#### 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).