On the unconditional subsequence property

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1 Journal of Functional Analysis ) On the unconditional subsequence property Edward Odell, Bentuo Zheng Department of Mathematics, The University of Texas at Austin, 1 University Station C100, Austin, TX , United States Received 18 April 009; accepted 5 July 009 Available online 7 August 009 Communicated by N. Kalton Abstract We show that a construction of Johnson, Maurey and Schechtman leads to the existence of a weakly null sequence f i ) in L pi ) l,wherep i 1, so that for all ε>0and1<q, every subsequence of f i ) admits a block basis 1 + ε)-equivalent to the Haar basis for L q. We give an example of a reflexive Banach space having the unconditional subsequence property but not uniformly so. Published by Elsevier Inc. Keywords: Unconditional subsequence property; Unconditionally saturated; Unconditional tree property 1. Introduction If every normalized weakly null sequence in a Banach space X has an unconditional subsequence, X is said to have the unconditional subsequence property USP). If, for some K<, every such sequence admits a K-unconditional subsequence, X has the K-USP). The first example of a space without the USP) was constructed in 1977 by Maurey and Rosenthal [16]. This construction later played a role in the work of Gowers and Maurey [4] where they gave an example of a reflexive space not containing an unconditional basic sequence. Subsequent work by S. Argyros and others has shown that such spaces are plentiful. Thus having the USP) is by no means automatic. In [16] it was asked if L 1 [0, 1] has the USP). In 007 Johnson, Maurey and Schechtman [8] showed that L 1 fails the USP). Moreover, for 1 p<, they constructed Edward Odell s research is supported in part by NSF grant DMS and Bentuo Zheng s research is supported in part by NSF grant DMS * Corresponding author. addresses: odell@math.utexas.edu E. Odell), btzheng@math.utexas.edu B. Zheng) /$ see front matter Published by Elsevier Inc. doi: /j.jfa

2 E. Odell, B. Zheng / Journal of Functional Analysis ) a weakly null sequence in L p [0, 1] so that for all ε>0, every subsequence admits a block basis 1+ε)-equivalent to the Haar basis for L p. Since the Haar basis for L p is unconditional for p>1 but the unconditional constant blows up as p 1, it follows that L p has the C p -USP) for all 1 <p<, but lim p 1 + C p =. The situation is different for p>. L p has the + ε-usp) for all p and ε>0see[9]forp N, [5] for the general case). In this note we present two examples. In Section we show how the [8] construction easily yields the following. Let >p 1 >p > with lim i p i = 1. Then X = L pi ) l contains a weakly null sequence such that for all ε>0 and 1 <q, every subsequence admits a block basis that is 1 + ε-equivalent to the Haar basis for L q [0, 1]. X is reflexive and for every infinite dimensional subspace of X and ε>0, some further subspace Z satisfies dz,l p )<1 + ε for some 1 <p. In Section 3 we construct a reflexive space X with the USP) which fails the K-USP) for all K. This solves problem 3 in [17]. X = X n ) l where each X n is isomorphic to l but fails the n 3 -USP). The X n s are a modification of an example in [16]. We will show that every normalized weakly null sequence in X admits an l subsequence. This example contrasts with the result that if every normalized weakly null sequence in a Banach space admits a subsequence equivalent to the unit vector basis of c 0, then this is uniformly so [1] see [3] for a more general uniformity theorem). It is worth mentioning that the USP) is a weak version of the UTP): X has the unconditional tree property if every normalized weakly null tree in X admits an unconditional branch see [10]). In this case it is automatic that X has the K-UTP) for some K [18]. The UTP), rather than the USP) is the property that ensures a space embeds into one with an unconditional basis if X is reflexive. A reflexive space with the UTP) embeds into a reflexive space with an unconditional basis [10]. If X is separable and X has the ω -UTP), i.e., every normalized weak* null tree in X admits an unconditional branch, then X embeds into a space with a shrinking unconditional basis [11]. The almost isometric version of this result is given in []. Results on the USP) date back to the 1970 s. In [6] and [17] it was proved that a quotient X of a space with a shrinking unconditional basis has the USP). From [11] we have more, namely X embeds into a space with a shrinking unconditional basis. We use standard Banach space notation [7]. X, Y, and Z will denote separable real infinite dimensional Banach spaces. S X and B X denote the unit sphere and unit ball, respectively, of X. [N] <ω denotes the finite subsets of X, [N] ={i, j): i<j,i,j N}.ForE,F [N] <ω, E<F means max E<min F and E is the cardinality of E. c 00 is the linear space of finitely supported sequences of reals. For x,y c 00 x<ydenotes suppx) < suppy) and we use the same notation for x,y spane i ) where e i) is a basic sequence. L p = L p [0, 1] and me) denotes the Lebesgue measure of E. We thank the referee for pointing out an embarrassing error in our original proof of Lemma A weakly null sequence in L pi ) l Example.1. Let >p 1 >p > with lim i p i = 1. There exists a weakly null sequence f i ) in X = L pi ) l with the following property. If 1 <p, ε>0 and f ni ) is any subsequence of f i ), then some block basis of f ni ) is 1 + ε)-equivalent to the Haar basis of L p.

