Modulated Lacunary Statistical and Strong-Cesàro Convergences

Abstract

Here, we continued the studies initiated by Vinod K. Bhardwaj and Shweta Dhawan which relate different convergence methods involving the classical statistical and the classical strong Cesàro convergences by means of lacunary sequences and measures of density in $\mathbb{N}$ modulated by a modulus function f. A method for constructing non-compatible modulus functions was also included, which is related to symmetries with respect to $y=x$.

1. Introduction

In Applied Science, sometimes limit processes do no fit to the usual concept of limit. However, for applications, in most cases it is sufficient to consider a different form of convergence. This paradigm appeared in mathematics for the first time with the study of the convergence of the Fourier series; sometimes they do not converge to a continuous function but their partial sums always converge in the Cesàro sense. This is what is known as summability methods or convergence methods, an alternative formulation of convergence. Since then, a theory has been developed that has attracted interest in its own right and this is a very active field of research with many contributors (see [1]).

For instance, we say a sequence $\left(x_n\right)\subset X$ converges statistically to L if, for any $\epsilon$, the subset $\{n:\parallel x_n-L\parallel >\epsilon \}$ has zero density on $\mathbb{N}$. We say that a sequence $\left(x_k\right)$ is strong-Cesàro convergent to L if ${lim}_{n\to \infty }\frac{1}{n}{\sum }_{k=1}^n\parallel x_k-L\parallel =0$. These seems to be the first examples of convergence methods. The notion of statistical convergence was first presented by Fast [2] and Steinhaus [3] independently in the same year, 1951. Hardy and Littlewood [4] and, independently, Fekete [5] discovered the strong-Cesàro convergence in connection with the convergence of Fourier series.

It is a little bit surprising how the above convergence methods, discovered at different times and which were developed independently, are basically the same concept. This fact was pointed out by Jeff Connor [6].

In many cases, applied problems required different convergence formulations and, from a theoretical point of view, it is very natural to ask if an alternative convergence method has really been introduced or, on the contrary, if it is the same as the usual one, or to ask what relationship exists between a new method and the existing ones. In this sense, we see here that applications are a good source of mathematical problems. On the other hand, classical results on convergence have different nuances from the point of view of a wide range of summability methods and this mathematical structure gives new meaning to the classical convergence results (see [7,8,9]).

Strong convergence modulated by a modulus function f (f-strong-Cesàro convergence) was presented by Nakano [10] in 1953. Independently, after many years, Aizpuru and coworkers, presented a form of density on the set of natural numbers, also modulated by a modulus function f, from which the concept of f-statistical convergence can be defined. Since then, a great effort has been made to link the two types of convergence methods. One of the main contributions was by V. K. Bhardwaj and S. Dhawan [11]. In [12,13], we fully solved several questions in connection with these convergence methods. For instance, we characterized the modulus functions f for which the f-statistical convergence and the statistical convergence are equivalent; namely, when f is compatible (see Definition 2 below).

Later, V. K. Bhardwaj and S. Dhawan [14] continued their work, relating both f-statistical and f-strong-Cesàro convergence in their lacunary versions, obtaining partial results. In this paper, we solved some questions posed by V. K. Bhardwaj and S. Dhawan [14], Remark 12, p. 4. Moreover, we completely characterized the relationship between the f-statistical convergence and the f-strong-Cesàro convergence in their lacunary versions. In summary, in [15] and in this paper, we see that the circle of ideas in [12,13] continues to be efficient for double sequences, lacunary convergences, etc.

2. Preliminaries

In 1978, Freeman et al., inspired by the classical studies on Cesàro convergence by Hardy-Littlewood and Fekete, studied the concept of strong lacunary convergence in [16]. By a lacunary sequence $\theta ={\left\{k_r\right\}}_{r\ge 0}\subset \mathbb{N}$ with $k_0=0$, we mean an increasing sequence of natural numbers such that $h_r=k_r-k_{r-1}\to \infty$ as $r\to \infty$. Let us denote the intervals $I_r=(k_{r-1},k_r]$. Assume that $\left(x_n\right)$ is a sequence on a normed space X. We say that sequence $\left(x_n\right)$ is strong lacunary convergent to L if

$$\underset{r\to \infty }{lim}\frac{1}{h_r}\sum _{k\in I_r}\parallel x_k-L\parallel =0.$$

Moreover, they proved in [16] that this new convergence method is exactly equal to the classical one introduced by Hardy-Littlewood and Fekete: strong-Cesàro convergence, if and only if the ratios $q_r=\frac{k_r}{k_{r-1}}$ satisfy $1<{lim\; inf}_r{q}_r\le {lim\; sup}_r{q}_r<\infty$.

Later, Fridy and Orhan [17] refined these studies by relating them to a new concept, lacunary statistical convergence, where it underlies new ways of measuring the density of subsets in $\mathbb{N}$. Specifically, given a lacunary sequence $\theta =\left\{k_r\right\}$, a sequence $\left(x_n\right)$ in a normed space X is said to be lacunary statistical convergent to L if, for any $\epsilon >0$, the subset $A_{\epsilon }=\{k\in \mathbb{N}:\parallel x_k-L\parallel >\epsilon \}$ has $\theta$-density equal 0. That is,

$$d_{\theta }\left(A_{\epsilon }\right)=\underset{r\to \infty }{lim}\frac{1}{h_r}\{k\in I_r:\parallel x_k-L\parallel >\epsilon \}=0.$$

Let us recall that $f:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is said to be a modulus function if it satisfies:

1.

