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On a class of Hilbert-type inequalities in the whole plane involving some classical kernel functions

Part of: Special classes of linear operators Inequalities

Published online by Cambridge University Press:  06 October 2022

Minghui You Open the ORCID record for Minghui You [Opens in a new window]
Minghui You*
Affiliation:
Department of Mathematics, Zhejiang Institute of Mechanical and Electrical Engineering, Hangzhou 310053, China (youminghui@zime.edu.cn)
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Abstract

In this paper, by the introduction of several parameters, we construct a new kernel function which is defined in the whole plane and includes some classical kernel functions. Estimating the weight functions with the techniques of real analysis, we establish a new Hilbert-type inequality in the whole plane, and the constant factor of the newly obtained inequality is proved to be the best possible. Additionally, by means of the partial fraction expansion of the tangent function, some special and interesting inequalities are presented at the end of the paper.

Keywords

Hilbert-type inequalityweight functionpartial fraction expansionwhole planeclassical kernel functions

MSC classification

Primary: 26D15: Inequalities for sums, series and integrals 26D10: Inequalities involving derivatives and differential and integral operators
Secondary: 47B38: Operators on function spaces (general)

Information

Type
Research Article
Information
Proceedings of the Edinburgh Mathematical Society , Volume 65 , Issue 3 , August 2022 , pp. 833 - 846
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Copyright
Copyright © The Author(s), 2022. Published by Cambridge University Press on Behalf of The Edinburgh Mathematical Society

1. Introduction

Let $S$ be a measurable set. $f(x)$ and $\mu (x)$ are two non-negative measurable functions defined on $S$. Define

\[ L_{\mu}^{p}(S):=\left\{f:\|f\|_{p,\mu}:=\left[{\displaystyle\int_{S}\mu(x)f^{p}(x)\mathrm{d}}x\right]^{\frac{1}{p}}<\infty\right\}. \]

Particularly, for $\mu (x)=1$, $\|f\|_{p,\mu }$ and $L^{p}_{\mu }(S)$ are abbreviated as $\|f\|_p$ and $L^{p}(S)$ respectively.

Suppose that $f$, $g$ are two real-valued functions, $f, g\geq 0$ and $f, g\in L^{2}(\mathbb {R}^{+})$. Then [Reference Hardy, Littlewood and Pólya3]

(1.1)\begin{equation} \displaystyle\int_0^{\infty}\int_0^{\infty} \displaystyle\frac{f(x)g(y)}{x+y} \mathrm{d}x\mathrm{d}y<\pi\|f\|_{2}\|g\|_{2}, \end{equation}

where the constant factor $\pi$ is the best possible. (1.1) is usually known as Hilbert inequality. Let $p>1$, and $\frac {1}{p}+\frac {1}{q}=1$, then we have a more general form of (1.1) as follows:

(1.2)\begin{equation} \displaystyle\int_0^{\infty}\int_0^{\infty} \displaystyle\frac{f(x)g(y)}{x+y} \mathrm{d}x\mathrm{d}y<\frac{\pi}{\sin\frac{\pi}{p}}\|f\|_{p}\|g\|_{q}. \end{equation}

In addition, it can also be found in [Reference Hardy, Littlewood and Pólya3] that some other inequalities similar to (1.1), such as

(1.3)\begin{align} \displaystyle\int_0^{\infty}\int_0^{\infty}\displaystyle\frac{f(x)g(y)}{\max \{x,y\}} \mathrm{d}x\mathrm{d}y& < pq\|f\|_{p}\|g\|_{q}, \end{align}
(1.4)\begin{align} \int_{0}^{\infty}\int_{0}^{\infty}\frac{\log\frac{x}{y}}{x-y}f(x)g(y)\mathrm{d} x\mathrm{d} y & <\left[\frac{\pi}{\sin\frac{\pi}{p}}\right]^{2}\|f\|_{p}\|g\|_{q}. \end{align}

Such inequalities as (1.2)–(1.4) are called Hilbert-type inequality. Since 1990s, by introducing parameters and special functions, researchers have established quite a lot of extensions and analogies of (1.2)–(1.4), such as the following one which was proved by Yang [Reference Yang26] in 2004:

(1.5)\begin{equation} \displaystyle\int_{0}^{\infty}\int_{0}^{\infty}\displaystyle\frac{f(x)g(y)}{x^{r}+y^{r}}\mathrm{d} x\mathrm{d} y<\frac{\pi}{r\sin{\frac{\pi}{s}}}\|f\|_{p,\mu}\|g\|_{q,\nu}, \end{equation}

where $r>0,$ $\mu (x)=x^{p(1-\frac {r}{s})-1},$ $\nu (y)=y^{q(1-\frac {r}{t})-1}$ and $\frac {1}{s}+\frac {1}{t}=1.$ Furthermore, Yang [Reference Yang28] established an analogy of (1.4) as follows:

(1.6)\begin{equation} \displaystyle\int_{0}^{\infty}\int_{0}^{\infty}\displaystyle\frac{\mid\log\frac{x}{y} \mid}{x+y}f(x)g(y)\mathrm{d} x\mathrm{d} y<8L_0\|f\|_{2}\|g\|_{2}, \end{equation}

where $L_0=\Sigma _{j=0}^{\infty }\frac {(-1)^{j}}{(2j+1)^{2}}=0.91596\cdots$ is the Catalan constant.

With regard to other extensions and analogies of (1.2)–(1.4), we can refer to [Reference Azar1,Reference Batbold, Krnić, Pečarić and Vuković2,Reference Krnić and Pečarić8,Reference Krnić, Pečarić, Perić and Vuković10,Reference Kuang12,Reference Li, Wang and He14,Reference Yang23–Reference Yang25,Reference Yang27,Reference Yang29,Reference Yang and Rassias31,Reference You36]. In addition, by the continuous innovation of kernel function, introducing parameters, studying the high-dimensional extension, coefficient refinement and the equivalent conditions of the best constant factor, and also considering the discrete and the half-discrete forms, a large number of new Hilbert-type inequalities have been established in recent years (see [Reference He, Yang and Chen4–Reference Hong, Liao, Yang and Chen6,Reference Krnić, Pečarić and Vuković9,Reference Krnić, Pečarić and Vuković11,Reference Rassias and Yang15,Reference Rassias and Yang19,Reference Rassias, Yang and Raigorodskii20,Reference Yang30,Reference Yang and Debnath32,Reference Yang, Wu and Wang34,Reference You35,Reference You38,Reference You and Yue39]). It should be pointed that the establishment of these new inequalities fully demonstrates the methods and techniques of analysis and proves to be of vital importance to the development of modern analysis.

