Introduction to Banach Spaces and Their Geometry

107

5 5

1. 2.

The Banach spaces

114

THE BANACH SPACES

L

P

(a,&, m

p)

(1 < p

<+

-

126 )

)

137

140 153 167

SOME METRIC PROPERTIES IN BANACH SPACES STRICT CONVEXITY AND SMOOTHNESS

175

1.

S t r i c t convexity

175

2.

Smoothness and G a t e a u x - d i f f e r e n t i a b i l i t y o f t h e norm

177

3.

D u a l i t y between s t r i c t c o n v e x i t y and smoothness

184

CHAPTER 11. 1. 2.

UNIFORM CONVEXITY AND UNIFORM SMOOTHNESS

Uniform convexity U n i f o r m smoothness and U n i f o r m F r e c h e t D i f f e r e n t i a b i l i t y

PART 4. CHAPTER I.

189 189 204

THE GEOMETRY OF SUPER-REFLEXIVE BANACH SPACES FINITE REPRESENTABILITY AND SUPER-PROPERTIES OF BANACH SPACES

5 5

123 123

V(K)

1. Subspaces o f L (1 G p < + P 2. The space L1 3. Banach valued f u n c t i o n s

CHAPTER I.

5 5

107

Extreme p o i n t s o f compact convex s e t s

PART 3.

5 5 5

AND co

EXTREME POINTS OF COMPACT CONVEX SETS AND THE BANACH SPACES W(K)

CHAPTER V I .

5 5 5

)

1. Subspaces o f 1 ( 1 < p Q t 0 0 ) and co P 2 . Some p a r t i c u l a r p r o p e r t i e s o f l1 and .Lm

CHAPTER V .

5 5

00

217

1.

F i n i t e r e p r e s e n t a b i l i t y and Ultrapowers o f Banach spaces

217

2.

S u p e r - p r o p e r t i e s o f Banach spaces ; s u p e r - r e f l e x i v e spaces

225

The Tree P r o p e r t i e s

226

$ 3.

TABLE OF CONTENTS

xi

CHAPTER 11.

B A S I C SEQUENCES I N SUPER-REFLEXIVE BANACH SPACES

243

CHAPTER 111.

UNIFORMLY NON-SQUARE AND J-CONVEX

255

5 5

1. U n i f o r m l y non-square Banach Spaces 2. J-convex Banach Spaces

CHAPTER I V .

5 5

BANACH SPACES

RENORMING SUPER-REFLEXIVE BANACH SPACES

1.

Banach-valued M a r t i n g a l e s

2.

A sequence o f norms on a s u p e r - r e f l e x i v e space

255 261 273

2 73 277

BIBLIOGRAPHY

301

INDEX

30 5

This Page Intentionally Left Blank

CHAPTER 0

NOTATIONS AND PRELIMINARIES

In t h i s chapter, we r e c a l l some elementary d e f i n i t i o n s and f i x some terminology. A topaLogicaR hpuce i s a s e t endowed with a topology.

-

0 A topaLvgicaR uectoh hpuce E i s a vector space, equipped w i t h a topology, such t h a t t h e mappings ( x , y ) x t y (from E x E i n t o E ) and ( x , x ) x x (from E x IK i n t o E ) a r e continuous.

-

I f E i s a vector space, i t s f i e l d of s c a l a r s will be denoted by I t w i l l always be IR o r d .

IK.

i s a topological space such t h a t t h e o r i g i n has a fundamental system of convex neighbourhoods. All t h e l o c a l l y convex spaces t h a t we s h a l l meet a r e moreover Huuhduh61;, t h a t i s , two d i s t i n c t points have neighbourhoods which do not meet. 0

A L v c ~ cunuex y t u p d v g i c d uectvh bpuce

E

finean dpzce is a topological vector space i n which t h e A topology i s given by a d i s t a n c e . 0

0

A hemi-nohm on a vector space

E

i s a mapping

p

from E

into

which s a t i s f i e s P(X

+

Y)

P(X) t P(Y)

for all

x,y E E

for all

x

(sub-addi t i v i t y ) PbX)

=

A nvmi on

bIP(X) E

, denoted by

llxil = 0 implies x A normed space

=

0

E

IK, all

xE E

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

.

i s a l o c a l l y convex topological vector space.

A Eunuch hwce i s a normed space which i s complete f o r the norm

topology

.

1

IR+,

B . BEAUZAMY

2

0

Y;

a normed space, BE w i l l be i t s c l o s e d u n i t b a l l , and

E

If

its

u n i t sphere :

E and F a r e two normed spaces, Y(E,F)

If

w i l l be t h e space o f

E i n t o F , A continuous l i n e a r mapping w i l l j u s t be c a l l e d an apehatatr. The gtruph o f a mapping u i s denoted a l l continuous l i n e a r mappings from by

G r u , and i s g i v e n b y Gru = {(x, u(x)), x E E I

If A x E BI

B a r e two s e t s , t h e complement o f

and

A \ B. The complement o f

i s denoted by

d if f e r e n e If

A A B

t h e convex h u l l o f the

N =icaixi 0

spanI(xn)nEIN 1

I ( xn) If

a sequence t h e xi

Is

E

Dv I ,

[A. The symmetric

; N>O,

xn I s , t h a t i s

ao,...,aN>O

and

N Zai 0

1I

t h e v e c t o r space spanned by t h e xn I s , t h a t i s

span{(xn)nEINl = I

E

is

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

( x ~ ) ~ ~ 1, , ,

conv

A

i n A = { x ; x E A,

B

(A U B ) \ (A n B ) .

is

('n)n E IN

conv

.

N

z

aixi ; N > 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

This Page Intentionally Left Blank

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?

<1

rl

and t h e r e f o r e belongs t o a l l

, s i n c e xi

Bo ( x n , n ) . Since

t o a point x

x

. Since

x

j

belong t o

E i s complete, t h e

{ ( x ~ ) ~ > 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



xo E E

x o + x @ UN

i E I, IIui(xo+

, that x)II

E

and

is

, its r

complement has

>0

such t h a t , f o r

q(xo+x) < N

< N , and

.

we o b t a i n , f o r

B. BEAUZAMY

10

llUi x)ll Q llUi

xo)II

X+

llxll Q 1 , t h e n

Ifnow

+ IIui(xo)ll

Q

2N

.

2N , from which we deduce b ) , w i t h

Ilui(x)ll

QT

M = - 2N r '

-

- I f a l l t h e Un's a r e dense i n E , so i s t h e i r i n t e r s e c t i o n , by { x E E, q ( x ) = + 3 , and we

Theorem 1. T h i s i n t e r s e c t i o n i s t h e s e t obtain a).

- l e t E be a Bamch bpUCe, F be a notrmed bpace, and f i n e m continuoub mappingd @om E i n t o F We anbume t h a t , ('n)nE IN doh a l l x E E , t h e bequence ( U , , ( X ) ) , , ~ ~ cunvehgen, when n -+ + , t o a Lltn-Lt which we c a l l u ( x ) Then u i d a con.tinuoub f i n e m mapping, llull Q sup llunll . sup llunll < + m , and COROLLARY 3.

-

.

.

nElN

PROOF.

nEIN

-

The l i m i t mapping

un(x) all

x E E

,

u

, then

u(x)

sup IIun(x)ll n€IN

i s obviously l i n e a r . If, f o r a l l IIun(x)ll


m

IN

Ilun(x)II

. For

x

every

Q M , and thus

and t h a t

llull < I 4

.

, and

x E E ,

therefore, f o r

,

By Theorem 2, t h e r e i s a number

nE

Ilu(x)II

M

such t h a t

i n the u n i t b a l l o f Ilu(x)ll d M

, which

E

, we

llunll GM

for all

have

u

proves t h a t

i s continuous,

As a f i r s t a p p l i c a t i o n o f Theorem 1, we s h a l l g i v e a r e s u l t concerning a l g e b r a i c bases i n Banach spaces.

E a s e t of v e c t o r s has a unique decomposition

We s h a l l c a l l a l g e b r a i c b a s i s i n a Banach space (ei)iE I

such t h a t any element

x

in

E

x = C a.e i i as a &h.iXe l i n e a r combination o f t h e

ei's

,

The e x i s t e n c e o f a l g e b r a i c bases can be e s t a b l i s h e d v e r y e a s i l y u s i n g Z o r n ' s Axiom. We s h a l l come back l a t e r (Second P a r t , Chapter 11) on t h e n o t i o n of b a s i s i n a Banach space. COROLLARY 4.

the

-

dement^)

PROOF.

-

Let

(ei)iEI

Fn

06

be an a l g e b r a i c b a s i s , and l e t us assume, on t h e

I i s i n f i n i t e and countable. Then we can assume I = IN. t h e v e c t o r space spanned by ( eo,...,en) Since i t i s f i n i t e -

contrary, t h a t We c a l l

In a Banach crpace t h e cmdinaLLty (ithat .in, t h e nwnbeh ad an dgebtlaic b u h .in d h e h d-inite oh uncauntablee.

.

11

B A I R E ' S PROPERTY

dimensional, each

Fn

i s closed, and, s i n c e

(ei)iEIN

i s an a l g e b r a i c

U F Since E i s a B a i r e Space, one o f t h e F n ' s , say n ' n >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


, and

E be a uectoh A cLoned in L (IR , d t ) and in m

leA

P by t h a n e Awo spaces c o i n c i d e .

t h e space E endowed w i t h t h e norm L , and E PROOF. - We c a l l E P P q into E w i t h t h e norm L We c a l l I t h e i d e n t i t y mapping from E q ' P q ' We s h a l l see t h a t I i s continuous, and, f o r t h i s , prove t h a t i t s graph i s closed. be a sequence o f f u n c t i o n s i n E ; we assume t h a t Let ( f n ) n > 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

<0 ,

-

f(xl)

, c G p(

X

+

; x1 E F I

we o b t a i n , d i v i d i n g by

e )

.

-A :

-

X

f ( T; )

,

,

IR

be w r i t t e n : t c

E

, that i s

X E

For

:

E spanned by

has a unique decomposition

T h i s i n e q u a l i t y , by assumption, i s v a l i d f( 5 X

a club-

p

that i s :

, i t can

20

6. BEAUZAMY f(x2)

-

-

c G p(x2

,

e)

with

x2 =

-X x y

and t h e r e f o r e : c 2 supIf(x2) The choice o f sup{f(x2)

p(x2

-

e ) ; x2 E F )

p(x2

-

f

to

. Note

G

+ e)

e ) ; x2 E F ) G i n f { p ( x l

and t h e r e f o r e t h e c h o i c e o f functional

.

w i l l be p o s s i b l e i f and o n l y i f :

c

-

-

c

-

f(xl)

; x, E F l

,

i s p o s s i b l e , and we can extend t h e

the choice o f

c

i s n o t unique i n general,

and so t h e e x t e n s i o n i s n o t unique, e i t h e r . We s h a l l now use Z o r n ' s Axiom. L e t us f i r s t r e c a l l t h e statement.

-

be an o t r d a e d neX, in which evehy Zo2dY.y otrdehed bubneX hao a majotrant (we nay ZhaA A b induct.LtiveJ. Then any d e m e n t 06 A b majonized by a maximat d e m e n t ( a n d e m e n t a d maximat .id a < b * a = b ) , ZORN'S A X I O M .

LeX

A

(H,fH)

L e t us c o n s i d e r t h e s e t o f couples subspace o f

E , containing F

, and

H2 3 H1

and

fH 2

H i s a vector

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

H , e x t e n d i n g f , and dominated by these couples by t h e r e l a t i o n :

on

if

, where

i s an e x t e n s i o n o f

p

on

fH 1

H

.

.

