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A Hilbert-type integral inequality in the whole plane with a non-homogeneous kernel and a few parameters

Abstract

By using the way of real analysis and estimating the weight functions, we build a new Hilbert-type integral inequality in the whole plane with a non-homogeneous kernel and a few parameters. The constant factor related to the beta function is proved to be the best possible. We also consider the equivalent forms, the reverses, and some particular cases.

Introduction

If \(f(x),g(y)\geq0\), satisfying \(0<\int_{0}^{\infty}f^{2}(x)\, dx<\infty\) and \(0<\int_{0}^{\infty}g^{2}(y)\, dy<\infty\), then we have (cf. [1])

$$ \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{x+y}\,dx\,dy< \pi \biggl( \int_{0}^{\infty}f^{2}(x) \,dx\int_{0}^{\infty}g^{2}(y)\,dy \biggr) ^{\frac {1}{2}}, $$
(1)

where the constant factor π is the best possible. Inequality (1) is known as Hilbert’s integral inequality, which is important in analysis and its applications (cf. [1, 2]).

In recent years, by using the way of weight functions, a number of extensions of (1) were given by Yang (cf. [3]). Noticing that inequality (1) is a homogeneous kernel of degree −1, in 2009, A survey of the study of Hilbert-type inequalities with the homogeneous kernels of degree negative numbers and some parameters is given by [4]. Recently, some inequalities with the homogeneous kernels of degree 0 and non-homogeneous kernels have been studied (cf. [510]). All of the above integral inequalities are built in the quarter plane.

In 2007, Yang [11] first gave a Hilbert-type integral inequality in the whole plane as follows:

$$\begin{aligned}& \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{f(x)g(y)}{(1+e^{x+y})^{\lambda}}\,dx\,dy \\& \quad < B\biggl(\frac{\lambda}{2},\frac{\lambda}{2}\biggr) \biggl(\int _{-\infty}^{\infty }e^{-\lambda x}f^{2}(x)\,dx\int _{-\infty}^{\infty}e^{-\lambda y}g^{2}(y)\,dy \biggr)^{\frac{1}{2}}, \end{aligned}$$
(2)

where the constant factor \(B(\frac{\lambda}{2},\frac{\lambda }{2})\) (\(\lambda >0\)) is the best possible, and

$$ B(u,v):=\int_{0}^{\infty}\frac{t^{u+1}}{(1+t)^{u+v}}\, dt \quad (u,v>0) $$
(3)

is the beta function (cf. [12]). He et al. [1324] also provided some Hilbert-type integral inequalities in the whole plane.

In this paper, by using the way of real analysis and estimating the weight functions, we build a new Hilbert-type integral inequality in the whole plane with the non-homogeneous kernel and a few parameters. The constant factor related to the beta function is proved to be the best possible. We also consider the equivalent forms, the reverses, and some particular cases.

Some lemmas

Lemma 1

Suppose that \(0<\alpha_{1}\leq\alpha_{2}<\pi\), \(\mu ,\sigma>0\), \(\mu+\sigma=\lambda\), \(\gamma\in\{\frac{1}{2k+1},2k-1\ (k\in \mathbf{N})\}\), \(\delta\in\{-1,1\}\). We define weight functions \(\omega (\sigma,y)\) (\(y\in\mathbf{R}\)), and \(\varpi(\sigma,x)\) (\(x\in \mathbf{R}\)) as follows:

$$\begin{aligned}& \omega(\sigma,y) : =\int_{-\infty}^{\infty}\min _{i\in\{1,2\}}\frac {|y|^{\sigma}|x|^{\delta\sigma-1}}{[|x^{\delta}y|^{\gamma }+(x^{\delta }y)^{\gamma}\cos\alpha_{i}+1]^{\lambda/\gamma}}\,dx, \end{aligned}$$
(4)
$$\begin{aligned}& \varpi(\sigma,x) : =\int_{-\infty}^{\infty}\min _{i\in\{1,2\}}\frac {|x|^{\delta\sigma}|y|^{\sigma-1}}{[|x^{\delta}y|^{\gamma }+(x^{\delta }y)^{\gamma}\cos\alpha_{i}+1]^{\lambda/\gamma}}\,dy. \end{aligned}$$
(5)

Then for \(y,x\in\mathbf{R}\backslash\{0\}\), we have

$$\begin{aligned} \omega(\sigma,y) =&\varpi(\sigma,x)=K(\sigma) \\ :=&\frac{1}{\gamma2^{\sigma/\gamma}} \biggl[ \biggl(\sec\frac{\alpha _{1}}{2}\biggr)^{\frac{2\sigma}{\gamma}}+ \biggl(\csc\frac{\alpha_{2}}{2}\biggr)^{\frac{2\sigma}{ \gamma}} \biggr] B\biggl( \frac{\mu}{\gamma},\frac{\sigma}{\gamma}\biggr)\in \mathbf{R}_{+}. \end{aligned}$$
(6)

Proof

(i) For \(\delta=1\), \(y\in\mathbf{R}\backslash\{0\}\), setting \(u=xy\), we find

