Banach Spaces of Vector-Valued Functions (Lecture Notes in Mathematics)

a copy of co if and only if X does. Theorem

( P i s i e r , 1978

[101]). For 1 < p < 0% LR(P,X) contains a

copy of ~1 if and only if X does. It should be mentioned that Kwapiefi's theorem lies very much on the Hoffmann-J0rgensen's preliminary work [73]. This chapter is devoted to study these two theorems, but in our way we will have occasion of learning also some other interesting results such as Bourgain's theorem 2.1.1 on averaging of co-sequences, Kadec-PetczyfiskiRosenthal l e m m a 2.1.3, or Maurey-Pisier-Bourgain theorem 2.2.1 on glsequences in Lp(#, X). Since we are interested in the search of copies of Co and ~1 in Lp(#, X), let us r e m e m b e r first which is the situation in the scalar case. We can make the following quite elementary analysis. For 1 < p < oc, Lp(#) spaces are reflexive, and so it is clear that they can not contain copies of any of the spaces we are considering: Co, ~1, or foo. It is also clear that LI(#) does not contain copies of Co, because L I ( # ) is weakly sequentially complete and Co does not enjoy this property. Of course, if it does not contain copies of Co, it can not contain the bigger space fo~. Notice that an idea behind Kwapiefi's theorem is the following: "Since Lp(#) has no copies of co, if co lies in Lp(#, X ) then it must lie in X " . And a similar philosofy is underlying Pisier's theorem, too. As we mentioned in the introduction these ideas have been very important and fruitful in the study of vector-valued function Banach spaces. Soon after the works of Hoffmann-Jorgensen, Kwapiefi and Pisier, Bourgain [14, 16] provided new proofs of the preceding theorems, giving results and introducing points of view which turned out to be very important in subsequent works. This is why Bourgain's ideas lie all over this chapter. In particular, we will follow his approach to get Kwapiefi's theorem because it

42

2. Copies of co and gl in L;(#, X)

contains a result (theorem 2.i.1) which will be crucial in C h a p t e r 3 to get Saab and Saab's t h e o r e m 3.1.4.

2.1

Copies

o f Co i n L p ( t t , X )

If one thinks a little bit how to show t h a t if Lp(#~X) contains a copy of Co then X must contain a copy of co, too, a first n a t u r a l question arises: If (rio is a Co-sequence in Lv(p,X), can we take ~o E ~2 such t h a t ( f . ( ~ 0 ) ) is a Co-sequence in X ? However, one immediately realizes t h a t this question is not properly formulated, since to find a point in a measure space is in general meaningless, because singletons are very often null sets. On the other hand, t r y i n g to make such an assertion a b o u t the sequence (fn(WO)) itself is clearly too much, even in the simplest situation. W h e n we have a c0-sequence (Xn, y~) in X • X , for instance in e0 • Co, we can not hope t h a t either (x~) or (y~) to be a co-sequence, we only know t h a t either (xn) or (Yn) have a Cosubsequence. Of course, this can be easily translated to Lp(#, X). We can give for instance the following quite trivial example: Let A1, A2 be two disjoint measurable sets with finite positive measure, and let (en) be the canonical basis of Co, then we define

for each n E N, where i = 1 if n is odd and i = 2, otherwise. It is immediate t h a t (fn) is a c0-sequence in Lp(#, co), but (fn(~)) is not a co-sequence for ,any point ~ C ~2. T h e preceding a r g u m e n t s lead to a more accurate formulation of the question: if (fn) is a Co-sequence in Lp(#, X), can we find a subsequence (fn~) of (f~) and measurable subset A of ~ , with positive measure, such t h a t (f~k (w)) is a co-sequence in X for each ~ C A? The following simple example shows t h a t the answer is "No".

Example 2.1.1. For 1 < p < + ~ there is a co-sequence (fn) in Lp([O, 1],Co) such that for each subsequence (fnk) of (fi~), (fnk (t)) is not a co-sequence for

(almost) any t C [0, 1]. It is enough to define for each n E N f,~: [0,1]

t

--+

Co

---+ (rn(t) + 1)en

where (e~) denotes the canonical basis of Co, and (r~) is the sequence of R a d e m a c h e r functions (remember t h a t R a d e m a c h e r functions (r~) are defined by

r~(t) = sign(sin(2%rt) ) for all $ C [0, 1] and all n C N).

2.1 Copies of co in Lv(p,X )

43

On one hand, it is clear t h a t for each subsequence (fnk) of (f,~), and for almost all t C [0, 1] the sequence ( f ~ (t)) vanishes for infinite n a t u r a l n u m b e r s /c, and so it can not be a Co-sequence. On the other hand, let us show t h a t (f~) is a Co-sequence in Lp([O, 1], co). Let ( ~ ) be a finite sequence of scalars. We have 1

m

=

mffx t A= I

=

max

fl

= =

Ir,~(t)+lldt

I An(r~(t) + 1) I dt

n Jo

<

j~0

infix I An(rn(t) + 1) l d t

/0

I1 E ~ f ~ ( t ) I I

dt

n

II Zxnf ?l,

<_

II E A ' ~ f ' ~ lip. n

And we also have

_-(/o 1 II

(ff -< (ff

=

=

(m~x I .~(r~(t) + 1)I)Pdt) 1/p (max 21An I)Pdt) 1/p

2maxlAn n

IIp dt)

I.

Therefore (fn) is indeed a c0-sequence in Lp([0, 1], co). If we look at the sequence (fn) of the preceding e x a m p l e we see t h a t a l t h o u g h (fn~(t)) is not a c0-sequence for (almost) any t E [0,1], (fi~(t)) does have a Co-subsequence for (almost) all t E [0, 1] (the point is t h a t the subsequence depends on the t). This leads us to a new question: if (fn) is a c0-sequence in Lp(tt, X), can we assure t h a t the set of all co E (2 for which (fi~(co)) has a co-subsequence is non null ? Bourgain showed t h a t the answer is "Yes" in the case p = 1, and we will see t h a t for 1 < p < co the answer is "Yes", too. Of course, this provides an answer to the p r o b l e m posed at the beginning of the section: this guarantees t h a t if Lp(/*, X ) contains a copy of Co then X has the s a m e property. It m i g h t be surprising t h a t some results in this section are stated for

seminormed spaces. However this is not a more or less superfluous search of

44

2. Copies of co and ~1 in L~(p,, X)

generality. We will see in T h e o r e m 2.1.2 and in Saab and Saab's t h e o r e m 3.1.4 t h a t this is actually convenient for our purposes. We will consider c0sequences in s e m i n o r m e d spaces. Denote by Coo the n o r m e d subspace of Co of all sequences of only finitely m a n y non-zero terms. We will say t h a t a sequence (x.,~) in a s e m i n o r m e d space (X, II 9 It) is a c0-sequence if there exist two positive n u m b e r s M, d such t h a t

for all (a~) E Coo. Of course this simply means that, by passing to the quotient n o r m e d space, the corresponding sequence is a c0-sequence in (the completion of) this n o r m e d space, too. In the proofs of the few results we will state for senfinormed spaces the only facts of the theory of n o r m e d spaces which are applied are Bessaga-Petczyfiski t h e o r e m 1.1.1 and Bessaga-Petczyfiski selection principle. It is easy to show t h a t the usual proofs of these t h e o r e m s also work for seminortned spaces or, better t h a n this, one can see in the s t a n d a r d way, by passing to the quotient n o r m e d space, t h a t Bessaga-Petczyfiski results for s e m i n o r m e d spaces follow immediately from the corresponding ones in n o r m e d spaces. Before going into Bourgain's a p p r o a c h we will s u m m a r i z e some i m p o r t a n t properties of R a d e m a c h e r functions (r~) in [0, 1], because they will play a very i m p o r t a n t role in our study. These properties are of course quite well known, but for the sake of completeness we will include proofs. Proposition 2.1.2 is just a version of the "contraction principle" [37, 12.2]. ' P r o p o s i t i o n 2.1.1. Let (X, ]] are elements of X , then

oll

II)

be a seminormed space. If x, xl, x 2 , . . . , x,~

dt=2,, 1

x + i=l

IIx+ E 01=~l

for any collection of natural numbers nx < n2 <

II x

+

(t)x i=1

tl dt =

.

.

.

/o1 II x +

'

Oixi

II

i=l

< rim. Consequently,

r (t)x

II dr.

i=1

Proof. T h e result is an i m m e d i a t e consequence of the fact t h a t rnl, rn2, rn~ are i n d e p e n d e n t (symmetric) r a n d o m variables which take the values 1 and - 1 with probability 1 In fact, if for each O = (01, .,0m) C { - 1 , 1 } "~ we put 9

-

9

,

7Tt

A e = {t e [0, 1] : r,~(t) = 0i for 1 < i < m} = N { t E [0,1] : rn,(t) = 0i}, i=1

it is clear t h a t this set has m e a s u r e 2A~. Since, for a given m, the A o ' s form a partition of [0, 1], we have

2.1 Copies of co in Lp (#, X ) 91

7T~

nt

= 9

i

45

~ OG{--1,J

1

} ....

1 ~ 2"~

-

II ~ + ~

~ O

llx+ E

Oi=•

r~,(t)zi It dt

i=1

Oia:i

II

"

i=1

P r o p o s i t i o n 2 . 1 . 2 . Let ( X , II . II) be a s e m i n o r m e d space. I f zo, z~, z 2 , . . . , ~:,,~ are e l e m e n t s of X arm ao, al, a,.2, 9 9 am are real numbers, then

/o1II ,~0,0 § ~

....

'

~ m ( t ) ~ II at <- - Onlax ] ai I
II x0 +

i=1

•

~ ( t ) ~ II at.

i=1

In particular

II xo II _<

In

.•01

II :~o + ~ ,.,(t)** II ~t,

(2.1)

i=l

and 91

9

'n~--i

.I

i = 1

,@

i=1

"

Proo./: By

h o m o g e n e i t y , we m a y a s s u m e I ai I -< ] fo~' g E {0, ] , . . . , - , , } L e t us c o n s i d e r t h e real f u n c t i o n defined on R - -(,g2 m + l ,, t h e u n i t b a h of era+; by '30 /,

Ca0,-, .... , ~-~)'

>/

,

.l

~7~

II ,,o~:o + ~ airi(t)zi It dt

'/ 0

i

1

This function is obviously continuous and convex. Therefore, it takes its m a y i m u m a m o n g the extreme points of" the. c o m p a c t set --,~ocR(Un+l~/,which are the points of the tbrm ( e 0 , q , . . . , e , , ) with q = :t:1 for i = 0, 1 , . . . ,m. Notice now t h a t it follows easily from the preceding proposition t h a t in all these points tile function takes the same value.. To be precise, 9l

II 60"s

9 )

+

~{r~(t):,:~ II dt =

fi i

1

/0 l

II 2:0 + ~n), ri(t):ci {I dt

'

i=1

tbr any choice of signs Co, q . . . . , e,,, = :t_1. Hence, for any (ao, al . . . . , a,~) C B (\f' ~, ,5,o+ l )/ W C h a v e

II 9

+ ~ a~.~(t):~,~ II dt _<

'

/=l

/o' II

"

+

•

rdt)z,z II at

i=]-

This completes the proof of the first inequality. The others are just particular cases of it.

46

2. Copies of co and gt in Lp(#, X)

Now, let us begin to study Bourgain's approach. The key of this approach is L e m m a 2.1.2, which is an "averaging version" of Bessaga-Petczyfiski classical theorem 1.1.1 on c0-sequences. In its proof we will use the following simple fact about sequences of positive numbers. L e m m a 2.1.1. Let (Ai) and (6.,) be sequences of positive numbers and ass u m e that (Ai) is bounded and non decreasing, then (Ai) has a subsequence ( Ai,~ ) such that ~i,.+l _< (1 + &7,);~i.~ for all m E N. P r o @ Observe first that the sequence (Ai) must be convergent. Let us denote by A its limit. Consider the sequence ( ~x) . It is clear that

lim~A = 1 < 1 +(~m for all m C N. So, (Ai) has a subsequence (Ai~) such that A - -

<

Ai,,

1+din

for all m E N. Since (Ai) is non decreasing, we have Ai < A for all i. Therefore, in particular, we have ,ki,,~ for all m. L e m m a 2.1.2. /f (X, II 9 II) is a seminormed space and (xi) is a non null sequence in X such that f 1

sup I ~z

dO

n

II

II dt < i=1

then (:ci) has a Co-subsequenee. Proof, Let (zi) be a sequence in X such that limsup ]1 xi I1> 0 and

sup n

II .

r~(t)x~

dt = B < +oe.

i=1

Observe first t h a t according to Propositions 2.1.1 and 2.1.2, for each subsequence (zk~) of (zk), we have

2.1 Copies of c0 in L~(#,X)

sup

II

7~

n(Oxk, II dt

=

II

sup

"k~(t)xk~ II dt

n

i= 1

i=1 1

s

<

47

rn

sup/ nz

[I E

Jo

ri(t)xi II d t = B < +oo,

i=l

and therefore,

sup n

II .

n(t)*k~ II dt ~ B < +<~

(2.2)

i=1

T h e n we can assume without loss of generality t h a t there exists 7 > 0 such t h a t II xi I1_> ~ > 0 for all i E N. Let us see now t h a t (xi) must be weakly null. Given x* E X*, t h a n k s to the Khinchine's inequalities, for a certain A > 0 we have

A(Z

'/'

<

i=1

I 9

<xi,x*>r~(t)

=

ri(t)xi,z* >t dt

l< i=l

9

.1

_< s

ldt

i=1

m

!tY~(Ox., II II~*lldt<_BIIx*ll i= 1

for all m E N. Thus, } - ~i=i I < x i x* >l 2 converges and so lim < :r~, :1:* > = 0. i

Take now a non increasing sequence (6.,,,) of positive n u m b e r s such t h a t o~

cA3

1-[(1 + < ) < +oo

Z2' 6,, < +oo

n

~=1

1

Using the Bessaga-Petczyfiski selection principle (see [35, page 42] or [93, pages 4-5] ), we can extract a subsequence (Y0 of (xi) in such a way t h a t (Yi) is a basic sequence and satisfies

I[

a~> I1_<(1 + &,)II ~ ~y~ bl i--1

i

(2.3)

1

if ~, < m and a , , a 2 , . . . , a,,, are scalars. By 2.2 and Proposition 2.1.2, (.1~ II Z*j=~ r~(t)y~ 11 dt)~~176 is a b o u n d e d non decreasing sequence of positive numbers, so, it follows from l e m m a 2.1.1 t h a t there is an increasing sequence (i,,~) on n a t u r a l n u m b e r s such t h a t

48

2. Copies of co and gl in Lp(p, X ) 1

i.~+1

fo IIE ri(t)yi IIdr<_ (l+5,,j~ i

1

{,~

I[ Er~(t)Y,

1

Ildt

(2.4)

i=1

for all m C N. Our aim now is to show that (yi~) is a c0-sequence. This is the only point which remains to be proved, however it is tile hardest part in the proof. Since (Yi,,) is a seminormalized basic sequence, by Theorem 1.1.1, we only need to show that ~ Yi., is a w.n.C, series, that is, we have to show that

sup{ll

}-~O,,y~,, I1:-,,

c H, 0n = +1} < +oc.

(2.5)

n=l

To prove this let us define some quantities which relate the expressions involved in 2.3 and 2.4. For any finite choice of signs (01,02,..., 0,~) put

p(Ol,02,... ,Om) ['1 II

o~y~ + ~ ~(t)y~ II dt

=

aO

n=l

iCGm

where ~,,~ = { 1 , 2 , 3 , . . . , i , ~ } \ {it,i~,...,i,~}. follows from 2.1 of Proposition 2.1.2 that

First. of all, observe that it

O,~yi,, IIs p(Ol,O2,...,Om)

II s rz: l

Therefore, in order to prove 2.5, it is enough to show that the numbers t)(01, Oe,..., Ore) are bounded. Second, notice that it follows immediately from 2.3 that for each choice of signs (01,09,..., 0m) we have

p(01,02,... ,Ore)

II

= 9

s 91

<

O,~yi~ § }7_, ~(t)y~ tl dt n

1

iEO'm

'nz+]

/ (1 + <,) I1 ~ <,w~ + ~, /0

9

,.~(t)y~II dt

i~o-,,t+l

n:l

<_ (l+*m)p(O1,...,Om,Om+l). Therefore,

p(<,<,...,

e,~) < (1 + &,)p(<,..., r

e,,~+~)

(2.6)

We will need another property of the numbers p(01,02,..., Ore) which is contained in the following Claim. We will prove it at the end of this proof.

Claim: .[01 II ~<"~=1 ri(tDyi II dt = .1o,1 p(r] (t), r 2 ( t ) , . . . , r',~(t))dt With these properties of the p(01,02,..., 0m)'S in mind we only have to do some computations to finish our proof. By our Claim and 2.4 we have

2.1 Copies of co in Lp(#, X) 1

./i I D(rl ( t ) , . . . , r m (t), rm+l ( t ) ) d t

im+l

= fo II ~ '

<

49

,.~(t)y~

II dt

/=1

It~-~.ri(t)yill dt

(l+6m)~o

i=1

/o

=

(1 + 5r~)

p(r~(t),...,r,~(t))dt,

and thus

./i ~ (p(r~(t),... ,,',,,,(t),r,-~+~(t)) - (1 + 5,~)p(rl(t),... ,r,~(t))) dt <_ O. Then, given a finite collection ( 0 1 , . . . , 0 .... 0re+l) of signs and putting S = {t G [0, 1]: (l'l(t),... ,rm(t),Fm_l_l(t)) r (01,.

9,0,~,om+~)},

we have

.fo I (p(rl (t),. . . ,r,,(t),rm+l (t)) - (1 + (~,~)p(rl(t), 1

-

-

2,,+](p(O,,...,O,,,Om+,)-

§

/ (p(,.,(t),

<

0.

1

(1+Sin)p(01,..

