Research Article | Open Access
New Difference Sequence Spaces Defined by Musielak-Orlicz Function
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 . It was further generalized by Et and Çolak  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  taking , , and .
Dutta  introduced the following difference sequence spaces using a new difference operator: where for all .
In , 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 [6–14] and references therein. Başar and Altay  introduced the generalized difference matrix for all , which is a generalization of -difference operator by
Başarir and Kayikçi  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  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  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.  as follows: The concept of 2-normed spaces was initially developed by Gähler  in the mid of 1960’s, while that of -normed spaces one can see in Misiak . Since then, many others have studied this concept and obtained various results; see Gunawan [23, 24] and Gunawan and Mashadi . For more details about sequence spaces see [26–33] 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.
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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