Keywords

1 Introduction and Preliminaries

The notion of statistical convergence of a sequence (of real or complex numbers) was defined by Fast [5] and Steinhaus [15] independently in 1951. After then this subject have been studied by various mathematicians (see [2, 4, 6, 7, 11,12,13,14]).

A sequence \(x=\left( x_{k}\right) \) of real (or complex) numbers is said to be statistically convergent to a number L if for every \(\varepsilon >0\)

$$ \underset{n\rightarrow \infty }{\lim }\frac{1}{n}\left| \left\{ k\le n:\left| x_{k}-L\right| \ge \varepsilon \right\} \right| =0. $$

Let \(\lambda =\left( \lambda _{n}\right) \) be a non-decreasing sequence of positive real numbers tending to \(\infty \) such that

$$ \lambda _{n+1}\le \lambda _{n}+1,\text { }\lambda _{1}=1. $$

The set of all such sequences will be denoted by \(\varLambda .\)

Let \(K\subset \mathbb {N} \), \(\lambda =\left( \lambda _{n}\right) \in \varLambda ,\) and define \(\lambda -\) density of K as

$$ \delta _{\lambda }\left( K\right) =\underset{n\rightarrow \infty }{\lim }\frac{1}{\lambda _{n}}\left| \left\{ k\in I_{n}:\text { }k\in K\right\} \right| $$

where \(I_{n}=\left[ n-\lambda _{n}+1,n\right] \) and \(\left| .\right| \) denotes the number of elements of the involved set. \(\lambda -\)density \(\delta _{\lambda }\left( K\right) \) reduces to the natural density \(\delta \left( K\right) \) in case \(\lambda _{n}=n\) [3].

The generalized de la Vallée-Poussin mean is defined by

$$ t_{n}\left( x\right) =\frac{1}{\lambda _{n}} {\textstyle \sum \limits _{k\in I_{n}}} x_{k}\text {.} $$

A sequence \(x=\left( x_{k}\right) \) is said to be \(\left( V,\lambda \right) \)-summable to a number L (see [10]) if

$$ t_{n}\left( x\right) \rightarrow L\text { as }n\rightarrow \infty \text {.} $$

If \(\lambda _{n}=n\) for each \(n\in \mathbb {N}\), then \(\left( V,\lambda \right) \)-summability reduces to \(\left( C,1\right) \)-summability. We write

$$ \left[ C,1\right] =\left\{ x=\left( x_{k}\right) :\underset{n\rightarrow \infty }{\lim }\frac{1}{n} {\textstyle \sum \limits _{k=1}^{n}} \left| x_{k}-L\right| =0\text { for some }L\right\} , $$
$$ \left[ V,\lambda \right] =\left\{ x=\left( x_{k}\right) :\underset{n\rightarrow \infty }{\lim }\frac{1}{\lambda _{n}} {\textstyle \sum \limits _{k\in I_{n}}} \left| x_{k}-L\right| =0\text { for some }L\right\} $$

for the sets of sequences \(x=\left( x_{k}\right) \) which are strongly Cesàro summable and strongly \(\left( V,\lambda \right) -\)summable, respectively.

The \(\lambda -\)statistical convergence was introduced by Mursaleen in [11] as follows for number sequences.

Let \(\lambda =\left( \lambda _{n}\right) \in \varLambda \). A sequence \(x=\left( x_{k}\right) \) is said to be \(\lambda -\) statistically convergent or \(S_{\lambda }\)-convergent to L if for every \(\varepsilon >0\)

$$ \underset{n\rightarrow \infty }{\lim }\frac{1}{\lambda _{n}}\left| \left\{ k\in I_{n}:\left| x_{k}-L\right| \ge \varepsilon \right\} \right| =0, $$

where \(I_{n}=\left[ n-\lambda _{n}+1,n\right] \). In this case we write \(S_{\lambda }-\lim x=L\) or \(x_{k}\rightarrow L(S_{\lambda })\), and \(S_{\lambda }=\left\{ x=\left( x_{k}\right) :S_{\lambda }-\lim x=L\text { for some number }L\right\} \).

In this study, we determine the relations between the sets \(S_{\lambda d}\) and \(S_{\mu d},\) \(S_{\mu d}\ \)and \(BS_{\lambda d},\) \(BS_{\mu d}\) and \(BS_{\lambda d}\), the sets \(S_{\lambda d}\) and \(\left[ V,\mu \right] _{d}\) for various sequences \(\lambda ,\mu \) in the class \(\varLambda \). Furthermore we determine the relations between the sets \(\left[ V,\lambda \right] _{d}\) and \(\left[ V,\mu \right] _{d}\) for various sequences \(\lambda ,\mu \) in the class \(\varLambda ^{*}\).

Throughout the paper \(c\left( X\right) \) and \(l_{\infty }\left( X\right) \) will denote the sets of convergent and bounded sequences in metric space (X, d),  respectively and by the statement “for all \(n\in \mathbb {N} _{n_{\circ }}\)” we mean “for all \(n\in \mathbb {N} \) except finite numbers of positive integers” where \( \mathbb {N} _{n_{\circ }}=\left\{ n_{\circ },n_{\circ }+1,n_{\circ }+2,...\right\} \) for some \(n_{\circ }\in \mathbb {N} =\left\{ 1,2,3,...\right\} \).

2 \(\lambda _{d}-\)Statistical Convergence and \(\lambda _{d}-\)statistical Boundedness in Metric Spaces

\(\lambda -\)statistical convergence of number sequences was introduced and studied by Mursaleen [11]. In this section we define and study \(\lambda _{d}-\)statistical convergence and \(\lambda _{d}-\)statistical boundedness of a sequence in a metric space and give the relations between the sets of \(\lambda _{d}-\)statistical convergent sequences and the sets of \(\lambda _{d}-\)statistical bounded sequences for various sequences \(\lambda =\left( \lambda _{n}\right) \) in \(\varLambda .\)

