A new Hilbert-type integral inequality in the whole plane with the non-homogeneous kernel

Abstract

By using the way of weight functions and the technique of real analysis, a new Hilbert-type integral inequality with the non-homogeneous kernel in the whole plane with the best constant factor is given. As applications, the equivalent inequalities with the best constant factors, the reverses and some particular cases are obtained.

2000 Mathematics Subject Classification

26D15

1 Introduction

If f(x), g(x) ≥ 0, such that $0<{\int }_0^{\infty }{f}^2\left(x\right)\mathsf{\text{d}}x<\infty$ and $0<{\int }_0^{\infty }{g}^2\left(x\right)\mathsf{\text{d}}x<\infty$, then we have (cf. [1]):

$$\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}\frac{f\left(x\right)g\left(y\right)}{x+y}\mathsf{\text{d}}x\mathsf{\text{d}}y<\pi {\left(\underset{0}{\overset{\infty }{\int }}{f}^2\left(x\right)\mathsf{\text{d}}x\underset{0}{\overset{\infty }{\int }}{g}^2\left(x\right)\mathsf{\text{d}}x\right)}^{\frac{1}{2}},$$

(1)

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

If p, r > 1, $\frac{1}{p}+\frac{1}{q}=1$, $\frac{1}{r}+\frac{1}{s}=1$ λ > 0, f(x), g(x) ≥ 0, such that $0<{\int }_0^{\infty }{x}^{p\left(1+\frac{\lambda }{r}\right)}{f}^p\left(x\right)\mathsf{\text{d}}x<\infty$ and, $0<{\int }_0^{\infty }{x}^{q\left(1+\frac{\lambda }{s}\right)}{g}^q\left(x\right)\mathsf{\text{d}}x<\infty$, then we have [3]:

$$\begin{array}{c}\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{\left(min\left\{x,y\right\}\right)}^{\lambda }f\left(x\right)g\left(y\right)\mathsf{\text{d}}x\mathsf{\text{d}}y\\ <\frac{rs}{\lambda }{\left(\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1+\frac{\lambda }{r}\right)-1}{f}^p\left(x\right)\mathsf{\text{d}}x\right)}^{\frac{1}{p}}{\left(\underset{0}{\overset{\infty }{\int }}{x}^{q\left(1+\frac{\lambda }{s}\right)-1}{g}^q\left(x\right)\mathsf{\text{d}}x\right)}^{\frac{1}{q}},\end{array}$$

(2)

where the constant factor $\frac{rs}{\lambda }$ is the best possible. By using the way of weight functions, we can get two Hilbert-type integral inequalities with non-homogeneous kernels similar to (1) and (2) as follows [4, 5]:

$$\begin{array}{c}\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}\frac{f\left(x\right)g\left(y\right)}{\mid 1+xy{\mid }^{\lambda }}\mathsf{\text{d}}x\mathsf{\text{d}}y\\ <B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right){\left\{\underset{-\infty }{\overset{\infty }{\int }}{x}^{p\left(1-\frac{\lambda }{2}\right)-1}{f}^p\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{p}}{\left\{\underset{-\infty }{\overset{\infty }{\int }}{x}^{q\left(1-\frac{\lambda }{2}\right)-1}{g}^q\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{q}}\left(\lambda >0\right),\end{array}$$

(3)

$$\begin{array}{c}\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{\left(min\left\{1,xy\right\}\right)}^{\lambda }f\left(x\right)g\left(y\right)\mathsf{\text{d}}x\mathsf{\text{d}}y\\ <\frac{\lambda }{\alpha \left(\lambda -\alpha \right)}{\left\{\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1+\alpha \right)-1}{f}^p\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{p}}{\left\{\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1+\alpha \right)-1}{g}^q\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{q}}\left(0<\alpha <\lambda \right)\end{array}.$$

(4)

Some inequalities with the non-homogenous kernels have been studied in [6–8].

In this paper, by using the way of weight functions and the technique of real analysis, a new Hilbert-type integral inequality in the whole plane with the non-homogenous kernel and a best constant factor is built. As applications, the equivalent forms, the reverses and some particular cases are obtained.