It is well known that the classical Hilbert-type inequalities are usually constructed in the first quadrant. In recent years, with more and more researchers participating in the study of Hilbert-type inequalities, some researchers extended the range of variables to the whole plane and established many valuable new results (see [Reference Huang and Yang7,Reference Rassias and Yang16–Reference Rassias and Yang18,Reference Rassias, Yang and Raigorodskii21,Reference Yang, Hauang and Zhong33,Reference You37,Reference Wang and Yang41]). Inspired by the related results in these literatures, we will construct the following new inequalities:

(1.7)\begin{align} & \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\displaystyle\frac{({\mid}\log\mid \frac{x}{y}\mid{\mid})^{\alpha}}{(\max\{{\mid} x\mid, \mid y\mid\})^{2} }f(x)g(y) \mathrm{d}x\mathrm{d}y <4\Gamma(\alpha+1)\| f \|_{p,\mu}\| g \|_{q, \nu}, \end{align}
(1.8)\begin{align} & \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{(\log\mid xy\mid)^{2k}}{ 1-xy+(xy)^{2} }f(x)g(y) \mathrm{d}x\mathrm{d}y <\frac{4\pi^{2k+1}}{6^{2k+1}}\phi^{(2k)}\left(\frac{\pi}{6}\right)\| f \|_{p,\mu}\| g \|_{q, \nu}, \end{align}

where $\phi (u)=\cot u$, $k\in \mathrm {N}$, $\mu (x)=\frac {1}{\mid x\mid }$, and $\nu (y)=\frac {1}{\mid y\mid }$.

Furthermore, we will construct a new kernel function which includes some classical kernel functions and is defined in the whole plane, and then establish a new Hilbert-type inequality. Interestingly, we also present some new Hilbert-type inequalities with special kernel functions at the end of the paper.

2. Some lemmas

Lemma 1 (H$\ddot {\text {o}}$lder's inequality [Reference Kuang13]) Let $p>1,$ $\frac {1}{p}+\frac {1}{q}=1$. Let $f(x), g(x)\geq 0,$ and $f\in L^{p}(\Omega ),$ $g\in L^{q}(\Omega )$. Then

(2.1)\begin{equation} \displaystyle\int_\Omega f(x)g(x)\mathrm{d}x \leq \| f\|_p\| g\|_q, \end{equation}

and (2.1) takes the form of equality if and only if $C_1f^{p}=C_2g^{q}$ a. e. in $\Omega$, where $C_1$ and $C_2$ are not both equal to zero.

Lemma 2 Let $\delta _1,\delta _2, \eta \in \{1,-1\}$. Let $\alpha \geq 0$ for $\delta _1\delta _2=1$ and $\alpha >0$ for $\delta _1\delta _2=-1$. Suppose that $\beta$, $\gamma$ are such that $0<\gamma <\beta +2m-2n,$ where $m,n\in \mathbb {N},$ $m\geq n$ for $\beta >0$ and $m> n$ for $\beta \geq 0$. Define

(2.2)\begin{equation} K(x,y):= \displaystyle\frac{\mid 1+\delta_2(xy^{\eta})^{2n+1}\mid({\mid}\log\mid xy^{\eta}\mid{\mid})^{\alpha}}{\mid 1+\delta_1(xy^{\eta})^{2m+1}\mid(\max\{1,\mid xy^{\eta}\mid\})^{\beta} }, \end{equation}

and

(2.3)\begin{align} \kappa(\delta_1,\delta_2, m, n, \alpha, \beta,\gamma) & :=2\displaystyle\sum_{j=0}^{\infty}\Bigg\{ \displaystyle\frac{1}{[(4m+2)j+\gamma]^{\alpha+1}}\\ & \quad-\displaystyle\frac{\delta_1\delta_2}{[(4m+2)j+2m+2n+2+\gamma]^{\alpha+1}}\nonumber\\ & \quad+\frac{1}{[(4m+2)j+2m-2n+\beta-\gamma]^{\alpha+1}}\nonumber\\ & \quad-\frac{\delta_1\delta_2}{[(4m+2)j+4m+2+\beta-\gamma]^{\alpha+1}} \Bigg\}. \nonumber \end{align}

Then

(2.4)\begin{equation} \displaystyle\int_{-\infty}^{\infty} K(u,1)\mid u\mid^{\gamma-1} \mathrm{d}u=\Gamma(\alpha+1)\kappa(\delta_1,\delta_2, m, n, \alpha, \beta,\gamma), \end{equation}

where $\Gamma (s)=\int _{0}^{\infty } u^{s-1}e^{-u} \mathrm {d}u (s>0)$ is the $\Gamma$-function [Reference Richard22,Reference Wang and Guo40], and $\Gamma (s)=(s-1)!$ for $s\in \mathbb {N}^{+}$.

Proof.

(2.5)\begin{align} & \displaystyle\int_{-\infty}^{\infty} K(u,1)\mid u\mid^{\gamma-1} \mathrm{d}u\\ & \quad=\displaystyle\int_{0}^{\infty} K(u,1) u^{\gamma-1}\mathrm{d}u+ \int_{0}^{\infty} K({-}u,1)\mid u\mid^{\gamma-1}\mathrm{d}u\nonumber\\ & \quad=\int_{0}^{\infty}\left[\displaystyle\frac{\mid 1+\delta_2u^{2n+1}\mid}{\mid 1+\delta_1u^{2m+1}\mid }+\frac{\mid 1-\delta_2u^{2n+1}\mid }{\mid 1-\delta_1u^{2m+1}\mid }\right]\frac{ u^{\gamma-1}\mid \log u\mid^{\alpha}}{(\max\{1,u\})^{\beta} }\mathrm{d}u\nonumber\\ & \quad=\int_{0}^{\infty}\left|\frac{ 1+\delta_2u^{2n+1}}{ 1+\delta_1u^{2m+1} }+\frac{ 1-\delta_2u^{2n+1} }{ 1-\delta_1u^{2m+1} }\right|\frac{ u^{\gamma-1}\mid \log u\mid^{\alpha}}{(\max\{1,u\})^{\beta} }\mathrm{d}u\nonumber\\ & \quad= 2\int_{0}^{\infty}\frac{\mid 1-\delta_1\delta_2u^{2m+2n+2}\mid }{\mid 1-u^{4m+2}\mid }\frac{\mid u^{\gamma-1}\log u\mid^{\alpha}}{(\max\{1,u\})^{\beta} }\mathrm{d}u\nonumber\\ & \quad= 2\int_{0}^{1}\frac{ (1-\delta_1\delta_2u^{2m+2n+2}) u^{\gamma-1}\mid\log u\mid^{\alpha}}{1-u^{4m+2} }\mathrm{d}u\nonumber\\ & \quad+ 2\int_{1}^{\infty}\frac{ (u^{2m+2n+2}-\delta_1\delta_2) u^{\gamma-\beta-1}\mid\log u\mid^{\alpha}}{u^{4m+2}-1 }\mathrm{d}u\nonumber\\ & \quad:=2J_1+2J_2. \nonumber \end{align}

