Characterizations of a Banach Space through the Strong Lacunary and the Lacunary Statistical Summabilities

Abstract

In this manuscript we characterize the completeness of a normed space through the strong lacunary ($N_{\theta }$) and lacunary statistical convergence ($S_{\theta }$) of series. A new characterization of weakly unconditionally Cauchy series through $N_{\theta }$ and $S_{\theta }$ is obtained. We also relate the summability spaces associated with these summabilities with the strong p-Cesàro convergence summability space.

1. Introduction

Let X be a normed space, a sequence $\left(x_k\right)\subset X$ is said to be strongly 1-Cesàro summable (briefly, $|{\sigma }_1|$-summable) to $L\in X$ if

$$\underset{n\to \infty }{lim}\frac{1}{n}\sum _{k=1}^n\parallel x_k-L\parallel =0.$$

Hardy-Littlewood [1] and Fekete [2] introduced this type of summability, which is related to the convergence of Fourier series (see [3,4]). The $|{\sigma }_1|$ summability along with the statistical convergence [5] started a very striking theory with important applications [6,7,8]. Some years later, the strong lacunary summability $N_{\theta }$ was presented by Freedman et al. [9] by introducing lacunary sequences and showed that $N_{\theta }$ is a larger class of $BK$-spaces which had many of the characteristics of $|{\sigma }_1|$. Later on, Fridy [10,11] showed the concept of statistical lacunary summability and they related it with the statistical convergence and the $N_{\theta }$ summability.

The characterization of a Banach space through different types of convergence has been dealt by authors such as Kolk [12], Connor, Ganichev, and Kadets [13],…

Consider X a normed space and $\sum x_i$ a series in X. In [14] the authors introduced the space of convergence $S(\sum x_i)$ associated with the series $\sum x_i$, which is defined as the space of sequences $\left(a_j\right)$ in ${\ell }_{\infty }$ such that $\sum a_i{x}_i$ converges. They also proved that the necessary and sufficient condition for X to be a complete space is that for every weakly unconditionally Cauchy series $\sum x_i$, the space $S(\sum x_i)$ is complete. Recall that $\sum x_i$ is a weakly unconditionally Cauchy (wuC) series if for every permutation $\pi$ of the set of natural numbers, the sequence $\left({\sum }_{i=1}^n{x}_{\pi \left(i\right)}\right)$ is a weakly Cauchy sequence. We will also rely on a powerful known result that states that a series $\sum x_i$ is wuC if and only if $\sum \left|f\right(x_i\left)\right|<\infty$ for all $f\in X^{*}$ (see [15] for Diestel’s complete monograph about series in Banach spaces).

In [16,17] a Banach space is characterized by means of the strong p-Cesàro summability (${\mathrm{w}}_p$) and ideal-convergence. In this manuscript, the $N_{\theta }$ and $S_{\theta }$ summabilities are used along with the concept of weakly unconditionally series to characterize a Banach space. In Section 2 we introduce these two kinds of summabilities which are regular methods and we recall some properties. In Section 3 and Section 4 we introduce the spaces $S_{S_{\theta }}\left({\sum }_i{x}_i\right)$ and $S_{N_{\theta }}\left({\sum }_i{x}_i\right)$ which will be used in Section 5 to characterize the completeness of a space.

2. Preliminaries

In this section, we present the definition of $N_{\theta }$ and $S_{\theta }$ summabilities for Banach spaces and the relations between them. First, we recall the concept of lacunary sequences.

Definition 1.

A lacunary sequence is an increasing sequence of natural numbers $\theta =\left(k_r\right)$ such that $k_0=0$ and $h_r=k_r-k_{r-1}$ tends to infinite as $r\to \infty$. The intervals determined by θ will be denoted by $I_r=(k_{r-1},k_r]$, the ratio ${\displaystyle \frac{k_r}{k_{r-1}}}$ will be denoted by $q_r$.

We now give the definition of strong lacunary summability for Banach spaces based on the one given by Freedman for real-valued sequences [9].

Definition 2.

Let X be a Banach space and $\theta =\left(k_r\right)$ a lacunary sequence. A sequence $x=\left(x_k\right)$ in X is lacunary strongly convergent or $N_{\theta }-\mathit{summable}$ to $L\in X$ if ${\displaystyle \underset{r\to \infty }{lim}\frac{1}{h_r}\sum _{k\in I_r}\parallel x_k-L\parallel =0}$, and we write $N_{\theta }$$\mathrm{-lim} x_k=L$ or $x_k\underset{N_{\theta }}{\to }L$.

Let $N_{\theta }$ be the space of all lacunary strongly convergent sequences,

$$N_{\theta }=\left\{\left(x_k\right)\subseteq X:\underset{r\to \infty }{lim}\frac{1}{h_r}\sum _{k\in I_r}\parallel x_k-L\parallel =0\mathrm{for}\mathrm{some}L\right\}.$$

The space $N_{\theta }$ is a BK−space endowed with the norm ${\displaystyle \parallel x_k{\parallel }_{\theta }=\underset{r}{sup}\frac{1}{h_r}\sum _{k\in I_r}\parallel x_k\parallel }$.

