1 Introduction
Suppose that \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(f(x),g(y)\geq0\), \(f\in L^{p}(\mathbf{R}_{+})\), \(g\in L^{q}(\mathbf{R}_{+})\), \(\|f\|_{p}=(\int_{0}^{\infty }f^{p}(x)\,dx)^{\frac{1}{p}}>0\), \(\|g\|_{q}>0\). We have the following well-known Hardy-Hilbert integral inequality (cf. [1]):
where the constant factor \(\frac{\pi}{\sin(\pi/p)}\) is best possible. If \(a_{m},b_{n}\geq0\), \(a=\{a_{m}\}_{m=1}^{\infty}\in l^{p}\), \(b=\{b_{n}\}_{n=1}^{\infty}\in l^{q}\), \(\|a\|_{p}=(\sum_{m=1}^{\infty }a_{m}^{p})^{\frac{1}{p}}>0\), \(\|b\|_{q}>0\), then we still have the discrete variant of the above inequality with the same best constant \(\frac{\pi }{\sin(\pi/p)}\) as follows:
Inequalities (1) and (2) are important in the analysis and its applications (cf. [1‐6]).
$$ \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{x+y}\,dx\,dy< \frac{\pi }{\sin(\pi/p)}\|f\|_{p}\|g\|_{q}, $$
(1)
$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{a_{m}b_{n}}{m+n}< \frac {\pi}{\sin(\pi/p)}\|a \|_{p}\|b\|_{q}. $$
(2)
In 1998, by introducing an independent parameter \(\lambda\in(0,1]\), Yang [7] gave an extension of (1) at \(p=q=2\) with the kernel \(\frac{1}{(x+y)^{\lambda}}\). In 2009 and 2011, Yang [3, 4] gave some best extensions of (1) and (2) as follows.
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If \(\lambda_{1},\lambda_{2},\lambda\in\mathbf{R}\), \(\lambda _{1}+\lambda _{2}=\lambda\), \(k_{\lambda}(x,y)\) is a non-negative homogeneous function of degree −λ, with
\(\phi(x)=x^{p(1-\lambda_{1})-1}\), \(\psi(y)=y^{q(1-\lambda _{2})-1}\), \(f(x),g(y)\geq0\),
\(g\in L_{q,\psi}(\mathbf{R}_{+})\), \(\|f\|_{p,\phi},\|g\|_{q,\psi}>0\), then we have the following inequality:
where the constant factor \(k(\lambda_{1})\) is best possible. Moreover, if \(k_{\lambda}(x,y)\) stays finite and \(k_{\lambda}(x,y)x^{\lambda _{1}-1}(k_{\lambda}(x,y)y^{\lambda_{2}-1})\) is decreasing with respect to \(x>0\) (\(y>0\)), then for \(a_{m},b_{n}\geq0\),
\(b=\{b_{n}\}_{n=1}^{\infty}\in l_{q,\psi}\), \(\|a\|_{p,\phi },\|b\|_{q,\psi }>0\), we have
where the constant factor \(k(\lambda_{1})\) is still best possible.
$$ k(\lambda_{1})=\int_{0}^{\infty}k_{\lambda}(t,1)t^{\lambda _{1}-1}\,dt \in \mathbf{R}_{+}, $$
$$ f\in L_{p,\phi}(\mathbf{R}_{+})= \biggl\{ f;\|f \|_{p,\phi }:=\biggl(\int_{0}^{\infty} \phi(x)\bigl|f(x)\bigr|^{p}\,dx\biggr)^{\frac{1}{p}}< \infty \biggr\} , $$
$$ \int_{0}^{\infty}\int_{0}^{\infty}k_{\lambda }(x,y)f(x)g(y)\,dx\,dy< k( \lambda _{1})\|f\|_{p,\phi}\|g\|_{q,\psi}, $$
(3)
$$ a\in l_{p,\phi}= \Biggl\{ a;\|a\|_{p,\phi}:=\Biggl(\sum _{n=1}^{\infty}\phi (n)|a_{n}|^{p} \Biggr)^{\frac{1}{p}}< \infty \Biggr\} , $$
$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}k_{\lambda }(m,n)a_{m}b_{n}< k( \lambda_{1})\|a\|_{p,\phi}\|b\|_{q,\psi}, $$
(4)
Clearly, for \(\lambda=1\), \(k_{1}(x,y)=\frac{1}{x+y}\), \(\lambda_{1}=\frac {1}{q}\), \(\lambda_{2}=\frac{1}{p}\), (3) reduces to (1), while (4) reduces to (2).