We o r d e r t h e s e t o f

21

INFINITE-DIMENSIONAL NORMED SPACES Under t h i s o r d e r i n g , t h i s s e t i s i n d u c t i v e : l e t a t o t a l l y o r d e r e d subset. The element

if

( 2 ,f2)

(Hi

, fHi)iEI

be

d e f i n e d by :

x E Hi

i s t h e m a j o r a n t we were l o o k i n g f o r . Consequently, t h e element called

(L

,

fL). I f

i s m a j o r i z e d by a maximal element,

(F,f)

L was n o t t h e whole space

E

, we

could,

f i r s t p a r t o f the proof, extend

L

,

and

-

by t h e

t o a subspace s t r i c t l y l a r g e r t h a n fL L would n o t be maximal. T h e r e f o r e L = E , and fL = f i s t h e

extension o f

f

t o t h e whole space

E

.

T h i s proves o u r theorem. I t s h o u l d

be observed t h a t no t o p o l o g y was i n t r o d u c e d , and, i n p a r t i c u l a r , assumed t o be c l o s e d o r

f

F

i s not

t o be continuous.

The p r e v i o u s theorem makes sense o n l y f o r r e a l v e c t o r spaces ( s i n c e we assume

f < p ) . We s h a l l now pass t o a c o r o l l a r y which i s v a l i d f o r

complex v e c t o r spaces as w e l l .

-

Let E be a veototr bpace on R otr C , p a bemi-notun o n E , F a ve.ctah bubbpace 06 E and f a fineah ~ u n c L i o n don F ,

THEOREM 2.

dominated by

PROOF. is

p

,in

mVdmb :

- Assume f i r s t t h a t E i s a r e a l v e c t o r space. By Theorem 1, t h e r e f , w i t h F ( y ) G p(y) f o r a l l y E E . This implies

?r , e x t e n d i n g

(changing y

into

1 3 Y ) I G P(Y)

*

NOW, if E

-

-y ) : f ( - y ) G p ( - y )

that i s

-

-f(y)
and

i s a complex v e c t o r space, I t i s , a f o r t i o r i , a r e a l

v e c t o r space. So, i f we p u t which we can e x t e n d i n t o

, we o b t a i n a [c(y) I < p(y) f o r a l l

g ( x ) = Re f ( x )

, with

real functional, y E E

, by

the

22

B. BEAUZAMY

p r e v i o u s argument. But we have 7m(x) = - g ( i x ) ,

for a l l

,

x E F

and so, i f we s e t 5

f(Y) = 3 Y ) - m y )

Y

we o b t a i n a complex l i n e a r f u n c t i o n a l t o show t h a t

p

dominates

171

,

which extends y E E

For t h i s , i f

We have : I?(Y)I = e

-f -

-is-

f ( y ) = die T(e-”y)

= g(e

-is

,

y ) Q p(e-iey)

. We

f

put

s t i l l have

.

0 = Arg 7 ( y )

= p(y)

,

and t h i s proves t h e theorem. From t h i s theorem r e s u l t s t h a t t h e dual o f a normed space c o n t a i n s non-zero elements. More p r e c i s e l y :

- 1eR E be a nohmed Apace, xo E E . ThQhe h a corttinuoub f i n e a t ~unc-tiond f , LLlith llfll = 1 , buch t h a t f ( x o ) = lIxoII . COROLLARY 3.

PROOF.

- If

,

xo = 0

on t h e subspace

F

spanned by

xo : Fx

.

T

the extension

1

< llyll

, and

:

satisfies for a l l

T h i s l a s t formula shows t h a t

IITII

IK 1 , by

x E F , we have I f ( x ) I = llxll We extend f t o t h e E , u s i n g theorem 2, t a k i n g t h e norm 11.11 as semi-norm p

and so, f o r a1 1

(F(y)1

E

,

f(XX ) = XIIXoII

whole space

,h

= {AX,

0

xO

0

f

t h e r e i s n o t h i n g t o prove. Otherwise, we define

IITll

y EE

< 1 , and,

. since

f ( x o ) = llxoll

, we

have

t h e c o r o l l a r y i s proved.

The l i n e a r f u n c t i o n a l thus c o n s t r u c t e d w i l l be c a l l e d “ l i n e a r f u n c t i o n a l norming t h e p o i n t

xo

‘I,

and w i l l be denoted by

f

XO

. This

linear

f u n c t i o n a l needs n o t be unique i n general ( s i n c e e x t e n s i o n s a r e n o t unique, as we a l r e a d y mentioned) ; we s h a l l come back on t h i s problem i n t h e T h i r d P a r t , Chapter I. On t h e same way, we o b t a i n :

-

LeA

be a nohmed Apace, F a vectoh bubbpace, f a continuvuo f i n e m &nca%tianal on F . Thefie LA an externion T LLlith

COROLLARY 4.

IITII

= llfll

.

E

23

INFINITE-DIMENSIONAL NORMED SPACES

REMARK.

-

The formula, symmetric t o f o r m u l a (1) :

(2)

llXllE =

sup

IIE II

* d1

IE(x)

I

E

i s a l s o a consequence o f Hahn-Banach Theorem We hirve :

I E ( x ) l d IlEII

sup I E ( x ) l d llxll

IIE II -*

. llxll

:

and t h u s :

,

E

b u t , i f we choose

E

SUP I E ( x )

IIE II

, we

obtain

IfxW

= llxll

= fx

,

E*

and t h e f o r m u l a i s proved. We s h a l l now i n v e s t i g a t e s e v e r a l t o p o l o g i e s which can be i n t r o d u c e d on a normed space o r on i t s d u a l .

5

E*

2. WEAK TOPOLOGY ON THE DUAL On

, we

E*

.

have a l r e a d y i n t r o d u c e d a norm, by t h e f o r m u l a (1) o f t h e

p r e v i o u s paragraph. But t h e r e i s a n o t h e r t o p o l o g y which can be d e f i n e d : t h i s i s t h 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

A fundamental system o f neighbourhoods o f an element t h i s t o p o l o g y i s g i v e n by t h e f o l l o w i n g s e t s

for all

E

>O

We c a l l

,

all

* o ( E ,E)

n

>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<

sup

Ilr:II

E*

<1

xo

, t >I

,

= IIxoII

which says t h a t t h e norm o f t h e element o f l a 1) i s equal t o t h e norm o f identifiy identify

xo E

w i t h a subspace o f

E

E

.

, and

E**

l a r g e r than E**

E

E

E** d e f i n e d by

. Therefore,

xo

( b y formu-

it i s legitimate t o

which i t d e f i n e s , t h a t i t : t o

i s normed, b u t n o t complete,

s i n c e i t i s complete. But even i f

On

E

E**

w i t h t h e element o f

i s t h e norm o f

If

,in

xo

E**

t h e norm induced on

E

by

i s c e r t a i n l y l a r g e r than

i s a Banach space

E**

**

E

E

i s generally

: we s h a l l come back on t h i s q u e s t i o n i n t h e n e x t c h a p t e r .

, just

r e p e a t i n g t h e d e f i n i t i o n s o f t h e p r e v i o u s paragraph,

we i n t r o d u c e a weak t o p o l o g y , c a l l e d p o i n t - w i s e convergence on t h e elements

** ,E*) * of E .

o(E

,

: i t i s the topology o f

B. BEAUZAMY

26 Calling instead o f

z t h e elements o f E** , and t a k i n g t h e n o t a t i o n < z , 5 > 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


for zo , t > .

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
<17

< E

*

* u ( E , E). . ,xn E E . That i s If(0l

or Max I t ( x k ) l < 1 7 l
*

Using t h e f a c t t h a t

f

constant

C

>0

If(t)I

< E

< E

f

were

be a l i n e a r

By d e f i n i t i o n , f o r e v e r y such t h a t , i f t o say : ;

.

i s homogeneous, we o b t a i n t h e e x i s t e n c e o f a

* ,

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

B. BEAUZAMY

28

L e t us c a l l

xl,

...,xn .

of

F

F

L e t us assume t h a t

. If the

...,xr

xl,

r r e l a t i o n s E(xi)

.

f(€) = 0

E

t h e v e c t o r subspace o f = 0

spanned by t h e p o i n t s

( r G n ) form an a l g e b r a i c b a s i s l,...,r a r e s a t i s f i e d , t h e n

, i

i = 1, ..., r , l e t us choose a l i n e a r f u n c t i o n a l t i on F , Ei(x.) = 0 i f i # j ( j = 1, ...,r ) , and e x t e n d i t J t o a continuous l i n e a r f u n c t i o n a l , E i , d e f i n e d on t h e whole space E . For each

w i t h ti(xi)

-

= 1

-

Y

F o r each € E E* equal t o z e r o on ai

=

Ci(x)

F

, the

linear functional

: if

x EF

f(€) =

,

f ( €I ) = 0

*zE(Xi)f(Fi)

1 =1

with

and r =

, for i =

Y

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

PROPOSITION 4.

-

1

is

i=l

w i t h pi = f(Ei)

PROOF.

1=1 p x = aixi

r

r

.

can be w r i t t e n

I

- . 2 €(Xi)Ei

z 'yiF(Xi)

Therefore,

-& E*

x

= €

, and

E'(x) =

f

,

€ I

,

-

1.6

r

= €(

L: PiC(Xi) i=1

z pixi)

i=l

,

.

l,...,r T h i s proves o u r p r o p o s i t i o n , s i n c e r Z Pixi E E . i=l

eadoued MlitcZ t h e topoeogy a ( E , E*)

E

*

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

, ~ X Adual

i s weaker t h a n t h e norm topology,

t h e r e f o r e , t h e r e a r e fewer c o n t i n u o u s l i n e a r f u n c t i o n a l s . B u t t h e elements of

E*

remain continuous : t h e r e f o r e , t h e p r o p o s i t i o n i s proved.

We see, from these l a s t two p r o p o s i t i o n s , t h a t t h e d u a l i t y i s p e r f e c t l y symmetric between (E, o(E, E*)) i n general , i f

E*

and

(E*,

o(E*,

E))

. This

i s n o t t h e case,

i s endowed w i t h i t s norm : i t s dual i s t h e n E**

u s u a l l y s t r i c t l y greater than

E

. We

,

s h a l l study t h i s question i n d e t a i l

i n t h e n e x t chapter, b u t l e t us a l r e a d y g i v e an example o f t h i s s i t u a t i o n :

w i t h t h e norm Il(x )II

= '0

sup nEN

IXn(

.

29

INFINITE-DIMENSIONAL NORMED SPACES Then

i s a Banach space, and has f o r dual

co

e1 = w i t h t h e norm :

11 (Yn ( t h e dua The dual o f

el

is

w i t h t h e norm II(Zn)llm =

lZnl

.

(co)**

=

sup n€lN

The norm o f

Lm has t h e same e x p r e s s i o n as t h e norm o f

co

,

i s a l a r g e r space : i t i s t h e space o f t h e bounded sequences o f

but

s c a l a r s , whereas

in f in i t y

.

was t h e space o f sequences t e n d i n g t o z e r o a t

co

These spaces w i l l be s t u d i e d i n g r e a t d e t a i l i n Second P a r t , c h a p t e r I V .

5

4. THE SEPARATION OF CONVEX SUBSETS : THE GEOMETRIC FORM OF HAHN-BANACH THEOREM, L e t us f i r s t r e c a l l a few f a c t s . I f

v e c t o r subspace o f the quotient

E/F

E

, we

E

i s a v e c t o r space and

c a l l cvdimeMnion o f

F

in

F

a

t h e dimension o f

E

( i t may o f course be i n f i n i t e ) . An hyperplane i s a

v e c t o r subspace o f codimension 1. I t i s t h e k e r n e l o f a l i n e a r f u n c t i o n a l : t h e hyperplane i s c l o s e d i f t h e l i n e a r f u n c t i o n a l i s continuous, i t i s dense i n t h e converse case. An addine subspace i s t h e t r a n s l a t e o f a v e c t o r subspace. I n p a r t i c u l a r , an a f f i n e hyperplane i s t h e s e t o f p o i n t s a t which a l i n e a r f u n c t i o n a l takes a prescribed value. THEOREM 1. (Hahn-Banach Theorem, geometric form).

