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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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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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