# A Hilbert-type operator with a symmetric homogeneous kernel of two parameters and its applications

## Abstract

We introduce a general homogeneous kernel whose degree is given by two parameters to establish the equivalent inequalities with the norm of a new Hilbert-type operator. As applications, we provide new extended Hilbert-type inequalities with the best possible constant factors.

## 1 Introduction

Let $$\{a_{n}\}$$ and $$\{b_{m}\}$$ be two sequences of nonnegative real numbers. The well-known Hilbert’s inequality says that if $$p > 1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$0< \sum_{m=1}^{\infty}a_{m} ^{p} < \infty$$ and $$0 < \sum_{n=1}^{\infty}b_{n} ^{q} < \infty$$, then

$$\sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{a_{m} b_{n}}{m+n} < \frac{\pi }{\sin(\frac{\pi}{p})} \Biggl(\sum_{m=1}^{\infty}a_{m} ^{p} \Biggr)^{1/p} \Biggl(\sum _{n=1}^{\infty}b_{n} ^{q} \Biggr)^{1/q},$$
(1)

where the constant factor $$\frac{\pi}{\sin(\frac{\pi}{p})}$$ is the best possible [1]. This inequality has been generalized in numerous ways with introducing suitable parameters and weight coefficients. (For example, see [213] and the references therein.) In particular, by introducing a Hilbert-type linear operator with a symmetric homogeneous kernel, one can obtain various Hilbert-type inequalities with the best constant factors. For this purpose, let $$k(x, y)$$ be a nonnegative symmetric function defined on $$(0,\infty )\times(0,\infty)$$, i.e., $$k(x,y)=k(y,x)$$. For $$p>1$$ and $$\frac {1}{p} +\frac{1}{q}=1$$, let $$\ell^{r}$$ ($$r=p, q$$) be two normed spaces. If T is a bounded self-adjoint semi-positive definite operator defined by

$$(Ta) (n):= \sum_{m=1}^{\infty}k(m,n) a_{m},\quad n \in\mathbb{N}$$

for $$a=\{a_{m}\}_{m=1}^{\infty}\in\ell^{p}$$, or similarly,

$$(Tb) (m):= \sum_{n=1}^{\infty}k(m,n) b_{n},\quad m \in\mathbb{N}$$

for $$b=\{b_{n}\}_{n=1}^{\infty}\in\ell^{q}$$. The operator T is called the Hilbert-type operator and the function $$k(x,y)$$ is called the symmetric kernel of T. In view of this point, Hilbert’s inequality (1) can be expressed by

$$(Ta, b)\leq\frac{\pi}{\sin(\frac{\pi}{p})} \|a\|_{p} \|b\|_{q},$$

where the kernel $$k(x,y)=\frac{1}{x+y}$$ and the formal inner product $$(Ta, b)$$ between Ta and b is given by $$(Ta, b):= \sum_{n=1}^{\infty}(Ta)(n)b_{n}$$. Motivated by this observation, Yang [14] defined a Hilbert-type linear operator $$T: \ell^{r} \rightarrow\ell^{r}$$ ($$r=p,q$$) with the kernel $$k(x,y)=\frac{(xy)^{\frac{\lambda -1}{2}}}{(x+y)^{\lambda}}$$ of degree −1. As a consequence, he was able to prove that if $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$a_{m}, b_{n} \geq0$$, $$1-2\min\{\frac{1}{p}, \frac{1}{q}\} <\lambda< 1+2\min\{\frac{1}{p}, \frac{1}{q}\}$$, then the following two inequalities are equivalent:

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{(mn)^{\frac{\lambda -1}{2}}a_{m}b_{n}}{m^{\lambda}+n^{\lambda}} < \frac{1}{\lambda}B \biggl(\frac {q(\lambda+1)-2}{2q\lambda}, \frac{p(\lambda+1)-2}{2p\lambda} \biggr)\| a\|_{p} \|b \|_{q} , \\& \Biggl\{ \sum_{n=1}^{\infty}\Biggl(\sum _{m=1}^{\infty}\frac{(mn)^{\frac {\lambda-1}{2}}a_{m}}{m^{\lambda}+n^{\lambda}} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} < \frac{1}{\lambda}B \biggl(\frac{q(\lambda+1)-2}{2q\lambda}, \frac {p(\lambda+1)-2}{2p\lambda} \biggr)\|a\|_{p} , \end{aligned}

where $$B(u,v)$$ denotes the beta function defined by

$$B(u,v):= \int_{0}^{\infty}\frac{t^{u-1}}{(1+t)^{u+v}}\,dt = B(u,v)\quad (u,v>0).$$

Moreover, the constant factor $$\frac{1}{\lambda}B (\frac{q(\lambda +1)-2}{2q\lambda}, \frac{p(\lambda+1)-2}{2p\lambda} )$$ is the best possible. In 2010, Jin and Debnath [15] generalized the Hilbert-type linear operator whose kernel is symmetric and homogeneous of degree −1. In fact, they obtained several extended Hilbert-type inequalities by using the kernel $$k(x,y)=\frac{1}{(x^{\frac{1}{\lambda }}+y^{\frac{1}{\lambda}})^{\lambda}}$$ ($$\lambda>0$$). For instance, they proved that if $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$\alpha, \beta>0$$, $$0 <\lambda\leq\min\{\frac{q}{\alpha}, \frac{p}{\beta}\}$$, then the following two inequalities are equivalent:

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m} b_{n}}{(m^{\alpha}+n^{\beta})^{\lambda}} < \frac{B(\frac{\lambda}{p}, \frac{\lambda}{q})}{\alpha^{\frac {1}{q}}\beta^{\frac{1}{p}}} \Biggl( \sum_{m=1}^{\infty}m^{(p-1)(1-\alpha \lambda)}|a_{m}|^{p} \Biggr)^{\frac{1}{p}} \Biggl( \sum_{n=1}^{\infty}n^{(q-1)(1-\beta\lambda)}|b_{n}|^{q} \Biggr)^{\frac {1}{q}}, \\& \Biggl\{ \sum_{n=1}^{\infty}n^{\beta\lambda-1} \Biggl(\sum_{m=1}^{\infty}\frac {a_{m} }{(m^{\alpha}+n^{\beta})^{\lambda}} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} < \frac{B(\frac{\lambda}{p}, \frac{\lambda}{q})}{\alpha^{\frac{1}{q}}\beta ^{\frac{1}{p}}} \Biggl( \sum_{m=1}^{\infty}m^{(p-1)(1-\alpha\lambda )}|a_{m}|^{p} \Biggr)^{\frac{1}{p}} , \end{aligned}

where the constant factor $$\frac{B(\frac{\lambda}{p}, \frac{\lambda }{q})}{\alpha^{\frac{1}{q}}\beta^{\frac{1}{p}}}$$ is the best possible. See [1623] for other Hilbert-type operators and the corresponding extended Hilbert-type inequalities with the best factors.

Most of the previous results were, however, obtained by using the Hilbert-type operator with the symmetric homogeneous kernel of −λ-order, which depends on a parameter $$\lambda>0$$. In this paper, we introduce a more general homogeneous kernel whose degree is given by two parameters (Definition 2.3). We establish the equivalent inequalities with the norm of a new Hilbert-type operator (Theorem 3.1). As applications, we provide new extended Hilbert-type inequalities with the best possible constant factors (Corollary 4.1 and Cases 1-3).

## 2 Hilbert-type operator with a symmetric homogeneous kernel whose degree is given by two parameters

For completeness, we begin with the following definitions and notations.

### Definition 2.1

Let $$p>1$$, $$n_{0} \in\mathbb{Z}$$, $$w(n)\geq0$$ ($$n \geq n_{0}$$, $$n \in\mathbb{Z}$$). Define the normed space $$\ell_{w,n_{0}}^{p}$$ by

$$\ell_{w,n_{0}}^{p} := \Biggl\{ a = \{a_{n} \}_{n=n_{0}}^{\infty}: \|a \|_{p,w} := \Biggl(\sum _{n=n_{0}}^{\infty}w(n)|a_{n}|^{p} \Biggr)^{1/p} < \infty \Biggr\} .$$

### Definition 2.2

Let $$\lambda_{1}, \lambda_{2}, \lambda>0$$ satisfying that $$\lambda= \lambda_{1}+\lambda_{2}$$. Denote by $$F_{n_{0}}(r)$$ ($$n_{0} \in\mathbb{Z}$$) the set of all real-valued $$C^{1}$$-functions $$\phi(x)$$ satisfying the following conditions:

1. (1)

$$\phi(x)$$ is strictly increasing in $$(n_{0}-1,\infty)$$ with $$\phi((n_{0}-1)+)=0$$, $$\phi(\infty) =\infty$$.