3 606 E. Odell, B. Zheng / Journal of Functional Analysis ) For p<q, L q embeds isometrically into L p. Moreover the Haar basis for L q is precisely reproducible [14]. This means that if L q Z, a space with a basis z i ), and ε>0 then some block basis of z i ) is 1+ε)-equivalent to the Haar basis for L q. Thus it will suffice to produce f i ) satisfying the desired property for each L p i. Let 1 <p 1 <. We begin by recalling some specific aspects of the [8] construction in L p1. Let A be the algebra generated by the dyadic subintervals of [0, 1]. LetE i ) be a listing of all elements in A so that for each E A, ME) {j N : E j = E} is infinite. In [8] a certain sequence k n ) n=1, of powers of, along with a sequence a n) n=1 0, ) and for n N, a sequence h i,n ) of functions on [0, 1] are constructed to satisfy the following: i) h i,n =1 Ai,n, A i,n E n ; ii) h i,n = 0; iii) ma i,n ) = me n )/k n ; iv) h i,n is A-measurable; v) h i,n ) are independent random variables in the probability space E 1 n, me n ) m); vi) n=1 a n h i,n p1 <. The desired sequence in L p1 [0, 1] is defined by f i = n=1 a n h i,n. Note that each sequence h i,n ) is weakly null in L p 1. Thus by passing to subsequences, using a diagonal argument, we may assume that h i,n ) i,n N is, in some order, a perturbation of a block basis of the Haar basis for L p1 and hence is unconditional. In particular there exists C = Cp 1 )< so that if g i ) is a disjointly supported sequence in span{h i,n: i, n N} w.r.t. the coordinates h i,n ) then see e.g. [1]) ) 1/ g i p 1 C g i..1) p1 The arguments of [8] yield the following. Let R be an infinite subsequence of N so that for all E A, ME) R is infinite. Let E A and ε>0. We shall say h is a Haar function on E if h =1 E, h = 0 and h is A-measurable. Then for all infinite M N there exists f spanf i : i M),sayf = b i f i, and a Haar function h on E so that f h p1 <ε. Moreover, writing f = i ) b i a n h i,n = a n n R i n=1 b i h i,n ) + ) a n b i h i,n fr)+ fn \ R) n/ R i we have fr) h p1 <εand fn\ R) p1 <ε..) It follows from this that f i ) has the property that every subsequence admits a block basis 1 + ε-equivalent to the Haar basis in L p1.

4 E. Odell, B. Zheng / Journal of Functional Analysis ) We next define f i ) X. First partition ME),forE A, into infinitely many infinite subsets M E,j ).Forj N let R j = E A M E,j. Leth j i,n be h i,n regarded as an element of L pj [0, 1].Fori N, define f i = Note that f i ) is seminormalized in X since f i = a n h j i,n n R j p j n R j a n h j i,n. ) 1/ a n h 1 i,n n R j p 1 ) 1/ C f j p1 by.1), since pj p1 for all j. f i ) is weakly null by vi) and the fact that each h i,n ) is weakly null. Let E A, ε>0 and let M be an infinite subsequence of N.Letj N and let f = b i f i spanf i : i M) and let h be a Haar function on E so that.) holds for R = R j.leth j be h regarded as an element of L pj [0, 1]. Set f = b i f i. Then f = n R j a n i Now fr j ) = fr j ). By.1) and.), fn \ R j ) X = ) b i h j i,n + b i a n h k i,n fr j ) + fn \ R j ). k j i n R k b i a n h k i,n k j n R k i p k ) 1/ C fn \ R j ) p1 <Cε. Thus f h j = fr j ) h j + fn \ R p j j ) X fr j ) h + C ε < 1 + C ) ε. p 1 As in [8], the fact that every subsequence of f i ) admits a block basis 1 + ε)-equivalent to the Haar basis of L p follows readily. Each L pj is stable and from this it is easy to check that X is stable. Thus for every infinite dimensional subspace of X and ε>0 some further subspace Z satisfies dz,l q )<1 + ε, for some 1 <q [13]. 3. The USP) does not imply the K-USP) Example 3.1. There exists a reflexive space X = n=1 X n ) l satisfying the following: i) For all i N, X n is isomorphic to l but fails the n 3 -USP).