2 Some lemmas

Lemma 1 If 0 < α₁ < α₂ < π, define the weight functions ω(y) and $\tilde{\omega }\left(x\right)$ as follow:

$$\omega \left(y\right):=\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}\frac{1}{\mid x\mid }\mathsf{\text{d}}x,\left(y\in \left(-\infty ,\infty \right)\right),$$

(5)

$$\tilde{\omega }\left(x\right):=\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}\frac{1}{\mid y\mid }\mathsf{\text{d}}y,\left(x\in \left(-\infty ,\infty \right)\right).$$

(6)

Then, we have $\omega \left(y\right)=\tilde{\omega }\left(x\right)=k\left(x,y\ne 0\right)$ where

$$k:=2ln\left[\left(1+sec\frac{{\alpha }_1}{2}\right)\left(1+csc\frac{{\alpha }_2}{2}\right)\right].$$

(7)

Proof. Setting u = x · |y| in (5), we find

$$\omega \left(y\right)=\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{\mid u\mid ,1\right\}}{\sqrt{u^2+2u\left(y∕\mid y\mid \right)cos{\alpha }_i+1}}\right\}\frac{1}{\mid u\mid }\mathsf{\text{d}}u.$$

(8)

For y ∈ (0, ∞), we have

$$\begin{array}{c}\omega (y)={\displaystyle \underset{-\infty }{\overset{-1}{\int }}\frac{1}{\sqrt{u^2+2ucos{\alpha }_2+1}}\frac{-1}{u}\text{d}u}+{\displaystyle \underset{-1}{\overset{0}{\int }}\frac{\text{d}u}{\sqrt{u^2+2ucos{\alpha }_2+1}}}\\ +{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\text{d}u}{\sqrt{u^2+2ucos{\alpha }_1+1}}}+{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{\sqrt{u^2+2ucos{\alpha }_1+1}}\frac{\text{d}u}{u}}.\end{array}$$

(9)

Setting

$$\begin{array}{c}{\omega }_1:=\underset{-\infty }{\overset{-1}{\int }}\frac{1}{\sqrt{u^2+2ucos{\alpha }_2+1}}\frac{-1}{u}\mathsf{\text{d}}u,\\ {\omega }_2:=\underset{-1}{\overset{0}{\int }}\frac{1}{\sqrt{u^2+2ucos{\alpha }_2+1}}\mathsf{\text{d}}u,\\ {\omega }_3:=\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{u^2+2ucos{\alpha }_1+1}}\mathsf{\text{d}}u,\\ {\omega }_4:=\underset{1}{\overset{\infty }{\int }}\frac{1}{\sqrt{u^2+2ucos{\alpha }_1+1}}\frac{1}{u}\mathsf{\text{d}}u,\end{array}$$

we find

$$\begin{array}{c}{\omega }_1\stackrel{v=-u}{=}\underset{1}{\overset{\infty }{\int }}\frac{\mathsf{\text{d}}v}{v\sqrt{v^2+2vcos\left(\pi -{\alpha }_2\right)+1}}\\ \stackrel{z=1∕v}{=}\underset{0}{\overset{1}{\int }}\frac{\mathsf{\text{d}}z}{\sqrt{z^2+2zcos\left(\pi -{\alpha }_2\right)+1}},\\ {\omega }_2\stackrel{v=-u}{=}\underset{0}{\overset{1}{\int }}\frac{\mathsf{\text{d}}v}{\sqrt{v^2+2vcos\left(\pi -{\alpha }_2\right)+1}}={\omega }_1,\\ {\omega }_4\stackrel{z=1∕u}{=}\underset{0}{\overset{1}{\int }}\frac{\mathsf{\text{d}}z}{\sqrt{z^2+2zcos{\alpha }_1+1}}={\omega }_3.\end{array}$$

Then, we have

$$\begin{array}{cc}\hfill \omega \left(y\right)& =2\left({\omega }_1+{\omega }_3\right)\hfill \\ =2\left\{\underset{0}{\overset{1}{\int }}\frac{\mathsf{\text{d}}u}{\sqrt{u^2+2ucos\left(\pi -{\alpha }_2\right)+1}}+\underset{0}{\overset{1}{\int }}\frac{\mathsf{\text{d}}u}{\sqrt{u^2+2ucos{\alpha }_1+1}}\right\}\hfill \\ =2\{\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{{\left[u+cos\left(\pi -{\alpha }_2\right)\right]}^2+{sin}^2\left(\pi -{\alpha }_2\right)}}\mathsf{\text{d}}u\hfill \\ +\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{{\left(u+cos{\alpha }_1\right)}^2+{sin}^2{\alpha }_1}}\mathsf{\text{d}}u\}\hfill \\ =2\left\{ln\left(1+csc\frac{{\alpha }_2}{2}\right)+ln\left(1+sec\frac{{\alpha }_1}{2}\right)\right\}\hfill \\ =2ln\left[\left(1+sec\frac{{\alpha }_1}{2}\right)\left(1+csc\frac{{\alpha }_2}{2}\right)\right]=k.\hfill \end{array}$$

