# Composition of Cesàro and backward difference operators

## Abstract

In this research, we combine the Cesàro and backward difference operators of different orders which results in introducing a matrix who has two different behaviors and includes several matrices. We also investigate the Köthe duals and inclusion relations of the associated sequence space of this new matrix. Moreover, we compute the norm of this matrix on some well-known sequence spaces.

## Introduction

Let ω be the space of all real-valued sequences. The space $$\ell _{p}$$ consists all real sequences $$u=(u_{k})_{k=0}^{\infty }\in \omega$$ such that $$\sum_{k=0}^{\infty }|u_{k}|^{p}<\infty$$, which is a Banach space with the norm

\begin{aligned}& \Vert u \Vert _{\ell _{p}}= \Biggl(\sum_{k=0}^{\infty } \vert u_{k} \vert ^{p} \Biggr)^{1/p}< \infty , \end{aligned}

where $$1< p<\infty$$.

The matrix domain of an infinite matrix T in a sequence space X is defined as

$$X_{T}=\{x\in \omega : Tx\in X\}$$

which is also a sequence space. By using matrix domains of special triangle matrices in classical spaces, many authors have introduced and studied new Banach spaces. For the relevant literature, we refer to the papers  and the textbooks  and .

The Köthe dual (α-, β-, γ-duals) of a sequence space X are defined by

\begin{aligned}& X^{\alpha }= \Biggl\{ a=(a_{k})\in \omega : \sum _{k=1}^{\infty } \vert a_{k}x_{k} \vert < \infty \text{ for all } x=(x_{k})\in X \Biggr\} , \\& X^{\beta }= \Biggl\{ a=(a_{k})\in \omega : \Biggl(\sum _{k=1}^{n} a_{k}x_{k} \Biggr)\in c \text{ for all } x=(x_{k})\in X \Biggr\} , \\& X^{\gamma }= \Biggl\{ a=(a_{k})\in \omega : \Biggl(\sum _{k=1}^{n} a_{k}x_{k} \Biggr)\in \ell _{\infty }\text{ for all } x=(x_{k})\in X \Biggr\} , \end{aligned}

respectively.

Consider the Hausdorff matrix $$H^{\mu }=(h_{jk})_{j,k=0}^{\infty }$$, with entries of the form

\begin{aligned}& h_{jk}=\textstyle\begin{cases} \binom{j}{k} \int _{0}^{1} \theta ^{k}(1-\theta )^{j-k}\,d\mu (\theta ) , & 0\le k \leq j, \\ 0 , & k>j, \end{cases}\displaystyle \end{aligned}

where μ is a probability measure on $$[0,1]$$. The Hausdorff matrix contains some famous classes of matrices. For positive integer n, by choosing $$d\mu (\theta )=n(1-\theta )^{n-1}\,d\theta$$ the Cesàro matrix $$C^{n} = (c^{n}_{jk})$$ of order n is defined as follows:

\begin{aligned}& c^{n}_{jk}=\textstyle\begin{cases} \frac{\binom{n+j-k-1 }{j-k}}{\binom{n+j }{j}} ,& 0\le k\le j, \\ 0 ,& \text{otherwise}. \end{cases}\displaystyle \end{aligned}
(1.1)

Hardy’s formula (, Theorem 216) states that the Hausdorff matrix is a bounded operator on $$\ell _{p}$$ if and only if $$\int _{0}^{1} \theta ^{\frac{-1}{p}}\,d\mu (\theta )<\infty$$ and

\begin{aligned}& \bigl\Vert H^{\mu } \bigr\Vert _{\ell _{p}}= \int _{0}^{1} \theta ^{\frac{-1}{p}}\,d\mu ( \theta ), \end{aligned}
(1.2)

hence the Cesàro matrix has the norm

\begin{aligned}& \bigl\Vert C^{n} \bigr\Vert _{\ell _{p}}= \frac{\Gamma (n+1)\Gamma (1/p^{*})}{\Gamma (n+1/p^{*})}, \end{aligned}
(1.3)

where $$p^{*}$$ is the conjugate of p i.e. $$\frac{1}{p}+\frac{1}{p^{*}}=1$$. Note that $$C^{1}$$ is the well-known Cesàro matrix C with $$\|C\|_{\ell _{p}}=p^{*}$$. The author has introduced the sequence spaces $$C^{n}_{p}$$ and $$C^{n}_{\infty }$$ as the set of all sequences whose $$C^{n}$$-transforms are in the spaces $$\ell _{p}$$ and $$\ell _{\infty }$$, respectively; that is,

\begin{aligned}& C^{n}_{p}= \Biggl\{ x=(x_{j})\in \omega :\sum _{j=0}^{\infty } \Biggl\vert \frac{1}{\binom{n+j }{j}} \sum_{k=0}^{j}\binom{n+j-k-1 }{j-k}x_{k} \Biggr\vert ^{p}< \infty \Biggr\} \end{aligned}
(1.4)

and

\begin{aligned}& C^{n}_{\infty }= \Biggl\{ x=(x_{j})\in \omega :\sup _{j} \Biggl\vert \frac{1}{\binom{n+j }{j}}\sum _{k=0}^{j}\binom{n+j-k-1 }{j-k}x_{k} \Biggr\vert < \infty \Biggr\} . \end{aligned}
(1.5)

Backward difference matrix. The backward difference matrix of order n, $$\Delta ^{n}=(\delta ^{n}_{jk})$$, is defined by

\begin{aligned}& \delta ^{n}_{jk}=\textstyle\begin{cases} (-1)^{j-k}\binom{n}{j-k} ,& k\le j \le k+n, \\ 0 ,& \text{otherwise}. \end{cases}\displaystyle \end{aligned}
(1.6)

This matrix has the inverse $$\Delta ^{-n}=(\delta ^{-n}_{jk})$$ which has the following entries:

\begin{aligned}& \delta ^{-n}_{jk}=\textstyle\begin{cases} \binom{n+j-k-1}{j-k} ,& j\ge k \ge 0, \\ 0 ,& \text{otherwise}. \end{cases}\displaystyle \end{aligned}
(1.7)

Motivation. The inverse of the backward difference matrix has some similarities in the definition to the Cesàro matrix, hence composing of the Cesàro matrix and the backward difference matrix results in two different cases: one acts like the backward and the other acts like the inverse of the backward difference operator. In this study, we try to discover some results of these combinations covering both types of operators. Therefore, the present study is a generalization of much research.

