On Zweier I-Convergent Double Sequence Spaces

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1 Filomat 3:12 (216), DOI /FIL K Published by Faculty of Sciences and Mathematics, University of Niš, Serbia Available at: On Zweier I-Convergent Double Sequence Spaces Vakeel A. Khan a, Nazneen Khan a a Department of Mathematics, Aligarh Muslim University, Aligarh-222 (INDIA) Abstract. In this article we introduce the Zweier I-convergent double sequence spaces 2 Z I, 2Z I and 2Z I. We prove the decomposition theorem and study topological properties, algebraic properties and some inclusion relations on these spaces. 1. Introduction Let N, R and C be the sets of all natural, real and complex numbers respectively. We write the space of all real or complex sequences. 2ω = x = (x ij ) : x ij R or C}, Let l, c and c denote the Banach spaces of bounded, convergent and null sequences respectively normed by x = sup x k. k At the initial stage the notion of I-convergence was introduced by Kostyrko,Šalát and Wilczyński[1]. Later on it was studied by Šalát, Tripathy and Ziman[2], Demirci [3] and many others. I-convergence is a generalization of Statistical Convergence. Now we have a list of some basic definitions used in the paper. Definition 1.1. [4,5] Let X be a non empty set. Then a family of sets I 2 X (2 X denoting the power set of X) is said to be an ideal in X if (i) I (ii) I is additive i.e A,B I A B I. (iii) I is hereditary i.e A I, B A B I. An Ideal I 2 X is called non-trivial if I 2 X. A non-trivial ideal I 2 X is called admissible if x} : x X} I. A non-trivial ideal I is maximal if there cannot exist any non-trivial ideal J I containing I as a subset. For each ideal I, there is a filter (I) corresponding to I. i.e (I) = K N : K c I}, where K c = N K. 21 Mathematics Subject Classification. 4C5; 46A45, 46E3, 46E3, 46E4, 46B2 Keywords. Ideal, filter, I-convergence field, I-convergent, monotone and solid double sequence spaces, Lipschitz function Received: 3 April 214 ; Accepted: 3 November 214 Communicated by Hari M. Srivastava addresses: vakhanmaths@gmail.com (Vakeel A. Khan), nazneenmaths@gmail.com (Nazneen Khan)

2 V. A. Khan, N. Khan / Filomat 3:12 (216), Definition 1.2. A double sequence of complex numbers is defined as a function x : N N C. We denote a double sequence as (x ij ), where the two subscripts run through the sequence of natural numbers independent of each other. A number a C is called a double limit of a double sequence (x ij ) if for every ɛ > there exists some N = N(ɛ) N such that (x ij ) a < ɛ, f or all i, j N (see [6, 7, 8]) Definition 1.3.[7] A double sequence (x ij ) 2 ω is said to be I-convergent to a number L if for every ɛ >, In this case we write I lim x ij = L. i, j N : x ij L ɛ} I. Definition 1.4.[7] A double sequence (x ij ) 2 ω is said to be I-null if L =. In this case we write I lim x ij =. Definition 1.5. [7] A double sequence (x ij ) 2 ω is said to be I-cauchy if for every ɛ > there exist numbers m = m(ɛ), n= n(ɛ) such that i, j N : x ij x mn ɛ} I. Definition 1.6.[7] A double sequence (x ij ) 2 ω is said to be I-bounded if there exists M > such that i, j N : x ij > M}. Definition 1.7.[7] A double sequence space E is said to be solid or normal if (x ij ) E implies (α ij x ij ) E for all sequence of scalars (α ij ) with α ij < 1 for all i,j N. Definition 1.8.[7] A double sequence space E is said to be monotone if it contains the canonical preimages of its stepspaces. Definition 1.9.[7] A double sequence space E is said to be convergence free if (y ij ) E whenever (x ij ) E and x ij = implies y ij =. Definition 1.1.[7] A double sequence space E is said to be a sequence algebra if (x ij.y ij ) E whenever (x ij ), (y ij ) E. Definition 1.11.[7] A double sequence space E is said to be symmetric if (x ij ) E implies (x π(ij) ) E, where π is a permutation on N. A sequence space λ ω with linear topology is called a K-space provided each of maps p i C defined by p i (x) = x i is continuous for all i N. Let λ and µ be two sequence spaces and A = (a nk ) be an infinite matrix of real or complex numbers a nk, where n, k N. Then we say that A defines a matrix mapping from λ to µ, and we denote it by writing A : λ µ. If for every sequence x = (x k ) λ the sequence Ax = (Ax) n }, the A transform of x is in µ, where (Ax) n = a nk x k, (n N). (1) k By (λ : µ), we denote the class of matrices A such that A : λ µ. Thus, A (λ : µ) if and only if series on the right side of (1) converges for each n N and every x λ. (see[9]). The approach of constructing the new sequence spaces by means of the matrix domain of a particular limitation method have been recently studied by Başar and Altay[1], Malkowsky[11], Ng and Lee[12] and