For y ∈ (-∞, 0), we can obtain

$$\begin{array}{c}\omega (y)={\displaystyle \underset{-\infty }{\overset{-1}{\int }}\frac{1}{\sqrt{u^2-2ucos{\alpha }_1+1}}\frac{-\text{d}u}{u}}+{\displaystyle \underset{-1}{\overset{0}{\int }}\frac{\text{d}u}{\sqrt{u^2-2ucos{\alpha }_1+1}}}\\ +{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\text{d}u}{\sqrt{u^2-2ucos{\alpha }_2+1}}}+{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{\sqrt{u^2-2ucos{\alpha }_2+1}}\frac{\text{d}u}{u}}\\ \text{= 2(}\omega \text{1 + }\omega \text{3) = }k\text{.}\end{array}$$

By the same way, we still can find that $\tilde{\omega }\left(x\right)=\omega \left(y\right)=k\left(x,y\ne 0\right)$. The lemma is proved.   □

Lemma 2 If p > 1, $\frac{1}{p}+\frac{1}{q}=1$, 0 < α₁ < α₂ < π, f (x) is a nonnegative measurable function in (-∞,∞), then we have

$$\begin{array}{l}J:={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}|y{|}^{-1}}{\left[{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \{1,2\}}{\min }}\left\{\frac{min\{1,|xy\left|\right\}}{\sqrt{1+2xycos{\alpha }_i+{(xy)}^2}}\right\}f(x)\text{d}x\right]}^p\text{d}y\\ \le k^p{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}|x{|}^{p-1}{f}^p(x)dx}.\end{array}$$

(10)

Proof. By Lemma 1 and Hölder's inequality [9], we have

$$\begin{array}{c}{\left(\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\left|xy\right|\right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}f\left(x\right)\mathsf{\text{d}}x\right)}^p\hfill \\ ={\left(\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\left|xy\right|\right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}\left[\frac{|x{|}^{1∕q}}{|y{|}^{1∕p}}f\left(x\right)\right]\left[\frac{|y{|}^{1∕p}}{|x{|}^{1∕q}}\right]\mathsf{\text{d}}x\right)}^p\hfill \\ \le \left[\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\left|xy\right|\right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}\frac{|x{|}^{p-1}}{\left|y\right|}{f}^p\left(x\right)\mathsf{\text{d}}x\right]\hfill \\ \times {\left[\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\left|xy\right|\right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}\frac{|y{|}^{q-1}}{\left|x\right|}\mathsf{\text{d}}x\right]}^{p-1}\hfill \\ =\omega {\left(y\right)}^{p-1}\left|y\right|\left[\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\left|xy\right|\right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}\frac{|x{|}^{p-1}}{\left|y\right|}{f}^p\left(x\right)\mathsf{\text{d}}x\right]\hfill \end{array}$$

(11)

Then, by (6), (11) and Fubini theorem [10], it follows

$$\begin{array}{cc}\hfill J& \le k^{p-1}\underset{-\infty }{\overset{\infty }{\int }}\left[\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}\frac{\mid x{\mid }^{p-1}}{\mid y\mid }{f}^p\left(x\right)\mathsf{\text{d}}x\right]\mathsf{\text{d}}y\hfill \\ =k^{p-1}\underset{-\infty }{\overset{\infty }{\int }}\left[\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}\frac{1}{\mid y\mid }\mathsf{\text{d}}y\right]\mid x{\mid }^{p-1}{f}^p\left(x\right)\mathsf{\text{d}}x\hfill \\ =k^{p-1}\underset{-\infty }{\overset{\infty }{\int }}\tilde{\omega }\left(x\right)\mid x{\mid }^{p-1}{f}^p\left(x\right)\mathsf{\text{d}}x\hfill \\ =k^p\underset{-\infty }{\overset{\infty }{\int }}\mid x{\mid }^{p-1}{f}^p\left(x\right)dx\hfill \end{array}$$

The lemma is proved.   □

3 Main results and applications

Theorem 3 If p > 1, $\frac{1}{p}+\frac{1}{q}=1$, 0 < α₁ < α₂ < π, f (x), g(x) ≥ 0, satisfying $0<{\int }_{-\infty }^{\infty }\mid x{\mid }^{p-1}{f}^p\left(x\right)\mathsf{\text{d}}x<\infty$ and $0<{\int }_{-\infty }^{\infty }\mid y{\mid }^{q-1}{g}^q\left(y\right)\mathsf{\text{d}}y<\infty$, then we have

$$\begin{array}{c}I:=\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}f\left(x\right)g\left(y\right)\mathsf{\text{d}}x\mathsf{\text{d}}y\\ <k{\left(\underset{-\infty }{\overset{\infty }{\int }}\mid x{\mid }^{p-1}{f}^p\left(x\right)\mathsf{\text{d}}x\right)}^{\frac{1}{p}}{\left(\underset{-\infty }{\overset{\infty }{\int }}\mid y{\mid }^{q-1}{g}^q\left(y\right)\mathsf{\text{d}}y\right)}^{\frac{1}{q}},\end{array}$$

