## 1 Introduction and preliminaries

Let w denote the space of all sequences. By $$\ell_{p}$$, $$\ell_{\infty }$$, c and $$c_{0}$$, we denote the spaces of p-absolutely summable, bounded, convergent and null sequences, respectively, where $$1\leq p<\infty$$. Let Z be a sequence space, then Kizmaz [1] introduced the following difference sequence spaces:

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

for $$Z=\ell_{\infty},c,c_{0}$$, where $$\Delta x_{k}=x_{k}-x_{k+1}$$ for each $$k\in\mathbb{N}=\{1,2,3,\ldots\}$$, the set of positive integers. Since then, many authors have studied further generalization of the difference sequence spaces [26]. Moreover, Altay and Polat [7], Başarir and Kara [812], Kara [13], Kara and İlkhan [14], Polat and Başar [15], and many others have studied new sequence spaces from a 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 $$Ax=\{(Ax)_{n}\}$$ and is supposed to be convergent for all $$n\in\mathbb{N}$$, where $$(Ax)_{n}=\sum_{k=0}^{\infty}a_{n,k}x_{k}$$. For two sequence spaces X, Y and an infinite matrix $$A=(a_{n,k})$$, the sequence space $$X_{A}$$ is defined by $$X_{A}=\{x=(x_{k})\in w:Ax \in X\}$$, which is called the domain of matrix A in the space X. 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

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

The Euler sequence spaces $$e^{r}_{p}$$ and $$e^{r}_{\infty}$$ were defined by Altay, Başar and Mursaleen [16] as follows:

$$e^{r}_{p}=\left \{x=(x_{k})\in w: \sum _{n}\left \vert \sum _{k=0}^{n} \left ( \begin{matrix} n\\ k \end{matrix} \right ) (1-r)^{n-k}r^{k}x_{k} \right \vert ^{p}< \infty \right \}$$

and

$$e^{r}_{\infty}=\left \{x=(x_{k})\in w: \sup _{n\in\mathbb{N}}\left \vert \sum_{k=0}^{n} \left ( \begin{matrix} n\\ k \end{matrix} \right ) (1-r)^{n-k}r^{k}x_{k} \right \vert < \infty \right \}.$$

Altay and Polat [7] defined further generalization of the Euler sequence spaces $$e^{r}_{0}(\nabla)$$, $$e^{r}_{c}(\nabla)$$ and $$e^{r}_{\infty}(\nabla)$$ by

\begin{aligned} &e^{r}_{0}(\nabla)= \bigl\{ x=(x_{k})\in w: ( \nabla x_{k})\in e^{r}_{0} \bigr\} , \\ &e^{r}_{c}(\nabla)= \bigl\{ x=(x_{k})\in w: ( \nabla x_{k})\in e^{r}_{c} \bigr\} \end{aligned}

and

\begin{aligned} e^{r}_{\infty}(\nabla)= \bigl\{ x=(x_{k})\in w: (\nabla x_{k})\in e^{r}_{\infty} \bigr\} , \end{aligned}

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. Many authors have used especially the Euler matrix for defining new sequence spaces, for instance, Kara and Başarir [17], Karakaya and Polat [18] and Polat and Başar [15].

Recently Bişgin [19, 20] defined another type of 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

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

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

1. (i)

$$\Vert B^{r,s} \Vert <\infty$$,

2. (ii)

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

3. (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 [20] defined the following spaces of binomial sequences:

$$b^{r,s}_{p}=\left \{x=(x_{k})\in w: \sum _{n}\left \vert \frac{1}{(s+r)^{n}}\sum _{k=0}^{n}\left ( \begin{matrix} n\\ k \end{matrix} \right )s^{n-k}r^{k}x_{k} \right \vert ^{p}< \infty \right \}$$

and

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

The main purpose of the present paper is to study the normed difference spaces $$b^{r,s}_{p}(\nabla)$$ and $$b^{r,s}_{\infty}(\nabla)$$ of the binomial sequence whose $$B^{r,s}(\nabla)$$-transforms are in the spaces $$\ell_{p}$$ and $$\ell_{\infty}$$, respectively. These new sequence spaces are the generalization of the sequence spaces defined in [7] and [20]. Also, we 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 $$b^{r,s}_{p}(\nabla)$$ and $$b^{r,s}_{\infty}(\nabla)$$ and prove that these sequence spaces are linearly isomorphic to the spaces $$\ell_{p}$$ and $$\ell_{\infty}$$, respectively.

