Binomial difference sequence spaces of order m

Abstract

In this paper, we introduce the binomial sequence spaces $b^{r,s}_{0}(\nabla^{(m)})$, $b^{r,s}_{c}(\nabla^{(m)})$ and $b^{r,s}_{\infty}(\nabla^{(m)})$ by combining the binomial transformation and mth order difference operator. We prove the BK-property and some inclusion relations. Also, we obtain the Schauder bases and compute the α-, β- and γ-duals of these sequence spaces.

1 Introduction and preliminaries

Let w denote the space of all sequences. By $\ell_{\infty}$, c and $c_{0}$, we denote the spaces of bounded, convergent and null sequences, respectively. We write bs, cs and $\ell_{p}$ for the spaces of all bounded, convergent and p-absolutely summable series, respectively; $1\leq p<\infty$. A Banach sequence space Z is called a BK-space [1] provided each of the maps $p_{n}:Z\rightarrow\mathbb{C}$ defined by $p_{n}(x)=x_{n}$ is continuous for all $n\in\mathbb{N}$, which is of great importance in the characterization of matrix transformations between sequence spaces. It is well known that the sequence spaces $\ell_{\infty },c$ and $c_{0}$ are BK-spaces with their usual sup-norm.

Let Z be a sequence space, then Kizmaz [2] introduced the following difference sequence spaces:

$$ Z(\Delta)=\bigl\{ (x_{k})\in w:(\Delta x_{k})\in Z\bigr\} $$

for $Z\in\{\ell_{\infty},c,c_{0}\}$, where $\Delta x_{k}=x_{k}-x_{k+1}$ for each $k\in\mathbb{N}$. Et and Colak [3] defined the generalization of the difference sequence spaces

$$ Z\bigl(\Delta^{m}\bigr)=\bigl\{ (x_{k})\in w:\bigl( \Delta^{m} x_{k}\bigr)\in Z\bigr\} $$

for $Z\in\{\ell_{\infty},c,c_{0}\}$, where $m\in\mathbb{N}$, $\Delta ^{0}x_{k}=x_{k}$, $\Delta^{m} x_{k}=\Delta^{m-1}x_{k}-\Delta ^{m-1}x_{k+1}$ for each $k\in\mathbb{N}$, which is equivalent to the binomial representation $\Delta^{m} x_{k}=\sum_{i=0}^{m}(-1)^{i}\binom {m}{i}x_{k+i}$. Since then, many authors have studied further generalization of the difference sequence spaces [4–8]. Moreover, Altay and Polat [9], Başarir [10], Başarir, Kara and Konca [11], Başarir and Kara [12–17], Başarir, Öztürk and Kara [18], Polat and Başarir [19] and many others have studied new sequence spaces from matrix point of view that represent difference operators.

For an infinite matrix $A=(a_{n,k})$ and $x=(x_{k})\in w$, the A-transform of x is defined by $(Ax)_{n}=\sum_{k=0}^{\infty }a_{n,k}x_{k}$ and is supposed to be convergent for all $n\in\mathbb {N}$. For two sequence spaces X, Y and an infinite matrix $A=(a_{n,k})$, the sequence space $X_{A}$ is defined by

$$ X_{A}=\bigl\{ x=(x_{k})\in w:Ax \in X\bigr\} , $$

(1.1)

which is called the domain of matrix A. By $(X : Y)$, we denote the class of all matrices such that $X \subseteq Y_{A}$.

The Euler means $E^{r}$ of order r is defined by the matrix $E^{r}=(e_{n,k}^{r})$, where $0< r<1$ and

$$e_{n,k}^{r}= \textstyle\begin{cases} \binom{n}{k}(1-r)^{n-k}r^{k}& \text{if }0\leq k\leq n, \\ 0& \text{if }k>n. \end{cases} $$

The Euler sequence spaces $e^{r}_{0}$, $e^{r}_{c}$ and $e^{r}_{\infty}$ were defined by Altay and Başar [20] and Altay, Başar and Mursaleen [21] as follows:

$$\begin{aligned}& e^{r}_{0}=\Biggl\{ x=(x_{k})\in w: \lim _{n\rightarrow\infty}\sum_{k=0}^{n} \binom{n}{k}(1-r)^{n-k}r^{k}x_{k}=0\Biggr\} , \\& e^{r}_{c}=\Biggl\{ x=(x_{k})\in w: \lim _{n\rightarrow\infty}\sum_{k=0}^{n} \binom{n}{k}(1-r)^{n-k}r^{k}x_{k} \text{ exists}\Biggr\} , \end{aligned}$$

and

$$ e^{r}_{\infty}=\Biggl\{ x=(x_{k})\in w: \sup _{n\in\mathbb{N}} \Biggl\vert \sum_{k=0}^{n} \binom{n}{k}(1-r)^{n-k}r^{k}x_{k} \Biggr\vert < \infty\Biggr\} . $$

Altay and Polat [9] defined further generalization of the Euler sequence spaces $e^{r}_{0}(\nabla)$, $e^{r}_{c}(\nabla)$ and $e^{r}_{\infty}(\nabla)$ by

$$ Z (\nabla)=\bigl\{ x=(x_{k})\in w: (\nabla x_{k})\in Z \bigr\} $$

for $Z\in\{e_{0}^{r}, e_{c}^{r}, e_{\infty}^{r}\}$, where $\nabla x_{k}=x_{k}-x_{k-1}$ for each $k\in\mathbb{N}$. Here any term with negative subscript is equal to naught.

