C OMPOSITIO M ATHEMATICA
J. G LOBEVNIK
Separability of analytic images of some Banach spaces
Compositio Mathematica, tome 38, n
o3 (1979), p. 347-354
<http://www.numdam.org/item?id=CM_1979__38_3_347_0>
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347
SEPARABILITY OF ANALYTIC IMAGES OF SOME BANACH SPACES
J. Globevnik
© 1979 Sijthoff & Noordhoff International Publishers-Alphen aan den Rijn
Printed in the Netherlands
Abstract
A Banach space contains a
nonseparable analytic image
of a ball inco(F)
iff it contains anisomorphic
copy ofco(B),
B uncountable.Let r be an uncountable set. It is known that the
(complex)
spaceco(T)
has someinteresting properties
withrespect
toanalytic
maps.For
instance,
every scalar-valuedanalytic
map onco(T)
factorsthrough
aseparable subspace
ofco(r) [8, 1].
Ailseparable complex
Banach
spaces X
and the spaces X =1P(B)
for anyB,
1 --5 p oo have theproperty
that everynonempty
open connected subset of X can be filleddensely
with ananalytic image
of a ball in X[4, 5],
while thespace
co(r)
does not have thisproperty [9].
No spacelp(B) (1 :5 p 00)
and no space with countable total set contains anonseparable analytic image
of a ball inco(r) [8, 6].
In thepresent
paper wesharpen
the last resultby proving
that a Banach space contains anonseparable analytic image
of a ball inco(F)
iff it contains anisomorphic
copy ofco(B),
B uncountable. This is known in the linearcase
(see
Remark 1below).
Preliminaries
The scalar field
(R
orC)
is the same for all Banach spaces considered. We denoteby
N the set of allpositive integers.
If A is amap 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
348
are
arbitrarily
small outside finite subsets ofF,
with sup norm. IfxEco(D
we write suppx y Ei F: x(-y) 0 01.
We denoteby fe(-y): y E F}
the standard basis inco(r): e(-y)(3) = 1(1’
=8), e(y)(8) _
0
(y 0 8).
Given a metric space M we denoteby
dens M thedensity
character of
M,
i.e. the smallest cardinal of a dense subset of M. Note that ifMi
is asubspace
of M then densMl _
dens M. Let X be aBanach space.
By BI(X)
we denote the open unit ball of X andby
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 abounded
symmetric
n-linear mapQ: X" ---> Y
such thatP(x)
=Q(x,
x, ...,
x) (x E X).
We use the term0-homogeneous polynomial
forconstant maps. A map
A: BI(X) ---->
Y is calledanalytic
ifgiven
anyxo E
B1(X)
there are an r > 0 and for each n a bounded n-homo- geneouspolynomial Pn : X - Y
such thatA(x) = 1.’=o Pn(x - xo)(flx - xofl r),
the seriesbeing uniformly convergent
forlix - xofl
r[11].
When the scalar field is C then A isanalytic
iff foreach x E
B 1(X )
the Fréchet derivative of A at x exists as a boundedcomplex-linear
map from X toY,
orequivalently,
if A isG-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 ananalytic
map Afrom
the openunit ball
of
someco(r)
to Y such that densR(A)
> d. Then Y containsan
isomorphic
copyof co(B)
where card B > d.REMARK 1: In the
special
case when A is bounded linear map theassumptions
aboveimply
that cardly e F;A(e(y» gé 01 > d
so forsome & > 0 card
ly
G r :))A(e(y)))) a 81
> d and the assertion followsby [12
p.30,
Rem.1 ] ;
see also[2, 3].
COROLLARY 1: A Banach space contains a
nonseparable analytic image of
a ball inco(r) iff it
contains anisomorphic
copyof co(B)
whereB is uncountable.
LEMMA 1: Let
X,
Y be two Banach spaces and let d be anyinfinite
cardinal.
Suppose
that there exists ananalytic
map A ::Bi(X)- Y
such that densR(A)
> d. Then there are an n E N and a boundedn-homogeneous polynomial
P : X - Y such that densR(P)
> d.PROOF: There is some r > 0 such that
where for each n,
Pn
is a boundedn-homogeneous polynomial.
