K-Β and K-Nearly Uniformly Convex Banach Spaces Denka Kutzarova

JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS vol. 162, No. 2, 1991 k-β and k-Nearly Uniformly Convex Banach Spaces Denka Kutzarova

Different uniform geometrical properties have been defined between the uniform convexity and the reflexivity of Banach spaces. In the present paper we introduce other properties of this type, namely k-β and k-nearly uniform convexity. The definitions, as well as some of the results presented here, are announced in [7]. Sullivan [24] has defined k-uniformly rotund Banach spaces which by a suggestion of Davis are all superreflexive. Fan and Glicksberg [1] have introduced fully k-convex Banach spaces. Let x1,...,xk+1 be vectors in a Banach space X. The k-dimensional volume enclosed by x1,...,xk+1 is given by

V (x1, x2,...,xk+1) 1 ··· 1 f (x ) ··· f (x )  1 1 1 k+1 ∗  = sup . . : fi ∈ X , kfik ≤ 1, i =1, 2,...,k .  . .    fk(x1) ··· fk(xk+1)

    A Banach space X is said to be k-uniformly rotund (k-UR), k ≥ 1, if for each ε > 0 there exists a δ > 0 such that for any xi ∈ X, kxik < 1, i = 1, 2,...,k + 1, k+1 with i=1 xi (k +1) ≥ 1−δ, then V (x1,...,xk+1) <ε. Clearly, 1-UR coincides with uniform convexity. . For equivalent definition of k-UR see [11]. P A Banach space X is said to be fully k-convex (kR), k ≥ 2, if for every sequence k {xi} in X such that limn1,...,nk→∞ i=1 xn k = 1, then {xi} is a Cauchy sequence in X. . P Another uniform property is the nearly uniform convexity, introduced by Huff [3]. He has proved that the class of nearly uniformly convexifiable spaces is strictly between superreflexive and reflexive Banach spaces. A Banach space X is called nearly uniformly convex (NUC) if for each ε > 0 there is a δ, 0 <δ< 1 such that for any sequence {xn} in the closed unit ball B with sep(xn) = inf{kxn − xmk : n =6 m} > ε, then conv({xn}) ∩ (1 − δ)= ∅. For equivalent definitions of NUC see [19].

1 The Kuratowski measure of non-compactness α(A)ofaset A in X is the infimum of those ε> 0 for which there is a covering of A by a finite number of sets Ai, such that diam(Ai) <ε. Let X be a Banach space with closed unit ball B. By the drop D(x, B) defined by an element x ∈ X \ B, we mean conv({x} ∪ B) and let R(x, B) = D(x, B) \ B. Rolewicz [18] has proved that X is uniformly convex if and only if for each ε > 0 there is a δ > 0 such that 1 < kxk < 1+δ implies diam(R(x, B)) <ε. In connection with this he has also introduced [19] the property (β). A Banach space X is called to have the property (β) if for each ε> 0 there is a δ > 0 such that 1 < kxk < 1+ δ implies α(R(x, B)) <ε. Rolewicz [19] has shown that UC ⇒ (β) ⇒ NUC. The class of Banach spaces with an equivalent norm with property (β) coincides neither with that of superreflexive spaces (independently done in [4] and [14]) nor with the class of nearly uniformly convexifiable spaces (see [5, 6]). Yu Xin-Tai [25] has proved that k-UR implies NUC. In connection with this result we may ask if k-UR implies (β). The answer is negative. For this purpose we may present first the following.

Proposition 1. Let X be the l1-direct sum of Banach spaces Y and Z, where Y is uniformly convex and Z is k-UR. Then X is (k + 1)-UR.

This is an addition to the results in [2, 13]. The author was informed by Prof. Bor-Luh Lin that the same result had been also obtained by Yu Xin-Tai.

Example 2. There is a 2-UR Banach space which does not possess the property (β).

1 Proof. Let Y = R and Z = l2. Denote by X the l1-direct sum of Y and Z. According to Proposition 1, X is 2-UR. On the other hand, it is easy to observe that X fails to have the property (β) (cf. [19, 14]).

