WEAKLY COMPACT SETS IN BANACH SPACES

Transcription

1 WEAKLY COMPACT SETS IN BANACH SPACES A dissertation submitted to the Kent State University Graduate College in partial fulfillment of the requirements for the degree of Doctor of Philosophy by John Alexopoulos August 992

2 Dissertation written by John Alexopoulos B.S. Clarion University of Pennsylvania, 985 M.A. Kent State University, 987 Ph.D Kent State University, 992 Approved by, Chair, Doctoral Dissertation Committee, Members, Doctoral Dissertation Committee,,, Accepted by, Chair, Department of Mathematics and Computer Science, Dean, Graduate College ii

3 Table of Contents AKNOWLEDGEMENTS v 0 INTRODUCTION DE LA VALLÉE POUSSIN S THEOREM REVISITED Uniform integrability and De La Vallée Poussin s Theorem Some facts about N-Functions Some facts about Orlicz Spaces De La Vallée Poussin s theorem revisited ORLICZ SPACES AND THE WEAK BANACH-SAKS PROPERTY A weak compactness result reminiscent of the Dunford-Pettis theorem An application in convex function theory REFLEXIVE SUBSPACES OF NON-REFLEXIVE ORLICZ SPACES Subspaces containing complemented l Some facts about Banach Spaces with type Subspaces of L F that have type A THE PRESENCE OF UNIFORM l n s IN L (µ) BIBLIOGRAPHY iii

4 John Alexopoulos, Ph.D., August, 992 Mathematics WEAKLY COMPACT SETS IN BANACH SPACES (44 pp.) Director of Dissertation: Dr. Joseph Diestel The classical theorem of Dunford and Pettis, identifies the bounded, uniformly integrable subsets of L (µ) with the relatively weakly compact sets. Another characterization of uniform integrability is given in a theorem of De La Vallée Poussin which states that a subset K of L (µ) is bounded and uniformly integrable if and only if it is a bounded subset of some Orlicz space L F. We refine and improve this theorem in several directions. The theorem of De La Vallée Poussin does not, for instance, specify just how well the function F can be chosen. It gives little additional information in case the set in question is relatively norm compact in L (µ). Finally it gives no information on the structure of the set in the corresponding Banach space of F -integrable functions. More specifically we establish the fact that a subset K of L is relatively compact if and only if there is an N-function F so that K is relatively compact in L F. Furthermore we prove that a subset K of L is relatively weakly compact if and only if there is an N-function F so that K is relatively weakly compact in L F. We then go on to show that a large class of non-reflexive Orlicz spaces has the weak Banach-Saks property, by establishing a result for these spaces, very similar to the Dunford-Pettis theorem for L. Finally we investigate some similarities of these spaces, with the space L (µ). Kadec and Pelczýnski have shown that every non-reflexive subspace of L (µ) contains a copy of l complemented in L (µ). On the other hand Rosenthal investigated the structure of reflexive subspaces of L (µ) and proved that such subspaces, have non-trivial type. We show the same facts to hold true, for the special class of non-reflexive Orlicz spaces, we have been investigating.

5 AKNOWLEDGEMENTS At this point I want to thank all the people that helped and encouraged me during the past years. I would like to express my most sincere gratitude to my advisor Joe Diestel. His enthusiasm, optimism, advice and support made an enormous difference in my life as a mathematician and as a person in general. I would also like to thank Richard Aron and Andrew Tonge for talking to me about a variety of problems as well as for conducting a multitude of stimulating courses and seminars, from which I benefited greatly. I take now this opportunity to also express my deepest appreciation to my colleague Paul Abraham who patiently listened to many arguments, offered many suggestions and taught me a great deal through seminars and private conversations. Thanks also to Miguel Lacruz, Chris Lennard, Tony Weston, Carmen Romero and Guillermo Curbera (Bill from Seville) for the many interesting discussions that made the environment in Kent so positive for mathematical development. My undergraduate teachers, Nicholas Bezak, Ben Freed, Stephen Gendler, Mike Osessia and especially Roger Engle for transforming my world of mathematics, from a necessary nuisance, to an everlasting passion. I cannot find the right words to thank my father, mother and brother for all they have done for me. So I will just say thanks and I hope that they understand. Special thanks also to Francie (Fritz) Gifford for her kindness compassion, understanding and encouragement during some very difficult moments of my life. Finally the support of the Department of Mathematics and Computer Science and Graduate College at Kent State University as well as that of Motor Oil Hellas is gratefully acknowledged. v

6 To Jane Elizabeth

7 Chapter 0 INTRODUCTION The notation used throughout this dissertation is fairly standard. A close model is the notation in Diestel [6]. (Ω, Σ, µ) will denote a non-atomic probability space and L p (µ) will denote the Banach space of (equivalence classes of) measurable, real-valued functions on Ω, whose p-th power is µ-integrable. IR denotes the set of real numbers. The symbol is used to denote a Banach space norm. Sometimes subscripts are placed in the norm symbol, in order to identify the space on which the norm is taken. The symbol χ ( ) is used to denote characteristic functions of sets. That is for a set A in Σ, χ A is a real valued function defined on Ω by if ω A χ A (ω) = 0 if ω / A Finally given a measurable function f on Ω and a real number a, the probabilistic notation [f > a], [f a], [f < a], [f a] and [f = a] is used to describe the sets of all elements ω Ω for which f(ω) > a, f(ω) a, f(ω) < a, f(ω) a and f(ω) = a respectively. For the various concepts in Banach space theory and measure theory, not defined explicitly, the reader should consult Banach [4], Dunford and Schwartz [9], Diestel [6], Diestel and Uhl [7], Rudin [28], [29] and Halmos []. Recall that a subset K of L (µ) is called uniformly integrable if given ε > 0 there is a δ > 0 so that sup { E f dµ : f K} < ε whenever µ(e) < δ. Alternatively K is bounded

8 2 and uniformly integrable if and only if given ε > 0 there is an N > 0 so that { } sup f dµ : f K < ε whenever c N. [ f >c ] The classical theorem of Dunford and Pettis [6, page 93], identifies the bounded, uniformly integrable subsets of L (µ) with the relatively weakly compact sets. Another characterization of uniform integrability is given in a theorem of De La Vallée Poussin [22, pages 9-20], which states that a subset K of L (µ) is bounded and uniformly integrable if and only if there is an N-function F so that sup{ F (f)dµ : f K} <. A function F : IR [0, ) is called an N-function, if it is continuous, even and convex with lim t F (t) t = and lim t 0 F (t) t = 0. Given an N-function F, the function G defined by G(x) = sup{t x F (t) : t 0} is an N-function, called the complement of F. De La Vallée Poussin s theorem is the focal point of Chapter and the main reason that the other chapters exist. We refine and improve this theorem in several directions. The theorem of De La Vallée Poussin does not, for instance, specify just how well the function F can be chosen. It gives little additional information in case the set in question is relatively norm compact in L (µ). Finally it gives no information on the structure of the set in the corresponding Banach space of F -integrable functions. Such a space is called an Orlicz space. Given an N-function F, the Orlicz space determined by F is defined by L F = {f measurable : c > 0 such that Ω F (cf(ω))dµ(ω) < }, where the usual identification of functions differing only on a set of measure zero, takes place. The norm of an element f L F is given by f F = inf{ c ( + F (cf)dµ) : c > 0}. It is worth mentioning at this stage that if < p < and F is defined by F (t) = t p, then L F is just the familiar Lp space. Most of the results in this dissertation deal with Orlicz spaces whose generating N-functions satisfy the 2 or conditions. We say that

