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# A study on Copson operator and its associated sequence space

## Abstract

In this research, we investigate two types of Copson matrices, the generalized Copson matrix and the Copson matrix of order n, and their associated sequence spaces generated by these matrices. We also investigate the topological properties, inclusions, and dual spaces of these new Banach spaces as well as compute the norm of Copson operators on the well-known matrix domains such as Hilbert and difference sequence spaces. Moreover, in a reverse manner, we investigate the norm of well-known operators on the Copson matrix domains generated with Copson matrices. Through this study we introduce several new inequalities, inclusions, and factorizations for well-known operators.

## Introduction

Let $$p\ge1$$ and ω denote the set of all real-valued sequences. The space $$\ell_{p}$$ is the set of all real sequences $$x=(x_{k})\in\omega$$ such that

$$\Vert x \Vert _{\ell_{p}}= \Biggl(\sum_{k=0}^{\infty} \vert x_{k} \vert ^{p} \Biggr)^{1/p}< \infty.$$

There are two different types of Copson matrices, the generalized Copson matrix and the Copson matrix of order n, we indicate them by the notations $$C^{N}$$ and $$C^{n}$$, respectively. In the sequel, we introduce these matrices and their differences as well as their associated matrix domains.

We say that $$A= (a_{n,k})$$ is a quasi-summability matrix if it is an upper-triangular matrix, i.e., $$a_{n,k}= 0$$ for $$n < k$$ and $$\sum_{n=0}^{k} a_{n,k}= 1$$ for all k.

Copson matrix. The Copson matrix is a quasi-summability matrix which is defined by

\begin{aligned} c_{j,k}=\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{1}{k+1}, & 0\le j \le k,\\ 0 & \text{otherwise} \end{array}\displaystyle \right . \end{aligned}

for all $$j,k \in\mathbb{N}_{0}$$, that is,

\begin{aligned} C= \begin{pmatrix} 1 & 1/2 & 1/3 & \cdots \\ 0 & 1/2 & 1/3 & \cdots \\ 0 & 0 & 1/3 & \cdots \\ \vdots& \vdots& \vdots& \ddots \end{pmatrix} , \end{aligned}

and has the $$\ell_{p}$$-norm $$\|C\|_{\ell_{p}}=p$$. This matrix is the transpose of the well-known Cesàro matrix.

Generalized Copson matrix. Suppose that $$N\ge1$$ is a real number. The generalized Copson matrix $$C^{N}=(c^{N}_{j,k})$$ is defined by

\begin{aligned} c^{N}_{j,k}=\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{1}{k+N}, & 0\le j \le k ,\\ 0 & \text{otherwise} \end{array}\displaystyle \right . \end{aligned}

for all $$j,k \in\mathbb{N}_{0}$$, has the $$\ell_{p}$$-norm $$\|C^{N}\|_{\ell _{p}}=p$$ (, Lemma 2.3), and the matrix representation

\begin{aligned} c^{N}= \begin{pmatrix} \frac{1}{N} &\frac{1}{1+N} & \frac{1}{2+N} & \cdots \\ 0 & \frac{1}{1+N}& \frac{1}{2+N} & \cdots \\ 0 & 0 & \frac{1}{2+N} & \cdots \\ \vdots& \vdots& \vdots& \ddots \end{pmatrix} . \end{aligned}

Note that $$C^{1}$$ is the Copson matrix C. For more examples,

\begin{aligned} C^{2}= \begin{pmatrix} 1/2 & 1/3 & 1/4 & \cdots \\ 0 & 1/3 & 1/4 & \cdots \\ 0 & 0 & 1/4 & \cdots \\ \vdots& \vdots& \vdots& \ddots \end{pmatrix} \quad \text{and}\quad C^{3}= \begin{pmatrix} 1/3 & 1/4 & 1/5 & \cdots \\ 0 & 1/4 & 1/5 & \cdots \\ 0 & 0 & 1/5 & \cdots \\ \vdots& \vdots& \vdots& \ddots \end{pmatrix} . \end{aligned}

Also note that, for $$N>1$$, the generalized Copson matrix is not a quasi-summability matrix.

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. The matrix domain associated with this matrix is the set $$\{ x = ( x_{k} )\in\omega: C^{N}x\in\ell_{p} \}$$, or

\begin{aligned} C^{N}_{p} =& \Biggl\{ x = ( x_{k} )\in \omega: \sum_{j=0}^{\infty}\Biggl\vert \sum_{k=j}^{\infty}\frac{x_{k}}{k+N} \Biggr\vert ^{p} < \infty \Biggr\} , \end{aligned}

which has the following norm:

\begin{aligned} \Vert x \Vert _{C^{N}_{p}} = \Biggl(\sum_{j=0}^{\infty}\Biggl\vert \sum_{k=j}^{\infty}\frac{x_{k}}{k+N} \Biggr\vert ^{p} \Biggr)^{\frac{1}{p}}. \end{aligned}

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 [25, 7, 9, 10, 12, 13, 16, 17, 21, 2325] and textbooks  and .

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

\begin{aligned} h_{j,k}=\left \{ \textstyle\begin{array}{l@{\quad}l} \int_{0}^{1} {j\choose k}\theta^{k}(1-\theta)^{j-k}\,d\mu(\theta) ,& j\ge k ,\\ 0, & j< k, \end{array}\displaystyle \right . \end{aligned}

where μ is a probability measure on $$[0,1]$$.

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}
(1.1)

In order to define and know the Copson matrix details, we need the following theorem also known as Hellinger–Toeplitz theorem.

### Theorem 1.1

(, Proposition 7.2)

Suppose that$$1 < p,q < \infty$$. A matrix A maps$$\ell_{p}$$into$$\ell _{q}$$if and only if the transposed matrix$$A^{t}$$maps$$\ell_{q^{*}}$$into$$\ell_{p^{*}}$$. We then have$$\|A\|_{\ell_{p},\ell_{q}}=\|A^{t}\|_{\ell _{q^{*}}, \ell_{p^{*}}}$$, where$$p^{*}$$is the conjugate ofp, i.e., $$\frac{1}{p}+\frac{1}{p^{*}}=1$$.

