MRI GIRRDI ii BSTRCT The interplay between the behavior of bounded linear operators from L 1 a Banach space X and the internal geometry of X has long

Transcription

1 DUNFORD-PETTIS OPERTORS ON L 1 ND THE COMPLETE CONTINUITY PROPERTY BY MRI KTHRYN GIRRDI B.S., Santa Clara University, 1984 DVISOR Dr. J.J. Uhl, Jr. THESIS Submitted in partial fulllment of the requirements for the degree of Doctor of Philosophy in Mathematics in the Graduate College of the University of Illinois at Urbana-Champaign, 1990 Urbana, Illinois i

2 MRI GIRRDI ii BSTRCT The interplay between the behavior of bounded linear operators from L 1 a Banach space X and the internal geometry of X has long been evident. into The Radon-Nikodym property (RNP) and strong regularity arose as operator theoretic properties but were later realized as geometric properties. nother operator theoretic property, the complete continuity property (CCP), is a weakening of both the RNP and strong regularity. Banach space X has the CCP if all bounded linear operators from L 1 into X are Dunford-Pettis (i.e. take weakly convergent sequences to norm convergent sequences). There are motivating partial results suggesting that the CCP also can be realized as a geometric property. This thesis provides such a realization. Our rst step is to derive an oscillation characterization of Dunford-Pettis operators. Using this oscillation characterization, we obtain a geometric description of the CCP; namely, we show that X has the CCP if and only if all bounded subsets of X are Bocce dentable, or equivalently, all bounded subsets of X are weak-normone dentable. This geometric description leads to yet another; X has the CCP if and only if no bounded separated -trees grow inx, or equivalently, no bounded -Rademacher trees grow inx. We also localize these results. We motivate these characterizations by the corresponding (known) characterizations of the RNP and of strong regularity.

3 THESIS iii CKNOWLEDGEMENTS The tree of wisdom and knowledge grows unboundedly in a nourishing environment. I wish to express my sincere gratitude to my thesis advisor, Professor J.J. Uhl, Jr., for providing the essentials for this growth through his generous help, guidance, encouragement, and patience during my life as a graduate student. More than a convivial advisor, he is above all else a true friend and exuberant teacher. Iwould like to thank the many professors who have encouraged me with their stimulating discussions and kind words. I am also grateful to my family for always believing in me. To my fellow chocolate and cappuccino connoisseurs, thank you for the comforting chats. To my racquetball partners, thank you for the many ne games in which we shared laughter and vented frustration. To my fellow baseball fans, maybe next year.

4 MRI GIRRDI iv TBLE OF CONTENTS CHPTER 1. INTRODUCTION :::::::::::::::::::::::::::::::::::::::::::::::::::::::1 2. OSCILLTION ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::6 3. DENTBILITY ::::::::::::::::::::::::::::::::::::::::::::::::::::::::19 4. BUSHES ND TREES :::::::::::::::::::::::::::::::::::::::::::::::::35 5. LOCLITION :::::::::::::::::::::::::::::::::::::::::::::::::::::::45 REFERENCES ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::49 VIT ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::51

5 THESIS 1 CHPTER 1 INTRODUCTION The interplay between the behavior of bounded linear operators from L 1 a Banach space X and the internal geometry of X has long been evident. into The Radon-Nikodym property (RNP) was originally dened in terms of the behavior of operators from L 1 into X. Rieel, Maynard, Hu, Davis, and Phelps provided the rst concrete realization of the RNP as a geometric property of a Banach space when they showed that X has the RNP if and only if all bounded subsets of X are dentable. From this geometric characterization follows another; namely, X has the RNP if and only if no bounded -bushes grow inx, and in the case that X is a dual space (Stegall), if and only if no bounded -trees grow inx. Similarly, the concept of strong regularity, aweakening of the RNP, arose as an operator theoretic property [B1] but was later realized as a geometric property [GGMS]. The complete continuity property (CCP), another operator theoretic property, has been around for many years. Dunford and Pettis showed that a Banach space with the RNP has the CCP; in fact, a strongly regular space also has the CCP. There are motivating partial results suggesting that the CCP also can be realized as a geometric property. Bourgain showed that X has the CCP if no bounded - trees grow inx. s for the case in which X is a dual space, work of Pelczynski, Hagler, and Bourgain gives that X has the CCP if and only if X is strongly regular. In this thesis, we shall complete the cycle by showing that the complete continuity property is indeed an internal geometric property of a Banach space.

6 MRI GIRRDI 2 Our rst step is to derive an oscillation characterization of Dunford-Pettis operators (Chapter 2). Using this oscillation characterization, we obtain a geometric description of the CCP; namely, we show that X has the CCP if and only if all bounded subsets of X are Bocce dentable (Chapter 3). This geometric description leads to yet another; X has the CCP if and only if no bounded separated -trees grow inx(chapter 4). We motivate these characterizations by the corresponding (known) characterizations of the Radon-Nikodym property and of strong regularity. Throughout this thesis, X denotes an arbitrary Banach space, X the dual space of X, B(X) the closed unit ball of X, and S(X) the unit sphere of X. The triple (; ;) refers to the Lebesgue measure space on [0; 1], + to the sets in with positive measure, L 1 to L 1 (; ;), and L 1 -compact to compact in the L 1 -norm topology. subset K of L 1 is uniformly bounded if sup f2k k f k L 1 is nite. ll notation and terminology, not otherwise explained, are as in [DU]. For clarity, known results are presented as facts while new results are presented as theorems, corollaries, lemmas, and observations. s for basic denitions, the RNP, strong regularity, and the CCP each have several equivalent formulations. We shall dene these properties by examining the behavior of bounded linear operators from L 1 into X. Recall that a bounded linear operator R T : L 1! X is (Bochner) representable if there is a g in L 1 (; X) such that Tf = fg d for each f in L 1(). Banach space X has the Radon-Nikodym property if all bounded linear operators from L 1 into X are Bochner representable. bounded linear operator T : L 1! X is strongly regular if for each bounded subset D of L 1 and >0there are nitely many slices S 1 ;S 2 ;::: ;S n of D satisfying TS1 +:::+TS n diam n <;

7 THESIS 3 where a slice of a subset D of Banach space Y is any non-empty set of the form fy 2 D : y (y) >gwhere y 2 Y and 2 R are xed. Banach space X is strongly regular if all bounded linear operators from L 1 into X are strongly regular. n operator is Dunford-Pettis if it maps weakly convergent sequences to norm convergent sequences. Banach space X has the complete continuity property (CCP) if all bounded linear operators from L 1 into X are Dunford-Pettis. It is well known that if a bounded linear operator T from L 1 into X is representable then T is strongly regular; and if T is strongly regular then T is Dunford- Pettis. Thus, if X has the Radon-Nikodym property then X is strongly regular, and if X is strongly regular then X has the CCP. The dual of the James tree space [cf. J] is strongly regular yet fails the RNP. Talagrand [cf. T1] has constructed a space that has the CCP yet is not strongly regular. The following fact provides several equivalent formulations of the CCP. Fact 1.1. For a bounded linear operator T from L 1 into X, the following statements are equivalent. (1) T is a Dunford-Pettis operator. (2) T maps weak compact sets to norm compact sets. (3) T (B(L 1 )) is a relatively norm compact subset of X. (4) The corresponding vector measure F :!Xgiven by F (E) =T( E ) has a relatively norm compact range in X. (5) The adjoint of the restriction of T to L 1 from X into L 1 is a compact operator. (6) s a subset of L 1, T (B(X )) is relatively L 1 -compact. (7) E T converges to T in the operator topology of L(X ;L 1 ). (8) TE converges to T in the operator topology of L(L 1 ; X).