Observing that $\frac {1}{1-u^{4m+2}} =\sum _{j=0}^{\infty } u^{(4m+2)j}$ for $0< u<1$, and using Fubini's theorem, we obtain

(2.6)\begin{align} J_1& =\displaystyle\sum_{j=0}^{\infty}\left[\displaystyle\int_{0}^{1}u^{(4m+2)j+\gamma-1}\mid \log u\mid^{\alpha}\mathrm{d}u\right.\\ & \quad\left.-\delta_1\delta_2\displaystyle\int_{0}^{1}u^{(4m+2)j+2m+2n+\gamma+1}\mid \log u\mid^{\alpha}\mathrm{d}u\right]\nonumber\\ & :=\displaystyle\sum_{j=0}^{\infty}(J_{11}-\delta_1\delta_2J_{12}). \nonumber \end{align}

Setting $\log u=\frac {-t}{(4m+2)j+\gamma }$, we get

(2.7)\begin{equation} J_{11}=\displaystyle\frac{1}{[(4m+2)j+\gamma]^{\alpha+1}}\displaystyle\int_{0}^{\infty}e^{{-}t}t^{\alpha} \mathrm{d}t=\frac{\Gamma(\alpha+1)}{[(4m+2)j+\gamma]^{\alpha+1}}. \end{equation}

Similarly, we have

(2.8)\begin{equation} J_{12}=\displaystyle\frac{\Gamma(\alpha+1)}{[(4m+2)j+2m+2n+\gamma+2]^{\alpha+1}}.\end{equation}

Plugging (2.7) and (2.8) into (2.6), we obtain

(2.9)\begin{equation} J_1=\displaystyle\sum_{j=0}^{\infty}\left\{\displaystyle\frac{\Gamma(\alpha+1)}{[(4m+2)j+\gamma]^{\alpha+1}}- \frac{\delta_1\delta_2\Gamma(\alpha+1)}{[(4m+2)j+2m+2n+\gamma+2]^{\alpha+1}}\right\}. \end{equation}

In addition, by variable substitution, we obtain

(2.10)\begin{align} J_2& =\displaystyle\int_{0}^{1}\displaystyle\frac{ (u^{2m-2n+\beta-\gamma-1}-\delta_1\delta_2u^{4m+\beta-\gamma+1})\mid\log u\mid^{\alpha}}{1-u^{4m+2} }\mathrm{d}u\\ & =\displaystyle\sum_{k=0}^{\infty}\Bigg\{\displaystyle\frac{\Gamma(\alpha+1)}{[(4m+2)j+2m-2n+\beta-\gamma]^{\alpha+1}}\nonumber\\ & \quad-\frac{\delta_1\delta_2\Gamma(\alpha+1)}{[(4m+2)j+4m+\beta-\gamma+2]^{\alpha+1}}\Bigg\}. \nonumber \end{align}

Applying (2.9) and (2.10) to (2.5), and using (2.3), we get (2.4).

Lemma 3 Let $\delta _1,\delta _2, \eta \in \{1,-1\}$. Let $\alpha \geq 0$ for $\delta _1\delta _2=1$ and $\alpha >0$ for $\delta _1\delta _2=-1$. Suppose that $\beta,$ $\gamma$ are such that $0<\gamma <\beta +2m-2n,$ where $m,n\in \mathbb {N},$ $m\geq n$ for $\beta >0$ and $m> n$ for $\beta \geq 0$. $K(x,y)$ and $\kappa (\delta _1,\delta _2, m, n, \alpha, \beta,\gamma )$ are defined via Lemma 2. Let $p>1,$ $\frac {1}{p}+\frac {1}{q}=1$, $E_{\eta }=\{y: \mid y\mid ^{\eta }<1\},$ and for arbitrary real number $\varepsilon (\varepsilon >0)$ which is sufficiently small, set

\[ \tilde{f}(x)= \begin{cases} 0, & x\in[{-}1, 1], \cr \mid x\mid^{{\gamma-1}-\frac{2\varepsilon}{p}}, & x\in\mathbb{R}\setminus [{-}1, 1],\end{cases}\quad \tilde{g}(y)= \begin{cases} \mid y\mid^{{\eta\gamma-1}+\frac{2\varepsilon\eta}{q}}, & y\in E_{\eta}, \cr 0, & y\in\mathbb{R}\setminus E_{\eta}. \end{cases} \]

Then

(2.11)\begin{equation} \varepsilon\displaystyle\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}K(x, y)\tilde{f}(x)\tilde{g}(y) \mathrm{d}x\mathrm{d}y=\Gamma(\alpha+1)\kappa(\delta_1,\delta_2, m, n, \alpha, \beta,\gamma)+o(1).\end{equation}

Proof. Write $E_{\eta }^{-}=\{y: y\in E_\eta, y<0\},$ $E_{\eta }^{+}=\{y: y\in E_\eta, y>0\}.$ Then

(2.12)\begin{align} & \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} K(x, y)\tilde{f}(x)\tilde{g}(y) \mathrm{d}x\mathrm{d}y\\ & \quad=2\displaystyle\int_1^{\infty}x^{{\gamma-1}-\frac{2\varepsilon}{p}} \int_{E_{\eta}^{+}} K(x, y)y^{\eta\gamma-1+\frac{2\eta\varepsilon}{q}} \mathrm{d}y\mathrm{d}x\nonumber\\ & \quad+2\int_1^{\infty}x^{{\gamma-1}-\frac{2\varepsilon}{p}} \int_{E_{\eta}^{-}} K(x, y)\mid y\mid^{\eta\gamma-1+\frac{2\eta\varepsilon}{q}} \mathrm{d}y\mathrm{d}x:=2L_1+2L_2. \nonumber \end{align}