If in the above mentioned characterization of uniform convexity by drops [18] we replace diam(R(x, B)) < ε by sup{diam(C) : C ⊂ R(x, B), C convex} < ε, obviously we have again UC. In contrast, if we do a similar substitution in the deﬁnition of (β), we obtain:

Theorem 3. A Banach space X is NUC if and only if for each ε > 0 there exists a δ > 0 such that 1 < kxk < 1+ δ implies

sup{α(C) : C ⊂ R(x, B), C convex} < ε, where B is the closed unit ball of X.

Proof. Necessity. Assume the contrary; i.e., there exists an ε > 0, elements xn ∈ X with 1 < kxnk < 1+1/n and convex sets Cn, Cn ⊂ R(xn, B), so that

2 α(Cn) ≥ ε, n = 1, 2,... . For every integer n we have that the convex set Cn ∗ is disjoint with B and thus, there is a functional fn ∈ X with kfnk = 1 which separates Cn and B, i.e.,

f(x) ≥ 1 for every x ∈ Cn.

Put En = Cn/(1+1/n). Clearly, En is convex and α(En) ≥ ε/2 for every integer n. By the choice of xn and Cn we obtain easily that f(x) > 1 − 1/n for every x ∈ En, i.e., En ∩ (1 − 1/n)B = ∅, n =1, 2,.... By [19], this means that the norm is not NUC which is a contradiction. Suﬃciency. Assume that the norm is not NUC. Then, following [19], we may ∗ ﬁnd an ε> 0 and a sequence {fn} ⊂ X with kfnk = 1 so that

α(S(fn, 1/n)) ≥ ε, n =1, 2,...,

∗ where S(f, δ) = {x ∈ B : f(x) ≥ 1 − δ} for f ∈ X and 0 <δ< 1. Take xn with 1+2/n ≤kxnk ≤ 1+3/n and fn(xn)=1+2/n, n =1, 2,... . Then, if we denote by Cn the set xn/2+ S(fn, 1/n)/2, we have that Cn is convex and Cn ⊂ R(xn, B). Moreover, α(Cn) ≥ ε/2 for every n, which is a contradiction.

We are now in a position to introduce the notions of k-β and k-nearly uniform convexity.

Definition 4. Let k ≥ 1 be an integer. A Banach space X with closed unit ball B is called to be k-β, provided for each ε > 0 there exists a δ > 0 so that 1 < kxk < 1+ δ implies sup{α(C)} < ε, where the supremum is taken over all subsets C of R(x, B) such that for every choice k k of elements {xi}i=1 ⊂ C and scalars γi ≥ 0, i = 1,...,k with i=1 γi = 1; then k γ x ∈ R(x, B). i=1 i i P P Clearly, 1-β coincides with the property (β).

Definition 5. Let k ≥ 2 be an integer. A Banach X is called to be k-nearly uniformly convex (k-NUC), provided for each ε> 0 there exists a δ,0 <δ< 1, such that for every sequence {xn} ⊂ X, with kxnk ≤ 1 and sep(xn) >ε, there are indices k k {ni}i=1 and scalars γi ≥ 0, i =1,...,k, with i=1 γi = 1 so that

k P γixn ≤ 1 − δ.

i=1 X

Evidently, k-NUC implies NUC.

3 Theorem 6. Let X be a Banach space and k ≥ 2 be an integer. Then the following are equivalent: (i) X is k-NUC; (ii) for each ε > 0 there is a δ, 0 <δ< 1 such that for every set E contained k is the closed unit ball B with α(E) > ε, there exist elements {xi}i=1 ⊂ E so that k conv({xi}i=1) ∩ (1 − δ) =6 ∅; (iii) for each ε> 0 there is a δ, 0 <δ< 1, such that for every sequence {xn} ⊂ B k with sep(xn) >ε, there are indices {ni}i=1 so that

xni k ≤ 1 − δ. , i=1 X

Proof. The equivalence between (i) and (ii) follows immediately from the properties of the Kuratowski measure of non-compactness. It remains to show that (i) ⇒ (iii). Let ε> 0 and choose δ > 0 according to Deﬁnition 5. Put η = δ/k. Let {xn} ⊂ B k be an arbitrary sequence with sep(xn) >ε. Then there exists a point y = i=1 γixni with γ ≥ 0, k γ = 1 so that kyk ≤ 1 − δ. Without loss of generality we may i i=1 i P assume that P γ1 = max{γi :1 ≤ i ≤ k}. We have the obvious representation

