Weakly null sequences with upper l p -estimates

Weakly null sequences with upper l p -estimates Helmut Knaust* and Edward Odell Department of Mathematics The University of Texas at Austin Austin, Texas 78712 1. Introduction A Banach space X has property (S) if every weakly null sequence in Ba X, the unit ball of X, has a subsequence which is C-dominated by the unit vector basis of c 0 for some constant C <. In [11] it was shown that if X has property (S), then the constant C can be chosen to be independent of the particular weakly null sequence in Ba X. Here we generalize this result to the case of upper l p -estimates. Definition 1.1. Let 1 < p <. A Banach space X has property (S p ) if every weakly null sequence (x n ) has a subsequence (y n ) such that for some constant C <, (1) α n y n C for all (α n ) IR with ( ) 1/p α n p 1. X has property (US p ), if there is a constant C such that every normalized weakly null sequence in Ba X admits a subsequence (y n ) so that (1) holds. We say that (y n ) has a C-upper l p -estimate, if (1) holds. Our main result is Theorem 1. A Banach space has property (S p ) if and only if it has property (US p ). Let us give some examples of Banach spaces which enjoy property (S p ): l p has property (S p ) (1 < p < ). L p [0, 1] has property (S r ), where r = min{2, p}. (More generally, every Banach space which has type p and can be embedded into a Banach space with an unconditional basis has property (S p ).) The James space J and its tree version JT have property (S 2 ) [1]. It follows from the results of James [8] (see also [7]) that every superreflexive Banach space has property (S p ) for some 1 < p <. Let us note that every subspace of a Banach space X with property (S p ) has property (S p ). If X is in addition reflexive, quotients of X have property (S p ) as well. A technique employed in the proof of Theorem 1 allows us to strengthen this last result in the following way: Corollary 1. Let X be a Banach space with property (S p ) and let Y be a subspace of X not containing l 1. Then the quotient space X/Y has property (US p ). Our proof of Theorem 1 is strongly motivated by the arguments in [11]. In fact, the proof we present here is valid for the case of property (S) as well (with the usual changes of notation). The key proposition in our proof (Proposition 3.4) is an improvement to our construction in [11]. Furthermore we no longer have Johnson s lemma [11, Proposition 3.4] at our disposal. The following remarks were made in [11] but seem worth recalling. * This is part of this author s Ph.D. Dissertation prepared at The University of Texas at Austin under the supervision of H.P. Rosenthal. Research partially supported by NSF Grant DMS-8601752. 1

One might ask whether a result like Theorem 1 remains true, if one considers the property that every normalized weakly null sequence in X admits a subsequence which is equivalent to the unit vector basis of l p. On the one hand Johnson and the second named author [9] have shown this to be false: for 1 < p < 2 they construct a subspace of L p [0, 1] where each normalized weakly null sequence has a subsequence equivalent to the unit vector basis of l p, but where the equivalence constant cannot be chosen uniformly for all sequences in question. Their construction can be carried out to produce counterexamples for p 2 as well (not within L p [0, 1]). We are indebted to H.P. Rosenthal for pointing out to us that under stronger conditions on the other hand, one obtains the following corollary of Theorem 1: Corollary 2. Let X be a Banach space such that X has property (S p ) and X has property (S q ) for some 1 < p, q < with 1 p + 1 q = 1. Then the following properties hold for some constant C < : (i) If l 1 does not embed in X, then every normalized weakly null sequence in X contains a subsequence which is C-equivalent to the unit vector basis of l p and whose closed linear span is C-complemented in X. (ii) If l 1 does not embed in X, then every normalized weakly null sequence in X contains a subsequence which is C-equivalent to the unit vector basis of l q and whose closed linear span is C-complemented in X. The proof of Theorem 1 will be presented in Section 3. In order to motivate the quite technical proof we present a version of Theorem 1 for spreading models in Section 2. We hope that the spreading-model version (which is quite easy and does not follow from Theorem 1) will give the reader some insight into our approach to the proof of Theorem 1. Section 4 contains the proof of the corollaries; we also state a weak Cauchy sequence criterion for property (S p ), due to C. Schumacher [16]. Our notation is standard as can be found in [5] or [12]. If F is a subset in a Banach space X, then [F ] denotes the closed linear span of F in X. If L is an infinite subsequence of IN, we denote by P (L) the set of all infinite subsequences of L. We would like to thank Haskell Rosenthal for his useful suggestions. 2. The spreading model case Let us recall that a semi-normalized basic sequence (x n ) in a Banach space X is said to have a spreading model (e n ), if (e n ) is basic in some Banach space such that for all k IN and for all ε > 0 there is an N IN such that for all N < n 1 < n 2 <... < n k and for all scalars (a i ) with sup i a i 1 k k a i e i a i x ni < ε. If (x n ) has a spreading model, the model is unique up to 1-equivalence. Every normalized weakly null sequence admits a subsequence with an unconditional spreading model [3], [4]. Definition 2.1. Let 1 < p <. A Banach space has property (M p ) if every normalized weakly null sequence in X admits a subsequence with a spreading model (e n ) which is C-dominated by the unit vector basis of l p, i.e., a i e i C for all (a i ) Ba l p. i We say X has property (UM p ), if the constant C can be chosen uniformly for all normalized weakly null sequences in X. We will show the following analogue of Theorem 1: 2

