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International Journal of Analysis
Volume 2013 (2013), Article ID 126163, 7 pages
On a New I-Convergent Double-Sequence Space
Department of Mathematics, A.M.U., Aligarh 202002, India
Received 26 November 2012; Revised 17 January 2013; Accepted 18 January 2013
Academic Editor: Wen Xiu Ma
Copyright © 2013 Vakeel A. Khan and Nazneen Khan. 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.
The sequence space was introduced and studied by Mursaleen (1983). In this article we introduce the sequence space 2 and study some of its properties and inclusion relations.
1. Introduction and Preliminaries
Let , , and be the sets of all natural, real, and complex numbers, respectively. We write showing the space of all real or complex sequences.
Definition 1. A double sequence of complex numbers is defined as a function . We denote a double sequence as where the two subscripts run through the sequence of natural numbers independent of each other . A number is called a double limit of a double sequence if for every there exists some such that (see ).
Let and denote the Banach spaces of bounded and convergent sequences, respectively, with norm . Let denote the space of sequences of bounded variation; that is, where is a Banach space normed by (see ).
Definition 2. Let be a mapping of the set of the positive integers into itself having no finite orbits. A continuous linear functional on is said to be an invariant mean or -mean if and only if(i) when the sequence has for all ;(ii), where ;(iii) for all .
In case is the translation mapping , a -mean is often called a Banach limit (see ), and , the set of bounded sequences all of whose invariant means are equal, is the set of almost convergent sequences (see ).
If , then . Then it can be shown that where , . Consider where denote the th iterate of at . The special case of (5) in which was given by Lorentz [5, Theorem 1], and that the general result can be proved in a similar way. It is familiar that a Banach limit extends the limit functional on .
Theorem 3. A -mean extends the limit functional on in the sense that for all if and only if has no finite orbits; that is to say, if and only if, for all , , (see ) Put assuming that . A straight forward calculation shows (see ) that
For any sequence , , and scalar , we have
Definition 4. A sequence is of -bounded variation if and only if (i) converges uniformly in ;(ii), which must exist, should take the same value for all .
We denote by , the space of all sequences of -bounded variation (see ):
Theorem 5. is a Banach space normed by (see ).
Subsequently, invariant means have been studied by Ahmad and Mursaleen , Mursaleen et al. [3, 6, 8, 10–14], Raimi , Schaefer , Savas and Rhoades , Vakeel et al. [18–20], and many others [21–23]. For the first time, I-convergence was studied by Kostyrko et al. . Later on, it was studied by Šalát et al. [25, 26], Tripathy and Hazarika , Ebadullah et al. [18–20, 28], and Vakeel et al. [1, 29].
An Ideal is called nontrivial if . A non-trivial ideal is called admissible if .
A non-trivial ideal is maximal if there cannot exist any non-trivial ideal containing as a subset.
For each ideal , there is a filter corresponding to . That is,
Definition 8 (see ). A double sequence is said to be -null if . In this case, we write
Definition 9. A double sequence is said to be -cauchy if for every there exist numbers , such that
Definition 10. A double sequence is said to be -bounded if there exists such that
Definition 11. A double-sequence space is said to be solid or normal if implies for all sequence of scalars with for all .
Definition 13. Let be a linear space. A function is called a paranorm, if for all ,(i) if ;(ii); (iii); (iv)if is a sequence of scalars with and with , in the sense that , in the sense that .
2. Main Results
In this paper, we introduce the sequence space
Theorem 14. is a linear space.
Proof. Let and , be two scalars in . Then for a given , we have Now let, be such that . Now consider this implies that the sequence space Hence, . Therefore, is a linear space.
Theorem 15. The space is a paranormed space, paranormed by
Proof. For , is trivial.
For , , we have(i) 0 for all .(ii) for all .(iii).(iv)Let be a sequence of scalars with and such that in the sense that Therefore, Hence, is a paranormed space.
Theorem 16. is a closed subspace of .
Proof. Let be a cauchy sequence in such that . We show that . Since , then there exists such that
We need to show that (i) converges to .(ii)If , then .
Since is a cauchy sequence in , then for a given , there exists such that
For a given , we have Then , and . Let where . Then . We choose , then for each , we have Then is a cauchy sequence of scalars in , so there exists a scalar such that , as .
For the next step, let be given. Then, we show that if then . Since , then there exists such that which implies that . The number , can be so chosen that together with (32), we have such that . Since , then we have a subset of such that , where Let , where
Therefore, for each , we have Hence, the result follows.
Theorem 17. The space is nowhere dense subsets of .
Proof. Proof of the result follows from the previous theorem.
Theorem 18. The space is solid and monotone.
Proof. Let and be a sequence of scalars with for all . Then, we have The space is solid follows from the following inclusion relation: Also a sequence space is solid implies monotone. Hence, the space is monotone.
Theorem 19. and the inclusions are proper.
Proof. Let . Then, we have . Since , implies
Now let, be such that . As , taking supremum over we get . Hence, .
Next we show that the inclusion is proper
(i) First for . Consider , then by the definition we have where Therefore, On solving, we get As is a translation map, that is, , we have Taking , we have Since , therefore which implies that . Hence, we get that the inclusion is proper.
(ii) Second for .
The result of this part follows from the proof of Theorem 18.
Theorem 20. and the inclusions are proper.
Proof. Let . Then, we have
Since , which implies implies
Now let, be such that . As taking over , we get . Hence,
Next, we show that the inclusion is proper
(i) First for . We show that .
Let , then by the definition We have, . We say that the .
Now considering the case when , then when , then we have . Therefore we get, Hence, . Hence, the inclusion is proper.
(ii) Second for .
The result follows from the proof of Theorem 18.
The authors would like to record their gratitude to the reviewer for his careful reading and making some useful corrections which improved the presentation of the paper.
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