1 Introduction
Suppose that \(a_{m},b_{n}\geq0\), \(a=\{a_{m}\}_{m=1}^{\infty }\), \(b=\{b_{n}\}_{n=1}^{\infty}\in l^{2}\) (\(l^{2}:=\{x=\{x_{k}\} _{k=1}^{\infty }:\sum_{k=1}^{\infty}x_{k}^{2}<\infty\}\)), \(\|a\|=\{\sum_{m=1}^{\infty }a_{m}^{2}\}^{\frac{1}{2}} >0\), \(\|b\|>0\), we have the following well-known discrete Hilbert inequality [1]:
where the constant factor π is the best possible. Moreover, for \(f(x),g(y)\geq0\), \(f,g\in L^{2}(\mathbf{R} _{+})\), \(\|f\|=(\int_{0}^{\infty}f^{2}(x)\,dx)^{\frac{1}{2}}>0\), \(\|g\|>0\), we still have the following Hilbert integral inequality:
with the same best constant factor π. Inequalities (1) and (2) are important in analysis and applications (cf. [2‐12]).
$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{a_{m}b_{n}}{m+n}< \pi \|a\|\|b\|, $$
(1)
$$ \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{x+y}\,dx\,dy< \pi\|f\|\|g\|, $$
(2)
All of the above integral inequalities are built in the quarter plane of the first quadrant. For some special kernel functions, it is meaningful to establish inequalities in the whole plane. In 2008, Yang [13] gave an inequality as follows:
where the constant factor \(k_{\lambda}:=B(\frac{\lambda}{2},\frac {\lambda }{2})+2B(1-\lambda,\frac{\lambda}{2})\) is the best possible and \(0<\lambda <1\), \(p>1\), \((1/p)+(1/q)=1\). In 2010, Zeng and Xie [14] got the following inequality: If \(p>1\), \((1/p)+(1/q)=1\), \(r\in (-1,0)\), \(0<\alpha <\beta <\pi\), \(f(x)\) and \(g(y)\) are nonnegative measurable functions, satisfying \(0<\int_{-\infty}^{\infty}|x|^{p(1-\frac{\lambda }{2})-1}f^{p}(x)\,dx<\infty\), \(0<\int_{-\infty}^{\infty}|y|^{q(1-\frac{\lambda }{2})-1}g^{q}(y)\,dy<\infty \), then
where the constant factor
is the best possible. Subsequently, many scholars began to study the inequalities of this type (cf. [15‐19]).
$$\begin{aligned} &\int_{-\infty}^{\infty}\int _{-\infty}^{\infty} \frac{f(x)g(y)}{|1+xy|^{\lambda}}\,dx\,dy \\ &\quad< k_{\lambda} \biggl\{ \int_{-\infty}^{\infty}|x|^{p(1-\frac{\lambda }{2})-1}f^{p}(x) \,dx \biggr\} ^{1/p} \biggl\{ \int_{-\infty}^{\infty }|y|^{q(1-\frac{\lambda}{2})-1}g^{q}(y) \,dy \biggr\} ^{1/q}, \end{aligned}$$
(3)
$$\begin{aligned} &\int_{-\infty}^{\infty}\int _{-\infty}^{\infty}f(x)g(y) \biggl| \ln\frac{x^{2}+2xy\cos\alpha+y^{2}}{x^{2}+2xy\cos\beta+y^{2}} \biggr|\,dx\,dy \\ &\quad< k \biggl\{ \int_{-\infty}^{\infty}|x|^{p(1+r)-1}f^{p}(x) \,dx \biggr\} ^{1/p} \biggl\{ \int_{-\infty}^{\infty}|y|^{q(1-r)-1}g^{q}(y) \,dy \biggr\} ^{1/q}, \end{aligned}$$
(4)
$$\begin{aligned} k=\frac{4\pi\sin[r(\beta-\alpha)/2]\cos [ (r/2)(\pi-\alpha -\beta) ] }{r\cos(r\pi/2)} \end{aligned}$$
Anzeige
In 1934, Hardy [1] published the following Hardy integral inequality: If \(p>1\), \(f(x)\geq0\), \(f\in L^{p}(\mathbf{R} _{+})\), \(\|f\|_{p}=\{\int_{0}^{\infty }f^{p}(x)\,dx\}^{1/p}>0\), \(F(x)=\int_{0}^{x}f(t)\,dt\), then
where the constant factor \([ p/ ( p-1 ) ] ^{p}\) is the best possible. From Theorem 328, [1], it follows that
where the constant factor \(p^{p}\) is still the best possible and \(F(x)=\int_{x}^{\infty}f(t)\,dt\). In 2009, Yang [20] gave the following best extensions of (6) by introducing an independent parameter λ.
$$ \int_{0}^{\infty} \biggl( \frac{F(x)}{x} \biggr) ^{p}\,dx< \biggl( \frac {p}{p-1} \biggr) ^{p}\|f \|_{p}^{p}, $$
(5)
$$ \int_{0}^{\infty}F^{p}(x) \,dx< p^{p} \int_{0}^{\infty} \bigl( xf(x) \bigr) ^{p}\,dx, $$
(6)
If \(0< p<1\), \(r>0\), \(f\geq0\), \(0<\int_{0}^{\infty}x^{p(1+\frac{1-\lambda}{r})-1}f^{p}(x)\,dx <\infty\) (\(\lambda\neq1\)), \(F(x)=\int_{0}^{x}f(t)\,dt\) (\(\lambda >1\)), \(F(x)=\int_{x}^{\infty}f(t)\,dt\) (\(\lambda<1\)), then
where the constant factor \(( \frac{r}{|1-\lambda|} ) ^{p}\) is the best possible.