5 608 E. Odell, B. Zheng / Journal of Functional Analysis ) ii) Every normalized weakly null sequence in X admits a subsequence equivalent to the unit vector basis of l. iii) For all ε>0 and subspace Y of X some subspace Z of Y satisfies dz,l )<1 + ε. The X n s are slight variants of spaces defined in [16] and attributed to W.B. Johnson). It is shown there that those spaces fail the n 3 -USP) and the same argument holds for our variants. Our work will be in establishing ii). We begin with the definition. Definition of X n. Let ε i 0 such that {i,j) [N] : i j} Let M = m i ) be a subsequence of N so that ε maxi,j) < ) for all i j, minm i,m j ) mi mj <ε maxi,j) 3.) and if E j [N] <ω with E j =m j for j N, then ) 1Ej is -equivalent to the unit vector basis of l in l. 3.3) Ej Let G ={E 1,...,E j ): j N, φ E i [N] <ω for i j and E 1 < <E j }. Choose an injection Φ : G M so that ΦE 1,...,E j )>ΦE 1,...,E j 1 ) for all E 1,...,E j ) G, j. For n N,let { 1 F n = n We also let n } 1 Ei Ei : E 1,...,E n ) G, E 1 =1, E j+1 =ΦE 1,...,E j ) for j<n. { 1E F 0 = : E [N] }. <ω E For x c 00 let X n is the completion of c 00 under x Fn = sup { f,x : f Fn }. x n = 1 n x l x Fn x F0.

6 E. Odell, B. Zheng / Journal of Functional Analysis ) Note that since F n S l,forx X n 1 n x l x n x l. Thus X n is isomorphic to l and the unit vector basis e i ) of c 00 is a normalized monotone basis for X n. X = X n ) l is then reflexive and has a monotone basis, namely the bases for each X n, properly ordered. The spaces defined in [16] were given by x = 1 n x l x Fn x. Our example requires the F0 term. We also have added the lacunary condition 3.3). Lemma 3.. See [16].) For n N, X n fails the n 3 -USP). The proof is the same as that given in [16]. The idea is to consider any subsequence e i ) i M of e i ) and form vectors y = 1 n x = 1 n n n 1) i+1 1 E i Ei 1 Ei Ei F n and with E i M for i n. One then shows that x 1 and y 1/ n using 3.1) and 3.). Also x F0 and y F0 are both of the order 1/ n so the proof remains valid in our modified space. We need 3.3) and F0 to prove that X has the USP) and, in fact, satisfies ii). Lemma 3.3. Let n N and ε>0. There exists Kn,ε) < so that if x l n, then {m j : there exists E [N] <ω, E =m j and 1 E mj,x ε} Kn,ε). Proof. Let E i =m ji for i K. Letg = 1 K K δ i 1 Ei mji Kε gx) n which yields K 4n. K ε for i K with j i j l if i l K and 1 E i mji,x = δ i 1 E i mji,x ε B l, by 3.3). Then gx) Kε K and x l n so Lemma 3.4. Let n N and let x i ) be a normalized block basis of X n. There exists a subsequence y i ) of x i) satisfying, for all a i) c 00, ) 1/ a i y i 4 a i Fn.