In 1993, Fridy and Orhan [11] introduced a generalization of the statistical convergence, the lacunary statistical convergence, using lacunary sequences. To accomplish this, they substituted the set $\{k:k\le n\}$ by the set $\{k:k_{r-1}<k\le k_r\}$. We recall now the definition of $\theta -$density of a subset $K\subset \mathbb{N}$.

Definition 3.

Let $\theta =\left(k_r\right)$ be a lacunary sequence. If $K\subset \mathbb{N}$, the $\theta -$density of K is denoted by ${\displaystyle d_{\theta }\left(K\right)=\underset{r}{lim}\frac{1}{h_r}\mathrm{card}\left(\{k\in I_r:k\in K\}\right),}$ whenever this limit exists.

It is easy to show that this density is a finitely additive measure and we can define the concept of lacunary statistically convergent sequences for Banach spaces.

Definition 4.

Let X be a Banach space and $\theta =\left(k_r\right)$ a lacunary sequence. A sequence $x=\left(x_k\right)$ is a lacunary statistically convergent sequence to $L\in X$ if given $\epsilon >0$,

$$d_{\theta }\left(\right\{k\in I_r:\parallel x_k-L\parallel \ge \epsilon \left\}\right)=0,$$

or equivalently,

$$d_{\theta }\left(\right\{k\in I_r:\parallel x_k-L\parallel <\epsilon \left\}\right)=1,$$

we say that $\left(x_k\right)$ is $S_{\theta }$-convergent and we write $x_k\underset{S_{\theta }}{\to }L$.

Theorem 1.

Let X be a Banach space and $\left(x_k\right)$ a sequence in X. Notice that $S_{\theta }$ and $N_{\theta }$ are regular methods.

Proof. 

• If $\left(x_k\right)\to L$, then $\left(x_k\right)\underset{N_{\theta }}{\to }L$.
  Let $\epsilon >0$, then there exists $k_0$ such that if $k\ge k_0$, then

  $$\parallel x_k-L\parallel <\epsilon .$$
  Hence there exists $r_0\in \mathbb{N}$ with $r_0\ge k_0$ such that if $r\ge r_0$ we have

  $$\begin{array}{c}\hfill \frac{1}{h_r}\sum _{k\in I_r}\parallel x_k-L\parallel <\frac{1}{h_r}\sum _{k\in I_r}\epsilon =\frac{h_r}{h_r}\epsilon =\epsilon \end{array}$$

  which implies that ${\displaystyle \underset{r\to \infty }{lim}\frac{1}{h_r}\sum _{k\in I_r}\parallel x_k-L\parallel =0}$.
• If $\left(x_k\right)\to L$, then $\left(x_k\right)\underset{S_{\theta }}{\to }L$.
  Simply observe that since $\left(x_k\right)\to L$, given $\epsilon >0$ there exists $k_0$ such that for every $k\ge k_0$ we get $\mathrm{card}\left(\right\{k\in I_r:\parallel x_k-L\parallel \ge \epsilon \left\}\right)=0$, which implies $d_{\theta }\left(\right\{k\in I_r:\parallel x_k-L\parallel \ge \epsilon \left\}\right)=0$ for every $k\ge k_0$. □

The reverse is not true, as we will show in Example 1, in which we introduce an unbounded sequence that is $N_{\theta }$-summable and Example 2 where an unbounded $S_{\theta }$ convergent sequence is presented.

Example 1.

There exist unbounded sequences which are $N_{\theta }$-summable.

Let $\theta =\left(k_r\right)$ be the lacunary sequence with $k_0=0$ and $k_r=2^r$. Notice that

• $h_1=k_1-k_0=2$ and $h_r=2^{r-1}$ for every $r\ge 2$.
• $I_1=(k_0,k_1]=(0,2]$ and $I_r=(2^{r-1},2^r]$ for every $r\ge 2$.

Consider the sequence defined by

$$x_k=\left\{\begin{array}{cc}0& ifk\ne 2^j\mathit{for}\mathit{all}j,\hfill \\ j-1& ifk=2^j\mathit{for}\mathit{some}j.\hfill \end{array}\right.$$

Notice that $\left(x_k\right)$ is unbounded and observe that

$$\frac{{\displaystyle \sum _{k\in I_r}|x_k-0|}}{h_r}=\left\{\begin{array}{cc}0& ifr=1,\hfill \\ {\displaystyle \frac{r-1}{2^{r-1}}}& ifr\ge 2,\hfill \end{array}\right\}\underset{r\to \infty }{⟶}0,$$

which implies that $x_k\underset{N_{\theta }}{\to }0$.

Fridy and Orhan [10] showed that $N_{\theta }$ and $S_{\theta }$ are equivalent for real-valued bounded sequences. This fact also holds for Banach spaces and we include the proof for the sake of completeness.

Theorem 2.