8 MRI GIRRDI 4 The equivalence of (2) and (3) follows from the fact that the subsets of L 1 that are relatively weakly compact are precisely those subsets that are bounded and uniformly integrable, which in turn, are precisely those subsets that can be uniformly approximated in L 1 -norm by uniformly bounded subsets. The other implications in Fact 1.1 are straightforward and easy to verify. Because of (4), the CCP is also referred to as the compact range property (CRP). Towards a martingale characterization of the CCP, x an increasing sequence f n g n0 of nite positive interval partitions of such that _ ( n ) = and 0 = fg. Let F n denote the sub--eld ( n ) of that is generated by n. For f in L 1 (X), let E n (f) denote the conditional expectation of f given F n. Denition 1.2. sequence ff n g n0 in L 1 (X)isanX-valued martingale with respect to ff n g if for each n we have that f n is F n -measurable and E n (f n+1 ) = f n. The martingale ff n g is uniformly bounded provided that sup n k f n k L 1 is nite. Often the martingale is denoted by ff n ; F n g in order to display both the functions and the sub--elds involved. There is a one-to-one correspondence between the bounded linear operators T from L 1 into X and the uniformly bounded X-valued martingales ff n ; F n g. This correspondence is obtained by taking T (g) = lim n!1 f n (!)g(!) d(!) if ff n g is the martingale, and f n (!) = X E2 n T( E ) (E) E(!) if T is the operator. It is easy to verify that a bounded linear operator from L 1 into X is representable by f 2 L 1 (; X) if and only if the corresponding martingale converges in L 1 (X)- norm to f. Fact implies that a bounded linear operator T from L 1 into X is

9 THESIS 5 Dunford-Pettis if and only if lim m;n!1 sup k E n (T x ), E m (T x ) k L 1=0: x 2B(X ) Since E n (T x )=x f n in L 1,wehave the following martingale characterization of Dunford-Pettis operators, and thus of the CCP. Fact 1.3. bounded linear operator from L 1 into X is Dunford-Pettis if and only if the corresponding martingale is Cauchy in the Pettis norm. Consequently, a Banach space X has the CCP if and only if all uniformly bounded X-valued martingales are Pettis-Cauchy.

10 MRI GIRRDI 6 CHPTER 2 OSCILLTION Fix a bounded linear operator T from L 1 into X and consider the uniformly bounded subset T (B(X )) = f T (x ) : k x k1gof L 1. In this chapter, we shall explore the oscillation behavior of elements in T (B(X )) in order to attain information about T, and thus also geometric information about X. We can determine if T is representable by examining T (B(X )) [cf. GGMS]. Fact 2.1. The following statements are equivalent. (1) T is representable. (2) T (B(X )) is equi-measurable, i.e. for each >0 there is an in + with () > 1, such that, as a subset of L 1, f(t x ) : k x k1gis relatively L 1 -norm-compact. (3) For each >0 and B in + there is a subset of B with positive measure satisfying diam n T (E ) (E) : E and E 2 + o <: Recall that for a bounded function f in L 1 and in +, the oscillation of f restricted to is given by osc f = ess sup f(!),!2 ess inf!2 f(!) : Condition (3) may be viewed as an oscillation condition on T (B(X )) since diam n T (E ) (E) : E and E 2 + o = sup osc (T x ) x 2B(X : )

11 THESIS 7 We can also determine whether T is strongly regular by examining the oscillation of elements in T (B(X )). Towards this, recall the following well-known fact [cf. GGMS]. Fact 2.2. If K is a uniformly bounded subset of L 1, then the following statements are equivalent. (1) K is a set of small oscillation, i.e. for each >0 there is a nite positive measurable partition P of such that for each f in K X () osc f < : 2P (2) K has the Bourgain property, i.e. for each >0 and B in + there is a nite collection F of subsets of B each with positive measure such that for each f in K there exists an in F satisfying osc f <: pplying this fact to the uniformly bounded subset T (B(X )) provides the following characterization of strongly regular operators [cf. GGMS]. Fact 2.3. The following statements are equivalent. (1) T is a strongly regular operator. (2) T (B(X )) is a set of small oscillation. (3) T (B(X )) has the Bourgain property. Clearly, the oscillation condition [Fact 2.1.3] on T (B(X )) that characterizes representable operators implies the oscillation condition [Fact 2.3.3] on T (B(X )) that characterizes strongly regular operators. The natural question to explore next

12 MRI GIRRDI 8 is what oscillation condition on T (B(X )) characterizes Dunford-Pettis operators. Clearly, the condition must be some weakening of the strongly regular conditions. Towards this weakening, we now present a form of a L 1 -oscillation that is a weakening of the usual L 1 -oscillation. Denition 2.4. For f in L 1 and in, the Bocce oscillation of f on is given by Bocce-osc f R R jf, fd jd () () ; where 0/0 is 0. Since Bocce-osc f osc f for any bounded f in L 1 and in +, the following denitions are weakenings of the corresponding conditions in Fact 2.2. Denition 2.5. subset K of L 1 is a set of small Bocce oscillation if for each >0 there is a nite positive measurable partition P of such that for each f in K X () Bocce-osc f < : 2P Denition 2.6. subset K of L 1 satises the Bocce criterion if for each >0 and B in + there is a nite collection F of subset of B each with positive measure such that for each f in K there is an in F satisfying Bocce-osc f <: We now state the result corresponding to Fact 2.2. Theorem 2.7. If K is a uniformly bounded subset of L 1, then the following statements are equivalent. (1) K is relatively L 1 -compact. (2) K is a set of small Bocce oscillation. (3) K satises the Bocce criterion.

13 THESIS 9 Before passing to the proof of Theorem 2.7, we note that a direct application of this theorem and Fact 1.1 to the uniformly bounded subset T (B(X )) of L 1 yields the following characterization of Dunford-Pettis operators, this chapter's main result. Corollary 2.8. If T is a bounded linear operator from L 1 into X, then the following statements are equivalent. (1) T is a Dunford-Pettis operator. (2) T (B(X )) is a set of small Bocce oscillation. (3) T (B(X )) satises the Bocce criterion. The remainder of this chapter is devoted to the proof of Theorem 2.7. Because of its length and complexity and also for the sake of the clarity of the exposition, we present the implications in this theorem as three separate theorems. Theorem 2.10 below shows the equivalence of (1) and (2) in Theorem 2.7. Remark 2.9. For a nite measurable partition of, let E (f) denote the conditional expectation of f relative to the sigma eld generated by. Thus E (f) = X B2 R B fd (B) B; observing the convention that 0=0 is 0. Theorem If K isabounded subset of L 1, then the following statements are equivalent. (1) K is relatively L 1 -compact. (2) For each >0there is a nite measurable partition of such that k f, E (f) k L 1 < for each f in K. (3) K is a set of small Bocce oscillation.

