Nonsquareness and Locally Uniform Nonsquareness in Orlicz-Bochner Function Spaces Endowed with Luxemburg Norm

Abstract

Criteria for nonsquareness and locally uniform nonsquareness of Orlicz-Bochner function spaces equipped with Luxemburg norm are given. We also prove that, in Orlicz-Bochner function spaces generated by locally uniform nonsquare Banach space, nonsquareness and locally uniform nonsquareness are equivalent.

1. Introduction

A lot of nonsquareness concepts in Banach spaces are known (see [1]). Nonsquareness are important notions in geometry of Banach space. One of reasons is that these properties are strongly related to the fixed point property (see [2]). The criteria for nonsquareness and locally uniform nonsquareness in the classical Orlicz function spaces have been given in [3, 4] already. However, because of the complicated structure of Orlicz-Bochner function spaces equipped with the Luxemburg norm, the criteria for nonsquareness and locally uniform nonsquareness of them have not been found yet. The aim of this paper is to give criteria for nonsquareness and locally uniform nonsquareness of Orlicz-Bochner function spaces equipped with Luxemburg norm.

Let be a real Banach space. and denote the unit sphere and unit ball, respectively. Let us recall some geometrical notions concerning nonsquareness. A Banach space is said to be nonsquare if for any we have . A Banach space is said to be uniformly nonsquare if there exists such that for any , . A Banach space is said to be locally uniformly nonsquare if for any , there exists such that , where .

Let be set of real numbers. A function is called an -function if is convex, even, , and , and .

Let be a nonatomic measurable space. denotes right derivative of . Moreover, for a given Banach space , we denote by the set of all strongly -measurable function from to , and for each , we define the modular of by

(1.1)

Put

(1.2)

The linear set endowed with the Luxemburg norm

(1.3)

is a Banach space. We say that an Orlicz function satisfies condition if there exist and such that

(1.4)

First let us recall a known result that will be used in the further part of the paper.

Lemma 1.1 (see [3]).

Suppose . Then

(1.5)

2. Main Results

Theorem 2.1.

is nonsquare if and only if

(a);

(b) is nonsquare.

In order to prove the theorem, we give a lemma.

Lemma 2.2.

If is nonsquare, then for any , we have

(2.1)

Proof.

Case 1.

If , then

(2.2)

or

(2.3)

Case 2.

If , then

(2.4)

or

(2.5)

This implies . This completes the proof.

Proof of Theorem 2.1.

• (a) Necessity. Suppose that , then there exist and such that . Pick such that is not a null set. Since , there exist sequence and disjont subsets of such that

  

  (2.6)

Therefore, if we define , then for any , we have

(2.7)

This yields

(2.8)

Hence, . But , and we deduce that . Moreover, we have and , a contradiction with nonsquareness of .

If (b) is not true, then there exist such that . Pick such that . Put

(2.9)

Then we have

(2.10)

It is easy to see . We know that

(2.11)

Hence, we have

(2.12)

It is easy to see , a contradiction!

Sufficiency. Suppose that there exists such that

(2.13)

We will derive a contradiction for each of the following two cases.

Case 1.

. Let . Hence, we have

(2.14)

Since , we have . Hence, . This implies , a contradiction!

Case 2.

. By Lemma 2.2, without loss of generality, we may assume that there exists such that , and . Therefore,

(2.15)

Since , we have . Hence, . This implies , a contradiction!

Theorem 2.3.

is locally uniformly nonsquare if and only if

(a);

(b) is locally uniformly nonsquare.

In order to prove the theorem, we give a lemma.

Lemma 2.4.

If is locally uniformly nonsquare, then

(a)For any , , we have

(2.16)

(b)If , then , where

(2.17)

Proof.

• (a) Since is locally uniformly nonsquare, we have and , where and

  

  (2.18)

In fact, since is locally uniformly nonsquare, we have

(2.19)

Case 1.

If , then

(2.20)

or

(2.21)

Case 2.

If , then

(2.22)

or

(2.23)

Therefore, we get, the following inequality

(2.24)

holds.

(b1) Suppose that , where . Then there exist and subsequence of , such that . By definition of , there exist such that

(2.25)

We will derive a contradiction for each of the following two cases.

Case 1.

. Since , there exists such that . Therefore,

(2.26)

This implies , a contradiction!

Case 2.

. Without loss of generality, we may assume , where . Since , there exists such that . Therefore, we have

(2.27)

This implies

(2.28)

Similarly, we have

(2.29)

Therefore, we have

(2.30)

This implies , a contradiction! Hence, .

(b2) Suppose that , where . Then there exist and subsequence of , such that . Since , then there exist such that , whenever . By definition of , there exist such that

(2.31)

Therefore, we have

(2.32)

whenever . Since , there exists such that , where

(2.33)

Hence, we have

(2.34)

This implies

(2.35)

which contradict (2.32). Hence, .

Combing (b1) with (b2), we get . This completes the proof.

Proof of Theorem 2.3.

Necessity. By Theorem 2.1, . If (b) is not true, then there exist , such that and as . Pick such that . Put

(2.36)

Then we have

(2.37)

It is easy to see . We know that

(2.38)

Moreover, we have , . By the dominated convergence theorem, we have

(2.39)

It is easy to see , as . By Lemma 1.1, we have and as , a contradiction with locally uniform nonsquareness of .

Sufficiency. Suppose that there exist , such that as . We will derive a contradiction for each of the following two cases.

Case 1.

There exist , such that , where . Put

(2.40)

We have

(2.41)

This implies . Hence, . We define a function

(2.42)

on , where . By Lemma 2.4, we have -a.e on . Let -a.e on , where is simple function. Hence,

(2.43)

is -measurable. By Lemma 2.4, we have -a.e on . Then is -measurable. Using

(2.44)

we get that there exists such that , where

(2.45)

Let , , . It is easy to see , and . If , by Lemma 2.4, we have

(2.46)

Without loss of generality, we may assume that there exists such that

(2.47)

Moreover, for any , we have

(2.48)

Hence, if , then

(2.49)

Let . Then . Therefore,

(2.50)

By Lemma 1.1, we have , as . This is in contradiction with .

Case 2.

For any , , there exists such that whenever . By the Riesz theorem, without loss of generality, we may assume that a.e on . Using

(2.51)

we get that there exist such that , where

(2.52)

Since is -function, we can choose such that . Since a.e on , by the Egorov theorem, there exists such that whenever , where , . Next, we will prove that if , then

(2.53)

In fact, we have

(2.54)

Moreover, we have

(2.55)

By (2.54) and (2.55), we have

(2.56)

This shows that if , then

(2.57)

It is easy to see . Therefore,

(2.58)

for large enough. By Lemma 1.1, we have , as , which contradicts , for large enough. This completes the proof.

Corollary 2.5.

The following statements are equivalent:

(a) is locally uniformly nonsquare if and only if is nonsquare;

(b) is locally uniformly nonsquare.