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

  1. (a)

    If all statistical convergent sequences are f-statistical convergent then f is a compatible modulus function.

  2. (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.