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A geometrical constant and normal normal structure in Banach Spaces

Journal of Inequalities and Applications20112011:16

https://doi.org/10.1186/1029-242X-2011-16

Received: 1 March 2011

Accepted: 23 June 2011

Published: 23 June 2011

Abstract

Recently, we introduced a new coefficient as a generalization of the modulus of smoothness and Pythagorean modulus such as J X , p (t). In this paper, We can compute the constant J X , p (1) under the absolute normalized norms on 2 by means of their corresponding continuous convex functions on [0, 1]. Moreover, some sufficient conditions which imply uniform normal structure are presented.

2000 Mathematics Subject Classification: 46B20.

Keywords

Geometrical constantAbsolute normalized normLorentz sequence spaceUniform normal structure

1. Introduction and preliminaries

We assume that X and X* stand for a Banach space and its dual space, respectively. By S X and B X we denote the unit sphere and the unit ball of a Banach space X, respectively. Let C be a non-empty bounded closed convex subset of a Banach space X. A mapping T : CC is said to be non-expansive provided the inequality

holds for every x, y C. A Banach space X is said to have the fixed point property if every non-expansive mapping T : CC has a fixed point, where C is a non-empty bounded closed convex subset of a Banach space X.

Recall that a Banach space X is called uniformly non-square if there exists δ > 0 such that ||x + y||/2 ≤ 1 - δ or ||x - y||/2 ≤ 1 - δ whenever x, y S X . A bounded convex subset K of a Banach space X is said to have normal structure if for every convex subset H of K that contains more than one point, there exists a point x0 H such that
A Banach space X is said to have uniform normal structure if there exists 0 < c < 1 such that for any closed bounded convex subset K of X that contains more than one point, there exists x0 K such that

It was proved by Kirk that every reflexive Banach space with normal structure has the fixed point property.

There are several constants defined on Banach spaces such as the James [1] and von Neumann-Jordan constants [2]. It has been shown that these constants are very useful in geometric theory of Banach spaces, which enable us to classify several important concept of Banach spaces such as uniformly non-squareness and uniform normal structure [38]. On the other hand, calculation of the constant for some concrete spaces is also of some interest [2, 5, 6, 9].

Recently, we introduced a new coefficient as a generalization of the modulus of smoothness and Pythagorean modulus such as J X , p (t).

Definition 1.1. Let x S X , y S X . For any t > 0, 1 ≤ p < ∞ we set

Some basic properties of this new coefficient are investigated in [6]. In particular, we compute the new coefficient in the Banach spaces l r , L r , l1, ∞ and give rough estimates of the constant in some concrete Banach spaces. In fact, the constant JX, p(1) is also important from the below Corollary in [6].

Corollary 1.2. If . Then R(X) < 2, where R(X) and ω(X) stand for García-Falset constant and the coefficient of weak orthogonality, respectively (see [10, 11]). It is well known that a reflexive Banach space X with R(X) < 2 enjoys the fixed property (see [10]).

In this paper, we compute the constant J X , p (1) under the absolute normalized norms on 2, and give exact values of the constant J X , p (1) in some concrete Banach spaces. Moreover, some sufficient conditions which imply uniform normal structure are presented.

Recall that a norm on 2 is called absolute if ||(z, w)|| = ||(|z|, |w|)|| for all z, w and normalized if ||(1,0)|| = ||(0,1)||. Let N α denote the family of all absolute normalized norms on 2, and let Ψ denote the family of all continuous convex functions on [0, 1] such that ψ (1) = ψ (0) = 1 and max{1 - s, s} ≤ ψ(s) ≤ 1(0 ≤ s ≤ 1). It has been shown that N α and Ψ are a one-to-one correspondence in view of the following proposition in [12].

Proposition 1.3. If ||·|| N α , then ψ(s) = ||(1 - s, s)|| Ψ. On the other hand, if ψ(s) Ψ, defined a norm ||·|| ψ as

then the norm ||·|| ψ N α .