Proposition 2.2. A Banach space with property (M p ) has property (UM p ). For its proof we will need some technical definitions. Let 1 < p < and let X be a Banach space. A normalized weakly null sequence in X is called good (respectively C-good), if it has a spreading model which satisfies an upper l p -estimate (respectively a C-upper l p -estimate). A good sequence in X is called M-bad, if its spreading model fails to have an M-upper l p -estimate. An array (x n i ) i, of elements in X is called a bad array, if each column (x n i ) i is an M n -bad sequence for all n IN and M n as n. An array (yi n) is called a subarray of an array (xn i ), if each column of (yn i ) is a subsequence of (xk n i ) i for some sequence k 1 < k 2 <.... Finally let us say a bad array (x n i ) satisfies the l p-array procedure (for spreading models) if it admits a subarray (yi n) and if there are positive scalars a n with a n 1 so that y i = a nyi n has no subsequence with a spreading model which has an upper l p -estimate. (Note that (y i ) is automatically weakly null.) Proof of Proposition 2.2. Suppose a Banach space X has property (M p ) but fails property (UM p ). Then X contains a bad array. We will show that every bad array satisfies the l p -array procedure for spreading models, thus obtaining a contradiction. Indeed, a bad array (x n i ) contains a (bad) subarray, which is basic in some order (see Lemma 3.7 below). The sequence (y n ) obtained from this subarray by the l p -array procedure is then seminormalized and weakly null. The sequence (y n /y n ) then has no spreading model with an upper l p -estimate. Let (x n i ) be a bad array in X. We claim that we can find a subarray (yn i ) of (xn i ), positive scalars a n with a n 1, k N IN and constants C N and M N such that the following properties hold for all N IN: (2) (y N i ) i is a C N -good sequence with its spreading model denoted by (e N i ) (3) (4) (5) (6) k N k N N 1 b N i e N i > 2M N for some (b N i ) Ba l p a N M N > N n=n+1 a n C n < 1 8 a N M N a n < 1 4 a N M N. Once the claim has been established we proceed as follows. We set y i = a ny n i and let (y i ) i L, L P (IN), be a subsequence of (y i ) with spreading model (e i ). Let C < be given. We have to show that (e i ) does not have a C-upper l p -estimate. By (4) we can choose N so that a N M N > 4C. Next we choose j 1 < j 2 <... < j kn in L with j 1 large enough, so that we obtain for all n N and all (c i ) IR with c i 1: (7) 2 k N c i e n i k N c i yj n i 1 2 k N c i e n i 3 and k N c i y ji 2 k N c i e i.

By (3) we can find (b N i )k N Ba l p with k N b N i y ji = k N b N i k N bn i en i ( ) a n yj n i > 2M N. Thus k N N 1 k N a N b N i yj N i a n b N i yj n i 1 2 a N k N b N i e N i N 1 k N 2a n b N i e n i N 1 a N M N 2 a n C N k N n=n+1 n=n+1 n=n+1 a n using (2) k N a n b N i yj n i k N a n b N i yj n i by (7) a N M N a NM N 4 a N M N 4 a NM N 2 2C by (5) and (6). Consequently k N bn i e i > C by (7). It is left to present the construction of the subarray (yi n). Let a 1 = 1/2. Since (x n i ) is a bad array we can find a column n such that the sequence (x n i ) i has a subsequence (z i ) which has a spreading model (e 1 i ) which fails to satisfy a 2M 1 -upper l p -estimate, where M 1 > 2, i.e., k 1 b1 i e1 i > 2M 1 for some k 1 IN and some (b 1 i )k1 Ba l p. We set yi 1 = z i. This defines the first column of the subarray (yi n) of (xn i ). By choosing C 1 large enough we can assure that (yi 1) is a C 1-good sequence. We pick a 2 < 1/4 such that a 2 < a 1 M 1 /8k 1, pick (yi 2), a subsequence of some (xn i ) i, such that this column is (2M 2 )-bad, where a 1 C 1 < a 2 M 2 /8 and a 2 M 2 > 2. We choose C 2 so that (yi 2) is a C 2-good sequence. If the first N 1 columns of the subarray have been chosen in the way just described, we choose { 0 < a N < min 2 N,,...,N 1 a n M n 4 2 N k n then we choose (yi N ), again a subsequence of some column of (xn i ), which is 2M N -bad where M N satisfies (5). Finally k N is chosen so that k N bn i en i > 2M N for some (b N i ) Ba l p. This completes the induction. It is straightforward to check that (2) (6) are indeed satisfied. }, 4

3. The proof of the main result The proof for the spreading model version is quite easy, since we can control the length of the sum we consider in the Nth column. This enables us to gain control over the behavior of the columns following the Nth one. Similarly it is straightforward to prove Theorem 1, when we assume additionally that X is embedded into a space with an unconditional basis. The projection onto the Nth column allows us then to preserve the badness of the N th column without disturbances from the other columns. In the general case we do not know an easy way to similarly gain easy access to the Nth column. One step in the proof will be considerations very similar to the ones employed in the proof of the spreading model case (see Lemma 3.6). We start with some technical definitions analogous to the ones employed in the proof in the last section. Definitions 3.1. Let X be a Banach space. (i) A sequence (x n ) in X is called a ul p -sequence, if x n 1 for all n IN, (x n ) converges weakly to 0 and (8) sup α n x n <. ( αn p ) 1/p 1 (x n ) is called C-ul p -sequence, if one can replace (8) by (9) sup α n x n C. ( αn p ) 1/p 1 (ii) A sequence (x n ) in X is called an M-bad ul p -sequence for a constant M <, if (x n ) is a ul p - sequence, and no subsequence of (x n ) is an M-ul p -sequence. Thus for all subsequences (y n ) there exist k IN and (α n ) k Ba l p with k α n y n > M. (iii) An array (x n i ) i, in X is called a bad ul p-array, if each sequence (x n i ) is an M n-bad ul p - sequence for some constants M n with M n as n. (iv) (yi k) is called a subarray of (xn i ), if there is a subsequence (n k) of IN such that every sequence (yi k) is a subsequence of (xn k i ). (v) A bad ul p -array (x n i ) i, is said to satisfy the l p-array procedure, if there exists a subarray (yi n) of (x n i ) and there exist (a n) IR + with a n 1 such that the (weakly null) sequence (y i ) with y i := a nyi n has no ul p -subsequence. An immediate consequence of the definitions are the following observations: A subarray of a bad ul p -array is a bad ul p -array. A bad ul p -array satisfies the l p -array procedure, if and only if it has a subarray satisfying the l p -array procedure. The global idea for the proof of Theorem 1 can be summarized as follows: Lemma 3.2. Let X have property (S p ). Then X has property (US p ) if and only if X satisfies the l p -array procedure. Proof. Clearly if X has property (US p ), X does not contain a bad ul p -array and so the l p -array procedure is satisfied. Let us now assume that X has property (S p ) and satisfies the l p -array procedure, but fails property (US p ). Thus X contains a bad ul p -array. Since X satisfies the l p -array procedure, we can find a sequence (y i ) in X (by the method described in the array procedure), which does not admit of a ul p - subsequence. Observing that (y i ) converges weakly to 0 by the conditions imposed on the a n s, we see that X fails property (S p ), a contradiction. The preceding lemma reduces the proof of Theorem 1 to the proof of 5