Polat and Başar [19] employed the technique matrix domain of triangle limitation method for obtaining the following sequence spaces:

$$ Z\bigl(\nabla^{(m)}\bigr)=\bigl\{ x=(x_{k})\in w: \bigl( \nabla^{(m)} x_{k}\bigr)\in Z\bigr\} $$

for $Z\in\{e_{0}^{r}, e_{c}^{r}, e_{\infty}^{r}\}$, where $\nabla ^{(m)}=(\delta_{n,k}^{(m)})$ is a triangle matrix defined by

$$\delta_{n,k}^{(m)}= \textstyle\begin{cases} (-1)^{n-k}\binom{m}{n-k}& \text{if }\max\{0,n-m\}\leq k\leq n, \\ 0& \text{if }0\leq k< \max\{0,n-m\}\mbox{ or }k>n, \end{cases} $$

for all $k,n,m\in\mathbb{N}$.

Recently Bişgin [22, 23] defined another generalization of the Euler sequence spaces and introduced the binomial sequence spaces $b^{r,s}_{0}$, $b^{r,s}_{c}$, $b^{r,s}_{\infty}$ and $b^{r,s}_{p}$. Let $r,s\in\mathbb{R}$ and $r+s\neq0$. Then the binomial matrix $B^{r,s}=(b_{n,k}^{r,s})$ is defined by

$$b_{n,k}^{r,s}= \textstyle\begin{cases} \frac{1}{(s+r)^{n}}\binom{n}{k} s^{n-k}r^{k}& \text{if }0\leq k\leq n, \\ 0& \text{if }k>n, \end{cases} $$

for all $k,n\in\mathbb{N}$. For $sr>0$ we have

• (i) $\| B^{r,s}\|<\infty$,

• (ii) $\lim_{n\rightarrow\infty}b_{n,k}^{r,s}=0$ for each $k\in \mathbb{N}$,

• (iii) $\lim_{n\rightarrow\infty}\sum_{k}b_{n,k}^{r,s}=1$.

Thus, the binomial matrix $B^{r,s}$ is regular for $sr>0$. Unless stated otherwise, we assume that $sr >0$. If we take $s+r =1$, we obtain the Euler matrix $E^{r}$. So the binomial matrix generalizes the Euler matrix. Bişgin defined the following spaces of binomial sequences:

$$\begin{aligned}& b^{r,s}_{0}=\Biggl\{ x=(x_{k})\in w: \lim _{n\rightarrow\infty}\frac {1}{(s+r)^{n}}\sum_{k=0}^{n} \binom{n}{k}s^{n-k}r^{k}x_{k}=0\Biggr\} , \\& b^{r,s}_{c}=\Biggl\{ x=(x_{k})\in w: \lim _{n\rightarrow\infty}\frac {1}{(s+r)^{n}}\sum_{k=0}^{n} \binom{n}{k}s^{n-k}r^{k}x_{k} \mbox{ exists} \Biggr\} , \end{aligned}$$

and

$$ b^{r,s}_{\infty}=\Biggl\{ x=(x_{k})\in w: \sup _{n\in\mathbb{N}} \Biggl\vert \frac {1}{(s+r)^{n}}\sum _{k=0}^{n}\binom{n}{k}s^{n-k}r^{k}x_{k} \Biggr\vert < \infty\Biggr\} . $$

The purpose of the present paper is to study the difference spaces $b^{r,s}_{0}(\nabla^{(m)})$, $b^{r,s}_{c}(\nabla^{(m)})$ and $b^{r,s}_{\infty}(\nabla^{(m)})$ of the binomial sequence whose $B^{r,s}(\nabla^{(m)})$-transforms are in the spaces $c_{0}$, c and $\ell_{\infty}$, respectively. These new sequence spaces are the generalization of the sequence spaces defined in [22, 23] and [19]. Also, we give some inclusion relations and compute the bases and α-, β- and γ-duals of these sequence spaces.

2 The binomial difference sequence spaces

In this section, we introduce the spaces $b^{r,s}_{0}(\nabla^{(m)})$, $b^{r,s}_{c}(\nabla^{(m)})$, $b^{r,s}_{\infty}(\nabla^{(m)})$ and prove the BK-property and inclusion relations.

