New Results on the Sequence Spaces Inclusion Equations of the Form $F\subset \mathcal{E}+F_x^{\prime }$ Where F, F′ ∈ {w₀, w, w_∞}

Abstract

We determine the multipliers $M(X,Y)$ where X, $Y\in \{w_0,w,w_{\infty }\}$, and we apply these results to the solvability of both the (SSIE) of the form $F\subset \mathcal{E}+w_x$, where $F\in \{w_0,w,w_{\infty }\}$, and the (SSIE) $w\subset \mathcal{E}+W_x^0$ and $w\subset \mathcal{E}+W_x$.

1. Introduction

The set of all complex sequences is denoted by $\omega$, and we use the standard notations for the classical sequence spaces. We write $yz$ for the termwise product sequence of $y,z\in \omega$. Also, U and $U^{+}$ denote the set of all sequences with no zero terms, and of real sequences with positive terms. If E is any subset of $\omega$, then $E^{+}=\{x\in E:x_k>0 \mathrm{for} \mathrm{all} k\}$. For any subset $\chi$ of $\omega$, we write $\overline{\chi }=\{x\in U^{+}:1/x=(1/x_k)\in \chi \}$.

Throughout the paper, we use the notations and concepts of [1], and the notation $M(X,Y)$ for the multiplier of $X,Y\subset \omega$ (p. 64, [2]).

The spaces $w_{\infty }$ and $w_0$ were introduced and studied by Maddox in [3,4,5]. The sets $W_a={(1/a)}^{-1}\ast w_{\infty }$, and $W_a^0={(1/a)}^{-1}\ast w_0$ are introduced in Chapter 4 [1].

We recall that an infinite matrix $\mathcal{T}$ is said to be a triangle if ${\mathcal{T}}_{nk}=0$ for $k>n$ and ${\mathcal{T}}_{nn}\ne 0$ for all n.

In this paper, we extend the results of Chapters 5 and 6 in [1] and [6], determine the multipliers $M(X,Y)$, where X, $Y\in \{w_0,w,w_{\infty }\}$, and study the solvability of each of the (SSIEs) of the form

(1)

$F\subset \mathcal{E}+w_x$, where $F\in \{w_0,w,w_{\infty }\}$, and

(2)

$w\subset \mathcal{E}+W_x^0$ and $w\subset \mathcal{E}+W_x$,

where $\mathcal{E}$ satisfies either of the conditions (i) $\mathcal{E}\subset c_0$ or (ii) $\mathcal{E}\subset W_a$ with $a\in c_0^{+}$.

We use results from the theory of $BK$ and $AK$ spaces, and refer the reader to the monographs [2,7,8,9]. We also refer the reader to [10,11] for notations and results from classical summability theory.

2. On the Triangle $\mathit{C}\mathbf{\left(}\mathit{\lambda }\mathbf{\right)}$ and the Sets ${\mathit{W}}_{\mathit{a}}$, ${\mathit{W}}_{\mathit{a}}^{\mathbf{0}}$, and ${\mathit{W}}_{\mathit{a}}^{\mathbf{\left(}\mathit{c}\mathbf{\right)}}$

For $\lambda \in U$, we define the triangles $C\left(\lambda \right)$ and $\Delta \left(\lambda \right)$ by ${\left[C\left(\lambda \right)\right]}_{nk}=1/{\lambda }_n$ for $k\le n$, and the non-zero entries of $\Delta \left(\lambda \right)$ by ${[\Delta \left(\lambda \right)]}_{nn}={\lambda }_n$ for all n and ${[\Delta \left(\lambda \right)]}_{n,n-1}=-{\lambda }_{n-1}$ for all $n\ge 2$. A prominent special case of $\Delta \left(\lambda \right)$ is the familiar matrix of the backward differences $\Delta \left(e\right)$, where $e=(1,1,\dots )$. The matrix domains of $\Delta$ in ${\ell }_{\infty },c,c_0$ were studied by Kızmaz [12].

We recall the definition of the sets (p. 209, [1])

$$\begin{array}{ccc}\hfill W_a& =& \left\{y\in \omega :\underset{n}{sup}\left(\frac{1}{n}\sum _{k=1}^n\frac{|y_k|}{a_k}\right)<\infty \right\}\mathrm{and}\hfill \\ \hfill W_a^0& =& \left\{y\in \omega :\underset{n\to \infty }{lim}\left(\frac{1}{n}\sum _{k=1}^n\frac{|y_k|}{a_k}\right)=0\right\},\hfill \end{array}$$

If $a={\left(r^n\right)}_{n\ge 1}$ with $r>0$, then we write $W_a=W_r$ and $W_a^0=W_r^0$. For $r=1$, we obtain the well-known sets $w_{\infty }$, $w_0$, and w ([3,4]). We have $y\in w_a$ if and only if there is a scalar L such that

$$\underset{n\to \infty }{lim}\left(\frac{1}{n}\sum _{k=1}^n\left|\frac{y_k}{a_k}-L\right|\right)=0.$$

It is easy to prove the following results.

Lemma 1. 