$$\begin{aligned} \omega(\sigma,y) =&\int_{-\infty}^{\infty}\min _{i\in\{1,2\}}\frac {1}{(|u|^{\gamma}+u^{\gamma}\cos\alpha_{i}+1)^{\lambda/\gamma }}|u|^{\sigma -1}\,du \\ =&\int_{0}^{\infty}\min_{i\in\{1,2\}} \frac{1}{[u^{\gamma}(1+\cos \alpha _{i})+1]^{\lambda/\gamma}}u^{\sigma-1}\,du \\ &{}+\int_{-\infty}^{0}\min_{i\in\{1,2\}} \frac{1}{[u^{\gamma}(-1+\cos \alpha_{i})+1]^{\lambda/\gamma}}(-u)^{\sigma-1}\,du \\ =&\int_{0}^{\infty}\min_{i\in\{1,2\}} \frac{1}{[u^{\gamma}(1+\cos \alpha _{i})+1]^{\lambda/\gamma}}u^{\sigma-1}\,du \\ &{}+\int_{0}^{\infty}\min_{i\in\{1,2\}} \frac{1}{[v^{\gamma}(1-\cos \alpha _{i})+1]^{\lambda/\gamma}}v^{\sigma-1}\,dv \\ =&\int_{0}^{\infty}\frac{u^{\sigma-1}\,du}{[u^{\gamma}(1+\cos\alpha _{1})+1]^{\lambda/\gamma}}+\int _{0}^{\infty}\frac{v^{\sigma-1}\,dv}{[v^{\gamma}(1-\cos\alpha_{2})+1]^{\lambda/\gamma}}. \end{aligned}$$
(7)

Setting \(t=u^{\gamma}(1+\cos\alpha_{1})\) (\(t=u^{\gamma}(1-\cos \alpha _{2})\)) in the above first (second) integral, by (3), it follows that

$$ \omega(\sigma,y)=\frac{1}{\gamma} \biggl[ \biggl(\frac{\sec^{2}\frac{\alpha _{1}}{2}}{2} \biggr)^{\frac{\sigma}{\gamma}}+\biggl(\frac{\csc^{2}\frac{\alpha _{2}}{2}}{2}\biggr)^{\frac{\sigma}{\gamma}} \biggr] \int_{0}^{\infty}\frac{t^{\frac {\sigma}{\gamma}-1}\,dt}{(t+1)^{\lambda/\gamma}}=K(\sigma). $$

(ii) For \(\delta=-1\), setting \(\frac{y}{x}\), we still can obtain \(\omega (\sigma,y)=K(\sigma)\).

Setting \(u=x^{\delta}y\), we also find

$$ \varpi(\sigma,x)=\int_{-\infty}^{\infty}\min _{i\in\{1,2\}}\frac{1}{ (|u|^{\gamma}+u^{\gamma}\cos\alpha_{i}+1)^{\lambda/\gamma }}|u|^{\sigma -1}\,du=K(\sigma). $$

Hence we have (6). □

Note

If we replace \(\min_{i\in\{1,2\}}\) by \(\max_{i\in\{ 1,2\}}\) in (4) and (5), then we may exchange \(\alpha_{1}\) and \(\alpha_{2}\) in (6).

Lemma 2

Suppose that \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(0<\alpha _{1}\leq\alpha_{2}<\pi\), \(\mu,\sigma>0\), \(\mu+\sigma=\lambda\), \(\gamma\in \{\frac{1}{2k+1},2k-1\ (k\in\mathbf{N})\}\), \(\delta\in\{-1,1\}\). If \(K(\sigma)\) is indicated by (6), \(f(x)\) is a non-negative measurable function in \((-\infty,\infty)\), then we have

$$\begin{aligned} J :=&\int_{-\infty}^{\infty}|y|^{p\sigma-1} \biggl\{ \int_{-\infty }^{\infty}\min_{i\in\{1,2\}} \frac{1}{[|x^{\delta}y|^{\gamma }+(x^{\delta }y)^{\gamma}\cos\alpha_{i}+1]^{\lambda/\gamma}}f(x)\,dx \biggr\} ^{p}\,dy \\ \leq&K^{p}(\sigma)\int_{-\infty}^{\infty}|x|^{p(1-\delta\sigma )-1}f^{p}(x) \,dx. \end{aligned}$$
(8)

Proof

We set

$$ k_{\lambda}^{(\delta)}(x,y):=\min_{i\in\{1,2\}} \frac{1}{[|x^{\delta }y|^{\gamma}+(x^{\delta}y)^{\gamma}\cos\alpha_{i}+1]^{\lambda /\gamma}} \quad (x,y\in\mathbf{R}). $$
(9)

By Hölder’s inequality (cf. [25]), we have

$$\begin{aligned}& \biggl( \int_{-\infty}^{\infty}k_{\lambda}^{(\delta )}(x,y)f(x) \,dx \biggr) ^{p} \\& \quad = \biggl\{ \int_{-\infty}^{\infty}k_{\lambda}^{(\delta)}(x,y) \biggl[ \frac{|x|^{(1-\delta\sigma)/q}}{|y|^{(1-\sigma)/p}}f(x) \biggr] \biggl[ \frac{|y|^{(1-\sigma)/p}}{|x|^{(1-\delta\sigma)/q}} \biggr] \,dx \biggr\} ^{p} \\& \quad \leq \int_{-\infty}^{\infty}k_{\lambda}^{(\delta)}(x,y) \frac{|x|^{(1-\delta\sigma)(p-1)}}{|y|^{1-\sigma}}f^{p}(x)\,dx \\& \qquad {}\times \biggl[ \int_{-\infty}^{\infty}k_{\lambda}^{(\delta )}(x,y) \frac{|y|^{(1-\sigma)(q-1)}}{|x|^{1-\delta\sigma}}\,dx \biggr] ^{p-1} \\& \quad =\bigl(\omega(\sigma,y)\bigr)^{p-1}|y|^{-p\sigma+1}\int _{-\infty}^{\infty }k_{\lambda}^{(\delta)}(x,y) \frac{|x|^{(1-\delta\sigma)(p-1)}}{|y|^{(1-\sigma)}}f^{p}(x)\,dx. \end{aligned}$$
(10)