Let X be a Banach space, $\left(x_k\right)$ a sequence in X and $\theta =\left(k_r\right)$ a lacunary sequence. Then:

1. 

$\left(x_k\right)\underset{N_{\theta }}{\to }L$ implies $\left(x_k\right)\underset{S_{\theta }}{\to }L$.

2. 

$\left(x_k\right)$ bounded and $\left(x_k\right)\underset{S_{\theta }}{\to }L$ imply $\left(x_k\right)\underset{N_{\theta }}{\to }L$.

Proof. 

1. If $\left(x_k\right)\underset{N_{\theta }}{\to }L$, then for every $\epsilon >0$,

$$\sum _{k\in I_r}\parallel x_k-L\parallel \ge \sum _{\begin{array}{c}k\in I_r\\ \parallel x_k-L\parallel \ge \epsilon \end{array}}\parallel x_k-L\parallel \ge \epsilon \mathrm{card}\left(\right\{k\in I_r:\parallel x_k-L\parallel \ge \epsilon \left\}\right),$$

which implies that $\left(x_k\right)\underset{S_{\theta }}{\to }L$.

2. Let us suppose that $\left(x_k\right)$ is bounded and $\left(x_k\right)\underset{S_{\theta }}{\to }L$. Since $\left(x_k\right)$ is bounded, there exists $M>0$ such that $\parallel x_k-L\parallel \le M$ for every $k\in \mathbb{N}$. Given $\epsilon >0$,

$$\begin{array}{cc}\hfill \frac{1}{h_r}\sum _{k\in I_r}\parallel x_k-L\parallel & =\frac{1}{h_r}\sum _{\begin{array}{c}k\in I_r\\ \parallel x_k-L\parallel \ge \epsilon \end{array}}\parallel x_k-L\parallel +\frac{1}{h_r}\sum _{\begin{array}{c}k\in I_r\\ \parallel x_k-L\parallel <\epsilon \end{array}}\parallel x_k-L\parallel \hfill \\ \hfill & \le \frac{M}{h_r}\mathrm{card}\left(\right\{k\in I_r:\parallel x_k-L\parallel \ge \epsilon \left\}\right)+\epsilon ,\hfill \end{array}$$

so we deduce that $\left(x_k\right)\underset{N_{\theta }}{\to }L$. □

Next, we give an example to illustrate that the hypothesis over the sequence to be bounded is necessary and cannot be removed.

Example 2.

There exist unbounded $S_{\theta }$-convergent sequences to L which are not $N_{\theta }$-summable to L.

Let $\theta =\left(k_r\right)$ be the lacunary sequence with $k_0=0$ and $k_r=2^r$. Notice that

• $h_1=k_1-k_0=2$ and $h_r=2^{r-1}$ for every $r\ge 2$.
• $I_1=(k_0,k_1]=(0,2]$ and $I_r=(2^{r-1},2^r]$ for every $r\ge 2$.

Consider the sequence defined by

$$x_k=\left\{\begin{array}{cc}0& ifk\ne 2^j\mathit{for}\mathit{all}j,\hfill \\ 2^j& ifk=2^j\mathit{for}\mathit{some}j.\hfill \end{array}\right.$$

Given $\epsilon >0$, it is easy to show that ${\displaystyle \frac{\mathrm{card}\left(\right\{k\in I_r:|x_k-0|\ge \epsilon \})}{h_r}\to 0}$ as $r\to \infty$, which implies that $\left(x_k\right)\underset{S_{\theta }}{\to }0$. Also, notice that $\left(x_k\right)$ is an unbounded sequence. However,

$$\frac{{\displaystyle \sum _{k\in I_r}|x_k-0|}}{h_r}=\left\{\begin{array}{cc}{\displaystyle \frac{2}{2}=1}& ifr=1,\hfill \\ {\displaystyle \frac{2^r}{2^{r-1}}=2}& ifr\ge 2,\hfill \end{array}\right\}\underset{r\to \infty }{⟶}2,$$

which implies that $x_k\underset{N_{\theta }}{\nrightarrow }0$.

We now give the definition of lacunary statistically Cauchy sequences in Banach spaces as a generalization of the definition for real-valued sequences by Fridy and Orhan in [11].

Definition 5.

Let X be a Banach space and $\theta =\left(k_r\right)$ a lacunary sequence. A sequence $x=\left(x_k\right)$ is a lacunary statistically Cauchy sequence if there exists a subsequence $x_{k^{\prime }\left(r\right)}$ of $x_k$ such that $k^{\prime }\left(r\right)\in I_r$ for every $r\in \mathbb{N}$, ${\displaystyle \underset{r\to \infty }{lim}{x}_{k^{\prime }\left(r\right)}=L}$ for some $L\in X$ and for every $\epsilon >0$,

$$\underset{r\to \infty }{lim}\frac{1}{h_r}\mathrm{card}\left(\right\{k\in I_r:\parallel x_k-x_{k^{\prime }\left(r\right)}\parallel \ge \epsilon \left\}\right)=0,$$

or equivalently,

$$\underset{r\to \infty }{lim}\frac{1}{h_r}\mathrm{card}\left(\right\{k\in I_r:\parallel x_k-x_{k^{\prime }\left(r\right)}\parallel <\epsilon \left\}\right)=1.$$

In this case, we say that $\left(x_k\right)$ is $S_{\theta }$-Cauchy.