,0,~))

.,,.,,~(t), I'm+ 1 (t)) - (1 + ~,dP(,'~ (t),

. , ,.,,~(t))) dt

Hence, using 2.2, 2.6 and the Claim, we have 1

2m,+1 (p(01,..., 0m, 0re+l) -- (1 -~- (~m)tO(0l,... ,0m)) ./S ((1 -I- (~m)p(rl(t),... ,r,n(t')) -- /9(rl(t),... ,rrn(t),"m+l(t)) ) d t

s

<_ .

(l+5,,)p(rl(t),.,.,r,,,,(t))

(5,, + 1

1 ~U(~).

,

l+6,

p(r](t),...,r,,,,(t))

p(ra(t'),... ,r,,,(t)) dt

.1

< 26,,,.~ p
=

2~,,~/i '

i,,~

II ~(t)w

II dt

< 25,,,.B

i=1

Therefore,

p(O,,.

,0,,,, 0,,~+1) _< (1 + (~m)p(01,..., 0m) + 4B2"~6,,~

)

dt

50

2.

Copies of co and

& in Lp(p,X)

Now, by induction, it is easy to see that

p(O1,...,Om,Om+l) <--n=lH(14-(~,z)p(01) +4B

nI~=l(l+~n)

(En=12nt~n)

and then

~(01,...,0,~) <

(p(1)+p(-1))H(I+6~)+4B

(1+6,~)

n=l

(

2~6~)<+oe n=l

for all m r IN and all finite collection of signs ( 0 1 , . . . , Ore). Hence we have shown that the numbers p(01,02,..., Om) are bounded. To finish the proof of the lemma it only remains to show the Claim.

Proof of the Claim: It is enough to observe that, using Proposition 2.1.2, we have .1

.~ p ( r , ( t ) , r 2 ( t ) , . . . , r , ~ ( t ) ) dt

=

j~0l(~l

~

r,~(t)yi. +

n=l

i~o-, n

2~m--.~----7 E

7~t II Er,~(t)yi.

el =-t- 1,iC or,.

:

=

1 1 2im-m 2m

1

2i,: II ~ 91

= This

e,:

E

•

/

+ E eiyi II at

n= l

0.=•

i E ~r.~

n=l

iCam

eiyi II

i=1 i.~

II ~ ~'i(t)y~ II dr.

s

'

i=1

completes the proof of the Claim and therefore, the proof of the

too.

It. is interesting to observe

i '

"n, i=1

that, since '17,

1

II ~0ix~ II, Oi 4zl

i=1

lemma,

2.1 Copies of c0 in L p ( # , X )

51

the preceding lelnma says that if (xn) is a non null sequence for which the set 1

{~

EII 0~=-F1

II: n E N, 0i = =t=1}

EOixi i=1

of averages is bounded, then it has a subsequence (Yi) for which the set n

{11 ~ O~y{ I1: ,~ E N, 0{

=

+1}

i=1

is bounded. Notice also that the preceding lemma provides the following particular version of one of the theorems we are trying to show (Theorem 2.1.2): Let (xi) be a sequence in X and assume that (ri(.)x{) is a cosequence in L1 ([0, 1], X ) , then (x~) has a co-subsequence.

We can finally state and prove Bourgain's theorem. T h e o r e m 2.1.1 ( B o u r g a i n [14], 1978). Let (f2, 2 , # ) be a finite measure ,space and suppose that for each co E f2, I . I~ is a s e m i n o r m on Coo such that co ~-~l x I~ is integrable on (f2, Z , p ) for each x E coo. Let us define the s e m i n o r m tl . IIo on Coo by

IIx Iio=/Ix

I~ d~(co)

for every x E coo. z] (xi) is ,~ sequence in Coo which is a co-sequence for II Ilo, then there exists a positive measure set A E ~ such that (xi) has a Co-subsequence for I. I~. whenever co E A. Proof. Assume that the hypothesis of the theorem holds. It is clear that thanks to the preceding lemma it, is enough to show that, the following is true: C l a i m : There exists a positive measure ,set A E Z such that .1

l i m s u P l x i I~> 0 and n,

sup/ 71,

.10

71,

I Zri(t)xi

I~ dt < +oo

i= 1

for all co E A.

Let M and 5 be two positive numbers such that 5max lai I < - II ~ - ~ a i x i IIo = / i l Z a i : r ' i I~; d#(w) <_ M m a x l a~ I

2. Copies of co and ~1 in Lp(tZ, X)

52

for all (ai) E coo. By our hypothesis, for each i E N, the function r defined by r = ]xi l~. for cc E g2, belongs to L1 (tL). Moreover, (r is a seminormalized sequence in L I ( # ) because

for all i E N. To prove t h a t our Claim is true we will show t h a t (a)

(r

(b)

p({w E ~2: l i m s u p i r

(c)

sup~z ~o1 I E i n

is uniformly integrable,

1

> 0}) > 0, and

ri(t)xi tw dt < + o c tbr almost all cv E f2.

If (a) does not hold, then it is well known (see [35, pages 93,89]) t h a t there exist a sequence (Ai) of pairwise disjoint measurable sets, a subsequence of (r which we will continue to denote in the same way, and an e > 0 such that

'

i

for all i E N. Now, Rosenthal's disjointification l e m m a 1.2.1 says t h a t there is an increasing sequence (ik) of n a t u r a l numbers for which

fbr all k E N. Now, we reach the tollowing contradiction: for each n E N we have

k:l

dg2

k=l

m=l

'

im

k:l

> nt

Z

'

i.~

'

im

k=l ,kern

d)~k(aJ)) dp(w)) II?,~1

k=l,k@m

>_ 'n~- ~ /A ( ~ ~rJ 1

=

~

-

k=l

r

i,,~ k = l , k T ~ m

m

~/c

l;

li.,

r (~)

d~(~)

) > .,e - ~

k=l

,~=1,.~r

A~,,,

= 7t~.

2.1 Copies of co in

Lp(~,X)

53

Thus (a) is proved. Now, let us note t h a t (b) is an i m m e d i a t e consequence of (a): If we suppose t h a t #({cJ E s : l i m s u p i @(co) > 0}) = 0 then lim~@(co) = 0 for almost all c~ E (2, because the functions @ are non negative. Hence Vitali's l e m m a [51, IV.10.9.] says us t h a t limi II r It~ = 0, which is of course a contradiction. In order to prove (c), for each n E N put ~(co) =

/01• ]

r~(t)z~ I~ dt

i=1

for all co E f2. Clearly ~ belongs to LI(#) since, by proposition 2.1.1,

Oi=:t:l i = l

and by our hypothesis, each function co ~1 ~'~in=l OiXi [w belongs to L I ( p ) . According to P r o p o s i t i o n 2.1.2 it is also clear t h a t ( ~ ) is an increasing sequence. On the other hand, since for each n E N

/; 9

0~=-4-1

-

1

i=1

II

fi

0 x ll0 <M,

Oi=-b_l i = I

we deduce from the Lebesgue's m o n o t o n e convergence t h e o r e m t h a t the function co ~ sup~ ~,,(co) is integrable, and therefore, sup{~(co) = sup rt

n

I dO

ri(t)xi Ix at < +oc i

1

tbr almost all co E ~2. This finishes the proof of (c). Once we have shown (a),(b) and (c), to complete our proof it is enough to observe t h a t (b) and (c) imply t h a t our claim is true. Now, the following two results are i m m e d i a t e consequences. T h e o r e m 2.1.2. Let (f2, ~ , p ) be a finite measure space and let (fn) be a co-sequence in L I ( # , X ) . Then, the set of all co E f2 such that (fn(co)) has a Co-subsequence is a measurable set with positive measure. Proof. Let (f,~) be a c0-sequence in L1 (#, X ) . T h a n k s to Proposition 1.6.3, we only have to show t h a t the set of all a~ E s such t h a t (f~(co)) has a c0-subsequence is a non null set. By our a s s u m p t i o n , there are two positive n u m b e r s M and ~ such t h a t

54

2. Copies of co and

5 m a x l a~ I < II ~ a ~ f ~

~1

in Lp(p,X)

II~ = / ~ If ~~aifi(c~)II d~(~) ~ M m a x l a~ I '

i

for all (ai) E Coo. This suggests to define for each w E f? I x I~ = II Z

aifi(w)II i

for all x = ~ aiei 6 Coo. Of course the function co ~-~l x I~ is integrable on (f2, Z , # ) for each x E coo, and so we can consider the s e m i n o r m II 9 Iio on coo as defined in the preceding theorem. Moreover, the canonical unit vector sequence (e~) in coo is a c0-sequence for 11 9 Ilo because

i=1

i=1

i=1

n

i=1

Therefbre, the preceding t h e o r e m implies t h a t the set of all points w E ~2 for which (e~) has a co-subsequence for I . I~ is a non null set. To complete the p r o o f it is enough to observe t h a t "(e~) has a e0-subsequence for I 9 I~" simply m e a n s t h a t (fn(c~)) has a c0-subsequence.

2.1.3. Let (f?, Z, #) be a finite measure space and let X be a Banach space, then L1 (#, X ) contains a copy of Co if and only if X does.

Theorem

Proof. T h e condition is of course sufficient. Necessity follows from the preceding theorem. Our aim now is to show t h a t the preceding results are also true for Lp(#, X ) with 1 < p < +oc. To prove this we will use the following "subsequence splitting l e m m a " for L1 (/L)-bounded sequences. It was discovered by Rosenthal [113], as a refinement of results and a r g u m e n t s due to K a d e c and Petczyfiski [77]. As far as we know it is the sharpest result of its kind. 2.1.3 (Kadee-Petczyfiski-Rosenthal, [77, 113]). Let (f2, Z , # ) be a finite measure space. If (fn) is a bounded sequence in L I ( # ) , then there exist a subsequence (f,~) of (fn) and a sequence (Ak) of pairwise disjoint measurable sets such that (X,\Ak f'~) is uniformly integrable.

Lemma

Proof. Of course if (f,~) is uniformly integrable the result is trivial: it is enough to take ( f ~ ) = (fn) and AN = O for all n E H. So, let us suppose t h a t (.f~) is not uniformly i n t e g r a b l e This means t h a t there exists e > 0 satisfying

2.1 Copies of co in L p ( f , X )

55

there exist a sequence (Ej) of measurable sets with limj p ( E j ) = O, and a subsequence (fnj) of (rio such that

i

for all j 6 N Let ~0 be the supremum of all ds for which (*) holds. It is clear that we can take a subsequence of (f.T,), which we continue to denote in the same way, and a sequence (E~) of measurable sets such that 1 #(E,J < ~

and

js

If n l d # > e o -

1

2~

n

for all n C N. Take nl = 1. Since lim p(U~_kE,, ) = 0, k

by the absolute continuity of the indefinite integral of ] f,~, 1, there exists -,2 > ",a such that

L

1

, "~\u~=-2E" [fn, I d# > eo - 2,---T. Put

Now, applying the same argument

to

Ifn2

I, we get n3 > n2 such that

~: \u~=~3E~ ]fn2 ] dtt > eo - ~2n. Put,

A2 = En: \ LJ?z~

T~ 3

E,~.

It is clear that in this way we obtain a sequence (Ak) of pairwise disjoint measurable sets, and a subsequence (f,~) of (fn) such that /A

I fnk I d# > eo - 2n-~1

for all k 6 N. Let us finish the proof integrable. Suppose it is not the case. of (f,~), which we continue to denote of measurable sets with limk #(Bk) =

showing that (Xn\a~f'~k) is uniformly Then there exist e' > 0, a subsequenee in the same way, and a sequence (Bk) 0, such that

B ]Xr~\Akf~k I d # > d k

56

2. Copies of co and gl in Lp(#, X)

for all k e N. But this implies that 1

Xa\ak ]fnk [ d# > eo - 2n k + e' for all k E N. Since e0 - ~ 1- +e' > ~0 for k large enough, this is a contradiction with the fact that e0 is the supremum of the e's satisfying (*). We can now begin our study of Lp(#, X) spaces. Next theorem is a good example of how the preceding lemma can be used to deal with vector-valued functions. T h e o r e m 2.1.4. Let (s Z , p ) be a finite measure space, let 1 < p < +oo, and let i : L p ( p , X ) ~ LI(p,,X) be the canonical embedding. Let H be a subspace of Lp(#, X ) which does not contain a complemented copy of gp, then the restriction i IH of i to H is an isomorphic embedding.

Pro@ If i ]H is not an isomorphic embedding, then not all II 9 [11-null sequences in H are [I 9 lip-null. Thus, using the continuity of i, it follows that there exists a II. II1-null sequence (f~) in H such that 11f~ lip = 1 for all n E N. Taking a subsequence if necessary, we may suppose that a2 \ (fT~(J), and therefore (11 f~(c~)If), goes to zero for almost all aJ E ~2. Now we can apply the preceding lemma to the sequence (11 f ~ ( - ) I f ) , and we get a further subsequence, which we still denote (f~), and a sequence (A,~) of pairwise disjoint measurable sets such that (X,\A~ II f~(.) IIp) is uniformly integrable. An appeal to Vitali's lemma [51, IV.10.9.] allows us to conclude that (X,\A~ II fn(.)II t') is null in LI(#), that is, (X,\Anf~) is null in Lp(#, X ) . Therefore, taking again a subsequence if necessary, we have that ( X ~ fn) = (f~ - Xa\A~ f~) is a seminormalized sequence in Lp(#, X). Since the functions are disjointly supported, (X~ fn) is a complemented gpsequence in Lp(#, X ) (Proposition 1.4.1). At this point, the classical result on perturbation of basis (see [93, 1.a.9] or [35, Chapter V, Theorem 12]), guarantees that (/n) = (X~o f~ + X,\~o/,~) has an/?p-subsequenee which is complemented in Lp(#, X), and therefore in H. This completes the proof.

Remark 2.1.1. Notice that it is possible to give more precise versions of the preceding theorem. For instance, it is also true for the canonical injection from Lp(#, X ) into L~(#, X), with p > r > 1. We may also deduce from the proof that it also holds for all subspaces H which do not contain sequences arbitrarily close to normalized disjointly supported sequences, and in particular, which does not contain (1 +()-complemented (1 + c)-/?p-sequences for arbitrarily small ~ > 0.

2.2 When does Lp(#, X) contain a copy of gt?

57

As an immediate consequence of the preceding theorem we have: C o r o l l a r y 2.1.1. Let (~2, Z, >) be a finite measure space and let 1 < p <

+oc. Then we have: (a) Every Co-sequence in L1,(p , X ) is a co-sequence in LI(#, X ) . (b) Every gl-sequence m Lp(p, X ) is an fl-.sequence in L1 (p: X ) . (c) Every g<sequence in L p ( # , X ) is art g,--sequence in L I ( # , X ) , whenever l <_r < +oo, r ' r (d) If Lp(p, X ) contains a copy of H and H does not contain complemented copies of gp, then L1 (#, X ) contains a copy of H. We can now extend Theorem 2.1.2 to Lv(#, X ) spaces with 1 < p < +co. In the extension we also remove the finiteness assumption on #. 2.1.5. Let (f2, 5 , p) be an arbitrary measure space, let 1 <_ p < +oc, and let (f~) be a co-sequence in Lp(#, X). Then, the set of all co C g? such that (f,~(oo)) has a Co-subsequence is a measurable set with positive

Theorem

7Tteasurc.

Proof. Let (f,~) be a c0-sequence in Lp(#, X). Notice first that if # is a finite measure, we can apply the preceding corollary, and we deduce that (fi~) is a c0-sequence in L1 (p, X ) , too. Now, the conclusion follows from Theorem 2.1.2. For the general case, we use the standard reduction argument we have a h e a d y given in Corollary 1.6.1. By L e m m a 1.6.1, there exists a a-finite measure space (f~l, ~1,#1) C (f~, E,p) such that (f~) lies in L p ( # I , X ) . Now, by Proposition 1.6.1, there exists a probability measure P0 on (/21, Z1) and an isometric isomorphism from LF(#I, X) onto Lp(po, X). Moreover, as we showed in the proof, this isometric isomorphism has the form f ----+ h f where h is a certain scalar function which does not vanish at any point. Therefore, it is cleat" that (h(co)fi~(w)) has a Co-Subsequence if an only if so does (f,~(co)). So the conclusion follows from the first part. Finally, we get the following

2.1.6 (Kwapiefi). Let (~2, ~ , # ) be an arbitrary measure space and let 1 < p < +0% then L p ( # , X ) contains a copy of Co if and only if X contains a copy of co.

Theorem

2.2 When

does

Lp(lz, X)

contain

a copy

of s

Our initial point of view here is very similar to the one in the preceding section. In this way we may see that the natural modification of Example 2.1.1 show's t h a t f o r 1 < p < +oo there are *?l-sequences (f~) in Lp([0, 1],~1) such

58

2. Copies of co and ~1 in L~,(p, X)

that for each subsequence (fi~.k) of (f,.), (fn~(t)) is not an gl-sequence for (almost) any t E [0, 1]. So, likewise in the preceding section, it is natural to ask whether given an tl-sequence (f,.) in Lp(#, X), we can assure that the set of all cJ E f2 such that (f,,(co)) has an fl-SUbsequence is a non null set. The main theorem of the section states that we can. We begin our work with quite a general result about sequences in Lp([0, 1],X) due to Batt and Hiermeyer. Recall that (r~) denotes the sequence of Rademacher functions in [0, 1]. P r o p o s i t i o n 2.2.1 ( P r o p o s i t i o n 4.1 of [3]). Let (x~) be a bounded sequence in X and let 1 < p < +oo. Then, the following are equivalent:

(a) (x~) has no e~-subsequences. (b) (r,~(.)x,~) is a weakly null sequence m Lp([0, 1], X). Proof9 Suppose that (a) holds. In order to conclude (b) we will see that every subsequence of (r,~(.)xn) has a further subsequence which is weakly null in L~([0, 1], X). But Rosenthal's theorem says us that each subsequence of (x~) has a weakly Cauchy subsequence. Thus, we only need to show (*) If (yk) is a weakly Cauehy sequence in X and (rnk) is a subsequenee of (r,.), then (r,,~ (.)Yk) is weakly null m Lp([0, 1], X). So, assume we are in the hypothesis of (*) and let F be a functional in Lp([0, 1],X)*. By theorem 1.5.4, there exists a w*-measurable function ~ : [0, 1] -+ X*, such that H~(.)I] is measurable, it belongs to s (as usual, q denotes the conjugate of p) and

_r'(f) =

< f(t), g,(t) > dt

for all f E Lp([0, 1], X). The sequence (< Yk,~P(t) >)k converges for almost all t E [0, 1] because (Yk) is weakly Cauchy. Let us denote by h the pointwise limit of ()k. Since

I< Yk,C'(t) >l <_ ,up,~ IlYnll Ilk(t) II for all t E [0, 1] and all k E N, we can use Lebesgue's dominated convergence theorem. Therefore, h belongs to Lq([O, 1]) and (< Yk,~(.) >)k converges to h in Lq([0, 1]). Now, note that

] F(rnk(.)yk) ] =

'01

< r,,~(t)yk,~P(t) > dt

=

fo

(t) < y k ,

>

~-

~o 1 r'~(t)(--h(t)) dt + ~oo1 rnk (t)h(t) dt

dt

2.2 When does Lp(#, X) contain a copy of &?

59

!

