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A universal infinite-dimensional modulus for normed spaces and applications
Nonlinear Analysis 58 (2004) 379 – 394
www.elsevier.com/locate/na
A universal in nite-dimensional modulus for normed spaces and applications T. Dom&'nguez Benavides∗ , B. Gavira Departamento de An alisis Matem atico, Facultad de Matem aticas, Universidad de Sevilla, P.O. Box 1160, 41080 Sevilla, Spain Received 13 January 2004; accepted 6 April 2004
Abstract We de ne an in nite-dimensional modulus which can be simultaneously considered as a measure of nearly uniform convexity and nearly uniform smoothness. We compute this modulus in some classical Banach spaces and we show some basic properties of this modulus and some applications to prove that a Banach space satis es the w- xed point property for non-expansive mappings. ? 2004 Elsevier Ltd. All rights reserved. MSC: 47H09; 47H10; 46B20 Keywords: Fixed point; Normal structure; Nearly uniform convexity; Nearly uniform smoothness
1. Introduction In 1995 Ben&'tez et al. [2] de ned a two-dimensional modulus for normed spaces. Given a normed space X , one observes that for any x; y ∈ X with y ¡ 1 ¡ x, there is a unique z=z(x; y) in the line segment [x; y] with z=1. They de ne X : [0; 1) → R by x − z(x; y) X ( ) = sup : y 6 ¡ 1 ¡ x : x − 1 They called modulus of squareness because its extreme values characterize nearly squareness (we recall that X is nearly square if for all ¿ 0 there exists Y subspace of X with dim Y = 2 such that d(Y; ‘1 (2)) ¡ 1 + , where d(E; F) is the ∗
Corresponding author. Tel.: +34-954-557982; fax: +34-954-557972. E-mail addresses: [email protected] (T.D. Benavides), [email protected] (B. Gavira).
0362-546X/$ - see front matter ? 2004 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2004.04.009
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Banach–Mazur distance between two normed spaces E and F). The behaviour of the modulus of squareness is strongly connected with the geometry of the space. In particular, the value of the modulus of squareness tells us whether or not a space is uniformly smooth, uniformly convex, uniformly non-square or an inner product space. Furthermore, this modulus can also be used to obtain uniform normal structure. The main properties of the modulus are collected in the following theorem: Theorem 1 (Benitez et al. [2]): Let X be any normed space and its modulus of squareness. (1) ( ) = sup{M ( ): M ⊂ X; dim M = 2}. (2) is strictly increasing and convex. (3) ( ) ¡ 1 ( ) for each ∈ (0; 1), unless X is nearly square, in that case ( ) = 1 ( ) for all ∈ (0; 1), where 1+ : 1 ( ) = 1− In particular, if X is non-re
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2. Preliminaries In this section we introduce some known results related to the existence of xed points for nonexpansive mappings which will be used through this paper. For more details the reader may consult, for instance [1,8]. Throughout this paper X will be a Banach space. Convergence of sequences in the weak topology is denoted by x n * x. We say that X has the weak xed point property (w-FPP) if every nonexpansive mapping T de ned from a weakly compact convex subset C of X into C has a xed point. We say that X has weak normal structure (w-NS) if every weakly compact convex subset of X with more than one member is not diametral. Theorem 2 (Goebel and Kirk [8]). If X has w-NS then X has the w-FPP. Associated to the weak normal structure of a Banach space, Bynum [4] de ned the weakly convergent sequence coeKcient. We shall use an equivalent de nition [1]. Let X be a Banach space without the Schur property. The weakly convergent sequence coeKcient of X is de ned by limn; m;n=m x n − xm ; WCS(X ) = inf limn x n where the in mum is taken over all weakly null sequences {x n } such that both limits exist and limn x n = 0. It must be noted that every bounded sequence contained in a metric space has a subsequence {x n } such that limn; m;n=m d(x n ; xm ) exists (see [1, Theorem III.1.5]). Theorem 3 (Bynum [4]). If WCS(X ) ¿ 1 then X has w-NS. We say that X has weak uniform normal structure (w-UNS) if WCS(X ) ¿ 1. In the following theorem we recall the value of WCS(X ) in some particular Banach spaces. Previously we recall the de nition of Bynum spaces. Let p ∈ [1; +∞), q ∈ [1; +∞]. The Bynum spaces ‘p; q are de ned as ‘p; q = (‘p ; · p; q ), where xp; q = (x+ qp + x− qp )1=q
if q ∈ [1; +∞);
xp; ∞ = max{x+ p ; x− p }; x+ ; x− denote the positive and the negative part of x respectively and xp denotes the norm of x in ‘p . Theorem 4 (Bynum [4]; Ayerbe et al. [1]): (1) For every real number p ¿ 1, we have WCS(‘p ) = 21=p . (2) WCS(c0 ) = 1. (3) For p ¿ 1, q ¿ 1, we have WCS(‘p; q ) = min{21=p ; 21=q }.
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Now we are going to consider some geometric properties connected with normal structure. The space X is said to be uniformly convex (UC) if for each ∈ (0; 2] there exists ¿ 0 such that for x; y ∈ X with x; y 6 1 and x−y ¿ then (x+y)=2 6 1−. The space X is said to be uniformly smooth (US) if !X (t) lim = 0; t→0+ t where
!X (t) = sup
x + ty + x − ty − 1: x 6 1; y 6 1 : 2
Next we recall the de nitions of nearly uniform convexity and nearly uniform smoothness which are in nite-dimensional generalizations of the notions of uniform convexity and uniform smoothness. The space X is said to be nearly uniformly convex (NUC) if X is reMexive and "X ( ) ¿ 0 for each ¿ 0, i.e. "0 (X ) = 0, where "X ( ) = inf {1 − x: {x n } ⊂ BX ; x n * x; lim inf x n − x ¿ }; n