An important result in [11] is the $S_{\theta }$-Cauchy Criterion and some of the next theorems in this work rely on it. This result can also be obtained for sequences in Banach spaces, and we include the proof for the sake of completeness.

Theorem 3.

Let X be a Banach space. A sequence $\left(x_k\right)$ in X is $S_{\theta }$-convergent if and only if it is $S_{\theta }$-Cauchy.

Proof. 

Let $\left(x_k\right)$ be an $S_{\theta }$-convergent sequence in X and for every $k\in \mathbb{N}$, we define $K_j=\{k\in \mathbb{N}:\parallel x_k-L\parallel <1/j\}$. Observe that $K_j\supseteq K_{j+1}$ and ${\displaystyle \frac{\mathrm{card}(K_j\cap I_r)}{h_r}\to 1}$ as $r\to \infty$.

Set $m_1$ such that if $r\le m_1$ then $\mathrm{card}(K_1\cap I_r)/h_r>0$, i.e., $K_1\cap I_r\ne \mathsf{⌀}$. Next, choose $m_2>m_1$ such that if $r\ge m_2$, then $K_2\cap I_r\ne \mathsf{⌀}$. Now, for each $m_1\le r\le m_2$, we choose $k_r^{\prime }\in I_r$ such that $k_r^{\prime }\in I_r\cap K_1$, i.e., $\parallel x_{k_r^{\prime }}-L\parallel <1$. Inductively, we choose $m_{p+1}>m_p$ such that if $r>m_{p+1}$, then $I_r\cap K_{p+1}\ne \mathsf{⌀}$. Thus, for all r such that $m_p\le r<m_{p+1}$, we choose $k_r^{\prime }\in I_r\cap K_p$, and we have $\parallel x_{k_r^{\prime }}-L\parallel <1/p$.

Therefore, we have a sequence $k_r^{\prime }$ such that $k_r^{\prime }\in I_r$ for every $r\in \mathbb{N}$ and ${lim}_{r\to \infty }{x}_{k_r^{\prime }}=L$. Finally,

$$\begin{array}{cc}\hfill \frac{1}{h_r}\mathrm{card}\left(\right\{k\in I_r:\parallel x_k-x_{k_r^{\prime }}\parallel \ge \epsilon \left\}\right)\le & \frac{1}{h_r}\mathrm{card}\left(\right\{k\in I_r:\parallel x_k-L\parallel \ge \epsilon /2\left\}\right)\hfill \\ \hfill & +\frac{1}{h_r}\mathrm{card}\left(\right\{k\in I_r:\parallel x_{k_r^{\prime }}-L\parallel \ge \epsilon /2\left\}\right).\hfill \end{array}$$

Since $\left(x_k\right)\underset{S_{\theta }}{\to }L$ and ${lim}_{r\to \infty }{x}_{k_r^{\prime }}=L$ we deduce that $\left(x_k\right)$ is $S_{\theta }$- Cauchy.

Conversely, if $\left(x_k\right)$ is a Cauchy sequence, for every $\epsilon >0$,

$$\begin{array}{cc}\hfill \mathrm{card}\left(\right\{k\in I_r:\parallel x_k-L\parallel \}\parallel \ge \epsilon \left\}\right)\le & \mathrm{card}\left(\right\{k\in I_r:\parallel x_k-x_{k_r^{\prime }}\parallel \ge \epsilon /2\left\}\right)\hfill \\ \hfill & +\mathrm{card}\left(\right\{k\in I_r:\parallel x_{k_r^{\prime }}-L\parallel \ge \epsilon /2\left\}\right).\hfill \end{array}$$

Since $\left(x_k\right)$ is $S_{\theta }$-Cauchy and ${lim}_{r\to \infty }{x}_{k_r^{\prime }}=L$, we deduce that $\left(x_k\right)\underset{S_{\theta }}{\to }L$. □

3. The Statistical Lacunary Summability Space

Let us consider X a real Banach space, ${\sum }_i{x}_i$ a series in X and $\theta =\left(k_r\right)$ a lacunary sequence. We define

$$S_{S_{\theta }}\left(\sum _i{x}_i\right)=\left\{{\left(a_i\right)}_i\in {\ell }_{\infty }:\sum _i{a}_i{x}_i\mathrm{is}{S}_{\theta }\mathrm{-summable}\right\}$$

endowed with the supremum norm. This space will be named as the space of $S_{\theta }$-summability associated with ${\sum }_i{x}_i$. We will characterize the completeness of the space $S_{S_{\theta }}\left({\sum }_i{x}_i\right)$ in Theorem 4, but first we need a lemma.