$$ \int_{0}^{\infty}x^{\frac{p}{r}(1-\lambda)-1}F^{p}(x) \,dx> \biggl( \frac {r}{|1-\lambda|} \biggr) ^{p}\int_{0}^{\infty}x^{\frac{p}{r}(1-\lambda )-1} \bigl( xf(x) \bigr) ^{p}\,dx, $$
(7)
The main objective of this paper is to establish inequalities in the whole plane, as well as the corresponding operator expression, the equivalent form, and the reverse inequality. Special emphasis is placed on a Hardy-type inequality with the best possible constant.
2 Some lemmas
In what follows, \(\alpha_{1}\), \(\alpha_{2}\) will be real numbers such that \(0<\alpha_{1}\leq\alpha_{2}<\pi\), \(\delta\in\{-1,1\}\), \(\lambda _{1}, \lambda_{2}<0\), \(\lambda_{1}+\lambda_{2}=-\lambda\), and
$$\begin{aligned}& h \bigl(x^{\delta}y \bigr):=\min_{i\in\{1,2\}}\min \bigl\{ 1, \bigl[ \bigl|x^{\delta }y\bigr|^{\gamma }+ \bigl( x^{\delta}y \bigr) ^{\gamma}\cos\alpha_{i} \bigr] ^{\lambda /\gamma} \bigr\} , \\& k_{\lambda}(x,y):=\min_{i\in\{1,2\}}\min \bigl\{ |x|, \bigl[ |y|^{\gamma }+y^{\gamma} \operatorname{sgn} x \cos\alpha_{i} \bigr] ^{\lambda/\gamma } \bigr\} . \end{aligned}$$
Anzeige
Lemma 1
If
\(C(\lambda):=\frac{\lambda}{\lambda_{1}\lambda _{2}}\cdot2^{\lambda_{2}/\gamma} [ ( \cos\frac{\alpha_{1}}{2} ) ^{2\lambda_{2}/\gamma}+ ( \sin\frac{\alpha_{2}}{2} ) ^{2\lambda_{2}/\gamma} ] \), the weight function
then for all
\(x,y\neq0\), we have
$$\begin{aligned}& \omega_{\delta}(\lambda_{2},y):=\int_{-\infty}^{\infty}h \bigl( x^{\delta}y \bigr) \frac{|y|^{\lambda_{2}}}{ |x|^{1-\delta\lambda_{2}}}\,dx, \end{aligned}$$
(8)
$$\begin{aligned}& \varpi_{\delta}(\lambda_{2},x):=\int_{-\infty}^{\infty}h \bigl( x^{\delta}y \bigr) \frac{|x|^{\delta\lambda_{2}}}{|y|^{1-\lambda_{2}}}\,dy, \end{aligned}$$
(9)
$$ \omega_{\delta}(\lambda_{2},y)=\varpi_{\delta}(\lambda _{2},x)=C(\lambda). $$
(10)
Proof
(i) For \(\delta=1\), setting \(u=xy\), we find for \(y\neq0\),
$$\begin{aligned} \omega_{1}(\lambda_{2},y) =&\int_{-\infty}^{\infty}h ( u ) |u|^{\lambda_{2}-1}\,du \\ =&\int_{0}^{\infty}\min_{i\in\{1,2\}}\min \bigl\{ 1,u^{\lambda}(1+\cos \alpha _{i})^{\lambda/\gamma} \bigr\} u^{\lambda_{2}-1}\,du \\ &{}+\int_{0}^{\infty}\min_{i\in\{1,2\}} \min \bigl\{ 1,u^{\lambda}(1-\cos \alpha _{i})^{\lambda/\gamma} \bigr\} u^{\lambda_{2}-1}\,du \\ =&\frac{1}{\lambda} \biggl[ \int_{0}^{\infty}\min _{i\in\{1,2\}}(1+\cos \alpha_{i})^{\lambda_{2}/\gamma}v^{\lambda_{2}/\lambda-1} \min \{1,v\}\,dv \\ &{} +\int_{0}^{\infty}\min_{i\in\{1,2\}}(1- \cos\alpha _{i})^{\lambda _{2}/\gamma}v^{\lambda_{2}/\lambda-1}\min\{1,v\}\,dv \biggr] \\ =&\frac{1}{\lambda} \bigl[ (1+\cos\alpha_{1})^{\lambda_{2}/\gamma }+(1- \cos \alpha_{2})^{\lambda_{2}/\gamma} \bigr] \int_{0}^{\infty }v^{\lambda_{2}/\lambda-1} \min\{1,v\}\,dv \\ =&\frac{\lambda}{\lambda_{1}\lambda_{2}}\cdot2^{\lambda_{2}/\gamma} \biggl[ \biggl( \cos \frac{\alpha_{1}}{2} \biggr) ^{2\lambda_{2}/\gamma }+ \biggl( \sin\frac{\alpha_{2}}{2} \biggr) ^{2\lambda_{2}/\gamma } \biggr] =C(\lambda). \end{aligned}$$
(ii) For \(\delta=-1\), setting \(u=y/x\), we still have \(\omega _{-1}(\lambda _{2},y)=C(\lambda)\).