Withno loss of
generality
assume thatPo
= 0.By
theanalyticity
of Agiven
any x EBl(X)
and any u E Y’ the scalar-valued mapt- F(t) = (A(tx)) u )
defined onI = (t : 0 = t S 1)
has an
analytic
extension to an open subset of Ccontaining
I soby
the
identity
theoremF( t )
= 0(0
tr) implies
thatF( 1 )
= 0.B y
theHahn-Banach theorem it follows that
A(x)
Esp{A(tx):O
tr}
soAssume that dens
R (Pn ) _ d
for all n and for each n letBn
be a dense subset ofR (Pn ) satisfying
cardBn
d. The set B of all vectorsy 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 sodens sp
B :5 D. On the otherhand, by (1)
and(2) R (A)
Csp B
so densR(A):5 d,
a contradiction which proves that for some n e N densR(Pn)
> d.Q.E.D.
PROOF OF THE THEOREM: Let r be an infinite set,
put
X =co(r)
and let A :
BI(X) -->
Y be ananalytic
mapsatisfying
densR (A)
> d.By
Lemma 1 there are an n E N and a bounded
n-homogeneous poly-
nomial P: X >Y such that dens
R(P) d.
LetQ: Xn ___> Y
be abounded
symmetric
m-linear map such thatP(x)
=Q(x,
x, ...,x) (x
EX).
Let d C rn be the set of all those a =(ai,
a2, ...,an)
for whichQ(e(al), e(a2),
...,e(an» 0
0. We prove that card A> d. To seethis,
assume that card A:5 d. For
i,
1 S i n writeAi = {B Er: 13
= ai forsome a E
A}. Clearly
carddi :5
card A d(1:5 i :5 n)
sowriting OU = U 1=1 di
we have card OU:5 ,doBy
the boundedness ofQ
it follows thatQ(e(y),
X2, x3, ...,xn)
= 0 for any y E r - OU and anyXi E X (2 S i :5 n)
soQ(y,
X2,X3, ..., xn)
= 0 for any Xi E X(2:5 i - n)
and any y E
X,
supp y nu =0.
SinceQ
issymmetric
it follows thatP(x + y) = Q(x + y, x + y, ..., x + y) = Q(x, x, ..., x) = P(x)
for any x, y EX, supp
y n OU= 13. Consequently
P = P - L where L is the pro-jection
from X ontoco(OU)
definedby
350
Now,
card ô/1 ::5 dimplies
that densco(’W):5 d
and it follows that densR(P) ::5 d,
a contradiction which proves that card s4 > d.By
Remark 1 theproof
will becomplete
once we haveproved
thefollowing
LEMMA 2: Let T be an
infinite
set andput X
=co(r).
Let Y be a Banach space, let m E N and let d be anyinfinite
cardinal.Suppose
that P X’ --> Y is a bounded m-linear map such that the set
satisfies
card A > d.Then there exist a set
D,
card D > d and a bounded linear map L :co(D)--->
Y such thatL(e(&» gé
0(8
ED).
PROOF: We prove the lemma
by
induction on m.Now
and the assertion of the lemma for m = n follows
by
the inductionhypothesis.
In the
sequel
we assume thatConsider the class 16 of all
nonempty
subsetsQ
c s4having
thefollowing property
Observe that (C is not
empty
since every setQ consisting
of oneelement
belongs
to ce.Partially
order (6by
inclusion. Let{Q( i),
i EIl
be a chain in ce. Put
Q
= U i,,:,Q(i),
let k E N and letaj(l :5 j
:5;k)
bedistinct elements of
Q.
Thereare c;
E 1(1 --5 j
:5k)
suchthat aj
EQ(ij) (1 :5 j :5; k).
SinceIQ(i),
i EIl
is a chain there is somejo : 1
:5;jo
:5; ksuch that
U f=l Q(ij)
=Q(iio).
SinceQ(iio) E ce
and since ai EQ(iio)
(1 :5; j :5; k)
it followsthat aj satisfy (4)
andconsequently 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% > d.Given
i,
1 s i :5 n denoteQi
=1,8
E f :/3
= ai for some a =(a}, a2, ..., an) E Q}.
Letb = (b}, b2, ..., bn) E /3.
Assume that for everydecomposition {l, 2, ..., n}
= A U B whereA,
B #0:
A n B =0,
implies that g = (gt, g2, ..., gn) e d.
