Banach spaces which are uniformly non-square over blocks and related properties

Nonlinear Analysis 34 (1998) 991 – 1004

Banach spaces which are uniformly non-square over blocks and related properties D. Kutzarova a;1 , T. Zachariades b;∗ b

a Institute of Mathematics, Bulgarian Academy of Sciences, 1090 So a, Bulgaria Department of Mathematics, University of Athens, Panepistemiopolis, 15784 Athens, Greece

Received 10 December 1996; received in revised form 6 March 1997; accepted 13 March 1997

Keywords: Schauder nite dimensional decomposition; Uniformly convex spaces; Spreading model; Weak Banach–Saks property; Alternate signs Banach–Saks property

Introduction In [1] the authors, using some ideas of Maurey [2], introduce orthogonally convex (O.C.) spaces and prove that uniformly convex spaces, spaces with the Schur property, the space c0 , James’ space J and others have this property. Also in that paper they prove that O.C. spaces have the xed point property (f.p.p.). In [3] it is proved that O.C. spaces have the weak Banach–Saks (w.B.S.) property, (that is, for every weakly 0 null sequence P n (x0 n )n∈N of X there exists a subsequence (xn )n∈N such that the Cesaro 1 means n i=1 xi ; n = 1; 2; : : : are norm convergent). Thus, if an O.C. space does not contain isomorphically the space l1 then it has the alternate signs Banach–Saks property, X there exists a subsequence (A.B.S.) (that is for every bounded sequence (xn )n∈N Pof n (xn0 )n∈N such that the alternate signs Cesaro means 1n i=1 (−1)i xi0 ; n = 1; 2; : : : are norm convergent). In this paper we introduce the subclass of Banach spaces with a Schauder nite dimensional decomposition (F.D.D) which are uniformly nonsquare over blocks. We prove that this property implies the A.B.S property and we give some examples of spaces which are such spaces. We also prove that the James tree spaces JT is uniformly nonsquare over blocks and give a sucient condition for a uniformly nonsquare over blocks space to be O.C. From this condition it follows that the JT space is O.C. Finally ∗ 1

Corresponding author. Research partially supported by the Bulgarian Ministry of Education and Science under contract MM213=92. 0362-546X/98/$19.00 ? 1998 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 7 ) 0 0 6 0 7 - X

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we prove a theorem which concerns the stability of this property for p-direct sums, with 1¡p¡∞. De nitions and notations. Let X be a Banach space and BX = {x ∈ X : kxk ≤ 1} be the unit ball of X . The space X is called uniformly nonsquare if there exists ¿0 such that there are no x; y ∈BX for which kx + yk¿2 −  and kx − yk¿2 −  [4]. If (xn )n∈N is a bounded sequence in X and ¿0 we set D[(xn )] = lim n sup(lim m sup kxn − xm k), and A [(xn )] = lim n sup(lim m sup |M (xn ; xm )|), where M (xn ; xm ) = {z ∈X : max{kz − xn k; kz − xm k} ≤ 12 (1 + )kxn − xm k} and |M (xn ; xm )| = sup{kzk : z ∈M (xn xm )}. The space X is called orthogonally convex (O.C.) if for every weakly null sequence (xn )n∈N in X , with D[(xn )]¿0, there exists ¿0 such that A [(xn )]¡D[(xn )] [1]. A sequence (Xn )n∈N of nite dimensional subspaces of X is called a Schauder nite dimensional decomposition (F.D.D.) of X if every x ∈X has a unique representation P∞ of the form x = n=1 xn with xn ∈ Xn for every n ∈N. It is clear that every Schauder basis (en ) in a space X forms (F.D.D.). Let (Xn )n∈N be a F.D.D. of X . For every P∞ Pk k ∈ N we set Pk ( n=1 xn ) = n=1 xn . We have that supk kPk k¡+ ∞ and this number is called the constant of the F.D.D. The F.D.D. is called bimonotone if kPk k ≤ 1 and kI − Pk k ≤ 1 for every k ∈ N, where I is the identity operator. For every x ∈ X the support of x is the set supp x = {n∈ N : xn 6= 0} and for x; y ∈X we write supp x¡supp y if max(supp x)¡ min(supp y). For further information on spaces with F.D.D. see [5]. Let (xn )n∈N be a bounded sequence in X . A spreading model (s.m.) of X , which is built on the sequence (xn )n∈N , is a Banach space F with a sequence (en )n∈N such that F = [(en )], that is F is the closed linear span of (en )n∈N , and for every ¿0; k ∈N and c1 ; : : : ; ck ∈R there exists N (k)∈N such that for every N (k) ≤ n1 ¡ · · · ¡nk



k k P

P

ci ei − ci xni ¡:



i=1

i=1

The sequence (en )n∈N is called the fundamental sequence (f.s.) of the s.m. F. In [6] it is proved that every bounded and no norm-convergent sequence (xn )n∈N has a subsequence with a s.m. If (xn )n∈N is a weakly null sequence with s.m. F then the f.s. (en )n∈N is an unconditional basis of F. For further information on s.m. see [7]. 1. Uniformly nonsquare over blocks spaces In the sequel by a space X we will mean a Banach space X . De nition 1.1. A space X is called uniformly nonsquare over blocks if there is an F.D.D. of X with the following property: there exists ¿0 such that there are no x; y ∈ BX , with supp x¡supp y, for which kx + yk¿2 −  and kx − yk¿2 − . For every space X and F.D.D. (Xn )n∈N of X we set b[X; (Xn )] = sup{min{kx + yk; kx − yk} : x; y ∈Bx and supp x¡supp y}:

D. Kutzarova, T. Zachariades / Nonlinear Analysis 34 (1998) 991 – 1004

993

The proof of the following property is easy. Proposition 1.2. Let X be a space. Then the following are equivalent; (i) X is uniformly nonsquare over blocks. (ii) There is an F.D.D. (Xn )n∈N of X such that b[X; (Xn )]¡2. (iii) There is an F.D.D. (Xn )n∈N of X such that X does not contain nearly isometrically over blocks of (Xn )n∈N the space l(2) 1 ; that is, there exists ¿0 such that there are no x; y ∈ X; with supp x¡supp y and (1 − )(|c1 | + |c2 |) ≤ kc1 x + c2 yk ≤ (1 + )(|c1 | + |c2 |) for every c1 ; c2 ∈ R. Remarks. (i) From Proposition 1.2 and from [4] we obtain that every uniformly nonsquare over blocks space does not contain isomorphically the space l1 . (ii) Let X be a space and (Xn )n∈N be an F.D.D. of X such that b[X; (Xn )]¡2. It is easy to see that if (Xn )n∈N is a weakly null sequence in X; with s.m. F and (en )n∈N is the f.s. of F; then b[F; (en )]¡2. So, from Proposition 1.2, if X is uniformly nonsquare over blocks then every s.m. F, which is built on a weakly null sequence of X; is uniformly nonsquare over blocks. From the above remarks and from [8] we obtain the next corollary. Corollary 1.3. Every space X which is uniformly nonsquare over blocks has the alternate signs Banach–Saks (A.B.S.) property; equivalently, no bounded sequence in X has a spreading model equivalent to the usual basis of l1 . Proof. From the Remark (i) above and from [8], it is enough to prove that no weakly null sequence in X has a s.m. which is isomorphic to the space l1 . This follows from Remarks (i), (ii) above. In [1] the authors de ne, for a space X and an F.D.D. (Xn )n∈N of X , the number p[X; (Xn )] = sup{kx + yk : x; y ∈BX with supp x¡supp y} and they prove that if p[X; (Xn )]¡2 and the F.D.D. is bimonotone then X is O.C. It is clear that if p[X; (Xn )] ¡2 then b[X; (Xn )]¡2 and so X is uniformly nonsquare over blocks. (We prove below that b[JT; (Xn )]¡2, for a suitable F.D.D. (Xn )n∈N of JT, but p[JT; (Yn )] = 2 for every F.D.D. (Yn )n of JT). Examples of uniformly nonsquare over blocks spaces are c0 , uniformly convex spaces with F.D.D. and others. Another interesting class of spaces which are uniformly nonsquare over blocks are dual spaces X ∗ such that X satis es (i) X has a shrinking F.D.D. (Xn )n∈N , and (ii) there exist 0¡c¡2 such that kxk + kyk ≤ c. kx + yk for every x; y ∈X , with supp x¡suppy. Examples of spaces X with the properties (i) and (ii), include among the others, are the space S of Schlumprecht [9] and the space GM of Gowers and Maurey [10]. (In fact these spaces have a stronger property). Spaces with property (ii) have weak normal structure (from [11]), which in re exive spaces is equivalent to having normal

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D. Kutzarova, T. Zachariades / Nonlinear Analysis 34 (1998) 991 – 1004