Definition 2.1

Let \(\left( X,d\right) \) be a metric space and let \(\lambda =\left( \lambda _{n}\right) \in \varLambda \) be given. A sequence \(x=\left( x_{k}\right) \) in metric space \(\left( X,d\right) \) is called \(\lambda _{d}-\)statistically convergent to a point \(x_{o}\in X\) if for every \(\varepsilon >0\),

$$ \underset{n\rightarrow \infty }{\lim }\frac{1}{\lambda _{n}}\left| \left\{ k\in I_{n}:x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} \right| =0, $$

where \(I_{n}=\left[ n-\lambda _{n}+1,n\right] \) and \(B_{\varepsilon }\left( x_{o}\right) =\left\{ x\in X:d\left( x,x_{o}\right) <\varepsilon \right\} \) is the open ball of radius \(\varepsilon \) and center \(x_{o}.\) The class of \(\lambda _{d}-\)statistically convergent sequences in the metric space \(\left( X,d\right) \) will be denoted by \(S_{\lambda d}\). If a sequence \(x=\left( x_{k}\right) \) in metric space \(\left( X,d\right) \) is \(\lambda _{d} -\) statistically convergent to the point \(x_{o}\in X\) then we write \(x_{k}\rightarrow x_{o}\left[ S_{\lambda d}\right] .\)

In case \(\lambda _{n}=n\), \(S_{\lambda d}\) reduces to the class \(S_{d}\) which is the set of those sequences such that for every \(\varepsilon >0\)

$$ \underset{n\rightarrow \infty }{\lim }\frac{1}{n}\left| \left\{ k\le n:x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} \right| =0 $$

for some \(x_{o}\in X\) [9].

Theorem 2.2

Let \(\left( X,d\right) \) be a metric space, \(\lambda =\left( \lambda _{n}\right) \) and \(\mu =\left( \mu _{n}\right) \) be two sequences in \(\varLambda \) such that \(\lambda _{n}\le \mu _{n}\) for all \(n\in \mathbb {N}_{n_{o}}\). Then in metric space X

\(\left( i\right) \) Every \(\mu _{d}-\)statistically convergent sequence is \(\lambda _{d}-\)statistically convergent, that is \(S_{\mu d}\subseteq S_{\lambda d}\) if

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim }\inf \frac{\lambda _{n}}{\mu _{n}}>0. \end{aligned}$$
(1)

\(\left( ii\right) \) Every \(\lambda _{d}-\)statistically convergent sequence is \(\mu _{d}-\)statistically convergent, that is \(S_{\lambda d}\subseteq \) \(S_{\mu d}\) if

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim }\frac{\lambda _{n}}{\mu _{n}}=1. \end{aligned}$$
(2)

In the following proof we will use the techniques given in [2].

Proof

\(\left( i\right) \) Suppose that \(\lambda _{n}\le \mu _{n}\) for all \(n\in \mathbb {N} _{n_{\circ }}\) and let (1) be satisfied. Then \(I_{n}\subset J_{n}\) and so that for every \(\varepsilon >0\) we may write

$$ \left| \left\{ k\in J_{n}:x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} \right| \ge \left| \left\{ k\in I_{n} :x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} \right| $$

and therefore we have

$$ \frac{1}{\mu _{n}}\left| \left\{ k\in J_{n}:x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} \right| \ge \frac{\lambda _{n}}{\mu _{n} }\frac{1}{\lambda _{n}}\left| \left\{ k\in I_{n}:x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} \right| $$

for all \(n\in \mathbb {N} _{n_{\circ }}\), where \(J_{n}=\left[ n-\mu _{n}+1,n\right] \). Now taking the limit as \(n\rightarrow \infty \) in the last inequality and using (1) we get \(x_{k}\rightarrow x_{o}\left[ S_{\mu d}\right] \Longrightarrow x_{k}\rightarrow x_{o}\left[ S_{\lambda d}\right] \) so that \(S_{\mu d}\subseteq S_{\lambda d}\).

\(\left( ii\right) \) Let \(\left( x_{k}\right) \in S_{\lambda d}\) and (2) be satisfied. Since \(I_{n}\subset J_{n}\), for every \(\varepsilon >0\) we may write

$$\begin{aligned} \frac{1}{\mu _{n}}\left| \left\{ k\in J_{n}:x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} \right|&=\frac{1}{\mu _{n}}\left| \left\{ n-\mu _{n}+1\le k\le n-\lambda _{n}:x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} \right| \\&+\frac{1}{\mu _{n}}\left| \left\{ k\in I_{n}:x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} \right| \\&\le \frac{\mu _{n}-\lambda _{n}}{\mu _{n}}+\frac{1}{\lambda _{n}}\left| \left\{ k\in I_{n}:x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} \right| \\&\le \left( 1-\frac{\lambda _{n}}{\mu _{n}}\right) +\frac{1}{\lambda _{n} }\left| \left\{ k\in I_{n}:x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} \right| \end{aligned}$$

for all \(n\in \mathbb {N} _{n_{\circ }}\).

Since \(\underset{n}{\lim }\frac{\lambda _{n}}{\mu _{n}}=1\) by (2) the first term and since \(x=\left( x_{k}\right) \in S_{\lambda d}\) the second term of right hand side of above inequality tend to 0 as \(n\rightarrow \infty \). This implies that \(\frac{1}{\mu _{n}}\left| \left\{ k\in J_{n}:x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} \right| \rightarrow 0\) as \(n\rightarrow \infty \) and so that \(x_{k} \rightarrow x_{o}\left[ S_{\lambda d}\right] \) \(\Longrightarrow x_{k}\rightarrow x_{o}\left[ S_{\mu d}\right] \). Therefore \(S_{\lambda d}\subseteq \) \(S_{\mu d}\).

From Theorem 2.2 we have the following result.

Corollary 2.3

Let \(\left( X,d\right) \) be a metric space, \(\lambda =\left( \lambda _{n}\right) \) and \(\mu =\left( \mu _{n}\right) \) be two sequences in \(\varLambda \) such that \(\lambda _{n}\le \mu _{n}\) for all \(n\in \mathbb {N}_{n_{o}}\). If (2) holds then \(S_{\lambda d}=S_{\mu d}\).

If we take \(\mu =\left( \mu _{n}\right) =\left( n\right) \) in Corollary 2.3 we have the following result.

Corollary 2.4

Let \(\left( X,d\right) \) be a metric space and \(\lambda =\left( \lambda _{n}\right) \) \(\in \varLambda \). If \(\underset{n\rightarrow \infty }{\lim }\frac{\lambda _{n}}{n}=1\) then \(S_{\lambda d}=S_{d}\).

