A Sequence Space and Uniform $(A,\varphi )$—Statistical Convergence

Abstract

In this, we introduce and study some properties of the new sequence space that is defined using the $\varphi$—function and de la Valée-Poussin mean. We also study some connections between $V_{\lambda }((A,\varphi ))$—strong summability of sequences and $\lambda$—strong convergence with respect to a modulus.

1 Introduction and Background

Let s denote the set of all real and complex sequences $x=(x_{k})$. By $l_{\infty }$ and c, we denote the Banach spaces of bounded and convergent sequences $x=(x_{k})$ normed by $||x||=\sup _{n}|x_{n}|$, respectively. A sequence $x\in l_{\infty }$ is said to be almost convergent if all of its Banach limits coincide. Let $\hat{c}$ denote the space of all almost convergent sequences. Lorentz [6] has shown that

$$\begin{aligned} \hat{c}=\left\{ x\in l_{\infty }: \lim _{m}t_{m,n}(x) \text{ exists } \text{ uniformly } \text{ in } n\right\} \end{aligned}$$

where

$$\begin{aligned} t_{m,n}(x)=\frac{x_{n}+x_{n+1} +x_{n+2}+\cdots + x_{n+m}}{m+1}. \end{aligned}$$

The space $[\hat{c}]$ of strongly almost convergent sequences was introduced by Maddox [7] and also independently by Freedman et al. [3] as follows:

$$\begin{aligned}{}[\hat{c}]=\left\{ x\in l_{\infty }: \lim _{m}t_{m,n}(|x-L |)= 0, \text{ uniformly } \text{ in } n, \text{ for } \text{ some } L\right\} . \end{aligned}$$

Let $\lambda = (\lambda _{i})$ be a nondecreasing sequence of positive numbers tending to $\infty$ such that

$$\begin{aligned} \lambda _{i+1}\le \lambda _{i} +1, \lambda _{1} =1. \end{aligned}$$

The collection of such sequence $\lambda$ will be denoted by $\varDelta .$

The generalized de la Valée-Poussin mean is defined as

$$\begin{aligned} T_{i}(x)= \frac{1}{\lambda _{i}}\sum _{k\in I_{i}}x_{k} \end{aligned}$$

where $I_{i}=[i-\lambda _{i} +1, i]$. A sequence $x=(x_{n})$ is said to be $(V,\lambda )$—summable to a number L, if $T_{i}(x)\rightarrow L \text{ as } i\rightarrow \infty$ (see [9]).

Recently, Malkowsky and Savaş [9] introduced the space $[V,\lambda ]$ of $\lambda$—strongly convergent sequences as follows:

$$\begin{aligned}{}[V,\lambda ]=\left\{ x=(x_{k}):\lim _{i}\frac{1}{\lambda _{i}} \sum _{k\in I_{i}}|x_{k} -L|=0, \text{ for } \text{ some } L \right\} . \end{aligned}$$

Note that in the special case where $\lambda _i=i$, the space $[V,\lambda ]$ reduces the space w of strongly Cesàro summable sequences which is defined as

$$\begin{aligned} w=\left\{ x=(x_{k}):\lim _{i}\frac{1}{i} \sum _{k=1}^i |x_k-L|)=0, \text{ for } \text{ some } L \right\} . \end{aligned}$$

More results on $\lambda$- strong convergence can be seen from [12, 20–24].

Ruckle [16] used the idea of a modulus function f to construct a class of FK spaces

$$\begin{aligned} L(f) = \left\{ x= (x_k) : \sum _{k=1}^{\infty } f\left( |x_k|\right) < \infty \right\} . \end{aligned}$$

The space L(f) is closely related to the space $l_1$, which is an L(f) space with $f(x) = x$ for all real $x\ge 0$.

Maddox [8] introduced and examined some properties of the sequence spaces $w_0 (f)$, w(f), and $w_{\infty } (f)$ defined using a modulus f, which generalized the well-known spaces $w_0$, w and $w_\infty$ of strongly summable sequences.

Recently, Savas [19] generalized the concept of strong almost convergence using a modulus f and examined some properties of the corresponding new sequence spaces.

Waszak [26] defined the lacunary strong $(A,\varphi )$—convergence with respect to a modulus function.

