Norm of Hilbert Operator’s Commutants

Abstract

In this study, we obtain the ${\ell }_p$-norms of six classes of operators that commute with the infinite Hilbert operators.

1. Introduction

We use the notation ${\parallel T\parallel }_p$ for the norm of a linear operator from the sequence space ${\ell }_p$ to itself. Several references have addressed the problem of finding the norm and lower bound of operators on matrix domains [1,2,3,4,5,6,7]. Our study considers infinite matrices ${\left[A\right]}_{j,k}$, where all the indices j and k are non-negative.

Definition 1 (Hilbert matrix).

If n is a non-negative integer, we define the Hilbert matrix of order n, $H_n$, as follows:

$$\begin{array}{c}\hfill {\left[H_n\right]}_{j,k}=\frac{1}{j+k+n+1}j,k=0,1,\dots .\end{array}$$

In the case of $n=0$, $H_0=H$ is the well-known Hilbert matrix, which was introduced by David Hilbert in 1894. According to [8] theorem 323, the Hilbert matrix is a bounded operator on ${\ell }_p$ and 

$$\begin{array}{c}\hfill {\parallel H\parallel }_p=\mathsf{\Gamma }(1/p)\mathsf{\Gamma }(1/p^{*})=\pi csc(\pi /p),\end{array}$$

where $p^{*}$ is the conjugate of p, i.e., $\frac{1}{p}+\frac{1}{p^{*}}=1$.

Definition 2 (Hausdorff matrices).

One of the best examples of summability matrices is $H_{\mu }$, which is defined as

$$\begin{array}{c}\hfill {\left[H_{\mu }\right]}_{j,k}=\left\{\begin{array}{cc}{\int }_0^1\left(\genfrac{}{}{0pt}{}{j}{k}\right){\theta }^k{(1-\theta )}^{j-k}d\mu \left(\theta \right)\hfill & 0\le k\le j,\hfill \\ & \\ 0\hfill & otherwise.\hfill \end{array}\right.\end{array}$$

where μ is a probability measure on $[0,1]$. Even though it is a difficult task to obtain the ${\ell }_p$-norm of operators, the Hausdorff matrices can be computed using Hardy’s formula [9], Theorem 216, which states that this matrix is a bounded operator on ${\ell }_p$, if and only if

$$\begin{array}{c}\hfill {\int }_0^1{\theta }^{\frac{-1}{p}}d\mu \left(\theta \right)<\infty ,1\le p<\infty .\end{array}$$

In fact,

$$\parallel H_{\mu }{\parallel }_p={\int }_0^1{\theta }^{\frac{-1}{p}}d\mu \left(\theta \right).$$

(1)

Hausdorff operators have the interesting norm-separating property.

Theorem 1

([7], Theorem 9). Let $p\ge 1$ and $H_{\mu }$, $H_{\phi }$ and $H_{\nu }$ be Hausdorff matrices such that $H_{\mu }=H_{\phi }{H}_{\nu }$. Then, $H_{\mu }$ is bounded on ${\ell }_p$ if and only if both $H_{\phi }$ and $H_{\nu }$ are bounded on ${\ell }_p$. Moreover, we have

$$\begin{array}{c}\hfill \parallel H_{\mu }{\parallel }_p=\parallel H_{\phi }{\parallel }_p{\parallel H_{\nu }\parallel }_p.\end{array}$$

For comprehensive information about the Hausdorff matrices, the enthusiastic reader can refer to [10,11].

Several famous matrices have been derived from the Hausdorff matrix. For positive integer n, the following are the two classes.

Definition 3 (Cesàro matrix).

The measure $d\mu \left(\theta \right)=n{(1-\theta )}^{n-1}d\theta$ gives the Cesàro matrix of order n, $C_n$, for which

$$\begin{array}{c}\hfill {\left[C_n\right]}_{j,k}=\left\{\begin{array}{cc}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{n+j-k-1}{j-k}\right)}{\left(\genfrac{}{}{0pt}{}{n+j}{j}\right)}}\hfill & 0\le k\le j,\hfill \\ & \\ 0\hfill & otherwise.\hfill \end{array}\right.\end{array}$$

Note that $C_0=I$, where I is the identity matrix, and $C_1=C$ is the classical Cesàro matrix. According to (1), $C_n$ has the ${\ell }_p$-norm

$$\begin{array}{c}\hfill \parallel C_n{\parallel }_p=\frac{\mathsf{\Gamma }(n+1)\mathsf{\Gamma }(1/p^{*})}{\mathsf{\Gamma }(n+1/p^{*})},\end{array}$$

which for the famous Cesàro matrix that is ${\parallel C\parallel }_p=\frac{p}{p-1}$.