In 2006, Hong [8] first published a multidimensional Hilbert integral inequality by using the transfer formula, which is an extension of (3). Some other related results are given by [9‐22], which provided some new methods to study these kinds of inequalities.
In this paper, by using the transfer formula and applying the method of weight functions and the technique of real analysis, we give a multidimensional Hilbert-type integral inequality with multi-parameters and the best possible constant factor related to the gamma function. The equivalent forms and the reverses are obtained. Furthermore, we also consider the operator expressions and a few particular results related to the kernels of non-homogeneous and homogeneous.
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2 Some lemmas
If \(m,n\in\mathbf{N}\) (N is the set of positive integers), \(\alpha ,\beta >0\), we set
$$\begin{aligned}& \|x\|_{\alpha} := \Biggl( \sum_{k=1}^{m}|x_{k}|^{\alpha} \Biggr) ^{\frac{1}{\alpha}}\quad\bigl(x=(x_{1},\ldots,x_{m})\in \mathbf{R}^{m}\bigr),\\& \|y\|_{\beta} := \Biggl( \sum_{k=1}^{n}|y_{k}|^{\beta} \Biggr) ^{\frac {1}{\beta}}\quad\bigl(y=(y_{1},\ldots,y_{n})\in \mathbf{R}^{n}\bigr). \end{aligned}$$
Lemma 1
If
\(s\in\mathbf{N}\), \(\gamma,M>0\), \(\Psi(u)\)
is a non-negative measurable function in
\((0,1]\), and
then we have the following transfer formula (cf. [6]):
where
\(\Gamma(\cdot)\)
is the gamma function defined by
$$ D_{M}:= \Biggl\{ x\in\mathbf{R}_{+}^{s};0< u= \sum_{i=1}^{s} \biggl( \frac {x_{i}}{M} \biggr) ^{\gamma}\leq1 \Biggr\} , $$
$$\begin{aligned} &\int\cdots\int_{D_{M}}\Psi \Biggl( \sum _{i=1}^{s} \biggl( \frac {x_{i}}{M} \biggr) ^{\gamma} \Biggr)\,dx_{1}\cdots \,dx_{s} \\ &\quad=\frac{M^{s}\Gamma^{s}(\frac{1}{\gamma})}{\gamma^{s}\Gamma(\frac {s}{\gamma})}\int_{0}^{1} \Psi(u)u^{\frac{s}{\gamma}-1}\,du, \end{aligned}$$
(5)
$$ \Gamma(t):=\int_{0}^{\infty}e^{-v}v^{t-1}\,dv\quad(t>0). $$
In view of (5), since \(\mathbf{R}_{+}^{s}=\lim_{M\rightarrow \infty }D_{M}\), we have
$$\begin{aligned} &\int\cdots\int_{\mathbf{R}_{+}^{s}}\Psi \Biggl( \sum _{i=1}^{s} \biggl( \frac{x_{i}}{M} \biggr) ^{\gamma} \Biggr)\,dx_{1}\cdots \,dx_{s} \\ &\quad=\lim_{M\rightarrow\infty}\frac{M^{s}\Gamma^{s}(\frac{1}{\gamma })}{\gamma^{s}\Gamma(\frac{s}{\gamma})}\int_{0}^{1} \Psi(u)u^{\frac {s}{\gamma }-1}\,du. \end{aligned}$$