Then by (6) and the Fubini theorem (cf. [26]), it follows that

$$\begin{aligned} J \leq&K^{p-1}(\sigma)\int_{-\infty}^{\infty} \biggl[ \int_{-\infty }^{\infty}k_{\lambda}^{(\delta)}(x,y) \frac{|x|^{(1-\delta\sigma )(p-1)}}{|y|^{(1-\sigma)}}f^{p}(x)\,dx \biggr] \,dy \\ =&K^{p-1}(\sigma)\int_{-\infty}^{\infty}\varpi( \sigma ,x)|x|^{p(1-\delta\sigma)-1}f^{p}(x)\,dx. \end{aligned}$$

Hence, still in view of (6), inequality (8) follows. □

Main results and applications

Theorem 1

Suppose that \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(0<\alpha _{1}\leq\alpha_{2}<\pi\), \(\mu,\sigma>0\), \(\mu+\sigma=\lambda\), \(\gamma\in \{\frac{1}{2k+1},2k-1\ (k\in\mathbf{N})\}\), \(\delta\in\{-1,1\}\). If \(K(\sigma)\) is indicated by (6), \(f(x),g(y)\geq0\), satisfying \(0<\int_{-\infty}^{\infty}|x|^{p(1-\delta\sigma)-1}f^{p}(x)\,dx<\infty\) and \(0<\int_{-\infty}^{\infty}|y|^{q(1-\sigma)-1}g^{q}(y)\,dy<\infty\), then we have the following equivalent inequalities:

$$\begin{aligned}& I : =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \min_{i\in\{ 1,2\}}\frac{1}{[|x^{\delta}y|^{\gamma}+(x^{\delta}y)^{\gamma}\cos\alpha _{i}+1]^{\lambda/\gamma}}f(x)g(y)\,dx\,dy \\& \hphantom{I} < K(\sigma) \biggl[ \int_{-\infty}^{\infty}|x|^{p(1-\delta\sigma )-1}f^{p}(x) \,dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{-\infty}^{\infty }|y|^{q(1-\sigma)-1}g^{q}(y) \,dy \biggr] ^{\frac{1}{q}}, \end{aligned}$$
(11)
$$\begin{aligned}& J : =\int_{-\infty}^{\infty}|y|^{p\sigma-1} \biggl\{ \int _{-\infty }^{\infty}\min_{i\in\{1,2\}} \frac{1}{[|x^{\delta}y|^{\gamma }+(x^{\delta }y)^{\gamma}\cos\alpha_{i}+1]^{\lambda/\gamma}}f(x)\,dx \biggr\} ^{p}\,dy \\& \hphantom{J} < K^{p}(\sigma)\int_{-\infty}^{\infty}|x|^{p(1-\delta\sigma )-1}f^{p}(x) \,dx, \end{aligned}$$
(12)

where the constant factors \(K(\sigma)\) and \(K^{p}(\sigma)\) are the best possible.

In particular, for \(\alpha_{1}=\alpha_{2}=\alpha\in(0,\pi)\), \(\gamma=1\) in (11) and (12), we find

$$ K(\sigma)=k(\sigma):=\frac{1}{2^{\sigma}}\biggl[\biggl(\sec\frac{\alpha}{2}\biggr)^{2\sigma}+\biggl(\csc\frac{\alpha}{2}\biggr)^{2\sigma} \biggr]B(\mu,\sigma), $$
(13)

and the following equivalent inequalities:

$$\begin{aligned}& \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{1}{(|x^{\delta }y|+x^{\delta}y\cos\alpha+1)^{\lambda}}f(x)g(y)\,dx\,dy \\& \quad < k(\sigma) \biggl[ \int_{-\infty}^{\infty}|x|^{p(1-\delta\sigma )-1}f^{p}(x) \,dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{-\infty}^{\infty }|y|^{q(1-\sigma)-1}g^{q}(y) \,dy \biggr] ^{\frac{1}{q}}, \end{aligned}$$
(14)
$$\begin{aligned}& \int_{-\infty}^{\infty}|y|^{p\sigma-1} \biggl[ \int _{-\infty }^{\infty}\frac{1}{(|x^{\delta}y|+x^{\delta}y\cos\alpha+1)^{\lambda }}f(x)\,dx \biggr] ^{p}\,dy \\& \quad < k^{p}(\sigma)\int_{-\infty}^{\infty}|x|^{p(1-\sigma)-1}f^{p}(x) \,dx. \end{aligned}$$
(15)

Proof

If (10) takes the form of equality for \(y\in(-\infty,0)\cup(0,\infty)\), then there exist constants A and B, such that they are not all zero, and

$$ A\frac{|x|^{(1-\delta\sigma)(p-1)}}{|y|^{(1-\sigma)}}f^{p}(x)=B\frac{ |y|^{(1-\sigma)(q-1)}}{|x|^{(1-\delta\sigma)}}\quad \mbox{a.e. in } (-\infty,\infty). $$

We suppose \(A\neq0\) (otherwise \(B=A=0\)). Then it follows that

$$ |x|^{p(1-\delta\sigma)-1}f^{p}(x)=|y|^{q(1-\sigma)}\frac{B}{A|x|}\quad \mbox{a.e. in }(-\infty,\infty), $$

which contradicts the fact that \(0<\int_{-\infty}^{\infty }|x|^{p(1-\delta \sigma)-1}f^{p}(x)\,dx<\infty\). Hence (10) takes the form of a strict inequality. So does (9), and we have (12).