2. (2)

For $$\alpha>0$$, $$\frac{\phi'(x)}{\phi(x)^{\alpha+1-\lambda _{i}}}$$ is decreasing in $$(n_{0}-1,\infty)$$.

Let $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$\lambda= \lambda _{1}+\lambda_{2}$$, $$\lambda_{1}, \lambda_{2}, \lambda>0$$. For $$\phi(x) \in F_{m_{0}}(r)$$ and $$\psi(y) \in F_{n_{9}}(s)$$, $$r,s>1$$, we define the following weight functions:

\begin{aligned}& w_{1}(m) := \frac{\phi(m)^{p(\alpha+1-\lambda_{2})-1}}{\phi'(m)^{p-1}},\qquad w_{2}(n) := \frac{\psi(n)^{q(\alpha+1-\lambda_{1})-1}}{\psi'(n)^{q-1}},\\& \widetilde{w}_{1}(n) := \frac{\psi'(n)}{\psi(n)^{p(\alpha-\lambda _{1})+1}}, \qquad\widetilde{w}_{2}(m) := \frac{\phi'(m)}{\phi(m)^{q(\alpha -\lambda_{2})+1}}. \end{aligned}

### Definition 2.3

Let $$\lambda_{1}, \lambda_{2}, \lambda>0$$ satisfying that $$\lambda= \lambda_{1}+\lambda_{2}$$. For $$\alpha>0$$ and $$x, y >0$$, $$K_{\alpha, \lambda}(x,y)$$ is a continuous real-valued function on $$(0, \infty) \times(0,\infty)$$ satisfying the following properties:

1. (1)

$$K_{\alpha, \lambda}(x,y)$$ is a symmetric homogeneous function of degree $$2\alpha-\lambda$$, that is,

\begin{aligned} &K_{\alpha, \lambda}(x,y) = K_{\alpha, \lambda}(y,x),\\ &K_{\alpha, \lambda}(tx,ty) = t^{2\alpha-\lambda} K_{\alpha, \lambda }(x,y) \quad\mbox{for any } t>0. \end{aligned}
2. (2)

$$K_{\alpha, \lambda}(x,y)$$ is decreasing with respect to x and y, respectively.

3. (3)

For sufficiently small $$\varepsilon\geq0$$, the following integral

$$\widetilde{K}_{\alpha, \lambda}(\lambda_{i},\varepsilon) := \int _{0}^{\infty}K_{\alpha, \lambda}(1,t)t^{-1+\lambda_{i}-\alpha-\varepsilon}\,dt$$

exists for $$i=1,2$$. Moreover, assume that $$\widetilde{K}_{\alpha, \lambda}(\lambda_{i},0):=K_{\alpha}(\lambda_{i})>0$$ and $$\widetilde{K}_{\alpha, \lambda}(\lambda_{i},\varepsilon) = K_{\alpha }(\lambda_{i}) + o(1)$$ as $$\varepsilon\rightarrow0+$$.

4. (4)

Given $$p>1$$, $$\phi(x) \in F_{m_{0}} (r)$$, and $$\psi(y) \in F_{n_{0}} (s)$$ ($$r,s>1$$),

$$\sum_{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi(n)^{1+\varepsilon}}\int _{0}^{\frac{\phi(m_{0})}{\psi(n)}} K_{\alpha, \lambda} (1, t) t^{-1+\lambda _{i}-\alpha-\frac{\varepsilon}{p}}\,dt=O(1)$$

as $$\varepsilon\rightarrow0+$$.

### Lemma 2.4

Let $$\lambda_{1}, \lambda_{2}, \lambda>0$$ satisfying that $$\lambda=\lambda _{1} + \lambda_{2}$$. For any $$\alpha>0$$, we have

$$K_{\alpha}(\lambda_{1}) = K_{\alpha}( \lambda_{2}).$$

### Proof

Since

\begin{aligned} K_{\alpha}(\lambda_{1}) &= \widetilde{K}_{\alpha,\lambda}( \lambda_{1},0) = \int_{0}^{\infty}K_{\alpha, \lambda}(1, t)t^{-1+\lambda_{1}-\alpha}\,dt, \end{aligned}

letting $$t=\frac{1}{s}$$ gives

$$K_{\alpha}(\lambda_{1}) = \int_{0}^{\infty} K_{\alpha, \lambda }(1,s)s^{-1+\lambda_{2}-\alpha}\,ds = K_{\alpha}( \lambda_{2}).$$

□

In view of Lemma 2.4, we may assume that

$$K_{\alpha}(\lambda) := K_{\alpha}(\lambda_{1}) = K_{\alpha}(\lambda_{2}).$$

### Lemma 2.5

Let $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$\lambda_{1}+\lambda_{2}=\lambda$$, $$\lambda_{1}, \lambda_{2} >0$$, $$\alpha>0$$. For $$\phi(x) \in F_{m_{0}}(r)$$ and $$\psi(y) \in F_{n_{0}}(s)$$, $$r,s >1$$, define the weight coefficients $$W_{1}(m)$$ and $$W_{2}(n)$$ by

\begin{aligned}& W_{1}(m) := \sum _{n=n_{0}}^{\infty}K_{\alpha, \lambda}\bigl(\phi(m),\psi(n) \bigr) \frac{\phi(m)^{\lambda_{2}-\alpha}}{\psi(n)^{\alpha+1-\lambda_{1}}} \psi '(n),\\& W_{2}(n) := \sum_{m=m_{0}}^{\infty}K_{\alpha, \lambda}\bigl(\phi(m),\psi(n)\bigr) \frac{\psi(n)^{\lambda_{1}-\alpha}}{\phi(m)^{\alpha+1-\lambda_{2}}} \phi '(m) . \end{aligned}

Then

\begin{aligned} W_{1}(m) < K_{\alpha}(\lambda) \quad\textit{and}\quad W_{2}(n) < K_{\alpha}(\lambda) \end{aligned}

for any $$m \geq m_{0}$$, $$n \geq n_{0}$$ ($$m,n \in\mathbb{Z}$$).

### Proof

We have

\begin{aligned} W_{1}(m) &= \sum _{n=n_{0}}^{\infty}K_{\alpha, \lambda} \biggl(1, \frac{\psi (n)}{\phi(m)} \biggr) \frac{\phi(m)^{\alpha-\lambda_{1}} }{\psi(n)^{\alpha +1-\lambda_{1}}} \psi'(n)\\ &< \int_{n_{0}-1}^{\infty}K_{\alpha, \lambda} \biggl(1, \frac{\psi(x)}{\phi (m)} \biggr) \frac{\psi'(x)}{\psi(x)^{\alpha+1-\lambda_{1}}} \phi (m)^{\alpha-\lambda_{1}}\,dx. \end{aligned}

Setting $$t=\frac{\psi(x)}{\phi(m)}$$, we get

$$W_{1}(m) < \int_{0}^{\infty}K_{\alpha, \lambda} (1,t) t^{-1+\lambda_{1}-\alpha }\,dt = K_{\alpha}(\lambda) .$$

Similarly, one can obtain $$W_{2}(n) < K_{\alpha}(\lambda)$$. □

### Lemma 2.6

Let $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$\lambda_{1}+\lambda_{2}=\lambda$$, $$\lambda_{1}, \lambda_{2} >0$$. For $$a_{m}, b_{n} \geq0$$ ($$m_{0}, n_{0} \in\mathbb{Z}$$), let $$a=\{a_{m}\} _{m=m_{0}}^{\infty}\in\ell_{w_{1},m_{0}}^{p}$$ and $$b=\{b_{n}\}_{n=n_{0}}^{\infty}\in\ell_{w_{2}, n_{0}}^{q}$$. Then, for $$\phi(x) \in F_{m_{0}}(r)$$ and $$\psi(y) \in F_{n_{0}}(s)$$ ($$r,s >1$$), we have

\begin{aligned} &\Biggl\Vert \sum_{m=m_{0}}^{\infty}K_{\alpha, \lambda} \bigl(\phi(m), \psi(n)\bigr) a_{m} \Biggr\Vert _{p,\widetilde{w}_{1}} \leq K_{\alpha}(\lambda) \|a\|_{p,w_{1}} \quad\textit{and} \\ &\Biggl\Vert \sum_{n=n_{0}}^{\infty}K_{\alpha, \lambda} \bigl(\phi(m), \psi(n)\bigr) b_{n} \Biggr\Vert _{q,\widetilde{w}_{2}} \leq K_{\alpha}(\lambda) \|b\|_{q,w_{2}}, \end{aligned}

and hence

\begin{aligned} &\Biggl\{ \sum_{m=m_{0}}^{\infty}K_{\alpha, \lambda}\bigl(\phi(m), \psi(n)\bigr) a_{m} \Biggr\} _{n=n_{0}}^{\infty}\in\ell_{\tilde{w_{1}},n_{0}}^{p} \quad\textit{and}\\ &\Biggl\{ \sum_{n=n_{0}}^{\infty}K_{\alpha, \lambda}\bigl(\phi(m), \psi(n)\bigr) b_{n} \Biggr\} _{m=m_{0}}^{\infty}\in\ell_{\tilde{w_{2}},m_{0}}^{q}. \end{aligned}

