Zweier I-Convergent Sequence Spaces

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1 Chapter 2 Zweier I-Convergent Sequence Spaces In most sciences one generation tears down what another has built and what one has established another undoes. In mathematics alone each generation builds a new story to the old structure. - Hankel.

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3 Chapter 2 Zweier I-Convergent Sequence Spaces 2.1 Introduction Let l, c and c 0 denote the Banach spaces of bounded,convergent and null sequences respectively normed by x = sup x k. k Each linear subspace of ω, for example, λ, µ ω is called a sequence space. A sequence space X with linear topology is called a K-space provided each of maps p i : X C defined by p i (x) = x i is continuous for all i N. A K-space λ is called an FK-space provided λ is a complete linear metric space. An FK-space whose topology is normable is called a BK-space. Let λ and µ be two sequence spaces and A = (a nk ) is 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 = k a nk x k, (n N). [2.1] 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 [2.1] converges for each n N and every x λ. The approach of constructing new sequence spaces by means of the Science Publishing Group 21

4 Zweier I-Convergent Sequence Spaces and Their Properties matrix domain of a particular limitation method have been recently employed by Altay,Başar and Mursaleen[1], Başar and Altay[3], Malkowsky[57], Ng and Lee[59], and Wang[74]. Şengönül[68] 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 = 0, 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), 0, otherwise. Following Başar and Altay[3], Şengönül[68] introduced the Zweier sequence spaces Z and Z 0 as follows Z = {x = (x k ) ω : Z p x c} Z 0 = {x = (x k ) ω : Z p x c 0 }. Here we list below some of the results of [68] which we will need as a reference in order to establish analogously some of the results of this article. Theorem [68, Theorem 2.1] The sets Z and Z 0 are the linear spaces with the co-ordinate wise addition and scalar multiplication which are the BK-spaces with the norm x Z = x Z0 = Z p x c. Theorem [68, Theorem 2.2] The sequence spaces Z and Z 0 are linearly isomorphic to the spaces c and c 0 respectively, i.e Z = c and Z 0 = c0 [See (Theorem 2.2.[18])] 22 Science Publishing Group

5 Chapter 2 Zweier I-Convergent Sequence Spaces Theorem [68, Theorem 2.3] The inclusions Z 0 Z strictly hold for p 1. Theorem [68, Theorem 2.6] Z 0 is solid. Theorem [68, Theorem 3.6] Z is not a solid sequence space. The following Lemmas will be used for establishing some results of this article. Lemma Let E be a sequence space. If E is solid then E is monotone. (see [20], page 53). Lemma If I 2 N and M N. If M / I, then M N / I. (see [71-72]). 2.2 Main Results In this chapter we introduce the following classes of sequence spaces. Z I = {x = (x k ) ω : {k N : I lim Z p x = L, for some L C} I} Z I 0 = {x = (x k ) ω : {k N : I lim Z p x = 0} I} Z I = {x = (x k ) ω : sup Z p x < }. k We also denote by m I Z = Z Z I and m I Z 0 = Z Z I 0 Throughout the article, for the sake of convenience now we will denote by Z p (x k ) = x /, Z p (y k ) = y /, Z p (z k ) = z / for x, y, z ω. Science Publishing Group 23

6 Zweier I-Convergent Sequence Spaces and Their Properties Theorem The classes of sequences Z I, Z0 I, m I Z and mi Z 0 are linear spaces. Proof. We shall prove the result for the space Z I. The proof for the other spaces will follow similarly. Let (x k ), (y k ) Z I and let α, β be scalars. Then I lim x / k L 1 = 0, for some L 1 C; I lim y / k L 2 = 0, for some L 2 C; That is for a given ɛ > 0, we have A 1 = {k N : x / k L 1 > ɛ } I, [2.2] 2 we have A 2 = {k N : y / k L 2 > ɛ } I. [2.3] 2 (αx / k + βy/ k ) (αl 1 + βl 2 ) α ( x / k L 1 ) + β ( y / k L 2 ) x / k L 1 + y / k L 2 Now, by [2.2] and [2.3], {k N : (αx / k + βy/ k ) (αl 1 + βl 2 ) > ɛ} A 1 A 2. Therefore (αx k + βy k ) Z I Hence Z I is a linear space. Theorem The spaces m I Z and mi Z 0 are normed linear spaces,normed by where x / k = Zp (x) x / k = sup Z p (x). [2.4] k Proof. It is clear from Theorem that m I Z and mi Z 0 are linear spaces. It is easy to verify that [2.4] defines a norm on the spaces m I Z and mi Z Science Publishing Group