Lemma 1.

Let X be a real Banach space and suppose that the series $\sum x_i$ is not wuC. Then there exist $f\in X^{*}$ and a null sequence ${\left(a_i\right)}_i\in c_0$ such that

$$\sum _i{a}_{i}f\left(x_i\right)=+\infty$$

and

$$a_{i}f\left(x_i\right)\ge 0.$$

Proof. 

Since ${\sum }_{i=1}^{\infty }\left|f\left(x_i\right)\right|=+\infty$, there exists $m_1$ such that ${\sum }_{i=1}^{m_1}\left|f\left(x_i\right)\right|>2·2$.

We define $a_i=\frac{1}{2}$ if $f\left(x_i\right)\ge 0$ and $a_i=-\frac{1}{2}$ if $f\left(x_i\right)<0$ for $i\in \{1,2,\cdots ,m_1\}$.

This implies that ${\sum }_{i=1}^{m_1}{a}_{i}f\left(x_i\right)>2$ and $a_{i}f\left(x_i\right)\ge 0$ if $i\in \{1,2,\cdots ,m_1\}$.

Let $m_2>m_1$ be such that ${\sum }_{i=m_1+1}^{m_2}\left|f\left(x_i\right)\right|>2^2·2^2$.

We define $a_i=\frac{1}{2^2}$ if $f\left(x_i\right)\ge 0$ and $a_i=-\frac{1}{2^2}$ if $f\left(x_i\right)<0$ for $i\in \{m_1+1,\cdots ,m_2\}$. Hence ${\sum }_{i=m_1+1}^{m_2}{a}_{i}f\left(x_i\right)>2^2$ and $a_{i}f\left(x_i\right)\ge 0$ if $i\in \{m_1+1,\cdots ,m_2\}$. So we have obtained a sequence ${\left(a_i\right)}_i\in c_0$ with the above properties. □

Theorem 4.

Let X be a real Banach space and $\theta =\left(k_r\right)$ a lacunary sequence. The following are equivalent:

(1) 

The series ${\sum }_i{x}_i$ is weakly unconditionally Cauchy (wuC).

(2) 

The space$S_{S_{\theta }}\left({\sum }_i{x}_i\right)$is complete.

(3) 

The space of all null sequences $c_0$ is contained in $S_{S_{\theta }}\left({\sum }_i{x}_i\right).$

Proof. 

(1)⇒(2): Since $\sum x_i$ is wuC, the following supremum is finite:

$$H=sup\left\{∥\sum _{i=1}^n{a}_i{x}_i∥:\left|a_i\right|\le 1,1\le i\le n,n\in \mathbb{N}\right\}<+\infty .$$

Let ${\left(a^m\right)}_m\subset S_{S_{\theta }}\left({\sum }_i{x}_i\right)$ such that ${\displaystyle \underset{m}{lim}{\parallel a^m-a^0\parallel }_{\infty }=0}$, with $a^0\in {\ell }_{\infty }$. We will show that $a^0\in S_{S_{\theta }}\left({\sum }_i{x}_i\right)$. Let us suppose without any loss of generality that $\parallel a^0{\parallel }_{\infty }\le 1$. Then, the partial sums $S_k^0={\sum }_{i=1}^k{a}_i^0{x}_i$ satisfy $\parallel S_k^0\parallel \le H$ for every $k\in \mathbb{N}$, i.e., the sequence $\left(S_k^0\right)$ is bounded. Then, $a^0\in S_{S_{\theta }}\left({\sum }_i{x}_i\right)$ if and only if $\left(S_k^0\right)$ is $S_{\theta }$-summable to some $L\in X$. According to Theorem 3, $\left(S_k^0\right)$ is lacunary statistically convergent to $L\in X$ if and only if $\left(S_k^0\right)$ is a lacunary statistically Cauchy sequence.

Given $\epsilon >0$ and $n\in \mathbb{N}$, we obtain statement (2) if we show that there exists a sub-sequence $\left(S_{k^{\prime }\left(r\right)}\right)$ such that $k^{\prime }\left(r\right)\in I_r$ for every r, ${\displaystyle \underset{r\to \infty }{lim}{S}_{k^{\prime }\left(r\right)}=L}$ and

$$d_{\theta }(\{k\in I_r:\parallel S_k^0-S_{k^{\prime }\left(r\right)}^0\parallel <\epsilon \})=1.$$

Since $a^m\to a^0$ in ${\ell }_{\infty }$, there exists $m_0>n$ such that ${\displaystyle \parallel a^m-a^0{\parallel }_{\infty }<\frac{\epsilon }{4H}}$ for all $m>m_0$, and since $S_k^{m_0}$ is $S_{\theta }$-Cauchy, there exists $k^{\prime }\left(r\right)\in I_r$ such that ${\displaystyle \underset{r\to \infty }{lim}{S}_{k^{\prime }\left(r\right)}^{m_0}=L}$ for some L and

$$d_{\theta }\left(\left\{k\in I_r:\parallel S_k^{m_0}-S_{k^{\prime }\left(r\right)}^{m_0}\parallel <\frac{\epsilon }{2}\right\}\right)=1.$$