-

LeL

c d vectotr space,

in

E

R

a nvn-empty cvnwex open

E

,

L

be a t v p v l v g i a nvn-empty

addine subspace 06 E , which dven no2 mee2 51 . Then, thehe a clvned a66ine hypehplane H , which c v n t a i ~ n L and dven n o t mee2 R .

6. BEAUZAMY

30

PROOF.

-

We may, making a t r a n s l a t i o n i f necessary, assume t h a t

contains

R

the o r i g i n . We f i r s t c o n s i d e r t h e case when E

i s a r e a l v e c t o r space ; L

i s then

a r e a l a f f i n e subspace. We d e f i n e t h e gauge

a

o f t h e convex s e t

by t h e formula :

.

IRt

j(x) = i n f h E

j

;X R ~ X I

Note t h a t t h i s d e f i n i t i o n makes sense, s i n c e neighbourhood o f

0

: therefore, f o r a l l

,

x

R

,

being open, i s a

there i s a

X

>0

such t h a t

XR3X.

I t i s obvious on t h e d e f i n i t i o n t h a t , i f

so

R

i s p o s i t i v e l y homogeneous. Since

j

,

xEs2

X

>0 ,

j ( a x ) = a j x ) , and

i s convex, i f

, then

if

x+y +cc

that i s

01

tive, that i s

E

(X,p

+

j(x

y)

> 0) .

<j(x)

From t h i s f o l l o w s t h a t

t j(y)

,

for a l l

We have one more p r o p e r t y o f t h e gauge

x,y E E

.

i s subaddi-

j

, due t o t h e f a c t t h a t R

j

i s open : LEMMA 2. R = Ix E E ; j(x)

-

PROOF OF LEMMA 2. R

.

< 11 .

It i t clear that

Conversely, l e t y E R

small enough t o have

. We

Ix E E

, j ( x ) < 11

can f i n d , s i n c e R

(1 + e ) y E R

; therefore

i s contained i n

i s open, an

<1

j(y)

. The

E

>

,

0

lemma i s

proved. L e t us c o n s i d e r now t h e v e c t o r subspace 0

assumption,

X

E

IK , x

i s not i n

, and

E L

L

i s a linear functional Consequently, we have every

(5

x

in

F

.

L

, every

F

spanned by

L

element o f

i s an a f f i n e hyperplane o f f

on

F

f(x) <j(x)

such t h a t

, for

L

. Since,

F

. Therefore,

L = { x E F ; f ( x ) = 11

every

x

by

i s o f t h e form Xx

in

L

, and

there

.

thus f o r

We may a p p l y t h e a n a l y t i c form o f Hahn-Banach Theorem

1, theorem l ) , s i n c e

,

j s a t i s f i e s t h e necessary c o n d i t i o n s ( p o s i t i v e

31

INFINITE-DIMENSIONAL NORMED SPACES homogeneous, and s u b a d d i t i v e ) . So t h e r e e x i s t s a l i n e a r f u n c t i o n a l

-

d e f i n e d on t h e whole space

, for

f ( z ) G p(z)

H

Let H

.

52

. Let

4CE

C

, extending

H

H

and dominated by

E ,

-f ( y )

p :

, and

L ;

therefore

E : but i t

i s c l o s e d o r everywhere dense i n

i s a complex v e c t o r space ( L

E

,

i s c l o s e d : from t h e i n c l u s i o n s

cannot be dense, s i n c e i t does n o t meet t h e open s e t I f now

f

11 : i t c o n t a i n s

=

= 1 if y E H

j ( y ) >?(y)

us observe f i n a l l y t h a t follows that

f

Iy E

be t h e a f f i n e hyperplane

and 52 a r e d i s j o i n t , s i n c e

y

H

E

z E E

all

4

52

.

b e i n g a complex a f f i n e

subspace), by t h e p r e v i o u s argument, t h e r e i s a r e a l c l o s e d a f f i n e hyperplane where

Hi

, containing

Ho

complex hyperplane, and which does n o t meet L'

L

, disjoint

from

52

H = a t H'

, and

52

H'

i s a r e a l v e c t o r hyperplane. Then

=

.

Ho

Say t h a t

HA

fl (iHA)

= a

+ HA ,

is a

i s a complex a f f i n e c l o s e d hyperplane,

which c o n t a i n s

L

since

i s a complex v e c t o r subspace, t h u s s a t i s f y i n g

L = a t L'

iL' = L'

.

where

T h i s proves

o u r theorem. We c o u l d argue d i r e c t l y , f o r complex Banach spaces, u s i n g theorem 2,

1, i n s t e a d o f theorem 1, 5 1, b u t t h i s would r e q u i r e a n o t h e r d e f i n i t i o n o f t h e gauge, namely t h e f o l l o w i n g :

5

j ( x ) = i n f f l x l ; A E IK

xa 3 XI ,

which would be a "symmetric" gauge. We now g i v e s e v e r a l a p p l i c a t i o n s o f Theorem 1 t o t h e s e p a r a t i o n o f convex s e t s . COROLLARY 3 .

-

1eX

E

; f(x)

be a heal! t o p o L o g L c d vectok space,

A

a non-empty apen canuex bubb&, B a non-empxy convex bubb&, which d o a n o t me& A . Then .thehe h a coM.tinuau fineah ~uncLLtianal! f and a heal! numbefi (Y buch that : A PROOF.

A

C

{x E

-

The s e t


B C {x ; f ( x ) >a1

U ( A - z ) = A - B ( s e t o f d i f f e r e n c e s o f elements o f ZEB B , n o t A n I B ! ) i s open, as a r e u n i o n o f open s e t s ,

and i t i s convex, s i n c e i t i s t h e image o f (y,z)

y-z

.

.

C =

and elements o f

-

E

A

x

B

by t h e a p p l i c a t i o n

0

I t i s non-empty, and does n o t c o n t a i n

t h e r e e x i s t s a c l o s e d hyperplane which does n o t meet

C

.

By t h e Theorem,

: one can f i n d a

32

B. BEAUZAMY

continuous l i n e a r f u n c t i o n a l Therefore, we have p u t LY =

, we

f(y) < a G f(z)

such t h a t

< f(z) , for

f(y)

i n f f(z) z€B

f

, for

f(x) < O y E A

all

, all

all

z E B

x E C

. So,

.

i f we

have

,

for all

, all

y E A

A

T h i s i s n o t y e t o u r statement, b u t s i n c e

.

z E B

i s open, we have

and t h e c o r o l l a r y i s proved.

I f we have A

C

I x E E ; f(x) Gal

I x ; f ( x ) = 00 bepuhuhu

say t h a t t h e hyperplane

them l i e s i n one c l o s e d h a l f - s p a c e . I f we have

B

C

, we

B C I x E E ; f(x) > a )

and

A

C

A

B

and

: each o f

{ x E E ; f ( x ) < a ) and

I x E E ; f ( x ) > a ) , we say t h a t t h i s hyperplane b,i7ulcLLg bepmateh

and

B

A

.

We s h a l l now e s t a b l i s h t h e analogue o f C o r o l l a r y 3 f o r complex v e c t o r

, i s ~-bymme,ttic i f , whenever 1x1 = 1 , Ax E A (such s e t s a r e

E

-

COROLLARY 4.

Let

c o n t a i n e d i n a complex v e c t o r space

x E A

,

then, f o r a l l

x

E

a: w i t h

a l s o c a l l e d "balanced").

be a compLex XopoLogicaf! vectvh Apace,

A

a non-

B a non-empty convex b d , w k i c h Then R h a e a (compLexJ continuouh f i n e a h 6unctionaL and a numbs a > 0 buch t h a t :

empty, open,

~-bymme,.7hic and convex be,t,

doen noX meet A f

E

,

A

spaces. We s h a l l say t h a t a s e t

.

PROOF.

-

on

and a number a E IR

E

By C o r o l l a r y 3, t h e r e i s a r e a l continuous l i n e a r f u n c t i o n a l

A c I x E E ; g(x) < D o

0 E A

Since f(x) = g(x)

-

I f ( x ) l >6& Take now

( A

such t h a t

,

B C I x E E ; g(x) > a )

i s convex and C-symmetric), we have

i g ( i x ) ; t h e n Re f = g f(x) > a

x E A , and e

g

if

x E B

such t h a t

.

, and

so

0
.

Put

I N F I N I T E - D I M E N S I ONAL NORMED S P A C E S (eisf(x))

/f(x)l =

A

Since

is

.

= g(eisx)

C-symmetric,

e

i sx

33

E

A

and

, which

/f(x)I
proves t h e

c o r o l 1a r y .

COROLLARY 5. - L& E be a fie&, Haundvfidd, LvcaUy cvnuex t v p v L v g i c d uectofi npace. le,t A be a nun-empty compact cvnuex A&, B a non-empty cloned conucx n&, wkich d v u n o t me& A . Then, t h m e & a cLvbed a d d i n e hypenplane wkcch b . t k h t . t y nepahate.6 A and B : t h e n e A a cvvttinuvud finean. &nc.tivn& f and a fie& a nuch t h d : A

C tx E

,

E ; f ( x )
PROOF. - I f A

B

C

I x E E ; f(x) > a )

i s a compact s e t and

B

.

a c l o s e d s e t i n a l o c a l l y convex

t o p o l o g i c a l v e c t o r space, which do n o t i n t e r s e c t , one can f i n d an open convex neighbourhood o f t h e o r i g i n ,

U , such t h a t A + U and B + U do n o t

i n t e r s e c t (see f o r i n s t a n c e Bourbaki [ l l ] ) .We can a p p l y c o r o l l a r y 3 t o

A t U

and

A+U

6 t

C {x E

U , and we o b t a i n

E

; f ( x )
B t U c I x E E ; f ( x ) >a1

the l a s t inclusion since and

B

.

B+U

, i s open.

A

f o r t i o r i , t h e same h o l d s f o r

A

The same way, we have :

COROLLARY 6 . - L e t uec.tofi npace, A E-dyrnmeLkic cloned complex cvn;tinuvud A

C

E be a comptex Haundvfidd S o c a U y convex X o p o L o g i c d a non-empty compact convex n&, B a nan-empty

cvnuex n e t , which d v u n o t me& A fineah dunc.tivn& f and a numbeh

IX E E ; / f ( x ) I >a1 , B

C

{x E

E

.

CY

; I f ( x ) I
'Thefie e x A & a > 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 )
set contains

, f(y)

f

does n o t c o n t a i n

C

for

u(E, E*)

xo

.

: t h i s proves t h a t

xo

cannot

I t f o l l o w s from t h i s p r o p o s i t i o n t h a t , f o r a convex s e t , t h e c l o s u r e s

f o r t h e norm and f o r

u

(E, E*) . a r e t h e same.

Recall t h a t a r e a l f u n c t i o n continuous i f , f o r every

h E

I f , moreover, t h e f u n c t i o n

q

q

,

, i s l o w e r semiE, ~ ( x 9) x 1 i s closed.

d e f i n e d on

IR , t h e

set

E

{x €

i s convex, t h e n these s e t s a r e a l s o convex,

and t h e p r e v i o u s p r o p o s i t i o n says t h a t i f a convex f u n c t i o n i s l o w e r semicontinuous f o r t h e norm, i t i s a l s o l o w e r semi-continuous f o r t h e weak topology

A

.

very i m p o r t a n t example i s t h e norm i t s e l f , which i s t h e r e f o r e l o w e r

semi-continuous f o r t h e weak t o p o l o g y of

E

are

a(E,

E*)-closed,

closed. I t i m p l i e s t h a t i f E

u ( E , E*)

.

o r t h a t the b a l l s o f (

x

~

, weakly converging t o an element x

T h i s means t h a t t h e b a l l s

E*

are

o(E*, E**)-

i s a sequence i n t h e u n i t b a l l o f ) ~ ~ ~ E , then :

E

INFINITE-DIMENSIONAL NORMED SPACES

35

11x11 G l i m i n f I I X ~ I I n-tt m

, converging

(and t h e same f o r a f a m i l y

-

REMARK.

