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 [2–13] 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 [16–23] 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) $\phi(x)$ is strictly increasing in $(n_{0}-1,\infty)$ with $\phi((n_{0}-1)+)=0$, $\phi(\infty) =\infty$.

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

• (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) 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 b̃ 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 $w_{2}(n)=n^{q(1-\lambda_{1}\gamma+\alpha\gamma)-1}$.

• (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 $w_{2}(n)=n^{(q-1)(1-\lambda\gamma)+q\alpha\gamma}$.

• (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 $w_{2}(n)=n^{q-1-\gamma\lambda+q\alpha\gamma}$.

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

• (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 $w_{2}(n)=n^{q-1}(\ln n)^{q(1-\lambda_{1} \gamma+\alpha\gamma)-1}$.

• (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 $w_{2}(n)=n^{q-1}(\ln n)^{(q-1)(1-\lambda\gamma)+q\alpha\gamma}$.

• (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 $w_{2}(n)=n^{q(1-\lambda_{1} \gamma+\alpha\gamma)-1}$.

• (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 $w_{2}(n)=n^{(q-1)(1-\lambda\gamma)+q\alpha\gamma}$.

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