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Research Article | Open Access

Volume 2011 |Article ID 361525 | https://doi.org/10.1155/2011/361525

Shaoqiang Shang, Yunan Cui, Yongqiang Fu, "Nonsquareness in Musielak-Orlicz-Bochner Function Spaces", Abstract and Applied Analysis, vol. 2011, Article ID 361525, 16 pages, 2011. https://doi.org/10.1155/2011/361525

# Nonsquareness in Musielak-Orlicz-Bochner Function Spaces

Revised03 Jan 2011
Accepted14 Feb 2011
Published06 Apr 2011

#### Abstract

The criteria for nonsquareness in the classical Orlicz function spaces have been given already. However, because of the complication of Musielak-Orlicz-Bochner function spaces, at present the criteria for nonsquareness have not been discussed yet. In the paper, the criteria for nonsquareness of Musielak-Orlicz-Bochner function spaces are given. As a corollary, the criteria for nonsquareness of Musielak-Orlicz function spaces are given.

#### 1. Introduction

A lot of nonsquareness concepts in Banach spaces are known. Nonsquareness are the important notion in geometry of Banach space. One of reasons is that the property is strongly related to the fixed point property (see ). The criteria for nonsquareness in the classical Orlicz function spaces have been given in  already. However, because of the complication of Musielak-Orlicz-Bochner function spaces, at present the criteria for nonsquareness have not been discussed yet. The aim of this paper is to give criteria nonsquareness of Musielak-Orlicz-Bochner function spaces. As a corollary, the criteria for nonsquareness of Musielak-Orlicz function spaces are given. The topic of this paper is related to the topic of .

Let be a real Banach space. and denote the unit sphere and unit ball, respectively. By , denote the dual space of . Let , and denote the set natural number, reals, and nonnegative reals, respectively. Let us recall some geometrical notions concerning nonsquareness. A Banach space is said to be nonsquare space if for any we have . A Banach space is said to be uniformly nonsquare space if for any , there exists such that .

Let be nonatomic measure space. Suppose that a function satisfies the following conditions:(1)for -a.e, , , and for some ,(2)for  -a.e, , is convex on with respect to ,(3)for each , is a -measurable function of on .

Let . It is well known that is -measurable (see ).

Moreover, for a given Banach space , we denote by , the set of all strongly -measurable function from to , and for each , define the modular of by Put It is well known that Musielak-Orlicz-Bochner function space is Banach spaces equipped with the Luxemburg norm or Orlicz’s norm In particular, and are said to be Musielak-Orlicz function space. Set In particular, the set can be nonempty. To show that, we give a proposition.

Proposition 1.1 (see ). If   μ-a.e.   , then for any .

We define a function Function will be used in the further part of the paper. Moreover, is -measurable. To show that, we give a proposition.

Proposition 1.2. Function is -measurable.

Proof. Pick a dense set in and set It is easy to see that for all ,   μ-a.e on . Hence, . For -a.e , arbitrarily choose . Then, there exists such that , that is, . Since is arbitrary, we find . Thus, .

It is easy to prove the following proposition.

Proposition 1.3. For any , if , then .

Proposition 1.4. For any , if , then .

Proposition 1.5. For any , if , then .

Definition 1.6 (see ). We say that satisfies condition if there exist and a measureable nonnegative function on such that and for almost all and all .

First, we give some results that will used in the further part of the paper.

Lemma 1.7 (see ). Suppose . Then .

Lemma 1.8 (see ). Let be Musielak-Orlicz-Bochner function spaces, then, if , one has , where .

#### 2. Main Results

Theorem 2.1. is nonsquare space if and only if (a),(b)for any , one has or ,(c) is nonsquare space.

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

Lemma 2.2. Let be nonsquare space, then for any , one has

Proof. For any , without loss of generality, we may assume . Since is nonsquare space, we have Therefore, by (2.2), we obtain or This implies . This completes the proof.