Definition 2.5

Let \(\left( X,d\right) \) be a metric space and let \(\lambda =\left( \lambda _{n}\right) \in \varLambda \) be given. A sequence \(x=\left( x_{k}\right) \) in metric space \(\left( X,d\right) \) is called \(\lambda _{d}-\)statistically bounded if there exist a point \(x\in X\) and a real number \(M>0\) such that

$$ \lim \limits _{n\rightarrow \infty }\frac{1}{\lambda _{n}}\left| \left\{ k\in I_{n}:d\left( x_{k},x\right) \ge M\right\} \right| =0. $$

The set of \(\lambda _{d}-\)statistically bounded sequences in the metric space \(\left( X,d\right) \) will be denoted by \(BS_{\lambda d}.\)

In case \(\left( \lambda _{n}\right) =\left( n\right) ,\) \(\lambda _{d}-\)statistical boundedness reduces to statistical boundedness and the set of statistically bounded sequences will be denoted by \(BS_{d}\) [8].

Theorem 2.6

Any bounded sequence in a metric space \(\left( X,d\right) \) is \(\lambda _{d}-\)statistically bounded for each \(\lambda \in \varLambda .\)

Proof

Assume that \(x=\left( x_{k}\right) \) is a bounded sequence in a metric space \(\left( X,d\right) \) and let \(\lambda \in \varLambda \) be any element. Since the sequence \(\left( x_{k}\right) \) is bounded then there exist a real number \(M>0\) and a point \(x\in X\) such that \(d\left( x_{k},x\right) <M\) for every \(k\in \mathbb {N=}\) \(\left\{ 1,2,3,...\right\} .\) Then

$$\begin{aligned} \left\{ k\le n:d\left( x_{k},x\right) \ge M\right\} =\emptyset \end{aligned}$$
(3)

and since the inclusion

$$ \left\{ k\in I_{n}:d\left( x_{k},x\right) \ge M\right\} \subset \left\{ k\le n:d\left( x_{k},x\right) \ge M\right\} $$

holds we have \(\left\{ k\in I_{n}:d\left( x_{k},x\right) \ge M\right\} =\emptyset \) for each \(n\in \mathbb {N}.\) Also we have

$$ \lim \limits _{n\rightarrow \infty }\left| \left\{ k\le n:d\left( x_{k},x\right) \ge M\right\} \right| =\left| \left\{ k\in \mathbb {N}:d\left( x_{k},x\right) \ge M\right\} \right| =0 $$

by (3). Now easily we have

$$ \lim \limits _{n\rightarrow \infty }\frac{1}{\lambda _{n}}\left| \left\{ k\in I_{n}:d\left( x_{k},x\right) \ge M\right\} \right| =0 $$

since \(\lambda _{n}\rightarrow \infty \) as \(n\rightarrow \infty .\) Therefore the sequence \(\left( x_{k}\right) \) is \(\lambda _{d}-\)statistically bounded. This completes the proof.

Remark 2.7

The invers of Theorem 2.6 may not be true. For this let us consider the following example:

Example: Let us take \(X=\mathbb {R}\) with usual metric. Then the sequence \(\left( x_{k}\right) \) defined by

$$ \begin{array} [c]{cc} &{} \\ x_{k} &{} =\\ &{} \end{array} \left\{ \begin{array} [c]{c} k,\text { }k=m^{2}\\ \\ \left( -1\right) ^{k},\text { }k\ne m^{2} \end{array} \right. $$

is not bounded. But since the inequality

$$\begin{aligned} \frac{1}{\lambda _{n}}\left| \left\{ k\in I_{n}:\left| x_{k} \right| \ge M\right\} \right|&\le \frac{\sqrt{n}-\sqrt{n-\lambda _{n}}}{\lambda _{n}}\\&=\frac{\lambda _{n}}{\lambda _{n}\left( \sqrt{n}+\sqrt{n-\lambda _{n} }\right) }\\&=\frac{1}{\sqrt{n}+\sqrt{n-\lambda _{n}}}\le \frac{1}{\sqrt{n}} \end{aligned}$$

is satisfied and the right side of this last inequality tends to 0 as \(n\rightarrow \infty ,\) we obtain that the sequence \(\left( x_{k}\right) \) is \(\lambda _{d}-\)statistically bounded.

Theorem 2.8

Let \(\left( X,d\right) \) be a metric space and \(\lambda =\left( \lambda _{n}\right) ,\mu =\left( \mu _{n}\right) \in \varLambda \) be two sequences such that \(\lambda _{n}\le \mu _{n}\) for all \(n\in \mathbb {N}_{n_{o}}.\)

\(\left( i\right) \) Suppose that the inequality (1) is satisfied. Then if a sequence \(x=\left( x_{k}\right) \) in X is \(\mu _{d} -\)statistically convergent, then it is \(\lambda _{d}-\) statistically bounded that is \(S_{\mu d}\subseteq BS_{\lambda d}.\)

\(\left( ii\right) \) Suppose that the equality (2) is satisfied. Then if a sequence \(x=\left( x_{k}\right) \) in X is \(\lambda _{d}-\)statistically bounded then it is \(\mu _{d}-\) statistically bounded that is \(BS_{\lambda d}\subseteq BS_{\mu d}.\)

Proof

\(\left( i\right) \) Suppose that \(\lambda _{n}\le \mu _{n}\) for all \(n\in \mathbb {N}_{n_{o}}\) and let (1) be satisfied. Assume that \(x=\left( x_{k}\right) \) is \(\mu _{d}\)-statistically convergent to \(x_{o}\in X.\) Then \(I_{n}\subset J_{n}\) and so that for \(\varepsilon >0\) and a large \(M>0\) we may write

$$ \left\{ k\in I_{n}:d\left( x_{k},x_{o}\right) \ge M\right\} \subset \left\{ k\in J_{n}:x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} . $$

From this inclusion we obtain the inequality

$$ \left| \left\{ k\in J_{n}:x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} \right| \ge \left| \left\{ k\in I_{n}:d\left( x_{k},x_{o}\right) \ge M\right\} \right| $$

and therefore we have

$$ \frac{1}{\mu _{n}}\left| \left\{ k\in J_{n}:x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} \right| \ge \frac{\lambda _{n}}{\mu _{n} }\frac{1}{\lambda _{n}}\left| \left\{ k\in I_{n}:d\left( x_{k} ,x_{o}\right) \ge M\right\} \right| $$

for all \(n\in \mathbb {N}_{n_{o}},\) where \(J_{n}=\left[ n-\mu _{n}+1,n\right] .\) Now taking the limit as \(n\rightarrow \infty \) in the last inequality and using (1) we get \(x=\left( x_{k}\right) \in BS_{\lambda d}\) so that \(S_{\mu d}\subseteq BS_{\lambda d}\).