14 MRI GIRRDI 10 Proof. For bounded subsets of L 1, the equivalence of (1) and (2) is well-known and easy to check. The equivalence of (2) and (3) follows directly from X R B k f, E (f) k L 1 = j f, fd (B) Bj d X B2 R B = j f, fd j d B2 B (B) = X B2 (B) Bocce-osc f B ; and the denition of a set of small Bocce oscillation. Viewed as a function from into the real numbers, Bocce-osc f () is not increasing (see Example 2.14); however, () Bocce-osc f () is increasing (see Remark 2.15). With this in mind, we now proceed with Theorem 2.11 which shows that (2) implies (3) in Theorem 2.7. Theorem If a subset of L 1 is a set of small Bocce oscillation, then it satises the Bocce criterion. Proof. Let the subset K of L 1 be a set of small Bocce oscillation. Fix >0 and B in +. Since K is a set of small Bocce oscillation, there is a positive measurable partition = f 1 ; 2 ;::: ; n g of such that for each f in K nx ( i ) (Bocce-osc f i ) < (B) : i=1 Set F = f i \ B : i 2 and ( i \ B) > 0g: Fix f in K. Since the function () Bocce-osc f () is increasing, if Bocce-osc f i \B for each set i \ B in F then we would have that (B) = nx i=1 ( i \ B)

15 THESIS 11 nx ( i \ B) Bocce-osc f i \B i=1 nx ( i ) Bocce-osc f i i=1 < (B) : This cannot be, so there is a set i \ B in F such that the Bocce-osc f i \B <. Thus the set K satises the Bocce criterion. Theorem 2.12 shows that (3) implies (2) in Theorem 2.7. We shall present the author's original proof of this implication. For a new short proof, we refer the reader to [G3]. Theorem If a uniformly bounded subset of L 1 satises the Bocce criterion, then it is relatively L 1 -compact. We need the following lemma which we will prove after the proof of Theorem Lemma Let the uniformly bounded subset K of L 1 satisfy the Bocce criterion. If >0and ff n g is a sequence ink, then there are disjoint sets B 1 ;::: ;B p in + and a subsequence fg n g of ff n g satisfying j g n, R px j=1 B j g n d Bj j d 2 : () (B j ) Proof of Theorem Let the uniformly bounded subset K of L 1 satisfy the Bocce criterion. Choose a sequence ff n g in K and a sequence f k g of positive real numbers decreasing to zero. It suces to nd a sequence f k g of nite measurable partitions of and a nested sequence fffn kg ng k of subsequences of ff n g such that for all positive integers n and k k f k n, E k (f k n) k L 1 2 k ; ()

16 MRI GIRRDI 12 where E k (f) denotes the conditional expectation of f relative to the -eld generated by k (see Remark 2.9). For then the set ff n n g is relatively L 1 -compact since it can be uniformly approximated in the L 1 -norm within 2 k by the relatively compact set fe k (f n n )g nk [ff n ng n<k ; hence, there is a L 1 -convergent subsequence of ff n g. Towards (), repeated applications of Lemma 2.13 yield for each positive integer k: (1) disjoint subsets B k 1 ;::: ;Bk n k of each with positive measure, and (2) a subsequence ff k n g n of ff k,1 n g n (where we write ff 0 n g for ff ng ) satisfying for each positive integer n Xn k j fn k, j=1 R B k j f k n d (B k j ) B k j j d k : Set B k 0 =n [ n k j=1 B k j and let k be the partition generated by B k 0 ;Bk 1 ;::: ;Bk n k. Still observing the convention that 0=0 is 0,two appeals to the above inequality yield for each positive integer n and k k f k n, E k(f k n ) k L1 = k + = k + Xn k j fn k, j=0 X j f k n, n k B k 0 B k 0 j=1 R B k j f k n d (Bj k) B k j j d R Bj k f n k d (Bj k) B k j j d + j f k n j d Xn k j fn k, j=1 R B k j f k n d (B k j ) B k j j d j R B0 k f n k d (B k) B k 0 j d 0

17 k + 2 k : THESIS 13 n k R X j fn k Bj, k f n k d (Bj k) B k j j d j=1 Thus the proof of Theorem 2.12 is nished as soon as we verify Lemma Proof of Lemma Let the uniformly bounded subset K of L 1 satisfy the Bocce criterion. Fix >0 and a sequence ff n g in K. The proof is an exhaustion-type argument. Let E 1 denote the collection of subsets B of such that there are disjoint subsets B 1 ;::: ;B p of B each with positive measure and a subsequence ff nk g of ff n g satisfying B j f nk, R px j=1 B j f nk d Bj j d (B) : (B j ) Since K satises the Bocce criterion, there is an in + and a subsequence ff nk g of ff n g satisfying j f nk, R f n k d () j d () : Thus the collection E 1 is not empty. If there is a set B in E 1 with corresponding subsets B 1 ;::: ;B p and a subsequence fg n g of ff n g satisfying j g n, R px j=1 then we are nished since (1) implies (). g B j n d Bj j d ; (1) (B j ) Otherwise, let j 1 be the smallest positive integer for which there is a C 1 in E 1 with j1 1 (C 1). ccordingly, there is a nite sequence fcj 1 g of disjoint subsets of C 1 each with positive measure and a subsequence ffn 1g of ff ng satisfying X R j fn 1, C 1 f n 1 d j C1 (C 1 j j ) C 1 j j d (C 1 ) : (2)

18 MRI GIRRDI 14 Note that since condition (1) was not satised, (C 1 ) <()=1. Let E 2 denote the collection of subsets B of n C 1 such that there are disjoint subsets B 1 ;::: ;B p of B each with positive measure and a subsequence ff nk g of ff 1 ng satisfying B j f nk, R px j=1 f B nk d j Bj j d (B) : (B j ) Since K satises the Bocce criterion and (nc 1 ) > 0, we see that E 2 is not empty. If there is a set B in E 2 with corresponding nite sequence fcj 2 g of subsets of B and a subsequence fg n g of ffng 1 satisfying X R C j g n, 2 g n d j nc1 (C 2 j j ) C 2 j j d ( n C 1 ) ; (3) then we are nished. For in this case, () holds since inequalities (2) and (3) insure that j g n, X k=1;2 j R C k j g n d (C k j ) C k j j d (C 1 )+( n C 1 ) = : Otherwise, let j 2 be the smallest positive integer for which there is a C 2 in E 2 with j2 1 (C 2). ccordingly, there is a nite sequence fcj 2 g of disjoint subsets of C 2 each with positive measure and a subsequence ffn 2g of ff n 1g satisfying X R j fn 2, C 2 f n 2 d j C2 (C 2 j j ) C 2 j j d (C 2 ) : Note that since condition (3) was not satised, (C 2 ) <( n C 1 ). Continue in this way. If the process stops in a nite number of steps then we are nished. If the process does not stop, then diagonalize the resultant sequence f ff k ng 1 n=1 g 1 k=1 of sequences to obtain the sequence ff n n g 1 n=1 and set C 1 = S 1 k=1 C k.

