Abstract and Applied Analysis

Volume 2010, Article ID 914183, 13 pages

http://dx.doi.org/10.1155/2010/914183

## Extreme Points and Rotundity in Musielak-Orlicz-Bochner Function Spaces Endowed with Orlicz Norm

^{1}Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China^{2}Department of Mathematics, Harbin University of Science and Technology, Harbin 150080, China

Received 17 June 2010; Accepted 27 August 2010

Academic Editor: Paul Eloe

Copyright © 2010 Shaoqiang Shang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

The criteria for extreme point and rotundity of Musielak-Orlicz-Bochner function spaces equipped with Orlicz norm are given. Although criteria for extreme point of Musielak-Orlicz function spaces equipped with the Orlicz norm were known, we can easily deduce them from our main results.

#### 1. Introduction

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.

A point is said to be extreme point of if and imply . The set of all extreme points of is denoted by . If , then is said to be rotund. A point is said to be strongly extreme point if for any with , and , there holds . If the set of all strongly extreme points of is equal to , then is said to be midpoint local uniform rotund.

The notion of extreme point plays an important role in some branches of mathematics. For example, the Krein-Milman theorem, Choquet integral representation theorem, Rainwater theorem on convergence in weak topology, Bessaga-Pelczynski theorem, and Elton test unconditional convergence are strongly connected with this notion. In [1], using the principle of locally reflexivity, a remarkable theorem describing connections between extreme points of and strongly extreme points of is proved. Namely, a Banach space is midpoint local uniformly rotund if and only if every point of is an extreme point in . Another proof of this theorem based on Goldstein's theorem is given in [2]. Analyzing the proof of this fact one can easily see its local version, namely, if is a strongly extreme point in , then is an extreme point in , where is the mapping of canonical embedding of into .

The criteria for extreme point and rotundity in the classical Musielak-Orlicz function spaces have been given in [3] already. However, because of the complication of Musielak-Orlicz-Bochner function spaces equipped with Orlicz norm, at present, the criteria for extreme point and rotundity have not been discussed yet. The aim of this paper is to give criteria for extreme point and rotundity of Musielak-Orlicz-Bochner function spaces equipped with Orlicz norm. By the result of this paper, it is easy to see that the result of [3] is true.

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

Let denote the right derivative of at (where if , let ) and let be the generalized inverse function of defined on by Then for any and -a.e. . It is well known that there holds the Young inequality for -a.e. . And or . Let

For fixed and , if there exists such that then we call a nonstrictly convex points of with respects to . The set of all nonstrictly convex point of with respect to is denoted by .

For fixed , if , then we call that is strictly convex with respect for .

Moreover, for a given Banach space , we denote by the set of all strongly -measurable functions from to , and for each , define the modular of by Put Then the Musielak-Orlicz-Bochner function space is Banach space. If , is 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. *If -a.e. , then for any .*

*Proof. *For any , there exists such that , where It is easy to see that , where
Noticing that , then . Hence there exists such that . This means that if , we have
This implies that if , then sequence is bounded. Without loss of generality, we may assume that . Without loss of generality, we may assume that or . If , by Levi theorem, we have
If , by dominated convergence theorem, we have
Therefore , namely, . This implies .

#### 2. Main Results

In order to obtain the main theorems of this paper, we first give some lemmas.

Lemma 2.1. *If , then , where .*

*Proof. *By proof of Proposition 1.1, we know that if , then there exists such that and as . Without loss of generality, we may assume that . By Levi theorem, we have
Hence the conclusion of the lemma is true.

Lemma 2.2. *If the set consists of one element from , then .*

*Proof. *Pick ; then we have
where
It follows that
Let ; then we obtain
where
If , then . Hence we have
Moreover, there exists such that , whenever . This means that function is nondecreasing, when is large enough. Pick sequence such that . By Levi theorem, we have
where . Hence the conclusion of the lemma is true.

Lemma 2.3 (see [3]). *Let be rotund, then is strictly convex with respect to for almost all . *

Theorem 2.4. *Let be Musielak-Orlicz-Bochner function spaces, then is an extreme point of if and only if *(a)*the set consists of one element from ; *(b)* with and that on implies , where ; *(c)*, where .*

*Proof. **Necessity*. (a1) Suppose that is an extreme point of the unit ball and , then by Lemma 2.1. Decompose into and such that . Pick . Put
Obviously, and . Moreover, we have
Similarly, we have . Hence . Therefore is not an extreme point of , a contradiction. Hence . Suppose that . Similarly, we get a contradiction.

The necessity of (b) is obvious.

(c) Set
where . Suppose that (c) does not hold. Then or .

If , then for any , by setting
we have and . Moreover, we have

Similarly, we have . Hence . Therefore is not an extreme point of , a contradiction.

If , it is easy to see that , where . Then there exists such that
is not a noll set. Decompose into and such that . Define
Then and . Furthermore, we have
Similarly, we have . Hence . Therefore is not an extreme point, a contradiction. Hence (c) is true.

(a2) If and is an extreme point, suppose that there exists satisfying . Define,
This implies that
(i.e., ),
Since on , then on , a contradiction. Therefore (a) is true. *Sufficiency*. We first prove that for with at least one of the sets or is nonempty. Suppose that and . Hence we have
a contradiction. This contradiction shows that or .

Now we will prove that and . Otherwise, we can assume without loss of generality that and . Put
Next we will prove that for all and for all . Assume first for the contrary that this is such that . Then there exists such that . Since , we have
Hence . Therefore
a contradiction.

Assume now for the contrary that this is such that . We can find such that and . Therefore there are and such that

By the convexity of the modular we have
Hence
Consequently, all inequalities from the last three lines are equalities in fact. Therefore and
for . By , it follows that for . And we have for . By (b), we have . Since , we get , which gives . This contradicts the inequality . Thus for any . Take for all . Then for all . Hence , and consequently for all . Note that and for . Since , with , we have
a contradiction. Therefore and . Now repeating the same procedure as above, putting and instead of and , respectively, we get
for . Hence, by the fact that , we have , and consequently, . Thus is an extreme point of .

Corollary 2.5 (see [3]). * is an extreme point of if and only if *(a)*the set consists of one element from ; *(b)*, where .*

Finally, we investigate the rotundity of .

Theorem 2.6. * is rotund if and only if: *(a)*for any , the set consists of one element from ; *(b)* is rotund; *(c)* is strictly convex with respect to for almost all .*

*Proof. *Sufficiency is obvious by Theorem 2.4.

Necessity. (a) is obvious by (a) of Theorem 2.4. is isometrically isomorphic to closed subspace of , thus is rotund. By Lemma 2.3, (c) is obvious.

If (b) is not true, then there exist with and . Pick , then there exists such that , where . Since , then there exists such that
Set
We have
This implies that . Similarly, we have . It is easy to see that . Then
However, for . By (b) of Theorem 2.4, we have . Hence . So is not an extreme point of . Contradicting the rotundity of .

#### Acknowledgment

The authors would like to thank the anonymous referees for some suggestions to improve the paper. This work was supported by Heilong Jiang Natural Science Fund A200902.

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