(6)
By (6), (i) for
setting \(\Psi(u)=0\) (\(u\in(0,\frac{1}{M^{\gamma}})\)), it follows that
(ii) for
setting \(\Psi(u)=0\) (\(u\in(\frac{1}{M^{\gamma}},\infty)\)), we have
$$\begin{aligned} &\bigl\{ x\in\mathbf{R}_{+}^{s};\|x \|_{\gamma}\geq1\bigr\} \\ &\quad=\lim_{M\rightarrow\infty} \Biggl\{ x\in\mathbf{R}_{+}^{s}; \frac{1}{M^{\gamma}}\leq u=\sum_{i=1}^{s} \biggl( \frac{x_{i}}{M} \biggr) ^{\gamma }\leq1 \Biggr\} , \end{aligned}$$
$$\begin{aligned} &\int\cdots\int_{\{x\in\mathbf{R}_{+}^{s};\|x\|_{\gamma }\geq 1\}}\Psi \Biggl( \sum _{i=1}^{s} \biggl( \frac{x_{i}}{M} \biggr) ^{\gamma } \Biggr)\,dx_{1}\cdots \,dx_{s} \\ &\quad=\lim_{M\rightarrow\infty}\frac{M^{s}\Gamma^{s}(\frac{1}{\gamma })}{\gamma^{s}\Gamma(\frac{s}{\gamma})}\int_{\frac{1}{M^{\gamma }}}^{1} \Psi (u)u^{\frac{s}{\gamma}-1}\,du; \end{aligned}$$
(7)
$$\begin{aligned} &\bigl\{ x\in\mathbf{R}_{+}^{s};\|x \|_{\gamma}\leq1\bigr\} \\ &\quad=\lim_{M\rightarrow\infty} \Biggl\{ x\in\mathbf{R}_{+}^{s};0< u=\sum_{i=1}^{s} \biggl( \frac{x_{i}}{M} \biggr) ^{\gamma}\leq\frac{1}{M^{\gamma}} \Biggr\} , \end{aligned}$$
$$\begin{aligned} &\int\cdots\int_{\{x\in\mathbf{R}_{+}^{s};\|x\|_{\gamma }\leq 1\}}\Psi \Biggl( \sum _{i=1}^{s} \biggl( \frac{x_{i}}{M} \biggr) ^{\gamma } \Biggr)\,dx_{1}\cdots \,dx_{s} \\ &\quad=\lim_{M\rightarrow\infty}\frac{M^{s}\Gamma^{s}(\frac{1}{\gamma })}{\gamma^{s}\Gamma(\frac{s}{\gamma})}\int_{0}^{\frac{1}{M^{\gamma }}} \Psi (u)u^{\frac{s}{\gamma}-1}\,du. \end{aligned}$$
(8)
Remark 1
For \(\delta\in\{-1,1\}\), \(s\in\mathbf{N}\), \(\gamma,M>0\), setting \(E_{\delta}:=\{u>0;u^{\delta}\geq\frac{1}{M^{\delta\gamma }}\}\), in view of (7) and (8), it follows that
$$\begin{aligned} &\int\cdots\int_{\{x\in\mathbf{R}_{+}^{s};\|x\|_{\gamma }^{\delta}\geq1\}}\Psi \Biggl( \sum _{i=1}^{s} \biggl( \frac {x_{i}}{M} \biggr) ^{\gamma} \Biggr)\,dx_{1}\cdots \,dx_{s} \\ &\quad=\lim_{M\rightarrow\infty}\frac{M^{s}\Gamma^{s}(\frac{1}{\gamma })}{\gamma^{s}\Gamma(\frac{s}{\gamma})}\int_{E_{\delta}} \Psi(u)u^{\frac {s}{\gamma}-1}\,du. \end{aligned}$$
(9)
Lemma 2
For
\(\delta\in\{-1,1\}\), \(s\in\mathbf{N}\), \(\gamma, \varepsilon>0\), we have