## Composition of the Cesàro and backward difference matrices

Suppose that n, m are two non-negative integers and $$1< p<\infty$$. Let us define the matrix $$S^{n,m}=(s^{n,m}_{jk})$$ by

\begin{aligned}& S^{n,m}=C^{n}\Delta ^{m}, \end{aligned}
(2.1)

where $$C^{n}$$ and $$\Delta ^{m}$$ are the Cesàro and backward difference matrices of order n and m, defined by Eqs. (1.1) and (1.6), respectively.

Note that $$S^{1,0}=C$$, $$S^{n,0}=C^{n}$$, $$S^{0,1}=\Delta$$ and $$S^{0,m}=\Delta ^{m}$$. The sequence space associated with this matrix, $$\ell _{p}(S^{n,m})$$, includes the following spaces:

• $$\ell _{p}(S^{1,0}):=C_{p}$$,

• $$\ell _{p}(S^{n,0}):=C^{n}_{p}$$,

• $$\ell _{p}(S^{0,1}):=bv_{p}$$,

• $$\ell _{p}(S^{0,m}):=\ell _{p}(\Delta ^{m})$$,

• $$\ell _{p}(S^{1,m}):=C_{p}(\Delta ^{m})$$;

these have been investigated in [7, 2026], respectively.

With regard to the cases $$n\ge m$$ or $$n\le m$$ we encounter two different types of matrix $$S^{n,m}$$, which we define by $$\Phi ^{n,m}$$ and $$\Psi ^{n,m}$$, respectively. They are

\begin{aligned}& S^{n,m}:=\Phi ^{n,m}, \quad n\ge m \ge 0, \end{aligned}

and

\begin{aligned}& S^{n,m}:=\Psi ^{n,m}, \quad 0\le n\le m. \end{aligned}

We can represent the matrices $$\Phi ^{n,m}=(\phi ^{n,m}_{jk})$$ and $$\Psi ^{n,m}=(\psi ^{n,m}_{jk})$$ by their entries as follows:

\begin{aligned}& \phi ^{n,m}_{jk}=\textstyle\begin{cases} \frac{\binom{n-m+j-k-1 }{j-k}}{\binom{n+j }{j}} ,& 0\le k\le j, \\ 0 ,& \text{otherwise}, \end{cases}\displaystyle \end{aligned}
(2.2)

and

\begin{aligned}& \psi ^{n,m}_{jk}=\textstyle\begin{cases} \frac{(-1)^{j-k}\binom{m-n }{j-k}}{\binom{n+j }{j}} ,& 0\le k\le j, \\ 0 ,& \text{otherwise}. \end{cases}\displaystyle \end{aligned}
(2.3)

### Lemma 2.1

The matrices $$\Phi ^{n,m}$$ and $$\Psi ^{n,m}$$ are invertible and their inverse $$(\Phi ^{n,m})^{-1}=((\phi ^{n,m})^{-1}_{jk})$$ and $$(\Psi ^{n,m})^{-1}=((\psi ^{n,m})^{-1}_{jk})$$ are defined by

\begin{aligned}& \bigl(\phi ^{n,m}\bigr)^{-1}_{jk}=\textstyle\begin{cases} (-1)^{j-k}\binom{n-m }{j-k}\binom{n+k }{k} ,& k\le j\le n-m+k, \\ 0 ,& \textit{otherwise}, \end{cases}\displaystyle \end{aligned}

and

\begin{aligned}& \bigl(\psi ^{n,m}\bigr)^{-1}_{jk}=\textstyle\begin{cases} \binom{m-n+j-k-1 }{j-k}\binom{n+k }{k} ,& j\ge k\ge 0, \\ 0 ,& \textit{otherwise}. \end{cases}\displaystyle \end{aligned}

### Proof

Since $$S^{n,m}=C^{n}\Delta ^{m}$$ we have $$(S^{n,m})^{-1}=\Delta ^{-m}C^{-n}$$ or

\begin{aligned}& \bigl(S^{n,m}\bigr)^{-1}_{jk}=\sum _{i} \delta ^{-m}_{ji} \delta ^{n}_{ik} \binom{n+k }{k}= \delta ^{n-m}_{jk} \binom{n+k }{k}. \end{aligned}

Now, by the definition of the backward difference operator, Eqs. (1.6) and (1.7)

\begin{aligned}& \bigl(\phi ^{n,m}\bigr)^{-1}_{jk}=(-1)^{j-k} \binom{n-m }{j-k}\binom{n+k }{k} \end{aligned}

and

\begin{aligned}& \bigl(\psi ^{n,m}\bigr)^{-1}_{jk}= \binom{m-n+j-k-1 }{j-k}\binom{n+k }{k}. \end{aligned}

□

### Theorem 2.2

Let n and m be two non-negative integers and $$x\in \ell _{p}$$. We have the following inequalities:

\begin{aligned}& \bigl\Vert S^{n,m}x \bigr\Vert _{\ell _{p}}\le \frac{\Gamma (n+1)\Gamma (1/p^{*})}{\Gamma (n+1/p^{*})} \bigl\Vert \Delta ^{m} x \bigr\Vert _{\ell _{p}}. \end{aligned}

In particular, we have the following.

• For $$m=0$$

\begin{aligned}& \bigl\Vert C^{n}x \bigr\Vert _{\ell _{p}}\le \frac{\Gamma (n+1)\Gamma (1/p^{*})}{\Gamma (n+1/p^{*})} \Vert x \Vert _{\ell _{p}}. \end{aligned}
• For $$n=1$$ and $$m=0$$ we have Hardy’s inequality,

\begin{aligned}& \Vert Cx \Vert _{\ell _{p}}\le \frac{p}{p-1} \Vert x \Vert _{\ell _{p}}. \end{aligned}

### Proof

By the definition of the matrix $$S^{n,m}$$ we have

\begin{aligned}& \bigl\Vert S^{n,m}x \bigr\Vert _{\ell _{p}}= \bigl\Vert C^{n}\Delta ^{m}x \bigr\Vert _{\ell _{p}}\le \bigl\Vert C^{n} \bigr\Vert _{\ell _{p}} \bigl\Vert \Delta ^{m}x \bigr\Vert _{\ell _{p}} = \frac{\Gamma (n+1)\Gamma (1/p^{*})}{\Gamma (n+1/p^{*})} \bigl\Vert \Delta ^{m} x \bigr\Vert _{\ell _{p}}. \end{aligned}

□

The following inclusions are the straightforward results of the above theorem.

