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

The Original Article was published on 23 May 2022


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


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


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


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

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

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.


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.


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

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

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.


  1. León-Saavedra, F., Listán-García, M.C., Romero de la Rosa, M.P.: On statistical convergence and strong Cesàro convergence by moduli for double sequences. J. Inequal. Appl. 2022, Paper No. 62 (2022). 14, MR4426807

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Correspondence to María del Carmen Listán-García.

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León-Saavedra, F., Listán-García, M.d.C. & Romero de la Rosa, M.d.P. Correction to: On statistical convergence and strong Cesàro convergence by moduli for double sequences. J Inequal Appl 2023, 111 (2023).

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  • Strong Cesàro convergence
  • f-density
  • f-statistical convergence
  • f-strong Cesàro convergence