# ON LACUNARY STATISTICAL CONVERGENCE AND LACUNARY STRONG CESÀRO CONVERGENCE
BY MODULI

María del Pilar Romero de la Rosa  M.P. Romero de la Rosa, University &
Departamento de Matemáticas, Universidad de Cádiz, Avenida de la Universidad,
11405 Jerez de la Frontera, Cádiz (Spain<EMAIL_ADDRESS>
(Date: November 2022)

###### Abstract.

Here we fully complete the studies initiated by Vinod K. Bhardwaj and Shweta
Dhawan in [hindawi] which relate different convergence methods which involves
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 $f$.

###### Key words and phrases:

Lacunary convergence; Statistical convergence ; Strong-Cesàro convergence;
Modulus function

###### 1991 Mathematics Subject Classification:

40H05; 40A35

##  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, appears 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 precisely what
is known as summability methods or convergence methods, an alternative
formulation of convergence. Since then, a theory has been developed that has
interest in its own right, moreover, this is a very active field of research
with many contributors (see [mursaleenbook]).

For instance, we say a sequence $(x_{n})\subset X$ is said to be statistically
convergent to $L$ if for any $\varepsilon$ the subset
$\\{n\,:\,\|x_{n}-L\|>\varepsilon\\}$ has zero density on $\mathbb{N}$. On the
other hand a sequence $(x_{k})$ on a normed space $(X,\|\cdot\|)$ is said to
be strong Cesàro convergent to $L$ if
$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\|x_{k}-L\|=0$. These seems to be
the first examples of convergence methods. The term statistical convergence
was firstly presented by Fast [Fast] and Steinhaus [Stein] independently in
the same year 1951. And the strong Cesàro convergence for real numbers was
introduced by Hardy-Littlewood [hardy] and Fekete [fekete] in connection with
the convergence of Fourier series. It is a little bit surprising how the above
convergence methods, discovered in different times and which were developed
independently, are basically the same concept thanks to a result discovered by
Jeff Connor [connor88].

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
it is the same as the usual one. Or what relationship exists between a new
method and the existing ones. In this sense, we see here that applications are
a good source for good mathematical problems.

Strong convergence modulated by a modulus function $f$ ($f$-strong Cesàro
convergence) were presented by Nakano [nakano] in 1953. And independently
after many years, Aizpuru et al. [aizpuru] presented a form of density on the
set of natural numbers, modulated also by a modulus function $f$, from which
can be defined the concept of $f$-statistical convergence. Since then, a great
effort has been made to link the two types of convergence methods. One of the
main contributions is due to V. K. Bhardwaj and S. Dhawan [jia0]. In [jia] we
fully solve several questions in connection with these convergence methods.
For instance, we characterized the modulus functions $f$ for which the
$f$-statistically convergence and the statistically convergence are
equivalents. Namelly, when $f$ is compatible (see Definition 2.2 below).

More later, V. K. Bhardwaj and S. Dhawan [hindawi], continue their work,
relating both convergences: $f$ -statistical and $f$-strong Cesàro, in their
lacunary versions, obtaining partial results. In this paper we solve some
questions posed by V. K. Bhardwaj and S. Dhawan [hindawi], Remark 12, p.4.
Moreover, we fully characterize the relationship between the $f$-statistical
convergence and the $f$-strong Cesàro convergence in their lacunary versions.

##  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 [sember]. By a lacunary sequence $\theta=\\{k_{r}\\}_{r\geq
0}\subset\mathbb{N}$ with $k_{0}=0$, we will means an increasing sequence of
natural numbers such that $h_{r}=k_{r}-k_{r-1}\to\infty$ as $r\to\infty$.
Denoting the intervals $I_{r}=(k_{r-1},k_{r}]$, a sequence $(x_{n})$ on a
normed space $X$ is said to be strong lacunary convergent to $L$ provided

$\lim_{r\to\infty}\frac{1}{h_{r}}\sum_{k\in I_{r}}^{\infty}\|x_{k}-L\|=0.$

Moreover, they proved in [sember] that this new convergence method is exactly
equal to 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 $1<\liminf_{r}q_{r}\leq\limsup_{r}q_{r}<\infty$.

Later, Fridy and Orhan [b5] 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 $\theta=\\{k_{r}\\}$, a sequence $(x_{n})$ in a normed space
$X$ is said to be lacunary statistical convergent to $L$ if for any
$\varepsilon>0$ the subset
$A_{\varepsilon}=\\{k\in\mathbb{N}\,:\,\|x_{k}-L\|>\varepsilon\\}$ has
$\theta$-density equal $0$. That is,

$d_{\theta}(A_{\varepsilon})=\lim_{r\to\infty}\frac{1}{h_{r}}\\{k\in
I_{r}\,:\,\|x_{k}-L\|>\varepsilon\\}=0.$

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

  1. (1)

$f(x)=0$ if and only if $x=0$.