<

(/o 1I

h(t)l q a t

+

/o 1r ~ ( t ) h ( t ) d t

Using the fact that (rnk) is weakly null in L~([0, 1]), it is clear that (F(rn~ (.)Yk)) is a null sequence. Thus we have shown that (a) implies (b). For the converse, it is enough to observe that, for 1 < p < +no, (r~(.)xn) is an gl-sequence in Lp([0, If, X) whenever (x,0 is an gl-sequence in X: if 5 > 0 satisfies

for all finite sequences ()~,~) of scalars, then we have 1

1/p

for every finite sequence (A,0 of scalars. This finishes the proof. Our next lemma provides a very easy sufficient condition for a sequence in L1 (#, X) to be weakly null. L e m m a 2.2.1. Let (9, ~U, #) be a finite measure space, let (f~) be a bounded sequence in L ~ ( # , X ) , and let us suppose that (fn(co)) is weakly null for almost all co E ~2. Then (f,~) is weakly null in L] (#, X). Proof. By Theorem 1.5.4 we only need to show that if we take a w*measurable function q~ : ~Q --+ X* such that the function If~P(.) Hiis measurable and belongs to s (#), then

/<

A(co),~(co) > d# -~ 0.

But this clearly follows from the Lebesgue's dominated convergence theorem, because lim,~ < .f,.(co), k~(co) > = 0 for almost atl ca r /2 and

I< .f,,(co),~'(co) >I _< liA(~)ll lie(co) li _< sup li A(co)II II ~'(co)li. We can now give the main result of the section.

Theorem 2.2.1 (Maurey-Pisier-Bourgain, 1978). Let (~2,~,#) be an arbitrary measure space, let 1 < p < +oo, and let (fn) be an gl-sequence in Lp(/t, X ) . Let us assume that one of the following two conditions holds (a) It is .finite p = 1 and (f,O is uniformly integrable.

(b) 1 < p <

+no.

Then, there the set of all co C I2 such that (f,~(co)) has art g~-subsequence is a measurable set with positive measure.

60

2. Copies of co and ~71 in Lp(p., X)

Proof. By Corollary 1.6.1 we only have to show that the set of all co E S? such that (fn(co)) has an ~.l-SUbsequence is non null. First, let us consider case (a). By the hypothesis, there exist two positive numbers M, ~ such that

<~Zl~,~,l<_llZ<,,,,.f,~ll,<_ M Z la,~l for all finite sequences (a,<) of scalars. Since (11in(.)II) is uniformly integrable, by standard arguments (see for instance [35, page 111]), it follows that there exist N > 0 and a sequence (An) of measurable sets such that for every n E N we have i ~ \A. IIf,*(co)II d#(co) < 7~

and

II f,~(co)It < N for all w C An.

Take g~ = )lAnfn for every n E N. Notice that (gn) is bounded in LI(v,X) (even in Loo(#, X)) and for every finite sequence (an) of scalars we have

II ~ a~.q~ I1, _> II ~ anf,~ II, -II ~

an(fi~ - g n ) I I ,

Hence, (g~,) is an ~71-sequence. Let us proceed now by contradiction. If (f~(w)) has no Q-subsequences for almost all co E $2, then the same holds for the sequence (9,~). But (9~(co)),~ is bounded in X for almost all co E s and so, according to Proposition 2.2.1, (rn(.)gn(co)) is a weakly null sequence in "LI([0, 1], X) for almost all co E .Q. Let us denote by r~g~(.) the function

> C,([0,1],X)

Notice that the sequence (rngn(.)) is bounded in Loo(#, L1 ([0, 1], X)) because

IIr~g~(.)ll~

--

ess s u p ~

IIr,m,~(co)ll,--

ess s u p ~

/0 '

IIr~(t)gn(~)ll dt

= esssup~en Ilgn(co)ll _< N Thus, by the preceding lemma, it follows that the sequence (rng~(.)) is weakly null in L1 (#, L1 ([0, 1], X)). Since this space is isometrically isomorphic in the natural way to L1 ([0, 1], L1 (#, X)), we deduce that (rn(.)g~) is weakly null in this last space, where of course (r,~(.)g,~) denote the functions [0,1]

t

)

>

L,(#,X) rn(t).qn

Again, an appeal to Proposition 2.2.1 says us that (gn) has no gl-subsequences in L1 (#, X). This contradiction finishes the proof in the case p = 1.

2.3 Notes and Remarks

61

For the case (b), assume first the measure it is finite. In this situation, notice that ( f , ) is uniformly integrable, just because it is bounded in Lp(it, X) with p > 1. On the other hand, it follows fl'om Corollary 2.1.1 that (f~) is an f,~-sequence in L 1 (it, X). Therefore, our sequence is actually in the hypothesis of case (a). So the conclusion holds. For an arbitrary measure we just have to follow the same procedure we used in Theorem 2.1.5. As an immediate consequence of this theorem we get T h e o r e m 2.2.2 ( P i s i e r [101], 1978). Let (~2, Z , it) be an arbitrary measure space and let 1 < p < +oc, then Lp(it, X ) contains a copy of fl if and only if X does.

Remark 2.2.1. One could think that the finiteness assumption on it in part (a) of Theorem 2.2.1 could be removed. However, this is not the case as the following simple example shows: Take any uniformly hounded sequence (h~) of norm one non negative functions in LI([0, +co)) such that ]l,,~(t) = 0

for all t 9 [0, n]

for all n E N. Let (e~) be the canonical fl-basis. It is not difficult to to show t h a t if we take L, = h,,(.)e,~, then (fi~) is an uniformly integrable g~-sequence in L1 ([0, +co), ~1). However, for each t E [0, +co) the sequence (f,.(t)) is eventually null and so, it can not have any fl-SUbsequence.

2.3 N o t e s a n d R e m a r k s Tile importance of Hoffmann-J0rgensen's paper [73] in Kwapiefi's theorem 2.1.6, the main result of section 2.1, can not be overestimated. Kwapiefi's short note [83] was published in the same issue in which the aforementioned paper appeared, and even its title ("On Banach spaces containing Co. A supplement to tile paper by J. Hoffmann-J0rgensen 'Sums of independent Banach space valued random variables' ") reveals how it depends oil it. Hoffmann-Jorgensen's fundamental paper [73] is one of the pioneering works on probability in Banach spaces. It is focused on generalizing several classical theorems about sums of independent real random variables to the case of Banach space valued random variables. Among other results, it is showed that for a Banach space X the following are equivalent: (a) X satisfies: "If ( ~ ) is a sequence of independent, symmetric, X-valued random variables and their partial sums (Sin) = (}--]~;~" 1 P~) are bounded ahnost everywhere, then (S,~) converges almost everywhere".

62

2. Copies of co and ~1 in L~(#, X)

(b) L~,([0, 1], X) does not contain a copy of co for some p E [1, +oc). (c) L~,([0, 1], X ) does not contain a copy of co for any p ~ [1, +oc). These equivalences led Hoffmann-J0rgensen to conjecture t h a t the preceding conditions would also be equivalent, to the following one (d) X does not contain a copy of co. Kwapiefi [83] showed that the conjecture was true. In 1978, four years after the publication of the papers of HoffrnannJergensen and of Kwapiefi, Bourgain [14] provided an alternative proof of Kwapiefi's theorem obtaining a more general result. Actually, we must say that Bourgain's results are formulated in a more general setting than required now, but, as we have already remarked, later on his solution will be crucial for us (see Saab and Saab's theorem 3.1.4). This is why we present here Bourgain's approach. We do not use the probability language and no particular knowledge of probability is assumed. However, its flavour is all over Section 2.1. This should not be surprising because, as we have explained, the results considered here had their roots precisely in one of the pioneering works on probability in Banach spaces. We knew about Kadec-Petczyfiski-Rosenthal's lemma 2.1.3 through an unpublished paper of Bourgain [17], and we learnt there how to apply it in the study of vector-valued functions. Although it is mentioned by passing in some places in the literature (see for instance [19]), we are afraid that it is widely ignored. However, we believe that it is a very important result which deserves a lot of attention. We have known through Bourgain and Rosenthal that this "subsequence splitting lemma" was formulated for the first time by the second n a m e d author in the spring of 1979, in the Topics Cmtrse at the University of Paris VI. T h e o r e m 2.1.4 is perhaps well known for many specialist but, as far as we know, it. appears explicitly here for the first time. The same can be said about the case p > 1 in Theorem 2.1.5. Some of the particular cases of T h e o r e m 2.1.4, which we give in its Corollary, may be found included in .~ome proofs in the literature (see, for instance, [108], [69] or [10]). Also in 1978, Pisier, answering a question of Diestel, showed that, for 1 < p < +oo, L v ( p , X ) contains a copy of gl if and only if X does. He remarked in his paper: "soon after m,.] proof, B. Maurey gave another prvof of the result giving the following improvement (he is referring to theorem 2.2.1); I could verify that my proof gave the same extension, too'. Since Bourgain also asserts in [16] that he got independently the same result, we have called it Maurey-Pisier-Bourgain theorem. Bourgain's proof is quite complicated, via an averaging result for gl-sequences. Our proof follows very much Pisier's approach. We must say that Batt and Hiermeyer's proposition 2.2.1 has a clear antecedent in Proposition 1 of [101], which in turn, in Pisier's own words, is

2.3 Notes and Remarks

63

"just a r'eJormulation of a remark of Stegall and Odell" [112, Addendum, p. 377]. It is one of the main tools used by Pisier to prove his theorem. In our fbrmulations of Bourgain's Theorem 2.1.5 and Maurey-PisierBourgain's Theorem 2.2.1, we have taken advantage of the measurability results obtained in Section 1.6. In fact, in the original formulations of these theorems, no assertion about the measurability of the the set of all points a~ C J? for which (f~,(co)) has either a c0-subsequence or an (1-subsequence was made. We thought natural to study this, and so we were led to Proposition 1.6.3 All Section 2.2 is clearly connected with the problems of weak compacthess, weak convergence, and gi-sequences in Lp(#,X) spaces. We have to point out that Talagrand's fundamental paper [127] must be studied by anyone interested in such problems. In particular in [127] one can find much stronger resiflts than our simple lemma 2.2.1. Besides Talagrand's paper, the following papers should also be mentioned: [129], [38], [9], [30] and [33]. Once we have touched on weak compactness in Lp(#, X) spaces one can not help wondering about stron9 compactness in these spaces. In this line we have some quite recent papers: [62, 63], [2] and [32]. Finally, we would like to call the attention here to an important line of research which has some connection with the problems studied in this Chapter: vector-valued versions of results on basis properties of Lp (#) spaces. Recall that ibr 1 < p < +oc and # finite: (a) Lp(#) has an unconditional basis. (b) Every sequence of non zero elements in Lp(#) which is a martingale differen(:e sequence is an unconditional basic sequence. Diestel and Uhl [39] asked if Lp(#, X) has an unconditional basis whenever X has, and in general they asked for vectorial versions of these results. The first question was soon negatively answered by Aldous [1], who found a necessary condition for L~)(#, X) to have an unconditional basis. We believe that at the moment it is unknown whether it is sufficient or not. Concerning ~he second question, we (:an say that the Banach spaces for which assertion (b) is true have been extensively studied (see [21, 22, 23]). They at'(; the so-called UMD spaces, which turned out to be the natural context in m a n y problems in analysis.

3. C (K, X) spaces

After the flmdamental results on Lp(p, X) spaces of the preceding chapter, let us see now which is the behavior of C(K, X) spaces. We will s t u d y the contributions of E. and P. Saab, P. C e m b r a n o s and F. Freniche, and L. Drewnowski. T h e y provide a complete answer to the problems we are dealing with. T h e questions: "does Lp(#) have a copy of co, g-1 or g~o'?", "does Lv(#) have a c o m p l e m e n t e d copy of Co or 61?" have quite an easy answer, and it does not depend on the measure space (~2, ~ , p) (remember we are always assuming Lp(p) infinite-dimensional). T h e situation is completely different for C(K) spaces. We know t h a t C(K) always contains a copy of co (see Section 1.4), and we also know t h a t it never contains a complemented copy of 61 (this is not difficult and it will be shown in the next section), but if we wish to know whether C(K) contains a copy of gl, or a copy of 6oc, or a complemented copy of Co, then the answer will be: "It depends on the c o m p a c t space K. For some of t h e m the answer is yes, and for some of t h e m the answer is no". Of course, we would like to know for which c o m p a c t spaces the answer is "yes", t h a t is, we would like to have a charaterization of these facts t h r o u g h topological properties of K . In this line we have a satisfactory answer for copies of 61 ( T h e o r e m 3.1.1), but for complemented copies of co and for copies of too, the problems are still open.

3.1 Copies and complemented copies of

E1

in

C(K, X)

Let us begin characterizing when C(K, X) contains a copy of fl. We should first know the situation in the scalar case. For this reason we have to recall quite a classical and well known result due to Pdczyfiski and Semadeni [100]. To u n d e r s t a n d it we need to remember t h a t a c o m p a c t Hausdorff space is said to be scattered (or dispersed) if each of its closed subsets have isolated points (see [84] or [124]). T h e simplest example of a scattered c o m p a c t space is the Alexandroff compactification of the natural numbers, and in general any countable and c o m p a c t space. A typical example of a non scattered c o m p a c t space is [0, 1]. T h e tbllowing t h e o r e m is just [84, Section 13, Corollary to T h e o r e m 4] (see also [124, Section 8.5]).

66

3. C(K,X) spaces

Theorem

[100], 1 9 5 9 ) . Let K be a corn-

3.1.1 ( P e 8 9

pact and Hausdorff space, then C(K) contains a copy of ~1 if and only if K is not scattered Now we can s t u d y the vector-valued case. The following result is p e r h a p s well known and very old, but we have only found it in [25]. 3.1.2. Let K be a compact and Hausdorff space, then C ( K , X ) contains a copy of ~1 if and only if either X contains a copy of ~1 Of [( iS not scattered (equivalently, C(K) contains a copy of gl).

Theorem

Pro@ Let us show the non trivial iinplication. Assume K scattered and let us suppose t h a t C(K, X ) contains a copy of fl. Since a Hausdorff space which is the range of a continuous m a p defined on a c o m p a c t scattered space is also a c o m p a c t scattered space [124, II.8.5.3.], an old technique (see for e x a m p l e the p r o o f of [51, VI.7.6.]) allows us to assume also t h a t K is metrizable. B u t it is well known t h a t a scattered metrizable c o m p a c t space must be countable (see [124, 8.5.5.] or [84, Section 5, Corollary 2]), so K must be countable. Then, p u t K = {tin : m E N} and let (r be an fl-sequence in C(K, X). If X does not contain ft then Rosenthal's t h e o r e m says that, for each m E N, (r has a weakly Cauchy subsequence. But then, by a s t a n d a r d diagonal a r g u m e n t , we can deduce t h a t (r has a subsequence, which we continue to denote in the same way, such t h a t (r is weakly C a u c h y for all m E N. Therefore, by Corollary 1.7.1, (r is weakly Cauchy in C(K, X). Of course, this contradicts the fact t h a t it is an t~l-sequenee. T h e p r o b l e m of characterizing when C ( K , X ) contains a complernented copy of 61 was solved by E. and P. Saab [116]. To u n d e r s t a n d E. and P. Saab's proof let us begin thinking how to prove t h a t C ( K ) does not contain a complemented copy of f-1 ( r e m e m b e r t h a t m a n y C(K) spaces do contain copies of ~ , for instance C([0, 1]), see also Theorein 3.1.1 above). P e r h a p s the simplest proof of this is the following: Let us suppose t h a t C(K) contains a c o m p l e m e n t e d copy of ~1. T h e n C(K)*, the dual of C(K), which we identify with the space rcabv(K) of all regular measures of b o u n d e d variation on the Borel subsets of K (see Section 1.7), contains a copy of f-oo, and theretbre it contains a Co-sequence (p,,~). Take

k= 1

where II tLa. II denotes the variation of #k. R a d o n - N i k o d y m t h e o r e m allows us to consider (#,~) as a sequence in LI(A) and so we would have a copy of co in L1 (A). Of course this is a contradiction (use for example that, L1 (A) is weakly sequentially complete).