3 V. A. Khan, N. Khan / Filomat 3:12 (216), Wang[13], Başar, Altay and Mursaleen[14]. For more information one can refer [15,16,17] Şengönül[18] defined the sequence y = (y i ) which is frequently used as the Z p transform of the sequence x = (x i ) i.e, y i = px i + (1 p)x i 1 where x 1 =, p 1, 1 < p < and Z p denotes the matrix Z p = (z ik ) defined by p, (i = k), z ik = 1 p, (i 1 = k); (i, k N),, otherwise. Following Basar,and Altay [1], Şengönül[18] introduced the Zweier sequence spaces Z and Z as follows Z = x = (x k ) ω : Z p x c} Z = x = (x k ) ω : Z p x c }. The following Lemmas will be used for establishing some results of this article. Lemma A sequence space E is solid implies that E is monotone.(see[19,2]) Lemma Let K (I) and M N. If M I, then M K I.(See[19,2]) Lemma If I 2 N and M N. If M I, then M K I.(See[21,22,23]) Main Results In this article we introduce the following classes of sequence spaces. We also denote by 2Z I = x = (x ij ) 2 ω : (i, j) N N : I lim Z p x = L for some L}} I 2Z I = x = (x ij) 2 ω : (i, j) N N : I lim Z p x = }} I 2Z I = x = (x ij ) 2 ω : (i, j) N N : sup Z p x < }} I i,j 2m I Z = 2Z 2 Z I and 2 m I Z = 2 Z 2 Z I Theorem 2.1. The classes of sequences 2 Z I, 2 Z I, 2m I Z and 2m I Z are linear spaces. Proof. We shall prove the result for the space 2 Z I. The proof for the other spaces will follow similarly. Let (x ij ), (y ij ) 2 Z I and let α, β be scalars. Then That is for a given ɛ >, we have I lim x ij L 1 =, for some L 1 C ; I lim y ij L 2 =, for some L 2 C ; A 1 = (i, j) N N : x ij L 1 > ɛ } I, (1) 2 we have A 2 = i, j N N : y ij L 2 > ɛ } I. (2) 2

4 V. A. Khan, N. Khan / Filomat 3:12 (216), Now, by (1) and (2), (αx ij + βy ij ) (αl 1 + βl 2 ) α ( x ij L 1 ) + β ( y ij L 2 ) x ij L 1 + y ij L 2. (i, j) N N : (αx ij + βy ij ) (αl 1 + βl 2 ) > ɛ} A 1 A 2. Therefore (αx ij + βy ij ) 2 Z I. Hence 2 Z I is a linear space. We state the following result without proof in view of Theorem 2.1. Theorem 2.2. The spaces 2 m I Z and 2m I Z are normed linear spaces, normed by x ij = sup x ij. (3) i,j Theorem 2.3. A sequence x = (x ij ) 2 m I Z I-converges if and only if for every ɛ > there exists N ɛ = (m, n) N N such that (i, j) N N : x ij x Nɛ < ɛ} 2 m I Z (4) Proof. Suppose that L = I lim x. Then B ɛ = (i, j) N N : x ij L < ɛ 2 } 2m I Z for all ɛ >. Fix an N ɛ = (m, n) B ɛ. Then we have which holds for all (i, j) B ɛ. Hence (i, j) N N : x ij x Nɛ < ɛ} 2 m I Z. x Nɛ x ij x Nɛ L + L x ij < ɛ 2 + ɛ 2 = ɛ Conversely, suppose that (i, j) N N : x ij x Nɛ < ɛ} 2 m I Z. That is (i, j) N N : x k x Nɛ < ɛ} 2 m I Z for all ɛ >. Then the set C ɛ = (i, j) N N : x ij [x Nɛ ɛ, x Nɛ + ɛ]} 2 m I Z for all ɛ >. Let J ɛ = [x Nɛ ɛ, x Nɛ + ɛ]. If we fix an ɛ > then we have C ɛ 2 m I Z as well as C ɛ 2 2m I Z. Hence C ɛ C ɛ 2 2m I Z. This implies that that is that is J = J ɛ J ɛ 2 φ (i, j) N N : x ij J} 2 m I Z diamj diamj ɛ where the diam of J denotes the length of interval J. In this way, by induction we get the sequence of closed intervals J ɛ = I I 1... I k...