We first define the binomial difference sequence spaces $b^{r,s}_{0}(\nabla^{(m)})$, $b^{r,s}_{c}(\nabla^{(m)})$ and $b^{r,s}_{\infty}(\nabla^{(m)})$ by

$$ Z\bigl(\nabla^{(m)}\bigr)=\bigl\{ x=(x_{k})\in w: \bigl( \nabla^{(m)} x_{k}\bigr)\in Z\bigr\} $$

for $Z\in\{b^{r,s}_{0}, b^{r,s}_{c}, b^{r,s}_{\infty}\}$. By using the notion of (1.1), the sequence spaces $b^{r,s}_{0}(\nabla ^{(m)})$, $b^{r,s}_{c}(\nabla^{(m)})$ and $b^{r,s}_{\infty}(\nabla ^{(m)})$ can be redefined by

$$ b^{r,s}_{0}\bigl(\nabla^{(m)}\bigr)= \bigl(b^{r,s}_{0}\bigr)_{\nabla^{(m)}},\qquad b^{r,s}_{c}\bigl(\nabla^{(m)}\bigr)= \bigl(b^{r,s}_{c}\bigr)_{\nabla^{(m)}},\qquad b^{r,s}_{\infty }\bigl(\nabla^{(m)}\bigr)= \bigl(b^{r,s}_{\infty}\bigr)_{\nabla^{(m)}}. $$

(2.1)

It is obvious that the sequence spaces $b^{r,s}_{0}(\nabla^{(m)})$, $b^{r,s}_{c}(\nabla^{(m)})$ and $b^{r,s}_{\infty}(\nabla^{(m)})$ may be reduced to some sequence spaces in the special cases of s, r and $m\in\mathbb{N}$. For instance, we take $m=0$, then obtain the spaces $b^{r,s}_{0}$, $b^{r,s}_{c}$ and $b^{r,s}_{\infty}$ defined by Bişgin [22, 23]. On taking $s+r=1$, we obtain the spaces $e^{r}_{0}(\nabla^{(m)})$, $e^{r}_{c}(\nabla^{(m)})$ and $e^{r}_{\infty}(\nabla^{(m)})$ defined by Polat and Başar [19].

Let us define the sequence $y=(y_{n})$ as the $B^{r,s}(\nabla ^{(m)})$-transform of a sequence $x=(x_{k})$ by

$$ y_{n}=\bigl[B^{r,s}\bigl(\nabla^{(m)} x_{k}\bigr)\bigr]_{n}=\frac{1}{(s+r)^{n}}\sum _{k=0}^{n}\binom{n}{k}s^{n-k}r^{k} \bigl(\nabla^{(m)} x_{k}\bigr) $$

(2.2)

for each $n\in\mathbb{N}$, where

$$ \nabla^{(m)} x_{k}=\sum_{i=0}^{m}(-1)^{i} \binom{m}{i}x_{k-i}=\sum_{i=\max\{0,k-m\}}^{m}(-1)^{k-i} \binom{m}{k-i}x_{i}. $$

Then the binomial difference sequence spaces $b^{r,s}_{0}(\nabla ^{(m)})$, $b^{r,s}_{c}(\nabla^{(m)})$ and $b^{r,s}_{\infty}(\nabla ^{(m)})$ can be redefined by all sequences whose $B^{r,s}(\nabla ^{(m)})$-transforms are in the spaces $c_{0}$, c and $\ell_{\infty}$.

Theorem 2.1

Let $Z\in\{b^{r,s}_{0}, b^{r,s}_{c}, b^{r,s}_{\infty}\}$. Then $Z(\nabla^{(m)})$ is a BK-space with the norm $\| x\|_{Z(\nabla^{(m)})}=\|(\nabla^{(m)} x_{k})\|_{Z}$.

Proof

The sequence spaces $b^{r,s}_{0}$, $b^{r,s}_{c}$ and $b^{r,s}_{\infty}$ are BK-spaces (see [22], Theorem 2.1 and [23], Theorem 2.1). Moreover, $\nabla^{(m)}$ is a triangle matrix and (2.1) holds. By using Theorem 4.3.12 of Wilansky [24], we deduce that the binomial sequence spaces $b^{r,s}_{0}(\nabla^{(m)})$, $b^{r,s}_{c}(\nabla^{(m)})$ and $b^{r,s}_{\infty}(\nabla^{(m)})$ are BK-spaces. □

Theorem 2.2

The sequence spaces $b^{r,s}_{0}(\nabla^{(m)})$, $b^{r,s}_{c}(\nabla ^{(m)})$ and $b^{r,s}_{\infty}(\nabla^{(m)})$ are linearly isomorphic to the spaces $c_{0}$, c and $\ell_{\infty}$, respectively.