Letting $u=xy^{\eta }$, and using Fubini's theorem, we have

(2.13)\begin{align} L_1& =\displaystyle\int_1^{\infty}x^{{-}1-2\varepsilon}\left[ \int_{0}^{x} K(u, 1)u^{\gamma-1+\frac{2\varepsilon}{q}} \mathrm{d}u\right]\mathrm{d}x\\ & =\displaystyle\int_1^{\infty}x^{{-}1-2\varepsilon}\left[ \int_{0}^{1} K(u, 1)u^{\gamma-1+\frac{2\varepsilon}{q}} \mathrm{d}u+\int_{1}^{x} K(u, 1)u^{\gamma-1+\frac{2\varepsilon}{q}} \mathrm{d}u\right] \mathrm{d}x \nonumber\\ & =\frac{1}{2\varepsilon}\int_{0}^{1} K(u, 1)u^{\gamma-1+\frac{2\varepsilon}{q}} \mathrm{d}u +\int_1^{\infty}K(u, 1)u^{\gamma-1+\frac{2\varepsilon}{q}} \int_{u}^{\infty} x^{{-}1-2\varepsilon} \mathrm{d}x\mathrm{d}u\nonumber\\ & =\frac{1}{2\varepsilon}\left[\int_{0}^{1} K(u, 1)u^{\gamma-1+\frac{2\varepsilon}{q}} \mathrm{d}u +\int_1^{\infty}K(u, 1)u^{\gamma-1-\frac{2\varepsilon}{p}} \mathrm{d}u\right]. \nonumber \end{align}

Similarly, letting $u=-xy^{\eta }$, it can be also obtained that

(2.14)\begin{equation} L_2=\displaystyle\frac{1}{2\varepsilon}\left[ \displaystyle\int_{0}^{1} K({-}u, 1)u^{\gamma-1+\frac{2\varepsilon}{q}} \mathrm{d}u +\int_1^{\infty}K({-}u, 1)u^{\gamma-1-\frac{2\varepsilon}{p}} \mathrm{d}u\right].\end{equation}

Applying (2.13) and (2.14) to (2.12), and letting $\varepsilon \to 0^{+}$, we get

\begin{align*} & \varepsilon\displaystyle\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}K(x, y)\tilde{f}(x)\tilde{g}(y) \mathrm{d}x\mathrm{d}y\\ & \quad=\displaystyle\int_0^{\infty} K(u, 1)u^{\gamma-1} \mathrm{d}u +\int_0^{\infty}K({-}u, 1)u^{\gamma-1} \mathrm{d}u+o(1)\\ & \quad=\int_{-\infty}^{\infty} K(u, 1)\mid u\mid^{\gamma-1} \mathrm{d}u+o(1)=\Gamma(\alpha+1)\kappa(\delta_1,\delta_2, m, n, \alpha, \beta,\gamma)+o(1). \end{align*}

Lemma 3 is proved.

Lemma 4 Let $0< z<1,$ $\phi (u)=\cot u$ and $k\in \mathbb {N}$. Then

(2.15)\begin{align} \phi^{(2k)}(z\pi)& =\displaystyle\frac{(2k)!}{\pi^{2k+1}} \displaystyle\sum_{j=0}^{\infty}\left[\frac{1}{(j+z)^{2k+1}}-\frac{1}{(j+1-z)^{2k+1}}\right], \end{align}
(2.16)\begin{align} \phi^{(2k+1)}(z\pi)& ={-}\frac{(2k+1)!}{\pi^{2k+2}}\sum_{j=0}^{\infty}\left[\frac{1}{(j+z)^{2k+2}}+\frac{1}{(j+1-z)^{2k+2}}\right]. \end{align}

Proof. Write $\psi (u)=\tan u$ in the form of rational fraction expansion [Reference Richard22,Reference Wang and Guo40]:

(2.17)\begin{equation} \psi(u)=\tan u=2\displaystyle\sum_{j=0}^{\infty}\left[\displaystyle\frac{1}{(2j+1)\pi-2u}-\frac{1}{(2j+1)\pi+2u}\right]. \end{equation}

Taking the $(2k)$th derivative of $\psi (u)=\tan u$, we get

(2.18)\begin{equation} \psi^{(2k)}(u)=2^{2k+1}(2k)!\displaystyle\sum_{j=0}^{\infty} \left\{\displaystyle\frac{1}{[(2j+1)\pi-2u]^{2k+1}} -\frac{1}{[(2j+1)\pi+2u]^{2k+1}}\right\}. \end{equation}

Observing that $\phi ^{(2k)}(z\pi )=\psi ^{(2k)}\left(\tfrac {\pi }{2}-z\pi \right)$, and setting $u=\frac {\pi }{2}-z\pi$ in (2.18), we obtain (2.15). Finding the first derivative of (2.18), and in view of $\phi ^{(2k+1)}(z\pi )=-\psi ^{(2k+1)} \left(\tfrac {\pi }{2}-z\pi \right)$, we arrive at (2.16). Lemma 4 is proved.

3. Main results

Theorem 1 Let $\delta _1,\delta _2, \eta \in \{1,-1\}$. Let $\alpha \geq 0$ for $\delta _1\delta _2=1$ and $\alpha >0$ for $\delta _1\delta _2=-1$. Suppose that $\beta,$ $\gamma$ are such that $0<\gamma <\beta +2m-2n,$ where $m,n\in \mathbb {N}$, $m\geq n$ for $\beta >0$ and $m> n$ for $\beta \geq 0$. Let $p>1,$ $\frac {1}{p}+\frac {1}{q}=1,$ $\mu (x)=\mid x\mid ^{p(1-\gamma )-1},$ and $\nu (y)= \mid y\mid ^{q(1-\eta \gamma )-1}$. Suppose that $f(x),$ $g(y)>0$ with $f(x)\in L_{\mu }^{p}(\mathbb {R})$ and $g(y)\in L_{\nu }^{q}(\mathbb {R})$. $K(x,y)$ and $\kappa (\delta _1,\delta _2, m, n, \alpha, \beta,\gamma )$ are defined via Lemma 2. Then

(3.1)\begin{equation} \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}K(x,y)f(x)g(y) \mathrm{d}x\mathrm{d}y <\Gamma(\alpha+1)\kappa(\delta_1,\delta_2, m, n, \alpha, \beta, \gamma)\| f \|_{p,\mu}\| g \|_{q, \nu}, \end{equation}

where the constant factor $\Gamma (\alpha )\kappa (\delta _1,\delta _2, m, n, \alpha, \beta,\gamma )$ is the best possible.