For nonnegative real number n, and by choosing $$d\mu(\theta )=n(1-\theta)^{n-1}\,d\theta$$ in the definition of Hausdorff matrix, we gain the Cesàro matrix of order n, which, according to Hardy’s formula, has the $$\ell_{p}$$-norm

\begin{aligned} \frac{\varGamma(n+1)\varGamma(1/p^{*})}{\varGamma(n+1/p^{*})}. \end{aligned}
(1.2)

Now, the Copson matrix of order n, $$C^{n}=(c^{n}_{j,k})$$, which is defined as the transpose of Cesàro matrix of order n, has the entries

\begin{aligned} c^{n}_{j,k}=\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{{n+k-j-1 \choose k-j}}{{n+k \choose k}}, & 0 \le j\le k,\\ 0 & \text{otherwise}, \end{array}\displaystyle \right . \end{aligned}
(1.3)

and, according to Hellinger–Toeplitz theorem, the $$\ell_{p}$$-norm

\begin{aligned} \bigl\Vert C^{n} \bigr\Vert _{\ell_{p}}= \frac{\varGamma(n+1)\varGamma(1/p)}{\varGamma(n+1/p)}. \end{aligned}
(1.4)

Note that $$C^{0}=I$$, where I is the identity matrix and $$C^{1}=C$$ is the well-known Copson matrix. For more examples,

\begin{aligned} C^{2}= \begin{pmatrix} 1 & 2/3 & 3/6 & \cdots\\ 0 & 1/3 & 2/6 & \cdots\\ 0 & 0 & 1/6 & \cdots\\ \vdots& \vdots& \vdots& \ddots \end{pmatrix} \quad \text{and} \quad C^{3}= \begin{pmatrix} 1 & 3/4 & 6/10 &\cdots\\ 0 & 1/4 & 3/10 & \cdots\\ 0 & 0 & 1/10 & \cdots\\ \vdots& \vdots& \vdots& \ddots \end{pmatrix} . \end{aligned}

We emphasize that the Copson matrix of order n is a quasi-summability matrix, while the generalized Copson matrix is not.

The Copson matrix domain $$C^{n}_{p}$$ is the set of all sequences whose $$C^{n}$$-transforms are in the space $$\ell_{p}$$; that is,

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

which is a Banach space with the norm

\begin{aligned} \Vert x \Vert _{C^{n}_{p}}= \Biggl(\sum_{j=0}^{\infty}\Biggl\vert \sum_{k=j}^{\infty}\frac{{n+k-j-1 \choose k-j}}{{n+k \choose k}}x_{k} \Biggr\vert ^{p} \Biggr)^{1/p}. \end{aligned}

Throughout this research, we use the notations $$\|\cdot\|_{A_{p},\ell_{p}}$$, $$\|\cdot\|_{\ell_{p},A_{p}}$$ and $$\|\cdot\|_{A_{p},B_{p}}$$ for the norm of operators from the matrix domain $$A_{p}$$ into sequence space $$\ell_{p}$$, from $$\ell_{p}$$ into the matrix domain $$A_{p}$$, and from the matrix domain $$A_{p}$$ into the matrix domain $$B_{p}$$, respectively.

Motivation. Although a lot of papers have been published on Cesàro matrix, Cesàro sequence spaces, and Cesàro function spaces and many mathematicians, like the pioneers Jagers, Bennett, Luxemburg, and Zaanen [1, 6, 14, 19], worked on that and later the work was continued by several mathematicians, there exist limited studies on the transpose of this matrix. Also, while we can extract many results of Copson matrix just by transposing the Cesàro matrix and applying some theorems like Hellinger–Toeplitz theorem, there are some special areas that work only for Copson matrices. We state one of these differences in this paper by computing the norm of Copson operators between difference sequence spaces, while we cannot do it for the Cesàro operator. Other examples introduce some topological properties, inequalities, and inclusions which are only applicable on Copson matrices.

## The Copson Banach spaces $$C^{n}_{p}$$ and $$C^{n}_{\infty}$$

In this section, the sequence spaces $$C^{n}_{p}$$ ($$1\leq p<\infty$$) and $$C^{n}_{\infty}$$ are introduced by using the Copson matrix of order n, and the inclusions, basis, and duals of this matrix domain will investigated.

### Lemma 2.1

The Copson matrix of ordern, $$C^{n}$$, is invertible and its inverse$$C^{-n}=(c^{-n}_{j,k})$$is defined by

\begin{aligned} c^{-n}_{j,k}=\left \{ \textstyle\begin{array}{l@{\quad}l} (-1)^{(k-j)}{n\choose k-j}{n+j \choose j}, & j\le k \le j+n, \\ 0 & \textit{otherwise}. \end{array}\displaystyle \right . \end{aligned}

### Proof

Let us recall the forward difference matrix of order n, $$\Delta ^{n}=(\delta^{n}_{j,k})$$, which is a lower triangle matrix with entries

\begin{aligned} \delta^{n}_{j,k}=\left \{ \textstyle\begin{array}{l@{\quad}l} (-1)^{(k-j)}{n\choose k-j}, & j\le k \le j+n, \\ 0 & \text{otherwise}. \end{array}\displaystyle \right . \end{aligned}

This matrix has the inverse $$\Delta^{-n}=(\delta^{-n}_{j,k})$$ with the following entries:

\begin{aligned} \delta^{-n}_{j,k}=\left \{ \textstyle\begin{array}{l@{\quad}l} {n+k-j-1\choose k-j} , & j\le k, \\ 0 & \text{otherwise}. \end{array}\displaystyle \right . \end{aligned}

From relation (1.3), one can see that the Copson matrix of order n and its inverse can be rewritten based on the forward difference operator and its inverse. For $$j\le k$$, we have

\begin{aligned} c^{n}_{j,k}=\frac{{n+k-j-1 \choose k-j}}{{n+k \choose k}} =\frac{\delta^{-n}_{j,k}}{{n+k \choose k}} \quad \text{and} \quad c^{-n}_{j,k}=\delta^{n}_{j,k} {n+j \choose j}. \end{aligned}

Now, by a simple calculation, we deduce that

\begin{aligned} \bigl(C^{-n} C^{n}\bigr)_{i,k}= \frac{{n+i \choose i}}{{n+k \choose k}}\sum_{j} \delta^{n}_{i,j}\delta^{-n}_{j,k}= \frac{{n+i \choose i}}{{n+k \choose k}}\bigl(\Delta^{n}\Delta^{-n} \bigr)_{i,k}=\frac{{n+i \choose i}}{{n+k \choose k}}I_{i,k}, \end{aligned}

which completes the proof. □

Now, we introduce 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 \sum _{k=j}^{\infty}\frac{{n+k-j-1 \choose k-j}}{{n+k \choose k}}x_{k} \Biggr\vert ^{p}< \infty \Biggr\} \end{aligned}

and

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

### Theorem 2.2

The spaces$$C^{n}_{p}$$and$$C^{n}_{\infty}$$are Banach spaces with the norms

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

and

\begin{aligned} \Vert x \Vert _{C^{n}_{\infty}}=\sup_{j\in\mathbb{N}_{0}} \Biggl\vert \sum_{k=j}^{\infty}\frac{{n+k-j-1 \choose k-j}}{{n+k \choose k}}x_{k} \Biggr\vert , \end{aligned}

respectively.

### Proof

We omit the proof which is a routine verification. □

### Theorem 2.3

The spaces$$C^{n}_{p}$$and$$C^{n}_{\infty}$$are linearly isomorphic to$$\ell _{p}$$and$$\ell_{\infty}$$, respectively.

### Proof

We only prove that $$C^{n}_{p}$$ is linearly isomorphic to $$\ell_{p}$$. Since $$C^{n}$$ is invertible, hence the map $$T:C^{n}_{p}\rightarrow\ell_{p}$$ as $$Tx=C^{n} x$$ for any $$x\in C^{n}_{p}$$ is bijective, which proves the isomorphism. □

### Theorem 2.4

The inclusion$$C^{n}_{p} \subset C^{n}_{q}$$is strict, where$$1\leq p< q<\infty$$.