"0 (X ) = sup{ ¿ 0: "X ( ) = 0}: The space X is said to be nearly uniformly smooth (NUS) if X is reMexive and for any ¿ 0 there exists $ ¿ 0 such that for any t ∈ (0; $) and any weakly null sequence {x n } ⊂ BX there exists k ¿ 1 such that x1 + txk 6 1 + t. Notice that every NUC space has normal structure ([1, Remark VI.4.7]). The situation for NUS spaces is diOerent. Indeed, ‘p; 1 is NUC, so its dual ‘q; ∞ is NUS but this space fails to have normal structure ([1, Example VI.2]). However NUS spaces have the xed point property [6]. Up to now we have deduced the existence of xed points for nonexpansive mappings as a consequence of normal structure. Now we recall other geometric constant of Banach spaces which give xed point results without normal structure. In 1991 Garc&'a Falset [6] de ned the following geometric coeKcient: R(X ) = sup lim inf x n + x: {x n }; x ∈ BX ; x n * 0 n→∞
and, later, he proved the following theorem. Theorem 5 (Garc&'a Falset [7]): A Banach space X has the w-FPP if R(X ) ¡ 2. 3. A universal innite-dimensional modulus The rst problem appearing when we try to de ne a modulus suitable for NUC and NUS is the non-existence of a relationship between these concepts for a xed space. Indeed, if X is at the worst situation for uniform convexity, i.e. 0 (X ) = 2, then X (0) = 2 and so X is not uniformly smooth (see [2, Theorem 2.4]). However there
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is no a similar behaviour for nearly uniform convexity and nearly uniform smoothness as the following example shows. Example 6. Let X be the Bynum space ‘2; ∞ . Since ‘2; 1 is NUC ([1, Example V.1]), its dual ‘2; ∞ is NUS. We recall that for any bounded set A ⊂ X , the HausdorO measure of noncompactness is de ned by ((A) = inf { ¿ 0: A has a nite -net}. The characteristic of noncompact convexity of X is de ned by ( (X )=sup{ ¿ 0: "( ( )=0} where "( ( ) = inf {1 − d(0; A): A ⊂ BX is convex, ((A) ¿ }. We are going to show that ( (X ) = 1, that is, the maximum value for ( . Let A be the convex hull of the sequence x n = e1 − en , which is w-convergent to e1 . Since limn x n − e1 exists and ‘2; ∞ satis es the non-strict Opial property [1] (that is, lim inf n x n 6 lim inf n x n + x for every w-null sequence {x n } and every x = 0 in X ), using [9] Lemma 1.1 we have (({x n }) = limx n − e1 = limen = 1: n
n
Since the measure ( satis es the invariance under passage to the convex hull [1], we have ((A) = 1 and ( (X ) = 1. Denition 7. Let X be a Banach space. For each ∈ (0; 1), the universal in nitedimensional modulus is de ned by x n − y ; X ( ) = sup lim inf n→∞ 1 − x where the supremum is taken over all sequences {x n } ⊂ B such that x n * x = 0, lim inf n x n − x 6 and y = x=x, where B denotes the closed ball centred at 0 with radius . Remark 8. It is easy to check that for each ∈ (0; 1) we have 1 : 1 6 X ( ) 6 1− In the following theorem we will show that both extreme values are attained in some spaces. Theorem 9. (1) If X satis>es Schur property, that is, every weakly convergent sequence is norm convergent, then X ( ) = 1 for any ∈ (0; 1). (2) ‘∞ ( ) = 1=(1 − ) for each ∈ (0; 1). (3) ‘2 ( ) = ‘2 ( ) = 1= 1 − 2 for each ∈ (0; 1). (4) If 1 ¡ p ¡ ∞, then ‘p ( )=((1− q )p−1 + p )1=p =(1− q )1=q for each ∈ (0; 1). (5) c0 ( ) = max(1; =(1 − )) for each ∈ (0; 1). Proof. (1) Obvious. ∞ (2) Denote 1= n=1 en and consider the sequence x n = (1−en ) ∈ B which converges weakly to 1 =: x with lim inf n x n − x 6 . Furthermore, x n − y = 1 for every n. Easily we deduce the claimed value of ‘∞ ( ).
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(3) The following result can be derived from [3]: Let {x n } be a w-null sequence in ‘p , 1 6 p ¡ + ∞. Then, lim inf x n − xp = lim inf x n p + xp n→∞
n→∞
(1)
for every x ∈ ‘p . Let {x n } be a sequence in B ⊂ ‘2 which is weakly convergent to x (which implies lim inf n x n − x 6 because ‘2 satis es the Opial property). We denote A = x and, using (1), we have 2 ¿ lim inf x n 2 = lim inf x n − x2 + A2 n
n
and lim inf n
x n − y2 x n − x2 2 − A2 = lim inf + 1 6 + 1: n (1 − x)2 (1 − x)2 (1 − A)2
Elementary calculus proves that the function f(A) = ( 2 − A2 )=(1 − A)2 attains its maximum in [0; ] at the point A = 2 . By substitution of this value we obtain 1 ‘2 ( ) 6 1 − 2 and this upper bound is attained for x n = 2 e1 + 1 − 2 en . (4) It is analogous to the case p = 2. (5) Let {x n } ⊂ B be a sequence which converges weakly to x (which implies lim inf n x n − x 6 because c0 satis es the non-strict Opial property). Taking a subsequence, if necessary, we can assume supp(x n − x) ∩ supp(x − y) = ∅; where for every x = (xk ) we denote supp x = {k: xk = 0}. We have lim inf x n − y = lim inf x n − x + x − y n
n
= max lim inf x n − x; x − y 6 max( ; 1 − x): n
Thus
max( ; 1 − x) co ( ) 6 6 max ;1 1 − x 1−
and this value is attained at x n = (e1 + en ). Now we study some basic properties of the new modulus. The following is obvious: Proposition 10. Let X be a Banach space. The function X (·) is increasing. In general, X (·) is not strictly increasing. Consider for example a Banach space X with Schur property or X = c0 (see Theorem 9). On the other hand this modulus is convex. This fact is noteworthy because the modulus of convexity does not need to be a convex function.
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Proposition 11. For every Banach space X , X (·) is a convex function and so X (·) is continuous in (0; 1). Proof. Consider 0 ¡ 1 ¡ 2 ¡ 1, 0 ¡ t ¡ 1. We have to prove X (t 1 + (1 − t) 2 ) 6 tX ( 1 ) + (1 − t)X ( 2 ): Let {x n } ⊂ Bt 1 +(1−t) 2 such that x n * x, lim inf n x n − x 6 t 1 + (1 − t) 2 and y = x=x. It is enough to nd two sequences {x1n } ⊂ B 1 , {x2n } ⊂ B 2 such that x2 − y2 x n − y x1 − y1 6 t lim inf n lim inf + (1 − t) lim inf n : 1 n→∞ 1 − x n n 1 − x 1 − x2 Consider for i = 1; 2 i i xn * x = xi xin = t 1 + (1 − t) 2 t 1 + (1 − t) 2
yi =
x xi = = y: i x x
Using the convexity of the norm, we have x n − y tx1n − y1 + (1 − t)x2n − y2 6 : 1 − x t(1 − x1 ) + (1 − t)(1 − x2 ) On the other hand if we consider zn = -x2 + (1 − -)x2n = .y + (1 − .)x1n with -=
1 2 x1
1− 1−
∈ (0; 1)
and
.=
1 2 x1
1− 1−
x2 ∈ (0; 1);
we have zn − y = (1 − .)x1n − y x2 = (1 − .)x1 + .y ⇒ 1 − x2 = (1 − .)(1 − x1 ): Furthermore, using that · is convex and w-slsc we obtain lim inf zn − y 6 -x2 − y + (1 − -) lim inf x2n − y 6 lim inf x2n − y n