$f\left(x\right)=0$ if and only if $x=0$;

2.

$f(x+y)\le f\left(x\right)+f\left(y\right)$ for every $x,y\in {\mathbb{R}}^{+}$;

3.

f is increasing;

4.

f is continuous from the right at 0.

To avoid trivialities, we will suppose that the modulus functions are unbounded.

In the 1990s, there was much interest in studying the classical Cesàro convergence, when it is modulated by a modulus function f. For instance, the studies started by Maddox [18] and continued later by Connor [19]. If one wants to find the “statistical” counterpart of the convergence methods studied by Maddox, we see that it is impossible [12]. That is, we cannot define a density in $\mathbb{N}$ that allows us to study the convergence introduced by Maddox by means of the statistical convergence defined by this density.

However, many years later, Aizpuru and coworkers, inspired in the logarithmic density concept of a subset, introduced a concept of density for sets of $\mathbb{N}$ modulated by a function f. Namely, given $A\subset \mathbb{N}$, the f-density is defined as:

$$d_f\left(A\right)=\underset{n\to \infty }{lim}\frac{1}{f\left(n\right)}f\left(\#\{k\le n:k\in A\}\right),$$

when this limit exists (here $\#C$ denotes the cardinal of a finite subset C).

As we showed in [12], a slight modification of the formula introduced by Madox gives us a concept of f-strong-Cesàro convergence that fits as a glove to the f-density. Now, it is possible to obtain the desired relationship between these two types of convergence methods.

A similar problem occurs with the lacunary version of the f-statistical convergence. Specifically, the lacunary f-statistical convergence introduced by Vinod K. Bhardwaj and Shweta Dhawanin [14] and a possible lacunary extension introduced by Pehlivan and Fisher [20] for the f-strong-Cesàro convergence.

The aim of this article was to explore the rich structure that exists between the lacunary statistical convergence and the strong-Cesàro convergence when modulated by a modulus function f. As it was defined in [14], given a lacunary sequence $\theta =\left(k_r\right)$ and a modulus function f, a sequence $\left(x_k\right)$ is said to be lacunary f-statistical convergent to L if, for any $\epsilon >0$ the subset $A_{\epsilon }=\{k\in \mathbb{N}:\parallel x_k-L\parallel >\epsilon \}$ has lacunary f-density equal to zero, that is

$$d_{f,\theta }\left(A_{\epsilon }\right)=\underset{t\to \infty }{lim}\frac{1}{f\left(h_t\right)}f\left(\#\{k\in I_r,:\parallel z_k-L\parallel >\epsilon \}\right)=0.$$

Now, let us see that the densities have two degrees of choice. The lacunary f-statistical convergence sheds light on how we can modify the term f-strong-Cesàro convergence studied by Pehlivan and Fisher.

Definition 1.

Given a modulus function f and $\theta =\left(k_r\right)$ a lacunary sequence, we say that a sequence $\left(x_n\right)$ is lacunary f-strong convergent to L if

$$\underset{t\to \infty }{lim}\frac{1}{f\left(h_t\right)}f\left(\sum _{k\in I_r}\parallel x_k-L\parallel \right)=0.$$

The notion of compatible modulus function, which is very important in the discussion, was introduced in [12]. We can see in [21,22,23] that this notion was very important for understanding the underlying structure between different convergence methods.

Definition 2.

For each $\epsilon >0$, set $\phi \left(\epsilon \right)={lim\; sup}_n\frac{f\left(n\epsilon \right)}{f\left(n\right)}$. We say that a modulus function f is compatible if ${lim}_{\epsilon \to 0}\phi \left(\epsilon \right)=0$.

Remark 1.

Most of the functions are compatible modulus function. For instance if $f\left(x\right)=x^p+x^q$, $0<p,q\le 1$, then f is compatible. However, if we add a non constant monomial to the logarithm, that is, $f\left(x\right)=x^p+log(x+1)$, then f is compatible. However, there are fewer non-compatible modulus functions. For instance, if we consider $f\left(x\right)=log(x+1)$ and $f\left(x\right)=W\left(x\right)$ (where W is the W-Lambert function restricted to ${\mathbb{R}}^{+}$, that is, the inverse of $x{e}^x$) then they are non-compatible modulus functions. In general, this is a good source of non-compatible modulus functions; we consider the inverse of functions which increase very quickly to infinite.

In general, it is very easy to check whether a modulus function is compatible or not. For instance, if we consider $f\left(x\right)=x+log(x+1)$, then

$$\underset{n\to \infty }{lim}\frac{f\left(n{\epsilon }^{\prime }\right)}{f\left(n\right)}=\underset{n\to \infty }{lim}\frac{n{\epsilon }^{\prime }+log(1+n{\epsilon }^{\prime })}{n+log(n+1)}={\epsilon }^{\prime },$$

which implies that it is compatible. Let us consider $f\left(x\right)=log(x+1)$; since

$$\underset{n\to \infty }{lim}\frac{log(1+n{\epsilon }^{\prime })}{log(1+n)}=1,$$

we deduce that $f\left(x\right)=log(x+1)$ is not compatible.