(ii) Let \(\left( x_{k}\right) \in BS_{\lambda d}\) and (2) be satisfied. Since \(I_{n}\subset J_{n},\) for a large number \(M>0\) we may write

$$\begin{aligned} \frac{1}{\mu _{n}}\left| \left\{ k\in J_{n}:d\left( x_{k},x_{o}\right) \ge M\right\} \right|&=\frac{1}{\mu _{n}}\left| \left\{ n-\mu _{n}+1\le k\le n-\lambda _{n}:d\left( x_{k},x_{o}\right) \ge M\right\} \right| \qquad \qquad \\&+\frac{1}{\mu _{n}}\left| \left\{ k\in I_{n}:d\left( x_{k} ,x_{o}\right) \ge M\right\} \right| \qquad \qquad \\&\le \frac{\mu _{n}-\lambda _{n}}{\mu _{n}}+\frac{1}{\lambda _{n}}\left| \left\{ k\in I_{n}:d\left( x_{k},x_{o}\right) \ge M\right\} \right| \\&\le \left( 1-\frac{\lambda _{n}}{\mu _{n}}\right) +\frac{1}{\lambda _{n} }\left| \left\{ k\in I_{n}:d\left( x_{k},x_{o}\right) \ge M\right\} \right| \end{aligned}$$

for all \(n\in \mathbb {N}_{n_{o}}.\) Since \(\lim \limits _{n}\frac{\lambda _{n}}{\mu _{n}}=1\) by (2) the first term and since \(x=\left( x_{k}\right) \in BS_{\lambda d}\) the second term of the right hand side of above inequality tend to 0 as \(n\rightarrow \infty \). This implies that

$$ \lim \limits _{n\rightarrow \infty }\frac{1}{\mu _{n}}\left| \left\{ k\in J_{n}:d\left( x_{k},x_{o}\right) \ge M\right\} \right| =0 $$

and so that the sequence \(\left( x_{k}\right) \) is \(\mu _{d}-\)statistically bounded, that is \(\left( x_{k}\right) \in BS_{\mu d}\). Therefore since \(\left( x_{k}\right) \in BS_{\lambda d}\) is an arbitrary element ve have \(BS_{\lambda d}\subseteq BS_{\mu d}\) and this completes the proof.

Theorem 2.9

Let \(\left( X,d\right) \) be a metric space and \(\lambda =\left( \lambda _{n}\right) \in \varLambda \) be given. Then every \(\lambda _{d}-\)statistically convergent sequence in X is \(\lambda _{d} -\)statistically bounded, that is \(S_{\lambda d}\subseteq BS_{\lambda d} \).

Taking \(\left( \mu _{n}\right) =\left( \lambda _{n}\right) \) the proof follows from Theorem 2.8 (i).

Corollary 2.10

Let \(\left( X,d\right) \) be a metric space and \(\lambda =\left( \lambda _{n}\right) \in \varLambda \) be given.

\(\left( i\right) \) Every statistically convergent sequence is \(\lambda _{d} -\)statistically bounded, that is \(S_{d}\subseteq BS_{\lambda d}\) if

$$\begin{aligned} \liminf \limits _{n\rightarrow \infty }\frac{\lambda _{n}}{n}>0. \end{aligned}$$
(4)

\(\left( ii\right) \) Every \(\lambda _{d}-\)statistically bounded sequence is statistically bounded, that is \(BS_{\lambda d}\subseteq BS_{d}\) if

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{\lambda _{n}}{n}=1. \end{aligned}$$
(5)

Taking \(\left( \mu _{n}\right) =\left( n\right) \) the proof follows from Theorem 2.8 (i) and (ii) respectively.

Remark 2.11

The invers of \(\left( i\right) \) in Theorem 2.8 may not be true. For example if we take \(X=\mathbb {R}\) with the usual metric, the sequence \(\left( x_{k}\right) \) defined by

$$ \begin{array} [c]{cc} &{} \\ x_{k} &{} =\\ &{} \end{array} \left\{ \begin{array} [c]{c} 0,\text { }k=2m+1\\ \\ 1,\text { }k=2m \end{array} \right. $$

is not \(\mu _{d}-\)statistically convergent but it is \(\lambda _{d} -\)statistically bounded for any \(\lambda ,\mu \in \varLambda .\) Note that in this example we do not need the restriction \(\lambda _{n}\le \mu _{n}\) for all \(n\in \mathbb {N}_{n_{o}}\).

3 Strong \(\left( V,\lambda \right) _{d}-\)summability in Metric Spaces

\(p-\)strong summability in metric spaces was studied by Bilalov and Nazarova [1]. In this section we introduce and study strong \(\left( V,\lambda \right) _{d}-\) summability and give the relations between the sets of strongly \(\left( V,\lambda \right) _{d}-\) summable sequences for various sequences \(\lambda =\left( \lambda _{n}\right) \) in \(\varLambda ^{*}\) in a metric space \(\left( X,d\right) ,\) where

$$ \varLambda ^{*}=\left\{ \lambda =\left( \lambda _{n}\right) :0<\lambda _{n} \le \lambda _{n+1},\text { for every }n\text { and }\lambda _{n}\rightarrow \infty \text { }\left( n\rightarrow \infty \right) \right\} . $$

Note that in order to obtain the class \(\varLambda ^{*}\) we remove the conditions \(\lambda _{n+1}\le \lambda _{n}+1\) and \(\lambda _{1}=1\) on the class \(\varLambda .\) It is clear that \(\varLambda \subset \varLambda ^{*}\) and the inclusion is strict. For example \(\lambda =\left( \lambda _{n}\right) =\left( n^{2}\right) \in \varLambda ^{*}-\varLambda .\)

In this chapter, we use the class \(\varLambda ^{*}\) instead of \(\varLambda .\)

Definition 3.1

Let \(\left( X,d\right) \) be a metric space and \(\lambda =\left( \lambda _{n}\right) \in \varLambda ^{*}\). A sequence \(x=\left( x_{k}\right) \subset X\) is said to be strongly \(\left( V,\lambda \right) _{d}-\) summable to \(x_{o}\in X\) if

$$ \underset{n\rightarrow \infty }{\lim }\frac{1}{\lambda _{n}}\underset{k\in I_{n} }{\sum }d\left( x_{k},x_{o}\right) =0. $$