9 3 an N-function F satisfies the 2 condition (F 2 ) if there is a constant K so that F (2x) KF (x) for large values of x. An N-function F satisfies the condition (F ) if there is a constant K so that F (xy) KF (x)f (y) for large values of x and y. More specifically in Section.4 we establish the fact that a subset K of L is relatively compact if and only if there is an N-function F so that K is relatively compact in L F (Theorem.4.3). Furthermore in the same Section we prove that a subset K of L is relatively weakly compact if and only if there is an N-function F so that K is relatively weakly compact in L F (Theorem.4.7). In establishing this last result, a weak compactness criterion for Orlicz spaces was used (Theorem.4.5). The technique employed to prove this criterion was mainly averaging. Thus the natural question of Orlicz spaces and their relationship to Banach-Saks types of properties arises. Recall that a Banach space X has the Banach-Saks (weak Banach-Saks) property if every bounded (weakly null) sequence in X has a subsequence, each subsequence of which, has norm convergent arithmetic means. Banach and Saks have shown in [5] that L p, for p >, has the Banach-Saks property, while Szlenk in [3] established the fact that L has the weak Banach-Saks property. Nishiura and Waterman showed in [24] that Banach spaces with the Banach-Saks property are reflexive. On the other hand Kakutani in [5] proved that uniformly convexifiable spaces have the Banach-Saks property. Baernstein in [3] gave the first example of a reflexive Banach space that fails the Banach Saks property. Furthermore Schreier in [30] established the fact that C[0, ], fails the weak Banach-Saks property. Akimovich in [] has shown that reflexive Orlicz spaces are uniformly convexifiable and so they have the Banach-Saks property. In Chapter 2 we show that a large class of non-reflexive Orlicz spaces has the weak Banach-Saks property, by establishing a result for these spaces, very similar to the Dunford- Pettis Theorem for L. Before we mention the results in Chapter 2, we need to recall, that The idea for studying this class comes from [8]

10 4 a subset K of an Orlicz space L F, has equi-absolutely continuous norms, if given ε > 0 there is a δ > 0 so that sup{ χ A f F : f K} < ε for all measurable sets A with µ(a) < δ. In Section 2. we show that if F 2 and its complement G satisfies lim t G(ct) G(t) = for some c > 0, then any weakly null sequence in L F has equi-absolutely continuous norms (Theorem 2..3). As a corollary to this theorem we have that if F is as above then a bounded set in L F is relatively weakly compact if and only if it has equi-absolutely continuous norms (Corollary 2..4). Furthermore, under the same hypothesis L F has the weak Banach-Saks property (Corollary 2..5). These results complement the ones of T. Ando in [2]. In Section 2.2 we give an application in convex function theory. Specifically we answer negatively the following question posed in [7, page 30]: Given an N-function F, is it possible to find an N-function H equivalent to F so that H satisfies the condition, for all real x, y? Having this Dunford-Pettis type of result for this special class of non-reflexive Orlicz spaces, we continue on to Chapter 3, where we investigate some similarities of these spaces, with the space L (µ). Kadec and Pelczýnski in [3] have shown that every non-reflexive subspace of L (µ) contains a copy of l complemented in L (µ). On the other hand Rosenthal in [27] investigated the structure of reflexive subspaces of L (µ) and proved that such subspaces, have non-trivial type. Recall that a Banach space X has type p for some < p 2, if there is a K > 0, so that ( 0 r i (t)x i p dt) p K ( x i p ) p, where (r i ) denotes the sequence of Rademacher functions 2 and x,..., x n are arbitrary 2 For a positive integer n, r n : [0, ] {, } is defined by { if t = r n(t) = ( ) i if t [ i i, ), where i =,..., 2 n 2 n 2 n

11 5 elements of the Banach space X. In Chapter 3 we show the same facts to hold true for the special class of non-reflexive Orlicz spaces we have been investigating. In particular, in Section 3. we show that if F is an N-function in 2 with its complement G satisfying lim t G(ct) G(t) non-reflexive subspace of L F, contains a copy of l complemented in L F = then every (Theorem 3..4). Furthermore in Section 3.3 we show that if F is an N-function in 2 with its complement G satisfying lim t G(ct) G(t) = then every reflexive subspace of L F has non trivial type (Theorem 3.3.3).

12 Chapter DE LA VALLÉE POUSSIN S THEOREM REVISITED. Uniform integrability and De La Vallée Poussin s Theorem. Definition.. A subset K of L (µ) is called uniformly integrable if lim sup{ f dµ : f K} = 0. c [ f c ] That is given ε > 0 there is a c ε > 0 so that for each f K and each c c ε we have [ f c ] f dµ < ε. Another way of defining uniform integrability is described in the following proposition: Proposition.. A subset K of L (µ) is uniformly integrable if and only if it is L - bounded and for each ε > 0 there is a δ > 0 so that sup{ A f dµ : f K} < ε for all A Σ with µ(a) < δ. Proof : A First note that for all measurable A, f K, c > 0 we have f dµ = A [ f <c ] f dµ + f dµ cµ(a) + f dµ. A [ f c ] [ f c ] Fix ε > 0 and choose c 0 > 0 so that sup{ [ f c ] f dµ : f K} < ε 2 whenever c c 0. Then for all f K we have and thus K is L bounded. Now let 0 < δ < and all f K we have A Ω f dµ c 0 µ(ω) + f dµ c 0 + ε [ f c 0 ] 2 ε 2c 0. Then for all measurable A with µ(a) < δ f dµ c 0 µ(a) + f dµ < ε [ f c 0 ] 2 + ε 2 = ε. 6