  2. (2)

$f(x+y)\leq f(x)+f(y)$ for every $x,y\in\mathbb{R}^{+}$.

  3. (3)

$f$ is increasing.

  4. (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 has been a lot of interest in studying the classical Cesàro
convergence, when it is modulated by a modulus function $f$. For instance,
these studies started by Maddox [maddox] and continued later by Connor
[canadian]. If one want to find the ”statistical” counterpart of the
convergence methods studied by Maddox, we see that it is impossible [jia].
That is, we cannot define a density in $\mathbb{N}$ that allows to study the
convergence introduced by Maddox by means of the statistical convergence
defined by this density.

However, many years later, Aizpuru et al. [aizpuru], introduced a concept of
density for sets of $\mathbb{N}$ modulated by a function $f$. Namelly, given
$A\subset\mathbb{N}$ the $f$ -density is defined as:

$d_{f}(A)=\lim_{n\to\infty}\frac{1}{f(n)}f\left(\\#\\{k\leq n\,:\,k\in
A\\}\right)$

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

As we show in [jia], 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. And 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 [hindawi] and a possible lacunary
extension introduced by Pehlivan and Fisher [fisher] for the $f$-strong Cesàro
convergence.

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

$d_{f,\theta}(A_{\varepsilon})=\lim_{t\to\infty}\frac{1}{f(h_{t})}f\left(\\#\\{k\in
I_{r},:\,\|z_{k}-L\|>\varepsilon\\}\right)=0.$

Now, let us see that the densities have two degrees of choice. Next we
introduce the slight modification of the lacunary $f$-strong Cesàro
convergence studied by Pehlivan and Fisher.

######  Definition 2.1.

Given a modulus function $f$ and $\theta=(k_{r})$ a lacunary sequence, we say
that a sequence $(x_{n})$ is lacunary $f$-strong convergent to $L$ if

$\lim_{t\to\infty}\frac{1}{f(h_{t})}f\left(\sum_{k\in
I_{r}}\|x_{k}-L\|\right)=0.$

In [jia] we introduced the notion of compatible modulus function that will
play a central role along the discussion.

######  Definition 2.2.

Let us denote $\varphi(\varepsilon)=\limsup_{n}\frac{f(n\varepsilon)}{f(n)}$.
A modulus function $f$ is said to be compatible if $\lim_{\varepsilon\to
0}\varphi(\varepsilon)=0$.

######  Remark 2.3.

The following functions $f(x)=x^{p}+x^{q}$, $0<p,q\leq 1$,
$f(x)=x^{p}+\log(x+1)$, $f(x)=x+\frac{x}{x+1}$ are modulus functions which are
compatible. And $f(x)=\log(x+1)$, $f(x)=W(x)$ (where $W$ is the $W$-Lambert
function restricted to $\mathbb{R}^{+}$, that is, the inverse of $xe^{x}$) are
modulus functions which are not compatible. Indeed, let us show that
$f(x)=x+\log(x+1)$ is compatible.

$\lim_{n\to\infty}\frac{f(n\varepsilon^{\prime})}{f(n)}=\lim_{n\to\infty}\frac{n\varepsilon^{\prime}+\log(1+n\varepsilon^{\prime})}{n+\log(n+1)}=\varepsilon^{\prime}$

On the other hand if $f(x)=\log(x+1)$, since

$\lim_{n\to\infty}\frac{\log(1+n\varepsilon^{\prime})}{\log(1+n)}=1$

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

######  Definition 2.4.

Let us denote
$\varphi_{\theta}(\varepsilon)=\limsup_{t\to\infty}\frac{f(h_{t}\varepsilon)}{f(h_{t})}$.
We will say that $f$ is $\theta$-compatible if if $\lim_{\varepsilon\to
0}\varphi_{\theta}(\varepsilon)=0$.

Clearly, if $f$ is compatible then $f$ is $\theta$-compatible for any lacunary
sequence $\theta$.