,a

E*

In

norm-closed, b u t n o t But, i f filter llfll

(fi)i

F o r example, i n

** , E

o(E

9,i n

o(~*,


i n f llfiII

x E

lim

, to

E)

BE

Ifi(x)

9

u n i t b a l l SE i s convex,

) - c l o s e d (see p r o p o s i t i o n 4, below).

an element

, we

f

BE*, c o n v e r g i n g

for a

a l s o have :

,

9

If(x)I =

** , t h e

E

i s a f a m i l y o f elements i n

EI

since, f o r every

9on I ) .

convex s e t may be c l o s e d f o r t h e n o r m - w i t h o u t b e i n g

* o(E , E) .

closed f o r

for a filter

Y


i n f llfiII

9

**

The same r e l a t i o n h o l d s a so i n

E

,

for

O(E**,

E*).

The f o l l o w i n g p r o p o s i t i o n i s a l s o due t o G. Mackey :

-

PROPOSITION 2. &o

The dubdetn

06

E

nohm-bounded (arid c o n v m d y ) .

w h i c h ahe bounded doh o ( E , E*)

PROOF. -We r e c a l l f i r s t t h a t , i n a t o p o l o g i c a l v e c t o r space, a s e t

bounded i f , f o r e v e r y neighbourhood h

such t h a t

-

hV 3

E

.

V

ahc

B

is

o f the origin, there i s a scalar

s e t E i n a n o m e d s p a c e E A bounded d o h o(E, E*) and o n l y 46 ewehy continuoub f i n e m I;unc,tiond i~ bounded on B ,RhaR LEMMA 3.

4 all

A

f E E*

dok euehy

i s a number

M

such t h a t

I f ( x ) 1 < bl ,

id,

604

x E B .

PROOF OF LEMMA 3. for

, there

46

o(E,

there i s a

E*)

x

-

Let

f E E*

, E > 0 , and V t h e neighbourhood o f 0 ,

, d e f i n e d by V = { x E E ; I f ( x ) I < e l . I f B i s bounded, such t h a t xV 3 B , t h a t i s , f o r a l l x E B , I f ( x ) ( < ( h ( e ,

which proves t h a t e v e r y continuous l i n e a r f u n c t i o n a l i s bounded on

B

.

Conversely, we assume t h a t e v e r y continuous l i n e a r f u n c t i o n a l i s bounded on

B

.

Let

VE ; t l , . . . ,tn

be a neighbourhood o f

0

for

a(E,

E*)

.

B. BEAUZAMY

36

Iti(X)I and so

B C

>0

M

There e x i s t s a number

QM

/ t Z ( x ) ( GMs...sltn(x)1

Q M s

MV

, and

x E B

such t h a t , f o r a l l

%

3

t h e lemma i s proved.

L e t us come back t o t h e p r o o f o f p r o p o s i t i o n 2. The u n i t b a l l gE* i s a m e t r i c space, f o r t h e induced norm, and complete, t h e r e f o r e a B a i r e n2 1:

space. We c o n s i d e r t h e c l o s e d subsets, f o r

tt

Fn =

E

LBE* ; I t ( x ) I ~n

,

x EB 1

for all

The lemma says t h a t t h e r e u n i o n o f

covers gE*, and so one o f

Fn's

them must have a non-empty i n t e r i o r : t h e r e must e x i s t a t o E E* , no E lN, E

: 0 , such

F,

that

T h i s means t h a t :

IIt

-

= I E ( x ) ( d no ,

toll < e

Therefore, i f

llfll

t o , w i t h radius e

contains the b a l l centered a t

0

,

< E

x E

for all

putting

f =

E - to ,

B

.

.

we o b t a i n

and thus

llfll

<1

2n0

]f(x)) Qy ,

=+

.

for all

x E B

for all

x E B ,

Consequently, Ilxll = and

B

2n0

sup I f ( x ) \ Q~ II f II < 1

,

i s bounded i n norm ; t h e converse b e i n g obvious, o u r p r o p o s i t i o n i s

proved. We have seen t h a t t h e s,pace

E

c o u l d be viewed as a subspace o f

E**

,

w i t h t h e induced norm. We s h a l l now l o o k more c l o s e l y a t t h i s q u e s t i o n . PROPOSITION 4.

-

L&t

E

be u ~ o m c dhpuce,

dn dud,

E*

F a t ,the RopoLogy o(E**, E*) , t h e b u l l BE= { x in t h e b& BE**= I x E E** ; llxll < 11

.

REMARK.

-

I n t h i s p r o p o s i t i o n , we c o n s i d e r

.

t h e r e f o r e , BEi s a subset o f gE**

E

hh b i d u u l . 11 A denne

E**

E E ; llxll Q

as a subspace o f

**

E

, and

37

INFINITE-DIMENSIONAL NORMED SPACES

It follows obviously for

E**

i s dense i n

, b u t , i f E i s a Banach space, i t i s c l o s e d i n E**

E*)

o(E**,

E

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

for

t h e norm-topology. PROOF.

-

o(E**,

E*))

Since BE** i s

i s a point llfll = 1

E*)-compact,

the set

.

TE.

i n BE** , n o t i n

xo

TEby

and

u(E**,

rE (closure

for

i s c o n t a i n e d i n BE** Conversely, l e t us assume t h a t t h e r e We can t h e n s e p a r a t e s t r i c t l y

a c l o s e d hyperplane : t h e r e i s an element

f

* E E

, with

, and a number a > 0 such t h a t

,

J f ( x o ) J> a ( b y c o r o l l a r y 6, a

other side,

<

sup I f ( x ) I < a

Jf(x)J< a

5

for a l l

x € T E

4). This holds a f o r t i o r i f o r a l l

< IIfll.IIxoll < 1 , and

/f(xo)I

< 1 , and

xo

so

a

x E BE

<1.

t h i s contradicts formula l . ( l ) ,

.

But, on t h e

Therefore, llfll = 1

since

.

X E B E

T h i s proves t h e p r o p o s i t i o n . We s h a l l now e x t e n d these n o t i o n s t o t h e o p e r a t o r s between normed spaces.

5

6. WEAK AND STRONG CONTINUITY. TRANSPOSITION.

u

C

Let

E

and

from

E

into

>0

F

be two normed spaces. We know t h a t a l i n e a r a p p l i c a t i o n

F

such t h a t

i s continuous i 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)llF

< CllxllE , and

we s e t

llull =

c a l l opehatoh a c o n t i n u o u s l i n e a r a p p l i c a t i o n . Let

,

E*

F*

be t h e d u a l s o f

w&.akLy continucub f r o m

,

u ( E , E*)

with

E

into

F

into

E

,

F

, endowed

and

F

.

We s h a l l say t h a t

i f i t i s continuous f r o m

with

. We

sup Ilu(x)llF llXllE <1

* .

E

u

is

, endowed

o(F, F )

F o r l i n e a r mappings, s t r o n g c o n t i n u i t y i m p l i e s weak c o n t i n u i t y : PROPOSITION 1.

-

-

VE ; E

PROOF. for

Let

o ( F , F*)

p u t v1 =

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

U

E

x1,.

.

*

$9

n

E

= {x E

E

i n t o F h weakLy cavLtinuoub. 0

be a neighbourhood o f

1’. * . ,En

: i t i s the set o f

. We

I t n & ) /< E

Evehy vpehatotr @om

,

y E u(E)

t10uy...,v since

E ; lvl(x)I

u

=

such t h a t

tno u

.

i n t h e image IE1(y)I

< E

u(E)

,...,

These a r e c o n t i n u o u s

i s continuous. T h e r e f o r e , t h e s e t < €

,...,

Iv,,(x)I

<el

,

B. BEAUZAMY

38 i s a neighbourhood o f contained i n

, the

image o f which, by

u

: t h i s proves t h e weak c o n t i n u i t y o f

u

for

0

V

u(E, E*)

; E l , . . . ,# n The &urnpabe o f u , denoted by E* , d e f i n e d by t h e f o r m u l a :

,is

.

E

tu

,is

an a p p l i c a t i o n f r o m

F*

into

T h i s a p p l i c a t i o n i s o b v i o u s l y l i n e a r , and has t h e f o l l o w i n g i n t e r e s t i n g continuity properties :

-

PROPOSITION 2.

L&

E i n t o F , tu

be an opehatoh d m m

u

Lth

ZkaMnpone. Then :

a)

tu

A2 continuoud ljhom F*

(we b h a l l bay

b)

tu

A2 CoPLtinuoub dhom F*

equipped &h PROOF.

b&ongBy

:

U ( E * ~E)

.the,.&

i n t o E* equipped &h t c o n f i n u o a ) , and I1 ull = 1lu11 . equipped w L t h

.

- L e t us show t h a t tu

u(?,

F)

i n t o E*

i s s t r o n g l y continuous. L e t 1 E

continuous l i n e a r f u n c t i o n a l , w i t h

IIvII = 1

,

Let

x E E

nohmb

*

F

llxll = 1

be a

.

We

have ItU(17)(X)l = l17(u(x))l 6 llull

.

We o b t a i n : t

II UII =

t

sup sup I u ( r ; ) ( x ) 111) llF* 61 llXllE <1

t h i s proves t h a t E

vo

> O , xo E

E E

* F , with

tu with

;

II t uII < IIuII . L e t now llu(xo)ll 2 (1 - e)llull , and l e t

i s continuous, and shows t h a t IIxoll = 1

, such

q o ( u ( x o ) ) = Ilu(xo)ll

and t h i s proves t h a t

I < llxll

II t uII = llull

that

. We

have :

,

The p r o o f o f t h e c o n t i n u i t y o f

tu

from

(F*, u(F*, F ) )

into

(E*, u(E*, E ) ) i s s i m i l a r t o t h e 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 l e a v e i t t o t h e reader. L e t us observe, moreover, t h a t , by p r o p o s i t i o n 1, tu i s a l s o continuous from (F*, O ( F * , $*I) i n t o (E*, U ( E * , E**)) .

INFINITE-DIMENSIONAL NORMED SPACES

39

We s h a l l now i n v e s t i g a t e t h e l i n k s between t h e i n j e c t i v i t y o r

u

surjectivity of

and t h e corresponding p r o p e r t i e s f o r

.

tu

L e t us f i r s t

r e c a l l some n o t i o n s .

M

Let

be a v e c t o r subspace o f

M'

; ~ ( x =)

E E*

=

o

E

*

M1 , t h e subspace o f

denote by

.

x E tJ}

for all

*

in

E*

M

, and

.

-

n + + m t and i f 1 [ ( x ) = 0 ) . A l s o , we can c o n s i d e r t h e o r t h o g o n a l o f M

T h i s i s o b v i o u s l y a c l o s e d subspace o f

, then

tn(x) = 0

We c a l l ohthogun& o f

d e f i n e d by :

E

endowed w i t h

u(E*,

.

E)

E

En

(if

Mil, c o n t a i n e d

This i s the s e t

E

in

,

,

and d e f i n e d b y : M"

o

E ; c(x) =

E

= IX

i d e n t i c a l , by

-

, and

M c 14"

M

5

for

* E )

o(E,

*

*

E

xo

xo

i s not i n

M

E

11

endowed w i t h

= {x E E ; c(x) =

.

o , for

c

all

E*

IE

=

E E* ; t ( x ) = N l l is

The subspace

is

u(E*,

o , for

E*)

o(E,

4,

(that i s :

all

, we

xo :

consider

E NI

and N"

5

t h e r e i s , by

M , and s t r i c t l y p o s i t i v e a t

i s a subspace o f

N

fl ,

does n o t belong t o

) , equal t o z e r o on

The same way, i f

'N

M1l

o r f o r t h e norm ( b o t h c l o s u r e s a r e

c o r o l l a r y 6, a l i n e a r f u n c t i o n a l on therefore

.