### Proof

Applying Hölder’s inequality, we observe

\begin{aligned} &\sum_{m=m_{0}}^{\infty}K_{\alpha, \lambda}\bigl( \phi(m),\psi(n)\bigr) a_{m} \\ &\quad= \sum_{m=m_{0}}^{\infty}\biggl(K_{\alpha, \lambda} \bigl(\phi(m),\psi(n)\bigr) \frac {\phi(m)^{\frac{\alpha+1-\lambda_{2}}{q}}}{\psi(n)^{\frac{\alpha+1-\lambda _{1}}{p}}} \frac{\psi'(n)^{\frac{1}{p}}}{\phi'(m)^{\frac{1}{q}}} a_{m} \biggr) \biggl( \frac{\psi(n)^{\frac{\alpha+1-\lambda_{1}}{p}}}{\phi(m)^{\frac {\alpha+1-\lambda_{2}}{q}}} \frac{\phi'(m)^{\frac{1}{q}}}{\psi'(n)^{\frac{1}{p}}} \biggr)\\ &\quad\leq \Biggl(\sum_{m=m_{0}}^{\infty}K_{\alpha, \lambda}\bigl(\phi(m),\psi(n)\bigr) \frac{\phi(m)^{(\alpha+1-\lambda_{2})(p-1)}}{\psi(n)^{\alpha+1-\lambda _{1}}} \frac{\psi'(n)}{\phi'(m)^{p-1}} a_{m}^{p} \Biggr)^{\frac{1}{p}}\\ &\qquad{} \times \Biggl(\sum_{m=m_{0}}^{\infty}K_{\alpha, \lambda}\bigl(\phi(m),\psi (n)\bigr) \frac{\psi(n)^{(\alpha+1-\lambda_{1})(q-1)}}{\phi(m)^{\alpha +1-\lambda_{2}}} \frac{\phi'(m)}{\psi'(n)^{q-1}} \Biggr)^{\frac{1}{q}}. \end{aligned}

By Definition 2.2, we get

\begin{aligned} &\sum_{m=m_{0}}^{\infty}K_{\alpha, \lambda}\bigl( \phi(m),\psi(n)\bigr) a_{m} \\ &\quad\leq \biggl(\int_{0}^{\infty}K_{\alpha, \lambda} (1, t) t^{-1+\lambda_{2}-\alpha}\,dt \biggr)^{\frac{1}{q}} \biggl( \frac{\psi(n)^{q(\alpha+1-\lambda _{1})-1}}{\psi'(n)^{q-1}} \biggr)^{\frac{1}{q}}\\ &\qquad{} \times \Biggl(\sum_{m=m_{0}}^{\infty}K_{\alpha, \lambda}\bigl(\phi(m),\psi (n)\bigr) \frac{\phi(m)^{(\alpha+1-\lambda_{2})(p-1)}}{\psi(n)^{\alpha +1-\lambda_{1}}} \frac{\psi'(n)}{\phi'(m)^{p-1}} a_{m}^{p} \Biggr)^{\frac{1}{p}}. \end{aligned}

Therefore, by using Lemma 2.5, we get

\begin{aligned} &\Biggl\Vert \sum_{m=m_{0}}^{\infty}K_{\alpha, \lambda} \bigl(\phi(m), \psi(n)\bigr) a_{m} \Biggr\Vert _{p,\widetilde{w}_{1}}\\ &\quad= \Biggl\{ \sum_{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi(n)^{p(\alpha-\lambda _{1})+1}} \Biggl(\sum_{m=m_{0}}^{\infty}K_{\alpha, \lambda}\bigl(\phi(m),\psi(n)\bigr) a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} \\ &\quad\leq K_{\alpha}(\lambda)^{\frac{1}{q}} \Biggl(\sum _{n=n_{0}}^{\infty}\sum_{m=m_{0}}^{\infty}K_{\alpha, \lambda} \bigl(\phi(m), \psi(n)\bigr) \frac{\phi (m)^{(\alpha+1-\lambda_{2})(p-1)}}{\psi(n)^{\alpha+1-\lambda_{1}}} \frac {\psi'(n)}{\phi'(m)^{p-1}} a_{m}^{p} \Biggr)^{\frac{1}{p}}\\ &\quad=K_{\alpha}(\lambda)^{\frac{1}{q}} \Biggl( \sum _{m=m_{0}}^{\infty}W_{1} (m) \frac{\phi(m)^{p(\alpha+1-\lambda_{2})-1}}{\phi'(m)^{p-1}} a_{m}^{p} \Biggr)^{\frac{1}{p}}\\ &\quad< K_{\alpha}(\lambda) \|a\|_{p,w_{1}}. \end{aligned}

In the same manner, one can obtain

$$\Biggl\Vert \sum_{n=n_{0}}^{\infty}K_{\alpha, \lambda} \bigl(\phi(m), \psi(n)\bigr) b_{n} \Biggr\Vert _{q,\widetilde{w}_{2}} \leq K_{\alpha}(\lambda) \|b\|_{q,w_{2}}.$$

□

In view of Lemma 2.6, we can define a Hilbert-type operator $$T: \ell_{w_{1}, m_{0}}^{p} \rightarrow\ell_{\widetilde {w}_{1}, n_{0}}^{p}$$ by

$$(Ta) (n):= \sum_{m=m_{0}}^{\infty}K_{\alpha, \lambda} \bigl(\phi(m), \psi(n)\bigr) a_{m},\quad n\geq n_{0}, n\in\mathbb{Z}.$$

Similarly, define $$T: \ell_{w_{2}, n_{0}}^{q} \rightarrow\ell_{\widetilde {w}_{2}, m_{0}}^{q}$$ by

$$(Ta) (m):= \sum_{n=n_{0}}^{\infty}K_{\alpha, \lambda} \bigl(\phi(m), \psi(n)\bigr) b_{n},\quad m\geq m_{0}, m\in\mathbb{Z}.$$

It immediately follows from Lemma 2.6 that

$$\|T\|_{p}:= \sup_{\|a\|_{p,\widetilde{w}_{1}}=1} \|Ta\|_{p,\widetilde{w}_{1}} \leq K_{\alpha}(\lambda)$$

and

$$\|T\|_{q}:= \sup_{\|a\|_{p,\widetilde{w}_{2}}=1} \|Tb\|_{q,\widetilde{w}_{2}} \leq K_{\alpha}(\lambda).$$

Hence the operator T is bounded. The formal inner product $$(Ta, b)$$ of Ta and b is defined by

$$(Ta, b):= \sum_{n=n_{0}}^{\infty}\sum _{m=m_{0}}^{\infty}K_{\alpha, \lambda} \bigl(\phi(m), \psi(n) \bigr) a_{m} b_{n}.$$

### Lemma 2.7

Let $$p>1$$, $$\frac{1}{p} +\frac{1}{q}=1$$. Let $$\widetilde{a}=\{\widetilde {a}_{m}\}_{m=m_{0}}^{\infty}$$ and $$\widetilde{b}=\{\widetilde{b}_{n}\} _{n=n_{0}}^{\infty}$$ with $$\widetilde{a}_{m}= \frac{\phi'(m)}{\phi(m)^{\alpha +1-\lambda_{2}+\frac{\varepsilon}{p}}}$$ and $$\widetilde{b}_{n}= \frac{\psi '(n)}{\psi(n)^{\alpha+1-\lambda_{1}+\frac{\varepsilon}{q}}}$$ for $$0<\varepsilon< p\lambda_{i}$$, $$i=1,2$$. Then, as $$\varepsilon \rightarrow0+$$,

$$K_{\alpha}(\lambda) \bigl(1-o(1)\bigr) \sum_{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi (n)^{1+\varepsilon}}< (T\widetilde{a}, \widetilde{b})< K_{\alpha}(\lambda ) \bigl(1+o(1)\bigr) \sum_{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi(n)^{1+\varepsilon}}.$$