1 k 1 1 k γ x = y + 1 − i x . k ni k γ γ ni i=1 1 i=2 1 ! X X By the choice of γ1, 1 − γi/γ1 ≥ 0, i =2,...,k, and thus

1 k 1 δ xni ≤ k − ≤ 1 − η, k k γ1 i=1 X which concludes the proof. Theorem 7. Let X be a Banach space with closed unit ball B and k ≥ 1 be an integer. Then the following are equivalent: (i) X is k-β; (ii) for each ε> 0 there is a δ > 0 so that 1 < kxk < 1+ δ implies

sup{sep(xn)} < ε, where the supremum is taken over all sequences {xn} ⊂ R(x, B) so that the convex k hull of every k elements {xni }i=1 also belongs to R(x, B); (iii) for each ε> 0 there exists a δ, 0 <δ< 1, so that for every element x ∈ B k and every sequence {xn} ⊂ B with sep(xn) > ε, there are indices {ni}i=1 so that k conv({x}∪{xni }i=1) ∩ (1 − δ)B =6 ∅;

4 (iv) for each ε> 0 there exists a δ, 0 <δ< 1, so that for every element x ∈ B n and every sequence {xn} ⊂ B with sep(xn) >ε, there are indices {ni}i=1 so that

x + xni (k + 1) ≤ 1 − δ. , i=1 X

Proof. The equivalence between (i) and (ii) follows from the properties of the measure of non-compactness.

(ii) ⇒ (iii). Let ε> 0. According to (ii), ﬁx for ε/2 a corresponding 0 <δ< 1. Put η = δ/4. Take an element x and a sequence {xn} in B. k Assume that for every choice of indices {ni}i=1, we have

k (1) conv {x}∪{xni }i=1 ∩ (1 − η)B = ∅.

Put y =(1+ δ)x and let yn=(y + xn)/2. Clearly, yn ∈ D(y, B). Moreover, δ 1+ δ 1 y = 1+ x + x n 2 2+ δ 2+ δ n k and hence for arbitrary choice {ni}i=1 and convex combination with coeﬃcients k γi ≥ 0, i =1,...,k, i=1 γi = 1, we get

k P δ 1+ δ 1 k γ y = 1+ x + γ x . i ni 2 2+ δ 2+ δ i ni i=1 i=1 ! X X Thus, by (1) k δ γ y > 1+ (1 − η) > 1. i ni 2 i=1 X k Therefore, conv({yn} ) ⊂ R( y, B) and 1 < kyk < 1+ δ. Then we get from (ii) i=1 that sep(yn) < ε/2, whence sep(xn) <ε.

(iii) ⇒ (ii). Let ε > 0. According to (iii), choose for ε/2 a corresponding 1 0 <δ< 2 . Take an x ∈ X with 1 < kxk < 1+ δ and a sequence {xn} ⊂ R(x, B) k as in (ii). By convexity we also get for every choice {ni}i=1 and scalars γi > 0, k i =0, 1,...,k, i=0 γi = 1; then P k

γ0x + γixni ∈ R(x, B). i=1 X Denote y = (1 − δ)x and yn = (1 − δ)xn. Evidently, y, yn ∈ B. Moreover,

γ0y + γiyni > 1 − δ.

k=1 X

5 Therefore, sep(yn) < ε/2 and thus, sep(xn) <ε. (iv) ⇒ (iii) is obvious. (iii) ⇒ (iv). Let ε> 0. Find a δ according to (iii). Put η = δ/(k +1). Take an k x ∈ B and {xn} ⊂ B with sep(xn) >ε. Then there are indices {ni}i=1 and scalars k γi ≥ 0, i =0, 1,...,k, i=0 γi = 1, so that

P k

γ0x + γixni ≤ 1 − δ.

i=1 X

As in the proof of Proposition 6, we get

x + xni (k + 1) ≤ 1 − η. , i=1 X

Theorem 8. Let X be a Banach space. Then, (i) k-β ⇒ (k + 1)-NUC for every k ≥ 1; (ii) k-NUC ⇒ k-β for every k ≥ 2.