3.1 Copies and complemented copies of fl in C(K, X)

67

Observe now t h a t the preceding proof generalizes easily to prove that if

C(K, X) contains a complemented copy of fl then X must contain a complemented copy of 61, for spaces X whose dual X* enjoys Radon-Nikodym property. It is enough to apply K w a p i e f i ' s Theorem 2.1.3 and the classical Bessaga-Petczyfiski theorem asserting that a Banach space X contains a complemented copy of f-1 if and only if X* contains a copy of co (Theorem 10, Chapter V of [35]). E. and P. Saab realized that it was possible to overcome the restriction on X following carefully the representation of the dual of Lp(p, X ) provided by lifting theory (see Section 1.5), and using Bourgain's Theorem 2.1.1 instead of K w a p i e f i ' s Theorem 2.1.3. We believe t h a t these are the main points in Saab & Saab's approach. We have summarized it in the following result, which is not explicitly stated in [116], but it is clearly contained in tile proof given there. At first., one might be surprised by the statement of the next theorem just because it seems it does not have much to do with C(K, X). However, notice that the dual of C(K, X), which we identify with rcabv(K, X*), is just a subspace of cabv(B(K), X*), a particular case of the space involved in the statement. On the other hand, we must say that the full generality of our statement will be used in Section 4.1 (Theorem 4.1.1). 3.1.3 (E. a n d P. S a a b [116], 1982). L e t X be a Banach space, then cabv( ~, X*) contains a copy of Co if and only if X* does9

Theorem

Proof. Assume cabv(~, X*) contains a copy of co. Let, (rnn) be a c0-sequence in cabv(~, X*). Then there are C1, C.) > 0 such t h a t C1 max l ai i

I <- - II ~

a~,,,,i

It <- - C., max l(~i I i

i ~c

1

fbr all (ai) E Coo. Take A = ~ , . l p: II ~,~ II. It is clear that (m,~) is a sequence in cabvx(i2, X*), and we can assume A to be complete (if necessary we A-complete de a-field 2 and we extend by zero the m~'s). Take a lifting p in Lo~(A), and use Theorem 1.5.2. We get a sequence ( ~ ) = (fi(m,.)) in s (.~, X*) satisfying:

i

9~ s z

'

i

i

for all (a i) C Coo. Then we deduce from Bourgain's Theorem 2.1.1 that X* must contain a copy of co (recall the proof of Theorem 2.1.2). An immediate consequence of the preceding result is the following

T h e o r e m 3.1.4 (E. and P. Saab [116], 1982). C ( K , X ) contains a complemented copy of ~,1 if and only if X does.

68

3.

C(K,X) spaces

Pro@ If C(K, X) contains a complemented copy of ~1 then rcabv(K, X*), the dual of C(K, X), contains a copy of ~oo, and therefore, a copy of Co. Of course, this implies that cabv(g(K), X*) (Drcabv(K,X*)) also contains a copy of co. The preceding theorem implies that X* contains a copy of co. But, by the classical result of Bessaga and Petczyfiski mentioned above (Theorem 10, Chapter V of [35]), this means that X contains a complemented copy of ~1.

3.2 Complemented

copies

o f Co i n

C(K, X)

We have seen that C(K) (and therefore C(K,X)) always contains m a n y copies of Co (see Section 1.4). However, what can be said about complernerzted copies of Co? Notice first that in general C(K) does not contain complemented copies of Co. For example take g~, which is just C(~N), where as usual, /3N denotes the Stone-Cech cotnpactification of the natural numbers. Phillips classical theorem (Corollary 1.3.2) guarantees that C(r does not contain complemented copies of Co. In general, to get a C(K) space with no complemented copies of Co, it is enough to take any extremally disconnected compact space K , because the corresponding C(K) is a Grothendieck space (see [35] and the Notes and Remarks of this Chapter). On the other hand, we m a y find lots of Banach spaces not containing e0 at all. However, we will see in the next theorem that C(K, X) always contains a complemented copy of co (remember that we are excluding the trivial cases of finite dimensional spaces C(K) or X). Of course, as we have already mentioned, this was quite surprising and provided the first negative answer to Problem 2 in tile Introduction. The result was obtained independently by Cembranos and Freniche. It is worth to mention that the main ingredient in its proof is Josefson-Nissenzweig Theorem [35, Chapter XII]. We have found the proof we give in Bombal's paper [5]. Theorem

3.2.1 ( C e m b r a n o s - F r e n i e h e

[26]

[59], 1984).

The Banach

space C(K', X) always contains a complemented copy of Co. Proof. By Josefson-Nissenzweig theorem, there exists a normalized weak*null sequence (.z.,~) " "* in X * , Take a bounded sequence (x,~) in X such t h a t z~.(:~:,) = 1, and with this sequence (z~), let us repeat the construction of a c0-sequence we (lid just before Proposition 1.4.2. We get a c0-sequence (g,~) in C(K, X) such that 11.9,~ II = II.q,~(tn) II tor all n E N, where (t,~) is a sequence in K such that

g~(t,~) =

z,,,

if'n=m

0

otherwise

3.3 Copies of ~o~ in C(K, X)

69

Since (x;.) is w*-null, it converges uniformly to zero on the compact subsets of X, and so

c(/(,x)

--~ c(/(, x)

r

--~ ~ x;(r

is a well defined continuous linear projection onto the closed linear span of the (g,~)'s. This completes the proof.

Remark ,?.2.1. It should be pointed out that if we wish to use the language of Proposition 1.1.2 in the preceding proof, we just have to realize that (~t~(.)x,*~ is a w*-null sequence in C ( K , X)* satisfying <.q~.~t.,(.)zm > = < g . ( t , . ) , z ; ~ >

= ~.m < z n , z , ~ >

= d,~.~

for all n , m 9 N, where of course the g.,~'s, xn's, t,,'s and x m*'s are the ones of the preceding proof.

3.3 Copies of s

in

C(K, X)

Finally, we wish t o know when C(K, X ) contains a copy of ~ . The problem was solved by Drewnowski. We will need to apply very carefully all the power of the results of Section 1.3. T h e o r e m 3.3.1 ( D r e w n o w s k i

[46], 1990). C ( K , X )

contains a copy of

~.oo if and only if either C ( K ) or' X does. Proo.f. The condition is of course sufficient. Let us show that it is also necessary. To do this let us observe first the following fact: Given a compact HausdorJf space K and a Banaeh space X , if T " goo --4 C ( K , X ) is a bounded linear operator such that T((
(~T(e,~)(t)

rl=l

for' all t E K and all ((n) 9 e~, then lira T(~.,~) = O,

where (e~) denotes the canonical unit vector sequence in too.

70

3.

C(K,X) spaces

This fact may be proved using Remark 1.3.1, but instead of this result we will use Orlicz-Pettis Theorem ( [35, page 24]). Let T : go~ --+ C(K, X) be a bounded linear operator such that oO

T((r

= ~

r

for all t E K and all (r E go~. Then for each ((~) E goo, the bounded k k sequence ( ~ 1r = (T(~,~=I r is pointwise convergent to T((r and by Proposition 1.7.1, it is weakly convergent, too. This means t h a t the series ~ (.~T(e,~) is weakly convergent for all ((,~) E g~. Therefore, by Orlicz-Pettis theorem ( [35, page 24]), the series ~ T(e~) is unconditionally convergent. Of course this implies that limn T(e,,) = 0. Our aim now is to show that. if C(K, X) contains a copy of go~ but neither C(K) nor X does then it is possible to find a closed subset K0 of K and a bounded linear operator T : g~ -4 C(Ko, X) such that

T((r

=~ r 7~-- I

for all t E K0 and all (r

E goo, but

IIZ(e~) 1174 0. This is a contradiction with the fact we have just shown and our proof will be complete. So, let us assume that C(K, X) contains a copy of go~ and suppose that, neither C(K) nor X contain a copy of g-oo. Let J : goo --+ C(K,X) be an isomorphic embedding, and let us denote f,~ = J(e~). For each t E K consider the bounded linear operator

Jr: go~ (r

--9 --+

X J((r

Since we are assuming that X does not contain g~, it follows that all these operators are weakly compact and that ~ Jt(en) = ~ f~(t) is an unconditionally convergent series for all t r K (see Corollary 1.3.1 and remark 1.3.2), that is,

Z r

is convergent

(3.1)

for all (r r goo. Let D be a countable subset of K such that for each n E N there exists t , r D such t h a t

3.3 Copies of goo in

C(K, X)

71

Since (Jt)eeD is a countable family of weakly compact operators defined on f ~ , it follows from Corollary 1.3.3 that there exists M E [N] (let us recall that [ ~ is the set of all infinite subsets of N) such that oo

71,=1

for all ((.,~) E f ~ ( M ) and all t E D. Of course, goo(M) is isometrically isomorphic to g~o. So, without loss of generality, we may assume that this equality holds tbr every ((,~) E ~oo and all t E D. But this means that

J ((r

(t) = ~ (,~f,~(t) n

(3.2)

1

whenever (<,~) E goo and t E D. Let, K0 be the closure of D in K and let Xo 00 be the dosed linear span of U~=lfk ( K 0)- Given ((~) E goo, notice that 3.2 implies t h a t J (((~)) (t,) E X0 for all t E D. Hence, by density and continuity, we have J ( ( ( , ) ) ( t ) E Xo

(3.3)

for all t E /40. Since Xo is a separable subspace of X , we can take now a w*-dense sequence (x~) in B(X~)). Thanks to the Hahn-Banach theorem we can extend these functionals and assume (x~) C B(X*). For each j let us consider the operator Jj:

t,~

)

(r

,

C(K) .T; o J ((r

By Corollary 1.3.1, all these operators are weakly compact. And, using again Corollary 1.3.3, we deduce that there exists M E [N] such that

J, (((~)) = ~ (,~4j(en) for all ((,~) E g~(M) and all j E N. As belbre, may assume M = N, and with this in mind the preceding equality means oo

xj o J ((.(~)) =

(,. (xj o J(e,~)) n

1

for all ((n) E g-oo and all j E N, where notice that the convergence of the series is in the norm of C(K). In particular, using 3.1, it follows that given ((n) E g ~ and t E Ko

72

3. C(K,X) spaces oo

.,~ X"~ (J ((r

(~))

=

~

=

Xj

(,:j * (J(~)(t))

n

en

for all j C N. Therefore, it follows from 3.3 and the w*-density of the x~'s that J (((,d) (t) = ~

(,~f,dt).

(3.4)

Now, let R be the restriction operator from C(K, X) in C(Ko, X) and take ,I:foo --+ C(Ko, X). We have

T = R o

IlT(en) ll

= =

l]Ro,l(e,,,)ll= IlRfnll IIf,~(t,,)ll = Ilfnll = IIJ(en) H ~ 0,

and by 3.4, for each ((n) E ~o and each t C K0 we also have (DO

T (((,,)) (t)

=

R o J(((,,))(:)

=

(,~R(f,d(t)

Z n=l

oo

=

Z

oc

(,,(R o ,J)(en)(t)

n=l

=

~

(.r(e,,)(t)

n=l

This is the desired contradiction and the proof is complete.

3.4 N o t e s a n d R e m a r k s Saab and Saab's theorem 3.1.3 is not true for cabv(5, X): Talagrand in [127] provides an example of a Banach lattice which does not contain co such that cabv(Z, X) even contains goo. Ferrando in [57] studies when cabv(5, X) contains goo. Recall that if the Banach space X is a dual then it contains ~oo if (and only if) it contains co [35, Chapter V, Theorem 10]. Therefore, Saab & Saab's Theorem 3.1.3 may also be read as follows: "if X is a dual space then cabv(Z, X) contains goo if and only if X does". Of course, Talagrand's example just mentioned shows that the preceding result is not true for general Banach spaces. Several results contained in this monograph have been extended to tensor products. This is specially true with Cembranos-Freniche's Theorem 3.2.1. Since it was published several extension have been given (see for instance [117]).

3.4 Notes and Remarks

73

We have to point out that in the proof of Drewnowski's Theorem 3.3.1 some ideas introduced by Rosenthal in [110] and by Kalton in his famous paper [78] on spaces of compact operators are crucial (and part of them had already been included in section 1.6). No good characterizations of when C(K) contains a copy of f~o or a complemented copy of co are known. But we have some important partial results. For instance we have Theorem

(Rosenthal

[39, T h e o r e m

V I . 2 . 1 0 ] ) . If K is extremally dis-

connected then C(K) contains a copy of ~o. Recall that K is extremally disconnected (or Stonean) if the clausure of each open is open. Easy examples of extremally disconnected compact Hausdorff spaces are the Stone-Cech compactification of discrete infinite topological spaces. These spaces are very important in Banach space theory an in particular they are important in the problems we are considering. Grothendieck proved t h a t if K is one of t h e m then each bounded linear operator from C ( K ) in co is weakly compact [39, Corollary VI.2.12]. As an immediate consequence of this result we obtain the following

If K is extremally disconnected then C(K) contains no complemented copies of Co.

Theorem.

The preceding result is also an easy corollary of another important fact [84, Section 11, Corollary 2]: C(K) is injective whenever K is extremally disconnected. It is enough to observe that complemented subspaces of an injective space must be injective, while Co is not. The aforementioned Grothendieck's result as well as some other related ones led to the introduction of the tbllowing definition [39, pages 156, 179]: we say t h a t a Banach space X is Grothendieck if weak and weak* convergence of sequences in X* coincide. Notice that thanks to the identification between w*-null sequences and Co-valued continuous linear operators (see Proposition 1.1.1 and [35, Chapter VII, Exercise 4]), this condition just means t h a t each bounded linear operator from X in co is weakly compact. So Grothendieck's result may be read as follows: Theorem.

If K is extremally disconnected then C(K) is a Grothendieck

space. Clearly there are m a n y Banach spaces with no complemented copies of co which are not Grothendieck spaces (think for instance in ~1). Nevertheless this is not the case for C(K) spaces because we have

74

3. C(K,X) spaces

C(K) is a Gvothendieck space if and only if it does not contain any complemented copy of Co.

Theorem.

Pro@ If C(K) contains a complemented copy of co, then it can not be Grothendieck, just because dearly Co is not Grothendieck. Conversely, if U(K) is not Grothendieck, then there exists a w*-null non w-null sequence (p,~) in C(K)*. But, thanks to the identification between w*-null sequences and c0-valued continuous linear operators we have mentioned above, this means t h a t the continuous linear operator T associated to (#~) is not weakly compact. So, by [39, VI.2.17.], there exists a bounded sequence (f~) in C(K) such t h a t f,~. f,~ = 0 for m r n, and lim~ T(f,~) r 0. It is immediate that the series ~ f,~ is w.u.C. (it is clear that (f~) is even a c0-sequence). On the other hand, it is easy to show that the condition limn T(f,~) r 0 implies that there are subsequences of (#,.) and (fn), which we continue to denote in the same way, such that #,,(fi~) 7# O. So, Theorem 1.1.2 guarantees that (f,,~) has a complemented co-subsequence. This completes the proof. Those compacts K such that C(K) is Grothendieck are some times called G-spaces. Information about these spaces may be found in [123]. For a time it was not known whether each Grothendieck C(K) space must contain a copy of g~. Haydon [68] constructed a G-space K such that C(K) does not contain copies of goo. However, this C(K) does admit goo as a quotient. Talagrand, using continuum hypothesis, constructed a G-space K such that the corresponding C(K) does not even admit ~oo as a quotient. Extremally disconnected compacts satisfy the following two properties: (a) T h e y do not contain copies of goo, and (b) They do not contain complemented copies of Co. So, one could thing there is some relationship between these two properties in C(K) spaces. This is not the case. We have the following examples: 1. If K = [0, 1], then C(K) Dc0 and C(K)73eoo. (,) 2. If K = / 3 N x/3N, then C(K) Dc0 and C(K)Dgcr (,) 3. If K is any of the compact spaces given by Haydon or Talagrand, then C(K) 73 co and C(K)73goo.

(c)

4. If K =/3N, then C(K) 73 Co and C(K) = g.~. (c)

4. Lp(tt, X)

s p a c e s

We have already studied in Chapter 2 when Lp(t.t , X ) , for 1 _< p < +oo, contains copies of co or gl. We wish to consider now complemented copies of these spaces, and also copies of f-oo (remember that for this last space copies and complemented copies are the same). The case p = oo is postponed until next Chapter. Let us begin with the gl case.

4.1 When o f ~l ?

does

Lp(tt, X)

contain

a complemented

The first result is a consequence of Saab & Saab's theorem provides an answer to our question in a particular case.

copy

3.1.3 which

[6]). Let (~, Z, >) be a finite measure space and let 1 <_ p < +oo. If Lp(>, X ) has a uniformly bounded complemented glsequence then X contains a complemented copy of gl.