19 THESIS 15 Note that ( n C 1 )=0. For if ( n C 1 ) > 0 then, since K satises the Bocce criterion, there is a subset B of n C 1 with positive measure and a subsequence fh n g of ff n n g satisfying B j h n, R B h n d (B) j d < (B): Since for each positive integer m mx k=1 1 j k ( m[ k=1 C k ) () and j m > 1, we can choose an integer m>1 such that But this implies that B is in E m since 1 j m, 1 (B) : B n C 1 n and fh n g 1 n=m,1 is a subsequence of ff n n g 1 n=m,1, which in turn is a subsequence of m,1 [ ff m,1 n g 1 n=1. This contradicts the choice of j m.thus ( n C 1 )=0. Since K is uniformly bounded, there is a real number M such that k f k 1 M for each f in K. Pick aninteger m so that M( n m[ k=1 k=1 C k C k ) : Since ffn n g 1 n=m is a subsequence of ffn m g 1 n=1, for each f in ffn n g 1 n=m we have X R Cj jf, kfd (Cj k) Cj kjd 1km j = mx k=1 mx k=1 C k j f, X j R C k j fd (C k ) + M( n + = 2 : (C k j ) C k j jd + n S m k=1 C k m[ k=1 C k ) jfj d

20 MRI GIRRDI 16 So the disjoint subsets fcj k : j 1 and k =1;::: ;mg along with the subsequence ffn n g 1 n=m of ff n g satisfy the conditions of the lemma. This completes the proof of Theorem 2.7. The proof that if K has the Bourgain property then K is a set of small oscillation uses the fact that the function osc f () from into the real numbers is increasing [cf. GGMS]. The proof of Theorem 2.12 is complicated by the fact that the function Bocce-osc f () does not enjoy this property, as our next example illustrates. Example Let =[0;1=4] and B =[0;1]. Dene the function f from [0,1] into the real numbers by f(t) = C (t) where C =[1=8;1]. It is straightforward to verify that Bocce-osc f =1=2 but Bocce-osc f B =7=32. Remark In the proof of Theorem 2.11, we used the fact that the function () Bocce-osc f () () j f, R () fd () j d is an increasing function from into the reals numbers. To see this, let be a subset of B with and B in +. Let R m = fd and m B = () R B fd (B) : Note that j f, m j d = B j f, m B j d + j f, m B j d, Bn j m B, m j d j f, m B j d + j m B, m j d : Thus j f, m j d B j f, m B j d

21 THESIS 17 since j m B, m j d = () j m B, m j = j () (B) =j () (B) =j =j Bn Bn B B fd, fd, B fdj fd, fd, (B),() (B) (f,m B )d j : B Bn fdj fd j Remark Fix f 2 L 1. Viewed as a function from into the real numbers, Bocce-osc f() is continuous, i.e. if a sequence f ng in converges to 2 in the sense that ( 4 n ) tends to 0, then Bocce-osc f n tends to Bocce-osc f. We can verify this using methods similar to the methods in Remark 2.15; or we can note that this is an easy consequence of the Lebesgue Dominated Convergence Theorem. Remark If a subset K of L 1 satises the Bocce criterion, then the translate of K by al 1 -function f also satises the Bocce criterion. If K is uniformly bounded and f is bounded, then this is an easy consequence of Theorem 2.7. For the general case, let the subset K of L 1 satisfy the Bocce criterion and f 2 L 1. We shall to show that the set K + f fg+f : g2kg satises the Bocce criterion. Towards this end, x >0 and B 2 +. Find B 0 B with B such that f is bounded on B 0. pproximate f B 0 in L 1-norm within 4 by a simple function ~ f. Find C B 0 with C 2 + such that ~ f is constant onc. Since K satises the Bocce criterion, we can nd a nite collection F of subsets corresponding to 2 and C. Fix g + f 2 K + f. Find 2Fsuch that Bocce-osc g < 2 : Note that since

22 MRI GIRRDI 18 ~f is constant on, Bocce-osc g = Bocce-osc (g + ~ f) : Compute, Bocce-osc (g + f) Bocce-osc (g + ~ f) + Bocce-osc ( ~ f, f) Bocce-osc g + 2 k ( ~ f, f) k L 1 < : Thus K + f satises the Bocce criterion.

23 THESIS 19 CHPTER 3 DENTBILITY In this chapter, we examine in which Banach spaces bounded subsets have certain dentability properties. Dentability characterizations of the RNP are well-known [cf. DU and GU]. Fact 3.1. The following statements are equivalent. (1) X has the RNP. (2) Every bounded subset D of X is dentable. Denition 3.2 D is dentable if for each >0 there is x in D such that x=2co (D n B (x)) where B (x) =fy2x:kx,yk<g. (3) Every bounded subset D of X is -dentable. Denition 3.3 D is -dentable if for each >0 there is an x in D such that if x has the form x = P n i=1 iz i with z i 2 D, 0 i, and P n i=1 i =1, then k x, z i k<for some i. (4) Every bounded subset D of X has slices of arbitrarily small diameter. weakening of condition (4) provides a dentability characterization of strongly regular spaces [cf. GGMS]. This characterization is often used as the denition. Fact 3.4. X is strongly regular if and only if for each bounded subset D of X and >0 there are nitely many slices S 1 ;::: ;S n of D such that diam n S1 + :::+S n n o <: The natural question to explore next is what dentability condition characterizes the CCP. Towards this, the next denition is a weakening of Denition 3.2.

24 MRI GIRRDI 20 Denition 3.5. subset D of X is weak-norm-one dentable if for each >0 there is a nite subset F of D such that for each x in S(X ) there is x in F satisfying x=2co fz 2 D : j x (z, x) jg co (D n V ;x (x)) : Petrakis and Uhl [PU] showed that if X has the CCP then every bounded subset of X is weak-norm-one dentable. For our characterization of the CCP, weintroduce the following variations of denition 3.3 that are useful in showing the converse of the above implication of [PU]. Denition 3.6. subset D of X is Bocce dentable if for each >0 there is a nite subset F of D such that for each x in S(X ) there is x in F satisfying: if x = nx i=1 i z i with z i 2 D; 0 i ; and nx i=1 i =1; then nx i=1 i j x (x, z i ) j <: Denition 3.7. subset D of X is midpoint Bocce dentable if for each >0 there is a nite subset F of D such that for each x in S(X ) there is x in F satisfying: if x = 1 2 z z 2 with z i 2 D then j x (x, z 1 ) jjx (x,z 2 )j<: We obtain equivalent formulations of the above denitions by replacing S(X ) with B(X ). The next theorem, this chapter's main result, shows that these dentability conditions provide an internal geometric characterization of the CCP.