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

$$b^{r,s}_{p}(\nabla)= \bigl\{ x=(x_{k})\in w: (\nabla x_{k})\in b^{r,s}_{p} \bigr\}$$

and

$$b^{r,s}_{\infty}(\nabla)= \bigl\{ x=(x_{k})\in w: (\nabla x_{k})\in b^{r,s}_{\infty} \bigr\} .$$

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

$$y_{n}= \bigl[B^{r,s}(\nabla x_{k}) \bigr]_{n}=\frac{1}{(s+r)^{n}}\sum_{k=0}^{n} \left ( \begin{matrix} n\\ k \end{matrix} \right )s^{n-k}r^{k}(\nabla x_{k})$$
(2.1)

for each $$n\in\mathbb{N}$$. Then the binomial difference sequence spaces $$b^{r,s}_{p}(\nabla)$$ or $$b^{r,s}_{\infty}(\nabla)$$ can be redefined by all sequences whose $$B^{r,s}(\nabla)$$-transforms are in the space $$\ell_{p}$$ or $$\ell_{\infty}$$.

### Theorem 2.1

The sequence spaces $$b^{r,s}_{p}(\nabla)$$ and $$b^{r,s}_{\infty}(\nabla )$$ are complete linear metric spaces with the norm defined by

$$f_{b^{r,s}_{p}(\nabla)}(x)= \Vert y \Vert _{p}= \Biggl(\sum _{n=1}^{\infty} \vert y_{n} \vert ^{p} \Biggr)^{\frac{1}{p}}$$

and

$$f_{b^{r,s}_{\infty}(\nabla)}(x)= \Vert y \Vert _{\infty}=\sup_{n\in\mathbb{N}} \vert y_{n} \vert ,$$

where $$1\leq p<\infty$$ and the sequence $$y=(y_{n})$$ is defined by the $$B^{r,s}(\nabla)$$-transform of x.

### Proof

The proof of the linearity is a routine verification. It is obvious that $$f_{b^{r,s}_{p}}(\alpha x)= \vert \alpha \vert f_{b^{r,s}_{p}}(x)$$ and $$f_{b^{r,s}_{p}}(x)=0$$ if and only if $$x=\theta$$ for all $$x\in b^{r,s}_{p}(\nabla)$$, where θ is the zero element in $$b^{r,s}_{p}$$ and $$\alpha\in\mathbb{R}$$. We consider $$x,z \in b^{r,s}_{p}(\nabla)$$, then we have

\begin{aligned} f_{b^{r,s}_{p}(\nabla)}(x+z)&= \biggl(\sum_{n} \bigl\vert \bigl(B^{r,s} \bigl[\nabla (x_{k}+z_{k}) \bigr] \bigr)_{n} \bigr\vert ^{p} \biggr)^{\frac{1}{p}} \\ &\leq \biggl(\sum_{n} \bigl\vert \bigl[B^{r,s}(\nabla x_{k}) \bigr]_{n} \bigr\vert ^{p} \biggr)^{\frac {1}{p}}+ \biggl(\sum_{n} \bigl\vert \bigl[B^{r,s}(\nabla z_{k}) \bigr]_{n} \bigr\vert ^{p} \biggr)^{\frac {1}{p}} = f_{b^{r,s}_{p}(\nabla)}(x)+f_{b^{r,s}_{p}(\nabla)}(z). \end{aligned}

Hence $$f_{b^{r,s}_{p}(\nabla)}$$ is a norm on the space $$b^{r,s}_{p}(\nabla)$$.