Theorem 2. Every Banach space satisfies the l p -array procedure. The proof of Theorem 2 will be broken up into two major steps. First we show that Theorem 2 is valid for the special case of C(K)-spaces where K is a countable compact metric space: Proposition 3.3. Let K be a countable compact metric space. Then C(K) satisfies the l p -array procedure. The second major step is to show that the case of a general Banach space can be reduced to the special case of Proposition 3.3: Proposition 3.4. Let (x n i ) i, be a bad ul p-array in a Banach space X. Then there exist a subarray (yi n) of (x n i ) and a countable ω -compact subset K of Ba Y, where we set Y = [yi n] i,, such that (yn i K) is a bad ul p -array in C(K). Let us observe that Theorem 2 is indeed an easy consequence of the Propositions 3.3 and 3.4. If (x n i ) is a bad ul p -array in X, Proposition 3.4 produces a subarray (yi n) and a countable ω -compact metric space K in Ba Y such that (yi n K) is a bad ul p -array in C(K). Thus there are (a n ) IR + with a n 1 so that the sequence (y i ) in C(K) defined by y i = a nyi n K has no ul p-subsequence. Since K Ba Y it follows that y i = a nyi n itself can have no ul p -subsequence in X. Thus (x n i ) satisfies the l p-array procedure. We now present the proof of Proposition 3.3. This follows rather easily by induction from the following result. Lemma 3.5. Let (X n ) be a sequence of Banach spaces each satisfying the l p -array procedure. ( X n) c0 satisfies the l p -array procedure. Before we present its proof we need another lemma: Lemma 3.6. Let (X n ) be a sequence of Banach spaces each satisfying the l p -array procedure and let (x n i ) be a bad ul p -array in some Banach space. Set X = [x n i ] i, and suppose that for all m IN there is a bounded linear operator T m : X X m with T m 1 such that (T m x m i ) is an m-bad ul p-sequence in X m. Then (x n i ) satisfies the l p-array procedure. Proof of Lemma 3.6. Let us first consider the (easy) case that there exist an m IN and a subarray (yi n) of (xn i ) such that (T myi n) i,n is a bad ul p -array in X m. Since X m satisfies the l p -array procedure, so does (T m yi n) i,n. It follows that (yi n) satisfies the l p-array procedure. Thus (x n i ) itself satisfies the l p-array procedure. Now let us assume that the easy case does not apply. By passing to a subarray, if necessary, we can then assume that for all m IN there is an integer M m such that (T m x n i ) i is an M m -ul p -sequence for all n IN. Inductively we will choose (m n ) P (IN), a subarray (yi n) = (xm n i ), a n > 0 with a n 1 and integers (N n ) such that the following properties hold for all n IN: Then (10) (T mn (y n i )) i is an m n -bad ul p -sequence in X mn (11) (y n i ) i is an N n -ul p -sequence (12) a n m n > n (13) (14) n 1 a j N j < a n m n /4 a j M mn < a n m n /4 j=n+1 6

(15) ( Tmn y l i ) i is an M m n -ul p -sequence for all l. Let us note that (10) is automatically fulfilled because of the hypotheses; (15) holds because of our assumption above. We start the induction as follows. Let a 1 = 1/2; choose m 1 such that a 1 m 1 > 1 (thus (12) is satisfied for ). Since (yi 1) i := (x m1 i ) i is a ul p -sequence, we can choose an N 1 so that (11) holds. If we require for future a j s that a j M m1 < 2 j a 1 m 1 /4 for all j > 1, condition (14) will hold for. The condition (13) does not apply for n = 1. Now let n > 1 and suppose that (a j ) n 1, (m j) n 1 and (N j) n 1 have been chosen such that (11) (13) hold for n 1 and additionally for all 2 j < n Choose a n such that { 0 < a j < min 2 j, 2 j a } km k : 1 k < j 4 M mk { 0 < a n < min 2 n, 2 n a } km k : 1 k < n. 4 M mk Then choose m n large enough to satisfy (12) and (13). This defines (yi n) i = (x mn i ) i. Next choose N n so that (11) is fulfilled. The induction is complete. Because of the conditions imposed on the a n s, (14) holds for all n IN and a n 1. We set y k = a jy j k. Let (y ) be a subsequence of (y k ). We have to show that (y ki ) is not a ul p -sequence, i.e., l sup sup α i y ki =. l IN (α i) Ba l (l) p To this end fix n and choose using (10) k IN and (β i ) Ba l p such that We obtain the following estimate: k k β i y ki = β i a j y j j=n k ( T mn βi yk n ) i > m n. k ( ) T mn β i a j y j k n 1 β i a j y j k ( a n T mn βi yk n ) i j=n+1 k ( ) n 1 a j T mn β i y j k a j β i y j a n m n j=n+1 n 1 a j M mn a j N j by using (10), (15) and (11) resp. a n m n a n m n /4 a n m n /4 by (13) and (14) = a n m n /2. 7

