Kadec-Klee Property in Orlicz Function Spaces Equipped with S-Norms

Abstract

Using some new techniques, the necessary and sufficient conditions for Kadec-Klee property of Orlicz function spaces equipped with s-norms are presented. An original method that was used in the process of inquiry and the obtained results also systematically complete and broaden the characterization of Kadec-Klee property of Orlicz spaces.

1. Introduction and Preliminaries

Orlicz spaces, introduced by W. Orlicz in 1932, form a wide class of Banach spaces of measurable functions (in the case of atomless measure) or sequences (in the case of counting measure) (see [1]). On Orlicz spaces of measurable functions, the classical Orlicz and Luxemburg norm can be defined by use of the Amemiya formula: and , respectively, where is an Orlicz function and . Based on these statements, H. Hudzik and L. Maligrada introduced Orlicz spaces equipped with the p-Amemiya norms, where and defined in 2000 (see [2, 3]). M. Wisła presented a universal and general method of introducing norms (s-norms) in Orlicz spaces in 2019, and the introduction of Orlicz spaces equipped with s-norms covers all the cases mentioned above.

1.1. Introduction

In the following, by we will denote the sets of natural numbers and and the sets of real and nonnegative real numbers, respectively. By and , we will denote the unit sphere and the unit ball of the Banach space , respectively.

Let be a finite nonatomic measure space, and be the set of all (-equivalence) classes of -measurable real functions defined on .

Definition 1. A continuous function is called convex iffor all . If, in addition, the two sides of formula (1) are not equal for all , then we say is strictly convex.

Definition 2. A function : goes by name of an Orlicz function if is nonnegative, even, convex satisfying and .
for all is called the complementary function of in the sense of Young. Obviously, is also an Orlicz function.

Definition 3. is introduced as a modular of byfor all .
Then, the Orlicz space and its subspace are generated by an Orlicz function and are linear spaces of measurable functions defined by the following formulas:For each , let Luxemburg norm be defined byand Orlicz norm by

Definition 4 (see [4]). A function will be called an outer function, if it is convex and

Definition 5 (see [4]). Let be an outer function and be an Orlicz function. Then, the functionis a norm that will be called s-norm on the Orlicz space .
Combing the definitions of various norms, the inequalityholds. Denote

Definition 6 (see [4]). is an outer function that is conjugate to s in the Hölder sense.
For any and the following Hölder inequality holds.

Definition 7 (see [4]). For an outer function and its right-hand derivative , definewhenever .
For all , , defineand for all ,

Lemma 1 (see [4]). For any , if and only if , where for short, we have

Definition 8. We say that an Orlicz function satisfies condition (, for short) if there exists and , such that

Definition 9. Let be a Banach space. If , , and imply , then we say that has the Kadec-Klee property (see [5–7]).

Lemma 2. For any has a unique decomposition,where F is the set of all singular functionals on .

Lemma 3 (see [8]). If is strictly convex, then for any and , there exists , such thatwhenever .

Lemma 4 (see [8]). A subset is weakly compact (i.e., there exists a subsequence and , such that for each ) if

Lemma 5 (see [8]). For any , assume that and ; if in addition, , then

2. Main Results

Theorem 1. Orlicz space has the Kadec-Klee property if and only if(1)(2) is strictly convex on R

Proof. Necessity. (i) If , there exists , such that , where and as .
For any , putUsing , we haveHence, for each , there exist satisfying . Define . Then,Let us prove that . For any , we have , thanks to andUsing and , the qualityholds.
Put . Then, , andthat is . But,this contract with that has the Kadec-Klee property.
(ii) If is not strictly convex, there exists , such that is not an extreme point, i.e., there exists , such that . Therefore, there exist , such thatfor some when .
PutThen, . Without loss of generality, we may assume that for each .
Therefore, there exists a , such that . Let . Then, . Put and . Then, we have for any and . Hence,i.e., . Then, we have that when and when . By formula (27), we haveDivide into two disjoint sets and , such thatand divide and into two disjoint sets , and , , respectively, such thatBy induction as above, we obtain a sequence , such thatfor every and . DefineSince, for , the following equalities hold:whence .
Let ; we get that , without loss of generality, using Lemma 4, we may assume that for some ; using , we have that holds. Put . Then, .
Let be a support functional of . Denote and . We get the following:whence .
On the other hand, take , satisfying . Then,Hence, we have that holds.
Since , , and , so .
Furthermore,It follows that is not a Cauchy sequence, a contradiction.
Sufficiency. Assume that s is an outer function, and Orlicz function is strictly convex satisfying the condition. Take satisfying and .
For any , let and for some with . By the definition of , we haveandSince , we obtain and , i.e., . Then, there exist , such that First, we will prove that in measure. Suppose that as , there exist , such that for any .(i)Assume that is bounded, and we have . Hence, Using , we have Then, there exists , such that . Set Then, . In virtue of Lemma 3, there exists , such that Whenever . Combining with the definition of the functions s and as well as with formulas (42) and (45), we obtain A contradiction. Consequently, as .(ii)If the sequence is not bounded and without loss of generality, we may assume that and set ; then, , , and the following inequalities hold. It is clear that the sequence is bounded, and Similar as the proof above, we have . Second, we will show that . We only need to prove that converges weakly star. For any and , using that has absolutely continuous norm, there exists , , and , such that By Yegolov’s theorem, there exists with for which uniformly . Hence, there is satisfying Whenever and . So, i.e., . By , we can get that holds. Finally, by formula (41), we have and . Hence, we get that holds. In virtue of Lemma 5, we obtain .

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that there are no conflicts of interest.

Acknowledgments

This work was supported by the National Natural Science Foundation of China (11871181) and the Natural Science Foundation of Heilongjiang Province (A2018006).