Distinguished subspaces in topological sequence spaces theory

1.   Introduction

Let $\omega$ denote the space of all real or complex valued sequences. An $FK$-space is a locally convex vector subspaces of $\omega$ which is also a Fréchet space (complete linear metric) with continuous coordinates. A $BK$-space is a normed $FK$-space^[1]. Some articles about $BK$-space and $FK$-space are studied in ^{[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]}. The definition of semiconservative $FK$-spaces and some properties of these spaces were given by Snyder and Wilansky in ^[9]. Ince ^[10] continued to work on Cesáro semiconservative $FK$-spaces and gave some characterizations. The main purpose of this paper is to introduce the concept of Reisz semiconservative $FK$-spaces, which contains the space strictly increasing sequence of positive integers lambda. We have proved several interesting conclusions about this concept in section 3. The results in this article are new, accurate and interesting. In addition, our work is extension of the works in ^{[6,9,10,12,16]}. The work brings innovations and information that attracts the attention of the mathematics reader.

2.   Preliminaries

Let $F$ be an infinite subset of $\mathbb{N}$ and $F$ as the range of a strictly increasing sequence of positive integers, say $F = \{\lambda(n)\}_{n = 1}^{\infty}.$ The Cesáro submethod $C_{\lambda}$ is defined as

where $\{x_{k}\}$ is a sequence of real or complex numbers ^[11]. The Riesz submethod is defined as the following; Let $(q_{k})$ be a positive sequence of real numbers.

where

and

Also,

In case $\lambda(n) = n,$ the method $R_{n} ^{\lambda}(f; x)$ give us classical known Riesz mean. Provided that $q_{n} = 1$ for all $(n\geq 0)$ Riesz mean yields

In addition to this, if $\lambda(n) = n$ for $\sigma_{n} ^{\lambda}(f; x),$ then it coincides with Cesáro method $C_{1}$ ^[12]. Let $q = (q_{k})$ and $(Q_{n})$ be given $q_{0} > 0$, $q_{k}\geq 0$ $(\forall k\in \mathbb{N})$, $Q_{n} = \sum_{k = 1}^{n} q_{k}\; \; (n\in \mathbb{N})$. The matrix $R = q_{nk}$ defined by

is called a Riesz matrix.

In this paper, The Riesz submethod is symbolized by $R_{n} ^{\lambda}(f; x)$ or, in short, $R_{\lambda}$. The sequences space

are $FK$ space with the classical norm. Note that $(cs)_{R} = rs (\text{or r}), (bs)_{R} = rb, (c_{0})_{R} = r_{0}$ are $FK$-spaces with the norms $\|x\|_{rb} = \sup_{n}\left|\frac{1}{Q_{n}}\sum_{k = 1}^{n}\sum_{j = 1}^{k}q_{j}x_{j} \right|$ and $\|x\|_{r_{o}} = \sup_{n}\frac{1}{Q_{n}}\left|\sum_{k = 1}^{n}q_{j}x_{j} \right|$ respectively ^[15]. Also,

$rb(\lambda) = \left\{x\in \omega: \sup_{n}\left|\frac{1}{Q_{\lambda(n)}}\sum_{k = 1}^{\lambda(n)}\sum_{j = 1}^{k}q_{j}x_{j} \right| < \infty\right\},$ $r_{0}(\lambda) = \left\{x\in \omega: \lim_{n}\left|\frac{1}{Q_{\lambda(n)}}\sum_{j = 1}^{\lambda(n)}q_{j}x_{j} \right| = 0\right\}$ are $FK$-spaces with the norms $\|x\|_{rb(\lambda)} = \sup_{n}\left|\frac{1}{Q_{\lambda(n)}}\sum_{k = 1}^{\lambda(n)}\sum_{j = 1}^{k}q_{j}x_{j} \right|, \|x\|_{r_{o}(\lambda)} = \sup_{n}\frac{1}{Q_{\lambda(n)}}\left|\sum_{k = 1}^{\lambda(n)}q_{j}x_{j} \right|,$ respectively. Throughout the paper, $\delta$ denotes sequences of the form $(1, 1, \dots, 1\dots)$. Let $\varphi = span\{\delta^{k}:k\in \mathbb{N}\}$ and $\varphi_{1} = \varphi \cup \{\delta\}$. The topological dual of $X$ is denoted by $X^{'}$. Let $(X, \tau)$ be a $K$-space with $\varphi\subset X$ and dual space $X^{'}$, and let $x = (x_{k})\in X$ be arbitrarily given. Define the $n^{th}$ section of $x$ to be sequence $x^{[n]} = \sum_{k = 1}^{n}x_{k}\delta^{k} = (x_{1}, x_{2}, \dots, x_{n}, 0, \dots)$, where $\delta^{k}$ denotes the sequence having $1$ in the j-th position and $0$'s elsewhere ^[13,14]. Also, $r^{[n]}x = \frac{1}{Q_{n}}\sum_{k = 1}^{n}q_{k}x_{k}\delta^{k}$ is called the $n^{th}$ Riesz section of $x$ ^[15]. This here $r$ is the set $\{r^{n}: n\in \mathbb{N}\}$. We define the following properties:

$x$ has ${AK}$ if $x^{[n]}\rightarrow x$ in $(X, \tau)$,

$x$ has ${SAK}$ if $x^{[n]}\rightarrow x$ in $(X, \sigma(X, X^{'}))$,

$x$ has ${FAK}$ if $\sum_{k}x_{k}f(\delta^{k})$ converges for all $f\in X^{'}$,

$x$ has ${AB}$ if $\{x^{[n]}: n\in \mathbb{N}\}$ is bounded in $(X, \tau)$,

$x$ has ${AD}$ if $X = \overline{\varphi}$ (closed of $\varphi$),

$x$ has ${rK(\lambda)(riesz\; \; sectional\; \; convergence)}$ if $\frac{1}{Q_{\lambda(n)}}\sum_{k = 1}^{\lambda(n)}q_{k}x^{(k)}\rightarrow x\; \; , {n\rightarrow\infty}$ ^[15]. Then, some duals of a subset $X$ are defined to be

By taking advantage of ^[1], we can easily see the following lemma:

Lemma 2.1. Let $X, X_{1}$ be sets of sequences. Then for $k = f, \beta, r, rb, r(\lambda), rb(\lambda)$

(1)$\; \; X\subset X^{kk},$

(2) $X^{kkk} = X^{k},$

(3) if $X\subset X_{1}\; \; then\; \; X_{1}^{k}\subset X^{k}$ holds.

Let $A = (a_{ij})$ be an infinite matrix. The matrix $A$ may be considered as a linear transformation of sequence by the formula $y = Ax,$ where $y_{i} = \sum_{j = 1}^{\infty}a_{ij}x_{j}$. For an $FK$-space $(\lambda, u)$ we consider the summability domain $\lambda_{A} = \{x\in \omega\ : Ax\in \lambda\}$. Then $\lambda_{A}$ is an $FK$-space under the semi-norms $p_{i} = |x_{i}|, \; \; (1, 2, \dots)$. A conservative matrix $A$, and the corresponding matrix method, is called conull if $\chi(A) = 0$, where $\chi(A) = \lim_{A}\delta-\sum_{k}\lim_{A}\delta^{k}$ ^[1].

Recall that, given a matrix $A$ with $\ell_{A}\supset \varphi$ is called $\ell$-replaceable if there is a matrix $B = (b_{nk})$ with $\ell_{B} = \ell_{A}$ and $\sum_{k = 1}^{\infty}b_{nk} = 1$ for all $k\in \mathbb{N}$ ^[16].

In addition an $FK$-space $X$ is called semiconservative if $X^{f}\subset cs$, this means that $X\supset \varphi$ and $\sum_{j = 1}^{\infty}f(\delta^{j})$ is convergent for each $f\in X^{'}$ ^[9].

3.   Main results

Firstly, we have defined the notations of $R_{\lambda}$-semiconservative $FK$-space in this section. Then, we investigate the properties of these spaces and we also give the relationship between $\ell$-reblaceable and $R_{\lambda}$-semiconservative $FK$-space. Note that it is accepted $Q_{n}\to \infty, (n\rightarrow \infty)$ in this paper.