Proof. By Lemma 1, we obtain

(3.2)\begin{align} & \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} K(x, y)f(x)g(y) \mathrm{d}x\mathrm{d}y\\ & \quad=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\left\{[ K(x, y)]^{\frac{1}{p}}\mid y\mid^{\frac{\eta\gamma-1}{p}}\mid x\mid^{\frac{1-\gamma}{q}}f(x)\right\}\nonumber\\ & \qquad\times\left\{[ K(x, y)]^{\frac{1}{q}}\mid x\mid^{\frac{\gamma-1}{q}}\mid y\mid^{\frac{1-\eta\gamma}{p}}g(y)\right\} \mathrm{d}x\mathrm{d}y\nonumber\\ & \quad\leq\left[\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} K(x, y)\mid y\mid^{\eta\gamma-1}\mid x\mid^{\frac{p(1-\gamma)}{q}}f^{p}(x) \mathrm{d}x\mathrm{d}y\right]^{\frac{1}{p}}\nonumber\\ & \qquad\times\left[\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} K(x, y)\mid x\mid^{\gamma-1} \mid y\mid^{\frac{q(1-\eta\gamma)}{p}}g^{q}(y) \mathrm{d}x\mathrm{d}y\right]^{\frac{1}{q}}\nonumber\\ & \quad=\left(\int_{-\infty}^{\infty} \omega(x)\mid x\mid^{\frac{p(1-\gamma)}{q}}f^{p}(x) \mathrm{d}x\right)^{\frac{1}{p}} \left(\int_{-\infty}^{\infty} \varpi(y)\mid y\mid^{\frac{q(1-\eta\gamma)}{p}}g^{q}(y)\mathrm{d}y\right)^{\frac{1}{q}}, \nonumber \end{align}

where $\omega (x)=\int _{-\infty }^{\infty } K(x, y)\mid y\mid ^{\eta \gamma -1}\mathrm {d}y,$ and $\varpi (y)=\int _{-\infty }^{\infty } K(x, y)\mid x\mid ^{\gamma -1} \mathrm {d}x.$

Letting $xy^{\eta }=u$, and using (2.4), we obtain

(3.3)\begin{align} \omega(x)& ={\mid} x\mid^{-\gamma}\displaystyle\int_{-\infty}^{\infty} K(u,1)\mid u\mid^{\gamma-1} \mathrm{d}u\\ & =\Gamma(\alpha+1)\kappa(\delta_1,\delta_2, m, n, \alpha, \beta,\gamma)\mid x\mid^{-\gamma},\quad (x\neq 0) \nonumber \end{align}

and

(3.4)\begin{align} \varpi(y)& ={\mid} y\mid^{-\eta\gamma}\displaystyle\int_{-\infty}^{\infty}K(u,1)\mid u\mid^{\gamma-1} \mathrm{d}u\\ & =\Gamma(\alpha+1)\kappa(\delta_1,\delta_2, m, n, \alpha, \beta,\gamma)\mid y\mid^{-\eta\gamma},\quad (y\neq 0). \nonumber \end{align}

Applying (3.3) and (3.4) to (3.2), it follows that

(3.5)\begin{equation} \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}K(x,y)f(x)g(y) \mathrm{d}x\mathrm{d}y \leq\Gamma(\alpha+1)\kappa(\delta_1,\delta_2, m, n, \alpha, \beta,\gamma)\| f \|_{p,\mu}\| g \|_{q, \nu}, \end{equation}

Assume that (3.5) takes the form of equality. By Lemma 1, it can be obtained that

(3.6)\begin{equation} C_1 K(x, y)\mid y\mid^{\eta\gamma-1}\mid x\mid ^{\frac{p(1-\gamma)}{q}}f^{p}(x)=C_2 K(x, y)\mid x\mid^{\gamma-1} \mid y\mid^{\frac{q(1-\eta\gamma)}{p}}g^{q}(y) \end{equation}

a. e. in $\mathbb {R}\times \mathbb {R}$, where $C_1$ and $C_2$ are not both equal to zero. Simplifying (3.6), we have

\[ C_1 \mid x\mid^{p(1-\gamma)}f^{p}(x)=C_2\mid y\mid^{q(1-\eta\gamma)}g^{q}(y), \]

a. e. in $\mathbb {R}\times \mathbb {R}$. It follows, therefore, that there exists a constant $C$ such that $C_1\mid x\mid ^{p(1-\gamma )}f^{p}(x)=C,$ a. e. in $\mathbb {R}$, and $C_2\mid y\mid ^{q(1-\eta \gamma )}g^{q}(y)=C,$ a. e. in $\mathbb {R}$. Assuming that $C_1\neq 0$, we obtain $\mid x\mid ^{p(1-\gamma )-1}f^{p}(x)=\frac {C}{C_1\mid x\mid }$ a.e. in $\mathbb {R},$ which is obviously in contradiction with the condition $f(x)\in L_{\mu }^{p}(\mathbb {R})$. Therefore, (3.5) takes the form of strict inequality, and (3.1)is proved.

In order to prove Theorem 1, it suffices to prove that the constant factor $\Gamma (\alpha +1) \kappa (\delta _1,\delta _2, m, n, \alpha, \beta,\gamma )$ is the best possible. Assume that there is a positive constant $c$ being such that $c<\Gamma (\alpha +1)\kappa (\delta _1,\delta _2, m, n, \alpha, \beta,\gamma )$, and (3.1) is still valid if $\Gamma (\alpha +1) \kappa (\delta _1,\delta _2, m, n, \alpha, \beta,\gamma )$ is replaced with $c$, that is,

(3.7)\begin{equation} \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} K(x, y)f(x)g(y)\mathrm{d}x\mathrm{d}y < c\| f \|_{p,\mu}\| g \|_{q, \nu}. \end{equation}

Replace $f$ and $g$ in (3.7) by $\tilde {f}$ and $\tilde {g}$ defined in Lemma 3, respectively. Then

(3.8)\begin{align} & \varepsilon\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} K(x,y)\tilde{f}(x)\tilde{g}(y) \mathrm{d}x\mathrm{d}y\\ & \quad< c\varepsilon \left(2\displaystyle\int_{1}^{\infty}x^{{-}2\varepsilon-1} \mathrm{d}x\right)^{\frac{1}{p}}\left(\int_{E_\eta}\mid y \mid^{2\eta\varepsilon-1}\mathrm{d}y\right)^{\frac{1}{q}}=c. \nonumber \end{align}

It follows from (2.11) and (3.8) that $\Gamma (\alpha +1)\kappa (\delta _1,\delta _2, m, n, \alpha, \beta,\gamma )\leq k,$ which contradicts the assumption obviously. Therefore, the constant factor in (3.1) is the best possible and the proof of Theorem 1 is completed.

