Strongly Extreme Points and Middle Point Locally Uniformly Convex in Orlicz Spaces Equipped with s-Norm

Abstract

As is well known, the extreme points and strongly extreme points play important roles in Banach spaces. In this paper, the criterion for strongly extreme points in Orlicz spaces equipped with s-norm is given. We complete solved criterionOrlicz space that generated by Orlicz function. And the sufficient and necessary conditions for middle point locally uniformly convex in Orlicz spaces equipped with s-norm are obtained.

1. Introduction

The extreme point set plays a crucial role in function analysis, convex analysis, and optimization. In fact, any compact convex set is the convex hull of its extreme point set, and the result is called Krein-Milman theorem. The notion of a dentable subset of a Banach space was introduced by Rieffel [1] in conjunction with a Radon-Nikodym theorem for Banach space-valued measures. Subsequent work by Maynard [2] and by Davis and Phelps [3] has shown that those Banach spaces in which Rieffels Radon-Nikodym theorem is valid are precisely the ones in which every bounded closed convex set is dentable. This is a real breakthrough in studying the nature of Radon-Nikodym as a geometric property. In 1988, Bor-Luh Lin, Pei-Kee Lin, and S. L. Troyanski [4] described the characteristic of denting points and obtained the notion that there is a close relationship between denting points and strongly extreme points. It is easy to see that every denting point of K is a strongly extreme point of K and it is known [Ken Kunen and Haskeil Rosenthal, Martingale proofs of some geometric results in Banach space theory, Pacific J. Math. 100 (1982), 153-175] that every strongly extreme point of K is a weak-extreme point of K. Orlicz space is a special kind of Banach space; it was introduced by the famous Polish mathematician W. Orlicz in 1932. The theory of Orlicz space [5, 6] has been greatly developed because of its important theoretical properties and application value. Up to now, the criterion of an element in the unit sphere of Orlicz spaces equipped with the Orlicz norm [5, 7], the Luxemburg norm [5], and p-Amemiya norm [8] has been given. In this paper, we use a new technique to study the strongly extreme point in Orlicz spaces generated by Orlicz function and equipped with a new norm, namely, s-norm. The criterion of strongly extreme points is given, and the sufficient and necessary conditions for middle point locally uniformly convex in Orlicz spaces equipped with s-norm are obtained.

2. Preliminaries

Throughout this paper, will denote a Banach space and stands for the dual space of . We denote by the nonatomic -measure finite space. By and we denote the unit ball and the unit sphere of , respectively. By and we will denote the sets of real and natural numbers, respectively.

A mapping is said to be an Orlicz function if it is even, continuous, convex, and , . Moreover, if satisfies and , is called -function. Let be the right-hand derivative of , where the function is defined by the formula and called complementary function to in the sense of Young.

We say that an Orlicz function satisfies -condition for large   ( for short) if there exist and such that whenever .

Let denote the set of all measure real functions on . For a given Orlicz function we define on a convex function (called a pseudomodular; see [6]) by It is clear that ; here

The Orlicz space generated by an Orlicz function is defined by the formula and its subspace is defined by

This space is usually equipped with the Luxemburg norm [5] or with the Orlicz norm (Amemiya norm) [5]

A function will be called an outer function, if it is convex and

In 2017, M.Wisła introduced -norm.

Definition 1. Let be an outer function. Denote -norm on Orlicz spaces by the formulaIt is easy to get if and if ([8]). Then we have .

In this paper, by we will denote an Orlicz space equipped with the -norm.

Definition 2. Let be the right-hand derivative of . For all , define

Definition 3. Let be an outer function. For all and , the function is nonincreasing. For any , define ([9]) Let . Then if and only if .

Definition 4. A point is said to be an extreme point of if for any with , then implies .
The set of all extreme points of the unit ball will be denoted by . is said to be strictly convex if .

Definition 5. A point is said to be a strongly extreme point of if for any with , and there holds as .
It is obvious that a strongly extreme point is an extreme point. is said a middle point locally uniformly convex Banach space if and only if each point on is a strongly extreme point.

Definition 6. Let . If for every such that and , we have , then is called to be a strictly convex point of . The set of all strictly convex points of will be denoted by .
For the results concerning strongly extreme points and convexities in Orlicz spaces which are generated by -function and equipped with the Orlicz norm, the Luxemburg norm, and p-Amemiya norm, we refer a reader to [10–17].

3. Main Theorem

Lemma 7. (1) If then for any ;
(2) If and then .

Proof. (1) Suppose . We have . Since for any , , then So , whence .
(2) By , we have , and then and Therefore .

Corollary 8. if and only if for any .

Proof. Necessity. We know that as . By , we can get . That is, . By , we have . Then for all . Since is nonincreasing, we have whence .
Here we infer that . If we have , a contradiction.
Sufficiency. By the definitions of and , and for any . Therefore for all whence , i.e., .

Theorem 9. Suppose that when and is an Orlicz function. A point is a strongly extreme point if and only if and for .

Proof. Necessity. As we know that a strongly extreme point is an extreme point, we only need to prove that implies for . Firstly, we will prove that if , then . If , we will have which implies that holds. There exists such that . Put and . Divide into two sets and with and . Take and define Then , . Moreover , . We have Similarly, we can get .
Next we will show that .
Suppose that for . There exists an interval such that , and is affine on : . Divide into two sets and with and . Define Then , . Thus whence . In the same way, we can prove . This contradicts the fact that is an extreme point of .
In order to complete this proof, we need to prove that if there are no strongly extreme points in .
Suppose . Then .
In fact, if , there exists such that holds for every . Then we have ; it implies , a contradiction.
For any , there exists such that Since , we can find such that . By , there exist and such that . We may assume that . Take with for any , satisfying . Define Then , , here , .
Notice that We have , that is, . And Consequently, . Hence . In the same way, we have . But , which implies , a contradiction.
Sufficiency. Let , with for . Take any with and .
Take sequences and of positive numbers such that Define then and .
Now we will prove that and . Otherwise, we can assume that ; there exist and such that, for every , Then a contradiction.
Since if and only if , we will consider the sequences and , where and in place of and .
Put and . Then and are bounded. Since , we have whence it follows that Analogously, Put . Assume that and as . Now we prove . Since then and if , consequently, as , a contradiction. Thus . Similarly, . Then we have .
Step 1. We will show that . In fact whence as .
Step 2. We will show that .
Firstly, we will prove that . Otherwise, there exist such that . Let and . Let . It can be easy to calculate that . In fact, since , is bounded in norm. Without loss of generality, we may assume that ; then whence . We have . Hence We know that is a close set. Let is a bounded close set. For every , the continuous function is Set maximum value equal to . For every , we have Since , we have for . Therefore, , i.e., for , and Hence Notice that Since , there holds Thus Similarly, we can get . Then as . By when , we have . The contradiction shows that .
Since s-norm is equivalent with the Luxemburg norm, their weak topology and weak star topology are all equivalent. Then is compact. Take , such that and . We can get . By where represents the unit ball of the dual space of , we have and For any , there exists such that Put Then and By the definition of “”, for any , there exists such that whenever .
Thus By arbitrariness of and combining with the above proof, we can obtain Therefore This shows .
As we know ; then . It implies that . Combining with the proof above and , we have . As a result, . So . We have as . Namely, By the proof above, we get ; thus Step 3. We will show that . In fact so . By the fact that and , now is strictly monotonous on . Hence, we have

Corollary 10. Let with and be an Orlicz function. is middle point locally uniformly convex if and only if and is strictly convex.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

Yunan Cui is supported by NSF of China (Grant no. 11871181).