Norm of Hilbert operator on sequence spaces

Hilbert matrices are frequently used both in mathematics and computational sciences. In image processing, for example, Hilbert matrices are commonly used. Any twodimensional array of natural numbers in the range [0, n] for all n ∈ N can be viewed as a gray-scale digital image. Cryptography is another example of applications of the Hilbert matrix. Cryptography is the science of using mathematics to encrypt and decrypt data. Cryptography enables you to store sensitive information or transmit it across insecure networks so that it cannot be

Hilbert matrices are frequently used both in mathematics and computational sciences. In image processing, for example, Hilbert matrices are commonly used. Any twodimensional array of natural numbers in the range [0, n] for all n ∈ N can be viewed as a gray-scale digital image.
Cryptography is another example of applications of the Hilbert matrix. Cryptography is the science of using mathematics to encrypt and decrypt data. Cryptography enables you to store sensitive information or transmit it across insecure networks so that it cannot be read by anyone except the intended recipient. The objectives of the proposed work are to propose a new cryptographic method based on the special matrix called the Hilbert matrix for authentication and confidentiality and to propose a model for confidentiality and authentication using a combination of symmetric and public cryptosystems. In some studies related to cryptographic methods, the Hilbert matrix is used for authentication and confidentiality [16]. It is well known that the Hilbert matrix is very unstable [15] and so it can be used in security systems.
In this paper we only focus on the infinite version of Hilbert matrix H = (h j,k ), which is defined by and is a bounded operator on p with p -norm H p = Γ (1/p)Γ (1/p * ) = π csc(π/p) by Theorem 323 in [8].
Matrix domain The matrix domain of an infinite matrix A in a sequence space X is defined as which is also a sequence space. It is easy to see that, for an invertible matrix A, the matrix domain A p is a normed space with x A p := Ax p . By using matrix domains of special triangle matrices in classical spaces, we can define more general sequence spaces than the space p . More recently, the author and some other mathematicians have investigated the problem of finding the norm of operators on several matrix domains [5, 6, 10-14, 17, 20, 21, 23].
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 , for the norm of operators from p into the matrix domain A p and for the norm of operators from matrix domain A p into the matrix domain B p , respectively.
Motivation Although a variety of research has been done on the finite Hilbert operator, see [1,4,9,22], and a lot of properties of this matrix have been discovered (determinant, inverse, . . . ) there exists a few information about the infinite version of Hilbert matrix, specially in the area of finding the norm of this operator on sequence spaces. Recently the author [18,19] has introduced some factorizations for the infinite Hilbert matrix based on the generalized Cesàro matrix and Cesàro and Gamma matrices of order n. Through this study the author has tried to compute the norm of Hilbert operator on several sequence spaces that have not been done before.

Norm of Hilbert operator on some sequence spaces
In this part of our study, we investigate the problem of finding the norm of well-known Hilbert operator on some sequence spaces. The following lemma plays a key role in finding the norm of operators between matrix domains.
Proof (i) Since A p and p are isomorphic, (ii) Again by isomorphism between A p and p we have which gives the desired result.

Norm of Hilbert operator on Cesàro and Copson sequence spaces
where μ is a probability measure on Cesàro matrix By letting dμ(θ ) = n(1θ ) n-1 dθ in the definition of the Hausdorff matrix, the Cesàro matrix of order n, C n = (c n j,k ), is defined by which according to Eq. (2.1) has the p -norm where p * is the conjugate of p i.e. 1 p + 1 p * = 1. Note that C 1 = C is the well-known Cesàro matrix which has the p -norm C p = p * .
Some more examples are The Cesàro 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 Transposing the Cesàro matrix of order n results the Copson matrix of order n, C nt = (c nt j,k ), which has the entries by the Hellinger-Toeplitz theorem, which is the following.
Consider that, for n = 0, B 0 = H, where H is the Hilbert matrix. Note that the matrix B n has also the representation where the β function is For computing the norm of Hilbert operator on Cesàro and Copson matrix domains we need the following lemma. Proof (i) By applying the identity ∞ j=0 n+j-1 j z j = (1z) -n for |z| < 1, we deduce that . The factorization will be obtained by the following calculations: β(i + k -2, n + 1) + (n + 2)!(i + k -2)! 2!(i + k + n -2)!(i + k + n + 1) . . . = λ n + i -1 i β(k + 1, n + 1) + (n + i -1)!(k + 1)! (i -1)!(k + n + 1)!(i + k + n + 1) (iii) This is obvious by parts (i) and (ii). (iv) For computing the p -norm of B n , we introduce a family of matrices, B(w), 0 < w ≤ 1, given by the row sums are all (1-w) n w and the column sums are all (1w) n-1 . Thus Schur's theorem results in On the other hand, Also the factorization H = B n C n results H p ≤ B n p C n p . Therefore