T h e o r e m 4.1.1 ( B o m b a l

Proof. Let (fn) be an uniformly bounded complemented gl-sequence in Lp(>, X). By the classical perturbation theorem (see [93, Proposition 1.a.9.] or [35, Theorem 12 of Chapter V]) we may assume that the f~'s are uniformly bounded simple functions. Let us denote f,~ = y'~.iXa~ (.)zp. We will assume the x?'s are in the unit ball of X. Let ~ Fn be a w.u.C, series in Lp(#, X)* such that F,~(f~) = 1 for all n EN. Let r be the canonical map defined in remark 1.5.3, and let us denote ~(F,~) = m~. Of course, y~. m,~ is a w.u.C, series in cabv(Z, X*), and for each n EN we have Ilrn'~ll

=

Ilrn~(A~)ll -> E

Ilrn,~H (D) >- E i

=

r 4

i =

=

mn(Ap)(xn)

76

4. Lp(#,X) spaces

Therefore c a b v ( Z , X * ) contains a copy of co. So, we deduce from E. and P. Saab's theorem 3.1.3 that X* contains a copy of co, too. Of course, by Bessaga-Petczyfiski's classical theorem (Theorem 10, Chapter V of [35]) this means that X contains a complemented copy of gl. Remark 4.1.1. It is interesting to know Bombal's original proof of the preceding result [6]. It is essentially the following: Assume our measure space is [0, 1] with the Lebesgue measure, and let (fn) be a a uniformly bounded complemented gl-sequence in Lp([0, 1], X). By Lusin's theorem and the classical perturbation of basis theorem, we may assume the f~'s continuous (and uniformly bounded). Considering the canonical injection from C([0, 1],X) into Lp([0, 1], X), it is clear that (fn) is also a complemented gl-sequence in C([0, 1], X). Now Saab & Saab's theorem 3.1.4 guarantees that X contains a complemented copy of gl, as desired. For simplicity, we have exposed the proof for Y2 = [0, 1], however we can use the same idea with any compact K instead of [0, 1], and in fact, using Kakutani's theorem [40, Section 18, Theorem 2], we have the result for all finite measure spaces. We have followed a different approach, because we have preferred to avoid such an structural theorem as Kakutani's. Now we wish to extend Bombal's theorem to all fl-sequences. To do this we will need the following lemma. L e m m a 4.1.1. Let (~7, Z , #) be a finite measure space and let 1 < p < oo. If (fn) is an gl-sequence in Lp(#, X ) then (fn) has a normalized block sequence (f88 such that sup IIf:,~(co)[I < oo ?lZ

for almost all co C ~2. Pro@ Let (fn) be an el-sequence in L v ( # , X ). The set {11 fn(.)I1: n e N} is bounded in Lp(#) and therefore it is relatively weakly compact. So, it is relatively weakly compact in LI(#), too. Let h E L I ( # ) be a w-cluster point of {li/n(-)ll: n e N}. Then h is in the w-closure of {11 f~(.)I1: n >_ m} for every m C N. But Mazur's theorem (see for instance Corollary 2, Ch. II of [35]) implies that for every m r N, h must be in the (norm) closed convex hull of {H f,~(.)II: n _> m}, too. Therefore, there are an increasing sequence 1 = k l , . . . , kin,.., of natural numbers and a sequence (r in LI(#) (norm) converging to h and such that 6m is a convex combination of {ll fkm (') l[, " " " , ll fkm+l-l

(') ll}

for each m E IN. Moreover, extracting a subsequence if necessary, we may suppose that (6~) converges to h almost everywhere. Hence sup I Cm(w ) 1< oo 79Z

4.1 When does Lp(#, X) contain a complemented copy of fl ?

77

for almost all co E .(2. Finally, if km + 1 -- 1

),~ It f{(.) II, X"~k''~+l-1

with Xi > 0, and z-,i=k,,~ _

Ai = 1, we define

kin+l--1

gm =

E

-Xifi

and

f,',~--

g,,~

i=k.~

for all ~rT.C N. Since (f,~) is an gl-sequence, it is clear t h a t Hgm lip is b o u n d e d away from 0, and so, it follows t h a t (f[~) satisfies the imposed requirements. At this m o m e n t we can answer the question of the Section. Theorem

4.1.2 ( M e n d o z a

[97], 1992). For 1 < p < oc, L p ( p , X ) con-

tains a complemented copy of gl if and only if X does. Pro@ By T h e o r e m 1.6.1, we m a y assume (~2, Z , #) is a finite m e a s u r e space. Let us prove the non trivial implication. Let (f~) be a c o m p l e m e n t e d flsequence in L~(t,,X ). By the preceding l e m m a and well known results on block basis (see for instance 2.a.1. of [93]) we m a y assume sup II f,~(co)l1 < c>o

(4.1)

?Z

for allnost all co E ~2. Since (f,~) is conlplemented, there is a w.u.C, series ~.,,. F,,. in Lj,(t*, X)* such t h a t < f,~,/2,, > = 1

(4.2)

for all 'H, E N. Using T h e o r e m 1.5.4 on the representation of Lv(#, X ) ~, we know t h a t there are w * - m e a s u r a b l e functions tp,~ : (2 --+ X* such t h a t for each 7~, E N, II~',,(.)II is measurable, it. belongs to 12q(#) and < . f , / ~ > = ./~ < f(co), ~P,~(co)> dp(c~) for all .f E L~,(I*, X). Therefore,

4.2 means t h a t

./~ < f,~(co),~,,(~) > dl~(w) = 1

(4.3)

ti)r all it, E N. Now we have two possibilities: (a) (< .f,~(.), q),,(.) >) C L 1 (tl.) is not uniformly integrable. In this ca.se, we deduce from 1.2.5. of [39] t h a t there exist e > 0, a sequence (A,.) of pairwise disjoint m e a s u r a b l e subsets of ~2, and a subsequence of (< f,,.(.), ~,,.(.) >), which we continue to denote in the s a m e way, such that.

4. Lp(p,X) spaces

78

fA

[.

< f~(~)' #~(~) > d , ( . )

=

[_ < XA. (~)f~(~), #.(~) > dp(w)

=

I< XA~(')fn('),Fn >l

n

d~2

> e

for all n E N. Since Y'~nFn is a w.u.C, series in Lv(#, X)*, the preceding inequality and Theorem 1.2.1 imply that (XA~ (.)fn(.)) has an gl-subsequence. But this is a contradiction because (XA~(.)f~(.)) is a seminormalized disjointly supported sequence in Lp(#,X) and therefore, it is an gp-sequence (Proposition 1.4.1). Thus we have to assume (b) (< f,~(.),kan(.) >) C LI(#) is uniformly integrable. Then, for each k E N put (2~ = { w E Q:llf,~(wDll_) it follows that there exists k0 9 N such that < f~(~), ~,~(~) > d#(cJ) > 1/2

/~ ko

for all n 9 N. Since ~ / 7 , ~ is a w.u.C, series in Lp(#, X)*, according to Theorem 1.2.1, (Xr~0 (.)f~(.)) has a complemented gl-subsequence in Lp(#, X). Now we deduce from Bombal's Theorem 4.1.1 that X has a complemented ~l-sequence.

Remark ~.1.2. We would like to point out that Bourgain's theorem 2.1.1 is on the basements of all this section. In fact, Theorem 4.1.1 (and therefore, Theorem 4.1.2) lays on Saab & Saab's theorem 3.1.3, but recall that this result in turn lays on the aforementioned Bourgain's theorem.

4.2 When

does

Lp(p, X)

contain a copy of ~?

The tbllowing result and its proof is strongly inspired by Drewnowski's Theorem 3.3.1 and so it borrows several Kalton's ideas. We will need again all the power of the results of Section 1.3. We will also use the preliminary work we have made in Section 1.6.

Theorem 4.2.1 (Mendoza

[96], 1990). If 1 <_ p < oc then Lp(p,X)

contains a copy of g.~ if and only if X does. Proof. The "if' part of the result is of course trivial. For the converse recall that we may assume (~?, ~ , p) separable and finite (Theorem 1.6.2). Let (Ak) be a sequence of measurable sets generating Z (see the beginning of Section 1.6). Let us suppose that Lp(#, X) contains a copy of goo and that X does not. Let ,] be an isomorphic embedding from go~ into Lp(p, X) and for each k E N define

Lp(tt, X)

4.2 W h e n does

Jk:

g~

---+

c o n t a i n a copy of g ~ ?

79

X

---+ /

J

a#

JA k

By Corollaries 1.3.1 and 1.3.3 there exists M r [N] such that, oo

Jk ((r

r

= E lzzl

tbr all ((~) C g ~ ( M ) and all k C N. We may assume, without loss of generality, M = N, and so we have oo

lZ z

i

for all ((n) E gee and all k C N. Let X0 be a closed separable subspace of X such that J(e,~)(t) C Xo #-almost everywhere for every n. Take ((.,~) r gee and k E N. We have ~J ((r '

Jk

~k

~'z=l

n=l

k

Therefore, L 9

E Xo

J ((~r~))d#

for all k 9 N

(4.4)

k

On the other hand, it is straightforward to show that r0={A 9

~: /' J((i,,,))dp 9 /A

is a sub-c,-algebra of 5 , and so, since 5 is generated by the Ak's, 4.4 means that, 50 = 5 . In other words, it follows that

./A J((C**))d# 9 Xo

for all A e Z.

Then we have (see for instance [85, X.5 Theorem 5]) J ((4,,,))(t) 9 X0

/t-almost everywhere,

and so

J

9 L,(#, Xo)

for every ((,,,) r foe. But this implies that J is an isomorphic embedding from ,0~ into the separable space Lp(#, Xo). This contradiction finishes the proof.

80

4. L v ( # , X ) spaces

4.3 Complemented

o f Co i n L p ( t t , X )

copies

In this section we wish to characterize when Lp(>, X ) contains complemented copies of e0. For the first time, it will be important to distinguish whether the measure space (27, ~ , >) is purely atomic or not. Let us begin studying the purely atomic case. L e m m a 4.3.1. Let 1 < p < +oc, let ~ xk be a w.u.C, series in [p(X), with (xk) = ((xkn)~)k, and let (%) = ((x~n),)k be a bounded sequence in eq(x*) = + 1 = 1). Then

lz - - T/~

Proof. If the conclusion were not true, then there exist e > O, an increasing sequence (m~:) of natural numbers and a subsequence of (Tk (Xk)), which we still denote in the same way, such that rnk+l

n~z

--i

k

for each k E N. Let us define the sequence (Zk)k = ((Zk~)n)k in [p(X) by

zk~ =

Xkn 0

if m k < n < mk+l otherwise

Notice that mA~+l --1

'lt:'fl~

,~,

Therefore the disjointly supported sequence (zk) is seminormalized. And so (see Proposition 1.4.1), it is an g~,-sequence. But Theorem 1.1.1 implies that (za:) has a e0-subsequence. Hence a contradiction.

Remark 4.3.1. Notice that the conclusion in the preceding lemma simply means that {(x;,.~(zk,~))~ : k r N} is a relatively compact subset of [l (see [51, IV.la.3.] or [35, Exercise 6, Chapter I]). T h e o r e m 4.3.1 ( B o m b a l [8], 1992). get (J2, Z , > ) be a purely atomic measure space and let 1 <_ p < oo, then L p ( # , X ) contains a complemented copy o/ Co if and only if X has the same property.

4.3 Complemented copies of co in Lv(#, X)

81

Proof. Let us show the non trivial implication. By L e m m a 1.6.1 we m a y assume (Y2, Z, #) or-finite, and by Proposition 1.6.4, Lp(#, X ) is isometrically isomorphic to gv(X). So all we have to show is the following assertion: "if (~v(X) contains a complemented copy of Co, then so does X " . Then, assume t h a t ~.p(X) contains a complemented copy of Co. In view of Proposition 1.1.2 there are a c0-basis (xk) in gv(X), where xk = (xk~)~ for each k E N, and a w * - n u l l sequence (%) in gq(X*) = gp(X)*, with rk = (x[m)n, such that X* rk(z.j) = E.,a:n(:r, jn) = 5kj

(4.5)

?Z

Notice that, *'or each n E N, ~ k xk,~ is a w.u.C, series in X and ( x ~ ) k is a w * - n u l l sequence in X*. So, thanks to Theorem 1.1..2, the only thing we must show is that limk xk, ~'* (zkm) # 0 for some m E N. But, if this is not the case, the preceding l e m m a implies that oo

lim%(xk)k

=

lip ~ z~.~(zk~) = 0, '~'z= 1

which clearly contradicts 4.5. This completes the proof. Let us see now which is the situation in the non purely atomic case. We have the following result of Emmanuele (see also [15]):

T h e o r e m 4.3.2 ( E m m a n u e l e [52], 1988). Let (g2, 5, p) be a non purely atomic measure space and let 1 <

= (Snm for all 7~, m E N, and let (r,~) be the sequence of Rademacher functions. It is immediate that (r,,(.)x,) is a c0-sequence in Lv(#, X ) , and in fact, in example 2.1.1 we studied a slightly more complicated situation. On the other hand, let us show t h a t (r',~(.)z*,.) is a w*-null sequence on L p ( # , X ) * . Notice first that if .f belongs to L1 ([0, 1], X ) , then / , .1

li~n / r.,~tt)ftt) dt = 0 /0

(4.6)

This is immediate *br simple functions because (r,~) is w*-null in L1 ([0, 1])* = s and then, by density, we can extend the result to all functions in Ll ([0, 1], X). Hence, for each .f E Lv(p, X ) we have

82

4. L~,(#,X) spaces

[ < f, r,~(.)z~*~ >1

=

Ii 1 < f(t), rn(t)z~ > dt

=

~

=

< /'01 rn(t)f(t) d t , z n* >

<

1

< r,~(t)f(t),z~ >

II:c,*~II

dt

fo 1 r~(t)f(t) dt

and the last sequence goes to zero, because of 4.6. Finally, it is clear that < r,~(.)x,,,,',,~(.)x;~ > = ~,,~ for all n, rn C N. So, it follows immediately from Proposition 1.1.2 that (rn(.)z,~) is complemented in Lp(#, X), and hence the conclusion. Of course, the preceding result leads to the following characterization. Theorem

4.3.3. Let (/?, Z, #) be a non purely atomic measure space and let

1 <_p < oo, then Lp(#, X ) contains a complemented copy of co if and only if

X contains a copy of Co. Proof. Sufficiency is the preceding theorem and necessity is just Kwapiefi's T h e o r e m 2.1.6 Finally, in the next theorem we summarize theorems 4.3.1 and 4.3.3. 4.3.4. Let 1 < p < co, ttten L p ( # , X ) contains a complemented copy of co ~f and only if one of the following two conditions holds:

Theorem

(a) (Y2, Z, #) is purely atomic and X contains a complemented copy of co. (b) (~?, Z, #) is not purely atomic aTtd X contains a copy of co. Re'mark ~.3.2. We will see in Section 5.1 that the preceding theorem is also true for p = +oo (at least tbr a-finite measures #).

5. The space L

(tt, X )

In the preceding chapters we have studied C ( K , X ) spaces and L v ( # , X ) spaces for 1 _< p < +oc. We devote this chapter to study Loo(#, X). For this space we restrict ourselves to consider a-finite measure spaces (#2, E, #). There are two important reasons which lead us to do this. The first one is the very definition of L ~ ( # , X). While for 1 <_ p < ec the definition of Lp(#, X) spaces is clear for arbitrary measure spaces and most (or all) authors give the same one, for Loo(#, X) the situation is completely different. There is no problem in the finite or a-finite case, but when we leave these cases some problems arise even when X is the scalar field: for Loo(#) spaces. We would like at least to attract the attention to the fact that the definitions of Loo(#) given by Dunford & Schwartz [51], A. and C. Ionescu Tulcea [75] or Rudin [115] do not coincide. Another important reason is that regarding the problems we are dealing with, for Loo(t,,X) we have a complete answer whenever the measure # is finite (or a-finite). However, for arbitrary measures, whichever definition of L~(I, , X) one takes, only quite partial answers are known at the moment. This is because the reductions to the a-finite case we made in Section 1.6 for Lv(t, , X) spaces, do not work for p = oo (or, at least, they do not work in an ingenuous way). We can add that some of the definitions of Loo(#, X) for arbitrary measure spaces involve some technicalities (for instance, the concept of locally/z-null sets) which, in our opinion, would hide the main ideas we are interested in. Therefore, in all this chapter (#2, 5,15) will be a a-finite measure space. In the Notes and Remarks we will mention some facts which are known in more general situations. Now, let us see which are our non trivial questions for L ~ ( # , X). We have already shown in Section 1.4 that Loo(t*, X) has many copies of foo, and so it has many copies of c0 a.nd fl, moreover, it has many copies of all separable spaces just because ~oo has this property. On the other hand, since t~ is a-finite, we know that Loo(#) is injective (see tor instance, [37, Theorem 4.14.]), and then it can not contain complemented copies of Co or g~, simply because they are not injective. We could also argue that Loo(#) is the dual of

84

5. The space

L~(p,X)

L I ( # ) and it is easy to deduce from Phillips' theorem (Corollary 1.3.2) that no dual can contain a complemented copy of Co. Alternatively, we could say that Loo(#) is isometrically isomorphic to a C(K) space with K extremally disconnected (see for instance [84, Section 11, Theorem 6]), and we have considered these compact spaces in Chapter 3. We mentioned there that the corresponding C(K) can not contain complemented copies of co. We also showed that no C(K) can contain a complemented copy of fl. Thus our goal here is to know when Loo(#,X) contains complemented copies of co or gl.

5.1 C o m p l e m e n t e d copies of Co in

Loo(tt, X)

When we studied complemented copies of Co in Lp(p,X) for 1 _< p < +oc (Section 4.3), we had to distinguish between purely and non purely atomic measure spaces. Now we have to do the same. Let us begin with the purely atomic case. For a purely atomic measure, Loo(#,X) is isometrically isomorphic to goo(X) (Proposition 1.6.4). Let us study this space. The first idea to do this is trying to apply the same arguments we used to study gv(X) for 1 <_ p < +oc (see Section 4.3). We will be able to apply some of them, however, it is plain that we will find extra difficulties. This is due to the fact that, likewise in the scalar case, for most purposes ~o~(X) is much more complicated than the others f.l,(X) spaces. This is particularly true when we deal with duals. In view of the results of section 1.1 (see Proposition 1.1.2 and Theorem 1.1.2), to characterize when goo(X) contains a complemented copy of Co it is convenient to know the w*-null sequences in foo(X)*. So, we need to work in g~(X)* and for this reason it is worth to recall first a few important facts about, the scalar case, that is, about g s We should remember that this space may be identified in a natural way with ha(N), the space of all scalar finitely additives nmasures with bounded variation defined on the a-algebra of all subsets of N. A particularly important subspace of ba(N) is ca(N), the subspaee of all countably additive measures. It is isometrically isomorphic to ~ and it, is norm-one complemented in t~o~ = ha(N). Thus we can decompose in a very nice way each member of ha(N) in sum of t w o measures, one of them countably additive, and the other one purely finitely additive. We will see that the same can be done in goo(X)* It is straightforward to show that Q (X*) is isometrically isomorphic in a canonical way to a subspace of ~ (X)*: each (x;) C f-1(X*) acts as a member of f ~ ( X ) * in the following way <

> =

fi

<

9>

5.1 Complemented copies of co in L~ (#, X)

85

for each (x,~) r g-oo(X). So, in all this section we will consider g-x(X*) as a subspace of g.co(X)*. Moreover, if i m : X --4 g.co(X) denotes the canonical injection given by ira(x) = ( 0 , . . . , 0 , x , 0 , 0 , . . ) m-1 times for each x e X, one can show easily that P : g-~(X)*

~

h(X*)

is a well defined norm-one projection onto h ( X * ) 9 This projection has an additional interesting property (see [102, Lemma 11.4] or [90, p. 53]): P r o p o s i t i o n 5.1.1. The projection P defined above is w*-sequentially continuous, that is, it transforms w*-null sequences in w*-null sequences (where we continue to consider g-l(X*) as a subspace of g-co(X)* in the canonical

way). Pro@ Let (Tk) be a w*-null sequence in fco(X)*. Take x = (xn)n E g-oo(X). We must show that < x, P(Tk) > --+ 0. Let us consider the continuous linear k

operator Tx : g-co

~

&o(x)

>

Since (Tk.) is w*-null in g-~(X)*, it is clear that (7k o T~)k is w*-null in g-~. Then, Phillips's lemma [35, p. 83] guarantees that oo

lira E k

<e'~'rk~

> =0,

n=]

where (en) is the canonical unit sequence in g-co (that is, the canonical c0basis). Therefore, using that T,(e~) = ( 0 , . . . , 0 , x , 0 , 0 , . . . )

= i~(x~)

~-1 times we have oo

< x, P(~-k) > rz~l

k

86

5. The space L ~ (p, X)

T h a n k s to the projection P defined above, gee(X)* is the direct sum of (in this space we have the "countably additive part" of the functionals) and k e r ( P ) , the kernel of P (in this space we have the "purely finitely additive part" of the functionals). Moreover, in view of the preceding l e m m a a. w*-null sequence in goo(X)* is sum of two w*-null sequences, one of t h e m in *71(X*) and the other one in ker(P). All we have to do is to study these two kinds of w*-null sequences. We will do this in the following two lemmas. We will see t h a t b o t h of t h e m have a "good" (although different) behavior. T h e first l e m m a m a y be found in [89, L e m m a 2.8] or [90, L e m m a 4]. T h e second one is implicitly contained in [90].