Setting \(u=x^{\delta}y\), we get
Hence we have (10). □
$$\begin{aligned} \varpi_{\delta}(\lambda_{2},x)=\int_{-\infty}^{\infty}h ( u ) |u|^{\lambda_{2}-1}\,du=C(\lambda). \end{aligned}$$
Remark 1
Let \(\alpha_{1}=\alpha_{2}=\alpha\), then \(h(u)=\min \{1, [ |u|^{\gamma}+ ( u ) ^{\gamma}\cos\alpha ] ^{\lambda/\gamma}\}\) and
$$\begin{aligned} C(\lambda)=\frac{\lambda}{\lambda_{1}\lambda_{2}}\cdot2^{\lambda _{2}/\gamma} \biggl[ \biggl( \cos \frac{\alpha}{2} \biggr) ^{2\lambda _{2}/\gamma}+ \biggl( \sin\frac{\alpha}{2} \biggr) ^{2\lambda _{2}/\gamma} \biggr] . \end{aligned}$$
Lemma 2
If
\(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(f(x)\)
is a nonnegative measurable function in
R, then
(i)
for
\(p>1\), we have
$$\begin{aligned} J :=&\int_{-\infty}^{\infty}|y|^{p\lambda_{2}-1} \biggl[ \int_{-\infty }^{\infty}h \bigl( x^{\delta}y \bigr) f(x) \,dx \biggr] ^{p}\,dy \\ \leq&C^{p}(\lambda)\int_{-\infty}^{\infty}|x|^{p(1-\delta\lambda _{2})-1}f^{p}(x) \,dx; \end{aligned}$$
(11)
Proof
For \(p>1\), by the Hölder inequality with weight [21], we have
Then by (10) and the Fubini theorem [22], it follows that
and so (11) is proved.
$$\begin{aligned} & \biggl[ \int_{-\infty}^{\infty}h \bigl( x^{\delta}y \bigr) f(x)\,dx \biggr] ^{p} \\ &\quad= \biggl[ \int_{-\infty}^{\infty}h \bigl( x^{\delta}y \bigr) \biggl( \frac{|x|^{(1-\delta\lambda_{2})/q}}{|y|^{(1-\lambda_{2})/p}}f(x) \biggr) \biggl( \frac{|y|^{(1-\lambda_{2})/p}}{|x|^{(1-\delta\lambda_{2})/q}} \biggr)\,dx \biggr] ^{p} \\ &\quad\leq\int_{-\infty}^{\infty}h \bigl( x^{\delta}y \bigr) \frac{|x|^{(1-\delta\lambda_{2})(p-1)}}{|y|^{1-\lambda_{2}}} f^{p}(x)\,dx \biggl[ \int_{-\infty}^{\infty}h \bigl( x^{\delta}y \bigr) \frac {|y|^{(1-\lambda _{2})(q-1)}}{|x|^{1-\delta\lambda_{2}}}\,dx \biggr] ^{p-1} \\ &\quad=\frac{[\omega_{\delta}(\lambda_{2},y)]^{p-1}}{|y|^{p\lambda_{2}-1}} \int_{-\infty}^{\infty}h \bigl( x^{\delta}y \bigr) \frac {|x|^{(1-\delta \lambda_{2})(p-1)}}{|y|^{1-\lambda_{2}}}f^{p}(x)\,dx. \end{aligned}$$
(12)
$$\begin{aligned} J \leq&C^{p-1}(\lambda)\int_{-\infty}^{\infty} \biggl[ \int_{-\infty }^{\infty}h \bigl( x^{\delta}y \bigr) \frac{|x|^{(1-\delta\lambda _{2})(p-1)}}{|y|^{1-\lambda_{2}}}f^{p}(x)\,dx \biggr]\,dy \\ =&C^{p-1}(\lambda)\int_{-\infty}^{\infty} \omega_{\delta}(\lambda _{2},x)|x|^{p(1-\delta\lambda_{2})-1}f^{p}(x) \,dx \\ =& C^{p}(\lambda)\int_{-\infty}^{\infty}|x|^{p(1-\delta\lambda _{2})-1}f^{p}(x) \,dx, \end{aligned}$$
3 Main results and applications
In this section, we set the functions and spaces as follows:
$$\begin{aligned}& \varphi(x) :=|x|^{p(1-\delta\lambda_{2})-1}\quad (x\in \mathbf{R} ),\qquad \psi(y):=|y|^{q(1-\lambda_{2})-1}\quad (y\in \mathbf{R} ), \\& L_{p,\varphi}(\mathbf{R} ) := \biggl\{ f:\|f\|_{p,\varphi}= \biggl\{ \int_{-\infty}^{\infty } \varphi (x)\bigl|f(x)\bigr|^{p}\,dx \biggr\} ^{1/p}< \infty \biggr\} , \\& L_{q,\psi}(\mathbf{R} ) := \biggl\{ g:\|g\|_{q,\psi}= \biggl\{ \int_{-\infty}^{\infty} \psi (y)\bigl|g(y)\bigr|^{q}\,dy \biggr\} ^{1/q}<\infty \biggr\} . \end{aligned}$$
Theorem 1
If
\(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(f(x),g(y) \geq0\), satisfying
\(f(x)\in L_{p,\varphi}(\mathbf{R} )\), \(g(y)\in L_{q,\psi}(\mathbf{R} )\), \(\|f\|_{p,\varphi},\|g\|_{q,\psi}>0\), then we have the following equivalent inequalities:
where the constant factors
\(C(\lambda)\)
and
\(C^{p}(\lambda)\)
in the above inequalities are the best possible.
