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Inclusions between weighted Orlicz spaces
Journal of Inequalities and Applications volume 2014, Article number: 390 (2014)
Abstract
Let Φ be a Young function and w be a weight. The weighted Orlicz space is a natural generalization of the weighted Lebesgue space () and a characterization of an inclusion between weighted Lebesgue spaces is well known. In this study, we will investigate the inclusions between weighted Orlicz spaces and with respect to Young functions , and weights , . Also, we give necessary and sufficient conditions for the equality of these two weighted Orlicz spaces under some conditions. Thereby, we obtain our result.
MSC:46E30.
1 Introduction and preliminaries
Generally Orlicz spaces are a natural generalization of the classical Lebesgue spaces , and there are many studies of Orlicz spaces in the literature (for example [1, 2]). In [3], the inclusion between spaces is investigated with respect to the measure space and in [4] inclusions between Orlicz spaces are examined for a finite measure space and in general [5, 6]. Also, inclusions between weighted spaces with respect to weights are studied in [7, 8] for a locally compact group with Haar measure. In this paper, we will investigate the inclusion between weighted Orlicz spaces with respect to a Young function Φ and a weight w for a general measure space. To this aim we will give the definition of a weighted Orlicz norm which depends on the usual Orlicz norm and we show that the inclusion map between the weighted Orlicz spaces is continuous. Also, we obtain the result that two weighted Orlicz spaces can be comparable with respect to Young functions for any measure space, although the weighted spaces are not comparable with respect to the numbers p. Moreover, in the case of we generalize some results in [7] to the weighted Orlicz spaces and we establish necessary and sufficient conditions on the weights and in order that .
A non-zero function is called a Young function if Φ is convex and satisfies the conditions and . We say that a Young function Φ satisfies the condition if there exists a such that for all . Also, for a Young function Φ, the complementary Young function Ψ of Φ is given by
for . If Ψ is the complementary function of Φ, then Φ is the complementary of Ψ and is called a complementary pair of Young functions. We have the Young inequality for the complementary functions Φ and Ψ,
Let be a measure space. We will assume that μ is a σ-finite measure. Given a Young function Φ, the Orlicz space or simply is defined by
where f shows μ-equivalence classes of measurable functions. Then the Orlicz space is a Banach space under the (Orlicz) norm defined for by
where Ψ is the complementary Young function of Φ.
For further information as regards Orlicz spaces, the reader is referred to [4–6].
Remark 1.1 By using the Young inequality and the definition of the norm , it is easy to see that a measurable function is in if and only if .
Now, let be a measure space and let Φ be a Young function. If w is a weight on X (i.e. is measurable function) then the weighted Orlicz spaces, denoted by , are defined as follows:
If we define
for all , then the function defines a norm on , and it is called a weighted Orlicz norm.
For the norm reduces to the usual Orlicz norm and we obtain the Orlicz space .
Let . Then, for the Young functions , the space becomes the weighted Lebesgue space and the norm is equivalent to the classical norm in . In particular, if then the complementary Young function of is
and in this case for all , since is true if and only if almost everywhere on X.
Also, if then, for the Young function Φ given by (1), the space is equal to the space . We have for all .
It can be shown that the weighted Orlicz space is also a Banach space by using the completeness of the usual Orlicz space.
Proposition 1.2 is a Banach space.
Proof To show that is a Banach space, take an arbitrary absolutely convergent series in . Then
Thus, is absolutely convergent in the Orlicz space . Since the Orlicz space is a Banach space, there exists a function such that , . If we set then and
where . So the space becomes a Banach space. □
2 Main result
Let be a measure space, and be two weights on X and let , be two Young functions. We will investigate the inclusion between the weighted Orlicz spaces and . For this investigation we need some definitions.
Let and be two weights on X. If there exists a such that
for all , then we write . If and then we say that and are equivalent and write . For example, and are weights on ℝ and it is clear that for .
Let and be two Young functions. Then we say that is stronger than , in symbols, if there exist a and (depending on c) such that for all . If then we write (). If and then we write . The same notation is valid for the case () [6].
Remark 2.1 It is clear that () implies . But is not sufficient to investigate the inclusion between weighted Orlicz spaces when the measure μ is not finite. So we need the condition () for infinite measures. Also, if () then () for the complementary Young functions and of and , respectively.
Example 2.2 , , and , , are Young functions and satisfy the inequality
for all . Thus, () for .
Now we can give the following theorem for the inclusion between the weighted Orlicz spaces and .
Theorem 2.3 If () and , then .
Proof Suppose that and . Let . Then there exists such that
On the other hand, since and () there exist numbers and such that
and
If we set then from (2) we get
for all , since the Young function is increasing. Then we obtain
for all from (3). So,
Thus . □
Remark 2.4 The converse of Theorem 2.3 is not true in general. The following example shows this.
Example 2.5 Let be a measure space with and let w be a weight on X. If , then for the Young functions
the weighted Orlicz spaces and become the weighted Lebesgue spaces and , respectively. Here, . But is not stronger than for . Indeed, if we assume that (), then there exists such that for all . This says that for all , and so for all . Then we get a contradiction, if we pass to the limit that x goes to zero, since .
