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`Abstract and Applied AnalysisVolume 2013 (2013), Article ID 371806, 7 pageshttp://dx.doi.org/10.1155/2013/371806`
Research Article

## An Extension of Modular Sequence Spaces

1Department of Mathematics, Gauhati University, Guwahati, Assam 781014, India
2Department of Mathematics, Faculty of Science, Taibah University, P.O. Box 30097, Almadinah Almunawwarah 41477, Saudi Arabia

Received 1 April 2013; Accepted 2 June 2013

Copyright © 2013 Hemen Dutta and Iqbal H. Jebril. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

The main aim of this paper is to present an extension of the modular sequence spaces by means of Cesàro mean of order one, to investigate several relevant algebraic and topological properties, and derive some other spaces in the sequel.

#### 1. Introduction

Throughout the paper will represent the spaces of all valued sequences spaces, where is a seminormed space, seminormed by . For , the space of complex numbers, these represent the corresponding scalar valued sequence spaces. The zero sequence is denoted by , where is the zero element of .

An Orlicz function is a function , which is continuous, nondecreasing, and convex with ,  , for and , as .

If convexity of Orlicz function is replaced by , then this function is called a modulus function introduced by Nakano [1].

Lindenstrauss and Tzafriri [2] used the idea of Orlicz function to construct sequence space

The space becomes a Banach space, with the norm which is called an Orlicz space. The space is closely related to the space which is an Orlicz sequence space with for .

Another generalization of Orlicz sequence spaces is due to Woo [3]. Let be a sequence of Orlicz functions. Define the vector space by and equip this space with the norm

Then becomes a Banach space and is called a modular sequence space. The space also generalizes the concept of modular sequence space introduced earlier by Nakano [4], who considered the space when , where for .

An Orlicz function is said to satisfy the -condition for all values of , if there exists a constant , such that . The -condition is equivalent to the satisfaction of inequality for all values of and for (see [5]).

The previous Δ2-condition implies , for all .

Bektaş and Altin [6], Parasar and Choudhary [7], Mursaleen et al. [8], Dutta and Başar [9], Dutta and Bilgin [10], Karakaya and Dutta [11], Tripathy and Dutta [12], Jebril [13], and many others have studied different summable spaces and other sequence spaces using Orlicz functions.

A -space (introduced by Zeller [14]) is a Banach space of complex sequences in which the coordinate maps are continuous; that is, , whenever as , where , for all and .

Let denote the set of all complex sequences which have only a finite number of nonzero coordinates, and let denote a -space of sequences which contains . An element of will be called sectionally convergent if where , where , for .

will be called -space if and only if each of its elements is sectionally convergent.

Let be a sequence of Orlicz functions, let be a seminormed space with seminorm , let be sequence of positive real numbers, and let be the Cesàro matrix of order one with if and , otherwise. Then for nonnegative real numbers we define

The following inequality will be used throughout the paper. Let be a positive sequence of real numbers with , . Then for all for all , we have

#### 2. Main Results

In this section we give the theorems that characterize the structure of the class of sequences and some other spaces which can be derived from this space.

Theorem 1. Let be bounded sequence of positive reals; then is a linear space over the field of complex numbers.

Proof. Let and . Then there exist some and such that We consider . Since each is non-decreasing and convex, and since is a seminorm, This completes the proof.

Theorem 2. is a paranormed space (need not total paranorm) space with paranorm , defined as follows where .

Proof. Clearly . Since , for all we get for .
Now let , and let us choose and such that Let . Then we have Hence .
Finally let be a given non-zero scalar; then the continuity of the scalar multiplication follows from the following equality This completes the proof.

The proof of the following theorem is easy, so omitted.

Theorem 3. Let and be sequences of Orlicz functions. For any two sequences and of bounded positive real numbers and for any two seminorms and one has(i)if is stronger than , then ,(ii),(iii), (iv), (v)If , then .

Theorem 4. Let and be sequences of Orlicz functions which satisfy -condition and , then

Proof. Let and . We choose such that each for . We write and consider where the first summation is over and the second is over . Now we have For , we use the fact that Since each is non-decreasing and convex, it follows that Since each satisfies -condition, we have Hence
Thus Hence .
This completes the proof.

Taking    and in , in Theorem 4, we get the next corollary.

Corollary 5. Let be any sequence of Orlicz functions which satisfy -condition and , then

We will write for non-negative functions and whenever for some .

Theorem 6. Let and be sequences of Orlicz functions. If for each , then .

Proof. The proof is obvious.

Theorem 7. Let be a sequence of Orlicz functions. If and , for each , then .

Proof. If the given conditions are satisfied, we have for each , and the proof follows from Theorem 6.

If we take , the sequence space reduces to the following sequence space:

Theorem 8. Let be bounded sequence of positive reals, and let be a complete seminormed space, then is a complete paranormed space paranormed by , defined by where .

Proof. Let be a Cauchy sequence in . Let be fixed, and let be such that for a given and . Then there exists a positive integer such that Hence we have It follows that For with , we have
Since is non-decreasing for each , we have Hence it follows that is a Cauchy sequence in for each . But is complete, and so is convergent in for each .
Let exists for each .
Now we have for all , Then we have Using the continuity of Orlicz functions, we have This implies It follows that .
Since and is a linear space, so we have .
This completes the proof.

If we take and , constant the sequence space reduces to the following sequence space:

Theorem 9. Let be a complete normed space; then is a Banach space normed by , defined by

Proof. We prove that is a norm on . The completeness part can be proved using similar arguments as applied to prove the previous theorem.
If , then it is obvious that . Conversely assume . Then using the definition of norm, we have This implies that for a given , there exists some such that It follows that Thus Suppose , for some . Let then .
It follows that as for some . This is a contradiction.
Therefore .
It follows that for all . Hence .
Again proof of the properties and for any scalar are similar to that of Theorem 2.

It is easy to see that implies that for each . Hence we have the following proposition.

Proposition 10. The space is a BK-space.

Now we study the AK-characteristic of the space . Before that we give a new definition and prove some results which will be required.

Definition 11. For any sequence of Orlicz functions , we define Clearly is a subspace of . The topology of is the one it inherits from .

Proposition 12. Let be a sequence of Orlicz functions which satisfy -condition. Then

Proof. It is enough to prove that .
Let , then for some , Choose an arbitrary . If then Let now and put .
Since each satisfies the -condition, there exist constants such that Let . Then for every This completes the proof.

Proposition 13. Let be a complete normed space, then is an AK-space.

Proof. Let . Then for each , we can find an such that Hence for , Thus we can conclude that is an AK space.

Combining Proposition 10 and Proposition 12, we have the following theorem.

Theorem 14. Let be a sequence of Orlicz functions which satisfy -condition, then is an AK-space.

Proposition 15. The space is a closed subspace of .

Proof. Let be a sequence in such that , where .
To complete the proof we need to show that ; that is, To there corresponds an such that . Then using convexity of each , Now from Theorem 9, using the definition of norm , we have It follows that Thus .

Hence we have the following corollary.

Corollary 16. The space is a BK-space.

#### Acknowledgment

This research is supported by Deanship of Scientific Research, Taibah University, Almadinah Almunawwarah, Saudi Arabia.

#### References

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