Proof

Similarly, we prove the theorem only for the space $b^{r,s}_{0}(\nabla^{(m)})$. To prove $b^{r,s}_{0}(\nabla ^{(m)})\cong c_{0}$, we must show the existence of a linear bijection between the spaces $b^{r,s}_{0}(\nabla^{(m)})$ and $c_{0}$.

Consider $T:b^{r,s}_{0}(\nabla^{(m)})\rightarrow c_{0}$ by $T(x)=B^{r,s}(\nabla^{(m)} x_{k})$. The linearity of T is obvious and $x=0$ whenever $T(x)=0$. Therefore, T is injective.

Let $y=(y_{n})\in c_{0}$ and define the sequence $x=(x_{k})$ by

$$ x_{k}=\sum_{i=0}^{k}(s+r)^{i} \sum_{j=i}^{k}\binom{m+k-j-1}{ k-j} \binom{j}{i}r^{-j}(-s)^{j-i}y_{i} $$

(2.3)

for each $k \in\mathbb{N}$. Then we have

$$ \lim_{n\rightarrow\infty}\bigl[B^{r,s}\bigl(\nabla^{(m)} x_{k}\bigr)\bigr]_{n}=\lim_{n\rightarrow\infty} \frac{1}{(s+r)^{n}}\sum_{k=0}^{n} \binom{n}{k} s^{n-k}r^{k}\bigl(\nabla^{(m)} x_{k}\bigr)=\lim_{n\rightarrow\infty}y_{n}=0, $$

which implies that $x\in b^{r,s}_{0}(\nabla^{(m)} )$ and $T(x)=y$. Consequently, T is surjective and is norm preserving. Thus, $b^{r,s}_{0}(\nabla^{(m)} )\cong c_{0}$. □

The following theorems give some inclusion relations for this class of sequence spaces. We have the well-known inclusion $c_{0}\subseteq c\subseteq\ell_{\infty}$, then the corresponding extended versions also preserve this inclusion.

Theorem 2.3

The inclusion $b^{r,s}_{0}(\nabla^{(m)})\subseteq b^{r,s}_{c}(\nabla ^{(m)})\subseteq b^{r,s}_{\infty}(\nabla^{(m)})$ holds.

Theorem 2.4

The inclusions $b^{r,s}_{0}(\nabla^{(m)})\subseteq b^{r,s}_{0}(\nabla ^{(m+1)})$, $b^{r,s}_{c}(\nabla^{(m)})\subseteq b^{r,s}_{c}(\nabla ^{(m+1)})$ and $b^{r,s}_{\infty}(\nabla^{(m)})\subseteq b^{r,s}_{\infty }(\nabla^{(m+1)})$ hold.

Proof

Let $x=(x_{k})\in b^{r,s}_{0}(\nabla^{(m)})$, then the inequality

$$\begin{aligned} \bigl\vert \bigl[B^{r,s}\bigl(\nabla^{(m+1)} x_{k} \bigr)\bigr]_{n} \bigr\vert =& \bigl\vert \bigl[B^{r,s} \bigl(\nabla ^{(m)}(\nabla x_{k})\bigr)\bigr]_{n} \bigr\vert \\ =& \bigl\vert \bigl[B^{r,s}\bigl(\nabla^{(m)} x_{k}\bigr)\bigr]_{n}-\bigl[B^{r,s}\bigl( \nabla^{(m)} x_{k}\bigr)\bigr]_{n-1} \bigr\vert \\ \leq& \bigl\vert \bigl[B^{r,s}\bigl(\nabla^{(m)} x_{k}\bigr)\bigr]_{n} \bigr\vert + \bigl\vert \bigl[B^{r,s}\bigl(\nabla ^{(m)} x_{k}\bigr) \bigr]_{n-1} \bigr\vert \end{aligned}$$

holds and tends to 0 as $n\rightarrow\infty$, which implies that $x\in b^{r,s}_{0}(\nabla^{(m+1)})$. □

Theorem 2.5

The inclusions $e_{0}^{r}(\nabla^{(m)})\subseteq b^{r,s}_{0}(\nabla ^{(m)})$, $e_{c}^{r}(\nabla^{(m)})\subseteq b^{r,s}_{c}(\nabla^{(m)})$ and $e_{\infty}^{r}(\nabla^{(m)})\subseteq b^{r,s}_{\infty}(\nabla ^{(m)})$ strictly hold.