The paper is structured as follows. In Section 3 we fix a lacunary sequence
$(\theta)$ and we will see 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 ($N_{\theta}^{f}$). Moreover,
we see also 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 Connor-Khan-Orhan’s type results by relating the
lacunary $f$ strong Cesàro convergence ($N_{\theta}^{f}$) with the lacunary
$f$-statistical convergence ($S_{\theta}^{f}$). Finally, in Section 5 we fix a
modulus function $f$ and we study 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 $\theta=(k_{r})$ and we move the
modulus functions $f$. We will see the relationship between lacunary
$f$-strong Cesàro convergence and the lacunary strong Cesàro convergence. Will
we show that both convergence methods are equivalents when the modulus
function $f$ is compatible.

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

Moreover, we fully solve a question posed by V. K. Bhardwaj and S. Dhawan
[hindawi] Remark 12, p.4. If we denote by $S_{\theta}^{f}$ the space of all
lacunary $f$-statistical convergence and by $S_{\theta}$ the set of all
lacunary statistical convergence sequences, by applying Theorem 3.2, and
Theorem 3.4 we have that $S_{\theta}^{f}=S_{\theta}$ if an only if $f$ is
$\theta$-compatible.

Let $f$ be a modulus function and $\theta=(k_{r})$ be a lacunary sequence. Let
us denote by $N^{f}_{\theta}$ the space of all lacunary $f$-strongly Cesàro
convergent sequences and by $N_{\theta}$ the space of all lacunary strong
Cesàro convergent sequences.

######  Theorem 3.1.

Let $f$ be a modulus function and let $\theta=(k_{t})$ be a lacunary sequence.
Then $N_{\theta}^{f}\subset N_{\theta}$.

###### Proof.

Indeed, for all $p\geq 1$, there exists $t_{0}$, such that:

$\frac{f\left(\sum_{k\in I_{t}}\|x_{k}-L\|\right)}{f(h_{t})}\leq\frac{1}{p}$

for all $t\geq t_{0}$. That is,

$f\left(\sum_{k\in I_{t}}\|x_{k}-L\|\right)\leq\frac{1}{p}f(h_{t})\leq
f\left(\frac{h_{t}}{p}\right)$

and since $f$ is increasing, we have that

$\sum_{k\in I_{t}}\|x_{k}-L\|\leq\frac{h_{t}}{p}$

for all $t\geq t_{0}$, that is, $(x_{n})$ is lacunary strong Cesàro convergent
to $L$. ∎

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

######  Theorem 3.2.

Let $f$ be a $\theta$-compatible modulus function and $\theta=(k_{r})$ a
lacunary sequence. Then $S_{\theta}=S_{\theta}^{f}$.

###### Proof.

It well know that for any modulus function $S_{\theta}^{f}\subset S_{\theta}$.
Let us fix $\varepsilon>0$ arbitrarily small. Since $f$ is
$\theta$-compatible, there exist $\varepsilon^{\prime}>0$ and
$r_{0}(\varepsilon)$ such that if $h_{r}\geq r_{0}$ then

$\frac{f(h_{r}\varepsilon^{\prime})}{f(h_{r})}<\varepsilon.$

Let $c>0$ and let us fix $\varepsilon^{\prime}>0$. Since $(x_{n})$ is lacunary
statistically convergent to $L$ then there exists $r_{1}(\varepsilon)$ (since
$\varepsilon^{\prime}$ depends on $\varepsilon$ we have that $r_{1}$ depends
actually on $\varepsilon$) such that if $h_{r}\geq r_{1}$:

$\\#\\{k\in I_{r}\,:\,\|x_{k}-L\|>c\\}\leq h_{r}\varepsilon^{\prime}.$

Since $f$ is increasing we have

$\frac{f(\\#\\{k\in
I_{r}\,\,:\,\|x_{k}-L\|>c\\})}{f(h_{r})}\leq\frac{f(h_{r}\varepsilon^{\prime})}{f(h_{r})}<\varepsilon.$

for all $h_{t}\geq\max\\{r_{0},r_{1}\\}$, which gives the desired result. ∎

######  Theorem 3.3.

Let $f$ be a $\theta$-compatible modulus function and $\theta=(k_{r})$ a
lacunary sequence. Then, $N_{f}^{\theta}=N_{\theta}$.

###### Proof.

Let us suppose that $(x_{n})$ is lacunary strong Cesàro convergent to $L$.
Then, for any $\varepsilon^{\prime}>0$ there exists $r_{0}$ such that if
$h_{r}\geq r_{0}$ then

$\sum_{k\in I_{r}}\|x_{k}-L\|\leq h_{r}\varepsilon^{\prime}$

and since $f$ is increasing, then

$f\left(\sum_{k\in I_{r}}\|x_{k}-L\|\right)\leq f(h_{r}\varepsilon^{\prime})$

thus

$\frac{f\left(\sum_{k\in
I_{r}}\|x_{k}-L\|\right)}{f(h_{r})}\leq\frac{f(h_{r}\varepsilon^{\prime})}{f(h_{r})}$

then by applying the same argument as above we get the desired result. ∎

######  Theorem 3.4.

Let $f$ be a modulus function and $\theta=(k_{r})$ a lacunary sequence.