This means thatQ U {b} E ce
which contradicts the
maximality
ofQ.
This proves thatgiven
any b EOO there is adecomposition {l, 2, ..., n}
= A UB, A,
B 00:
A f1B =
0,
such that there is some g E dsatisfying gi
=bi (i
EA)
andg; E
Qi (i
EB).
Since the set of allpossible decompositions
is finiteand since card 00 >A there is some fixed
decomposition {1,2,.... ni =
A
U B, A,
B 00;
A fl B =0
and someset 001
c00,
card001
> d such that forevery b G 2%i
1 there is some g E dsatisfying gj = b; ( j
EA) and g;
EQj (j
EB).
Write each b EôA in
the form b =PA(b)EBPB(b)
where
PA(b)
EIIjEAr
andPB(b)
EIIjEBr
are definedby (PA(b»j
=bj (j E A)
and(PB(b» = bj (j E B).
We show that cardPA(OOI)
> d. Tosee
this,
assume that cardPA(-O-e 1) - d.
Sincecard 00 1 > d
it follows that there is some002 C OO},
card002> d
and some u ElljEAr
such thatu =
PA(b)
for all b E002.
Inparticular,
there are an i E A and a y E rsuch that y =
bi
for all b E002
which contradicts(3)
since card002>
d.This proves that card
PA(,-9-el) > d.
For eachuEPA(OOI)
choose an element fromPÃl(u)
nB1 1 and
denote the set of all these elementsby S2- Clearly
card002
> d andRecall that for every b E
9JJ2
there is some g G s4 such thatPA(b) =
PA(g)
and such thatgj E Qj (j e B).
Since cardQcard Q - d
(1 Ç j n)
it follows that cardIIjE!I Qj :5 d.
Since card9JJ2
> d it fol- lows that there is some’,»3
C9JJ2,
card9JJ3>
d and some v EIIjEB r
352
such that for
every b
EB3
there is some g E dsatisfying PA(b) = PA(g)
andPB(g)
= v.By (5)
it follows that there are an i E B and a y E r such thatcard{a
EE 4: ai =y}
> d which contradicts(3).
Thuswe have
proved
that cardQ
> d.Since
Q
E Cf¿ it follows thatPut D =
Q
and define themap .0
from the basisand it follows that
By (7)
it follows thatConsequently 0
admits a bounded linear extension L to allco(D).
Since card D > d this
completes
theproof
for m = n.Q.E.D.
COROLLARY 2: Let X =
co(F)
where r is aninfinite
set and let Y bea Banach space.
Suppose
that the rangeof
every bounded linear mapfrom
X to Y isseparable.
Then the rangeof
everyanalytic
mapfrom BI(X) to
Y isseparable.
PROOF: If r is countable there is
nothing
to prove so assume that r is uncountable.Suppose
that there is ananalytic
map fromB1(X)
toY with
nonseparable
range.By
Theorem there are an uncountable set 4 and a bounded linear mapA: co(,à) --->
Y which mapsco(,à)
isomor-phically
ontoR(A).
SincecoC1)
is up toisometry
determinedby
cardL1 assume with no loss of
generality
that either 4 C r or r C L1. IfL1 C F define B : X ---> Y
by
B = AoP where P is theprojection
from Xonto
co(L1)
definedby (Px) (y) = x (y) (y e à; x e X).
B : X - Y is abounded linear map whose range
R (B )
=R (A)
isnonseparable,
a contradiction. Let r C 4. Since r is uncountable X is anonseparable subspace
ofco(,à)
andby
theproperties
ofA, A(X)
isnonseparable.
Consequently AIX: X ---> 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 from12(T)
to
l’(T)
iscompact [10]
it follows that the range of every bounded linear map from12(T)
to1’(F)
isseparable.
On the otherhand,
therange of the bounded
2-homogeneous polynomial
P:1’(F) ---> 11(r)
defined
by P(x) = y
wherey( 1’) = x( 1’)2 (1’ E r, x E 12(r»
is non-separable
since P issurjective.
This shows thatCorollary
2 does not hold ingeneral.
We ask under which conditions on a Banach space X does the assertion ofCorollary
2 hold.REFERENCES
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(Oblatum 15-11-1978 & 21-VIII-1978) Institute of Mathematics Physics and Mechanics University of Ljubljana Ljubljana, Yugoslavia