By Hölder’s inequality (cf. [25]), we find

$$\begin{aligned} I =&\int_{-\infty}^{\infty} \biggl( |y|^{\sigma-\frac{1}{p}}\int _{-\infty }^{\infty}k_{\lambda}^{(\delta)}(x,y)f(x) \,dx \biggr) \bigl(|y|^{\frac{1}{p}-\sigma}g(y)\,dy\bigr) \\ \leq&J^{\frac{1}{p}} \biggl[ \int_{-\infty}^{\infty}|y|^{q(1-\sigma )-1}g^{q}(y) \,dy \biggr] ^{\frac{1}{q}}. \end{aligned}$$
(16)

Then by (12), we have (11). On the other hand, suppose that (11) is valid. Setting

$$ g(y):=|y|^{p\sigma-1} \biggl( \int_{-\infty}^{\infty}k_{\lambda }^{(\delta )}(x,y)f(x) \,dx \biggr) ^{p-1} ,\quad y\in\mathbf{R}, $$

then it follows that \(J=\int_{-\infty}^{\infty}|y|^{q(1-\sigma )-1}g^{q}(y)\,dy\). By (9), we have \(J<\infty\). If \(J=0\), then (12) is obviously of value; if \(0< J<\infty\), then by (11), we obtain

$$\begin{aligned}& \int_{-\infty}^{\infty}|y|^{q(1-\sigma)-1}g^{q}(y) \,dy \\& \quad =J=I < K(\sigma) \biggl[ \int_{-\infty}^{\infty}|x|^{p(1-\delta\sigma )-1}f^{p}(x) \,dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{-\infty}^{\infty }|y|^{q(1-\sigma)-1}g^{q}(y) \,dy \biggr] ^{\frac{1}{q}}, \end{aligned}$$
(17)
$$\begin{aligned}& J^{\frac{1}{p}}= \biggl[ \int_{-\infty}^{\infty}|y|^{q(1-\sigma )-1}g^{q}(y) \,dy \biggr] ^{\frac{1}{p}} \\& \hphantom{J^{\frac{1}{p}}}< K(\sigma) \biggl[ \int_{-\infty }^{\infty}|x|^{p(1-\delta\sigma)-1}f^{p}(x) \,dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$
(18)

Hence we have (12), which is equivalent to (11).

We set \(E_{\delta}:=\{x\in\mathbf{R};|x|^{\delta}\geq1\}\), and \(E_{\delta}^{+}:=E_{\delta}\cap\mathbf{R}_{+}=\{x\in\mathbf{R}_{+};x^{\delta}\geq1\}\). For \(\varepsilon>0\), we define functions \(\tilde{f}(x)\), \(\tilde{g}(y)\) as follows:

$$\begin{aligned}& \tilde{f}(x) :=\left \{ \textstyle\begin{array}{l@{\quad}l} |x|^{\delta(\sigma-\frac{2\varepsilon}{p})-1}, & x\in E_{\delta} , \\ 0, & x\in\mathbf{R}\backslash E_{\delta}, \end{array}\displaystyle \right . \\& \tilde{g}(y) :=\left \{ \textstyle\begin{array}{l@{\quad}l} 0, & y\in(-\infty,-1)\cup(1,\infty), \\ |y|^{\sigma+\frac{2\varepsilon}{q}-1}, & y\in[-1,1].\end{array}\displaystyle \right . \end{aligned}$$

Then we obtain

$$\begin{aligned} \tilde{L} :=& \biggl[ \int_{-\infty}^{\infty}|x|^{p(1-\delta\sigma )-1}\tilde{f}^{p}(x)\,dx \biggr] ^{\frac{1}{p}} \biggl[ \int _{-\infty}^{\infty }|y|^{q(1-\sigma)-1}\tilde{g}^{q}(y) \,dy \biggr] ^{\frac{1}{q}} \\ =&2 \biggl( \int_{E_{\delta}^{+}}x^{-2\delta\varepsilon-1}\,dx \biggr) ^{ \frac{1}{p}} \biggl( \int_{0}^{1}y^{2\varepsilon-1} \,dy \biggr) ^{\frac {1}{q}} \\ =&\frac{1}{\varepsilon}. \end{aligned}$$

We find

$$ h(x):=\int_{-1}^{1}\min_{i\in\{1,2\}} \frac{|y|^{\sigma+\frac {2\varepsilon }{q}-1}}{[|x^{\delta}y|^{\gamma}+(x^{\delta}y)^{\gamma}\cos\alpha _{i}+1]^{\lambda/\gamma}}\,dy=h(-x). $$

In fact, setting \(Y=-y\), we obtain

$$\begin{aligned} h(-x) =&\int_{-1}^{1}\min_{i\in\{1,2\}} \frac{|y|^{\sigma+\frac{2\varepsilon}{q}-1}}{[|-x^{\delta}y|^{\gamma}+(-x^{\delta}y)^{\gamma }\cos\alpha_{i}+1]^{\lambda/\gamma}}\,dy \\ =&\int_{-1}^{1}\min_{i\in\{1,2\}} \frac{|Y|^{\sigma+\frac {2\varepsilon}{q}-1}}{[|x^{\delta}Y|^{\gamma}+(x^{\delta}Y)^{\gamma}\cos\alpha _{i}+1]^{\lambda/\gamma}}\, dY=h(x). \end{aligned}$$

It follows that

$$\begin{aligned} \tilde{I} =&\int_{-\infty}^{\infty}\int _{-\infty}^{\infty }k_{\lambda }^{(\delta)}(x,y) \tilde{f}(x)\tilde{g}(y)\,dx\,dy \\ =&\int_{E_{\delta}}|x|^{\delta(\sigma-\frac{2\varepsilon}{p})-1}h(x)\,dx=2\int _{E_{\delta}^{+}}x^{\delta(\sigma-\frac{2\varepsilon }{p})-1}h(x)\,dx \\ \stackrel{u=x^{\delta}y}{=}&2\int_{E_{\delta}^{+}}x^{-2\delta \varepsilon -1} \biggl\{ \int_{-x^{\delta}}^{x^{\delta}}\min_{i\in\{1,2\}} \frac{|u|^{\sigma+\frac{2\varepsilon}{q}-1}}{[|u|^{\gamma}+u^{\gamma}\cos \alpha_{i}+1]^{\lambda/\gamma}}\,du \biggr\} \,dx. \end{aligned}$$