structure. So they have the f.p.p. For the dual space X ∗ of a space X satisfying (i) and (ii) it is easy to see that p[X ∗ ; (Xn∗ )]¡2. Thus X ∗ is uniformly nonsquare over blocks and, if moreover the F.D.D. is bimonotone, then X ∗ is O.C. So the spaces S ∗ and GM ∗ are O.C., uniformly nonsquare over blocks spaces but they do not have an equivalent uniformly nonsquare norm since they are not super re exive [12]. Pei-kee Lin has shown that S has a bounded sequence with s.m. isomorphic to l1 . From this we obtain that GM also has the same property. So S and GM do not have the A.B.S. property and thus, from [3], they are not O.C.; also, from Corollary 1.3, they are not uniformly nonsquare over blocks. The famous space JT (James tree) [13] was the rst example of a space with nonseparable dual having no subspace isomorphic to l1 . We will prove that JT is uniformly nonsquare over blocks, which, as we saw in the remarks after Proposition 1.2, implies that l1 does not embedd in JT. Before the proof we repeat the de nition of the space JT. Let T be the set T = {(n; i) : n = 0; 1; : : : ; i = 0; 1; : : : ; 2n − 1}. For every t = (n; i)∈T we put |t| = n. Let (n; i) and (m; j) be in T . We put (n; i) ≤ (m; j) if n = m and i = j or n¡m and there exist integers i0 = i; i1 ; : : : ; ik = j, with k = m − n and il ∈{2il−1 ; 2il−1 + 1} for l = 1; : : : ; k. A set S = {(n; i0 ); (n + 1; i1 ); : : : ; (n + k; ik )} is called a segment. For every n ∈ N and s ∈T we put Tn = {t ∈ T : n ≤ |t|} and Ts = {t ∈T : s ≤ t}. It is easy to see that, if s = (m; i)∈T then Ts = {(m + n; 2n i + k) : n = 0; 1; : : : ; and k = 0; 1; : : : ; 2n − 1}. The space JT is the completion of the space of all sequences of real numbers x = (xt )t∈T , with nite support, and with norm   !2 1=2    P  n P   kxk = sup xt : S1 ; : : : ; Sn pairwise disjoint segments :    i=1 t∈Si  For every s ∈T we put es = (xt )t∈T with xs = 1 and xt = 0 for t 6= s. It is straightforward to show that (es )s∈T is a Schauder basis of JT (enumerated e0;0 ; e1;0 ; e1;1 ; : : : ; en;0 ; en;1 ; : : : ; en+2n −1 ; en+1;0 ; : : :). For every n∈ N let Xn be the nite dimensional subspace of JT which is generated by {et : |t| = n}. It is clear that the sequence (Xn )n∈N forms an F.D.D. of JT. In the sequel, when we write supp x, for x ∈ JT, we mean the support of x related to the basis (es )s∈T . We will prove that b[JT; (es )s∈T ] = b[JT; (Xn )]¡2. Proposition 1.4. b[JT; (es )s∈T ] = b[JT; (Xn )]. Proof. It is obvious that b[JT; (Xn )] ≤ b[JT; (es )s∈T ]. Let x = (xt )t∈T and y = (yt )t∈T in BJT with supp x¡supp y. If there exists l∈N such that supp x ⊂ T \Tl+1 and supp y ⊂ Tl+1 then min{kx + y}; kx − yk} ≤ b[JT; (Xn )]. We suppose that there exist l ∈N and 0¡k ≤ 2l − 1 such that supp x ⊂ (T \T1 ) ∪ {(l; i) : 0 ≤ i¡k}

and

suppy ⊂ Tl+1 ∪ {(l; i) : k ≤ i ≤2l − l}: For every k ≤ i ≤ 2l − 1 we set ’i : T(l; i) → T(l+1; 2i) ; with ’i ((l + n; 2n i + k)) = (l + n + 1; 2n+1 i + k) for n = 0; 1; : : : ; and k = 0; : : : ; 2n − 1: It is clear that ’i is 1 − 1, onto and

D. Kutzarova, T. Zachariades / Nonlinear Analysis 34 (1998) 991 – 1004

995

if (m1 ; j1 ); (m2 ; j2 ) ∈ T(l; i) with (m1 ; j1 )≤(m2 ; j2 ) then ’i ((m1 ; j1 ))≤ ’i ((m2 ; j2 )): We consider the element z = (zt )t∈T ∈ JT with for t = (l; i) with k ≤ i ≤ 2l − 1

z t = − yt

for t ∈T(l; i) with k ≤ i ≤ 2l − 1;

z’i (t) = yt − y’i (t) zt = 0

otherwise

and we put y0 = y + z: Then we have (i) kyk = ky0 k≤1 (ii) supp y0 ⊂Tl+1 , and (iii) kx + y0 k = kx + yk; kx − y0 k = kx − yk: So, since supp x ⊂ T \Tl+1 , we obtain min{kx + yk; kx − yk} ≤ b[JT; (Xn )]: Thus b[JT; (es )s∈T ]≤ b[JT; (Xn )]: 2

Lemma 1.5. Let ¿0 and x = (xt )t∈T ; y = (yt )t∈T in BJT with kx + yk ¿4 − . We suppose that there exists l∈ N such that supp x ⊂ T \Tl+1 and supp y ⊂ Tl+1 . Then there n pairwise disjoint segments of T such that exist n ∈N and (Si )i=1 (i) Si; 1 = Si P ∩ (T \TP l+1 ) 6= ∅ and Si; 2 = Si ∩ Tl+1 6= ∅ for i = 1; : : : ; n; n (ii) 1 −  ≤ i=1 ( t∈Si; 1 xt )2 ≤1; Pn P (iii) 1 −  ≤ i=1 ( t∈Si; 2 yt )2 ≤1; Pn P P (iv) 1 − 2 ≤ i=1 ( t∈Si; 1 xt ) · ( t∈Si; 2 yt )≤ 1; and P Pn P 2 (v) t∈Si; 1 xt − t∈Si; 2 yt ) ≤: i=1 ( n Proof. From the hypothesis it follows that there exist n ≥1; m; k ≥ 0 and (Si )i=1 , m k (Mj )j=1 (if m = 0 there are no Mj ) and (Kr )r=1 (if k = 0 there are no Kr ) pairwise disjoint segments of T such that