Proof

Similarly, we only prove the inclusion $e_{0}^{r}(\nabla ^{(m)})\subseteq b^{r,s}_{0}(\nabla^{(m)})$. If $r+s=1$, we have $E^{r}=B^{r,s}$. So $e_{0}^{r}(\nabla^{(m)})\subseteq b^{r,s}_{0}(\nabla ^{(m)})$ holds. Take $0< r<1$ and $s=4$. We define a sequence $x=(x_{k})$ by

$$ x_{k}=\sum_{j=0}^{k} \binom{m+k-j-1}{ k-j} \biggl(-\frac{3}{r}\biggr)^{j} $$

for all $m, k\in\mathbb{N}$. It is clear that $[E^{r}(\nabla^{(m)} x_{k})]_{n}=((-2-r)^{n})\notin c_{0}$ and $[B^{r,s}(\nabla^{(m)} x_{k})]_{n}=((\frac{1}{4+r})^{n})\in c_{0}$. So, we have $x\in b^{r,s}_{0}(\nabla^{(m)})\setminus e_{0}^{r}(\nabla^{(m)})$. This shows that the inclusion $e_{0}^{r}(\nabla^{(m)})\subseteq b^{r,s}_{0}(\nabla ^{(m)})$ strictly holds. □

3 The Schauder basis and α-, β- and γ-duals

For a normed space $(X, \|\cdot\|)$, a sequence $\{ x_{k}:x_{k}\in X\}_{k\in\mathbb{N}}$ is called a Schauder basis [1] if for every $x\in X$, there is an unique scalar sequence $(\lambda_{k})$ such that $\| x-\sum_{k=0}^{n}\lambda _{k}x_{k}\|\rightarrow0$ as $n\rightarrow\infty$. We shall construct the Schauder bases for the sequence spaces $b_{0}^{r,s}(\nabla ^{(m)})$ and $b_{c}^{r,s}(\nabla^{(m)})$.

We define the sequence $g^{(k)}(r,s)=\{g^{(k)}_{i}(r,s)\}_{i \in\mathbb {N}}$ by

$$g^{(k)}_{i}(r,s)= \textstyle\begin{cases} 0& \text{if }0\leq i < k, \\ (s+r)^{k}\sum_{j=k}^{i}\binom{m+i-j-1}{i-j}\binom{j}{k} r^{-j}(-s)^{j-k}& \text{if }i\geq k, \end{cases} $$

for each $k\in\mathbb{N}$.

Theorem 3.1

The sequence $(g^{(k)}(r,s))_{k\in\mathbb{N}}$ is a Schauder basis for the binomial sequence space $b_{0}^{r,s}(\nabla^{(m)})$ and every $x=(x_{i})\in b_{0}^{r,s}(\nabla^{(m)})$ has an unique representation by

$$ x=\sum_{k} \lambda_{k}(r,s) g^{(k)}(r,s), $$

(3.1)

where $\lambda_{k}(r,s)= [B^{r,s}(\nabla^{(m)} x_{i})]_{k}$ for each $k\in\mathbb{N}$.

Proof

Obviously, $B^{r,s}(\nabla^{(m)} g^{(k)}_{i}(r,s))=e_{k}\in c_{0}$, where $e_{k}$ is the sequence with 1 in the kth place and zeros elsewhere for each $k\in\mathbb{N}$. This implies that $g^{(k)}(r,s)\in b_{0}^{r,s}(\nabla^{(m)})$ for each $k\in\mathbb{N}$.

For $x \in b_{0}^{r,s}(\nabla^{(m)})$ and $n\in\mathbb{N}$, we put

$$ x^{(n)}=\sum_{k=0}^{n} \lambda_{k}(r,s) g^{(k)}(r,s). $$

By the linearity of $B^{r,s}(\nabla^{(m)})$, we have

$$ B^{r,s}\bigl(\nabla^{(m)} x^{(n)}_{i}\bigr)= \sum_{k=0}^{n}\lambda _{k}(r,s)B^{r,s} \bigl(\nabla^{(m)} g^{(k)}_{i}(r,s)\bigr)=\sum _{k=0}^{n}\lambda _{k}(r,s)e_{k} $$

and

$$\bigl[B^{r,s}\bigl(\nabla^{(m)}\bigl(x_{i}-x_{i}^{(n)} \bigr)\bigr)\bigr]_{k}= \textstyle\begin{cases} 0& \text{if }0\leq k < n, \\ [B^{r,s}(\nabla^{(m)} x_{i})]_{k}& \text{if }k\geq n, \end{cases} $$

for each $k\in\mathbb{N}$.

For every $\varepsilon>0$, there is a positive integer $n_{0}$ such that

$$ \bigl\vert \bigl[B^{r,s}\bigl(\nabla^{(m)} x_{i} \bigr)\bigr]_{k} \bigr\vert < \frac{\varepsilon}{2} $$

for all $k\geq n_{0}$. Then we have

$$ \bigl\Vert x-x^{(n)} \bigr\Vert _{b_{0}^{r,s}(\nabla^{(m)})}=\sup _{k\geq n} \bigl\vert \bigl[B^{r,s}\bigl( \nabla^{(m)} x_{i}\bigr)\bigr]_{k} \bigr\vert \leq \sup_{k\geq n_{0}} \bigl\vert \bigl[B^{r,s}\bigl( \nabla^{(m)} x_{i}\bigr)\bigr]_{k} \bigr\vert < \frac{\varepsilon}{2}< \varepsilon, $$

which implies $x \in b_{0}^{r,s}(\nabla^{(m)})$ is represented as in (3.1).