We write

$$ \left[ V,\lambda \right] _{d}=\left\{ x=\left( x_{k}\right) :\text { }\underset{n\rightarrow \infty }{\lim }\frac{1}{\lambda _{n}}\underset{k\in I_{n} }{\sum }d\left( x_{k},x_{o}\right) =0\text { for some }x_{o}\in X\right\} $$

for the set of the sequences which are strongly \(\left( V,\lambda \right) _{d}-\)summable in metric space X with the metric d. If a sequence \(x=\left( x_{k}\right) \) in metric space \(\left( X,d\right) \) is strongly \(\left( V,\lambda \right) _{d}-\)summable to the point \(x_{o}\in X\) then we write \(x_{k}\rightarrow x_{o}\left[ V,\lambda \right] _{d}.\)

Strong \(\left( V,\lambda \right) _{d}-\)summability reduces to strong \(\left( C,1\right) _{d}\) -summability in case \(\lambda _{n}=n\) [1].

Theorem 3.2

Let \(\left( X,d\right) \) be a metric space and \(\lambda =\left( \lambda _{n}\right) ,\) \(\mu =\left( \mu _{n}\right) \) \(\in \varLambda ^{*}\) and suppose that \(\lambda _{n}\le \mu _{n}\) for all \(n\in \mathbb {N} _{n_{\circ }}\).

\(\left( i\right) \) If (1) holds then a strongly \(\left( V,\mu \right) _{d}-\)summable sequence in the metric space X is also strongly \(\left( V,\lambda \right) _{d}-\)summable, that is \(\left[ V,\mu \right] _{d}\subseteq \left[ V,\lambda \right] _{d}\),

\(\left( ii\right) \) Suppose (2) holds. If \(x=\left( x_{k}\right) \subset X\) is bounded and \(x_{k}\rightarrow x_{o}\left[ V,\lambda \right] _{d}\) then \(x_{k}\rightarrow x_{o}\left[ V,\mu \right] _{d} \).

Proof

\(\left( i\right) \) Let \(\left( X,d\right) \) be a metric space and suppose that \(\lambda _{n}\le \mu _{n}\) for all \(n\in \mathbb {N} _{n_{\circ }}\). Then \(I_{n}\subseteq J_{n}\) and so that we may write

$$ \frac{1}{\mu _{n}}\underset{k\in J_{n}}{\sum }d\left( x_{k},x_{o}\right) \ge \frac{1}{\mu _{n}}\underset{k\in I_{n}}{\sum }d\left( x_{k},x_{o}\right) $$

for all \(n\in \mathbb {N} _{n_{\circ }}\) and hence we may write the inequality

$$ \frac{1}{\mu _{n}}\underset{k\in J_{n}}{\sum }d\left( x_{k},x_{o}\right) \ge \frac{\lambda _{n}}{\mu _{n}}\frac{1}{\lambda _{n}}\underset{k\in I_{n}}{\sum }d\left( x_{k},x_{o}\right) . $$

Then taking limit as \(n\rightarrow \infty \) in the last inequality and using (1) we obtain \(x_{k}\rightarrow x_{o}\left[ V,\mu \right] _{d}\Longrightarrow x_{k}\rightarrow x_{o}\left[ V,\lambda \right] _{d}\). Since \(x=\left( x_{k}\right) \in \left[ V,\mu \right] _{d}\) is an arbitrary sequence we obtain that \(\left[ V,\mu \right] _{d}\subseteq \left[ V,\lambda \right] _{d}\).

\(\left( ii\right) \) Let the sequence \(x=\left( x_{k}\right) \subset X\) be bounded and \(x_{k}\rightarrow x_{o}\left[ V,\lambda \right] _{d}\). Suppose (2) holds. Since the sequence \(x=\left( x_{k}\right) \ \)is bounded then there exists some open ball \(B_{r}\left( x^{\prime }\right) \) such that \(x_{k}\in B_{r}\left( x^{\prime }\right) \) for all \(k\in \mathbb {N},\) \(r>0\) and \(x^{\prime }\in X\). Hence we may write

$$ d\left( x_{k},x_{\circ }\right) \le d\left( x_{k},x^{\prime }\right) +d\left( x^{\prime },x_{\circ }\right) <r+d\left( x^{\prime },x_{\circ }\right) =M. $$

Now since \(\lambda _{n}\le \mu _{n}\) and so that \(\frac{1}{\mu _{n}}\le \frac{1}{\lambda _{n}}\), and \(I_{n}\subset J_{n}\) for all \(n\in \mathbb {N} _{n_{\circ }}\), we may write

$$\begin{aligned} \frac{1}{\mu _{n}}\underset{k\in J_{n}}{\sum }d\left( x_{k},x_{\circ }\right)&=\frac{1}{\mu _{n}}\underset{k\in J_{n}-I_{n}}{\sum }d\left( x_{k},x_{\circ }\right) +\frac{1}{\mu _{n}}\underset{k\in I_{n}}{\sum }d\left( x_{k} ,x_{\circ }\right) \\&\le \frac{\mu _{n}-\lambda _{n}}{\mu _{n}}M+\frac{1}{\mu _{n}}\underset{k\in I_{n}}{\sum }d\left( x_{k},x_{\circ }\right) \\&\le \left( 1-\frac{\lambda _{n}}{\mu _{n}}\right) M+\frac{1}{\lambda _{n} }\underset{k\in I_{n}}{\sum }d\left( x_{k},x_{\circ }\right) \end{aligned}$$

for every \(n\in \mathbb {N} _{n_{\circ }}\). Since \(\underset{n}{\lim }\frac{\lambda _{n}}{\mu _{n}}=1\) by (2) the first term and since \(x_{k}\rightarrow x_{\circ }\) \(\left[ V,\lambda \right] _{d}\) the second term of rigt hand side of above inequality tend to 0 as \(n\rightarrow \infty \). Hence we get \(x_{k}\rightarrow x_{\circ }\left[ V,\lambda \right] _{d}\Longrightarrow x_{k}\rightarrow x_{\circ }\left[ V,\mu \right] _{d}\).

If we take discrete metric instead of any metric in Theorem 3.2 we have the following result.