$$\begin{aligned}& I:=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}h \bigl( x^{\delta }y \bigr) f(x)g(y)\,dx\,dy< C(\lambda)\|f\|_{p,\varphi} \|g \|_{q,\psi}, \end{aligned}$$
(13)
$$\begin{aligned}& J=\int_{-\infty}^{\infty}|y|^{p\lambda_{2}-1} \biggl[ \int _{-\infty }^{\infty}h \bigl( x^{\delta}y \bigr) f(x)\,dx \biggr] ^{p}\,dy< C^{p}(\lambda )\|f\|_{p,\varphi}^{p}, \end{aligned}$$
(14)
Proof
If (12) takes the form of equality for a \(y\neq 0\), then there exist constants A and B, such that they are not all zero, and
We confirm that \(A\neq0\) (otherwise \(B=A=0\)), from which
in view of \(\int_{-\infty}^{\infty}\frac{1}{|x|}\,dx=\infty\), which contradicts the fact that \(\|f\|_{p,\varphi}>0\). Hence (12) takes a strict inequality, so does (11), thus (14) is valid.
$$\begin{aligned} A\frac{|x|^{(1-\delta\lambda_{2})(p-1)}}{|y|^{1-\lambda _{2}}}f^{p}(x)=B\frac{|y|^{(1-\lambda_{2})(q-1)}}{|x|^{1-\delta\lambda_{2}}} \quad \mbox{a.e. in } \mathbf{R}. \end{aligned}$$
$$\begin{aligned} |x|^{p(1-\delta\lambda_{2})-1}f^{p}(x)=|y|^{q(1-\lambda_{2})}\frac {B}{A|x|}\quad \mbox{a.e. in } \mathbf{R}, \end{aligned}$$
By the Hölder inequality with weight [21], we obtain
By (14), we have (13). On the other hand, suppose that (13) is valid, setting
then we find \(J=\|g\|_{q,\psi}^{q}\). By (11), we have \(J<\infty \). If \(J=0\), then (14) is valid trivially; if \(0< J<\infty\), then by (13), it follows that
and then (14) follows, which is equivalent to (13).
$$ I=\int_{-\infty}^{\infty} \biggl[ |y|^{\lambda_{2}-\frac{1}{p}}\int _{-\infty}^{\infty}h \bigl( x^{\delta}y \bigr) f(x)\,dx \biggr] \bigl( |y|^{\frac{1}{p}-\lambda_{2}}g(y)\,dy \bigr) \leq J^{1/p}\|g \|_{q,\psi}. $$
(15)
$$\begin{aligned} g(y):=|y|^{p\lambda_{2}-1} \biggl[ \int_{-\infty}^{\infty}h \bigl( x^{\delta }y \bigr) f(x)\,dx \biggr] ^{p-1} \quad\bigl(y\in \mathbf{R} \backslash\{0\} \bigr), \end{aligned}$$
$$\begin{aligned}& 0< \|g\|_{q,\psi}^{q}=J=I<C(\lambda)\|f\|_{p,\varphi}\|g \|_{q,\psi }<\infty, \end{aligned}$$
(16)
$$\begin{aligned}& J^{1/p}=\|g\|_{q,\psi}^{q/p}< C(\lambda)\|f \|_{p,\varphi}, \end{aligned}$$
(17)
For any \(n>\frac{1}{p|\lambda_{2}|}\) (\(n\in \mathbf{N} \)), setting \(E_{\delta}:=\{x\in \mathbf{R} :|x|^{\delta}\geq1\}\) and
then
Suppose that \(I(x):=\int_{-1}^{1}h ( x^{\delta}y ) |y|^{\lambda _{2}+\frac{1}{nq}-1}\,dy\), we confirm that
In fact, for \(x>0\),
Similarly, we get
for \(x<0\).
