P. PROPOSITION.

; E)

and Lp' (~2,d,cc ; E*)

i s related

as f o l l o w s (see L. SCHWARTZ 1461 f o r a d e t a i l e d s t u d y )

-

The d u d

L p ' (n,d,cr ; E*)

06

:

Lp ( n , d , p ; E) c o i n c i d a u l i t h i6 and o n l y .id E* b R.N.P.

We s h a l l r e l a t e R.N.P.

.

w i t h m a r t i n g a l e ' s convergence i n t h e complements

t o t h e f o u r t h chapter, f o u r t h p a r t .

PART 3 SOME METRIC PROPERTIES IN BANACH SPACES

We have a l r e a d y made t h e d i s t i n c t i o n between a " t o p o l o g i c a l p r o p e r t y " and a " m e t r i c p r o p e r t y " . The f i r s t depends o n l y on t h e t o p o l o g y o f t h e space, and w i l l t h e r e f o r e be conserved i f we r e p l a c e t h e norm by an e q u i v a l e n t one. The SecGnd i s a c h a r a c t e r i s t i c o f t h e g i v e n m e t r i c , and may disappear i n an e q u i v a l e n t change o f norm.

As examples o f p r o p e r t i e s o f t h e f i r s t type, we met r e f l e x i v i t y , t h e e x i s t e n c e o f a subspace isomorphic t o

Ll

or

c

0

,

the existence o f a

Schauder b a s i s , and s e v e r a l o t h e r s . We have n o t seen so many p r o p e r t i e s o f t h e second t y p e : one i s t h e e x i s t e n c e o f a monotone Schauder b a s i s ( i f t h e norm changes, t h e b a s i s remains a b a s i s , b u t n o t n e c e s s a r i l y monotone). O f course, t h e P a r a l l e l o g r a m I d e n t i t y , which c h a r a c t e r i z e s p r e h i l b e r t i a n spaces, i s a m e t r i c p r o p e r t y . I n t h i s t h i r d p a r t , we s h a l l now d e s c r i b e several i m p o r t a n t m e t r i c p r o p e r t i e s which can be s a t i s f i e d i n a Banach space.

CHAPTER

I

STRICT CONVEXITY AND SMOOTHNESS

I n t h i s chapter, we i n v e s t i g a t e two m e t r i c p r o p e r t i e s : s t r i c t convexit y and smoothness. These p r o p e r t i e s a r e " l o c a l " ,

i n t h e sense t h a t t h e y a r e

n o t u n i f o r m on t h e u n i t sphere. The u n i f o r m analogues w i l l be s t u d i e d i n the next chapter.

5

1. STRICT CONVEXITY. Let

E

whenever

be a Banach space. We say t h a t x

and

y

s t r i c t l y convex. The spaces

,

second, t a k e

x = el

basis), then

Ilxlll = llylll y = e

i s clL%iDtey convex i f ,

are not colinear, then :

L

We s h a l l see l a t e r t h a t t h e

x=el+e2,

E

-

co

spaces (1 < p < t m ) a r e P L1 a r e n o t s t r i c t l y convex : i n t h e

,L

( t h e f i r s t two v e c t o r s o f t h e c a n o n i c a l

y = e2 = 1 ;

e2

P and

Ilx + yII1 = 2

: then

; i n the f i r s t ,

llxllo = IlyIlo = 1

, Ilx + yll,

take =

2

.

We s h a l l g i v e two e q u i v a l e n t c h a r a c t e r i z a t i o n s o f s t r i c t c o n v e x i t y , which, v e r y o f t e n , prove t o be more c o n v e n i e n t f o r computations : PROPOSITION 1. a)

(2)

E

id

iA n&~LcteyConvex id and o n l y -id, llxll = llyll = 1

,

then

It

(3) 175

y It
E

6

( E )

:

= 1

-

47.

Uniform convexity imp i e s s t r i c t convexity, and, a t l e a s t formally, i s s t r i c t l y s t r o n g e r , since t assumes t h e differences 1 t o be

1 y I(

uniformly bounded from be ow, f o r a l l x , y , IlxII = IIyII = 1 , IIx - YII > I n f a c t , uniform convexity i s r e a l l y s t r o n g e r , as t h e following example shows :

E .

E n ) p , w i t h 1 < p < t 00 , where a l l t h e E n ' s a r e n€IN uniformly convex. Then, by chap. I , § 3 , prop. 3, ( En)p i s s t r i c t l y n€IN convex. B u t , i f the E n ' s a r e l e s s and l e s s uniformly convex, t h a t i s , i f f o r every E > 0 , 6 ( E ) 0 , then ( En) will not be uniEn n€lN P formly convex. As examples o f E n ' s , we s h a l l s e e l a t e r t h a t we can take En = 1 , where 1 < p, < t m , and pn tends t o 1 o r t o t m , Pn when n t = .

Take

(

-f

The conditions llxll = llyll = 1 may be replaced, in the d e f i n i t i o n , by llxll < 1 and llyll Q 1 This w i l l allow us t o give homogeneous c h a r a c t e r i zations of uniform convexity :

.

PROPOSITION 1. onLq

.id,

604

-

Let

evmy

E

p , 1 < p < t m . E .i~ unidohmdq > 0 , thme b a rumba S ( E ) > 0

P

convex

.id and

nuch t h a t ,

UNIFORM CONVEXITY AND UNIFORM SMOOTHNESS

PROOF.

-

LEMMA 2.

We s h a l l f i r s t prove a lemma o f geometric n a t u r e .

- LeR

cunwex. SeR

X,

y

-

= IIx

E ’

be Awu clhfinc& poi&

.

yll

Then, doh all t

+ t(l 19 1) O , that is, for a l l

conv(xk+l,...))

>e

i s u n i f o r m l y convex, we o b t a i n , f o r a l l

And, s i n c e x3 (1 2 x1 x2 2 4 II>e +

+

,

we g e t :

which imp1 i e s

or

Using t h i s procedure again, we g e t

and a l s o :

. n 2 1:

(

x

k 2 1 :

~

)

~

~

B. BEAUZAMY

198

and t h e r e f o r e :

But s ( 0 )

i s s t r i c t l y p o s i t i v e , and t h i s i n e q u a l i t y i s i m p o s s i b l e f o r

n

1arge enough. I n a u n i f o r m l y convex space, weak convergence p l u s convergence o f norms i m p l y s t r o n g convergence. Indeed, j u s t r e p e a t i n g t h e arguments o f t h e f i r s t p r o o f o f t h e p r e v i o u s p r o p o s i t i o n , we o b t a i n immediately :

-

be a damdy 0 6 &emen& 06 E , u n i d o m ~ L y 9 be u &Xtehon I . We Mbume that convehgu weakLy t o an eRemmevLt x E E , and ,that ( I I X ~ I I ) ConuehgeA ~ ~ ~ to PROPOSITION 7.

LeZ

convex pace. L e t

IIxII

.

Then

ConvehgeA t o

x

i n nohm.

