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THE COMPLETE CONTINUITY PROPERTY AND

arXiv:math/9302204v1 [math.FA] 2 Feb 1993THE COMPLETE CONTINUITY PROPERTY ANDFINITE DIMENSIONAL DECOMPOSITIONSMaria GirardiandWilliam B. JohnsonAbstract. A Banach space X has the complete continuity property (CCP) if eachbounded linear operator from L1 into X is completely continuous (i.e., maps weaklyconvergent sequences to norm convergent sequences).

The main theorem shows thata Banach space failing the CCP (resp., failing the CCP and failing cotype) has asubspace with a finite dimensional decomposition (resp., basis) which fails the CCP.1. introductionGiven a property of Banach spaces which is hereditary, it is natural to askwhether a Banach space has the property if every subspace with a basis (or with afinite dimensional decomposition) has the property.

The motivation for such ques-tions is of course that it is much easier to deal with Banach spaces which have abasis (or at least a finite dimensional decomposition) than with general spaces. Inthis note we consider these questions for the complete continuity property (CCP),which means that each bounded linear operator from L1 into the space is com-pletely continuous (i.e., carries weakly convergent sequences into norm convergentsequences).The CCP is closely connected with the Radon-Nikod´ym property (RNP).

Sincea representable operator is completely continuous, the RNP implies the CCP; how-ever, the Bourgain-Rosenthal space [BR] has the CCP but not the RNP. Bourgain[B1] showed that a space failing the RNP has a subspace with a finite dimensionaldecomposition which fails the RNP.

Wessel [W] showed that a space failing the CCPhas a subspace with a basis which fails the RNP. It is open whether a space has theRNP (respectively, CCP) if every subspace with a basis has the RNP (respectively,CCP).Our main theorem shows that if X fails the CCP, then there is an operatorT : L1 →X that behaves like the identity operator I : L1 →L1 on the Haar func-tions {hj}.Specifically, there is a sequence {x∗n} in the unit ball of X∗suchthat x∗n keeps the image of each Haar functions along the nth-level large (i.e.1991 Mathematics Subject Classification.

46B22, 46B20, 46B28, 46G99.The first author was supported in part by NSF DMS-9204301.The second author was supported in part by NSF DMS-9003550.Typeset by AMS-TEX1

x∗n(Th2n+k) > δ > 0 ) and the natural blocking {sp(Th2n+k : k = 1, . .

. , 2n)}nof the images of the Haar functions is a finite dimensional decomposition for somesubspace X0.

Note that X0 fails the CCP since T is not completely continuous (Tkeeps the Rachemacher functions larger than δ in norm). Thus a space failing theCCP has a subspace with a finite dimensional decomposition which fails the CCP.In the language of Banach space geometry, the theorem says that in any Banachspace which fails the CCP grows a separated δ-tree with a difference sequence nat-urally blocking into a finite dimensional decomposition.

If furthermore the spacealso fails cotype, then modifications produce a separated δ-tree growing inside asubspace with a basis.Throughout this paper, X denotes an arbitrary Banach space, X∗the dual spaceof X, and S(X) the unit sphere of X. The triple (Ω, Σ, µ) refers to the Lebesguemeasure space on [0, 1], Σ+ to the sets in Σ with positive measure, and L1 toL1(Ω, Σ, µ).

All notation and terminology, not otherwise explained, are as in [DU].The authors are grateful to Michel Talagrand and Peter Casazza for helpfuldiscussions.2. operator view-pointA system A = {Ank ∈Σ: n = 0, 1, 2, .

. .

and k = 1, . .

. , 2n} is a dyadic splittingof A01 ∈Σ+ if each Ank is partitioned into the two sets An+12k−1 and An+12kof equal mea-sure for each admissible n and k .

Thus the collection πn = {Ank : k = 1, . .

. , 2n} ofsets along the n th-level partition A01 with πn+1 refining πn and µ(Ank) = 2−nµ(A01).To a dyadic splitting corresponds a (normalized) Haar system {hj}j≥1 whereh1 =1µ(A01) 1A01andh2n+k =2nµ(A01) (1An+12k−1 −1An+12k )for n = 0, 1, 2, .

. .

and k = 1, . .