5 V. A. Khan, N. Khan / Filomat 3:12 (216), with the property that diami k 1 2 diami k 1 for (k=2,3,4,...) and (i, j) N N : x ij I k } m I Z for (k=1,2,3,4,...). Then there exists a ξ I k where (i, j) N N such that ξ = I lim x, that is L = I lim x. Theorem 2.4. Let I be an admissible ideal. Then the following are equivalent. (a) (x ij ) 2 Z I ; (b) there exists (y ij ) 2 Z such that x ij = y ij, for a.a.k.r.i; (c) there exists(y ij ) 2 Z and (z ij ) 2 Z I such that x ij = y ij + z ij for all (i, j) N N and (i, j) N N : y ij L ɛ} I; (d) there exists a subset K = k 1 < k 2...} of N such that K (I) and lim n x kn L =. Proof. (a) implies (b). Let (x ij ) 2 Z I. Then there exists L C such that (i, j) N N : x ij L ɛ} I. Let (m t, n t ) be an increasing sequence with (m t, n t ) N N such that Define a sequence (y ij ) as For (m t, n t ) < (i, j) (m t+1, n t+1 ) f or t N. (i, j) (m t, n t ) : x ij L 1 t } I. y ij = x ij, for all (i, j) (m 1, n 1 ). y ij = Then (y ij ) 2 Z and form the following inclusion We get x ij = y ij, for a.a.k.r.i. xij, if x ij L < t 1, L, otherwise. (i, j) (m t, n t ) : x ij y ij } (i, j) (m t, n t ) : x ij L ɛ} I. (b) implies (c). For (x ij ) 2 Z I, there exists (y ij ) 2 Z such that x ij = y ij, for a.a.k.r.i. Let K = (i, j) N N : x ij y ij }, then K I. Define a sequence (z ij ) as xij y z ij = ij, if (i, j) K,, otherwise. Then z ij 2 Z I and y ij 2 Z. (c)implies (d). Let P 1 = (i, j) N N : z ij ɛ} I and Then we have lim n x (in,j n ) L =. K = P c 1 = (i 1, j 1 ) < (i 2, j 2 ) <...} (I). (d) implies (a). Let K = (i 1, j 1 ) < (i 2, j 2 ) <...} (I) and lim n x (in,j n ) L =. Then for any ɛ >, and Lemma 1.17, we have (i, j) N N : x ij L ɛ} K c (i, j) K : x ij L ɛ}.

6 V. A. Khan, N. Khan / Filomat 3:12 (216), Thus (x ij ) 2 Z I. Theorem 2.5. The inclusions 2 Z I 2Z I 2 Z I hold and are proper. Proof. Let (x ij ) 2 Z I. Then there exists L C such that I lim x ij L = We have x ij 1 2 x ij L L. Taking the supremum over (i, j) on both sides we get (x ij ) 2 Z I. The inclusion 2 Z I 2Z I is obvious. The strict inclusion is also trivial. Theorem 2.6. The function : 2m I R is the Lipschitz function,where Z 2m I Z = 2Z I 2 Z, and hence uniformly continuous. Proof. Let x, y 2 m I, x y. Then the sets Z Thus the sets, Hence also B = B x B y (I), so that B φ. Now taking (i,j) in B, Thus is a Lipschitz function. For 2 m I Z the result can be proved similarly. A x = (i, j) N N : x ij (x) x y } I, A y = (i, j) N N : y ij (y) x y } I. B x = (i, j) N N : x ij (x) < x y } (I) B y = (i, j) N N : y ij (y) < x y } (I). (x) (y) (x) x ij + x ij y ij + y (y) 3 x y. Theorem 2.7. If x, y 2 m I Z, then (x.y) 2m I and (xy) = (x) (y). Z Proof. For ɛ > Now, B x = (i, j) N N : x (x) < ɛ} (I), B y = (i, j) N N : y (y) < ɛ} (I). x.y (x) (y) = x.y x (y) + x (y) (x) (y) x y (y) + (y) x (x) (5) As 2 m I Z Z, there exists an M R such that x < M and (y) < M. Using eqn(5) we get x.y (x) (y) Mɛ + Mɛ = 2Mɛ For all (i, j) B x B y (I). Hence (x.y) 2 m I and (xy) = (x) (y). Z For 2 m I the result can be proved similarly. Z Theorem 2.8. The spaces 2 Z I and 2m I Z are solid and monotone.

