Generalized statistical convergence of difference sequences

Abstract

In this paper we define the $\lambda (u)$-statistical convergence that generalizes, in a certain sense, the notion of λ-statistical convergence. We find some relations with sets of sequences which are related to the notion of strong convergence.

MSC:40A05, 40H05.

1 Introduction and preliminaries

The notion of statistical convergence (see Fast [1]) has been studied in various setups, and its various generalizations, extensions and variants have been studied by various authors so far. For example, lacunary statistical convergence [2], A-statistical convergence [3, 4], statistical summability $(C,1)$ [5, 6], statistical λ-summability [7], statistical σ-convergence [8], statistical A-summability [9], λ-statistical convergence with order α [10], lacunary and λ-statistical convergence in a solid Riesz space [11, 12], lacunary statistical convergence and ideal convergence in random 2-normed spaces [13, 14], generalized weighted statistical convergence [15]etc. In this paper we define the notion of λ-statistical convergence as a matrix domain of a difference operator [16], which is obtained by replacing the sequence x by $u\mathrm{\Delta }x$, where $\mathrm{\Delta }x={(x_k-x_{k+1})}_{k=1}^{\mathrm{\infty }}$ and $u={(u_k)}_{k=1}^{\mathrm{\infty }}$ is another sequence with $u_k\ne 0$ for all k. We find some relations with sets of sequences which are related to the notion of strong convergence [17].

Let K be a subset of the set of natural numbers ℕ. Then the asymptotic density of K denoted by $\delta (K)$ is defined as $\delta (K)={lim}_n\frac{1}{n}|\{k\le n:k\in K\}|$, where the vertical bars denote the cardinality of the enclosed set.

A number sequence $x=(x_k)$ is said to be statistically convergent to the number L if for each $ϵ>0$, the set $K(ϵ)=\{k\le n:|x_k-L|>ϵ\}$ has asymptotic density zero, i.e.,

$$\underset{n}{lim}\frac{1}{n}\left|\{k\le n:|x_k-L|\}\right|=0.$$

In this case, we write $S\text{-}limx=L$.

Let $\lambda =({\lambda }_n)$ be a non-decreasing sequence of positive numbers tending to ∞ such that

$${\lambda }_{n+1}\le {\lambda }_n+1,{\lambda }_1=0.$$

The generalized de la Vallée-Poussin mean is defined by

$$t_n(x)=:\frac{1}{{\lambda }_n}\sum _{j\in I_n}{x}_j,$$

where $I_n=[n-{\lambda }_n+1,n]$.

A sequence $x=(x_j)$ is said to be $(V,\lambda )$-summable to a number L if

$$t_n(x)\to L\text{as }n\to \mathrm{\infty }.$$

In this case, L is called the λ-limit of x.

Let $K\subseteq \mathbb{N}$. Then the λ-density of K is defined by

$${\delta }_{\lambda }(K)=\underset{n}{lim}\frac{1}{{\lambda }_n}|\{n-{\lambda }_n+1\le j\le n:j\in K\}|.$$

In case ${\lambda }_n=n$, λ-density reduces to the asymptotic density. Also, since $({\lambda }_n/n)\le 1$, $\delta (K)\le {\delta }_{\lambda }(K)$ for every $K\subseteq \mathbb{N}$.

A sequence $x=(x_k)$ is said to be λ-statistically convergent (see [12]) to L if for every $ϵ>0$ the set $K_{ϵ}:=\{k\in \mathbb{N}:|x_k-L|\ge ϵ\}$ has λ-density zero, i.e., ${\delta }_{\lambda }(K_{ϵ})=0$. That is,

$$\underset{n}{lim}\frac{1}{{\lambda }_n}\left|\{k\in I_n:|x_k-L|\ge ϵ\}\right|=0.$$

In this case, we write $S_{\lambda }\text{-}limx=L$ and we denote the set of all λ-statistically convergent sequences by $S_{\lambda }$.

2 $\lambda (u)$-Statistical convergence

We consider the infinite matrix of first difference $\mathrm{\Delta }={(a_{nm})}_{n,m\ge 1}$ defined by $a_{nn}=1$, $a_{n,n+1}=-1$ and $a_{nm}=0$ otherwise. Let $D_u$ be the diagonal matrix defined by ${[D_u]}_{nn}=u_n$ for all n and consider the set U of all sequences such that $u_n\ne 0$ for all n. Then we write $\mathrm{\Delta }(u)=D_u\mathrm{\Delta }$ for $u\in U$.