$$ \int_{\{x\in\mathbf{R}_{+}^{s};\|x\|_{\gamma}^{\delta}\geq 1\}}\|x\|_{\gamma}^{-s-\delta\varepsilon}\,dx= \frac{\Gamma^{s}(\frac {1}{\gamma})}{\varepsilon\gamma^{s-1}\Gamma(\frac{s}{\gamma})}. $$
(10)
Proof
By (9), for \(\delta\in\{-1,1\}\), it follows that
Hence, we have (10). □
$$\begin{aligned} &\int_{\{x\in\mathbf{R}_{+}^{s};\|x\|_{\gamma}^{\delta}\geq 1\}}\|x\|_{\gamma}^{-s-\delta\varepsilon}\,dx \\ &\quad=\int\cdots\int_{\{x\in\mathbf{R}_{+}^{s};\|x\|_{\gamma}^{\delta }\geq 1\}} \Biggl\{ M \Biggl[ \sum _{i=1}^{s}\biggl(\frac{x_{i}}{M} \biggr)^{\gamma} \Biggr] ^{\frac{1}{\gamma}} \Biggr\} ^{-s-\delta\varepsilon}\,dx_{1} \cdots \,dx_{s} \\ &\quad=\lim_{M\rightarrow\infty}\frac{M^{s}\Gamma^{s}(\frac{1}{\gamma })}{\gamma^{s}\Gamma(\frac{s}{\gamma})}\int_{E_{\delta}} \bigl(Mu^{1/\gamma }\bigr)^{-s-\delta\varepsilon}u^{\frac{s}{\gamma}-1}\,du \\ &\quad=\lim_{M\rightarrow\infty}\frac{M^{-\delta\varepsilon}\Gamma^{s}( \frac{1}{\gamma})}{\gamma^{s}\Gamma(\frac{s}{\gamma})}\int_{E_{\delta }}u^{\frac{-\delta\varepsilon}{\gamma}-1}\,du\\ &\quad\overset{v=M^{\gamma}u}{=}\frac{M^{-\delta\varepsilon}\Gamma ^{s}(\frac{1}{\gamma})}{\gamma^{s}\Gamma(\frac{s}{\gamma})}\int_{\{ v>0;v^{\delta }\geq1\}} \bigl(M^{-\gamma}v\bigr)^{\frac{-\delta\varepsilon}{\gamma }-1}M^{-\gamma }\,dv\\ &\quad=\frac{\Gamma^{s}(\frac{1}{\gamma})}{\gamma^{s}\Gamma(\frac {s}{\gamma})}\int_{\{v>0;v^{\delta}\geq1\}}v^{\frac{-\delta\varepsilon}{\gamma }-1}\,dv= \frac{\Gamma^{s}(\frac{1}{\gamma})}{\varepsilon\gamma ^{s-1}\Gamma(\frac{s}{\gamma})}. \end{aligned}$$
Definition 1
For \(m,n\in\mathbf{N}\), \(\alpha,\beta,\lambda _{1},\lambda_{2}>0\), \(\lambda_{1}+\lambda_{2}=\lambda\), \(\eta>-1\), \(\delta \in\{-1,1\}\), \(x=(x_{1},\ldots,x_{m})\in\mathbf{R}_{+}^{m}\), \(y=(y_{1},\ldots,y_{n})\in\mathbf{R}_{+}^{n}\), we define two weight functions \(\omega(\lambda_{1},y)\) and \(\varpi(\lambda_{2},x)\) as follows:
$$\begin{aligned}& \omega(\lambda_{1},y) :=\|y\|_{\beta}^{\lambda_{2}}\int _{\mathbf {R}_{+}^{m}}\frac{|\ln(\|y\|_{\beta}/\|x\|_{\alpha}^{\delta})|^{\eta }}{(\max\{\|x\|_{\alpha}^{\delta},\|y\|_{\beta}\})^{\lambda}}\frac{1}{ \|x\|_{\alpha}^{m-\delta\lambda_{1}}}\,dx, \end{aligned}$$
(11)