Consider $r\in \mathbb{N}$ and fix $k\in I_r$ such that

$$\begin{array}{c}\hfill {\displaystyle \parallel S_k^{m_0}-S_{k^{\prime }\left(r\right)}^{m_0}\parallel <\frac{\epsilon }{2}.}\end{array}$$

(1)

We will show that $\parallel S_k^0-S_{k^{\prime }\left(r\right)}^0\parallel <\epsilon$, and this will prove that

$$\left\{k\in I_r:\parallel S_k^{m_0}-S_{k^{\prime }\left(r\right)}^{m_0}\parallel <\frac{\epsilon }{2}\right\}\subset \{k\in I_r:\parallel S_k^0-S_{k^{\prime }\left(r\right)}^0\parallel <\epsilon \}.$$

Since the first set has density 1, the second will also have density 1 and we will be done.

Let us observe first that for every $j\in \mathbb{N}$,

$$∥\sum _{i=1}^j\frac{4H}{\epsilon }(a_i^m-a_i^{m_0})x_i∥\le H,$$

for every $m>m_0$, therefore

$$\begin{array}{c}\hfill ∥S_j^0-S_j^{m_0}∥=∥\sum _{i=1}^j(a_i^0-a_i^{m_0})x_i∥\le \frac{\epsilon }{4}.\end{array}$$

(2)

Then, by applying the triangular inequality,

$$\begin{array}{cc}\hfill \parallel S_k^0-S_{k^{\prime }\left(r\right)}^0\parallel & \le \parallel S_k^0-S_k^{m_0}\parallel +\parallel S_k^{m_0}-S_{k^{\prime }\left(r\right)}^{m_0}\parallel +\parallel S_{k^{\prime }\left(r\right)}^{m_0}-S_{k^{\prime }\left(r\right)}^0\parallel \hfill \\ \hfill & <\frac{\epsilon }{4}+\frac{\epsilon }{2}+\frac{\epsilon }{4}=\epsilon .\hfill \end{array}$$

where the last inequality follows by applying (1) and (2), which yields the desired result.

$\left(2\right)\Rightarrow \left(3\right)$: Let us observe that if $S_{S_{\theta }}\left({\sum }_i{x}_i\right)$ is complete, then it contains the space of eventually zero sequences $c_{00}$ and therefore the thesis comes, since the supremum norm completion of $c_{00}$ is $c_0$.

$\left(3\right)\Rightarrow \left(1\right)$: By way of contradiction, suppose that the series $\sum x_i$ is not wuC. Therefore there exists $f\in X^{*}$ such that ${\displaystyle \sum _{i=1}^{\infty }\left|f\left(x_i\right)\right|=+\infty }$. By Lemma 1 we can construct inductively a sequence ${\left(a_i\right)}_i\in c_0$ such that

$$\sum _i{a}_{i}f\left(x_i\right)=+\infty$$

and

$$a_{i}f\left(x_i\right)\ge 0.$$

Now we will prove that the sequence $S_k={\sum }_{i=1}^k{a}_{i}f\left(x_i\right)$ is not $S_{\theta }$-summable to any $L\in \mathbb{R}$. By way of contradiction, suppose that it is $S_{\theta }$-summable to $L\in \mathbb{R}$, then we have

$$\frac{1}{h_r}\mathrm{card}\left(\right\{k\in I_r:|S_k-L|\ge \epsilon \})=\frac{1}{h_r}\sum _{\begin{array}{c}k=k_{r-1}\\ |S_k-L|\ge \epsilon \end{array}}^{k_r}1\underset{r\to \infty }{\to }0.$$

Since $S_k$ is an increasing sequence and $S_k\to \infty$, there exists $k_0$ such that $|S_k-L|\ge \epsilon$ for every $k\ge k_0$. Let us suppose that $k_r>k_0$ for every r. Hence,

$$\frac{1}{h_r}\sum _{\begin{array}{c}k=k_{r-1}\\ |S_k-L|\ge \epsilon \end{array}}^{k_r}1=\frac{h_r}{h_r}=1\underset{r\to \infty }{\nrightarrow }0,$$

which is a contradiction. This implies that $S_k$ is not $S_{\theta }$-convergent and this is a contradiction with (3). □

4. The Strong Lacunary Summability Space

Let X be a real Banach space, ${\sum }_i{x}_i$ a series in X and $\theta =\left(k_r\right)$ a lacunary sequence. We define

$$S_{N_{\theta }}\left(\sum _i{x}_i\right)=\left\{{\left(a_i\right)}_i\in {\ell }_{\infty }:\sum _i{a}_i{x}_{i}is{N}_{\theta }\mathrm{-summable}\right\}$$

endowed with the supremum norm. This will be named as the space of $N_{\theta }$-summability associated with the series ${\sum }_i{x}_i$. We can now present a theorem very similar to that of Theorem 4 but for the case of $N_{\theta }$-summability. Indeed Theorem 5 characterizes the completeness of the space $S_{N_{\theta }}\left({\sum }_i{x}_i\right)$.