7 610 E. Odell, B. Zheng / Journal of Functional Analysis ) As above we will use the notation f = 1 n nl=1 f i F n to signify that f i = 1 E i Ei, E 1 < <E n, E 1 =1 and E j+1 =ΦE 1,...,E j ) for j<n. We will write fx)for f,x. Proof of Lemma 3.4. Given x c 00,letx denote the decreasing rearrangement of xi) ). Passing to a subsequence, and relabeling, we may assume that xi ) converges pointwise to some x 0. x 0 could be identically 0 but, in any event, x 0 nb l. By passing to a further subsequence, and perturbing, we may assume, in the case x 0 0, that x i = x i 1) + x i ) where x i 1) = x 0 [1,mi ],forsomem i, x i ) 0asi and x i 1) x i ) =0. Fix ε>0 and define Aε) ={i, j) [N] : there exists f = 1 nl=1 n f i F n with f l1 x i ) ε and f l x j ) ε for some <l n}. Passing again to a subsequence, using Ramsey s Theorem, we may assume Aε) = or Aε) =[N]. We will show that the latter is impossible. Assume Aε) =[N]. Passing to a further subsequence, using Ramsey s Theorem, we may assume the integers <l in the definition of Aε) are fixed, independently of i, j). Moreover, we may assume we have one of three cases. Case 1): For all i<jthere exists f i,j) = 1 n nl=1 f i,j) l f i,j) l x j ) ε. F n with f i,j) x i 1)) ε/ and Case, s): for s {1, }: For all i<j there exists f i,j) = 1 nl=1 n f i,j) l F n with f i,j) x i )) ε/ and f i,j l x j s)) ε/. Assume Case 1) holds. Now {j: there exists E [N] <ω, E =m j and 1 E mj,x 0 ε/} < by Lemma 3.3. Thus, passing to a subsequence and using Ramsey s Theorem, we may assume that for some fixed j 0, f i,j) = 1 E i,j) mj0, where E i,j) =m j0,fori<j. We claim that if j is sufficiently large then at least Kn,ε) + 1 of the functions {f i,j) : i<j} are distinct. Hence at least Kn,ε)+ 1 of the functions {f i,j) l : i<j} are distinct and this contradicts Lemma 3.3. Each set E i,j) could intersect the support of at most m j0 x i s. Thus the claim is evident and Case 1) is impossible. Next assume Case, 1) holds. As we argued in Case 1), we may similarly assume that for some fixed j 1, 1 f i,j) i,j) E l l =, mj1 E i,j) l = m j1, for i<j. Since x i ) 0 it follows that for f i,j) = E i,j) l 1 i,j) E l 1E 1 i,j), E i,j) as i. This forces as i which is a contradiction. Finally we assume Case, ) holds. By passing to a subsequence we may assume for all f = 1 nl=1 n f l F n, for all l<n, for all j N,ifmaxsupp f l ) max suppx j )) then i=j+1 fl+1 xi ) ) ε < and f1 xi ) ) ε <. 3.4)

8 E. Odell, B. Zheng / Journal of Functional Analysis ) Indeed suppf 1 ) =1, suppf l+1 ) is determined by f 1,...,f l ) and x i ) 0. Thus the last statement is achieved by taking the subsequence so that x i ) <ε/ for all i. Now fixing x 1 ) we can consider all f s and l<nso that max suppf l ) max supx 1 )) and pass to a subsequence to achieve 3.4) for j = 1. Let x ) be the first term of this new subsequence and repeat, and so on. Fix j 0 and let i 1 <i <j 0. By 3.4) f i 1,j 0 ) f i,j 0 ) since max supp f i 1,j 0 ) 1 max suppx i1 ). Hence f i 1,j 0 ) l f i,j 0 ) l as well for i 1 <i <j 0. Thus if j 0 >Kn,ε)+ 1 we again contradict Lemma 3.3. Let δ>0 and let ε i 0 with nεi <δ. 3.5) We use the above to inductively pass to a subsequence y i ) of x i), so that for all m N, Am) Aε m ) {i, j) [N] : i m}= where Aε m ) is defined as above with respect to the sequence y i ). Let f = 1 nl=1 n f l F n and a i ) c 00. We shall estimate f a j y j ) by breaking it into 3 sums. First note that if g = 1 E E F 0, then gy j ) ) B l. 3.6) Indeed, let E =m and E suppy j ) =k j for j N. Thus k j m. Moreover, since y j F0 1 for all j, 1 E,y j k j. Hence gy j ) ) is coordinatewise dominated by 1 kj m ).But ) 1 kj m = 1 ) 1/ m k j = 1. l m m Let K ={i: f l y i ) 0 for at most one l n} and let L = N \ K. We split L into two sets. If L ={i 1,i,...} in increasing order, let L 0 ={i 1,i 3,...} and L e ={i,i 4,...}. Thus ) f ) a i y i f ) a i y i + f ) a i y i + f a i y i. i L 0 i L e i K From the definition of K and 3.6), the vectors f i y j )) j K form a block basis in B K l, over the index i, and thus f y j )) j K B K l. We then have ) 1/ fa j y j ) aj. 3.7) j K j K