By (12) a n m n as n and the proof is complete. Proof of Lemma 3.5. Let (x n i ) be a bad ul p-array in X = ( X n ) c0. We denote by R m the natural projections R m : X X m. Lemma 3.5 is an easy consequence of the following claim: For all M < there are n, m IN and a subsequence (y i ) of (x n i ) such that (R my i ) i is an M-bad ul p -sequence. Assume the claim is false. Thus we can find M < such that for all m, n IN every subsequence of (x n i ) contains a further subsequence (y i ) such that (R m y i ) i is an M-ul p -sequence. Fix n such that (x n i ) is an (M + 3)-bad ul p -sequence. We can find a subsequence (y i ) of (x n i ) and (m i) P (IN) such that for all i IN (16) sup m>m i R m y i i 1 (17) (R m y j ) j=i+1 is an M-ul p-sequence for all m m i. Indeed, set y 1 = x n 1 and choose m 1 (= 1) such that R m y 1 1 for all m > m 1. Now pass to a subsequence (y 1; i ) of (xn i ) i=2 such that (R my 1; j ) is an M-ul p-sequence for all m m 1. Set y 2 = y 1; 1. Choose m 2 such that R m y 2 2 1 for m > m 2. Again by our assumption, we can find a subsequence (y 2; i ) of (y 1; i ) i=2 such that (R my 2; j ) is an M-ul p-sequence for all m m 2. Set y 3 = y 2; 1 and continue in the obvious fashion. The sequence (y i ) we have constructed clearly satisfies (16) and (17). Since (x n i ) is an (M + 3)-bad ul p-sequence, we can find (α j ) Ba l p with (18) α j y j > M + 3. On the other hand, by (16) and (17) we obtain the following estimate for i IN and m (m i 1, m i ] (where m 0 := 0) : i 1 R m (α j y j ) R m (α j y j ) + R m(α i y i ) + R m (α j y j ) j=i+1 (i 1) (i 1) 1 + 1 + M = M + 2. Thus we obtain that α j y j = sup m R m α j y j M + 2, which contradicts (18). The claim allows us to choose integers N(1) < N(2) <... and (M(n)), and subsequences (yi n) of (x N(n) i ) i, such that ( ) R M(n) yi n is an n-bad ul i p-sequence for all n IN. We let T n = R M(n) [y r i ] i,r=1 and apply Lemma 3.6. This finishes the proof. Proof of Proposition 3.3. Recall that for every countable limit ordinal α we can find a sequence of ordinals β n < α, β n α such that C(α) is isomorphic to ( C(β n )) c0. Using induction and Lemma 3.5 we obtain that all C(α)-spaces, where α is a countable limit ordinal, satisfy the l p -array procedure. Thus, in view of the isomorphic classification of C(K)-spaces for countable compact metric spaces K (see [2]), all C(K)-spaces for countable compact metric spaces K satisfy the l p -array procedure. 8

This completes the first major step. In order to complete the proof of Theorem 2 (and thus the proof of Theorem 1) we have to prove Proposition 3.4. What is the idea behind its proof? Let (x n i ) i, be the given bad ul p-array. Let us focus on the nth column, (x n i ) i. Since (x n i ) is a M n-bad ul p -sequence for some constant M n, we can find for each subsequence (y i ) of (x n i ) integers k 1 < k 2 <... such that α iy ki > M n for some choice of (α i ) with ( α i p ) 1/p 1. This badness can be witnessed by a functional f Ba X : there is an f Ba X such that f( α iy ki ) > M n. The countable ω -compact metric space K will be formed from a subcollection of the functionals f, which witness the badness of a subsequence of the n-th column of a subarray (yi n) of (x n i ) for some n IN. While it is not hard to make sure that only countably many functionals are required to sufficiently witness the badness of the array (x n i ), major difficulties arise from the restriction that K has to be ω -compact. The construction of the right functionals will be given in the proof of Proposition 3.8. Before we state this result and present the construction of the subarray (yi n ) and K needed for Proposition 3.4 we will change the shape of the array (x n i ) to facilitate the quite technical proofs which follow. Instead of the square array (x n i ) i, we will use its triangulated version (zn i ) i,, where zn i = x n i if n i and zi n = 0 otherwise. It is easy to see that a square array satisfies the l p -array procedure if and only if its triangulated version does. From now on we will drop the zero-entries and just label (x n i ) in a triangular fashion (x n i ) 1 n i<. We use the triangular labeling to improve the structure of the given bad ul p -array (x n i ) : Lemma 3.7. A triangular bad ul p -array (x n i ) n i admits a triangular subarray (y n i ) n i, which is a basic sequence in its lexicographical order (where i is the first letter and n is the second letter ): y 1 1, y 1 2, y 2 2, y 1 3, y 2 3, y 3 3, y 1 4,.... Note that we can assume that each column of (x n i ) is semi-normalized. The proof is then an easy adaptation of the the proof that a normalized weakly null sequence has a basic subsequence, using each column of (x n i ) is weakly null. We leave the details of the proof of this lemma to the reader. In the sequel we will assume that the given bad ul p -array (x n i ) is labeled in a triangular fashion and that it is a bimonotone basic sequence in its lexicographical order. (Note that we can renorm the underlying Banach space X, since both properties being a bad ul p -array and satisfying the l p -array procedure are invariant under isomorphisms.) We also assume that (x n i ) i is a semi-normalized M n -bad ul p -sequence where M n. We will also have occasion to use arrays which are labeled in a trapezoidal manner. For this purpose we define the index set T n0 = {(i, n) n i and i n 0 } Proposition 3.8. Given n 0 IN and a trapezoidal subarray of (x n i ) indexed by T n 0 (and again denoted by (x n i ) ), there exist (l i) i=n 0 P (IN) and finite sets F (i, n) [ 1, 1] for (i, n) T n0 such that the following holds: If yi n = x n l i for (i, n) T n0 and if n 0 k 1 < k 2 <... < k q are given such that q β iy n o > M n0 for some (β i ) Ba l p, then there exists an f 3 Ba Y (where Y = [yi n] (i,n) T n0 ) such that (19) q f(α i y n0 ) > M n0 /4 for some (α i ) Ba l p (20) f(y n i ) F (i, n) for all (i, n) T n0 (21) f(y n i ) = 0 for all (i, n) T n0 with i {k 1, k 2,..., k q }. We are now able to complete the proof of Theorem 1 modulo the proof of Proposition 3.8. Proof of Proposition 3.4. We will define the subarray (yi n) n i by a diagonal procedure using Proposition 3.8 repeatedly. First we apply Proposition 3.8 for n 0 = 1 and the array (x n i ) n i and obtain a certain subarray (zi n) (i,n) T 1 ; we set y1 1 = z1. 1 Next we apply the proposition to (zi n) (i,n) T 2. In return we get a new subarray, which we denote by (zi n) (i,n) T 2. The second row of (yi n) is defined by setting y1 2 = z2 1 and y2 2 = z2. 2 If the lth row of (yi n) has been defined (via the array (zn i ) (i,n) T l ), we apply Proposition 3.8 to the array (zi n) (i,n) T l+1. The first row of the array we obtain will be the (l + 1)st row of (yi n ). This completes the induction. Clearly (yi n) n i is a subarray of (x n i ) n i. Moreover (yi n ) in its lexicographical order is a 9