Following Ruckle [16], a modulus function f is a function from $[0, \infty )$ to $[0, \infty )$ such that

• (i) $\displaystyle {f(x) = 0}$ if and only if $x=0$,

• (ii) $\displaystyle {f(x+y) \le f(x) + f(y)}$ for all $x,y \ge 0$,

• (iii) f increasing,

• (iv) f is continuous from the right at zero.

Since $\displaystyle {\left| f(x) - f(y)\right| \le f\left( |x-y|\right) }$, it follows from condition $(\textit{iv})$ that f is continuous on $[0,\infty )$.

If $x=(x_k)$ is a sequence and $A=(a_{nk})$ is an infinite matrix, then Ax is the sequence whose nth term is given by $A_{n}(x) = \sum _{k=0}^{\infty } a_{nk}x_{k}$. Thus we say that x is A-summable to L if $\lim _{n\rightarrow \infty }A_{n}(x) = L$. Let X and Y be two sequence spaces and $A=(a_{nk})$ an infinite matrix. If for each $x\in X$ the series $A_n(x) = \sum _{k=0}^{\infty } a_{nk}x_{k}$ converges for each n and the sequence $Ax= A_n(x) \in Y$ we say that A maps X into Y. By (X, Y) we denote the set of all matrices which maps X into Y, and in addition if the limit is preserved then we denote the class of such matrices by $(X,Y)_{reg}$.

A matrix A is called regular , i.e., $A\in (c,c)_{reg}.$ if $A\in (c,c)$ and $lim_{n}A_n(x) = lim_{k}x_{k}$ for all $x\in c$.

In 1993, Nuray and Savas [14] defined the following sequence spaces:

Definition 1

Let f be a modulus and A a nonnegative regular summability method. We let

$$ w(\hat{A},f) = \left\{ x : \textit{lim}_{n}\sum _{k=1}^{\infty }a_{nk}f(|x_{k+m}-L|)=0, \text{ for } \text{ some } \text{ L, } \text{ uniformly } \text{ in } m \right\} $$

and

$$ w(\hat{A},f)_0 = \left\{ x : \textit{lim}_{n}\sum _{k=1}^{\infty }a_{nk}f(|x_{k+m}|)=0, \text{ uniformly } \text{ in } m \right\} . $$

If we take $A=(a_{nk})$ as

$$\begin{aligned} a_{nk}:=\{ \begin{array}{ccc} \frac{1}{n}, &{} \text{ if } &{} n\ge k , \\ 0, &{} &{} \text{ otherwise. } \end{array} \end{aligned}$$

Then the above definitions are reduced to $[\hat{c}(f)]$ and $[ \hat{c}(f)]_0$ which were defined and studied by Pehlivan [15].

If we take $A=(a_{nk})$ is a de la Valée poussin mean, i.e.,

$$\begin{aligned} a_{nk}:=\{ \begin{array}{ccc} \frac{1}{\lambda _n}, &{} \text{ if } &{} k \in I_n = [ n-\lambda _{n}+1,n] , \\ 0, &{} &{} \text{ otherwise. } \end{array} \end{aligned}$$

Then these definitions are reduced to the following sequence spaces which were defined and studied by Malkowsky and Savas [9].

$$\begin{aligned} w(\hat{V},\lambda ,f) = \left\{ x : lim_{j}\frac{1}{\lambda _j}\sum _{k \in I_j}f(|x_{k+m}-L|)=0, \text{ for } \text{ some } \text{ L, } \text{ uniformly } \text{ in } m \right\} \end{aligned}$$

and

$$\begin{aligned} w(\hat{V},\lambda ,f)_0 = \left\{ x : lim_{j}\frac{1}{\lambda _j}\sum _{k \in I_j}f(|x_{k+m}|)=0, \text{ uniformly } \text{ in } m \right\} \end{aligned}$$

When $\lambda _j = j$ the above sequence spaces become $[\hat{c}(f)]_0$ and $[\hat{c}(f)]$.

By a $\varphi$-function we understand a continuous nondecreasing function $\varphi (u)$ defined for $u\ge 0$ and such that $\varphi (0)=0, \varphi (u)>0$, for $u>0$ and $\varphi (u)\rightarrow \infty$ as $u\rightarrow \infty$, (see, [26]).

A $\varphi$-function $\varphi$ is called non-weaker than a $\varphi$-function $\psi$ if there are constants $c, b, k,l>0$ such that $c\psi (lu)\le b\varphi (ku)$, (for all large u) and we write $\psi \prec \varphi$.