Theorem 2 Let $\delta _1,\delta _2, \eta \in \{1,-1\}$. Let $\alpha \geq 0$ for $\delta _1\delta _2=1$ and $\alpha >0$ for $\delta _1\delta _2=-1$. Suppose that $\beta,$ $\gamma$ are such that $0<\gamma <\beta +2m-2n,$ where $m,n\in \mathbb {N},$ $m\geq n$ for $\beta >0$ and $m> n$ for $\beta \geq 0$. Let $p>1,$ $\frac {1}{p}+\frac {1}{q}=1,$ and $\mu (x)=\mid x\mid ^{p(1-\gamma )-1}$. Suppose that $f(x)\geq 0$ with $f(x)\in L_{\mu }^{p}(\mathbb {R})$. $K(x,y)$ and $\kappa (\delta _1,\delta _2, m, n, \alpha, \beta,\gamma )$ are defined via Lemma 2. Then

(3.9)\begin{align} & \displaystyle\int_{-\infty}^{\infty}\mid y\mid^{p\eta\gamma-1}\left[\int_{-\infty}^{\infty}K(x,y)f(x) \mathrm{d}x\right]^{p}\mathrm{d}y\\ & \quad<[\Gamma(\alpha+1)\kappa(\delta_1,\delta_2, m, n, \alpha, \beta,\gamma)\| f \|_{p,\mu}]^{p}, \nonumber \end{align}

where $[\Gamma (\alpha +1)\kappa (\delta _1,\delta _2, m, n, \alpha, \beta,\gamma )]^{p}$ is the best possible, and (3.9) is equivalent to (3.1).

Proof. Let $\hat g(y)$ be an arbitrary measurable function with $\hat g\in L_{\hat \nu }^{q}(\mathbb {R})$, $\|\hat g\|_{q,\hat \nu }=1$, where $\hat \nu (y):=\mid y\mid ^{p\eta \gamma -1}$. Let $g(y):=\hat g(y)\hat \nu (y)$, $\nu (y):=\mid y\mid ^{q(1-\eta \gamma )-1}$. Then we obtain $g\in L_{\nu }^{q}(\mathbb {R})$, and $\| g\|_{q,\nu }=\|\hat g\|_{q,\hat \nu }=1$. By the use of Theorem 1, we have

\[ \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} K(x,y)f(x)g(y) \mathrm{d}x\mathrm{d}y<\Gamma(\alpha+1)\kappa(\delta_1,\delta_2, m, n, \alpha, \beta,\gamma)\| f \|_{p,\mu}, \]

that is,

(3.10)\begin{align} & \displaystyle\int_{-\infty}^{\infty}\left[(\hat\nu(y))^{\frac{1}{p}}\int_{-\infty}^{\infty} K(x,y)f(x)\mathrm{d}x\right]\hat g(y) (\hat\nu(y))^{\frac{1}{q}} \mathrm{d}y\nonumber\\& \quad <\Gamma(\alpha+1)\kappa(\delta_1,\delta_2, m, n, \alpha, \beta,\gamma)\| f \|_{p,\mu}. \end{align}

Therefore, it follows from $\|\hat g\|_{q,\hat \nu }=1$ and (3.10) that

(3.11)\begin{align} & \left\{\displaystyle\int_{-\infty}^{\infty} \mid y\mid^{p\eta\gamma-1}\left[\int_{-\infty}^{\infty} K (x,y)f(x) \mathrm{d}x\right]^{p}\mathrm{d}y\right\}^{\frac{1}{p}}\nonumber\\ & \quad=\sup_{\hat g}\displaystyle\int_{-\infty}^{\infty}\left[(\hat\nu(y))^{\frac{1}{p}}\int_{-\infty}^{\infty} K(x,y)f(x)\mathrm{d}x\right]\hat g(y) (\hat\nu(y))^{\frac{1}{q}} \mathrm{d}y\nonumber\\ & \quad<\Gamma(\alpha+1)\kappa(\delta_1,\delta_2, m, n, \alpha, \beta,\gamma)\| f \|_{p,\mu}. \end{align}

By (3.11), (3.9) follows naturally. On the contrary, assuming that (3.9) holds true, and using H$\ddot {o}$lder's inequality, we have

(3.12)\begin{align} & \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} K(x,y)f(x)g(y) \mathrm{d}x\mathrm{d}y\\ & \quad=\displaystyle\int_{-\infty}^{\infty} \left[{\mid} y\mid^{-\left(1-\eta\gamma-\frac{1}{q}\right)}\int_{-\infty}^{\infty} K(x,y)f(x) \mathrm{d}x\right]\left[{\mid} y\mid ^{1-\eta\gamma-\frac{1}{q}}g(y)\right]\mathrm{d}y\nonumber\\ & \quad\leq\left\{\int_{-\infty}^{\infty} \mid y\mid^{p\eta\gamma-1}\left[\int_{-\infty}^{\infty} K(x,y)f(x) \mathrm{d}x\right]^{p}\mathrm{d}y\right\}^{\frac{1}{p}} \| g\|_{q,\nu}. \nonumber \end{align}

Applying (3.9) to (3.12), we obtain (3.1). Therefore, (3.1) and (3.9) are equivalent. It follows from the equivalence of (3.1) and (3.9) that the constant factor on the right side of (3.9) is also the best possible. Theorem 2 is proved.

4. Applications

Letting $m=n$, $\delta _1\delta _2=1$ in Theorem 1, we can easily get

\[ \kappa(\delta_1,\delta_2, m, n, \alpha, \beta,\gamma)=\frac{2}{\gamma^{\alpha+1}}+\frac{2}{(\beta-\gamma)^{\alpha+1}}, \]

and Theorem 1 is transformed into the following corollary.