  1. (1)

If $S_{\theta}=S^{f}_{\theta}$ then $f$ must be $\theta$-compatible.

  2. (2)

If $N_{\theta}=N_{\theta}^{f}$ then $f$ must be $\theta$-compatible.

###### Proof.

Assume that $f$ is not a $\theta$-compatible modulus function. Let us
construct on the normed space $X=\mathbb{R}$ a sequence $(x_{n})\in
S_{\theta}\setminus S_{\theta}^{f}$. An easy modification provides a
counterexample on any normed space.

Since $\varphi_{\theta}(\varepsilon)$ is an increasing function, if $f$ is not
$\theta$-compatible then there exists $c>0$ such that for any $\varepsilon>0$:
$\varphi_{\theta}(\varepsilon)>c$.

Let us fix $\varepsilon_{k}\to 0$. Thus, for each $k$ there exists $h_{r_{k}}$
such that $f(h_{r_{k}}\varepsilon_{k})\geq cf(h_{r_{k}})$. Moreover, since
$h_{r}$ is increasing we can suppose that a

$h_{r_{k}}(1-\varepsilon_{k})-1>0.$ (3.1)

Let us denote by $\lfloor x\rfloor$ the integer part of $x$. We set
$n_{k}=\lfloor h_{r_{k}}\varepsilon_{k}\rfloor+1$. According to equation
(3.1), we get that $h_{r_{k}}-n_{k}>0$. Let us define the subset
$A_{k}=[k_{r_{k}}-n_{k},k_{r_{k}}]\cap\mathbb{N}\subset I_{r_{k}}$, and
$A=\bigcup_{k}A_{k}$. If we denote by $\chi_{A}(\cdot)$ the characteristic
function of $A$, we claim that the sequence $x_{n}=\chi_{A}(n)$, is
$\theta$-statistically convergent to $0$ but not $f$-statistically convergent,
a contradiction.

Indeed, if $r\neq r_{k}$ for any $k$, then

$\frac{\\#\\{l\in
I_{r}\,:\,|x_{l}|>\varepsilon\\}}{h_{r}}\leq\frac{0}{h_{r}}=0.$

And for $r=r_{k}$:

$\frac{\\#\\{l\in
I_{r_{k}}\,:\,|x_{l}|>\varepsilon\\}}{h_{r_{k}}}=\frac{n_{k}}{h_{r_{k}}}\to 0$

as $k\to\infty$.

On the other hand,

$\frac{f(\\#\\{l\in
I_{r_{k}}\,:\,|x_{l}|>\varepsilon\\})}{f(h_{r_{k}})}=\frac{f(n_{k})}{f(h_{r_{k}})}\geq\frac{f(h_{r_{k}}\varepsilon_{k})}{f(h_{r_{k}})}\geq
c$

which yields part (1). To show part (2), let us see that if $f$ is not
$\theta$-compatible the sequence $(x_{n})$ constructed before satisfies that
$(x_{n})\in N_{\theta}$ but $(x_{n})\notin N_{\theta}^{f}$ a contradiction.

∎

##  4\. Lacunary $f$-strong Cesàro convergence vs Lacunary $f$-statistical
convergence

In this section we will obtain Connor-Khan-Orhan’s type results by relating
the lacunary $f$ strong Cesàro convergence ($N_{\theta}^{f}$) with the
lacunary $f$-statistical convergence ($S_{\theta}^{f}$).

######  Theorem 4.1.

Assume that $\theta=(k_{t})$ is a lacunary sequence and $f$ is a modulus
function. Then $N_{\theta}^{f}\subset S_{\theta}^{f}$.