Proof of Theorem 2.1. Necessity. (a) If is nonsquare space, then is nonsquare space, because is isometrically embedded into . Since is nonsquare space, then which follows from the theorem proved in more general case (see [10, 11]). Namely, if , then contains isometric copy of .
If (b) is not true, then there exist such that and . Let . We have By and , we obtain that two inequalities of (2.5) are equations. This implies . By Lemma 1.7, we have . Thus, . Similarly, we have , a contradiction!
(c) Pick , then there exists such that , where . Put , where . It is easy to see that . Hence, there exists such that . By Lemma 1.7, we have Let . Then, . The necessity of (c) follows from the fact that is isometrically embedded into . Namely, defining the operator by Hence, for any , we have By Lemma 1.7, we have , hence .
Sufficiency. The proof requires the consideration of two cases separately.
Case 1. . Without loss of generality, we may assume . Let . Put Since is nonsquare space, we have or by Lemma 2.2. Without loss of generality, we may assume . Moreover, we have Let , . Then, or . By , we obtain whenever . By , we obtain whenever . This means that one of three inequalities of (2.10) is strict inequality. By , we have . By Lemma 1.7, we have .Case 2. . By (b), we have . Let . Put Since is nonsquare space, we have or by Lemma 2.2. Without loss of generality, we may assume . Hence, Therefore, by (2.14), we have By , we have . By Lemma 1.7, we have . This completes the proof.

Corollary 2.3. is nonsquare space if and only if (a),(b)for any , one has or .

Theorem 2.4. Let   μ-a.e on . Then, is nonsquare space if and only if (a),(b),(c) is nonsquare space.

Proof. Necessity. By Theorem 2.1, (a) and (c) are obvious. Suppose that . Then, there exists such that , and , where . Set where . It is easy to see that , and . Contradicting Theorem 2.1.
Sufficiency. We only need to prove that for any , if , then . Suppose that there exist such that , . By , we have . This implies or . Hence, or , a contradiction! This completes the proof.

Theorem 2.5. is nonsquare space if and only if (a)for any , one has ,(b)at least one of the conditions (b1), (b2), (b3)is true, where , and ,(c) is nonsquare space.

Proof. Necessity. (a) Suppose that there exists such that , then by Lemma 1.8. Decompose into disjoint sets and such that . Put Obviously, . Pick sequence such that as . Let . By Levi theorem, we have Therefore, Similarly, we have . By , we obtain . By , we have . Therefore, This implies a contradiction!
If (b) is not true, then there exist such that , and , where , . Let . It is easy to see that if , then on . This implies Therefore, by (2.22), we have By , we have By (2.23) and (2.24), we have By , we have Therefore, by (2.25) and (2.26), we have Similarly, we have . This implies , a contradiction!
(c) Pick , then there exists such that , where . Put , where . It is easy to see that . Hence, there exists such that . The necessity of (c) follows from the fact that is isometrically embedded into . Namely, defining the operator by It is easy to see that . Hence, for any , we have
Sufficiency. Suppose that there exists such that . Let , . We will derive a contradiction for each of the following two cases.
Case 1. . By Lemma 2.2, we have . Put Moreover, we have By , three inequalities of (2.31) are equation. This implies Next, we will prove . Suppose that . Without loss of generality, we may assume . Therefore, by (2.32), we have   μ-a.e. on . Hence, . Since three inequalities of (2.31) are equation, we deduce Moreover, it is easy to see on . Therefore, by (2.33) and (2.34), we have -a.e. on . Since is nonsquare space, we have on . Thus, -a.e. on . Therefore, by (2.34) and (2.37), we have This implies , a contradiction! Hence, . Since three inequalities of (2.31) are equation, we deduce and . By (b), we have or . If , then This implies , a contradiction! If , then This implies , a contradiction!Case 2. . Let . Put Then, or . If , then Therefore, by (2.42), we have This implies , a contradiction! If , then or by Lemma 2.2, where Without loss of generality, we may assume . Hence, Therefore, by (2.45), we have This implies , a contradiction! This completes the proof.

Corollary 2.6. is nonsquare space if and only if (a)for any , one has ,(b)at least one of the conditions (b1), (b2),(b3)is true, where , and .

#### Acknowledgments

The authors would like to thank the anonymous referee for the suggestions to improve the manuscript. This work was supported by Academic Leaders Fund of Harbin University of Science and Technology and supported by China Natural Science Fund 11061022.

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