A study on Copson operator and its associated sequence space

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 ≥ 1 and ω denote the set of all real-valued sequences. The space p is the set of all real sequences x = (x k ) ∈ ω such that 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.
and has the p -norm C p = p. This matrix is the transpose of the well-known Cesàro matrix. Generalized Copson matrix. Suppose that N ≥ 1 is a real number. The generalized Copson matrix C N = (c N j,k ) is defined by for all j, k ∈ N 0 , has the p -norm C N p = p ( [8], Lemma 2.3), and the matrix representation 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 which is also a sequence space. The matrix domain associated with this matrix is the set which has the following norm: 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 [2-5, 7, 9, 10, 12, 13, 16, 17, 21, 23-25] and textbooks [2] and [20].
Copson matrix of order n. Consider the Hausdorff matrix H μ = (h j,k ) ∞ j,k=0 , with entries of the form: where μ is a probability measure on In order to define and know the Copson matrix details, we need the following theorem also known as Hellinger-Toeplitz theorem.
For nonnegative real number n, and by choosing dμ(θ ) = n(1θ ) n-1 dθ in the definition of Hausdorff matrix, we gain the Cesàro matrix of order n, which, according to Hardy's formula, has the p -norm 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 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 p ; that is, which is a Banach space with the norm Throughout this research, we use the notations · A p , p , · p ,A p and · A p ,B p for the norm of operators from the matrix domain A p into sequence space p , from 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.

∞
In this section, the sequence spaces C n p (1 ≤ p < ∞) and C n ∞ 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 order n, C n , is invertible and its inverse C
Proof Let us recall the forward difference matrix of order n, n = (δ n j,k ), which is a lower triangle matrix with entries This matrix has the inverse -n = (δ -n j,k ) with the following entries: 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 ≤ k, we have Now, by a simple calculation, we deduce that which completes the proof. Now, we introduce the sequence spaces C n p and C n ∞ as the set of all sequences whose C n -transforms are in the spaces p and ∞ , respectively; that is,

Theorem 2.2
The spaces C n p and C n ∞ are Banach spaces with the norms respectively.
Proof We omit the proof which is a routine verification.

Theorem 2.3
The spaces C n p and C n ∞ are linearly isomorphic to p and ∞ , respectively.
Proof We only prove that C n p is linearly isomorphic to p . Since C n is invertible, hence the map T : C n p → p as Tx = C n x for any x ∈ C n p is bijective, which proves the isomorphism.

Theorem 2.4 The inclusion C n
Proof Let x ∈ C n p . Then we have C n x ∈ p . Since the inclusion p ⊂ q holds for 1 ≤ p < q < ∞, we have C n x ∈ q , which implies that x ∈ C n q . Hence, we conclude that the inclusion C n p ⊂ C n q holds. Now, since the inclusion p ⊂ q is strict, we can choose y = (y j ) ∈ q \ p . By defining x = C -n y, we have C n x = y, which results in C n x ∈ q \ p . Hence, we conclude that x ∈ C n q \ C n p , and so the inclusion C n p ⊂ C n q is strict.
Proof Choose any x ∈ C n p . Then we have C n x ∈ p . Since the inclusion p ⊂ ∞ holds for 1 ≤ p < ∞, we have C n x ∈ ∞ . This implies that x ∈ C n ∞ . Hence, we conclude that the inclusion C n p ⊂ C n ∞ holds. Similar to the proof of the previous theorem, we can choose x such that C n x = ((-1) j ) ∈ ∞ \ p , and consequently it results in x ∈ C n ∞ \ C n p . Therefore, the inclusion C n p ⊂ C n ∞ is strict.
It is known from Theorem 2.3 of Jarrah and Malkowsky [15] that if T is a triangle then the domain λ 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.
Then the sequence (h (k) ) is a basis for the space C n p and every sequence x ∈ C n p has a unique representation of the form x = k (C n x) k h (k) .
The following lemma is essential to determine the dual spaces. Throughout the paper, N is the collection of all finite subsets of N.

Lemma 2.7 ([26])
The following statements hold: Definition The α-, β-, and γ -duals of a sequence space X are defined by respectively. In the following, we find the Köthe dual of the Copson sequence space.

Theorem 2.8
The α-duals of the spaces C n 1 , C n p (1 < p < ∞), and C n ∞ are as follows: otherwise.
Given any x = (x j ) ∈ C n p (1 ≤ p ≤ ∞), we have b j x j = (Ay) j for all j ∈ N. This implies that bx ∈ 1 with x ∈ C n p if and only if Ay ∈ 1 with y ∈ p . Hence, we conclude that b ∈ (C n p ) α if and only if A ∈ ( p , 1 ). This completes the proof by Lemma 2.7.