$$\begin{aligned}& \widetilde{f}(x):= \left \{ \begin{array}{@{}l@{\quad}l} |x|^{\delta(\lambda_{2}-\frac{1}{np})-1},& x\in E_{\delta}, \\ 0,& x\in \mathbf{R} \backslash E_{\delta}, \end{array} \right . \qquad \widetilde{g}(y):= \left \{ \begin{array}{@{}l@{\quad}l} |y|^{\lambda_{2}+\frac{1}{nq}-1},& y\in [-1,1], \\ 0,& y\in(-\infty,-1)\cup(1,\infty), \end{array} \right . \end{aligned}$$
$$\begin{aligned} \widetilde{L} :=&\|\widetilde{f}\|_{p,\varphi}\|\widetilde {g} \|_{q,\psi }= \biggl\{ \int_{E_{\delta}}|x|^{-\frac{\delta}{n}-1}\,dx \biggr\} ^{1/p} \biggl\{ \int_{-1}^{1}|y|^{\frac{1}{n}-1} \,dy \biggr\} ^{1/q} \\ =&2 \biggl\{ \int_{\{x>0:x^{\delta}\geq1\}}x^{-\frac{\delta}{n}-1}\,dx \biggr\} ^{1/p} \biggl\{ \int_{0}^{1}y^{\frac{1}{n}-1} \,dy \biggr\} ^{1/q}=2n. \end{aligned}$$
$$\begin{aligned} I(x)=|x|^{-\delta\lambda_{2}-\frac{\delta}{nq}}\int_{0}^{|x|^{\delta }} \bigl(h(v)+h(-v) \bigr)v^{\lambda_{2}+\frac{1}{nq}-1}\,dv. \end{aligned}$$
$$\begin{aligned}[b] I(x) &=\int_{-1}^{0}h \bigl(x^{\delta}y \bigr) (-y)^{\lambda_{2}+\frac{1}{nq}-1}\,dy+\int_{0}^{1}h \bigl(x^{\delta}y \bigr)y^{\lambda_{2}+\frac{1}{nq}-1}\,dy \\ &=x^{-\delta\lambda_{2}-\frac{\delta}{nq}}\int_{0}^{x^{\delta }}h(-v)v^{\lambda_{2}+\frac{1}{nq}-1} \,dv+x^{-\delta\lambda_{2}-\frac{\delta}{nq}} \int_{0}^{x^{\delta}}h(v)v^{\lambda_{2}+\frac {1}{nq}-1} \,dv \\ &=x^{-\delta\lambda_{2}-\frac{\delta}{nq}}\int_{0}^{x^{\delta }} \bigl(h(v)+h(-v) \bigr)v^{\lambda_{2}+\frac{1}{nq}-1}\,dv. \end{aligned} $$
$$\begin{aligned} I(x) =&\int_{-1}^{0}h \bigl(x^{\delta}y \bigr) (-y)^{\lambda_{2}+\frac{1}{nq}-1}\,dy+\int_{0}^{1}h \bigl(x^{\delta}y \bigr)y^{\lambda_{2}+\frac{1}{nq}-1}\,dy \\ =&(-x)^{-\delta\lambda_{2}-\frac{\delta}{nq}}\int_{0}^{(-x)^{\delta }} \bigl( h(v)+h(-v) \bigr) v^{\lambda_{2}+\frac{1}{nq}-1}\,dv \end{aligned}$$
Applying the Fubini theorem [22],
$$\begin{aligned} \widetilde{I} :=&\int_{-\infty}^{\infty}\int _{-\infty}^{\infty }h \bigl( x^{\delta}y \bigr) \widetilde{f}(x)\widetilde{g}(y)\,dx\,dy \\ =&\int_{E_{\delta}}|x|^{\delta(\lambda_{2}-\frac{1}{np})-1} \biggl( \int _{-1}^{1}h \bigl(x^{\delta}y \bigr)|y|^{\lambda_{2}+\frac{1}{nq}-1}\,dy \biggr)\,dx \\ =&\int_{E_{\delta}}|x|^{-\frac{\delta}{n}-1} \biggl( \int _{0}^{|x|^{\delta }} \bigl(h(u)+h(-u) \bigr)u^{\lambda_{2}+\frac{1}{nq}-1} \,du \biggr)\,dx \\ =&\int_{E_{\delta}}|x|^{-\frac{\delta}{n}-1} \biggl( \int _{0}^{1} \bigl(h(u)+h(-u) \bigr)u^{\lambda_{2}+\frac{1}{nq}-1} \,du \biggr)\,dx \\ &{}+\int_{E_{\delta}}|x|^{-\frac{\delta}{n}-1} \biggl( \int _{1}^{|x|^{\delta }} \bigl(h(u)+h(-u) \bigr)u^{\lambda_{2}+\frac{1}{nq}-1} \,du \biggr)\,dx \\ =&2n\int_{0}^{1} \bigl(h(u)+h(-u) \bigr)u^{\lambda_{2}+\frac{1}{nq}-1}\,du \\ &{}+\int_{1}^{\infty} \biggl( \int _{\{x:|x|^{\delta}\geq u\} }^{1}|x|^{-\frac{\delta}{n}-1}\,dx \biggr) \bigl( h(u)+h(-u) \bigr) u^{\lambda_{2}+\frac {1}{nq}-1}\,du \\ =& 2n \biggl[ \int_{0}^{1} \bigl(h(u)+h(-u) \bigr)u^{\lambda_{2}+\frac{1}{nq}-1}\,du+\int_{1}^{\infty} \bigl( h(u)+h(-u) \bigr) u^{\lambda_{2}-\frac {1}{np}-1}\,du \biggr] . \end{aligned}$$
If there exists a constant \(K\leq C(\lambda)\), such that (13) is still valid when replacing \(C(\lambda)\) by K, and \(f(x)\), \(g(y)\) by \(\widetilde{f}(x)\), \(\widetilde{g}(y)\), respectively, then it gives
Since \(\{ u^{\lambda_{2}+\frac{1}{nq}-1} \} \) (\(0< u\leq1\)) and \(\{ u^{\lambda_{2}-\frac{1}{np}-1} \} \) (\(u>1\)) are nonnegative, monotonically increasing, by the Levi theorem [22],
Hence \(C(\lambda)=K\) in (13) is the best possible.