25 THESIS 21 Theorem 3.8. The following statements are equivalent. (1) X has the CCP. (2) Every bounded subset of X is weak-norm-one dentable. (3) Every bounded subset of X is midpoint Bocce dentable. (4) Every bounded subset of X is Bocce dentable. The remainder of this chapter is devoted to the proof of Theorem 3.8. Because of its length and complexity and also for the sake of clarity of the exposition, we present the implications as separate theorems. It is clear from the denitions that (2) implies (3) and that (4) implies (3). For completeness sake, Fact 3.14 presents the proof of [PU] that (1) implies (2). Using Corollary 2.8, Theorem 3.11 shows that (3) implies (1). That (1) implies (4) follows from Theorem 3.9 and the martingale characterization (Fact 1.3) of the CCP. Theorem 3.9. If a subset D of X is not Bocce dentable, then there is an increasing sequence f n g of partition of [0; 1) and a D-valued martingale ff n ;( n )g that is not Cauchy in the Pettis norm. Moreover, f n g can be chosen so that _ ( n )=, 0 =fgand each n partitions [0; 1) into a nite number of half-open intervals. Proof. Let D be a non-bocce-dentable subset of X. ccordingly, there is > 0 satisfying: for each nite subset F of D there is x F in S(X ) such that each x in F has the form x = with mx i=1 mx i=1 i j x F (x, z i) j > i z i for a suitable choice of z i 2 D and i > 0 with mx i=1 i =1. () We shall use property () to construct an increasing sequence f n g n0 of nite

26 MRI GIRRDI 22 partitions of [0; 1), a martingale ff n ;( n )g n0, and a sequence fx n g n1 in S(X ) such that for each nonnegative integer n: P (1) R f n has the form f n = E2 n x E E where x E is in D, (2) j x n+1(f n+1, f n ) j d, (3) if E is in n, then E has the form [a; b) and (E) < 1=2 n and (4) 0 = fg. Condition (3) guarantees that _ ( n ) = while condition (2) guarantees that ff n g is not Cauchy in the Pettis norm. Towards the construction, pick an arbitrary x in D. Set 0 = fg and f 0 = x. Fix n 0. Suppose that a partition P n of consisting of intervals of length at most 1/2 n and a function f n = E2 n x E E with x E 2 D have been constructed. We now construct f n+1, n+1 and x n+1 satisfying conditions (1), (2) and (3). pply () tof=fx E : E2 n gand nd the associated x F = x n+1 in S(X ). Fix an element E =[a; b) of n. We rst dene f n+1 E. Property () gives that x E has the form x E = with mx i=1 mx i=1 i x i i j x n+1(x, x i ) j > for a suitable choice of x i 2 D and positive real numbers 1 ;::: ; m whose sum is one. Using repetition, we arrange to have i < 1=2 n+1 for each i. It follows that there are real numbers d 0 ;d 1 ;::: ;d m such that a = d 0 <d 1 <::: <d m,1 <d m =b and d i, d i,1 = i (b, a) for i =1;::: ;m :

27 Set f n+1 E = THESIS 23 mx i=1 x i [di,1 ;d i ) : Dene f n+1 on all of similarly. Let n+1 be the partition consisting of all the intervals [d i,1 ;d i ) obtained from letting E range over n. Clearly, f n+1 and n+1 satisfy conditions (1) and (3). Condition (2) is also satised since for each E =[a; b) in n wehave, using the above notation, E j x n+1(f n+1, f n ) j d = mx i=1 di =(b,a) (E) : d i,1 j x n+1(x i, x E ) j d mx i=1 i j x n+1 (x i, x E ) j To insure that ff n g is indeed a martingale, we need to compute E n (f n+1 ). Fix E =[a; b) in n. Using the above notation, we have for almost all t in E, E n (f n+1 )(t) = 1 b b, a = 1 b, a = = mx mx i=1 i=1 f n+1 d a mx di i=1 d i, d i,1 b, a i x i = f n (t) : d i,1 f n+1 d x i = x E Thus E n (f n+1 )=f n a.e., as needed. This completes the necessary constructions. We need the following lemma which we will prove after the proof of Theorem Lemma If is in + and f in L 1 () is not constant a.e. on, then there is an increasing sequence f n g of positive nite measurable partitions of such that

28 MRI GIRRDI 24 _ ( n )=\and for each n [ n E : E 2 n ; R E fd (E) R fd () o= () 2 ; and so [ n E : E 2 n ; R E fd (E) < R fd () o= () 2 : Theorem If all bounded subsets of X are midpoint Bocce dentable, then X has the complete continuity property. Proof. Let all bounded subsets of X be midpoint Bocce dentable. Fix a bounded linear operator T from L 1 into X. We shall show that the subset T (B(X )) of L 1 satises the Bocce criterion. Then an appeal to Corollary 2.8 shows that X has the complete continuity property. To this end, x >0 and B in +. Let F denote the vector measure from into X given by F (E) =T( E ). Since the subset f F (E) (E) : E B and E 2 + g of X is bounded, it is midpoint Bocce dentable. ccordingly, there is a nite collection F of subsets of B each in + such that for each x in the unit ball of X there is a set in F such that if F () () = 1 2 F (E 1 ) (E 1 ) F (E 2 ) (E 2 ) for some subsets E i of B with E i 2 +, then 1 2 x F (E 1 ) (E 1 ), x F () () x F (E 2 ) (E 2 ), x F () () < : (1) Fix x in the unit ball of X and nd the associated in F. By denition, the set T (B(X )) will satisfy the Bocce criterion provided that Bocce-osc (T x ).

29 THESIS 25 If the L 1 -function T x is constant a.e. on, then the Bocce-osc (T x ) is zero and we are nished. So assume T x is not constant a.e. on. For a nite positive measurable partition of, denote f = X E2 F (E) (E) E and E + = [ E 2 : x F (E) (E) x F () () and E, = [ E 2 : x F (E) (E) < x F () () : Note that for E in x F (E) = E (x T )d : Compute = () X E2 x f, x F () d X () = x F (E) E2 E (E) (E) () = () " (E + ) (), x F () () x F (E) (E) x F (E + ) (E + ) d, x F () (), x F () () (2) + (E, ) () x F (E, ) (E, ), x F () () Since the L 1 -function T x is bounded, for now wemay view T x as an element in L 1. Lemma 3.10 allow us to apply property (1) to equation (2). For applying Lemma 3.10 to with f T x produces an increasing sequence f n g of positive measurable partitions of satisfying # : _ ( n )=\ and (E + n )= () 2 =(E, n ) :