Let $p\ge 1$. We have

(i) 

(a) $M(c,c_0)=M({\ell }_{\infty },c)=M({\ell }_{\infty },c_0)=c_0$ and $M(c,c)=c$,

(b) $M(E,{\ell }_{\infty })=M(c_0,F)={\ell }_{\infty }$ for E, $F=c_0$, c or ${\ell }_{\infty }$,

(c) $M(c_0,{\ell }^p)=M(c,{\ell }^p)=M({\ell }_{\infty },{\ell }^p)={\ell }^p$,

(d) $M({\ell }^p,F)={\ell }_{\infty }$ for $F\in \{c_0,c,s_1,{\ell }^p\}$;

(ii) 

(a) $M(w_0,F)=s_{{(1/n)}_{n\ge 1}}$ for $F=c_0$, c or ${\ell }_{\infty }$,

(b) $M(w_{\infty },c_0)=M(w_{\infty },c)=s_{{(1/n)}_{n\ge 1}}^0$,

(c) $M({\ell }^1,w_{\infty })=s_{{\left(n\right)}_{n\ge 1}}$ and $M({\ell }^1,w_0)=s_{{\left(n\right)}_{n\ge 1}}^0$,

(d) $M(E,w_0)=w_0$ for $E=s_1$ or c,

(e) $M(E,w_{\infty })=w_{\infty }$ for $E=c_0$, $s_1$ or c;

(iii) 

$M(w,{\ell }_{\infty })=M(w,c)=M(w,c_0)=s_{{(1/n)}_{n\ge 1}}$.

Remark 1. 

It can easily be shown that $M(w_0,w_{\infty })=M(w_{\infty },w_{\infty })={\ell }_{\infty }$. Since $w_0\subset w\subset w_{\infty }$, this implies $M(w,w_{\infty })={\ell }_{\infty }$.

3. On the Multipliers Involving the Sets ${\mathit{w}}_{\mathbf{0}}$, $\mathit{w}$, and ${\mathit{w}}_{\infty }$

In this section, we determine the multipliers $M(X,Y)$, where $X\in \{w_{\infty },w\}$ and $Y\in \{w_0,w,w_{\infty }\}$. We state the following lemma.

Lemma 2. 

We have

$$\left(i\right)M(w_{\infty },Y)=\left\{\begin{array}{cc}{c}_0\hfill & ifY=w_0\mathrm{or}w\hfill \\ {\ell }_{\infty }\hfill & ifY=w_{\infty };\hfill \end{array}\right.$$

$$\left(ii\right)M(w,Y)=\left\{\begin{array}{cc}{w}_0\cap {\ell }_{\infty }\hfill & ifY=w_0\hfill \\ w\cap {\ell }_{\infty }\hfill & ifY=w\hfill \\ {\ell }_{\infty }\hfill & ifY=w_{\infty }.\hfill \end{array}\right.$$

Proof. 

(i) First we show $M(w_{\infty },w_0)=M(w_{\infty },w)=c_0$.

Since $w\subset s_{{\left(n\right)}_{n\ge 1}}^0$, we have $M(w_{\infty },w)\subset M(w_{\infty },s_{{\left(n\right)}_{n\ge 1}}^0)$. It follows that $\alpha \in M(w_{\infty },s_{{\left(n\right)}_{n\ge 1}}^0)$ if and only if ${({\alpha }_n/n)}_{n\ge 1}\in M(w_{\infty },c_0)=s_{{(1/n)}_{n\ge 1}}^0$, that is, $\alpha \in c_0$. So, we have shown the inclusion $M(w_{\infty },w)\subset c_0$.

Now, we show $c_0\subset M(w_{\infty },w_0)$.

Let $\alpha \in c_0$. Then, for any given $y\in w_{\infty }$ and $\epsilon >0$, there is an integer $N_0$ such that $n^{-1}{sup}_{k\le N_0}|{\alpha }_k|{\sum }_{k=1}^{N_0}\left|y_k\right|\le \epsilon /2$ for all $n\ge N_0$ and $|{\alpha }_k{|·\Vert y\Vert }_{w_{\infty }}\le \epsilon /2$ for $k\ge N_0+1$. We have

$$\begin{array}{ccc}\hfill \frac{1}{n}\sum _{k=1}^n\left|{\alpha }_k{y}_k\right|& =& \frac{1}{n}\left(\sum _{k=1}^{N_0}\left|{\alpha }_k{y}_k\right|+\sum _{k=N_0+1}^n\left|{\alpha }_k{y}_k\right|\right)\hfill \\ & \le & \frac{1}{n}\underset{k\le N_0}{sup}\left|{\alpha }_k\right|\sum _{k=1}^{N_0}\left|y_k\right|+\left(\underset{k\ge N_0+1}{sup}\left|{\alpha }_k\right|\right){∥y∥}_{w_{\infty }}\hfill \\ & \le & \epsilon /2+\epsilon /2=\epsilon \mathrm{for}\mathrm{all}n\ge N_0.\hfill \end{array}$$

We obtain $\alpha y\in w_0$ and $c_0\subset M(w_{\infty },w_0)$. Since $w_0\subset w$, we obtain the inclusions

$$c_0\subset M(w_{\infty },w_0)\subset M(w_{\infty },w)\subset c_0$$

and we conclude $M(w_{\infty },w_0)=M(w_{\infty },w)=c_0$. As we have seen in Remark 1, we have $M(w_{\infty },w_{\infty })={\ell }_{\infty }$. This concludes the proof of Part (i).

(ii) First, we show the identity $M(w,w_0)=w_0\cap {\ell }_{\infty }$.