A $\varphi$-function $\varphi$ and $\psi$ are called equivalent and we write $\varphi \sim \psi$ if there are positive constants $b_1,b_2, c, k_1, k_2 , l$ such that $b_ 1\varphi (k_1u) \le c\psi (lu)\le b_2\varphi (k_2u)$, (for all large u ), (see, [26]).

A $\varphi$-function $\varphi$ is said to satisfy $(\varDelta _2)$-condition, (for all large u) if there exists constant $K>1$ such that $\varphi (2u) \le K\varphi (u)$.

In this paper, we introduce and study some properties of the following sequence space that is defined using the $\varphi$- function and de la Valée-Poussin mean and some known results are also obtained as special cases.

2 Main Results

Let $\varLambda =(\lambda _j)$ be the same as above, $\varphi$ be given $\varphi$-function, and f be given modulus function, respectively. Moreover, let $\mathbf {A}=(a_{nk}(i))$ be the generalized three-parametric real matrix. Then we define

$$\begin{aligned} V_{\lambda }^{0}((A,\varphi ),f)= \left\{ x=(x_{k}): \lim _{j}\frac{1}{\lambda _{j}}\sum _{n\in I_{j}}f\Big (\left| \sum _{k=1}^{\infty }a_{nk}(i)\varphi (|x_{k}|)\right| \Big )= 0, \text{ uniformly } \text{ in } i \right\} . \end{aligned}$$

If $\lambda _j = j,$ we have

$$\begin{aligned} V_{\lambda }^{0}((A,\varphi ),f)= \left\{ x=(x_{k}): \lim _{j}\frac{1}{j}\sum _{n=1}^{j}f\Big (\left| \sum _{k=1}^{\infty }a_{nk}(i)\varphi (|x_{k}|)\right| \Big )= 0, \text{ uniformly } \text{ in } i \right\} . \end{aligned}$$

If $x\in V_{\lambda }^{0}((A,\varphi ),f)$, the sequence x is said to be $\lambda$—strong $(A,\varphi )$—convergent to zero with respect to a modulus f. When $\varphi (x)= x$ for all x, we obtain

$$\begin{aligned} V_{\lambda }^{0}((A),f)= \left\{ x=(x_{k}): \lim _{j}\frac{1}{\lambda _{j}}\sum _{n\in I_{j}}f\Big (\left| \sum _{k=1}^{\infty }a_{nk}(i)(|x_{k}|)\right| \Big )= 0, \text{ uniformly } \text{ in } i \right\} . \end{aligned}$$

If $f(x) = x$, we write

$$\begin{aligned} V_{\lambda }^{0}(A,\varphi )= \left\{ x=(x_{k}): \lim _{j}\frac{1}{\lambda _{j}}\sum _{n\in I_{j}}\Big (\left| \sum _{k=1}^{\infty }a_{nk}(i)\varphi (|x_{k}|)\right| \Big )= 0, \text{ uniformly } \text{ in } i \right\} . \end{aligned}$$

If we take $A=I$ and $\varphi (x)=x$ respectively, then we have

$$\begin{aligned} V_{\lambda }^{0}(I,f) = \left\{ x=(x_{k}): \lim _{j}\frac{1}{\lambda _{j}}\sum _{k\in I_{j}}f \Big (\left| x_{k}\right| \Big )=0 \right\} . \end{aligned}$$

If we take $A=I$, $\varphi (x)=x$ and $f(x)=x$ respectively, then we have

$$\begin{aligned} V_{\lambda }^{0}((I)) = \left\{ x=(x_{k}): \lim _{j}\frac{1}{\lambda _{j}}\sum _{k\in I_{j}}|x_{k}|=0 \right\} , \end{aligned}$$

which was defined and studied by Savaş and Savaş [18].