The symmetry plays some role in order to obtain non compatible modulus functions. For instance, we considered a function that grows very quickly. For instance, $x^2{e}^x,x^3{e}^x,\cdots e^{x^2}$ and so on. These functions stand well above the identity function $Id\left(x\right)=x$. In most cases, their inverses, which are symmetric with respect the the axe $y=x$, are non-compatible modulus functions.

Definition 3.

Set ${\phi }_{\theta }\left(\epsilon \right)={lim\; sup}_{t\to \infty }\frac{f\left(h_t\epsilon \right)}{f\left(h_t\right)}$. A function f is said to be θ-compatible provided ${lim}_{\epsilon \to 0}{\phi }_{\theta }\left(\epsilon \right)=0$.

Of course, the notion of compatibility on f is stronger than $\theta$-compatible for any lacunary sequence.

The paper is structured as follows. In Section 3, we fix a lacunary sequence $\theta$ and we will discuss the relationship between the space of sequences which are lacunary strong-Cesàro convergent ($N_{\theta }$) and the space of lacunary f-strong-Cesàro convergent sequences ($N_{\theta }^f$). We also investigated the relationship between the spaces $S_{\theta }$ and $S_{\theta }^f$, that is, the space of lacunary statistical convergent sequences and the space of lacunary f-statistical convergent sequences (respectively).

In Section 4, we will obtain a relationship between the spaces $N_{\theta }^f$ and $S_{\theta }^f$, that is, respectively, the space of all lacunary f-strong-Cesàro convergence and the space of all lacunary f-statistical convergence. These kinds of results are called Connor–Khan–Orhan’s type. Finally, in Section 5, we fixed a modulus function f and we studied the relationship between the lacunary f-strong-Cesàro convergence and the f-strong-Cesàro convergence, for any lacunary sequence $\theta$.

3. Lacunary Convergence vs. Lacunary Convergence Modulated by a Function f

In this section, we fix a lacunary sequence $\theta =\left(k_r\right)$ and we move the modulus functions f. We aimed to see the relation between sequences that are lacunary f-strong-Cesàro convergent and those which are lacunary strong-Cesàro convergent. Will we show that both convergence spaces are identical when f is compatible.

A similar characterization was obtained for the lacunary f-statistical convergence and the lacunary statistical convergence. If the modulus function f is compatible, both convergence methods are equivalent.

Moreover, we fully solved a question posed by V. K. Bhardwaj and S. Dhawan [14] (Remark 12, p. 4). If we denote by $S_{\theta }^f$ the space of all lacunary f-statistical convergences and by $S_{\theta }$ the set of all lacunary statistical convergence sequences, by applying Theorem 2 and Theorem 4, we have that $S_{\theta }^f=S_{\theta }$ if and only if fis θ-compatible.

Set f as modulus function a and $\theta =\left(k_r\right)$ as a lacunary sequence. Let us denote by $N_{\theta }^f$ the space of all lacunary f-strongly Cesàro convergent sequences and by $N_{\theta }$ the space of all lacunary strong-Cesàro convergent sequences.

Theorem 1.

Assume that f is a modulus function. If $\theta =\left(k_t\right)$ is a lacunary sequence, then $N_{\theta }^f\subset N_{\theta }$.

Proof. 

Indeed, for all $p\ge 1$, there exists $t_0$, such that:

$$\frac{f\left({\sum }_{k\in I_t}\parallel x_k-L\parallel \right)}{f\left(h_t\right)}\le \frac{1}{p},$$

for all $t\ge t_0$. That is,

$$f\left(\sum _{k\in I_t}\parallel x_k-L\parallel \right)\le \frac{1}{p}f\left(h_t\right)\le f\left(\frac{h_t}{p}\right)$$

and since f is increasing, we have that

$$\sum _{k\in I_t}\parallel x_k-L\parallel \le \frac{h_t}{p}$$

for all $t\ge t_0$, that is, $\left(x_n\right)$ is lacunary strong-Cesàro convergent to L. □

A similar result can be obtained for the lacunary statistical convergence.

Theorem 2.

Assume that f is a θ-compatible modulus function. If $\theta =\left(k_r\right)$ is a lacunary sequence then $S_{\theta }=S_{\theta }^f$.

Proof. 