From the generalized de la Vallée-Poussin mean defined by

$$t_n(x)=\frac{1}{{\lambda }_n}\sum _{k\in I_n}{x}_k\text{for }x={(x_k)}_k,$$

we are led to define the following sets:

$$\begin{array}{r}{[V,\lambda ]}_0(\mathrm{\Delta }(u))=\{x=(x_k):\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{\lambda }_n}\sum _{k\in I_n}|\mathrm{\Delta }(u)x_k|=0\}\\ \phantom{{[V,\lambda ]}_0(\mathrm{\Delta }(u))}=\{x=(x_k):\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{\lambda }_n}\sum _{k\in I_n}|u_k(x_k-x_{k+1})|=0\},\\ {[V,\lambda ]}_{\mathrm{\infty }}(\mathrm{\Delta }(u))=\{x=(x_k):\underset{n}{sup}\frac{1}{{\lambda }_n}\sum _{k\in I_n}|\mathrm{\Delta }(u)x_k|<\mathrm{\infty }\}\\ \phantom{{[V,\lambda ]}_{\mathrm{\infty }}(\mathrm{\Delta }(u))}=\{x=(x_k):\underset{n}{sup}\frac{1}{{\lambda }_n}\sum _{k\in I_n}|u_k(x_k-x_{k+1})|=0\}.\end{array}$$

In the case when ${\lambda }_n=n$, we write the previous sets ${[V]}_0(\mathrm{\Delta }(u))$ and ${[V]}_{\mathrm{\infty }}(\mathrm{\Delta }(u))$, respectively. Now we can state the definition of $\lambda (u)$-statistical convergence to zero.

A sequence $x={(x_k)}_{k\ge 1}$ is said to be $\lambda (u)$-statistically convergent to zero if for every $\epsilon >0$,

$$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{\lambda }_n}\left|\{k\in I_n:|\mathrm{\Delta }(u)x_k|\ge \epsilon \}\right|=0.$$

In this case, we write $x_k\to 0S_{\lambda }(\mathrm{\Delta }(u))$. If ${\lambda }_n=n$ for all n, we then write $x_k\to 0S(\mathrm{\Delta }(u))$.

3 Main results

We are ready to prove the following result.

Theorem 1 Let $u\in U$. Then

• (a) ${[V,\lambda ]}_0(\mathrm{\Delta }(u))\subset S_{\lambda }^0(\mathrm{\Delta }(u))$ and the inclusion is proper,

• (b) if $x\in l_{\mathrm{\infty }}$ and $x_k\to 0S_{\lambda }(\mathrm{\Delta }(u))$, then $x\in {[V,\lambda ]}_0(\mathrm{\Delta }(u))$,

• (c) $S_{\lambda }^0(\mathrm{\Delta }(u))\cap l_{\mathrm{\infty }}={[V,\lambda ]}_0(\mathrm{\Delta }(u))\cap l_{\mathrm{\infty }}$.

Proof (a) Let $\epsilon >0$ be given and $x\in {[V,\lambda ]}_0(\mathrm{\Delta }(u))$. Then we have

$$\frac{1}{{\lambda }_n}\sum _{k\in I_n}|\mathrm{\Delta }(u)x_k|\ge \frac{1}{{\lambda }_n}\underset{|x_k-L|\ge \epsilon }{\sum _{k\in I_n}}|\mathrm{\Delta }(u)x_k|\ge \frac{\epsilon }{{\lambda }_n}\left|\{k\in I_n:|\mathrm{\Delta }(u)x_k|\ge \epsilon \}\right|.$$

Therefore $x\in S_{\lambda }^0(\mathrm{\Delta }(u))$. The following example shows that the inclusion is proper: Let $x=(x_k)$ be defined by

$$x_k=\{\begin{array}{ll}{\sum }_{j=k}^{\mathrm{\infty }}j,& \text{for }n-[\sqrt{{\lambda }_n}]+1\le k\le n,\\ 0,& \text{otherwise.}\end{array}$$

Then $x\notin l_{\mathrm{\infty }}$ and for $0<\epsilon \le 1$,

$$\frac{1}{{\lambda }_n}\left|\{k\in I_n:|\mathrm{\Delta }(u)x_k|\ge \epsilon \}\right|=\frac{[\sqrt{{\lambda }_n}]}{{\lambda }_n}\to 0(n\to \mathrm{\infty }),$$

i.e., $x\in S_{\lambda }^0(\mathrm{\Delta }(u))$. But

$$\frac{1}{{\lambda }_n}\sum _{k\in I_n}|\mathrm{\Delta }(u)x_k|\nrightarrow 0(n\to \mathrm{\infty }),$$

i.e., $x\notin {[V,\lambda ]}_0(\mathrm{\Delta }(u))$.

• (b) Let $x\in l_{\mathrm{\infty }}$ and $x_k\to 0S_{\lambda }(\mathrm{\Delta }(u))$. Then $|\mathrm{\Delta }(u)x_k|\le M$ for all k, where $M>0$. For $\epsilon >0$, we have

  $$\begin{array}{rcl}\frac{1}{{\lambda }_n}\sum _{k\in I_n}|\mathrm{\Delta }(u)x_k|& =& \frac{1}{{\lambda }_n}\underset{|x_k-L|\ge ϵ}{\sum _{k\in I_n}}|\mathrm{\Delta }(u)x_k|+\frac{1}{{\lambda }_n}\underset{|x_k-L|<ϵ}{\sum _{k\in I_n}}|\mathrm{\Delta }(u)x_k|\\ \le & \frac{M}{{\lambda }_n}\left|\{k\in I_n:|\mathrm{\Delta }(u)x_k|\ge \epsilon \}\right|+\epsilon .\end{array}$$

Hence, $x\in {[V,\lambda ]}_0(\mathrm{\Delta }(u))$.