Setting \(v=x^{\delta}\) in the above integral, by the Fubini theorem (cf. [26]), we find

$$\begin{aligned} \tilde{I} =&2\int_{1}^{\infty}v^{-2\varepsilon-1} \biggl\{ \int_{-v}^{v}\min_{i\in\{1,2\}} \frac{|u|^{\sigma+\frac{2\varepsilon }{q}-1}}{[|u|^{\gamma}+u^{\gamma}\cos\alpha_{i}+1]^{\lambda/\gamma}}\,du \biggr\} \,dv \\ =& 2\int_{1}^{\infty}v^{-2\varepsilon-1}\biggl\{ \int _{0}^{v}\biggl[ \min_{i\in\{1,2\}} \frac{1}{[u^{\gamma}(1+\cos\alpha_{i})+1]^{\lambda /\gamma}} \\ &{} +\min_{i\in\{1,2\}}\frac{1}{[u^{\gamma}(1-\cos\alpha _{i})+1]^{\lambda/\gamma}}\biggr] u^{\sigma+\frac{2\varepsilon}{q}-1}\,du \biggr\} \,dv \\ =& 2\int_{1}^{\infty}v^{-2\varepsilon-1}\biggl\{ \int _{0}^{v}\biggl[ \frac{1}{[u^{\gamma}(1+\cos\alpha_{1})+1]^{\lambda/\gamma}} +\frac{1}{[u^{\gamma}(1-\cos\alpha_{2})+1]^{\lambda /\gamma}}\biggr] u^{\sigma+\frac{2\varepsilon}{q}-1}\,du\biggr\} \,dv \\ =&2\int_{1}^{\infty}v^{-2\varepsilon-1} \biggl\{ \int _{0}^{1}\biggl[\frac {u^{\sigma +\frac{2\varepsilon}{q}-1}}{[u^{\gamma}(1+\cos\alpha_{1})+1]^{\frac{ \lambda}{\gamma}}}+ \frac{u^{\sigma+\frac{2\varepsilon }{q}-1}}{[u^{\gamma }(1-\cos\alpha_{2})+1]^{\frac{\lambda}{\gamma}}}\biggr]\,du \biggr\} \,dv \\ &{}+2\int_{1}^{\infty}v^{-2\varepsilon-1} \biggl\{ \int _{1}^{v}\biggl[\frac {u^{\sigma +\frac{2\varepsilon}{q}-1}}{[u^{\gamma}(1+\cos\alpha_{1})+1]^{\frac{ \lambda}{\gamma}}}+ \frac{u^{\sigma+\frac{2\varepsilon }{q}-1}}{[u^{\gamma }(1-\cos\alpha_{2})+1]^{\frac{\lambda}{\gamma}}}\biggr]\,du \biggr\} \,dv \\ =& \frac{1}{\varepsilon}\int_{0}^{1} \biggl\{ \frac{u^{\sigma+\frac{2\varepsilon}{q}-1}}{[u^{\gamma}(1+\cos\alpha_{1})+1]^{\lambda /\gamma}}+\frac{u^{\sigma+\frac{2\varepsilon}{q}-1}}{[u^{\gamma}(1-\cos\alpha _{2})+1]^{\lambda/\gamma}} \biggr\} \,du \\ &{}+2\int_{1}^{\infty}\biggl(\int _{u}^{\infty}v^{-2\varepsilon-1}\,dv\biggr) \\ &{}\times \biggl\{ \frac{u^{\sigma+\frac{2\varepsilon}{q}-1}}{[u^{\gamma }(1+\cos\alpha_{1})+1]^{\lambda/\gamma}}+\frac{u^{\sigma+\frac{2\varepsilon}{q}-1}}{[u^{\gamma}(1-\cos\alpha_{2})+1]^{\lambda /\gamma}} \biggr\} \,du \\ =& \frac{1}{\varepsilon}\biggl\{ \int_{0}^{1} \biggl[ \frac{u^{\sigma +\frac{2\varepsilon}{q}-1}}{[u^{\gamma}(1+\cos\alpha_{1})+1]^{\lambda /\gamma}}+\frac{u^{\sigma+\frac{2\varepsilon}{q}-1}}{[u^{\gamma}(1-\cos\alpha _{2})+1]^{\lambda/\gamma}} \biggr] \,du \\ &{} +\int_{1}^{\infty} \biggl[ \frac{u^{\sigma-\frac{2\varepsilon }{p}-1}}{[u^{\gamma}(1+\cos\alpha_{1})+1]^{\lambda/\gamma}}+ \frac {u^{\sigma -\frac{2\varepsilon}{p}-1}}{[u^{\gamma}(1-\cos\alpha _{2})+1]^{\lambda /\gamma}} \biggr] \,du\biggr\} . \end{aligned}$$

If the constant factor \(K(\sigma)\) in (11) is not the best possible, then there exists a positive number k, with \(K(\sigma)< k\), such that (11) is valid when replacing \(K(\sigma)\) by k. Then we have \(\varepsilon \tilde{I}<\varepsilon k\tilde{L}\), and

$$\begin{aligned} \begin{aligned}[b] &\int_{0}^{1} \biggl\{ \frac{u^{\sigma+\frac{2\varepsilon}{q}-1}}{[u^{\gamma}(1+\cos\alpha_{1})+1]^{\lambda/\gamma}}+ \frac{u^{\sigma +\frac{2\varepsilon}{q}-1}}{[u^{\gamma}(1-\cos\alpha_{2})+1]^{\lambda /\gamma}} \biggr\} \,du \\ &\qquad {} +\int_{1}^{\infty} \biggl\{ \frac{u^{\sigma-\frac{2\varepsilon }{p}-1}}{[u^{\gamma}(1+\cos\alpha_{1})+1]^{\lambda/\gamma}}+\frac{u^{\sigma -\frac{2\varepsilon}{p}-1}}{[u^{\gamma}(1-\cos\alpha_{2})+1]^{\lambda /\gamma}} \biggr\} \,du \\ &\quad = \varepsilon\tilde{I}< \varepsilon k\tilde{L}=k. \end{aligned} \end{aligned}$$
(19)