It is well known that, for any modulus function, $S_{\theta }^f\subset S_{\theta }$. Set $\epsilon >0$ small enough. By applying that f is $\theta$-compatible, we find that ${\epsilon }^{\prime }>0$ and $r_0\left(\epsilon \right)$ satisfy

$$\frac{f\left(h_r{\epsilon }^{\prime }\right)}{f\left(h_r\right)}<\epsilon ,$$

provided that $h_r\ge r_0$. Let us fix $c>0$ and ${\epsilon }^{\prime }>0$. If $\left(x_n\right)$ is lacunary statistically convergent to L then there exists $r_1\left(\epsilon \right)$ (in fact, we can suppose that $r_1$ depends actually on $\epsilon$) such that if $h_r\ge r_1$:

$$\#\{k\in I_r:\parallel x_k-L\parallel >c\}\le h_r{\epsilon }^{\prime }.$$

However, f is increasing, therefore

$$\frac{f(\#\{k\in I_r:\parallel x_k-L\parallel >c\left\}\right)}{f\left(h_r\right)}\le \frac{f\left(h_r{\epsilon }^{\prime }\right)}{f\left(h_r\right)}<\epsilon ,$$

provided $h_t\ge max\{r_0,r_1\}$, which gives the desired result. □

Theorem 3.

Assume that f is θ-compatible. If $\theta =\left(k_r\right)$ is a lacunary sequence, then $N_{\theta }^f=N_{\theta }$.

Proof. 

If $\left(x_n\right)$ is lacunary strong-Cesàro convergent to L, then for any ${\epsilon }^{\prime }>0$ there exists $r_0$ such that if $h_r\ge r_0$ then

$$\sum _{k\in I_r}\parallel x_k-L\parallel \le h_r{\epsilon }^{\prime };$$

by applying that f is increasing, we get:

$$f\left(\sum _{k\in I_r}\parallel x_k-L\parallel \right)\le f\left(h_r{\epsilon }^{\prime }\right),$$

thus

$$\frac{f\left({\sum }_{k\in I_r}\parallel x_k-L\parallel \right)}{f\left(h_r\right)}\le \frac{f\left(h_r{\epsilon }^{\prime }\right)}{f\left(h_r\right)},$$

which gives the desired result t by applying the same argument as above. □

Theorem 4.

Assume that f is a modulus function and let us suppose that $\theta =\left(k_r\right)$ is a lacunary sequence.

1.

f is θ-compatible provided $S_{\theta }=S_{\theta }^f$;

2.

f is θ-compatible provided $N_{\theta }=N_{\theta }^f$.

Proof. 

We proceed by means of contradiction, that is, if f is not $\theta$-compatible, we will construct a sequence $\left(x_n\right)$ on $\mathbb{R}$ such that $\left(x_n\right)\in S_{\theta }\setminus S_{\theta }^f$. In fact, a slight modification of the construction provides a sequence on any normed space.

If f is not compatible, by applying that ${\phi }_{\theta }\left(\epsilon \right)$ is increasing we can find $c>0$, such that for any $\epsilon >0$: ${\phi }_{\theta }\left(\epsilon \right)>c$.

If $\left({\epsilon }_k\right)$ is a sequence converging to zero, then for each k there exists $h_{r_k}$ such that $f\left(h_{r_k}{\epsilon }_k\right)\ge cf\left(h_{r_k}\right)$. Moreover, since $h_r$ is increasing, we can suppose that

$$h_{r_k}(1-{\epsilon }_k)-1>0.$$

(1)

Set $\lfloor x\rfloor$ as the integer part of x. We denote $n_k=\lfloor h_{r_k}{\epsilon }_k\rfloor +1$. According to Equation (1), we get that $h_{r_k}-n_k>0$. We consider the subsets $A_k=[k_{r_k}-n_k,k_{r_k}]\cap \mathbb{N}\subset I_{r_k}$, and $A={\bigcup }_k{A}_k$. Denoting by ${\chi }_A(·)$ the characteristic function of A, we claim that the sequence $x_n={\chi }_A\left(n\right)$, is the sequence that we are looking for that is $\left(x_n\right)$ is $\theta$-statistically convergent to 0 but not f-statistically convergent.

Effectively, if $r\ne r_k$ for any k, then

$$\frac{\#\{l\in I_r:|x_l|>\epsilon \}}{h_r}\le \frac{0}{h_r}=0.$$

If $r=r_k$, then:

$$\frac{\#\{l\in I_{r_k}:|x_l|>\epsilon \}}{h_{r_k}}=\frac{n_k}{h_{r_k}}\to 0$$

as $k\to \infty$.

On the other hand,

$$\frac{f(\#\{l\in I_{r_k}:|x_l|>\epsilon \})}{f\left(h_{r_k}\right)}=\frac{f\left(n_k\right)}{f\left(h_{r_k}\right)}\ge \frac{f\left(h_{r_k}{\epsilon }_k\right)}{f\left(h_{r_k}\right)}\ge c,$$

which yields part (1). Part (2) is similar. We can see that the sequence $\left(x_n\right)$ constructed before satisfies that $\left(x_n\right)\in N_{\theta }$ but $\left(x_n\right)\notin N_{\theta }^f$, which proves (2). □

4. Lacunary f-Strong-Cesàro Convergence vs. Lacunary f-Statistical Convergence

In this section, we aim to relate the spaces $N_{\theta }^f$ and $S_{\theta }^f$, that is, respectively the space of all lacunary f-strong-Cesàro convergent sequences and the space of all lacunary f-statistical convergent sequences.

Theorem 5.

If $\theta =\left(k_t\right)$ is a lacunary sequence and f is any modulus function, then $N_{\theta }^f\subset S_{\theta }^f$.