~1 ( X * )

L e m m a 5 . 1 . 1 . Let (Tk) = ((X~.,~),~)k be a w*-null sequence in goo(X)* which is contained in gl(X*) , that is,

,-k =

r el(X*)

lirn

[lxk~][ = 0

for" all k E N. Then

uniformly in k E N. Proof. If the conclusion were not, true, then there exist e > O, a subsequence of (rk), which we still denote in the same way, and an increasing sequence (ink) of n a t u r a l numbers such t h a t 77~k+ 1 --1

75= g/~,k

for each k C N. Therefore, there exists a vector (xn) in the unit ball of t?oo(X) such t h a t rnk+l--1

< x , ~ , x k*n > > e

E

(5.1)

for each k E N. On the other hand, since (rk) is w*-null, we have t h a t < t,~x,~, Xkn > = lim

lim < (t,,z~), rk > = lim k

k

k

n=l

t~ < x~, xkn > = 0 n=l

for each (tn) E t.oo. In other words, we deduce t h a t ((< x,,,x[.~ >)n)k is a weakly null sequence in gl. Hence, we deduce from Schur t h e o r e m [35, p. 85] t h a t ((<xn,:r:*k~ >)~)k is n o r m null, and so

lim Z

I<

>1 -- 0,

rz=l

which contradicts

5.1. This completes the proof.

5.1 Complemented copies of co in L~o (#, X)

87

5 . 1 . 2 . Let ~ Xk be a w.u.C, series in go~(X) and assume that (~-k) is a w*-null sequence in g~(X)* which lies in the kernel of P (t:' is the projection defined at the beginning of the Section), then Lemma

lira <xk,~-k > = 0 k

Pro@ Let ~ xk be a w.u.C, series in g ~ ( X ) and assume t h a t (vk) is a w*-null sequence in go~(X)* which lies in the kernel of P. We will suppose t h a t <xk,~-k > 74 0, and we will reach a contradiction. Since the elements of ker(P) vanish in every finitely s u p p o r t e d vector of goo(X), we have <Xk,Tk >

--~

< (Xkl,Xk2,...,Xkk--l,0,

O . . . ) , T k :>

+ < (O,...,O,xkk,Xkk+l,Xkk+2,...),Tk

>

k-~ times

=

< (O,...,O,xkk,Xkk+l,Xkk+2,...),Tk > k--~ times

Now, let us define

~:~oo

~ g~(X) n

(tk)

)

( E tkXkn)n k=l

Since ~ xk is a w.u.C, series, it is very easy to show t h a t r is a well defined continuous linear operator. Notice also t h a t if we denote by (ek) the canonical unit sequence in go~, then

~(ek) = ( 0 , . . . , 0, zkk, xkk+l, Zkk+2,...) k-1 times for all k E N., and so

< ek,Tk o4~ > = <xk,~-k > ~ 0 On the other hand, (rk o ~5) is a w*-null sequence in gc~. Therefore, we deduce from T h e o r e m 1.1.2 t h a t (ek) has a complemented c0-subsequence. But this is a well known contradiction (see Corollary 1.3.2). 5.1.1 ( L e u n g - R i i b i g e r , 1 9 9 0 ) . goo(X) contains a complemented copy of Co if and only if X contains a complemented copy of co.

Theorem

88

5. The space L~ (#, X)

Pro@ Let us show the non trivial implication. Assume that g~ (X) contains a complemented copy of co. Then, there exists a c0-sequence (xk) in go~(X) and a w*-null sequence (~-k) in g~(X)* such that <Xk,Tk > = 1 We can write and by Proposition 5.1.1 and the preceding lemma we have lim < x~,, Tk - P(~-k) > = 0, k

therefore, we conclude lim <xk, P(Tk) > = 1 k

(5.2)

Let us denote P(wk) = ( x ~ ) ~ E gl(X*) and xk = (Xk~)o C g o ( x ) . Since x~ is a w.u.C, series, it follows very easily that

Z Xnk k is a w.u.C, series for every n C N. Besides, by the w*-sequential continuity of P we have that (x~)~ is a w*-null sequence in X* for all n E N. On the other hand, it follows from Lemma 5.1.1 that

lira in

Ilzk, II

0

uniformly in k E N, and so

lira

x* > 1 = 0 ??,~m

uniformly in k E N. Therefore, we deduce from 5.2 that there exists no such that lim<xkn0, X*k~o > ~ 0 k

Now, it is enough a look to Theorem 1.1.2 to get the conclusion. Of course, another way to state the preceding theorem is the following: T h e o r e m 5 . 1 . 2 ( L e u n g - R ~ i b i g e r , 1990). Let ( ~2, Z , # ) be a a-finite purely

atomic measure space, then L ~ ( # , X ) contains a complemented copy of co if and only if X does.

5.1 Complemented copies of co in Loo(#,X)

89

Once we have solved the problem for purely atomic measures, let us consider the non purely atomic case. We will only need two results which are true for general Lo~(#, X) spaces. They were proved by Dfaz [27]. The first one is a characterization of w.u.C, series in Loo(#, X). Roughly speaking, it says that a series ~ f~ in L~(#, X) is w.u.C, if and only if it is "(essentially) uniformly w.u.C." L e m m a 5.1.3. Let (f2, X,p) be a or-finite measure space, then a series ~ f,~

in Lo~(p,X) is w.u.C, if and only if there exist a null subset Z of f2 and a positive number M such that

II E t,~f~(w)II <- M rt~ l

for all

all m

N and allc~E [ 2 \ Z .

Proof. The condition is clearly sufficient [35, Chapter V, Theorem 6]. Let us show necessity. Assume ~ fn is a w.u.C, series in L~(#, X). By [35, Chapter V, Theorem 6], there exists M > 0 such that

n=l

for all m C N. But the preceding inequality means that for each m C N and each (t~) belonging to B ( g ~ ) , the unit ball of g~, there exists a null subset Z(,~,(t~)) of f2 such that II ~

t~I~(co) II-< M

for all a~ E ~2 \ Z(,,~,(t~)). To get the conclusion it is enough to take Z = UZ(,~,(t~)), where the union runs over the denumerable set of all (m, (tn)) with m C N and (t~) C B ( g ~ ) with rational coordinates. T h e o r e m 5.1.3 (Dfaz, 1994). Let (fJ, Z , # ) be a a-finite measure space.

If Loo(#, X) contains a complemented copy of co, then X contains a copy of C0

.

Proof. Assume that Loo(#, X) contains a complemented copy of Co. Let (f~) be a co-basis in Lo~(#, X) and let (z,~) be a w*-null sequence in L~(#, X)* such that = 1 for all n C N. By the preceding lemma, there exist a null set Z of s and a positive number M such that

II E t~f,~(a~) II < M

(5.3)

90

5. The space L ~ ( p , X )

for all (t,~) E B ( f ~ ) , all m E N and all w E ~2 \ Z. Obviously, the series f,~(w) is w.u.C, for all c~ E ~? \ Z. If for some of these ~ ' s the series f,.(w) is not unconditionally convergent, then we deduce from BessagaPe~czyfiski's classical theorem [35, Chapter V, Theorem 8] that X contains a copy of co. Otherwise, let us show that we are led to a contradiction. We must assume that for all w E ~2 \ Z the series ~ f.,~(w) is unconditionally convergent and therefore [93, page 16],

Z t,~f~(w) ~,~1

is convergent for all (t~) E fo~. Thus, if (t~) E ~ the series ~ tnf,~ is pointwise convergent almost everywhere, although it is not in general convergent in L ~ ( p , X ) because Ilf~ II~ = 1 for all n. Then, for each (t~) E f ~ we can define

s((t,,)):~

x

-~

~=ltnfn(w) w

--+

0

ifw E ~ 2 \ Z otherwise

Now, 5.3 says that the measurable function S((tn)) is bounded, and moreover, it implies that

S :~ (t,~)

~ L~(it, X) ~

S((tn))

is a well defined continuous linear operator. Therefore, (~-n o S) is a w*-null sequence in g ~ , and, if we denote by (en) the canonical unit sequence in e ~ , we have <en, TnoS> = = 1 So, we deduce from Theorem 1.1.2 that ~o~ contains a complemented copy of co, a contradiction (see Corollary 1.3.2) which has appeared quite often here. C o r o l l a r y 5.1.1. If (f2, Z , it) is a a-finite non purely atomic measure space, then L ~ ( i t , X ) contains a complemented copy of co if and only if X contains a copy of Co.

Proof. Necessity is the preceding theorem. Sufficiency follows from Theorem 4.3.2 We can finally give a theorem summarizing the results of the section: T h e o r e m 5.1.4. Let (~2, E , # ) be a a-finite measure space, then L ~ ( # , X ) contains a complemented copy of co if and only if one of the following two conditions holds:

(a) (f2, Z, #) is purely atomic and X contains a complemented copy of co.

5.2 Complemented copies of ~1 in Loo(#, X)

91

(b) ([2, Z, #) is not purely atomic and X contains a copy of Co. Putting together the preceding theorem and Theorem 4.3.4 we get the following: T h e o r e m 5.1.5. Let 1 <_ p <_ ec and let (52, Z , # ) be a a-finite measure

space, then we have 1. If the measure p is purely atomic then Lp(p, X ) D c 0 if and only if X D co. (~) (c) 2. If the measure p is not purely atomic then Lp(p, X ) D Co if and only if (c) X Dco.

5.2 Complemented

copies

of

~1

in Lcc(tt, X)

Since Lc~(#) is quite different to the other Lp(#) spaces, it is not surprising we find a great difference between Loo(p,X) and L p ( # , X ) , too. However, taking in account tile previous results in the monograph, one could expect that, one of the following results be true: (a) Loo(#, X ) contains a complemented copy of 61 if and only if X does. (b) Loo(>, X ) contains a complemented copy of 61 if and only if X contains a copy of glThis is not the case. It was first noticed by Montgomery-Smith (p. 389 of [117]). He realized that an example given by Johnson [76, Remark to Theorem 1] provided a Banach space X0 with no copies of ~1 such that L ~ ( # , X0) contains a complemented copy of gl. Let us begin giving Johnson's example.

Example 5.2.1 ( J o h n s o n / 7 @ .

contains a complemented copy of

(~|

fl. In fact, if we denote by 5 the subspace of ( ~ | the forrn

of all sequences of

((a,),(al,a.2),..., ( a , , a 2 , . . . , a , , ) , . . . ) with (at.) belonging to ~1, we will show that ~ is a 1-complemented subspace

of (~-~.Q .1 )oo isometrically isomorphic to 61. It, is straightforward to show that .7" is isometrically isomorphic to t~l. To find a norm one projection from (~@f~)oo onto ff take a norm one contimmus linear functional L on foo such that k

for each convergent, sequence (at.). Now, let us define L1 : ( E @~r)oo ((a]),(a 2 ' a 2 ) , . . . , ( a 1., n -.,an),-..)n

> IK > L((al,al,a3,..))l 2

5. The space Loo(l~,X)

92

a n d for m _> 2,

L~: (E .e?)~ ((al), (a~, a ~ ) , . . . , (a~,

a'~

9 " " ~

n)~

K m

---+

" 9 ")

m+l

L((O,...,O,a,~,a

m

arn+2

....

...))

m-1 t i m e s C l e a r l y t h e L,~'s are n o r m one c o n t i n u o u s linear f u n c t i o n a l s a n d for any ((al),(al,a2),...,(al,a2,...,an),...) e 7=

.~/:, =

L ~ ( x ) = L(( O , . . . , O , a m , a , ~ , a , ~ , . . . ) ) m-- 1

= a,~

times

Now for each k ~ N define t h e c o n t i n u o u s linear o p e r a t o r

Pk: ( E e e ~ ) ~

~

x

e~~

~ (Ll(x),L2(x),...,Lk(x))

To c o m p u t e t h e n o r m of Pk t a k e

= (@), (~, ~),..., (a?,..., aD,...) 9 ( E ~e~)~, a n d d e n o t e Oj = s g n ( L j ( x ) ) for 1 < j < k. We have

II P~(x)I1~= [I (Ll(x),... ,Lk(x))I1~= I L~(x) I § 2 4 7 =

01Ll(X) §

(a 11 , a 21, a13,...)) + 02L ((0, a~,a3,...)) + . . .

01L

... + O k L ( ( O , . . . , O

=

<

Lk(x) l

L

,k ~kq-1 ,~k,'*k ,'''))

k-~ t i m e s

(

01(al a2,a 3,

.) +

+ Ok(O '

(I

~k ~kq-1

k-~ t i m e s

/)

II 01(al,a~,a~,...) § 02(O,a~,a*~,...) § ... + 0~(0,... ,0, a~, ~+1,...)I1~ k-1 t i m e s

=

II (01a~, 01a~ + 0 2 G . . . , . . . + Oka~ +1 ,

=

01~ + . . . + ok~L < ~ + 1

01a~ +~ + . . .

+ 0 ka k k+2

+...

,...)11~

sup{I Oxa{ l, I O~a~ + 6~'~ I,..., I 01ak q-'" @ Okak I, I eX~ +~ + . . . 9..+Oka~

+ l l l ,o l a k l + 2 +

...

+ 0 k a k+2 ~ I,...}

5.2 Complemented copies of gl in Loo(#, X)

<

sup{I a{ I, I a~ I + la~ I,.-., l a~ 1+... +la~ I, I a~+ll + ... 9 -.+1 ak+l k I, l a lk+2 I+ 9+la~ +2 I, .-}

<

sup{ll (al)I1~,1t (a~, aF,)II,,--. ..., I[ (a~:,... ,a~)I1~, I1 (a~+~,-

=

9

~k+l Ill, ,(*~:+I)

(ak+2,

.

'

9 , a kk++ 22, ~

93

II1,.-'}

II x 11~

Hence, we have shown that

Ilrkll _< 1 for all k E N. Now, it, is immediate that

P : (E~> 1)~ z

>

>

f

(PI(X),P2(z),P3(z),...)

is a well defined norm one projection onto 5c.

R e m a r k 5.2.1. Actually, in the preceding example it would have been natural to take as L a generalized Banach limit (see [51, II.4.22.]). In this case we could simply have taken m + l ,a,,~ m+2 , . . . ) ) L((a,~,a,~

instead of L ( ( O , . . . , O , a , ~.m, a .,,.,+i ... am m + i ,. ..~3 ]/

,,~

i

times

because both values coincide [51, II.4.22.].

R e m a r k 5.2.2. In [76] Johnson mentioned Example 5.2.1 while working in very general conditions. We have preferred to study the example from a very direct and elementary point of view. So we give an explicit form of the subspace isometrically isomorphic to ~1 and an explicit form of the projection, too. However we must point out that, the space ( ~ ~g~)oo even contains a complemented copy of L~([0, 1]). This is an immediate consequence of the following result of Hagler and StegalI [67]: A Ba'aach space X contains a copy of ( ~ (i~f~).~ iJ and only if X * contains a complemented copy of L1 ([0, 1])9

94

5. The space L~(#,X)

Remark 5.2.3. (Montgomery-Smith, 1992) There exists a Banach space Xo with no copies off1, such that Lo~(#, Xo) has a complemented copy off1. Consider X0 = ( ~ | that is the direct sum of (f'l~),~ in the sense of co ([931). On one hand it is clear that X0 has no copies of fl (notice for instance that Xg = (Y~'.| is separable). On the other hand, if we take f ~ ( X o ) , it is immediate t h a t this space contains a complemented copy of (~ | the space of the preceding example. Therefore, taking in account that L~(p, Xo) contains a complemented copy of foo(X0) (this is a very easy fact, and it is shown in the proof of Theorem 5.2.3), we have the following chain:

Lo~(#,Xo) (~)Deoo(Xo) (D) (~-~|

(c)Del

So, we conclude that Loo(p, X0) contains a complemented copy of gl. Montgomery-Smith's remark was very important because it showed what one can not expect. However, it seemed the example involved was a very particular case and it made difficult to give a reasonable conjecture concerning which Banach spaces X provide Lo~(p, X) with complemented copies of fl. In fact, some people thought it was not possible to give a useful condition describing the Banach spaces X enjoying this property. For this reason, it was surprising when S. Diaz [28] realized that the crucial condition which is satisfied by the space X0 in Montgomery-Smith's remark is quite simple. Dfaz discovered that the point was that X0 c o n t a i n s / ~ ' s uniformly complemented (see the definition below). This was a completely new point of view, because it related for the first time the theory we are considering with the local theory of Banach spaces (the theory mainly concerned with the structure of finite dimensional subspaces of the Banach spaces). Let us give a look to a few well known and important results on the finite dimensional structure of Banach spaces. Our aim is to understand the main facts about the Banach spaces X which c o n t a i n / ~ ' s uniformly complemented. Given a Banach space X, we will denote by idx the identity operator on X. If X and Y are Banach spaces and a bounded linear operator T : X --+ Y is an isomorphism onto T(X), we will denote by T -1 de inverse of T, defined on

T(X).