Definition 4 (Gamma matrix).

The measure $d\mu \left(\theta \right)=n{\theta }^{n-1}d\theta$ gives the Gamma matrix of order n, $G_n$, for which

$$\begin{array}{c}\hfill {\left[G_n\right]}_{j,k}=\left\{\begin{array}{cc}\frac{\left(\genfrac{}{}{0pt}{}{n+k-1}{k}\right)}{\left(\genfrac{}{}{0pt}{}{n+j}{j}\right)}\hfill & 0\le k\le j,\hfill \\ & \\ 0\hfill & otherwise.\hfill \end{array}\right.\end{array}$$

Hence, by Hardy’s formula, $G_n$ has the ${\ell }_p$-norm

$$\begin{array}{c}\hfill \parallel G_n{\parallel }_p=\frac{np}{np-1}.\end{array}$$

You should note that $G_1$ is the classical Cesàro matrix C.

A well-known property of Hausdorff means is that products are determined by the diagonal elements. Specifically, if A, B, and C are Hausdorff means and ${\left[A\right]}_{j,j}{\left[B\right]}_{j,j}={\left[C\right]}_{j,j}$ for all j, then $AB=C$. (This is proved in [9], Section 11.3, though in different notation.) The following result is also known from [9]:

Theorem 2.

$C_{n-1}{G}_n=C_n$, hence $C_n=S_{n,m}{C}_m$ for $m<n$, where $S_{n,m}=G_{m+1}\cdots G_n$.

Proof. 

The diagonal elements of $C_{n-1}{G}_n$ are

$$\begin{array}{c}\hfill {\left[C_{n-1}{G}_n\right]}_{j,j}=\frac{j!}{n(n+1)\cdots (n+j-1)}\frac{n}{n+j}={\left[C_n\right]}_{j,j}.\end{array}$$

Hence, the stated identity. □

The following result is known as the Hellinger–Toeplitz theorem.

Theorem 3

([1], Proposition 7.2). Let $1<p,q<\infty$. The matrix M maps ${\ell }_p$ into ${\ell }_q$ if and only if the transposed matrix, $M^t$, maps ${\ell }_{q^{*}}$ into ${\ell }_{p^{*}}$. Then, we have

$$\begin{array}{c}\hfill {\parallel M\parallel }_{{\ell }_p\to {\ell }_q}={\parallel M^t\parallel }_{{\ell }_{q^{*}}\to {\ell }_{p^{*}}}.\end{array}$$

As an example of the Hellinger–Toeplitz theorem, the transposed Cesàro matrix of order n has the ${\ell }_p$-norm

$$\begin{array}{c}\hfill \parallel C_n^t{\parallel }_p=\frac{\mathsf{\Gamma }(n+1)\mathsf{\Gamma }(1/p)}{\mathsf{\Gamma }(n+1/p)}.\end{array}$$

Motivation. Hilbert operators are used in a wide range of fields including approximation theory, cryptography, image processing, functional analysis, representation theory, and noncommutative geometry. The estimates of the norm of this operator and the study of its properties in various spaces are of considerable interest and have a long history. Recently, ref. [12] has introduced some classes of Hilbert’s commutators mostly based on Cesàro and Gamma matrices. In this study, we establish the ${\ell }_p$ norm of these operators.

For non-negative integers $n,j$ and k, let us define the matrix $B_n$ by

$$\begin{array}{c}\hfill {\left[B_n\right]}_{j,k}=\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)\beta (j+k+1,n+1)=\frac{(k+1)\cdots (k+n)}{(j+k+1)\cdots (j+k+n+1)},\end{array}$$

where the $\beta$ function is

$$\begin{array}{c}\hfill \beta (m,n)={\int }_0^1{x}^{m-1}{(1-x)}^{n-1}dz(m,n=1,2,\dots ).\end{array}$$

$B_0=H$ where H represents Hilbert’s matrix.

We need the following lemma before we can discuss the Hilbert operator’s commutants, which reveals the relationship between the Hilbert operator and the Cesàro and Gamma matrices.