subsequence of (x n i ) in its lexicographical order and thus also bimonotone. Furthermore (yn i ) i is a subsequence of (x n i ) i for all n IN. The set K is constructed next. Let Y = [y n i ] n i and m IN, and define K m = { (k 1, k 2,..., k q ) m k 1 < k 2 <... < k q, r α i yk m i M m for all (α i ) Ba l p for all r < q and } q α i yk m i > M m for some (α i ) Ba l p. Whenever k = (k 1,..., k q ) K m, our application of Proposition 3.8 in the definition of the subarray (yi n) yields a functional f of norm not greater than 3, which is defined on a certain subspace of X, spanned by some elements of (x n i ). By first restricting to [(yn i )] (i,n) T n0 and then extending to Y in a trivial manner, we may assume that f 3 Ba Y with q f(α iyki n ) > M n/4 for some (α i ) Ba l p (among its other properties (20) and (21)). It is here, where we use that (yi n ) is a bimonotone basic sequence. We denote the functional f/3 by f k and let K n = {Q mf k m IN, k K n }. Here Q m denotes the natural projection of norm 1 from Y onto [y n i ] 1 n i m. Finally we define K = K n {0}. Let us first note that (yi n K) n i is a bad ul p -array. Indeed, fix a column n 0. By our construction (y n 0 i ) is an M n0 -bad ul p -sequence. Consequently, given a subsequence (y n0 ) of (y n0 i ), k := (k 1,..., k q ) K n0 for some q IN. By (21) f k = Q mf k for m large enough and thus f k K n0 K. By (19) we obtain that (y n0 i K) is an M n 0 /12-bad ul p -sequence in C(K). By the construction of K it is obvious that K is a countable subset of Ba Y. Since Y is separable, K is ω -metrizable. It remains to show that K is ω -closed. Let (g j ) K and assume that (g j ) converges ω to some g Ba Y. We have to show that g K. Every g j is of the form Q m j f kj for some m j IN and some k j K nj for some n j IN. By passing to a subsequence of the sequence (g j ) we can assume that either n j as j, or there is an n IN such that n j = n for all j IN. Let us first deal with the first alternative. Let i j be the first element of k j. Since i j n j, i j as j. Moreover f kj (yi n) = 0 for all n i < i j by (21). Thus we obtain that f kj 0 in the ω -topology as j, i. e. g = 0 K. From now on we assume that there is an n IN such that k j K n for all j IN. K n is relatively sequentially compact, if endowed with the relative product topology on {0, 1} IN ; thus we can assume, by passing to a subsequence of (g j ) if necessary, that k j k for some k K n, the closure of K n. We claim that s finite. Suppose to the contrary that k = ( ). Since k K n we can find for every q IN an element in K n of the form (k 1,..., k q, l 1,..., l r ). By the definition of K n, q α iyk n i M n for all (α i ) Ba l p. Thus (yk n i ) is an M n-ul p -sequence, a contradiction to the fact that (yi n) is an M n-bad ul p -sequence. Since Ba Y is ω -sequentially compact, we may additionally assume that f kj converges in the ω -sense to some f Ba Y. We claim that f K. First we observe that Q mf K for all m IN. Why? By (20) and (21) the set {Q mf kj (yi n) j IN, 1 n i} has only finitely many elements. Since Q mf kj ω Q mf as j, we obtain that Q mf kj = Q mf for 10

j IN large enough; in particular Q mf K. Next let q = max k. Since k j k and s finite, Q qf = f. Thus f K. We want to show that g K. By passing to yet another subsequence of the (g j ) we can assure that either m j as j, or there is an m IN such that m j = m for all j IN. If the first case occurs or if m q in the second case, g j = Q m j f kj converges ω to f, hence g = f K. If on the other hand the second case applies and m < q, g j = Q m j f kj converges ω to Q mf. Since f K, g = Q mf K. The major tool for the construction of the subarray (yi n ) in Proposition 3.8 is the next lemma, which is a variation (and generalization) of a lemma employed by J. Elton in his thesis [6]. It allows us to almost achieve the condition in (21) for one row preceding the s. Its versatility will allow repeated application to construct the array (yi n ) row by row, carefully preserving properties obtained in previous steps. Lemma 3.9. Let n 0 IN and let (x n i ) (i,n) T n0 be our bad ul p -array indexed trapezoidally. Let B 2 Ba X, C > 0, ε > 0, δ > 0, N P (IN) and n n 0 be given. Then there exists L P (N) such that if (l i ) q i=0 L with n l 0 < l 1 <... < l q is given and if there exists an f B with (22) q f(α j x n 0 l j ) > C for some (α j ) with q α j p 1/p δ then there exists a g B with q g(β j x n 0 l (23) j ) > C for some (β j ) with and g(x m l 0 ) < ε for 1 m n. q β j p 1/p δ Proof. We let A q = { I = (l j ) j=0 P (N) l 0 n and if there is an f B satisfying (22), then there is a g B satisfying (23) }. Let A = q=1 A q. Every set A q (and hence A) is a Ramsey set, since the conditions imposed on I involve only the first (q + 1) elements in I. By Ramsey theory (see e.g., [14] for a discussion of Ramsey theory) we can thus find an L P (N) such that either P (L) A or P (L) P (N) \ A. In the first case the proof is finished; we show that the second Ramsey alternative leads to a contradiction. We write L = {l 0, l 1,...}. For fixed q IN and 1 r q we let L r = {l r, l q+1,, l q+2...}. By our assumption L r A, thus L r A sr for some s r IN. Consequently there exist (αj r) with ( s r αr j p ) 1/p δ and a function f r B such that f r ( sr α r jx n 0 l q+j ) > C and whenever g( s r β jx n0 l q+j ) > C for some function g B and some (β j ) with ( s r β j p δ, then g(x m l r ) ε for some 1 m n. Let r 0 be chosen such that s r0 = min {s r 1 r q}. We obtain for each 1 r q ( s r0 ) C < f r0 α r 0 j xn 0 l q+j = f r0 ( sr ) α r 0 j xn 0 l q+j, where we set α r 0 j = 0 for s r0 < j s r. It follows that for each r we can find an 1 m r n with f r0 (x m r l r ) ε. Now we let q vary. Set g q = f r0 and let g be a ω -clusterpoint of {g q } q IN. Due to our construction we can find for each r IN a column m r, 1 m r n, with g(x m r l r ) ε. Thus one of the sequences (x m l i ), 1 m n, is not weakly null, a contradiction. 11