Si; 1 = Si ∩ (T \Tl+1 ) 6= ∅; Mj ⊂ T \Tl+1

Si; 2 = Si ∩ Tl+1 6= ∅

for j = 1; : : : ; m;

Kr ⊂Tl+1

for r = 1; : : : ; k; and !2 n m P P P P xt + yt + 4 − ≤ i=1

t∈Si; 1

n P

P

i=1

t∈Si; 1

t∈Si; 2

So we have 4− ≤

+

for i = 1; : : : ; n;

!2 xt

m P

P

j=1

t∈Mj

+ !2 xt

P

i=1

t∈Si; 2

+

k P r=1

t∈Mj

j=1

n P



!2 yt

P t∈Kr

!2

P

+2

xt

+

P

i=1

t∈Si; 1

≤2 + 2



2

P t∈Kr

r=1

n P

2 yt

k P

! xt

·

n P

P

i=1

t∈Si; 1

yt

:

P t∈Si; 2

!

xt

·

! yt P t∈Si; 2

! yt :

996

D. Kutzarova, T. Zachariades / Nonlinear Analysis 34 (1998) 991 – 1004

From the above and the inequality of Cauchy–Schwarz it follows that ! ! n P P P  xt · yt 1− ≤ 2 i=1 t∈Si; 1 t∈Si; 2  !2 1=2  n n P P P xt  ·  ≤ i=1

t∈Si; 1

i=1

P t∈Si; 2

!2 1=2 yt  ≤ 1:

From the last inequalities the lemma follows. Theorem 1.6. b[JT; (Xn )]¡2: Proof. We suppose that b[JT; (Xn )] = 2: Then for every ¿0 there exist x = (xt )t∈T ; 2 2 y = (yt )t∈T in BJT , with supp x¡supp y, such that kx + yk ¿4 −  and kx − yk ¿4 − . Let ¿0 and x; y in BJT be as above. Let l ∈ N be such that supp x ⊂T \Tl+1 and supp y ⊂ Tl+1 : From Lemma 1.5 there exist segments (Si )ni=1 of T, pairwise disjoint, which satisfy m of T, pairwise disjoint, the conclusions of Lemma 1.4 for x; y and segments (Sj0 )j=1 which satisfy the conclusions of Lemma 1.4 for x;−y. We set  - m [ Sj;0 2  for i = 1; : : : ; n; Li; 2 = Si; 2 

and

j=1

L0j; 2

= Sj;0 2

m [

! Si; 2

for j = 1; : : : ; m:

i=1 m it follows that Then, from the properties of (Si )ni=1 and (Sj0 )j=1 n P

P

i=1

t∈Li; 2

!2 yt

≤

and

m P

P

j=1

t∈L0j; 2

!2 yt

≤

(1)

We set ti = min Si and tj = min Sj0 for i = 1; : : : ; n and j = 1; : : : ; m and clearly we can suppose xti 6= 0 and xtj 6= 0. We de ne I1 = {1 ≤ j ≤ m : there exists (exactly one) 1≤ i( j)≤ n such that Sj;0 2 ∩Si( j); 2 6= ∅ and ti( j) ≤tj } I2 = {1 ≤ j ≤ m : there exists (exactly one) 1≤ i( j)≤ n such that Sj;0 2 ∩Si( j); 2 6= ∅ and tj ¡ti( j) } and I3 = {1 ≤ j ≤ m : for every 1≤i ≤n Sj;0 2 ∩Si; 2 = ∅}:

D. Kutzarova, T. Zachariades / Nonlinear Analysis 34 (1998) 991 – 1004

997

It is clear that I1 ; I2 ; I3 are pairwise disjoint and I1 ∪I2 ∪I3 = {1; : : : ; m}: For every j ∈I1 we set Lj; 1 = Si( j); 1 \Sj;0 1 and we have P

P

j∈I1

t∈Sj;0 1

=

=

P

"

t∈Sj;0 2

P

j∈I1

t∈Si( j); 1

P

P

j∈I1

t∈Si( j); 1

+

j∈I1

t∈Lj;0 2

−2 −2

t∈Si( j); 1

P

P

j∈I1

t∈Si( j); 1

P

P

j∈I1

t∈Lj; 1

P

P

j∈I1

t∈Li( j); 2

j∈I1

t∈Si( j); 1

xt − ! !

j∈I1

t∈L0j; 2

P

j∈I1

t∈Si( j); 1

+2 

P

P

j∈I1

t∈Si( j); 1

+2 

P

P

j∈I1

t∈Lj; 1

+2

yt

t∈Li( j); 2

P

!

yt

t∈Si( j); 2



xt −

xt −

j∈I1

t∈Li( j); 2

yt

P

·

t∈Lj; 1

yt

!

xt

yt

yt

P

P

j∈I1

t∈Lj; 1

!