Lemma 1

(Lemmas 2.3 and 3.1 of [13,14]). Hilbert matrices satisfy the following identities for positive integer n:

• $H=B_n{C}_n$
• $H_n=C_n{B}_n$
• $H_n{C}_n=C_{n}H$
• $H_n{G}_n=G_n{H}_{n-1}$
• $B_n$ is a bounded operator that has the ${\ell }_p$-norm

  $$\begin{array}{c}\hfill \parallel B_n{\parallel }_p=\frac{\mathsf{\Gamma }(n+1/p^{*})\mathsf{\Gamma }(1/p)}{\mathsf{\Gamma }(n+1)}.\end{array}$$
• ${\parallel H\parallel }_p=\parallel B_n{\parallel }_p{\parallel C_n\parallel }_p$,

where $C_n$ and $G_n$ are the Cesàro and Gamma matrices of order n and $B_n$ is the matrix, which was defined earlier.

Commutants of the infinite Hilbert operator. Assume that n is a non-negative integer, and define the symmetric matrix as follows:

$$\begin{array}{c}\hfill {\mathsf{\Phi }}_n^b=B_n^t{B}_n{\mathsf{\Psi }}_n^b=B_n{B}_n^t\end{array}$$

$$\begin{array}{c}\hfill {\mathsf{\Phi }}_n^c=C_n^t{C}_n{\mathsf{\Psi }}_n^c=C_n{C}_n^t\end{array}$$

and for $n\ge 1$

$$\begin{array}{c}\hfill {\mathsf{\Phi }}_n^g=G_n^t{G}_n{\mathsf{\Psi }}_n^g=G_n{G}_n^t,\end{array}$$

Note that for $n=1$,

$$\begin{array}{c}\hfill \mathsf{\Psi }:={\mathsf{\Psi }}_1^c={\mathsf{\Psi }}_1^g=C{C}^{t}and\mathsf{\Phi }:={\mathsf{\Phi }}_1^c={\mathsf{\Phi }}_1^g=C^{t}C.\end{array}$$

In [12] Theorems 11.2.2 and 11.2.4, the author has proved that the above matrices are commutants of Hilbert operators. We present those theorems with their proofs.

Theorem 4.

The operators ${\mathsf{\Phi }}_n^c$ and ${\mathsf{\Psi }}_n^b$ are commutants of H.

Proof. 

By applying Lemma 1 twice, we have

$$\begin{array}{ccc}\hfill {\mathsf{\Phi }}_n^{c}H& =& C_n^t{H}_n{C}_n={\left(H_n{C}_n\right)}^t{C}_n\hfill \\ & =& {\left(C_{n}H\right)}^t{C}_n=H{C}_n^t{C}_n=H{\mathsf{\Phi }}_n^c.\hfill \end{array}$$

It can easily be seen from Lemma 1 that $H{B}_n=B_n{H}_n$. Now,

$$\begin{array}{ccc}\hfill {\mathsf{\Psi }}_n^{b}H& =& B_n{\left(H{B}_n\right)}^t=B_n{\left(B_n{H}_n\right)}^t\hfill \\ & =& B_n{H}_n{B}_n^t=H{B}_n{B}_n^t=H{\mathsf{\Psi }}_n^b.\hfill \end{array}$$

 □

Theorem 5.

The operators ${\mathsf{\Phi }}_n^b$, ${\mathsf{\Phi }}_{n+1}^g$, ${\mathsf{\Psi }}_n^c$ and ${\mathsf{\Psi }}_n^g$ are commutants of the Hilbert operator of order n.

Proof. 

By applying Lemma 1 twice, we have

$$\begin{array}{ccc}\hfill {\mathsf{\Psi }}_n^c{H}_n& =& C_n{\left(H_n{C}_n\right)}^t=C_n{\left(C_{n}H\right)}^t\hfill \\ & =& C_{n}H{C}_n^t=H_n{C}_n{C}_n^t=H_n{\mathsf{\Psi }}_n^c.\hfill \end{array}$$

Additionally, applying Lemma 1 results in

$$\begin{array}{ccc}\hfill {\mathsf{\Psi }}_n^g{H}_n& =& G_n{\left(H_n{G}_n\right)}^t=G_n{\left(G_n{H}_{n-1}\right)}^t\hfill \\ & =& G_n{H}_{n-1}{G}_n^t=H_n{G}_n{G}_n^t=H_n{\mathsf{\Psi }}_n^g.\hfill \end{array}$$