Proof of Proposition 3.8. In preparation for the diagonal procedure which follows we will introduce the following quantities and sets. Let ε = min {1, M n0 /4}. Let (b n i ) be the biorthogonal functionals associated with the bimonotone basic sequence (x n i ) (i,n) T n0 in its lexicographical order. For (i, n) T n0 we choose (ε n i ) > 0 such that (24) We fix for each (i, n) T n0 i ε n i b n i < ε. i=n 0 a finite ε n i /2-net in [ 1, 1] denoted by Hn i. Let B 1 := {f 2 Ba X f(x n i ) Hi n for all (i, n) T n0 } Let us observe that by (24), whenever we can find a g Ba X, (α i ) Ba l p, and (l i ), l 1 > n 0, with g( α ix n0 l i ) > M n0, then there is an f B 1 with f( α ix n0 l i ) > 3 4 M n 0. Next we choose ε m > 0 for m n 0 so that (25) m=n 0 mε m sup {b n i (i, n) T n0, n m} < ε. The suprema above are finite, since each of the sequences (x n i ) was assumed to be semi-normalized. For m n 0 we let Γ m be a finite ε m -net in the interval (0, M] and we let m be a set of positive reals such that the set {δ p : δ m } is a (2 mp )-net for [0, 1], which contains 1. Furthermore we require that m m+1 for all m n 0. We are ready to start the induction. Choose C 1 Γ n0 and δ 1 n0 arbitrarily, apply Lemma 3.9 to (B 1, ε n0, C 1, δ 1, n 0, IN) and obtain L 1 1 P (IN). We pick δ 2 n0, δ 1 δ 2, and apply the lemma to (B 1, ε n0, C 1, δ 2, n 0, L 1 1), obtaining a new infinite subset L 1 2. We continue applying Lemma 3.9 successively until we have exhausted all combinations for which (C, δ) Γ n0 n0. If L 1 is the last infinite subset of IN thus obtained, we let l 1 = min L 1. This defines the first row of the (trapezoidal) subarray (yi n): yj n 0 = x j l 1 for 1 j n 0. We set F (n 0, j) = H j l 1 for 1 j n 0. For the second step of the induction we first partition the set B 1 into finitely many sets as follows: For t = (t 1,..., t n0 ) n 0 F (n 0, j) we let B 2 t = { f B 1 f(y j n 0 ) = t j for all 1 j n 0 }. Similarly to what we did in the first step we apply Lemma 3.9 successively to (B t 2, ε n 0 +1, C, δ, n 0 + 1, ) beginning with the infinite set L 1, until we have exhausted all combinations ( t, C, δ) n 0 (F (n 0, j) Γ n0+1 n0+1). Let L 2 denote the last sequence thus obtained. We choose as l 2 an element in L 2 with l 2 > l 1. This defines the second row of the subarray: y j n 0 +1 = xj l 2 for 1 j n 0 +1. We set F (n 0 +1, j) = H j l 2 for 1 j n 0 + 1. For the general induction step let us assume that l 1 < l 2 <... < l m and L m have been found in the way now described. This defines the first m rows of (yi n) (i,n) T n0. We set F (m, j) = H j l m for 1 j m, where m = n 0 + m 1. We partition B 1 as before into finitely many sets: for t = (t n i ) {F (i, n) (i, n) T n0 and i m } we let B m+1 t = { f B 1 f(y n i ) = t n i for all (i, n) T n0 with i m }. Now we apply Lemma 3.9, starting with the sequence L m, successively to (B m+1, ε t m +1, C, δ, m +1, ) where ( t, C, δ) ranges over all possible combinations in (F (i, n) Γ m +1 m +1). If L m+1 is the last sequence we obtain, we choose l m+1 L m+1 with l m+1 > l m. The induction is complete. From now on we will identify the functionals f B 1 with their restrictions to Y = [yi n] (i,n) T n0. The subarray (yi n) (i,n) T n0 of (x n i ) (i,n) T n0 has been chosen such that the following holds: 12

Lemma 3.10. Let n 0 n < k 1 < k 2 <... < k r be given. If there exists an f B 1 such that r f(α iy n0 ) > C for some (α i ) with ( r α i p ) 1/p δ, δ n, C Γ n, then there exists g B 1 with (26) (27) (28) r g(β i y n0 ) > C for some (β i ) with g(y m i ) = f(y m i ) F (i, m) for all (i, m) T n0 g(y m n ) < ε n for all 1 m n. ( r ) 1/p β i p δ with i < n We claim that the proof of Proposition 3.8 is complete, once we prove Lemma 3.11. Let n 0 < k 1 <... < k q be given. If q α iy n 0 > M n0 for some (α i ) with ( q α i p ) 1/p 1, then there exists an h B 1 such that (29) q h(γ i y n0 ) > M n0 /4 for some (γ i ) with ( q ) 1/p γ i p 1 (30) h(y n i ) = 0 for all n n 0, if i > k q (31) h(y n i ) < ε i for all n n 0, if i {k 1, k 2,..., k q }. Completion of the proof of Proposition 3.8. Indeed, by perturbing h, we can obtain an f Y such that for (i, n) T n0, f(yi n) = h(yn i ) if i {k 1, k 2,..., k q } and f(yi n ) = 0 otherwise. Thus (21) is satisfied. Moreover it follows that q f(γ iy n0 ) = q h(γ iy n0 ) > M n0 /4 and f(yi n) F (i, n) for all (i, n) T n 0. Thus (19) and (20) hold. By using (25) we can estimate f as follows: f h k q ε i i=n 0 i b n l i m=n 0 m ε m sup {b n i (i, n) T n0, n m} < ε 1. Since h 2, we have f 3. This ends the proof of Proposition 3.8. Proof of Lemma 3.11. If q α iy n0 > M n0 for some q IN, n 0 k 1 <... < k q and some (α i ) with ( q α i p ) 1/p 1, we can find by our earlier observation a function g B 1 with g( q α iy n 0 ) > 3 4 M n 0. We will apply Lemma 3.10 (at most) (k q n 0 + 1) times beginning with the function g and the row n = n 0. We choose C n0 Γ n0 such that 0 < 3 4 M n 0 C n0 < ε n0. If n 0 = k 1, we set h n0 = g and α i,n0 = α i for 1 i q. We set γ 1 = α 1,n0, if h n0 (α 1,n0 y n 0 n 0 ) 0, and γ 1 = α 1,n0 otherwise. We let β n0 = h n0 (γ 1 yn n0 0 ) and choose a δ 1 n0 such that δ p 1 2 p q j=2 α j,n 0 p δ p 1. If on the other hand n 0 < k 1, we apply Lemma 3.10 to g, n 0 = n < k 1 <... < k q, δ = 1 ( n0) and C = C n0 Γ n0. Note that C < 3 4 M n 0 and that therefore the premise of the lemma is indeed fulfilled. The application yields a new functional h n0 B 1 and a new (α i,n0 ) q Ba lq p with q h n 0 (α i,n0 y n0 ) > C n0 and h n0 (yn m 0 ) < ε n0 for 1 m n 0. We let β n0 = 0 and δ 1 = 1. 13