### Corollary 2.3

Let n and m are two non-negative integers and $$x\in \ell _{p}$$. Then

\begin{aligned}& \ell _{p}\bigl(\Delta ^{m}\bigr)\subset \ell _{p} \bigl(S^{n,m}\bigr). \end{aligned}
(2.4)

In particular,

• $$\ell _{p}(\Delta ^{m})\subset C_{p}(\Delta ^{m})$$,

• $$\ell _{p}\subset C^{n}_{p}$$.

### Theorem 2.4

The spaces $$\ell _{p}(S^{n,m})$$ and $$\ell _{\infty }(S^{n,m})$$ are linearly isomorphic to $$\ell _{p}$$ and $$\ell _{\infty }$$, respectively. In particular, we have the following.

• The spaces $$C^{n}_{p}$$ and $$C^{n}_{\infty }$$ are linearly isomorphic to $$\ell _{p}$$ and $$\ell _{\infty }$$, respectively (, Theorem 2.4).

• The spaces $$\ell _{p}(\Delta ^{n})$$ and $$\ell _{\infty }(\Delta ^{n})$$ are linearly isomorphic to $$\ell _{p}$$ and $$\ell _{\infty }$$, respectively.

### Proof

Let us define the map $$T:\ell _{p}(S^{n,m})\rightarrow \ell _{p}$$ with $$T(u)=S^{n,m}u$$ for any $$u\in \ell _{p}(S^{n,m})$$. It is clear that T is linear and one-to-one. Also, since $$S^{n,m}$$ is invertible, T is onto. Now, since $$\|u\|_{\ell _{p}(S^{n,m})}=\|S^{n,m}u\|_{\ell _{p}}$$ holds, we find that T preserves the norms. This completes the proof. □

### Theorem 2.5

The inclusion $$\ell _{p}(S^{n,m}) \subset \ell _{q}(S^{n,m})$$ strictly holds, where $$1< p<q<\infty$$. In particular, we have the following.

• The inclusion $$C^{n}_{p} \subset C^{n}_{q}$$ strictly holds.

• The inclusion $$\ell _{p}(\Delta ^{n}) \subset \ell _{q}(\Delta ^{n})$$ strictly holds.

### Proof

Let $$u\in \ell _{p}(S^{n,m})$$. Then we have $$S^{n,m}u\in \ell _{p}$$. Since the inclusion $$\ell _{p}\subset \ell _{q}$$ holds for $$1< p<q<\infty$$, we have $$S^{n,m} u\in \ell _{q}$$. This implies that $$u\in \ell _{q}(S^{n,m})$$. Hence, we conclude that the inclusion $$\ell _{p}(S^{n,m}) \subset \ell _{q}(S^{n,m})$$ holds.

Now, we show that the inclusion is strict. Since the inclusion $$\ell _{p}\subset \ell _{q}$$ is strict, we can choose $$v\in \ell _{q}\backslash \ell _{p}$$. Define the sequence $$u=(S^{n,m})^{-1}v$$, which means $$S^{n,m}u=v$$ and so $$S^{n,m}u\in \ell _{q}\backslash \ell _{p}$$. Hence, we conclude that $$u\in \ell _{q}(S^{n,m}) \backslash \ell _{p}(S^{n,m})$$ and so the inclusion $$\ell _{p}(S^{n,m}) \subset \ell _{q}(S^{n,m})$$ is strict. □

### Theorem 2.6

The inclusion $$\ell _{p}(S^{n,m}) \subset \ell _{\infty }(S^{n,m})$$ strictly holds. In particular, we have the following.

• The inclusion $$C^{n}_{p} \subset C^{n}_{\infty }$$ strictly holds.

• The inclusion $$\ell _{p}(\Delta ^{n}) \subset \ell _{\infty }(\Delta ^{n})$$ strictly holds.

### Proof

Let $$u\in \ell _{p}(S^{n,m})$$. Then we have $$S^{n,m}u\in \ell _{p}$$. Since the inclusion $$\ell _{p}\subset \ell _{\infty }$$ holds for $$1< p<\infty$$, we have $$S^{n,m}u\in \ell _{\infty }$$. This implies that $$u\in \ell _{\infty }(S^{n,m})$$. Hence, we conclude that the inclusion $$\ell _{p}(S^{n,m}) \subset \ell _{\infty }(S^{n,m})$$ holds.

Now, we show that the inclusion is strict. Consider the sequence $$v=(v_{j})=(-1)^{j}$$ and let $$u=(S^{n,m})^{-1}v$$. We deduce that $$S^{n,m}u=((-1)^{j})\in \ell _{\infty }\backslash \ell _{p}$$, we obtain $$u\in \ell _{\infty }(S^{n,m}) \backslash \ell _{p}(S^{n,m})$$. Consequently, the inclusion $$\ell _{p}(S^{n,m}) \subset \ell _{\infty }(S^{n,m})$$ is strict. □

### Lemma 2.7

(Theorem 20.3, )

Let $$1< p<\infty$$ and $$\alpha >\beta \ge 0$$. The Cesàro matrix of order α, $$C^{\alpha }$$, has a factorization of the form

\begin{aligned}& C^{\alpha }=\mathcal{R}^{\alpha ,\beta } C^{\beta }=C^{\beta }\mathcal{R}^{ \alpha ,\beta }, \end{aligned}
(2.5)

where $$\mathcal{R}^{\alpha ,\beta }$$ is a bounded operator on $$\ell _{p}$$ and

\begin{aligned}& \bigl\Vert \mathcal{R}^{\alpha ,\beta } \bigr\Vert _{\ell _{p}}= \frac{\Gamma (\alpha +1)\Gamma (\beta +1/p^{*})}{\Gamma (\beta +1)\Gamma (\alpha +1/p^{*})}. \end{aligned}

### Corollary 2.8

For $$\alpha >\beta \ge 0$$, the inclusion $$\ell _{p}(S^{\beta ,m})\subset \ell _{\infty }(S^{\alpha ,m})$$ holds. In particular, for $$m=0$$, we have $$C^{\beta }_{p} \subset C^{\alpha }_{p}$$.