The proof of the other items is similar. □

2. Main Results

For non-negative integers m and n, let us define the following matrices:

$$\begin{array}{c}\hfill {\mathsf{\Phi }}_{m,n}^b=B_m^t{B}_n{\mathsf{\Psi }}_{m,n}^b=B_m{B}_n^t\end{array}$$

$$\begin{array}{c}\hfill {\mathsf{\Phi }}_{m,n}^c=C_m^t{C}_n{\mathsf{\Psi }}_{m,n}^c=C_m{C}_n^t\end{array}$$

and for $m,n\ge 1$

$$\begin{array}{c}\hfill {\mathsf{\Phi }}_{m,n}^g=G_m^t{G}_n{\mathsf{\Psi }}_{m,n}^g=G_m{G}_n^t.\end{array}$$

Note that for $m=n$, all the above matrices are reduced to the Hilbert operator’s commutators that we introduced earlier. Through this section, we will prove the norm-separating property for the Cesàro and Gamma matrices of the form:

$$\begin{array}{c}\hfill \parallel C_m{C}_n^t{\parallel }_p=\parallel C_m{\parallel }_p{\parallel C_n^t\parallel }_p,\end{array}$$

$$\begin{array}{c}\hfill \parallel C_m^t{C}_n{\parallel }_p=\parallel C_m^t{\parallel }_p{\parallel C_n\parallel }_p,\end{array}$$

$$\begin{array}{c}\hfill \parallel G_m{G}_n^t{\parallel }_p=\parallel G_m{\parallel }_p{\parallel G_n^t\parallel }_p,\end{array}$$

$$\begin{array}{c}\hfill \parallel G_m^t{G}_n{\parallel }_p=\parallel G_m^t{\parallel }_p{\parallel G_n\parallel }_p.\end{array}$$

Theorem 6.

For non-negative integers m and n, matrices ${\mathsf{\Psi }}_{m,n}^c$ and ${\mathsf{\Phi }}_{m,n}^c$ are bounded operators on ${\ell }_p$ and

$$\begin{array}{c}\hfill \parallel {\mathsf{\Psi }}_{m,n}^c{\parallel }_p=\frac{\mathsf{\Gamma }(m+1)\mathsf{\Gamma }(n+1)}{\mathsf{\Gamma }(m+1/p^{*})\mathsf{\Gamma }(n+1/p)}\pi csc(\pi /p)\end{array}$$

$$\begin{array}{c}\hfill \parallel {\mathsf{\Phi }}_{m,n}^c{\parallel }_p=\frac{\mathsf{\Gamma }(m+1)\mathsf{\Gamma }(n+1)}{\mathsf{\Gamma }(m+1/p)\mathsf{\Gamma }(n+1/p^{*})}\pi csc(\pi /p).\end{array}$$

In particular, the matrices ${\mathsf{\Psi }}_n^c$ and ${\mathsf{\Phi }}_n^c$ are bounded operators on ${\ell }_p$ and

$$\begin{array}{c}\hfill \parallel {\mathsf{\Psi }}_n^c{\parallel }_p={\parallel {\mathsf{\Phi }}_n^c\parallel }_p=\frac{{\mathsf{\Gamma }}^2(n+1)}{\mathsf{\Gamma }(n+1/p)\mathsf{\Gamma }(n+1/p^{*})}\pi csc(\pi /p).\end{array}$$

Theorem 7.

For positive integers m and n, matrices ${\mathsf{\Psi }}_{m,n}^g$ and ${\mathsf{\Phi }}_{m,n}^g$ are bounded operators on ${\ell }_p$ and

$$\begin{array}{c}\hfill \parallel {\mathsf{\Psi }}_{m,n}^g{\parallel }_p=\frac{mnp{p}^{*}}{(mp-1)(n{p}^{*}-1)}\end{array}$$

$$\begin{array}{c}\hfill \parallel {\mathsf{\Phi }}_{m,n}^g{\parallel }_p=\frac{mnp{p}^{*}}{(m{p}^{*}-1)(np-1)}.\end{array}$$

In particular, the matrices ${\mathsf{\Psi }}_n^g$ and ${\mathsf{\Phi }}_n^g$ are bounded operators on ${\ell }_p$ and

$$\begin{array}{c}\hfill \parallel {\mathsf{\Psi }}_n^g{\parallel }_p={\parallel {\mathsf{\Phi }}_n^g\parallel }_p=\frac{n^{2}p{p}^{*}}{(np-1)(n{p}^{*}-1)}.\end{array}$$

Theorem 8.