Corollary 1 Let $\eta \in \{1,-1\},$ $\alpha \geq 0,$ $\beta >\gamma >0,$ $p>1$ and $\frac {1}{p}+\frac {1}{q}=1$. Suppose that $\mu (x)=\mid x\mid ^{p(1-\gamma )-1},$ $\nu (y)=\mid y\mid ^{q(1-\eta \gamma )-1},$ and $f(x),$ $g(y)>0$ with $f(x)\in L_{\mu }^{p}(\mathbb {R}),$ $g(y)\in L_{\nu }^{q}(\mathbb {R})$. Then

(4.1)\begin{align} & \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\displaystyle\frac{({\mid}\log\mid xy^{\eta}\mid{\mid})^{\alpha}}{(\max\{1,\mid xy^{\eta}\mid\})^{\beta} } f(x)g(y) \mathrm{d}x\mathrm{d}y\\ & \quad<\Gamma(\alpha+1)\left[\displaystyle\frac{2}{\gamma^{\alpha+1}}+\frac{2}{(\beta-\gamma)^{\alpha+1}}\right]\| f \|_{p,\mu}\| g \|_{q, \nu}. \nonumber \end{align}

Let $\eta =-1$ in (4.1), and replacing $g(y)\mid y\mid ^{\beta }$ by $g(y),$ We obtain the following Hilbert-type inequality with a homogeneous kernel:

(4.2)\begin{align} & \quad\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\displaystyle\frac{({\mid}\log\mid \frac{x}{y}\mid{\mid})^{\alpha}}{(\max\{{\mid} x\mid, \mid y\mid\})^{\beta} }f(x)g(y) \mathrm{d}x\mathrm{d}y\\ & \quad<\Gamma(\alpha+1)\left[\frac{2}{\gamma^{\alpha+1}}+\frac{2}{(\beta-\gamma)^{\alpha+1}}\right]\| f \|_{p,\mu}\| g \|_{q, \nu}, \nonumber \end{align}

where $\mu (x)=\mid x\mid ^{p(1-\gamma )-1}$, $\nu (y)=\mid y\mid ^{q(1+\gamma -\beta )-1}$.

Let $\beta =2$, $\gamma =1$ in (4.2), then it follows (1.7).

Let $\alpha =0$, $\beta =1$, $\gamma =\frac {1}{q}$ in (4.2), then we have the following inequality which is the extension of (1.3) in the whole plane:

\[ \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\displaystyle\frac{f(x)g(y)}{\max \{{\mid} x\mid, \mid y\mid\}} \mathrm{d}x\mathrm{d}y<2pq\|f\|_{p}\|g\|_{q}. \]

Letting $\delta _1=\delta _2=1$, $\beta =n=0$, $\alpha =2k$, $k\in \mathbb {N}$ in Theorem 1, and using (2.15), we obtain the following corollary.

Corollary 2 Let $\eta \in \{1,-1\},$ $k\in \mathbb {N},$ $m\in \mathbb {N}^{+}$ and $0<\gamma <2m$. Suppose that $p>1,$ $\frac {1}{p}+\frac {1}{q}=1,$ $\mu (x)=\mid x\mid ^{p(1-\gamma )-1}$ and $\nu (y)=\mid y\mid ^{q(1-\eta \gamma )-1}$. Let $\phi (u)=\cot u,$ and $f(x),$ $g(y)>0$ with $f(x)\in L_{\mu }^{p}(\mathbb {R})$, $g(y)\in L_{\nu }^{q}(\mathbb {R})$. Then

(4.3)\begin{align} & \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\displaystyle\frac{(\log\mid xy^{\eta}\mid)^{2k}}{\mid 1-xy^{\eta}+(xy^{\eta})^{2}-\dots+(xy^{\eta})^{2m}\mid }f(x)g(y) \mathrm{d}x\mathrm{d}y\\ & \quad<\displaystyle\frac{2\pi^{2k+1}}{(4m+2)^{2k+1}}\left[\phi^{(2k)}\left(\frac{\gamma\pi}{4m+2}\right)+\phi^{(2k)}\left(\frac{(2m-\gamma)\pi}{4m+2}\right)\right]\| f \|_{p,\mu}\| g \|_{q, \nu}. \nonumber \end{align}

Let $\eta =m=\gamma =1$ in (4.3), then it follows (1.8).

Let $\eta =-1$, $m=\gamma =1$ in (4.3), and replacing $g(y)y^{2}$ by $g(y)$. Then

(4.4)\begin{equation} \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\displaystyle\frac{(\log\mid \frac{x}{y}\mid)^{2k}}{ x^{2}-xy+y^{2} }f(x)g(y) \mathrm{d}x\mathrm{d}y <\frac{4\pi^{2k+1}}{6^{2k+1}}\phi^{(2k)}\left(\frac{\pi}{6}\right)\| f \|_{p,\mu}\| g \|_{q, \nu}, \end{equation}

where $\mu (x)=\frac {1}{\mid x\mid }$ and $\nu (y)=\frac {1}{\mid y\mid }$.

Letting $\delta _1=\delta _2=\delta$, $\delta \in \{1,-1\}$, $\alpha =\beta =0$ in Theorem 1, and using (2.15) again, we obtain Corollary 3.

Corollary 3 Let $\delta, \eta \in \{1,-1\},$ $m, n\in \mathbb {N},$ $m>n,$ and $0<\gamma <2m-2n$. Suppose that $p>1,$ $\frac {1}{p}+\frac {1}{q}=1,$ $\mu (x)=\mid x\mid ^{p(1-\gamma )-1}$ and $\nu (y)=\mid y\mid ^{q(1-\eta \gamma )-1}$. Let $\phi (u)=\cot u,$ and $f(x),$ $g(y)>0$ with $f(x)\in L_{\mu }^{p}(\mathbb {R}),$ $g(y)\in L_{\nu }^{q}(\mathbb {R})$. Then

(4.5)\begin{align} & \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\displaystyle\frac{\mid 1+\delta(xy^{\eta})^{2n+1}\mid}{\mid 1+\delta(xy^{\eta})^{2m+1}\mid }f(x)g(y) \mathrm{d}x\mathrm{d}y\\ & \quad<\displaystyle\frac{\pi}{2m+1}\left[\phi\left(\frac{\gamma\pi}{4m+2}\right)+\phi\left(\frac{(2m-2n-\gamma)\pi}{4m+2}\right)\right]\| f \|_{p,\mu}\| g \|_{q, \nu}. \nonumber \end{align}

Setting $\gamma =m-n$, $\eta =-1$, and replacing $g(y)\mid y\mid ^{2m-2n}$ by $g(y)$, we obtain

(4.6)\begin{equation} \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\displaystyle\frac{\mid x^{2n+1}+\delta y^{2n+1}\mid}{\mid x^{2m+1}+\delta y^{2m+1}\mid }f(x)g(y) \mathrm{d}x\mathrm{d}y <\frac{2\pi}{2m+1}\phi\left(\frac{(m-n)\pi}{4m+2}\right)\| f \|_{p,\mu}\| g \|_{q, \nu},\end{equation}

where $\mu (x)=\mid x\mid ^{p(1+n-m)-1}$ and $\nu (y)=\mid y\mid ^{q(1+n-m)-1}$.