We s h a l l say t h a t a Banach space i s u n i f o r m l y c o n v e x i f i a b l e i f i t i s isomorphic t o a u n i f o r m l y convex space, t h a t i s , i f i t can be endowed w i t h an e q u i v a l e n t u n i f o r m l y convex norm. The p r e v i o u s p r o p o s i t i o n says t h a t a u n i f o r m l y c o n v e x i f i a b l e space must be r e f l e x i v e . But r e f l e x i v i t y i s n o t s u f f i c i e n t : we s h a l l i n v e s t i g a t e t h i s q u e s t i o n i n d e t a i l i n t h e f o u r t h p a r t . We s h a l l now g i v e some examples o f u n i f o r m l y convex spaces. PROPOSITION 8. convex

-

1")

16

1< p

E

we o b t a i n

that i s

which g i v e s

and t h i s l a s t q u a n t i t y i s e q u i v a l e n t t o

l e p p(z)

I t i s easy t o see t h a t one has i n f a c t

i f s2 = [0,11

, take

f(t) = 1

for all

g(t) = 1

if

0

t

€ P < t < 1 - (7)

when

6 ( ~ )= 1

E

-

.

[1 - (;)'q -f

0

I/P

:

B. BEAUZAMY

200

and

-1 i f

1

-

.

E P (7) l ..

E i s o f no importance,

would l e a d t o t h e same r e s u l t . But,

L2-norm, o r , more g e n e r a l l y , w i t h an L -norm, (1 < p < + m ) , P i s r e f l e x i v e ( f i r s t p a r t , chap. 111, e x e r c i s e 3 ) . We o b t a i n , t h e r e f o r e , ~

with the E

an example o f a non r e f l e x i v e space (here,

Ll ), which i s f i n i t e l y

r e p r e s e n t a b l e i n a r e f l e x i v e one. R e f l e x i v i t y i s n o t p r e s e r v e d under f i n i t e representability. For uniform convexity, t h e s i t u a t i o n i s d i f f e r e n t :

-

PROPOSITION 1.

in

E

PROOF. q

16

unidahmLy cvnuex and

E

-

Let

>0 , let

x,y E F

X'

= TX

,

, with

llxll = llyll = 1

spantx,yl

y ' = Ty

onto

E", w i t h

.

Then : 11x'll Q 1 + 9

or

E

,

IIx

be a two-dimensional subspace o f

Eo

isomorphism f r o m

Since

~ i n i t e R yfiepkenentable

F

, ,then F h unid0hmLy cvnuex.

y

Ily'II

Q

1+

1)

i s u n i f o r m l y convex, we o b t a i n :

IIT-lII

-

yll

E

> E

and

0 ,

1) 91) '1 7 1 '( l x' t y'

and t h i s proves t h a t

-

q)(l

'E(

& ))

9

i s u n i f o r m l y convex, and t h a t

F

(The p r e v i o u s l i m i t i s n o t n e c e s s a r i l y equal t o t i E ( € )

, since

n o t be continuous. But i t i s c e r t a i n l y g r e a t e r t h a n 6 E ( every a

>0

&)

needs

6E

, for

) . Our p r o p o s i t i o n i s proved.

We s h a l l now g i v e a g e n e r i c way o f producing Banach spaces which a r e

.

f i n i t e l y r e p r e s e n t a b l e i n a g i v e n Banach space E The key f o r t h i s c o n s t r u c t i o n i s t h e n o t i o n o f U l t r a p o w e r o f a Banach space. Let 4

be a n o n - t r i v i a l u l t r a f i l t e r on

IN

( r e c a l l t h a t an u l t r a -

f i l t e r i s s a i d t o be MUMi f i t c o n s i s t s i n a l l subsets o f contain a given integer

IN

which

ko ) .

We c o n s i d e r t h e p r o d u c t

and, i n i t , t h e subspace

EN

9 o f bounded

sequences :

9={ i € E I N, x-

= (

x

~ ; ) sup ~ IlxnII ~ < ~ t =1 n€lN

.

On t h i s subspace, t h e a p p l i c a t i o n

-

x =

L i m /IxnII

4

d e f i n e s a semi-norm. ( I f t h e u l t r a f i l t e r was t r i v i a l , i t would j u s t be The k e r n e l o f t h i s semi-norm i s t h e subspace

.N=C x

E EI

N, x- = (

We c o n s i d e r t h e q u o t i e n t

x

JV

~ ; ) Lim~ llxnll~ = 0) ~

4

91M',which

(xn),,€,,,

-

llxk

0

1

).

:

.

we c a l l E INla o r , more simply,

-E .

SUPER-PROPERTIES OF BANACH SPACES

221

On t h i s q u o t i e n t , t h e a p p l i c a t i o n

-x if

-+

-x

Lim IlxnII

4

, , i s a norm.

i s the class o f

-x -E , t h e n

We r e c a l l t h a t (by d e f i n i t i o n o f a q u o t i e n t ) , i f a r e two r e p r e s e n t a n t s o f a c l a s s

E

(

Lirn IIxn

4

-

x

~

~ A l l=

, )( 0 ,

X ~ A ~) ~ ~ ~ ~ and

Lirn Ilxnll = Lirn l l x ~ l l( t h e f i r s t f a c t i m p l i e s t h e second ! ) .

4

4

If

E

i s f i n i t e dimensional, t h e c l o s e d b a l l s a r e compact, and f o r

e v e r y bounded sequence i n f o r t h e norm. I f to

E

,

(

x

x = Lirn xn , t h e n

4

~

, )t h e

~l i m i~t

Lim ~ xn

exists i n E

a!

llxll = Lirn llxnll , and

i s isometric

4

E : one g e t s n o t h i n g more by t h i s process when E i s f i n i t e d i a e n s i o n a

-

PROPOSITION 2. PROOF.

h a Banach hpace.

E

- L e t (f(")),

be a Cauchy sequence i n

5

E

.

I n o r d e r t o show

t h a t i t converges, i t s u f f i c e s t o show t h a t i t has a c o n v e r g i n g subsequence, T h e r e f o r e , we may assume

then

f(n) =

n

Z u(j) j=l

show t h a t t h e s e r i e s For every

llf(n)

-

f(n-l)ll-

2

u ( ~ ), we have

0 , vne can bind a closed cvnuex hubs& C 06 K , nvt equal t o K ,. huch t h a t ,the diame2ten 06 K \ C -LA d m a l l e ~ than E

.

.

(K\C X,Y

sup EK \ C

IIX

i s the complement of - yll ).

C

in

K

; i t s diameter i s

PROOF OF THE LEMMA. - We c a l l b(K) the s e t of extreme points o f K (See second p a r t , chap. V ) , and D the closure of b(K) i n o ( F , F*) . D i s a compact set f o r t h i s topology, therefore a Baire space ( f i r s t p a r t , chapter I , prop. 1 ) .

Let E > 0 family of b a l l s

. Since

K i s separable, i t may be covered by a countable * Bn , of radius These b a l l s a r e a l s o o ( F , F )-closed,

5.

SUPER-PROPERTIES OF BANACH SPACES they cover D

, and

D

has non-empty i n t e r i o r

, open

set

0

Put

U = D\O

for

o(F

, because

D

0

i s open, meets

PO

nt of

t h e weak t o p o l o g y ) : t h e r e i s a

nD

D , and

K1

0 1, prop. 4 ) , b u t t h i s i s n o t p o s s i b l e :

t h e r e f o r e meets

K which i s n o t i n

.