To show the uniqueness of this representation, we assume that

$$ x=\sum_{k} \mu_{k}(r,s) g^{(k)}(r,s). $$

Then we have

$$ \bigl[B^{r,s}\bigl(\nabla^{(m)} x_{i}\bigr) \bigr]_{k}=\sum_{k}\mu_{k}(r,s) \bigl[B^{r,s}\bigl(\nabla ^{(m)} g^{(k)}_{i}(r,s) \bigr)\bigr]_{k}=\sum_{k} \mu_{k}(r,s) (e_{k})_{k}=\mu_{k}(r,s), $$

which is a contradiction with the assumption that $\lambda _{k}(r,s)=[B^{r,s}(\nabla^{(m)} x_{i})]_{k}$ for each $k \in\mathbb {N}$. This shows the uniqueness of this representation. □

Theorem 3.2

We define $g=(g_{n})$ by

$$ g_{n}=\sum_{k=0}^{n}(s+r)^{k} \sum_{j=k}^{n}\binom{m+n-j-1}{n-j} \binom{j}{k}r^{-j}(-s)^{j-k} $$

for all $n\in\mathbb{N}$ and $\lim_{k\rightarrow\infty}\lambda _{k}(r,s)=l$. The set $\{g, g^{(0)}(r,s), g^{(1)}(r,s),\ldots ,g^{(k)}(r,s),\ldots\}$ is a Schauder basis for the space $b_{c}^{r,s}(\nabla^{(m)})$ and every $x\in b_{c}^{r,s}(\nabla^{(m)})$ has an unique representation by

$$ x=lg+\sum_{k} \bigl[ \lambda_{k}(r,s)-l\bigr] g^{(k)}(r,s). $$

(3.2)

Proof

Obviously, $B^{r,s}(\nabla^{(m)} g^{k}_{i}(r,s))=e_{k}\in c_{0}\subseteq c$ and $g\in b_{c}^{r,s}(\nabla^{(m)})$. For $x \in b_{c}^{r,s}(\nabla^{(m)})$, we put $y=x-lg$ and we have $y\in b_{0}^{r,s}(\nabla^{(m)})$. Hence, we deduce that y has an unique representation by (3.1), which implies that x has an unique representation by (3.2). Thus, we complete the proof. □

From Theorem 2.1, we know that $b_{0}^{r,s}(\nabla^{(m)})$ and $b_{c}^{r,s}(\nabla^{(m)})$ are Banach spaces. By combining this fact with Theorem 3.1 and Theorem 3.2, we can give the following corollary.

Corollary 3.3

The sequence spaces $b_{0}^{r,s}(\nabla^{(m)})$ and $b_{c}^{r,s}(\nabla ^{(m)})$ are separable.

Köthe and Toeplitz [25] first computed the dual whose elements can be represented as sequences and defined the α-dual (or Köthe-Toeplitz dual). Chandra and Tripathy [26] generalized the notion of Köthe-Toeplitz dual of sequence spaces. Next, we compute the α-, β- and γ-duals of the sequence spaces $b_{0}^{r,s}(\nabla^{(m)})$, $b_{c}^{r,s}(\nabla^{(m)})$ and $b_{\infty }^{r,s}(\nabla^{(m)})$.

For the sequence spaces X and Y, define multiplier space $M(X,Y)$ by

$$ M(X,Y)=\bigl\{ u=(u_{k})\in w:ux=(u_{k}x_{k})\in Y \mbox{ for all } x=(x_{k})\in X\bigr\} . $$

Then the α-, β- and γ-duals of a sequence space X are defined by

$$ X^{\alpha}=M(X,\ell_{1}),\qquad X^{\beta}=M(X,cs)\quad \mbox{and}\quad X^{\gamma}=M(X, bs), $$

respectively.

Let us give the following properties:

$$\begin{aligned}& \sup_{K\in\Gamma} \sum_{n} \biggl\vert \sum_{k\in K} a_{n,k} \biggr\vert < \infty, \end{aligned}$$

(3.3)

$$\begin{aligned}& \sup_{n\in\mathbb{N}} \sum_{k} \vert a_{n,k} \vert < \infty, \end{aligned}$$

(3.4)

$$\begin{aligned}& \lim_{n\rightarrow\infty}a_{n,k}=a_{k}\quad \mbox{for each } k\in\mathbb {N}, \end{aligned}$$

(3.5)

$$\begin{aligned}& \lim_{n\rightarrow\infty}\sum_{k}a_{n,k}=a, \end{aligned}$$

(3.6)

$$\begin{aligned}& \lim_{n\rightarrow\infty}\sum_{k} \vert a_{n,k} \vert =\sum_{k} \Bigl\vert \lim_{n\rightarrow\infty}a_{n,k} \Bigr\vert , \end{aligned}$$

(3.7)

where Γ is the collection of all finite subsets of $\mathbb{N}$.