I

5, p r o p o s i t i o n 1). Indeed, one checks immediately t h a t

i f a point

an element o f

E M

E , which i s o ( E , E ) - c l o s e d . I n f a c t ,

T h i s i s a subspace o f the closure o f

1

c

for all

x E N

E)-closed, j u s t a

1

1

.

M l l was

*

o(E, E ) - c l o s e d :

t h i s i s n o t a new n o t i o n , b u t j u s t a consequence o f t h e f a c t t h a t t h e d u a l i t y i s p e r f e c t l y symmetric between B u t one s h o u l d observe t h a t

NL1

whereas N

in

(though norm-closed i n

.

E*

PROPOSITION

3.

&Laupuse. Then a)

( I m u)'

-

L&

t

u ;

and

norm-closure o f

(E*, u(E*,

M

in

E

E))

,

E*) needs n o t be t h e norm-closure o f

be un o p m h a t o t d t o m E i n t o

u

:

= Ker

(E, u(E, E*))

M l l i s the

F , tu

.

B. BEAUZAMY

40 Ker u = (Im(tu))'

b)

;

t

c)

K

d)

(Ker u)* = Im(tu)

PROOF.

-

-

= ( K e r ( u))'

b)

t

*t(y)

=

*

(closure f o r

L e t us show a) :

1; E ( I m u f

x E E

;

o

for a l l

u(.E)(x) = 0 , f o r a l l

x

E)).

o(E

-

y E I m u I+g(u(x)) t

E

in

I s o b t a i n e d t h e same way.

=

u(E) = 0 - E

o

for all t E Ker u .

Is o b t a i n e d f r o m a ) by t a k i n g o r t h o g o n a l s , and d) f r o m b ) . The c) p r o p o s i t i o n i s proved. Consequently, i t i s e q u i v a l e n t t o say t h a t and t h a t

( r e s p . tu) i s i n j e c t i v e

u

( r e s p . u) has a dense image. One w i l l observe t h a t t h i s l a s t

tu

p r o p e r t y d i f f e r s from s u r j e c t i v i t y . We s h a l l now d e s c r i b e t h e c o n d i t i o n s on u

t o o b t a i n the s u r j e c t i v i t y o f

-

tu

A

and c o n v e r s e l y :

.id

h a weak Aomotrpkinm, ( t h a t A, a bijecti-on wktch h bicantinuoun 604 t h e weak * topologienl, dhom F o n t o tu(F*) . A h v , u a weak Avmu4pkinm 6aom E onto u(E) .id and o n l y id tu iA auhjecLLtive. PROPOSITION 4.

PROOF.

-

The opehato4

u

nuhjecL&e id and o n l y

tu

L e t us show 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 . We assume

s u r j e c t i v e . Then, by a) of t h e p r e v i o u s p r o p o s i t i o n , i s a b i j e c t i o n , weakly continuous, from

F*

onto

u

t o be

tu

i s injective : i t

%(F*)

We have t o show

.

t h a t t h e i n v e r s e b i j e c t i o n i s a l s o weakly continuous. F o r t h i s , l e t

V

f

;y1 Y . .

. ,Yn

Vf ;y1 Y

* *.

be a neighbourhood o f ,Yn

=

{v

E F* ; h ( y l ) l

0

in

< f Y . . . ¶

F*

for

o(F*,

F)

.

We have

lv(yn)1

and thus tU(VE ;yl'.

. . ,y n 1

= ctu(v') =

i f we denote by

; IS(Y1)l

t

{E E u(F*) ; I t ( x , ) l

X ~ , . . . ~ X ~t h e

points o f )

0

for

o(E*,

E)

and

tu

IS(Yn)l < f l

< f Y . . . ,


...,

IE(xn)l < e l

,

E such t h a t u(xl) = yl, ..., i s t h e r e f o r e a neighbourhood o f

i s a weak isomorphism.

41

INFINITE-DIMENSIONAL NORMED SPACES

Conversely, l e t us assume y E F

. Then

continuous. I f we c o n s i d e r

, which

E*

is

t~ t o be a weak isomorphism, and l e t

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

y

, we

y o (tu)-l

F*

*

, which i s o ( F , F ) -

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

E)-continuous : such a f u n c t i o n a l , as we know

o(E*,

prop. 3 ) , i s g i v e n by an element

. Therefore,

x E E

(5

3,

we have, f o r e v e r y

q E F * :

< u(x), and

> = < x,

1)

. This

u(x) = y

tu(s)

>=
0

(tu)-l,

tU(l?)

> = < y,

q

>

y

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 ; t h e second

i s proved analogously, and i s l e f t t o t h e r e a d e r .

5

7. DUALITY BETWEEN SUBSPACES AND QUOTIENTS FOR WEAK AND STRONG TOPOLOGIES. I n t h i s paragraph, we s h a l l i n v e s t i g a t e t h e f o l l o w i n g q u e s t i o n :

If

F

E

F

i s a normed space and

and o f t h e q u o t i e n t

E/F ?

a subspace o f

E

, what

are the duals o f

The answer i s g i v e n i n t h e f o l l o w i n g

proposition :

-

be a notrmed hpace, with t h e induced nohm. Then : PROPOSITION 1.

LeL

E

F

nubnpace o d

E

, endowed

a) The d u d F* 0 6 F can be hvme,Ltic&y i d e n 2 4 i e d with t h e dedined b y : cjuoLLent E * l F I ; t h e duaLLty mapping

(1) ( ( x ) = E(x) i n the d u h f , i n

, E*

.

id

x E F

, f

.

Let a) functional 5

115 llF* (if t

<

=

f

E E*lFI on

SUP

IIXllF
F

h any d e m e n t

F

by

; i t i s obvious t h a t f o r m u l a (1) d e f i n e s a l i n e a r

, and one has :

It(X)I =

SUP

It x I

It(x)I

<1

i s any element i n t h e c l a s s sup I t ( x ) I = lltllE* ; llXllE <1

therefore

, and !.

h t h e topology induced on

The weak t o p o l o g y o(F, E*jF1) b) o(E, E*) PROOF.

E E*iF1

f

),

42

B. BEAUZAMY

Conversely, i f t

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

F

.?

u s i n g Hahn-Banach Theorem, extend i t t o a l i n e a r f u n c t i o n a l

,x

w i t h t h e same norm, and t h e v a l u e .?(x) of

r

modulo

F'.

f

Calling

E

the class o f

F

r

, depends o n l y , we o b t a i n :

, we on

can,

,

E

on t h e c l a s s

5

llt !IF* = 115' llE* 2 lli "E* IIt llF

which proves t h a t

'

/F1

IlsllE*

=

L e t ve;g 1'. . * ,En neighbourhood o f 0 f o r

Ix

b)

gl,...,g

i n the classes

=

t l , . . . ,c; n

"e;

n

o(F,

.

/F' E

'

F ; Icl(x)I

<e,

...,

Ign(x)I < e l

E IF1), We choose E1,...,En

IX E F ; Itl(x)I

if v€;E l , . . . , E

F n Ve ;E

have

l""'t n

<e,...,lEn(x)1

<el

nF

.

i s a neighbourhood o f 0 f o r o(E, E*)

n

= I x E F ; 15'1(x)) < ~ , . . . , l g ~ ( x ) \ < e l

a r e c l a s s e s o f ~l,...,En

fly...,~,.,

any elements

We can w r i t e :

= I x E E ; IE1(x)I < e , . . . , ~ E n ( x ) ~

Conversely,

be a

modulo

F*

,

, we

, where

and t h e p r o p o s i t i o n i s

proved. The dual o f a subspace o f a quofient o f

E*

.

E

i s t h e r e f o r e n o t a bubnpace o f

come back l a t e r ) , where t h e dual o f

but

*.

can be viewed as a bubbpace o f

F

E i s a normed space and F a c l o s e d subspace o f

If

E*

There i s , however, an i m p o r t a n t case (on which we s h a l l E

, we

E

s h a l l say

t h a t F i s compLeme&ed i n E i f t h e r e e x i s t s a continuous l i n e a r p r o j e c t i o n P from E o n t o F (we r e c a l l t h a t a p r o j e c t i o n s a t i s f i e s 2 P = P ) . From now on, we s h a l l j u s t say " p r o j e c t i o n " , and o m i t t h e words "continuous 1 i n e a r " . PROPOSITION 2. - L e t E be a notmed bpace, F a c h e d bubdpace 06 We ahnume t h a t thehe O a phujecZLun P dhom E onto F Then t h e

.

Ztannpobe PROOF.

-

a n Oomohpkinm dhom F*

tP

Since

P

i s s u r j e c t i v e from

dense range, and t h e r e f o r e

tP

, from

E

* F

o n t o a d o b e d bubdpace 06 onto into

F , E*

E E*

i t has a f o r t i o r i a

,

i s injective

. .

INFINITE-DIMENSIONAL NORMED SPACES

(5

6, prop. 3). We p u t

Y

We s h a l l p r o v e t h a t that

Y

Y = Im

,

P

and l o o k a t

i s closed i n

E*

tP

,

from

F*

Y

into

.

( f o r t h e norm) : t h i s w i l l prove

i s a Banach space, and, by t h e open mapping theorem ( c h a p t e r I,

Theorem 8 ) , w i l l prove t h a t We c a l l

i;

5

P = io j

E ) . We a l s o p u t

P

i s bicontinuous.

tP

the operator

itself (that is, into

t

43

considered as an o p e r a t o r from

P

into

E

, where i i s t h e c a n o n i c a l embedding from t = i , and we have t h e f o l l o w i n g diagrams :

F

5

We have from E*

-P -P

F* =

TOt p = ti; , and

i; , t h e r e f o r e

o n t o a c l o s e d subspace of i t . The embedding

on i t s image, t h e r e f o r e a weak isomorphism ( t h a t i s , o(E, E*) and

j

restricted t o i s surjective

F

( 0 6,

i s a projection

i i s an isomorphism o ( F , F*)

and

co'incide ; see b ) o f t h e p r e v i o u s p r o p o s i t i o n ) , t prop. 4 ) . So Y = I m P = I m ti; , and t h i s

l a s t space i s closed, which proves o u r a s s e r t i o n . T h i s p r o p o s i t i o n means a l s o t h a t subspace

of

E*

*

F

can be i d e n t i f i e d w i t h a c l o s e d

, by means o f t h e f o r m u l a :

[ E E * - [ ~ P E E *

,

b u t one s h o u l d observe t h a t

tp

needs n o t be an i s o m e t r y .

We s h a l l now s t u d y t h e dual o f q u o t i e n t s o f PROPOSITION 3. - LQR E be a nokmed npace, lclith t h e induced nvfim. Then :

E

F u nubspace

06

The d u d 0 6 E l F can be hvmeXhicaeRy idcnti6ied wi t h a) d u U y mapping being given b y t h e ~ o ~ o w i n6ohmuLa g : (2)

< i ,[ > =

[ (x)

,

c l a n i , x E E , and [ E F

06

The t v p o l o g y u ( E I F b) o(E, E*) b y F .

i6 1

.

i

1

E

ElF

E

, endOUJed F*

, the

, x being any element i n t h c

, F ) h i d e n t i c a l uLith Xhe quoLLent t o p o l o g y

44

B. BEAUZAMY

PROOF.

,E

I f I E F1

a)

E(i) = F ( x ) , i f x If

i

if

F

(E

l,... ;F

, we

E (E/,)* We c a l l

ve

a .

x

1

. * Y E

n

E/F ; IEl(x)l

{x

=

E

p

W

Conversely, l e t

E ; IE,(x)l

f

and

G

o(E,

by

E

. We

,F

o(EiF

F1

in

for

(E/F)*. EjF .

onto

<el

o ( E / ~, F

E*)lF

<el

1

) . Then

G

q u o t i e n t by

0

in

F)

for

E

s h a l l show t h a t we can f i n d

)

1

,

(that i s :

be a neighbourhood o f

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

a supplement o f

Go

E E )

0

<e,..,,ltl(x)l

= p ( % ;E 1,. . . ,F

.."f n

and t h i s w i l l prove t h a t Let

.,.En

;t 1Y . .