Let us now assume that s n 0 and we have thus far constructed h s B 1, α i,s for i = 1,..., q and for n 0 r s, scalars C r Γ r, δ r r and β r 0, and furthermore for each 1 s we have chosen γ i IR such that the following conditions hold: (32) 0 < (C r 1 β r 1 ) C r < ε r for all n 0 r s (33) ( {i >r} α i,r p ) 1/p δ j for all n 0 r s with k j r < k j+1 (34) δ p j 2 jp (35) q i=j+1 {i s} α i,kj p for all n 0 k j s h s (α i,s y n 0 ) > C s (36) β r = h s (γ i y n 0 ) if r = for some 1 i q, and β r = 0 otherwise (37) h s (y m r ) < ε r for all 1 m r s with r {k 1, k 2,..., k q }. Let us observe that all these conditions are satisfied for s = n 0, if we let C n0 1 = 3 4 M n 0, β n0 1 = 0, k 0 = n 0 and δ 0 = 1. (We also set α i,n0 1 = α i for 1 i q.) Next we choose C s+1 Γ s+1 with 0 < (C s β s ) C s+1 < ε s+1 to satisfy (32) for s + 1. Note that C s+1 < C s. (If C s β s < ε s+1, we quit the procedure and set h = Q sh s. Estimates below will show that h satisfies the conclusion of Lemma 3.11. We set γ j = 0 for k j > s.) If s + 1 = k j for some 1 j q we set h s+1 = h s and α i,s+1 = α i,s for 1 i q. We set γ j = ±α j,s+1 so that β s+1 := h s+1 (γ j y n 0 k j ) 0 and choose δ j s+1 so that (33) and (34) are satisfied for s + 1. (36) is satisfied for s + 1 by our construction; in (37) no new condition is imposed, so it remains to check (35). Indeed we have {i s+1} h s+1 (α i,s+1 y n0 ) = {i s+1} {i s} h s (α i,s y n0 ) h s (α i,s y n 0 ) β s by (36) > C s β s > C s+1 using (35) for r = s and (32) for r = s + 1. If s + 1 {k 1, k 2,..., k q }, say k j 1 < s + 1 < k j for some j, we apply Lemma 3.10 to h s, n = s + 1 < k j <... < k q, C s+1, δ j 1 and (α i,s ), j i q. Let us check that the lemma applies for these parameters: Clearly C s+1 Γ s+1 and δ j 1 s s+1. Moreover Finally, as above, we obtain q α i,s p i=j q i=j 1/p = h s (α i,s y n0 ) = {i >s} {i s+1} α i,s p 14 1/p δ j 1 by (33). h s (α i,s y n0 ) > C s+1.

The application of Lemma 3.10 yields a new functional h s+1 B 1, and new (α i,s+1 ) for j i q with ( q i=j α i,s+1 p ) 1/p δ j 1, q i=j h s+1(α i,s+1 y n 0 ) > C s+1 and h s+1 (ys+1) m < ε s+1 for all 1 m s + 1; hence (37) is satisfied for r = s + 1. Since h s+1 preserves the values of h s on the rows prior to the (s + 1)st row, h s+1 satisfies (37) also for 1 r s. We set β s+1 = 0. By our construction (33), (35) and (36) are satisfied for s + 1; (34) does not impose a new condition. Unless we stopped the construction earlier, we quit after we have obtained h kq and let h = Q k q h kq. Observing that the conclusions (30) and (31) in Lemma 3.11 hold, it remains to check (29). If h is defined to be h = Q k q h kq we obtain q h(γ i y n 0 ) = q β ki by (36) = β kq + β kq + (Note that β kq C kq 0 by (35) and (36).) A similar estimate holds if h = Q sh s for some s < k q : k q r=n 0 β r 1 k q r=n 0 (C r 1 C r ε r ) (β kq C kq ) + 3 4 M n 0 3 4 M n 0 ε 1 2 M n 0. k q r=n 0 ε r q h(γ i y n0 ) β s + s r=n 0 (C r 1 C r ε r ) ( 3 (β s C s ) + 4 M n 0 s r=n 0 ε r ) 3 4 M n 0 ε 1 2 M n 0. (This time β s C s ε s+1 by our stopping condition.) The proof is complete, once we show that ( q γ j p ) 1/p 2. Indeed, if h = Q k q h kq, we have for 1 j < q, q q γ j p = α j,kj 1 p α i,kj 1 p α i,kj p i=j i=j+1 δ p j 1 (δp j 2 jp ) by (33) and (34); Thus we obtain γ q p = α q,kq 1 p δ p q 1 by (33). q q 1 q 1 γ j p (δ p j 1 δp j ) + δp q 1 + 2 jp q 1 1 + 2 jp 2. We obtain the same estimate, if we stopped the procedure before reaching the row k q. 15