From $M(w,w_0)\subset M(c,w_0)=w_0$ and $M(w,w_0)\subset M(w,w_{\infty })={\ell }_{\infty }$, we obtain

$$M(w,w_0)\subset w_0\cap {\ell }_{\infty }.$$

Now, we show the inclusion $w_0\cap {\ell }_{\infty }\subset M(w,w_0)$.

Let $\alpha \in w_0\cap {\ell }_{\infty }$. For every $y\in w$, there is a scalar l such that $n^{-1}{\sum }_{k=1}^n|y_k-l|=o\left(1\right)$ $(n\to \infty )$, and we have

$$\begin{array}{ccc}\hfill \frac{1}{n}\sum _{k=1}^n\left|{\alpha }_k{y}_k\right|& =& \frac{1}{n}\sum _{k=1}^n|{\alpha }_k|·|y_k-l+l|\hfill \\ & \le & \frac{1}{n}\sum _{k=1}^n|{\alpha }_k|·|y_k-l|+\frac{1}{n}\left|l\right|\sum _{k=1}^n\left|{\alpha }_k\right|\hfill \\ & \le & {\Vert \alpha \Vert }_{{\ell }_{\infty }}o\left(1\right)+o\left(1\right)=o\left(1\right)(n\to \infty ).\hfill \end{array}$$

Thus, we have shown $w_0\cap {\ell }_{\infty }\subset M(w,w_0)$, and we conclude $M(w,w_0)=w_0\cap {\ell }_{\infty }$.

Finally, we show the identity $M(w,w)=w\cap {\ell }_{\infty }$.

First, it can easily be seen that the inclusions $M(w,w)\subset M(c,w)\subset w$ and $M(w,w)\subset M(w,w_{\infty })={\ell }_{\infty }$ hold, which imply the inclusion $M(w,w)\subset w\cap {\ell }_{\infty }$. It remains to show the inclusion $w\cap {\ell }_{\infty }\subset M\left(w,w\right)$. Let $\alpha \in w\cap {\ell }_{\infty }$. Then, there is a scalar $\chi$ such that $\alpha -\chi e\in w_0$. For any given $y\in w$, we need to show $\alpha y\in w$. We have $y-le\in w_0$ for some scalar l, and we may write

$${\alpha }_k{y}_k-l\chi =({\alpha }_k-\chi )(y_k-l)+({\alpha }_k-\chi )l+\chi (y_k-l).$$

If we put ${\rho }_n=n^{-1}{\sum }_{k=1}^n|{\alpha }_k{y}_k-l\chi |$, then we have

$$\begin{array}{ccc}\hfill {\rho }_n& \le & \frac{1}{n}\sum _{k=1}^n|{\alpha }_k-\chi |·|y_k-l|+|l|\frac{1}{n}\sum _{k=1}^n|{\alpha }_k-\chi |+|\chi |\frac{1}{n}\sum _{k=1}^n|y_k-l|\hfill \\ & \le & O\left(1\right)\frac{1}{n}\sum _{k=1}^n|y_k-l|+|l|o\left(1\right)+|\chi |\frac{1}{n}\sum _{k=1}^n|y_k-l|=o\left(1\right)(n\to \infty ).\hfill \end{array}$$

We deduce ${\rho }_n=o\left(1\right)$$(n\to \infty )$ and $\alpha \in M(w,w)$. So, we have shown $w\cap {\ell }_{\infty }\subset M(w,w)$, and $M(w,w)=w\cap {\ell }_{\infty }$.

Finally, by Remark 1, the identity $M(w,w_{\infty })={\ell }_{\infty }$ follows from the inclusions

$$M(w_{\infty },w_{\infty })\subset M(w,w_{\infty })\subset M(w_0,w_{\infty }),$$

where $M(w_0,w_{\infty })=M(w_{\infty },w_{\infty })={\ell }_{\infty }$.

This concludes the proof. □

4. Application to the Solvability of the (SSIE) of the Form $\mathit{F}\mathbf{\subset }\mathcal{E}\mathbf{+}{\mathit{w}}_{\mathit{x}}$, Where $\mathit{F}\mathbf{\in }\mathbf{\{}{\mathit{w}}_{\mathbf{0}}\mathbf{,}\mathit{w}\mathbf{,}{\mathit{w}}_{\mathbf{\infty }}\mathbf{\}}$, and of the (SSIE) $\mathit{w}\mathbf{\subset }\mathcal{E}\mathbf{+}{\mathit{F}}_{\mathit{x}}^{\mathbf{\prime }}$, Where ${\mathit{F}}^{\mathbf{\prime }}\mathbf{\in }\mathbf{\{}{\mathit{w}}_{\mathbf{0}}\mathbf{,}{\mathit{w}}_{\mathbf{\infty }}\mathbf{\}}$

Now, we are interested in the study of the set of all positive sequences x that satisfy the (SSIE) $F\subset \mathcal{E}+F_x^{\prime }$, where $\mathcal{E}$, F and $F^{\prime }$ are linear spaces of sequences. If the inclusion $F\subset \mathcal{E}$ holds, then the set of all the solutions of this (SSIE) is equal to $U^{+}$. We have seen in Chapter 5 [1] that the solvability of most (SSIEs) is connected to the cases when e does or does not belong to F. We may consider this study as a perturbation problem stated as follows. If we know the set $M(F,F^{\prime })$, then the solutions of the elementary inclusion $F_x^{\prime }\supset F$ are determined by $1/x\in M(F,F^{\prime })$. Then, we consider a linear space $\mathcal{E}\subset \omega$ and deal with the solvability of the perturbed inclusion $F_x^{\prime }+\mathcal{E}\supset F$. So, we are led to study the case when the elementary and the perturbed inclusions have the same set solutions.

