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

Abstract

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).

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 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.

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

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

(1.1)

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 f-statistical 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 $\varepsilon >0$:

$$\begin{aligned} \frac{\#\{l\leq m: \vert x_{l} \vert >\varepsilon \}}{m}&\leq \frac{\#\{l\leq m_{k}: \vert x_{k} \vert >\varepsilon \}}{m_{k}}+ \frac{n_{k+1}-n_{k}}{m_{k+1}-n_{k+1}+n_{k}} \\ &\leq \frac{n_{k}}{m_{k}}+\frac{1}{\frac{m_{k+1}}{n_{k+1}-n_{k}}-1} \to 0 \end{aligned}$$

as $k\to \infty$. On the other hand, since $\varepsilon _{k+1}<\frac{n_{k+1}}{m_{k+1}}$

$$\begin{aligned} \frac{f (\#\{n< m_{k+1}: \vert x_{n} \vert >1/2\} )}{f(m_{k+1})} = \frac{f(n_{k+1})}{f(m_{k+1})} \geq \frac{f(m_{k+1}\varepsilon _{k+1})}{f(m_{k+1})}>c, \end{aligned}$$

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 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 $(\varepsilon _{k})$, $(m_{k})$ 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

$$\begin{aligned} r_{k}= \frac{m_{k+1}\varepsilon _{k+1}-m_{k}\varepsilon _{k}}{m_{k+1}-m_{k}} \end{aligned}$$

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 f-statistically 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 f-strong Cesàro convergent, as we desired. □

The corrections have been indicated in this article and the original article [1] has been corrected.