K would be

o t h e r w i s e a l l extreme p o i n t s o f

Kl

. Let

D

.

xo

be an extreme

K1

L e t us observe, a t t h i s p o i n t o f t h e p r o o f , t h a t t h e s e t

cannot

C , because n o t h i n g says t h a t t h e d i a m e t e r o f

be taken as t h e d e s i r e d

K \ K1

with

, F*) , such t h a t 0 n D i s c o n t a i n e d i n B

(Second p a r t , chap. V,

U

Bn

intersection o f

0

, for

D

(in

, the

: t h i s i s a weakly c l o s e d s e t . I t s c l o s e d convex h u l l

i s n o t equal t o in

no 2 1

so, f o r some

233

i s small. T h i s i s f a l s e i n general, as t h e f o l l o w i n g p i c t u r e shows :

K2 = G ( 0 n D)

We p u t

K

extreme p o i n t o f conv(Kl U K2) image o f

. Then

i s either i n

i s closed, s i n c e

K1 x K 2

x

[0,11

K = conv(Kl U K2) K1

or i n and

K1

(1)

k = Xkl t (1

We c a l l

Ar

-

h)k2

,

x E K1

>r

We s h a l l see t h a t , i f

r

k E K

0 <X 4 1

the set o f points o f

t h e f o r m (1) f o r some X looking for.

a r e compact : i t i s t h e

i n the application

Ax t (1 - X ) y ) . Consequently, e v e r y p o i n t

(0

every

(recall that

K2

K2

, because

,

,y

E K2

,h

E

[0,11 +.

has a decomposition

kl E K1

,

k2 E K2

.

K which a d m i t a decomposition o f

< r < 1) .

i s small enough,

A,

i s the set

C we a r e

234

B. BEAUZAMY

A,

a)

i s convex : i f

+

k = Akl

-

-k -+ k '

2

with -

-

T

-

A

-

+

b)

& + A ' Akl + A u k ; 2 h+X' + (1

X+X'

-72-1

Ar

1- A

(1-A)k2 EK1

1- A

Y

A,A'

E r

,

and

(1 - A ) k 2 + (1 - a ' ) k ;

+xi

1

Akl + A ' k i

A'

:

k ' = A ' k ' + (1 - A ' ) k ; , w i t h 1

,

(1 - A)k2

, then

k,k' E Ar

+ +

(1

+

I -A'

- A')k;

1- A '

K2

i s weakly c l o s e d ( o r s t r o n g l y : t h i s i s t h e same, s i n c e A,

is

convex). Let

,

k E K

be a sequence o f p o i n t s i n Ar

k(n)

f o r t h e norm. L e t

be t h e corresponding decompositions. Since by f i r s t p a r t , chap. 111, (nj) which kl j + + m b '1 9 k = Akl

converging t o an element

+

(1

-

X)k2

,

5

K1

and

K2

a r e weakly compact,

3, prop. 1 , t h e r e i s a sequence (nj)jc,,, (nj) An h , and k2 j + +m+ k2

for

--+

with

A

>r

j ( i n t h i s p r o o f , one c o u l d as w e l l use

f i l t e r s , and a v o i d any appeal t o t h e mentioned p r o p o s i t i o n ) . c)

Ar

i s n o t equal t o

Indeed, t h e p o i n t x0 = O a k l t x0

of

.

I f M = sup{IIxII,

d) than

xo

E

, if

r

0,

E

F;

),

= TIE;

U and V , from E i o n t o Eo and f r o m F i o n t o F"

two isomorphisms

r e s p e c t i v e l y , w i t h , on

IlUll

IIu-lI1

1,

there i s , f o r every

K

>1.

n 2 1

E : t h i s w i l l prove t h a t

E

m

,

>1,

nuch

u l i t h b a n h cvnb-tant

had

(The b a s i s c o n s t a n t o f a b a s i c sequence i s d e f i n e d p. PROOF.

0

buch t h a t

:

(yl,..

. ,y

2

N)

dvtrm a

244

B. BEAUZAMY

-

PROOF OF LEMMA 2.

k

< ZN , e v e r y

By d e f i n i t i o n o f t h e b a s i s c o n s t a n t , we have, f o r e v e r y

sequence

,. .. ,a

o f scalars :

ZN

< k' < 2N ,

1X(ffl/S

245

,

a r e p o s i t i v e numbers,

Min

C .

m

Z laiI > M ( Z lai

Observe t h a t

1

’ill

K

X

, from

n .

Now, t h e f o r m u l a

implies

+- 1I amXm 1I
1 E

such

i s super-

r e f l e x i v e . The r e s u l t f o l l o w s . We now g i v e t h e second James'Theorem,

-

THEOREM 4.

LeA E

concerning l o w e r e s t i m a t e s :

be a hupeh-hed&xive hpace. Fok evehy c and evehy K

1 hatisdying 0 < 2c < K < 1 , thehe e x h a 2 a numbeh r , with 1 < r < -+ buck that, doh evehy n o m f i z e d banic hequence ( x ~ ulith ) ~b a n d~ con,5Xatant K , doh evehy d i n i t e hequence 06 h c d a h (ai) , one h a ~

PROOF.

-

Let

c

, K , with

0

< 2c < K1 < 1 .

00,

~

We s h a l l show t h a t , i f t h e

c o n c l u s i o n f a i l s , t h e assumptions o f t h e f o l l o w i n g lemma a r e s a t i s f i e d w i t h a' = 4K2c2 , 0' = K ,

-

Ahhwne that thehe me p o h X v e numbm 01' , 0' huch that, 6ok evehy n 2 1 , one can dind poi& yl,. ..,yn i n E huch that, doh aU , a U hequenceb 06 h c d m k , 1< k Q n (al,.. . ,an) :

LEMMA 5.

Then E h not hupeh-hedeexive. PROOF OF LEMMA 5.

Hzkll Q 1

. On

-

We p u t

span{zl,

zk =

TJT

yi

...,z n 1 = spanIyl,

j = 1, ...,n , a l i n e a r f u n c t i o n a l 0 if i # j Then

.

1 k A

g

j

by

, for

k = l,...,n

...,y ,I

g.(x.) J

1

, we

. Then

define, f o r

=a' i f

i= j

,

BASIC SEQUENCES

f o r e v e r y sequence Therefore

g

j

(al

251

,. . . ,an) .

can be extended t o t h e whole space, w i t h

Ilg.(( J

1

K

,

f o r which

252

B. BEAUZAMY

I

(we have used the f a c t t h a t s i n c e

and t h e r e f o r e m

>n

.

NOW, we choose al,...,am

with

has basis constant K

m

1I.Z aixill = 1 1=1

,

, and

(The Infimum i n ( 4 ) i s a t t a i n e d by compactness). For each

we have :

k = l,...,m,

and thus laklr

< KC)^

m

I: lai 1

Ir

0 , one can bind 6-hquaUb 06 E , equipped &h t h A nokm.

i n the unit b a l l

A s u p e r - r e f l e x i v e spa e can v e r y w e l l have squares. T h i s i s t h e case, f o r example, o f t h e space

d2

, equipped

which i s o b v i o u s l y e q u i v a l e n t t o t h e points

x = ( l , O , O ,...)

llxll = llyll = 1

3

,

w i t h t h e norm :

12-norm. I n t h i s space, t h e two

and y = (0,1,0 ,...) IIX

yll = 2

satisfy

.