### Proof

We have

\begin{aligned} (T\widetilde{a}, \widetilde{b}) &= \sum_{n=n_{0}}^{\infty}\sum_{m=m_{0}}^{\infty}K_{\alpha, \lambda} \bigl( \phi(m), \psi(n)\bigr) \frac{\phi '(m)}{\phi^{(}m)^{\alpha+1-\lambda_{2}+\frac{\varepsilon}{p}}} \frac{\psi '(n)}{\psi(n)^{\alpha+1-\lambda_{1}+\frac{\varepsilon}{q}}} \\ &< \sum_{n=n_{0}}^{\infty}\int _{m_{0} -1}^{\infty}K_{\alpha, \lambda} \bigl(\phi (x), \psi(n)\bigr) \frac{\phi'(x)}{\phi(x)^{\alpha+1-\lambda_{2}+\frac {\varepsilon}{p}}} \frac{\psi'(n)}{\psi(n)^{\alpha+1-\lambda_{1}+\frac {\varepsilon}{q}}}\,dx. \end{aligned}

Setting $$t=\frac{\phi(x)}{\psi(n)}$$, we get

\begin{aligned} (T\widetilde{a}, \widetilde{b}) &< \sum_{n=n_{0}}^{\infty}\biggl( \int_{0}^{\infty}K_{\alpha, \lambda} (1, t) t^{-1+\lambda_{2}-\alpha-\frac {\varepsilon}{p}}\,dt \biggr) \frac{\psi'(n)}{\psi(n)^{1+\varepsilon}} \\ &=K_{\alpha}(\lambda) \bigl(1+o(1)\bigr) \sum _{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi (n)^{1+\varepsilon}}. \end{aligned}

Moreover,

\begin{aligned} (T\widetilde{a}, \widetilde{b}) &> \sum_{n=n_{0}}^{\infty}\biggl(\int_{\frac {\phi(m_{0})}{\psi(n)}}^{\infty}K_{\alpha, \lambda} (1, t) t^{-1+\lambda _{2}-\alpha-\frac{\varepsilon}{p}}\,dt \biggr)\frac{\psi'(n)}{\psi (n)^{1+\varepsilon}} \\ &=\sum_{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi(n)^{1+\varepsilon}} \biggl(\int_{0}^{\infty}K_{\alpha, \lambda} (1, t) t^{-1+\lambda_{2}-\alpha-\frac {\varepsilon}{p}}\,dt - \int_{0}^{\frac{\phi(m_{0})}{\psi(n)}} K_{\alpha, \lambda} (1, t) t^{-1+\lambda_{2}-\alpha-\frac{\varepsilon}{p}}\,dt \biggr). \end{aligned}

Note that the definition of $$K_{\alpha, \lambda} (x,y)$$ implies that

$$\int_{0}^{\infty}K_{\alpha, \lambda} (1, t) t^{-1+\lambda_{2}-\alpha-\frac {\varepsilon}{p}}\,dt = K_{\alpha}(\lambda_{2}) + o(1)$$

and

$$\sum_{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi(n)^{1+\varepsilon}}\int _{0}^{\frac{\phi(m_{0})}{\psi(n)}} K_{\alpha, \lambda} (1, t) t^{-1+\lambda _{2}-\alpha-\frac{\varepsilon}{p}}\,dt=O(1).$$

Thus, using the fact that for $$a>0$$,

$$\sum_{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi(n)^{1+\varepsilon}} = \frac {1}{\varepsilon}\bigl(1+o(1)\bigr) \quad\mbox{and}\quad \sum _{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi(n)^{1+a+\frac{\varepsilon }{q}}}=O(1)$$

as $$\varepsilon\rightarrow0+$$, we obtain

\begin{aligned} (T\widetilde{a}, \widetilde{b}) &> K_{\alpha}(\lambda) \bigl(1+ o(1) \bigr) \Biggl( \sum_{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi(n)^{1+\varepsilon}}-O(1) \Biggr) \\ &=K_{\alpha}(\lambda) \Biggl[1+ o(1) - O(1) \sum _{n=n_{0}}^{\infty}\biggl(\frac{\psi'(n)}{\psi(n)^{1+\varepsilon}} \biggr)^{-1} \Biggr] \sum_{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi(n)^{1+\varepsilon}} \\ &=K_{\alpha}(\lambda) \bigl(1-o(1)\bigr) \sum _{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi (n)^{1+\varepsilon}}, \end{aligned}

which completes the proof. □

### Theorem 2.8

Let $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$\lambda_{1}+\lambda_{2}=\lambda$$, $$\lambda_{1}, \lambda_{2} >0$$. For $$a_{m}, b_{n} \geq0$$ ($$m_{0}, n_{0} \in\mathbb{Z}$$), let $$a=\{a_{m}\} _{m=m_{0}}^{\infty}\in\ell_{w_{1},m_{0}}^{p}$$ and $$b=\{b_{n}\}_{n=n_{0}}^{\infty}\in\ell_{w_{2}, n_{0}}^{q}$$. Then, for $$\phi(x) \in F_{m_{0}}(r)$$ and $$\psi(y) \in F_{n_{0}}(s)$$ ($$r,s >1$$),

$$\|T\|_{p} =\|T\|_{q} =K_{\alpha}(\lambda).$$

### Proof

Suppose that $$\|T\|_{p}< K_{\alpha}(\lambda)$$. Consider $$\widetilde{a}_{m} = \phi' (m)\phi(m)^{-1+\lambda_{2}-\alpha-\frac{\varepsilon}{p}}$$ and $$\widetilde{b}_{n} = \phi' (n) \psi(n)^{-1+\lambda_{1}-\alpha-\frac {\varepsilon}{q}}$$, where $$m\geq m_{0}$$, $$n\geq n_{0}$$, $$m,n\in\mathbb {Z}$$, $$0<\varepsilon<p\lambda_{i}$$, $$i=1,2$$. A simple computation shows that $$\widetilde{a} \in\ell_{w_{1},m_{0}}^{p}$$ and $$\widetilde{b} \in\ell _{w_{2}, n_{0}}^{q}$$ with $$\|\widetilde{a}\|_{p, w_{1}}>0$$ and $$\|\widetilde {b}\|_{q,w_{2}}>0$$. Then

\begin{aligned} \|T\widetilde{a}\|_{p,\widetilde{w}_{1}} &= \Biggl\{ \sum_{n=n_{0}}^{\infty}\psi'(n)\psi(n)^{p(\lambda_{1}-\alpha)-1} \Biggl(\sum _{m=m_{0}}^{\infty}K_{\alpha, \lambda} \bigl(\phi(m), \psi(n) \bigr) \widetilde{a}_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} \\ &\leq\|T\|_{p} \|\widetilde{a}\|_{p, w_{1}}. \end{aligned}

Moreover, we have

\begin{aligned} (T\widetilde{a}, \widetilde{b}) &= \sum_{n=n_{0}}^{\infty}\sum_{m=m_{0}}^{\infty}K_{\alpha, \lambda} \bigl( \phi(m), \psi(n)\bigr) \widetilde{a}_{m} \widetilde{b}_{n} \\ &= \sum_{n=n_{0}}^{\infty}\Biggl\{ \psi'(n) \psi(n)^{p(\lambda_{1}-\alpha)-1} \Biggl(\sum _{m=m_{0}}^{\infty}K_{\alpha, \lambda} \bigl(\phi(m), \psi(n) \bigr) \widetilde{a}_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} \|\widetilde{b}\| _{q,w_{2}} \\ &\leq\|T\|_{p} \|\widetilde{a}\|_{p,w_{1}} \|\widetilde{b}\| _{q,w_{2}} \\ &= \|T\|_{p} \Biggl(\sum_{m=m_{0}}^{\infty}\frac{\phi'(m)}{\phi (m)^{1+\varepsilon}} \Biggr)^{\frac{1}{p}} \Biggl(\sum _{n=n_{0}}^{\infty}\frac {\psi'(n)}{\psi(n)^{1+\varepsilon}} \Biggr)^{\frac{1}{q}}. \end{aligned}
(2)

On the other hand, from Lemma 2.7 it follows

\begin{aligned} K_{\alpha}(\lambda) \bigl(1-o(1)\bigr) \sum_{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi (n)^{1+\varepsilon}}< (T\widetilde{a}, \widetilde{b}). \end{aligned}
(3)

Therefore, combining these inequalities (2) and (3),

\begin{aligned} K_{\alpha}(\lambda) \bigl(1-o(1)\bigr) \Biggl(\sum _{n=n_{0}}^{\infty}\frac{\psi '(n)}{\psi(n)^{1+\varepsilon}} \Biggr)^{\frac{1}{p}}\leq\|T\|_{p} \Biggl(\sum_{m=m_{0}}^{\infty}\frac{\phi'(m)}{\phi(m)^{1+\varepsilon}} \Biggr)^{\frac{1}{p}} . \end{aligned}

Since

$$\sum_{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi(n)^{1+\varepsilon}} = \frac {1}{\varepsilon}\bigl(1+o(1)\bigr) \quad\mbox{and}\quad \sum _{m=m_{0}}^{\infty}\frac{\phi '(m)}{\phi(m)^{1+\varepsilon}} = \frac{1}{\varepsilon} \bigl(1+o(1)\bigr)$$

as $$\varepsilon\rightarrow0+$$, we obtain that $$K_{\alpha}(\lambda) \leq\|T\|_{p}$$, which is a contradiction. Thus we conclude that $$\|T\|_{p} = K_{\alpha}(\lambda)$$. Applying the same argument, we have $$\|T\|_{q} = K_{\alpha}(\lambda)$$, which completes the proof. □

## 3 Two equivalent inequalities for the Hilbert-type operator

Equipped with the Hilbert-type operator defined as above, we have the following theorem.