### Proof

Let $$x\in C^{n}_{p}$$. Then we have $$C^{n} x\in\ell_{p}$$. Since the inclusion $$\ell_{p}\subset\ell_{q}$$ holds for $$1\leq p< q<\infty$$, we have $$C^{n} x\in\ell_{q}$$, which implies that $$x\in C^{n}_{q}$$. Hence, we conclude that the inclusion $$C^{n}_{p} \subset C^{n}_{q}$$ holds. Now, since the inclusion $$\ell_{p}\subset\ell_{q}$$ is strict, we can choose $$y=(y_{j})\in\ell _{q}\setminus\ell_{p}$$. By defining $$x=C^{-n}y$$, we have $$C^{n} x=y$$, which results in $$C^{n} x\in\ell_{q}\setminus\ell_{p}$$. Hence, we conclude that $$x\in C^{n}_{q} \setminus C^{n}_{p}$$, and so the inclusion $$C^{n}_{p} \subset C^{n}_{q}$$ is strict. □

### Theorem 2.5

The inclusion$$C^{n}_{p} \subset C^{n}_{\infty}$$is strict, where$$1\leq p<\infty$$.

### Proof

Choose any $$x\in C^{n}_{p}$$. Then we have $$C^{n} x\in\ell_{p}$$. Since the inclusion $$\ell_{p}\subset\ell_{\infty}$$ holds for $$1\leq p<\infty$$, we have $$C^{n} x\in\ell_{\infty}$$. This implies that $$x\in C^{n}_{\infty}$$. Hence, we conclude that the inclusion $$C^{n}_{p} \subset C^{n}_{\infty}$$ holds. Similar to the proof of the previous theorem, we can choose x such that $$C^{n} x=((-1)^{j})\in\ell_{\infty}\setminus\ell_{p}$$, and consequently it results in $$x\in C^{n}_{\infty}\setminus C^{n}_{p}$$. Therefore, the inclusion $$C^{n}_{p} \subset C^{n}_{\infty}$$ is strict. □

It is known from Theorem 2.3 of Jarrah and Malkowsky  that if T is a triangle then the domain $$\lambda_{T}$$ of T in a normed sequence space λ has a basis if and only if λ has a basis. As a direct consequence of this fact, we have the following.

### Corollary 2.6

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

$$\bigl(h^{(k)}\bigr)_{j}=\left \{ \textstyle\begin{array} [c]{l@{\quad}l}(-1)^{k-j}{n+j \choose j}{n\choose k-j} , & k\geq j, \\ 0 , & k< j \end{array}\displaystyle \right .\quad (j\in\mathbb{N}_{0}).$$
(2.1)

Then the sequence$$(h^{(k)})$$is a basis for the space$$C^{n}_{p}$$and every sequence$$x \in C^{n}_{p}$$has a unique representation of the form$$x=\sum_{k}(C^{n} x)_{k}h^{(k)}$$.

The following lemma is essential to determine the dual spaces. Throughout the paper, $$\mathcal{N}$$ is the collection of all finite subsets of $$\mathbb{N}$$.

### Lemma 2.7

()

The following statements hold:

1. (i)

$$T=(t_{j,k})\in(\ell_{1},\ell_{1})$$if and only if

$$\sup_{k}\sum_{j=0}^{\infty} \vert t_{j,k} \vert < \infty.$$
2. (ii)

$$T=(t_{j,k})\in(\ell_{p},\ell_{1})$$if and only if

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

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

3. (iii)

$$T=(t_{j,k})\in(\ell_{\infty},\ell_{1})$$if and only if

$$\sup_{K\in\mathcal{N}}\sum_{j=0}^{\infty}\biggl\vert \sum_{k\in K}t_{j,k} \biggr\vert < \infty.$$
4. (iv)

$$T=(t_{j,k})\in(\ell_{1},c)$$if and only if

$$\lim_{j\rightarrow\infty}t_{j,k} \quad\textit{exists for each } k\in \mathbb{N}$$
(2.2)

and

$$\sup_{j,k} \vert t_{j,k} \vert < \infty.$$
(2.3)
5. (v)

$$T=(t_{j,k})\in(\ell_{p},c)$$if and only if (2.2) holds and

$$\sup_{j}\sum _{k=0}^{\infty} \vert t_{j,k} \vert ^{p*}< \infty,$$
(2.4)

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

6. (vi)

$$T=(t_{j,k})\in(\ell_{\infty},c)$$if and only if (2.2) holds and

$$\lim_{j\rightarrow\infty}\sum _{k=0}^{\infty} \vert t_{j,k} \vert = \sum_{k=0}^{\infty}\Bigl\vert \lim _{j\rightarrow\infty}t_{j,k} \Bigr\vert .$$
7. (vii)

$$T=(t_{j,k})\in(\ell_{1},\ell_{\infty})$$if and only if (2.3) holds.

8. (viii)

$$T=(t_{j,k})\in(\ell_{p},\ell_{\infty})$$if and only if (2.4) holds, where$$1< p<\infty$$.

9. (ix)

$$T=(t_{j,k})\in(\ell_{\infty},\ell_{\infty})$$if and only if

$$\sup_{j}\sum_{k=0}^{\infty} \vert t_{j,k} \vert < \infty.$$

### Definition

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

\begin{aligned} \begin{gathered} 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{gathered} \end{aligned}

respectively. In the following, we find the Köthe dual of the Copson sequence space.

### Theorem 2.8

Theα-duals of the spaces$$C^{n}_{1}$$, $$C^{n}_{p}$$ ($$1< p<\infty$$), and$$C^{n}_{\infty}$$are as follows:

$$\begin{gathered} \bigl(C^{n}_{1}\bigr)^{\alpha}= \Biggl\{ b=(b_{j})\in\omega:\sup_{k}\sum _{j=0}^{\infty}\biggl\vert (-1)^{k-j} {n+j\choose j}{n\choose k-j}b_{j} \biggr\vert < \infty \Biggr\} , \\ \bigl(C^{n}_{p}\bigr)^{\alpha}= \Biggl\{ b=(b_{j})\in\omega:\sum_{k=0}^{\infty}\Biggl(\sum_{j=0}^{\infty}\biggl\vert (-1)^{k-j}{n+j\choose j}{n\choose k-j}b_{j} \biggr\vert \Biggr)^{p*}< \infty \Biggr\} ,\end{gathered}$$

and

$$\bigl(C^{n}_{\infty}\bigr)^{\alpha}= \Biggl\{ b=(b_{j})\in\omega:\sup_{K\in\mathcal {N}}\sum _{j=0}^{\infty}\biggl\vert \sum _{k\in K}(-1)^{k-j}{n+j\choose j} {n\choose k-j}b_{j} \biggr\vert < \infty \Biggr\} .$$

### Proof

Let $$b=(b_{j})\in\omega$$. Consider the matrix $$A=(a_{j,k})$$ defined by

$$a_{j,k}=\left \{ \textstyle\begin{array} {l@{\quad}l}(-1)^{k-j}{n+j\choose j}{n\choose k-j}b_{j} , & 0\leq j\leq k, \\ 0 , & \text{otherwise.} \end{array}\displaystyle \right .$$

