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C OMPOSITIO M ATHEMATICA

J. G LOBEVNIK

Separability of analytic images of some Banach spaces

Compositio Mathematica, tome 38, n

o

3 (1979), p. 347-354

<http://www.numdam.org/item?id=CM_1979__38_3_347_0>

© Foundation Compositio Mathematica, 1979, tous droits réservés.

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347

SEPARABILITY OF ANALYTIC IMAGES OF SOME BANACH SPACES

J. Globevnik

© 1979 Sijthoff &#x26; Noordhoff International Publishers-Alphen aan den Rijn

Printed in the Netherlands

Abstract

A Banach space contains a

nonseparable analytic image

of a ball in

co(F)

iff it contains an

isomorphic

copy of

co(B),

B uncountable.

Let r be an uncountable set. It is known that the

(complex)

space

co(T)

has some

interesting properties

with

respect

to

analytic

maps.

For

instance,

every scalar-valued

analytic

map on

co(T)

factors

through

a

separable subspace

of

co(r) [8, 1].

Ail

separable complex

Banach

spaces X

and the spaces X =

1P(B)

for any

B,

1 --5 p oo have the

property

that every

nonempty

open connected subset of X can be filled

densely

with an

analytic image

of a ball in X

[4, 5],

while the

space

co(r)

does not have this

property [9].

No space

lp(B) (1 :5 p 00)

and no space with countable total set contains a

nonseparable analytic image

of a ball in

co(r) [8, 6].

In the

present

paper we

sharpen

the last result

by proving

that a Banach space contains a

nonseparable analytic image

of a ball in

co(F)

iff it contains an

isomorphic

copy of

co(B),

B uncountable. This is known in the linear

case

(see

Remark 1

below).

Preliminaries

The scalar field

(R

or

C)

is the same for all Banach spaces considered. We denote

by

N the set of all

positive integers.

If A is a

map we denote its

image by R(A).

Let r be an infinite set.

By co(r)

we denote the Banach space of all scalar-valued functions on r which

AMS Subject Classification 46G20

This work was supported in part by the Boris Kidric Fund, Ljubljana, Yugoslavia 0010-437X/79/03/0347-08 $00.20/0

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348

are

arbitrarily

small outside finite subsets of

F,

with sup norm. If

xEco(D

we write supp

x y Ei F: x(-y) 0 01.

We denote

by fe(-y): y E F}

the standard basis in

co(r): e(-y)(3) = 1(1’

=

8), e(y)(8) _

0

(y 0 8).

Given a metric space M we denote

by

dens M the

density

character of

M,

i.e. the smallest cardinal of a dense subset of M. Note that if

Mi

is a

subspace

of M then dens

Ml _

dens M. Let X be a

Banach space.

By BI(X)

we denote the open unit ball of X and

by

X’

we denote the dual of X. If S C X we write sp S for the closed linear span of S. Let Y be another Banach space and let n E N. A map P : X- Y is called a bounded

n-homogeneous polynomial

if there is a

bounded

symmetric

n-linear map

Q: X" ---&#x3E; Y

such that

P(x)

=

Q(x,

x, ...,

x) (x E X).

We use the term

0-homogeneous polynomial

for

constant maps. A map

A: BI(X) ----&#x3E;

Y is called

analytic

if

given

any

xo E

B1(X)

there are an r &#x3E; 0 and for each n a bounded n-homo- geneous

polynomial Pn : X - Y

such that

A(x) = 1.’=o Pn(x - xo)(flx - xofl r),

the series

being uniformly convergent

for

lix - xofl

r

[11].

When the scalar field is C then A is

analytic

iff for

each x E

B 1(X )

the Fréchet derivative of A at x exists as a bounded

complex-linear

map from X to

Y,

or

equivalently,

if A is

G-analytic

and continuous on

BI(X) [7].

Our main result is the

following

THEOREM: Let Y be a Banach space and let d be any

infinite

cardinal.

Suppose

that there exists an

analytic

map A

from

the open

unit ball

of

some

co(r)

to Y such that dens

R(A)

&#x3E; d. Then Y contains

an

isomorphic

copy

of co(B)

where card B &#x3E; d.

REMARK 1: In the

special

case when A is bounded linear map the

assumptions

above

imply

that card

ly e F;A(e(y» gé 01 &#x3E; d

so for

some &#x26; &#x3E; 0 card

ly

G r :

))A(e(y)))) a 81

&#x3E; d and the assertion follows

by [12

p.