* E ) ( w i t h E l , ... ,Fn . ,fn E F1 such t h a t

P(W,;fl,

t ( x ) = E(i) , IIEIIE* = IltII

E

F1 , and

E

.

o f t h e neighbourhood

.

fly..

F

<ey...,ltn(x)1

be a neighbourhood o f

and t h i s proves t h a t t h e t o p o l o g y 1 i s stronger than u ( E l F , F )

a(E,

So

Q llEIIE*

t h e canonical s u r j e c t i v e mapping from

{i E

=

)

.

F

modulo

E l F by

llEll(EIF)*

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

i s t h e image by

,tn

;El'.

p

- .,F n

,.En E F

1" we

1'.

i , and

i s any element i n

i s the class o f

b) Let

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

I

)

*

i s s t r o n g e r t h a n o(E, E )/F E*

spanned by

, which

F1 and E l , .

.

. .,En

,

i s o f f i n i t e dimension ( a t most

.

The r e s t r i c n ) . L e t q l , . . . , q m (m Q n) be an a l g e b r a i c b a s i s o f Go t i o n s t o F o f t h e l i n e a r f u n c t i o n a l s v l , . . . , v m a r e l i n e a r l y independent ( i f a l i n e a r combination

would

belong t o

a1ql

, and

F1

+

... t amvm was

equal t o zero on

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

F , it Go

).

x E E , t h e r e i s a y E F , such t h a t v . ( x ) = q . ( y ) , f o r j = 1,... ,m. For t h i s , l e t el,...,e be p o i n t s i n F J J m m) . For a such t h a t s i ( e . ) = 0 i f i # j , = 1 i f i = j (i,j = 1, We s h a l l see t h a t , f o r a l l

given

J x E E , we s e t

j = l,...,m all

q E Go

.

,

...,

n

: we have y E F , and TI . ( y ) = s j ( x ) , 1 J j J NOW, f o r x E E , we s e t x ' = x - y : we have q ( x ' ) = 0 f o r and t h e r e f o r e , we have a decomposition x = x ' t y , where

x ' E G i and y E F

y = Z q.(x)e

. We

a l s o decompose, f o r

i = l,...,n

:

INFINITE-DIMENSIONAL NORMED SPACES

ti

+

= fi

,

f' i

fi E F1

where

.

f; E Go

and

45

We o b t a i n : Fi(X')

+

= fi(X')

-

x

.. ,'!n )

P(W€ ;'! 1,.

1

-

COROLLARY 4.

x E Weif

, and

iA abubbpace 0 6

G

16

o(E/Gl

F = G1

-

*

G ) = o(E, E ) J G l

9

such

ends t h e p r o o f o f o u r p r o p o s i t i o n .

have :

PROOF.

1

, t h e r e i s an x ' E WE ;'! 1,.. . ,c 1I . . . 'fn x ' E F : t h i s proves t h a t p(W, ;f ) i s contained i n 1'. * . 'fn

Therefore, f o r a l l that

.

f ! ( x ' ) = f . ( x ' ) = fi(X)

,

E

which iA

o(E*,

E)-cLubed, we

*

We come back t o t h e s i t u a t i o n o f t h e p r e v i o u s p r o p o s i t i o n , p u t t i n g : we have

F

1

= G

,

G

since

is

, E)-closed.

o(E

EXERCISES ON CHAPTER 11.

-

EXERCISE 1.

Show t h a t i f two Banach spaces E*

t h e i r duals

and

F*

, defined

x(s)

-

s

+

+

c

a)

Show t h a t

b)

We d e f i n e on

k=l

m

p

E

Let

[O,

on

l n

p ( x ) = i n f I l i m sup

x(s

+

m[

.

+ ak)

For a l l ; n E

E

a linear functional

f u n c t i o n a l dominated by

call

p

.

We

2)

Lim x ( s ) 2 0

3)

Lim x ( s

4)

Lim 1 = 1

+

b y ( s ) ) = a Lim x ( s )

,

if

x(s) 2 0

s o ) = Lim x ( s )

.

, we p u t :

x E E

IN ,

al,...

,a n E I R + l

+

F

, the

x(s)

the value o f t h i s

m

.

b Lim y ( s )

.

f o r every

f o l l o w i n g way : on

, and we extend F t o a l i n e a r

Lim S++

e x t e n s i o n . Show t h a t :

+

a r e isomorphic,

i s s u b - a d d i t i v e and p o s i t i v e l y homogeneous.

F(1) = 1

Lim(ax(s)

F

be t h e space o f r e a l bounded

t h e c o n s t a n t f u n c t i o n 1, we p u t

1)

and

a r e a l s o isomorphic.

EXERCISE 2. (Banach's L i m i t ) . functions

E

so 2 0

.

B. BEAUZAMY

46

< Lim

Lim i n f x ( s )

5)

S ' t -

< Lim

x(s)

( f o r t h e 3 " ) , one may choose t h e (n

-

sup x ( s )

S + t m

a1 = 0, a2 = so,...,

ak's with

W0).

c)

Let

be any bounded sequence o f r e a l s . We p u t :

x(s)=En

n - l < s < n ,

if

Lim t n =

and

Lim x ( s ) n+t-

n+tm

Lim(a E n

2)

Lim t n 2 0

3)

Lim

4)

Lim 1 = 1

.

5)

Lim i n f E n n + t -

<

t

n > l .

for

Show t h a t :

tn2 0

if

= Lim

-

.

b v n ) = a Lim E n

1)

EXERCISE 3.

an =

t

b Lim

.

vn

n2 0

for a l l

.

.

En

-

Lim E n n+t

< Lim

sup t n

n + t m

.

Let

be a sequence o f l i n e a r f u n c t i o n a l s on (fn)n E N ( r e c a l l t h a t f, i s g i v e n by an element o f L q ' P 1 1 ( - t - 1). Show t h a t fn f f o r o ( L , L ) i f and o n l y i f one p .q q P has s i m u l t a n e o u s l y :

L

(1 d p

a)

< +m )

3M

(that i s :

>0 ,

IIfnllLq

{(fn)nEml

QM ,

i s bounded i n L

b) F o r a l l m 2 0 , fn(m) t h e mt_h c o o r d i n a t e s o f f, and

EXERCISE 4. Lp (1 Q p Lq' ? 1

-


Let m

t q 1 =1

(fn)nEIN

)

b)

).

9

f(m)

.

IN

where

fn(m), f(m)

(recall that

fn i s g i v e n by an element o f

).

fn n3+m- f simultaneously : 3M

f

n E

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

Show t h a t a)

for all

>0 ,

For a l l

u

IIfnIIL

,

9 with

for

QM 0

o(L

q

for a l l


, LP )

I

U

0

n

i f and o n l y i f one has

E

IN ,

fn(t)dt

I

U

--+

0

f(t)dt

.

are

INFINITE-DIMENSIONAL NORMED SPACES

-

EXERCISE 5. Show t h a t

-

EXERCISE 6. on

E*

BE*

Let

E

be a Banach space,

, endowed

E/F

Let

E

F

47

E

a c l o s e d subspace o f

.

w i t h t h e q u o t i e n t norm, i s a Banach space.

be a Banach space. Show t h a t a l i n e a r f u n c t i o n a l

i s continuous f o r

* , E)

o(E

z

i f and o n l y i f i t s r e s t r i c t i o n t o

i s continuous f o r t h i s topology.

REFERENCES ON CHAPTER 11. The elements o f F u n c t i o n a l A n a l y s i s which we g i v e i n t h i s c h a p t e r can be found i n e v e r y book d e a l i n g w i t h these t o p i c s . L e t us mention, i n p a r t i c u l a r , G. CHOQUET D21

, G.

KOTHE 1311 and N. BOURBAKI [ill. Our

p r e s e n t a t i o n s f o l l o w s more c l o s e l y these l a s t two. The

5

6 follows also

I.EKELAND 1171. E x e r c i s e s 2, 3, 4 come f r o m Banach's book [51.

COMPLEMENTS ON CHAPTER I 1 : HAHN-BANACH THEOREMS FOR OPERATORS. The Hahn-Banach Theorem g i v e n i n

5

1 i s valid only f o r linear

f u n c t i o n a l s . One can wonder i f such a theorem can e x i s t f o r o p e r a t o r s w i t h values i n a n o t h e r Banach space. One i s l e d t o t h e f o l l o w i n g d e f i n i t i o n :

-

DEFINITION.

A Banach bpace

vne v d t h e dvUv&ng

E

.b h a i d .to be a

(.YX)-bpace (X

eqLLivdent phvpQhtiM hoedb :

a) Fvh evehy Banach bpace Z , cvntaining E a phvjectivn P 6 4 v m Z v n t v E , d t h IlPll < A

Fvh evehy Banach Apace Z , cvntaining E Banach bpace Y , evehy apehatoh T @am E i n t v ex.teMivn T , d4vm z i n t v Y , ulith IITII < XIITII b)

> 1) id

-

M

.

a bubbpace, .thehe

a bubbpace, evehy Y , thehe .b an

M

.

Fvh evehy Banach b p U C e Y , evehy opehatoh T ,@om Y intv E , evehy Banach bpace Z c v k n i n g Y an a bubbpace, Ahehe .b an exAenAivn T , @am Z i n t v E , d t h IIYII < XIITII . c)

-

The p r o o f o f t h e e q u i v a l e n c e o f these p r o p e r t i e s can be found i n M.M. D A Y ' S book [131. The analogue o f Hahn-Banch Theorem f o r o p e r a t o r s corresponds t o t h e case

X = 1

,

that is, t o

(Y1)-spaces.

These spaces were c h a r a c t e r i z e d

6. BEAUZAMY

48 by

L

.

NACHBIN 1371 , i n terms o f i n t e r s e c t i o n p r o p e r t i e s o f b a l l s .

We say t h a t

E

has t h e b i n a r y i n t e r s e c t i o n p r o p e r t y ( i n s h o r t 6.1.)

i n E , i f Bi n B i f , f o r any c o l l e c t i o n o f c l o s e d b a l l s (Bi)iEI j non-empty f o r a l l i, j E I , then n Bi i s non-empty. I n 1371 , iG1 L. proved t h e f o l l o w i n g r e s u l t : THEOREM.

-

E

h a

( P1)-npace

i6

and ovCey

4

E

h a

€5. 7 .

The spaces h a v i n g t h i s p r o p e r t y were c o m p l e t e l y determined by L . NACHBIN, D.B.

THEOREM.

-

E

GOODER and J.L. KELLEY (see [13] f o r a p r o o f ) :

h a B . I . id and o n l y id

E

i~ t h e hpace

06

cuntinuouh

~uncLLom on an e x ~ e r n a l l ydincannecZed compact A&. ( A s e t i s e x t r e m a l l y disconnected i f t h e c l o s u r e o f any open s e t i s open).

is

CHAPTER I 1 1 REFLEXIVE BANACH SPACES ; SEPARABLE BANACH SPACES

5

1.

REFLEXIVITY.

E

We have seen t h a t t h e elements o f

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

E* , by t h e f o r m u l a t ---t < x, t > , 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 ~

<1

. So

y

does

More g e n e r a l l y , i t has been proved by R.C. James [231 t h a t , i f e v e r y l i n e a r f u n c t i o n a l a t t a i n s i t s supremum on t h e u n i t b a l l , t h e space must be r e f 1e x i ve. As o t h e r examples o f n o n - r e f l e x i v e spaces, we s h a l l meet

W ( [0,1])

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

[0,11 ) ,

Lm

L1 ,

: we s h a l l see l a t e r t h a t

REFLEXIVE BANACH SPACES

these spaces have a subspace i s o m o r p h i c t o

L1

51

, and

so by c o r o l l a r y 3,

t h e y cannot be r e f l e x i v e . We s h a l l now g i v e a geometric c h a r a c t e r i z a t i o n o f n o n - r e f l e x i v i t y . We c a l l i t geometric, opposed t o t o p o l o g i c a l , because, whereas o u r p r e v i o u s c h a r a c t e r i z a t i o n used weak compactness o f t h e u n i t b a l l ( w h i c h i s pure t o p o l o g y ) , t h i s one w i l l use t h e s t r u c t u r e o f a sequence o f p o i n t s ( o r o f two sequences, one o f p o i n t s , one o f l i n e a r f u n c t i o n a l s ) . T h i s r e s u l t , which w i l l be o f g r e a t h e l p f o r t h e sequel, i s due t o R.C. THEOREM 6.