Given any $$x=(x_{j})\in C^{n}_{p}$$ ($$1\leq p\leq\infty$$), we have $$b_{j}x_{j}=(Ay)_{j}$$ for all $$j\in\mathbb{N}$$. This implies that $$bx\in \ell_{1}$$ with $$x\in C^{n}_{p}$$ if and only if $$Ay\in\ell_{1}$$ with $$y\in \ell_{p}$$. Hence, we conclude that $$b\in(C^{n}_{p})^{\alpha}$$ if and only if $$A\in(\ell_{p},\ell_{1})$$. This completes the proof by Lemma 2.7. □

### Theorem 2.9

Let us define the following sets:

\begin{aligned} \begin{gathered} B_{1}= \Biggl\{ b=(b_{k})\in\omega:\lim _{j\rightarrow\infty}\sum_{k=i}^{j}(-1)^{k-i} {n+i\choose i}{n\choose k-i} b_{i} \textit{ exists for each } k\in\mathbb{N} \Biggr\} , \\ B_{2}= \Biggl\{ b=(b_{k})\in\omega:\sup _{j,k} \Biggl\vert \sum_{k=i}^{j}(-1)^{k-i} {n+i\choose i}{n\choose k-i} b_{i} \Biggr\vert < \infty \Biggr\} , \\ B_{3}= \Biggl\{ b=(b_{k})\in\omega:\sup _{j}\sum_{k=0}^{\infty}\Biggl\vert \sum_{k=i}^{j}(-1)^{k-i} {n+i\choose i}{n\choose k-i} b_{i} \Biggr\vert ^{p*}< \infty \Biggr\} ,\end{gathered} \end{aligned}

and

\begin{aligned} \begin{aligned} B_{4}&= \Biggl\{ b=(b_{k})\in\omega:\\&\quad\lim _{j\rightarrow\infty}\sum_{k=0}^{\infty}\Biggl\vert \sum_{k=i}^{j}(-1)^{k-i} {n+i\choose i}{n\choose k-i} b_{i} \Biggr\vert =\sum_{k=0}^{\infty}\Biggl\vert \sum_{k=i}^{\infty}(-1)^{k-i} {n+i\choose i}{n\choose k-i} b_{i} \Biggr\vert \Biggr\} .\end{aligned} \end{aligned}

Then we have$$(C^{n}_{1})^{\beta}=B_{1}\cap B_{2}$$, $$(C^{n}_{p})^{\beta}=B_{1}\cap B_{3}$$ ($$1< p<\infty$$), and$$(C^{n}_{\infty})^{\beta}=B_{1}\cap B_{4}$$.

### Proof

$$b=(b_{k})\in(C^{n}_{1})^{\beta}$$ if and only if the series $$\sum_{k=0}^{\infty}b_{k}x_{k}$$ is convergent for all $$x=(x_{k})\in C^{n}_{1}$$. The equality

\begin{aligned} \sum_{k=0}^{j}b_{k}x_{k} &=\sum_{k=0}^{j}b_{k} \Biggl(\sum_{i=0}^{k}(-1)^{k-i} {n+i\choose i}{n\choose k-i}y_{i} \Biggr) \\ &=\sum_{k=0}^{j} \Biggl(\sum _{i=k}^{j}(-1)^{k-i} {n+i\choose i}{n\choose k-i} b_{i} \Biggr)y_{k} \end{aligned}

implies that $$b=(b_{k})\in(C^{n}_{1})^{\beta}$$ if and only if the matrix $$B=(b_{j,k})$$ is in the class $$(\ell_{1},c)$$, where

$$b_{j,k}=\left \{ \textstyle\begin{array} {l@{\quad}l}\sum_{i=k}^{j}(-1)^{k-i}{n+i\choose i}{n\choose k-i} b_{i} , & 0\leq j\leq k,\\ 0 , & \text{otherwise}. \end{array}\displaystyle \right .$$

Hence, we deduce from Lemma 2.7 that

$$\lim_{j\rightarrow\infty}\sum_{i=k}^{n}(-1)^{k-i} {n+i\choose i}{n\choose k-i} b_{i} \quad\text{exists for each }k\in\mathbb{N}$$

and

$$\sup_{j,k} \Biggl\vert \sum_{i=k}^{j}(-1)^{k-i} {n+i\choose i}{n\choose k-i} b_{i} \Biggr\vert < \infty,$$

which means $$b=(b_{k})\in B_{1}\cap B_{2}$$, and so we have $$(C^{n}_{1})^{\beta}=B_{1}\cap B_{2}$$. The other results can be proved similarly. □

### Theorem 2.10

Theγ-duals of the spaces$$C^{n}_{1}$$, $$C^{n}_{p}$$ ($$1< p<\infty$$), and$$C^{n}_{\infty}$$are as follows:

$$\begin{gathered} \bigl(C^{n}_{1}\bigr)^{\gamma}= \Biggl\{ b=(b_{k})\in\omega:\sup_{j,k} \Biggl\vert \sum _{i=k}^{j}(-1)^{k-i} {n+i\choose i}{n\choose k-i} b_{i} \Biggr\vert < \infty \Biggr\} , \\ \bigl(C^{n}_{p}\bigr)^{\gamma}= \Biggl\{ b=(b_{k})\in\omega:\sup_{j}\sum _{k=0}^{\infty}\Biggl\vert \sum _{i=k}^{j}(-1)^{k-i} {n+i\choose i}{n\choose k-i} b_{i} \Biggr\vert ^{p*}< \infty \Biggr\} ,\end{gathered}$$

and

$$\bigl(C^{n}_{\infty}\bigr)^{\gamma}= \Biggl\{ b=(b_{k})\in\omega:\sup_{j}\sum _{k=0}^{\infty}\Biggl\vert \sum _{i=k}^{j}(-1)^{k-i} {n+i\choose i}{n\choose k-i} b_{i} \Biggr\vert < \infty \Biggr\} .$$

### Proof

It follows with the same technique as that in the proof of Theorem 2.9. □

## Norm of operators on Copson matrix domain

In this section we intend to compute the norm of well-known operators, such as Hilbert, Hausdorff, and Copson operators, on the Copson matrix domain. In so doing, we need the following lemma.

### Lemma 3.1

LetUbe a bounded operator on$$\ell_{p}$$and$$A_{p}$$and$$B_{p}$$be two matrix domains such that$$A_{p}\simeq\ell_{p}$$.

1. (i)

IfBTis a bounded operator on$$\ell_{p}$$, thenTis a bounded operator from$$\ell_{p}$$into$$B_{p}$$and

\begin{aligned} \Vert T \Vert _{\ell_{p}, B_{p}}= \Vert BT \Vert _{\ell_{p}}. \end{aligned}
2. (ii)

IfThas a factorization of the form$$T=UA$$, thenTis a bounded operator from the matrix domain$$A_{p}$$into$$\ell_{p}$$and

\begin{aligned} \Vert T \Vert _{A_{p}, \ell_{p}}= \Vert U \Vert _{\ell_{p}}. \end{aligned}
3. (iii)

If$$BT=UA$$, thenTis a bounded operator from the matrix domain$$A_{p}$$into$$B_{p}$$and

\begin{aligned} \Vert T \Vert _{A_{p}, B_{p}}= \Vert U \Vert _{\ell_{p}}. \end{aligned}

In particular, if$$AT=UA$$, thenTis a bounded operator from the matrix domain$$A_{p}$$into$$A_{p}$$and

\begin{aligned} \Vert T \Vert _{A_{p}}= \Vert U \Vert _{\ell_{p}}. \end{aligned}

Also, ifTandAcommute, then$$\|T\|_{A_{p}}=\|T\|_{\ell_{p}}$$.