If we define the matrix $A=(a_{nk}(i))$ as follows: for all i

$$\begin{aligned} a_{nk}(i):=\{ \begin{array}{ccc} \frac{1}{n}, &{} \text{ if } &{} n\ge k , \\ 0, &{} &{} \text{ otherwise. }\end{array}\end{aligned}$$

then we have,

$$\begin{aligned} V_{\lambda }^{0}(\mathbf {C},\varphi ,f)= \left\{ x=(x_{k}): \lim _{j}\frac{1}{\lambda _{j}}\sum _{n\in I_{j}}f\Big (\left| \frac{1}{n}\sum _{k=1}^{n}\varphi (|x_{k}|)\right| \Big )= 0 \right\} . \end{aligned}$$

If we define

$$\begin{aligned} a_{nk}(i):=\{ \begin{array}{ccc} \frac{1}{n}, &{} \text{ if } &{} i\le k\le i+n-1 , \\ 0, &{} &{} \text{ otherwise. } \end{array}\end{aligned}$$

then we have,

$$\begin{aligned} V_{\lambda }^{0}(\hat{c},\varphi ,f)= \left\{ x=(x_{k}): \lim _{j}\frac{1}{\lambda _{j}}\sum _{n\in I_{j}}f \Big (\left| \frac{1}{n}\sum _{k=i}^{i+n}\varphi (|x_{k}|)\right| \Big )= 0, \text{ uniformly } \text{ in } i \right\} . \end{aligned}$$

We now have:

Theorem 1

Let $\mathbf A =(a_{nk}(i))$ be the generalized three parametric real matrix and let the $\varphi$—function $\varphi (u)$ satisfy the condition $(\varDelta _2)$. Then the following conditions are true:

(a) If $x = (x_k)\in w((\mathbf A ,\varphi ),f)$ and $\alpha$ is an arbitrary number, then $\alpha x\in w((\mathbf A ,\varphi ),f).$

(b) If $x,y\in w((\mathbf A ,\varphi ),f)$ where $x = (x_k)$, $y=(y_k)$ and $\alpha , \beta$ are given numbers, then $\alpha x + \beta y\in w((\mathbf A ,\varphi ),f).$

The proof is a routine verification by using standard techniques and hence is omitted.

Theorem 2

Let f be a modulus function.

$$V_{\lambda }^{0}(A,\varphi )\subseteq V_{\lambda }^{0}((A,\varphi ),f).$$

Proof

Let $x\in V_{\lambda }^{0}(A,\varphi )$. For a given $\varepsilon >0$ we choose $0<\delta <1$ such that $f(x)<\varepsilon$ for every $x\in [0,\delta ]$. We can write for all i

$$\frac{1}{\lambda _{j}}\sum _{n\in I_{j}}f\Big (\left| \sum _{k=1}^{\infty }a_{nk}(i)\varphi (| x_{k}|)\right| \Big ) = S_1 + S_2 ,$$

where $S_1=\frac{1}{\lambda _{j}}\sum _{n\in I_{j}}f\Big (\left| \sum _{k=1}^{\infty }a_{nk}(i)\varphi (| x_{k}|)\right| \Big )$ and this sum is taken over

$$\sum _{k=1}^{\infty }a_{nk}(i)\varphi (|x_{k}|)\le \delta $$

and

$$ S_2=\frac{1}{\lambda _{j}}\sum _{n\in I_{j}}f\Big (\left| \sum _{k=1}^{\infty }a_{nk}(i)\varphi (| x_{k}|)\right| \Big ) $$

and this sum is taken over

$$ \sum _{k=1}^{\infty }a_{nk}(i)\varphi (| x_{k}|)> \delta . $$

By definition of the modulus f we have $S_1=\frac{1}{\lambda _{j}}\sum _{n\in I_{j}}f\Big (\delta \Big ) = f(\delta )<\varepsilon$ and moreover

$$S_2=f(1)\frac{1}{\delta }\frac{1}{\lambda _{j}}\sum _{n\in I_{j}}\sum _{k=1}^{\infty }a_{nk}(i)\varphi (|x_{k}|).$$

Thus we have $x\in V_{\lambda }^{0}((A,\varphi ),f)$.

This completes the proof.

3 Uniform $(A,\varphi )$—Statistical Convergence

The idea of convergence of a real sequence was extended to statistical convergence by Fast [2] (see also Schoenberg [25]) as follows: If $\mathbb {N}$ denotes the set of natural numbers and $K\subset \mathbb {N}$ then K(m, n) denotes the cardinality of the set $K\cap [m,n],$ the upper and lower natural densities of the subset K are defined as

$$\begin{aligned} \overline{d}(K)= \displaystyle {\lim _{n\rightarrow \infty }} \sup \frac{K(1,n)}{n} ~~\text{ and } ~~\underline{d}(K)=\displaystyle {\lim _{n\rightarrow \infty }}\inf \frac{K(1,n)}{n}. \end{aligned}$$

If $\overline{d}(K)=\underline{d}(K)$ then we say that the natural density of K exists and it is denoted simply by d(K). Clearly $d(K)= \displaystyle {\lim _{n\rightarrow \infty }} \frac{K(1,n)}{n}.$

A sequence $(x_{k})$ of real numbers is said to be statistically convergent to L if for arbitrary $\epsilon >0,$ the set $K(\epsilon )=\{k\in \mathbb {N}: |x_k-L|\ge \epsilon \}$ has natural density zero.