Corollary 3.3

Let \(\left( X,d\right) \) be a metric space with discrete metric, \(\lambda =\left( \lambda _{n}\right) \) and \(\mu =\left( \mu _{n}\right) \) be two sequences in \(\varLambda ^{*}\) such that \(\lambda _{n}\le \mu _{n}\) for all \(n\in \mathbb {N}_{n_{o}}\). If (2) holds then \(\left[ V,\lambda \right] _{d}=\left[ V,\mu \right] _{d} \).

Proof

If (2) holds then \(\underset{n\rightarrow \infty }{\lim }\frac{\lambda _{n}}{\mu _{n}}=1>0\) so that (1) holds, too. Then from Theorem 3.2 \(\left( i\right) \) we have \(\left[ V,\mu \right] _{d}\subseteq \left[ V,\lambda \right] _{d}\). Since any sequence in discrete metric space is bounded then any sequence in \(\left[ V,\lambda \right] _{d}\) is bounded and using (2) from Theorem 3.2 \(\left( ii\right) \) we get \(\left[ V,\lambda \right] _{d} \subseteq \left[ V,\mu \right] _{d}\). Both inclusions give the equality \(\left[ V,\lambda \right] _{d}=\left[ V,\mu \right] _{d}\).

4 Relations Between \(\lambda _{d}-\)statistical Convergence and Strong \(\left( V,\lambda \right) _{d}-\)summability in Metric Spaces

In this section we give the relations between the sets of \(\lambda _{d} -\)statistically convergent sequences and the sets of strongly \(\left( V,\lambda \right) _{d}-\)summable sequences for various sequences \(\lambda =\left( \lambda _{n}\right) \) belong to \(\varLambda \) in metric spaces.

Theorem 4.1

Let \(\left( X,d\right) \) be a metric space and \(\lambda =\left( \lambda _{n}\right) \in \varLambda \). Then

\(\left( i\right) \) \(x_{k}\rightarrow x_{o}\left[ V,\lambda \right] _{d}\Longrightarrow x_{k}\rightarrow x_{o}\left[ S_{\lambda d}\right] . \)

\(\left( ii\right) \) If \(\left( x_{k}\right) \) is bounded and \(x_{k}\rightarrow x_{o}\left[ S_{\lambda d}\right] \) then \(x_{k}\rightarrow x_{o}\left[ V,\lambda \right] _{d}\).

Proof

\(\left( i\right) \) Let \(\varepsilon >0\) and \(x_{k}\rightarrow x_{o}\left[ V,\lambda \right] _{d}\). We may write

$$ \sum _{k\in I_{n}}d\left( x_{k},x_{o}\right) \ge {\displaystyle \sum \limits _{\overset{k\in I_{n}}{d\left( x_{k},x_{o}\right) \ge \varepsilon }}} d\left( x_{k},x_{o}\right) \ge \varepsilon .\left| \left\{ k\in I_{n}:x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} \right| \text {} $$

and so that

$$ \frac{1}{\lambda _{n}}\sum _{k\in I_{n}}d\left( x_{k},x_{o}\right) \ge \frac{1}{\lambda _{n}} {\displaystyle \sum \limits _{\overset{k\in I_{n}}{d\left( x_{k},x_{o}\right) \ge \varepsilon }}} d\left( x_{k},x_{o}\right) \ge \frac{1}{\lambda _{n}}\left| \left\{ k\in I_{n}:x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} \right| .\varepsilon \text {.} $$

Hence we obtain that \(x_{k}\rightarrow x_{o}\left[ S_{\lambda d}\right] \) whenever \(x_{k}\rightarrow x_{o}\left[ V,\lambda \right] _{d}\).

\(\left( ii\right) \) Let \(\left( x_{k}\right) \) be a bounded sequence and \(x_{k}\rightarrow x_{o}\left[ S_{\lambda d}\right] \) in metric space \(\left( X,d\right) \). Then there is an open ball \(B_{r}\left( x^{\prime }\right) \subset X\) such that \(x_{k}\in B_{r}\left( x^{\prime }\right) \) for every \(k\in \mathbb {N} \) since \(\left( x_{k}\right) \) is bounded, where \(r>0\) and \(x^{\prime }\in X. \)

Now we may write

$$ d\left( x_{k},x_{o}\right) \le d\left( x_{k},x^{\prime }\right) +d\left( x^{\prime },x_{o}\right) <r+d\left( x^{\prime },x_{o}\right) =M $$

and since \(x_{k}\rightarrow x_{o}\left[ S_{\lambda d}\right] {\text { for every }} \varepsilon > 0 \) we have

$$ \underset{n\rightarrow \infty }{\lim }\frac{1}{\lambda _{n}}\left| \left\{ k\in I_{n}:x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} \right| =0\text {.} $$

Thus we obtain

$$\begin{aligned} \frac{1}{\lambda _{n}}\underset{k\in I_{n}}{\sum }d\left( x_{k},x_{o}\right)&=\frac{1}{\lambda _{n}}\underset{\overset{k\in I_{n}}{d\left( x_{k},x_{\circ }\right) \ge \varepsilon }}{\sum }d\left( x_{k},x_{o}\right) +\frac{1}{\lambda _{n}}\underset{\overset{k\in I_{n}}{d\left( x_{k} ,x_{o}\right)<\varepsilon }}{\sum }d\left( x_{k},x_{o}\right) \\&<\frac{M}{\lambda _{n}}\left| \left\{ k\in I_{n}:x_{k}\notin B_{\varepsilon }\left( x_{o}\right) \right\} \right| +\varepsilon \text {.} \end{aligned}$$

This means that \(x_{k}\rightarrow x_{o}\left[ V,\lambda \right] _{d}\).

Theorem 4.2

Let \(\left( X,d\right) \) be a metric space and \(\lambda =\left( \lambda _{n}\right) ,\) \(\mu =\left( \mu _{n}\right) \) \(\in \varLambda \) such that \(\lambda _{n}\le \mu _{n}\) for all \(n\in \mathbb {N} _{n_{\circ }}\).

\(\left( i\right) \) If (1) holds then

$$ x_{k}\rightarrow x_{\circ }\left[ V,\mu \right] _{d}\Longrightarrow x_{k}\rightarrow x_{\circ }\left[ S_{\lambda d}\right] $$

and the inclusion \(\left[ V,\mu \right] _{d}\subset S_{\lambda d}\) is strict for some \(\lambda ,\mu \in \varLambda \),

\(\left( ii\right) \) If \(\left( x_{k}\right) \ \)is bounded and \(x_{k}\rightarrow x_{\circ }\left[ S_{\lambda d}\right] \) then \(x_{k} \rightarrow x_{\circ }\left[ V,\mu \right] _{d}\), whenever (2) holds.