Lemma 3.4

[27]

Let $A=(a_{n,k})$ be an infinite matrix, then:

• (i) $A\in(c_{0}:\ell_{1})=(c:\ell_{1})=(\ell_{\infty}:\ell_{1})$ if and only if (3.3) holds.

• (ii) $A\in(c_{0}:c)$ if and only if (3.4) and (3.5) hold.

• (iii) $A\in(c:c)$ if and only if (3.4), (3.5) and (3.6) hold.

• (iv) $A\in(\ell_{\infty}:c)$ if and only if (3.5) and (3.7) hold.

• (v) $A\in(c_{0}:\ell_{\infty})=(c:\ell_{\infty})=(\ell_{\infty}:\ell _{\infty})$ if and only if (3.4) holds.

Theorem 3.5

The α-dual of the spaces $b_{0}^{r,s}(\nabla^{(m)})$, $b_{c}^{r,s}(\nabla^{(m)})$ and $b_{\infty}^{r,s}(\nabla^{(m)})$ is the set

$$ U^{r,s}_{1}=\Biggl\{ u=(u_{k})\in w:\sup _{K\in\Gamma}\sum_{k} \Biggl\vert \sum _{i\in K} (s+r)^{i}\sum _{j=i}^{k}\binom{m+k-j-1}{ k-j}\binom{j}{i}r^{-j}(-s)^{j-i}u_{k} \Biggr\vert < \infty\Biggr\} . $$

Proof

Let $u=(u_{k})\in w$ and $x=(x_{k})$ be defined by (2.3), then we have

$$ u_{k}x_{k}=\sum_{i=0}^{k}(s+r)^{i} \sum_{j=i}^{k}\binom{m+k-j-1}{ k-j} \binom{j}{i}r^{-j}(-s)^{j-i}u_{k}y_{i}= \bigl(G^{r,s}y\bigr)_{k} $$

for each $k\in\mathbb{N}$, where $G^{r,s}=(g^{r,s}_{k,i})$ is defined by

$$g^{r,s}_{k,i}= \textstyle\begin{cases} (s+r)^{i}\sum_{j=i}^{k}\binom{m+k-j-1}{k-j} \binom{j}{i} r^{-j}(-s)^{j-i}u_{k}& \text{if }0\leq i\leq k, \\ 0& \text{if }i>k. \end{cases} $$

Therefore, we deduce that $ux= (u_{k}x_{k})\in\ell_{1}$ whenever $x\in b_{0}^{r,s}(\nabla^{(m)})$, $b_{c}^{r,s}(\nabla^{(m)})$ or $b_{\infty }^{r,s}(\nabla^{(m)})$ if and only if $G^{r,s}y\in\ell_{1}$ whenever $y\in c_{0}, c$ or $\ell _{\infty}$, which implies that $u=(u_{k})\in[b_{0}^{r,s}(\nabla ^{(m)})]^{\alpha}, [b_{c}^{r,s}(\nabla^{(m)})]^{\alpha} \mbox{ or } [b_{\infty}^{r,s}(\nabla^{(m)})]^{\alpha}$ if and only if $G^{r,s}\in (c_{0}:\ell_{1})=(c:\ell_{1})=(\ell_{\infty}:\ell_{1})$. By Lemma 3.4(i), we obtain

$$ u=(u_{k})\in\bigl[b_{0}^{r,s}\bigl( \nabla^{(m)}\bigr)\bigr]^{\alpha}=\bigl[b_{c}^{r,s} \bigl(\nabla ^{(m)}\bigr)\bigr]^{\alpha} =\bigl[b_{\infty}^{r,s} \bigl(\nabla^{(m)}\bigr)\bigr]^{\alpha} $$

if and only if

$$ \sup_{K\in\Gamma}\sum_{k} \Biggl\vert \sum_{i\in K}(s+r)^{i}\sum _{j=i}^{k}\binom{m+k-j-1}{k-j} \binom{j}{i}r^{-j}(-s)^{j-i}u_{k} \Biggr\vert < \infty. $$

Thus, we have $[b_{0}^{r,s}(\nabla^{(m)})]^{\alpha}=[b_{c}^{r,s}(\nabla ^{(m)})]^{\alpha} =[b_{\infty}^{r,s}(\nabla^{(m)})]^{\alpha}=U^{r,s}_{1}$. □