$$\begin{aligned} &\int_{0}^{1} \bigl(h(u)+h(-u) \bigr)u^{\lambda_{2}+\frac{1}{nq}-1}\,du+\int_{1}^{\infty } \bigl( h(u)+h(-u) \bigr) u^{\lambda_{2}-\frac{1}{np}-1}\,du \\ &\quad=\frac{1}{2n}\widetilde{I}< \frac{1}{2n}k\widetilde{L}=K. \end{aligned}$$
(18)
$$\begin{aligned} C(\lambda) =&\int_{0}^{1} \lim _{n\rightarrow\infty} \bigl(h(u)+h(-u) \bigr)u^{\lambda_{2}+\frac{1}{nq}-1}\,du +\int_{1}^{\infty} \lim_{n\rightarrow\infty} \bigl( h(u)+h(-u) \bigr) u^{\lambda_{2}-\frac{1}{np}-1}\,du \\ =& \lim_{n\rightarrow\infty} \biggl[ \int_{0}^{1} \bigl(h(u)+h(-u) \bigr)u^{\lambda_{2}+\frac{1}{nq}-1}\,du +\int_{1}^{\infty} \bigl( h(u)+h(-u) \bigr) u^{\lambda _{2}-\frac{1}{np}-1}\,du \biggr] \leq K. \end{aligned}$$
Remark 2
(i) With the assumptions of Theorem 1, we define a Hilbert-type integral operator \(T:L_{p,\varphi}(\mathbf{R} )\rightarrow L_{p,\psi^{1-p}}(\mathbf{R} )\) as follows: For any \(f\in L_{p,\varphi}(\mathbf{R} )\), there exists a T, satisfying, for any \(y\neq0\),
By (14), we obtain
from which \(Tf\in L_{p,\psi^{1-p}}(\mathbf{R} )\). Hence T is a bounded linear operator with \(\|T\|\leq C(\lambda)\). Since the constant factor in (14) is the best possible, we have
$$\begin{aligned} (Tf) (y)=\int_{-\infty}^{\infty}h \bigl(x^{\delta}y \bigr)f(x)\,dx. \end{aligned}$$
$$\begin{aligned} \|Tf\|_{p,\psi^{1-p}}\leq C(\lambda)\|f\|_{p,\varphi}, \end{aligned}$$
$$\begin{aligned} \|T\|:=\sup_{f(\neq\theta)\in L_{p,\varphi}(\mathbf{R})} \frac{\|Tf\|_{p,\psi^{1-p}}}{\|f\|_{p,\varphi}}=C(\lambda). \end{aligned}$$
(ii) For \(\delta=1\), (13) and (14) are reduced to the following inequalities with non-homogeneous kernel and the best possible constant factors:
$$\begin{aligned}& \begin{aligned}[b] &\int_{-\infty}^{\infty}\int _{-\infty}^{\infty}h ( xy ) f(x)g(y)\,dx\,dy \\ &\quad< C(\lambda) \biggl\{ \int_{-\infty}^{\infty}|x|^{p(1-\lambda _{2})-1}f^{p}(x) \,dx \biggr\} ^{1/p} \biggl\{ \int_{-\infty}^{\infty }|y|^{q(1-\lambda_{2})-1}g^{q}(y) \,dy \biggr\} ^{1/q}, \end{aligned} \\& \int_{-\infty}^{\infty}|y|^{p\lambda_{2}-1} \biggl[ \int _{-\infty }^{\infty }h ( xy ) f(x)\,dx \biggr] ^{p} \,dy<C^{p}(\lambda)\int_{-\infty }^{\infty}|x|^{p(1-\lambda_{2})-1}f^{p}(x) \,dx. \end{aligned}$$
(iii) For \(\delta=-1\) in (13) and (14), replacing \(|x|^{\lambda }f(x)\) by \(f(x)\) gives the following inequalities with homogeneous kernel and the best possible constant factors (\(k_{\lambda}(x,y):=\min_{i\in \{1,2\}}\min\{|x|, [ |y|^{\gamma}+y^{\gamma} \operatorname{sgn} x \cos \alpha _{i} ] ^{\lambda/\gamma}\}\)):
$$\begin{aligned}& \begin{aligned}[b] &\int_{-\infty}^{\infty}k_{\lambda}(x,y)f(x)g(y) \,dx\,dy \\ &\quad< C(\lambda) \biggl\{ \int_{-\infty}^{\infty}|x|^{p(1-\lambda _{1})-1}f^{p}(x) \,dx \biggr\} ^{1/p} \biggl\{ \int_{-\infty}^{\infty }|y|^{q(1-\lambda_{2})-1}g^{q}(y) \,dy \biggr\} ^{1/q}, \end{aligned} \\& \int_{-\infty}^{\infty}|y|^{p\lambda_{2}-1} \biggl[ \int _{-\infty }^{\infty }k_{\lambda}(x,y)f(x)\,dx \biggr] ^{p}\,dy<C^{p}(\lambda)\int_{-\infty }^{\infty}|x|^{p(1-\lambda_{1})-1}f^{p}(x) \,dx. \end{aligned}$$
Theorem 2
With the assumptions of Theorem
1, replacing
\(p>1\)
by
\(0< p<1\), we have the following equivalent reverse inequalities:
where the constant factors
\(C(\lambda)\)
and
\(C^{p}(\lambda)\)
in the above inequalities are the best possible.
