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Correction to: On statistical convergence and strong Cesàro convergence by moduli for double sequences
Journal of Inequalities and Applications volume 2023, Article number: 111 (2023)
Abstract
We correct a logic mistake in our paper “On statistical convergence and strong Cesàro convergence by moduli for double sequences” (LeónSaavedra 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 fstrong 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
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 fstatistically 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 \(\varepsilon >0\):
which goes to zero as \(r\to \infty \) as desired. On the other hand, if we set
we shall show that there exists ε such that the limit
is not zero. Indeed,
which gives that
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 \(\delta >0\)
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 fstrong Cesàro convergent double sequences are fstatistically 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
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 fstatistically convergent to zero, but not fstrong 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 2compatible, provided \(\lim_{\varepsilon \to 0}\limsup_{n} \frac{f(n\varepsilon )}{f(n^{2})}=0\). Clearly, if f is compatible, then f is 2compatible. The next result correct Theorem 2.6 in [1].
Proposition 1.3
Assume that for any fstatistical convergent double sequence \((x_{i,j})\) we have that for any \(\varepsilon >0\)
then f must be 2compatible.
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
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 fstatistical 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 2compatible function that is not compatible, and to improve Proposition 1.3 replacing 2compatibility by compatibility.
The corrections have been indicated in this article and the original article [1] has been corrected.
References
LeónSaavedra, F., ListánGarcía, 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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LeónSaavedra, F., ListánGarcía, 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). https://doi.org/10.1186/s13660023029871
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DOI: https://doi.org/10.1186/s13660023029871