13 7 We now prove the converse. Fix ε > 0 and choose δ > 0 so that sup{ A f dµ : f K} < ε whenever A is measurable with µ(a) < δ. Let M = sup{ Ω f dµ : f K} and choose c 0 > 0 so that M c 0 < δ. Then for all f K and all c c 0 we have µ([ f c ]) f dµ M < δ. c [ f c ] c 0 So [ f c ] f dµ < ε and so we are done. The following well known theorem of Dunford and Pettis, gives some more insight to the notion of uniform integrability. Theorem..2 (Dunford-Pettis) : A subset K of L (µ) is uniformly integrable if and only if it is relatively weakly compact. A proof of this theorem can be found in [6, page 93]. Yet another characterization of uniformly integrable sets is an old theorem that finds its roots in Harmonic Analysis and Potential theory. It is due to De La Vallée Poussin. Since it is this theorem that we deal with in this chapter, we state and prove this result in detail (see [22, pages 9 20]). Theorem..3 (De La Vallée Poussin) A subset K of L (µ) is uniformly integrable if and only if there is a non-negative and convex function Q with lim t Q(t) t sup{ Q( f )dµ : f K} <. Ω = so that Proof : Suppose that K is a uniformly integrable subset of L (µ). We will construct a non-negative and non-decreasing function q that is constant on [n, n + ) for n = 0,,... with lim t q(t) = and we will set Q(x) = x 0 q(t)dt for x > 0. Use the hypothesis to choose a subsequence (c n ) of the positive integers so that sup{ [ f c n ] f dµ : f K} < n =, 2,.... 2n

14 8 Then for each f K and all n =, 2,... we have f dµ = [ f c n ] m=c n [ m f <m+ ] f dµ mµ([ m f < m + ]) m=c n µ([ f m ]). m=c n So for all f K we have n= m=c n µ([ f m ]). Now for m =, 2,... let q m be the number of the positive integers n, for which c n m. Then q m. Furthermore observe that n= m=c n µ([ f m ]) = q k µ([ f k ]). Let q 0 = 0 and define q(t) = q n if t [n, n + ) for n = 0,, 2,.... Then if Q(x) = x 0 q(t)dt we have Ω Q( f )dµ = n=0 ( n=0 m=0 [ n f <n+ ] k= Q( f )dµ q m ) µ([ n f < n + ]) = q 0 µ([ 0 f < ]) + (q 0 + q ) µ([ f < 2 ]) + = q n µ([ f n ]) n=0. So sup{ Ω Q( f )dµ : f K} <. To see that Q is convex, fix 0 x < x 2. We then have Q( 2 (x + x 2 )) = = 2 (x +x 2 ) 0 x 0 q(t)dt q(t)dt + x 2 (x +x 2 ) q(t)dt

15 9 = 2 x 0 x 0 q(t)dt + 2 q(t)dt (x +x 2 ) x q(t)dt + 2 x2 = 2 (Q(x ) + Q(x 2 )). 0 q(t)dt x2 2 (x +x 2 ) q(t)dt Finally observe that Q(x) = x and thus Q(x) x 2 q( x 2 ) as x. 0 q(t)dt x x 2 q(t)dt x 2 q(x 2 ) We now prove the converse. Let M = sup{ Ω Q( f )dµ : f K}. Let ε > 0 and choose c 0 > 0 so that Q(t) t > M ε whenever t c 0. Then for f K and c c 0 we have that f < ε M Q( f ) on the set [ f c ]. Thus and so we are done. [ f c ] f dµ ε M [ f c ] Q( f )dµ ε M M = ε.2 Some facts about N-Functions In this section we will summarize the necessary facts about a special class of convex functions called N-functions. For a detailed account of these facts, the reader could consult the first chapter in [7]. Definition.2. Let p : [0, ) [0, ) be a right continuous, monotone increasing function with. p(0) = 0; 2. lim t p(t) = ; 3. p(t) > 0 whenever t > 0; then the function defined by F (x) = x 0 p(t)dt

16 0 is called an N-function. The following proposition gives an alternative view of N-functions. Proposition.2. The function F is an N-function if and only if F is continuous, even and convex with. lim x 0 F (x) x = 0; 2. lim x F (x) x = ; 3. F (x) > 0 if x > 0. Definition.2.2 For an N-function F define G(x) = sup{t x F (t) : t 0}. Then G is an N-function and it is called the complement of F. Observe that F is the complement of its complement G. Theorem.2.2 (Young s Inequality) If F and G are two mutually complementary N- functions then xy F (x) + G(y) x, y IR. Proposition.2.3 The composition of two N-functions is an N-function. Conversely every N-function can be written as a composition of two other N-functions. The following material deals with the comparative growth of N-functions. Definition.2.3 For N-functions F, F 2 we write F F 2 if there is a K > 0 so that F (x) F 2 (Kx) for large values of x. If F F 2 and F 2 F then we say that F and F 2 are equivalent.

17 Proposition.2.4 If F F 2 then G 2 G, where G i is the complement of F i. In particular if F (x) F 2 (x) for large values of x then G 2 (x) G (x) for large values of x. Definition.2.4 A convex function Q is called the principal part of an N-function F, if F (x) = Q(x) for large x. Q(x) Proposition.2.5 If Q is convex with lim = then Q is the principal part of some x x N-function. Definition.2.5 An N-function F is said to satisfy the 2 condition (F 2 ) if lim sup x F (2x) F (x) of x. <. That is, there is a K > 0 so that F (2x) KF (x) for large values Definition.2.6 An N-Function F is said to satisfy the condition (F ) if there is a K > 0 so that F (xy) KF (x)f (y) for large values of x and y. Definition.2.7 An N-function F is said to satisfy the 3 condition (F 3 ) if there is a K > 0 so that xf (x) F (Kx) for large values of x. Definition.2.8 An N-function F is said to satisfy the 2 condition (F 2 ) if there is a K > 0 so that (F (x)) 2 F (Kx) for large values of x. Theorem.2.6 Let F be an N-function and let G be its complement; then the following hold. If F then F 2. If F 3 then its complement G 2. If F 2 then its complement G. If F 2 then there is a p > so that if H(x) = x p then F H. Finally the classes, 2, 3 and 2 are preserved under equivalence of N-functions.

18 2.3 Some facts about Orlicz Spaces In this section we summarize the necessary definitions and results about Orlicz spaces. A detailed account can be found in chapter two of [7]. Definition.3. For an N-function F and a measurable f define F(f) = F (f)dµ. Let L F = {f measurable : F(f) < }. If G denotes the complement of F let L F = {f measurable : fgdµ < g L G }. The collection L F is then a linear space. For f L F define f F = sup{ fgdµ : G(g) }. Then (L F, F ) is a Banach space, called an Orlicz space. The following theorem establishes the fact that an Orlicz space is a dual space. Theorem.3. Let F be an N-function and let E F be the closure of the bounded functions in L F. Then the conjugate space of E F is L G, where G is the complement of F. Theorem.3.2 Let F be an N-function and G be its complement. Then the following statements are equivalent:. L F = E F. 2. L F = L F. 3. The dual of L F is L G. 4. F 2.