n
n
and we deduce x1 − y zn − y x2n − y lim inf n 1 = lim inf 6 lim inf : n n n 1 − x 1 − x2 1 − x2 Thus, using [2, Lemma 3.1] we have the desired inequality. Next we obtain the following relationship between modulus and modulus . As a consequence, the new modulus is a re nement of the previous one. Theorem 12. For any normed space X and any ∈ (0; 1), X ( ) 6 X ( ). Proof. Let {x n } be a sequence in B which converges weakly to x = 0 with lim inf n x n − x 6 . Choose / ¿ 1 arbitrary and zn ∈ [x n ; /y] such that zn = 1. There exists
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-n ∈ (0; 1) such that zn = -n /y + (1 − -n )x n . Thus -n /y + (1 − -n )x n = 1. We can assume, without loss of generality, that {-n } is convergent to some -. We obtain -/y + (1 − -)x 6 1, which implies -/ x + 1 − - 6 1 x and we deduce -(/ − x) 6 1 − x: So we have 1−-¿
/−1 : / − x
Choose c ¡ 1 arbitrary. For n large enough (n ¿ n0 (c; /)) we have /−1 : 1 − -n ¿ c / − x Thus, X ( ) ¿
/y − zn (1 − -n )/y − x n /y − x n = ¿c : /−1 /−1 / − x
Hence, X ( ) ¿ c lim inf n
y − x/n /y − x n : = c lim inf n / − x 1 − x /
Taking supremum we obtain X ( ) ¿ cX ( =/). Since X (·) is continuous and c is arbitrary, letting / → 1+ , we obtain X ( ) ¿ X ( ). Theorem 13. Let X and Y be two isomorphic Banach spaces whose Banach–Mazur distance is less than 1 + , where 6 1. Then, for all ∈ (0; 1), |X ( ) − Y ( )| 6
2 + 2 : (1 − )2
Proof. Our hypothesis implies that we may regard X and Y as the same vector space equipped with two equivalent norms, · and | · | respectively, such that x 6 |x| 6 (1 + )x for every x ∈ X . Let {x n } ⊂ BX (0; ) be a sequence such that x n * x with lim inf n x n − x 6 and y = x=x. Consider the sequence xn = [1=(1 + )]x n ∈ BY (0; ) which converges weakly to x =[1=(1+)]x with lim inf n |xn −x | 6 and y =x =|x |=x=|x|. We have x n − y − |xn − y | 6 |x n − y| − |xn − y | 6 |x n − y − xn + y |
T.D. Benavides, B. Gavira / Nonlinear Analysis 58 (2004) 379 – 394
6 |x n − 6 |x n |
xn |
387
1 1 1 + |x| − + |y − y| = |x n | 1 − 1 + |x| x
1 + |x| 6 + (1 + ) = 2 + (1 + ): 1+ x
On the other hand we have 1 x = 1 − x + x 6 1 − x + : 1 − |x | 6 1 − 1+ 1+ 1+ Thus, x n − y |xn − y | x n − y x n − y − [2 + (1 + )] − 6 − 1 − x 1 − |x | 1 − x 1 − x + 1+
=
+ [2 + (1 + )](1 − x) x n − y 1+
(1 − x)(1 − x +
1+ )
2 + (1 + ) x n − y 1+ + 6 1 − x 1 − x 1 − x
6
1 + 1+ 2 + (1 + ) + 1− 1− 1−
=
(1 + ) 1+ + [2 + (1 + )](1 − ) (1 − )2
=
− 2 ) + 2 + − ) + ( 1+ 2 ( 1+ (1 − )2
6 ¡
1+
+ 2 + (1 − )2
1+
+ 2 + 2 + 2 6 2 (1 − ) (1 − )2
which implies 2 + 2 |xn − y | x n − y − : ¿ 1 − |x | 1 − x (1 − )2 Hence, Y ( ) ¿ X ( ) −
2 + 2 : (1 − )2
A symmetric argument yields X ( ) ¿ Y ( ) − (2 + 2)=(1 − )2 . Next we give the characterizations of nearly uniformly convex and nearly uniformly smooth spaces.
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Lemma 14. Let X be a Banach space. Then, ( =2)"0 (X ) + − 1 X ( ) ¿ : 1− Proof. The inequality is obvious if "0 (X ) = 0. If "0 (X ) ¿ 0, assume 0 ¡ c ¡ "0 (X ). For any $ ¿ 0 there exists a sequence {zn } in BX weakly convergent, say to z, such that zn − z ¿ c and z ¿ 1 − $. Consider the sequence x n = 2 (zn + z) ∈ B which is weakly convergent to x = z with lim inf n x n − x 6 . Then, we have x n − y ¿ x n − x − x − y ¿ c + x − 1 ¿ c + (1 − $) − 1: 2 2 Hence, ( =2)c + (1 − $) − 1 x n − y ¿ : 1 − x 1 − (1 − $) Letting $ → 0 and later c → "0 (X ) we obtain the inequality as stated. Lemma 15. Let X be a Banach space. Then, 1 "0 (X ) 6 lim inf (1 − )X ( ) 6 lim sup(1 − )X ( ) 6 2"0 (X ): →1 2 →1 Proof. From Lemma 14, we know that lim inf →1 (1 − )X ( ) ¿ 1=2"0 (X ). On the other hand, consider any sequence {x n } in B w-convergent to x with lim inf n x n − x 6 . Taking a subsequence, if necessary, we can assume that limn x n −y, limn x n = 0 6 and limn x n − x = 6 20 do exist. For an arbitrary $ ∈ (0; 1 − ) we can assume |x n − x − | ¡ $ and |x n − 0| ¡ $ for every n. Choose p ∈ (0; 1) and de ne for ∈ (0; 1) rp ( ) = sup{ ¿ 0: "X ( ) ¡ (1 − )p }: It is clear that rp (·) is nonincreasing and we claim that lim →1− rp ( ) 6 "0 (X ). Indeed, otherwise there is a number c such that rp ( ) ¿ c ¿ "0 (X ) for any ¡ 1. Choosing ( ) ∈ (c; rp ( )) such that "X ( ( )) ¡ (1 − )p , we obtain "X (c) 6 (1 − )p for any ¡ 1, which implies "X (c) = 0 and this is a contradiction because c ¿ "0 (X ). We have the following inequalities: x n − y 6 x n − x + x − y 6 + $ + 1 − x
6 (0 + $)(1 − "X ( )) x 6 (0 + $) 1 − "X 0+$ x ¿ x n − x n − x ¿ 0 − 2$ − : Thus, x n − y 1 − 0 + 3$ + 2 6 : 1 − x 1 − 0(1 − "X ( )) If ¿ rp ( ), then "X ( ) ¿ (1 − )p and x n − y 1 + 3 + 3$ 1 − 0 + 3$ + 40 1 + 30 + 3$ 6 : 6 6 p 1 − x 1 − (1 − "X ( )) 1 − + (1 − ) 1 − + (1 − )p
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If 6 rp ( ), we have 1 − 0 + 3$ + 2rp ( ) 1 − + 3$ + 2rp ( ) x n − y 6 6 : 1 − x 1−0 1− In all cases, since $ is arbitrary, we have lim sup(1 − )X ( ) →1
1 + 3 ; lim sup(1 − + 2rp ( )) 6 max lim sup(1 − ) 1 − + (1 − )p →1 →1
6 max 0; lim sup 2rp ( ) 6 2"0 (X ): →1
Theorem 16. Let X be a Banach space. Then X is NUC if and only if X is renition X (0) = 1. Proof. Assume that X is NUS and choose ¿ 0 arbitrary and $ = $( ) given by the de nition of NUS. Take ¡ $=(1 + $). If {x n } is a sequence in B which converges weakly to x with lim inf n x n − x 6 ¡ $. We can assume, without loss of generality, that limn x − x n =(1 − x) =: t exists. We have y−x x − xn y − x n = y − x + y − x y − x y−x x − x n x − x n = (1 − x) + y − x 1 − x x − x n y−x x − xn x − x n 6 (1 − x) : +t (1 − x) t − y − x x − x n 1 − x Since, t = lim n
x − x n 6 ¡ $; 1 − x 1−
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we have
x − x n y − x n 6 (1 − x)(1 + t) + (1 − x) t − 1 − x
x − x n 6 (1 − x) 1 + + t − 1− 1 − x
for in nitely many n. Hence, X ( ) 6 1 + 1− which implies X ( ) − 1 06 6 1−
for ¡
$ : 1+$
Thus, 0 6 lim inf →0
X ( ) − 1 X ( ) − 1 6 lim sup 6 : →0