(12)

$$\begin{array}{cc}\hfill J& =\underset{-\infty }{\overset{\infty }{\int }}\mid y{\mid }^{-1}{\left[\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}f\left(x\right)\mathsf{\text{d}}x\right]}^p\mathsf{\text{d}}y\hfill \\ <k^p\underset{-\infty }{\overset{\infty }{\int }}\mid x{\mid }^{p-1}{f}^p\left(x\right)dx,\hfill \end{array}$$

(13)

where the constant factors k and k^p are the best possible (k is defined by (7)). Inequality (12) and (13) are equivalent.

Proof. If (11) takes the form of equality for a y ∈ (-∞, 0) ∪ (0, ∞), then there exists constants M and N, such that they are not all zero, and

$$M\frac{\mid x{\mid }^{p∕q}}{\mid y\mid }{f}^p\left(x\right)=N\frac{\mid y{\mid }^{q∕p}}{\mid x\mid }a.e.\mathsf{\text{in }}\left(-\infty ,\infty \right).$$

Hence, there exists a constant C, such that

$$M\mid x{\mid }^p{f}^p\left(x\right)=N\mid y{\mid }^q=Ca.e.in\left(-\infty ,\infty \right).$$

We suppose M ≠ 0 (otherwise N = M = 0). Then, it follows

$$\mid x{\mid }^{p-1}{f}^p\left(x\right)=\frac{C}{M\mid x\mid }a.e.in\left(-\infty ,\infty \right),$$

which contradicts the fact that $0<{\int }_{-\infty }^{\infty }\mid x{\mid }^{p-1}{f}^p\left(x\right)\mathsf{\text{d}}x<\infty$. Hence, (11) takes the form of strict sign-inequality; so does (10), and we have (13).

By Hö lder's inequality [9], we have

$$\begin{array}{cc}\hfill I& =\underset{-\infty }{\overset{\infty }{\int }}\left[\mid y{\mid }^{\frac{-1}{p}}\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}f\left(x\right)\mathsf{\text{d}}x\right]\left[\mid y{\mid }^{\frac{1}{p}}g\left(y\right)\mathsf{\text{d}}y\right]\hfill \\ \le J^{\frac{1}{p}}{\left(\underset{-\infty }{\overset{\infty }{\int }}\mid y{\mid }^{q-1}{g}^q\left(y\right)\mathsf{\text{d}}y\right)}^{\frac{1}{q}}.\hfill \end{array}$$

(14)

By (13), we have (12). On the other hand, suppose that (12) is valid. Setting

$$g\left(y\right)=\mid y{\mid }^{-1}{\left[\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}f\left(x\right)\mathsf{\text{d}}x\right]}^{p-1},$$

then $J=\underset{-\infty }{\overset{\infty }{\int }}\mid y{\mid }^{q-1}{g}^q\left(y\right)dy$. By (10), it follows J < ∞. If J = 0, then (13) is naturally valid. Assuming that 0 < J < ∞, by (12), we obtain

$$\underset{-\infty }{\overset{\infty }{\int }}\mid y{\mid }^{q-1}{g}^q\left(y\right)\mathsf{\text{d}}y=J=I<k{\left(\underset{-\infty }{\overset{\infty }{\int }}\mid x{\mid }^{p-1}{f}^p\left(x\right)\mathsf{\text{d}}x\right)}^{\frac{1}{p}}{\left(\underset{-\infty }{\overset{\infty }{\int }}\mid y{\mid }^{q-1}{g}^q\left(y\right)\mathsf{\text{d}}y\right)}^{\frac{1}{q}},$$

(15)

$$J^{\frac{1}{p}}={\left(\underset{-\infty }{\overset{\infty }{\int }}\mid y{\mid }^{q-1}{g}^q\left(y\right)\mathsf{\text{d}}y\right)}^{\frac{1}{p}}<k{\left(\underset{-\infty }{\overset{\infty }{\int }}\mid x{\mid }^{p-1}{f}^p\left(x\right)\mathsf{\text{d}}x\right)}^{\frac{1}{p}}.$$

(16)

Hence, we have (13), which is equivalent to (12).