For non-negative integers m and n, the matrices ${\mathsf{\Psi }}_{m,n}^b$ and ${\mathsf{\Phi }}_{m,n}^b$ are bounded operators on ${\ell }_p$ and

$$\begin{array}{c}\hfill \parallel {\mathsf{\Psi }}_{m,n}^b{\parallel }_p=\frac{\mathsf{\Gamma }(m+1/p^{*})\mathsf{\Gamma }(n+1/p)}{\mathsf{\Gamma }(m+1)\mathsf{\Gamma }(n+1)}\pi csc(\pi /p)\end{array}$$

$$\begin{array}{c}\hfill \parallel {\mathsf{\Phi }}_{m,n}^b{\parallel }_p=\frac{\mathsf{\Gamma }(m+1/p)\mathsf{\Gamma }(n+1/p^{*})}{\mathsf{\Gamma }(m+1)\mathsf{\Gamma }(n+1)}\pi csc(\pi /p).\end{array}$$

In particular, the matrices ${\mathsf{\Psi }}_n^b$ and ${\mathsf{\Phi }}_n^b$ are bounded operators on ${\ell }_p$ and

$$\begin{array}{c}\hfill \parallel {\mathsf{\Psi }}_n^b{\parallel }_p={\parallel {\mathsf{\Phi }}_n^b\parallel }_p=\frac{\mathsf{\Gamma }(n+1/p)\mathsf{\Gamma }(n+1/p^{*})}{{\mathsf{\Gamma }}^2(n+1)}\pi csc(\pi /p).\end{array}$$

3. Proof of Theorems

In this section, we focus on proving our claims, but first, we need the following lemmas.

Lemma 2.

For the Hilbert operator, we have $\parallel H^2{\parallel }_p={\parallel H\parallel }_p^2$.

Proof. 

Let H be the Hilbert operator with matrix entries $1/(j+k)(j,k\ge 1)$, and write $M_r=\pi /sin\left(r\pi \right)$. It is well known that ${\parallel H\parallel }_p\le M_{1/p}$ for $p>1$. Here, we show that ${\parallel H\parallel }_p\ge M_{1/p}$ and $\parallel H^2{\parallel }_p\ge M_{1/p}^2$ (so that equality holds in both cases). The same statements hold for the alternative Hilbert operator with matrix entries $\frac{1}{j+k-1}$.

Choose r with $rp>1$, and let $x_k=1/k^r$ for $k\ge 1$. Let $y=Hx$ and $z=Hy$. Then,

$$\begin{array}{c}\hfill y_j=\sum _{k=1}^{\infty }\frac{1}{(j+k)k^r}\ge {\int }_1^{\infty }\frac{1}{(t+j)t^r}dt.\end{array}$$

Now,

$$\begin{array}{c}\hfill {\int }_0^{\infty }\frac{1}{(t+j)t^r}dt=\frac{M_r}{j^r},\end{array}$$

and

$$\begin{array}{c}\hfill {\int }_0^1\frac{1}{(t+j)t^r}dt\le {\int }_0^1\frac{1}{j{t}^r}dt=\frac{1}{(1-r)j}.\end{array}$$

so

$$\begin{array}{c}\hfill y_j\ge \frac{M_r}{j^r}-\frac{1}{(1-r)j}.\end{array}$$

(2)

Informally, $y_j$ is approximately $M_r{x}_j$, so ${\parallel y\parallel }_{{\ell }_p}$ is approximately $M_r{\parallel x\parallel }_{{\ell }_p}$. For $0<x<a$, we have ${(1-\frac{x}{a})}^p\ge 1-\frac{px}{a}$, hence ${(a-x)}^p\ge a^p-p{a}^{p-1}x$. Hence,

$$\begin{array}{c}\hfill y_j^p\ge \frac{M_r^p}{j^{rp}}-\frac{p}{1-r}\frac{M_r^{p-1}}{j^{rp-r+1}},\end{array}$$

so,

$$\begin{array}{c}\hfill \sum _{j=1}^{\infty }{y}_j^p\ge M_r^p\zeta \left(rp\right)-\frac{p}{1-r}{M}_r^{p-1}\zeta (rp-r+1),\end{array}$$

while ${\sum }_{k=1}^{\infty }{x}_k^p=\zeta \left(rp\right)$. Now, let $r\to 1/p$ from above. Then, $\zeta \left(rp\right)\to \infty$, while $\zeta \left(rp-r+1\right)\to \zeta (2-1/p)$. Hence, $\frac{{\parallel y\parallel }_{{\ell }_p}}{{\parallel x\parallel }_{{\ell }_p}}$ tends to $M_{1/p}$.