30,

Rem.

1 ] ;

see also

[2, 3].

COROLLARY 1: A Banach space contains a

nonseparable analytic image of

a ball in

co(r) iff it

contains an

isomorphic

copy

of co(B)

where

B is uncountable.

LEMMA 1: Let

X,

Y be two Banach spaces and let d be any

infinite

cardinal.

Suppose

that there exists an

analytic

map A :

:Bi(X)- Y

such that dens

R(A)

&#x3E; d. Then there are an n E N and a bounded

n-homogeneous polynomial

P : X - Y such that dens

R(P)

&#x3E; d.

(4)

PROOF: There is some r &#x3E; 0 such that

where for each n,

Pn

is a bounded

n-homogeneous polynomial.

With

no loss of

generality

assume that

Po

= 0.

By

the

analyticity

of A

given

any x E

Bl(X)

and any u E Y’ the scalar-valued map

t- F(t) = (A(tx)) u )

defined on

I = (t : 0 = t S 1)

has an

analytic

extension to an open subset of C

containing

I so

by

the

identity

theorem

F( t )

= 0

(0

t

r) implies

that

F( 1 )

= 0.

B y

the

Hahn-Banach theorem it follows that

A(x)

E

sp{A(tx):O

t

r}

so

Assume that dens

R (Pn ) _ d

for all n and for each n let

Bn

be a dense subset of

R (Pn ) satisfying

card

Bn

d. The set B of all vectors

y E Y of the form Y

= 1 lc=

1 yi where y;

e Bi ( 1

i

:5 n)

and n E N satisfies card B _ d so

dens sp

B :5 D. On the other

hand, by (1)

and

(2) R (A)

C

sp B

so dens

R(A):5 d,

a contradiction which proves that for some n e N dens

R(Pn)

&#x3E; d.

Q.E.D.

PROOF OF THE THEOREM: Let r be an infinite set,

put

X =

co(r)

and let A :

BI(X) --&#x3E;

Y be an

analytic

map

satisfying

dens

R (A)

&#x3E; d.

By

Lemma 1 there are an n E N and a bounded

n-homogeneous poly-

nomial P: X &#x3E;Y such that dens

R(P) d.

Let

Q: Xn ___&#x3E; Y

be a

bounded

symmetric

m-linear map such that

P(x)

=

Q(x,

x, ...,

x) (x

E

X).

Let d C rn be the set of all those a =

(ai,

a2, ...,

an)

for which

Q(e(al), e(a2),

...,

e(an» 0

0. We prove that card A&#x3E; d. To see

this,

assume that card A:5 d. For

i,

1 S i n write

Ai = {B Er: 13

= ai for

some a E

A}. Clearly

card

di :5

card A d

(1:5 i :5 n)

so

writing OU = U 1=1 di

we have card OU:5 ,do

By

the boundedness of

Q

it follows that

Q(e(y),

X2, x3, ...,

xn)

= 0 for any y E r - OU and any

Xi E X (2 S i :5 n)

so

Q(y,

X2,

X3, ..., xn)

= 0 for any Xi E X

(2:5 i - n)

and any y E

X,

supp y nu =

0.

Since

Q

is

symmetric

it follows that

P(x + y) = Q(x + y, x + y, ..., x + y) = Q(x, x, ..., x) = P(x)

for any x, y E

X, supp

y n OU

= 13. Consequently

P = P - L where L is the pro-

jection

from X onto

co(OU)

defined

by

(5)

350

Now,

card ô/1 ::5 d

implies

that dens

co(’W):5 d

and it follows that dens

R(P) ::5 d,

a contradiction which proves that card s4 &#x3E; d.

By

Remark 1 the

proof

will be

complete

once we have

proved

the

following

LEMMA 2: Let T be an

infinite

set and

put X

=

co(r).

Let Y be a Banach space, let m E N and let d be any

infinite

cardinal.

Suppose

that P X’ --&#x3E; Y is a bounded m-linear map such that the set

satisfies

card A &#x3E; d.

Then there exist a set

D,

card D &#x3E; d and a bounded linear map L :

co(D)---&#x3E;

Y such that

L(e(&#x26;» gé

0

(8

E

D).

PROOF: We prove the lemma

by

induction on m.