Now, let μ be a finite measure. If we take instead of () in Theorem 2.3, then by using similar techniques as in [[5], Theorem 3.17.1], we get the following proposition.
Proposition 2.6 Let . If and ; then .
By considering Theorem 2.3 we derive the following corollary.
Corollary 2.7 If () (or when ) and then .
Before we investigate the inclusions between the weighted Orlicz spaces with respect to the Young function and weight, respectively, we will show that, if then the inclusion map is continuous.
Proposition 2.8 Let be a measure space. If , are Young functions and , are weights, then if and only if there exists a such that for all .
Proof For the sufficiency part of the proof assume that the condition given in the theorem is true. Conversely, suppose that is true. Let . Then and . Then we can define
for all , and hence becomes a normed space.
Moreover, the norm convergence of in the Orlicz space implies the convergence almost everywhere of some subsequence on X [6]. So, a similar assertion holds for the weighted Orlicz space since a weight w is strictly positive. Then, by using similar techniques as in the Lebesgue space case, it can be shown that the normed space is a Banach space.
If we consider the mapping , for all . Then from the closed graph theorem there exists a such that
for all . □
We can summarize the above results in the following corollary.
Corollary 2.9 Let be a measure space, let , be two Young functions, and let , be two weights. If we have the conditions
-
(i)
() and ,
-
(ii)
,
-
(iii)
there exists such that for all ,
then we have (i) ⇒ (ii) ⇔ (iii).
Remark 2.10 Note that a similar result can be shown if we change () by for a finite measure space.
Now, let w be a weight on X and fix . We will investigate the inclusions between the weighted Orlicz spaces and with respect to Young functions and .
Since a weight w satisfies , we get the following consequences of Theorem 2.3.
Corollary 2.11 Let and be two Young functions and let w be a weight. If (), then .
Remark 2.12 Corollary 2.11 shows that the weighted Orlicz spaces and can be comparable with respect to Young functions and , although the weighted Lebesgue spaces and cannot be comparable with respect to in general (for instance with Lebesgue measure μ).
Remark 2.13 The converse of the Corollary 2.11 is not true in general (see Example 2.5).
On the other hand, if we consider the finite measure space then by using a similar method as in [5] we can derive the following corollary.
Corollary 2.14 Let and w be a fixed weight on X. If , are Young functions, then if and only if .
Remark 2.15 If we combine this corollary with the result mentioned in Remark 2.10 then we get the following.
Corollary 2.16 Let , be two Young functions. If w is a weight then the following statements are equivalent for a finite measure space.
-
(i)
.
-
(ii)
.
-
(iii)
There exists such that for all .
Let be two numbers. If we take , in Corollary 2.16, we get the following well-known result.
Corollary 2.17 Let . The following statements are equivalent for a finite measure space.
-
(i)
.
-
(ii)
.
-
(iii)
There exists such that for all .
Remark 2.18 If we combine this corollary with the result mentioned in Remark 2.10 then we get the following.
Now, let Φ be a Young function and let , be two weights. Fixing , we will study the inclusion between the weighted Orlicz spaces and with respect to the weights and .
We get the following corollaries of Theorem 2.3, since .
Corollary 2.19 If , then .
Corollary 2.20 If then .
We will show that the converse of the Corollary 2.19 is true when and μ is the Lebesgue measure on . To do this, we will assume
for the weight w, then by using similar techniques in [7, 8], it is easy to see that the weighted Orlicz spaces have the following properties.
Lemma 2.21 Let w be a weight on satisfying the condition (4). If Φ is a continuous Young function satisfying the condition, then:
-
(i)
For all and for all and .
-
(ii)
If , then the map from to is continuous.
-
(iii)
If and , then there exists a (depends on f) such that
Now, we can give the necessary and sufficient conditions for the inclusion between the weighted Orlicz spaces and .
Theorem 2.22 Let Φ be a continuous Young function satisfying condition and , be two weights on satisfying the condition (4). Then if and only if .
Proof If , then it is clear that from Corollary 2.19. Conversely, assume that . Then, by Proposition 2.8, there exists a such that for all . If we fix then from Lemma 2.21(iii), there exist such that
and
for all . So, we get the inequality
for all . This shows that for all where . Thus . □
The following is an easy consequence of Theorem 2.22.
Corollary 2.23 if and only if .
Also, if we combine Theorem 2.22 and Corollary 2.19 we obtain the following corollary.
Corollary 2.24 Let Φ be a continuous Young function satisfying the condition. If and are weights satisfying condition (4), then the following statements are equivalent.
-
(i)
.
-
(ii)
.
-
(iii)
There exists a such that for all .
Let . If we take in Corollary 2.24, then Φ satisfies the condition so we get the following well-known result.
Corollary 2.25 Let and , be two weights satisfying condition (4), then the following statements are equivalent.
-
(i)
.
-
(ii)
.
-
(iii)
There exists a such that for all .
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Acknowledgements
The author would like to thank S Öztop for her helpful suggestions. This research is supported by Scientific Research Projects Coordination Unit of Istanbul University with project number 14671.
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Osançlıol, A. Inclusions between weighted Orlicz spaces. J Inequal Appl 2014, 390 (2014). https://doi.org/10.1186/1029-242X-2014-390
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DOI: https://doi.org/10.1186/1029-242X-2014-390