Now, we define the sets $U_{2}^{r,s}$, $U_{3}^{r,s}$, $U_{4}^{r,s}$ and $U_{5}^{r,s}$ by

$$\begin{aligned}& U_{2}^{r,s}=\biggl\{ u=(u_{k})\in w: \sup _{n\in\mathbb{N}}\sum_{k} \vert u_{n,k} \vert < \infty\biggr\} , \\& U_{3}^{r,s}=\Bigl\{ u=(u_{k})\in w: \lim _{n\rightarrow\infty} u_{n,k} \mbox{ exists for each } k \in\mathbb{N} \Bigr\} , \\& U_{4}^{r,s}=\biggl\{ u=(u_{k})\in w: \lim _{n\rightarrow\infty}\sum_{k} \vert u_{n,k} \vert =\sum_{k} \Bigl\vert \lim_{n\rightarrow\infty}u_{n,k} \Bigr\vert \biggr\} , \end{aligned}$$

and

$$ U_{5}^{r,s}=\biggl\{ u=(u_{k})\in w: \lim _{n\rightarrow\infty}\sum_{k}u_{n,k} \mbox{ exists}\biggr\} , $$

where

$$ u_{n,k}=(s+r)^{k}\sum_{i=k}^{n} \sum_{j=k}^{i}\binom{m+i-j-1}{ i-j} \binom{j}{k} r^{-j}(-s)^{j-k}u_{i}. $$

Theorem 3.6

The following equations hold:

• (i) $[b_{0}^{r,s}(\nabla^{(m)})]^{ \beta}=U_{2}^{r,s}\cap U_{3}^{r,s}$,

• (ii) $[b_{c}^{r,s}(\nabla^{(m)})]^{ \beta}=U_{2}^{r,s}\cap U_{3}^{r,s}\cap U_{5}^{r,s}$,

• (iii) $[b_{\infty}^{r,s}(\nabla^{(m)})]^{ \beta}=U_{3}^{r,s}\cap U_{4}^{r,s}$.

Proof

Since the proof may be obtained in the same way for (ii) and (iii), we only prove (i). Let $u=(u_{k})\in w$ and $x=(x_{k})$ be defined by (2.3), then we consider the following equation:

$$\begin{aligned} \sum_{k=0}^{n}u_{k}x_{k} =& \sum_{k=0}^{n}u_{k}\Biggl[\sum _{i=0}^{k}(s+r)^{i}\sum _{j=i}^{k}\binom{m+k-j-1}{ k-j}\binom{j}{i} r^{-j}(-s)^{j-i}y_{i}\Biggr] \\ =&\sum_{k=0}^{n}\Biggl[(s+r)^{k} \sum_{i=k}^{n}\sum _{j=k}^{i}\binom {m+i-j-1}{ i-j} \binom{j}{k} r^{-j}(-s)^{j-k}u_{i}\Biggr]y_{k} \\ =&\bigl(U^{r,s}y\bigr)_{n}, \end{aligned}$$

where $U^{r,s}=(u^{r,s}_{n,k})$ is defined by

$$u_{n,k}= \textstyle\begin{cases} (s+r)^{k}\sum_{i=k}^{n}\sum_{j=k}^{i}\binom{m+i-j-1}{i-j} \binom{j}{k} r^{-j}(-s)^{j-k}u_{i}& \text{if }0\leq k \leq n, \\ 0& \text{if }k> n. \end{cases} $$

Therefore, we deduce that $ux= (u_{k}x_{k})\in cs$ whenever $x\in b_{0}^{r,s}(\nabla^{(m)})$ if and only if $U^{r,s}y\in c$ whenever $y\in c_{0}$, which implies that $u=(u_{k})\in[b_{0}^{r,s}(\nabla^{(m)})]^{ \beta}$ if and only if $U^{r,s}\in(c_{0}:c)$. By Lemma 3.4(ii), we obtain $[b_{0}^{r,s}(\nabla^{(m)})]^{ \beta}=U_{2}^{r,s}\cap U_{3}^{r,s}$. □

Theorem 3.7

The γ-dual of the spaces $b_{0}^{r,s}(\nabla^{(m)})$, $b_{c}^{r,s}(\nabla^{(m)})$ and $b_{\infty}^{r,s}(\nabla^{(m)})$ is the set $U_{2}^{r,s}$.

Proof

Using Lemma 3.4(v) instead of (ii), the proof can be given in a similar way. So, we omit the details. □

4 Conclusion

By considering the definitions of the binomial matrix $B^{r,s}=(b^{r,s}_{n,k})$ and mth order difference operator, we introduce the sequence spaces $b_{0}^{r,s}(\nabla^{(m)})$, $b_{c}^{r,s}(\nabla^{(m)})$ and $b_{\infty}^{r,s}(\nabla^{(m)})$. These spaces are the natural continuation of [3, 19, 22, 23]. Our results are the generalization of the matrix domain of the Euler matrix of order r.