2 . J-CONVEX BANACH SPACES. We s h a l l now e x t e n d t h e r e s u l t s o f t h e p r e v i o u s

paragraph, s t i l l u s i n g

t h e same techniques. P r e v i o u s l y , we have shown, i n a n o n - r e f l e x i v e space the

e x i s t e n c e o f two p o i n t s f o r m i n g a square. We s h a l l now show t h e

e x i s t e n c e o f an a r b i t r a r y number o f p o i n t s h a v i n g a s p e c i f i c g e o m e t r i c property.

26 2

B. BEAUZAMY

THEOREM 1.

-

doh euehy K doh

16

>1

all k G K , llxl

t

... t

[doh

k = K

PROOF.

-

number 6

.

.

E h not hedlexive, one can dind, doh euehy

xk

paid xl,..

-

(xktl

t

. ,xK

... t

i n t h e u n i t b a l l od E

xK))) > K 6

ttkin h intended an llxl

t

.

6

>0 ,

buch that,

.

... t

xKI/ 1.

We use t h e same n o t a t i o n s as i n t h e p r o o f o f theorem 1. The

>0

r

being given,

E

,m

a r e chosen as p r e v i o u s l y , and we

have

We a l s o t a k e a sequence that

We now f i n d

2Km

integers

the following order :

such t h a t : a)

for

k = l,.,.,K

o f norm-one l i n e a r f u n c t i o n a l s such

(gj)j

:

p( k ) j

.

k = 1,

....K

.

j = 1,

....Zm

.

in

UNIFORMLY NON-SQUARE BANACH SPACES

Then, we have, f o r

263

k = l,...,K

q3 < (-1)i - 1gh(uk) < 1 , i f

p2i-l (k)

and t h e r e f o r e on t h e i n t e r s e c t i o n

< h
1 , and

, conv(xktl,...,xK))

d i s t ( c o n v ( x l,...,xk)

i s not super-reflexive.

by

T h i s ends t h e p r o o f

We s h a l l now g i v e a homogeneous c h a r a c t e r i z a t i o n o f

J-convexity, j u s t

as i n t h i r d p a r t , c h a p t e r 11, we gave a homogeneous c h a r a c t e r i z a t i o n o f uniform convexity. K

For a given signs, o f l e n g t h PROPOSITION 4.

-

> 2 , we K

. Thus

Fot a Banach bbace E , t h e do.UoLng am equivalent

1)

E

2)

Thehe e x h t an integeh

evehy

t h e s e t of a d m i s s i b l e sequences of IK c a r d bK = K .

call

:

i~ J-convex, xl,..

. ,xK

i n

E

, one

El,..

., f K

3)

Thehe e x h t an intcgeh

evehy

xl,..

PROOF.

-

K2 2

,

a pobi,tive r u m b a 6

' ,

duch t h a t , doh

can dind an a d m i d b i b l e nequence ad bigisnn

WiAh:

. ,xK

K2 2

, a pobiXive

numbeh 6 "

in E ,

We show f i r s t t h a t 2) and 3) a r e e q u i v a l e n t .

buch t h a t

,

doh

267

UNIFORMLY NON-SQUARE BANACH SPACES

2) * 3 ) .

We assume t h a t , f o r an a d m i s s i b l e sequence o f s i g n s

(€9 j = I , . . . ,K

9

F o r t h e o t h e r a d m i s s i b l e sequences o f s i g n s

(E.

J

...,K , we

). J=1,

just write :

So we o b t a i n :

and

with

S" =

3 ) * 2).

2s' - s I 2 > 0 , and 3 ) i s proved If

then, f o r a t l e a s t one a d m i s s i b l e sequence o f s i g n s

(E;

)j=l,...,K

, we

get

or

and 2 ) f o l l o w s . 2)

* 1). I s obvious

s h a l l now prove t h a t 1)

: one j u s t takes

* 2).

xl,

...,xK

i n t h e u n i t b a l l . We

6. BEAUZAMY

268 So we assume t h e r e e x i s t

x1

,. . . ,xK

(‘j ) j = l , . . . , ~

If

Ilxlll

IIx .II J

with

llxKll

Now, t a k e any write, f o r every

x

6

>0

such t h a t f o r e v e r y sequence

i s an a d m i s s i b l e sequence o f s i g n s

such t h a t :

... =

=

,

K 2 2

< 1 , there

, we

obtain

...,xK

x1,x2,

in E

,a

j = l,...,K

,all

d i f f e r e n t from

0

. We

decomposition

= x! t x” J j y

j

where

We s e t m =

min

l Q i < K

So we have

0

Ilx’!ll = m

J

llxiII

.

< X.J < 1 , and ,

for

j = l,...,K

I

Using ( 2 ) , we know t h a t t h e r e i s an a d m i s s i b l e sequence o f s i g n s (‘j)j=l,...,~

1; K

ejx;jll

9

Q

such t h a t (1 - 6 )

K

c

1

IIxjII

.

can

UNIFORMLY NON-SQUARE BANACH SPACES So we o b t a i n , f o r t h i s sequence

K

c

1

Ilxjll

i.

(1

-

(ej

)j=l,e..,K

26 9

:

K

6)

c

1

""jl

By Cauchy-Schwarz i n e q u a l i t y , we o b t a i n :

But f o r one other

j 's

, we

j

a t l e a s t , we have X

have

A. 2 0

J

.

= 1

j We o b t a i n :

(say

j = jo ).

For the

I

and

with

and 2 ) i s proved. T h i s p r o p o s i t i o n w i l l be used t o show t h a t t h e space i s s u p e r - r e f l e x i v e i f and o n l y i f E i s s u p e r - r e f l e x i v e . graph w i l l be devoted t o t h e s t u d y o f t h i s space.

L2(Q,d,p

; E)

The n e x t para-

270

B. BEAUZAMY

We r e c a l l t h a t L2(52,d,cl ; E) i s t h e space o f measurable f u n c t i o n s on (52,d) , w i t h values i n E , such t h a t

( /llf(t)l12dp ( t ) ) l / 2

-

PROOF.

no C 52

First,

, with

.

~ ( 5 2 ~ )1 ; t h e a p p l i c a t i o n

sometry. So, i f

s u p e r - r e f l e x i ve , E

L ( f L d , p ; E)

2

L ( 5 2 , d , p ; E) : take 2

( i n short,

, we

L2(n ; E ) ) i s

has t h e same p r o p e r t y .

E t o be s u p e r - r e f l e x i v e . L e t fl,...,fK

Now, assume w E 52

-

i s i s o m e t r i c t o a subspace o f

E

i s the requi red

each

0 . n

UNIFORMLY NON-SQUARE BANACH SPACES

P u t E = 2 - e , E ' = 5 . Since E i s n o t n X ! ~ ) , . . . , X ( ~ ) i n t h e u n i t b a l l , such t h a t , f o r n every admissible choice o f signs ( E ~ ) ~ , < ,,, PROOF.

-

271

Let

n

>1, e >O .