Let $$(x_{m})$$ be a Cauchy sequence in $$b^{r,s}_{p}(\nabla)$$, where $$x_{m}=(x_{m_{k}})_{k=1}^{\infty}\in b^{r,s}_{p}(\nabla)$$ for each $$m\in\mathbb{N}$$. For every $$\varepsilon>0$$, there is a positive integer $$m_{0}$$ such that $$f_{b^{r,s}_{p}(\nabla)}( x_{m}-x_{l})<\varepsilon$$ $$\text{for } m,l\geq m_{0}$$. Then we get

$$\bigl\vert \bigl(B^{r,s} \bigl[\nabla(x_{m_{k}}-x_{l_{k}}) \bigr] \bigr)_{n} \bigr\vert \leq \biggl(\sum _{n} \bigl\vert \bigl(B^{r,s} \bigl[ \nabla(x_{m_{k}}-x_{l_{k}}) \bigr] \bigr)_{n} \bigr\vert ^{p} \biggr)^{\frac {1}{p}}< \varepsilon$$

for $$m,l\geq m_{0}$$ and each $$k\in\mathbb{N}$$. So $$(B^{r,s}(\nabla x_{m_{k}}))_{m=1}^{\infty}$$ is a Cauchy sequence in the set of real numbers $$\mathbb{R}$$. Since $$\mathbb{R}$$ is complete, we have $$\lim_{m\rightarrow\infty}B^{r,s}(\nabla x_{m_{k}})=B^{r,s}(\nabla x_{k})$$ for each $$k\in\mathbb{N}$$. We compute

\begin{aligned} \sum_{n=0}^{i} \bigl\vert \bigl(B^{r,s} \bigl[\nabla(x_{m_{k}}-x_{l_{k}}) \bigr] \bigr)_{n} \bigr\vert \leq f_{b^{r,s}_{p}(\nabla)}( x_{m}-x_{l})< \varepsilon \end{aligned}
(2.2)

for $$m>m_{0}$$. We take i and l →∞, then the inequality (2.2) implies that

$$f_{b^{r,s}_{p}(\nabla)}( x_{m}-x)\rightarrow0.$$

We have

$$f_{b^{r,s}_{p}(\nabla)}(x)\leq f_{b^{r,s}_{p}(\nabla )}(x_{m}-x)+f_{b^{r,s}_{p}(\nabla)}(x_{m})< \infty,$$

that is, $$x\in b^{r,s}_{p}(\nabla)$$. Thus, the space $$b^{r,s}_{p}(\nabla )$$ is complete. For the space $$b^{r,s}_{\infty}(\nabla)$$, the proof can be completed in a similar way. So, we omit the detail. □

### Theorem 2.2

The sequence spaces $$b^{r,s}_{p}(\nabla)$$ and $$b^{r,s}_{\infty}(\nabla )$$ are linearly isomorphic to the spaces $$\ell_{p}$$ and $$\ell_{\infty}$$, respectively, where $$1\leq p< \infty$$.

### Proof

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

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

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

\begin{aligned} x_{k}=\sum_{i=0}^{k}(s+r)^{i} \sum_{j=i}^{k} \left ( \begin{matrix} j\\ i \end{matrix} \right )r^{-j}(-s)^{j-i}y_{i} \end{aligned}
(2.3)

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

\begin{aligned} f_{b^{r,s}_{p}(\nabla)}(x) =& \bigl\Vert \bigl[B^{r,s}(\nabla x_{k}) \bigr]_{n} \bigr\Vert _{p} \\ =&\left (\sum_{n=1}^{\infty} \left \vert \frac{1}{(s+r)^{n}}\sum_{k=0}^{n} \left ( \begin{matrix} n\\ k \end{matrix} \right )s^{n-k}r^{k}( \nabla x_{k}) \right \vert ^{p} \right )^{\frac{1}{p}} \\ =& \Biggl(\sum_{n=1}^{\infty} \vert y_{n} \vert ^{p} \Biggr)^{\frac{1}{p}} = \Vert y \Vert _{p}< \infty, \end{aligned}