19 3 Theorem.3.3 (Hölder s Inequality) For f L F and g L G we have fg dµ f F g G. Theorem.3.4 If f L F then f F = inf { k ( + F(kf)) : k > 0 }. It follows then that f L F if and only if there is c > 0 so that F(cf) <. Proposition.3.5 If f F then f L F and F(f) f F. Comparison of N-functions, gives rise to the following result concerning their corresponding Orlicz spaces. Proposition.3.6 If F F 2 then L F 2 L F and the inclusion mapping is continuous. Definition.3.2 We say that a collection K L F if has equi-absolutely continuous norms ε > 0 δ > 0 so that sup{ χ E f F : f K} < ε whenever µ(e) < δ. For f L F we say that f has absolutely continuous norm if {f} has equi-absolutely continuous norms. The following two results deal with the equi-absolute continuity of the norms. Theorem.3.7 A function f L F has absolutely continuous norm if and only if f E F. Theorem.3.8 If K L F, K has equi-absolutely continuous norms and K is relatively compact in the topology of convergence in measure, then K is relatively (norm) compact in L F.

20 4.4 De La Vallée Poussin s theorem revisited We state and prove the following lemma which can be found in [7, page 62]. Lemma.4. Given an N-function F, there is an N-function H so that H (H(x)) F (x) for large values of x. Proof : Write F = F F 2, where F, F 2 are N-functions and let G i be the complement of F i. Let Q(x) = e G (x)+g 2 (x) Q(x). The function Q is convex, with lim =. Hence there x x is an N-function K whose principal part is Q. Clearly K 2 and G i (x) K(x) for large x. So if H is complementary to K, we must have H and H(x) F i (x) for large x. Thus H (H(x)) F (F 2 (x)) = F (x) for large values of x. Lemma.4.2 If F 2 and K L F then the following statements are equivalent: I) The set K has equi-absolutely continuous norms. II)The collection {F (f) : f K} is uniformly integrable in L. Proof : The implication (I) (II) follows directly from the fact that F (f)dµ = E F (χ E f)dµ = F(χ E f) χ E f F whenever χ E f F. Next suppose {F (f) : f K} is uniformly integrable. Let ε > 0 and choose n IN so that 2 n < ε. Since F 2, there are K > 0, c > 0 so that F (2 n x) KF (x) for x c. Choose 0 < δ < 2F (c) so that Then for µ(e) < δ, f K we have { } sup F (f)dµ : f K < E 2K whenever µ(e) < δ. E F (2 n f)dµ F (c)dµ + F (2 n f)dµ E E [ f c ] < 2 + K F (f)dµ <. E [ f c ]

21 5 Thus 2 n fχ E F F (2 n fχ E )dµ + < 2. So fχ E F < 2 n < ε. From these two lemmas we obtain the following characterization of norm compact subsets of L. Theorem.4.3 A subset K of L (µ) is relatively compact if and only if there is an N- function F so that K is relatively compact in L F. Proof : Since the inclusion map L F L is continuous, necessity follows. Suppose K is relatively compact in L. Then K is also relatively weakly compact in L and so by the theorem of De La Vallée Poussin there is an N-function H so that sup { H(f)dµ : f K} <. By Lemma (.4.) there is an N-function F with F (F (x)) H(x) for large values of x. Thus sup { F (F (f)) dµ f K} < and by De La Vallée Poussin s theorem again, we have that {F (f) f K} is uniformly integrable in L. So by Lemma (.4.2) K has equi-absolutely continuous norms in L F. Since K is relatively compact in L, it is also relatively compact in the topology of convergence in measure. Hence K is relatively compact in L F. The following results deal with relative weak compactness in L and L F. We begin by mentioning a remarkable theorem of J. Komlós [6]. Theorem.4.4 (Komlós) If (f n ) is bounded in L then there is a subsequence (f nk ) of (f n ) and a function f L so that each subsequence of (f nk ) has arithmetic means µ-a.e. convergent to f. Definition.4. A subset S of a Banach space X is a Banach-Saks set if every sequence in S has a subsequence, each subsequence of which has norm convergent arithmetic means. The space X is said to have the Banach-Saks property if every bounded set of X is a Banach-Saks set. Similarily X is said to have the weak Banach-Saks property, if every weakly compact set in X is a Banach-Saks set.

22 6 It is an easy consequence of the Hahn-Banach theorem, that Banach-Saks sets are weakly compact. So we are now ready for the next theorem. Theorem.4.5 Let K L F. If K has equi-absolutely continuous norms and it is norm bounded, then K is a Banach-Saks set in L F. In particular K is relatively weakly compact in L F. Proof : Since K has equi-absolutely continuous norms, K E F. Let (f n ) be a sequence in K. Since (f n ) is bounded in L F -norm, it is also bounded in L -norm. Hence by Komlós s theorem, there is a subsequence (f nk ) of (f n ) and a function f L so that any subsequence of (f nk ) has µ a.e. convergent arithmetic means to f. Let G denote the complement of F. Note that for any measurable E and any g L G with g G we have gχ E f dµ gχ E f dµ lim inf n sup n sup n n n gχ E f nk dµ n k= gχ E f nk dµ k= k= g G χ E f nk F sup{ χ E h F : h K}. Thus χ E f F sup{ gχ E fdµ : g G } sup{ χ E h F : h K}. So f L F and f has absolutely continuous norm. Let (h k ) be any subsequence of (f nk ) and let a n = n nk= h k. We now claim that a n f in L F norm. Since the inclusion map L G L is continuous, there is a K > 0 so that g K g G for all g L G. Fix ε > 0 and choose δ > 0 so that sup{ χ A h F : h K} < ε 3 whenever µ(a) < δ. By Egorov s theorem, there is a measurable set E with µ(ω \ E) < δ so that a n f uniformly on E. Choose N IN so that χ E (a n f) < ε 3K whenever n N. Then for

23 7 any g L G with g G and n N we have g(a n f)dµ = E g a n f dµ g a n f dµ + Ω\E g a n f dµ g χ E (a n f) + g G χ Ω\E (a n f) F ε K g G 3K + g G( a n χ Ω\E F + fχ Ω\E F ) < ε ( ) 3 + h k χ n Ω\E F + ε 3 2ε 3 + n k= k= < 2ε 3 + ε 3 = ε. h k χ Ω\E So the claim is established. Thus K is a Banach-Saks set in L F. It also follows that f n k f weakly in L F and so K is relatively weakly compact in L F thanks to the Eberlein-Smulian theorem. A. Grothendieck has shown that if p < and X is a closed subspace of L p (µ) contained in L (µ), then X is finite dimensional (see [0] and [28, ch. 5]). We generalize this result as follows. Theorem.4.6 Suppose that X L (µ) and suppose that X is a closed subspace of an Orlicz space L F. Then X is finite dimensional. Proof : Let i : X L (µ) and i 2 : L (µ) L F be the natural inclusion maps, with X having the topology inherited from L F. Let (f n) be a sequence in X and assume that f n f F 0 for some f X. Also assume that f n g 0 for some g L. The first assumption yields a subsequence (f nk ) of (f n ) with f nk f µ a.e.. Since f n g uniformly µ a.e. we have that f = g µ a.e.. Thus by the closed graph theorem i is continuous.