J-convex, t h e r e a r e p o i n t s

NOW, l e t

k

Z: a . =

1

'

,

k n

Z ai

ktl

1 ,< k ,< n

= 1

. We

,

al

,... ,ak

,aktl

,...,a n

p o s i t i v e numbers w i t h

get

>n(l-e')

-

k t l

- (nk)tl

= 2 - n ~ ='

e ,

and t h e p r o p o s i t i o n i s proved.

REFERENCES ON CHAPTER 111. The p r o o f t h a t a n o n - r e f l e x i v e space has squares i s t a k e n f r o m R.C.

JAMES 1261. The statement i s a l s o g i v e n i n [261 f o r

three points ;

f o r h i g h e r number o f p o i n t s , t h e a d a p t a t i o n i s immediate ; i t was made by J. SCHAFFER and SUNDARESAN i n 1441. P r o p o s i t i o n 4 (homogeneous c h a r a c t e r i zation o f

J - c o n v e x i t y ) i s i n t h e paper o f G. P I S I E R 1401

.

This Page Intentionally Left Blank

CHAPTER I V RENORMING SUPER-REFLEXIVE BANACH SPACES

T h i s c h a p t e r w i l l be devoted t o t h e p r o o f o f a renorming theorem f o r s u p e r - r e f l e x i v e Banach spaces. The f a c t t h a t e v e r y s u p e r - r e f l e x i v e space admits an e q u i v a l e n t u n i f o r m l y convex norm was f i r s t proved by P. ENFLO

1201. The r e s u l t we p r e s e n t here i s more p r e c i s e and i s due t o G. P I S I E R

1401.

S(E) > C

E'

6a/r evetry

E

>O

.

The p r o o f w i l l be d i v i d e d i n t o s e v e r a l s t e p s :

5

1. BANACH-VALUED MARTINGALES. I n c h a p t e r I , we have e x p l a i n e d how s u p e r - r e f l e x i v i t y was connected t o

t h e F i n i t e Tree P r o p e r t y . I n c h a p t e r 11, we have g i v e n e s t i m a t e s ( o n b o t h s i d e s ) f o r b a s i c sequences. We s h a l l now connect F i n i t e Trees i n L 2 ( n ; E)

b a s i c sequences i n a space e s t i m a t e s , s i n c e we know t h a t We c o n s i d e r measure P

= C-1,

+I}'

i s t h e Haar measure

L (a; E ) 2

. The P

with

E

: t h i s w i l l a l l o w us t o use these i s s u p e r - r e f l e x i v e when

a-field

E

is.

d i s P ( n ) , and t h e

on f2 ( f2 i s a compact g r o u p ) . T h i s

measure i s a p r o b a b i l i t y , which can a l s o be w r i t t e n

where

6-1

, S1

a r e t h e D i r a c masses a t 273

-1

, +

1 r e s p e c t i v e l y . So we

B. BEAUZAMY

274

have f o r example

PI ( 1,-1 ,E 3 4 ,. .. )

; ~k = '1,

,f

n

In

, we

= {-1, tl}'

1 k 231 =-. 4

may c o n s i d e r t h e f o l l o w i n g sequence o f

a - f i e l d s (each o f them has o n l y f i n i t e l y many atoms) : go = I4

,

, and

f o r every

by t h e p r o j e c t i o n on t h e f i r s t

Bn ( n

and more g e n e r a l l y ,

n

> 1)

n 2 1

,

gn i s t h e

o - f i e l d generated

coordinates. That i s :

c o n s i s t s i n $J

,

, and

2n

atoms,

which a r e :

BE l y . . . , E

for

el =

= { ( E l ~ - - . ~ En+ly...) f ~ ~

n

f l y ...,E n = '1

We a l s o c a l l

Q,

;

ek

=

+1 f o r

k >n tl}

.

the p r o j e c t i o n onto the f i r s t

n

coodinates, t h a t

is :

BE1

=

)..

.,E

Q, -1{ ( E ~ , . . . , E ~ ) ~ .

n

The sequence

(3?n),n>o

Therefore, one can d e f i n e t h e

. By

(ay (

i s an i n c r e a s i n g sequence o f

E-valued m a r t i n g a l e s corresponding t o

d e f i n i t i o n , such a m a r t i n g a l e i s a sequence

(Mn)nEm)

of

-

Mn

gn-measurable f o r e v e r y

-

E

is

9

"M,+~

where EBn

o-fields.

E-valued random v a r i a b l e s w i t h t h e f o l l o w i n g p r o p e r t i e s :

= M~

,

f o r every

n

n

>0 ,

o ,

denotes t h e c o n d i t i o n a l e x p e c t a t i o n on t h e

o-field

Bn

.

275

RENORMING SUPER-REFLEXIVE BANACH SPACES Here, due t o the special form o f

!2

and B n ’ s

, the

corresponding

martingales are very simple : into a f u n c t i o n from R = { - I , t1)’ ‘(‘k ) k >1 E The f u n c t i o n s which we consider w i l l depend o n l y on a f i n i t e number o f , t h e r e i s an index k, 2 1 coordinates : f o r each such f u n c t i o n ‘(‘k ) k > l such t h a t

.

We c a l l

f o r every

(‘k)k>l

€k0+l

€kotl

‘kot2

$ot2

7..

. .

A f u n c t i o n w i t h t h i s p r o p e r t y w i l l be c a l l e d b W o m g .

For every

n

> 1 , we

euentu&y

c c ~ n ~ t a not ,r

can d e f i n e : n 2 k,

if

el

,...y~ny~n+ly

..., e

,. .

if n0

.A

consequence i s t h a t t h e

sequence (An& >o i s a monotone b a s i c sequence i n L2(S2 ; E) , Indeed, f o r e v e r y m , n w i t h m < n , e v e r y sequence o f s c a l a r s ao,. ,an ,

..

RENORMING SUPER-REFLEXIVE BANACH SPACES

277

we have :

(An)n >o

which proves t h a t

L2(n ;

x E E

, we

, with

(n , ( a n ) n > o ) x

,

in

E)

For every have

1

i s basic, w i t h basis constant

call M ( x )

values i n

E

t h e s e t o f m a r t i n g a l e s on

, which

as " s t a r t i n g p o i n t " , t h a t i s

are stationary,

EMn = x

(n

> 0)

.(

and which

JE

i s the

e x p e c t a t i o n ) . T h i s l a s t c o n d i t i o n s means a l s o t h a t x

xtl

=

x-l

2

We a l s o c a l l

.

Mm

the terminal value o f the martingale

since i t i s stationary,

we know t h a t , f o r some

ko 2 1

(Mn),,>,

;

,

I t s h o u l d be observed t h a t we have developed t h e language o f v e c t o r -

valued m a r t i n g a l e s s i n c e , as w i l l be seen, i t i s q u i t e w e l l adapted t o o u r aim, b u t t h i s language i s b y no means necessary, and c o u l d be c o m p l e t e l y avoided, s i n c e t h e p r o p e r t i e s o f m a r t i n g a l e s which we need a r e elementary, and can be e s t a b l i s h e d d i r e c t l y f o r t r e e s .

