Weighted almost convergence and related infinite matrices

The purpose of this paper is to introduce the notion of weighted almost convergence of a sequence and prove that this sequence endowed with the sup-norm \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}${\Vert \cdot \Vert } _{\infty}$\end{document}∥⋅∥∞ is a BK-space. We also define the notions of weighted almost conservative and regular matrices and obtain necessary and sufficient conditions for these matrix classes. Moreover, we define a weighted almost A-summable sequence and prove the related interesting result.


Introduction and preliminaries
Let ω denote the space of all complex sequences s = (s j ) ∞ j=0 (or simply write s = (s j )). Any vector subspace of ω is called a sequence space. By N we denote the set of natural numbers, and by R the set of real numbers. We use the standard notation ∞ , c and c 0 to denote the sets of all bounded, convergent and null sequences of real numbers, respectively, where each of the sets is a Banach space with the sup-norm . ∞ defined by s ∞ = sup j∈N |s j |. We write the space p of all absolutely p-summable series by Clearly, p is a Banach space with the following norm: For p = 1, we obtain the set l 1 of all absolutely summable sequences. For any sequence s = (s j ), let s [n] = n j=0 s j e j be its n-section, where e j is the sequence with 1 in place j and 0 elsewhere and e = (1, 1, 1, . . . ).
A sequence space X is called a BK-space if it is a Banach space with continuous coordinates p j : X → C, the set of complex fields, and p j (s) = s j for all s = (s j ) ∈ X and every j ∈ N. A BK-space X ⊃ ψ, the set of all finite sequences that terminate in zeros, is said to have AK if every sequence s = (s j ) ∈ X has a unique representation s = ∞ j=0 s j e j .
Let X and Y be two sequence spaces, and let A = (a n,k ) be an infinite matrix. If, for each s = (s k ) in X, the series A n s = k a n,k s k = ∞ k=0 a n,k s k (1) converges for each n ∈ N and the sequence As = (A n s) belongs to Y , then we say that matrix A maps X into Y . By the symbol (X, Y ) we denote the set of all such matrices which map X into Y . The series in (1) is called A-transform of s whenever the series converges for n = 0, 1, . . . . We say that s = (s k ) is A-summable to the limit λ if A n s converges to λ (n → ∞).
The sequence s = (s k ) of ∞ is said to be almost convergent, denoted by f , if all of its Banach limits [1] are equal. We denote such a class by the symbol f , and one writes f -lim s = λ if λ is the common value of all Banach limits of the sequence s = (s k ). For a bounded sequence s = (s k ), Lorentz [2] proved that f -lim s = λ if and only if lim k→∞ s m + s m+1 + · · · + s m+k k + 1 = λ uniformly in m. This notion was later used to (i) define and study conservative and regular matrices [3]; (ii) introduce related sequence spaces derived by the domain of matrices [4][5][6]; (iii) study some related matrix transformations [7][8][9]; (iv) define related sequence spaces derived as the domain of the generalized weighted mean and determine duals of these spaces [10,11]. As an extension of the notion of almost convergence, Kayaduman and Şengönül [12,13] defined Cesàro and Riesz almost convergence and established related core theorems. The almost strongly regular matrices for single sequences were introduced and characterized [14], and for double sequences they were studied by Mursaleen [15] (also refer to [16][17][18][19]). As an application of almost convergence, Mohiuddine [20] proved a Korovkin-type approximation theorem for a sequence of linear positive operators and also obtained some of its generalizations. Başar and Kirişçi [21] determined the duals of the sequence space f and other related spaces/series and investigated some useful characterizations. We now recall the following result. We shall use the notation f (N) for the space of all sequences which are f (N)-convergent, that is,

Weighted almost convergence
We remark that if we take t k = 1 for all k, then (2) is reduced to the notion of almost convergence introduced by Lorentz [2]. Clearly, a convergent sequence is f (N)-convergent to the same limit, but its converse is not always true.