### Proof

By multiplying both sides of Eq. (2.5) in terms of $$\Delta ^{m}$$ we obtain the equality

\begin{aligned}& S^{\alpha ,m}=\mathcal{R}^{\alpha ,\beta } S^{\beta ,m}. \end{aligned}

Now, let $$x \in \ell _{p}(S^{\beta ,m})$$. Since

\begin{aligned}& \bigl\Vert S^{\alpha ,m}x \bigr\Vert _{\ell _{p}}\le \bigl\Vert \mathcal{R}^{\alpha ,\beta } \bigr\Vert _{ \ell _{p}} \bigl\Vert S^{\beta ,m}x \bigr\Vert _{\ell _{p}}, \end{aligned}

hence $$x \in \ell _{p}(S^{\alpha ,m})$$, which results in the inclusion. □

### Corollary 2.9

Let α and β be two non-negative integers that $$\alpha \ge \beta$$. The backward difference matrix, $$\Delta ^{\alpha -\beta }$$, is a bounded operator from $$\ell _{p}$$ into $$\ell _{p}(S^{\alpha ,\beta })$$ and

\begin{aligned}& \bigl\Vert \Delta ^{\alpha -\beta } \bigr\Vert _{\ell _{p},\ell _{p}(S^{\alpha ,\beta })}=1. \end{aligned}

In particular, the backward difference matrix of order α, $$\Delta ^{\alpha }$$, is a bounded operator from $$\ell _{p}$$ into $$C^{\alpha }_{p}$$ and

\begin{aligned}& \bigl\Vert \Delta ^{\alpha } \bigr\Vert _{\ell _{p},C^{\alpha }_{p}}=1. \end{aligned}

### Proof

According to the definition of matrix $$S^{\alpha ,\beta }$$ it is not difficult to prove that $$S^{\alpha ,\alpha }=D^{\alpha }$$, where $$D^{\alpha }=(d^{\alpha }_{jk})$$ is a diagonal matrix defined by

\begin{aligned}& d^{\alpha }_{jk}=\textstyle\begin{cases} \frac{1}{\binom{\alpha +j }{j}} ,& k= j, \\ 0 ,& \text{otherwise}. \end{cases}\displaystyle \end{aligned}

Now, we have

\begin{aligned} \bigl\Vert \Delta ^{\alpha -\beta } \bigr\Vert _{\ell _{p},\ell _{p}(S^{\alpha ,\beta })} =&\sup _{x \in \ell _{p}} \frac{ \Vert \Delta ^{\alpha -\beta }x \Vert _{\ell _{p}(S^{\alpha ,\beta })}}{ \Vert x \Vert _{\ell _{p}}}= \sup_{x\in \ell _{p}} \frac{ \Vert S^{\alpha ,\beta }\Delta ^{\alpha -\beta } x \Vert _{\ell _{p}}}{ \Vert x \Vert _{\ell _{p}}} \\ =& \bigl\Vert S^{\alpha ,\alpha } \bigr\Vert _{\ell _{p}}=\sup _{j} d^{\alpha }_{jj}=1. \end{aligned}

In particular, by letting $$\beta =0$$, $$S^{\alpha ,\beta }=C^{\alpha }$$, we have the desired result. □

### Corollary 2.10

Let α and β be two non-negative integers such that $$\alpha >\beta \ge 0$$ and $$t+k=r$$. The backward difference matrix, $$\Delta ^{k}$$, is a bounded operator from $$\ell _{p}(S^{\beta ,r})$$ into $$\ell _{p}(S^{\alpha ,t})$$ and

\begin{aligned}& \bigl\Vert \Delta ^{k} \bigr\Vert _{\ell _{p}(S^{\beta ,r}),\ell _{p}(S^{\alpha ,t})}= \frac{\Gamma (\alpha +1)\Gamma (\beta +1/p^{*})}{\Gamma (\beta +1)\Gamma (\alpha +1/p^{*})}. \end{aligned}

In particular, the identity matrix, I, is a bounded operator from $$C^{\beta }_{p}$$ into $$C^{\alpha }_{p}$$ and

\begin{aligned}& \Vert I \Vert _{C^{\beta }_{p},C^{\alpha }_{p}}= \frac{\Gamma (\alpha +1)\Gamma (\beta +1/p^{*})}{\Gamma (\beta +1)\Gamma (\alpha +1/p^{*})}. \end{aligned}

### Proof

According to Theorem 2.4, $$\ell _{p}(S^{\beta ,r})$$ and $$\ell _{p}$$ are isomorphic spaces. Hence by applying Lemma 2.7 we deduce that

\begin{aligned} \bigl\Vert \Delta ^{k} \bigr\Vert _{\ell _{p}(S^{\beta ,r}),\ell _{p}(S^{\alpha ,t})} =&\sup _{x \in \ell _{p}(S^{\beta ,r})} \frac{ \Vert \Delta ^{k}x \Vert _{\ell _{p}(S^{\alpha ,t})}}{ \Vert x \Vert _{\ell _{p}(S^{\beta ,r})}}= \sup_{x\in \ell _{p}(S^{\beta ,r})} \frac{ \Vert S^{\alpha ,t}\Delta ^{k} x \Vert _{\ell _{p}}}{ \Vert S^{\beta ,r}x \Vert _{\ell _{p}}} \\ =&\sup_{S^{\beta ,r} x\in \ell _{p}} \frac{ \Vert \mathcal{R}^{\alpha ,\beta }S^{\beta ,r} x \Vert _{\ell _{p}}}{ \Vert S^{\beta ,r}x \Vert _{\ell _{p}}}= \sup_{y\in \ell _{p}} \frac{ \Vert \mathcal{R}^{\alpha ,\beta }y \Vert _{\ell _{p}}}{ \Vert y \Vert _{\ell _{p}}} \\ =& \bigl\Vert \mathcal{R}^{\alpha ,\beta } \bigr\Vert _{\ell _{p}}= \frac{\Gamma (\alpha +1)\Gamma (\beta +1/p^{*})}{\Gamma (\beta +1)\Gamma (\alpha +1/p^{*})}. \end{aligned}

Now, by letting $$k=r=t=0$$ we find the desired result. □

## Two special cases

The sequence spaces $$\ell _{p}(S^{n,m})$$ and $$\ell _{\infty }(S^{n,m})$$ are introduced as the set of all sequences whose $$S^{n,m}$$-transforms are in the spaces $$\ell _{p}$$ and $$\ell _{\infty }$$, respectively; that is,

\begin{aligned}& \ell _{p}\bigl(S^{n,m}\bigr)= \Biggl\{ u=(u_{j}) \in \omega :\sum_{j=0}^{\infty } \Biggl\vert \sum_{k=0}^{j}s^{n,m}_{jk}u_{k} \Biggr\vert ^{p}< \infty \Biggr\} \end{aligned}

and

\begin{aligned}& \ell _{\infty }\bigl(S^{n,m}\bigr)= \Biggl\{ u=(u_{j}) \in \omega :\sup_{j} \Biggl\vert \sum _{k=0}^{j}s^{n,m}_{jk}u_{k} \Biggr\vert < \infty \Biggr\} . \end{aligned}

Now, regarding the double reaction of the matrix $$S^{n,m}$$ there are two separate sequence spaces $$\ell _{p}(\Phi ^{n,m})$$ and $$\ell _{p}(\Psi ^{n,m})$$ that have different bases and Köthe duals. In this section, we intend to investigate both these spaces.