If the constant factor k in (12) is not the best possible, then there exists a positive constant K with K < k, such that (12) is still valid as we replace k by K, then we have

$$I<K{\left(\underset{\infty }{\overset{\infty }{\int }}\mid x{\mid }^{p-1}{f}^p\left(x\right)\mathsf{\text{d}}x\right)}^{\frac{1}{p}}{\left(\underset{\infty }{\overset{\infty }{\int }}\mid y{\mid }^{q-1}{g}^q\left(y\right)\mathsf{\text{d}}y\right)}^{\frac{1}{q}}.$$

(17)

For ε > 0, define functions $\tilde{f}\left(x\right)$,$\tilde{g}\left(y\right)$ as follows:

$$\begin{array}{cc}\hfill \tilde{f}\left(x\right):& =\left\{\begin{array}{cc}\hfill x^{-\frac{2\epsilon }{p}-1},\hfill & \hfill x\in \left(1,\infty \right),\hfill \\ \hfill 0,\hfill & \hfill x\in \left[-1,1\right],\hfill \\ \hfill {\left(-x\right)}^{-\frac{2\epsilon }{p}-1},\hfill & \hfill x\in \left(-\infty ,-1\right),\hfill \end{array}\right.\hfill \\ \hfill \tilde{g}\left(y\right):& =\left\{\begin{array}{cc}\hfill \begin{array}{c}{y}^{\frac{2\epsilon }{q}-1},\\ 0,\end{array}\hfill & \hfill \begin{array}{c}y\in \left(0,1\right),\\ y\in \left(-\infty ,-1\right]\cup \left[1,\infty \right),\end{array}\hfill \\ \hfill {\left(-y\right)}^{\frac{2\epsilon }{q}-1},\hfill & \hfill y\in \left(-1,0\right).\hfill \end{array}\right.\hfill \end{array}$$

Replacing f(x), g(y) by $\tilde{f}\left(x\right)$, $\tilde{g}\left(y\right)$ in (17), we obtain

$$\begin{array}{cc}\hfill Ĩ:& =\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}\tilde{f}\left(x\right)\tilde{g}\left(y\right)\mathsf{\text{d}}x\mathsf{\text{d}}y\hfill \\ <K{\left(\underset{\infty }{\overset{\infty }{\int }}\mid x{\mid }^{p-1}{\tilde{f}}^p\left(x\right)\mathsf{\text{d}}x\right)}^{\frac{1}{p}}{\left(\underset{\infty }{\overset{\infty }{\int }}\mid y{\mid }^{q-1}{\tilde{g}}^q\left(y\right)\mathsf{\text{d}}y\right)}^{\frac{1}{q}}\hfill \\ =\frac{K}{\epsilon },\hfill \end{array}$$

(18)

$$Ĩ=\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}\tilde{f}\left(x\right)\tilde{g}\left(y\right)\mathsf{\text{d}}x\mathsf{\text{d}}y=\sum _{i=1}^4{I}_i,$$

(19)

where,

$$\begin{array}{cc}\hfill I_1:& =\underset{-1}{\overset{0}{\int }}{\left(-y\right)}^{\frac{2\epsilon }{q}-1}\left[\underset{-\infty }{\overset{-1}{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}{\left(-x\right)}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}x\right]\mathsf{\text{d}}y,\hfill \\ \hfill I_2:& =\underset{-1}{\overset{0}{\int }}{\left(-y\right)}^{\frac{2\epsilon }{q}-1}\left[\underset{1}{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}{x}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}x\right]\mathsf{\text{d}}y,\hfill \\ \hfill I_3:& =\underset{0}{\overset{1}{\int }}{y}^{\frac{2\epsilon }{q}-1}\left[\underset{-\infty }{\overset{-1}{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}{\left(-x\right)}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}x\right]\mathsf{\text{d}}y,\hfill \\ \hfill I_4:& =\underset{0}{\overset{1}{\int }}{y}^{\frac{2\epsilon }{q}-1}\left[\underset{1}{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_i+{\left(xy\right)}^2}}\right\}{x}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}x\right]\mathsf{\text{d}}y.\hfill \end{array}$$