In the following, we use the notation

$$\mathcal{I}(\mathcal{E},F,F^{\prime })=\{x\in U^{+}:F\subset \mathcal{E}+F_x^{\prime }\},$$

where $\mathcal{E}$, F, and $F^{\prime }$ are linear spaces of sequences and $a\in U^{+}$ (p. 236, [1]).

In this section, we study the solvability of each of the (SSIEs): (1) $w_0\subset \mathcal{E}+w_x$, (2) $w\subset \mathcal{E}+W_x^0$, (3) $w\subset \mathcal{E}+w_x$, (4) $w\subset \mathcal{E}+W_x$, and (5) $w_{\infty }\subset \mathcal{E}+w_x$.

In all that follows, we say that a linear space $\mathcal{E}$ of sequences satisfies the condition (CW0) if either of the conditions (i) or (ii) holds, where

$$\left(\mathrm{i}\right)\mathcal{E}\subset c_0\mathrm{and}\left(\mathrm{ii}\right)\mathcal{E}\subset W_a\mathrm{with}a\in c_0^{+}.$$

(1)

It can easily be seen that $c_0⊈W_a$ for all $a\in c_0^{+}$, and $W_a⊈c_0$ for all $a\in c_0^{+}\setminus s_{{(1/n)}_{n\ge 1}}^{0+}$.

4.1. On the (SSIE) $w_0\subset \mathcal{E}+w_x$

In this subsection, we study the solvability of the (SSIE)

$$w_0\subset \mathcal{E}+w_x.$$

(2)

First, we need the next lemma, which follows from Proposition 13, p. 61 in [13].

Lemma 3. 

The solutions of each of the (SSIEs) $w_0\subset \mathcal{E}+W_x^0$ and $w_0\subset \mathcal{E}+W_x$, where $\mathcal{E}\subset s_{\lambda }$ with ${\lambda }_n/n\to 0$$(n\to \infty )$ is a linear space of sequences, are determined by $\mathcal{I}(\mathcal{E},w_0,w_0)=\mathcal{I}(\mathcal{E},w_0,w_{\infty })=\overline{s_1}$.

We obtain the following theorem on the solvability of the (SSIE) in (2).

Theorem 1. 

Let $\lambda \in U^{+}$ and let $\mathcal{E}\subset s_{\lambda }$ be a linear space of sequences, where $\lambda {\in }_{{\left(n\right)}_{n\ge 1}}^0$. Then, the set of all the positive solutions of the (SSIE) in (2) is determined by

$$\mathcal{I}(\mathcal{E},w_0,w)=\overline{s_1}.$$

(3)

Proof. 

Using Lemma 3, we obtain $\mathcal{I}(\mathcal{E},w_0,w)\subset \mathcal{I}(\mathcal{E},w_0,w_{\infty })=\overline{s_1}$ and $\mathcal{I}(\mathcal{E},w_0,w)\subset \overline{s_1}$. Then we have $M(w_0,w)=s_1$ by Lemma 2, and conclude (3). □

We deduce the next corollary.

Corollary 1. 

Let $\mathcal{E}$ be a linear space of sequences satisfying condition (CW0). Then the identity in (3) holds.

Proof. 

This result follows from the inclusions $c_0\subset s_{\lambda }$ for some $\lambda \in s_{{\left(n\right)}_{n\ge 1}}^0$, and $W_a\subset s_{\lambda }$ for some $\lambda \in s_{{\left(n\right)}_{n\ge 1}}^0$ and $a\in c_0$. Indeed, the inclusion $W_a\subset s_{\lambda }$ holds if and only if ${(a_n/{\lambda }_n)}_{n\ge 1}\in M(w_{\infty },s_1)=s_{{(1/n)}_{n\ge 1}}$. Then, if we take ${\lambda }_n=n{a}_n$, we have $\lambda \in s_{{\left(n\right)}_{n\ge 1}}^0$, since $a\in c_0$. This completes the proof. □

4.2. On the (SSIE) $w\subset \mathcal{E}+W_x^0$

We state the following result:

Theorem 2. 

Let $\mathcal{E}$ be a linear space of sequences that satisfies the condition in (CW0). Then, the set of all positive sequences x that satisfy the (SSIE) $w\subset \mathcal{E}+W_x^0$ is determined by

$$\mathcal{I}(c_0,w,w_0)=\overline{w_0\cap s_1}.$$

Proof. 

By Lemma 10 (ii) (p. 11, [6]), it is enough to show the result in each of the cases $\mathcal{E}=c_0$ and $\mathcal{E}=W_a$ with $a\in c_0^{+}$.