. , 2n.A set N in the unit sphere of the dual of a Banach space X is said to norm asubspace X0 within τ > 1 if for each x ∈X0 there is x∗∈N such that ||x|| ≤τx∗(x).It is well known and easy to see that a sequence {Xj} of subspaces of X forms afinite dimensional decomposition with constant at most τ provided that for eachn ∈N the space generated by {X1, .

. .

, Xn} can be normed by a set from S(X⊥n+1)within τn > 1 where Πτn ≤τ.Theorem 1. If the bounded linear operator T : L1 →X is not completely continu-ous and {τn}n≥0 is a sequence of numbers larger than 1, then there exists(A) a dyadic splitting A = {Ank}(B) a sequence {x∗tn}n≥0 in S(X∗)(C) a finite set {y∗n,i}pni=1 in S(X∗) for each n ≥0such that for the Haar system {hj}j≥1 corresponding to A, for some δ > 0, andeach n ≥0(1) x∗tn(Th2n+k) > δ for k = 1, .

. ., 2n(2) {y∗n,i}pni=1 norms sp(Thj : 1 ≤j ≤2n) within τn(3) y∗n,i(Th2n+k) = 0 for k = 1, .

. .

, 2n and i = 1, . .

., pn.2

Note that if Πτn is finite, then conditions (2) and (3) guarantee that the naturalblocking {sp(Thj : 2n−1 < j ≤2n)}n≥0 forms a finite dimensional decompositionwith constant at most Πτn.The proof uses the following standard lemma which, for completeness, we shallprove later.Lemma 2. If A ∈Σ+ and {gi}ni=1 is a finite collection of L1 functions, then anextreme point u of the set C≡{f ∈L1 : |f| ≤1A andRA fgi dµ = 0 for i =1, .

. .

, n} has the form |u| = 1A.Proof of Theorem 1. Let T : L1 →X be a norm one operator that is not completelycontinuous.

Then there is a sequence {rt} in L1 and a sequence {x∗t } in S(X∗)satisfying:(a) ||rt||L∞≤1(b) rt converges to 0 weakly in L1(c) 4δ ≤x∗t T rt for some δ > 0 .Consider T ∗x∗t ∈L∞. Since ||rt(T ∗x∗t)||L∞is at most 1, by passing to a subsequencewe may assume that {rt(T ∗x∗t )} converges to some function h in the weak-startopology on L∞.

SinceRh dµ ≥4δ the set A01 ≡[h ≥4δ] is in Σ+. (Compare thiswith [B2, proposition 5]).We shall construct, by induction on the level n, a dyadic splitting of A01 alongwith the desired functional.

Fix n ≥0.Suppose we are given a finite dyadic splitting {Amk : m = 0, . .

. , n and k =1, .

. .

, 2m} of A01 up to n th-level.This gives the corresponding Haar functions{hj : 1 ≤j ≤2n}. For each 1 ≤k ≤2n, we shall partition Ank into 2 sets An+12k−1 andAn+12kof equal measure (thus finding h2n+k) and find x∗tn ∈S(X∗) and a sequence{y∗n,i}pni=1 in S(X∗) such that conditions (1), (2), and (3) hold.Find a finite set {y∗n,i}pni=1 in S(X∗) that norms sp(Thj : 1 ≤j ≤2n) within τn .LetCnk ≡{f ∈L1 : |f| ≤1Ank ,ZAnkf dµ = 0 andZAnk(T ∗y∗n,i)f dµ = 0 for 1 ≤i ≤pn}.Note that each Cnk is a convex weakly compact subset of L1.Since {rt} tends weakly to 0, for large t there is a small perturbation ˜rt of rt sothat ˜rt1Ank is in Cnk for each k. To see this, putF = sp{1Ank } ∪{(T ∗y∗n,i)1Ank : k = 1, .

. .

, 2n and i = 1, . .

. , pn}⊂L∞≃L∗1 .Now pick tn ≡t so large that for k = 1, .

. .

, 2n and i = 1, . .