By Fubini theorem [10], we obtain

$$\begin{array}{cc}\hfill I_1& =I_4=\underset{0}{\overset{1}{\int }}{y}^{\frac{2\epsilon }{q}-1}\left[\underset{1}{\overset{\infty }{\int }}\frac{min\left\{1,xy\right\}}{\sqrt{1+2xycos{\alpha }_1+{\left(xy\right)}^2}}{x}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}x\right]\mathsf{\text{d}}y\hfill \\ \stackrel{u=xy}{=}\underset{0}{\overset{1}{\int }}{y}^{2\epsilon -1}\left[\underset{y}{\overset{\infty }{\int }}\frac{min\left\{u,1\right\}}{\sqrt{u^2+2ucos{\alpha }_1+1}}{u}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}u\right]\mathsf{\text{d}}y\hfill \\ =\underset{0}{\overset{1}{\int }}{y}^{2\epsilon -1}\left[\underset{y}{\overset{1}{\int }}\frac{u^{-\frac{2\epsilon }{p}}}{\sqrt{u^2+2ucos{\alpha }_1+1}}\mathsf{\text{d}}u\right]\mathsf{\text{d}}y\hfill \\ +\underset{0}{\overset{1}{\int }}{y}^{2\epsilon -1}\left[\underset{1}{\overset{\infty }{\int }}\frac{u^{-\frac{2\epsilon }{p}-1}}{\sqrt{u^2+2ucos{\alpha }_1+1}}\mathsf{\text{d}}u\right]\mathsf{\text{d}}y\hfill \\ =\underset{0}{\overset{1}{\int }}\left(\underset{0}{\overset{u}{\int }}{y}^{2\epsilon -1}\mathsf{\text{d}}y\right)\frac{u^{-\frac{2\epsilon }{p}}\mathsf{\text{d}}u}{\sqrt{u^2+2ucos{\alpha }_1+1}}+\frac{1}{2\epsilon }\underset{1}{\overset{\infty }{\int }}\frac{u^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}u}{\sqrt{u^2+2ucos{\alpha }_1+1}}\hfill \\ =\frac{1}{2\epsilon }\left[\underset{0}{\overset{1}{\int }}\frac{u^{\frac{2\epsilon }{q}}}{\sqrt{u^2+2ucos{\alpha }_1+1}}\mathsf{\text{d}}u+\underset{1}{\overset{\infty }{\int }}\frac{u^{-\frac{2\epsilon }{p}-1}}{\sqrt{u^2+2ucos{\alpha }_1+1}}\mathsf{\text{d}}u\right],\hfill \\ \hfill I_2& =I_3=\underset{0}{\overset{1}{\int }}{y}^{\frac{2\epsilon }{q}-1}\left[\underset{1}{\overset{\infty }{\int }}\frac{min\left\{1,xy\right\}}{\sqrt{1-2xycos{\alpha }_2+{\left(xy\right)}^2}}{x}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}x\right]\mathsf{\text{d}}y\hfill \\ =\frac{1}{2\epsilon }\left[\underset{0}{\overset{1}{\int }}\frac{u^{\frac{2\epsilon }{q}}}{\sqrt{u^2-2ucos{\alpha }_2+1}}\mathsf{\text{d}}u+\underset{1}{\overset{\infty }{\int }}\frac{u^{-\frac{2\epsilon }{p}-1}}{\sqrt{u^2-2ucos{\alpha }_2+1}}\mathsf{\text{d}}u\right].\hfill \end{array}$$

In view of the above results, by using (18) and (19), it follows

$$\begin{array}{c}\underset{0}{\overset{1}{\int }}\frac{u^{\frac{2\epsilon }{q}}}{\sqrt{u^2+2ucos{\alpha }_1+1}}\mathsf{\text{d}}u+\underset{1}{\overset{\infty }{\int }}\frac{u^{-\frac{2\epsilon }{p}-1}}{\sqrt{u^2+2ucos{\alpha }_1+1}}\mathsf{\text{d}}u\\ +\underset{0}{\overset{1}{\int }}\frac{u^{\frac{2\epsilon }{q}}}{\sqrt{u^2-2ucos{\alpha }_2+1}}\mathsf{\text{d}}u+\underset{1}{\overset{\infty }{\int }}\frac{u^{-\frac{2\epsilon }{p}-1}}{\sqrt{u^2-2ucos{\alpha }_2+1}}\mathsf{\text{d}}u\\ =\epsilon Ĩ<\epsilon \cdot \frac{K}{\epsilon }=K.\end{array}$$

(20)