We now turn to $H^2$. We require the following:

Let $u_k=1/k$ for $k\ge 1$. Then,

$$\begin{array}{ccc}\hfill {\left(Hu\right)}_j& =& \sum _{k=1}^{\infty }\frac{1}{(j+k)k}=\frac{1}{j}\sum _{k=1}^{\infty }\left(\frac{1}{k}-\frac{1}{j+k}\right)\hfill \\ & =& \frac{1}{j}\left(1+\frac{1}{2}+\cdots +\frac{1}{j}\right)=L_j/j,\hfill \end{array}$$

where $L_j={\sum }_{i=1}^j\frac{1}{i}$. By (2), $y\ge M_{r}x-u/(1-r)$, so

$$\begin{array}{c}\hfill z\ge M_r\left(Hx\right)-\frac{Hu}{1-r}=M_{r}y-\frac{Hu}{1-r}.\end{array}$$

So, again by (2),

$$\begin{array}{c}\hfill z_j\ge \frac{M_r^2}{j^r}-\frac{M_r}{(1-r)j}-\frac{L_j}{(1-r)j}.\end{array}$$

Hence,

$$\begin{array}{c}\hfill z_j^p\ge \frac{M_r^{2p}}{j^{rp}}-p\frac{M_r^{2p-2}}{j^{r(p-1)}}\frac{M_r+L_j}{(1-r)j}.\end{array}$$

Write $\eta \left(s\right)={\sum }_{j=1}^{\infty }\frac{L_j}{j^s}$: this is convergent for $s>1$. Then,

$$\begin{array}{c}\hfill \sum _{j=1}^{\infty }{z}_j^p\ge M_r^{2p}\zeta \left(rp\right)-\frac{p}{(1-r)}{M}_r^{2p-2}(M_r\zeta (rp-r+1)+\eta (rp-r+1)).\end{array}$$

When $r\to 1/p$ from above, $\eta (rp-r+1)$ tends to the finite limit $\eta (2-1/p)$. So ${\parallel z\parallel }_{{\ell }_p}/{\parallel x\parallel }_{{\ell }_p}$ tends to $M_{1/p}^2$. □

Lemma 3.

For the Hilbert operator of order n, we have $\parallel H_n^2{\parallel }_p={\parallel H_n\parallel }_p^2$.

Proof. 

With x and z as defined in the previous lemma, it shows that, given $ϵ>0$, there exists $\delta >0$ such that if $r<\frac{1}{p}+\delta$, then ${\parallel z\parallel }_{{\ell }_p}\ge (M_{1/p}^2-ϵ){\parallel x\parallel }_{{\ell }_p}$. Now, let $u=(H-H_n)x$ and $w=(H^2-H_n^2)x=(H+H_n)u$. Then, ${\parallel w\parallel }_{{\ell }_p}\le 2M_{1/p}{\parallel u\parallel }_{{\ell }_p}$. We show that for r close enough to $\frac{1}{p}$, ${\parallel u\parallel }_{{\ell }_p}\le ϵ{\parallel x\parallel }_{{\ell }_p}$. Now, for any $r>\frac{1}{p}$,

$$\begin{array}{c}\hfill u_j=\sum _{k=1}^{\infty }\left(\frac{1}{j+k}-\frac{1}{j+k+n}\right)\frac{1}{k^r}<\sum _{k=1}^{\infty }\frac{n}{{(j+k)}^2{k}^{1/p}}.\end{array}$$

Hence,

$$\begin{array}{c}\hfill {\parallel u\parallel }_{{\ell }_p}\le {\parallel u\parallel }_{{\ell }_1}=\sum _{j=1}^{\infty }{u}_j<n\sum _{k=1}^{\infty }\frac{1}{k^{1/p}}\sum _{j=1}^{\infty }\frac{1}{{(j+k)}^2}<n\zeta (1+\frac{1}{p}),\end{array}$$

since ${\sum }_{j=1}^{\infty }\frac{1}{{(j+k)}^2}<\frac{1}{k}$. Meanwhile, ${\parallel u\parallel }_{{\ell }_p}=\zeta \left(rp\right)\to \infty$ as $r\to \frac{1}{p}$. Hence, for r close enough to $\frac{1}{p}$, ${\parallel u\parallel }_{{\ell }_p}\le ϵ{\parallel x\parallel }_{{\ell }_p}$, as required. □

Proof of Theorem 6.