A simple example of absolute normalized norm is usual l r (1 ≤ r ≤ ∞) norm. From Proposition 1.3, one can easily get the corresponding function of the l r norm:

Also, the above correspondence enable us to get many non-l r norms on 2. One of the properties of these norms is stated in the following result.

Proposition 1.4. Let ψ, φ Ψ and φψ. Put , then

The Cesà ro sequence space was defined by Shue [13] in 1970. It is very useful in the theory of matrix operators and others. Let l be the space of real sequences.

For 1 < p < ∞, the Cesà ro sequence space ces p is defined by
The geometry of Cesà ro sequence spaces have been extensively studied in [1416]. Let us restrict ourselves to the two-dimensional Cesà ro sequence space which is just 2 equipped with the norm defined by

2. Geometrical constant J X, p (1) and absolute normalized norm

In this section, we give a simple method to determine and estimate the constant J X, p (1) of absolute normalized norms on 2. For a norm || · || on 2, we write J X, p (1)(|| · ||) for J X, p (1)(2, || · ||). The following is a direct result of Proposition 2.4 in [6].

Proposition 2.1. Let X be a non-trivial Banach space. Then
Proposition 2.2. Let X be the space l r or L r [0, 1] with dimX ≥ 2 (see [6])
  1. (1)

    Let 1 < r ≤ 2 and 1/r + 1/r' = 1. Then for all t > 0

     

if 1 < p < r' then .

if r' ≤ p < ∞ then , for some K ≥ 1.
  1. (2)
    Let 2 ≤ r < ∞, 1 ≤ p < ∞ and h = max{r, p}. Then
     
Proposition 2.3. Let φ Ψ and ψ(s) = φ (1 - s). Then
Proof. For any x = (a, b) 2 and a ≠ 0, b ≠ 0, put . Then
Consequently, we have
We now consider the constant J X, p (1) of a class of absolute normalized norms on 2. Now let us put
Theorem 2.4. Let ψ Ψ and ψψ r (2 ≤ r < ∞). If the function attains its maximum at s = 1/ 2 and rp, then
Proof. By Proposition 1.4, we have || · || ψ ≤ || · || r M1|| · || ψ . Let x, y X, (x, y) ≠ (0, 0), where X = 2. Then
from the definition of J X, p (t), implies that
Note that rp and the function attains its maximum at s = 1/ 2, i.e., . From Proposition 2.2, implies that
(1)
On the other hand, let us put x = (a, a), y = (a, -a), where . Hence ||x|| ψ = ||y|| ψ = 1, and
(2)
From (1) and (2), we have
Theorem 2.5. Let ψ Ψ and ψψ r (1 ≤ r ≤ 2). If the function attains its maximum at s = 1/ 2 and 1 ≤ p < r', then
Proof. By Proposition 1.4, we have || · || r ≤ || · || ψ M2|| · || r . Let x, y X, (x, y) ≠ (0, 0), where X = 2. Then
From the definition of J X, p (t), it implies that
note that 1 ≤ p < r' and the function attains its maximum at s = 1/ 2, i.e., . From Proposition 2.2, it implies that
(3)
On the other hand, let us put x = (1, 0), y = (0, 1). Then ||x|| ψ = ||y|| ψ = 1, and
(4)
From (3) and (4), we have
Lemma 2.6 (see [6]). Let || · || and |.| be two equivalent norms on a Banach space. If a|.| ≤ || · || ≤ b|.| (ba > 0), then
Example 2.7. Let X = 2 with the norm
Then
Proof. It is very easy to check that ||x|| = max{||x||2, λ||x||1} α and its corresponding function is
Therefore,
Since ψ2(s) attains minimum at s = 1/ 2 and hence attains maximum at s = 1/ 2. Therefore, from Theorem 2.5, we have
Example 2.8. Let X = 2 with the norm
Then
Proof. It is obvious to check that the norm ||x|| = max{||x||2, λ||x||} is absolute, but not normalized, since ||(1, 0)|| = ||(0, 1)|| = λ. Let us put
Then |.| α and its corresponding function is
Then
Consider the increasing continuous function . Because g(0) = 1 and , there exists a unique 0 ≤ a ≤ 1 such that g(a) = λ. In fact g(s) is symmetric with respect to s = 1/ 2. Then we have
Obviously, g(s) attains its maximum at s = 1/ 2. Hence, from Theorem 2.4 and Lemma 2.6, we have
Example 2.9. Let X = 2 with the norm
Then
Proof. It is obvious to check that the norm is absolute, but not normalized, since ||(1, 0)|| = ||(0, 1)|| = (1 + λ)1/ 2. Let us put
Therefore, |.| α and its corresponding function is
Obvious ψ(s) ≤ ψ2(s). Since is symmetric with respect to s = 1/ 2, it suffices to consider for s [0, 1/ 2]. Note that, for any s [0, 1/ 2], put . Taking derivative of the function g(s), we have
We always have g'(s) ≥ 0 for 0 ≤ s ≤ 1/ 2. This implies that the function g(s) is increased for 0 ≤ s ≤ 1/2. Therefore, the function attains its maximum at s = 1/2. By Theorem 2.4 and Lemma 2.6, we have
Example 2.10. (Lorentz sequence spaces). Let ω1ω2> 0, 2 ≤ r < ∞. Two-dimensional Lorentz sequence space, i.e. 2 with the norm
where is the rearrangement of (|z|, |ω|) satisfying , then
Proof. It is obvious that , and the corresponding convex function is given by
Obviously ψ(s) ≤ ψ r (s) and . It suffices to consider Φ(s) for s [0, 1/2] since Φ(s) is symmetric with respect to s = 1/2. Note that for s [0, 1/2]
Some elementary computation shows that u(s) - v(s) = (1-(ω2/ω1))s r attains its maximum and v(s) attains its minimum at s = 1/2. Hence,
attains its maximum at s = 1/2 and so does Φ(s). Then by Theorem 2.4 and Lemma 2.6, we have
Example 2.11. Let X be two-dimensional Cesà ro space , then
Proof. We first define
for (x, y) 2. It follows that is isometrically isomorphic to (2, |.|) and |.| is an absolute and normalized norm, and the corresponding convex function is given by
Indeed, defined by is an isometric isomorphism. We prove that ψ(s) ≥ ψ2(s). Note that
Consequently,
Some elementary computation shows that attains its maximum at s = 1/2. Therefore, from Theorem 2.5, we have