Statistical convergence turned out to be one of the most active areas of research in summability theory after the work of Fridy [4] and Šalát [17].

In another direction, a new type of convergence called $\lambda$-statistical convergence was introduced in [13] as follows.

A sequence $(x_{k})$ of real numbers is said to be $\lambda$- statistically convergent to L (or, $S_\lambda$-convergent to L) if for any $\epsilon >0,$

$$\begin{aligned} \displaystyle {\lim _{ j\rightarrow \infty }}~\frac{1}{\lambda _{j}}|\{k\in I_{j}:|x_{k}-L|\ge \epsilon \}|=0 \end{aligned}$$

where |A| denotes the cardinality of $A\subset \mathbb {N}.$ In [13] the relation between $\lambda$-statistical convergence and statistical convergence was established among other things.

Recently, Savas [20] defined almost $\lambda$-statistical convergence using the notion of $(V, \lambda )$-summability to generalize the concept of statistical convergence.

Assume that A is a nonnegative regular summability matrix. Then the sequence $x=(x_n)$ is called statistically convergent to L provided that, for every $\varepsilon >0$, (see, [5])

$$\begin{aligned} lim_{j} \sum _{n: |x_{n}-L|\ge \varepsilon }a_{jn} =0. \end{aligned}$$

Let $\mathbf {A}=(a_{nk}(i))$ be the generalized three parametric real matrix and the sequence $x=(x_k)$, the $\varphi$-function $\varphi (u)$ and a positive number $\varepsilon > 0$ be given. We write, for all i

$$\begin{aligned} K_{\lambda }^j((A,\varphi ),\varepsilon )= \{ n \in I_j : \sum _{k=1}^{\infty }a_{nk}(i)\varphi (|x_{k}|)\ge \varepsilon \}. \end{aligned}$$

The sequence x is said to be uniform $(A,\varphi )$—statistically convergent to a number zero if for every $\varepsilon > 0$

$$\begin{aligned} lim_j \frac{1}{\lambda _j}\mu (K_{\lambda }^j((A,\varphi ),\varepsilon ))=0, \text{ uniformly } \text{ in } i \end{aligned}$$

where $\mu (K_{\lambda }^j((A,\varphi ),\varepsilon ))$ denotes the number of elements belonging to $K_{\lambda }^j((A,\varphi ),\varepsilon )$. We denote by $S_{\lambda }^0((A,\varphi )),$ the set of sequences $x=(x_k)$ which are uniform $(A,\varphi )$—statistical convergent to zero.

If we take $A= I$ and $\varphi (x) = x$ respectively, then $S_{\lambda }^0((A,\varphi ))$ reduce to $S_{\lambda }^0$which was defined as follows, (see, Mursaleen [13]).

$$\begin{aligned} S_{\lambda }^0= \left\{ x=(x_k): lim_j\frac{1}{\lambda _j}|\{k\in I_j:|x_{k}|\ge \varepsilon \}| =0 \right\} . \end{aligned}$$

Remark 1

(i) If for all i,

$$ a_{nk}:=\{ \begin{array}{ccc} \frac{1}{n}, &{} \text{ if } &{}n\ge k ,\\ 0, &{} &{}\text{ otherwise. } \end{array} $$

then $S_{\lambda }((A,\varphi ))$ reduce to $S_{\lambda }^0((C,\varphi ))$, i.e., uniform $(C,\varphi )$— statistical convergence. (ii) If for all i, (see, [1]),

$$ a_{nk}:=\{ \begin{array}{ccc} \frac{p_k}{P_n}, &{} \text{ if } &{}n\ge k ,\\ 0, &{} &{}\text{ otherwise. } \end{array} $$

then $S_{\lambda }((A,\varphi ))$ reduce to $S_{\lambda }^0((N,p),\varphi ))$, i.e., uniform $((N,p),\varphi )$— statistical convergence, where $p=p_k$ is a sequence of nonnegative numbers such that $p_{0}> 0$ and

$$P_i = \sum _{k=0}^{n}p_k \rightarrow \infty ( n\rightarrow \infty ). $$

We are now ready to state the following theorem.