$$\begin{aligned}& \varpi(\lambda_{2},x) :=\|x\|_{\alpha}^{\delta\lambda_{1}}\int _{\mathbf{R}_{+}^{n}}\frac{|\ln(\|y\|_{\beta}/\|x\|_{\alpha}^{\delta })|^{\eta}}{(\max\{\|x\|_{\alpha}^{\delta},\|y\|_{\beta}\} )^{\lambda}}\frac{1}{\|y\|_{\beta}^{n-\lambda_{2}}}\,dy. \end{aligned}$$
(12)
By (6), we find
Setting \(t=v^{\delta_{1}}\) in (13), for \(\delta=\pm1\), by simplification, it follows that
$$\begin{aligned} \omega(\lambda_{1},y) &=\|y\|_{\beta}^{\lambda_{2}}\int _{\mathbf{R} _{+}^{m}}\frac{|\ln(\frac{\|y\|_{\beta}}{M^{\delta}[\sum_{i=1}^{m} ( \frac{x_{i}}{M} ) ^{\alpha}]^{\delta/\alpha}})|^{\eta}}{(\max \{M^{\delta}[\sum_{i=1}^{m} ( \frac{x_{i}}{M} ) ^{\alpha }]^{\delta/\alpha},\|y\|_{\beta}\})^{\lambda}} \\ &\hphantom{=}{}\times\frac{1}{M^{m-\delta\lambda_{1}}[\sum_{i=1}^{m} ( \frac {x_{i}}{M} ) ^{\alpha}]^{\frac{m-\delta\lambda_{1}}{\alpha }}}\,dx_{1}\cdots \,dx_{m} \\ &=\|y\|_{\beta}^{\lambda_{2}}\lim_{M\rightarrow\infty} \frac {M^{m}\Gamma ^{m}(\frac{1}{\alpha})}{\alpha^{m}\Gamma(\frac{m}{\alpha})} \\ &\hphantom{=}{}\times\int_{0}^{1}\frac{|\ln(\frac{\|y\|_{\beta}}{M^{\delta }u^{\delta /\alpha}})|^{\eta}}{(\max\{M^{\delta}u^{\delta/\alpha },\|y\|_{\beta }\})^{\lambda}} \frac{u^{\frac{m}{\alpha}-1}\,du}{M^{m-\delta\lambda _{1}}u^{\frac{m-\delta\lambda_{1}}{\alpha}}} \\ &=\|y\|_{\beta}^{\lambda_{2}}\lim_{M\rightarrow\infty} \frac {M^{\delta \lambda_{1}}\Gamma^{m}(\frac{1}{\alpha})}{\alpha^{m}\Gamma(\frac {m}{\alpha})}\int_{0}^{1}\frac{|\ln(\frac{\|y\|_{\beta}}{M^{\delta }u^{\delta /\alpha}})|^{\eta}u^{\frac{\delta\lambda_{1}}{\alpha}-1}}{(\max \{M^{\delta}u^{\delta/\alpha},\|y\|_{\beta}\})^{\lambda}}\,du \\ &\overset{v=M\|y\|^{\frac{-1}{\delta}}u^{1/\alpha}}{=}\frac{\Gamma ^{m}(\frac{1}{\alpha})}{\alpha^{m-1}\Gamma(\frac{m}{\alpha})}\int _{0}^{\infty }\frac{|\ln v^{\delta}|^{\eta}v^{\delta\lambda_{1}-1}}{(\max\{ v^{\delta },1\})^{\lambda}}\,dv. \end{aligned}$$
(13)
$$\begin{aligned} \omega(\lambda_{1},y) &=\frac{\Gamma^{m}(\frac{1}{\alpha})}{\alpha ^{m-1}\Gamma(\frac{m}{\alpha})}\int_{0}^{\infty} \frac{|\ln t|^{\eta }t^{\lambda_{1}-1}}{(\max\{t,1\})^{\lambda}}\,dt \\ &=\frac{\Gamma^{m}(\frac{1}{\alpha})}{\alpha^{m-1}\Gamma(\frac{m}{\alpha})} \biggl[ \int_{0}^{1}(-\ln t)^{\eta}t^{\lambda _{1}-1}\,dt+\int_{1}^{\infty} \frac{(\ln t)^{\eta}t^{\lambda_{1}-1}}{t^{\lambda}}\,dt \biggr] \\ &=\frac{\Gamma^{m}(\frac{1}{\alpha})}{\alpha^{m-1}\Gamma(\frac{m}{\alpha})}\int_{0}^{1}(-\ln