$$\begin{aligned}& I =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}h \bigl( x^{\delta }y \bigr) f(x)g(y)\,dx\,dy >C(\lambda)\|f\|_{p,\varphi} \|g \|_{q,\psi}, \end{aligned}$$
(19)
$$\begin{aligned}& J =\int_{-\infty}^{\infty}|y|^{p\lambda_{2}-1} \biggl[ \int _{-\infty }^{\infty}h \bigl( x^{\delta}y \bigr) f(x)\,dx \biggr] ^{p}\,dy > C^{p}(\lambda)\|f\|_{p,\varphi}^{p}, \end{aligned}$$
(20)
Proof
The proof is similar to the proof of Theorem 1 and hence by Lemma 2 with \(0< p<1\), we have the reverses of (13), (14), and (15), thus (19) and (20) are valid. On the other hand, suppose that (19) is valid. Let \(g(y)\) be the same as in Theorem 1, then \(J>0\) with the reverse of (11). If \(J=\infty\), then (20) is obviously valid; if \(0< J<\infty\), then by (19), we obtain the reverses of (16) and (17). Hence we have (20), which is equivalent to (19).
If the constant factor \(C(\lambda)\) in (19) is not the best possible, then there exists a positive constant k, with \(k\geq C(\lambda)\), such that (19) is still valid when replacing \(C(\lambda)\) by k. There exists \(0< a<2\lambda_{2}\), such that \(n\geq\frac{2}{a|q|}\), we have the reverse of (18):
Since
then by the Levi theorem,
In view of \(q<0\) and \(n\geq\frac{2}{a|q|}\), we have
By the Lebesgue dominated convergence theorem [22], we have
Therefore, we get \(C(\lambda)\geq K\). Hence \(C(\lambda)=K\) is the best possible constant factor of (19).
$$\begin{aligned} \int_{0}^{1} \bigl(h(u)+h(-u) \bigr)u^{\lambda_{2}+\frac{1}{nq}-1}\,du+\int_{1}^{\infty } \bigl( h(u)+h(-u) \bigr) u^{\lambda_{2}-\frac{1}{np}-1}\,du>k. \end{aligned}$$
$$\begin{aligned} 0\leq \bigl(h(u)+h(-u) \bigr)u^{\lambda_{2}-\frac{1}{np}-1}\leq \bigl(h(u)+h(-u) \bigr)u^{\lambda _{2}-\frac{1}{(n+1)p}-1}\quad \bigl(u\in{}[1,\infty )\bigr), \end{aligned}$$
$$\begin{aligned}& \int_{1}^{\infty} \bigl( h(u)+h(-u) \bigr) u^{\lambda_{2}-\frac {1}{np}-1}\,du \\& \quad= \int_{1}^{\infty} \bigl( h(u)+h(-u) \bigr) u^{\lambda _{2}-1}\,du+o(1) \quad(n\rightarrow\infty). \end{aligned}$$
$$\begin{aligned}& \bigl(h(u)+h(-u) \bigr)u^{\lambda_{2}+\frac{1}{nq}-1} \leq \bigl(h(u)+h(-u) \bigr)u^{\lambda _{2}-\frac{a}{2}-1}\quad \bigl(u \in(0,1]\bigr), \\& \begin{aligned}[b] 0 &\leq\int_{0}^{1} \bigl(h(u)+h(-u) \bigr)u^{\lambda_{2}-\frac{a}{2}-1}\,du\leq \int_{0}^{\infty} \bigl(h(u)+h(-u) \bigr)u^{\lambda_{2}-\frac{a}{2}-1}\,du \\ &=\frac{-4\lambda}{ ( a+2\lambda_{1} ) ( a-2\lambda _{2} ) }\cdot2^{\lambda_{2}/\gamma} \biggl[ \biggl( \cos \frac {\alpha _{1}}{2} \biggr) ^{2\lambda_{2}/\gamma}+ \biggl( \sin\frac{\alpha _{2}}{2} \biggr) ^{2\lambda_{2}/\gamma} \biggr] < \infty. \end{aligned} \end{aligned}$$
$$\begin{aligned} \int_{0}^{1} \bigl(h(u)+h(-u) \bigr)u^{\lambda_{2}+\frac{1}{nq}-1}\,du=\int_{0}^{1} \bigl( h(u)+h(-u) \bigr) u^{\lambda_{2}-1}\,du+o(1)\quad (n\rightarrow\infty). \end{aligned}$$
As an application of Hilbert-type inequality, suppose that \(\delta =1\),
\(h(u)=0\) (\(|u|>1\)), i.e.