30 For = n, condition (2) becomes x f n, x F () d () " 1 = () 2 x F (E + n ) (E + n ) Since F () =() has the form MRI GIRRDI 26, x F () () x F (E, n ) (E, n ), x F () () # : (3) F () () = (E+ n ) () F (E + n ) (E + n ) + (E, n ) F (E, n ) () (E, n ) = 1 2 F (E + n ) (E + n ) F (E, n ) (E, n ) ; applying property (1) to equation (3) yields that for each n Since _ ( n )=\and (x f n ) = X x F (E) (E) E2 n Bocce-osc (T x ) x f n, x F () d < () : () X R E E = (T x ) d E = E n (T x ) (E) ; E2 n we have that (x f n ) converges to (T x ) in L 1 -norm. Hence, R (T x ), d R () (T x ) d () : Thus T (B(X )) satises the Bocce criterion and so, as needed, X has the complete continuity property. We now verify Lemma Proof of Lemma Fix in + and f in L 1 (). Without loss of generality, f R is not constant a.e. on and fd= 0. Find P and N in satisfying = P [ N (P )= () 2 =(N) P\N=; and P fd 2M>0 N fd,2m<0 :

31 THESIS 27 pproximate f by a simple function f() ~ P = i i () satisfying (1) k f, f ~ kl 1 <M, (2) [ i = and the i are disjoint, (3) i P if i m and i N if i>m for some positive integer m. Note that [ [ P = i and N = i : im i>m To nd the sequence f n g,we shall rst nd an increasing sequence f P n g of partitions of P and an increasing sequence f N n g of partitions of N. Then n will be the union of P n and N n.to this end, for each i obtain an increasing sequence of partitions of i : i E i0 1. & E i1 1 E i1 2. &. & E i2 1 E i2 2 E i2 3 E i2 4 such that for n =0;1;2;::: and k =1;::: ;2 n E in+1 2k,1 [ Ei n+1 2k = E in k E in+1 2k,1 \ Ei n+1 2k = ; (E in k )= ( i) 2 n : For each positive integer n, let P n be the partition of P given by P n = fp n k : k =1;::: ;2 n g where P n k = [ im E in k ; N n be the partition of N given by N n = fn n k : k =1;::: ;2 n g where N n k = [ i>m E in k ;

32 and n be the partition of given by MRI GIRRDI 28 n = P n [ N n : The sequence f n g has the desired properties. Since (P n k ) = X im ( i ) = (P ) = () 2 n 2 n 2 n+1 and (N n k ) = X i<m ( i ) 2 n = (N) 2 n = () 2 n+1 ; any element in n has measure () = 2 n+1.thus _ ( n )=\. s for the other properties, since ~ f takes the value i on E in k i we have P n k ~fd = X im E in k ~f d X = i (Ek in) im = 1 X 2 n i ( i ) im = 1 2 n P > 0 ~f d and likewise N n k ~f d = 1 2 n N ~fd < 0: Wechose ~ f close enough to f so that the above inequalities still hold when we

33 THESIS 29 replace ~ f by f, P n k fd P n k = 1 2 n P ( ~ f,m)d ~f d, M(P n k ) 1 2 P n (f,m)d, M() 2 n+1 = 1 2 P n fd, M() > M 2 n, M() 2 n = 0, M() 2 n+1 2 n+1 M [1, ()] 2 n and likewise N n k fd < M [(),1] 2 n 0 : Thus the other properties of the lemma are satised since for each n, [ n E : E 2 n ; E fd0 o = = (P) = () 2 [ n E : E2 P n o and so [ n E : E 2 n ; E fd<0 o= () 2 : Note that the partitions f n g are nested by construction. [PU, Theorem II.7] shows that (1) implies (2) in Theorem 3.8 by constructing, in a bounded non-weak-norm-one dentable subset D, a(co D)-valued martingale that is not Cauchy in the Pettis norm. Since quasi-martingales enter into their proof, let

34 MRI GIRRDI 30 us pause to recall a few denitions and facts. Fix the notation as in Denition 1.2. Let f n g n0 be a summable sequence of non-negative numbers. Denition sequence ff n g n0 in L 1 (X) isax-valued quasi-martingale with respect to ff n g and corresponding to f n g if for each n we have that f n is F n - measurable and k E n (f n+1 ), f n k L 1 n : self-contained presentation of quasi-martingales along with the decomposition theorem below may be found in [KR]. Fact 3.13 quasi-martingale decomposition theorem. Let K be a closed bounded convex subset of X. If ff n ; F n g is a K-valued quasi-martingale corresponding to f n g, then there is a K-valued martingale fg n ; F n g satisfying k f n, g n k L 1 1X j=n j : Note that a quasi-martingale is Pettis-Cauchy if and only if the corresponding martingale from the decomposition theorem is Pettis-Cauchy. Fact 3.14 [PU, Theorem II.7]. Let f n g n0 be a summable sequence of positive numbers. If a subset D of X is not weak-norm-one dentable, then there is an increasing sequence f n g of partition of [0; 1) and a D-valued quasi-martingale ff n ;( n )gcorresponding to f n g that is not Cauchy in the Pettis norm. Moreover, f n g can be chosen so that _ ( n )=, 0 =fgand each n partitions [0; 1) into a nite number of half-open intervals. Consequently, if there is a bounded nonweak-norm-one dentable subset D of X, then there is a (co D)-valued martingale that is not Pettis-Cauchy. Proof. Fix a summable sequence f n g n0 of positive numbers. Let D be subset of

35 THESIS 31 X that is not weak-norm-one dentable. ccordingly, there is an >0 satisfying: for each nite subset F of D there is an x F in S(X ) such that if x is in F then x 2 co, z 2 D : j x F (z, x) j : () We shall use property () to construct an increasing sequence f n g n0 of nite partitions of [0; 1), a quasi-martingale ff n ;( n )g n0 corresponding to f n g, and a sequence fx ng n1 in S(X ) such that for each nonnegative integer n P (1) f n has the form f n = E2 n x E E where x E is in D, (2) j x n+1(f n+1, f n ) j a.e., (3) if E is in n, then E has the form [a; b) and (E) < 1=2 n and (4) 0 = fg. Condition (3) guarantees that _ ( n ) = while condition (2) guarantees that ff n g is not Cauchy in the Pettis norm. Towards the construction, pick an arbitrary x in D. Set 0 = fg and f 0 = x. Fix n 0. Suppose P that a partition n of into intervals of length at most 1/2 n and f n = E2 n x E E with x E 2 D have been dened. We now construct f n+1, n+1 and x n+1 satisfying conditions (1), (2) and (3). To this end, apply () tof =fx E : E2 n gand nd the associated x F = x n+1 in S(X ). Fix an element E =[a; b) of n. We rst dene f n+1 E. Property () gives that for a suitable choice of x 1 ;::: ;x m in D and positive real numbers 1 ;::: ; m whose sum is one, k x E, mx i=1 i x i k n and for each i j x n+1 (x i, x E ) j : () Using repetition, we arrange to have i < 1=2 n+1 for each i. It follows that there are real numbers d 0 ;d 1 ;::: ;d m such that a = d 0 <d 1 <::: <d m,1 <d m =b