9 61 E. Odell, B. Zheng / Journal of Functional Analysis ) Let L 0 ={j 1,j,...} in increasing order. Note that if for some l n, f l y ji ) 0 then for i i, f l y ji ) = 0. Let m 1 = inf { m: f l y jm ) ε jm for some l n }. If no such m exists, then for all i fy ji ) 1 n nε ji and so fy ji ) nεji <δ, by 3.5). Otherwise, fy ji ) nε ji for i<m 1 and fy jm1 ) 1. Furthermore, since Aj m1 ) =, fy jm ) nε jm1 for m>m 1. Let m = inf{m>m 1 : f l y jm ) ε jm for some l n}. Ifm does not exist, we argue as in the case above where m 1 does not exist. We have for m 1 <m<m that fyjm ) nεjm and fyjm ) nεjm1. Continuing in this fashion we obtain fyj ) = fyji ) 1 + nεji < 1 + δ. 3.8) j L 0 A similar estimate holds for j L e fy j ). Thus from 3.7) and 3.8) we have ) f aj y j j K a j ) 1/ δ) a j ). Taking δ< 1 yields the lemma. Corollary 3.5. Let n N, δ>0 and let x i ) be a normalized block sequence in X n. There exists a subsequence y i ) of x i) satisfying for all a i ) c 00. a i y i 4 a i ) 1/ Proof. Note that 1 a i x i = n l 1 ) 1/ 1/ a n i x ) 1/ i l ai x i n) = ai. Thus the corollary follows from Lemma 3.4, 3.6) and the definition of n.

10 E. Odell, B. Zheng / Journal of Functional Analysis ) Remark 3.6. Corollary 3.5 can be improved in the case x i 0. Given δ>0 we can choose y i ) to satisfy ) 1/ a i y i 1 + δ) a i for all scalars a i ). To do this we choose a suitable δ i 0 and then choose y i ) so that if f l y i ) δ i for some l n, then j>i f ly i ) <δ i. Thus the argument reduces to that used in the estimation of f j K a j y j ). Proposition 3.7. Let x i ) be a normalized weakly null sequence in X and let δ>0. a) If x i 0, there exists a subsequence y i ) of x i) satisfying ) 1/ a i y i 1 + δ) a i for all a i ) c 00. b) There exists a subsequence y i ) of x i ) satisfying a i y i 5 for all scalars a i ). Proof. We prove a) only, using Remark 3.6. The proof of b) is similar, using Corollary 3.5. Let P n : X X n be the natural restriction projection. If I is an interval in N,letP I = n I P n. Passing to a subsequence and perturbing we may assume that x i ) is a block basis of the basis for X, and, for n N, lim i P n x i λ n exists. We have that n=1 λ n 1 since x i =1. Choose λ 0 0 so that n=0 λ n = 1. Let δ>0. Passing to a subsequence, and perturbing, we may assume that we have ε, 0< δ<δ, and integers 0 = n 0 <n 1 <n <, so that ε + λ 0 + ε1 + δ) < λ δ), if λ 0 > 0, 3.9) 1 + δ) + ε + ε1 + δ) ) <1 + δ), 3.10) λ n < ε, 3.11) n=n 1 +1 P n x j ) = λ n for all j N, j i and n n i, 3.1) x i = P [1,ni+1 ]x i ) and Pni,n i+1 ]x i ) λ0 < ε for all i, 3.13) ) 1/ for j N, n j 1 <n n j and a i ) c 00 a i P n x i ) 1 + δ)λ n ai. 3.14) i=j Eq. 3.14) comes from applying Corollary 3.5 and 3.1). a i ) 1/ i=j