!

P

xt

+

j∈I1

t∈Lj; 1

j∈I1

t∈Si( j); 1

xt −

t∈L0j; 2

yt

2

t∈Si( j); 2

P t∈Si( j); 2

yt 

·

j∈I1

t∈Li( j); 2

j∈I1

t∈Li( j); 2

P

!2 1=2  P yt  ·  2

P

j∈I1

t∈Li( j); 2

P t∈L0j; 2

!2 1=2 yt 

!2 yt

j∈I1

! 1=2

P

j∈I1

P

P

+

!2 1=2  P yt  · 

P

P

xt

t∈Si( j); 2

! 1=2 

P

!2

P

P

·

!2

P

!

P

P

!2 1=2 

yt

yt

!2

P

t∈L0j; 2

+2

t∈L0j; 2

#2

P

!

t∈Si( j); 2

P

!

P

!

+

P

t∈Li( j); 2

yt

t∈L0j; 2

xt 

·

yt

xt

P

!

t∈Si( j); 2

+ 2

P

!

P

!2

t∈Lj; 1

xt −

yt −

t∈Li( j); 2

j∈I1

t∈Si( j); 1

t∈Si( j); 2

xt + P

j∈I1

P

P

P

P

yt

yt

+

t∈Lj; 1

P

!2

P

P

−

yt

P

xt

xt −

P 

xt −

yt !2

t∈Si( j); 2

−2

j∈I1

P

t∈Si( j); 2

P

yt

P

P

+

xt −

P

!

P

!2

P

−2

yt

xt −

P

+2

≤

xt −

!2

P

yt 

!2 1=2 yt 

P t∈Lj; 1

!2 1=2 xt 

998

D. Kutzarova, T. Zachariades / Nonlinear Analysis 34 (1998) 991 – 1004



P

P

j∈I1

t∈Lj; 1

+2 

P

j∈I1

t∈Li( j); 2

P

P

j∈I1

t∈Si( j); 1

+

xt −

P

j∈I1

t∈L0j; 2

t∈L0j; 2

!2 1=2  P yt  ·  !2

P

yt

t∈Si( j); 2



+

t∈L0j; 2

P

j∈I1

t∈Lj; 1

P

P

j∈I1

t∈Si( j); 1

+ 6

P

P

j∈I1

t∈Li( j); 2

 !2 1=2 P yt  + 4 

+2

!2 1=2 yt 

P

yt



!2 1=2 yt 

P

j∈I1

!2

P

P

j∈I1

P

+2

≤

!2 1=2  P xt  · 

xt −

!2 xt P

t∈Si( j); 2

P

j∈I1

t∈L0j; 2

+

P

P

j∈I1

t∈Li( j); 2

!2 1=2

!2 yt

yt 

!2 1=2 yt  :

So from the above inequalities, from Lemma 1.5 and from (1) we obtain !2 !2 √ √ P P P P P xt − yt ≤3 + 12  + xt ≤ 3 + 12  + 1: j∈I1

t∈Sj;0 1

t∈Sj;0 2

j∈I1

t∈Lj; 1

P

P

j∈I2

t∈L0j; 1

(2)

Similarly we have P

P

j∈I2

t∈Sj;0 1

xt −

!2

P t∈Sj;0 2

yt

√

≤3 + 12  +

!2 xt

√ ≤ 3 + 12  + 1

(3)

where L0j; 1 = Sj;0 1 \Si( j); 1 : For every j ∈I3 we have that Sj;0 2 = L0j; 2 and it is easy to see that P

P

j∈I3

t∈Sj;0 1

xt −

!2

P t∈Sj;0 2

yt

√ ≤1 +  + 2 :

From (2), (3), (4) it follows that !2 m √ P P P xt − yt ≤3 + 7 + 26 : j=1

t∈Sj;0 1

t∈Sj;0 2

(4)

(5)

m it follows that But from the properties of (Sj0 )j=1

4 − 3 ≤

m P

P

j=1

t∈Sj;0 1

xt −

P t∈Sj;0 2

!2 yt

:

Choosing  small enough (5) and (6) give a contradiction.