t)^{\eta}\bigl(t^{\lambda_{1}-1}+t^{\lambda _{2}-1}\bigr)\,dt \\ &\overset{u=-\ln t}{=}\frac{\Gamma^{m}(\frac{1}{\alpha})}{\alpha ^{m-1}\Gamma(\frac{m}{\alpha})}\int_{\infty}^{0}u^{\eta} \bigl[e^{-u(\lambda _{1}-1)}+e^{-u(\lambda_{2}-1)}\bigr]\bigl(-e^{-u}\bigr)\,du \\ &=\frac{\Gamma^{m}(\frac{1}{\alpha})}{\alpha^{m-1}\Gamma(\frac{m}{\alpha})}\int_{0}^{\infty} \bigl(e^{-\lambda_{1}u}+e^{-\lambda _{2}u}\bigr)u^{\eta }\,du \\ &=\frac{\Gamma^{m}(\frac{1}{\alpha})}{\alpha^{m-1}\Gamma(\frac{m}{\alpha})} \biggl( \frac{1}{\lambda_{1}^{\eta}}+\frac{1}{\lambda _{2}^{\eta}} \biggr) \int _{0}^{\infty}e^{-v}v^{(\eta+1)-1}\,dv \\ &=\frac{\Gamma^{m}(\frac{1}{\alpha})\Gamma(\eta+1)}{\alpha ^{m-1}\Gamma (\frac{m}{\alpha})} \biggl( \frac{1}{\lambda_{1}^{\eta}}+\frac {1}{\lambda _{2}^{\eta}} \biggr) . \end{aligned}$$
(14)
Lemma 3
For
\(m,n\in\mathbf{N}\), \(\alpha,\beta,\lambda _{1},\lambda _{2},\widetilde{\lambda}_{1},\widetilde{\lambda}_{2}>0\), \(\lambda _{1}+\lambda_{2}=\widetilde{\lambda}_{1}+\widetilde{\lambda }_{2}=\lambda\), \(\eta>-1\), \(\delta\in\{-1,1\}\), we have
$$\begin{aligned}& \omega(\lambda_{1},y) =K_{\alpha}(\lambda_{1}):= \frac{\Gamma^{m}( \frac{1}{\alpha})\Gamma(\eta+1)}{\alpha^{m-1}\Gamma(\frac {m}{\alpha})} \biggl( \frac{1}{\lambda_{1}^{\eta}}+\frac{1}{\lambda_{2}^{\eta }} \biggr)\quad \bigl(y\in\mathbf{R}_{+}^{n}\bigr), \end{aligned}$$
(15)
$$\begin{aligned}& \varpi(\lambda_{2},x) =K_{\beta}(\lambda_{1}):= \frac{\Gamma ^{n}(\frac{1}{\beta})\Gamma(\eta+1)}{\beta^{n-1}\Gamma(\frac{n}{\beta})} \biggl( \frac{1}{\lambda_{1}^{\eta}}+\frac{1}{\lambda_{2}^{\eta}} \biggr) \quad\bigl(x \in \mathbf{R}_{+}^{m}\bigr), \end{aligned}$$
(16)
$$\begin{aligned}& \begin{aligned}[b] w(\widetilde{\lambda}_{1},y) &:=\|y\|_{\beta}^{\widetilde{\lambda}_{2}} \int_{\{x\in\mathbf{R}_{+}^{m};\|x\|_{\alpha}^{\delta}\geq1\} }\frac{|\ln(\|y\|_{\beta}/\|x\|_{\alpha}^{\delta})|^{\eta}}{(\max \{\|x\|_{\alpha}^{\delta},\|y\|_{\beta}\})^{\lambda}}\frac{1}{\|x\|_{\alpha}^{m-\delta\widetilde{\lambda}_{1}}}\,dx \\ &=K_{\alpha}(\widetilde{\lambda}_{1}) \bigl(1- \theta_{\widetilde{\lambda} _{1}}(y)\bigr), \end{aligned} \end{aligned}$$
(17)
