Correction to: On statistical convergence and strong Cesàro convergence by moduli for double sequences

Abstract

We correct a logic mistake in our paper “On statistical convergence and strong Cesàro convergence by moduli for double sequences” (León-Saavedra et al. in J. Inequal. Appl. 2022:62, 2022).

1 Introduction

It has come to our attention that there is a logic mistake with the converse of some results in our paper [1]. These converse of these results are not central in the papers, but they could be interested in its own right.

Let us denote $\lfloor x\rfloor$ the integer part of $x\in \mathbb{R}$. The following result correct Theorem 3.5 in [1].

Proposition 1.1

a):

If all statistical convergent double sequences $(x_{i,j})$ are f statistical convergent then f must be compatible.

b):

If all strong Cesàro convergent double sequences $(x_{i,j})$ are f-strong Cesàro convergent then f is compatible.

Proof

If f is not compatible, then there exists $c>0$ such that $\limsup_{n}\frac{f(n\varepsilon )}{f(n)}>c$. Assume that $\varepsilon _{k}$ is a decreasing sequence converging to 0, for each k we can construct inductively an increasing sequence $m_{k}$ satisfying $f(m_{k}\varepsilon _{k})>cf(m_{k})$ and

$$ \varepsilon _{k+1}-\frac{m_{k}\varepsilon _{k}-1}{m_{k+1}}< \biggl(1- \sqrt{ \frac{m_{k}}{m_{k+1}}}-\frac{2}{\sqrt{m_{k+1}}} \biggr)^{2}. $$

(1.1)

Let us define by $n_{k}=\lfloor m_{k}\varepsilon _{k}\rfloor +1$, and we set $\ell _{k}=\lfloor \sqrt{m_{k}}\rfloor$. An easy check using (1.1) yields $(\ell _{k+1}-\ell _{k})^{2}>n_{k+1}-n_{k}$.

Let us fix $A_{k+1} \subset [\ell _{k+1}-\lfloor \sqrt{n_{k+1}-n_{k}}\rfloor -1] \times [\ell _{k+1}-\lfloor \sqrt{n_{k+1}-n_{k}}\rfloor -1]$ a subset of $\mathbb{N}\times \mathbb{N}$ such that $\#(A_{k+1})=n_{k+1}-n_{k}$. Let us denote $A=\bigcup_{k}A_{k}$ and set $x_{i,j}=\chi _{A}(i,j)$. Let us see that $x_{i,j}$ is statistically convergent to 0, but not f-statistically convergent. Indeed, for any $m, n\in \mathbb{N}$, there exist p and q, such that $\ell _{p}< m\leq \ell _{p+1}$ and $\ell _{q}< n\leq \ell _{q+1}$. Set $k=\min \{p,q\}$. We can suppose without loss that $k\geq \ell _{k+1}-\lfloor \sqrt{n_{k+1}-n_{k}}\rfloor -1$. Hence, since $(\sqrt{m_{k+1}}-1)^{2}\leq \ell _{k+1}^{2}\leq m_{k}$, for any $\varepsilon >0$:

$$\begin{aligned} \frac{\#\{(i,j) : i\leq m, j\leq n \text{ and } \vert x_{i,j} \vert >\varepsilon \}}{mn} \leq & \frac{n_{k}}{\ell _{k}^{2}}+ \frac{n_{k+1}-n_{k}}{[\ell _{k+1}-\lfloor \sqrt{n_{k+1}-n_{k}}\rfloor -1 ]^{2}} \\ \leq & \frac{n_{k}}{m_{k}}+ \frac{n_{k+1}-n_{k}}{(\sqrt{m_{k+1}}-\sqrt{n_{k+1}-n_{k}}-2)^{2}} \\ \leq & \frac{n_{k}}{m_{k}}+ \frac{\frac{n_{k+1}-n_{k}}{m_{k+1}}}{(1-\sqrt{\frac{n_{k+1}-n_{k}}{m_{k+1}}}-\frac{2}{\sqrt{m_{k+1}}})^{2}} \end{aligned}$$

which goes to zero as $r\to \infty$ as desired. On the other hand, if we set

$$ A_{\varepsilon}(p,q,m,n)= \bigl\{ (i,j) \in \mathbb{N} \times \mathbb{N} : p\leq i\leq m , q\leq j\leq n , \vert x_{ij} \vert > \varepsilon \bigr\} , $$