(6)

D. Kutzarova, T. Zachariades / Nonlinear Analysis 34 (1998) 991 – 1004

999

From Theorem 1.6, Propositions 1.4, and 1.2 we obtain the next corollaries. Corollary 1.7. b[JT; (es )s∈T ]¡2: Corollary 1.8. JT is uniformly nonsquare over blocks. Remark. Let (xn ) be a normalized weakly null sequence in JT with s.m. F and (en ) be the f.s. of F. From the remarks after Proposition 1.2; we obtain that there exists a constant 0¡b¡2 such that for every w1 ; w2 ∈BF we have kw1 + w2 k¡b

or

kw1 − w2 k¡b:

From this it follows that JT does not have s.m. isomorphic to l1 or equivalently JT has the A.B.S. property. This also follows from the fact that (en )n∈N is 2-equivalent to the usual basis of l2 [14]. √ As Professor E. Odell informs us; the best constant for the above equivalence is 2: This means that for every k ∈N and c1 ; : : : ; ck ∈R we have

 k  k 1=2 k 1=2

P

√ P P 2 2

cn ≤ cn en ≤ 2 cn : n=1

n=1

n=1

2. Uniformly nonsquare over blocks and orthogonally convex spaces The next theorem gives a sucient condition for a uniformly nonsquare over blocks space to be O.C. Below (after Theorem 3.2) we will give an example of uniformly nonsquare over blocks space which is not O.C. Let 1 ≤p¡ + ∞. For a space X and an F.D.D. (Xn )n∈N of X we put p p p [X; (Xn )] = inf {c ∈R : (kxk + kyk )1=p ≤c: kx + yk for every x; y ∈ X with supp x¡ supp y} [11]. Theorem 2.1. Let X be a space with a bimonotone F.D.D. (Xn )n∈N such that n )] )]1=p for some 1 ≤ p¡ + ∞: Then X is b[X; (Xn )]¡2 and p [X; (Xn )]¡[1 + (1− b[X;(X 2 O.C. Proof. Let X be a space with the properties of the theorem. We set b = b[X; (Xn )] and = p [X; (Xn )]: From the hypothesis, we can choose ; , ¿0 and k ∈N such that 1+ 2 2 − ¿b, (1 − 2 )(1 +  + )¡1; N1 + 1++ ¡1 and 2( p − (1 − )p )1=p + b + N2 ¡ 1++  and we set M = (1 − 2 )(1 +  + ). Let (xn )n∈N be a weakly null sequence of X with D[(xn )]¿0: It is obvious that |M (xn ; xm )| ≤M kxn − xm k

(7)

for every n∈ N; with xn = 0, and m∈ N: Let n ∈ N such that xn 6= 0: Since w−limm (xn −xm ) = xn we have kxn k ≤ limm inf kxn − xm k: So there exists mn ∈ N such that 12 kxn k ≤ kxn − xm k for every m ≥ mn . 1 We choose k ∈ N with k(I −Pk )(xn )k ≤ 4N (1++)kxn k and, since w −limm xm = 0, 1 we can suppose that kPk (xm )k ≤ 4N (1 +  + )kxn k for every m ≥ mn .

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D. Kutzarova, T. Zachariades / Nonlinear Analysis 34 (1998) 991 – 1004

So we have k(I − Pk )(xn )k ≤ and kPk (xm )k ≤

1 (1 +  + )kxn − xm k 2N

1 (1 +  + )kxn − xm k 2N

for m ≥ mn :

Let m ≥ mn and z ∈ M (xn ; xm ), that is 1 kz − xn k ≤ (1 + )kxn − xm k 2 and

1 kz − xm k ≤ (1 + )kxn − xm k: 2

From the above, since the F.D.D. is bimonotone, we have 1 kPk (z)k ≤ (1 +  + )kxn − xm k 2 and

1 k(I − Pk )(z)k ≤ (1 +  + )kxn − xm k: 2

We set y = 2(1 +  + )−1 kxn − xm k−1 z; un = 2(1 +  + )−1 kxn − xm k−1 xn ; and um = 2(1 +  + )−1 kxn − xm k−1 xm : So it follows that kPk (y)k; k(I − Pk )(y)k ≤ 1; k(I − Pk )(un )k ≤ and kPk (um )k ≤

1 ; N

1 : N

Claim. kyk ≤ 2 − . We suppose kyk¿2 − ¿b. Then we have kPk (y) − (I − Pk )(y)k ≤ b; 1 −  ≤ k(I − Pk )(y)k ≤ 1;

D. Kutzarova, T. Zachariades / Nonlinear Analysis 34 (1998) 991 – 1004

1001

and 1 −  ≤ kPk (y)k ≤ 1: Also, from the choice of numbers ,  and N , we have ky − Pk (un )k ≤ ky − un k + k(I − Pk )(un )k ≤ 1; and ky − (I − Pk )(um )k ≤ ky − um k + kPk (um )k ≤ 1: So from the above we obtain p ≥ p ky − Pk (un )kp ≥ kPk (y − un )kp + k(I − Pk )(y)kp ≥ kPk (y − un )kp + (1 − )p and so kPk (y − un )kp ≤ p − (1 − )p : Similarly we have k(I − Pk )(y − um )kp ≤ p − (1 − )p : Thus kun − um k ≤ kPk (un − y)k + k(I − Pk )(un )k + kPk (y) − (I − Pk )(y)k + k(I − Pk )(y − um )k + kPk (um )k ≤ 2( p − (1 − )p )1=p 2 2 = kun − um k +b + ¡ N 1++ which is a contradiction. It follows from the claim that |M (xn ; xm )| ≤ M kxn − xm k