Since is arbitrary we obtain X (0) = 0. Conversely, assume that X is not NUS. There exists 0 ∈ (0; 1) such that for every $ ∈ (0; 1) there exists t ∈ (0; $) and a weakly null sequence {un } in BX satisfying u1 + tun ¿ 1 + 0 t for all n. Choose $ ¿ 0 arbitrary and a = 0 =2 with small enough such that t = a=(1 − a). Consider the sequence x n = a(u1 − un )=u1 which converges weakly to au1 =u1 =: x. Since, a a a 6 u1 + 1 + 0 ¡ u1 + un 1−a 1−a 1−a and a=(1 − a) ¡ 1 we obtain a u1 ¿ 1 + ( 0 − 1) ¿ 0 : 1−a Thus x n 6 2a= 0 = , lim inf n x n − x 6 and u1 y−x un x − xn a = y − x y − x n = y − x + + y − x y − x u1 1 − a u1
y − x a a ¿ ¿ (1 − x) (1 + 0 : (1 + 0 u1 1−a 1−a Hence X ( ) ¿ 1 + 0 a=(1 − a) which implies (X ( ) − 1)= ¿ 02 =(2 − 0 ) and X ( ) − 1 2 lim inf ¿ 0 ¿ 0: →0 2 We can give a lower estimate for WCS(X ) depending on the value of X ( ). Proposition 18. Let X be a Banach space. Then, (2 + X ( )) WCS(X ) ¿ sup : ∈(0;1) X ( ) + 1
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Proof. Denote w=WCS(X ). For any $ ¿ 0 there exists a weakly null sequence {zn } in BX such that zn − zm 6 1 + $ and zn ¿ 1=w(1 − $). Fix k ∈ N and for any ∈ (0; 1) we consider the sequence x n = (zk − zn )=(1 + $) ∈ B which converges weakly to zk =(1 + $) =: x with lim inf n x n − x 6 . Then we have x n − y ¿ 1 − x
1+$
zn − zk ( z1k − 1−
1+$ )
w(1−$2 )
¿
2 w(1−$2 )
1−
−1
w(1−$2 )
:
Letting $ → 0 we obtain X ( ) ¿
2 − w w−
which is equivalent to w¿
(2 + X ( )) : X ( ) + 1
Remark 19. The above lower bound for WCS(X ) is not sharp. Indeed, for X = ‘1 we obtain (2 + X ( )) 3 = 2 ∈(0;1) X ( ) + 1
2 = WCS(X ) ¿ sup
and for X = ‘2 elementary calculus proves that √ (2 + X ( )) 2 = WCS(X ) ¿ sup ≈ 1:19: ∈(0;1) X ( ) + 1 However, for X = c0 we have the equality WCS(X ) = 1 = sup
∈(0;1)
(2 + X ( )) : X ( ) + 1
As a consequence of the previous proposition and having in mind Theorem 3 and Theorem 2 we obtain a suKcient condition so that a Banach space X has w-UNS. Corollary 20. Let X be a Banach space. If X ( ) ¡
2 − 1 1−
for some ∈ (0; 1)
in particular if lim inf →1 (1 − )X ( ) ¡ 1, then X has w-UNS and, so, X has the w-FPP. Remark 21. Note that the estimate in Corollary 20 is not sharp. For example, the space X = (R ⊕ l2 )∞ satis es X ( ) ¿
2 − 1 ¿ 1− 1−
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√ for every ∈ (0; 1) and WCS(X ) = 2. Furthermore, since X ( ) ¿ 1, this inequality only can be satis ed for ¿ 2=3. This fact is not surprising because we have shown that the “good” behaviour of for near zero is related to nearly uniform smoothness (Theorem 17) and this property does not imply weak normal structure (see [1]). On the other hand, we cannot hope the estimation X ( ) ¡ 1=(1− ) for some as a warranty for weak uniform normal structure as it happened with the nite-dimensional modulus [2, Proposition 2.9]. Indeed, the space X = ‘2; ∞ does not have w-UNS and satis es X ( ) ¡ 1=(1 − ) for some ∈ (0; 1) because otherwise X ( ) = 1=(1 − ) for all which implies X (0) = 1 and this is a contradiction because X is NUS. If X satis es the non-strict Opial property, then the above estimate can be improved. Proposition 22. Let X be a Banach space satisfying the non-strict Opial property. Then (1) WCS(X ) ¿ sup ∈(0; 1) (1 + X ( ))=X ( ). (2) If X ( ) ¡ =(1 − ) for some ∈ (0; 1), then X has w-UNS. Proof. Denote w = WCS(X ). Let {zn } and {x n } be as in Proposition 18. Since X satis es the non-strict Opial property we have lim inf x n − y ¿ lim inf x n − x = n
n
lim inf zn ¿ 1+$ n w(1 − $2 )
and lim inf ¿ n→∞
=w(1 − $2 ) : 1 − =w(1 − $2 )
Letting $ → 0 we obtain the desired estimation. Finally, we give a suKcient condition for Banach space X to have the w-FPP in absence of w-NS. We shall use the coeKcient R(X ). Theorem 23. Let X be a Banach space. If X ( ) ¡ 1 + for some ∈ (0; 1), in particular if X (0) ¡ 1, then R(X ) ¡ 2 and, so, X has the w-FPP. Proof. Assume R(X ) = 2 and choose $ ¿ 0 arbitrary. There exists a weakly null sequence {un } in BX and a point u ∈ BX such that lim inf n un + u ¿ 2 − $. We can assume un +u ¿ 2−$ for all n. Thus, un ¿ 1−$ for every n and u ¿ 1−$. It is clear that -un + .u ¿ (- + .)(1 − $) for every -; . ∈ (0; +∞) and every n. Consider the sequence xn =
a 1 u− un 1+a 1+a
*
u=x 1+a
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with a ¿ 0 arbitrary. We have
1 a − u+ un y − x n = u 1 + a 1+a
1 a a ¿ − + (1 − $) ¿ 1 − + (1 − $) u 1 + a 1 + a 1+a 1+a u ¿ x = (1 − $): 1+a 1+a Hence a (1 − 1+a + 1+a )(1 − $) x n − y ¿ : 1 − x 1 − 1+a (1 − $)
Letting $ → 0 and a → ∞, we obtain X ( ) ¿ 1 + for all ∈ (0; 1). Remark 24. (1) If X is a reMexive Banach space with X (0) ¡ 1 then, as a consequence of the previous theorem and [10] Corollary 4.3.6, we have 5X (0) ¡ 1, where 5X (·) denotes the modulus of NUS of X (see [5]). (2) We do not know if R(X ) can be less than 2 for some space X satisfying X ( ) ¿ 1 + for all . (3) The estimate in Theorem √ 23 does not imply new xed point results for close to 1. Indeed, if ¿ − 1 + 3 we have 2 − 1 ¿ 1 + : 1− In this sense, we can say that Corollary 20 is useful for close to 1 and Theorem 23 for close to 0. It would be interesting to obtain xed point results from the modulus X ( ) for medium values of in some spaces where WCS(X ) = 1 and R(X ) = 2. Acknowledgements This research was partially supported by DGES Grant BFM2000-0344-C02-01 and Junta de Andaluc&' a Grant FQM-127. The authors are very grateful to S. Prus and an anonymous referee for some useful suggestions. References [1] J.M. Ayerbe, T. Dom&'nguez, G. L&opez, Measures of Noncompactness in Metric Fixed Point Theory, BirkhTauser, Basel, 1997. [2] C. Ben&itez, K. Przeslawski, D. Yost, A universal modulus for normed spaces, Studia Math. 127 (1) (1998) 21–46. [3] H. Brezis, E. Lieb, A relation between pointwise convergence of functionals, Proc. Amer. Math. Soc. 88 (3) (1983) 486–490. [4] W.L. Bynum, Normal structure coeKcient for Banach spaces, Pac. J. Math. 86 (2) (1980) 427–436.