### Fractional Cesàro spaces

By assuming $$n\ge m$$ the matrix $$S^{n,m}=\Phi ^{n,m}$$ and the associated sequence spaces $$\ell _{p}(\Phi ^{n,m})$$ and $$\ell _{\infty }(\Phi ^{n,m})$$ are introduced routinely as the set of all sequences whose $$\Phi ^{n,m}$$-transforms are in the spaces $$\ell _{p}$$ and $$\ell _{\infty }$$, respectively; that is,

\begin{aligned}& \ell _{p}\bigl(\Phi ^{n,m}\bigr)= \Biggl\{ u=(u_{j})\in \omega :\sum_{j=0}^{\infty } \Biggl\vert \frac{1}{\binom{n+j }{j}}\sum_{k=0}^{j} \binom{n-m+j-k-1 }{j-k}u_{k} \Biggr\vert ^{p}< \infty \Biggr\} \end{aligned}

and

\begin{aligned}& \ell _{\infty }\bigl(\Phi ^{n,m}\bigr)= \Biggl\{ u=(u_{j})\in \omega :\sup_{j} \Biggl\vert \frac{1}{\binom{n+j }{j}}\sum_{k=0}^{j} \binom{n-m+j-k-1 }{j-k}u_{k} \Biggr\vert < \infty \Biggr\} . \end{aligned}

### Theorem 3.1

The spaces $$\ell _{p}(\Phi ^{n,m})$$ and $$\ell _{\infty }(\Phi ^{n,m})$$ are Banach spaces with the norms

\begin{aligned}& \Vert u \Vert _{\ell _{p}(\Phi ^{n,m})}= \Biggl(\sum_{j=0}^{\infty } \Biggl\vert \frac{1}{\binom{n+j }{j}}\sum_{k=0}^{j} \binom{n-m+j-k-1 }{j-k} u_{k} \Biggr\vert ^{p} \Biggr)^{1/p} \end{aligned}

and

\begin{aligned}& \Vert u \Vert _{\ell _{\infty }(\Phi ^{n,m})}=\sup_{j} \Biggl\vert \frac{1}{\binom{n+j }{j}}\sum_{k=0}^{j} \binom{n-m+j-k-1 }{j-k}u_{k} \Biggr\vert , \end{aligned}

respectively.

### Proof

We omit the proof which is a routine verification. □

### Remark 3.2

By choosing $$m=0$$ in the above theorem, we obtain the Cesàro sequence spaces $$\ell _{p}(C^{n})=C^{n}_{p}$$ and $$\ell _{\infty }(C^{n})=C^{n}_{\infty }$$, defined in , which are Banach spaces endowed with the norms

\begin{aligned}& \Vert u \Vert _{C^{n}_{p}}= \Biggl(\sum_{j=0}^{\infty } \Biggl\vert \frac{1}{\binom{n+j }{j}}\sum_{k=0}^{j} \binom{n+j-k-1 }{j-k} u_{k} \Biggr\vert ^{p} \Biggr)^{1/p} \end{aligned}

and

\begin{aligned}& \Vert u \Vert _{C^{n}_{\infty }}=\sup_{j} \Biggl\vert \frac{1}{\binom{n+j }{j}}\sum_{k=0}^{j} \binom{n+j-k-1 }{j-k}u_{k} \Biggr\vert , \end{aligned}

respectively.

### Theorem 3.3

Define the sequence $$(b^{(k)})=(b^{(k)}_{j})$$ for each $$k\in \mathbb{N}$$ by

$$\bigl(b^{(k)}\bigr)_{j}=\textstyle\begin{cases} (-1)^{j-k}\binom{n-m }{j-k}\binom{n+k }{k} ,& j\geq k, \\ 0 ,& j< k, \end{cases}\displaystyle \quad (j\in \mathbb{N}_{0}).$$

Then the sequence $$(b^{(k)})$$ is a basis for the space $$\ell _{p}(\Phi ^{n,m})$$, and each $$u \in \ell _{p}(\Phi ^{n,m})$$ has a unique representation of the form $$u=\sum_{k}(\Phi ^{n,m} u)_{k}b^{(k)}$$.

### Proof

Let A be a triangle. By Theorem 2.3 of Jarrah and Malkowsky , the matrix domain $$U_{A}$$ has a basis if and only if the normed sequence space U has a basis. Hence the proof follows immediately. □

We use the following lemma to compute the dual spaces. By $$\mathcal{N}$$, we denote the family of all finite subsets of $$\mathbb{N}$$.

### Lemma 3.4

()

The following statements hold:

$$(i)$$ $$A=(a_{jk})\in (\ell _{1},\ell _{1})$$ if and only if

$$\sup_{k}\sum_{j=0}^{\infty } \vert a_{jk} \vert < \infty .$$

$$(ii)$$ $$A=(a_{jk})\in (\ell _{p},\ell _{1})$$ if and only if

$$\sum_{k=0}^{\infty } \Biggl(\sum _{j=0}^{\infty } \vert a_{jk} \vert \Biggr)^{p*}< \infty ,$$

where $$1< p<\infty$$.

$$(iii)$$ $$A=(a_{jk})\in (\ell _{\infty },\ell _{1})$$ if and only if

$$\sup_{K\in \mathcal{N}}\sum_{j=0}^{\infty } \biggl\vert \sum_{k\in K}a_{jk} \biggr\vert < \infty .$$

$$(iv)$$ $$A=(a_{jk})\in (\ell _{1},c)$$ if and only if

$$\lim_{j\rightarrow \infty }a_{jk} \textit{ exists for each } k\in \mathbb{N}$$
(3.1)

and

$$\sup_{jk} \vert a_{jk} \vert < \infty .$$
(3.2)

$$(v)$$ $$A=(a_{jk})\in (\ell _{p},c)$$ if and only if (3.1) holds and

$$\sup_{j}\sum_{k=0}^{\infty } \vert a_{jk} \vert ^{p*}< \infty ,$$
(3.3)

where $$1< p<\infty$$.

