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

Abstract

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]. If all statistical convergent sequences aref-statistical convergent thenfis a compatible modulus function. If all strong Cesàro convergent sequences aref-strong Cesàro convergent thenfis a compatible modulus function. Let (Formula presented.) be a decreasing sequence converging to 0. Since f is not compatible, there exists (Formula presented.) such that, for each k, there exists (Formula presented.) such that (Formula presented.). Moreover, we can select (Formula presented.) inductively satisfying (Formula presented.) Now we use an standard argument used to construct subsets with prescribed densities. Let us denote (Formula presented.) the integer part of (Formula presented.). Set (Formula presented.). And extracting a subsequence if it is necessary, we can assume that (Formula presented.) , (Formula presented.) . Thus, set (Formula presented.). Condition (1.1) guarantee that (Formula presented.). Let us denote (Formula presented.) , and (Formula presented.). Let us prove that (Formula presented.) is statistical convergent to 0, but not f-statistical convergent, a contradiction. Indeed, for any m, there exists k such that (Formula presented.). Moreover, we can suppose without loss that (Formula presented.) , that is, (Formula presented.). Thus for any (Formula presented.) : (Formula presented.) as (Formula presented.). On the other hand, since (Formula presented.) (Formula presented.) which yields (a) as promised. The part (b) is same proof. Indeed, for the sequence (Formula presented.) defined in part (a), we have that (Formula presented.). □ The following result corrects the converse of Theorem 3.4 in [1]. If allf-strong Cesàro convergent sequences aref-statistically and uniformly bounded thenfmust be compatible. Assume that f is not compatible. Thus, as in the proof in Proposition 1.1 we can construct sequences (Formula presented.) , (Formula presented.) such that (Formula presented.) for some (Formula presented.). Moreover, we can construct (Formula presented.) inductively, such that the sequence (Formula presented.) is decreasing and converging to 0. Let us consider (Formula presented.). Since (Formula presented.) is decreasing, (Formula presented.) if f-statistically convergent to 0. On the other hand (Formula presented.) , which gives that (Formula presented.) 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.