Theorem 5.

Let X be a real Banach space and $\theta =\left(k_r\right)$ a lacunary sequence. The following are equivalent:

(1) 

The series ${\sum }_i{x}_i$ is weakly unconditionally Cauchy (wuC).

(2) 

The space$S_{N_{\theta }}\left({\sum }_i{x}_i\right)$is complete.

(3) 

The space of all null sequences $c_0$ is contained in $S_{N_{\theta }}\left({\sum }_i{x}_i\right).$

Proof. 

(1) ⇒ (2): Since $\sum x_i$ is wuC, the following supremum is finite

$$H=sup\left\{∥\sum _{i=1}^n{a}_i{x}_i∥:\left|a_i\right|\le 1,1\le i\le n,n\in \mathbb{N}\right\}<+\infty .$$

Let ${\left(a^m\right)}_m\subset S_{N_{\theta }}\left({\sum }_i{x}_i\right)$ such that ${\displaystyle \underset{m}{lim}{\parallel a^m-a^0\parallel }_{\infty }=0}$, with $a^0\in {\ell }_{\infty }$.

We will show that $a^0\in S_{N_{\theta }}\left({\sum }_i{x}_i\right)$.

Without loss of generality we can suppose that $\parallel a^0{\parallel }_{\infty }\le 1$. Therefore the partial sums $S_k^0={\sum }_{i=1}^k{a}_i^0{x}_i$ satisfy $\parallel S_k^0\parallel \le H$ for every $k\in \mathbb{N}$, i.e., the sequence $\left(S_k^0\right)$ is bounded. Hence $a^0\in S_{N_{\theta }}\left({\sum }_i{x}_i\right)$ if and only if $\left(S_k^0\right)$ is $N_{\theta }$-summable to some $L\in X$. Since $\left(S_k^0\right)$ is bounded, it is sufficient to show that $\left(S_k\right)$ is $S_{\theta }$-convergent, thanks to to Fridy and Orhan’s Theorem ([10], Theorem 2.1) (see Theorem 2). The result follows analogously as in Theorem 4.

$\left(2\right)\Rightarrow \left(3\right)$: It is sufficient to notice that $S_{S_{\theta }}\left({\sum }_i{x}_i\right)$ is a complete space and it contains the space of eventually zero sequences $c_{00}$, so it contains the completion of $c_{00}$ with respect to the supremum norm, hence it contains $c_0$.

$\left(3\right)\Rightarrow \left(1\right)$: By way of contradiction, suppose that the series $\sum x_i$ is not wuC. Therefore there exists $f\in X^{*}$ such that ${\displaystyle \sum _{i=1}^{\infty }\left|f\left(x_i\right)\right|=+\infty }$. By Lemma 1 we can construct inductively a sequence ${\left(a_i\right)}_i\in c_0$ such that ${\sum }_i{a}_{i}f\left(x_i\right)=+\infty$ and $a_{i}f\left(x_i\right)\ge 0$.

The sequence $S_k={\sum }_{i=1}^k{a}_{i}f\left(x_i\right)$ is not $N_{\theta }$-summable to any $L\in \mathbb{R}$.

As $S_k\to \infty$, for every $A>0$, there exists $k_0$ such that $|S_k|>A$ if $k\ge k_0$. Then we have

$$\frac{1}{h_r}\sum _{k\in I_r}\left|S_k\right|>\frac{h_{r}A}{h_r}=A.$$

Hence $S_k$ is not $N_{\theta }$-summable to any $L\in \mathbb{R}$, otherwise

$$\infty \leftarrow \frac{1}{h_r}\sum _{k\in I_r}|S_k|\le |L|+\frac{1}{h_r}\sum _{k\in I_r}|S_k-L|\to |L|$$

We can conclude that $S_k$ is not $N_{\theta }$-convergent, a contradiction with (3). □

5. Characterizations of the Completeness of a Banach Space

A Banach space X can be characterized by the completeness of the space $S_{N_{\theta }}\left({\sum }_k{x}_k\right)$ for every wuC series ${\sum }_k{x}_k$, as we will show next.

Theorem 6.

Let X be a normed real vector space. Then X is complete if and only if$S_{N_{\theta }}\left({\sum }_k{x}_k\right)$is a complete space for every weakly unconditionally Cauchy series (wuC) ${\sum }_k{x}_k$.

Proof. 

Thanks to Theorem 4, the condition is necessary.

Now suppose that X is not complete, hence there exists a series $\sum x_k$ in X such that ${\displaystyle \parallel x_k\parallel \le \frac{1}{k{2}^k}}$ and $\sum x_k=x^{**}\in X^{**}\backslash X$.

We will provide a wuC series ${\sum }_k{y}_k$ such that $S_{N_{\theta }}\left({\sum }_k{y}_k\right)$ is not complete, a contradiction.