4. Proofs of the corollaries Proposition 3.8 can be phrased for a single weakly null sequence as follows: Corollary 3. Let (x n ) be a semi-normalized weakly null sequence in a Banach space X. Assume none of its subsequences has a C-upper l p -estimate. Then there exists a subsequence (y n ) (x n ), such that for all (m n ) P (IN) there are a functional f Ba([y n ] ), an l IN and (α i ) l Ba l p such that ( l ) f α i y mi > C/12 and f(y i ) = 0, if i / {m 1,..., m l }. C. Schumacher [16] uses this corollary to deduce the following weak-cauchy criterion for property (S p ), which is a generalization of an analogous result in the c 0 -case in [11]: Proposition 4.1. Let X be a Banach space. The following are equivalent: (i) X has property (S p ). (ii) Every weak Cauchy-sequence in X subsequences (y n) of (y n ) satisfy (Here y 0 = 0.) has a subsequence (y n ) such that, for some constant C <, all a n (y n y n 1) < C for all (a n) Ba l p. Corollary 1 is an easy consequence of this proposition: Proof of Corollary 1. Let (z n ) be a weak Cauchy-sequence in Ba(X/Y ) and let q : X X/Y denote the quotient map. By a result of R.H. Lohman [13] we can find a weak Cauchy-sequence (x n ) in X such that its image under q is some subsequence of (z n ). Since X has property (S p ), (x n ) satisfies the conclusion of (ii) of Proposition 4.1; so does its image under q, which is still a subsequence of (z n ). Next we present the proof of the second corollary: Proof of Corollary 2. We will only prove the second statement. The proof of the first statement is quite similar and will be left to the reader. By Theorem 1 we can find constants C p and C q such that every weakly null sequence in Ba X (resp. in Ba X ) admits a subsequence with a C p -upper l p -estimate (resp. C q -upper l q -estimate). Let (f n ) be a normalized weakly null sequence in X. By passing to a subsequence we may assume that (f n ) has a C q -upper l q -estimate. We choose a separable subspace Y of X which isometrically norms all the f n s and denote by g n the restriction of f n to Y. Using a result due to W.B. Johnson and H.P. Rosenthal [10], we can find a basic sequence (x n ) Y with x n 3 for all n, such that a subsequence of (g n ), which we still denote by (g n ), is biorthogonal to (x n ). Since l 1 does not embed into Y, we can assume that (x n ) is a weak Cauchy sequence. Furthermore, since Y has property (S p ) we may assume, by passing to a subsequence of (x n ), that y n := x 2n x 2n 1 has a 6C p -ul p -estimate. We claim that (f 2n ) has a lower l q -estimate. Indeed, let (b n ) be given with ( b n q ) 1/q = 1. Choose (a n ) with ( a n p ) 1/p = 1 and a nb n = 1. We obtain the following estimate: ( ) 1 ( ) ( ) b n f 2n a m y m b n f 2n a m y m m=1 (6C p ) 1 a n b n = (6C p ) 1. 16 m=1

The projection Q : X [f 2n ] is defined by Qf = f(y n)f 2n. Note that f is only applied to the ul p -sequence (y n ) in Y ; thus Q is well defined. It is easy to check that Q is the identity map on [f 2n ] and that Q 6C p C q. J. Elton obtained in [6] the following characterization: Let (x n ) be a semi-normalized weakly null sequence in a Banach space without a subsequence equivalent to the unit vector basis of c 0. Then (x n ) has a subsequence (y n ) such that k a ny n as k, whenever (a n ) / c 0. We conclude by showing that an analogous result fails in a strong way for the l p -case: Proposition 4.2. Let 1 < p <. There is a Banach space X with a 1-symmetric basis (e n ) such that the following properties hold: (i) (e n ) does not have an upper l p -estimate. (ii) There is a sequence (α n ) / l p such that a ne n converges. Proof. Choose 1 < p 0 < p < p 1. It is easy to construct a concave increasing function λ : IN IR + with λ(1) = 1, such that both B := {n : λ(n) n 1/p0 } and L := {n : λ n n 1/p1 } are infinite subsets of IN. Let X be the Lorentz sequence space d(w, 1), where w 1 = λ(1) and w n = λ(n) λ(n 1) for n > 1. X has a 1-symmetric basis (e n ) with the property λ(n) = n e i (see [12, I, p.120]). Since #B =, (e n ) does not have an upper l p -estimate. To see (ii) we proceed as follows: we choose an increasing sequence (l k ) L with (37) l 1 (p/p 1) k > k 2p. Set m 0 = 0, m j = j k=1 l k and F j = (m j 1, m j ] for j 1. We consider a ne n, where Since l j L, l 1/p 1 j n F j e n 1, and hence On the other hand, for N IN, a n = j 2 l 1/p 1 j, if n F j. ( a n e n = j 2 l 1/p 1 j n F j e n ) < j 2 <. m N N a n p = l j (j 2 l 1/p 1 j ) p = N l 1 (p/p 1) j j 2p N by (37). Thus (a n ) / l p. References 1. I. Amemiya and T. Ito, Weakly null sequences in James spaces on Trees, Kodai Math. J., 4 (1981), 418 25. 2. C. Bessaga and A. Pe lczyński, Spaces of continuous functions IV, Studia Math., 19 (1960), 53 62. 3. A. Brunel and L. Sucheston, On B-convex Banach spaces, Math. Systems Th., 7 (1974), 294 9. 4. A. Brunel and L. Sucheston, On J-convexity and some ergodic super-properties of Banach spaces, Trans. AMS, 204 (1975), 79 90. 5. J. Diestel, Sequences and Series in Banach spaces, Springer-Verlag, N.Y., 1984. 6. J. Elton, Weakly null normalized sequences in Banach spaces, Dissertation, Yale University, 1978. 7. V.I. Gurariǐ and N.I. Gurariǐ, Bases in uniformly convex and uniformly flattened Banach spaces, Math. USSR Izv., 5 (1971), 220 5, English translation. 8. R.C. James, Super-reflexive spaces with bases, Pacific J. Math., 41 (1972), 409 19. 9. W.B. Johnson and E. Odell, Subspaces of L p which embed into l p, Compos. Math., 28 (1974), 37 49. 17

10. W.B. Johnson and H.P. Rosenthal, On ω -basic sequences and their applications to the study of Banach spaces, Studia Math., 43 (1972), 77 92. 11. H. Knaust and E. Odell, On c 0 -sequences in Banach spaces, to appear in Israel J. Math. 12. J. Lindenstrauss and L. Tzafriri, Classical Banach Spaces, 2 vols., Springer-Verlag, Berlin, 1977, 1979. 13. R.H. Lohman, A note on Banach spaces containing l 1, Canad. Math. Bull., 19 (1976), 365 7. 14. E. Odell, Applications of Ramsey theorems to Banach space theory, in: H.E. Lacey (ed.), Notes in Banach Spaces, University of Texas Press, Austin and London, 1981, 379 404. 15. H.P. Rosenthal, A characterization of Banach spaces containing l 1, Proc. Nat. Acad. Sci. (USA), 71 (1974), 2411 3. 16. C. Schumacher, Dissertation, The University of Texas at Austin, in preparation. 18