Theorem 3

If $\psi \prec \varphi$ then $S_{\lambda }^0((A,\psi ))\subset S_{\lambda }^0((A,\varphi )).$

Proof

By our assumptions we have $\psi (|x_{k}|)\le b\varphi (c|x_{k}|)$ and we have for all i,

$$ \sum _{k=1}^{\infty }a_{nk}(i)\psi (|x_{k}|) \le b \sum _{k=1}^{\infty }a_{nk}(i)\varphi (c|x_{k}|)\le K\sum _{k=1}^{\infty }a_{nk}(i)\varphi (|x_{k}|)$$

for $b,c > 0$, where the constant K is connected with properties of $\varphi$. Thus, the condition $\sum _{k=1}^{\infty }a_{nk}(i)\psi (|x_{k}|)\ge \varepsilon$ implies the condition $\sum _{k=1}^{\infty }a_{nk}(i)\varphi (|x_{k}|)\ge \varepsilon$ and in consequence we get

$$\mu (K_{\lambda }^j((A,\varphi ),\varepsilon ))\subset \mu (K_{\lambda }^j((A,\psi ),\varepsilon ))$$

and

$$lim_j \frac{1}{\lambda _j}\mu \Big (K_{\lambda }^j((A,\varphi ),\varepsilon ))\le lim_j \frac{1}{\lambda _j}\mu (K_{\lambda }^j ((A,\psi ),\varepsilon ))\Big ).$$

This completes the proof.

Theorem 4

(a) If the matrix A, functions f, and $\varphi$ are given, then

$$V_{\lambda }^{0}((A,\varphi ),f)\subset S_{\lambda }^{0}(A,\varphi ).$$

(b) If the $\varphi$- function $\varphi (u)$ and the matrix A are given, and if the modulus function f is bounded, then

$$S_{\lambda }^{0}(A,\varphi )\subset V_{\lambda }^{0}(A,\varphi ),f).$$

(c) If the $\varphi$- function $\varphi (u)$ and the matrix A are given, and if the modulus function f is bounded, then

$$S_{\lambda }^{0}(A,\varphi )= V_{\lambda }^{0}(A,\varphi ),f).$$

Proof

(a) Let f be a modulus function and let $\varepsilon$ be a positive number. We write the following inequalities:

$$\begin{aligned} \frac{1}{\lambda _{j}}\sum _{n\in I_{j}}&f\Big (\left| \sum _{k=1}^{\infty }a_{nk}(i)\varphi (| x_{k}|)\right| \Big )\\&\ge \frac{1}{\lambda _{j}} \sum _{n\in I_{j}^1}f\Big (\left| \sum _{k=1}^{\infty }a_{nk}(i)\varphi (|x_{k}|)\right| \Big )\\&\ge \frac{1}{\lambda _{j}} f(\varepsilon ) \sum _{n\in I_{j}^1}1\\&\ge \frac{1}{\lambda _{j}}f(\varepsilon )\mu (K_{\lambda }^j(A,\varphi ),\varepsilon ),\\ \end{aligned}$$

where

$$ I_j^1 = \left\{ n \in I_j: \sum _{k=1}^{\infty }a_{nk}(i)\varphi (|x_{k}|)\ge \varepsilon \right\} . $$

Finally, if $x\in V_{\lambda }^{0}((A,\varphi ),f)$ then $x\in S_{\lambda }^{0}(A,\varphi )$.

(b) Let us suppose that $x\in S_{\lambda }^{0}(A,\varphi ).$ If the modulus function f is a bounded function, then there exists an integer M such that $f(x)<M$ for $x\ge 0$. Let us take

$$ I_j^2 = \left\{ n \in I_j: \sum _{k=1}^{\infty }a_{nk}(i)\varphi (|x_{k}|)<\varepsilon \right\} . $$