Proof. (i) Let ε> 0. Select a corresponding δ > 0 according to Theorem 7(iv). Take {xn} ⊂ B with sep(xn) > ε. Then for {xn} and x = x1 there are indices k+1 k+1 {ni}i=2 so that the norm of the arithmetic mean of xi and {xn}i=2 is less than or equal to 1 − δ; i.e., denoting n1 = 1 we get

k+1

xn (k + 1) ≤ 1 − δ. , i=1 X

(ii) Let ε > 0. Choose δ > 0 according to Theorem 6(iii). Then the condition (iv) of Theorem 7 is fulfilled for δ1 = kδ/(k + 1). We shall prove in the sequel that the converse implications are not true; i.e., all the notions of k-β and k-NUC are isometrically different. For this purpose we investigate first their relation to the other uniform geometrical properties. A Banach space X is said to be uniformly Kadec-Klee (UKK) if for every ε> 0 there is a δ > 0 such that for every sequence {xn}, kxnk ≤ 1 which converges weakly to x and sep(xn) > ε, we have kxk ≤ 1 − δ. X is NUC if and only if it is UKK and reflexive [3]. X is said to be weakly uniformly Kadec-Klee (wUKK or UKKε) if there exists an 0 <ε< 1 so that there is a δ > 0 for which kxk ≤ 1 − δ if xn → x weakly, kxnk ≤ 1, sep(xn) > ε. X is NUCε for some 0 <ε< 1; i.e., there is a 0 <δ< 1 such that for every sequence {xn} ⊂ B with sep(xn) > ε we have conv({xn}) ∩ (1 − δ)B, it and only if X is wUKK and reflexive (cf., e.g., [10]). Similarly, we may define k-NUCε for 0 <ε< 1, k ≥ 2. Clearly, k-NUCε ⇒ NUCε.

6 A Banach space X has the Banach-Saks property (BS) whenever every bounded sequence in X has a subsequence whose arithmetic means converge in norm. In [6] it is proved that every space with property (β) (i.e., 1-β) has (BS). Partington [16] has proved that Baernstein’s example of a reﬂexive space without (BS) is NUC. Since (BS) is invariant under isomorphisms, (β) and NUC are isomorphically diﬀerent [6].

Theorem 9. Let k ≥ 2 be an integer. If X is k-NUCε for some 0 <ε< 1 then X possesses the Banach-Saks property.

Since X is reflexive, it suffices to show that X does not have a spreading model isomorphic to l1. The last fact can be easily proved as in [6]. We prefer to give here an alternative argument based on a result of Partington [15] because it clarifies also the relation of k-NUC to the properties Ak defined in [15]. Let k ≥ 2. A Banach space X has property Ak if it is reflexive and there exists a number η, 0 <η< 1, such that, whenever xn → 0 weakly, kxnk < 1, then there exist n1,...,nk with k i=1 xnk/ k ≤ 1 − η. Partington has proved that Ak ⇒ (BS) and every superreflexive space has Ak for some k ≥ 2. P Proposition 10. If X is k-NUCε for some 0 <ε< 1 then X has the property Ak.

Proof. Let δ > 0 be chosen according to the deﬁnition of k-NUCε. Fix ν, ε<ν< 1. Put η = min{1 − ν, δ}. Take now an arbitrary sequence {xn}, kxnk ≤ 1 with xn → 0 weakly. The interesting case is when kxnk > ν for inﬁnitely many n.

For brevity let kxnk > ν for all n. Put n1 = 1. Having chosen xn1 ,...,xnk with kxni − xnj k > ν whenever i =6 j, 1 ≤ i, j ≤ m, select supporting functionals fi at xni , i =1,...,m, i.e., kfik = 1, fi(xni )= kxni k. Since xn → 0 weakly, ﬁnd xnm+1 so that |fi(xnm+1 )| < kxik − ν for every i =1,...,m. Thus for every 1 ≤ i ≤ m,

kxni − xnm+1 k ≥ fi(xni ) −|fi(xnm+1 )| > ν.

∞ Hence, for the above constructed subsequence {xnm }m=1 we get sep(xnm ) > ε.

Therefore, there are indices ji = nmi , i =1,...,k so that

xji k ≤ 1 − δ, , i=1 X which completes the proof. The converse implication of Proposition 10 is not fulﬁlled.

