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Correction to: On statistical convergence and strong Cesàro convergence by moduli
Journal of Inequalities and Applications volume 2023, Article number: 110 (2023)
Abstract
We correct a logic mistake in our paper “On statistical convergence and strong Cesàro convergence by moduli” (LeónSaavedra et al. in J. Inequal. Appl. 23:298, 2019).
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. The next result correct Proposition 2.9 in [1].
Proposition 1.1

(a)
If all statistical convergent sequences are fstatistical convergent then f is a compatible modulus function.

(b)
If all strong Cesàro convergent sequences are fstrong Cesàro convergent then f is a compatible modulus function.
Proof
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. Let us denote \(\lfloor x\rfloor \) the integer part of \(x\in \mathbb{R}\). 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 (1.1) guarantee that \(A_{k}\subset [m_{k},m_{k+1}]\).
Let us denote \(A=\bigcup_{k}A_{k}\), and \(x_{n}=\chi _{A}(n)\). Let us prove that \(x_{n}\) is statistical convergent to 0, but not fstatistical convergent, a contradiction. Indeed, for any m, there exists k such that \(m_{k}< m\leq m_{k+1}\). Moreover, we can suppose without loss that \(m\in A\), that is, \(m_{k+1}n_{k+1}+n_{k}\leq m\). Thus for any \(\varepsilon >0\):
as \(k\to \infty \). On the other hand, since \(\varepsilon _{k+1}<\frac{n_{k+1}}{m_{k+1}}\)
which yields (a) as promised. The part (b) is same proof. Indeed, for the sequence \((x_{n})\) defined in part (a), we have that \(\frac{f(\sum_{k=1}^{n}x_{n})}{f(n)}= \frac{f(\{k\leq n x_{k}>\varepsilon \})}{f(n)}\). □
The following result corrects the converse of Theorem 3.4 in [1].
Proposition 1.2
If all fstrong Cesàro convergent sequences are fstatistically and uniformly bounded then f must be compatible.
Proof
Assume that f is not compatible. Thus, as in the proof in Proposition 1.1 we can construct sequences \((\varepsilon _{k})\), \((m_{k})\) such that \(f(m_{k}\varepsilon _{k})\geq c f(m_{k})\) for some \(c>0\). Moreover, we can construct \((m_{k})\) inductively, such that the sequence
is decreasing and converging to 0. Let us consider \(x_{n}=\sum_{k=0}^{\infty}r_{k+1}\chi _{(m_{k},m_{k+1}]}(n)\). Since \((x_{n})\) is decreasing, \((x_{n})\) if fstatistically convergent to 0. On the other hand \(f(\sum_{l=1}^{m_{k}} x_{l})=f(m_{k}\varepsilon _{k})\geq cf(m_{k})\), which gives that \((x_{n})\) is not fstrong Cesàro convergent, as we desired. □
The corrections have been indicated in this article and the original article [1] has been corrected.
References
LeónSaavedra, F., ListánGarcía, M.C., Pérez Fernández, F.J., Romero de la Rosa, M.P.: On statistical convergence and strong Cesàro convergence by moduli. J. Inequal. Appl. 12, Article ID 298 (2019)
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LeónSaavedra, F., ListánGarcía, M.d.C., Pérez Fernández, F.J. et al. Correction to: On statistical convergence and strong Cesàro convergence by moduli. J Inequal Appl 2023, 110 (2023). https://doi.org/10.1186/s13660023029880
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DOI: https://doi.org/10.1186/s13660023029880