7 Chapter 2 Zweier I-Convergent Sequence Spaces Theorem A sequence x = (x k ) m I Z for every ɛ > 0 there exists N ɛ N such that I-converges if and only if {k N : x / k x/ N ɛ < ɛ} m I Z [2.5] Proof. Suppose that L = I lim x /. Then B ɛ = {k N : x / k L < ɛ 2 } mi Z for all ɛ > 0. Fix an N ɛ B ɛ. Then we have x / N ɛ x / k x/ N ɛ L + L x / k < ɛ 2 + ɛ 2 = ɛ which holds for all k B ɛ. Hence {k N : x / k x/ N ɛ < ɛ} m I Z. Conversely, suppose that {k N : x / k x/ N ɛ < ɛ} m I Z. That is {k N : x / k x/ N ɛ < ɛ} m I Z for all ɛ > 0. Then the set C ɛ = {k N : x / k [x/ N ɛ ɛ, x / N ɛ + ɛ]} m I Z for all ɛ > 0. Let J ɛ = [x / N ɛ ɛ, x / N ɛ + ɛ]. If we fix an ɛ > 0 then we have C ɛ m I Z as well as C ɛ mi 2 Z. Hence C ɛ C ɛ mi 2 Z. This implies that that is that is J = J ɛ J ɛ 2 φ {k N : x / k J} mi 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 0 I 1... I k... Science Publishing Group 25

8 Zweier I-Convergent Sequence Spaces and Their Properties with the property that diami k 1 2 diami k 1 for (k=2,3,4,...) and {k N : x / k I k} m I Z for (k=1,2,3,4,...). Then there exists a ξ I k where k N such that ξ / = I lim x /, that is L = I lim x /. Theorem Let I be an admissible ideal. Then the following are equivalent. (a) (x k ) Z I ; (b) there exists (y k ) Z such that x k = y k, for a.a.k.r.i; (c) there exists (y k ) Z and (z k ) Z I 0 such that x k = y k + z k for all k N and {k N : y k L ɛ} I ; (d) there exists a subset K = {k 1 < k 2...} of N such that K (I) and lim x k n L = 0. n Proof. (a) implies (b). Let (x k ) Z I. Then there exists L C such that {k N : x / k L ɛ} I. Let (m t ) be an increasing sequence with m t N such that {k m t : x / k L 1 t } I. Define a sequence (y k ) as y k = x k, for all k m 1. For m t < k m t+1, t N. { x k, if x / k y k = L < t 1, L, otherwise. 26 Science Publishing Group

9 Chapter 2 Zweier I-Convergent Sequence Spaces Then (y k ) Z and form the following inclusion {k m t : x k y k } {k m t : x / k L ɛ} I. We get x k = y k, for a.a.k.r.i. (b) implies (c). For (x k ) Z I. Then there exists (y k ) Z such that x k = y k, for a.a.k.r.i. Let K = {k N : x k y k }, then K I. Define a sequence (z k ) as { x k y k, if k K, z k = 0, otherwise. Then z k Z0 I and y k Z. (c) implies (d). Let P 1 = {k N : z k ɛ} I and K = P1 c = {k 1 < k 2 < k 3 <...} (I). Then we have lim n x kn L = 0. (d) implies (a). Let K = {k 1 < k 2 < k 3 <...} (I) and lim n x kn L = 0. Then for any ɛ > 0, and Lemma, we have {k N : x / k L ɛ} Kc {k K : x / k L ɛ}. Thus (x k ) Z I. Theorem The inclusions Z0 I Z I Z I are proper. Proof. Let (x k ) Z I. Then there exists L C such that I lim x / k L = 0 We have x / k 1 2 x/ k L + 1 L. Taking the supremum over k on both 2 sides we get (x k ) Z. I The inclusion Z0 I Z I is obvious. Science Publishing Group 27

10 Zweier I-Convergent Sequence Spaces and Their Properties Theorem The function : m I Z where m I Z = ZI Z, and hence uniformly continuous. Proof. Let x, y m I Z, x y. Then the sets R is the Lipschitz function, A x = {k N : x / k (x/ ) x / y / } I, A y = {k N : y / k (y/ ) x / y / } I. Thus the sets, B x = {k N : x / k (x/ ) < x / y / } m I Z, B y = {k N : y / k (y/ ) < x / y / } m I Z. Hence also B = B x B y m I Z, so that B φ. Now taking k in B, (x / ) (y / ) (x / ) x / k + x/ k y/ k + y/ (y / ) 3 x / y /. Thus is a Lipschitz function. For m I Z 0 the result can be proved similarly. Theorem If x, y m I Z, then (x.y) mi Z and (xy) = (x) (y). Proof. For ɛ > 0 B x = {k N : x / (x / ) < ɛ} m I Z, B y = {k N : y / (y / ) < ɛ} m I Z. Now, x /.y / (x / ) (y / ) = x /.y / x / (y / ) + x / (y / ) (x / ) (y / ) x / y / (y / ) + (y / ) x / (x / ) [2.6] 28 Science Publishing Group