5

2. A SEQUENCE OF NORMS ON A SUPER-REFLEXIVE SPACE. We have seen t h a t , i f

E was s u p e r - r e f l e x i v e , so was L 2 ( n

Therefore, by c h a p t e r 11, t h e r e i s a c o n s t a n t

C

> 1 , and

t h a t , f o r e v e r y n o r m a l i z e d monotone b a s i c sequence

(f,)

one has, f o r e v e r y f i n i t e sequence o f s c a l a r s

:

(a,)

a in

p

; E)

>2

. such

L 2 ( n ; E)

,

2 78

6. BEAUZAMY

T h i s i m p l i e s , f o r e v e r y monotone b a s i c sequence

go, ...,gN

o f length

N t l :

By H o l d e r ' s i n e q u a l i t y , we have :

And so we g e t :

We p u t

c

J3

N

1 = T C.(Ntl)' N 21

For every

(XE1

T

'

, we

define, i f

From t h e Tree

-

,. .

= '1)

.,E

n

(ei

)

n-branches ( f o r a l l i

x = * 2n

p o i n t o f view, we know t h a t

. We may c o n s i d e r M ( x )

n 2'1 )

" s t a r t i n g " from

n E ,

L'1 =

A .

E

n

=

'1

x E E :

XE l,...,E

n

'

Mn

defines a

n-branch

as t h e s e t o f a l l

x , that is, satisfying

RENORMI NG SUPER-REFLEX IVE BANACH SPACES

279

I f we s e t : 6+1 = xtl

and, f o r 6

-x

y

6 - 1 = x-l

-x

,..., E n -

XE1

n 2 1 :

.

El'.

.,E

=

XE1

,. .

.,€

n- 1 An ) , we can a l s o w r i t e ( 3 ) under t h e

( t h i s corresponds t o t h e d i f f e r e n c e s form :

E

,N

all

n

all

A C {l,...,nl

with

n

n

= k1

'j =

>N ,all with

xE

]A/ = N

1.

We now g i v e some p r o p e r t i e s o f a)

II.IIN

llxll

b)

IIAj12

L2(s2 ; E )


o

A

c IN*

with

get

/A1 = N

2

' (Aj)j

, we

j=o" J"L2(f2 ; E)

1

but since

x

*

I f we a p p l y ( 2 ) , we o b t a i n , i f

j €A

s t a r t i n g from

.

I f we t a k e t h e m a r t i n g a l e c o n s t a n t l y equal t o

IIXNIl

z

, n-branches

1 ,...YE n

+'

-

1

2

i s monotone b a s i c :

I/ ,!

j = oAJ. # 'L

2

(a; E ) '

,

x

,

B. BEAUZAMY

280

1 --

L

C

A 2 d C2 ( N t l ) j € A J1'L2(R ; E )

(4)

'I

By d e f i n i t i o n o f t h e norm c) we can f i n d a m a r t i n g a l e (Mn)nEN

IAl

= N

, such

P

2 "Mm11L2(Sl; E)

, every

II*IIN , f o r e v e r y x E E i n M ( x ) and a s e t

A

C

01

>0 ,

lN* , w i t h

that

Using ( 4 ) , we g e t :

Also, we have

x = EMm , and t h e r e f o r e

which g i v e s

and s i n c e t h i s i s t r u e f o r e v e r y

We s h a l l now show tha.t

01

> 0 , we

obtain f i n a l l y :

II-IIN i s a norm on E

. This

w i l l be done i n

two steps :

PROOF OF LEMMA 2. (Mn)n '1 sets

A

and

, A'

-

(M,!)n>l

Let

x,y E E

,

, belonging

contained i n

01

>0

. We

can f i n d two m a r t i n g a l e s

r e s p e c t i v e l y t o M ( x ) and M ( y )

IN* , w i t h

I A l = I A ' I = N , two numbers

, two

RENORMING SUPER-REFLEXIVE BANACH SPACES

A = A'

We s h a l l f i r s t reduce t o t h e case where (Mn)n >1

martingales Let

kl

- . Then we

A'

(Ml;In>l

3

-M ' = M ' n

-M,',

replace the martingale

for

n

n

5

-Md, ,

(M,',)n>l

Md, =

-A ' = I m E I N

-

belongs t o A ( y ) A! = A! J-1 J

; m E A '

and now, we have a s e t

x'

, or

in

A'

.

m kl,

,

Ai

we have a l s o :

m-1EA'

kl

A

= M.

, we r e p l a c e j < k2-1 ,

s h a l l f i n a l l y o b t a i n a s e t , which we c a l l

A' = A

if

m>klluIkll.

.

-

M'

or

J Mk2 = Since t h e r e a r e o n l y f i n i t e l y many p o i n t s i n A

t a k e ( 7 ) ( a ) and ( b ) w i t h

i s now equal

1

i s t h e n e x t i n t e g e r which i s i n

but not i n

m a r t i n g a l e which s a t i s f i e s

> k2

if

j

,

which c o n t a i n s

We go on t h e same way : i f

j

and n o t i n

< kl-1

n>kl.

=

A'

A

by t h e m a r t i n g a l e

with

t o zero, s i n c e

not i n

by " s l o w i n g down" t h e

i s done t h e f o l l o w i n g way :

be t h e f i r s t i n d e x which i s , f o r example, i n

O f course,

where

. This

and

281

.

A

, common

M

Mk2-l and

-

A

but

by a

' MJ. A'

= Mj - 1

, we

f o r both. So we j u s t

B. BEAUZAMY

282 We c o n s i d e r t h e s h i f t

on

S

L2(s2 ; E )

, that

i s the application

d e f ined by

We now d e f i n e a m a r t i n g a l e MI1

0

=

, using

(Mi)n2l

( 7 ) ( a ) and ( b ) , by :

'Y

2'

M"(f 1 1) = x

if

el

=

= y

if

el

= -1

and, if n

t1

>1 ,

T h i s c o n s t r u c t i o n can be more e a s i l y understood i n terms o f n-branches: i f we s t a r t from

ztl = x

3

xe1

¶ a *

2-1 = Y

.YE

and Y,

n

1s..

.,, n

we d e f i n e

z,

1 Y...,,

Y

and Z1.e 2 , . . . , E

n

= xE2

,.

Obviously, we have :

and a l s o , f o r

For

j = 1

,

j

>1

we have

:

..,€

n

'

2

-lyE2

,...ye

n

= ye2

,..., n E

*

n

by :

RENORMING SUPER-REFLEXIVE BANACH SPACES

and f o r

,

j = 0

A , , ---x + y

2

0

.

O b v i o u s l y a l s o , we have

B

and t h e s e t

= A

II*II

-

A E IN*

with

F o r any increments o f

Mn

Let

IAl = N

u E E

. Therefore,

Mn

+

(Mn u

we g e t :

> 0 , the

(Y

lly 1;

x

€

E

OM

E ,

(Y

we g e t :

lemma i s proved.

II*llN i s continuous on

COW%LUU~

N

PROOF OF LEMMA 3.

I f we use t h i s m a r t i n g a l e ,

+ 1 , t o e s t i m a t e t h e norm

We s h a l l now see t h a t

-

y ).

E &(

(M;)n>O

and s i n c e t h i s h o l d s f o r e v e r y

LEMMA 3.

283

E

.

.

>0 .