Thus we have

$$\begin{aligned} \frac{1}{\lambda _{j}}\sum _{n\in I_{j}}&f\Big (\left| \sum _{k=1}^{\infty }a_{nk}(i)\varphi (| x_{k}|)\right| \Big )\\&\le \frac{1}{\lambda _{j}} \sum _{n\in I_{j}^1}f\Big (\left| \sum _{k=1}^{\infty }a_{nk}(i)\varphi (|x_{k}|)\right| \Big )\\&+\frac{1}{\lambda _{j}} \sum _{n\in I_{j}^2}f\Big (\left| \sum _{k=1}^{\infty }a_{nk}(i) \varphi (|x_{k}|)\right| \Big )\\&\le \frac{1}{\lambda _{j}}M \mu (K_{\lambda }^j((A,\varphi ),\varepsilon ) + f(\varepsilon ).\\ \end{aligned}$$

Taking the limit as $\varepsilon \rightarrow 0$, we obtain that $x\in V_{\lambda }^{0}(A,\varphi ,f).$

The proof of (c) follows from (a) and (b).

This completes the proof.

In the next theorem we prove the following relation.

Theorem 5

If a sequence $x=(x_{k})$ is $S(A,\varphi )$—convergent to L and

$$ \lim inf_{j} \Big (\frac{\lambda _{j}}{j} \Big ) > 0 $$

then it is $S_{\lambda }(A,\varphi )$ convergent to L, where

$$ S(A,\varphi )= \{ x=(x_{k}): lim_{j} \frac{1}{j}\mu (K(A,\varphi ,\varepsilon ))=0 \}.$$

Proof

For a given $\varepsilon >0$, we have, for all i

$$\{ n\in I_{j}: \sum _{k=0}^{\infty }a_{nk}(i)\varphi ({|x_{k}-L|}) \ge \varepsilon \} \subseteq \{ n\le j: \sum _{k=0}^{\infty }a_{nk}(i)\varphi ({|x_{k}-L|}) \ge \varepsilon \}.$$

Hence we have,

$$K_{\lambda }(A,\varphi ,\varepsilon )\subseteq K(A,\varphi ,\varepsilon ).$$

Finally the proof follows from the following inequality:

$$ \frac{1}{j}\mu (K(A,\varphi ,\varepsilon ))\ge \frac{1}{j}\mu (K_{\lambda }(A,\varphi ,\varepsilon )) = \frac{\lambda _{j}}{j}\frac{1}{\lambda _{j}}\mu (K_{\lambda }(A,\varphi ,\varepsilon )) .$$

This completes the proof.

Theorem 6

If $\lambda \in \triangle$ be such that $lim_j \frac{\lambda _j}{j}=1$ and the sequence $x=(x_{k})$ is $S_{\lambda }(A,\varphi )$—convergent to L then it is $S(A,\varphi )$ convergent to L,

Proof

Let $\delta >0$ be given. Since $lim_j \frac{\lambda _j}{j}=1$, we can choose $m \in N$ such that $|\frac{\lambda _j}{j}-1 |<\frac{\delta }{2},$ for all $j \ge m .$ Now observe that, for $\varepsilon >0$

$$\begin{aligned}&\frac{1}{j}\left| \left\{ n\le j:\sum _{k=0}^{\infty }a_{nk}(i)\varphi ({|x_{k}-L|}) \ge \varepsilon \right\} \right| \\&= \frac{1}{j}\left| \left\{ k\le j-\lambda _j:\sum _{k=0}^{\infty }a_{nk}(i)\varphi ({|x_{k}-L|}) \ge \epsilon \right\} \right| \\&+ \frac{1}{j}\left| \left\{ n \in I_j:\sum _{k=0}^{\infty }a_{nk}(i)\varphi ({|x_{k}-L|}) \ge \epsilon \right\} \right| \\&\le \frac{j-\lambda _j}{j} + \frac{1}{j}\left| \left\{ n \in I_j:\sum _{k=0}^{\infty }a_{nk}(i)\varphi ({|x_{k}-L|}) \ge \epsilon \right\} \right| \\&\le 1-(1-\frac{\delta }{2})+\frac{1}{j}\left| \left\{ n \in I_j:\sum _{k=0}^{\infty }a_{nk}(i)\varphi ({|x_{k}-L|}) \ge \epsilon \right\} \right| \\&= \frac{\delta }{2}+ \frac{1}{j}\left| \left\{ n \in I_j:\sum _{k=0}^{\infty }a_{nk}(i)\varphi ({|x_{k}-L|}) \ge \epsilon \right\} \right| , \end{aligned}$$

This completes the proof.