### Proof

(i) For every $$x\in\ell_{p}$$,

\begin{aligned} \Vert T \Vert _{\ell_{p}, B_{p}} =&\sup_{x\in\ell_{p} } \frac{ \Vert T x \Vert _{B_{p}}}{ \Vert x \Vert _{\ell_{p}}}=\sup_{x\in\ell_{p}}\frac{ \Vert BT x \Vert _{\ell_{p}}}{ \Vert x \Vert _{\ell_{p}}}= \Vert BT \Vert _{\ell_{p}}. \end{aligned}

(ii) Since $$A_{p}$$ and $$\ell_{p}$$ are isomorphic, hence

$$\Vert T \Vert _{A_{p}, \ell_{p}}=\sup_{x\in A_{p} } \frac{ \Vert T x \Vert _{\ell_{p}}}{ \Vert x \Vert _{A_{p}}} =\sup_{x\in A_{p}}\frac{ \Vert UA x \Vert _{\ell_{p}}}{ \Vert A x \Vert _{\ell_{p}}} =\sup_{y\in\ell_{p}}\frac{ \Vert U y \Vert _{\ell_{p}}}{ \Vert y \Vert _{\ell_{p}}}= \Vert U \Vert _{\ell_{p}},$$

which gives the desired result. Part (iii) has a similar proof. □

### Norm of Hilbert operator on Copson matrix domain

Recall the definition of the well-known Hilbert matrix $$H=(h_{j,k})$$, which was introduced (1894) by David Hilbert to study a question in approximation theory. For $$j,k=0,1,\dots$$, the Hilbert matrix is defined by

\begin{aligned} h_{j,k}=\frac{1}{j+k+1},\qquad H= \begin{pmatrix} 1 & 1/2 & 1/3 & \cdots \\ 1/2 & 1/3 & 1/4 & \cdots \\ 1/3 & 1/4 & 1/5 & \cdots \\ \vdots& \vdots& \vdots& \ddots \end{pmatrix} , \end{aligned}

which is a bounded operator on $$\ell_{p}$$ with $$\ell_{p}$$-norm $$\|H\|_{\ell_{p}}=\varGamma(1/p)\varGamma(1/p^{*})=\pi\csc(\pi/p)$$.

For a nonnegative integer n, we define the Hilbert matrix of order n, $$H^{n}=(h^{n}_{j,k})$$, by

\begin{aligned} h^{n}_{j,k}=\frac{1}{j+k+n+1} \quad(j,k=0,1,\ldots). \end{aligned}
(3.1)

Note that for $$n=0$$, $$H^{0}=H$$ is the Hilbert matrix. For more examples:

\begin{aligned} H^{1}= \begin{pmatrix} 1/2 & 1/3 & 1/4 & \cdots \\ 1/3 & 1/4 & 1/5 & \cdots \\ 1/4 & 1/5 & 1/6 & \cdots \\ \vdots& \vdots& \vdots& \ddots \end{pmatrix} , \qquad H^{2}= \begin{pmatrix} 1/3 & 1/4 & 1/5 & \cdots \\ 1/4 & 1/5 & 1/6 & \cdots \\ 1/5 & 1/6 & 1/7 & \cdots \\ \vdots& \vdots& \vdots& \ddots \end{pmatrix} . \end{aligned}

For nonnegative integers n, j, and k, let us define the matrix $$B^{n}=(b^{n}_{j,k})$$ by

\begin{aligned} b^{n}_{j,k}=\frac{(j+1)\cdots(j+n)}{(j+k+1)\cdots(j+k+n+1)}. \end{aligned}

Consider that, for $$n=0$$, $$B^{0}=H$$, where H is the Hilbert matrix.

Note that the matrix $$B^{n}$$ has also the representation

\begin{aligned} b^{n}_{j,k}={n+j \choose j}\beta(j+k+1, n+1) \quad(j,k=0,1,\ldots), \end{aligned}

where the β function is

\begin{aligned} \beta(m,n)= \int_{0}^{1} z^{m-1}(1-z)^{n-1}\,dz \quad(m,n=1,2,\ldots). \end{aligned}

For computing the norm of Hilbert operator on the domain of Copson matrix, we need the following lemmas.

### Lemma 3.2

For$$|z|<1$$, we have

\begin{aligned} (1-z)^{-n}=\sum_{j=0}^{\infty}{n+j-1 \choose j}z^{j}. \end{aligned}

### Proof

By differentiating $$n-1$$ times the identity $$(1-z)^{-1}= \sum_{j=0}^{\infty}z^{j}$$, we obtain the result. □

### Lemma 3.3

(, Lemma 2.3)

The Hilbert matrixHand the Hilbert matrix of ordern, $$H^{n}$$, have the following factorizations:

1. (i)

$$H= C^{n}B^{n}$$,

2. (ii)

$$H^{n}= B^{n}C^{n}$$,

3. (iii)

$$HC^{n}= C^{n} H^{n}$$,

4. (iv)

$$B^{n}$$is a bounded operator on$$\ell_{p}$$and

\begin{aligned} \bigl\Vert B^{n} \bigr\Vert _{\ell_{p}}= \frac{\varGamma(n+1/p)\varGamma(1/p^{*})}{\varGamma(n+1)}. \end{aligned}

### Corollary 3.4

Let$$H^{n}$$be the Hilbert operator of ordern. Then

1. (i)

$$H^{n}$$is a bounded operator from$$C^{n}_{p}$$into$$\ell_{p}$$and

\begin{aligned} \bigl\Vert H^{n} \bigr\Vert _{C^{n}_{p}, \ell_{p}}= \frac{\varGamma(n+1/p)\varGamma (1/p^{*})}{\varGamma(n+1)}. \end{aligned}
2. (ii)

$$H^{n}$$is a bounded operator from$$C^{n}_{p}$$into$$C^{n}_{p}$$and

\begin{aligned} \bigl\Vert H^{n} \bigr\Vert _{C^{n}_{p}}=\pi\csc(\pi/p). \end{aligned}

### Proof

According to Lemma 3.3, we have $$H^{n}=B^{n}C^{n}$$ and $$C^{n}H^{n}=HC^{n}$$. Now, by applying Lemma 3.1 parts (ii) and (iii), we gain the result. □

As an application of Lemma 3.3, we are ready to generalize the inequality

\begin{aligned} \Vert H x \Vert _{\ell_{p}}\le\pi\csc(\pi/p) \Vert x \Vert _{\ell_{p}}, \end{aligned}

also known as Hilbert’s inequality.