Example 11. There exists a Banach space which has the property An for some n ≥ 2 and fails to be k-NUCε for any integer k ≥ 2 and 0 <ε< 1.

7 Proof. Let X be the space l2 supplied with the emquivalent norm k·ka considered in [26]. It has been shown there that (l2, k·ka) is non-wUKK and hence X fails to be even NUCε for any 0 <ε< 1. On the other hand, X is superreﬂexive and by the result of Partington, it has An for some n ≥ 2.

Actually, the properties k-NUC are isomorphically stronger than Ak, as we can see in the next example.

Example 12. There exists a Banach space which has the property A2 but fails to have an equivalent NUC norm.

Proof. Let X be the l2-direct sum of the spaces Xn = ln, n ≥ 2. It is not diﬃcult to show that X has the property A2. On the other hand, X fails to have an equivalent NUC norm (cf. [3, 19]).

As a consequence of Proposition 10 we obtain the following.

Corollary 13. The properties k-β and k-NUC are isomorphically diﬀerent from NUC.

Bor-Luh Lin and Yu Xin-Tai [13] have proved that every strictly convex k-UR space is (k + 1)R. We improve this result in the following way.

Theorem 14. Let X be a Banach space. Then, (i) if X is k-UR for some integer k ≥ 1, then X is k-β; (ii) if X is strictly convex and k-NUC for some integer k ≥ 2, then X is kR. Proof. (i) Assume the contrary; i.e., there exists an ε > 0 such that for every (n) (n) ∞ (n) integer n there is an element x0 and a sequence {xm }m=1 with kxm k ≤ 1, m = (n) 1, 2,... , n = 1, 2,... and sep(xm ) > ε, n = 1, 2,... , so that for arbitrary choice k {mi}i=1 of indices we get

k (n) (n) (2) x0 + xmi (k + 1) > 1 − 1/n. , i=1 X

It is easily seen that for every bounded set E with α(E) >ε,

(3) sup inf kx − yk > ε/2, y∈L x∈E for any ﬁnite-dimensional subspace L. (n) k Put m0 = 0. By (3), we may ﬁnd successively elements {xmi }i=1, so that for every n =1, 2,... ,

(n) (n) (n) (4) dist(xmi+1 , [xm0 ,...,xmi ]) > ε/2, i =1, 2,...,K,

8 (n) (n) (n) i where [xm0 ,...,xmi ] means the closed convex of {xmj }j=1. It follows from (4) and [2] that

(n) (n) k (5) V (xm0 ,...,xmk ) > (ε/2) , n =1, 2,.... On the other hand, (2) implies that

k (n) xmi (k + 1) → 1, as n →∞, , i=0 X which according to k-UR contradicts (5).

(ii) Let {xn} be a sequence for which

(6) lim xni k =1. n1,...,nk→∞ , i=1 X

Without affecting the generality we may suppose that kxnk = 1 for every n. We prove first that every subsequence of {xn} has Cauchy subsequence. Take an arbitrary subsequence of {xn}, for brevity denote it again by {xn}, and assume that it does not have a Cauchy subsequence. Then, α({xn}) > 0. Hence, one can find a subsequence {xm} with sep(xm) > α({xn})/3= ε. For this ε> 0 there exists by k-NUC a δ > 0 so that for every sequence {yn} in the closed unit ball with k sep(yn) >ε, there are indices {ni}i=1

(7) yni k ≤ 1 − δ. , i=1 X

It follows from (6) that there is a number N so that for mi ≥ N,

(8) xmi k > 1 − δ. , i=1 X

Since the separation of {xm}m ≥N is also greater than ε, then (8) contradicts (7), which proves our claim.

It remains to show that {xn} has unique cluster point. Assume that there are subsequences {xni } → x, {xmi } → y. By (6) and the triangle inequality, limi→∞ kxni + xmi k/2 = 1, whence kx + yk/2 = 1. Since X is strictly convex, this implies x = y, which completes the proof.

Remark 15. In view of Corollary 12, Theorem 14(i) isomorphically improves the implication k-UR ⇒ NUC of Yu Xin-Tai [25]. As we mentioned in the beginning,

9 k-UR spaces are superreflexive [24] and not every space with property (β) is super- reﬂexive [4, 14]. Bor-Luh Lin and Pei-Kee Lin [12] have shown that the Baerstein’s space which fails to have (BS), admits an equivalent 2R norm. Since k-β and k-NUC spaces have (BS), this shows that Theorem 14 isomorphically improves the result of Bor-Luh Lin and Yu Xin-Tai [13].

