Correction to: On statistical convergence and strong Cesàro convergence by moduli

We correct a logic mistake in our paper “On statistical convergence and strong Cesàro convergence by moduli” (León-Saavedra et al. in J. Inequal. Appl. 23:298, 2019).


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].Proof Let ε n be a decreasing sequence converging to 0. Since f is not compatible, there exists c > 0 such that, for each k, there exists Now we use an standard argument used to construct subsets with prescribed densities.Let us denote x the integer part of Let us prove that x n is statistical convergent to 0, but not f -statistical convergent, a contradiction.Indeed, for any m, there exists k such that m k < m ≤ m k+1 .Moreover, we can suppose without loss that m ∈ A, that is, © The Author(s) 2023.Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made.The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material.If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder.To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.m k+1n k+1 + n k ≤ m.Thus for any ε > 0: which yields (a) as promised.The part (b) is same proof.Indeed, for the sequence (x n ) defined in part (a), we have that .
The following result corrects the converse of Theorem 3.4 in [1].
Proposition 1.2 If all f -strong Cesàro convergent sequences are f -statistically 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 (ε k ), (m k ) such that f (m k ε k ) ≥ cf (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 = ∞ k=0 r k+1 χ (m k ,m k+1 ] (n).Since (x n ) is decreasing, (x n ) if f -statistically convergent to 0. On the other hand f ( , which gives that (x n ) is not f -strong Cesàro convergent, as we desired.
The corrections have been indicated in this article and the original article [1] has been corrected.

Proposition 1. 1
(a) If all statistical convergent sequences are f -statistical convergent then f is a compatible modulus function.(b) If all strong Cesàro convergent sequences are f -strong Cesàro convergent then f is a compatible modulus function.