Set ${\displaystyle S_N=\sum _{k=1}^N{x}_k}$. As $X^{**}$ is a Banach space endowed with the dual topology, ${\displaystyle \underset{\parallel y^{*}\parallel \le 1}{sup}|y^{*}\left(S_N\right)-x^{**}\left(y^{*}\right)|}$ tends to 0 as $N\to \infty$, i.e.,

$$\begin{array}{c}\hfill \underset{N\to +\infty }{lim}{y}^{*}\left(S_N\right)=\underset{N\to +\infty }{lim}\sum _{k=1}^N{y}^{*}\left(x_k\right)=x^{**}\left(y^{*}\right),\mathrm{for}\mathrm{every}\parallel y^{*}\parallel \le 1.\end{array}$$

(3)

Put $y_k=k{x}_k$ and let us observe that $\parallel y_k\parallel <\frac{1}{2^k}$. Therefore $\sum y_k$ is absolutely convergent, thus it is unconditionally convergent and weakly unconditionally Cauchy.

We claim that the series ${\displaystyle \sum _k\frac{1}{k}{y}_k}$ is not $N_{\theta }$-summable in X.

By way of contradiction suppose that $S_N={\sum }_{k=1}^N\frac{1}{k}{y}_k$ is $N_{\theta }$-summable in X, i.e., there exists L in X such that ${\displaystyle \underset{r\to \infty }{lim}\frac{1}{h_r}\sum _{k\in I_r}\parallel S_k-L\parallel =0}$. This implies that

$$\begin{array}{c}\hfill \underset{r\to +\infty }{lim}\frac{1}{h_r}\sum _{k\in I_r}{y}^{*}\left(S_k\right)=y^{*}\left(L\right),\mathrm{for}\mathrm{every}\parallel y^{*}\parallel \le 1.\end{array}$$

(4)

From Equations (3) and (4), the uniqueness of the limit and since $N_{\theta }$ is a regular method, we have $x^{**}\left(y^{*}\right)=y^{*}\left(L\right)$ for every $\parallel y^{*}\parallel \le 1$, so we obtain $x^{**}=L\in X$, a contradiction. Hence $S_N={\sum }_{k=1}^N\frac{1}{k}{y}_k$ is not $N_{\theta }$-summable to any $L\in X$.

Finally, let us observe that since ${\sum }_k{y}_k$ is a weakly unconditionally Cauchy series and $S_N={\sum }_{k=1}^N\frac{1}{k}{y}_k$ is not $N_{\theta }$-summable, we have $\left(\frac{1}{k}\right)\notin S_{N_{\theta }}\left({\sum }_k{y}_k\right)$ and this means that $c_0⊈S_{N_{\theta }}\left({\sum }_k{y}_k\right)$ which is a contradiction with Theorem 5(3), so the proof is complete. □

By a similar argument and taking into account Theorem 2, we have also the characterization for the $S_{\theta }$-summability:

Theorem 7.

Let X be a normed real vector space. Then X is complete if and only if$S_{S_{\theta }}\left({\sum }_i{x}_i\right)$is a complete space for every weakly unconditionally Cauchy series (wuC) ${\sum }_i{x}_i$.

Let $0<p<+\infty$, the sequence $\left(x_n\right)$ is said to be strongly p-Cesàro or ${\mathrm{w}}_p$-summable if there is $L\in X$ such that

$$\underset{n}{lim}\frac{1}{n}\sum _{i=1}^n{\parallel x_i-L\parallel }^p=0;$$

in this case we will write $\left(x_k\right)\underset{{\mathrm{w}}_p}{\to }L$ and $L={\mathrm{w}}_p-{lim}_n{x}_n.$ Let $\sum x_i$ be a series in a real Banach space X, let us define

$$S_{{\mathrm{w}}_p}\left(\sum _i{x}_i\right)=\left\{{\left(a_i\right)}_i\in {\ell }_{\infty }:\sum _i{a}_i{x}_i\mathrm{is}{\mathrm{w}}_p-summable\right\}$$

endowed with the supremum norm. We refer to [16] for other properties of the space $S_{{\mathrm{w}}_p}\left({\sum }_i{x}_i\right)$.

Finally, from Theorem 6, Theorem 7 and ([16], Theorem 3.5), we derive the following corollary.

Corollary 1.

Let X be a normed real vector space and $p\ge 1$. The following are equivalent:

1. 

X is complete.

2. 

$S_{N_{\theta }}\left({\sum }_k{x}_k\right)$ is complete for every weakly unconditionally Cauchy series (wuC) ${\sum }_k{x}_k$.

3. 

$S_{S_{\theta }}\left({\sum }_k{x}_k\right)$ is complete for every weakly unconditionally Cauchy series (wuC) ${\sum }_k{x}_k$.

4. 

$S_{{\mathrm{w}}_p}\left({\sum }_k{x}_k\right)$ is complete for every weakly unconditionally Cauchy series (wuC) ${\sum }_k{x}_k$.