Definition 3.1. An $FK$-space $X$ is called $R_{\lambda}$-semiconservative if $X^{f}\subset rs(\lambda)$. It is obvious that $X^{f}\subset rs(\lambda)$ if and only if $\left\{\frac{1}{Q_{\lambda(n)}}\sum_{k = 1}^{\lambda(n)}q_{k}f(\delta^{k})\right\}$ is convergent for each $f\in X'$ equivalently

exists.

Definition 3.2. An $FK$-space containing $\varphi_{1}$ is called $R_{\lambda}$-conull if

for all, $f\in X^{'}$.

Lemma 3.3. Let $X$ be an $FK$-space containing $\varphi$. Then

(1)$X^{\beta}\subset X^{r(\lambda)}\subset X^{rb(\lambda)}\subset X^{f},$

(2)If $X$ is a $rK(\lambda)$ space then $X^{f} = X^{r(\lambda)},$

(3)If $X$ is an $AD$ space then $X^{r(\lambda)} = X^{r b(\lambda)}$.

Proof. (2) Let $u\in X^{r(\lambda)}$ and define

for $x\in X$. Then $f\in X^{'}$; by the Banach-Steinhaus Theorem 1.0.4 of ^[2]. Also $f(\delta^{p}) = \lim_{n}\frac{1}{Q_{\lambda(n)}}(\lambda(n)-(p-1))q_{p}u_{p} = u_{p}, (p < \lambda(n))$ so $u\in X^{f}$. Thus $X^{r(\lambda)}\subset X^{f}$. Now we show that $X^{f}\subset X^{r(\lambda)}$. Let $u\in X^{f}$. Since $X$ is $rK(\lambda)$ space

for $x\in X,$ then $u\in X^{r(\lambda)}$. Hence $X^{f} = X^{r(\lambda)}$.

(3) Let $u\in X^{rb(\lambda)}$ and define $f_{n}(x) = \frac{1}{Q_{\lambda(n)}}\sum_{k = 1}^{\lambda(n)}\sum_{j = 1}^{k}q_{j}x_{j}u_{j}$ for $x\in X.$ Then $\{f_{n}\}$ is pointwise bounded, hence equicontinuous by [2,Theorem 7.0.2]. Since $\lim_{n}f(\delta^{p}) = u_{p}$ $(p < \lambda(n))$ then $\varphi\subset \{x:lim_{n}f_{n}(x) \; \; exists\}.$ Hence $\{x:lim_{n}f_{n}(x)\; \; exists\}$ is closed subspace of $X$ by the Convergence Lemma [2,Theorem 1.0.5,7.0.3]. Since $X$ is an $AD$ space then $X = \{x:lim_{n}f_{n}(x)\; \; exists\} = \bar{\varphi}$ and then $\lim_{n}f_{n}(x)$ exists for all $x\in X$. Thus $u\in X^{r(\lambda)}$. The opposite inclusion is trivial.

(1) $\bar{\varphi}\subset X$ by the hypothesis. Since $\bar{\varphi}$ is $AD$ space, then $X^{r b(\lambda)}\subset (\bar{\varphi})^{r b(\lambda)} = (\bar{\varphi})^{r(\lambda)} = (\bar{\varphi})^{f} = X^{f}$ by (2), (3) and [2,Theorem 7.2.4].

Theorem 3.4. If a matrix $A$ is $\ell$-replaceable then $\ell_{A}$ is not $R_{\lambda}$-semiconservative $FK$-space.

Proof. If $A$ is $\ell$-replaceable then there is $f\in \ell_{A}^{'}$ such that $f(\delta^{j}) = 1$ for all $j\in \mathbb{N}$ ^[16]. Hence

does not exist, so $\ell_{A}$ is not $R_{\lambda}$-semiconservative space.

Theorem 3.5. If $X_{A}$ is $R_{\lambda}$-conull $FK$-space then it is $R_{\lambda}$-semiconservative space.

Proof. Suppose that $X_{A}$ is $R_{\lambda}$-conull. Then

for all $f\in X_{A}^{'}.$ Hence $X_{A}^{f}\subset rs(\lambda).$

Theorem 3.6. (1) A closed subspace, containing $\varphi$, of a $R_{\lambda}$-semiconservatif space is a $R_{\lambda}$-semiconservative space.