By (7) and the Levi theorem (cf. [26]), we have

$$\begin{aligned} K(\sigma) =&\int_{0}^{\infty}\frac{u^{\sigma-1}\,du}{[u^{\gamma }(1+\cos \alpha_{1})+1]^{\lambda/\gamma}}+\int _{0}^{\infty}\frac{u^{\sigma -1}\,du}{[u^{\gamma}(1-\cos\alpha_{2})+1]^{\lambda/\gamma}} \\ =&\int_{0}^{1}\lim_{\varepsilon\rightarrow0^{+}} \biggl\{ \frac {u^{\sigma+\frac{2\varepsilon}{q}-1}}{[u^{\gamma}(1+\cos\alpha_{1})+1]^{\lambda /\gamma}}+\frac{u^{\sigma+\frac{2\varepsilon}{q}-1}}{[u^{\gamma }(1-\cos \alpha_{2})+1]^{\lambda/\gamma}} \biggr\} \,du \\ &{}+\int_{1}^{\infty}\lim_{\varepsilon\rightarrow0^{+}} \biggl\{ \frac{u^{\sigma-\frac{2\varepsilon}{p}-1}}{[u^{\gamma}(1+\cos\alpha _{1})+1]^{\lambda/\gamma}}+\frac{u^{\sigma-\frac{2\varepsilon }{p}-1}}{[u^{\gamma}(1-\cos\alpha_{2})+1]^{\lambda/\gamma}} \biggr\} \,du \\ =&\lim_{\varepsilon\rightarrow0^{+}}\biggl\{ \int_{0}^{1} \biggl[ \frac{ u^{\sigma+\frac{2\varepsilon}{q}-1}}{[u^{\gamma}(1+\cos\alpha _{1})+1]^{\lambda/\gamma}}+\frac{u^{\sigma+\frac{2\varepsilon }{q}-1}}{[u^{\gamma}(1-\cos\alpha_{2})+1]^{\lambda/\gamma}} \biggr] \,du \\ &{} +\int_{1}^{\infty} \biggl[ \frac{u^{\sigma-\frac{2\varepsilon }{p}-1}}{[u^{\gamma}(1+\cos\alpha_{1})+1]^{\lambda/\gamma}}+ \frac {u^{\sigma-\frac{2\varepsilon}{p}-1}}{[u^{\gamma}(1-\cos\alpha_{2})+1]^{\lambda /\gamma}} \biggr] \,du\biggr\} \leq k, \end{aligned}$$

which contradicts the fact that \(k< K(\sigma)\). Hence the constant factor \(K(\sigma)\) in (11) is the best possible.

If the constant factor in (12) is not the best possible, then by (16), we may get a contradiction: that the constant factor in (11) is not the best possible. □

Theorem 2

As the assumptions of Theorem  1, replacing \(p>1\) by \(0< p<1\), we have the equivalent reverses of (11) and (12) with the same best constant factors.

Proof

By the reverse Hölder’s inequality (cf. [25]), we have the reverses of (9) and (16). It is easy to obtain the reverse of (12). In view of the reverses of (12) and (16), we obtain the reverse of (11). On the other hand, suppose that the reverse of (11) is valid. Setting the same \(g(y)\) as Theorem 1, by the reverse of (9), we have \(J>0\). If \(J=\infty\), then the reverse of (12) is obviously value; if \(J<\infty\), then by the reverse of (11), we obtain the reverses of (17) and (18). Hence we have the reverse of (12), which is equivalent to the reverse of (11).

If the constant factor \(K(\sigma)\) in the reverse of (11) is not the best possible, then there exists a positive constant k, with \(k>K(\sigma)\), such that the reverse of (11) is still valid when replacing \(K(\sigma)\) by k. By the reverse of (19), we have

$$\begin{aligned}& \int_{0}^{1} \biggl\{ \frac{u^{\sigma+\frac{2\varepsilon }{q}-1}}{[u^{\gamma }(1+\cos\alpha_{1})+1]^{\lambda/\gamma}}+ \frac{u^{\sigma+\frac{2\varepsilon}{q}-1}}{[u^{\gamma}(1-\cos\alpha_{2})+1]^{\lambda /\gamma}} \biggr\} \,du \\& \quad {}+\int_{1}^{\infty} \biggl\{ \frac{u^{\sigma-\frac{2\varepsilon}{p}-1}}{ [u^{\gamma}(1+\cos\alpha_{1})+1]^{\lambda/\gamma}}+\frac{u^{\sigma -\frac{2\varepsilon}{p}-1}}{[u^{\gamma}(1-\cos\alpha_{2})+1]^{\lambda /\gamma}} \biggr\} \,du>k. \end{aligned}$$
(20)

For \(\varepsilon\rightarrow0^{+}\), by the Levi theorem (cf. [26]), we find that

$$\begin{aligned}& \int_{1}^{\infty} \biggl\{ \frac{u^{\sigma-\frac{2\varepsilon}{p}-1}}{[u^{\gamma}(1+\cos\alpha_{1})+1]^{\lambda/\gamma}}+ \frac{u^{\sigma -\frac{2\varepsilon}{p}-1}}{[u^{\gamma}(1-\cos\alpha_{2})+1]^{\lambda /\gamma}} \biggr\} \,du \\& \quad \rightarrow\int_{1}^{\infty} \biggl\{ \frac{u^{\sigma-1}}{[u^{\gamma}(1+\cos \alpha _{1})+1]^{\lambda/\gamma}}+\frac{u^{\sigma-1}}{[u^{\gamma}(1-\cos \alpha _{2})+1]^{\lambda/\gamma}} \biggr\} \,du. \end{aligned}$$
(21)