We first compute the ${\ell }_p$-norm of ${\mathsf{\Psi }}_{m,n}^c$. Obviously,

$$\begin{array}{ccc}\hfill \parallel {\mathsf{\Psi }}_{m,n}^c{\parallel }_p& \le & \parallel C_m{\parallel }_p{\parallel C_n^t\parallel }_p.\hfill \end{array}$$

According to the Lemma 2, we also have ${\parallel H\parallel }_p=\parallel B_n^t{\parallel }_p{\parallel C_n^t\parallel }_p$, which results in

$$\begin{array}{c}\hfill {\parallel H\parallel }_p^2=\parallel B_m{\parallel }_p\parallel C_m{\parallel }_p\parallel C_n^t{\parallel }_p{\parallel B_n^t\parallel }_p.\end{array}$$

Now, regarding the identity $H^2=B_m{\mathsf{\Psi }}_{m,n}^c{B}_n^t$ and Lemma 1,

$$\begin{array}{c}\hfill {\parallel H\parallel }_p^2=\parallel H^2{\parallel }_p\le \parallel B_m{\parallel }_p\parallel {\mathsf{\Psi }}_{m,n}^c{\parallel }_p{\parallel B_n^t\parallel }_p.\end{array}$$

Hence,

$$\begin{array}{c}\hfill \parallel {\mathsf{\Psi }}_{m,n}^c{\parallel }_p\ge \parallel C_m{\parallel }_p{\parallel C_n^t\parallel }_p,\end{array}$$

which completes the proof.

For computing the norm of ${\mathsf{\Phi }}_{m,n}^c$, we suppose that $m\ge n$. The other case $m<n$ has a similar proof. In this case, regarding Lemma 1 and Theorem 2, we have

$$\begin{array}{c}\hfill H_m^2={\left(C_m{B}_m\right)}^t{C}_m{B}_m=B_m^t{C}_m^t{C}_n{S}_{m,n}{B}_m=B_m^t{\mathsf{\Phi }}_{m,n}^c{S}_{m,n}{B}_m.\end{array}$$

However, by applying Lemma 3,

$$\begin{array}{ccc}\hfill \parallel H_m{\parallel }_p^2& =& \parallel B_m^t{\parallel }_p\parallel C_m^t{\parallel }_p\parallel C_m{\parallel }_p{\parallel B_m\parallel }_p\hfill \\ & =& \parallel B_m^t{\parallel }_p\parallel C_m^t{\parallel }_p\parallel C_n{\parallel }_p\parallel S_{m,n}{\parallel }_p{\parallel B_m\parallel }_p\hfill \\ & =& \parallel H_m^2{\parallel }_p\le \parallel B_m^t{\parallel }_p\parallel {\mathsf{\Phi }}_{m,n}^c{\parallel }_p\parallel S_{m,n}{\parallel }_p{\parallel B_m\parallel }_p,\hfill \end{array}$$

which shows

$$\begin{array}{c}\hfill \parallel {\mathsf{\Phi }}_{m,n}^c{\parallel }_p\ge \parallel C_m^t{\parallel }_p{\parallel C_n\parallel }_p.\end{array}$$

The other side of the above inequality is obvious, so the proof is complete. □

Proof of Theorem 7.

First, $\parallel {\mathsf{\Psi }}_{m,n}^g{\parallel }_p\le \parallel G_m{\parallel }_p{\parallel G_n^t\parallel }_p$. From the definition and relation $C_n=C_{n-1}{G}_n=G_n{C}_{n-1}$, we have

$$\begin{array}{c}\hfill {\mathsf{\Psi }}_{m,n}^c=C_{m-1}{G}_m{\left(C_{n-1}{G}_n\right)}^t=C_{m-1}{\mathsf{\Psi }}_{m,n}^g{C}_{n-1}^t,\end{array}$$