We are ready now to distinguish isometrically k-β and k-NUC.

Theorem 16. For the properties k-β and k-NUC the following hold: (i) for every k ≥ 2, there exists a strictly convex Banach space Xk, isomorphic to l2, which is k-β but is not k-NUC; (ii) for every k ≥ 1, there exists a Banach space Yk, isomorphic to l2, which is (k + 1)-NUC but is not k-β.

Proof. (i) Let Xk be the example of a strictly convex Banach space which is k-UR but is not kR, given in [13] (it is a modiﬁcation of an example in [23]). From Theorem 14 (i) and (ii) it follows immediately that Xk is k-β but is not k-NUC.

(ii) Put X1 = l2 endowed with its usual norm and let for k ≥ 2, Xk be as in (ii). 1 Denote by Yk, k = 1, 2,... the l1-direct sum of Xk and R . By Theorem 8(i), Xk is (k + 1)-NUC and thus, by [8], Yk is also (k + 1)-NUC. The space Y1 is not 1-β (see Example 2). Let k ≥ 2. Since Xk is not k-NUC, there exists an ε > 0 so that for each δ > 0 there is a sequence {xn(δ)}, kxn(δ)k ≥ 1 with sep(xn(δ)) >ε and for k every choice of indices {ni}i=1,

xni (δ) k> 1 − δ. , i=1 X R1 k Let x be the unit of considered as element of Yk. Then, for every choice {ni}i=1 we get

k k

x + xni (δ) (k +1) = xni (δ) (k +1)+1/(k + 1) > 1 − kδ/(k + 1) , , i=1 i=1 X X which means that Yk is not k-β . In connection with Theorems 8 and 16, Corollary 13, and Remark 15 one can ask about the isomorphic relationship between the properties k-β, k ≥ 1 and k-NUC, k ≥ 2. In this direction we show a representative of k-NUC spaces which fails to have an equivalent 1-β norm. For this purpose we use an example of Schachermayer [20, 21]. In the above sense Schachermayer’s space is a typical example of a k-NUC space. The deﬁnition of Schachermayer’s space E [21] is as follows.

10 Let γ = {n1, n2,...,nm} be an increasing ﬁnite sequence of natural numbers. ui ui Write ni =2 + vi where this expression is unique, if we require that 0 ≤ vi < 2 . ui Associate to every ni the real number t(ni)= vi/2 ∈ [0, 1) and call γ admissible if (i) m ≤ n1, ui+1 u1+1 (ii) for every 0 ≤ j < 2 there is at most one i such that t(ni) ∈ [j/2 , (j + 1)/2u1+1).

(N) For an admissible γ = {n1,...,nm} and x ∈ R the space of ﬁnite sequences, deﬁne m

kxkγ = |xni |. i=1 X Put ∞ 1/2 kxk = sup kxk2 ,  γj  j=1 !  X  where the supremum is taken over all increasing sequences {γ }∞ of admissible sets   j j=1 (i.e., the last member of γj is smaller than the first member of γj+1). R(N) ∞ By (E, k·k) denote the completion of with respect to this norm. Let {en}n=1 be its natural basis. Theorem 17. Schachermayer’s space E is 8-NUC and it does not admit an equivalent 1-β norm. Proof. In [5] we have shown that the example from [9] of a reflexive Banach space which does not admit an equivalent norm, uniformly differentiable in every direction, is NUC but it fails to have an equivalent norm with property (β), i.e., 1-β. The proof of the last fact was based on Day’s technique. After showing that (β) ⇒ (BS), we managed to give a simpler proof of the fact that (β) and NUC are isomorphically different [6], using Baernstein’s space. In view of Theorem 9 such a method is not applicable to distinguish isomorphically k-NUC and 1-β. Still, one can repeat almost literally the proof in [5] to show that Schachermayer’s space E fails to have an equivalent norm with property 1-β. Let us mention that the original norm k·k is not 2-NUC. Indeed, one can easily ∞ construct a sequence of integers {ni}i=1 so that all the two-point sets {ni, nj} are ∞ admissible, and then consider in E the sequence {eni }i=1. Thus, it remains to see that E is 8-NUC. Let ε > 0. Consider an arbitrary sequence {yn}, kynk ≤ 1 with sep(yn) >ε. Passing to a subsequence (denote again by {yn}) we may assume that yn is weakly convergent because of reflexivity of E. Put yn = x0 + xn, where {xn} is weakly null sequence. Observe by a standard ∞ perturbation argument that there is no loss of generality to suppose that {xn}n=0 ∞ are supported by finite increasing sets {In}n=0 (i.e., the last member of In is less than the first member of In+1). Thus, (n) (9) xn = λi ei, n =0, 1, 2,.... Xi∈In 11 ∞ Passing to a subsequence assume that {kxnk}n=1 is convergent, moreover without affecting the generality we may suppose that kxnk = b, n = 1, 2,... . By (9) and the definition of the norm in E, we obtain