\(h(xy)=0\) (\(|x|>\frac{1}{|y|}\)), then we have
and we get Hardy-type inequality with the non-homogeneous kernel as follows.
$$\begin{aligned} h(xy)=\min_{i\in\{1,2\}}\min \bigl\{ 1, \bigl[ |xy|^{\gamma}+ ( xy ) ^{\gamma}\cos\alpha_{i} \bigr] ^{\lambda/\gamma} \bigr\} , \end{aligned}$$
$$\begin{aligned} C(\lambda)=C_{1}(\lambda):=\int_{-1}^{1}h ( u ) |u|^{\lambda _{2}-1}\,du=\int_{0}^{1} \bigl( h ( u ) +h ( -u ) \bigr) u^{\lambda_{2}-1}\,du, \end{aligned}$$
Corollary 1
If
\(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(C_{1}(\lambda)=\int_{-1}^{1}h ( u ) |u|^{\lambda_{2}-1}\,du\), \(f(x),g(y)\geq0\), satisfying
\(0<\int_{-\infty}^{\infty }|x|^{p(1-\lambda_{2})-1}f^{p}(x)\,dx<\infty\), \(0<\int_{-\infty }^{\infty }|y|^{q(1-\lambda_{2})-1}g^{q}(y)\,dy<\infty\), then we have the following equivalent inequalities:
where the constant factors
\(C_{1}(\lambda)\)
and
\(C_{1}^{p}(\lambda)\)
in the above inequalities are the best possible. If
\(0< p<1\), then we have the reverse equivalent inequalities with the same best possible constant factors.
$$\begin{aligned}& \begin{aligned}[b] &\int_{-\infty}^{\infty} \biggl( \int _{\frac{-1}{|y|}}^{\frac {1}{|y|}}h ( xy ) f(x)\,dx \biggr) g(y)\,dy=\int _{-\infty}^{\infty} \biggl( \int_{\frac{-1}{|x|}}^{\frac{1}{|x|}}h ( xy ) g(y)\,dy \biggr) f(x)\,dx \\ &\quad< C_{1}(\lambda) \biggl\{ \int_{-\infty}^{\infty}|x|^{p(1-\lambda _{2})-1}f^{p}(x) \,dx \biggr\} ^{1/p} \biggl\{ \int_{-\infty}^{\infty }|y|^{q(1-\lambda_{2})-1}g^{q}(y) \,dy \biggr\} ^{1/q}, \end{aligned} \end{aligned}$$
(21)
$$\begin{aligned}& \int_{-\infty}^{\infty}|y|^{p\lambda_{2}-1} \biggl( \int _{\frac {-1}{|y|}}^{\frac{1}{|y|}}h ( xy ) f(x)\,dx \biggr) ^{p} \,dy< C_{1}^{p}(\lambda )\int_{-\infty}^{\infty}|x|^{p(1-\lambda_{2})-1}f^{p}(x) \,dx, \end{aligned}$$
(22)
Similarly, if \(h(u)=0\) (\(|u|<1\)), then \(h(xy)=0\) (\(|x|<\frac{1}{|y|}\)). Setting
and we get another Hardy-type inequality with the non-homogeneous kernel as follows:
$$\begin{aligned} E_{x^{-1}}= \biggl\{ y\in \mathbf{R} :y \leq-\frac{1}{|x|} \mbox{ or } y\geq\frac{1}{|x|} \biggr\} , \end{aligned}$$
Corollary 2
If
\(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\),
\(f(x),g(y)\geq0\), such that
\(0<\int_{-\infty}^{\infty }|x|^{p(1-\lambda_{1})-1}f^{p}(x)\,dx<\infty\),
then we have the following equivalent inequalities with the best possible constant factors:
$$\begin{aligned} C_{2}(\lambda)=\int_{1}^{\infty} \bigl(h ( u ) +h(-u) \bigr)u^{\lambda _{2}-1}\,du, \end{aligned}$$
$$\begin{aligned} 0< \int_{-\infty}^{\infty}|y|^{q(1-\lambda_{2})-1}g^{q}(y) \,dy< \infty, \end{aligned}$$
$$\begin{aligned}& \begin{aligned}[b] &\int_{-\infty}^{\infty} \biggl( \int _{E_{y^{-1}}}h(xy)f(x)\,dx \biggr) g(y)\,dy \\ &\quad=\int_{-\infty}^{\infty} \biggl( \int _{E_{x^{-1}}}h(xy)g(y)\,dy \biggr) f(x)\,dx \\ &\quad< C_{2}(\lambda) \biggl\{ \int_{-\infty}^{\infty}|x|^{p(1-\lambda _{1})-1}f^{p}(x) \,dx \biggr\} ^{1/p} \biggl\{ \int_{-\infty}^{\infty }|y|^{q(1-\lambda_{2})-1}g^{q}(y) \,dy \biggr\} ^{1/q}, \end{aligned} \end{aligned}$$
(23)
$$\begin{aligned}& \begin{aligned}[b] &\int_{-\infty}^{\infty}|y|^{p\lambda_{2}-1} \biggl( \int_{E_{y^{-1}}}k_{\lambda}(x,y)f(x)\,dx \biggr) ^{p}\,dy \\ &\quad< C_{2}^{p}(\lambda )\int_{-\infty}^{\infty} |x|^{p(1-\lambda_{1})-1}f^{p}(x)\,dx. \end{aligned} \end{aligned}$$
(24)
If
\(0< p<1\), then we get the reverse equivalent inequalities with the same best possible constant factors.
Acknowledgements
This work is supported by the National Natural Science Foundation of China (No. 11301090), Training Plan Fund of Outstanding Young Teachers of Higher Learning Institutions of Guangdong Province in China (No. Yq20145084602) and the Natural Science Foundation of Guangdong University of Education (No. 2014yjxm05).
Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
XH carried out the mathematical studies, participated in the sequence alignment and drafted the manuscript. JC participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.