Proof. 

To prove that $\left(x_n\right)\in S_{\theta }^f$, it is sufficient to show that, for all $m\in \mathbb{N}$,

$$\underset{t\to \infty }{lim}\frac{f(\#\{k\in I_t:\parallel x_k-L\parallel >\frac{1}{m}\left\}\right)}{f\left(h_t\right)}=0.$$

(2)

Fix $\epsilon >0$ and we consider m such that $\frac{1}{m+1}\le \epsilon \le \frac{1}{m}$. Hence,

$$\#\{k\in I_t:\parallel x_k-L\parallel >\epsilon \}\le \#\left\{k\in I_t:\parallel x_k-L\parallel >\frac{1}{m+1}\right\},$$

by applying that f is increasing

$$\underset{t\to \infty }{lim}\frac{f\left(\#\{k\in I_t:\parallel x_k-L\parallel >\epsilon \}\right)}{f\left(h_t\right)}\le \underset{t\to \infty }{lim}\frac{f\left(\#\{k\in I_t:\parallel x_k-L\parallel >\frac{1}{m+1}\}\right)}{f\left(h_t\right)},$$

when $t\to \infty$, we get the desired result.

Thus, set $m\in \mathbb{N}$ and let us show Equation (2).

$$\begin{array}{ccc}\hfill f\left(\sum _{k\in I_t}\parallel x_k-L\parallel \right)& \ge & f\left(\sum _{\begin{array}{c}k\in I_t\\ \parallel x_k-L\parallel \ge \frac{1}{m}\end{array}}\parallel x_k-L\parallel \right)\hfill \end{array}$$

(3)

$$\begin{array}{ccc}& \ge & f\left(\sum _{\begin{array}{c}k\in I_t\\ \parallel x_k-L\parallel \ge \frac{1}{m}\end{array}}\frac{1}{m}\right)\hfill \end{array}$$

(4)

$$\begin{array}{ccc}& \ge & \frac{1}{m}f\left(\sum _{\begin{array}{c}k\in I_t\\ \parallel x_k-L\parallel \ge \frac{1}{m}\end{array}}1\right)\hfill \end{array}$$

(5)

$$\begin{array}{ccc}& =& \frac{1}{m}f\left(\#\left\{k\in I_t:\parallel x-L\parallel >\frac{1}{m}\right\}\right).\hfill \end{array}$$

(6)

Since $\left(x_n\right)\in N_{\theta }^f$:

$$\underset{t\to \infty }{lim}\frac{f\left({\sum }_{k\in I_t}\parallel x_k-L\parallel \right)}{f\left(h_t\right)}=0;$$

therefore, dividing by $f\left(h_t\right)$ Equation (3), when $t\to \infty$, we obtain that for each $m\in \mathbb{N}$

$$\underset{t\to \infty }{lim}\frac{f(\#\{k\in I_t:\parallel x_k-L\parallel >\frac{1}{m}\left\}\right)}{f\left(h_t\right)}=0,$$

which implies that $\left(x_n\right)$ is lacunary f-statistically convergent to L as we desired. □

If a sequence $\left(x_n\right)$ is bounded and statistically convergent then $\left(x_n\right)$ is strong-Cesàro convergent; this is the result of Connor [6]. Later, Khan and Orhan [24] sharpened this result by proving that a sequence $\left(x_n\right)$ is statistically convergent if and only if $\left(x_n\right)$ is strong-Cesàro convergent and uniformly integrable.

The following question arises. It is true Connor–Khan and Orhan true for modulated lacunary statistical convergence? The answer is true; however, it is necessary to measure the integrability of a sequence in a lacunary form.

Definition 4.

Assume that $\theta =\left(k_t\right)$ is a lacunary sequence. We say that a sequence $\left(x_n\right)$ is lacunary uniformly integrable if

$$\underset{M\to \infty }{lim}\underset{t}{sup}\sum _{\begin{array}{c}k\in I_t\\ \parallel x_k\parallel \ge M\end{array}}\parallel x_k\parallel =0.$$

We denote the space of all lacunary uniform integrable sequences by $I_{\theta }$. Let us observe that ${\ell }_{\infty }\left(X\right)\subset I_{\theta }$. Moreover, if $\left(x_n\right)\in I_{\theta }$ then $(x_n-L)\in I_{\theta }$ for every $L\in X$.

Theorem 6.

If $\theta =\left(k_r\right)$ is a lacunary sequence and f is θ-compatible, then $S_{\theta }^f\cap I_{\theta }\subset N_{\theta }^f$. Conversely, if $S_{\theta }^f\cap I_{\theta }\subset N_{\theta }^f$ for some modulus function f, then f is θ-compatible.

Proof. 

Set $\left(x_n\right)\in S_{\theta }^f\cap I_{\theta }$. Set $\epsilon >0$; by applying that f is $\theta$-compatible we find ${\epsilon }^{\prime }>0$ such that

$$\frac{f\left(h_t{\epsilon }^{\prime }\right)}{f\left(h_t\right)}<\frac{\epsilon }{3}$$

(7)

for all $t\ge t_0\left(\epsilon \right)$.