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

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

For a normed space $$(X, \Vert \cdot \Vert )$$, a sequence $$\{ x_{k}:x_{k}\in X\}_{k\in\mathbb{N}}$$ is called a Schauder basis [21] if for every $$x\in X$$, there is a unique scalar sequence $$(\lambda_{k})$$ such that $$\Vert x-\sum_{k=0}^{n}\lambda _{k}x_{k} \Vert \rightarrow0 \text{ as } n\rightarrow\infty$$. Next, we shall give a Schauder basis for the sequence space $$b_{p}^{r,s}(\nabla)$$.

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

$$g^{(k)}_{i}(r,s)= \textstyle\begin{cases} 0& \text{if 0\leq i < k},\\ (s+r)^{k}\sum_{j=k}^{i}\bigl({\scriptsize\begin{matrix}{} j\cr k\end{matrix}} \bigr)r^{-j}(-s)^{j-k}& \text{if i\geq k}, \end{cases}$$

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 spaces $$b_{p}^{r,s}(\nabla)$$ and every $$x=(x_{i})\in b_{p}^{r,s}(\nabla)$$ has a unique representation by

$$x=\sum_{k} \lambda_{k}(r,s) g^{(k)}(r,s),$$
(3.1)

where $$1\leq p<\infty$$ and $$\lambda_{k}(r,s)= [B^{r,s}(\nabla x_{i})]_{k}$$ for each $$k\in\mathbb{N}$$.

### Proof

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

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

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

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

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

and

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

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

For any given $$\varepsilon>0$$, there is a positive integer $$m_{0}$$ such that

$$\sum_{k=m_{0}+1}^{\infty} \bigl\vert \bigl[B^{r,s}(\nabla x_{i}) \bigr]_{k} \bigr\vert ^{p}< \biggl(\frac {\varepsilon}{2} \biggr)^{p}$$

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

\begin{aligned} f_{b^{r,s}_{p}(\nabla)} \bigl( x-x^{(m)} \bigr) =& \Biggl(\sum _{k=m+1 }^{\infty} \bigl\vert \bigl[B^{r,s}(\nabla x_{i}) \bigr]_{k} \bigr\vert ^{p} \Biggr)^{\frac{1}{p}} \\ \leq& \Biggl(\sum_{k=m_{0}+1 }^{\infty} \bigl\vert \bigl[B^{r,s}(\nabla x_{i}) \bigr]_{k} \bigr\vert ^{p} \Biggr)^{\frac {1}{p}} \\ < & \frac{\varepsilon}{2}< \varepsilon, \end{aligned}

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

To prove the uniqueness of this representation, we assume that

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

Then we have

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

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

### Corollary 3.2

The sequence space $$b_{p}^{r,s}(\nabla)$$ is separable, where $$1\leq p<\infty$$.

For the duality theory, the study of sequence spaces is more useful when we investigate them equipped with linear topologies. Köthe and Toeplitz [22] first computed duals whose elements can be represented as sequences and defined the α-dual (or Köthe-Toeplitz dual).

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

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

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

$$X^{\alpha}=M(X,\ell_{1}),\qquad X^{\beta}=M(X,c)\quad \text{and}\quad X^{\gamma }=M(X,\ell_{\infty}),$$

respectively.