Proof

\(\left( i\right) \) Let \(\varepsilon >0\) be given and let \(x_{k}\rightarrow x_{\circ }\left[ V,\mu \right] _{d}\). Then for every \(\varepsilon >0\) we may write

$$ \underset{k\in J_{n}}{\sum }d\left( x_{k},x_{\circ }\right) \ge \underset{k\in I_{n}}{\sum }d\left( x_{k},x_{\circ }\right) \ge {\textstyle \sum \limits _{\underset{d\left( x_{k},x_{\circ }\right) \ge \varepsilon }{k\in I_{n}}}} d\left( x_{k},x_{\circ }\right) \ge \varepsilon .\left| \left\{ k\in I_{n}:d\left( x_{k},x_{\circ }\right) \ge \varepsilon \right\} \right| $$

and so that

$$ \frac{1}{\mu _{n}}\underset{k\in J_{n}}{\sum }d\left( x_{k},x_{\circ }\right) \ge \frac{\lambda _{n}}{\mu _{n}}\frac{1}{\lambda _{n}}\left| \left\{ k\in I_{n}:d\left( x_{k},x_{\circ }\right) \ge \varepsilon \right\} \right| .\varepsilon $$

for all \(n\in \mathbb {N} _{n_{\circ }}\). Then taking limit as \(n\rightarrow \infty \) in the last inequality and using (1) we obtain that \(x_{k}\rightarrow x_{\circ }\left[ V,\mu \right] _{d}\Longrightarrow x_{k}\rightarrow x_{\circ }\left[ S_{\lambda d}\right] \). Since \(x=\left( x_{k}\right) \in \left[ V,\mu \right] _{d}\) is an arbitrary sequence we obtain \(\left[ V,\mu \right] _{d}\subseteq S_{\lambda d}\).

To show that the inclusion \(\left[ V,\mu \right] _{d}\subset S_{\lambda d}\) is strict for some \(\lambda ,\mu \in \varLambda \) we take \(X=\mathbb {R}\), \(d\left( x,y\right) =\left| x-y\right| \) and \(\lambda _{n}=\frac{n+1}{2},\) \(\mu _{n}=n\) for all \(n\in \mathbb {N} \). Then \(\underset{n}{\lim }\frac{\lambda _{n}}{\mu _{n}}=\frac{1}{2}>0\) and hence \(\left[ V,\mu \right] _{d}\subseteq S_{\lambda d}\).

Define \(x=\left( x_{k}\right) \) as

$$ x_{k}=\left\{ \begin{array} [c]{ccc} \frac{1}{k}, &{} &{} k\ne m^{3}\\ &{} &{} \\ k, &{} &{} k=m^{3} \end{array} \right. . $$

Let \(\varepsilon >0\) be given. Then there exists \(k_{\circ }\in \mathbb {N} \) such that \(\left| x_{k}\right| <\varepsilon \) for all \(k>k_{\circ }\) and \(k\ne m^{3}\). Now since

$$\begin{aligned} \frac{1}{\lambda _{n}}\left| \left\{ k\in I_{n}:\left| x_{k} \right| \ge \varepsilon \right\} \right|&\le \frac{1}{\lambda _{n} }\left( k_{\circ }+\root 3 \of {n}-\root 3 \of {\frac{n-1}{2}}\right) \\&=\frac{2}{n+1}\left( k_{\circ }+\root 3 \of {n}-\root 3 \of {\frac{n-1}{2}}\right) \rightarrow 0 \end{aligned}$$

as \(n\rightarrow \infty \) we have \(x_{k}\rightarrow 0\left[ S_{\lambda }\right] \left( \mathbb {R}\right) \). On the other hand we know that the equality

$$ 1+2^{3}+3^{3}+4^{3}+...+n^{3}=\frac{n^{2}\left( n+1\right) ^{2}}{4} $$

is satisfied for every \(n\in \mathbb {N} \). Considering this equality, since \(\root 3 \of {n}<\left[ \root 3 \of {n}\right] +1\) and so that \(\frac{1}{n}>\frac{1}{\left( \left[ \root 3 \of {n}\right] +1\right) ^{3}}\) we have

$$\begin{aligned} \frac{1}{\mu _{n}}\underset{k\in J_{n}}{\sum }\left| x_{k}\right|&=\frac{1}{n}\sum _{k=1}^{n}x_{k}=\frac{1}{n}\sum _{\overset{k=1}{k=m^{3}}} ^{n}x_{k}+\frac{1}{n}\sum _{\overset{k=1}{k\ne m^{3}}}^{n}x_{k}>\frac{1}{n}\sum _{\overset{k=1}{k=m^{3}}}^{n}x_{k}=\frac{1}{n}\sum _{\overset{k=1}{k=m^{3}}}^{n}k\\&=\frac{1}{n}\left( 1+2^{3}+3^{3}+4^{3}+...+\left[ \root 3 \of {n}\right] ^{3}\right) \\&=\frac{\left[ \root 3 \of {n}\right] ^{2}\left( \left[ \root 3 \of {n}\right] +1\right) ^{2}}{4n}>\frac{\left[ \root 3 \of {n}\right] ^{2}\left( \left[ \root 3 \of {n}\right] -1\right) ^{2}}{4\left( \left[ \root 3 \of {n}\right] +1\right) ^{3}}\rightarrow \infty \text { }\left( n\rightarrow \infty \right) . \end{aligned}$$

Therefore \(x=\left( x_{k}\right) \notin \left[ V,\mu \right] \left( \mathbb {R}\right) \). Thus the inclusion \(\left[ V,\mu \right] _{d}\subset S_{\lambda d}\) is strict.