11 614 E. Odell, B. Zheng / Journal of Functional Analysis ) Let a i ) S l. Then, using 3.13), n 1 a i x i = a i P n x i ) + n=1 n 1 n=1 + a i n j j= n=n j 1 +1 ) λ n 1 + δ) n j j= n=n j 1 +1 a j 1P n x j 1 ) + a i P n x i ) i=j [ a j 1 P n x j 1 ) )] + P n a i x i, by 3.14). Now the right most term above is, by 3.13) and the triangle inequality in l, by 3.13) and 3.14), [ a j P nj,n j+1 ]x j ) ) 1/ n j + j= [ ) 1/ aj λ 0 + ε) + n j [ λ 0 + ε) δ) n=n 1 +1 j= n=n j 1 +1 λ n ) 1/ ] [ λ 0 + ε) δ) ε ], by 3.11). i=j n=n j 1 +1 i=j a i ) P ) 1/ ] n a i x i i=j )λ n 1 + δ) ) 1/ ] We thus have, if λ 0 > 0, n 1 a i x i 1 + δ) λ n + λ δ), by 3.9), <1 + δ). n=1 If λ 0 = 0 then n 1 a i x i 1 + δ) λ n + [ ε + ε1 + δ) ] n=1 <1 + δ) by 3.10). Corollary 3.8. Let x i ) be a normalized weakly null sequence in X. Then some subsequence is equivalent to the unit vector basis of l.

12 E. Odell, B. Zheng / Journal of Functional Analysis ) Proof. By Proposition 3.7 we may assume that x i ) satisfies an upper l estimate. If for some n, lim i P n x i ) > 0 then we easily obtain a lower l estimate for some subsequence. If this never happens then some subsequence y i ) of x i ) satisfies lim P ni,n i+1 ]y i ) y i = 0 for somen 1 <n <, i and so we also easily obtain a lower l estimate. Corollary 3.9. Let Y be a subspace of X and ε>0. There exists a subspace Z of Y with dz,l )<1 + ε. Proof. By Corollary 3.8 we may assume Y contains a basic sequence y i ) which is equivalent to the unit vector basis of l. Replacing y i ) by a suitable block sequence of long averages we may assume that y i ) is a block basis of X with y i 0. By James argument that is not distortable see e.g. [15]) some normalized block basis z i ) of y i ) satisfies a i z i 1 + ε) 1 ai )1/. Passing to a subsequence of z i ), using Proposition 3.7a), we obtain the corollary. References [1]D.Alspach,E.Odell,L p spaces, in: Handbook of the Geometry of Banach Spaces, vol. I, North-Holland, Amsterdam, 001, pp [] S.R. Cowell, N.J. Kalton, Asymptotic unconditionality, Q. J. Math. 009), in press, doi: /qmath/han036. [3] D. Freeman, Weakly null sequences with upper estimates, Studia Math ) 008) [4] W.T. Gowers, B. Maurey, The unconditional basic sequence problem, J. Amer. Math. Soc. 6 4) 1993) [5] S. Guerre, Types and symmetric sequences in L p,1 p<+, p, Israel J. Math. 53 ) 1986) [6] W.B. Johnson, On quotients of L p which are quotients of l p, Compos. Math. 34 1) 1977) [7] W.B. Johnson, J. Lindenstrauss, Basic concepts in the geometry of Banach spaces, in: Handbook of the Geometry of Banach Spaces, vol. I, North-Holland, Amsterdam, 001, pp [8] W.B. Johnson, B. Maurey, G. Schechtman, Weakly null sequences in L 1, J. Amer. Math. Soc. 0 1) 007) [9] W.B. Johnson, B. Maurey, G. Schechtman, L. Tzafriri, Symmetric structures in Banach spaces, Mem. Amer. Math. Soc ) 1979) v+98. [10] W.B. Johnson, B. Zheng, A characterization of subspaces and quotients of reflexive Banach spaces with unconditional bases, Duke Math. J ) 008) [11] W.B. Johnson, B. Zheng, Subspaces and quotients of Banach spaces with shrinking unconditional bases, preprint. [1] H. Knaust, E. Odell, On c 0 sequences in Banach spaces, Israel J. Math. 67 ) 1989) [13] J.-L. Krivine, B. Maurey, Espaces de Banach stables, Israel J. Math. 39 4) 1981) [14] J. Lindenstrauss, A. Pełczyński, Contributions to the theory of the classical Banach spaces, J. Funct. Anal ) [15] J. Lindenstrauss, L. Tzafriri, Classical Banach Spaces I: Sequence Spaces, Ergeb. Math. Grenzgeb., vol. 9, Springer-Verlag, New York, [16] B. Maurey, H.P. Rosenthal, Normalized weakly null sequence with no unconditional subsequence, Studia Math ) [17] E. Odell, On quotients of Banach spaces having shrinking unconditional bases, Illinois J. Math ) [18] E. Odell, Th. Schlumprecht, A. Zsák, On the structure of asymptotic l p spaces, Quart. J. Math ) 85 1.