(2) An $FK$-space that contains a $R_{\lambda}$-semiconservative space must be a $R_{\lambda}$-semiconservative space.

(3) A countable intersection of $R_{\lambda}$-semiconservative space is a $R_{\lambda}$-semiconservative space.

The proof is easily obtained from elementary properties of $FK$-spaces (see ^[2]).

Theorem 3.7. If $z^{r(\lambda)}$ is a $R_{\lambda}$-semiconservative space then $z\in rs(\lambda)$.

Proof. Let $z^{r(\lambda)}$ be a $R_{\lambda}$-semiconservative space. Then $z^{r(\lambda)f}\subset rs(\lambda)$. Since $z^{r(\lambda)}$ is a $rK(\lambda)$ space, we have $z^{r(\lambda)f} = z^{r(\lambda)r(\lambda)}$. Since $\{z\}\subset z^{r(\lambda)r(\lambda)}\subset rs(\lambda)$, we get $z\in rs(\lambda)$.

Now we give study a new the subspaces which are called $R_{\lambda}S, R_{\lambda}W, R_{\lambda}F^{+}$ and $R_{\lambda} B^{+}$.

Definition 3.8. Let $X$ be an $FK$-space containing $\varphi$. Then, the following definitions hold.

Also, $R_{\lambda}F = R_{\lambda}F^{+}\cap X$ and $R_{\lambda}B = R_{\lambda}B^{+}\cap X.$

Definition 3.9. Sequence sets of above definitions show that:

1. $X_{rK{(\lambda)}} = R_{\lambda}S = \{x\in X: x {{\; \; has\; \; }} rK{(\lambda)}\} \subset X$

2. $X_{SrK{(\lambda)}} = R_{\lambda}W = \{x\in X: x {{\; \; has\; \; }} SrK{(\lambda)}\} \subset X$

3. $X_{FrK{(\lambda)}} = R_{\lambda}F = \{x\in X: x {{\; \; has\; \; }} FrK{(\lambda)}\} \subset X$

4. $X_{rB{(\lambda)}} = R_{\lambda}B = \{x\in X: x {{\; \; has\; \; }} rB{(\lambda)}\} \subset X$

Corollary 3.1. By definition 3.8 we obtain from following results:

1. $X {{\; \; has\; \; }} FrK{(\lambda)} {{\; \; iff\; \; }} X\subset R_{\lambda}F {{\; \; , i.e., \; \; }} X = R_{\lambda}F,$

2. $X {{\; \; has\; \; }} rB{(\lambda)} {{\; \; iff\; \; }} X\subset R_{\lambda}B {{\; \; , i.e., \; \; }} X = R_{\lambda}B.$

Theorem 3.10. Let $X$ be an $FK$-space containing $\varphi$ and $z\in \omega$. Then $z\in R_{\lambda}F^{+}$ if and only if $Y = z^{-1}X = \{x: zx = \{z_{n}x_{n}\}\in X\}$ is a $R_{\lambda}$-semiconservative $FK$-space.

Proof. Let $(z^{-1}X)$ be an $R_{\lambda}$-semiconservative space. Hence $f\in (z^{-1}X)^{'}$. Then $f(x) = \alpha x+g(zx)$, $\alpha\in \varphi, g\in Y^{'}$, by [2,Theorem 4.4.10] and $f(\delta^{n}) = \alpha_{n}+ g(z \delta ^{n}) = \alpha_{n}+ g(z_{n}\delta^{n}) = \alpha_{n}+z_{n}g(\delta^{n})$. Hence, since $\alpha \in \varphi\subset r s(\lambda)$ then $\{f(\delta^{n})\}\in r s(\lambda)$ if and only if $\{z_{n}g(\delta^{n})\}\in r s(\lambda)$, i. e., $z\in R_{\lambda}F^{+}$.

An $FK$-space is called bounded convex $R_{\lambda}$-semiconservative if it is $R_{\lambda}$-semiconservative space and includes $\delta$.