$$(vi)$$ $$A=(a_{jk})\in (\ell _{\infty },c)$$ if and only if (3.1) holds and

$$\lim_{j\rightarrow \infty }\sum_{k=0}^{\infty } \vert a_{jk} \vert =\sum_{k=0}^{\infty } \Bigl\vert \lim_{j\rightarrow \infty }a_{jk} \Bigr\vert .$$

$$(vii)$$ $$A=(a_{jk})\in (\ell _{1},\ell _{\infty })$$ if and only if (3.2) holds.

$$(viii)$$ $$A=(a_{jk})\in (\ell _{p},\ell _{\infty })$$ if and only if (3.3) holds, where $$1< p<\infty$$.

$$(ix)$$ $$A=(a_{jk})\in (\ell _{\infty },\ell _{\infty })$$ if and only if

$$\sup_{j}\sum_{k=0}^{\infty } \vert a_{jk} \vert < \infty .$$

Dual spaces. The α-dual of a sequence space U consists of all sequences $$a=(a_{k})\in \omega$$ such that $$au=(a_{k} u_{k})\in \ell _{1}$$ for all $$u=(u_{k})\in U$$.

### Theorem 3.5

The α-duals of the spaces $$\ell _{1}(\Phi ^{n,m})$$, $$\ell _{p}(\Phi ^{n,m})$$ $$(1< p<\infty )$$ and $$\ell _{\infty }(\Phi ^{n,m})$$ are as follows:

\begin{aligned}& \bigl[\ell_{1}(\Phi^{n,m})\bigr]^{\alpha }:= \Biggl\{ a=(a_{j})\in \omega : \sup _{k}\sum_{j=0}^{\infty } \biggl\vert (-1)^{j-k}\binom{n-m }{j-k}\binom{n+k }{k}a_{j} \biggr\vert < \infty \Biggr\} , \\& \bigl[\ell_{p}(\Phi^{n,m})\bigr]^{\alpha }:= \Biggl\{ a=(a_{j})\in \omega : \sum _{k=0}^{\infty } \Biggl(\sum_{j=0}^{\infty } \biggl\vert (-1)^{j-k}\binom{n-m }{j-k}\binom{n+k }{k}a_{j} \biggr\vert \Biggr)^{p*}< \infty \Biggr\} , \end{aligned}

and

\begin{aligned}& \bigl[\ell _{\infty }(\Phi ^{n,m})\bigr] ^{\alpha }:= \Biggl\{ a=(a_{j})\in \omega : \sup _{K\in \mathcal{N}}\sum_{j=0}^{\infty } \biggl\vert \sum_{k\in K}(-1)^{j-k} \binom{n-m }{j-k}\binom{n+k }{k}a_{j} \biggr\vert < \infty \Biggr\} . \end{aligned}

### Proof

Let $$a=(a_{j})\in \omega$$ and define the matrix $$D=(d_{jk})$$ as

\begin{aligned}& d_{jk}=\textstyle\begin{cases} (-1)^{j-k}\binom{n-m }{j-k}\binom{n+k }{k}a_{j} ,& 0\leq k\leq j, \\ 0 ,& \text{otherwise.} \end{cases}\displaystyle \end{aligned}

For any $$u=(u_{j})\in \ell _{p}(\Phi ^{n,m})$$ $$(1< p<\infty )$$, we have $$a_{j}u_{j}=(Dv)_{j}$$ for all $$j\in \mathbb{N}$$. Thus $$au\in \ell _{1}$$ with $$u\in \ell _{p}(\Phi ^{n,m})$$ if and only if $$Dv\in \ell _{1}$$ with $$v\in \ell _{p}$$. Hence, we conclude that $$a\in [\ell _{p}(\Phi ^{n,m})]^{\alpha }$$ if and only if $$D\in (\ell _{p},\ell _{1})$$. This completes the proof by part $$(ii)$$ of Lemma 3.4. The other cases can be proved similarly. □

The β-dual of a sequence space U consists of all sequences $$a=(a_{k})\in \omega$$ such that $$(\sum_{k=1}^{n} a_{k} u_{k} )\in c$$ for all $$u=(u_{k})\in U$$.

### Theorem 3.6

Let us define the following sets:

\begin{aligned}& P_{1}:= \Biggl\{ a=(a_{k})\in \omega : \lim _{j\rightarrow \infty } \sum_{i=k}^{j}(-1)^{i-k} \binom{n-m }{i-k}\binom{n+k }{k} a_{i} \textit{ exists for each } k \in \mathbb{N} \Biggr\} , \\& P_{2}:= \Biggl\{ a=(a_{k})\in \omega : \sup _{jk} \Biggl\vert \sum_{i=k}^{j}(-1)^{i-k} \binom{n-m }{i-k}\binom{n+k }{k} a_{i} \Biggr\vert < \infty \Biggr\} , \\& P_{3}:= \Biggl\{ a=(a_{k})\in \omega : \sup _{j}\sum_{k=0}^{\infty } \Biggl\vert \sum_{i=k}^{j}(-1)^{i-k} \binom{n-m }{i-k}\binom{n+k }{k} a_{i} \Biggr\vert ^{p*}< \infty \Biggr\} , \end{aligned}

and

\begin{aligned}& \begin{aligned} P_{4}&:= \Biggl\{ a=(a_{k})\in \omega :\\ &\quad \lim _{j\rightarrow \infty } \sum_{k=0}^{\infty } \Biggl\vert \sum_{i=k}^{j}(-1)^{i-k} \binom{n-m }{i-k}\binom{n+k }{k} a_{i} \Biggr\vert =\sum _{k=0}^{\infty } \Biggl\vert \sum _{i=k}^{\infty }(-1)^{i-k}\binom{n-m }{i-k} \binom{n+k }{k} a_{i} \Biggr\vert \Biggr\} . \end{aligned} \end{aligned}

Then $$[\ell _{1}(\Phi ^{n,m})]^{\beta }=P_{1}\cap P_{2}$$, $$[\ell _{p}(\Phi ^{n,m})]^{\beta }=P_{1}\cap P_{3}$$ $$(1< p<\infty )$$ and $$[\ell _{\infty }(\Phi ^{n,m})]^{\beta }=P_{1}\cap P_{4}$$ hold.