There exists a constant \(\delta_{0}>0\), such that \(\sigma-\frac{1}{2}\delta_{0}>0\), and then \(K(\sigma-\frac{\delta_{0}}{2})<\infty\). For \(0<\varepsilon<\frac{\delta_{0}|q|}{4}\) (\(q<0\)), since \(u^{\sigma+\frac{2\varepsilon}{q}-1}\leq u^{\sigma-\frac{\delta_{0}}{2}-1}\), \(u\in(0,1]\), and

$$\begin{aligned} 0 < &\int_{0}^{1} \biggl\{ \frac{u^{\sigma-\frac{\delta_{0}}{2}-1}}{[u^{\gamma}(1+\cos\alpha_{1})+1]^{\lambda/\gamma}}+ \frac{u^{\sigma -\frac{\delta_{0}}{2}-1}}{[u^{\gamma}(1-\cos\alpha_{2})+1]^{\lambda /\gamma}} \biggr\} \,du \\ \leq& K\biggl(\sigma-\frac{\delta_{0}}{2}\biggr)< \infty, \end{aligned}$$

then by the Lebesgue control convergence theorem (cf. [26]), for \(\varepsilon\rightarrow0^{+}\), we have

$$\begin{aligned}& \int_{0}^{1} \biggl\{ \frac{u^{\sigma+\frac{2\varepsilon}{q}-1}}{[u^{\gamma}(1+\cos\alpha_{1})+1]^{\lambda/\gamma}}+ \frac{u^{\sigma +\frac{2\varepsilon}{q}-1}}{[u^{\gamma}(1-\cos\alpha_{2})+1]^{\lambda /\gamma}} \biggr\} \,du \\& \quad \rightarrow \int_{0}^{1} \biggl\{ \frac{u^{\sigma-1}}{[u^{\gamma }(1+\cos \alpha_{1})+1]^{\lambda/\gamma}}+\frac{u^{\sigma-1}}{[u^{\gamma }(1-\cos \alpha_{2})+1]^{\lambda/\gamma}} \biggr\} \,du. \end{aligned}$$
(22)

By (20), (21), and (22), for \(\varepsilon\rightarrow 0^{+}\), we find \(K(\sigma)\geq k\), which contradicts the fact that \(k>K(\sigma)\). Hence, the constant factor \(K(\sigma)\) in the reverse of (11) is the best possible.

If the constant factor in reverse of (12) is not the best possible, then by the reverse of (16), we may get a contradiction that the constant factor in the reverse of (11) is not the best possible. □

Remarks

For \(\delta=-1\) in (11) and (12), replacing \(|x|^{\lambda}f(x)\) by \(f(x)\), we obtain the following equivalent inequalities with the homogeneous kernel and the best possible constant factors:

$$\begin{aligned}& \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \min_{i\in\{1,2\} }\frac{1}{(|y|^{\gamma}+\operatorname{sgn}(x)y^{\gamma}\cos\alpha_{i}+|x|^{\gamma })^{\lambda /\gamma}}f(x)g(y)\,dx\,dy \\& \quad < K(\sigma) \biggl[ \int_{-\infty}^{\infty}|x|^{p(1-\mu)-1}f^{p}(x) \,dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{-\infty}^{\infty}|y|^{q(1-\sigma )-1}g^{q}(y) \,dy \biggr] ^{\frac{1}{q}}, \end{aligned}$$
(23)
$$\begin{aligned}& \int_{-\infty}^{\infty}|y|^{p\sigma-1} \biggl[ \int _{-\infty }^{\infty }\min_{i\in\{1,2\}} \frac{1}{(|y|^{\gamma}+\operatorname{sgn}(x)y^{\gamma}\cos\alpha _{i}+|x|^{\gamma})^{\lambda/\gamma}}f(x)\,dx \biggr] ^{p}\,dy \\& \quad < K^{p}(\sigma)\int_{-\infty}^{\infty}|x|^{p(1-\mu)-1}f^{p}(x) \,dx. \end{aligned}$$
(24)

In particular, for \(\alpha_{1}=\alpha_{2}=\alpha\in(0,\pi)\), \(\gamma=1\) in (23) and (24), we obtain the following equivalent inequalities:

$$\begin{aligned}& \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{1}{(|y|+\operatorname{sgn}(x)y\cos\alpha+|x|)^{\lambda}}f(x)g(y)\,dx\,dy \\& \quad < k(\sigma) \biggl[ \int_{-\infty}^{\infty}|x|^{p(1-\mu)-1}f^{p}(x) \,dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{-\infty}^{\infty}|y|^{q(1-\sigma )-1}g^{q}(y) \,dy \biggr] ^{\frac{1}{q}}, \end{aligned}$$
(25)
$$\begin{aligned}& \int_{-\infty}^{\infty}|y|^{p\sigma-1} \biggl[ \int _{-\infty }^{\infty}\frac{1}{(|y|+\operatorname{sgn}(x)y\cos\alpha+|x|)^{\lambda}}f(x)\,dx \biggr] ^{p}\,dy \\& \quad < k^{p}(\sigma)\int_{-\infty}^{\infty}|x|^{p(1-\mu)-1}f^{p}(x) \,dx, \end{aligned}$$
(26)

where \(k(\sigma)\) is indicated by (13).

References

  1. 1.

    Hardy, GH, Littlewood, JE, Pólya, G: Inequalities. Cambridge University Press, Cambridge (1934)

    Google Scholar 

  2. 2.

    Mitrinović, DS, Pečarić, JE, Fink, AM: Inequalities Involving Functions and Their Integrals and Derivatives. Kluwer Academic, Boston (1991)

    Google Scholar 

  3. 3.