Theorem 3.11. Let $X$ be an $FK$-space containing $\varphi$ and $z\in \omega$. Then $z\in R_{\lambda}F$ if and only if $z^{-1}X$ is bounded convex $R_{\lambda}$-semiconservative $FK$-space.

Proof. Let $z\in R_{\lambda}F$. Since $R_{\lambda}F = R_{\lambda}F^{+}\cap X$ then $z\in X$ so $\delta\in z^{-1}X$ and since $z\in R_{\lambda}F^{+}$, $z^{-1}X$ is $R_{\lambda}$-semiconservative $FK$-space by Theorem 3.10. Thus $z^{-1}X\in X$ is bounded convex $R_{\lambda}$-semiconservative $FK$-space. Contrary, let $z^{-1}X\in X$ is bounded convex $R_{\lambda}$-semiconservative $FK$-space. Then $z^{-1}X$ is $R_{\lambda}$-semiconservative $FK$-space and $\delta\in z^{-1}X$ so $z\in X$. Since $z\in R_{\lambda}F^{+}$ by Theorem 3.10 and $z\in X$, we get the result $z\in R_{\lambda}F$.

Theorem 3.12. Let $X$ be an $FK$-space containing $\varphi.$ Then

$\varphi\subset R_{\lambda}S\subset R_{\lambda}W\subset R_{\lambda}F\subset R_{\lambda}B\subset X$ and $\varphi\subset R_{\lambda}S\subset R_{\lambda}W\subset \bar{\varphi}.$

Proof. First conclusion is obvious by Definition 3.8. We show that the inclusion $R_{\lambda}W\subset \bar{\varphi}$. Let $f \in X^{'}$ and $f = 0$ on $\varphi.$ The definition of $R_{\lambda}W$ shows that $f = 0$ on $R_{\lambda}W$. Thus, the Hanh-Banach theorem gives the result.

Theorem 3.13. The subspaces $E = R_{\lambda}S, R_{\lambda}W, R_{\lambda}F, R_{\lambda}B, R_{\lambda}F^{+}$ and $R_{\lambda}B^{+}$ of $X$ are monotone, i. e., if $X\subset Y$ then $E(X)\subset E(Y).$

Proof. The proof is obtained from elementary properties of FK-spaces (see ^[2]) and Definition 3.8.

Theorem 3.14. Let $X$ be an $FK$-space containing $\varphi$. Then,

(1) $R_{\lambda}B^{+} = (X^{f})^{rb(\lambda)}.$

(2) $R_{\lambda}B^{+}$ is the same for all $FK$-spaces $Y$ between $\bar{\varphi}$ and $X$; i. e., $\bar{\varphi}\subset Y\subset X$ implies $R_{\lambda}B^{+}(Y) = R_{\lambda}B^{+}(X)$. Here the closure of $\varphi$ is calculated in $X$.

Proof. (1) By Definition 3.8, $z\in R_{\lambda}B^{+}$ if and only if $zu\in rb(\lambda)$ for each $u\in X^{f}.$ Hence $R_{\lambda}B^{+}\subset (X^{f})^{rb(\lambda)}$ holds. The converse inclusion is trivial. This is precisely the claim.

(2) By Theorem 3.13, we have $R_{\lambda}B^{+}(\bar{\varphi})\subset R_{\lambda}B^{+}(Y)\subset R_{\lambda}B^{+}(X).$ By Theorem 3.14 (1), the first and last are equal.

Theorem 3.15. Let $X$ be an $FK$-space containing $\varphi$. Then

(1) $R_{\lambda}F^{+} = (X^{f})^{r(\lambda)}.$

(2) $R_{\lambda} F^{+}$ is the same for all $FK$-spaces $Y$ between $\bar{\varphi}$ and $X$; i. e., $\bar{\varphi}\subset Y\subset X$ implies $R_{\lambda}F^{+}(Y) = R_{\lambda}F^{+}(X)$. Here the closure of $\varphi$ is calculated in $X$.

Proof. This can be proved as in Theorem 3.14, with $rs(\lambda)$ instead of $rb(\lambda).$

Theorem 3.16. Let $X$ be an $FK$-space in which $\bar{\varphi}$ has $rK{(\lambda)}$. Then

(1) $R_{\lambda}F^{+} = (\bar{\varphi})^{r(\lambda)r(\lambda)}$.