Now

and the assertion of the lemma for m = n follows

by

the induction

hypothesis.

In the

sequel

we assume that

Consider the class 16 of all

nonempty

subsets

Q

c s4

having

the

following property

(6)

Observe that (C is not

empty

since every set

Q consisting

of one

element

belongs

to ce.

Partially

order (6

by

inclusion. Let

{Q( i),

i E

Il

be a chain in ce. Put

Q

= U i,,:,

Q(i),

let k E N and let

aj(l :5 j

:5;

k)

be

distinct elements of

Q.

There

are c;

E 1

(1 --5 j

:5

k)

such

that aj

E

Q(ij) (1 :5 j :5; k).

Since

IQ(i),

i E

Il

is a chain there is some

jo : 1

:5;

jo

:5; k

such that

U f=l Q(ij)

=

Q(iio).

Since

Q(iio) E ce

and since ai E

Q(iio)

(1 :5; j :5; k)

it follows

that aj satisfy (4)

and

consequently Q E

le.

By

Zorn lemma there exists a maximal element

Q

in ce.

Assume first that card

Q :5

d. Write 00 = d -

Q. Clearly

card 2% &#x3E; d.

Given

i,

1 s i :5 n denote

Qi

=

1,8

E f :

/3

= ai for some a =

(a}, a2, ..., an) E Q}.

Let

b = (b}, b2, ..., bn) E /3.

Assume that for every

decomposition {l, 2, ..., n}

= A U B where

A,

B #

0:

A n B =

0,

implies that g = (gt, g2, ..., gn) e d.

This means that

Q U {b} E ce

which contradicts the

maximality

of

Q.

This proves that

given

any b EOO there is a

decomposition {l, 2, ..., n}

= A U

B, A,

B 0

0:

A f1

B =

0,

such that there is some g E d

satisfying gi

=

bi (i

E

A)

and

g; E

Qi (i

E

B).

Since the set of all

possible decompositions

is finite

and since card 00 &#x3E;A there is some fixed

decomposition {1,2,.... ni =

A

U B, A,

B 0

0;

A fl B =

0

and some

set 001

c

00,

card

001

&#x3E; d such that for

every b G 2%i

1 there is some g E d

satisfying gj = b; ( j

E

A) and g;

E

Qj (j

E

B).

Write each b E

ôA in

the form b =

PA(b)EBPB(b)

where

PA(b)

E

IIjEAr

and

PB(b)

E

IIjEBr

are defined

by (PA(b»j

=

bj (j E A)

and

(PB(b» = bj (j E B).

We show that card

PA(OOI)

&#x3E; d. To

see

this,

assume that card

PA(-O-e 1) - d.

Since

card 00 1 &#x3E; d

it follows that there is some

002 C OO},

card

002&#x3E; d

and some u E

lljEAr

such that

u =

PA(b)

for all b E

002.

In

particular,

there are an i E A and a y E r

such that y =

bi

for all b E

002

which contradicts

(3)

since card

002&#x3E;

d.

This proves that card

PA(,-9-el) &#x3E; d.

For each

uEPA(OOI)

choose an element from

PÃl(u)

n

B1 1 and

denote the set of all these elements

by S2- Clearly

card

002

&#x3E; d and

Recall that for every b E

9JJ2

there is some g G s4 such that

PA(b) =

PA(g)

and such that

gj E Qj (j e B).

Since card

Qcard Q - d

(1 Ç j n)

it follows that card

IIjE!I Qj :5 d.

Since card

9JJ2

&#x3E; d it fol- lows that there is some

’,»3

C

9JJ2,

card

9JJ3&#x3E;

d and some v E

IIjEB r

(7)

352

such that for

every b

E

B3

there is some g E d

satisfying PA(b) = PA(g)

and

PB(g)

= v.

By (5)

it follows that there are an i E B and a y E r such that

card{a

EE 4: ai =

y}

&#x3E; d which contradicts

(3).

Thus

we have

proved

that card

Q

&#x3E; d.

Since

Q

E Cf¿ it follows that

Put D =

Q

and define the

map .0

from the basis

and it follows that

By (7)

it follows that

(8)

Consequently 0

admits a bounded linear extension L to all

co(D).

Since card D &#x3E; d this

completes

the

proof

for m = n.

Q.E.D.

COROLLARY 2: Let X =

co(F)

where r is an

infinite

set and let Y be

a Banach space.