3. Constant and uniform normal structure

First, we recall some basic facts about ultrapowers. Let l(X) denote the subspace of the product space IInX equipped with the norm ||(x n )|| := supn||x n || < ∞. Let be an ultrafilter on and let

The ultrapower of X, denoted by , is the quotient space equipped with the quotient norm. Write to denote the elements of the ultrapower. Note that if is non-trivial, then X can be embedded into isometrically. We also note that if X is super-reflexive, that is , then X has uniform normal structure if and only if has normal structure (see [17]).

Theorem 3.1. Let X be a Banach space with

for some t (0, 1]. Then X has uniform normal structure.

Proof. Observe that X is uniform non-square (see [6]) and then X is super-reflexive, it is enough to show that X has normal structure. Suppose that X lacks normal structure, then by Saejung [18, Lemma 2], there exist and satisfying:
  1. (1)
    and for all ij.
     
  2. (2)
    for i = 1, 2, 3.
     
  3. (3)
    .
     

Let and consider three possible cases.

First, if . In this case, let us put and . It follows that , and
Secondly, if and . In this case, let us put and . It follows that , and
Thirdly, and . In this case, let us put and . It follows that , and
Then, by definition of J X, p (t) and the fact ,

This is a contradiction and thus the proof is complete.

Declarations

Acknowledgements

The author wish to express their heartfelt thanks to the referees for their detailed and helpful suggestions for revising the manuscript.

Authors’ Affiliations

(1)
Department of Mathematics and Computer Science, Chongqing Three Gorges University, Wanzhou, China

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Copyright

© Zuo; licensee Springer. 2011

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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