We give the following properties:

\begin{aligned} &\sup_{n\in\mathbb{N}} \sum_{k} \vert a_{n,k} \vert ^{q}< \infty, \end{aligned}
(3.2)
\begin{aligned} &\sup_{k\in\mathbb{N}} \sum_{n} \vert a_{n,k} \vert < \infty, \end{aligned}
(3.3)
\begin{aligned} &\sup_{n,k\in\mathbb{N}} \vert a_{n,k} \vert < \infty, \end{aligned}
(3.4)
\begin{aligned} &\lim_{n\rightarrow\infty}a_{n,k}=a_{k} \quad\text{for each } k \in \mathbb{N}, \end{aligned}
(3.5)
\begin{aligned} &\sup_{K\in\Gamma} \sum_{k} \biggl\vert \sum_{n\in K} a_{n,k} \biggr\vert ^{q}< \infty , \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 $$\mathbb{N}$$, $$\frac{1}{p}+\frac{1}{q}=1$$ and $$1< p\leq\infty$$.

### Lemma 3.3

[23]

Let $$A=(a_{n,k})$$ be an infinite matrix. Then the following statements hold:

1. (i)

$$A\in(\ell_{1}:\ell_{1})$$ if and only if (3.3) holds.

2. (ii)

$$A\in(\ell_{1}:c)$$ if and only if (3.4) and (3.5) hold.

3. (iii)

$$A\in(\ell_{1}:\ell_{\infty})$$ if and only if (3.4) holds.

4. (iv)

$$A\in(\ell_{p}:\ell_{1})$$ if and only if (3.6) holds with $$\frac {1}{p}+\frac{1}{q}=1$$ and $$1< p\leq\infty$$.

5. (v)

$$A\in( \ell_{p}:c)$$ if and only if (3.2) and (3.5) hold with $$\frac{1}{p}+\frac{1}{q}=1$$ and $$1< p<\infty$$.

6. (vi)

$$A\in( \ell_{p}:\ell_{\infty} )$$ if and only if (3.2) holds with $$\frac{1}{p}+\frac{1}{q}=1$$ and $$1< p<\infty$$.

7. (vii)

$$A\in( \ell_{\infty}:c )$$ if and only if (3.5) and (3.7) hold with $$\frac{1}{p}+\frac{1}{q}=1$$ and $$1< p<\infty$$.

8. (viii)

$$A\in( \ell_{\infty}:\ell_{\infty} )$$ if and only if (3.2) holds with $$q=1$$.

### Theorem 3.4

We define the set $$U_{1}^{r,s}$$ and $$U_{2}^{r,s}$$ by

$$U_{1}^{r,s}=\left \{u=(u_{k})\in w:\sup _{i\in\mathbb{N}}\sum_{k}\left \vert (s+r)^{i}\sum_{j=i}^{k}\left ( \begin{matrix} j\\ i \end{matrix} \right )r^{-j}(-s)^{j-i}u_{k} \right \vert < \infty \right \}$$

and

$$U_{2}^{r,s}=\left \{u=(u_{k})\in w:\sup _{K\in\Gamma}\sum_{i}\left \vert \sum_{k\in K}(s+r)^{i}\sum _{j=i}^{k}\left ( \begin{matrix} j\\ i \end{matrix} \right )r^{-j}(-s)^{j-i}u_{k} \right \vert ^{q}< \infty \right \}.$$

Then $$[b^{r,s}_{1}(\nabla)]^{\alpha}=U_{1}^{r,s}$$ and $$[b^{r,s}_{p}(\nabla)]^{\alpha}=U_{2}^{r,s}$$, where $$1< p\leq\infty$$.

### Proof

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

$$u_{k}x_{k}=\sum_{i=0}^{k}(s+r)^{i} \sum_{j=i}^{k}\left ( \begin{matrix} j\\ i \end{matrix} \right )r^{-j}(-s)^{j-i}u_{k}y_{i}= \bigl(G^{r,s}y \bigr)_{k}$$

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

$$g^{r,s}_{k,i}= \textstyle\begin{cases} (s+r)^{i}\sum_{j=i}^{k}\left({\scriptsize\begin{matrix}{} j\cr i \end{matrix}} \right)r^{-j}(-s)^{j-i}u_{k}& \text{if 0\leq i\leq k},\\ 0& \text{if i>k}. \end{cases}$$