24 8 Now by Theorem (.4.5) i 2 is weakly compact and as L (µ) has the Dunford-Pettis property, i 2 is completely continuous. Hence i 2 i is weakly compact and completely continuous. But i 2 i is the identity on X. Now it is not hard to see that the identity on X is compact and hence X is finite dimensional. We now prove the following stronger version of De La Vallée Poussin s theorem. Theorem.4.7 A set K is relatively weakly compact in L if and only if there is F so that K is relatively weakly compact in L F. Proof : Since the inclusion map L F L is continuous and thus weak-to-weak continous, necessity follows. So suppose that K is relatively weakly compact in L. By De La Vallée Poussin s theorem, there is an N-function H with sup{ H(f)dµ : f K} <. By Lemma (.4.), there is F with F (F (x)) H(x) for large x. So sup{ F (F (f)) dµ : f K} <, and by De La Vallée Poussin s theorem once more, we have that {F (f) : f K} is relatively weakly compact in L. Hence by Lemma (.4.2), K has equi-absolutely continuous norms in L F. Since K is obviously bounded in L F, we then have that K is relatively weakly compact in L F, thanks to Theorem (.4.5). Remark: If K L and if there is an N-function F with its complementary G 2 so that sup{ F (f)dµ : f K} < then K is a bounded subset of L p for some p >. Indeed, if G 2 then there is q > so that L q L G. Let T : Lq L G denote the natural inclusion map. Then if p + q = the adjoint operator T : L F Lp is also a natural inclusion map. Since T is continuous so is T. Hence K bounded in L F, implies that K is also bounded in L p.

25 Chapter 2 ORLICZ SPACES AND THE WEAK BANACH-SAKS PROPERTY 2. A weak compactness result reminiscent of the Dunford-Pettis theorem In this section we deal with a special class of Orlicz spaces, namely those spaces whose generating N-function F satisfies 2 and the function G complementary to F satisfies lim t G(ct) G(t) = for some c > 0. This class of spaces has been examined by D. Leung and in [8] they have been shown to satisfy the weak Dunford-Pettis property 2 while they fail the Dunford-Pettis property. This fact shows that such spaces are not isomorphic to L (p) for any probability p. Nonetheless they exhibit some striking similarities with L spaces. Some of these similarities are discussed in Chapters 2 and 3. At this point we should mention that V. A. Akimovich has shown in [] that every reflexive Orlicz space over a probability is isomorphic to a uniformly convex Orlicz space. Combining this result with Kakutani s result in [5] that states that uniformly convex spaces have the Banach-Saks property, one can immediately conclude that reflexive Orlicz spaces have the Banach-Saks property. Lemma 2.. Let K L F where F 2. Suppose that K fails to have equi-absolutely continuous norms. Then there is an ε 0 > 0, a sequence (f n ) K and a sequence (E n ) of The following question remains unresolved. Given an N-function F 2 with its complement G / 2 Ψ(ct) does there exist an N-function Φ equivalent to F so that its complement Ψ satisfies lim t = for Ψ(t) some c > 0? 2 A Banach lattice X has the weak Dunford-Pettis property if any weakly compact operator from X into any Banach space maps disjoint, weakly null sequences onto norm null sequences. 9

26 20 pairwise disjoint measurable sets, so that χ En f n F > ε 0 for all positive integers n. Proof : Since K does not have equi-absolutely continuous norms, there is an η 0 > 0 and sequences (k n ) K, (A n ) Σ, with µ(a n ) < 2 n, so that χ An k n F > η 0 for all positive integers n. For each n let B n = j=n A j. Then B n B n+. Furthermore µ(b n ) = µ( A j ) j=n µ(a j ) j=n j=n 2 j 0 as n, with χ Bn k n F χ An k n F > η 0 for all positive integers n. Since F 2 we have that each f L F has absolutely continuous norm. So if n = then there is n 2 > n so that χ Bn \B n2 k n F > η 0 2 (After all µ(b n ) 0 ). Let E = B n \ B n2 and let f = k n. Now choose n 3 > n 2 so that χ Bn2 \B n3 k n2 F > η 0 2. Let E 2 = B n2 \ B n3 and let f 2 = k n2. Continue on. The result is now established if we take ε 0 = η 0 2. We next present a Rosenthal s Lemma type of result. (cf. [6, page 82].) Lemma 2..2 Let X be a Banach space. Suppose that (x n ) X is weakly null and (x n) X is weak null. Then for each ε > 0 there is a subsequence (n k ) of the positive integers, so that, for each positive integer k we have < x n j, x nk > < ε. j k Proof : Let ε > 0. Let n =. Since x n 0 weak there is an infinite subset A of the positive integers so that j A < x j, x n > < ε 2. Since x n 0 weakly and since A is infinite, we can find n 2 > n with n 2 A, so that < x n, x n2 > < ε 2. Similarly there is an infinite subset A 2 of A so that j A 2 < x j, x n 2 > < ε 2. Again choose n 3 > n 2 with n 3 A 2 so that < x n, x n3 > < ε 4 and < x n 2, x n3 > < ε 4. There is an infinite subset A 3 of A 2 so that j A 3 < x j, x n 3 > < ε 2. Choose n 4 > n 3 with n 4 A 3 so that < x n i, x n4 > < ε 6 for i = Continue inductively to construct a sequence of infinite subsets of the positive integers, A A 2 A k and a sequence n < n 2 < of

27 2 positive integers with (i) n k+ A k for all k. (ii) (iii) Now for fixed positive integer k we have < x j, x nk+ > < ε for all k. 2 j A k < x n j, x nk+ > < ε for all k and for j =, 2,..., k. 2k < x n j, x nk > = j k < k j= < x n j, x nk > + j=k+ < x n j, x nk > ε (k ) + < x n 2(k ) j, x nk > j A k < ε 2 + ε 2 = ε. And so we are done. Now we are ready for the main result of this section. Theorem 2..3 Suppose that F 2 and that its complement G satisfies G(ct) lim = for some c > 0. t G(t) Then any weakly null sequence in L F has equi-absolutely continuous norms. Proof : Suppose not. Then there is a weakly null sequence (f n ) L F that fails to have equi-absolutely continuous norms. Using Lemma (2..) we may assume that there is an ε 0 > 0 and a sequence (E n ) of pairwise disjoint measurable sets so that χ En f n F > ε 0 for all positive integers n. Now choose a sequence (g n ) L G so that each g n is supported on E n with G(g n ) dµ and so that g n f n dµ > ε 0. For a fixed f L F Hölder s Inequality yields fg n dµ = χ En fg n dµ χ En f F g n G. But since (E n ) are pairwise disjoint and µ is finite we have that µ(e n ) 0. Furthermore since F 2 and f L F, f has absolutely continuous norm. Thus χ E n f F 0. As