Example 2.2
Consider a sequence s = (s k ) defined by s k = 1 if k is odd and 0 for even k. Also, let t k = 1 for all k. Then we see that s = (s k ) is f (N)-convergent to 1/2 but not convergent. Letting i → ∞, one obtains mr s ks i < /3 and |λ k -λ| < /3 (4) for each m, r and k > M. Now, for fixed k, the above inequality holds. Since s k ∈ f (N), for fixed k, we get lim m→∞ mr s k = λ k uniformly in r.
For given > 0, there exists positive integers M 0 (independent of r, but dependent upon ) such that for m > M 0 and for all r. It follows from (4) and (5) that This proves that f (N) is a Banach space normed by (3).
Since c ⊂ f (N) ⊂ l ∞ , there exist positive real numbers α and β with α < β such that α s ∞ ≤ s f (N) ≤ β s ∞ . That is to say, two norms · ∞ and · f (N) are equivalent. It is well known that the spaces c and l ∞ endowed with the norm · ∞ are BK-spaces, and hence the space f (N) endowed with the norm · ∞ is also a BK-space.
We prove the following characterization of weighted almost conservative matrices. t n a n,k = λ exists uniformly in r. ( 8 ) Proof Necessity. Let A ∈ (c, f (N)). Since the sequences e and e k both are convergent, so A-transforms of the sequences e k and e belong to f (N) and exist uniformly in r. It follows that (7) and (8)  It is clear that ( mr (s)) is bounded for s = (s k ) ∈ c and fixed r. Hence, by the uniform boundedness principle, ( mr ) is bounded. For each p ∈ Z + (the positive integers), the sequence x = (x k ) is defined by Then a sequence x ∈ c, x = 1 and r+m-1 n=r t n a n,k .
Therefore, we obtain Equations (9) and (10) give that 1 T m p k=0 r+m-1 n=r t n a n,k ≤ mr < ∞, it follows that (6) is valid. Sufficiency. Let conditions (6)-(8) hold. Let r be any nonnegative integer, and let s k ∈ c. Then t n a n,k s k r+m-1 n=r t n a n,k s k , which gives r+m-1 n=r t n a n,k s .
It follows from hypothesis (6) that | mr (s)| ≤ B r s , where B r is a constant independent of r. Thus we have mr ∈ c for each m ≥ 1, which gives that a sequence ( mr ) is bounded for each nonnegative integer r. Hypotheses (7) and (8) imply that the limit of mr (e k ) and mr (e) must exist for all nonnegative integers k and r. Since {e, e 0 , e 1 , . . . } is a fundamental set in c, it follows from [23, p. 252] that lim m mr (s) = r (s) exists and r ∈ c . Therefore r has the following form (see [23, p. 205]): where ξ = lim s k . From (7) and (8), we see that r (e k ) = λ k for a nonnegative integer k and r (e) = λ. Therefore, for each s ∈ c and a nonnegative integer r, we have lim m→∞ mr (s) = (s) with the following expression: Since mr ∈ c , so it has the representation We observe from (11) and (12) that the convergence of mr (s) to (s) is uniform since lim m→∞ mr (e k ) = λ k and lim m→∞ mr (e) = λ uniformly in r. Hence, A is a weighted almost conservative matrix.
In the following theorem, we obtain the characterization of weighted almost regular matrices. t n a n,k = 1 uniformly in r.
Proof Necessity. Let A ∈ (c, f (N)) R . We see that condition (13) holds by using the fact that A is also weighted almost conservative. Take e k , e ∈ c. Then A-transforms of the sequences e k and e are weighted almost convergent to 0 and 1, respectively, since e k → 0 and e → 1.
Hence e k ∈ c gives condition (14) and e ∈ c proves the validity of (15). Sufficiency. Let conditions (13)-(15) hold. It is easy to see that A is weighted almost conservative. So, for each (s k ) ∈ c, lim m→∞ mr (s) = (s) uniformly in r. Thus we obtain from (11) and our hypotheses (13)-(15) that (s) = ξ = lim s k . This yields A is weighted almost regular.
We now obtain necessary and sufficient conditions for the matrix A which transform the absolutely convergent series into the space of weighted almost convergence.
lim m→∞ 1 T m r+m-1 n=r t n a n,k = λ k exists for each k ∈ N 0 uniformly in r.
Proof Necessity. Let A ∈ (l 1 , f (N)). Condition (17) follows since e k ∈ l 1 . Let mr be a continuous linear functional on l 1 defined by We obtain from (18) and (19) that r+m-1 n=r t n a n,k .