36 MRI GIRRDI 32 and Set d i, d i,1 = i (b, a) for i =1;::: ;m : f n+1 E = mx i=1 x i [di,1 ;d i ) : Dene f n+1 on all of similarly. Let n+1 be the partition consisting of all the intervals [d i,1 ;d i ) obtained from letting E range over n. Clearly, f n+1, n+1 and x n+1 satisfy conditions (1), (2) and (3). Towards insuring that ff n g is indeed a quasi-martingale corresponding to f n g, x an E =[a; b) in n. Using the above notation, we have that for almost all t in E, f n (t) = x E and E n (f n+1 )(t) = 1 b b, a = 1 b, a = = mx mx i=1 i=1 f n+1 d a mx di i=1 d i, d i,1 b, a i x i : d i,1 f n+1 d n appeal to condition () yields the necessary inequality, x i k E n (f n+1 ), f n k L 1 n : This completes the necessary constructions. Remark For a convex set D, Bocce dentability and midpoint Bocce dentability coincide since, for a xed x 2 S(X ), if x 2 D satises the condition if x = 1 2 z z 2 with z i 2 D then j x (x, z 1 ) jjx (x,z 2 )j<; then x 2 D also satises the condition if x = nx i=1 i z i with z i 2 D; 0 i ; and nx i=1 i =1;

37 then nx i=1 THESIS 33 i j x (x, z i ) j < 2: Remark It is easy to verify that B(L 1 ) is not weak-norm-one dentable, Bocce dentable, nor midpoint Bocce dentability but that B + (L 1 ) ff 2 L 1 : k f k L 1 1 and f 0 a.e. g is weak-norm-one dentable, Bocce dentable, and midpoint Bocce dentability. Remark For a bounded subset D and > 0, a point x 2 D is called an -strong extreme point of D if there is a > 0 such that if x 1 ;x 2 belong to D and there is a point u on the line segment joining x 1 and x 2 with k u, x k<, then k u, x 1 k< or k u, x 2 k<. closed bounded convex set has the pproximate Krein-Milman property (KMP) if each of its nonempty subsets has an -strong extreme point for every >0. It is clear from the denitions that if a closed bounded convex set has the KMP, then it is midpoint Bocce dentable. Remark nalogous to the RNP case, in Theorem 3.8 we would like to reduce our test sets from bounded sets to closed bounded convex sets. However, the author is not certain whether this is possible. close look at the proofs reveal that we only need that, for each bounded linear operator T from L 1 into X, the sets of the form B n T ( ) () : B and 2 + o where B 2 + to be midpoint Bocce dentable. But such sets need not be closed nor convex. If T is the identity operator on L 1, then is closed but is not convex. If T : L 1! L 2 is given by (Tf)() R 0 fd, then is neither closed nor convex.

38 MRI GIRRDI 34 If T is representable by a simple function f and B 2 +, then B is the convex hull of the values that f takes on B. Thus, if T is a bounded linear operator into a Banach space with the RNP and B 2 +, then the B is convex. If T is a bounded linear operator into a Banach space with the CCP and B 2 +, then we have already seen that B need not be closed nor convex; however, B is convex. This follows from the fact that for each >0 there is an operator T (Fact 1.1.8) that is representable by a simple function such that sup k (T, T )(f)kx < : f2b(l1)

39 THESIS 35 CHPTER 4 BUSHES ND TREES In this chapter, we examine which Banach spaces allow certain types of bushes and trees to grow in them. First let us review some known implications. Banach space X fails the RNP precisely when a bounded -bush grows in X. Thus if a bounded -tree grows in X then X fails the RNP. The converse is false; the Bourgain-Rosenthal space [BR] fails the RNP yet has no bounded -trees. However, if X is a dual space then the converse does hold. Bourgain [B2] showed that if X fails the CCP then a bounded -tree grows in X. The converse is false; the dual of the James Tree space has a bounded -tree and the CCP. Itiswell-known that if a bounded -Rademacher tree grows in X then X fails the CCP. Riddle and Uhl [RU] showed that the converse holds in a dual space. This chapter's main theorem, Theorem 4.1 below, makes precise exactly which types of bushes and trees grow in a Banach space failing the CCP. Theorem 4.1. The following statements are equivalent. (1) X fails the CCP. (2) bounded separated -tree grows in X. (3) bounded separated -bush grows in X. (4) bounded -Rademacher tree grows in X. That (1) implies (2) will follow from Theorem 4.2 below. ll the other implications are straightforward and will be veried shortly. s usual, we start with denitions. While typing this thesis, I learned that H.P. Rosenthal has also recently obtained the result that if X fails the CCP then a bounded -Rademacher tree grows in X.

40 MRI GIRRDI 36 Perhaps it is easiest to dene a bush via martingales. If f n g n0 is an increasing sequence of nite positive interval partitions of [0; 1) with _ ( n )=and 0 =fgand if ff n ;( n )g n0 is an X-valued martingale, then each f n has the form and the system f n = X E2 n x n E E fx n E : n =0;1;2;::: and E 2 n g is a bush in X. Moreover, every bush is realized this way. bush is a -bush if the corresponding martingale satises for each positive integer n k f n (t), f n,1 (t) k >: (i) bush is a separated -bush if there exists a sequence fx n g n1 in S(X ) such that the corresponding martingale satises for each positive integer n j x n (f n (t), f n,1 (t)) j >: (ii) In this case, we say that the bush is separated by fx ng. Clearly a separated -bush is also a -bush. Observation that (3) implies (1) in Theorem 4.1. If a bounded separated -bush grows in a subset D of X, then condition (ii) guarantees that the corresponding D-valued martingale ff n ;( n )g is not Pettis-Cauchy since k f n, f n,1 k P ettis j x n (f n (t), f n,1 (t)) j d > : Thus, if a bounded separated -bush grows in X then X fails the CCP (Fact 1.3). If each n is the n th dyadic partition then we call the bush a (dyadic) tree. Let us rephrase the above denitions for this case, without the help of martingales. tree

41 THESIS 37 in X is a system of the form fx n k : n =0;1;::: ; k =1;::: ;2n g satisfying for n =1;2;::: and k =1;::: ;2 n,1 x n,1 k = xn 2k,1 +xn 2k 2 : (iii) Condition (iii) guarantees that ff n g is indeed a martingale. It is often helpful to think of a tree diagrammatically: x 0 1 x 1 1 x 1 2 x 2 1 x 2 2 x 2 3 x 2 4 x 3 1 x 3 2 x3 3 x 3 4 x 3 5 x 3 6 x3 7 x 3 8 It is easy to see that (iii) is equivalent to x n 2k,1,xn 2k = 2(xn 2k,1,xn,1 k ) = 2(x n,1 k,x n 2k ): (iii') tree fx n kg is a -tree if for n =1;2;::: and k =1;::: ;2n,1 k x n 2k,1,x n,1 k kkx n 2k,x n,1 k k> : (iv) n appeal to (iii') shows that (iv) is equivalent to kx n 2k,1,x n 2kk>2: (iv') tree fx n k g is a separated -tree if there exists a sequence fx ng n1 in S(X ) such that for n =1;2;::: and k =1;::: ;2 n,1 j x n(x n 2k,1,x n,1 k )jjx n(x n 2k,x n,1 k )j> : (v)