For two isomorphic Banach spaces X and I/ we denote by d(X,Y) the Banach-Mazur distance between X and Y, that is

d(X, Y) = inf{I IT IIII T -1 I1: :F is an isomorphism from X onto Y}. Let 1 _< p _< +oc. We say that a Banach space X contains fpn ~ s uniformly if there exists a sequence (E~) of finite dimensional subspaces of X such that

5.2 Complemented copies of gi in Loo(#, X )

95

s,~p d(E,,, e~) < + c ~ If there are also projections P,~ from X onto E~ with sup IIP~,II < +o~ ?z

we say that X contains g.p'S uniformly complemented. Thanks to the injectivity of all g~'s it is clear that in the case p = +oc both definitions coincide. Notice that another way of saying that X contains g~'s uniformly complemented is to say that for each n C N there are bounded linear operators Cvn

I. >

X Q% ~pn

satisfying Q,~oI,~=ide?,

and

sup ItQ,~It {lI,~ II < +oo lz

If' one takes adjoints, it is clear that if X contains g~'s uniformly complemented then X* contains gq'S uniformly complemented, where, as usual, q is the conjugate of p, that is, 71 + 1 = 1. Therefore, applying the argument again, we deduce that X** contains g'~"s uniformly complemented, too. The converse is also true, that is, if X** contains g~'s uniformly complemented then X has the same property. This follows easily from the tollowing version of the Local Reflexivity Principle [92, Theorem II.5.1]: T h e o r e m 5 . 2 . 1 ( L o c a l R e f l e x i v i t y P r i n c i p l e ) . Let X be a Banaeh space and let E , G be finite dimensional subspaces of X** and X * , respectively, and let e > 0. A s s u m e that P is a projection from X** onto E. Then there are (zn i'rl:'jective linear operator T : E > X and a projection Po from X onto T ( E ) such that 1. T I~':nx = i d E n x . 2. < T ( x * * ) , x * > = < x * , x * * > for allx** C E and allx* E G. 3. IITll IIT-111 < 1 + e (and so we can assume IITII <_ 1 + e), and

~. lIPs, II _< (1 +~-)IIPII. T h e following p r o p o s i t i o n sumInarizes our remarks given above.

P r o p o s i t i o n 5 . 2 . 1 . Let X be a Ba'naeh, ,~pace, let 1 < p <_ + o o and let "as 1 = 1). Then the following are &:'note by q the co'r~:j'u,gate of p (that is, ~1 + 71 equivalent: (a) X eontain.s ~"~i's '~mi,formly complemented. (b) X* contains fl~ 's "wrhformly complemented. (c) X** contains f;~ 's uniformly complemented.

96

5. The space Loo(#,X)

Now, let us concentrate our attention in the case p = 1 (and by duality, in the case p = +oc, too). We need a good characterization of the Banach spaces containing f~'s uniformly complemented. This will follow from the following particular version of the "Great Theorem" of Maurey-Pisier (remember that the "only if' part in the theorem is trivial, the difficult part is the converse):

T h e o r e m 5.2.2 (Maurey-Pisier, T h e o r e m 14.1 of [37] ). X contains f ~ 's uniformly if and only if X does not have finite cotype. With the two preceding results in mind the following characterization fbllows immediately. As usual, let us denote by co(X) the Banach space of all null sequences in X, endowed with the supremum norm. We only have to recall that f l ( X * ) is the dual of co(X) and that X and gl(X) has the same cotype [37, Theorem 11.12]. P r o p o s i t i o n 5.2.2. The following are equivalent: (a) X contains PZ's uniformly complemented. (b) co(X) contains e[~'s uniformly complemented. (c) X* does not have f n i t e cotype.

(d) e~(X*) does not have finite eotype.

Once we know the main facts concerning Banach spaces which contain f~'s uniformly complemented, we need two more ingredients to get the characterization of Banach spaces X such that Lo~ (#, X) contains a complemented copy of fl. The first one is a very well known perturbation lemma. We will include the proof for the sake of completeness. As a consequence of it we will show the close relationship between the finite-dimensional subspaces of L ~ (p, X) and foo(X). The second ingredient is Kalton's contribution. It is a lemma which relates g~(X**) and f~o(X)**. It is crucial in the proof of the main theorem 5.2.3 (see also the Notes and Remarks).

L e m m a 5.2.1 ( P e r t u r b a t i o n l e m m a ) . Let F be a finite dimensional sub.space of X and let P be a projection from X onto F. Suppose d i m ( F ) = n and let { x t , . . . , x T ~ } be a unitary basis of F. Then, given c > 0 there exists 5 > 0 such that if {Yl, 999 Yr~} are vectors in X satisfying max II

l
z~ -

:~;~ II

< ~,

and we denote by G the linear span of the Yi 's, we have (i) d( F,G) < 1 + e, and (ii) There exists a projection Po from X onto G with IIPo II <- (1 + e)IIPll.

5.2 Complemented copies of gl in Loo (#, X)

97

Pro@ First, r e m e m b e r t h a t i f Y is a subspace of X and T : X + Y is a b o u n d e d linear o p e r a t o r which satisfies II idy - T II<_ r / < 1, then T is an isomorphism onto its range T(Y). Moreover, since for any y E Y II Z(y) II _< II Y - r ( y ) II + II y It _< (1 § ~j) II Y II, and 11T(y) II > II y II - II y - r ( : A II > (1 -,7) II y II, it, follows t h a t

II T II I1 T -~ II _< 1 + ~ l-r/

Suppose now the assumptions of our l e m m a are satisfied. Then, there exists M > 0 such t h a t

M 3-~ I Xi I_< II i=1

Aix~ II i--1

for all scalars (ki). Given e > 0 take rl ~ (0, 1) such t h a t

l+r]

--
11

M(IIPII § satisfies the required properties. Take { Y l , - . . , Y,~} vectors in X verifying

max II x~ - :,j~ II < d,

l
n

and denote by G the linear span of the yi's. For x = }-~i=1/~ixi E F, define S(z) = ~ i = 1 Aiyi E G. Since

II : , : - S(x) I1 i=1

<

i

1

i=1

i=1

a M II * II,

we have

II'idr

_

S{I<_ a M

rl

_

< 7 / < 1

(IIPII +1)

Therefore, S is an isomorphism from F onto G witch

IlSll tlS < II < l § -

Thus, d(F,G) < IlSlt IIS-' II ~< 1 §

l-r/-

< 1+~

98

5. The space Loo(p,X)

In order to get the projection P0, consider the operator T : X --+ X defined by T = idx + (S o P) - P. It verifies r] II r - i d y II = II (S - i d a ) o P II <

II P II

(11 P II §

< ,1 < 1.

Hence, it is well known that T is an automorphism on X. Moreover,

11T II II T - ~ II _<

1 + r/ < 1 + e. 1 - 7/

Since T ( F ) = G, it is clear that P,, = T o P o T -1 is a projection from X onto G with H P0 II _< (1 + e) II P II, as required. P r o p o s i t i o n 5.2.3. Let ($2, 5, p) be a c,-fi'nite measure space, let F be a finite dimensional subspace of Lar and let P be a projection from L o c ( # , X ) onto F. Then, given e > 0 there exist a finite dimensional subspace G of g.oo(X) and a projection Po from goo(X) onto Cj such that

d(Jz, G) <_ 1 § e arm

II 5, II _< (1 +

e)

II P II

Proof. Assume the hypothesis of the proposition holds, let e > 0, and apply the preceding l e m m a to the finite dimensional subspace r of Loo(>, X ) , taking any unitary basis { f l , . . . , f,~} of S. Given the positive number 6 provided by the lemma, take {g~,... ,g,~} in L ~ ( p , X ) countably valued such that max

l
II .f,

- .~t~ I1~ <

Let (A,,~) be a sequence of palrwise disjoint measurable sets with positive measure such that each gi is constant in the Am's and >(F2 \ u,,A,~) = 0. Let us denote by 7 / t h e subspace of Loo(p, X ) of all functions which are constant in tile A,,,'s. Of course, ~ is isometrically isomorphic to ,~ If we denote by g tile finite dimensional subspace of L ~ ( p , X ) spanned by the 9i's, we have

g

c ~ c

Loo(#,X),

and tile preceding l e m m a guarantees that d(m,g)_< l + e , and t h a t there is a projection P0 flom Loo(#, X ) onto g such that

II Po II _< (1 + ~) II P II. If we consider the restriction of Po to 7{, we get the desired conclusion.

5.2 Complemented copies of g-1 in Loo(#, X)

99

As usual, in the next l e m m a we identify in the canonical way X as a subspace of X**. In this way g.oo(X) is a subspace of goo(X**) and a subspace

of ~oo(X)**, too. Lemma

Given e > 0 there exists a continuous linear op-

5.2.2 ( K a l t o n ) .

eratoF

J: e~(x**) ~

:~(x)**

such that It J II _< 1 + e and Jle~(x)= ide~(x). Pry@ Let ~: be a finite dimensional subspace of goo(X**). Then there exists a sequence (F,-~) of finite dimensional subspaees of X** such that

By the local reflexivity principle (Theorem 5.2.1), tbr each n E N there exists an injective operator u~. : F , -+ X such that

u,dxnv~= i d x n f ,

and

II u~ II < 1 + e.

Now we can define

Ia:

a ---+ (.,~) ---+

:oo(x) (~,.(~,~))

It is clear that I a is a well defined continuous linear operator which satisfies II Ia II -< 1 + e and /,xlane~(x)= idane~(x), Now we will use a classical Lindenstrauss compactness argument. Denote by B(goo(X**)) and B(foo(X)**) the unit balls of foo(X**) and goo(X)** respectively, and define .:a : B(goo(X**))

x

>

>

:o~(X) Ia(x) 0

ifzE otherwise

Of course Ja neither is linear nor continuous, but notice t h a t

.Ia(B(.:~(X**)) ) C (1 + e)B(g~(X)) C (1 + e)B(goo(X)**)

(5.4)

Now we can consider the net (Ja)a, where ~ runs on the directed set of all finite dimensional subspaces of goo (X**). By 5.4, this is a net in tile compact space

( (1 + ~)B(e.~(x)**), ,,,:).le~(x"ll Therefore immediate to denote immediate

(Ja)a has an accumulation point J in this compact space. It is that J can be extended to a linear operator, which we continue J, fl'om the whole space foo(X**) into goo(X)**, and it is also that this linear operator J has tee required properties.

5. The space L~o(#,X)

100

We can finally state and prove the main result of the Section. T h e o r e m 5.2.3 ( D f a z - K a l t o n ) . Let (Y2, ~, >) be a or-finite measure space,

then the following are equivalent: (a) (b) (c) (d) (e)

Loo(p,X) contains a complemented copy of gl. L~o(p,X) contains g[~'s uniformly complemented. ~.oo(X) contains a complemented copy of gl. ~.oo(X) contains g.[~'s uniformly complemented. X contains g.~ 's uniformly complemented.

Proof. We will show the following chain of implications: (a) ~ (b) ~ (d) (~-) ~ (6 ~ (a). T h a t (a) implies (b) is trivial, and (b) implies (d) is an immediate consequence of Proposition 5.2.3. Let us show now that (d) implies (e). We assume that goo(X) contains g~"s uniformly complemented. So there exist M > 0, a sequence (F~) of finite dimensional subspaces of goo(X) and a sequence of projections ~

: coo(x) + G

such that

d(G,q~) < M and for all

n C N.

IIP~II-<M

Kalton's lemma 5.2.2 provides a continuous linear operator J : Go(X**)

go~(X)**

such that

Jigs(x) = idg~(x) Of course we may consider the F~,'s as subspaces of goo (X**), and if we denote by P,~* the second adjoint of P,~, it is clear that the operators

R,F o j : e~(x**) + G provide a bounded sequence of projections. Therefore, goo(X**) = co(X)** contains g~'s uniformly complemented. Then, by Proposition 5.2.1 this means that c0(X) contains f~'s uniformly complemented, too. But now, thanks to Proposition 5.2.2, we deduce that (e) holds. To see that (e) implies (c) is very easy. Assume that there exist M > 0, a sequence (F,,,) of finite dimensional subspaces of X and a sequence (P~) of projections from X onto F~ such that

d ( G , q ~) <_ M and

IIP,~II<M

for all ,~. C N. It is clear that (}-~ (~F,,)oo is isomorphic to (y~. Og.~)oc, and we know (,Johnson's example 5.2.1) that, ( ~ Og.~)~o contains a complemented

5.3 Notes and Remarks

101

copy of gl. Therefore (}-~.| contains a complemented copy of gl, too. On the other hand, it is clear that

(P,~) : e~(x)

--+

x

---+

( ~ eF,,)oo (p~(x))

is a projection onto (y~ | So, it follows that goo(X) contains a complemented copy of g~. Finally, note that goo (X) is isometrically isomorphic to a complemented subspace of Loo(p, X). Indeed, likewise in the proof of proposition 5.2.3, take a sequence (Am) of pairwise disjoint measurable sets with finite and positive measure, and consider the subspace 7 / o f Loo(#, X) of all functions which are :onstant in the A,~'s. Then the map

Loo(#,X) f

~

7/

1(s

> ~-~.X,~(.),-777-77 ?1,

fd#) n

ts a projection (it is a conditional expectation). Since 7t is clearly isometrically isomorphic to goo(X), our statement is proved. Hence (c) implies (a), obviously. This completes the proof.

Remark 5.2.4. It should be pointed out that the preceding result provides immediately examples of Banach spaces X which are even separable and reflexive for which L ~ ( p , X) contains a complemented copy of ~1. Take for instance X = ( ~ | )2. 5.3 N o t e s a n d R e m a r k s As we have already mentioned, in this chapter we have considered only orfinite measure spaces, while in the rest of the monograph we have been working in arbitrary measure spaces. In the introduction of the chapter we have explained why, but we would like to call the attention here to a significant fact. At the moment, Banach space specialists usually study almost exclusively L ~ ( # ) spaces for finite (or ~-finite) measures >, with only one i m p o r t a n t exception: spaces of type goo(F). Here, as usual, s denotes the Banach space of all bounded scalar functions defined on a certain set F, endowed with the supremum norm. The vectorial version of this space is goo(F, X), which is d e f n e d in the natural way. We believe t h a t while considering the finite measure case, we have exposed most of the main ideas related with our problems, but we would like to mention also the results which are known in more general situations. Concerning the search of complemented copies of co in Loo(/~, X ) there are two results. One for the purely atomic case, and the other one for the non purely atomic one:

102

5. The space Loo(#,X)

(a) Leung and Rgbiger [90] showed that if the set F of indices is measurable (which includes of course the denumerable case), then Coo(F, X) contains a complemented copy of c0 if and only if X does. They left open the general problem for arbitrary sets of indices. (b) Dfaz proved in [27] his theorem 5.1.3 (and corollary 5.1.1) for arbitrary measure spaces. About complemented copies of Cl in Loo(/~, X) we can say that if X contains C~"s uniformly complemented then Loo(/z, X) contains a complemented copy of g~ (this is just because if X contains C~'s uniformly complemented, by theorem 5.2.3, g~(X) contains a complemented copy of Cl, and this space is clearly complemented in Loo (p, X)). However, we do not know if the converse is true or not. We must say that actually Leung and Rgbiger [90] work in spaces more general than Coo(F, X). They study sums in the sense of Coo of families of Banach spaces. However, we would add that, concerning the problem we are dealing with, these two kind of spaces present a completely analogous behavior. Now, let us give a look at the results of the chapter. At the beginning of Section 5.1 we recall that we can decompose in a very nice way each member of ba(~ -- go~ in sum of two measures, one of them countably additive, and the other one purely finitely additive. Of course, this is just a particular case of Yosida-Hewitt decomposition theorem [51, III.5.8]. When we move the above decomposition from C~ to Coo (X)* we introduce the projection P in t~oo(X)* which appears at the beginning of the section, and which is crucial to get Leung and RSbiger' theorem 5.1.1. It is worth to observe that this projection could also be viewed from a more general point of view. One can easily show that if X is a closed subspace of Y and there exists a bounded linear operator S : Y -+ X** such that S Ix = idx (where, as usual, we identify X with a closed subspace of X** in the canonical way), then there exists a continuous linear projection P : Y* -+ Y* whose kernel is X • = {x* E Y* : x*(X) = {0}}, and therefore, X* = Y*/X • is complemented in Y*. In particular, our assumption is satisfied whenever X C Y C X**, taking as S the natural embedding. If we apply this to

co(X) c coo(x) c coo(x**) = co(X)**, we get a pro.jection P from g.oo(X)* onto its subspace co(X)* = Cx(X*). However, we do not conclude from the general theory that this projection is w*-sequentially continuous, which is very important for us. This is true in our particular situation, as we show in Proposition 5.1.1. In this monograph we have hardly dealt with (~-~| spaces, but one should notice that Johnson's example also shows that the behavior of ( ~ @X,,.)oo is quite different to the one of ( ~ | for 1 < p < +oo.