Theorem 2.5 Let H n be the Hilbert matrix of order n. Then (i) H is a bounded operator from C n p into p and
H C n p , p = Γ (n + 1/p * )Γ (1/p) Γ (n + 1) .
(ii) H is a bounded operator from C n p into C n p and H C n p = π csc(π/p).
(iii) H n is a bounded operator from C nt p into p and H n C nt p , p = Γ (n + 1/p)Γ (1/p * ) Γ (n + 1) .
(iv) H n is a bounded operator from C nt p into C nt p and H n C nt p = π csc(π/p).
Consider that, for n = 0, C 0 = I and we have the Hilbert inequality. Also by applying Lemma 2.3, H n = B n C n , hence we have Consider that, for n = 0, C 0 = I and we have the Hilbert inequality.
Let H n p be the sequence space associated with the Hilbert matrix H n , which is and has the norm x k j + k + n + 1 As another application of Lemma 2.3, we have the following inclusions.

Norm of Hilbert operator on the generalized Cesàro matrix domain
Suppose that N ≥ 1 is a real number. The generalized Cesàro matrix, Note that C 1 is the well-known Cesàro matrix C. Some more examples are The matrix domain associated with this matrix is the set which has the following norm:

Theorem 2.8 The Hilbert operator H is a bounded operator from C N p into p and
In particular, the Hilbert operator H is a bounded operator from C p into p and H C p , p = π p * csc(π/p).
Proof The author in [18] has proved that the Hilbert matrix admits a factorization of the form H = B N C N , where B N is a bounded operator on p and π p * csc(π/p) ≤ B N p ≤ πN p * csc(π/p).
Now, according to Lemma 2.1 we have the result.

Norm of Hilbert operator on Gamma sequence spaces
By letting dμ(θ ) = nθ n-1 dθ in the definition of the Hausdorff matrix, the Gamma matrix of order n, Γ n = (γ n j,k ), is The sequence space associated with Γ n , is the set which is called the Gamma space of order n.

Theorem 2.9 The Hilbert operator H is a bounded operator from Γ n p into p and
In particular, the Hilbert operator H is a bounded operator from C p into p and H C p , p = π p * csc(π/p).
Proof The author in [19] has proved that the Hilbert matrix has a factorization of the form H = S n Γ n , where the matrix S n has the p -norm S n p = π 1 -1 np csc(π/p). Now, by applying part (ii) of Lemma 2.1 the proof is obvious.

Norm of Hilbert operator on difference sequence spaces
Let n ∈ N and n F = (δ n F j,k ) be the forward difference operator of order n with entries Note that this function will not be 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 p ( n F ), except · p ( n B ) is a norm.
Theorem 2.10 The Hilbert matrix H, is a bounded operator from p ( n B ) into p ( n F ) and H p ( n B ), p ( n F ) = π sin(π/p) .
Proof Let A = n F H. By using the identity n j=0 (-1) j n j z j = (1z) n , we have Obviously, A is a symmetric matrix which implies that n F H = H n B . Now by using Lemma 2.1 part (ii), H p ( n B ), p ( n F ) = H p = π csc(π/p).