We begin with the case $\mathcal{E}=c_0$. Let $x\in \mathcal{I}(c_0,w,w_0)$. Then we have

$$w\subset c_0+W_x^0\subset s_{{(1+n{x}_n)}_{n\ge 1}}^0,$$

which implies

$${\left(\frac{1}{1+n{x}_n}\right)}_{n\ge 1}\in M(w,c_0),$$

where $M(w,c_0)=s_{{(1/n)}_{n\ge 1}}$. Then, there is $K>0$ such that

$$\frac{1+n{x}_n}{n}=\frac{1}{n}+x_n\ge K\mathrm{for}\mathrm{all}n.$$

We conclude $x\in \overline{s_1}$ and $\mathcal{I}(c_0,w,w_0)\subset \overline{s_1}$. Then, we have $e\in w$ and since $w\subset c_0+W_x^0$, and there are $\epsilon \in c_0$ and $\nu \in w_0$ such that

$$1={\epsilon }_n+x_n{\nu }_n,$$

and $(1-{\epsilon }_n)/x_n={\nu }_n$ for all n. Since ${lim}_{n\to \infty }(1-{\epsilon }_n)=1$, we obtain $x\in \overline{w_0}$. So, we have shown $\mathcal{I}(c_0,w,w_0)\subset \overline{s_1}\cap \overline{w_0}=\overline{w_0\cap s_1}$. Since $M(w,w_0)=w_0\cap s_1$, we conclude $\mathcal{I}(c_0,w,w_0)=\overline{w_0\cap s_1}$.

Now, we consider the case $\mathcal{E}=W_a$ with $a\in c_0^{+}$. We show the inclusion $\mathcal{I}(W_a,w,w_0)\subset \overline{w_0\cap {\ell }_{\infty }}$. Let $x\in \mathcal{I}(W_a,w,w_0)$. Then, we have

$$w\subset W_a+W_x^0\subset W_a+W_x=W_{a+x},$$

which implies $1/(a+x)\in M(w,w_{\infty })=s_1$. Since $a\in c_0$, this implies $x\in \overline{s_1}$ and $\mathcal{I}(W_a,w,w_0)\subset \overline{s_1}$. It remains to show $\mathcal{I}(W_a,w,w_0)\subset \overline{w_0}$. For this, we let $x\in \mathcal{I}(W_a,w,w_0)$ and we begin to show the inclusion $W_a+W_x^0\subset W_{a+x}^0$. Since $\mathcal{I}(W_a,w,w_0)\subset \overline{s_1}$, we have $x\in \overline{s_1}$, and since $a\in c_0^{+}$, these two conditions imply

$$a/(a+x)\in c_0.$$

Now, we have $c_0\subset w_0\cap s_1=M(w,w_0)$, which implies $a/(a+x)\in M(w,w_0)$ and $w_a\subset W_{a+x}^0$. Then, we have $x/(a+x)\in s_1$. Since $s_1=M(w_0,w_0)$, we obtain $W_x^0\subset W_{a+x}^0$. Thus, we have shown the inclusion $w\subset W_{a+x}^0$. Then, we have ${(a+x)}^{-1}\in M(w,w_0)$. By Lemma 2, where $M(w,w_0)=w_0\cap s_1$, we deduce ${(a+x)}^{-1}\in w_0$. Now, we show $1/x\in w_0$. From the identity

$$x^{-1}={(a+x)}^{-1}(e+a{x}^{-1}),$$

where $e+a{x}^{-1}\in c$, and by the inclusion $c\subset M(w_0,w_0)$, we deduce that the condition ${(a+x)}^{-1}\in w_0$ implies

$${(a+x)}^{-1}(e+a{x}^{-1})\in w_0$$

and $x^{-1}\in w_0$. We conclude $\mathcal{I}(W_a,w,w_0)\subset \overline{s_1}\cap \overline{w_0}$. Again, from the identity $M(w,w_0)=w_0\cap s_1$ and by Lemma 10 in [6], we obtain $\mathcal{I}(W_a,w,w_0)=\overline{w_0\cap s_1}$. This completes the proof. □

Let $R>0$. We may illustrate this result with the solvability of each of the (SSIEs) (1) $w\subset s_R^0+W_x^0$, (2) $w\subset {\ell }_R^p+W_x^0$, and (3) $w\subset {\left(s_R^0\right)}_{\mathsf{\Sigma }}+W_x^0$.

Example 1. 

The set ${\mathcal{I}}_{w,R}^0$ of all positive sequences x that satisfy the (SSIE) $w\subset s_R^0+W_x^0$ is given by

$${\mathcal{I}}_{w,R}^0=\left\{\begin{array}{cc}\overline{w_0\cap s_1}\hfill & ifR\le 1\hfill \\ U^{+}\hfill & ifR>1.\hfill \end{array}\right.$$

(4)

The case $R\le 1$ follows from Theorem 2.

In the case $R>1$, by Lemma 1, we have $M(w,c_0)=s_{{(1/n)}_{n\ge 1}}$, and the condition ${lim}_{n\to \infty }n{R}^{-n}=0$ implies ${\left(R^{-n}\right)}_{n\ge 1}\in M(w,c_0)$ and $w\subset s_R^0$. We conclude ${\mathcal{I}}_{w,R}^0=U^{+}$.

Example 2. 

Let ${\mathcal{I}}_{w,R}^p$$(p\ge 1)$ be the set of all positive sequences x that satisfy the (SSIE) $w\subset {\ell }_R^p+W_x^0$. We show that the set ${\mathcal{I}}_{w,R}^p$ is determined by (4) in Example 1. Indeed, the solvability of this (SSIE) in the case $R\le 1$ follows from Theorem 2.

