Open Access

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

Journal of Inequalities and Applications20152015:266

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

Received: 10 July 2015

Accepted: 17 August 2015

Published: 2 September 2015

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.

Keywords

Hilbert-type operator Hilbert-type inequality beta function kernel norm

MSC

47A07 26D15

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

Declarations

Acknowledgements

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

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, Sookmyung Women’s University

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