7 V. A. Khan, N. Khan / Filomat 3:12 (216), Proof. We shall prove the result for 2 Z I. Let (x ij ) Z I. Then I lim k x ij = (6) Let (α ij ) be a sequence of scalars with α ij 1 for all (i, j) N N. Then the result follows from (6) and the following inequality α ij x ij α ij x ij x ij for all (i, j) N N. That the space 2 Z I is monotone follows from the Lemma For 2 m I the result can be proved similarly. Z Theorem 2.9. If I is not maximal, then the space 2 Z I is neither solid nor monotone. Proof. Here we give a counter example. Let (x ij ) = 1 for all (i, j) N N. Then (x ij ) 2 Z I. Let K N N be such that K I and N N K I Define the sequence (xij ), if (i, j) K, (y ij ) =, otherwise. Then (y ij ) belongs to the canonical preimage of K-step space of 2 Z I but (y ij ) 2 Z I. Hence 2 Z I is not monotone. Theorem 2.1. The spaces 2 Z I and 2 Z I are sequence algebras. Proof. We prove that 2 Z I is a sequence algebra. Let (x ij ), (y ij ) 2 Z I. Then I lim x ij = and I lim y ij = Then we have I lim (x ij.y ij ) =. Thus (x ij.y ij ) 2 Z I Hence 2 Z I is a sequence algebra. For the space 2 Z I, the result can be proved similarly. Theorem The spaces 2 Z I and 2 Z I are not convergence free in general. Proof. Here we give a counter example. Let I = I f. Consider the sequence (x ij ) and (y ij ) defined by x ij = 1 i.j and y ij = i.j for all (i,j) N N Then (x ij ) 2 Z I and 2 Z I, but (y ij) 2 Z I and 2 Z I. Hence the spaces 2 Z I and 2 Z I are not convergence free. Theorem If I is not maximal and I I f, then the spaces 2 Z I and 2 Z I are not symmetric. Proof. Let A I be infinite. If x ij = 1, for i, j A,, otherwise. Then x ij 2 Z I 2Z I let K N be such that K I and N K I. Let φ : K A and ψ : N K N A be bijections, then the map π : N N defined by φ(k), for k K, π(k) = ψ(k), otherwise.

8 V. A. Khan, N. Khan / Filomat 3:12 (216), is a permutation on N, but x (π(m)π(n)) Z I and x (π(m)π(n)) 2 Z I. Hence 2Z I and 2 Z I are not symmetric. Theorem The sequence spaces 2 Z I and 2 Z I are linearly isomorphic to the spaces 2c I and 2 c I respectively, i.e 2 Z I 2 c I and 2 Z I 2c I. Proof. We shall prove the result for the space 2 Z I and 2 c I. The proof for the other spaces will follow similarly. We need to show that there exists a linear bijection between the spaces 2 Z I and c I. Define a map T : 2Z I 2 c I such that x x = Tx T(x ij ) = px ij + (1 p)x (i 1)(j 1) = x ij where x 1 =, p 1, 1 < p <. Clearly T is linear. Further, it is trivial that x = = (,,,...) whenever Tx = and hence injective. Let x ij 2c I and define the sequence x = x ij by for (i, j) N N and where M = 1 p x ij = M i r= j ( 1) (i r)(j s) N (i r)(j s) x ij s= 1 p and N = p. Then we have lim px ij + (1 p)x (i 1)(j 1) = p lim M (i,j) (i,j) r= i r= j ( 1) (i r)(j s) N (i r)(j s) x ij i 1 j 1 + (1 p) lim M ( 1) (i 1 r)(j 1 s) N (i 1 r)(j 1 s) x (i,j) (i 1)(j 1) s= = lim (i,j) x ij s= which shows that x 2 Z I. Hence T is a linear bijection. Also we have x = Z p x c Therefore Hence 2 Z I 2 c I. x = = sup pm (i,j) N N sup px ij + (1 p)x (i 1)(j 1) (i,j) N N i r= j ( 1) (i r)(j s) N (i r)(j s) x ij s= i 1 j 1 + (1 p)m ( 1) (i 1 r)(j 1 s) N (i 1 r)(j 1 s) x (i 1)(j 1) r= s= = sup x ij = x 2 c I. (i,j) N N Acknowledgments. 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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