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Research Article | Open Access

Volume 2014 |Article ID 691632 | https://doi.org/10.1155/2014/691632

M. Mursaleen, Sunil K. Sharma, S. A. Mohiuddine, A. Kılıçman, "New Difference Sequence Spaces Defined by Musielak-Orlicz Function", Abstract and Applied Analysis, vol. 2014, Article ID 691632, 9 pages, 2014. https://doi.org/10.1155/2014/691632

# New Difference Sequence Spaces Defined by Musielak-Orlicz Function

Revised11 Jul 2014
Accepted11 Jul 2014
Published22 Jul 2014

#### Abstract

We introduce new sequence spaces by using Musielak-Orlicz function and a generalized -difference operator on -normed space. Some topological properties and inclusion relations are also examined.

#### 1. Introduction and Preliminaries

The notion of the difference sequence space was introduced by Kızmaz [1]. It was further generalized by Et and Çolak [2] as follows: for , and , where is a nonnegative integer and or equivalent to the following binomial representation: These sequence spaces were generalized by Et and Basarir [3] taking , , and .

Dutta [4] introduced the following difference sequence spaces using a new difference operator: where for all .

In [5], Dutta introduced the sequence spaces , , , , and , where and and for all , which is equivalent to the following binomial representation: The difference sequence spaces have been studied by authors [614] and references therein. Başar and Altay [15] introduced the generalized difference matrix for all , which is a generalization of -difference operator by

Başarir and Kayikçi [16] defined the matrix which reduced the difference matrix in case , . The generalized -difference operator is equivalent to the following binomial representation:

Let be a sequence of nonzero scalars. Then, for a sequence space , the multiplier sequence space , associated with the multiplier sequence , is defined as An Orlicz function is a function, , which is continuous, nondecreasing, and convex with , for , and as .

We say that an Orlicz function satsfies the -condition if there exists and such that for all . The -condition is equivalent to for all and for .

Lindenstrauss and Tzafriri [17] used the idea of Orlicz function to define the following sequence space: which is called an Orlicz sequence space. The space is a Banach space with the norm It is shown in [17] that every Orlicz sequence space contains a subspace isomorphic to .

A sequence of Orlicz function is called a Musielak-Orlicz function; see [18, 19]. A sequence defined by is called the complimentary function of a Musielak-Orlicz function . For a given Musielak-Orlicz function , the Musielak-Orlicz sequence space and its subspace are defined as follows: where is a convex modular defined by We consider equipped with the Luxemburg norm or equipped with the Orlicz norm By a lacunary sequence , , where , we will mean an increasing sequence of nonnegative integers . The intervals determined by are denoted by and the ratio will be denoted by . The space of lacunary strongly convergent sequences was defined by Freedman et al. [20] as follows: The concept of 2-normed spaces was initially developed by Gähler [21] in the mid of 1960’s, while that of -normed spaces one can see in Misiak [22]. Since then, many others have studied this concept and obtained various results; see Gunawan [23, 24] and Gunawan and Mashadi [25]. For more details about sequence spaces see [2633] and references therein. Let and be linear space over the field , where is the field of real or complex numbers of dimension , where .

A real valued function on satisfying the following four conditions:(1) if and only if are linearly dependent in ;(2) is invariant under permutation;(3) for any ;(4)is called an -norm on and the pair is called an -normed space over the field . For example, we may take being equipped with the Euclidean -norm the volume of the -dimensional parallelepiped spanned by the vectors which may be given explicitly by the formula where for each and denotes the Euclidean norm. Let be an -normed space of dimension and linearly independent set in . Then the following function on defined by defines an norm on with respect to .

A sequence in an -normed space is said to converge to some if A sequence in a normed space is said to be Cauchy if If every Cauchy sequence in converges to some then is said to be complete with respect to the -norm. Any complete -normed space is said to be -Banach space.

Let be an -normed space and let denote the space of -valued sequences. Let be any bounded sequence of positive real numbers and a Musielak-Orlicz function. We define the following sequence spaces in this paper: when , we get when , for all , we get The following inequality will be used throughout the paper. If , , then for all and . Also for all .