### Corollary 3.5

Let$$p>1$$and$$x\in\ell_{p}$$. Then

\begin{aligned} \bigl\Vert H^{n} x \bigr\Vert _{\ell_{p}}\le \frac{\varGamma(n+1/p)\varGamma (1/p^{*})}{\varGamma(n+1)} \bigl\Vert C^{n} x \bigr\Vert _{\ell_{p}} \end{aligned}

or

\begin{aligned} \sum_{j=0}^{\infty}\Biggl\vert \sum _{k=0}^{\infty}\frac {x_{k}}{j+k+n+1} \Biggr\vert ^{p} \le \biggl(\frac{\varGamma(n+1/p)\varGamma (1/p^{*})}{\varGamma(n+1)} \biggr)^{p} \sum_{j=0}^{\infty}\Biggl\vert \sum _{k=j}^{\infty}\frac{{n+k-j-1 \choose k-j}}{{n+k \choose k}}x_{k} \Biggr\vert ^{p}. \end{aligned}

In particular, for$$n=0$$, Hilbert’s inequality occurs, and for$$n=1$$, we have the inequality

\begin{aligned} \bigl\Vert H^{1} x \bigr\Vert _{\ell_{p}}\le\pi/p \csc( \pi/p) \Vert C x \Vert _{\ell_{p}} \end{aligned}

or

\begin{aligned} \sum_{j=0}^{\infty}\Biggl\vert \sum _{k=0}^{\infty}\frac{x_{k}}{j+k+2} \Biggr\vert ^{p} \le \bigl(\pi/p \csc(\pi/p) \bigr)^{p} \sum _{j=0}^{\infty}\Biggl\vert \sum _{k=j}^{\infty}\frac{x_{k}}{1+k} \Biggr\vert ^{p}. \end{aligned}

### Proof

According to Lemma 3.3, $$H^{n}=B^{n} C^{n}$$, hence

\begin{aligned} \bigl\Vert H^{n} x \bigr\Vert _{\ell_{p}}= \bigl\Vert B^{n}C^{n} x \bigr\Vert _{\ell_{p}}\le \frac{\varGamma (n+1/p)\varGamma(1/p^{*})}{\varGamma(n+1)} \bigl\Vert C^{n} x \bigr\Vert _{\ell_{p}}. \end{aligned}

Consider that, for $$n=0$$, $$C^{0}=I$$, and we have Hilbert’s inequality. □

### Lemma 3.6

Letαandnbe two nonnegative integers that$$\alpha\ge n \ge0$$. For$$j=1, 2, \dots$$, we have

\begin{aligned} \sum_{k=0}^{j} (-1)^{k} {n \choose k}{\alpha+j-k-1 \choose j-k}=\left \{ \textstyle\begin{array}{l@{\quad}l} {\alpha-n+j-1 \choose j},& \alpha> n,\\ 1, &\alpha=n. \end{array}\displaystyle \right . \end{aligned}

### Proof

Since $$\alpha-n \ge0$$, hence Lemma 3.2 results in

\begin{aligned} (1-z)^{-(\alpha-n)} =&\sum_{j=0}^{\infty}{\alpha-n+j-1 \choose j}z^{j}. \end{aligned}

On the other hand,

\begin{aligned} (1-z)^{-(\alpha-n)} =&(1-z)^{n}(1-z)^{-\alpha} \\ =&\sum_{j=0}^{n}(-1)^{j} {n \choose j}z^{j} \sum_{j=0}^{\infty}{\alpha +j-1 \choose j}z^{j} \\ =&\sum_{j=0}^{\infty}\sum _{k=0}^{j}(-1)^{k} {n\choose k}{\alpha +j-k-1\choose j-k}z^{j}. \end{aligned}

Now, the result is obvious. □

### Theorem 3.7

Letα, nbe two nonnegative integers that$$\alpha\ge n\ge0$$. The Copson matrix of orderαhas a factorization of the form$$C^{\alpha}=C^{n}S^{\alpha,n}=S^{\alpha,n} C^{n}$$, where$$C^{n}$$is the Copson matrix of ordern, $$S^{\alpha ,n}=(s^{\alpha,n}_{j,k})$$is a bounded operator on$$\ell_{p}$$with the entries

\begin{aligned} s^{\alpha,n}_{j,k}=\frac{{n+j \choose j}{\alpha-n+k-j-1 \choose k-j}}{{\alpha+k\choose k}} \quad(j,k=0,1,\ldots), \end{aligned}

and$$\ell_{p}$$-norm

\begin{aligned} \bigl\Vert S^{\alpha,n} \bigr\Vert _{\ell_{p}}= \frac{\varGamma(\alpha+1)\varGamma (n+1/p)}{\varGamma(n+1)\varGamma(\alpha+1/p)}. \end{aligned}

### Proof

For obtaining the matrix $$S^{\alpha,n}$$, it is sufficient to compute $$C^{-n} C^{\alpha}$$. By applying Lemma 3.6, we gain

\begin{aligned} \bigl(C^{-n}C^{\alpha}\bigr)_{i,k} =& \frac{{n+i \choose i}}{{\alpha+k \choose k}}\sum_{j=i}^{k}(-1)^{j-i} {n \choose j-i}{\alpha+k-j-1 \choose k-j} \\ =&\frac{{n+i \choose i}}{{\alpha+k \choose k}}\sum_{j=0}^{k-i}(-1)^{j} {n \choose j}{\alpha+k-i-j-1 \choose k-i-j} \\ =&\frac{{n+i \choose i}{\alpha-n+k-i-1 \choose k-i}}{{\alpha+k \choose k}}=s^{\alpha,n}_{i,k}, \end{aligned}

which proves the identity

\begin{aligned} C^{\alpha}=C^{n} S^{\alpha,n}. \end{aligned}
(3.2)

Now, by letting $$n=0$$ and $$n=\alpha$$ in relation (3.2), we gain $$S^{\alpha,0}=C^{\alpha}$$ and $$S^{n,n}=I$$, respectively.

For computing the $$\ell_{p}$$-norm of $$S^{\alpha,n}$$, by inserting $$n=0$$ in identity (3.2), we gain

\begin{aligned} \bigl\Vert S^{\alpha,n} \bigr\Vert _{\ell_{p}}= \frac{\varGamma(\alpha+1)\varGamma (1/p)}{\varGamma(\alpha+1/p)} g(n), \quad g(0)\neq0. \end{aligned}
(3.3)

Now, inserting $$n=\alpha$$ in (3.3) results in

\begin{aligned} \bigl\Vert S^{\alpha,n} \bigr\Vert _{\ell_{p}}= \frac{\varGamma(\alpha+1)\varGamma (n+1/p)}{\varGamma(n+1)\varGamma(\alpha+1/p)}, \end{aligned}

which completes the proof. □

### Corollary 3.8

Let$$p>1$$, $$\alpha\ge n \ge0$$, and$$x\in\ell_{p}$$. Then

1. (i)

$$\|C^{\alpha}x\|_{\ell_{p}}\le \frac{\varGamma(\alpha +1)\varGamma(n+1/p)}{\varGamma(n+1)\varGamma(\alpha+1/p)} \|C^{n} x\|_{\ell_{p}}$$,

2. (ii)

$$\|C x\|_{\ell_{p}}\le p \| x\|_{\ell_{p}}$$,

3. (iii)

$$C^{n}_{p} \subset C^{\alpha}_{p}$$.

