Journal of Function Spaces

Journal of Function Spaces / 2020 / Article

Research Article | Open Access

Volume 2020 |Article ID 8512636 | https://doi.org/10.1155/2020/8512636

Xinran Bai, Yunan Cui, Joanna Kończak, "Monotonicities in Orlicz Spaces Equipped with Mazur-Orlicz -Norm", Journal of Function Spaces, vol. 2020, Article ID 8512636, 7 pages, 2020. https://doi.org/10.1155/2020/8512636

Monotonicities in Orlicz Spaces Equipped with Mazur-Orlicz -Norm

Academic Editor: Gen Q. Xu
Received25 Dec 2019
Revised29 Apr 2020
Accepted04 May 2020
Published31 May 2020

Abstract

Some basic properties in Orlicz spaces and Orlicz sequence spaces that are generated by monotone function equipped with the Mazur-Orlicz -norm are studied in this paper. We give some relationships between the modulus and the Mazur-Orlicz -norm. We obtain an interesting result that the norm of an element in line segments is formed by two elements on the unit sphere less than or equal to 1 if and only if that the monotone function is a convex function. The criterion that Orlicz spaces and Orlicz sequence spaces that are generated by monotone function equipped with the Mazur-Orlicz -norm are strictly monotone or lower locally uniform monotone is presented.

1. Introduction

Geometry of Banach space has important applications in the control theory, fixed-point theory, ergodic theory, probability theory, and vector analytic function theory. Recall that the monotonicity properties are restrictions of appropriate rotundity properties to the set of couples of comparable elements in the positive cone of a Banach lattice (see [15]). Consequently, then in many cases, good convex properties can be replaced successfully by respective monotonicity properties (see [68]). It is well known that monotonicity properties (strict and uniform monotonicity) play an analogous role in the best dominated approximation problems in Banach lattices as the respective rotundity properties (strict and uniform rotundity) in the best approximation problems in Banach spaces (see [9, 10]). Moreover, monotonicity properties are applicable in the ergodic theory, since they provide a tool for estimating a norm (see [1113]). In this paper, based on the paper in 2019 by Cui et al., after removing the continuous condition of generating function (see [6]), we found some conclusions completely different from those in the Banach spaces we studied in the past, which provided us with a new understanding and enlightenment for studying geometric properties of seminormed linear spaces in the future.

2. General Auxiliary and New Results

Definition 1. Letbe a-finite measure space andbe the space of all (equivalence) classes of-measurable real-valued funcions defined on. A functionis called an Orlicz function iffor all , and evenis the nondecreasing function and .Any Orlicz functiondetermines a functionaldefined by the formulaand called the modular. The order idealin is called an Orlicz space.
Let us define a lattice ideal in as The space is linear with respect to the following lattice -norm, called the Mazur-Orlicz -norm (see [6]): Note that if is a convex function, then , endowed with the equivalent lattice norm (called the Luxemburg norm) , becomes a Banach lattice. If is right continuous and is a Banach space, increasing function on such that . Define The properties of yield that if for , we put then satisfies the axioms of a -norm in (see [14]). Moreover, is equivalent to an -norm under which is complete, whence is an -space.
In the case of the counting measure space, the space is always nontrivial and

Remark 2. Ifis not right continuous at zero, then. If we take, thenis convergent to 0 uniformly, butfor all. This means that the case. We do not even have the fact that the uniform convergence of the sequenceimplies its modular convergence.

Lemma 3. is a linear space.

Proof. Obviously, if is a monotone increasing function, then for all and we have Take any , by the definition of the ; there exists such that We choose , that is, Assume now that or , and we conclude that Hence, is a linear space.

Definition 4 (see [6]). An-normed Orlicz spaceis said to be strictly monotone (we write) if for anysuch that, we havewhenever(or equivalentlywhennever).

Definition 5 (see [6]). An -normed Orlicz spaceis said to be lower locally uniformly monotone (we write) if for anyandinsuch thatfor allandas, the conditionasholds.

Definition 6. We say thatsatisfies-condition (for short) whenever there are constantsand such that (see [15])

Definition 7. We say thatsatisfies-condition (for short) whenever there are constantsandsuch that

Lemma 8. if and only iffor all.

Proof. Sufficiency: given that , choose an integer such that . Then, by (12), we have Necessity: now choose an integer such that . Then, by (11) we get which ends the proof.

Definition 9. We say thatsatisfies-condition (for short) whenever there are constantsandsuch that

Theorem 10. Ifis an Orlicz function such that, then for any, there existssuch thatfor all .

Proof. By , there exists and such that (). Let us assume that and for any and , we get We choose . We can hence easily get that for any . By , for any , there exists such that , whenever . Defining , we obtain

Theorem 11. For any,if and only ifis right continuous and.