Let

(Mn)n>O

E

&(x)

and

such t h a t :

+ u for

)

n

belongs ~ t~o d ( x~ + u )

> 1 , are

,

and t h e

t h e same as t h e increments o f

284

B. BEAUZAMY

So, s i n c e t h i s i s v a l i d f o r e v e r y

> 0 , we

a

get

which i m p l i e s

and t h e lemma i s proved. I t i s now c l e a r t h a t

Il*IIN i s a norm on E

: i t is obviously

p o s i t i v e l y homogeneous, and, from lemma 2, f o l l o w s t h a t

More g e n e r a l l y ,

and i f A

can be w r i t t e n

Since t h e s e t o f such

A's

X =

n 2 a j=1 J/$

, for

i s dense i n

[0,1]

some

, and

n2 1 since

continuous, t h e same r e l a t i o n h o l d s f o r e v e r y X E [0,11

{x E E

, llxllN < 11

i s convex, and t h e r e f o r e

t h i s norm i s e q u i v a l e n t t o t h e norm o f

E

modulus o f c o n v e x i t y , which we c a l l

.

hN

. We

,

al,..,,anEIN.

II-IIN i s

: t h i s says t h a t

II*llN i s a norm. By ( 6 ) , s h a l l now i n v e s t i g a t e i t s

285

RENORMING SUPER-REFLEXIVE BANACH SPACES

-

LEMMA 4.

16

ahe Awa pointn in E ouch tthat

x, y

then

whme

t h e camtatant i n ( 2 )

C

PROOF OF LEMMA 4.

a

>0 .

Let

-

x, y

Let

in

, with

E

1 (11x11;

be m a r t i n g a l e s i n & ( x )

,

M;l)n >O r e s p e c t i v e l y , s a t i s f y i n g ( 7 ) ( a ) and ( 7 ) ( b ) . L e t (Mn)n>O

We have seen t h a t , f o r

' CN

j

>1

:

IAl = N

Z A 2 (J

be t h e

c o n s t r u c t e d i n t h e p r o o f o f lemma 2.

m a r t i n g a l e i n A(

Thus, i f

t llyllN) 2 =

Let

B. BEAUZAMY

286

B u t a l s o , we know t h a t :

L e t us assume t h a t

we o b t a i n :

We d e f i n e t h e s e t have :

B = ( ( A \ { j o l )t 1) u 111

. Then

IBI = N

, and

we

RENORMING SUPER-REFLEXIVE BANACH SPACES Since t h i s h o l d s f o r e v e r y and

c;

I/

o(

> 0 , we

obtain that, i f

287 llxllN 2

+

(Iy(IN 2 = 2

So i f

then

and t h e lemma i s proved. T h i s i s n o t y e t uniform c o n v e x i t y o f t h e norm

II.llN , f o r two reasons : II-II , and n o t II-II N

f i r s t , t h e d i s t a n c e i n (10) i s measured w i t h t h e norm

( b u t t h i s i s n o t a s e r i o u s o b j e c t i o n , because o f t h e e q u i v a l e n c e o f t h e two norms), and, m a i n l y , because we have a c o n c l u s i o n o n l y f o r 3

1

, and

n o t f o r a r b i t r a r i l y small

>0 , in

E

o f (10). One can say t h a t each o f t h e norms

II.llN

t h e second member

has t h e u n i f o r m

c o n v e x i t y p r o p e r t y o n l y f o r p o i n t s which a r e a t d i s t a n c e a t l e a s t each o t h e r . NOW, by m i x i n g up t h e norms

II*llN

, we

eN

i s r e a l l y u n i f o r m l y convex.

LEMMA 5.

-

LeA

01

> 1 , P > 1 , and le,t

nohm on E

, Lclith

1")

< llxllN < llxll ,

llxll

(Il*IIN)N21

be a nequence

:

dolr

& x E E

,

from

s h a l l b u i l d a norm which

& N 2 1

,

06

288

B.

t h e n , do4 evehy

p'

>p

BEAUZAMY

thehhe e x d l 2 a comtartt C '

, and

a nOhm

I I

E , ndtin6ying

and wkich i~ uvLi6om.Ly convex, and ha4 a r n o d u h 06 convexLty wLth

> C'

6(E)

EP'

604

aee

E

>o .

PROOF OF L E M M A 5. - We p u t

T h i s d e f i n e s a norm on

Take now

IIx

-

yll

> E

which s a t i s f i e s o b v i o u s l y

1

2 (x (1x1' t ( y ( ) = 1 N 2 1 by ( 1 2 ) . We have a l s o , f o r a l l

x,y E E

,

E

with

-

y( > e

. Then

and so :

We t a k e E

9"-

CY

N = Zk

, where

k

i s t h e s m a l l e s t i n t e g e r such t h a t

that i s

2k/P (where Then we have

1-1

i s the e n t i r e part).

on

RENORMING SUPER-REFLEXIVE BANACH SPACES By assumption 2), t h i s i m p l i e s , f o r t h i s

For every

n

> 1 , we

have, s i n c e

ll*1/2n

k

289

:

i s a norm :

So we o b t a i n :

1

2

G T (1'1

+

I y I 2 - m6 TP 71 a k 2

2 (l/X1/2k

+

2 \/Y112k)

but, u s i n g a g a i n I " ) and (12), we have :

and t h e r e f o r e : X + Y

2 23 kP + 3 n k 2

'

since

1x1'

+

So we have o b t a i n e d t h e f o l l o w i n g : f o r e v e r y

with

then

1x1'

+

IyI

2

= 2

, Ix

-

yI > e

,

i f we s e t

ly12 = 2 E

>0 ,

. every

x,y E E

2 90

B. BEAUZAMY T h i s shows indeed t h a t t h e norm

1.1

i s u n i f o r m l y convex, b u t we s t i l l

have t o compute i t s modulus o f c o n v e x i t y . We have

and t h e r e f o r e

we have :

which i m p l i e s , a f o r t i o r i ,

C'

>0

s(e)

that for

very

I

> p , there

exists

constant

such t h a t

2

c

€PI ,

f o r every e

>o

and o u r theorem i s proved, So we have shown t h a t e v e r y s u p e r - r e f l e x i v e space c o u l d be endowed w i t h an e q u i v a l e n t norm f o r which i t was u n i f o r m l y convex. Moreover, t h i s norm hasa modulus o f c o n v e x i t y which i s o f "power-type", f o r some

q

>2

. The

norm

1.1

that i s

which we have b u i l t s a t i s f i e s

8 ( ~ 2) C e q

291

RENORMING SUPER-REFLEXIVE BANACH SPACES

1 Tllxll

< 1x1 < Hxll

f o r every

x E

E ,

b u t we can replace i t by another one, c l o s e r t o

1.1

Then, 6vtl each

PROOF.

\*I7 .

Let

:

be a nupeh-&e@exive Banach bpace, be a unidah.mly convex n o m v n E , Iclith

PROPOSITION 6.

L&

-

II-II

-

y

, 0

E

E 7

we have :

Also :

which gives

.

B. BEAUZAMY

292

By u n i f o r m c o n v e x i t y , we o b t a n, p u t t i n g

I 7l2 X + Y

(1

Q

e l

=

;

- 6 (el))

(

Since we have

we g e t

But

and so f i n a l l y

and t h e p r o p o s i t i o n i s proved. T h i s p r o p o s i t i o n shows t h a t i f 6 ( E )

,C -r

0

-to u n i f o r m l y convex. PROPOSITION 7. nmooth n o m nome

C"

P(T)

,

QCI' T

- E u e ~ [email protected]? npace I

r

, then

(7 1

6(e)

.