In the case that $R>1$, we use Part (i) (c) of Lemma 1 and have $s_{{\left(n\right)}_{n\ge 1}}^0\subset {\ell }_R^p$, if and only if ${(n/R^n)}_{n\ge 1}\in M(c_0,{\ell }^p)={\ell }^p$. Since $w\subset s_{{\left(n\right)}_{n\ge 1}}^0$, we conclude $w\subset {\ell }_R^p$ and ${\mathcal{I}}_{w,R}^p=U^{+}$. This completes the proof.

Example 3. 

The set ${\mathcal{I}}_{w,R}^{\prime 0}$ of all positive sequences x that satisfy the (SSIE)

$$w\subset {\left(s_R^0\right)}_{\mathsf{\Sigma }}+W_x^0$$

is given by (4) in Example 1. Indeed, let $R>1$. Then we have $R^{-n}{\sum }_{k=1}^{n}k=O\left(1\right)$$(n\to \infty )$, which successively implies $D_{1/R}\mathsf{\Sigma }{D}_{{\left(k\right)}_{k\ge 1}}\in (c_0,c_0)$, $D_{1/R}\mathsf{\Sigma }\in (s_{{\left(n\right)}_{n\ge 1}}^0,c_0)$, $D_{1/R}\mathsf{\Sigma }\in (w,c_0)$, and $w\subset {\left(s_R^0\right)}_{\mathsf{\Sigma }}$. So, if $R>1$, we have ${\mathcal{I}}_{w,R}^0=U^{+}$.

The case $R\le 1$ follows from Theorem 2, since ${\left(s_R^0\right)}_{\mathsf{\Sigma }}\subset c_0$.

4.3. On the (SSIE) $w\subset \mathcal{E}+F_x^{\prime }$, Where $F^{\prime }\in \{w,w_{\infty }\}$

In this subsection, we study each of the (SSIEs) $w\subset \mathcal{E}+w_x$ and $w\subset \mathcal{E}+W_x$, where $\mathcal{E}$ satisfies (CW0). Solving the (SSIE) $w\subset c_0+w_x$ consists of determining the set of all positive sequences $x={\left(x_n\right)}_{n\ge 1}$ that satisfy the next statement. For every $y\in \omega$ that satisfies the condition ${lim}_{n\to \infty }{n}^{-1}{\sum }_{k=1}^n|y_k-l|=0$ for some scalar l, there are sequences $u,v\in \omega$ with $y=u+v$ such that ${lim}_{n\to \infty }{u}_n=0$ and ${lim}_{n\to \infty }{n}^{-1}{\sum }_{k=1}^n|v_k/x_k-l^{\prime }|=0$ for some scalar $l^{\prime }$.

We state the following result.

Theorem 3. 

Let $\mathcal{E}$ be a linear space of sequences that satisfies the condition in (CW0). Then, the set of all positive sequences $x={\left(x_n\right)}_{n\ge 1}$ that satisfy the (SSIE) $w\subset \mathcal{E}+w_x$ is given by

$$\mathcal{I}(\mathcal{E},w,w)=\overline{w\cap s_1}.$$

Proof. 

Since $w_0\subset w$, we have by Lemma 9 in [6] and Proposition 1

$$\mathcal{I}(c_0,w,w)\subset \mathcal{I}(c_0,w_0,w)\subset \overline{s_1}.$$

It remains to show the inclusion $\mathcal{I}(\mathcal{E},w,w)\subset \overline{w}$ with $\mathcal{E}=c_0$. For this, we let $x\in \mathcal{I}(c_0,w,w)$. Since $e\in w$, there are $\epsilon \in c_0$ and $\xi \in w$ such that $e=\epsilon +x\xi$ and $(e-\epsilon )/x=\xi$. Then, we have

$$\frac{1}{x}=\frac{1}{e-\epsilon }\xi ,$$

where $1-{\epsilon }_n\to 1$$(n\to \infty )$ and $1/(e-\epsilon )\in c$. So, by the inclusion $c\subset M(w,w)$, we obtain ${(e-\epsilon )}^{-1}\xi \in w$ and $1/x\in w$. We conclude $\mathcal{I}(c_0,w,w)\subset \overline{w}$. Now, as we have seen above, we have $\mathcal{I}(c_0,w,w)\subset \overline{s_1}$ and the inclusion

$$\mathcal{I}(c_0,w,w)\subset \overline{s_1}\cap \overline{w}=\overline{w\cap s_1},$$

holds. Finally, by the identity $M(w,w)=w\cap s_1$, we obtain $\mathcal{I}(c_0,w,w)=\overline{w\cap s_1}$ and $\mathcal{I}(\mathcal{E},w,w)=\overline{w\cap s_1}$, for any linear space $\mathcal{E}\subset$$c_0$.

Now, we consider the case $\mathcal{E}=W_a$ with $a\in c_0^{+}$. We show the inclusion $\mathcal{I}(W_a,w,w)\subset \overline{w\cap s_1}$. For this, let $x\in \mathcal{I}(W_a,w,w)$. Then, we have

$$w\subset W_a+w_x\subset W_a+W_x=W_{a+x}$$

and $w\subset W_{a+x}$. This implies $1/(a+x)\in M(w,w_{\infty })=s_1$. Since $a\in c_0$, we deduce $x\in \overline{s_1}$, and we conclude $\mathcal{I}(W_a,w,w)\subset \overline{s_1}$. It remains to show $\mathcal{I}(W_a,w,w)\subset \overline{w}$. For this, we begin to show the inclusion $W_a+w_x\subset w_{a+x}$. As we have seen above, the conditions $x\in \overline{s_1}$ and $a\in c_0$ imply $a/(a+x)\in c_0$. Since $c_0=M(w_{\infty },w)$, we obtain $W_a\subset w_{a+x}$. Then, we have