Theorem 2.9
Let us define the following sets: The other results can be proved similarly.

Theorem 2.10
The γ -duals of the spaces C n 1 , C n p (1 < p < ∞), and C n ∞ are as follows: 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 Let U be a bounded operator on p and A p and B p be two matrix domains such that A p p . (i) If BT is a bounded operator on p , then T is a bounded operator from p into B p and
(ii) If T has a factorization of the form T = UA, then T is a bounded operator from the matrix domain A p into p and (iii) If BT = UA, then T is a bounded operator from the matrix domain A p into B p and

In particular, if AT = UA, then T is a bounded operator from the matrix domain A p into A p and
Also, if T and A commute, then T A p = T p .
(ii) Since A p and p are isomorphic, hence 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, . . . , the Hilbert matrix is defined by For nonnegative integers n, j, and k, let us define the matrix B n = (b n j,k ) by b n j,k = (j + 1) · · · (j + n) (j + k + 1) · · · (j + k + n + 1) .
Consider that, for n = 0, B 0 = H, where H is the Hilbert matrix.
Note that the matrix B n has also the representation b n j,k = n + j j β(j + k + 1, n + 1) (j, k = 0, 1, . . .), where the β function is 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
Proof By differentiating n -1 times the identity (1z) -1 = ∞ j=0 z j , we obtain the result. (ii) H n is a bounded operator from C n p into C n p and H n C n p = π csc(π/p).
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 Hx p ≤ π csc(π/p) x p , also known as Hilbert's inequality.
Theorem 3.7 Let α, n be two nonnegative integers that α ≥ n ≥ 0. The Copson matrix of order α has a factorization of the form C α = C n S α,n = S α,n C n , where C n is the Copson matrix of order n, S α,n = (s α,n j,k ) is a bounded operator on p with the entries and p -norm S α,n p = Γ (α + 1)Γ (n + 1/p) Γ (n + 1)Γ (α + 1/p) .
Proof For obtaining the matrix S α,n , it is sufficient to compute C -n C α . By applying Lemma 3.6, we gain which proves the identity Now, by letting n = 0 and n = α in relation (3.2), we gain S α,0 = C α and S n,n = I, respectively.
For computing the p -norm of S α,n , by inserting n = 0 in identity (3.2), we gain which completes the proof.
(ii) Consider that, for α = 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).
Proof According to Lemma 3.1 and Theorem 3.7, we have

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 ∈ N and n F = (δ n F j,k ) be the forward difference operator of order n with entries δ n F j,k = (-1) k-j n k-j , j ≤ k ≤ n + j, 0 o t h e r w i s e .
We define the sequence space p ( n F ) as the set {x = (x k ) : with semi-norm · p ( n F ) , which is defined by Note that this function will be not a norm since if x = (1, 1, 1, . . .), then x p ( n F ) = 0 while x = 0. The definition of backward difference sequence space p ( n B ) is similar to that of For special case n = 1, we use the notations B and F to indicate the backward and forward difference matrices of order 1, respectively. These matrices are defined by and their associated sequence spaces p ( B ) and p ( F ) are respectively. The idea of difference sequence spaces was introduced by Kizmaz [18] 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 p ( F ), p ( B ), and p ( n F ) in [9,21,24].

Theorem 4.1
The Copson matrix of order n, C n , is a bounded operator from p into p ( n F ) and C n p , p ( n F ) = 1.
In particular, the Copson matrix is a bounded operator from p into p ( F ) and C p , p ( F ) = 1.
Proof Let n F C n = D n . By the definition of Copson matrix, the matrix D n = (d n i,j ) has the entries If i = j, then ji = 0, which results in k = 0, hence d n j,j = 1/ n+j 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 j , where I is the identity matrix. Now, since D n is diagonal, Lemma 3.1 results in C n p , p ( n F ) = D n p = sup j d n j,j = 1.

Theorem 4.2 The generalized Copson matrix C N is a bounded operator from p into matrix domain p ( F ) and
In particular, the Copson matrix is a bounded operator from p into p ( F ) and C p , p ( F ) = 1.
Proof It is not difficult to verify the identity F C N = D N , where the diagonal matrix D N = (d N j,k ) has the entries
In particular, the Copson matrix is a bounded operator from p ( B ) into p ( F ) and C p ( B ), p ( F ) = p * .
Proof The fact that n B is the transpose of n F and n F C n is a diagonal matrix, according to Theorem 4.1, results in the identity n F C n = C nt n B . Now, Lemma 3.1 and relation (1.2) complete the proof.