By Fatou lemma [10] and (20), we find

$$\begin{array}{c}k=\omega \left(y\right)=\underset{0}{\overset{\infty }{\int }}\frac{\min \left\{u,1\right\}\mathsf{\text{d}}u}{u\sqrt{u^2+2ucos{\alpha }_1+1}}+\underset{0}{\overset{\infty }{\int }}\frac{min\left\{u,1\right\}\mathsf{\text{d}}u}{u\sqrt{u^2-2ucos{\alpha }_2+1}}\\ =\underset{0}{\overset{1}{\int }}\underset{\epsilon \to 0^{+}}{lim}\frac{u^{\frac{2\epsilon }{q}}\mathsf{\text{d}}u}{\sqrt{u^2+2ucos{\alpha }_1+1}}+\underset{1}{\overset{\infty }{\int }}\underset{\epsilon \to 0^{+}}{lim}\frac{u^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}u}{\sqrt{u^2+2ucos{\alpha }_1+1}}\\ +\underset{0}{\overset{1}{\int }}\underset{\epsilon \to 0^{+}}{lim}\frac{u^{\frac{2\epsilon }{q}}\mathsf{\text{d}}u}{\sqrt{u^2-2ucos{\alpha }_2+1}}+\underset{1}{\overset{\infty }{\int }}\underset{\epsilon \to 0^{+}}{lim}\frac{u^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}u}{\sqrt{u^2-2ucos{\alpha }_2+1}}\\ \le {\underset{¯}{\lim }}_{\epsilon \to 0^{+}}[\underset{0}{\overset{1}{\int }}\frac{u^{\frac{2\epsilon }{q}}\mathsf{\text{d}}u}{\sqrt{u^2+2ucos{\alpha }_1+1}}+\underset{1}{\overset{\infty }{\int }}\frac{u^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}u}{\sqrt{u^2+2ucos{\alpha }_1+1}}\\ +\underset{0}{\overset{1}{\int }}\frac{u^{\frac{2\epsilon }{q}}\mathsf{\text{d}}u}{\sqrt{u^2-2ucos{\alpha }_2+1}}+\underset{1}{\overset{\infty }{\int }}\frac{u^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}u}{\sqrt{u^2-2ucos{\alpha }_2+1}}]\le K,\end{array}$$

(21)

which contradicts the fact that K < k. Hence, the constant factor k in (12) is the best possible.

If the constant factor in (13) is not the best possible, then by (14), we may get a contradiction that the constant factor in (12) is not the best possible. Thus, the theorem is proved.   □

Theorem 4 As the assumptions of Theorem 3, replacing p > 1 by 0 < p < 1, we have the equivalent reverse of (12) and (13) with the best constant factors.

Proof. The way of proving of Theorem 4 is similar to Theorem 3. By the reverse Hö lder's inequality [9], we have the reverse of (10) and (14). It is easy to obtain the reverse of (13). In view of the reverses of (13) and (14), we obtain the reverse of (12). On the other hand, suppose that the reverse of (12) is valid. Setting the same g(y) as theorem 3, by the reverse of (10), we have J > 0. If J = ∞, then the reverse of (13) is obvious value; if J < ∞, then by the reverse of (12), we obtain the reverses of (15) and (16). Hence, we have the reverse of (13), which is equivalent to the reverse of (12).

If the constant factor k in the reverse of (12) is not the best possible, then there exists a positive constant $\tilde{K}\left(\mathsf{\text{with }}\tilde{K}>k\right)$, such that the reverse of (12) is still valid as we replace k by $\tilde{K}$. By the reverse of (20), we have

$$\begin{array}{c}\underset{0}{\overset{1}{\int }}\left[\frac{1}{\sqrt{u^2+2ucos{\alpha }_1+1}}+\frac{1}{\sqrt{u^2-2ucos{\alpha }_2+1}}\right]u^{\frac{2\epsilon }{q}}\mathsf{\text{d}}u\\ +\underset{1}{\overset{\infty }{\int }}\left[\frac{1}{\sqrt{u^2+2ucos{\alpha }_1+1}}+\frac{1}{\sqrt{u^2-2ucos{\alpha }_2+1}}\right]u^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}u\\ >\tilde{K}.\end{array}$$

(22)

For $0<{\epsilon }_0<\frac{\mid q\mid }{2}$, we have $\frac{2{\epsilon }_0}{q}>-1$. For 0 < ε ≤ ε₀, we obtain $u^{\frac{2\epsilon }{q}}\le u^{\frac{2{\epsilon }_0}{q}}\left(u\in \left(0,1\right]\right)$ and

$$\begin{array}{c}\underset{0}{\overset{1}{\int }}\left[\frac{1}{\sqrt{u^2+2ucos{\alpha }_1+1}}+\frac{1}{\sqrt{u^2-2ucos{\alpha }_2+1}}\right]u\frac{2{\epsilon }_0}{q}\mathsf{\text{d}}u\\ \le \left(\frac{1}{sin{\alpha }_1}+\frac{1}{sin{\alpha }_2}\right)\underset{0}{\overset{1}{\int }}u\frac{2{\epsilon }_0}{q}\mathsf{\text{d}}u=\left(\frac{1}{sin{\alpha }_1}+\frac{1}{sin{\alpha }_2}\right)\frac{1}{1+\left(2{\epsilon }_0\right)∕q}<\infty .\end{array}$$