Suppose

that the range

of

every bounded linear map

from

X to Y is

separable.

Then the range

of

every

analytic

map

from BI(X) to

Y is

separable.

PROOF: If r is countable there is

nothing

to prove so assume that r is uncountable.

Suppose

that there is an

analytic

map from

B1(X)

to

Y with

nonseparable

range.

By

Theorem there are an uncountable set 4 and a bounded linear map

A: co(,à) ---&#x3E;

Y which maps

co(,à)

isomor-

phically

onto

R(A).

Since

coC1)

is up to

isometry

determined

by

card

L1 assume with no loss of

generality

that either 4 C r or r C L1. If

L1 C F define B : X ---&#x3E; Y

by

B = AoP where P is the

projection

from X

onto

co(L1)

defined

by (Px) (y) = x (y) (y e à; x e X).

B : X - Y is a

bounded linear map whose range

R (B )

=

R (A)

is

nonseparable,

a contradiction. Let r C 4. Since r is uncountable X is a

nonseparable subspace

of

co(,à)

and

by

the

properties

of

A, A(X)

is

nonseparable.

Consequently AIX: X ---&#x3E; Y

is a bounded linear map with nonsepar-

able range, a contradiction.

Q.E.D.

REMARK 2: Let r be an uncountable set and let

1 p (T) ( 1 _ p oo)

be the Banach space of all scalar-valued functions x on r such that

llxll

=

(Y,,.rlx(Y)lp)"p

-. Since every bounded linear map from

12(T)

to

l’(T)

is

compact [10]

it follows that the range of every bounded linear map from

12(T)

to

1’(F)

is

separable.

On the other

hand,

the

range of the bounded

2-homogeneous polynomial

P:

1’(F) ---&#x3E; 11(r)

defined

by P(x) = y

where

y( 1’) = x( 1’)2 (1’ E r, x E 12(r»

is non-

separable

since P is

surjective.

This shows that

Corollary

2 does not hold in

general.

We ask under which conditions on a Banach space X does the assertion of

Corollary

2 hold.

REFERENCES

[1] S. DINEEN: Growth properties of pseudoconvex domains and domains of holomorphy in locally convex linear topological vector spaces. Math. Ann. 226 (1977) 229-236.

[2] L. DREWNOWSKI: An extension of a theorem of Rosenthal on operators acting from 1~(0393). Studia Math. 57 (1976) 209-215.

[3] L. DREWNOWSKI: Un théoreme sur les opérateurs de 1~(0393). C. R. Acad. Sc.

Paris, Ser A, 281 (1975) 967-969.

[4] J. GLOBEVNIK: On the range of analytic functions into a Banach space. Infinite Dim. Holomorphy Appl. (Matos Ed.) North Holland Math. Studies 12 (1977)

pp. 201-209.

(9)

354

[5] J. GLOBEVNIK: On the ranges of analytic maps in infinite dimensions. (To appear in Advances in Holomorphy, Barroso Ed., North Holland).

[6] J. GLOBEVNIK: On the range of analytic maps on c0(0393). (To appear in Boll. Un. Mat.

Ital.)

[7] E. HILLE, R.S. PHILLIPS: Functional Analysis and semi-groups. Amer. Math.

Soc. Colloq. Publ. 31 (1957).

[8] B. JOSEFSON: A counterexample in the Levi problem. Proc. Inf. Dim. Holomor- phy. Lecture Notes in Math. 364, pp. 168-177, Springer 1974.

[9] B. JOSEFSON: Some remarks on Banach valued polynomials on c0(A). Infinite Dim. Holomorphy Appl. (Matos Ed.) North Holland Math. Studies 12 (1977) pp.

231-238.

[10] E. LACEY, R.J. WHITLEY: Conditions under which all the bounded linear maps

are compact. Math. Ann. 158 (1965) 1-5.

[11] L. NACHBIN: Topology on spaces of holomorphic mappings. Erg. der Math., Bd.

47, Springer 1969.

[12] H.P. ROSENTHAL: On relatively disjoint families of measures, with some ap-

plications to Banach space theory. Studia Math. 37 (1970) 13-16.

(Oblatum 15-11-1978 &#x26; 21-VIII-1978) Institute of Mathematics Physics and Mechanics University of Ljubljana Ljubljana, Yugoslavia

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