###### Proof.

First of all, in order to prove that $(x_{n})$ is lacunary $f$-statistically
convergent to $L$, it is sufficient to show that for all $m\in\mathbb{N}$,

$\lim_{t\to\infty}\frac{f(\\#\\{k\in
I_{t}\,\,:\,\,\|x_{k}-L\|>\frac{1}{m}\\})}{f(h_{t})}=0.$ (4.1)

Indeed, let us fix $\varepsilon>0$ and let us consider $m$ such that
$\frac{1}{m+1}\leq\varepsilon\leq\frac{1}{m}$. Then, we get

$\\#\\{k\in I_{t}\,:\|x_{k}-L\|>\varepsilon\\}\leq\\#\left\\{k\in
I_{t}\,:\|x_{k}-L\|>\frac{1}{m+1}\right\\},$

therefore, since $f$ is increasing

$\lim_{t\to\infty}\frac{f\left(\\#\\{k\in
I_{t}\,:\|x_{k}-L\|>\varepsilon\\}\right)}{f(h_{t})}\leq\lim_{t\to\infty}\frac{f\left(\\#\\{k\in
I_{t}\,:\|x_{k}-L\|>\frac{1}{m+1}\\}\right)}{f(h_{t})}$

thus taking limits on $t\to\infty$ we obtain what we desired.

Therefore, let $m\in\mathbb{N}$, and let us show equation (4.1).

$\displaystyle f\left(\sum_{k\in I_{t}}\|x_{k}-L\|\right)$ $\displaystyle\geq
f\left(\sum_{\begin{subarray}{c}k\in I_{t}\\\
\|x_{k}-L\|\geq\frac{1}{m}\end{subarray}}\|x_{k}-L\|\right)$ (4.2)
$\displaystyle\geq f\left(\sum_{\begin{subarray}{c}k\in I_{t}\\\
\|x_{k}-L\|\geq\frac{1}{m}\end{subarray}}\frac{1}{m}\right)\geq\frac{1}{m}f\left(\sum_{\begin{subarray}{c}k\in
I_{t}\\\ \|x_{k}-L\|\geq\frac{1}{m}\end{subarray}}1\right)$ (4.3)
$\displaystyle=\frac{1}{m}f\left(\\#\left\\{k\in
I_{t}\,\,:\,\,\|x-L\|>\frac{1}{m}\right\\}\right).$ (4.4)

Since $(x_{n})$ is lacunary $f$-strongly Cesàro convergent to $L$, we have
that

$\lim_{t\to\infty}\frac{f\left(\sum_{k\in
I_{t}}\|x_{k}-L\|\right)}{f(h_{t})}=0,$

therefore, dividing by $f(h_{t})$ equation (4.2), and taking limit as
$t\to\infty$, we obtain that for each $m\in\mathbb{N}$

$\lim_{t\to\infty}\frac{f(\\#\\{k\in
I_{t}\,\,:\,\,\|x_{k}-L\|>\frac{1}{m}\\})}{f(h_{t})}=0$

which implies that $(x_{n})$ is lacunary $f$-statistically convergent to $L$
as we desired. ∎

Statistical convergence implies strong Cesàro convergence in the realm of all
bounded sequences. This is the 1988 Connor’s result [connor88]. This result
was sharpened by Khan and Orhan [orhan] by proving that a sequence is
statistically convergent if and only if it is Strong Cesàro convergent and
uniformly integrable.

In the following statement we can obtain a result analogous to that of Khan
and Orhan, which is even new for the lacunary statistical convergence. Here,
is crucial to measure optimally the integrability (in a lacunary form) of a
sequence.

######  Definition 4.2.

Let $\theta=(k_{t})$ be a lacunary sequence. A sequence $(x_{n})$ is said to
be lacunary uniformly integrable if

$\lim_{M\to\infty}\sup_{t}\sum_{\begin{subarray}{c}k\in I_{t}\\\ \|x_{k}\|\geq
M\end{subarray}}\|x_{k}\|=0.$

Let us denote by $I_{\theta}$ the space of all lacunary uniform integrable
sequences. Let us observe that if a sequence $(x_{n})$ is bounded then
$(x_{n})$ is lacunary uniformly integrable, that is, $\ell_{\infty}(X)\subset
I_{\theta}$. On the other hand, if a sequence $(x_{n})$ is uniformly
integrable then $(x_{n}-L)$ is also uniformly integrable for every $L\in X$.

######  Theorem 4.3.

Assume that $\theta=(k_{r})$ is a lacunary sequence and $f$ is a
$\theta$-compatible modulus function. Then $S_{\theta}^{f}\cap
I_{\theta}\subset N_{\theta}^{f}$. Moreover, if for some modulus function $f$
we have $S_{\theta}^{f}\cap I_{\theta}\subset N_{\theta}^{f}$ then the modulus
$f$ must be $\theta$-compatible.