(2) $X$ has $FrK{(\lambda)}$ if and only if $X\subset (\bar{\varphi})^{r(\lambda)r(\lambda)}$.

(3) If the inclusion $R_{\lambda}B\supset \bar{\varphi}$ holds, $R_{\lambda}S = R_{\lambda}W = \bar{\varphi}$.

Proof. (1) It is obvious that $R_{\lambda}F^{+} = (X^{f})^{r(\lambda)}$. Since $X^{f} = (\bar{\varphi})^{f}$ by ^[2], we have $(X^{f})^{r(\lambda)} = (\bar{\varphi}^{f})^{r(\lambda)}$. Thus by Lemma 3.3 the result follows.

(2) Firstly, suppose that $X$ has $FrK{(\lambda)}$. $X$ has $rB{(\lambda)}$ since $R_{\lambda}F\subset R_{\lambda}B$. If $\bar{\varphi}$ has $rK{(\lambda)}$ then $X\subset R_{\lambda}F^{+}$. Hence $X\subset (\bar{\varphi})^{r(\lambda)r(\lambda)}$. Sufficiency is given by Theorem 3.16 (1).

(3) By Theorem 3.12, $\varphi\subset R_{\lambda}S\subset R_{\lambda}W\subset \bar{\varphi}\subset R_{\lambda}B.$ Firstly, suppose that $X$ has $R_{\lambda}B$. Define $f_{n}:X\rightarrow X$ by $x\rightarrow f_{n}(x) = x-\frac{1}{Q_{\lambda(n)}}\sum_{k = 1}^{\lambda(n)}q_{k}x^{(k)}.$ Then $\{f_{n}\}$ is pointwise bounded, hence equicontinuous by ^[2]. Since $f_{n}\rightarrow 0$ on $\varphi$ then also $f_{n}\rightarrow 0$ on $\bar{\varphi}$ by ^[2]. By $\bar{\varphi}$ has $rK{(\lambda)}$, the proof is complete.

Theorem 3.17. Let $X$ be an $FK$-space containing $\varphi$. Then $X$ has

(1) $rB(\lambda)$ property if and only if $X^{f}\subset X^{rb(\lambda)}$.

(2) $rF(\lambda)$ property if and only if $X^{f}\subset X^{r(\lambda)}$.

Proof. Necessity; Let $X$ be $rB(\lambda)$ property. Then $X\subset R_{\lambda}B^{+} = (X^{f})^{rb(\lambda)}$ and $X^{rb(\lambda)}\supset (X^{f})^{rb(\lambda)rb(\lambda)}\supset X^{f}$. Sufficiency is clear. The proof of (2) is similar to that of (1).

Theorem 3.18. Let $Y$ be a $R_{\lambda}$-semiconservative $FK$-space and $Z$ an $AD$-space. Suppose that for an $FK$-space $X$, $X\supset Y.Z$. Then $X\supset Z$, where $Y.Z = \{y.z : y\in Y, z\in Z\}$.

Proof. Let $z\in Z$. Then, since $X\supset Y.Z$, $z^{-1}.X\supset Y$. Thus, since $Y$ is $R_{\lambda}$-semiconservative space then $z^{-1}.X$ is $R_{\lambda}$-semiconservative space by Theorem 3.6 and so $z\in R_{\lambda} F^{+}$ by Theorem 3.10. Hence $Z\subset R_{\lambda} F^{+} = (X^f)^{r(\lambda)}$. Thus $X^{f}\subset X^{fr(\lambda)r(\lambda)}\subset Z^{r(\lambda)}\subset Z^{f}$ and so $X\supset Z$ by Theorem 8.6.1 of ^[2].

Theorem 3.19. Let $X$ be an $FK$-space containing $\varphi$. The following statements are equivalent:

(1) $X$ has $FrK{(\lambda)}$,

(2) $X\subset (R_{\lambda}S)^{r(\lambda)r(\lambda)}$,

(3) $X\subset (R_{\lambda}W)^{r(\lambda)r(\lambda)}$,

(4) $X\subset(R_{\lambda}F)^{r(\lambda)r(\lambda)}$,

(5) $X^{r(\lambda)} = (R_{\lambda}S)^{r(\lambda)}$,

(6) $X^{r(\lambda)} = (R_{\lambda}F)^{r(\lambda)}$.