- Fvh a Banach Apace

equiwaeent

:

a)

E

b)

in E ,

E

James 1251.

, t h e d a U v w i n g cvnd&vMn

me

iA not h ed lex ive

Fvh ewmq 0 , 0 < 0 < 1 , t h a t iA a bequence 0 6 p o i n t 2 ( x ~ ) ~ , - ~ Luith llxkll = 1 dvh a&? k , and a bequence 0 4 LLngm duncLbn& , Lclith llf,II = 1 doh n , buch t h a t :

( f n ) n E IN

f (x,) n

= O

FOR all K me

n>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

1

k C ai

1 conv(x l,...,xk)

= 1

, then

= 1

)

:

1 i s t h e convex

, this

l a s t condi-

t i o n can be w r i t t e n i n s h o r t : for all

K

dist(conv(xl,.

and we s h a l l c a l l i t

1 , all

. . ,xk) “cvnd&on

k < K ,

, conv(xk+l,. J

I’

..)) >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
.

E

<1

, a ,

. We

of (x~)~,-,,, o f l i n e a r f u n c t i o n a l s , such t h a t :

We s h a l l show by i n d u c t i o n t h a t t h e r e e x i s t a sequence points

-

n E

a sequence ,,€,,,n)f(

xkll = 1

,

IIfnII

for all

F(fn) = 0 fn(xk

1 , for all n E IN

= B

if

n & k

= O

if

n > k .

n, k E IN

,

F o r t h i s , we s h a l l use t h e f o l l o w i n g lemma, which i s c a l l e d "He1

Y'S

Condition" :

- LeA E be a nomed Apace, f l y ..,fn be continuoub f i n e a h &ncfion&, M > 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


K

z: ai

ktl

= 1

* b).

a)

al

,... ,ak

.

We have :

,.,. ,aK

,aktl

be

and c ) i s proved. c ) +. a ) . prove t h a t

E

r e f l e x i v e . For The s e t s

,

Assume t h a t c ) h o l d s f o r some 0

0

<0 < 1 .

i s n o t r e f l e x i v e . Assume on t h e c o n t r a r y t h a t

k E

IN , we

rk( c l o s u r e

p u t Ak = I x k

E

, ... 1 , and

, xktl

f o r t h e norm, o r f o r

o(E,

We s h a l l is

Bk = conv Ak

*

.

E ) : b o t h co’incide)

form a decreasing sequence o f compact s e t s , s i n c e t h e b a l l BC i s o(E, E*) L compact. Therefore, n Bk must be non-empty. L e t yo be a p o i n t i n i t . kElN I n p a r t i c u l a r , xo belongs t o t h e norm-closure o f Bo : t h e r e i s a

... t

aLxL , w i t h a i l s 2 0

( f i n i t e ) l i n e a r combination alxl

t

= 1 , such t h a t Z aixi‘1 ’ i=l index i n t h e c o m b i n a t i o n ) . Since

yo// < $ (we have c a l l e d

e

I/

L: a.

L

the l a s t BL+l

,

We o n l y used, i n t h i s l a s t p r o o f , t h e f a c t t h a t c ) h o l d s f o r some 0

,

yo

i s a l s o i n t h e norm-closure o f

t h e r e i s another l i n e a r combination aetl xLtl ails

L

,

>0 ,

obtain :

L

‘

Z a. = 1

Lt1

,

such-that

1

L

Z: aixi

i=Ltl

t

-

... t aLxL

yell G e3

.

, with

F i n a l l y , we

2e

7 ’ and t h i s c o n t r a d i c t s ( J ) , and proves t h e theorem. REMARKS.

1)

REFLEXIVE BANACH SPACES

0
<1.

Therefore, c ) i s a l s o equivalent t o :

,

: F o r some 0

(c')

55

< 1 , there

0
i s a n o r m a l i z e d sequence

such t h a t ( J ) h o l d s .

('k)k E IN

T h i s l a s t p r o o f shows t h a t a sequence s a t i s f y i n g c o n d i t i o n ( J ) cannot 2) be weakly convergent. Since ( J ) a p p l i e s a l s o t o t h e subsequences o f t h e g i v e n sequence, i t shows i n f a c t t h a t a sequence s a t i s f y i n g c o n d i t i o n ( J ) cannot have any weakly convergent subsequence. So we o b t a i n : COROLLARY 8.

- ld, i n

a Banach

E ,

bpUCL

w e a k l y c o w a g e n t bubbequence, t h e n

e v m y bounded

b&qUeMU

hu a

Lb hedlexiue.

E

A v e r y s i m p l e example o f a sequence s a t i s f y i n g ( J ) i s t h e c a n o n i c a l

ll : so we have a new p r o o f ( s l i g h t l y l o n g e r ! ) o f t h e f a c t t h a t

basis o f

ll

i s n o t r e f l e x i v e . 3 u t we s h a l l have many o p p o r t u n i t i e s o f u s i n g

James'Theorem i n i t s f u l l s t r e n g t h . We s h a l l now t u r n t o t h e s t u d y o f t h e Banach spaces which, though t h e y

are infinite-dimensional

5

,

do n o t have a t o o l a r g e c a r d i n a l i t y .

2 . SEPARABLE BANACH SPACES. We say t h a t a normed space

E

E

i s separable i f t h e r e i s a subset

which i s c o u n t a b l e and dense i n

.

E

A

of

I t i s t h e case, as one sees e a s i l y

ll , L2 , l ( 1 < p < + 00 ) : t h e f i n i t e s e q u e n c e s , w i t h P r a t i o n a l c o e f f i c i e n t s , form a dense and c o u n t a b l e s e t . I t i s n o t t h e case f o r t h e spaces for

L,

( c f . exercise 4).

More g e n e r a l l y , we s h a l l say t h a t a s e t E

i s separable i f t h e r e e x i s t s a subset

dense i n

M

(that is,

PROPOSITION 1.

-

M

A cM

c o n t a i n e d i n a normed space

, which

i s c o u n t a b l e and

M CA).

Fvh a nohmed space

E

, t h e 6aMowing

UAbehtiVnA

eqLLivaeent :

1)

E Lb

2)

The u n i t ball BE Lb nepahable,

3)

The u n i t dphme YE = IX E E ;

sepmable, llxll = 11 hi nepmable.

me

B. BEAUZAMY

56

- 1) * 2 ) .

PROOF.

2) * 3 ) .

If A

i s dense i n

E

, A n BE

i s a dense sequence i n BE ( t h e p r e v i o u s s e t ,

If ( u n l n E

which was countable, and has been enumerated), t h e n dense i n YE .

3)

* 1).

A = {Xu,

Let

-

PROPOSITION 2. PROOF.

-

Let

(Un)nEM

IN , X

; n E

E

i s dense i n BE . Un ( llunll ) n E

be a dense sequence i n YE

. Then

QtI i s c o u n t a b l e and i s dense i n E

IN

is

the set

.

14 E Lb nepahabte., nu m e afY -La2 nubnpaced.

(

u

~

be ) a ~dense ~ sequence ~ in

. We c o n s i d e r t h e $ , for m 2 1 : l e t

E

f a m i l y o f the b a l l s centered a t un's , with r a d i i 1 1 B(un , ) be these b a l l s . L e t An,,, = F n B(un , ) The r e u n i o n o f t h e s e t s

An,,,

,n

E

IN , m

2 1

covers

A

, we

course, a r e empty). I n each o f t h e non-empty

,

n E F

IN,

m 2 1

.

( b u t some, o f

choose a p o i n t

n .m : we o b t a i n t h i s way a c o u n t a b l e f a m i l y which, as one checks n ,m immediately, i s dense i n F .

x

REMARK. - The d e f i n i t i o n we have g i v e n f o r s e p a r a b i l i t y extends, more g e n e r a l l y , t o t o p o l o g i c a l v e c t o r spaces. B u t t h e p r e v i o u s p r o p o s i t i o n does

n o t h o l d i n t h i s l a r g e r frame. PROPOSITION 3. - 14 E Lb n e p m b t e and i n d i n i t e d i m e n n i o n d , thehe Lb a devLcle and f i n e a h l y independent nequence ( v ~ IN) in ~ E .

PROOF.

-

F i r s t r e c a l l t h a t a sequence i s s a i d t o be l i n e a r l y independent

i f e v e r y &b&te s u b - f a m i l y i s l i n e a r l y independent.

.

.

be a dense sequence i n E We t a k e vo = u Assume Let ( q n E m 0 vl, ,vk have been chosen, i n o r d e r t o s a t i s f y : 1 1 . Then we choose 1 llvk - ukll llVl - ulII < 2 , IIv2 - u211 < v ~ ,+ 3 y . - * y 1 -U~+~II This independent of v ~ , v ~ , . . . , v ~ , and s a t i s f y i n g

...


i s p o s s i b l e , s i n c e t h e open b a l l o f c e n t e r

uktl

<m.

and r a d i u s $€

has

non-empty i n t e r i o r , and t h e r e f o r e cannot been c o n t a i n e d i n spanIvo,vl,...,

vk}

, which

i s f i n i t e dimensional. Therefore, we b u i l d t h e

sequence ( V n I n E IN by i n d u c t i o n : i t i s independent, by c o n s t r u c t i o n , and i t i s dense because, f o r every x E E , t h e r e i s a subsequence which converges t o x , and t h e subsequence (vn ) j a1 so ( ' n . ) j E IN j J converges t o x

.

~

57

REFLEXIVE BANACH SPACES

T h i s p r o p o s i t i o n , which i s v e r y easy t o p r o v e ( a s t h e r e a d e r has seen) i l l u s t r a t e s w e l l t h e d i f f e r e n c e w i t h t h e f i n i t e - d i m e n s i o n a l case, where such a t h i n g i s o b v i o u s l y f a l s e : i n a sense, one has much more freedom when t h e space i s i n f i n i t e - d i m e n s i o n a l . The main i n t e r e s t o f separable spaces i s t h a t we s h a l l be allowed, i n them, t o use sequences t o t e s t c o n t i n u i t y o r convergence ( i n s t e a d o f f i l t e r s , which a r e u s u a l l y r e q u i r e d ) . One knows t h a t sequences s u f f i c e i f t h e space i s m e t r i z a b l e . T h i s i s o f course so when we c o n s i d e r t h e norms,

E

in

or

, b u t i s n o t so when we t u r n t o weak t o p o l o g i e s u(E, E*) E o r E* , a r e n o t

E*

E)

u(E*,

or

: these t o p o l o g i e s , on t h e whole spaces

m e t r i z a b l e i f t h e space i s i n f i n i t e dimensional. But, i f we r e s t r i c t o u r s e l v e s t o b a l l s , t h e s i t u a t i o n may be d i f f e r e n t :

- 78

E iA bepahable, t h e 6 a . U ~ u6 E* ah& m&zable doh E ) , and t h e m e x h h a cuuntable beZ which iA devme Ru ( E , E ) (one bOm&eA Aayb t h a t E* iA weahey bepahable).

PROPOSITION 4.

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

in E*

~ U

-

PROOF.

Let

(un),-

be a dense sequence i n

E

.

We s h a l l show t h a t

t h e u n i t b a l l gE* i s m e t r i z a b l e . F o r t h i s , we c o n s i d e r t h e f o l l o w i n g f a m i l y o f semi-distances, d e f i n e d on

-

d n ( f y g ) = 1f(un)

g(un)I

for

( t h i s i s o n l y a semi-distance,

since

*

E

by :

f,g E E* = 0

dn(f,g)

does n o t i m p l y

f = g ).