• (c) This immediately follows from (a) and (b).

This completes the proof of the theorem. □

Theorem 2 $S^0(\mathrm{\Delta }(u))\subseteq S_{\lambda }^0(\mathrm{\Delta }(u))$ if and only if

where by $x\in S^0(\mathrm{\Delta }(u))$ (or $x\in S_{\lambda }^0(\mathrm{\Delta }(u))$) we mean $x_k\to 0S(\mathrm{\Delta }(u))$ (or $x_k\to 0S_{\lambda }(\mathrm{\Delta }(u))$).

Proof For $\epsilon >0$, we have

$$\{k\in I_n:|\mathrm{\Delta }(u)x_k|\ge \epsilon \}\subset \{k\le n:|\mathrm{\Delta }(u)x_k|\ge \epsilon \}.$$

Therefore

$$\begin{array}{rcl}\frac{1}{n}\left|\{k\le n:|\mathrm{\Delta }(u)x_k|\ge \epsilon \}\right|& \ge & \frac{1}{n}\left|\{k\in I_n:|\mathrm{\Delta }(u)x_k|\ge \epsilon \}\right|\\ \ge & \frac{{\lambda }_n}{n}\cdot \frac{1}{{\lambda }_n}\left|\{k\in I_n:|\mathrm{\Delta }(u)x_k|\ge \epsilon \}\right|.\end{array}$$

Taking the limit as $n\to \mathrm{\infty }$ and using (∗), we get the inclusion.

Conversely, suppose that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{{\lambda }_n}{n}=0.$$

Choose a subsequence ${(n(j))}_{j\ge 1}$ such that $\frac{{\lambda }_{n(j)}}{n(j)}<\frac{1}{j}$. Define a sequence $x={(x_k)}_{k\ge 1}$ such that

$$\mathrm{\Delta }{x}_k=\{\begin{array}{ll}1,& \text{for }k\in I_{n(j)},j=1,2,3,\dots ,\\ 0,& \text{otherwise.}\end{array}$$

Then $\mathrm{\Delta }x\in [C,1]$ and hence, by Theorem 2.1 of [18], $x\in S^0(\mathrm{\Delta }(u))$. On the other hand, $x\notin {[V,\lambda ]}_0(\mathrm{\Delta }(u))$ and Theorem 1(b) implies that $x\notin S_{\lambda }^0(\mathrm{\Delta }(u))$. Hence, (∗) is necessary.

This completes the proof of the theorem. □

Presently, for the reverse inclusion, we have only one way condition.

Theorem 3 If $lim{sup}_n(n-{\lambda }_n)<\mathrm{\infty }$, then $S_{\lambda }^0(\mathrm{\Delta }(u))\subseteq S^0(\mathrm{\Delta }(u))$.

Proof Let $lim{sup}_n(n-{\lambda }_n)<\mathrm{\infty }$. Then there exists $M>0$ such that $n-{\lambda }_n\le M$ for all n. Since $\frac{1}{n}\le \frac{1}{{\lambda }_n}$ and $\{1\le k\le n:|\mathrm{\Delta }(u)x_k|\ge \epsilon \}$ $\subseteq \{k\in I_n:|\mathrm{\Delta }(u)x_k|\ge \epsilon \}\cup \{1\le k\le n-{\lambda }_n:|\mathrm{\Delta }(u)x_k|\ge \epsilon \}$, we have

$$\begin{array}{r}\frac{1}{n}\left|\{1\le k\le n:|\mathrm{\Delta }(u)x_k|\ge \epsilon \}\right|\\ \le \frac{1}{{\lambda }_n}\left|\{1\le k\le n:|\mathrm{\Delta }(u)x_k|\ge \epsilon \}\right|\\ \le \frac{1}{{\lambda }_n}\left|\{k\in I_n:|\mathrm{\Delta }(u)x_k|\ge \epsilon \}\right|+\frac{1}{{\lambda }_n}\left|\{k\le n-{\lambda }_n:|\mathrm{\Delta }(u)x_k|\ge \epsilon \}\right|\\ \le \frac{1}{{\lambda }_n}\left|\{k\in I_n:|\mathrm{\Delta }(u)x_k|\ge \epsilon \}\right|+\frac{M}{{\lambda }_n}.\end{array}$$

Now, taking the limit as $n\to \mathrm{\infty }$, we get $S_{\lambda }^0(\mathrm{\Delta }(u))\subseteq S^0(\mathrm{\Delta }(u))$.

This completes the proof of the theorem. □