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[5] T. Dom&'nguez Benavides, Modulus of nearly uniform smoothness and Lindenstrauss formulae, Glasgow Math. J. 37 (1995) 143–153. [6] J. Garc&'a-Falset, Stability and xed points for nonexpansive mappings, Houston J. Math. 20 (3) (1994) 495–506. [7] J. Garc&'a-Falset, The xed point property in Banach spaces with NUS-property, J. Math. Anal. Appl. 215 (1997) 532–542. [8] K. Goebel, W.A. Kirk, Topics in Metric Fixed Point Theory, Cambridge University Press, Cambridge, 1990. [9] M.A. Jap&on Pineda, Stability of the xed point property for nonexpansive mappings, Ph.D. Thesis, Seville, 1998. [10] E.M. Mazcu˜na& n Navarro, Geometry of Banach spaces in metric xed point theory, Ph.D. Thesis, Valencia, 2003.
www.elsevier.com/locate/na
A universal in nite-dimensional modulus for normed spaces and applications T. Dom&'nguez Benavides∗ , B. Gavira Departamento de An alisis Matem atico, Facultad de Matem aticas, Universidad de Sevilla, P.O. Box 1160, 41080 Sevilla, Spain Received 13 January 2004; accepted 6 April 2004
Abstract We de ne an in nite-dimensional modulus which can be simultaneously considered as a measure of nearly uniform convexity and nearly uniform smoothness. We compute this modulus in some classical Banach spaces and we show some basic properties of this modulus and some applications to prove that a Banach space satis es the w- xed point property for non-expansive mappings. ? 2004 Elsevier Ltd. All rights reserved. MSC: 47H09; 47H10; 46B20 Keywords: Fixed point; Normal structure; Nearly uniform convexity; Nearly uniform smoothness
1. Introduction In 1995 Ben&'tez et al. [2] de ned a two-dimensional modulus for normed spaces. Given a normed space X , one observes that for any x; y ∈ X with y ¡ 1 ¡ x, there is a unique z=z(x; y) in the line segment [x; y] with z=1. They de ne X : [0; 1) → R by x − z(x; y) X ( ) = sup : y 6 ¡ 1 ¡ x : x − 1 They called modulus of squareness because its extreme values characterize nearly squareness (we recall that X is nearly square if for all ¿ 0 there exists Y subspace of X with dim Y = 2 such that d(Y; ‘1 (2)) ¡ 1 + , where d(E; F) is the ∗
Corresponding author. Tel.: +34-954-557982; fax: +34-954-557972. E-mail addresses: [email protected] (T.D. Benavides), [email protected] (B. Gavira).
0362-546X/$ - see front matter ? 2004 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2004.04.009
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Banach–Mazur distance between two normed spaces E and F). The behaviour of the modulus of squareness is strongly connected with the geometry of the space. In particular, the value of the modulus of squareness tells us whether or not a space is uniformly smooth, uniformly convex, uniformly non-square or an inner product space. Furthermore, this modulus can also be used to obtain uniform normal structure. The main properties of the modulus are collected in the following theorem: Theorem 1 (Benitez et al. [2]): Let X be any normed space and its modulus of squareness. (1) ( ) = sup{M ( ): M ⊂ X; dim M = 2}. (2) is strictly increasing and convex. (3) ( ) ¡ 1 ( ) for each ∈ (0; 1), unless X is nearly square, in that case ( ) = 1 ( ) for all ∈ (0; 1), where 1+ : 1 ( ) = 1− In particular, if X is non-re
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2. Preliminaries In this section we introduce some known results related to the existence of xed points for nonexpansive mappings which will be used through this paper. For more details the reader may consult, for instance [1,8]. Throughout this paper X will be a Banach space. Convergence of sequences in the weak topology is denoted by x n * x. We say that X has the weak xed point property (w-FPP) if every nonexpansive mapping T de ned from a weakly compact convex subset C of X into C has a xed point. We say that X has weak normal structure (w-NS) if every weakly compact convex subset of X with more than one member is not diametral. Theorem 2 (Goebel and Kirk [8]). If X has w-NS then X has the w-FPP. Associated to the weak normal structure of a Banach space, Bynum [4] de ned the weakly convergent sequence coeKcient. We shall use an equivalent de nition [1]. Let X be a Banach space without the Schur property. The weakly convergent sequence coeKcient of X is de ned by limn; m;n=m x n − xm ; WCS(X ) = inf limn x n where the in mum is taken over all weakly null sequences {x n } such that both limits exist and limn x n = 0. It must be noted that every bounded sequence contained in a metric space has a subsequence {x n } such that limn; m;n=m d(x n ; xm ) exists (see [1, Theorem III.1.5]). Theorem 3 (Bynum [4]). If WCS(X ) ¿ 1 then X has w-NS. We say that X has weak uniform normal structure (w-UNS) if WCS(X ) ¿ 1. In the following theorem we recall the value of WCS(X ) in some particular Banach spaces. Previously we recall the de nition of Bynum spaces. Let p ∈ [1; +∞), q ∈ [1; +∞]. The Bynum spaces ‘p; q are de ned as ‘p; q = (‘p ; · p; q ), where xp; q = (x+ qp + x− qp )1=q
if q ∈ [1; +∞);
xp; ∞ = max{x+ p ; x− p }; x+ ; x− denote the positive and the negative part of x respectively and xp denotes the norm of x in ‘p . Theorem 4 (Bynum [4]; Ayerbe et al. [1]): (1) For every real number p ¿ 1, we have WCS(‘p ) = 21=p . (2) WCS(c0 ) = 1. (3) For p ¿ 1, q ¿ 1, we have WCS(‘p; q ) = min{21=p ; 21=q }.