Since $\left(x_n\right)\in I_{\theta }$, there exists $M\in \mathbb{N}$ satisfying $\frac{1}{M}<{\epsilon }^{\prime }$ and

$$\frac{1}{h_t}\sum _{\begin{array}{c}k\in I_t\\ \parallel x_k-L\parallel \ge M\end{array}}\parallel x_k-L\parallel <{\epsilon }^{\prime }.$$

(8)

for all $t\in \mathbb{N}$. On the other hand, since $\left(x_n\right)\in S_{\theta }^f$, there exists a natural number $t_0\left(\epsilon \right)$, such that if $t\ge t_0\left(\epsilon \right)$, then

$$\frac{1}{f\left(h_t\right)}f(\#\{k\in I_t:\parallel x_k-L\parallel >{\epsilon }^{\prime }\left\}\right)<\frac{\epsilon }{3M}.$$

(9)

for some L. Therefore,

$$\begin{array}{c}\hfill \frac{f\left({\sum }_{k\in I_t}\parallel x_k-L\parallel \right)}{f\left(h_t\right)}\le \frac{1}{f\left(h_t\right)}f\left(\sum _{\begin{array}{c}k\in I_t\\ M>\parallel x_k-L\parallel \ge {\epsilon }^{\prime }\end{array}}\parallel x_k-L\parallel \right)+\\ \hfill \frac{1}{f\left(h_t\right)}f\left(\sum _{\begin{array}{c}k\in I_t\\ \parallel x_k-L\parallel \ge M\end{array}}\parallel x_k-L\parallel \right)+\\ \hfill \frac{1}{f\left(h_t\right)}f\left(\sum _{\begin{array}{c}k\in I_t\\ \parallel x_k-L\parallel <{\epsilon }^{\prime }\end{array}}\parallel x_k-L\parallel \right).\end{array}$$

(10)

By using that f is increasing, according to (9), we get that for all $t\ge t_0\left(\epsilon \right)$, the first term of (10) is:

$$\begin{array}{cc}\hfill \frac{1}{f\left(h_t\right)}f\left(\sum _{\begin{array}{c}k\in I_t\\ M>\parallel x_k-L\parallel \ge {\epsilon }^{\prime }\end{array}}\parallel x_k-L\parallel \right)& <\frac{f\left(\#\{k\in I_t:\parallel x_k-L\parallel >{\epsilon }^{\prime }\}·M\right)}{f\left(h_t\right)}\hfill \\ \hfill & \le M\frac{1}{f\left(h_t\right)}f\left(\#\{k\in I_t:\parallel x_k-L\parallel >{\epsilon }^{\prime }\}\right)\hfill \\ \hfill & <M\frac{\epsilon }{3M}=\frac{\epsilon }{3}.\hfill \end{array}$$

(11)

Now, we estimate the second term of the inequality (10). By applying that f is increasing and using the inequalities (8) and (7), we get that for $t\ge t_0\left(\epsilon \right)$:

$$\begin{array}{ccc}\hfill \frac{1}{f\left(h_t\right)}f\left(\sum _{\begin{array}{c}k\in I_t\\ \parallel x_k-L\parallel \ge M\end{array}}\parallel x_k-L\parallel \right)& =& \frac{1}{f\left(h_t\right)}f\left(h_t\frac{1}{h_t}\sum _{\begin{array}{c}k\in I_t\\ \parallel x_k-L\parallel \ge M\end{array}}\parallel x_k-L\parallel \right)\hfill \\ & \le & \frac{1}{f\left(h_t\right)}f\left(h_t{\epsilon }^{\prime }\right)\le \frac{\epsilon }{3}.\hfill \end{array}$$

(12)

Finally, we consider the third summand in (10) and we apply inequality (7). Hence, if $t\ge t_0\left(\epsilon \right)$, then we get

$$\begin{array}{ccc}\hfill \frac{1}{f\left(h_t\right)}f\left(\sum _{\begin{array}{c}k\in I_t\\ \parallel x_k-L\parallel \le {\epsilon }^{\prime }\end{array}}\parallel x_k-L\parallel \right)& \le & \frac{1}{f\left(h_t\right)}f\left(h_t\frac{1}{M}\right)<\frac{\epsilon }{3}.\hfill \end{array}$$

(13)

Now, by using (11)–(13) into (10), we obtain that if $t\ge t_0\left(\epsilon \right)$

$$\frac{f\left({\sum }_{k\in I_t}\parallel x_k-L\parallel \right)}{f\left(h_t\right)}\le \epsilon ,$$

that is, $\left(x_n\right)\in N_{\theta }^f$ as we desired.

Conversely, assume that f is not $\theta$-compatible. If ${\epsilon }_k\to 0$, then there exists a subsequence $r_k$ such that $f\left(h_{r_k}{\epsilon }_k\right)\ge cf\left(h_{r_k}\right)$, for some $c>0$.