### Theorem 3.1

Let $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$\lambda_{1}+\lambda_{2}=\lambda$$, $$\lambda _{1}, \lambda_{2} >0$$. For $$a_{m}, b_{n} \geq0$$ ($$m_{0}, n_{0} \in\mathbb{Z}$$), let $$a=\{a_{m}\} _{m=m_{0}}^{\infty}\in\ell_{w_{1},m_{0}}^{p}$$, $$b=\{b_{n}\}_{n=n_{0}}^{\infty}\in \ell_{w_{2}, n_{0}}^{q}$$ $$\|a\|_{p,w_{1}}>0$$, $$\|b\|_{q,w_{2}}>0$$. Then, for $$\phi (x) \in F_{m_{0}}(r)$$ and $$\psi(y) \in F_{n_{0}}(s)$$ ($$r,s >1$$), we have the following equivalent inequalities:

\begin{aligned} &(Ta, b)= \sum_{n=n_{0}}^{\infty}\sum _{m=m_{0}}^{\infty}K_{\alpha, \lambda} \bigl(\phi(m), \psi(n)\bigr) a_{m} b_{n} < K_{\alpha}(\lambda) \|a \|_{p,w_{1}} \|b\| _{q,w_{2}}, \end{aligned}
(4)
\begin{aligned} &\|Ta\|_{p, \widetilde{w}_{1}} < K_{\alpha}(\lambda) \|a\|_{p,w_{1}}. \end{aligned}
(5)

Furthermore, the constant factor $$K_{\alpha}(\lambda)$$ is the best possible.

### Proof

It follows from Hölder’s inequality that

\begin{aligned} (Ta, b)={}& \sum_{n=n_{0}}^{\infty}\sum _{m=m_{0}}^{\infty}K_{\alpha, \lambda} \bigl(\phi(m), \psi(n) \bigr) \biggl( \frac{\phi(m)^{\frac{\alpha+1-\lambda _{2}}{q}}}{\psi(n)^{\frac{\alpha+1-\lambda_{1}}{p}}} \frac{\psi'(n)^{\frac {1}{p}}}{\phi'(m)^{\frac{1}{q}}} a_{m} \biggr)\\ &{}\times\biggl( \frac{\psi(n)^{\frac {\alpha+1-\lambda_{1}}{p}}}{\phi(m)^{\frac{\alpha+1-\lambda_{2}}{q}}} \frac {\phi'(m)^{\frac{1}{q}}}{\psi'(n)^{\frac{1}{p}}} b_{n} \biggr) \\ \leq{}& \Biggl( \sum_{n=n_{0}}^{\infty}\sum _{m=m_{0}}^{\infty}K_{\alpha, \lambda } \bigl( \phi(m), \psi(n)\bigr) \frac{\phi(m)^{(\alpha+1-\lambda_{2})(p-1)}}{\psi (n)^{\alpha+1-\lambda_{1}}} \frac{\psi'(n)}{\phi'(m)^{p-1}} a_{m}^{p} \Biggr)^{\frac{1}{p}} \\ &{} \times \Biggl( \sum_{n=n_{0}}^{\infty}\sum _{m=m_{0}}^{\infty}K_{\alpha, \lambda} \bigl( \phi(m), \psi(n)\bigr) \frac{\psi(n)^{(\alpha+1-\lambda _{1})(q-1)}}{\phi(m)^{\alpha+1-\lambda_{2}}} \frac{\phi'(m)}{\psi '(n)^{q-1}} b_{n}^{q} \Biggr)^{\frac{1}{q}} \\ ={}& \Biggl( \sum_{m=m_{0}}^{\infty}W_{1} (m) \frac{\phi(m)^{p(\alpha+1-\lambda _{2})-1}}{\phi'(m)^{p-1}} a_{m}^{p} \Biggr)^{\frac{1}{p}} \Biggl( \sum_{n=n_{0}}^{\infty}W_{2} (n) \frac{\psi(n)^{q(\alpha+1-\lambda _{1})-1}}{\psi'(n)^{q-1}} b_{n}^{q} \Biggr)^{\frac{1}{q}}. \end{aligned}

Applying Lemma 2.5, we see that

$$(Ta, b) < K_{\alpha}(\lambda) \|a\|_{p,w_{1}} \|b \|_{q,w_{2}}.$$

In order to prove that inequality (4) implies inequality (5), we define as follows:

$$\widetilde{b}_{n}:= \frac{\psi'(n)}{\psi(n)^{p(\alpha-\lambda_{1})+1}} \Biggl(\sum _{m=m_{0}}^{\infty}K_{\alpha, \lambda} \bigl(\phi(m), \psi(n) \bigr) \Biggr)^{p-1}$$

for $$n\geq n_{0}$$, $$n\in\mathbb{Z}$$. Then we see that $$\widetilde{b} \in \ell_{w_{2},n_{0}}^{q}$$ and $$\|\widetilde{b}\|_{q,w_{2}}>0$$ as before. Thus using inequality (4) shows that

\begin{aligned} \|\widetilde{b}\|_{q,w_{2}}^{q} &= \sum _{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi(n)^{p(\alpha-\lambda_{1})+1}} \Biggl(\sum _{m=m_{0}}^{\infty}K_{\alpha, \lambda}\bigl(\phi(m),\psi(n) \bigr) a_{m} \Biggr)^{p} \\ &= \sum_{n=n_{0}}^{\infty}\sum _{m=m_{0}}^{\infty}K_{\alpha, \lambda} \bigl(\phi (m), \psi(n)\bigr) a_{m} \widetilde{b}_{n} < K_{\alpha}(\lambda) \|a\|_{p,w_{1}} \|\widetilde{b} \|_{q,w_{2}}, \end{aligned}

which gives $$\|Ta\|_{p,\widetilde{w}_{1}} = \|\widetilde{b}\|_{q,w_{2}}^{q-1}< K_{\alpha}(\lambda) \|a\|_{p,w_{1}}$$. Hence inequality (4) implies inequality (5).

Now suppose that inequality (5) holds for any $$a \in\ell _{w_{1},m_{0}}^{p}$$.

\begin{aligned} (Ta, b)&= \sum_{n=n_{0}}^{\infty}\sum _{m=m_{0}}^{\infty}K_{\alpha, \lambda} \bigl(\phi(m), \psi(n) \bigr) a_{m} b_{n} \\ &= \sum_{n=n_{0}}^{\infty}\Biggl( \frac{\psi'(n)^{\frac{1}{p}}}{\psi(n)^{\alpha -\lambda_{1}+\frac{1}{p}}}\sum_{m=m_{0}}^{\infty}K_{\alpha, \lambda} \bigl(\phi (m), \psi(n)\bigr) a_{m} \Biggr) \biggl( \frac{\psi(n)^{\alpha-\lambda_{1}+\frac {1}{p}}}{\psi'(n)^{\frac{1}{p}}} b_{n} \biggr) \\ &\leq \Biggl\{ \sum_{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi(n)^{p(\alpha -\lambda_{1})+1}} \Biggl(\sum_{m=m_{0}}^{\infty}K_{\alpha, \lambda} \bigl(\phi(m), \psi(n)\bigr) a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} \|b\|_{q,w_{2}} \\ &< K_{\alpha}(\lambda) \|a\|_{p,w_{1}} \|b\|_{q,w_{2}}, \end{aligned}

which means that inequality (5) implies inequality (4). Therefore inequality (4) is equivalent to inequality (5). Furthermore, Theorem 2.8 implies that the constant factor $$K_{\alpha}(\lambda)$$ in inequalities (4) and (5) is the best possible, which completes the proof. □