28 22 (g n ) is norm bounded, we can conclude that χ En f F g n G 0 and so fg n dµ 0. Hence (g n ) is weak null. By Lemma (2..2) there is a subsequence (n k ) of the positive integers so that for each k we have j k g nj f nk dµ < ε 0 2. We now claim that G( gn c )dµ 0. Fix ε > 0. Since lim t G(ct) G(t) lim t G(t/c) G(t) = then = 0. Choose t 0 > 0 so that G(t/c) G(t) < ε 2 whenever t t 0. Since µ(e n ) 0, there is a positive integer N so that µ(e n ) < ε 2G(t 0 /c) whenever n N. Hence if n N we have G(g n /c)dµ = G(g n /c)dµ + G(g n /c)dµ [ g n <t 0 ] [ g n t 0 ] ε G(t 0 /c)µ(e n ) + 2 G(g n)dµ < ε 2 + ε 2 = ε. So the claim is established. Now choose a subsequence (n km ) of (n k ) so that m= G( g n km c )dµ <. Let g = m= g nkm. Then g is well defined and g L G, since G(g/c)dµ <. Since (f n ) is weakly null, we must have gf nkm dµ 0 as m. But for each positive integer m we have gf nkm dµ = ( g nkj )f nkm dµ j= g nkm f nkm dµ g nkj f nkm dµ j m j k m g nkm f nkm dµ > ε 0 ε 0 2 = ε 0 2, g nj f nkm dµ which is a contradiction. As a corollary to the theorem above, we get the following result that resembles the Dunford-Pettis theorem for L.

29 23 Corollary 2..4 Let F 2 and suppose that its complement G satisfies G(ct) lim t G(t) = for some c > 0. Then a bounded set K L F is relatively weakly compact if and only if K has equi-absolutely continuous norms. Proof : Suppose that K L F is relatively weakly compact. If K fails to have equi-absolutely continuous norms then there is an ε 0 > 0, a sequence (f n ) K and a sequence (E n ) of measurable sets with µ(e n ) 0 so that χ En f n F > ε 0, for each positive integer n. By the Eberlein-Smulian theorem, there is an f L F and a subsequence (f n k ) of (f n ) so that f nk f weakly in L F. So by Theorem (2..3), (f n k f) has equi-absolutely continuous norms. Thus χ Enk (f nk f) F 0 as k. As F 2 and f L F, f has absolutely continuous norm. Hence χ Enk f F 0 as k. But ε 0 < χ Enk f nk F χ Enk f F + χ Enk (f nk f) F which is a contradiction. The converse is just Theorem (.4.5). Corollary 2..5 Under the hypothesis of Corollary (2..4), L F has the weak Banach-Saks property. Proof : It follows directly from Corollary (2..4) and Theorem (.4.5). 2.2 An application in convex function theory Recall that an N-function G satisfies the 3 condition if there is c > 0 so that tg(t) G(ct) for large values of t. If G 3 then its complement F 2 [7, pages 29 30]. Furthermore it is clear that lim t G(ct) G(t) =. Also note that if G 2 then G 3. In [7, page 30] the following question is posed: Given an N-function F is it possible to find an N-function H, equivalent to F so that for some K > 0 H(xy) K H(x) H(y) x, y IR?

30 24 The following theorem answers this question in the negative. Theorem 2.2. Suppose that G 2 and let F denote the complement of G. Then there is no N-function H equivalent to F which satisfies the following condition: There is a K > 0 so that H(t t 2 ) K H(t ) H(t 2 ) for all real t and t 2. Proof : Suppose that such an H existed. Let µ denote Lebesgue measure on the interval [0, ]. Since µ is non-atomic, we can find a sequence (E n ) of pairwise disjoint measurable sets, each of which has positive measure. For each positive integer n, let h n = H ( µ(e n) )χ E n. Then h n L H with H(h n )dµ = for all positive integers n. It follows from Lemma (.4.2) that no subsequence of (h n ) has equi-absolutely continuous norms. We now claim that (h n ) is weakly null. Let (h nk ) be any subsequence of (h n ). Then for any positive integer N we have H( N N h nk )dµ K H( N N ) H( h nk )dµ k= = K H( N ) N k= = K H( N ) N k= = K H( N ) N. k= H(h nk )dµ µ(e nk ) µ(e n k ) Since H is an N-function, lim t 0 H(t) t = 0. Thus lim N K H( N ) N = 0. But since H then H 2. So N Nk= h nk H 0. To summarize, every subsequence of (h n ) has norm null arithmetic means and so (h n ) is weakly null as we claimed. Now since F is equivalent to H, there are constants λ > 0 and λ 2 > 0 so that λ f F f H λ 2 f F for all f L H(= L F ). By Theorem(2..3), (h n ) has equi-absolutely continuous F-norms and thus, by the inequality above, (h n ) also has equi-absolutely continuous H-norms.

31 25 But this is clearly a contradiction. Remark: The same result can be obtained from the work of T. Ando in [2]. Specifically it follows directly from [2, Theorem ], that given F 2, a subset K of L F weakly compact, if and only if is relatively lim (sup{f(tf) : f K}) = 0 t 0 t With this fact in hand, we can easily prove the following theorem. Theorem Let F be an N-function satisfying the condition for all real x, y. That is there is K > 0 so that F (xy) K F (x) F (y) for all x, y IR. Then L F is reflexive. Proof: Since F then f 2. Furthermore lim (sup{f(tf) : f B L t 0 t F }) = lim t 0 (sup{ lim (sup{ t 0 Ω F (tf(ω))dµ(ω) t Ω K F (t) = lim = 0. t 0 t : f B L F }) K F (t) F (f(ω))dµ(ω) t : f B L F }) Thus B L F is relatively weakly compact and so L F is reflexive. Now it is easy to see that given any N-function F so that its complement G / 2, then there is no N-function H equivalent to F so that, H satisfies for all real x, y.