I I , ulith a madukkn 06

> 0 , nome

C eq

c,

q But o f course t h e c o n s t a n t ,C depends on Y i f t h e o r i g i n a l norm II*II on E was n o t a l r e a d y

f o r t h e same exponent and

yo

r

,

1

q

q'

r and

are not therefore d i r e c t l y r e l a t e d

t o t h e exponents i n James' e s t i m a t e s f o r t h e b a s i c sequences.

EXERCISES ON CHAPTER I V . EXERCISE 1. E

-

Let

E

be a Banach space and

: there i s a constant

q(x)

< Cllxll

C

>0

for all

q(x1

-

x-1)

> E

q(xel

points

(xE 1Y . . ,+1

and i f t h e

2"'

(n

- 1,

form a

-

Put

property

.

6

(n

- 1,

)

,E

q-branch

)

.,E

n e.=fl

e ) q-branch has been defined.

1

form a

XE1

Y

. ..,

Etl-l

,-I)

(n

,E )

q-branch

for a l l

> E

E

> 0 , there E

, we

el

i s , f o r every n 2 1 , a

say t h a t

E

has t h e f i n i t e

be another Banach space, and T

q ( x ) = IlTxII

exercise

(1

in

if

E

We say t h a t t h e if

,. .

= *1

,

e ) q-branch.

I f , f o r some

F

a

points

i n the u n i t b a l l o f Let

, x - ~ form

x1

1

,. . .;En-l

.

E

in

*

Assume t h a t a 2"

a continuous semi-norm on

such t h a t

x

We say t h a t two p o i n t s

q

.

Show t h a t

i f and o n l y i f

,E

(n

)

q-branch

q-Tree p r o p e r t y .

an o p e r a t o r f r o m

E

into

F

i s u n i f o r m l y c o n v e x i f y i n g (see c h a p t e r I ,

T E

does n o t have t h e f i n i t e

q-Tree

.

RENORMING SUPER-REFLEXIVE BANACH SPACES EXERCISE 2.

-

Let

295

be a continuous semi-norm, as i n e x e r c i s e 1. Assume

q

E

t h a t t h e r e e x i s t s on

1.1 , s a t i s f y i n g

an e q u i v a l e n t norm,

the following

property : For every

1x191

Show t h a t EXERCISE 3.

is a

(1

E )

E Let

> 0 , there

E

l Y l 9 1

Y

6

>0

cannot have t h e f i n i t e

E

>0

x+l x-l =

,

Assume t h a t a

(n

such t h a t , f o r a l l

q(x-Y)>E

Y

and

q-partition o f

+

is a

z E E z

llx+lll =

x,y E E

Y

q-Tree p r o p e r t y .

. We

say t h a t t h e couple

( x + ~, x - ~ )

if IIX-lII

and

that

( xE

- 1,

,. .., E n ) E i. = + 1

form a

(n-I,

E )

E )

q-partition o f

is a

(n

,E

)

q-partition o f

,

z

has been d e f i n e d . We say

q - p a r t i t i o n of

z

.

z

if

296

B. BEAUZAMY We assume t h a t

1")

E

does n o t have t h e f i n i t e

Show t h a t t h e r e e x i s t an i n t e g e r

f o r every

z E E

, every

(n

,E)

>0

n

and a 6

,

z

q-partition o f

>0

such t h a t ,

( xEl,...,€

n

one has

2")

Let

E

>O

Put, f o r

m

for a l l

,

n ,6

and

>1,

c,

g i v e n by 1 " ) . Assume

= (1

+

6

(1

1

+

> 1 , and

m

< E

4m

x E E :

t h e infimum b e i n g taken o v e r a l l

0

8 ( ~ ) C ep

. These

8(e)

like

e s t i m a t e s were g i v e n by G. P I S I E R [401 , who f i r s t used

m a r t i n g a l e s i n s t e a d o f t r e e s , and c o u l d make use o f James' e s t i m a t e s f o r b a s i c sequences i n

L2(i2 ; E)

. Pisier's

p r o o f depended upon deep proper-

t i e s o f m a r t i n g a l e s , much more t h a n t h e p r o o f we p r e s e n t here, which i s , from t h i s p o i n t o f view, q u i t e elementary. The p r e s e n t p r o o f i s due t o B. MAUREY, and i s reproduced here w i t h h i s k i n d p e r m i s s i o n ( i t has n o t been p u b l i s h e d elsewhere). The d e f i n i t i o n i n terms o f m a r t i n g a l e s a l l o w s t o make use of James' e s t i m a t e s . The o t h e r i d e a s o f t h e p r o o f ( l i k e " s l o w i n g down" t h e m a r t i n g a l e s , m i x i n g t h e sequence o f norms, and so on) were a l r e a d y i n ENFLO's paper

POI , and

are j u s t

adapted

from P O I . I t seems t o us t h a t

t h e p r e s e n t proof i s "minimal", i n t h e sense t h a t one cannot a v o i d t h e d i f f e r e n t steps i t contains. S u p e r - r e f l e x i v i t y and u n i f o r m c o n v e x i t y can a l s o be d e f i n e d f o r 181 ,

o p e r a t o r s between Banach spaces : t h i s was done b y t h e a u t h o r i n [71,

and t h e e x e r c i s e s i n t h i s c h a p t e r a r e t a k e n f r o m t h i s work. So, some o f t h e r e s u l t s o f chapters I , 11, 111, I V can be extended t o t h i s new frame, b u t James' e s t i m a t e s f o r b a s i c sequences do n o t h o l d anymore. There i s , however, a renorming theorem ( e x e r c i s e 3 above, taken

rom [71), which

g i v e s a new norm on t h e space on which t h e o p e r a t o r i s defined, b u t t h e modulus o f c o n v e x i t y does n o t s a t i s f y i n general be, f o r example

l i k e ST(€)

yo

C e

4€.

"(e

> C ep

, but

can

We s h a l l f i n i s h w i t h some complements about Banach-valued m a r t i n g a l e s . COMPLEMENTS ON CHAPTER I V . The m a r t i n g a l e s used i n t h i s c h a p t e r a r e o f v e r y s p e c i a l type, because o f t h e v e r y p e c u l i a r d e f i n i t i o n o f s2 , an , P "Walsh-Paley m a r t i n g a l e s " by G. PISIER i n [401.

. They

are c a l l e d

More g e n e r a l l y , i f (Li3a)aEI i s a monotone i n c r e a s i n g n e t o f sub o - f i e l d s o f d , one says t h a t a- f a m i l y (fa)cuEI o f f u n c t i o n s d e f i n e d on (a,&,P) , w i t h values i n E , i s a m a r t i n g a l e i f each fa i s

RENORMING SUPER-REFLEXIVE BANACH SPACES

a'-measurable for

fl < a

(see second p a r t , c h a p t e r V I ,

J

fBdP

A

The m a r t i n g a l e i s c a l l e d that i s sup

a

An

fa E

I

n

3 ) , and i f

i s t h e c o n d i t i o n a l e x p e c t a t i o n on 93

(Ego

JA fadP =

also that

5

Ll(n, Ba , P