so $\parallel {\mathsf{\Psi }}_{m,n}^c{\parallel }_p\le \parallel G_{m-1}{\parallel }_p\parallel {\mathsf{\Psi }}_{m,n}^g{\parallel }_p{\parallel C_{n-1}^t\parallel }_p$. By Theorem 6 and $\parallel C_m{\parallel }_p=\parallel C_{m-1}{\parallel }_p{\parallel G_m\parallel }_p$

$$\begin{array}{c}\hfill \parallel {\mathsf{\Psi }}_{m,n}^c{\parallel }_p=\parallel C_m{\parallel }_p\parallel C_n^t{\parallel }_p=\parallel C_{m-1}{\parallel }_p\parallel G_m{\parallel }_p\parallel G_n^t{\parallel }_p{\parallel C_{n-1}^t\parallel }_p,\end{array}$$

hence $\parallel {\mathsf{\Psi }}_{m,n}^g{\parallel }_p\ge \parallel G_m{\parallel }_p{\parallel G_n^t\parallel }_p$.

Similarly, using $C_m^t{C}_n=C_{m-1}^t{\mathsf{\Phi }}_{m,n}^g{C}_{n-1}$ and Theorem 6. □

Proof of Theorem 8.

Using the identities $H=B_n{C}_n=C_m^t{B}_m^t$, we obtain $H^2=C_m^t{\mathsf{\Phi }}_{m,n}^b{C}_n$. Reasoning as in the proof of Theorem 6, we obtain

$$\begin{array}{c}\hfill \parallel {\mathsf{\Phi }}_{m,n}^b{\parallel }_p\ge \parallel B_m^t{\parallel }_p{\parallel B_m^t\parallel }_p,\end{array}$$

hence equality. Similarly, $\parallel {\mathsf{\Psi }}_{m,n}^b{\parallel }_p=\parallel B_m{\parallel }_p{\parallel B_n^t\parallel }_p$. □

4. Conclusions

The author, in his previous work, has introduced the following symmetric matrices:

$$\begin{array}{c}\hfill {\mathsf{\Phi }}_n^b=B_n^t{B}_n{\mathsf{\Psi }}_n^b=B_n{B}_n^t\end{array}$$

$$\begin{array}{c}\hfill {\mathsf{\Phi }}_n^c=C_n^t{C}_n{\mathsf{\Psi }}_n^c=C_n{C}_n^t\end{array}$$

$$\begin{array}{c}\hfill {\mathsf{\Phi }}_n^g=G_n^t{G}_n{\mathsf{\Psi }}_n^g=G_n{G}_n^t,\end{array}$$

as the Hilbert operators commutants. In this study, we have obtained the ${\ell }_p$-norm of these operators as:

• $\parallel {\mathsf{\Psi }}_n^b{\parallel }_p={\parallel {\mathsf{\Phi }}_n^b\parallel }_p=\frac{\mathsf{\Gamma }(n+1/p)\mathsf{\Gamma }(n+1/p^{*})}{{\mathsf{\Gamma }}^2(n+1)}\pi csc(\pi /p)$
• $\parallel {\mathsf{\Psi }}_n^c{\parallel }_p={\parallel {\mathsf{\Phi }}_n^c\parallel }_p=\frac{{\mathsf{\Gamma }}^2(n+1)}{\mathsf{\Gamma }(n+1/p)\mathsf{\Gamma }(n+1/p^{*})}\pi csc(\pi /p)$
• $\parallel {\mathsf{\Psi }}_n^g{\parallel }_p={\parallel {\mathsf{\Phi }}_n^g\parallel }_p=\frac{n^{2}p{p}^{*}}{(np-1)(n{p}^{*}-1)}$

Through this research, we have also proved the norm-separating property for the Cesàro and Gamma matrices of the form:

$$\begin{array}{c}\hfill \parallel C_m{C}_n^t{\parallel }_p=\parallel C_m{\parallel }_p{\parallel C_n^t\parallel }_p\end{array}$$

$$\begin{array}{c}\hfill \parallel C_m^t{C}_n{\parallel }_p=\parallel C_m^t{\parallel }_p{\parallel C_n\parallel }_p\end{array}$$

$$\begin{array}{c}\hfill \parallel G_m{G}_n^t{\parallel }_p=\parallel G_m{\parallel }_p{\parallel G_n^t\parallel }_p\end{array}$$

$$\begin{array}{c}\hfill \parallel G_m^t{G}_n{\parallel }_p=\parallel G_m^t{\parallel }_p{\parallel G_n\parallel }_p\end{array}$$