32 Chapter 3 REFLEXIVE SUBSPACES OF NON-REFLEXIVE ORLICZ SPACES. 3. Subspaces containing complemented l In this section we derive a theorem similar to the one of Kadec and Pelczýnski, about L in [3]. The proofs are modeled after the ones in [6, pages 94-98]. Lemma 3.. Let (f n ) be a normalized disjointly supported sequence in L F, where F 2 and its complement G satisfies lim x G(cx) G(x) =, for some c > 0. Then there is a subsequence (f nk ) of (f n ) so that i. (f nk ) is equivalent to l s unit vector basis. ii. The closed linear span of (f nk ) is complemented in L F by means of a projection of norm less than or equal to 4c. iii. The coefficient functionals (φ k ) extend to all of the dual of L F and φ k 4 for all positive integers k. Proof : Let E n denote the support of f n. For each positive integer n choose g n L G with G(gn )dµ so that g n f n dµ 2. There is no harm in assuming that each g n is also supported on E n. Claim that G(g n /c)dµ 0 as n. Fix ε > 0. Since lim x G(cx) G(x) lim x G(x/c) G(x) = 0. So we can choose x 0 > 0 so that G(x/c) G(x) = then < ε 2 whenever x x 0. Since the E n s are pairwise disjoint and µ is a probability, we have that µ(e n ) 0 as n. 26

33 27 So there is a positive integer N so that µ(e n ) < we have ε 2G(x 0 /c) whenever n N. So for n N G(g n /c)dµ = [ g n <x 0 ] G(g n /c)dµ + G(x 0 /c)µ(e n ) + ε 2 [ g n x 0 ] G(g n )dµ G(g n /c)dµ < ε 2 + ε 2 = ε and so the claim is established. Now choose a subsequence (n k ) of the positive integers so that k= G( g nk c )dµ. For any sequence of signs σ = (ε k ) define g σ = k= ε k g nk. Since the g nk s are disjointly supported, g σ is well defined. Furthermore G( g σ c )dµ = = = k= k= k=. G( g σ E nk c )dµ G( ε kg nk )dµ E nk c E nk G( g n k c )dµ So g σ L G. Recall that the norm of g σ in L G, is given by g σ G = inf{ k ( + G(kg σ )dµ) : k > 0} and so it is easy to see that g σ G remains constant as σ varies. Denote this constant by M and observe that M = g σ G c( + G( g σ c )dµ) = c( + k= G( g n k )dµ) 2c. c Now for (a k ) l let σ = (sign(a k )). Then a k f nk F k= (g σ a k f nk )dµ g σ G k= = ( a k g nk f nk )dµ M k=

34 28 = M 2M k= k= a k a k. g nk f nk dµ Hence (i) is established. Now define for each k, a functional φ k on all of L F by φ k (f) = gnk f nk dµ g nk fdµ and define P : L F L F by Then for k =, 2,... P (f) = φ k (f)f nk. k= φ k 2 g nk G 2 ( + G(g nk )) 4. Furthermore P is a projection of L F onto the closed linear span of (f n k ) with P = sup f F 2 sup k= f F k= 2 sup f F = 2M 4c. ( gnk fdµ gnk f nk dµ f n k F k= g nk f dµ g nk ) G f F And so our proof is complete. 50]. We state now the following result in form of a lemma. Its proof can be found in [6, page Lemma 3..2 Let (z n ) be a basic sequence in the Banach space X with coefficient functionals (z n). Suppose that there is a bounded linear projection P : X X onto the closed

35 29 linear span [z n ] of (z n ). If (y n ) is any sequence in X for which P zn z n y n <, n= then (y n ) is a basic sequence equivalent to (z n ) and the closed linear span [y n ] of (y n ) is also complemented in X. Lemma 3..3 Let (f n ) be a sequence in L F where F 2 and its complement G satisfies lim x G(cx) G(x) = for some c > 0. Suppose that for each ε > 0 there is a positive integer n ε so that µ([ f nε ε f nε F ]) < ε. Then there is a subsequence (r n ) of (f n ) so that ( rn r n F ) is equivalent to l s unit vector basis. Furthermore the closed linear span [r n ] of (r n ) is complemented in L F. Proof : First observe that if f L F, E = [ f ε f F ] and K is the norm of the inclusion map L G L then f χ E F f f F χ E c F f F f F f F = sup{ gχ E cfdµ : g L G and G(g) } f F f F sup{ g χ E cf : g L G and G(g) } K f F χ E cf K f F f F ε = Kε. So using the hypothesis there is a measurable set E and a positive integer n so that µ(e ) < 6c 4 2 K and χ f n E F f n F 6c 4 2. Since F 2 then each f L F has an absolutely continuous norm. This fact together with the hypothesis again, yields a measurable E 2 and a positive integer n 2 > n so that µ(e 2 ) < 6c 4 3 K,

36 30 χ E2 f n2 f n2 F F > 6c 4 3 and χ E2 f n f n F F < 6c 4 3. Continue inductively to construct a subsequence (g n ) of (f n ) and a sequence of measurable sets (E n ) so that and Now let Then χ En n k= A n = E n \ µ(e n ) < g n 6c 4 n+ K, g n F F > χ En k=n+ g k g k F F < 6c 4 n+ 6c 4 n+. E k and h n = g n g n F χ An. g n g n h n F = χ A c g n n F F g n F g n g n χ E c n F + χ g n En\An F F g n F 6c 4 n+ + χ g n k=n+ E k F g n F < 6c 4 n+ + 6c 4 n+ + 6c 4 n+ + 6c 4 n. χ Ek k=n+ χ Ek k=n+ k=n+ g n g n F F g n g n F F 6c 4 k+ Thus h n F

37 3 = χ An g n g n F F χ En g n g n F F χ > 6c 4 n+ 6c 4 n+ 6c 4 n. k=n+ k=n+ k=n+ E k χ Ek g n g n F F g n g n F F 6c 4 k+ And so g n g n F h n h n F F g n g n F h n F + h n = 6c 4 n + ( h n F ) 6c 4 n + ( + 6c 4 n ) 2 6c 4 n. h n h n F F By Lemma (3..), there is a subsequence (n k ) of the positive integers so that hn k ( h nk F ) is equivalent to l s unit vector basis. The closed linear span [h nk ] of (h nk ) is complemented in L F by means of a projection P, of norm less than or equal to 4c. The coefficient functionals φ k extend to all of L G with φ k G 4 for all k. So we have that if r k = g nk then P φ k G r k h n k F 6c r k= k F h nk F g n k g k= nk F 6c g n g n= n F 2 6c 6c 4 n= n 2 = 4 n= n <. h n k h nk F F h n h n F F