Let us consider the sequence $x_n={\sum }_{k=1}^{\infty }{\epsilon }_k{\chi }_{(k_{r_k}-1,k_{r_k}]}\left(n\right)$. The sequence $\left(x_n\right)$ is clearly bounded and, since ${\epsilon }_k$ is decreasing, an easy check shows that $\left(x_n\right)$ is lacunary f-statistically convergent to 0. On the other hand:

$$\frac{1}{f\left(r_k\right)}f\left(\sum _{n\in I_{r_k}}\left|x_n\right|\right)=\frac{f\left(h_{r_k}{\epsilon }_k\right)}{f\left(h_{r_k}\right)}\ge c,$$

which proves that $\left(x_n\right)$ is not lacunary f-strong-Cesàro convergent, as we desired to prove. □

5. Lacunary f-Strong-Cesàro Convergence vs. Lacunary Strong Cesàro Convergence

Now, let us fix a modulus function f, and we studied the relationship between the set of all lacunary f-strong-Cesàro convergent sequences and the set of all f-strong-Cesàro convergent sequences. We obtained results in terms of the properties of a lacunary sequence $\theta$. Assume that $\theta =\left(k_r\right)$ is a lacunary sequence and f is a modulus function. Let us denote by $N_{\theta }^f$ the space of all lacunary f-strong-Cesàro convergent sequences in the normed space X. We denote by $N^f$ the space of all f-strong-Cesàro convergent sequences in the normed space X. When $f\left(x\right)=x$, then $N^x=N$ and $N_{\theta }^x=N_{\theta }$ will denote, respectively, the space of strong-Cesàro convergent sequences and the space of lacunary strong-Cesàro convergent sequences.

Proposition 1.

Assume that $\theta =\left(k_r\right)$ is a lacunary sequence and let f be a modulus function:

(i) 

If $1<lim\; inf{q}_r$ then $N^f\subset N_{\theta }^f$;

(ii) 

Assume that f is compatible. If $N^f\subset N_{\theta }=N_{\theta }^f$ then $1<lim\; inf{q}_r$.

Proof. 

To prove (i), assume that $lim\; inf{q}_r>1$. Thus, there exists $\delta >0$ such that $1+\delta <q_r$ for all r. Let $\left(x_n\right)\in N^f$, such that $x_n\stackrel{N^f}{⟶}L$ for some $L\in X$, we will show that $x_n\stackrel{N_{\theta }^f}{⟶}L$. Indeed, set

$${\tau }_r=\frac{1}{f\left(h_r\right)}f\left(\sum _{k\in I_r}\parallel x_k-L\parallel \right).$$

Then,

$$\begin{array}{ccc}\hfill {\tau }_r& =& \frac{1}{f\left(h_r\right)}f\left(\sum _{k=k_{r-1}-1}^{k_r}\parallel x_k-L\parallel \right)\hfill \\ & \le & \frac{1}{f\left(h_r\right)}f\left(\sum _{k=1}^{k_r}\parallel x_k-L\parallel \right)+\frac{1}{f\left(h_r\right)}f\left(\sum _{k=1}^{k_{r-1}}\parallel x_k-L\parallel \right)\hfill \\ & =& \frac{f\left(k_r\right)}{f\left(h_r\right)}\frac{1}{f\left(k_r\right)}f\left(\sum _{k=1}^{k_r}\parallel x_k-L\parallel \right)+\frac{f\left(k_{r-1}\right)}{f\left(h_r\right)}\frac{1}{f\left(k_{r-1}\right)}f\left(\sum _{k=1}^{k_{r-1}}\parallel x_k-L\parallel \right)\hfill \\ & \le & C\left[\frac{1}{f\left(k_r\right)}f\left(\sum _{k=1}^{k_r}\parallel x_k-L\parallel \right)+\frac{1}{f\left(k_{r-1}\right)}f\left(\sum _{k=1}^{k_{r-1}}\parallel x_k-L\parallel \right)\right],\hfill \end{array}$$

where the last inequality follows from the fact that $\frac{k_r}{h_r}\le \frac{1+\delta }{\delta }$ and $\frac{k_{r-1}}{h_r}<\frac{1}{\delta }$ and using that f is increasing, we get that $\frac{f\left(k_r\right)}{f\left(h_r\right)}$ and $\frac{f\left(k_{r-1}\right)}{f\left(h_r\right)}$ are bounded. The result follows, taking limits as $r\to \infty$.

We prove (ii) by contradiction. Let us assume that $lim\; inf{q}_r=1$, and we will construct a sequence $\left(x_n\right)\in N^f$ such that $\left(x_n\right)\notin N_{\theta }^f$. Indeed, since $lim\; inf{q}_r=1$ by Freedman-Sember’s result (see [16] (Lemma 2.1)), there exists $\left(x_n\right)\in N$ but $\left(x_n\right)\notin N_{\theta }$. However, since f is compatible, in particular f is $\theta$-compatible, therefore $N_{\theta }^f=N_{\theta }$. On the other hand, since f is compatible, $N=N^f$. Hence, there exists $\left(x_n\right)\in N^f=N$ such that $\left(x_n\right)\notin N_{\theta }^f=N_{\theta }$ as we desired to prove. □

Proposition 2.

Assume that $\theta =\left(k_r\right)$ is a lacunary sequence and let f be a modulus function.

1. 

Assume that f is compatible. If $lim\; sup{q}_r<\infty$ then $N_{\theta }^f\subset N^f$;

2. 