5.3 Notes and Remarks

103

We would like to point out that the classical perturbation lemma about, basis ( [93, 1.a.9.] or [35, Chapter V, Theorem 12]) is just an infinitedimensional version of the well known iemma 5.2.1. In Section 5.2, we explain in general when we say that a Banach space X contains ~l]"s uniformly (complemented) for 1 _< p _< +oc. Actually we are only interested in the case p = 1 (and by duality, in the case p = +oo), but, we believe that to consider the general situation helps to understand better these concepts. Notice that Theorem 5.2.3, the main result of Section 2, lies on the "Great Theorem" of Maurey-Pisier (Theorem 5.2.2). This is because this theorem provides a good characterization of when X contains ~ ' s uniformly complemented, and by duality and the Local Reflexivity Principle (Theorem 5.2.1) this provides a good characterization of when X contains g~'s uniformly complemented. We do not know of any "elementary proof'. Theorem 5.2.3 is due to S. Dfaz [28] and N.J. Kalton. S. Dfaz in [28] proved that conditions (a), (b), (c) and (d) are equivalent, and that they hold whenever (e) is satisfied. He also showed that the five condition (a), (b), (c), (d) and (e) are equivalent in quite general conditions (for instance, for Banach lattices). Therefore, the only open problem left by him was whether (a) (or any of the other equivalent conditions (b), (c) or (d)) implied (e) or not. In the Spring of 1996 we asked N.J. Kalton about this. He showed us how to prove that (d) indeed implies (e), and kindly allowed us to include his solution here. We have summarized it in lemma 5.2.2 which is a reformulation of Kalton's ideas contained in [79]. We believe that the lemma is much more subtle than it seems and provides a very interesting connection between 5oo (X**) and f-oo(X)**. In fact, we have seen several times people proclaiming that necessity of condition (e) was very easy and later on they were not able to give a proof. We may add that we. have read several wrong proofs of the assertion (some of these wrong proofs were made by us). To show necessity of (e) in Kalton-Dfaz Theorem 5.2.3 one tbllows more or less the following idea: if too(X) contains g)'s uniformly complemented, then its second dual, g~(X)**, does either. From this fact we wish to conclude that go~(X**) = CI(X*)* = co(X)** has the same property, but these spaces are quite well known, and thanks to Maurey-Pisier theorem we may conclude our thesis. The point is that although too(X)** and goo(X**) might seem close in a first view, a careful look shows us that they are indeed very different. This is the problem solved by Kalton's lemma. We wish to point out that Kalton's lemma is closely related with the important notion of locally complemented subspace. From this point of view it just says that too(X) is locally complemented in too(X**). The notion of locally complemented subspace has been studied by several authors and has found many interesting applications in Banach space theory (see [55], [79]

104

5. The space L ~ ( p , X )

and [64]). Its isometric version has been considered recently in [106].

[91] and

It is worth noticing that Diaz-Kalton Theorem 5.2.3 and Hagler-Stegall result mentioned in remark 5.2.2 can be combined fruitfully to obtain a significant consequence: T h e o r e m . Let (Y?, Z, #) be a a-finite measure space, then the following are equivalent:

(a) (b) (c) (d) (e)

L ~ ( # , X ) contains a complemented copy of L1 ([0, 1]). got(X) contains a complemented copy of Ll([O, 1]). L ~ ( p , X ) contains a complemented copy of ~1. go~(X) contains a complemented copy off1. X contains g~l~'s uniformly complemented.

Pro@ We know by Theorem 5.2.3 that (c), (d) and (e) are equivalent. On the other hand, goo(X) contains a complemented copy of (y~ | whenever (e) holds, as shown in the proof of Dfaz-Kalton Theorem 5.2.3. Thus, HaglerStegall's result mentioned in remark 5.2.2 proves that (e) implies (b). To see that (b) implies (a), it is enough to observe that L ~ ( p , X ) contains a complemented copy of ~o~(X) (this easy fact was shown in the proof of Theorem 5.2.3). Finally, (a) implies (c) is trivial. Dfaz [28] has also proved that for 1 < p < +o% Loo(p,X) contains a complemented copy of gp whenever X contains g~ 's uniformly complemented.

6. T a b u l a t i o n o f R e s u l t s

Here is a t a b l e w h i c h s u m m a r i z e s all t h e r e s u l t s of o u r m o n o g r a p h .

C(K,X)

LI(/Z, X )

Lp(It, X )

Lo~(#,X)

l
D co

D co (c)

always

XDco

XDco

always

(trivial)

(2.1.6)

(2.1.6)

(trivial)

X~co (~)

always

X Dec (c) or XDco and # is not purely atomic

# is not purely atomic

X Dco (c) or XDco and p is not purely atomic

(4.3.4)

(4.3.4)

(5.1.4)

or XDgl

always

XDCl

always .

(3.1.2)

(trivial)

(2.2.2)

(trivial)

XDgi (~)

always

XDg~ (~)

xJ~ uniformly complemented

(3.1.4)

(trivial)

(4.1.2)

(5.2.3)

or XDg~

XDgoo

XDg~

always

(3.3.1)

(4.2,1)

(4.2.1)

(trivial)

(3.2.1)

or

XDco and

C(K)~el D gl

D gi

(A

D g~

T h e m e a n i n g of t h e T a b l e is clear. In t h e first row we h a v e t h e spaces we a r e s t u d y i n g , in t h e first c o l u m n we see t h e different c o n t a i n m e n t s we a r e

106

6. Tabulation of Results

considering, and in each box we put the condition which must be satisfied and the n u m b e r of the corresponding theorem if it is not trivial. "Trivial" means t h a t it follows from the following easy containments

C(K, X) D C(K) D co (c) (c)

L1 (#, X) D L1 (i.t) D el (~) (c) (c)

Recall t h a t we are assuming C ( K ) , Lp(#) and X infinite-dimensional. For 1 <_ p < + c o the measure # in Lp(#, X) spaces is arbitrary, in the case p = + o c it is assumed to be ~-finite. A look at the table shows t h a t in most cases we get the natural answers, t h a t is, we get a ~ r m a t i v e answers to P r o b l e m 2 in the Introduction. There are only the following exceptions: (a) The different containments of co as a c o m p l e m e n t e d subspace, and (b) C o m p l e m e n t e d copies of gl in Loo(#, X). In b o t h cases we see t h a t vector valued function spaces contain complemented copies of Co or ~1 more often t h a n expected. Notice once more t h a t case (b) provides the only answer which depends exclusively on the local structure of the B a n a c h space X.

7. Some Related Open Problems

To finish our work we would like to mention a few lines of research and open problems which are closely related with the results of this monograph. I. If we remember Problem 1 posed in the Introduction, we see that we have a complete solution to it with only one exception: For L ~ (#, X) we know the answer for a-finite spaces, but for arbitrary positive measure spaces we do not. Then we have the following P r o b l e m 7.1. Assume (f2, &, #) is a non a-finite measure space, when does L ~ ( p , X ) have a complemented copy of co!, when does it have a complemented copy of ~,1 "~ It would be particularly interesting to answer these questions for the space L ~ ( p , X ) = eo~ (C, X). Leung and R~biger [90] already posed this problem for complemented copies of co in the sum in the sense of goo of an arbitrary family {Xi : i E F} of Banach spaces. See Section 5.3, and in general Chapter 5. II. In order to consider analogous problems to the ones studied here, a natural thing is to add to our list of spaces (Co, el and too) the fr spaces (1 < r < oe), that is, we can ask: P r o b l e m 7.2. When does C(K, X) or Lp(#, X) have copies or complemented copies of ~r (1 < r < oo)? Tile general problem seems to be very difficult because no good characterization of spaces containing ~,. (1 < r < oc) is known. However, some partial answers have already been given (see [11], [28], [71], [107], [108], [119], [120]). Let us mention at least that Lp(>, X) can contain copies of gr, while neither Lv(#) nor X contain a copy of g, (this is an old known result, see [107], [108] or [94, Remark to Lemma 1.f.8]). And in the same way we can consider LI([0, 1]):

108

7. Some Related Open Problems

P r o b l e m 7.3. If L p ( p , X ) (1 < t) < oc) contains a copy of L1([0,1]) must X contain a copy of LI([0, 1]) ? or, in general, when does L p ( # , X ) have a copy of Cl([0, 1]) These questions were posed to the authors by J. Diestel and they can also be found in [117, Question 10]. The answer to the first one is yes if X is a dual space [117], but we do not know a complete answer even for p = 2. W h a t about complemented copies of L1 ([0, 1]) ? In this case the question is natural for L ~ ( p , X) and C(K, X), too. We have just shown in the Notes and Remarks of Chapter 5 a complete characterization for Lo~(#,X). But concerning the other spaces we do not know anything. 7.4. When does C(K,X) or Lp(p,X) (1 < p < oc) contain a complemented copy of LI([0, 1]) ?

Problem

For C(K,X) this has already been asked by E. and P. Saab in [117, Question 12]. An anonymous referee pointed out to us that Hagler and Stegall paper [67] might help to give a solution to this problem. Actually, it was this remark t h a t led us to formulate the solution tbr Loo(p, X). I I I . All problems considered here can also be posed in more general situations, for instance in locally convex spaces and, in particular in Frdchet spaces. This last case is probably the most important. Of course, some work in this direction has already been done. For instance, in connection with J. Hoffmann-Jorgensen and Kwapiefi's work see [56]. I V . Theorems studied in this monograph have already inspired the search of copies of sequence spaces in several different situations. Let us mention some of them: (a) Tensor products (see [117]). (b) Function spaces more general than Lp(#,X) spaces, like Orlicz vector valued function spaces L~(X) (see, for instance [6] or [7]), or even the more general E(X) spaces, where E is an order continuous Banach lattice which has weak unit as considered in [127] (see for instance [53] or [97]). (c) Vector measure spaces or operator spaces (as in [46] , [47] or [121]). (d) Vector-valued Hardy spaces (as in [44]). Nevertheless, we believe t h a t in these directions many things remain to be done. V . It is well known that vector-valued function spaces may be viewed as operator spaces. For example, the space I~(X, C(K)) of all compact operators from X into C(K) is isometrically isomorphic to C(K, X*) [51, VI.7.1], and Loo(p, X) is very useful in the representation of operators from L I ( # ) into

7. Some Related Open Problems

109

X [39, Chapter III]. For this reason the results given here have inspired and have found applications in the study of several operator spaces (see [12], [13], [43], [48], and see also [61] and [24] in connection with [48]). Among these applications are some complementation results more or less related to the famous open problem of the uncomplementability of the space K2(X, Y), of compact operators from X to Y, in the space s Y ) of all operators from X to Y (see [48]). This should not be surprising because it was already shown in [78, Theorem 6] that such a problem is related to the presence of copies of co or goo in operator spaces. VI. We have so far been mainly interested in isomorphic copies of spaces. Another natural point of view is to consider isometric copies. Some results in this line have been recently obtained by Koldobsky [81] (see also [65], I103], [108] and [66]). V I I . Instead of asking about subspaces we could also ask about quotients. That is, P r o b l e m 7.5. Wizen do the spaces considered here have quotients isomorphic to any of th,e classical sequence spaces? In fact Dfaz [29] has already got an interesting answer to the preceding question. He has shown that: "For" 1 <_p <_ oc, if the measure Ix is not purely atomic and X* contains a copy of ~1, then Lp(Ix, X ) has a quotient isomorphic to e0 ". This result has allowed him to get a striking characterization of Grothendieck Lp(Ix, X) spaces. The study of quotients was continued by Dfaz with Schliichternann in [34]. The result we have .just mentioned recalls us that many properties considered in Banach space theory are connected with the presence of copies or quotients of certain spaces. For instance, property (V*) of Petczynski is connected with the existence of complemented/ql-sequences [7], and Grothendieck property with the lack of" complemented copies of Co [26]. Of course these connections haw', already been exploited ([7], [25], [26], [29], [59], [104]), tout it seems natural to think that they could be used further. V I I I . Let us finally consider the tbllowing problem: It is well known that ~ and Loo([0, 1]) are isomorphic [93, p. 111]. However, let us suppose that X has a copy but not a complemented copy of co. For instance X may be ~'.~ or (qoo/co (a well studied space, see for example [49]), or more generally X may be any Grothendieck C ( K ) space or a quotient of it. It, follows from Theorem 5.1.4 that Loo([0, 1],X) does contain a complemented copy of co, but f,~(X) does ',tot. Hence L~o([0, 1],X) and g ~ ( X ) are not isomorphic. Then the tbllowing natural question arises: P r o b l e m 7 . 6 . / n which conditions aT'e Loo ([0, 1], X ) and goo(X) isomorphic ? We do not know the answer even for separable spaces X.

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116

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121. G. Schliichtermann, Properties of Operator-Valued Functions and Applications to Banach Spaces and Linear Semigroups, Habilitation Thesis. Ludwig Maximilians Universit~t, Miinchen 1994. 122. Th. Schlumprecht, Limitierte Mengen in Banachriiume, Thesis. Ludwig Maximilians Universit/it, Miinchen 1988. 123. G.L. Seever, Measures on F-spaces, Trans. Amer. Math. Soc. 133 (1968), 267-280. 124. Z. Semadeni, Banach Spaces of Continuous Functions, PWN-Polish Scientific Publishers, \u 1971. 125. I. Singer, Best Approximation in Normed Linear Spaces by Elements of Linear Snbspaces, Die Grundlehren der mathematischen Wissenschaften in Einzeldarstellungen vol. 171, Springer-Verlag, Berlin Heidelberg New-York 1970. 126. M. Talagrand, Un nouveau C(K) qui poss~de la propridtd de Grothendieck, Israel J. of Math. 37 (1980), 181-191. 127. IvI. Talagrand, Weak Cauchy sequences in LI(E) , Amer. J. of Math. 106 (1984), 703-7'24. 128. M. Talagrand, Quand l'espace des measures it variation bornde est-il faiblemente s~quentiellement complet ?, Proc. Amer. Math. Soc. 90 (1984), 285-288. 129. A. Ulger, Weak compactness in L1 (#, X), Proc. Amer. Math. Soc. 113 (1991), 143-149.

Index

(~-2 | (&), 12 (r~), 42

102

C(K, X), 8 E(X), 108 G-space, 74 Lq~. (#, X*), 26 L1 ([0, +o0)), 61 Lp(~,X), 8 L~(X), 108 X DY, 8 X DY, 8 (c) [A], 13 [N], 13

Lp(#), 8 /3N, 68, 74 &, 35 5~m, 10 gp(X), 32

& ( M ) , 13

K(X, C(K)), 108

c ( x , Y), 109 s (A, x*), 19 s s

s

(#, X*), 26 18

18

P(N), 11 m} 8 (s.s.) sequence, 37 B(K), 34 C(K, X)*, 35

C(K), 8 i~(~,X)*, 17

cabv(Z), 8 cabv(Z, X ), 8 co-sequence, 9 gp-sequence, 9 &-sequence, 9 rcabv(K, X*), 35 ( E eX~)p, 39, 102

&~(X), 32 &o(F), 101

Bessaga-Petczyfiski theorem 9

&o(F,X), 101, 107

Caratheodory theorem, 40 compact space - extremally disconnected (or Stonean), 73 scattered (or dispersed), 65 complemented copy, 8 complemented sequence, 9 conditional expectation operator, 29 contains/~'s uniformly, 94 contains g~'s uniformly complemented, 95 contains ~p'S uniformly, 94 contains gp'S uniformly complemented, 95 contraction principle, 44 copy, 8 - complemented, 8 counting measure, 32

II. I1~, 8 II. I1r1628 II ~ il, 8

p, 18 C (for measure spaces), 27 ba(N), 84 c0(X), 96 coo, 44

ca(N), 84

cabv(t3(K), X*), 35 cabv~(Z, X*) , 19 d(X, Y), 94 rn,~, 35

rcabv(K), 66 w*-measurable, 18 fi, 19 ~ ( X , Y), 109

118

Index

density function, 17 dispersed (or scattered) compact space, 65 Dunford-Pettis, 34 extremally disconnected, 73, 74, 84 Fr6chet spaces, 108 function - w*-measurable, 18 Grothendieck property (or space), 68, 74, 109 Hardy spaces, 108 injective, 30 isometric copy, 15, 109 isometric theory of Lp(#) spaces, 32, 40 lifting, 17, 18 linear, 23 Local Reflexivity Principle, 95 locally complemented subspace, 103

-

martingale difference sequence, 63 Maurey-Pisier Great Theorem, 96 measure - w*-countably additive, 35 w*-regular, 35 regular, 35 strongly additive, 14 measure algebra, 40 non-trivial weak-Cauchy sequence, 37 operation (A), see Souslin operation, 30 Orlicz vector valued function spaces, 108 Phillips Theorem, 13 probability in Banach spaces, 61, 62 property (V*) of Petczynski, 109 purely atomic, 28, 32 Rademacher functions, 42 Radon-Nikod3~m property, 18 Radon-Nikod3~m type theorem, 19, 22 random variables, 61 independent, 61 symmetric, 61 Riesz-Singer theorem, 35 Rosenthal's Disjointification lemma, 11

scattered (or dispersed) compact space, 65 seminormalized, 9 semivariation, 8 separable measure space, 27 sequence (s.s.), 37 co-sequence, 9 - gv-sequence, 9 ~?l-sequence, 9 complemented, 9 - non-trivial weak-Cauchy, 37 seminormalized, 9 strongly summing, 37 Souslin operation, 5, 30, 31, 39 Stone-Cech compactification of N, 68 Stonean, 73 strong compactness in Lp(#, X), 63 strongly summing, 37 subsequence splitting lemma, 54, 62 UMD spaces, 63 unconditional basic sequence (in Lp(#)), 63 unconditional basis, 63 unconditionally converging operator, 12 uniform integrability, 27, 33, 34, 40 uniformly integrable, 34, 52, 54 56, 59-61, 77, 78 variation, 8 w.u.C., 9 weak compactness in Lv(#, X), 63 weak convergence in LB(#, X), 63 weakly compact operator, 12 weakly unconditionally Cauchy, 9 weakly unconditionally convergent (or weakly unconditionally Cauchy) series, 9