    Yang, B: The Norm of Operator and Hilbert-Type Inequalities. Science Press, Beijing (2009)

    Google Scholar 

  4. 4.

    Yang, B: A survey of the study of Hilbert-type inequalities with parameters. Adv. Math. 38(3), 257-268 (2009)

    MathSciNet  Google Scholar 

  5. 5.

    Yang, B: On the norm of an integral operator and applications. J. Math. Anal. Appl. 321, 182-192 (2006)

    MATH  MathSciNet  Article  Google Scholar 

  6. 6.

    Xu, J: Hardy-Hilbert’s inequalities with two parameters. Adv. Math. 36(2), 63-76 (2007)

    Google Scholar 

  7. 7.

    Yang, B: On the norm of a Hilbert’s type linear operator and applications. J. Math. Anal. Appl. 325, 529-541 (2007)

    MATH  MathSciNet  Article  Google Scholar 

  8. 8.

    Xin, D: A Hilbert-type integral inequality with the homogeneous kernel of zero degree. Math. Theory Appl. 30(2), 70-74 (2010)

    MathSciNet  Google Scholar 

  9. 9.

    Yang, B: A Hilbert-type integral inequality with the homogeneous kernel of degree 0. J. Shandong Univ. Nat. Sci. 45(2), 103-106 (2010)

    MathSciNet  Google Scholar 

  10. 10.

    Debnath, L, Yang, B: Recent developments of Hilbert-type discrete and integral inequalities with applications. Int. J. Math. Math. Sci. 2012, Article ID 871845 (2012)

    MathSciNet  Google Scholar 

  11. 11.

    Yang, B: A new Hilbert-type integral inequality. Soochow J. Math. 33(4), 849-859 (2007)

    MATH  MathSciNet  Google Scholar 

  12. 12.

    Wang, Z, Guo, D: Introduction to Special Functions. Science Press, Beijing (1979)

    Google Scholar 

  13. 13.

    He, B, Yang, B: On a Hilbert-type integral inequality with the homogeneous kernel of 0-degree and the hypergeometric function. Math. Pract. Theory 40(18), 105-211 (2010)

    MathSciNet  Google Scholar 

  14. 14.

    Yang, B: A new Hilbert-type integral inequality with some parameters. J. Jilin Univ. Sci. Ed. 46(6), 1085-1090 (2008)

    Google Scholar 

  15. 15.

    Yang, B: A Hilbert-type integral inequality with a non-homogeneous kernel. J. Xiamen Univ. Nat. Sci. 48(2), 165-169 (2008)

    Google Scholar 

  16. 16.

    Zeng, Z, Xie, Z: On a new Hilbert-type integral inequality with the homogeneous kernel of degree 0 and the integral in whole plane. J. Inequal. Appl. 2010, Article ID 256796 (2010)

    MathSciNet  Article  Google Scholar 

  17. 17.

    Yang, B: A reverse Hilbert-type integral inequality with some parameters. J. Xinxiang Univ. Nat. Sci. Ed. 27(6), 1-4 (2010)

    MATH  Google Scholar 

  18. 18.

    Wang, A, Yang, B: A new Hilbert-type integral inequality in whole plane with the non-homogeneous kernel. J. Inequal. Appl. 2011, 123 (2011)

    Article  Google Scholar 

  19. 19.

    Xin, D, Yang, B: A Hilbert-type integral inequality in whole plane with the homogeneous kernel of degree −2. J. Inequal. Appl. 2011, Article ID 401428 (2011)

    MathSciNet  Article  Google Scholar 

  20. 20.

    He, B, Yang, B: On an inequality concerning a non-homogeneous kernel and the hypergeometric function. Tamsui Oxf. J. Inf. Math. Sci. 27(1), 75-88 (2011)

    MATH  MathSciNet  Google Scholar 

  21. 21.

    Yang, B: A reverse Hilbert-type integral inequality with a non-homogeneous kernel. J. Jilin Univ. Sci. Ed. 49(3), 437-441 (2011)

    Google Scholar 

  22. 22.

    Xie, Z, Zeng, Z, Sun, Y: A new Hilbert-type inequality with the homogeneous kernel of degree −2. Adv. Appl. Math. Sci. 12(7), 391-401 (2013)

    MATH  MathSciNet  Google Scholar 

  23. 23.

    Zeng, Z, Raja Rama Gandhi, K, Xie, Z: A new Hilbert-type inequality with the homogeneous kernel of degree −2 and with the integral. Bull. Math. Sci. Appl. 3(1), 11-20 (2014)

    Google Scholar 

  24. 24.

    Yang, B, Chen, Q: Two kinds of Hilbert-type integral inequalities in the whole plane. J. Inequal. Appl. 2015, 21 (2015)

    Article  Google Scholar 

  25. 25.

    Kuang, J: Applied Inequalities. Shangdong Science and Technology Press, Jinan (2004)

    Google Scholar 

  26. 26.

    Kuang, J: Introduction to Real Analysis. Hunan Education Press, Changsha (1996)

    Google Scholar 

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Acknowledgements

The authors wish to express their thanks to the referees for their careful reading of the manuscript and for their valuable suggestions. This work is supported by the National Natural Science Foundation (No. 61370186), and 2013 Knowledge Construction Special Foundation Item of Guangdong Institution of Higher Learning College and University (No. 2013KJCX0140).

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Correspondence to Zhaohui Gu.

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BY carried out the mathematical studies, participated in the sequence alignment and drafted the manuscript. ZG participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.

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Gu, Z., Yang, B. A Hilbert-type integral inequality in the whole plane with a non-homogeneous kernel and a few parameters. J Inequal Appl 2015, 314 (2015). https://doi.org/10.1186/s13660-015-0844-8

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MSC

  • 26D15

Keywords

  • Hilbert-type integral inequality
  • weight function
  • equivalent form
  • beta function
  • reverse