### Proof

(i) Since $$C^{\alpha}=S^{\alpha,\beta} C^{n}$$, according to Theorem 3.7, we have

\begin{aligned} \bigl\Vert C^{\alpha}x \bigr\Vert _{\ell_{p}}= \bigl\Vert S^{\alpha, n} C^{n} x \bigr\Vert _{\ell_{p}}\le \frac {\varGamma(\alpha+1)\varGamma(n+1/p)}{\varGamma(n+1)\varGamma (\alpha+1/p)} \bigl\Vert C^{n} x \bigr\Vert _{\ell_{p}}. \end{aligned}

(ii) Consider that, for $$\alpha=1$$ and $$n=0$$, $$C^{1}=C$$ and $$C^{0}=I$$ in part (i), hence we have the inequality.

(iii) is a straightforward result of part (i). □

### Theorem 3.9

Let$$\alpha\ge n \ge0$$. The Copson operator of orderα, $$C^{\alpha}$$, is a bounded operator from$$C^{n}_{p}$$into$$\ell_{p}$$and

\begin{aligned} \bigl\Vert C^{\alpha}\bigr\Vert _{C^{n}_{p},\ell_{p}}= \frac{\varGamma(\alpha+1)\varGamma (n+1/p)}{\varGamma(n+1)\varGamma(\alpha+1/p)}. \end{aligned}

### Proof

According to Lemma 3.1 and Theorem 3.7, we have

\begin{aligned} \bigl\Vert C^{\alpha}\bigr\Vert _{C^{n}_{p},\ell_{p}}= \bigl\Vert S^{\alpha,n}C^{n} \bigr\Vert _{C^{n}_{p},\ell_{p}}= \bigl\Vert S^{\alpha,n} \bigr\Vert _{\ell_{p}}=\frac{\varGamma(\alpha+1)\varGamma (n+1/p)}{\varGamma(n+1)\varGamma(\alpha+1/p)}. \end{aligned}

□

## Norm of Copson operators on some sequence spaces

In this section, we investigate the problem of finding the norm of Copson operators on several sequence spaces.

### Norm of Copson operators on difference sequence spaces

In this part of study, we investigate the norm of both the generalized Copson matrix and the Copson matrix of order n on the difference sequence spaces. In so doing we need the following preliminaries.

Let $$n\in\mathbb{N}$$ and $$\Delta^{n_{F}}=(\delta^{n_{F}}_{j,k})$$ be the forward difference operator of order n with entries

\begin{aligned} \delta^{n_{F}}_{j,k}=\left \{ \textstyle\begin{array}{l@{\quad}l} (-1)^{k-j}{n \choose k-j}, & j\le k\le n+j, \\ 0 & \text{otherwise}. \end{array}\displaystyle \right . \end{aligned}

We define the sequence space $$\ell_{p} (\Delta^{n_{F}})$$ as the set $$\{ x = ( x_{k} ): \Delta^{n_{F}}x\in\ell_{p} \}$$ or

\begin{aligned} \ell_{p}\bigl(\Delta^{n_{F}}\bigr) = \Biggl\{ x = ( x_{k} ):\sum_{j=0}^{\infty}\Biggl\vert \sum_{k=0}^{n}(-1)^{k} {n \choose k} x_{k+j} \Biggr\vert ^{p} < \infty \Biggr\} , \end{aligned}

with semi-norm $$\|\cdot\|_{\ell_{p}(\Delta^{n_{F}})}$$, which is defined by

\begin{aligned} \Vert x \Vert _{\ell_{p}(\Delta^{n_{F}})} = \Biggl(\sum_{j=0}^{\infty}\Biggl\vert \sum_{k=0}^{n}(-1)^{k} {n \choose k} x_{k+j} \Biggr\vert ^{p} \Biggr)^{\frac{1}{p}}. \end{aligned}

Note that this function will be not a norm since if $$x=(1,1,1,\ldots )$$, then $$\|x\|_{\ell_{p}(\Delta^{n_{F}})}=0$$ while $$x \ne0$$. The definition of backward difference sequence space $$\ell_{p}(\Delta ^{n_{B}})$$ is similar to that of $$\ell_{p}(\Delta^{n_{F}})$$, except $$\|\cdot\| _{\ell_{p}(\Delta^{n_{B}})}$$ is a norm.

For special case $$n=1$$, we use the notations $$\Delta^{B}$$ and $$\Delta ^{F}$$ to indicate the backward and forward difference matrices of order 1, respectively. These matrices are defined by

\begin{aligned} \delta^{B}_{j,k}=\left \{ \textstyle\begin{array}{l@{\quad}l} 1, & k=j, \\ -1, & k=j-1, \\ 0 & \text{otherwise}, \end{array}\displaystyle \right . \quad \text{and} \quad \delta^{F}_{j,k}= \left \{ \textstyle\begin{array}{l@{\quad}l} 1, & k= j, \\ -1, & k=j+1, \\ 0 & \text{otherwise}, \end{array}\displaystyle \right . \end{aligned}

and their associated sequence spaces $$\ell_{p}(\Delta^{B})$$ and $$\ell _{p}(\Delta^{F})$$ are

\begin{aligned} \ell_{p}\bigl(\Delta^{B}\bigr)= \Biggl\{ x = ( x_{n} ) : \sum_{n = 1}^{\infty} \vert x_{n} - x_{n - 1} \vert ^{p} < \infty \Biggr\} \end{aligned}

and

\begin{aligned} \ell_{p}\bigl(\Delta^{F}\bigr)= \Biggl\{ x = ( x_{n} ) : \sum_{n = 1}^{\infty} \vert x_{n} - x_{n +1} \vert ^{p} < \infty \Biggr\} , \end{aligned}

respectively.

The idea of difference sequence spaces was introduced by Kizmaz  in 1981. Although topological properties and inclusion relations of these spaces have been studied till now, the problem of finding the norm of operators on difference sequence spaces has not been studied extensively. More recently, Roopaei and Foroutannia investigated this problem for the difference sequence spaces $$\ell _{p}(\Delta^{F})$$, $$\ell_{p}(\Delta^{B})$$, and $$\ell_{p}(\Delta^{n_{F}})$$ in [9, 21, 24].

### Theorem 4.1

The Copson matrix of ordern, $$C^{n}$$, is a bounded operator from$$\ell_{p}$$into$$\ell_{p}(\Delta^{n_{F}})$$and

\begin{aligned} \bigl\Vert C^{n} \bigr\Vert _{\ell_{p},\ell_{p}(\Delta^{n_{F}})}=1. \end{aligned}

In particular, the Copson matrix is a bounded operator from$$\ell_{p}$$into$$\ell_{p}(\Delta^{F})$$and$$\|C\|_{\ell_{p}, \ell_{p}(\Delta^{F})}=1$$.