$$\begin{aligned}& \theta_{\widetilde{\lambda}_{1}}(y):=\frac{\widetilde{\lambda }_{1}^{-\eta }+\widetilde{\lambda}_{2}^{-\eta}}{\Gamma(\eta+1)}\int_{0}^{\|y\|_{\beta }^{-1}} \frac{|\ln t|^{\eta}t^{\widetilde{\lambda}_{1}-1}}{(\max \{t,1\})^{\lambda}}\,dt=O\bigl(\|y\|_{\beta}^{-\frac{\widetilde{\lambda }_{1}}{2}}\bigr) \quad\bigl(y\in \mathbf{R}_{+}^{n}\bigr). \end{aligned}$$
(18)
Proof
In view of (9) and (13), we find
$$\begin{aligned} w(\widetilde{\lambda}_{1},y) &=\|y\|_{\beta}^{\widetilde{\lambda}_{2}} \lim_{M\rightarrow\infty}\frac{M^{\delta\widetilde{\lambda}_{1}}\Gamma^{m}(\frac{1}{\alpha})}{\alpha^{m}\Gamma(\frac{m}{\alpha})} \int_{\{u>0;u^{\delta}\geq\frac{1}{M^{\delta\alpha}}\} }\frac{|\ln(\frac{\|y\|_{\beta}}{M^{\delta}u^{\delta/\alpha}})|^{\eta}u^{ \frac{\delta\widetilde{\lambda}_{1}}{\alpha}-1}}{(\max\{M^{\delta }u^{\delta/\alpha},\|y\|_{\beta}\})^{\lambda}}\,du \\ &\overset{v=M\|y\|^{\frac{-1}{\delta}}u^{\frac{1}{\alpha}}}{=}\frac{\Gamma^{m}(\frac{1}{\alpha})}{\alpha^{m-1}\Gamma(\frac{m}{\alpha})} \int_{\{v>0;v^{\delta}\geq\|y\|_{\beta}^{-1}\}}\frac{|\ln v^{\delta }|^{\eta}v^{\delta\widetilde{\lambda}_{1}-1}}{(\max\{v^{\delta },1\})^{\lambda}}\,dv \\ &\overset{t=v^{\delta}}{=}\frac{\Gamma^{m}(\frac{1}{\alpha})}{\alpha ^{m-1}\Gamma(\frac{m}{\alpha})}\int_{\|y\|_{\beta}^{-1}}^{\infty } \frac{|\ln t|^{\eta}t^{\widetilde{\lambda}_{1}-1}}{(\max\{t,1\})^{\lambda}}\,dt \\ &=\frac{\Gamma^{m}(\frac{1}{\alpha})}{\alpha^{m-1}\Gamma(\frac{m}{\alpha})} \biggl[ \int_{0}^{\infty} \frac{|\ln t|^{\eta}t^{\widetilde{\lambda}_{1}-1}\,dt}{(\max\{t,1\})^{\lambda}}-\int_{0}^{\|y\|_{\beta }^{-1}}\frac{|\ln t|^{\eta}t^{\widetilde{\lambda}_{1}-1}\,dt}{(\max \{t,1\})^{\lambda}} \biggr] \\ &=K_{\alpha}(\widetilde{\lambda}_{1}) \bigl(1- \theta_{\widetilde{\lambda} _{1}}(y)\bigr). \end{aligned}$$
Setting \(F(u):=\int_{0}^{u}\frac{|\ln t|^{\eta}t^{\widetilde{\lambda}_{1}-1}}{(\max\{t,1\})^{\lambda}}\,dt\) (\(u\in(0,\infty)\)), it follows that \(F(u)\) is continuous in \((0,\infty)\). Since
there exists a constant \(L>0\) such that
Then we have
namely, \(\theta_{\widetilde{\lambda}_{1}}(y)=O(\|y\|_{\beta}^{-\frac{ \widetilde{\lambda}_{1}}{2}})\) (\(y\in\mathbf{R}_{+}^{n}\)). Hence, we have (17) and (18). □