Therefore, we deduce that $$ux= (u_{k}x_{k})\in\ell_{1}$$ whenever $$x\in b_{1}^{r,s}(\nabla)$$ or $$b_{p}^{r,s}(\nabla)$$ if and only if $$G^{r,s}y\in\ell_{1}$$ whenever $$y\in \ell_{1}$$ or $$\ell_{p}$$, which implies that $$u=(u_{k})\in[b_{1}^{r,s}(\nabla )]^{\alpha} \text{ or } [b_{p}^{r,s}(\nabla)]^{\alpha}$$ if and only if $$G^{r,s}\in(\ell_{1}:\ell_{1})$$ and $$G^{r,s}\in(\ell_{p}:\ell_{1})$$ by parts (i) and (iv) of Lemma 3.3, we obtain $$u=(u_{k})\in [b_{1}^{r,s}(\nabla)]^{\alpha}$$ if and only if

$$\sup_{i\in\mathbb{N}}\sum_{k}\left \vert (s+r)^{i}\sum_{j=i}^{k} \left ( \begin{matrix} j\\ i \end{matrix} \right )r^{-j}(-s)^{j-i}u_{k} \right \vert < \infty$$

and $$u=(u_{k})\in[b_{p}^{r,s}(\nabla)]^{\alpha}$$ if and only if

$$\sup_{K\in\Gamma}\sum_{i}\left \vert \sum_{k\in K}(s+r)^{i}\sum _{j=i}^{k}\left ( \begin{matrix} j\\ i \end{matrix} \right )r^{-j}(-s)^{j-i}u_{k} \right \vert ^{q}< \infty.$$

Thus, we have $$[b^{r,s}_{1}(\nabla)]^{\alpha}=U_{1}^{r,s}$$ and $$[b^{r,s}_{p}(\nabla)]^{\alpha}=U_{2}^{r,s}$$, where $$1< p\leq\infty$$. □

Now, we define the sets $$U_{3}^{r,s}$$, $$U_{4}^{r,s}$$, $$U_{5}^{r,s}$$, $$U_{6}^{r,s}$$ and $$U_{7}^{r,s}$$ by

\begin{aligned} &U_{3}^{r,s}=\left \{u=(u_{k})\in w: \lim _{n\rightarrow\infty} (s+r)^{k}\sum_{i=k}^{n} \sum_{j=k}^{i}\left ( \begin{matrix} j\\ k \end{matrix} \right )r^{-j}(-s)^{j-k}u_{i} \text{ exists for each } k \in\mathbb {N} \right \}, \\ &U_{4}^{r,s}=\left \{u=(u_{k})\in w: \sup _{n,k\in\mathbb{N}}\left \vert (s+r)^{k}\sum _{i=k}^{n}\sum_{j=k}^{i} \left ( \begin{matrix} j\\ k \end{matrix} \right )r^{-j}(-s)^{j-k}u_{i} \right \vert < \infty \right \}, \\ &U_{5}^{r,s}= \left\{u=(u_{k})\in w: \lim _{n\rightarrow\infty}\sum_{k}\left \vert (s+r)^{k}\sum_{i=k}^{n}\sum _{j=k}^{i}\left ( \begin{matrix} j\\ k \end{matrix} \right )r^{-j}(-s)^{j-k}u_{i} \right \vert \right. \\ &\left.\phantom{U_{5}^{r,s}=}=\sum_{k}\left \vert \lim _{n\rightarrow\infty}(s+r)^{k}\sum_{i=k}^{n} \sum_{j=k}^{i}\left ( \begin{matrix} j\\ k \end{matrix} \right )r^{-j}(-s)^{j-k}u_{i} \right \vert \right\}, \\ &U_{6}^{r,s}=\left \{u=(u_{k})\in w: \sup _{n\in\mathbb{N}}\sum_{k=0}^{n}\left \vert (s+r)^{k}\sum_{i=k}^{n} \sum_{j=k}^{i}\left ( \begin{matrix} j\\ k \end{matrix} \right )r^{-j}(-s)^{j-k}u_{i} \right \vert ^{q}< \infty \right \},\quad 1< q< \infty, \end{aligned}