Proof. Sufficiency: for any the function is right continuous on the interval . By the definition of , we have and for any , we have . Using and the Lebesgue dominated convergence theorem, we obtain which finishes the proof of the implication .
Necessity: assume that does not satisfy the -condition. This yields that there exists a strictly increasing sequence of position numbers such that and Next, choose a sequence of pairwise disjoint sets in such that . Defining , we have On the other hand, we have that for any , there exists such that for any . Hence, Consequently, this shows that is a necessary condition.
Next, we will prove the necessary of the right-continuous function of . If is not right continuous on , then there exists a point such that ; that is, for any , we get . Take any with . Let us take large enough. Since , we have Since is nonatomic, we can find such that Let us define Then, , whence we obtain that and . On the other hand, for any , we have which gives that yields , a contradiction, which finishes the proof of necessity.

For convenience, from now on, we write .which contradicts the assumption . Therefore, .

Theorem 12. Assume thatand. Then,if and only ifis a convex function on.

Proof. Sufficiency: for any , we have and . Assume that . Then, . Hence, . This finishes the proof.
Necessity: if is not a convex function, then there exists , such that . Let us choose such that and . Take any such that Define We obtain Consequently, is a contradiction.

Example 13. PutThen, satisfies the -condition and is right continuous. Define Then, and and . Moreover, . Now let us define Then, . This means that .

Remark 13. This example shows that the norm of an element in a line segmentconsisting of two elementsandin a unit sphere may be greater than 1. This is essentially different from Banach space. For Banach space ,for anyand for any ,we get that.Using Theorem 12, ifis a convex function, we conclude that if, , then.

Theorem 14. The Orlicz spaceequipped with the Mazur-Orlicz-normis strictly monotone if and only ifis strictly increasing onandsatisfies the-condition andis right continuous.

Proof. Sufficiency: let , , then there exists and such that . Assuming for the contrary that , we have by Theorem 11,

Necessity: if is not right continuous on , then there exists a point such that ; that is,for any , we get . Take any with . Let us take large enough. Since , we obtain

Since is nonatomic, we can find such that

Moreover, take , where is a fixed positive number such that

Let us define

Then, , whence we obtain that and . On the other hand, we have that for any , which gives that yields . By (36), we obtain . Since and , this means that the space is not strictly monotone.

We will prove the necessity of the -condition. Assume first that does not satisfy the -condition. This yields that there exists a strictly increasing sequence of positive numbers such that and . Let us choose first such that . We have . Since is nonatomic, we can find such that . Continuing this procedure by induction, we can find a sequence of measurable sets such that and , for any , . Let us define

Then, , and , whence we obtain that and . On the other hand, we have that for any , there exists , , such that for any , whence which gives that and by the arbitrariness of , we get . The inequalities , , and yield . This equality shows that space is not strictly monotone.

Now, we will prove the necessity of strict monotonicity of on . Assume for the contrary that is constant on some interval , where , so we have . Take such that , and find a positive number such that ; using the measure space which is nonatomic, there exists such that

Define

We obviously have , , which yields ; this means that is not strictly monotone.

Theorem 15. is lower locally uniformly monotone if and only ifis strictly monotone and a right continuous function andsatisfies the-condition.

Proof. In order to prove the sufficiency of the theorem, assume that and as . We need to prove that as .

First, we will prove that in measure. Assuming that in measure, without loss of generality, then there exists , such that . Define and for each . It is clear that hold. Using condition , we know that and then, there exists such that

Therefore, by conditions (43) and (44), we get that

Next, we will prove that there exists such that

It is obvious that , since is strictly increasing. If (46) does not hold, then , such that

By the compactness theorem, assume that . If , since is right continuous, we get , a contradiction. If , since , we have

Moreover, ; we have known that there exists such that , which gives . This shows that a contradiction.

Put for each . From condition (44) there exists a such that . Put for each . Then,

Therefore,

Without loss of generality, we may assume that for each . Put . Applying condition (46), there exists a such that when .

Using and as , we have thanks to that is right continuous. Then, there exists a constant such that for all and . Since , we have . By the Lebesgue dominated convergence theorem, we obtain

Consequently, which contradicts with the assumption . Therefore, is convergent to in measure.

Hence, . By the fact that there is absolute continuity of integral, for any , there exists such that for every measureable set with .

Consequently, this implies . By the Egorov theorem, for the above , there exists such that and

Since , we get

Consequently,

By the arbitrariness of , we obtain that as holds.

In virtue of Theorem 14, we have a necessity that is obvious.

Theorem 16. For any, if and only if(i)(ii) is right continuous

Proof. Necessity: (i)If , then there exists an element such that and for any . Hence, and (ii)If there exists a point such that , that is, for any , we get

Put . Then, and for any , we have . Hence, and .

The sufficiency of the proof is similar as Theorem 11.

Theorem 17. The following statements are equivalent:(1)The Orlicz sequence space equipped with the Mazur-Orlicz F-norm is lower locally uniformly monotone(2)The Orlicz sequence space equipped with the Mazur-Orlicz F-norm is strictly monotone(3)(a) is strictly increasing on (b)
(c) is right continuous

Proof. The implication (1)(2) is clear.
(2)(3) The proof is similar as the proof of Theorem 15. We only give the proof that is right continuous, if there exists a point such that . That is, for any , we get . Put . Then, , Since , there exists a such that . Put . Then, , and