42 nother appeal to (iii') shows that (v) is equivalent to MRI GIRRDI 38 jx n(x n 2k,1,xn 2k)j>2: (v') Furthermore, by switching x n 2k,1 and xn 2k when necessary, wemay assume that (v') is equivalent to x n(x n 2k,1,xn 2k) > 2: (v") Since a separated -tree is also a separated -bush, (2) implies (3) in Theorem 4.1. tree fx n k : n =0;1;::: ; k =1;::: ;2n g is called a -Rademacher tree [RU] if for each positive integer n k 2X n,1 k=1 (x n 2k,1, xn 2k ) k > 2n : Perhaps a short word on the connection between the Rademacher functions fr n g and Rademacher trees is in order. In light of our discussion in Chapter 1, there is a one-to-one correspondence between all bounded trees in X and all bounded linear operators from L 1 into X. If fx n kg is a bounded tree in X with associated operator T, then it is easy to verify that fx n k g is a -Rademacher tree precisely when k T (r n ) k >for all positive integers n. Fact that (4) implies (1) in Theorem 4.1 [RU]. Let ff n g be the (dyadic) martingale associated with a -Rademacher tree fx n k g.if x is in X and I n k j x (f n, f n,1 ) j d = = 2X n,1 " k=1 = 1 2 n is the dyadic interval [ (k, 1)=2n ; k=2 n ) then I2k,1 n 2 n,1 Xh k=1 2X n,1 k=1 I n,1 k j x (x n 2k,1, x n,1 k ) j d + j x (f n, f n,1 ) j d I n 2k j x (x n 2k,1, x n,1 k ) j + j x (x n 2k, x n,1 k ) j j x (x n 2k, x n,1 k ) j d i #

43 = 1 2X n,1 2 n j x (x n 2k,1, x n 2k) j k=1 1 2 n,1 X 2 n j x (x n 2k,1, x n 2k) k=1 THESIS 39 by (iii') j : From this we see that ff n g is not Cauchy in the Pettis norm since k f n, f n,1 k P ettis = sup x 2B(X ) sup x 2B(X ) X = 1 2 n,1 2 n k k=1 > 1 2 n 2n = : j x (f n, f n,1 ) j d X 1 2 n,1 2 n j x k=1 (x n 2k,1, xn 2k ) k (x n, 2k,1 xn 2k ) j Thus if a bounded -Rademacher tree grows in a subset D of X, then there is a bounded D-valued martingale that in not Pettis-Cauchy and so X fails the CCP (Fact 1.3). Observation that (2) implies (4) in Theorem 4.1. separated -tree can easily be reshued so that it is a -Rademacher tree. For if fx n k g is a separated -tree then we may assume, by switching xn 2k,1 and xn 2k when necessary, that there is a sequence fx ng in S(X ) satisfying x n(x n 2k,1, x n 2k) > 2 : With this modication fx n kg is a -Rademacher tree since X 2 n,1 k (x n 2k,1, x n 2k) kj k=1 2X n,1 k=1 X x n (x n 2k,1, x n 2k) j = 2X n,1 k=1 2 n,1 > 2 = 2 n : k=1 x n (x n 2k,1, x n 2k)

44 MRI GIRRDI 40 To complete the proof of Theorem 4.1, we need only to show that (1) implies (2). Towards this end, let X fail the CCP. n appeal to Theorem 3.8 gives that there is a bounded non-midpoint-bocce-dentable subset of X. In such a set, we can construct a separated -tree. This construction is made precise in the following theorem. Theorem 4.2. separated -tree grows in a non-midpoint-bocce-dentable set. Proof. Let D be a subset of X that is not midpoint Bocce dentable. ccordingly, there is a > 0 satisfying: for each nite subset F of D there is x F 2 S(X ) such that each x in F has the form x = x 1 + x 2 2 for a suitable choice of x 1 and x 2 in D. with jx F (x 1, x 2 )j > () We shall use the property () to construct a tree fx n k : n =0;1;::: ;k =1;::: ;2n g in D that is separated by a sequence fx n g n=1 of norm one linear functionals. Towards this construction, let x 0 1 be any element ofd. pply () with F = fx0 1 g and nd x F = x 1. Property () provides x1 1 and x1 2 in D satisfying x 0 1 = x1 1 + x1 2 2 and jx 1 (x1, 1 x1 2 )j >: Next apply () with F = fx 1 1 ;x1g and nd 2 x F = x 2.For k = 1 and 2, property () provides x 2 2k,1 and x2 2k in D satisfying x 1 k = x2 2k,1 + x2 2k 2 and jx 2(x 2 2k,1, x 2 2k)j >: Instead of giving a formal inductive proof we shall be satised by nding x 3 along with x 3 1 ;x3 2 ;::: ;x3 8 in D. pply () with F = fx 2 1 ;x2 2 ;x2 3 ;x2 4g and nd x F = x 3. For k =1;2;3 and 4, property () provides x 3 2k,1 and x3 2k in D satisfying x 2 k = 1 2 (x3 2k,1 + x 3 2k) and jx 3(x 3 2k,1, x 3 2k)j >:

45 THESIS 41 It is now clear that a separated -tree grows in such a set D. Remark 4.3. Theorem 3.8 presents several dentability characterizations of the CCP. Our proof that (1) implies (2) in Theorem 4.1 uses part of one of these characterizations; namely, ifxfails the CCP then there is a bounded non-midpoint-boccedentable subset of X. If X fails the CCP, then there is also a bounded non-weaknorm-one-dentable subset of X (Theorem 3.8). In the closed convex hull of such a set we can construct a martingale that is not Pettis-Cauchy [Fact 3.14]; furthermore, the bush associated with this martingale is a separated -bush. However, it is unclear whether this martingale is a dyadic thus the separated -bush may not be a tree. If X fails the CCP, then there is also a bounded non-bocce-dentable subset of X (Theorem 3.8). In such a set we can construct a martingale that is not Pettis-Cauchy (Theorem 3.9), but it is unclear whether the bush associated with this martingale is a separated -bush. Remark 4.4. The Rademacher functions fr n g may be viewed as a test sequence for the CCP. For if X has the CCP and T is a bounded linear operator from L 1 into X, then ft (r n )g converges to 0 in norm. If X fails the CCP, then there is a bounded linear operator T from L 1 into X such that inf n k T (r n ) k > some >0. Remark 4.5. Recall that a separated -tree is also a -tree and, after a reshuing, a -Rademacher tree. However, a bounded -Rademacher ~ tree need neither be a - tree nor a separated -tree. For example, consider the c 0 -valued dyadic martingale ff n (s 0 ;::: ;s n ;0;0;:::)g where the function s n from [0; 1] into [,1; 1] is dened by (,1) k 2,n if! 2 Ik n with k 2; s n = (,1) k if! 2 Ik n with k>2: The tree associated with ff n g is a bounded 1 -Rademacher tree but is neither a 4 -tree nor a separated -tree for any positive. Thus, since a -tree grows in a