2 2 2 (10) kx0 + xnk ≥kx0k + b .

It follows from sep(xn) = sep(yn) >ε that

(11) b>ε/2.

∞ Schachermayer [21] has proved that every normalized weakly null sequence {zn}n=1 ∞ k has a subsequence {zni }i=1 so that for every ﬁnite sequence of scalars {ai}i=1

k k 1/2 2 aizni ≤ 6+1/2 ai ! i=1 i=1 X p X ∞ holds. Hence, we may choose a subsequence {xni }i=1 so that for every integer k,

k 1 1/2 (12) xni ≤ 6+1/2b/k . k i=1 X p

For the sake of brevity we denote the above subsequence by {xm}. Write i ∈ I0 in the form i =2u + v, 0 ≤ v < 2u. Put

r = max{u + 1 : i ∈ I0}.

r r r r Let ∆j = [j/2 , (j + 1)/2 ), 0 ≤ j < 2 . Consider for 0 ≤ j < 2 and m =1, 2,...

(m) (m) µj = max{|λi | : t(i) ∈ ∆j}

(write 0 if {i ∈ Im : t(i) ∈ ∆j} = ∅). (m) ∞ r Since {xm} is bounded, the sequences {µj }m=1, j =0,..., 2 are also bounded. Thus, passing to a subsequence, we may suppose that for every 0 ≤ j < 2r,

(m) lim µj = µj. m→∞ Fix k. Then for m large enough, say m>s, we get

(m) 2 r (13) |µj − µj| <ε /64k, 0 ≤ j < 2 .

Consider s+k 1 s+k 1 y = y = x + x . k m 0 k m m=s+1 m=s+1 X X

12 q Let {γj}j=1 be an increasing sequence of admissible sets such that

q 1/2 2 kyk = kykγj . j=1 ! X s+k Note that if there is no γj which intersects simultaneously I0 and I = ∪m=s+1Im then it is easy to see that

s+k 2 2 2 1 kyk = kx0k + xm k m=s+1 X and thus by (12), 2 2 2 kyk ≤kx0k +(6+1/2)b /k. 2 2 It follows from (10) that kx0k ≤ 1 − b and therefore by (11),

kyk2 ≤ 1 − (k − 6 − 1/2)ε2/4k.

Let there exist a γp = {i1 < i2 < ··· < ip} which intersects both I0 and I. Then u1 ′ r1 r1 r1 i1 =2 +v1 ∈ I0 whence r1 = u1 +1 ≤ r. Put ∆j = [j/2 , (j +1)/2 ), 0 ≤ j < 2 . (m) (m) ′ Clearly, for vj = max{|λi | : t(i) ∈ ∆j} we have that

(m) (m) r−r1 r−r1 r1 vj = max{µl :2 j ≤ l< 2 (j + 1)}, 0 ≤ j < 2 . Evidently, (13) implies that for every m>s,

(m) 2 r1 (14) |vj − vj| <ε /64k, 0 ≤ j < 2

(m) where vj = limm→∞ vj . Denote

r−1 ′ A = {0 ≤ j < 2 : there is an i(j) ∈ γp ∩ I0 with t(i(j)) ∈ ∆j}, r−1 ′ B = {0 ≤ j < 2 : there is an i(j) ∈ γp ∩ I with t(i(j)) ∈ ∆j}. Note that for every j ∈ A ∪ B the element i(j) is unique. We write