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Now we are going to consider some geometric properties connected with normal structure. The space X is said to be uniformly convex (UC) if for each ∈ (0; 2] there exists ¿ 0 such that for x; y ∈ X with x; y 6 1 and x−y ¿ then (x+y)=2 6 1−. The space X is said to be uniformly smooth (US) if !X (t) lim = 0; t→0+ t where
!X (t) = sup
x + ty + x − ty − 1: x 6 1; y 6 1 : 2
Next we recall the de nitions of nearly uniform convexity and nearly uniform smoothness which are in nite-dimensional generalizations of the notions of uniform convexity and uniform smoothness. The space X is said to be nearly uniformly convex (NUC) if X is reMexive and "X ( ) ¿ 0 for each ¿ 0, i.e. "0 (X ) = 0, where "X ( ) = inf {1 − x: {x n } ⊂ BX ; x n * x; lim inf x n − x ¿ }; n
"0 (X ) = sup{ ¿ 0: "X ( ) = 0}: The space X is said to be nearly uniformly smooth (NUS) if X is reMexive and for any ¿ 0 there exists $ ¿ 0 such that for any t ∈ (0; $) and any weakly null sequence {x n } ⊂ BX there exists k ¿ 1 such that x1 + txk 6 1 + t. Notice that every NUC space has normal structure ([1, Remark VI.4.7]). The situation for NUS spaces is diOerent. Indeed, ‘p; 1 is NUC, so its dual ‘q; ∞ is NUS but this space fails to have normal structure ([1, Example VI.2]). However NUS spaces have the xed point property [6]. Up to now we have deduced the existence of xed points for nonexpansive mappings as a consequence of normal structure. Now we recall other geometric constant of Banach spaces which give xed point results without normal structure. In 1991 Garc&'a Falset [6] de ned the following geometric coeKcient: R(X ) = sup lim inf x n + x: {x n }; x ∈ BX ; x n * 0 n→∞
and, later, he proved the following theorem. Theorem 5 (Garc&'a Falset [7]): A Banach space X has the w-FPP if R(X ) ¡ 2. 3. A universal innite-dimensional modulus The rst problem appearing when we try to de ne a modulus suitable for NUC and NUS is the non-existence of a relationship between these concepts for a xed space. Indeed, if X is at the worst situation for uniform convexity, i.e. 0 (X ) = 2, then X (0) = 2 and so X is not uniformly smooth (see [2, Theorem 2.4]). However there
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is no a similar behaviour for nearly uniform convexity and nearly uniform smoothness as the following example shows. Example 6. Let X be the Bynum space ‘2; ∞ . Since ‘2; 1 is NUC ([1, Example V.1]), its dual ‘2; ∞ is NUS. We recall that for any bounded set A ⊂ X , the HausdorO measure of noncompactness is de ned by ((A) = inf { ¿ 0: A has a nite -net}. The characteristic of noncompact convexity of X is de ned by ( (X )=sup{ ¿ 0: "( ( )=0} where "( ( ) = inf {1 − d(0; A): A ⊂ BX is convex, ((A) ¿ }. We are going to show that ( (X ) = 1, that is, the maximum value for ( . Let A be the convex hull of the sequence x n = e1 − en , which is w-convergent to e1 . Since limn x n − e1 exists and ‘2; ∞ satis es the non-strict Opial property [1] (that is, lim inf n x n 6 lim inf n x n + x for every w-null sequence {x n } and every x = 0 in X ), using [9] Lemma 1.1 we have (({x n }) = limx n − e1 = limen = 1: n
n
Since the measure ( satis es the invariance under passage to the convex hull [1], we have ((A) = 1 and ( (X ) = 1. Denition 7. Let X be a Banach space. For each ∈ (0; 1), the universal in nitedimensional modulus is de ned by x n − y ; X ( ) = sup lim inf n→∞ 1 − x where the supremum is taken over all sequences {x n } ⊂ B such that x n * x = 0, lim inf n x n − x 6 and y = x=x, where B denotes the closed ball centred at 0 with radius . Remark 8. It is easy to check that for each ∈ (0; 1) we have 1 : 1 6 X ( ) 6 1− In the following theorem we will show that both extreme values are attained in some spaces. Theorem 9. (1) If X satis>es Schur property, that is, every weakly convergent sequence is norm convergent, then X ( ) = 1 for any ∈ (0; 1). (2) ‘∞ ( ) = 1=(1 − ) for each ∈ (0; 1). (3) ‘2 ( ) = ‘2 ( ) = 1= 1 − 2 for each ∈ (0; 1). (4) If 1 ¡ p ¡ ∞, then ‘p ( )=((1− q )p−1 + p )1=p =(1− q )1=q for each ∈ (0; 1). (5) c0 ( ) = max(1; =(1 − )) for each ∈ (0; 1). Proof. (1) Obvious. ∞ (2) Denote 1= n=1 en and consider the sequence x n = (1−en ) ∈ B which converges weakly to 1 =: x with lim inf n x n − x 6 . Furthermore, x n − y = 1 for every n. Easily we deduce the claimed value of ‘∞ ( ).
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(3) The following result can be derived from [3]: Let {x n } be a w-null sequence in ‘p , 1 6 p ¡ + ∞. Then, lim inf x n − xp = lim inf x n p + xp n→∞
n→∞
(1)
for every x ∈ ‘p . Let {x n } be a sequence in B ⊂ ‘2 which is weakly convergent to x (which implies lim inf n x n − x 6 because ‘2 satis es the Opial property). We denote A = x and, using (1), we have 2 ¿ lim inf x n 2 = lim inf x n − x2 + A2 n
n
and lim inf n
x n − y2 x n − x2 2 − A2 = lim inf + 1 6 + 1: n (1 − x)2 (1 − x)2 (1 − A)2
Elementary calculus proves that the function f(A) = ( 2 − A2 )=(1 − A)2 attains its maximum in [0; ] at the point A = 2 . By substitution of this value we obtain 1 ‘2 ( ) 6 1 − 2 and this upper bound is attained for x n = 2 e1 + 1 − 2 en . (4) It is analogous to the case p = 2. (5) Let {x n } ⊂ B be a sequence which converges weakly to x (which implies lim inf n x n − x 6 because c0 satis es the non-strict Opial property). Taking a subsequence, if necessary, we can assume supp(x n − x) ∩ supp(x − y) = ∅; where for every x = (xk ) we denote supp x = {k: xk = 0}. We have lim inf x n − y = lim inf x n − x + x − y n
n
= max lim inf x n − x; x − y 6 max( ; 1 − x): n
Thus
max( ; 1 − x) co ( ) 6 6 max ;1 1 − x 1−
and this value is attained at x n = (e1 + en ). Now we study some basic properties of the new modulus. The following is obvious: Proposition 10. Let X be a Banach space. The function X (·) is increasing. In general, X (·) is not strictly increasing. Consider for example a Banach space X with Schur property or X = c0 (see Theorem 9). On the other hand this modulus is convex. This fact is noteworthy because the modulus of convexity does not need to be a convex function.