11 Chapter 2 Zweier I-Convergent Sequence Spaces As m I Z Z, there exists an M R such that x / < M and (y / ) < M. Using eqn[2.6] we get x /.y / (x / ) (y / ) Mɛ + Mɛ = 2Mɛ For all k B x B y m I Z. Hence (x.y) mi Z For m I Z 0 the result can be proved similarly. and (xy) = (x) (y). Theorem The spaces Z I 0 and m I Z 0 are solid and monotone. Proof. We shall prove the result for Z I 0. Let (x k ) Z I 0. Then I lim k x / k = 0 [2.7] Let (α k ) be a sequence of scalars with α k 1 for all k N. Then the result follows from [2.7] and the following inequality α k x / k α k x / k x / k for all k N. That the space ZI 0 is monotone follows from the Lemma For m I Z 0 the result can be proved similarly. Theorem The spaces Z I and m I Z are neither monotone nor solid, if I is neither maximal nor I = I f in general. Proof. Here we give a counter example. Let I = I δ. Consider the K-step space X K of X defined as follows, Let (x k ) X and let (y k ) X K be such that (y / k ) = { (x / k ), if k is odd, 1, otherwise. Consider the sequence (x / k ) defined by (x/ k ) = k 1 for all k N. Then (x k ) Z I but its K-stepspace preimage does not belong to Z I. Thus Z I is not monotone. Hence Z I is not solid. Theorem The spaces Z I and Z0 I are sequence algebras. Science Publishing Group 29

12 Zweier I-Convergent Sequence Spaces and Their Properties Proof. We prove that Z0 I is a sequence algebra. Let (x k ), (y k ) Z0 I. Then I lim x / k = 0 and I lim y / k = 0 Then we have I lim (x / k.y/ k ) = 0 Thus (x k.y k ) Z I 0. Hence Z I 0 is a sequence algebra. For the space Z I, the result can be proved similarly. Theorem The spaces Z I and Z0 I general. are not convergence free in Proof. Here we give a counter example. Let I = I f. Consider the sequence (x / k ) and (y/ k ) defined by x / k = 1 k and y/ k = k for all k N Then (x k ) Z I and Z I 0, but (y k ) / Z I and Z I 0. Hence the spaces Z I and Z I 0 are not convergence free. Theorem If I is not maximal and I I f, then the spaces Z I and Z I 0 are not symmetric. Proof. Let A I be infinite. If x / k = { 1, for k A, 0, otherwise. Then by lemma x k Z I 0 Z 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, 30 Science Publishing Group

13 Chapter 2 Zweier I-Convergent Sequence Spaces then the map π : N N defined by { φ(k), for k K, π(k) = ψ(k), otherwise. is a permutation on N, but x π(k) / Z I and x π(k) / Z I 0. Hence Z I and Z I 0 are not symmetric. Theorem The sequence spaces Z I and Z I 0 are linearly isomorphic to the spaces c I and c I 0 respectively, i.e Z I = c I and Z I 0 = c I 0. Proof. We shall prove the result for the space Z I and c I. The proof for the other spaces will follow similarly. We need to show that there exists a linear bijection between the spaces Z I and c I. Define a map T : Z I c I such that x x / = T x T (x k ) = px k + (1 p)x k 1 = x / k where x 1 = 0, p 1, 1 < p <. Clearly T is linear. Further, it is trivial that x = 0 = (0, 0, 0,...) whenever T x = 0 and hence injective. Let x / k ci and define the sequence x = x k by x k = M k ( 1) k i N k i x / i. (i N) i=0 where M = 1 p 1 p and N =. Then we have p lim px k + (1 p)x k 1 = p lim M k k which shows that x Z I. k ( 1) k i N k i x / i + i=0 k 1 (1 p) lim M ( 1) k i N k i x / i = lim k k x/ k i=0 Science Publishing Group 31

14 Zweier I-Convergent Sequence Spaces and Their Properties Hence T is a linear bijection. Also we have x = Z p x c. Therefore = sup pm k N Hence Z I = c I. x = sup px k + (1 p)x k 1 k N k k 1 ( 1) k i N k i x / i + (1 p)m ( 1) k i N k i x / i i=0 = sup x / k = x/ c I k N i=0 32 Science Publishing Group