Then, by Lebesgue control convergence theorem [10], we have for ε → 0⁺ that

$$\begin{array}{c}\underset{0}{\overset{1}{\int }}\left[\frac{1}{\sqrt{u^2+2ucos{\alpha }_1+1}}+\frac{1}{\sqrt{u^2-2ucos{\alpha }_2+1}}\right]u^{\frac{2\epsilon }{q}}\mathsf{\text{d}}u\\ =\underset{0}{\overset{1}{\int }}\left[\frac{1}{\sqrt{u^2+2ucos{\alpha }_1+1}}+\frac{1}{\sqrt{u^2-2ucos{\alpha }_2+1}}\right]\mathsf{\text{d}}u+o\left(1\right).\end{array}$$

(23)

By Levi's theorem [10], we find for ε → 0⁺ that

$$\begin{array}{c}\underset{1}{\overset{\infty }{\int }}\left[\frac{1}{\sqrt{u^2+2ucos{\alpha }_1+1}}+\frac{1}{\sqrt{u^2-2ucos{\alpha }_2+1}}\right]u^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}u\\ =\underset{1}{\overset{\infty }{\int }}\left[\frac{1}{\sqrt{u^2+2ucos{\alpha }_1+1}}+\frac{1}{\sqrt{u^2-2ucos{\alpha }_2+1}}\right]u^{-1}\mathsf{\text{d}}u+\tilde{o}\left(1\right).\end{array}$$

(24)

By (22), (23) and (24), for ε → 0⁺ in (22), we have $k\ge \tilde{K}$, which contradicts the fact that $k<\tilde{K}$. Hence, the constant factor k in the reverse of (12) is the best possible.

If the constant factor in reverse of (13) is not the best possible, then by the reverse of (14), we may get a contradiction that the constant factor in the reverse of (12) is not the best possible. Thus, the theorem is proved.   □

Remark 1 For α₁ = α₂ = α ∈ (0, π) in (12) and (13), we have the following equivalent inequalities:

$$\begin{array}{c}\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos\alpha +{\left(xy\right)}^2}}f\left(x\right)g\left(y\right)\mathsf{\text{d}}x\mathsf{\text{d}}y\\ <k_0{\left(\underset{-\infty }{\overset{\infty }{\int }}\mid x{\mid }^{p-1}{f}^p\left(x\right)\mathsf{\text{d}}x\right)}^{\frac{1}{p}}{\left(\underset{-\infty }{\overset{\infty }{\int }}\mid x{\mid }^{q-1}{g}^q\left(y\right)\mathsf{\text{d}}y\right)}^{\frac{1}{q}},\end{array}$$

(25)

$$\begin{array}{c}\underset{-\infty }{\overset{\infty }{\int }}\mid y{\mid }^{-1}{\left[\underset{-\infty }{\overset{\infty }{\int }}\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos\alpha +{\left(xy\right)}^2}}f\left(x\right)\mathsf{\text{d}}x\right]}^p\mathsf{\text{d}}y\\ <k_0^p\underset{-\infty }{\overset{\infty }{\int }}\mid x{\mid }^{p-1}{f}^p\left(x\right)dx,\end{array}$$

(26)

where the constant factors $k_0:=2ln\left[\left(1+sec\frac{\alpha }{2}\right)\left(1+csc\frac{\alpha }{2}\right)\right]$ and $k_0^p$ are the best possible.

Remark 2 For ${\alpha }_1={\alpha }_2=\frac{\pi }{3}$, p = q = 2 in (12) and (13), we have the following equivalent inequalities:

$$\begin{array}{c}\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+xy+{\left(xy\right)}^2}}f\left(x\right)g\left(y\right)\mathsf{\text{d}}x\mathsf{\text{d}}y\\ <2ln\left(3+2\sqrt{3}\right){\left(\underset{-\infty }{\overset{\infty }{\int }}\mid x\mid f^2\left(x\right)\mathsf{\text{d}}x\underset{-\infty }{\overset{\infty }{\int }}\mid y\mid g^2\left(y\right)\mathsf{\text{d}}y\right)}^{\frac{1}{2}},\end{array}$$

(27)

$$\begin{array}{c}\underset{-\infty }{\overset{\infty }{\int }}\mid y{\mid }^{-1}{\left[\underset{-\infty }{\overset{\infty }{\int }}\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+xy+{\left(xy\right)}^2}}f\left(x\right)\mathsf{\text{d}}x\right]}^2\mathsf{\text{d}}y\\ <2^2{\left[ln\left(3+2\sqrt{3}\right)\right]}^2\underset{-\infty }{\overset{\infty }{\int }}\mid x\mid f^2\left(x\right)dx.\end{array}$$

(28)