and

$$U_{7}^{r,s}=\left \{u=(u_{k})\in w: \sup _{n\in\mathbb{N}}\sum_{k=0}^{n}\left \vert (s+r)^{k}\sum_{i=k}^{n} \sum_{j=k}^{i}\left ( \begin{matrix} j\\ k \end{matrix} \right )r^{-j}(-s)^{j-k}u_{i} \right \vert < \infty \right \}.$$

### Theorem 3.5

We have the following relations:

1. (i)

$$[b_{1}^{r,s}(\nabla)]^{ \beta}=U_{3}^{r,s}\cap U_{4}^{r,s}$$,

2. (ii)

$$[b_{p}^{r,s}(\nabla)]^{ \beta}=U_{3}^{r,s}\cap U_{6}^{r,s}$$, where $$1< p<\infty$$,

3. (iii)

$$[b_{\infty}^{r,s}(\nabla)]^{ \beta}=U_{3}^{r,s}\cap U_{5}^{r,s}$$,

4. (iv)

$$[b_{1}^{r,s}(\nabla)]^{ \gamma}=U_{4}^{r,s}$$,

5. (v)

$$[b_{p}^{r,s}(\nabla)]^{ \gamma}=U_{6}^{r,s}$$, where $$1< p<\infty$$,

6. (vi)

$$[b_{\infty}^{r,s}(\nabla)]^{ \gamma}=U_{7}^{r,s}$$.

### Proof

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}\left [ \sum_{i=0}^{k}(s+r)^{i}\sum _{j=i}^{k}\left ( \begin{matrix} j\\ i \end{matrix} \right )r^{-j}(-s)^{j-i}y_{i} \right ] \\ &=\sum_{k=0}^{n}\left [(s+r)^{k} \sum_{i=k}^{n}\sum _{j=k}^{i}\left ( \begin{matrix} j\\ k \end{matrix} \right )r^{-j}(-s)^{j-k}u_{i} \right ]y_{k} = \bigl(U^{r,s}y \bigr)_{n}, \end{aligned}

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

$$u_{n,k}= \textstyle\begin{cases} (s+r)^{k}\sum_{i=k}^{n}\sum_{j=k}^{i}\bigl({\scriptsize\begin{matrix}{} j\cr k\end{matrix}} \bigr)r^{-j}(-s)^{j-k}u_{i}&\text{if 0\leq k \leq n},\\ 0&\text{if k> n}. \end{cases}$$

Therefore, we deduce that $$ux= (u_{k}x_{k})\in c$$ whenever $$x\in b_{1}^{r,s}(\nabla)$$ if and only if $$U^{r,s}y\in c$$ whenever $$y\in\ell_{1}$$, which implies that $$u=(u_{k})\in[b_{1}^{r,s}(\nabla)]^{ \beta}$$ if and only if $$U^{r,s}\in(\ell_{1}:c)$$. By Lemma 3.3(ii), we obtain $$[b_{1}^{r,s}(\nabla)]^{ \beta}=U_{3}^{r,s}\cap U_{4}^{r,s}$$. Using Lemma 3.3(i) and (iii)-(viii) instead of (ii), the proof can be completed 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 the difference operator, we introduce the sequence spaces $$b^{r,s}_{p}(\nabla)$$ and $$b^{r,s}_{\infty}(\nabla)$$. These spaces are the natural continuations of [1, 7, 20]. Our results are the generalizations of the matrix domain of the Euler matrix of order r. In order to give fully inform the reader on related topics with applications and a possible line of further investigation, the e-book [24] is added to the list of references.