## 4 Applications to various Hilbert-type inequalities

In this section, we apply our previous theorems to obtain several Hilbert-type inequalities. Recall that the beta function $$B(u,v)$$ is defined by

$$B(u,v):= \int_{0}^{\infty}\frac{t^{u-1}}{(1+t)^{u+v}}\,dt = B(u,v) \quad(u,v>0).$$

Define the function $$K_{\alpha,\lambda} (x,y)$$ by

$$K_{\alpha,\lambda} (x,y):= \frac{(xy)^{\alpha}}{(x+y)^{\lambda}}$$

for $$\lambda>\alpha\geq0$$. Then $$K_{\alpha,\lambda} (x,y)$$ is a symmetric homogeneous function of degree $$2\alpha-\lambda$$ and is decreasing with respect to x and y, respectively. Moreover,

$$\sum_{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi(n)^{1+\varepsilon}}\int _{0}^{\frac{\phi(m_{0})}{\psi(n)}} K_{\alpha, \lambda} (1, t) t^{-1+\lambda _{2}-\alpha-\frac{\varepsilon}{p}}\,dt=O(1).$$

To see this, for $$0< \varepsilon< p\lambda_{2}$$,

\begin{aligned} \sum_{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi(n)^{1+\varepsilon}}\int _{0}^{\frac{\phi(m_{0})}{\psi(n)}} \frac{t^{-1+\lambda_{2}-\frac{\varepsilon }{p}}}{(1+t)^{\lambda}}\,dt &\leq\sum _{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi (n)^{1+\varepsilon}}\int _{0}^{\frac{\phi(m_{0})}{\psi(n)}} t^{-1+\lambda _{2}-\frac{\varepsilon}{p}}\,dt \\ &=\sum_{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi(n)^{1+\varepsilon}} \frac {1}{\lambda_{2}-\frac{\varepsilon}{p}} \biggl(\frac{\phi(m_{0})}{\psi (n)} \biggr)^{\lambda_{2}-\frac{\varepsilon}{p}} \\ &=\frac{\phi(m_{0})^{\lambda_{2}-\frac{\varepsilon}{p}}}{\lambda_{2}-\frac {\varepsilon}{p}} \sum_{n=n_{0}}^{\infty}\frac{\psi'(n)}{\psi(n)^{1+\lambda _{2}+\frac{\varepsilon}{q}}} \\ &=O(1). \end{aligned}

Note that since

\begin{aligned} \widetilde{K}_{\alpha, \lambda}(\lambda_{i},\varepsilon) &:= \int _{0}^{\infty}K_{\alpha, \lambda}(1,t)t^{-1+\lambda_{i}-\alpha-\varepsilon}\,dt\\ &= \int_{0}^{\infty}\frac{t^{-1+\lambda_{i}-\varepsilon}}{(1+t)^{\lambda}}\,dt, \end{aligned}

we see that

\begin{aligned} \widetilde{K}_{\alpha, \lambda}(\lambda_{i},\varepsilon) \rightarrow \int_{0}^{\infty}\frac{t^{\lambda_{i}-1}}{(1+t)^{\lambda}}\,dt = B( \lambda_{1}, \lambda_{2})=K_{\alpha}( \lambda_{i}) = K_{\alpha}(\lambda) \end{aligned}

as $$\varepsilon\rightarrow0+$$. Therefore from Theorem 3.1 we observe the following.

### Corollary 4.1

Let $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$\lambda_{1}+\lambda_{2}=\lambda$$, $$\lambda_{1}, \lambda_{2} >0$$, $$\lambda>\alpha\geq0$$. For $$a_{m}, b_{n} \geq0$$ ($$m_{0}, n_{0} \in\mathbb{Z}$$), let $$a=\{a_{m}\} _{m=m_{0}}^{\infty}\in\ell_{w_{1},m_{0}}^{p}$$, $$b=\{b_{n}\}_{n=n_{0}}^{\infty}\in \ell_{w_{2}, n_{0}}^{q}$$ and $$\|a\|_{p,w_{1}}>0$$, $$\|b\|_{q,w_{2}}>0$$. Then, for $$\phi(x) \in F_{m_{0}}(r)$$ and $$\psi(y) \in F_{n_{0}}(s)$$ ($$r,s >1$$), we have the following equivalent inequalities:

\begin{aligned}& \sum_{n=n_{0}}^{\infty}\sum _{m=m_{0}}^{\infty}\frac{\phi(m)^{\alpha}\psi (n)^{\alpha}a_{m} b_{n}}{(\phi(m)+\psi(n))^{\lambda}} < B( \lambda_{1}, \lambda _{2}) \|a\|_{p,w_{1}} \|b \|_{q,w_{2}},\\& \Biggl\{ \sum_{n=n_{0}}^{\infty}\psi'(n)\psi(n)^{p(\lambda_{1}-\alpha)-1} \Biggl(\sum _{m=m_{0}}^{\infty}\frac{\phi(m)^{\alpha}\psi(n)^{\alpha}a_{m}}{(\phi (m)+\psi(n))^{\lambda}} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} < B(\lambda_{1}, \lambda_{2}) \|a\|_{p,w_{1}}. \end{aligned}

Furthermore, the constant factor $$B(\lambda_{1}, \lambda_{2})$$ is the best possible.

As applications, we have the following.

Case 1. Let $$\phi(x)=x^{\beta}$$ and $$\psi(x)=x^{\gamma}$$ ($$\beta, \gamma>0$$) for $$m_{0}=n_{0}=1$$. For $$0<\lambda_{i} <\alpha+\min\{ \frac{1}{\beta}, \frac{1}{\gamma}\}$$ and $$0\leq\alpha<\lambda$$, one has the following equivalent inequalities:

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{(m^{\beta}n^{\gamma})^{\alpha}}{(m^{\beta}+ n^{\gamma})^{\lambda}} a_{m} b_{n} < \frac{B(\lambda_{1}, \lambda _{2})}{\beta^{\frac{1}{q}}\gamma^{\frac{1}{p}}} \|a\|_{p,w_{1}} \|b \|_{q,w_{2}}, \\& \Biggl\{ \sum_{n=1}^{\infty}n^{\gamma p(\lambda_{1}-\alpha)-1} \Biggl(\sum_{m=1}^{\infty}\frac{(m^{\beta}n^{\gamma})^{\alpha}}{(m^{\beta}+ n^{\gamma})^{\lambda}}a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} < \frac{B(\lambda_{1}, \lambda _{2})}{\beta^{\frac{1}{q}}\gamma^{\frac{1}{p}}} \|a \|_{p,w_{1}}, \end{aligned}

where $$w_{1} (m)=m^{p(1-\lambda_{2}\beta+\alpha\beta)-1}$$ and $$w_{2}(n)=n^{q(1-\lambda_{1}\gamma+\alpha\gamma)-1}$$.

1. (I)

For $$\lambda_{1}=\frac{\lambda}{p}$$ and $$\lambda _{2}=\frac{\lambda}{q}$$ with $$0<\lambda_{i}<\alpha+\min\{\frac{1}{\beta}, \frac{1}{\gamma}\}$$ and $$0 \leq\alpha< \lambda$$, one has the following equivalent inequalities:

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{(m^{\beta}n^{\gamma})^{\alpha}}{(m^{\beta}+ n^{\gamma})^{\lambda}} a_{m} b_{n} < \frac{B(\frac{\lambda }{p},\frac{\lambda}{q})}{\beta^{\frac{1}{q}}\gamma^{\frac{1}{p}}} \|a\| _{p,w_{1}} \|b \|_{q,w_{2}},\\& \Biggl\{ \sum_{n=1}^{\infty}n^{\gamma(\lambda-p\alpha)-1} \Biggl(\sum_{m=1}^{\infty}\frac{(m^{\beta}n^{\gamma})^{\alpha}}{(m^{\beta}+ n^{\gamma})^{\lambda}}a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} < \frac{B(\frac{\lambda }{p},\frac{\lambda}{q})}{\beta^{\frac{1}{q}}\gamma^{\frac{1}{p}}} \|a \|_{p,w_{1}}, \end{aligned}

where $$w_{1} (m)=m^{(p-1)(1-\lambda\beta)+p\alpha\beta}$$ and $$w_{2}(n)=n^{(q-1)(1-\lambda\gamma)+q\alpha\gamma}$$.