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Proposition 11. For every Banach space X , X (·) is a convex function and so X (·) is continuous in (0; 1). Proof. Consider 0 ¡ 1 ¡ 2 ¡ 1, 0 ¡ t ¡ 1. We have to prove X (t 1 + (1 − t) 2 ) 6 tX ( 1 ) + (1 − t)X ( 2 ): Let {x n } ⊂ Bt 1 +(1−t) 2 such that x n * x, lim inf n x n − x 6 t 1 + (1 − t) 2 and y = x=x. It is enough to nd two sequences {x1n } ⊂ B 1 , {x2n } ⊂ B 2 such that x2 − y2 x n − y x1 − y1 6 t lim inf n lim inf + (1 − t) lim inf n : 1 n→∞ 1 − x n n 1 − x 1 − x2 Consider for i = 1; 2 i i xn * x = xi xin = t 1 + (1 − t) 2 t 1 + (1 − t) 2
yi =
x xi = = y: i x x
Using the convexity of the norm, we have x n − y tx1n − y1 + (1 − t)x2n − y2 6 : 1 − x t(1 − x1 ) + (1 − t)(1 − x2 ) On the other hand if we consider zn = -x2 + (1 − -)x2n = .y + (1 − .)x1n with -=
1 2 x1
1− 1−
∈ (0; 1)
and
.=
1 2 x1
1− 1−
x2 ∈ (0; 1);
we have zn − y = (1 − .)x1n − y x2 = (1 − .)x1 + .y ⇒ 1 − x2 = (1 − .)(1 − x1 ): Furthermore, using that · is convex and w-slsc we obtain lim inf zn − y 6 -x2 − y + (1 − -) lim inf x2n − y 6 lim inf x2n − y n
n
n
and we deduce x1 − y zn − y x2n − y lim inf n 1 = lim inf 6 lim inf : n n n 1 − x 1 − x2 1 − x2 Thus, using [2, Lemma 3.1] we have the desired inequality. Next we obtain the following relationship between modulus and modulus . As a consequence, the new modulus is a re nement of the previous one. Theorem 12. For any normed space X and any ∈ (0; 1), X ( ) 6 X ( ). Proof. Let {x n } be a sequence in B which converges weakly to x = 0 with lim inf n x n − x 6 . Choose / ¿ 1 arbitrary and zn ∈ [x n ; /y] such that zn = 1. There exists
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-n ∈ (0; 1) such that zn = -n /y + (1 − -n )x n . Thus -n /y + (1 − -n )x n = 1. We can assume, without loss of generality, that {-n } is convergent to some -. We obtain -/y + (1 − -)x 6 1, which implies -/ x + 1 − - 6 1 x and we deduce -(/ − x) 6 1 − x: So we have 1−-¿
/−1 : / − x
Choose c ¡ 1 arbitrary. For n large enough (n ¿ n0 (c; /)) we have /−1 : 1 − -n ¿ c / − x Thus, X ( ) ¿
/y − zn (1 − -n )/y − x n /y − x n = ¿c : /−1 /−1 / − x
Hence, X ( ) ¿ c lim inf n
y − x/n /y − x n : = c lim inf n / − x 1 − x /
Taking supremum we obtain X ( ) ¿ cX ( =/). Since X (·) is continuous and c is arbitrary, letting / → 1+ , we obtain X ( ) ¿ X ( ). Theorem 13. Let X and Y be two isomorphic Banach spaces whose Banach–Mazur distance is less than 1 + , where 6 1. Then, for all ∈ (0; 1), |X ( ) − Y ( )| 6
2 + 2 : (1 − )2
Proof. Our hypothesis implies that we may regard X and Y as the same vector space equipped with two equivalent norms, · and | · | respectively, such that x 6 |x| 6 (1 + )x for every x ∈ X . Let {x n } ⊂ BX (0; ) be a sequence such that x n * x with lim inf n x n − x 6 and y = x=x. Consider the sequence xn = [1=(1 + )]x n ∈ BY (0; ) which converges weakly to x =[1=(1+)]x with lim inf n |xn −x | 6 and y =x =|x |=x=|x|. We have x n − y − |xn − y | 6 |x n − y| − |xn − y | 6 |x n − y − xn + y |
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6 |x n − 6 |x n |
xn |
387
1 1 1 + |x| − + |y − y| = |x n | 1 − 1 + |x| x
1 + |x| 6 + (1 + ) = 2 + (1 + ): 1+ x
On the other hand we have 1 x = 1 − x + x 6 1 − x + : 1 − |x | 6 1 − 1+ 1+ 1+ Thus, x n − y |xn − y | x n − y x n − y − [2 + (1 + )] − 6 − 1 − x 1 − |x | 1 − x 1 − x + 1+
=
+ [2 + (1 + )](1 − x) x n − y 1+
(1 − x)(1 − x +
1+ )
2 + (1 + ) x n − y 1+ + 6 1 − x 1 − x 1 − x
6
1 + 1+ 2 + (1 + ) + 1− 1− 1−
=
(1 + ) 1+ + [2 + (1 + )](1 − ) (1 − )2
=
− 2 ) + 2 + − ) + ( 1+ 2 ( 1+ (1 − )2
6 ¡
1+
+ 2 + (1 − )2
1+
+ 2 + 2 + 2 6 2 (1 − ) (1 − )2
which implies 2 + 2 |xn − y | x n − y − : ¿ 1 − |x | 1 − x (1 − )2 Hence, Y ( ) ¿ X ( ) −
2 + 2 : (1 − )2
A symmetric argument yields X ( ) ¿ Y ( ) − (2 + 2)=(1 − )2 . Next we give the characterizations of nearly uniformly convex and nearly uniformly smooth spaces.
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Lemma 14. Let X be a Banach space. Then, ( =2)"0 (X ) + − 1 X ( ) ¿ : 1− Proof. The inequality is obvious if "0 (X ) = 0. If "0 (X ) ¿ 0, assume 0 ¡ c ¡ "0 (X ). For any $ ¿ 0 there exists a sequence {zn } in BX weakly convergent, say to z, such that zn − z ¿ c and z ¿ 1 − $. Consider the sequence x n = 2 (zn + z) ∈ B which is weakly convergent to x = z with lim inf n x n − x 6 . Then, we have x n − y ¿ x n − x − x − y ¿ c + x − 1 ¿ c + (1 − $) − 1: 2 2 Hence, ( =2)c + (1 − $) − 1 x n − y ¿ : 1 − x 1 − (1 − $) Letting $ → 0 and later c → "0 (X ) we obtain the inequality as stated. Lemma 15. Let X be a Banach space. Then, 1 "0 (X ) 6 lim inf (1 − )X ( ) 6 lim sup(1 − )X ( ) 6 2"0 (X ): →1 2 →1 Proof. From Lemma 14, we know that lim inf →1 (1 − )X ( ) ¿ 1=2"0 (X ). On the other hand, consider any sequence {x n } in B w-convergent to x with lim inf n x n − x 6 . Taking a subsequence, if necessary, we can assume that limn x n −y, limn x n = 0 6 and limn x n − x = 6 20 do exist. For an arbitrary $ ∈ (0; 1 − ) we can assume |x n − x − | ¡ $ and |x n − 0| ¡ $ for every n. Choose p ∈ (0; 1) and de ne for ∈ (0; 1) rp ( ) = sup{ ¿ 0: "X ( ) ¡ (1 − )p }: It is clear that rp (·) is nonincreasing and we claim that lim →1− rp ( ) 6 "0 (X ). Indeed, otherwise there is a number c such that rp ( ) ¿ c ¿ "0 (X ) for any ¡ 1. Choosing ( ) ∈ (c; rp ( )) such that "X ( ( )) ¡ (1 − )p , we obtain "X (c) 6 (1 − )p for any ¡ 1, which implies "X (c) = 0 and this is a contradiction because c ¿ "0 (X ). We have the following inequalities: x n − y 6 x n − x + x − y 6 + $ + 1 − x
6 (0 + $)(1 − "X ( )) x 6 (0 + $) 1 − "X 0+$ x ¿ x n − x n − x ¿ 0 − 2$ − : Thus, x n − y 1 − 0 + 3$ + 2 6 : 1 − x 1 − 0(1 − "X ( )) If ¿ rp ( ), then "X ( ) ¿ (1 − )p and x n − y 1 + 3 + 3$ 1 − 0 + 3$ + 40 1 + 30 + 3$ 6 : 6 6 p 1 − x 1 − (1 − "X ( )) 1 − + (1 − ) 1 − + (1 − )p
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389
If 6 rp ( ), we have 1 − 0 + 3$ + 2rp ( ) 1 − + 3$ + 2rp ( ) x n − y 6 6 : 1 − x 1−0 1− In all cases, since $ is arbitrary, we have lim sup(1 − )X ( ) →1
1 + 3 ; lim sup(1 − + 2rp ( )) 6 max lim sup(1 − ) 1 − + (1 − )p →1 →1
6 max 0; lim sup 2rp ( ) 6 2"0 (X ): →1
Theorem 16. Let X be a Banach space. Then X is NUC if and only if X is re
x − x n 6 ¡ $; 1 − x 1−
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we have
x − x n y − x n 6 (1 − x)(1 + t) + (1 − x) t − 1 − x