\(\left( ii\right) \) Suppose that \(x_{k}\rightarrow x_{\circ }\left[ S_{\lambda d}\right] \) and \(x=\left( x_{k}\right) \ \)is bounded. Then there exist a number \(r>0\) and \(x^{\prime }\in X\) such that \(x_{k}\in B_{r}\left( x^{\prime }\right) \) for all \(k\in \mathbb {N} \). Hence we may write

$$ d\left( x_{k},x_{\circ }\right) \le d\left( x_{k},x^{\prime }\right) +d\left( x^{\prime },x_{\circ }\right) <r+d\left( x^{\prime },x_{\circ }\right) =M. $$

Also since \(\frac{1}{\mu _{n}}\le \frac{1}{\lambda _{n}}\) then for every \(\varepsilon >0\) we may write

$$\begin{aligned} \frac{1}{\mu _{n}}\underset{k\in J_{n}}{\sum }d\left( x_{k},x_{\circ }\right)&=\frac{1}{\mu _{n}}\underset{k\in J_{n}-I_{n}}{\sum }d\left( x_{k},x_{\circ }\right) +\frac{1}{\mu _{n}}\underset{k\in I_{n}}{\sum }d\left( x_{k} ,x_{\circ }\right) \\&\le \frac{\mu _{n}-\lambda _{n}}{\mu _{n}}M+\frac{1}{\mu _{n}}\underset{k\in I_{n}}{\sum }d\left( x_{k},x_{\circ }\right) \\&\le \left( 1-\frac{\lambda _{n}}{\mu _{n}}\right) M+\frac{1}{\lambda _{n} }\underset{k\in I_{n}}{\sum }d\left( x_{k},x_{\circ }\right) . \end{aligned}$$

\(\left( ii\right) \) Suppose that \(x_{k}\rightarrow x_{\circ }\left[ S_{\lambda d}\right] \) and \(x=\left( x_{k}\right) \ \)is bounded. Then there exist a number \(r>0\) and \(x^{\prime }\in X\) such that \(x_{k}\in B_{r}\left( x^{\prime }\right) \) for all \(k\in \mathbb {N} \). Hence we may write

$$ d\left( x_{k},x_{\circ }\right) \le d\left( x_{k},x^{\prime }\right) +d\left( x^{\prime },x_{\circ }\right) <r+d\left( x^{\prime },x_{\circ }\right) =M. $$

Also since \(\frac{1}{\mu _{n}}\le \frac{1}{\lambda _{n}}\) then for every \(\varepsilon >0\) we may write

$$\begin{aligned} \frac{1}{\mu _{n}}\underset{k\in J_{n}}{\sum }d\left( x_{k},x_{\circ }\right)&\le \left( 1-\frac{\lambda _{n}}{\mu _{n}}\right) M+\frac{1}{\lambda _{n} }\underset{\overset{k\in I_{n}}{d\left( x_{k},x_{\circ }\right) \ge \varepsilon }}{\sum }d\left( x_{k},x_{\circ }\right) \\&+\frac{1}{\lambda _{n}}\underset{\overset{k\in I_{n}}{d\left( x_{k},x_{\circ }\right) <\varepsilon }}{\sum }d\left( x_{k},x_{\circ }\right) \\&\le \left( 1-\frac{\lambda _{n}}{\mu _{n}}\right) M+\frac{M}{\lambda _{n} }\left| \left\{ k\in I_{n}:d\left( x_{k},x_{\circ }\right) \ge \varepsilon \right\} \right| +\varepsilon \end{aligned}$$

for all \(n\in \mathbb {N} _{n_{\circ }}\). Using (2) we obtain that \(x_{k}\rightarrow x_{\circ }\left[ V,\mu \right] _{d}\) whenever \(x_{k}\rightarrow x_{\circ }\left[ S_{\lambda d}\right] \).

Corollary 4.3

If \(\underset{n\rightarrow \infty }{\lim } \inf \frac{\lambda _{n}}{\mu _{n}}>0\) then \(S_{\mu d}\cap \left[ V,\mu \right] _{d}\subset S_{\lambda d}\).

If we take \(\mu _{n}=n\) for all n in Theorem 4.2 then we have the following results. Because \(\underset{n\rightarrow \infty }{\lim }\frac{\lambda _{n}}{\mu _{n}}=1\) implies that \(\underset{n\rightarrow \infty }{\lim }\inf \frac{\lambda _{n}}{\mu _{n}}=1>0\), that is (2) \(\Longrightarrow \) (1).

Corollary 4.4

If \(\underset{n\rightarrow \infty }{\lim }\frac{\lambda _{n}}{n}=1\) then

\(\left( i\right) \) If \(\left( x_{k}\right) \ \)is bounded and \(x_{k}\rightarrow x_{\circ }\left[ S_{\lambda d}\right] \) then \(x_{k} \rightarrow x_{\circ }\left[ C,1\right] \left( X_{d}\right) \),

\(\left( ii\right) \) If \(x_{k}\rightarrow x_{\circ }\left[ C,1\right] \left( X_{d}\right) \) then \(x_{k}\rightarrow x_{\circ }\left[ S_{\lambda d}\right] \).

Remark 4.5

Let \(\left( X,d\right) \) be a metric space, \(\lambda =\left( \lambda _{n}\right) \in \varLambda ^{*}\) and \(0<p<\infty \). Define

$$ \left[ V,\lambda \right] _{dp}=\left\{ x=\left( x_{k}\right) :\underset{n\rightarrow \infty }{\lim }\frac{1}{\lambda _{n}}\underset{k\in I_{n} }{\sum }\left[ d\left( x_{k},x_{\circ }\right) \right] ^{p}=0\text { for some }x_{o}\in X\right\} . $$

Then Theorem 4.2 is satisfied for \(\left[ V,\lambda \right] _{dp}\) and \(\left[ V,\mu \right] _{dp}\), if we take \(\left[ V,\lambda \right] _{dp}\) instead of \(\left[ V,\lambda \right] _{d}\) and \(\left[ V,\mu \right] _{dp}\) instead of \(\left[ V,\mu \right] _{d}\).

5 Conclusion

We have introduced and studied \(\lambda _{d}-\)statistical convergence, \(\lambda _{d}-\)statistical boundedness and strong \(\left( V,\lambda \right) _{d}-\)summability for a sequence in a metric space \(\left( X,d\right) \). Furthermore we have established some inclusion relations between the sets \(S_{\lambda d}\) and \(S_{\mu d},\) between the sets \(BS_{\lambda d}\) and \(BS_{\mu d},\) between the sets \(\left[ V,\lambda \right] _{d}\) and \(\left[ V,\mu \right] _{d}\) and between the sets \(S_{\lambda d}\) and \(\left[ V,\mu \right] _{d}\) under some conditions for \(\lambda ,\mu \in \varLambda \) in a metric space \(\left( X,d\right) \).