#### 2. Main Results

Theorem 1. Let be a Musielak-Orlicz function and a bounded sequence of positive real numbers; the spaces , , and are linear over the field of complex numbers .

Proof. Let , , and . Then there exist positive real numbers and such that Define . Since is an -norm on and are nondecreasing and convex functions so by using inequality (23) we have Thus, we have . Hence is a linear space. Similarly, we can prove that and are linear spaces. This completes the proof of the theorem.

Theorem 2. Let be a Musielak-Orlicz function and a bounded sequence of positive real numbers; the space is a topological linear space paranormed by where .

Proof. Clearly for . Since , we get . Again, if , then This implies that, for a given , there exist some such that
Thus for each , and suppose that for each . This implies that for each . Let , then which is a contradiction. Therefore, for each and thus for each . Let and be such that Let ; then by using Minkowski’s inequality, we have
Since are nonnegative, we have Therefore, .
Finally, we prove that the scalar multiplication is continuous. Let be any complex number. By definition, Then where . Since , we have So, the fact that scalar multiplication is continuous follows from the above inequality. This completes the proof of theorem.

Theorem 3. Let be a Musielak-Orlicz function. If for all fixed , then .

Proof. Let . Then there exists some positive number such that Define . Since is nondecreasing and convex and by using inequality (23), we have Hence . This completes the proof of the theorem.

Theorem 4. Let be a Musielak-Orlicz function and . Then if and only if

Proof. Let . Suppose (40) does not hold. Therefore there are a subinterval of the set of intervals and a number , where for all , such that Let us define as follows: Thus by (41), . But . Hence (40) must hold.
Conversely, suppose that (40) holds and let . Then, Suppose that . Then for some number , , there is a number such that, for a subinterval of the set of intervals , We have , which contradicts (40) by using (43). Hence we get This completes the proof.

Theorem 5. Let . For any Musielak-Orlicz function which satisfies -condition, one has(i)(ii)(iii).

Proof. (i) Let . Then, we have Let , and choose with such that for . We can write For the first summation above, we can write By using continuity of , for the second summation we can write Since each is nondecreasing and convex and satisfies -condition, it follows that Taking limit as and , it follows that . Hence . Similarly, we can prove (ii) and (iii). This completes the proof of the theorem.

Theorem 6. Let be a Musielak-Orlicz function. Then the following statements are equivalent: (i); (ii); (iii) for all , where .

Proof. (i) (ii) Suppose (i) holds. In order to prove (ii) we have to show that Let . Then for a given there exists such that Hence there exists such that This shows that .
(ii) (iii) Suppose (ii) holds and (iii) fails to hold. Then for some , and, therefore, we can find a subinterval of the set of intervals such that Let us define as follows: Thus . But by (55), which contradicts (ii). Hence (iii) must hold.
(iii) (i) Let (iii) hold. Suppose that . Then for . Let for each , and then by (57) , which contradicts (iii). Hence (i) must hold. This completes the proof of the theorem.

Theorem 7. Let be a Musielak-Orlicz function. Then the following statements are equivalent: (i); (ii); (iii) for all .

Proof. (i) (ii) is obvious
(ii) (iii) Let (ii) hold and let (iii) fail to hold. Then and we can find a subinterval of the set of intervals such that Let us define as follows: Thus by (iii), . But which contradict (ii). Hence (iii) must hold.
(iii) (i) Let (iii) hold. Suppose that . Therefore, Again suppose for some number and a subinterval of the set of intervals , we have Then, from properties of the Orlicz function, we can write Consequently, by (61), we have , which contradicts (iii). Hence (i) must hold. This completes the proof of the theorem.

Theorem 8. (i) If for all , then .
(ii) If , then .

Proof. (i) Let . Since , we get and hence .
(ii) and . Then for each there exists a positive integer such that
This implies that Therefore . This completes the proof of the theorem.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The authors gratefully acknowledge that this research was partially supported by the University Putra Malaysia under the ERGS Grant Scheme having Project no. ERGS 1-2013/5527179. The authors are grateful also to the anonymous referees for a careful checking of the details and for helpful comments that improved the paper.

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