we shall show that there exists ε such that the limit

$$ \lim_{p,q}\lim_{m,n} \frac{f(\#A_{\varepsilon}(p,q,m,n))}{f(mn)} $$

is not zero. Indeed,

$$ \frac{f(\#\{(i,j) :i\leq \ell _{k}, j\leq \ell _{k}, \vert x_{i,j} \vert >1/2\})}{f(\ell _{k}^{2}) } \geq \frac{f(n_{k})}{f(m_{k})}\geq \frac{f(m_{k}\varepsilon _{k})}{f(m_{k})}\geq c, $$

which gives that

$$ \lim_{m,n}\frac{f(\#A_{1/2}(0,0,m,n))}{f(mn)}\geq c. $$

On the other hand, for each p, q there exists $p'$, $q'$ such that $\ell _{p}'\leq p\leq \ell _{p'+1}$ and $\ell _{q}'\leq q\leq \ell _{q'+1}$. Set $s=\max \{p',q'\}$. Since $A_{1/2}(0,0,m,n)\subset A_{1/2}(p,q,m,n)\cup A_{1/2}(s,s,m,n)$, we get that for any $\delta >0$

$$ \lim_{m,n}\frac{A(p,q,m,n)}{f(mn)} \geq \lim_{m,n} \frac{f(A_{1/2}(0,0,m,n))}{f(mn)}-\frac{f(n_{s})}{f(m,n)}\geq c- \delta $$

which yields part a). Again the part b) is the same proof. □

The next result fixed the converse of Theorem 3.7 in [1].

Proposition 1.2

If all f-strong Cesàro convergent double sequences are f-statistically and bounded then f must be compatible.

Proof

If f is not compatible then there exist two sequences $(\varepsilon _{k})$, $(m_{k})$ satisfying $f(m_{k}\varepsilon _{k})\geq cf(m_{k})$ for some $c>0$. We set $\ell _{k}=\lfloor \sqrt{m_{k}}\rfloor$, we can select $m_{k}$ inductively, such that the sequence

$$ r_{k+1}= \frac{m_{k+1}\varepsilon _{k+1}-m_{k}\varepsilon _{k}}{(\ell _{k+1}-\ell _{k})^{2}} $$

is decreasing and converges to zero. Again it is direct to show that $x_{i,j}=\sum_{i,j}r_{k+1}\chi _{(\ell _{k},\ell _{k+1}]}(i,j)$ is f-statistically convergent to zero, but not f-strong Cesàro convergent. □

Let us recall that f is a compatible modulus function provided $\lim_{\varepsilon \to 0}\limsup_{n}\frac{f(n\varepsilon )}{f(n)}=0$. We will say that a modulus function f is compatible of second order or 2-compatible, provided $\lim_{\varepsilon \to 0}\limsup_{n} \frac{f(n\varepsilon )}{f(n^{2})}=0$. Clearly, if f is compatible, then f is 2-compatible. The next result correct Theorem 2.6 in [1].

Proposition 1.3

Assume that for any f-statistical convergent double sequence $(x_{i,j})$ we have that for any $\varepsilon >0$

$$ \lim_{m,n}\frac{f(\#A_{\varepsilon}(0,0,m,n))}{f(mn)}=0 $$

then f must be 2-compatible.

Proof

Indeed, assume that f is not compatible. Let $\varepsilon _{n}$ be a decreasing sequence converging to 0. Since f is not compatible, there exists $c>0$ such that, for each k, there exists $m_{k}$ such that $f(m_{k}\varepsilon _{k})>cf(m_{k})$. Moreover, we can select $m_{k}$ inductively satisfying

$$ 1-\varepsilon _{k+1}-\frac{1}{m_{k+1}}> \frac{(1-\varepsilon _{k})m_{k}}{m_{k+1}}. $$

(1.2)

Now we use an standard argument used to construct subsets with prescribed densities. Set $n_{k}=\lfloor m_{k}\varepsilon _{k}\rfloor +1$. And extracting a subsequence if it is necessary, we can assume that $n_{1}< n_{2}<\cdots$ , $m_{1}< m_{2}<\cdots$ . Thus, set $A_{k}=[m_{k+1}-(n_{k+1}-n_{k})]\cap \mathbb{N}$. Condition (1.2) guarantee that $A_{k}\subset [m_{k},m_{k+1}]$.

Let us denote $A=\bigcup_{k}A_{k}$, and $x_{n}=\chi _{A}(n)$.

An easy check show that the sequence $x_{1,n}=x_{n}$ is f-statistical convergent to zero, but $\frac{f(\#A_{\varepsilon}(0,0,m_{k},m_{k}))}{f(m_{k}^{2})}\geq c$ which yields the desired result. □

It is worthy to find a 2-compatible function that is not compatible, and to improve Proposition 1.3 replacing 2-compatibility by compatibility.

The corrections have been indicated in this article and the original article [1] has been corrected.