. , pn(d)RAnk rt(T ∗x∗t ) dµ ≥2δαn(e)RΩrtf dµ ≤δ2αn||f||for all f in F3

where αn = 2−n µ(A01) ≡µ(Ank). Condition (d) follows from the definition of A01and the weak-star convergence of {rt(T ∗x∗t )} to h while condition (e) follows from(b) and the fact that F is finite dimensional.Thus the L1-distance from rt to ⊥F ≡{g ∈L1 :RΩfg dµ = 0 for each f ∈F}is at most δ αn2 .

So there is ˜rt ∈⊥F such that ||˜rt −rt||L1 is less than δαn. Clearly˜rt1Ank ∈Cnk for each k = 1, .

. .

, 2n.The functional T ∗x∗t ∈L∗1 attains its maximum on Cnk at an extreme point unkof Cnk . By the lemma, unk = 1An+12k−1 −1An+12kfor 2 disjoint sets An+12k−1 and An+12kwhose union is Ank.Furthermore, An+12k−1 and An+12kare of equal measure sinceRAnk unk dµ = 0.Condition (3) holds since for i = 1, .

. .

, pn and k = 1, . .

. , 2ny∗n,i(Th2n+k) = α−1nZAnk(T ∗y∗n,i)unk dµ = 0 .Condition (1) follows from the observations thatx∗tn(Th2n+k) = α−1n(T ∗x∗tn)unk ≥α−1n(T ∗x∗tn)(˜rt1Ank )and|(T ∗x∗tn)(˜rt1Ank ) −(T ∗x∗tn)(rt1Ank )| ≤||˜rt −rt||L1 < δ αnand(T ∗x∗tn)(rt1Ank ) ≥2δαn .Proof of Lemma 2.

Fix a function f of C such that |f| ̸= 1A. Find a positive αand a subset B of A with positive measure such that |f1B| < 1 −α.Let ˜Σ = B ∩Σ.

Consider the measures λi : ˜Σ →R given by λi(E) ≡RE gi dµ.Define the measure λ: ˜Σ →Rn+1 byλ(E) = (λ1(E), . .

. , λn(E), µ(E)) .Liapounoff’s Convexity Theorem gives a subset B1 of B satisfyingλ(B1) = 12λ(B) + 12λ(∅).

Set B2 = B \ B1. Note thatλi(B1) = 12λi(B) = λi(B2)andµ(B1) = 12µ(B) = µ(B2)for i = 1, .

. .

, n. Setf1 = f + α (1B1 −1B2)andf2 = f + α (1B2 −1B1) .Clearly f1 and f2 are in C and f = 12f1 + 12f2. Thus f is not an extreme point ofC.4

3. geometric view-pointConsider a non-completely-continuous operator T : L1 →X along with the corre-sponding Haar system {hj} from Theorem 1. LetInk = [ k−12n , k2n )n,k be the usualdyadic splitting of [0, 1] with corresponding Haar functions {˜hj}j≥1.

Consider themap ˜T ≡T ◦S where S : L1 →L1 is the isometry that takes ˜hj to hj. Theorem 1gives that there is a sequence {x∗n}n≥0 in S(X∗) and a subspace X0 of X such that(1) x∗n( ˜T˜h2n+k) > δ for some δ > 0(2) {sp( ˜T˜hj : 2n−1 < j ≤2n)}n≥0 is a finite dimensional decomposition of X0with constant at most 1 + τ.The next corollary follows from the observation that ˜T is not completely continuousand ˜T L1 ⊂X0.Corollary 3.

A Banach space failing the CCP has a subspace with a finite dimen-sional decomposition (with constant arbitrarily close to 1) that fails the CCP.A tree in X is a system of the form {xnk : n = 0, 1, . .

. ; k = 1, .

. .

, 2n} satisfyingxnk = xn+12k−1 + xn+12k2.Associated to a tree is its difference system {dj}j≥1 where d1 = x01 andd2n+k = xn+12k−1 −xn+12k2.There is a one-to-one correspondence between the bounded linear operators T fromL1 into X and bounded trees {xnk} growing in X. This correspondence is realizedby T(˜hj) = dj.A tree is a δ-Rademacher tree if || P2nk=1 d2n+k|| ≥2nδ.

A tree is a separated δ-tree if there exists a sequence {x∗n}n≥0 in S(X∗) such that x∗n(d2n+k) >δ. Clearly,a separated δ-tree is also a δ-Rademacher tree.