### Proof

$$a=(a_{k})\in [\ell _{1}(\Phi ^{n,m})]^{\beta }$$ if and only if the series $$\sum_{k=0}^{\infty }a_{k}u_{k}$$ is convergent for all $$u=(u_{k})\in \ell _{1}(\Phi ^{n,m})$$. From the equality

\begin{aligned} \sum_{k=0}^{j}a_{k}u_{k} &=\sum_{k=0}^{j}a_{k} \Biggl( \sum_{i=0}^{k}(-1)^{k-i} \binom{n-m }{k-i}\binom{n+i }{i}v_{i} \Biggr) \\ &=\sum_{k=0}^{j} \Biggl(\sum _{i=k}^{j}(-1)^{i-k}\binom{n-m }{i-k} \binom{n+k }{k} a_{i} \Biggr)v_{k}, \end{aligned}

it follows that $$a=(a_{k})\in [\ell _{1}(\Phi ^{n,m})]^{\beta }$$ if and only if the matrix $$P=(p_{jk})$$ is in $$(\ell _{1},c)$$, where

\begin{aligned}& p_{jk}=\textstyle\begin{cases} \sum_{i=k}^{j}(-1)^{i-k}\binom{n-m }{i-k}\binom{n+k }{k} a_{i} ,& 0 \leq k\leq j, \\ 0 ,& \text{otherwise}. \end{cases}\displaystyle \end{aligned}

Hence, by part $$(iv)$$ of Lemma 3.4 we conclude that

\begin{aligned}& \lim_{j\rightarrow \infty }\sum_{i=k}^{n}(-1)^{i-k} \binom{n-m }{i-k}\binom{n+k }{k} a_{i} \text{ exists for each }k\in \mathbb{N} \end{aligned}

and

\begin{aligned}& \sup_{jk} \Biggl\vert \sum_{i=k}^{j}(-1)^{i-k} \binom{n-m }{i-k}\binom{n+k }{k} a_{i} \Biggr\vert < \infty , \end{aligned}

which means $$a=(a_{k})\in P_{1}\cap P_{2}$$ and so we have $$[\ell _{1}(\Phi ^{n,m})]^{\beta }=P_{1}\cap P_{2}$$. The other cases can be proved similarly. □

The γ-dual of a sequence space U consists of all sequences $$a=(a_{k})\in \omega$$ such that $$(\sum_{k=1}^{n} a_{k} u_{k} )\in \ell _{\infty }$$ for all $$u=(u_{k})\in U$$.

### Theorem 3.7

The γ-duals of the spaces $$\ell _{1}(\Phi ^{n,m})$$, $$\ell _{p}(\Phi ^{n,m})$$ $$(1< p<\infty )$$ and $$\ell _{\infty }(\Phi ^{n,m})$$ are as follows:

\begin{aligned}& \bigl[\ell_{1}(\Phi^{n,m})\bigr]^{\gamma }:= \Biggl\{ a=(a_{k})\in \omega : \sup _{jk} \Biggl\vert \sum_{i=k}^{j}(-1)^{i-k} \binom{n-m }{i-k}\binom{n+k }{k} a_{i} \Biggr\vert < \infty \Biggr\} , \\& \bigl[\ell_{p}(\Phi^{n,m})\bigr]^{\gamma }:= \Biggl\{ a=(a_{k})\in \omega : \sup _{j}\sum_{k=0}^{\infty } \Biggl\vert \sum_{i=k}^{j}(-1)^{i-k} \binom{n-m }{i-k}\binom{n+k }{k} a_{i} \Biggr\vert ^{p*}< \infty \Biggr\} , \end{aligned}

and

\begin{aligned}& \bigl[\ell _{\infty }(\Phi ^{n,m})\bigr] ^{\gamma }:= \Biggl\{ a=(a_{k})\in \omega : \sup _{j}\sum_{k=0}^{\infty } \Biggl\vert \sum_{i=k}^{j}(-1)^{i-k} \binom{n-m }{i-k}\binom{n+k }{k} a_{i} \Biggr\vert < \infty \Biggr\} . \end{aligned}

### Proof

By using the same technique as in the proof of Theorem 3.6, we obtain the gamma duals. □

### Fractional difference spaces

The sequence spaces $$\ell _{p}(\Psi ^{n,m})$$ $$(1< p<\infty )$$ and $$\ell _{\infty }(\Psi ^{n,m})$$ are introduced similarly by

\begin{aligned}& \ell _{p}\bigl(\Psi ^{n,m}\bigr)= \Biggl\{ u=(u_{j})\in \omega :\sum_{j=0}^{\infty } \Biggl\vert \frac{1}{\binom{n+j }{j}}\sum_{k=0}^{j}(-1)^{j-k} \binom{m-n }{j-k} u_{k} \Biggr\vert ^{p}< \infty \Biggr\} \end{aligned}

and

\begin{aligned}& \ell _{\infty }\bigl(\Psi ^{n,m}\bigr)= \Biggl\{ u=(u_{j})\in \omega :\sup_{j} \Biggl\vert \frac{1}{\binom{n+j }{j}}\sum_{k=0}^{j}(-1)^{j-k} \binom{m-n }{j-k}u_{k} \Biggr\vert < \infty \Biggr\} . \end{aligned}

### Theorem 3.8

The spaces $$\ell _{p}(\Psi ^{n,m})$$ and $$\ell _{\infty }(\Psi ^{n,m})$$ are Banach spaces with the norms

\begin{aligned}& \Vert u \Vert _{\ell _{p}(\Psi ^{n,m})}= \Biggl(\sum_{j=0}^{\infty } \Biggl\vert \frac{1}{\binom{n+j }{j}}\sum_{k=0}^{j}(-1)^{j-k} \binom{m-n }{j-k} u_{k} \Biggr\vert ^{p} \Biggr)^{1/p} \end{aligned}

and

\begin{aligned}& \Vert u \Vert _{\ell _{\infty }(\Psi ^{n,m})}=\sup_{j} \Biggl\vert \frac{1}{\binom{n+j }{j}}\sum_{k=0}^{j}(-1)^{j-k} \binom{m-n}{j-k}u_{k} \Biggr\vert , \end{aligned}

respectively.