x − x n 6 (1 − x) 1 + + t − 1− 1 − x
for in nitely many n. Hence, X ( ) 6 1 + 1− which implies X ( ) − 1 06 6 1−
for ¡
$ : 1+$
Thus, 0 6 lim inf →0
X ( ) − 1 X ( ) − 1 6 lim sup 6 : →0
Since is arbitrary we obtain X (0) = 0. Conversely, assume that X is not NUS. There exists 0 ∈ (0; 1) such that for every $ ∈ (0; 1) there exists t ∈ (0; $) and a weakly null sequence {un } in BX satisfying u1 + tun ¿ 1 + 0 t for all n. Choose $ ¿ 0 arbitrary and a = 0 =2 with small enough such that t = a=(1 − a). Consider the sequence x n = a(u1 − un )=u1 which converges weakly to au1 =u1 =: x. Since, a a a 6 u1 + 1 + 0 ¡ u1 + un 1−a 1−a 1−a and a=(1 − a) ¡ 1 we obtain a u1 ¿ 1 + ( 0 − 1) ¿ 0 : 1−a Thus x n 6 2a= 0 = , lim inf n x n − x 6 and u1 y−x un x − xn a = y − x y − x n = y − x + + y − x y − x u1 1 − a u1
y − x a a ¿ ¿ (1 − x) (1 + 0 : (1 + 0 u1 1−a 1−a Hence X ( ) ¿ 1 + 0 a=(1 − a) which implies (X ( ) − 1)= ¿ 02 =(2 − 0 ) and X ( ) − 1 2 lim inf ¿ 0 ¿ 0: →0 2 We can give a lower estimate for WCS(X ) depending on the value of X ( ). Proposition 18. Let X be a Banach space. Then, (2 + X ( )) WCS(X ) ¿ sup : ∈(0;1) X ( ) + 1
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391
Proof. Denote w=WCS(X ). For any $ ¿ 0 there exists a weakly null sequence {zn } in BX such that zn − zm 6 1 + $ and zn ¿ 1=w(1 − $). Fix k ∈ N and for any ∈ (0; 1) we consider the sequence x n = (zk − zn )=(1 + $) ∈ B which converges weakly to zk =(1 + $) =: x with lim inf n x n − x 6 . Then we have x n − y ¿ 1 − x
1+$
zn − zk ( z1k − 1−
1+$ )
w(1−$2 )
¿
2 w(1−$2 )
1−
−1
w(1−$2 )
:
Letting $ → 0 we obtain X ( ) ¿
2 − w w−
which is equivalent to w¿
(2 + X ( )) : X ( ) + 1
Remark 19. The above lower bound for WCS(X ) is not sharp. Indeed, for X = ‘1 we obtain (2 + X ( )) 3 = 2 ∈(0;1) X ( ) + 1
2 = WCS(X ) ¿ sup
and for X = ‘2 elementary calculus proves that √ (2 + X ( )) 2 = WCS(X ) ¿ sup ≈ 1:19: ∈(0;1) X ( ) + 1 However, for X = c0 we have the equality WCS(X ) = 1 = sup
∈(0;1)
(2 + X ( )) : X ( ) + 1
As a consequence of the previous proposition and having in mind Theorem 3 and Theorem 2 we obtain a suKcient condition so that a Banach space X has w-UNS. Corollary 20. Let X be a Banach space. If X ( ) ¡
2 − 1 1−
for some ∈ (0; 1)
in particular if lim inf →1 (1 − )X ( ) ¡ 1, then X has w-UNS and, so, X has the w-FPP. Remark 21. Note that the estimate in Corollary 20 is not sharp. For example, the space X = (R ⊕ l2 )∞ satis es X ( ) ¿
2 − 1 ¿ 1− 1−
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√ for every ∈ (0; 1) and WCS(X ) = 2. Furthermore, since X ( ) ¿ 1, this inequality only can be satis ed for ¿ 2=3. This fact is not surprising because we have shown that the “good” behaviour of for near zero is related to nearly uniform smoothness (Theorem 17) and this property does not imply weak normal structure (see [1]). On the other hand, we cannot hope the estimation X ( ) ¡ 1=(1− ) for some as a warranty for weak uniform normal structure as it happened with the nite-dimensional modulus [2, Proposition 2.9]. Indeed, the space X = ‘2; ∞ does not have w-UNS and satis es X ( ) ¡ 1=(1 − ) for some ∈ (0; 1) because otherwise X ( ) = 1=(1 − ) for all which implies X (0) = 1 and this is a contradiction because X is NUS. If X satis es the non-strict Opial property, then the above estimate can be improved. Proposition 22. Let X be a Banach space satisfying the non-strict Opial property. Then (1) WCS(X ) ¿ sup ∈(0; 1) (1 + X ( ))=X ( ). (2) If X ( ) ¡ =(1 − ) for some ∈ (0; 1), then X has w-UNS. Proof. Denote w = WCS(X ). Let {zn } and {x n } be as in Proposition 18. Since X satis es the non-strict Opial property we have lim inf x n − y ¿ lim inf x n − x = n
n
lim inf zn ¿ 1+$ n w(1 − $2 )
and lim inf ¿ n→∞
=w(1 − $2 ) : 1 − =w(1 − $2 )
Letting $ → 0 we obtain the desired estimation. Finally, we give a suKcient condition for Banach space X to have the w-FPP in absence of w-NS. We shall use the coeKcient R(X ). Theorem 23. Let X be a Banach space. If X ( ) ¡ 1 + for some ∈ (0; 1), in particular if X (0) ¡ 1, then R(X ) ¡ 2 and, so, X has the w-FPP. Proof. Assume R(X ) = 2 and choose $ ¿ 0 arbitrary. There exists a weakly null sequence {un } in BX and a point u ∈ BX such that lim inf n un + u ¿ 2 − $. We can assume un +u ¿ 2−$ for all n. Thus, un ¿ 1−$ for every n and u ¿ 1−$. It is clear that -un + .u ¿ (- + .)(1 − $) for every -; . ∈ (0; +∞) and every n. Consider the sequence xn =
a 1 u− un 1+a 1+a
*
u=x 1+a
T.D. Benavides, B. Gavira / Nonlinear Analysis 58 (2004) 379 – 394
393
with a ¿ 0 arbitrary. We have
1 a − u+ un y − x n = u 1 + a 1+a
1 a a ¿ − + (1 − $) ¿ 1 − + (1 − $) u 1 + a 1 + a 1+a 1+a u ¿ x = (1 − $): 1+a 1+a Hence a (1 − 1+a + 1+a )(1 − $) x n − y ¿ : 1 − x 1 − 1+a (1 − $)
Letting $ → 0 and a → ∞, we obtain X ( ) ¿ 1 + for all ∈ (0; 1). Remark 24. (1) If X is a reMexive Banach space with X (0) ¡ 1 then, as a consequence of the previous theorem and [10] Corollary 4.3.6, we have 5X (0) ¡ 1, where 5X (·) denotes the modulus of NUS of X (see [5]). (2) We do not know if R(X ) can be less than 2 for some space X satisfying X ( ) ¿ 1 + for all . (3) The estimate in Theorem √ 23 does not imply new xed point results for close to 1. Indeed, if ¿ − 1 + 3 we have 2 − 1 ¿ 1 + : 1− In this sense, we can say that Corollary 20 is useful for close to 1 and Theorem 23 for close to 0. It would be interesting to obtain xed point results from the modulus X ( ) for medium values of in some spaces where WCS(X ) = 1 and R(X ) = 2. Acknowledgements This research was partially supported by DGES Grant BFM2000-0344-C02-01 and Junta de Andaluc&' a Grant FQM-127. The authors are very grateful to S. Prus and an anonymous referee for some useful suggestions. References [1] J.M. Ayerbe, T. Dom&'nguez, G. L&opez, Measures of Noncompactness in Metric Fixed Point Theory, BirkhTauser, Basel, 1997. [2] C. Ben&itez, K. Przeslawski, D. Yost, A universal modulus for normed spaces, Studia Math. 127 (1) (1998) 21–46. [3] H. Brezis, E. Lieb, A relation between pointwise convergence of functionals, Proc. Amer. Math. Soc. 88 (3) (1983) 486–490. [4] W.L. Bynum, Normal structure coeKcient for Banach spaces, Pac. J. Math. 86 (2) (1980) 427–436.
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