38 32 Hence the result is established by an appeal to Lemma (3..2). Theorem 3..4 Let F 2 with its complement G satisfying G(cx) lim = for some c > 0. x G(x) If X is any non-reflexive subspace of L F then X contains an isomorphic copy of l that is complemented in L F. Proof : Since X is not reflexive, then the ball B X of X is not relatively weakly compact. Hence by Theorem (2..4), B X does not have equi-absolutely continuous norms. So by Lemma (.4.2), the set {F (f) : f B X } is not uniformly integrable in L. Thus there is a δ > 0 so that lim sup{ F (f)dµ ; f B X } = δ. a [ f a ] Keeping in mind that the above limit is actually an infimum we can find an increasing sequence (a n ) of positive reals, with a n as n so that δ sup{ F (f)dµ ; f B X } < δ + [ f a n ] n, for each positive integer n. It follows then, that there is a sequence (f n ) in B X so that δ n < [ f n a n ] F (f n )dµ < δ + n for all positive integers n. Now let g n = f n χ [ fn a n ] and h n = f n g n. Observe that for each ε > 0 we have µ([ g n ε g n F ]) µ([ g n > 0 ]) µ([ f n a n ]) f n dµ a n [ f n a n ] F (f n )dµ a n [ f n a n ], a n

39 33 provided that n is large enough. Since a n 0 as n then µ([ g n ε g n F ]) < ε for even larger n. So by Lemma (3..3), (g n ) has a subsequence that spans a complemented l in L F. We now show that (h n ) has equi-absolutely continuous norms. Note that if m n then [ h m a n ] = while if m > n then [ h m a n ] F (h m )dµ = = [ f m <a m ] [ f m a n ] [ f m a n ] F (f m )dµ F (f m )dµ [ f m a m ] F (f m )dµ sup{ F (f)dµ : f B X } δ + [ f a n ] m δ + n δ + n = 2 n. So for each positive integer n we have sup F (h m )dµ = sup F (h m )dµ 2 m [ h m a n ] m>n [ h m a n ] n. It follows then that {F (h m ) : m } is uniformly integrable in L and so by Lemma (.4.2), (h n ) has equi-absolutely continuous norms as we claimed. Hence by Corollary (2..4), (h n ) is relatively weakly compact in L F. So by passing to appropriate subsequences, we can assume that (g n ) spans a complemented l in L F and (h n) is weakly convergent in L F. Thus (h 2n h 2n+ ) is weakly null. So by Mazur s theorem, there is an increasing sequence (n k ) of positive integers and a sequence (a k ) of non-negative reals so that n k+ j=n k + a j =. The sequence (w k ) defined by w k = n k+ j=n k + a j(h 2j h 2j+ ) is norm-null in L F. Let u k = n k+ j=n k + a j (f 2j f 2j+ )

40 34 and v k = n k+ j=n k + a j (g 2j g 2j+ ). Then u k = v k + w k and u k v k F = w k F 0 as k. By selection, ( gn g n F ) was equivalent to l s unit vector basis with complemented span in L F. As g n F F (g n )dµ δ n, (g n) itself is equivalent to l s unit vector basis. A little thought convinces us that this is also the case with (v k ), with the closed linear span of (v k ) still complemented in L F of course. By passing to a subsequence to ensure that u k v k F converges to zero fast enough to apply Lemma (3..2), the result is finished. 3.2 Some facts about Banach Spaces with type In this section, we denote by (r n ), the sequence of Rademacher functions. Recall that for a positive integer n, r n : [0, ] {, } is defined by r n () =. r n (t) = ( ) (i ) for t [ i 2 n, i 2 n ), where i =,..., 2 n. Definition 3.2. A Banach space X is said to have type p, for some < p 2, if there is a constant K so that for any x,..., x n X. ( 0 r i (t)x i p dt) p K( x i p ) p, The following result allows some computational freedom : Theorem 3.2. (Kahane s inequality) A Banach space X has type < p 2 if and only if for each q < there is a constant K q > 0 such that for any x,..., x n X. ( 0 r i (t)x i q dt) q K q ( x i p ) p

41 35 It turns out that type s presence in a Banach space, is ultimately connected with the space s finite dimensional structure. To be more specific, we need the following notion. Definition Let λ and X be a Banach space. We say that X contains l n s λ-uniformly if for each positive integer n there is an isomorphism T : l n X so that T T λ. It is easy to see from the definition above that X contains l n s λ-uniformly if and only if for each positive integer n, x,..., x n B X such that for all choices of scalars a,..., a n. a i x i a i, λ Theorem (Pisier) The following are equivalent for a Banach space X :. For each λ >, X does not contain l n s λ-uniformly. 2. For some λ >, X does not contain l n s λ-uniformly. 3. The space X has type p for some < p 2. For a proof of this theorem as well as a more detailed account and bibliography, the reader should consult [26] and [25, pages 3-40]. 3.3 Subspaces of L F that have type The work of Kadec and Pelczýnski in [3], finds its natural continuation in the work of Rosenthal. In [27], Rosenthal shows that a subspace of L is reflexive if and only if it has non-trivial type. In this section, we follow his lead, to show that the same fact holds true for the special class of Orlicz spaces, we have been considering. The following result, mentioned in the form of a lemma, is due to Dor and Kauffman ( see appendix and [8] ).

42 36 Lemma 3.3. Suppose f,..., f n B L (µ) satisfy a i f i θ a i. for any a,..., a n, where 0 < θ <. Then there exist pairwise disjoint measurable sets A,..., A n such that f i dµ θ 2. A i We now adapt that lemma to our purposes. Lemma Suppose f,..., f n B L F (µ) satisfy a i f i F θ a i, for any a,..., a n, where 0 < θ <. Then there exist pairwise disjoint measurable sets A,..., A n such that χ Ai f i F θ 2. Proof : There is no loss in assuming that n a i f i F > θ n a i, provided that not all of a,..., a n are zero. Choose now g B L G, where G is the complement of F, so that g( a i f i )dµ > θ a i. Then a i (gf i ) dµ > θ a i and so by Lemma (3.3.) there is a collection of measurable and pairwise disjoint sets A,..., A n so that A i gf i dµ θ 2 i =,..., n.

43 37 By Hölder s inequality we then have that for each i =,..., n χ Ai f i F g G χ Ai f i F gf i dµ A i θ 2, which is what we wanted. The following theorem, characterizes reflexive subspaces of L F, for F 2, with complement G satisfying lim t G(mt) G(t) = Theorem Let F 2, with its complement G satisfying G(mt) lim = t G(t) for some m > 0. Let X be a subspace of L F. Then the following are equivalent :. The space X is not reflexive. 2. The space X contains a copy of l complemented in L F. 3. The space X contains l n s uniformly. 4. The space X fails to have non-trivial type. Proof : The implication 2 is just theorem (3..4). As for 2 3 it follows directly from the definitions. The double implication 3 4 is Pisier s theorem. So we will only show 3. Suppose that X contains l n s uniformly. Then there is a 0 < θ < so that for each positive integer n, there are functions f,..., f n B X satisfying a i f i F θ a i,