2. (II)

For $$\lambda_{1}=\frac{\lambda}{q}$$ and $$\lambda _{2}=\frac{\lambda}{p}$$ with $$0<\lambda_{i}<\alpha+\min\{\frac{1}{\beta}, \frac{1}{\gamma}\}$$ and $$0 \leq\alpha< \lambda$$, one has the following equivalent inequalities:

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{(m^{\beta}n^{\gamma})^{\alpha}}{(m^{\beta}+ n^{\gamma})^{\lambda}} a_{m} b_{n} < \frac{B(\frac{\lambda }{p},\frac{\lambda}{q})}{\beta^{\frac{1}{q}}\gamma^{\frac{1}{p}}} \|a\| _{p,w_{1}} \|b \|_{q,w_{2}},\\& \Biggl\{ \sum_{n=1}^{\infty}n^{\gamma\lambda(p-1)-p\alpha\gamma-1} \Biggl(\sum_{m=1}^{\infty}\frac{(m^{\beta}n^{\gamma})^{\alpha}}{(m^{\beta}+ n^{\gamma})^{\lambda}}a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} < \frac{B(\frac{\lambda }{p},\frac{\lambda}{q})}{\beta^{\frac{1}{q}}\gamma^{\frac{1}{p}}} \|a \|_{p,w_{1}}, \end{aligned}

where $$w_{1} (m)=m^{p-1-\beta\lambda+p\alpha\beta}$$ and $$w_{2}(n)=n^{q-1-\gamma\lambda+q\alpha\gamma}$$.

3. (III)

Let $$\lambda_{1}=\frac{p+\lambda-2}{p}$$, $$\lambda _{2}=\frac{q+\lambda-2}{q}$$, $$\lambda>\max\{2-p, 2-q\}$$, $$0 < \beta< \frac{p}{p+\lambda-2-p\alpha}$$, $$0 < \gamma< \frac{q}{q+\lambda -2-q\alpha}$$, $$0\leq\alpha< \min\{\frac{p+\lambda-2}{p}, \frac {q+\lambda-2}{q}\}$$. Then one has the following equivalent inequalities:

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{(m^{\beta}n^{\gamma})^{\alpha}}{(m^{\beta}+ n^{\gamma})^{\lambda}} a_{m} b_{n} < \frac{B(\frac{p+\lambda -2}{p}, \frac{q+\lambda-2}{q})}{\beta^{\frac{1}{q}}\gamma^{\frac{1}{p}}} \| a\|_{p,w_{1}} \|b \|_{q,w_{2}},\\& \Biggl\{ \sum_{n=1}^{\infty}n^{\gamma(p+\lambda-2)-p\alpha\gamma-1} \Biggl(\sum_{m=1}^{\infty}\frac{(m^{\beta}n^{\gamma})^{\alpha}}{(m^{\beta}+ n^{\gamma})^{\lambda}}a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} < \frac{B(\frac{p+\lambda -2}{p}, \frac{q+\lambda-2}{q})}{\beta^{\frac{1}{q}}\gamma^{\frac{1}{p}}} \| a \|_{p,w_{1}}, \end{aligned}

where $$w_{1} (m)=m^{(p-1)(1-\beta(q+\lambda-2))+p\alpha\beta}$$ and $$w_{2}(n)=n^{(q-1)(1-\gamma(p+\lambda-2))+q\alpha\gamma}$$.

4. (IV)

Let $$\lambda_{1}=\frac{q+\lambda-2}{q}$$, $$\lambda _{2}=\frac{p+\lambda-2}{p}$$, $$\lambda>\max\{2-p, 2-q\}$$, $$0 < \beta< \frac{q}{q+\lambda-2-q\alpha}$$, $$0 < \gamma< \frac{p}{p+\lambda -2-p\alpha}$$, $$0\leq\alpha< \min\{\frac{p+\lambda-2}{p}, \frac {q+\lambda-2}{q}\}$$. Then one has the following equivalent inequalities:

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{(m^{\beta}n^{\gamma})^{\alpha}}{(m^{\beta}+ n^{\gamma})^{\lambda}} a_{m} b_{n} < \frac{B(\frac{p+\lambda -2}{p}, \frac{q+\lambda-2}{q})}{\beta^{\frac{1}{q}}\gamma^{\frac{1}{p}}} \| a\|_{p,w_{1}} \|b \|_{q,w_{2}},\\& \Biggl\{ \sum_{n=1}^{\infty}n^{\gamma(p-1)(q+\lambda-2)-p\alpha\gamma-1} \Biggl(\sum_{m=1}^{\infty}\frac{(m^{\beta}n^{\gamma})^{\alpha}}{(m^{\beta}+ n^{\gamma})^{\lambda}}a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} < \frac{B(\frac {p+\lambda-2}{p}, \frac{q+\lambda-2}{q})}{\beta^{\frac{1}{q}}\gamma^{\frac {1}{p}}} \|a \|_{p,w_{1}}, \end{aligned}

where $$w_{1} (m)=m^{p-1-\beta(p+\lambda-2)+p\alpha\beta}$$ and $$w_{2}(n)=n^{q-1-\gamma(q+\lambda-2)+q\alpha\gamma}$$.

Case 2. For $$A, B>0$$, let $$\phi(x)=A(\ln x)^{\beta}$$ and $$\psi(x)=B(\ln x)^{\gamma}$$ ($$\beta, \gamma>0$$), $$m_{0}=n_{0}=2$$. For $$0<\lambda_{i}< \alpha+\min\{\frac{1}{\beta}, \frac{1}{\gamma}\}$$ and $$0\leq\alpha< \lambda$$, one has the following equivalent inequalities:

\begin{aligned}& \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{((\ln m)^{\beta}(\ln n)^{\gamma})^{\alpha}}{(A(\ln m)^{\beta}+ B(\ln n)^{\gamma})^{\lambda}} a_{m} b_{n} < \frac{B(\lambda_{1}, \lambda_{2})}{A^{\lambda_{2}}B^{\lambda_{1}}\beta^{\frac {1}{q}}\gamma^{\frac{1}{p}}} \|a\|_{p,w_{1}} \|b \|_{q,w_{2}},\\& \Biggl\{ \sum_{n=2}^{\infty}\frac{1}{n}(\ln n)^{p\gamma(\lambda_{1}-\alpha )-1} \Biggl(\sum _{m=2}^{\infty}\frac{((\ln m)^{\beta}(\ln n)^{\gamma})^{\alpha}}{(A(\ln m)^{\beta}+ B(\ln n)^{\gamma})^{\lambda}} a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} < \frac{B(\lambda_{1}, \lambda_{2})}{A^{\lambda_{2}}B^{\lambda _{1}}\beta^{\frac{1}{q}}\gamma^{\frac{1}{p}}} \|a \|_{p,w_{1}}, \end{aligned}

where $$w_{1} (m)=m^{p-1}(\ln m)^{p(1-\lambda_{2}\beta+\alpha\beta)-1}$$ and $$w_{2}(n)=n^{q-1}(\ln n)^{q(1-\lambda_{1} \gamma+\alpha\gamma)-1}$$.

1. (I)

For $$\lambda_{1}=\frac{\lambda}{p}$$ and $$\lambda _{2}=\frac{\lambda}{q}$$ with $$0<\lambda_{i}<\alpha+\min\{\frac{1}{\beta}, \frac{1}{\gamma}\}$$ and $$0 \leq\alpha< \lambda$$, one has the following equivalent inequalities:

\begin{aligned}& \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{((\ln m)^{\beta}(\ln n)^{\gamma})^{\alpha}}{(A(\ln m)^{\beta}+ B(\ln n)^{\gamma})^{\lambda}} a_{m} b_{n} < \frac {B(\frac{\lambda}{p}, \frac{\lambda}{q})}{A^{\lambda_{2}}B^{\lambda _{1}}\beta^{\frac{1}{q}}\gamma^{\frac{1}{p}}} \|a\|_{p,w_{1}} \|b \|_{q,w_{2}},\\& \Biggl\{ \sum_{n=2}^{\infty}\frac{1}{n}(\ln n)^{\gamma-p\alpha\gamma-1} \Biggl(\sum _{m=2}^{\infty}\frac{((\ln m)^{\beta}(\ln n)^{\gamma})^{\alpha}}{(A(\ln m)^{\beta}+ B(\ln n)^{\gamma})^{\lambda}} a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} \\& \quad< \frac{B(\frac{\lambda}{p}, \frac{\lambda}{q})}{A^{\lambda _{2}}B^{\lambda_{1}}\beta^{\frac{1}{q}}\gamma^{\frac{1}{p}}} \|a \|_{p,w_{1}}, \end{aligned}

where $$w_{1} (m)=m^{p-1}(\ln m)^{(p-1)(1-\lambda\beta)+p\alpha\beta}$$ and $$w_{2}(n)=n^{q-1}(\ln n)^{(q-1)(1-\lambda\gamma)+q\alpha\gamma}$$.