###### Proof.

Assume that $(x_{n})$ is a sequence such that $(x_{n})$ is lacunary
$f$-statistically convergent to $L$ and lacunary uniformly integrable. Let us
consider $\varepsilon>0$. Since $f$ is $\theta$-compatible there exists
$\varepsilon^{\prime}>0$ such that

$\frac{f\left(h_{t}\varepsilon^{\prime}\right)}{f(h_{t})}<\frac{\varepsilon}{3}$
(4.5)

for all $t\geq t_{0}(\varepsilon)$.

Now, since $(x_{n})$ is lacunary uniformly integrable, there exists a natural
number $M>0$ large enough satisfying $\frac{1}{M}<\varepsilon^{\prime}$ and
for all $t\in\mathbb{N}$

$\frac{1}{h_{t}}\sum_{\begin{subarray}{c}k\in I_{t}\\\ \|x_{k}-L\|\geq
M\end{subarray}}\|x_{k}-L\|<\varepsilon^{\prime}.$ (4.6)

And since $(x_{n})$ is lacunary $f$-statistically convergent to $L$, there
exists a natural number, which we abusively denote by $t_{0}(\varepsilon)$,
such that for all $t\geq t_{0}(\varepsilon)$:

$\frac{1}{f(h_{t})}f(\\#\\{k\in
I_{t}\,:\,\,\|x_{k}-L\|>\varepsilon^{\prime}\\})<\frac{\varepsilon}{3M}.$
(4.7)

Therefore

$\displaystyle\frac{f\left(\sum_{k\in
I_{t}}\|x_{k}-L\|\right)}{f(h_{t})}\leq\frac{1}{f(h_{t})}f\left(\sum_{\begin{subarray}{c}k\in
I_{t}\\\
M>\|x_{k}-L\|\geq\varepsilon^{\prime}\end{subarray}}\|x_{k}-L\|\right)+$
$\displaystyle\frac{1}{f(h_{t})}f\left(\sum_{\begin{subarray}{c}k\in I_{t}\\\
\|x_{k}-L\|\geq M\end{subarray}}\|x_{k}-L\|\right)+$
$\displaystyle\frac{1}{f(h_{t})}f\left(\sum_{\begin{subarray}{c}k\in I_{t}\\\
\|x_{k}-L\|<\varepsilon^{\prime}\end{subarray}}\|x_{k}-L\|\right)$ (4.8)

Since $f$ is increasing, according to (4.7) we get that for all $t\geq
t_{0}(\varepsilon)$ the first term of (4) is:

$\displaystyle\frac{1}{f(h_{t})}f\left(\sum_{\begin{subarray}{c}k\in I_{t}\\\
M>\|x_{k}-L\|\geq\varepsilon^{\prime}\end{subarray}}\|x_{k}-L\|\right)$
$\displaystyle<\frac{f\left(\\#\\{k\in
I_{t}\,\,:\,\,\|x_{k}-L\|>\varepsilon^{\prime}\\}\cdot M\right)}{f(h_{t})}$
$\displaystyle\leq M\frac{1}{f(h_{t})}f\left(\\#\\{k\in
I_{t}\,\,:\,\,\|x_{k}-L\|>\varepsilon^{\prime}\\}\right)$
$\displaystyle<M\frac{\varepsilon}{3M}=\frac{\varepsilon}{3}.$ (4.9)

On the other hand, let us estimate the second summand of the inequality (4).
Using that $f$ is increasing and by applying firstly the inequality (4.6) and
later inequality (4.5) we have that for $t\geq t_{0}(\varepsilon)$:

$\displaystyle\frac{1}{f(h_{t})}f\left(\sum_{\begin{subarray}{c}k\in I_{t}\\\
\|x_{k}-L\|\geq M\end{subarray}}\|x_{k}-L\|\right)$
$\displaystyle=\frac{1}{f(h_{t})}f\left(h_{t}\frac{1}{h_{t}}\sum_{\begin{subarray}{c}k\in
I_{t}\\\ \|x_{k}-L\|\geq M\end{subarray}}\|x_{k}-L\|\right)$
$\displaystyle\leq\frac{1}{f(h_{t})}f\left(h_{t}\varepsilon^{\prime}\right)\leq\frac{\varepsilon}{3}.$
(4.10)

Finally, for the third summand in (4) by applying inequality (4.5) we obtain
that if $t\geq t_{0}(\varepsilon)$:

$\frac{1}{f(h_{t})}f\left(\sum_{\begin{subarray}{c}k\in I_{t}\\\
\|x_{k}-L\|\leq\varepsilon^{\prime}\end{subarray}}\|x_{k}-L\|\right)\leq\frac{1}{f(h_{t})}f\left(h_{t}\frac{1}{M}\right)<\frac{\varepsilon}{3}.$
(4.11)

Thus, by using inequalities (4),(4), and (4.11) into inequality (4) we obtain
that if $t\geq t_{0}(\varepsilon)$

$\frac{f\left(\sum_{k\in I_{t}}\|x_{k}-L\|\right)}{f(h_{t})}\leq\varepsilon,$

that is, $(x_{n})$ is lacunary $f$-strong Cesàro convergent to $L$ as we
desired.