2 p−1 c = |λ(0)| and a = |λ(0) |.  i  i(j) j=1 i∈γ X Xj Xj∈A   By the deﬁnition of he norm, we have for every m

2 (m) 2 c + a + vj ≤kymk , j∈B ! X 13 whence

(15) c + a + vj ≤ 1. ! Xj∈B It follows from (14) and the deﬁnition of the norm that

2 s+k 2 2 2 kyk ≤ c + a + vj/k + xm/k + ε /16k. ! j∈B m=s+1 X X 2 2 Clearly, (15) yields 2a j∈B vj ≤ 1 − c − a − j∈B vj and thus, P 2 P 1 k − 1 c + a + v /k ≤ + (c + a2). j k k j∈B ! X 2 2 2 Taking into account (12), kx0k + b ≤ 1 and c + a ≤kx0k , we obtain

1 k − 1 1 b2 ε2 kyk2 ≤ + (1 − b2)+ 6+ + k k 2 k 16k k − 7 − 1/2 ε2 =1 − b2 + . k 16k Let now k = 8. Because of (11), we have

kyk2 ≤ 1 − ε2/16k.

Evidently, this implies that the norm is 8-NUC. Moreover, the following holds true.

Proposition 18. The dual E∗ of Schachermayer’s space does not admit an equivalent NUC norm.

Proof. Huﬀ [3] has introduced a property (∗), suggested by J. Bourgain. Let X be a Banach space. Given a set A ⊂ X and ε> 0, deﬁne the ε-derived set of A to be the set

ηε(A)= {x : there exists {xn} ⊂ A with sep(xn) >ε and xn → x weakly}.

Denote by B the closed unit ball in X. In [3] it is proved that if X has an equivalent norm which is UKK, then

for every ε> 0 there exists n such that (∗) η(n)(B)= ∅. ε 14 ∗ ∗ 1 Let B be the closed unit ball of E . Consider ε = 2 . We show that for every n, (n) ∗ ∗ η1/2(B ) =6 ∅, which implies that E fails to have an equivalent NUC norm. Clearly, it is enough to take n = 2j, j = l, 2,... . For every admissible set γ consider its ∗ characteristic function χγ as an element of E , evidently kχγ k = 1 (here k·k stands for the dual norm). Fix n = 2j. Consider admissible sets γ whose first element k n1 is greater or equal to n. Let γ = {n1 < n2 < ··· < n2k }. Fix the first 2 − 1 elements and vary n2k = m. Obviously, there are infinitely many permissible m. For ′ ′′ ′ ′ the characteristic functions χ and χ of the admissible sets γ = {n1,...,n2k−1, m } ′′ ′′ ′ ′′ and γ = {n1,...,n2k−1, m } with m =6 m , we get

kχ′ − χ′′k ≥ 1.

(m) (m) So, if χ is the characteristic function of {n1,...,n2k−1, m}, then sep(χ ) ≥ 1. ∗ Since the biorthogonal basis {en} tends weakly to zero, we obtain that

m χ −→ χ 2k−1 weakly, m {n1,...,n }

(1) ∗ whence χ{n1,...,n2k−1} belongs to η1/2(B ). Repeating the argument, we get

w- lim (··· (w- lim (χ{n1,...,n2k })) ··· )=0. n1→∞ n2k →∞

(n) ∗ Thus, 0 ∈ η1/2(B ), which ends the proof.

Remark 19. Sekowski and Stachura [22] and Prus [17] have deﬁned the notion of nearly uniform smoothness (NUS). They have proved that a Banach space X (resp. X∗) is NUS if and only if X∗ (resp. X) is NUC. Prus has characterized the existence of an equivalent NUS or NUC norm in Banach spaces with countable basis. Using his results, we may, give an alternative proof of Proposition 18, Theorem 17 and Proposition 18 show that Schachermayer’s space is an example of a k-NUC space which does not admit an equivalent NUS norm.

To conclude we mention the following simple statement.

Proposition 20. Let X be a Banach space and Y be its subspace. If X is k-β (resp. k-NUC), then (i) Y is k-β (resp. k-NUC); (ii) the quotient space X/Y is k-β (resp. k-NUC).

The proof is evident in view of Theorem 7.

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