(8)

for every n ∈ N, with xn 6= 0, and m ≥ mn . Therefore, from (7), (8), we have     A [(xn )] = lim sup lim sup |M (xn ; xm )| ≤ M lim sup lim sup kxn − xm k n

m

n

m

= M:D[(xn )]¡D[(xn )]: So the space X is O.C. From the above we have Corollary 2.2. If a space X has a bimonotone F.D.D. (Xn )n∈N such that b[X; (Xn )]¡2 and p [X; (Xn )] = 1 for some 1 ≤ p¡+∞ then X is O.C.

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D. Kutzarova, T. Zachariades / Nonlinear Analysis 34 (1998) 991 – 1004

Corollary 2.3. The space JT is O.C. Proof. Since 2 [JT; (Xn )] = 1, this follows from Theorem 1.6 and Corollary 2.2. Remarks. 1. It is easy to prove that if p[X; (Xn )]¡2 then the F.D.D. (Xn )n∈N of X is shrinking. So JT is O.C. but p[JT; (Yn )] = 2 for every F.D.D. (Yn )n∈N of JT. 2. If a space X with an F.D.D. (Xn )n∈N is re exive and p[X; (Xn )]¡2 then it is easy to see that there exists 0¡c¡1 such that for every basic sequence (xn )n∈N in Bx there exist k¿1 for which kx1 + xk k ≤ 2 − c, that is, X is weakly nearly uniformly smooth [3]. So as in [15] it follows that X has an equivalent nearly uniformly smooth norm (the dual notion of nearly uniform convexity).

3. p-direct sums of uniformly nonsquare over blocks spaces It is clear that if (Zn )n∈N is a sequence of Banach spaces with F.D.D. and Mn is thePconstant of F.D.D., for n = 1; 2; : : : ; if supn Mn ¡+∞ then the p-direct sum ∞ Z = [ n=1 ⊕Zn ]p , for 1 ≤ p¡+∞ or p = 0, has an F.D.D. with the property: If x = (xn )n∈N , y = (yn )n∈N are in Z with supp x¡supp y then supp xn ¡supp yn for every n ∈ N. Our nal result is a theorem which concerns the p-direct sum of a sequence of uniformly nonsquare over blocks spaces for 1¡p¡+∞. The proof of the following lemma is easy. Lemma 3.1. Let 1¡p¡+∞ and f : [0; 1] → R, with f(t) = (1 + t)1=p + a(1 − t)1=p , where a ∈ R. Then f(t) ≤ 21=p (1 + a q )1=q for every t ∈ [0; 1], where p1 + q1 = 1. Theorem 3.2. Let Zn be a space with an F.D.D. (Xkn )k and Mn the constant of this F.D.D., P for n = 1; 2; : : : . If supn Mn ¡+∞ and b = supn b[Zn ; (Xkn )k ]¡2 then the space ∞ Z = [⊕ n=1 Zn ]p is uniformly nonsquare over blocks for every 1¡p¡+∞. Proof. We suppose that the space Z is not uniformly nonsquare over blocks. Then for every ¿0 there exist x = (xn )n∈N , y = (yn )n∈N in BZ , with supp x¡supp y, kx + yk ≥ 2 −  and kx − yk ≥ 2 − . (In the space Z we consider the F.D.D. of the above remark). For every n ∈ N we set n = max{kxn k; kyn k} and n = min{kxn k; kyn k}. Since supp x¡suppy we have supp xn ¡supp yn . So it is easy to see that either kxn +yn k ≤ (b− 1) n + n or kxn − yn k ≤ (b − 1) n + n for every n ∈ N. So for every n ∈ N we have (kxn + yn kp + kxn − yn kp )1=p ≤ [((b − 1) n + n ))p + ( n + n )p ]1=p ≤ [((b − 1)p np + np ]1=p + (2np )1=p = [((b − 1)p + 1]1=p : n + 21=p n and thus kxn + yn kp + kxn − yn kp ≤ [((b − 1)p + 1)1=p n + 21=p n ]p :

D. Kutzarova, T. Zachariades / Nonlinear Analysis 34 (1998) 991 – 1004

1003

From the above we obtain (kx + ykp + kx − ykp )1=p ∞ 1=p P = kxn + yn kp + kxn − yn kp n=1

 ≤

∞ P n=1

 ≤

∞ P n=1

1=p

[((b − 1)p + 1)1=p n + 21=p n ]p ((b − 1)p + 1) np

"