### Proof

Let $$\Delta^{n_{F}}C^{n}=D^{n}$$. By the definition of Copson matrix, the matrix $$D^{n}=(d^{n}_{i,j})$$ has the entries

\begin{aligned} d^{n}_{i,j} =&\frac{1}{{n+j \choose j}}\sum _{k=i}^{j} (-1)^{k-i} {n \choose k-i}{n+j-k-1 \choose j-k} \\ =&\frac{1}{{n+j \choose j}}\sum_{k=0}^{j-i} (-1)^{k} {n \choose k}{n+j-i-k-1 \choose j-i-k}. \end{aligned}

If $$i=j$$, then $$j-i=0$$, which results in $$k=0$$, hence $$d^{n} _{j,j}=1/{n+j \choose j}$$. If $$j>i$$, then Lemma 3.6 will result in $$d^{n} _{i,j}=0$$. Therefore $$d^{n}_{i,j}=I_{i,j}/{n+j \choose j}$$, where I is the identity matrix. Now, since $$D^{n}$$ is diagonal, Lemma 3.1 results in

\begin{aligned} \bigl\Vert C^{n} \bigr\Vert _{\ell_{p},\ell_{p}(\Delta^{n_{F}})}= \bigl\Vert D^{n} \bigr\Vert _{\ell_{p}}=\sup_{j} d^{n} _{j,j}=1. \end{aligned}

□

### Theorem 4.2

The generalized Copson matrix$$C^{N}$$is a bounded operator from$$\ell_{p}$$into matrix domain$$\ell_{p}(\Delta^{F})$$and

\begin{aligned} \bigl\Vert C^{N} \bigr\Vert _{\ell_{p}, \ell_{p}(\Delta^{F})}= \frac{1}{N}. \end{aligned}

In particular, the Copson matrix is a bounded operator from$$\ell_{p}$$into$$\ell_{p}(\Delta^{F})$$and$$\|C\|_{\ell_{p}, \ell_{p}(\Delta^{F})}=1$$.

### Proof

It is not difficult to verify the identity $$\Delta^{F}C^{N}=D^{N}$$, where the diagonal matrix $$D^{N}=(d^{N}_{j,k})$$ has the entries

\begin{aligned} D^{N}= \begin{pmatrix} \frac{1}{N} & 0 & 0 & \cdots \\ 0 & \frac{1}{N+1} & 0 & \cdots \\ 0 & 0 & \frac{1}{N+2} & \cdots \\ \vdots& \vdots& \vdots& \ddots \end{pmatrix} , \end{aligned}
(4.1)

and the $$\ell_{p}$$-norm $$1/N$$. Now, by applying Lemma 3.1, we have

\begin{aligned} \bigl\Vert C^{N} \bigr\Vert _{\ell_{p}, \ell_{p}(\Delta^{F})}= \bigl\Vert \Delta^{F} C^{N} \bigr\Vert _{\ell_{p}}= \bigl\Vert D^{N} \bigr\Vert _{\ell_{p}}=\frac{1}{N}. \end{aligned}

□

### Theorem 4.3

The Copson matrix of ordern, $$C^{n}$$, is a bounded operator from$$\ell_{p}(\Delta^{n_{B}})$$into$$\ell_{p}(\Delta ^{n_{F}})$$and

\begin{aligned} \bigl\Vert C^{n} \bigr\Vert _{\ell_{p}(\Delta^{n_{B}}),\ell_{p}(\Delta^{n_{F}})}= \frac {\varGamma(n+1)\varGamma(1/p^{*})}{\varGamma(n+1/p^{*})}. \end{aligned}

In particular, the Copson matrix is a bounded operator from$$\ell _{p}(\Delta^{B})$$into$$\ell_{p}(\Delta^{F})$$and

\begin{aligned} \Vert C \Vert _{\ell_{p}(\Delta^{B}), \ell_{p}(\Delta^{F})}=p^{*}. \end{aligned}

### Proof

The fact that $$\Delta^{n_{B}}$$ is the transpose of $$\Delta^{n_{F}}$$ and $$\Delta^{n_{F}}C^{n}$$ is a diagonal matrix, according to Theorem 4.1, results in the identity $$\Delta^{n_{F}}C^{n}=C^{nt}\Delta^{n_{B}}$$. Now, Lemma 3.1 and relation (1.2) complete the proof. □

### Theorem 4.4

The generalized Copson matrix$$C^{N}$$is a bounded operator from$$\ell_{p}(\Delta^{B})$$into matrix domain$$\ell_{p}(\Delta^{F})$$and

\begin{aligned} \bigl\Vert C^{N} \bigr\Vert _{\ell_{p}(\Delta^{B}), \ell_{p}(\Delta^{F})}=p^{*}. \end{aligned}

In particular, the Copson matrix is a bounded operator from$$\ell _{p}(\Delta^{B})$$into$$\ell_{p}(\Delta^{F})$$and

\begin{aligned} \Vert C \Vert _{\ell_{p}(\Delta^{B}), \ell_{p}(\Delta^{F})}=p^{*}. \end{aligned}

### Proof

Through the proof of Theorem 4.2 we showed that $$\Delta^{F} C^{N}=D^{N}$$, where $$D^{N}$$ is a diagonal matrix as defined by relation (4.1). Similar to the proof of Theorem 4.3, $$\Delta^{F} C^{N}=C^{Nt} \Delta^{B}$$, where $$C^{Nt}$$ is the transpose of the generalized Copson matrix. Now, according to Lemma 3.1, we have

\begin{aligned} \bigl\Vert C^{N} \bigr\Vert _{\ell_{p}(\Delta^{B}), \ell_{p}(\Delta^{F})}= \bigl\Vert C^{Nt} \bigr\Vert _{\ell_{p}}=p^{*}. \end{aligned}

This completes the proof. □

### Norm of Copson operator on the Hilbert matrix domain

Let $$H^{n}_{p}$$ be the sequence space associated with Hilbert matrix $$H^{n}$$, which is

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

and has the norm

\begin{aligned} \Vert x \Vert _{H^{n} _{p}} = \Biggl(\sum_{j=0}^{\infty}\Biggl\vert \sum_{k=0}^{\infty}\frac{x_{k}}{j+k+n+1} \Biggr\vert ^{p} \Biggr)^{\frac{1}{p}}. \end{aligned}

### Corollary 4.5

The Copson operator of ordern, $$C^{n}$$, is a bounded operator from$$H^{n}_{p}$$into$$H_{p}$$and

\begin{aligned} \bigl\Vert C^{n} \bigr\Vert _{H^{n}_{p}, H_{p}}= \frac{\varGamma(n+1/p)\varGamma (1/p)}{\varGamma(n+1)}. \end{aligned}

### Proof

According to Lemma 3.3, we have $$HC^{n}=C^{n}H^{n}$$. Now, Lemma 3.1 completes the proof. □

As another application of Lemma 3.3, we have the following inclusions.

### Corollary 4.6

Let$$p>1$$. Then

1. (i)

$$C^{n}_{p} \subset H^{n}_{p}$$,

2. (ii)

$$H^{n}_{p} \subset H^{n-1}_{p} \subset\cdots\subset H_{p}$$.

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