The operator corresponding to aδ-Rademacher tree is not completely continuous since the image of the Rademacherfunctions stay large in norm. Thus if a bounded δ-Rademacher tree (or separatedδ-tree) grows in X, then X fails the CCP.In any Banach space failing the CCP, a bounded δ-Rademacher tree grows (see[G1] for a direct proof); in fact, even a bounded separated δ-tree grows (see [G2]for an indirect proof).

The proof of Theorem 1 is a direct proof that if X fails CCPthen a bounded separated δ-tree, with a difference sequence naturally blocking intoa finite dimensional decomposition, grows in X.As previously mentioned, we do not know whether a space failing CCP must havea subspace with a basis which fails CCP. However, if the space also fails cotype (i.e.the space contains ℓn∞uniformly for all n), then Theorem 1 can be modified to showthis is so.5

The main point is that if X fails cotype, W is a finite dimensional subspace ofX, and Z is a finite codimensional subspace of X, then there is a finite dimensionalsubspace Y of Z such that W + Y has a basis with basis constant less than, say,10. To see this, use the fact ([P], [JRZ]) that W is (1 + ǫ)-complemented in afinite dimensional space which has a basis with basis constant less than 1 + ǫ andembed the complement to W in that superspace into Z ∩⊥F, where F is a finitesubset of X∗which (1 + ǫ)-norms W. This is possible because finite codimensionalsubspaces of X must contain ℓn∞uniformly for all n and hence [J] contain even(1 + ǫ)-isomorphs of ℓn∞for all n.Corollary 4.

A Banach space failing the CCP and failing cotype has a subspacewith a basis that fails the CCP.To see this, it is enough by the argument for Corollary 3 to observe that whenX fails cotype Theorem 1 can be improved by adding:(D) a finite dimensional subspace Yn of X for each n ≥0 with Y0 = ∅,changing (2) and (3) to:(2′) {y∗n,i}pni=1 norms sp(∪nk=0Yk ∪{Thj : 1 ≤j ≤2n}) within τn(3′) y∗n,i(y) = 0 for y ∈Yn+1 ∪{Th2n+k}2nk=1 and 1 ≤i ≤pn,and adding:(4) sp (Yn+1 ∪{Th2n+k}2nk=1) has a basis with basis constant less than 10.Condition (2′) is easily arranged. To achieve (D), (3′), and (4), at the inductivestep in the proof of Theorem 1, after selecting An+1k(thereby defining hj for j =2n + 1, .

. ., 2n+1), choose a finite dimensional space Yn+1 ⊂⊥{y∗n,i}pni=1 so thatYn+1 + sp{Th2n+k}2nk=1 has a basis with basis constant less than 10.References[B1].

J. Bourgain, Dentability and finite-dimensional decompositions, Studia MathematicaLXVII (1980), 135–148.[B2]. J. Bourgain, Dunford-Pettis operators on L1 and the Radon-Nikod´ym property, Israel J.Math.

37 (1980), 34–47.[BR]. J. Bourgain and H. P. Rosenthal, Martingales valued in certain subspaces of L1, IsraelJ.

Math. 37 (1980), 54–75.[C].

P. G. Casazza, Finite dimensional decompositions in Banach spaces, ContemporaryMathematics 52 (1986), 1 –31.[G1]. Maria Girardi, Dunford-Pettis operators on L1 and the complete continuity property,Thesis (1990).[G2].

Maria Girardi, Dentability, trees, and Dunford-Pettis operators on L1, Pacific J. Math.148 (1991), 59–79.[DU]. J. Diestel and J. J. Uhl, Jr., Vector Measures, Math.

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80 (1964), 542–550.[JRZ]. W. B. Johnson, H. P. Rosenthal, and M. Zippin, On bases, finite dimensional decompo-sitions, and weaker structures in Banach spaces, Israel J.

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[W]. Alan Wessel, S´eminaire d’Analyse Fonctionnelle (Paris VII–VI, 1985–1986), Publica-tions Math´ematiques de l’Universit´e Paris VII, Paris.University of South Carolina, Department of Mathematics, Columbia, SC 29208.E-mail: girardi@math.scarolina.eduTexas A&M University, Department of Mathematics, College Station, TX 77843.E-mail: WBJ7835@venus.tamu.edu7

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