$$x{(a+x)}^{-1}=e-a{(a+x)}^{-1}\in c.$$

Since $c\subset M(w,w)$, we obtain $w_x\subset w_{a+x}$. Thus, we have shown the inclusions

$$W_a+w_x\subset w_{a+x}$$

and $w\subset w_{a+x}$. The last inclusion implies ${(a+x)}^{-1}\in M(w,w)$. Since $M(w,w)=w\cap s_1\subset w$, we obtain ${(a+x)}^{-1}\in w$. It remains to show $1/x\in w$. From the identity

$$x^{-1}={(a+x)}^{-1}(e+a{x}^{-1}),$$

where $e+a{x}^{-1}\in c$, and since $c\subset M(w,w)$, we deduce that the condition ${(a+x)}^{-1}\in w$ implies

$${(a+x)}^{-1}(e+a{x}^{-1})\in w$$

and $x^{-1}\in w$. We conclude $\mathcal{I}(W_a,w,w)\subset \overline{s_1}\cap \overline{w}$. Again, from the identity $M(w,w)=w\cap s_1$ and by Lemma 10 in [6], we conclude $\mathcal{I}(W_a,w,w)=\overline{w\cap s_1}$. This completes the proof. □

We obtain the next application.

Corollary 2. 

The sets of all positive sequences $x={\left(x_n\right)}_{n\ge 1}$ that satisfy each of the (SSIEs)

(1) $w\subset c_0+w_x$ and

(2) $w\subset w_a+w_x$, with $a\in c_0$,

are determined by

$$\mathcal{I}(c_0,w,w)=\mathcal{I}(w_a,w,w)=\overline{w\cap s_1}.$$

Proof. 

This result follows from Theorem 3. □

By similar arguments to those in Theorem 3, we obtain the following result on the solvability of the (SSIE) $w\subset \mathcal{E}+W_x$.

Theorem 4. 

Let $\mathcal{E}$ be a linear space of sequences that satisfies the condition in (CW0). Then the set of all positive sequences $x={\left(x_n\right)}_{n\ge 1}$ that satisfy the (SSIE) $w\subset \mathcal{E}+W_x$ is given by $\mathcal{I}(\mathcal{E},w,w_{\infty })=\overline{s_1}$.

4.4. Application to the Solvability of the (SSE) $w_{\infty }\subset \mathcal{E}+w_x$

We obtain the next result on the (SSIE) $w_{\infty }\subset \mathcal{E}+w_x$.

Theorem 5. 

Let $\mathcal{E}\subset s_{{\left(n\right)}_{n\ge 1}}^0$ be a linear space of sequences. Then, the set of all the positive solutions of the (SSIE) $w_{\infty }\subset \mathcal{E}+w_x$ is determined by

$$\mathcal{I}(\mathcal{E},w_{\infty },w)=\overline{c_0}.$$

Proof. 

We have $w_{\infty }\subset s_{{\left(n\right)}_{n\ge 1}}^0+w_x\subset s_{{(n+n{x}_n)}_{n\ge 1}}^0$ and

$${\left(\frac{1}{n(1+x_n)}\right)}_{n\ge 1}\in M(w_{\infty },c_0)=s_{{(1/n)}_{n\ge 1}}^0.$$

This implies $1/(1+x)\in c_0$, and $x\in \overline{c_0}$. So, we have $\mathcal{I}(\mathcal{E},w_{\infty },w)\subset \overline{c_0}$. We conclude this by Lemma 2, where $M(w_{\infty },w)=c_0$. □

We state the next corollary, where we solve some particular (SSIEs) involving classical sets of sequences.

Corollary 3. 

Let $p\ge 1$. The sets of all the positive solutions of each of the (SSIEs) (1) $w_{\infty }\subset s_{{\left(n\right)}_{n\ge 1}}^0+w_x$, (2) $w_{\infty }\subset c_{C_1}+w_x$, (3) $w_{\infty }\subset w+w_x$, (4) $w_{\infty }\subset {\left(c_0\right)}_{C_1}+w_x$, (5) $w_{\infty }\subset w_0+w_x$, (6) $w_{\infty }\subset {\left({\ell }^p\right)}_{C_1}+w_x$, (7) $w_{\infty }\subset {\left(c_0\right)}_{\Delta }+w_x$, and $w_{\infty }\subset {\ell }_{\Delta }^p+w_x$ are determined by $\mathcal{I}=\overline{c_0}$.

Proof. 