2. (II)

Let $$\lambda_{1}=\frac{p+\lambda-2}{p}$$, $$\lambda _{2}=\frac{q+\lambda-2}{q}$$, $$\lambda>\max\{2-p, 2-q\}$$, $$0 < \beta< \frac{p}{p+\lambda-2-p\alpha}$$, $$0 < \gamma< \frac{q}{q+\lambda -2-q\alpha}$$, $$0\leq\alpha< \min\{\frac{p+\lambda-2}{p}, \frac {q+\lambda-2}{q}\}$$. Then one has the following equivalent inequalities:

\begin{aligned}& \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{((\ln m)^{\beta}(\ln n)^{\gamma})^{\alpha}}{(A(\ln m)^{\beta}+ B(\ln n)^{\gamma})^{\lambda}} a_{m} b_{n} < \frac {B(\frac{p+\lambda-2}{p}, \frac{q+\lambda-2}{q})}{A^{\lambda _{2}}B^{\lambda_{1}}\beta^{\frac{1}{q}}\gamma^{\frac{1}{p}}} \|a\|_{p,w_{1}} \| b \|_{q,w_{2}},\\& \Biggl\{ \sum_{n=2}^{\infty}\frac{1}{n}(\ln n)^{\gamma(p+\lambda-2)-p\alpha \gamma-1} \Biggl(\sum _{m=2}^{\infty}\frac{((\ln m)^{\beta}(\ln n)^{\gamma})^{\alpha}}{(A(\ln m)^{\beta}+ B(\ln n)^{\gamma})^{\lambda}} a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} \\& \quad< \frac{B(\frac{p+\lambda-2}{p}, \frac{q+\lambda-2}{q})}{A^{\lambda _{2}}B^{\lambda_{1}}\beta^{\frac{1}{q}}\gamma^{\frac{1}{p}}} \|a\|_{p,w_{1}}, \end{aligned}

where $$w_{1} (m)=m^{p-1}(\ln m)^{(p-1)(1-\beta(q+\lambda-2))+p\alpha\beta }$$ and $$w_{2}(n)=n^{q-1}(\ln n)^{(q-1)(1-\gamma(p+\lambda-2))+q\alpha \gamma}$$.

Case 3. For $$A, B>0$$, let $$\phi(x)=A(\ln x)^{\beta}$$ and $$\psi(x)=Bx^{\gamma}$$ ($$\beta, \gamma>0$$), $$m_{0}=2$$, $$n_{0}=1$$. For $$0<\lambda_{i}< \alpha+\min\{\frac{1}{\beta}, \frac{1}{\gamma}\}$$ and $$0\leq\alpha< \lambda$$, one has the following equivalent inequalities:

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=2}^{\infty}\frac{((\ln m)^{\beta}n^{\gamma})^{\alpha}}{(A(\ln m)^{\beta}+ Bn^{\gamma})^{\lambda}} a_{m} b_{n} < \frac {B(\lambda_{1}, \lambda_{2})}{A^{\lambda_{2}}B^{\lambda_{1}}\beta^{\frac {1}{q}}\gamma^{\frac{1}{p}}} \|a\|_{p,w_{1}} \|b \|_{q,w_{2}},\\& \Biggl\{ \sum_{n=1}^{\infty}n^{p\gamma(\lambda_{1}-\alpha)-1} \Biggl(\sum_{m=2}^{\infty}\frac{((\ln m)^{\beta}n^{\gamma})^{\alpha}}{(A(\ln m)^{\beta}+ Bn^{\gamma})^{\lambda}} a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} < \frac{B(\lambda _{1}, \lambda_{2})}{A^{\lambda_{2}}B^{\lambda_{1}}\beta^{\frac{1}{q}}\gamma ^{\frac{1}{p}}} \|a\|_{p,w_{1}}, \end{aligned}

where $$w_{1} (m)=m^{p-1}(\ln m)^{p(1-\lambda_{2}\beta+\alpha\beta)-1}$$ and $$w_{2}(n)=n^{q(1-\lambda_{1} \gamma+\alpha\gamma)-1}$$.

1. (I)

For $$\lambda_{1}=\frac{\lambda}{p}$$ and $$\lambda _{2}=\frac{\lambda}{q}$$ with $$0<\lambda_{i}<\alpha+\min\{\frac{1}{\beta}, \frac{1}{\gamma}\}$$ and $$0 \leq\alpha< \lambda$$, one has the following equivalent inequalities:

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=2}^{\infty}\frac{((\ln m)^{\beta}n^{\gamma})^{\alpha}}{(A(\ln m)^{\beta}+ Bn^{\gamma})^{\lambda}} a_{m} b_{n} < \frac{B(\frac {\lambda}{p}, \frac{\lambda}{q})}{A^{\lambda_{2}}B^{\lambda_{1}}\beta^{\frac {1}{q}}\gamma^{\frac{1}{p}}} \|a\|_{p,w_{1}} \|b \|_{q,w_{2}}, \\& \Biggl\{ \sum_{n=1}^{\infty}n^{\gamma(1-p\alpha)-1} \Biggl(\sum_{m=2}^{\infty}\frac{((\ln m)^{\beta}n^{\gamma})^{\alpha}}{(A(\ln m)^{\beta}+ Bn^{\gamma})^{\lambda}} a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} < \frac{B(\frac {\lambda}{p}, \frac{\lambda}{q})}{A^{\lambda_{2}}B^{\lambda_{1}}\beta^{\frac {1}{q}}\gamma^{\frac{1}{p}}} \|a\|_{p,w_{1}}, \end{aligned}

where $$w_{1} (m)=m^{p-1}(\ln m)^{(p-1)(1-\lambda\beta)+p\alpha\beta}$$ and $$w_{2}(n)=n^{(q-1)(1-\lambda\gamma)+q\alpha\gamma}$$.

2. (II)

Let $$\lambda_{1}=\frac{p+\lambda-2}{p}$$, $$\lambda _{2}=\frac{q+\lambda-2}{q}$$, $$\lambda>\max\{2-p, 2-q\}$$, $$0 < \beta< \frac{p}{p+\lambda-2-p\alpha}$$, $$0 < \gamma< \frac{q}{q+\lambda -2-q\alpha}$$, $$0\leq\alpha< \min\{\frac{p+\lambda-2}{p}, \frac {q+\lambda-2}{q}\}$$. Then one has the following equivalent inequalities:

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=2}^{\infty}\frac{((\ln m)^{\beta}n^{\gamma})^{\alpha}}{(A(\ln m)^{\beta}+ Bn^{\gamma})^{\lambda}} a_{m} b_{n} < \frac{B(\frac {p+\lambda-2}{p}, \frac{q+\lambda-2}{q})}{A^{\lambda_{2}}B^{\lambda _{1}}\beta^{\frac{1}{q}}\gamma^{\frac{1}{p}}} \|a\|_{p,w_{1}} \|b \|_{q,w_{2}},\\& \Biggl\{ \sum_{n=1}^{\infty}n^{\gamma(p+\lambda-2)-p\alpha\gamma-1} \Biggl(\sum_{m=2}^{\infty}\frac{((\ln m)^{\beta}n^{\gamma})^{\alpha}}{(A(\ln m)^{\beta}+ Bn^{\gamma})^{\lambda}} a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} < \frac {B(\frac{p+\lambda-2}{p}, \frac{q+\lambda-2}{q})}{A^{\lambda _{2}}B^{\lambda_{1}}\beta^{\frac{1}{q}}\gamma^{\frac{1}{p}}} \|a\|_{p,w_{1}}, \end{aligned}

where $$w_{1} (m)=m^{p-1}(\ln m)^{(p-1)(1-\beta(q+\lambda-2))+p\alpha\beta }$$ and $$w_{2}(n)=n^{(q-1)(1-\gamma(p+\lambda-2))+q\alpha\gamma}$$.

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## Acknowledgements

This research was supported by the Sookmyung Women’s University Research Grants 2012.

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Correspondence to Keomkyo Seo.

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Seo, K. A Hilbert-type operator with a symmetric homogeneous kernel of two parameters and its applications. J Inequal Appl 2015, 266 (2015). https://doi.org/10.1186/s13660-015-0788-z

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• DOI: https://doi.org/10.1186/s13660-015-0788-z