Proof. Since $R_{\lambda}S\subset R_{\lambda}W\subset R_{\lambda}F\subset X$, it is trivial that (2) $\Rightarrow$ (3) and (3) $\Rightarrow$ (4). If (4) is true, then

so (1) is true by Lemma 3.3. If (1) holds, then Theorem 3.16 implies that $\bar{\varphi} = R_{\lambda}S$ which means (2) holds. The equivalence of (5), (6) with others is clear.

Theorem 3.20. Let $X$ be an $FK$-space containing $\varphi$. The following are equivalent:

(1) $X$ has $SrK{(\lambda)}$,

(2) $X$ has $rK{(\lambda)}$,

(3) $X^{r(\lambda)} = X^{'}.$

Proof. By Theorem 3.12, it is clear (2) implies (1). Conversely if $X$ has $SrK(\lambda)$ it must have $AD$ for $R_{\lambda}W\subset \bar{\varphi}$ by Theorem 3.12. It also has $rB(\lambda)$ since $R_{\lambda}W\subset R_{\lambda}B$. Thus $X$ has $rK(\lambda)$ by Theorem 3.16, this proves that (1) and (2) are equivalent. Assume that (3) holds. Let $f\in X'$, then there exists $u\in X^{r(\lambda)}$ such that

for $x\in X.$ Since $u_{j} = f(\delta^{j})$, it follows that each $x\in R_{\lambda}W$ which shows that (3) implies (1). Let $X$ has $rK(\lambda)$, then by Theorem 3.12 it has $SrK(\lambda)$. So, by Definition 3.8, for all $f\in X'$ there is

such that $u\in X^{r(\lambda)}$ which is $u_{j} = f(\delta^{j}), (\forall x\in X).$ This shows that (2) implies (3).

Theorem 3.21. Let $X$ be an $FK$-space containing $\varphi$. The following are equivalent:

(1) $R_{\lambda}W$ is closed in $X$,

(2) $\bar{\varphi} \subset R_{\lambda}B$,

(3) $\bar{\varphi} \subset R_{\lambda}F$,

(4) $\bar{\varphi} = R_{\lambda}W$,

(5) $\bar{\varphi} = R_{\lambda}S$,

(6) $R_{\lambda}S$ is closed in $X$.

Proof. (2) $\Rightarrow$ (5): By Theorem 3.16, $\bar{\varphi}$ has $rK(\lambda)$, i.e., $\bar{\varphi} \subset R_{\lambda}S$. The opposite inclusion is Theorem 3.12. Note that (5) implies (4), (4) implies (3) and (3) implies (2) because

(1)$\Rightarrow$ (4) and (6)$\Rightarrow$ (5) since $\varphi \subset R_{\lambda}S\subset R_{\lambda}W\subset \bar{\varphi}.$ Finally (4) implies (1) and (5) implies (6).

4.   Conclusion

The main motivation of this article is to develop some distinguished subspaces in the theory of topological sequence spaces. These subspaces are called as $R_{\lambda}S, R_{\lambda}W, R_{\lambda}F^{+}$ and $R_{\lambda}B^{+}$ for a locally convex $FK$-space X containing $\varphi$. Moreover, we study $R_{\lambda}$-semiconservative $FK$-spaces for Riesz method defined by the Riesz matrix ($R$) and give some characterizations. In addition, we determine a new $r(\lambda)$ and $rb(\lambda)$ type duality of a sequence space $X$ containing $\varphi$ and we examine monotone of the distinguished subspaces. Finally, we prove some theorems related to the $f$-, $r(\lambda)$- and $rb(\lambda)$- duality of a sequence spaces $X$. Our main results give information that holds the mathematics reader's attention.

Authors' contributions

Authors contributed to each part of this work equally, and they read and approved the final manuscript.

Conflict of interest

The authors declare no conflict of interest.