First, we have the equivalence of $c_{C_1}\subset s_{{\left(n\right)}_{n\ge 1}}^0$ and $D_{{(1/n)}_{n\ge 1}}{C}_1^{-1}\in (c,c_0)$. Then, the non-zero entries of the triangle $D_{{(1/n)}_{n\ge 1}}{C}_1^{-1}$ are determined by ${\left[D_{{(1/n)}_{n\ge 1}}{C}_1^{-1}\right]}_{nn}=1$ and ${\left[D_{{(1/n)}_{n\ge 1}}{C}_1^{-1}\right]}_{n,n-1}=-(n-1)/n$ for all n. By the characterization of $(c,c_0)$ (p. 23, [1]), we conclude $c_{C_1}\subset s_{{\left(n\right)}_{n\ge 1}}^0$. Then, we obtain the next elementary inclusions ${\left({\ell }^p\right)}_{C_1}\subset {\left(c_0\right)}_{C_1}\subset c_{C_1}$ and $w\subset$$c_{C_1}$, $w_0\subset$${\left(c_0\right)}_{C_1}$. From Theorem 5 andLemma 10 (ii) in [6], we obtain the solvability of the (SSIE) in (1), (2),…, (6). By similar arguments to those used above, we obtain the inclusion ${\left(c_0\right)}_{\Delta }\subset s_{{\left(n\right)}_{n\ge 1}}^0$, since this inclusion is equivalent to $D_{{(1/n)}_{n\ge 1}}\mathsf{\Sigma }\in (c_0,c_0)$, and the solvability of each of the (SSIEs) in (7) and (8) follows from the inclusion ${\ell }_{\Delta }^p\subset {\left(c_0\right)}_{\Delta }$. □

Remark 2. 

The result for the solvability of the (SSIE) in (4) of Corollary 3 may be extended to that of the (SSIE) $w_{\infty }\subset {\left(s_r^0\right)}_{C_1}+w_x$ with $r>0$, and the set of all the solutions of this (SSIE) is given by

$${\mathcal{I}}_{w_{\infty },r}^0=\left\{\begin{array}{cc}\overline{c_0}\hfill & ifr\le 1\hfill \\ U^{+}\hfill & ifr>1.\hfill \end{array}\right.$$

(5)

We obtain a similar result for the (SSIE) $w_{\infty }\subset X_{\mathsf{\Sigma }}+w_x$, where $X=s_r^0$, $s_r^{\left(c\right)}$ or $s_r$.

We can also state the next corollary.

Corollary 4. 

Let $\mathcal{E}$ be a linear space of sequences that satisfies the condition in (CW0). Then, we have $\mathcal{I}(\mathcal{E},w_{\infty },w)=\overline{c_0}$.

Proof. 

This result follows from the inclusions $c_0\subset s_{{\left(n\right)}_{n\ge 1}}^0$ and $W_a\subset s_{{\left(n\right)}_{n\ge 1}}^0$ for $a\in c_0$. The first inclusion is trivial, and the second inclusion follows from the equivalence of $W_a\subset s_{{\left(n\right)}_{n\ge 1}}^0$ and ${(a_n/n)}_{n\ge 1}\in M(w_{\infty },c_0)=s_{{(1/n)}_{n\ge 1}}^0$. This concludes the proof. □

Example 4. 

By Corollary 4, the set of all positive sequences $x={\left(x_n\right)}_{n\ge 1}$ that satisfy the (SSIE) $w_{\infty }\subset w_a+w_x$, where $a\in c_0^{+}$, is given by $\mathcal{I}(w_a,w_{\infty },w)=\overline{c_0}$.

Similarly, for any given $r>0$, the set of all positive solutions of the (SSIE) $w_{\infty }\subset w_r+w_x$ is given by

$${\mathcal{I}}_{w_{\infty },r}^w=\left\{\begin{array}{cc}\overline{c_0}\hfill & ifr<1\hfill \\ U^{+}\hfill & ifr\ge 1.\hfill \end{array}\right.$$

5. Conclusions

In this article, we have dealt with the solvability of some (SSIEs) involving the spaces $w_0$, w, and $w_{\infty }$. In this way, we solved the (SSIE) of the form $F\subset \mathcal{E}+w_x$, where $F\in \{w_0,w,w_{\infty }\}$, and the (SSIEs) $w\subset \mathcal{E}+W_x^0$ and $w\subset \mathcal{E}+W_x$ for a particular class of linear spaces $\mathcal{E}$ of sequences. We can gather some of the previous results, and state that the positive solutions $x={\left(x_n\right)}_{n\ge 1}$ of the (SSIE) $F\subset c_0+w_x$, where $F\in \{w_0,w,w_{\infty }\}$, are given by

$$\mathcal{I}(c_0,F,w)=\left\{\begin{array}{cc}\overline{s_1}\hfill & \mathrm{if}F=w_0\hfill \\ \overline{c_0}\hfill & \mathrm{if}F=w_{\infty }\hfill \\ \overline{w\cap {\ell }_{\infty }}\hfill & \mathrm{if}F=w.\hfill \end{array}\right.$$

In future, these results could be extended to the study of each of the (SSIEs) $w\subset w_0+Y_x$ and $w_{\infty }\subset w+Y_x$, where Y is any of the spaces $c_0$, c, ${\ell }_{\infty }$, $w_0$, w, or $w_{\infty }$. Some (SSIEs) of the form ${\left(w_0\right)}_{\Delta -\lambda I}\subset Y_x$, $w_{\Delta -\lambda I}\subset Y_x$, and ${\left(w_{\infty }\right)}_{\Delta -\lambda I}\subset Y_x$, where Y is any of the spaces $c_0$, c, ${\ell }_{\infty }$, $w_0$, w, and $w_{\infty }$, and $\Delta$ is the operator of the first difference, should also be solved. Their solutions should involve the fine spectrum of the operator $\Delta$ considered as an operator from X to itself, where X is successively equal to $w_0$, w, and $w_{\infty }$. Finally, using the spectrum of the band matrix $B(r,s)$ on $w_0$, it should be interesting to solve the (SSE) $W_{B(r,s)-\lambda I}^0+W_x=W_b$.