=2

=2

1=p

"

+

∞ (b − 1)p + 1 P np 2 n=1

(b − 1)p + 1 2



1=p

1=p 

∞ P n=1

1=p + ∞ P n=1

np

1=p

2np

∞ P n=1

1=p

np

1=p

1=p #

 +

∞ P n=1

np

1=p # :

(9)

P∞ P∞ We set  = n=1 np and = n=1 np . Since 0 ≤  + ≤ 2 and 0 ≤ ≤ , there exists 0 ≤ t ≤ 1 such that  ≤ 1 + t and ≤ 1 − t. So, from (9), we have " # 1=p (b − 1)p + 1 p p 1=p 1=p 1=p 1=p (1 − t) + (1 + t) (kx + yk + kx − yk ) ≤ 2 2 and from Lemma 3.1 it follows that " p

p 1=p

(kx + yk + kx − yk ) where

1 p

≤2

2=p

 1+

(b − 1)p + 1 2

q=p #1=q (10)

+ q1 = 1. But

(kx + ykp + kx − ykp )1=p ≥ 21=p (2 − ):

(11)

Since b¡0 choosing  small enough (10) and (11) give a contradiction. The next example was constructed in [3]. In this paper the authors proved, among other properties, that this space is not O.C. Using Theorem 3.1 we see that this is uniformly nonsquare over blocks. Example ([3]). For every (; ) in R2 we set   "  2 #1=2   9 1 k(; )k0 = max  + ; t  − ; 2 +  10  10 10

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D. Kutzarova, T. Zachariades / Nonlinear Analysis 34 (1998) 991 – 1004

2 where t = 216 217 . Let  = (n )n∈N be in l . For every k = 1; 2; : : : we set Sk () = (k+n )n∈N and we de ne  " 2 #1=2    1 kSk ()k2 : k¿1 : kk = sup k(1 ; k )k20 +   10

P∞ Then, k·k is a norm on l2 equivalent to k·k2 . We set E = (l2 ; k·k) and Z = [ n=1 ⊕Zn ]8 where Zn = E for every n = 1; 2; : : : . In [3] it is proved that Z is not O.C. For the usual basis (en )n∈N of l2 it is easy to see that p[E; (en )]¡2. Thus E is uniformly nonsquare over blocks and so, from Theorem 3.2, Z is uniformly nonsquare over blocks. Acknowledgements We wish to express our thanks to Professor E. Odell for bringing to our attention certain information related to the JT space. This paper was prepared during the visit of the second named author in So a. He would like to express his thanks to the Institute of Mathematics at the Bulgarian Academy of Sciences and to the Department of Mathematics at the University of So a, for their hospitality. References [1] A. Jimenez-Melado, E. Llorens-Fuster, A geometric property of Banach spaces which implies the xed point property for nonexpansive mappings, preprint. [2] B. Maurey, Points xes des contractions sur un convexe borne de L1 , Seminaire d’Analyse Fonctionelle 1980 –1981, Ecole Polytechnique, Palaiseau VIII, pp. 1–18. [3] D. Kutzarova, S. Prus, B. Sims, Remarks on Orthogonal Convexity of Banach spaces, Houston Journal of Mathematics 19 (1993) 603 – 614. [4] R.C. James, Uniformly nonsquare Banach spaces, Ann. of Math. 80 (1964) 542–550. [5] J. Lindenstrauss, L. Tzafriri, Classical Banach Spaces I Sequence Spaces, Springer-Verlag, Berlin, 1979. [6] A. Brunel, L. Sucheston, On B-convex Banach spaces, Math. System Theory 7 (1974) 294 –329. [7] B. Beauzamy, J.-T. Lapreste, Modeles e tales des espaces de Banach, Travaux en Cours, Hermann, Paris, 1984. [8] B. Beauzamy, Banach–Saks properties and spreading models, Math. Scand. 44 (1979) 357–384. [9] T. Schlumprecht, An arbitrarily distortable Banach space, Israel J. Math. 76 (1991) 81–95. [10] W.T. Gowers, B. Maurey, The unconditional basic sequence problem, Journal Amer. Math. Soc. 6 (1993) 851– 874. [11] M.A. Khamsi, Normal structure for Banach spaces with Schauder decomposition, Canad. Math. Bull. 32 (3) (1989) 344 –350. [12] D.P. Giesy, R.C. James, Uniformly Non-l(1) and B-convex Banach spaces. Studia Mathematica XLVIII (1973) 61–69. [13] R.C. James, A separable somewhat re exive Banach space with nonseparable dual, Bull. Amer. Math. Soc. 80 (1974) 738 –743. [14] I. Amemiya, T. Ito, Weakly null sequences in James spaces on trees, Kodai Math. J. 4 (1981) 418 – 425. [15] S. Prus, Nearly uniformly smooth Banach spaces, Boll. U.M.I. 3-B7 (1987) 507–521.