### Proof

We omit the proof, which is a routine verification. □

### Theorem 3.9

Define the sequence $$(d^{(k)})=(d^{(k)}_{j})$$ for each $$k\in \mathbb{N}$$ by

$$\bigl(d^{(k)}\bigr)_{j}=\textstyle\begin{cases} \binom{m-n+j-k-1 }{j-k}\binom{n+k }{k} ,& j\geq k, \\ 0 , & j< k, \end{cases}\displaystyle \quad (j\in \mathbb{N}_{0}).$$

Then the sequence $$(d^{(k)})$$ is a basis for the space $$\ell _{p}(\Psi ^{n,m})$$, and each $$u \in \ell _{p}(\Psi ^{n,m})$$ has a unique representation of the form $$u=\sum_{k}(\Psi ^{n,m} u)_{k}d^{(k)}$$.

### Theorem 3.10

The α-duals of the spaces $$\ell _{1}(\Psi ^{n,m})$$, $$\ell _{p}(\Psi ^{n,m})$$ $$(1< p<\infty )$$ and $$\ell _{\infty }(\Psi ^{n,m})$$ are as follows:

\begin{aligned}& \bigl[\ell_{1}(\Psi^{n,m})\bigr]^{\alpha }:= \Biggl\{ a=(a_{j})\in \omega : \sup _{k}\sum_{j=0}^{\infty } \biggl\vert \binom{m-n+j-k-1 }{j-k}\binom{n+k }{k}a_{j} \biggr\vert < \infty \Biggr\} , \\& \bigl[\ell_{p}(\Psi^{n,m})\bigr]^{\alpha }:= \Biggl\{ a=(a_{j})\in \omega : \sum _{k=0}^{\infty } \Biggl(\sum_{j=0}^{\infty } \biggl\vert \binom{m-n+j-k-1 }{j-k}\binom{n+k }{k}a_{j} \biggr\vert \Biggr)^{p*}< \infty \Biggr\} , \end{aligned}

and

\begin{aligned}& \bigl[\ell _{\infty }(\Psi ^{n,m})\bigr] ^{\alpha }:= \Biggl\{ a=(a_{j})\in \omega : \sup _{K\in \mathcal{N}}\sum_{j=0}^{\infty } \biggl\vert \sum_{k\in K}\binom{m-n+j-k-1 }{j-k} \binom{n+k }{k}a_{j} \biggr\vert < \infty \Biggr\} . \end{aligned}

### Proof

The proof is similar to the proof of Theorem 3.5. □

### Theorem 3.11

Let us define the following sets:

\begin{aligned}& Q_{1}:= \Biggl\{ a=(a_{k})\in \omega : \lim _{j\rightarrow \infty } \sum_{i=k}^{j} \binom{m-n+i-k-1 }{i-k}\binom{n+k }{k} a_{i} \textit{ exists for each } k \in \mathbb{N} \Biggr\} , \\& Q_{2}:= \Biggl\{ a=(a_{k})\in \omega : \sup _{jk} \Biggl\vert \sum_{i=k}^{j} \binom{m-n+i-k-1 }{i-k}\binom{n+k }{k} a_{i} \Biggr\vert < \infty \Biggr\} , \\& Q_{3}:= \Biggl\{ a=(a_{k})\in \omega : \sup _{j}\sum_{k=0}^{\infty } \Biggl\vert \sum_{i=k}^{j} \binom{m-n+i-k-1 }{i-k}\binom{n+k }{k} a_{i} \Biggr\vert ^{p*}< \infty \Biggr\} , \end{aligned}

and

\begin{aligned}& \begin{aligned} Q_{4}&:= \Biggl\{ a=(a_{k})\in \omega : \\ &\quad \lim _{j\rightarrow \infty }\sum_{k=0}^{\infty } \Biggl\vert \sum_{i=k}^{j} \binom{m-n+i-k-1 }{i-k} \binom{n+k }{k} a_{i} \Biggr\vert \\ &\qquad =\sum _{k=0}^{\infty } \Biggl\vert \sum _{i=k}^{\infty }\binom{m-n+i-k-1 }{i-k}\binom{n+k }{k} a_{i} \Biggr\vert \Biggr\} . \end{aligned} \end{aligned}

Then $$[\ell _{1}(\Psi ^{n,m})]^{\beta }=Q_{1}\cap Q_{2}$$, $$[\ell _{p}(\Psi ^{n,m})]^{\beta }=Q_{1}\cap Q_{3}$$ $$(1< p<\infty )$$ and $$[\ell _{\infty }(\Psi ^{n,m})]^{\beta }=Q_{1}\cap Q_{4}$$ hold.

### Proof

The proof is similar to the proof of Theorem 3.6. □

### Theorem 3.12

The γ-duals of the spaces $$\ell _{1}(\Psi ^{n,m})$$, $$\ell _{p}(\Psi ^{n,m})$$ $$(1< p<\infty )$$ and $$\ell _{\infty }(\Psi ^{n,m})$$ are as follows:

\begin{aligned}& \bigl[\ell_{1}(\Psi^{n,m})\bigr]^{\gamma }:= \Biggl\{ a=(a_{k})\in \omega : \sup _{jk} \Biggl\vert \sum_{i=k}^{j} \binom{m-n+i-k-1 }{i-k}\binom{n+k }{k} a_{i} \Biggr\vert < \infty \Biggr\} , \\& \bigl[\ell_{p}(\Psi^{n,m})\bigr]^{\gamma }:= \Biggl\{ a=(a_{k})\in \omega : \sup _{j}\sum_{k=0}^{\infty } \Biggl\vert \sum_{i=k}^{j} \binom{m-n+i-k-1 }{i-k} \binom{n+k }{k} a_{i} \Biggr\vert ^{p*}< \infty \Biggr\} , \end{aligned}

and

\begin{aligned}& \bigl[\ell _{\infty }(\Psi ^{n,m})\bigr] ^{\gamma }:= \Biggl\{ a=(a_{k})\in \omega : \sup _{j}\sum_{k=0}^{\infty } \Biggl\vert \sum_{i=k}^{j}\binom{m-n+i-k-1 }{i-k} \binom{n+k }{k} a_{i} \Biggr\vert < \infty \Biggr\} . \end{aligned}

### Proof

The proof is similar to the proof of Theorem 3.7. □

## Availability of data and materials

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