For the converse, assume that $f$ is not $\theta$-compatible. We can construct
without loss a counterexample normed space $X=\mathbb{R}$. Since $f$ is not
$\theta$-compatible given $(\varepsilon_{k})$ a decreasing sequence converging
to zero, there exists a subsequence $r_{k}$ such that
$f(h_{r_{k}}\varepsilon_{k})\geq cf(h_{r_{k}})$, for some $c>0$. If we
consider the sequence
$x_{n}=\sum_{k=1}^{\infty}\varepsilon_{k}\chi_{(k_{r_{k}}-1,k_{r_{k}}]}(n)$.
The sequence $(x_{n})$ is clearly bounded and since $\varepsilon_{k}$ is
decreasing an easy check show that $(x_{n})$ is lacunary $f$-statistically
convergent to $0$. On the other hand:

$\frac{1}{f(r_{k})}f\left(\sum_{n\in
I_{r_{k}}}|x_{n}|\right)=\frac{f(h_{r_{k}}\varepsilon_{k})}{f(h_{r_{k}})}\geq
c$

which proves that $(x_{n})$ 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 study the relationship between
the lacunary $f$-strong Cesàro convergence and the $f$-strong Cesàro
convergence in terms of the properties of a lacunary sequence $\theta$. Let
$f$ be a modulus function and $\theta=(k_{r})$ a lacunary sequence. Let us
denote by $N_{\theta}^{f}$ the space of all lacunary $f$-strong Cesàro
convergent sequences in the normed space $X$. And we denote by $N^{f}$ the
space of all $f$-strong Cesàro convergent sequences in the normed space $X$.
When $f(x)=x$, then $N^{x}=N$ will denote the space of strong Cesàro
convergent sequences.

######  Proposition 5.1.

Let $f$ be a modulus function and let $\theta=(k_{r})$ be a lacunary sequence.

i):

If $1<\liminf q_{r}$ then $N^{f}\subset N_{\theta}^{f}$.

ii):

Assume that $f$ is compatible. If $N^{f}\subset N_{\theta}=N_{\theta}^{f}$
then $1<\liminf q_{r}$.

###### Proof.

To prove i) assume that $\liminf q_{r}>1$. Thus there exists $\delta>0$ such
that $1+\delta<q_{r}$ for all $r$. Let $(x_{n})\in N^{f}$, such that
$x_{n}\overset{N^{f}}{\longrightarrow}L$ for some $L\in X$, we will show that
$x_{n}\overset{N^{f}_{\theta}}{\longrightarrow}L$. Indeed, set

$\tau_{r}=\frac{1}{f(h_{r})}f\left(\sum_{k\in I_{r}}\|x_{k}-L\|\right).$

Then

$\displaystyle\tau_{r}$ $\displaystyle=$
$\displaystyle\frac{1}{f(h_{r})}f\left(\sum_{k=k_{r-1}-1}^{k_{r}}\|x_{k}-L\|\right)$
$\displaystyle\leq$
$\displaystyle\frac{1}{f(h_{r})}f\left(\sum_{k=1}^{k_{r}}\|x_{k}-L\|\right)+\frac{1}{f(h_{r})}f\left(\sum_{k=1}^{k_{r-1}}\|x_{k}-L\|\right)$
$\displaystyle=$
$\displaystyle\frac{f(k_{r})}{f(h_{r})}\frac{1}{f(k_{r})}f\left(\sum_{k=1}^{k_{r}}\|x_{k}-L\|\right)+\frac{f(k_{r-1})}{f(h_{r})}\frac{1}{f(k_{r-1})}f\left(\sum_{k=1}^{k_{r-1}}\|x_{k}-L\|\right)$
$\displaystyle\leq$ $\displaystyle
C\left[\frac{1}{f(k_{r})}f\left(\sum_{k=1}^{k_{r}}\|x_{k}-L\|\right)+\frac{1}{f(k_{r-1})}f\left(\sum_{k=1}^{k_{r-1}}\|x_{k}-L\|\right)\right],$

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

We prove ii) by contradiction. Let us suppose that $\liminf q_{r}=1$, and we
will construct a sequence $(x_{n})\in N^{f}$ such that $(x_{n})\notin
N^{f}_{\theta}$. Indeed, since $\liminf q_{r}=1$ by Freedman-Sember’s result,
there exists $(x_{n})\in N$ but $(x_{n})\notin N_{\theta}$. 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 $(x_{n})\in N^{f}=N$ such that $(x_{n})\notin
N_{\theta}^{f}=N_{\theta}$ as we desired to prove.