Let $m=1,$ $n=0$ in (4.6), then we obtain

(4.7)\begin{equation} \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\displaystyle\frac{f(x)g(y)}{ x^{2}\pm xy+y^{2} } \mathrm{d}x\mathrm{d}y <\frac{2\sqrt{3}\pi}{3}\| f \|_{p,\mu}\| g \|_{q, \nu}.\end{equation}

Let $m=7,$ $n=2$ in (4.6), then we obtain

(4.8)\begin{equation} \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\displaystyle\frac{f(x)g(y)}{ x^{10}\pm x^{5}y^{5}+y^{10} } \mathrm{d}x\mathrm{d}y <\frac{2\sqrt{3}\pi}{15}\| f \|_{p,\mu}\| g \|_{q, \nu}.\end{equation}

Letting $\delta _1=\delta$, $\delta _2=-\delta$, $\delta \in \{1,-1\}$, $\beta =0$, $\alpha =2k+1$, $k\in \mathbb {N}$ in Theorem 1, and using (2.16), we obtain the following corollary.

Corollary 4 Let $\delta, \eta \in \{1,-1\},$ $m, n, k\in \mathbb {N},$ $m>n,$ and $0<\gamma <2m-2n$. Suppose that $p>1,$ $\frac {1}{p}+\frac {1}{q}=1,$ $\mu (x)=\mid x\mid ^{p(1-\gamma )-1}$ and $\nu (y)=\mid y\mid ^{q(1-\eta \gamma )-1}$. Let $\phi (u)=\cot u$, and $f(x),$ $g(y)>0$ with $f(x)\in L_{\mu }^{p}(\mathbb {R}),$ $g(y)\in L_{\nu }^{q}(\mathbb {R})$. Then

(4.9)\begin{align} & \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\displaystyle\frac{\mid 1-\delta(xy^{\eta})^{2n+1}\mid}{\mid 1+\delta(xy^{\eta})^{2m+1}\mid }\mid\log\mid xy^{\eta}\mid{\mid}^{2k+1}f(x)g(y) \mathrm{d}x\mathrm{d}y\\ & \quad<\displaystyle\frac{-2\pi^{2k+2}}{(4m+2)^{2k+2}}\left[\phi^{(2k+1)}\left(\frac{\gamma\pi}{4m+2}\right)+\phi^{(2k+1)}\left(\frac{(2m-2n-\gamma)\pi}{4m+2}\right)\right]\| f \|_{p,\mu}\| g \|_{q, \nu}. \nonumber \end{align}

Setting $\delta =\eta =-1$, $n=k=0$, $m=\gamma =1$ in (4.9), and replacing $g(y)\mid y\mid ^{2}$ by $g(y)$, we obtain

(4.10)\begin{equation} \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\displaystyle\frac{\mid\log\mid\frac{x}{y}\mid{\mid}}{\mid x^{3}-y^{3}\mid }\mid x+y\mid f(x)g(y) \mathrm{d}x\mathrm{d}y <\frac{4\pi^{2}}{9}\| f \|_{p,\mu}\| g \|_{q, \nu}. \end{equation}

Letting $\delta _1=\delta _2=-1$, $\beta =n=0$, $\alpha =2k$, $k\in \mathbb {N}$ in Theorem 1, and using (2.15), we get the following corollary.

Corollary 5 Let $\eta \in \{1,-1\},$ $k\in \mathbb {N},$ $m\in \mathbb {N}^{+}$ and $0<\gamma <2m$. Suppose that $p>1,$ $\frac {1}{p}+\frac {1}{q}=1,$ $\mu (x)=\mid x\mid ^{p(1-\gamma )-1}$ and $\nu (y)=\mid y\mid ^{q(1-\eta \gamma )-1}$. Let $\phi (u)=\cot u,$ and $f(x),$ $g(y)>0$ with $f(x)\in L_{\mu }^{p}(\mathbb {R}),$ $g(y)\in L_{\nu }^{q}(\mathbb {R})$. Then

(4.11)\begin{align} & \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\displaystyle\frac{(\log\mid xy^{\eta}\mid)^{2k}}{\mid 1+xy^{\eta}+(xy^{\eta})^{2}+\dots+(xy^{\eta})^{2m}\mid }f(x)g(y) \mathrm{d}x\mathrm{d}y\\ & \quad<\displaystyle\frac{2\pi^{2k+1}}{(4m+2)^{2k+1}}\left[\phi^{(2k)}\left(\frac{\gamma\pi}{4m+2}\right)+\phi^{(2k)}\left(\frac{(2m-\gamma)\pi}{4m+2}\right)\right]\| f \|_{p,\mu}\| g \|_{q, \nu}. \nonumber \end{align}

Let $\eta =-1$, $m=\gamma =1$ in (4.11), and replace $g(y)y^{2}$ by $g(y)$. Then

(4.12)\begin{equation} \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\displaystyle\frac{(\log\mid \frac{x}{y}\mid)^{2k}}{ x^{2}+xy+y^{2} }f(x)g(y) \mathrm{d}x\mathrm{d}y <\frac{4\pi^{2k+1}}{6^{2k+1}}\phi^{(2k)}\left(\frac{\pi}{6}\right)\| f \|_{p,\mu}\| g \|_{q, \nu}, \end{equation}

where $\mu (x)=\frac {1}{\mid x\mid }$ and $\nu (y)=\frac {1}{\mid y\mid }$.

Let $k=1$ in (4.12), then it follows that

\[ \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\displaystyle\frac{(\log\mid \frac{x}{y}\mid)^{2}}{ x^{2}+xy+y^{2} }f(x)g(y) \mathrm{d}x\mathrm{d}y <\frac{4\sqrt{3}\pi^{3}}{27}\| f \|_{p,\mu}\| g \|_{q, \nu}. \]

Acknowledgements

The author is indebted to the anonymous referees for their valuable suggestions and comments that helped improve the paper significantly.

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