T h i s f a m i l y i s countable, and t h e r e f o r e d e f i n e s a m e t r i z a b l e t o p o l o g y , which we c a l l -G E)

u(E*,

Let

.

VE ;xl,.

,...,urn

Uml IIu,

i

-xill

. We

s h a l l now show t h a t , on gE*,

. . ,xn

be a neighbourhood o f

be p o i n t s o f t h e f a m i l y

< -2 , f o r n E

. We

i = l,...,n

(

u

Z co'incides w i t h

0

for

~

such ) ~t h a t~

, and

u(E*, E ) ~

now d e f i n e a neighbourhood o f

0

for

T;,by: V ' = i f E BE* ; dm (f,O)

i

V'

Then

C

VE;x

1Y *

<; , and I f ( X i ) I < If(u,J

If(u,.)I

-

*

:

,xn

<; ,

if

d

t If(Xi - u 1

.;

(f,O)

mi

thus :

1

)I

i

for

mi

<;

t

,

.

l,...,n}

=

then,

llfll

for

IIX-u

i = 1,..., n ,

II < E "'i

58

B. BEAUZAMY Conversely, l e t

a neighbourhood o f

f o r some

0

0

of

-e

; i t contains

o f t h e form :

and some i n d i c e s

>O

B

be a neighbourhood o f

W

W ' = {feaE* ; If(um ) I < E i and t h i s i s a neighbourhood o f 0

ml, ...,mn for

. Therefore,

i = 1,

for

o(E*,

...,n~

E)

we have :

,

. This

proves t h e f i r s t

part o f our proposition. I n o r d e r t o prove t h e second, we s h a l l f i r s t observe t h a t t h e r e s t r i c tion o f

E) t o gE* has a c o u n t a b l e b a s i s . Indeed, t h i s i s t h e

o(E*,

case, s i n c e t h e s e t s o f t h e form ( i f E

i s real)

4, ,... ,mn ; I~,...,In = IfE dE*; f ( um l ) E where

n

E

11,

... , f ( u

"n

1E

In]

lN , ml, ...,mn E IN , and I1,...,I and open i n t e r v a l s w i t h n

r a t i o n a l e x t r e m i t i e s , c o n s t i t u t e a b a s i s f o r t h i s topology, and t h e f a m i l y o f these s e t s i s countable. I f f(u

) E I1 x ml

iIi , and

E i s complex, we r e p l a c e t h i s c o n d i t i o n b y

so on, where

I and

I' have r a t i o n a l end p o i n t s .

So, i n o r d e r t o end t h e p r o o f o f o u r p r o p o s i t i o n , a l l we have t o do i s t o prove t h e f o l l o w i n g lemma :

- L&t G a me.txiza6Le t v p d v g i c d Apace. 1-12 topvLogy ha^ a cvuvLta6Le 6 a L 4 .id and v n t y 4 4 t h a e exha2 a cvunta6Le A & .in G , d m e LEMMA 5. .in

G

,

-

PROOF OF THE LEMMA.

The c o n d i t i o n i s necessary, because i f

(Bn)neIN i s i f one chooses bn E Bn f o r a l l n , i s dense. Conversely, i f t h e r e e x i s t s i n G a dense

a basis o f the topology o f the s e t

{(bn)nEMl

sequence

-1 ( n m

E

,if

(u~),,~,,,

IN , m > 1) , t h e

PROPOSITION 6.

-

16

G

0

Bm,n balls

, and

i s t h e open b a l l o f c e n t e r 0

Bm,n

un

, radius

a r e a b a s i s o f t h e topology.

the d u d E* LA depaha6Le, t h e Apace E &t?L.ed i~ n v t .t)Lue, a t h e examp& 06 El , Lm , A h V W A ) .

oepaha6Le. (The c v n v m e PROOF.

-

If

E*

i s metrizable f o r

i s separable, by t h e p r e v i o u s p r o p o s i t i o n , t h e b a l l BE** o(E**,

E*)

, and

t h i s topology, r e s t r i c t e d t o BE** ,

59

REFLEXIVE BANACH SPACES

, restricted

o(E, E )

has a c o u n t a b l e b a s i s . T h e r e f o r e , t h e t o p o l o g y

BE (which i s t h e t o p o l o g y induced on BE by u(E**,

E*))

to

has a

c o u n t a b l e b a s i s . The p r e v i o u s lemma t h e n says t h a t t h e r e i s a c o u n t a b l e family The

conv{(un)nEN I

convex h u l l

.

o(E, E )

for

BE

which i s dense i n BE f o r

(UnInEM

norm-topology.

A =

i s t h e r e f o r e , a f o r t i o r i , dense i n

5 , prop. 1, i t i s a l s o dense f o r t h e

f i n i t e sequence o f p o s i t i v e r a t i o n a l s , w i t h

i i ; (Xi)

L: Xi

< 11 ,

we o b t a i n a c o u n t a b l e s e t . T h i s s e t i s dense i n BE s u f f i c e s t o show t h a t i t i s dense i n x = E

N

, ai

I: aiui i=1

E

> 0 , rationals , and

i = l,...,N

-

IIX

XIII

.

E*)

I f now we c o n s i d e r :

CX u

{

5

By Chapter 11,

o(E,

N


IR , ai > 0 , 0 <Xi

with

Xi

we have, i f (ai

-

Xi)

< E

.

conv{(un)nEN3

N

C a 1.

i=l

= 1

, we

< a i and ai

To see t h i s , i t

.

But, i f

choose, f o r a g i v e n

- A I. < NE ,

for

,

x ' = ZXiui

.

T h i s proves t h a t BE i s separable, and t h e r e f o r e

E

i s , by p r o p o s i -

t i o n 1.

5

3. THE UTILIZATION OF CONVERGENT SEQUENCES I N ORDER TO DEFINE A TOPOLOGY ; WEAK COMPACTNESS. We have seen t h a t , i f

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

E

was separable, t h e u n i t b a l l BE* was a

* a ( E , E) .

On t h i s b a l l , we s h a l l t h e r e f o r e be

a l l o w e d t o use sequences, i n s t e a d o f f i l t e r s , t o d e f i n e t h e t o p o l o g y . F o r

, d e f i n e d on BE* , and f o r o(E*, E ) , we o n l y have converging t o xo , f o r

example, i f we want t o show t h a t a f u n c t i o n i s continuous a t

xo E BE*

show t h a t , f o r e v e r y sequence

( x ~ ) ~

real-valued, E) If o(E**,

3

E*

E*)

b u t now i n

4 x n ) -IP(x)

IP

to

*

i s separable, t h e b a l l s o f

E**

, and t h e r e f o r e t h e b a l l s o f E , one can use sequences.

E

w i l l be m e t r i z a b l e f o r for

o(E, E*)

: here again,

6. BEAUZAMY

60

I f one wants t o s t u d y t h e weak compactness o f a subset o f

, on

E

can

a l s o use sequences. T h i s i s a much deeper f a c t , which does n o t depend on the s e p a r a b i l i t y o f

E

- Le,t

PROPOSITION 1.

or E

,

E*

t o which we s h a l l now t u r n o u r a t t e n t i o n .

be u Bafiach Apace,

A

a bubn&t 0 6

aLeRatiwelq weakly compact (that -in, id t h e ~ L V A W LA ~ ad

. 16

E

A

iA

A

doh o(E, E*) h compact doh t h h topology), ewmy nequence 0 6 pointn i n A covLtdinb a oubnequence which ,LA weakly conwmgent ( t o a p o i n t 0 6 t h e ceonwle A I * PROOF.

-

Let

(

x

~

c l o s e d subspace spanned by t h e consequently, i t s dual

* Eo

. This

xnls

, dense

Eo

*

in

Eo

The s e t in

hl(A)

. We

; n E IN1

thl(xn)

, which

t h e n enumerate

. The

converges i n

IK

contained i n

(xn( 1 )

(

x

~

>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

En's

i s separable and

1

- L e t E be a Banach space,

-

En)p

can be

i f a l l the

An's

Show t h a t , i f

E

k1

, show

that

(An)nEm

,

m

,

n k e r fi

=

I01

Cm i s

a bquence o f subspaces

a r e separable, t h e n

U-A, n

i s separable, f r o m any f a m i l y

i s also

(fi)iEI

norm-one l i n e a r f u n c t i o n a l s s a t i s f y i n g

i€ 1

00

i s separable.

separable. EXERCISE 6.

nEN

i s separable.

By c o n s i d e r i n g t h e sequences made o f

. Show t h a t

(

i s reflexive.

n o t separable. Show t h a t W ( [0,11) EXERCISE 5 .

, the

-t - = 1

E*) n q '

Show t h a t i f each o f t h e

(

then

(


Show t h a t i f each of t h e

b) then

1< p

Show t h a t i f

a)

,

one can e x t r a c t a c o u n t a b l e s u b f a m i l y which has t h e same p r o p e r t y .

of

REFLEXIVE BANACH SPACES (One may i n t r o d u c e t h e s e t

-

EXERCISE 7.

T : E

An o p e r a t o r

--+

T(BE)

i s weakly r e l a t i v e l y compact

T(BE)

for

*

Prove t h a t

2)

Prove t h a t i f

from

E

F

i s s a i d t o be weakly compact i f n

F

(i.e.

: the closure o f

i s weakly compact i f and o n l y i f

T

into

F

consider

i s compact f o r t h i s t o p o l o g y ) .

a(F, F )

1)

< 11 , and

n I x E E ; Ifi(x)I iE1

C =

C ).

t h e gauge o f

63

(Tn)n>l

c F

ttT(E**)

i s a sequence o f weakly compact o p e r a t o r s ,

, converging

to

T

T i s weakly

i n o p e r a t o r norm, t h e n

compact.

3)

Prove t h a t (F*,

from

4)

T F))

O(F*,

Prove t h a t

EXERCISE 8.

-

T

E**

,

C nE = C

2)

Show t h a t , i f

. Let

A = dist(F,E)

F E E**

,

O

, F E C , but

0 <8
Choose

, I<

= { Z E E**

*

fl E E

T , show

> - < z,f

F,f

that, f o r

Assume t h a t

X~,...,X~-~ E

C

n,k

< p-1

fn(xk) > B

if

n

fn(xk) = 0

if

n

been chosen so t h a t , i f

,

is.

F

C, then

IlFll > O

F

>[<el

< F,fl > > O , Ilflll every E > 0 , V (F) flYE x1 E C

tT

that

such t h a t

Deduce t h a t one can f i n d a p o i n t

4)

. Show

Consider t h e f o l l o w i n g neighbourhood o f Vf,€(F)

i s continuous

C=

. .

O(E**, E*)

Show t h a t

FE

*

tT

a bounded s e t ,

C

o(E, E )-compact. We p u t

i s not

1)

3)

i s weakly compact i f and o n l y i f

C

for

.

(E*, O(E*, E**))

into

L e t E b e a Banach space,

We assume t h a t in

i s weakly compact i f and o n l y i f

for

*

E )-closed.

closure o f

F g E

.

. Put

E*)

o(E**,

:

= 1

. Using t h e

fact that

contains a p o i n t o f fl(xl)

*

E E

>O

with

IIfiII

k

F(fn) > O dist(F,E) > O

, to

C

.

.

:

Use H e l l y ’ s c o n d i t i o n , and t h e f a c t t h a t

,

C

.

such t h a t

fly...yfp-l

o(E,

find

= 1 have

64 f

P

5)

B. BEAUZAMY

,

EE*

, with

f (x ) = 0 P k

,

k
Use t h e f a c t t h a t t h e neighbourhood o f

meets

(F)

... , f p ; E

V fly

6)

IIfpII = 1

C

**

= {Z E E

t o find a point

x

;

in

P

F

, and

F(fp) > B

:

) - < F,fl > I < €

,..., [< z,f P > - < F,f P >I<€}

C

such t h a t

f (x ) > B n P

,

n