∎

######  Proposition 5.2.

Let $f$ be a modulus function and let $\theta=(k_{r})$ be a lacunary sequence.

  1. (1)

Assume that $f$ is compatible. If $\limsup q_{r}<\infty$ then
$N_{\theta}^{f}\subset N^{f}$.

  2. (2)

Assume that $f$ is $\theta$-compatible. If $N_{\theta}^{f}\subset N^{f}$ then
$\limsup q_{r}<\infty$.

###### Proof.

Let us shown $i)$. Since $f$ is compatible, in particular $f$ is
$\theta$-compatible, therefore $N_{\theta}^{f}=N_{\theta}$. On the other hand,
since $f$ is compatible $N=N^{f}$. Then, the result follows by applying
Freemand-Sember’s result (see [sember]). That is, if $1<\liminf q_{r}$ then
$N_{\theta}=N_{\theta}^{f}=\subset N=N^{f}$ as we desired.

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

$x_{k}=\begin{cases}x_{0}\neq 0&k_{r(j-1)}<k<2k_{r(j-1)}\quad
j=1,2,3,\cdots\\\ 0&\textrm{otherwise}.\end{cases}$

We wish to prove that $(x_{k})\in N_{\theta}^{f}$. Let us fix $\varepsilon>0$.
Since $f$ is $\theta$-compatible, there exists $\varepsilon^{\prime}>0$ and
$t_{0}$ such that if $j\geq j_{0}$ then
$\frac{f(h_{r(j)}\varepsilon^{\prime})}{f(h_{r(j)})}<\varepsilon$. Since
$q(r(j))>j$, we get

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

We consider $r(j)$ large enough such that $1/j-1<\varepsilon^{\prime}$. Thus
since $f$ is increasing for $r(j)\geq r(j_{0})$ we get

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

Dividing by $f(h_{r(j)})$ we get that for $r(j)\geq r(j_{0})$

$\frac{f(k_{r(j-1)})}{f(h_{r(j)})}<\frac{f(\varepsilon^{\prime}h_{r(j)})}{f(h_{r(j)})}<\varepsilon.$

Therefore for $r(j)\geq r(j_{0})$

$\displaystyle\frac{1}{f(h_{r(j)})}f\left(\sum_{k\in
I_{r(j)}}\|x_{k}\|\right)$ $\displaystyle=$
$\displaystyle\frac{f(k_{r(j-1)})}{f(h_{r(j)})}\|x_{0}\|$ $\displaystyle\leq$
$\displaystyle\varepsilon\|x_{0}\|$

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

$\displaystyle\frac{1}{k_{r(j)}}\sum_{i=1}^{k_{r(j)}}\|x_{i}-x_{0}\|$
$\displaystyle\geq$
$\displaystyle\frac{k_{r(j)}-2k_{r(j-1))}}{k_{r(j)}}\|x_{0}\|$
$\displaystyle\geq$ $\displaystyle\left(1-\frac{2}{j}\right)\|x_{0}\|$

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

$\displaystyle\frac{1}{2k_{r(j-1)}}\sum_{i=1}^{2k_{r(j-1)}}\|x_{i}\|$
$\displaystyle\geq$ $\displaystyle\frac{k_{r(j-1)}}{2k_{r(j-1)}}\|x_{0}|$
$\displaystyle\geq$ $\displaystyle\frac{\|x_{0}\|}{2}$

which does not converge to zero. Therefore $(x_{k})\notin N^{f}$ which gives
the desired result. ∎

######  Corollary 5.3.

Let $f$ be a compatible modulus function and let $\theta=(k_{r})$ be a
lacunary sequence. Then $N_{\theta}^{f}=N^{f}$ if and only if $1<\liminf
q_{r}\leq\limsup q_{r}<\infty$.

## Availability of data and material

Not applicable.

## Acknowledgements

## Author’s contribution

The author contributed significantly to analysis and manuscript preparation
and she helped perform the analysis with constructive discussions.

## Competing interests

The author declare that they have no competing interests.

## Funding

## References