Assume that f is θ-compatible. If $N_{\theta }^f\subset N^f$ then $lim\; sup{q}_r<\infty$.

Proof. 

Let us show $\left(i\right)$. If f is compatible, then f is $\theta$-compatible. Therefore, $N_{\theta }^f=N_{\theta }$. On the other hand, since f is compatible, $N=N^f$. Then, the result follows by applying Freemand-Sember’s result (see [16] (Lemma 2.2)). That is, if $lim\; sup{q}_r<\infty$, then $N_{\theta }^f=N_{\theta }=\subset N=N^f$ as we desired.

We prove the necessity by contradiction. Let us suppose that $lim\; sup{q}_r=\infty$. Following as in Lemma 2 of [16], we can select a sequence $k_{r\left(j\right)}\subset \theta$ satisfying $q_{r\left(j\right)}>j$ and let us consider the sequence

$$x_k=\left\{\begin{array}{cc}{x}_0\ne 0\hfill & k_{r(j-1)}<k<2k_{r(j-1)}j=1,2,3,\cdots \hfill \\ 0\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

We wish to prove that $\left(x_k\right)\in N_{\theta }^f$. Set $\epsilon >0$. By using that f is $\theta$-compatible, there exists ${\epsilon }^{\prime }>0$ and $t_0$ such that if $j\ge j_0$ then $\frac{f\left(h_{r\left(j\right)}{\epsilon }^{\prime }\right)}{f\left(h_{r\left(j\right)}\right)}<\epsilon$. Since $q\left(r\right(j\left)\right)>j$, we get

$$\frac{k_{r(j-1)}}{k_{r\left(j\right)}-k_{r(j-1)}}=\frac{1}{q\left(r_j\right)-1}<\frac{1}{j-1}.$$

We consider $r\left(j\right)$ large enough such that $1/j-1<{\epsilon }^{\prime }$. Thus, since f is increasing for $r\left(j\right)\ge r\left(j_0\right)$ we get

$$f\left(k_{r(j-1)}\right)\le f\left(\frac{1}{j-1}(k_{r\left(j\right)}-k_{r(j-1)})\right)<f\left({\epsilon }^{\prime }{h}_{r\left(j\right)}\right).$$

Dividing by $f\left(h_{r\left(j\right)}\right)$, we get that, for $r\left(j\right)\ge r\left(j_0\right)$,

$$\frac{f\left(k_{r(j-1)}\right)}{f\left(h_{r\left(j\right)}\right)}<\frac{f\left({\epsilon }^{\prime }{h}_{r\left(j\right)}\right)}{f\left(h_{r\left(j\right)}\right)}<\epsilon .$$

Therefore, for $r\left(j\right)\ge r\left(j_0\right)$

$$\begin{array}{ccc}\hfill \frac{1}{f\left(h_{r\left(j\right)}\right)}f\left(\sum _{k\in I_{r\left(j\right)}}\parallel x_k\parallel \right)& =& \frac{f\left(k_{r(j-1)}\right)}{f\left(h_{r\left(j\right)}\right)}\parallel x_0\parallel \hfill \\ & \le & \epsilon \parallel x_0\parallel ,\hfill \end{array}$$

which gives that $\left(x_k\right)\in N_{\theta }^f$. Now, let us see that $\left(x_k\right)\notin N^f$. Indeed, since the coordinates of $\left(x_k\right)$ are $x_0$ or 0 the possible strong limits for $\left(x_k\right)$ are 0 and $x_0$. Let us see that $\left(x_k\right)\stackrel{N^f}{\nrightarrow }{x}_0$. Indeed, for $i=1,\cdots ,k\left(r\right(j\left)\right)$:

$$\begin{array}{ccc}\hfill \frac{1}{k_{r\left(j\right)}}\sum _{i=1}^{k_{r\left(j\right)}}\parallel x_i-x_0\parallel & \ge & \frac{k_{r\left(j\right)}-2k_{r(j-1))}}{k_{r\left(j\right)}}\parallel x_0\parallel \hfill \\ & \ge & \left(1-\frac{2}{j}\right)\parallel x_0\parallel ,\hfill \end{array}$$

which does not converge to zero. Now, let us see that $\left(x_k\right)\stackrel{N^f}{\nrightarrow }0$. Indeed, for $i=1,\cdots ,2k_{r(j-1)}$:

$$\begin{array}{ccc}\hfill \frac{1}{2k_{r(j-1)}}\sum _{i=1}^{2k_{r(j-1)}}\parallel x_i\parallel & \ge & \frac{k_{r(j-1)}}{2k_{r(j-1)}}\parallel x_0|\hfill \\ & \ge & \frac{\parallel x_0\parallel }{2},\hfill \end{array}$$

which does not converge to zero. Therefore, $\left(x_k\right)\notin N^f$, which gives the desired result. □

Corollary 1.

Let f be a compatible modulus function and let $\theta =\left(k_r\right)$ be a lacunary sequence. Then, $N_{\theta }^f=N^f$ if and only if $1<lim\; inf{q}_r\le lim\; sup{q}_r<\infty$.