$$\begin{aligned} &\lim_{u\rightarrow0^{+}}\frac{1}{u^{\widetilde{\lambda}_{1}/2}}\int _{0}^{u}\frac{|\ln t|^{\eta}t^{\widetilde{\lambda}_{1}-1}}{(\max \{t,1\})^{\lambda}}\,dt \\ &\quad=\lim_{u\rightarrow0^{+}}\frac{2}{\widetilde{\lambda }_{1}u^{(\widetilde{\lambda}_{1}/2)-1}}\frac{|\ln u|^{\eta}u^{\widetilde{\lambda }_{1}-1}}{(\max\{u,1\})^{\lambda}}=\lim _{u\rightarrow0^{+}}\frac{2(-\ln u)^{\eta }u^{\widetilde{\lambda}_{1}/2}}{\widetilde{\lambda}_{1}}=0, \\ &\lim_{u\rightarrow\infty}\frac{1}{u^{\widetilde{\lambda}_{1}/2}}\int _{0}^{u}\frac{|\ln t|^{\eta}t^{\widetilde{\lambda}_{1}-1}}{(\max \{t,1\})^{\lambda}}\,dt=0, \end{aligned}$$
$$ 0< \int_{0}^{u}\frac{|\ln t|^{\eta}t^{\widetilde{\lambda}_{1}-1}}{(\max \{t,1\})^{\lambda}}\,dt\leq Lu^{\frac{\widetilde{\lambda}_{1}}{2}}\quad\bigl(u\in (0,\infty)\bigr). $$
$$ 0< \theta_{\widetilde{\lambda}_{1}}(y)\leq\frac{\widetilde{\lambda}_{1}^{-\eta}+\widetilde{\lambda}_{2}^{-\eta}}{\Gamma(\eta+1)}L\|y \|_{\beta}^{-\frac{\widetilde{\lambda}_{1}}{2}}, $$
Lemma 4
As the assumptions of Definition
1, if
\(p\in\mathbf {R}\backslash\{0,1\}\), \(\frac{1}{p}+\frac{1}{q}=1\), \(f(x)=f(x_{1},\ldots ,x_{m})\geq0\), \(g(y)=g(y_{1},\ldots,y_{n})\geq0\), then (i) for
\(p>1\), we have the following inequality:
(ii) for
\(0< p<1\), or
\(p<0\), we have the reverse of (19).
$$\begin{aligned} J_{1}&:= \biggl\{ \int_{\mathbf{R}_{+}^{n}}\frac{\|y\|_{\beta}^{p\lambda _{2}-n}}{(\omega(\lambda_{1},y))^{p-1}} \biggl[ \int_{\mathbf {R}_{+}^{m}}\frac{|\ln(\|y\|_{\beta}/\|x\|_{\alpha}^{\delta})|^{\eta }f(x)}{(\max \{\|x\|_{\alpha}^{\delta},\|y\|_{\beta}\})^{\lambda}}\,dx \biggr] ^{p}\,dy \biggr\} ^{\frac{1}{p}} \\ &\leq \biggl[ \int_{\mathbf{R}_{+}^{m}}\varpi(\lambda _{2},x)\|x \|_{\alpha }^{p(m-\delta\lambda_{1})-m}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}}; \end{aligned}$$
(19)
Proof
(i) For \(p>1\), by Hölder’s inequality with weight (cf. [23]), it follows that
Then by Fubini’s theorem (cf. [24]), we have
Hence, (19) follows.
$$\begin{aligned} &\int_{\mathbf{R}_{+}^{m}}\frac{|\ln(\|y\|_{\beta}/\|x\|_{\alpha }^{\delta})|^{\eta}}{(\max\{\|x\|_{\alpha}^{\delta},\|y\|_{\beta }\})^{\lambda}}f(x)\,dx \\ &\quad=\int_{\mathbf{R}_{+}^{m}}\frac{|\ln(\|y\|_{\beta}/\|x\|_{\alpha }^{\delta})|^{\eta}}{(\max\{\|x\|_{\alpha}^{\delta},\|y\|_{\beta }\})^{\lambda}} \biggl[ \frac{\|x\|_{\alpha}^{(m-\delta\lambda _{1})/q}}{\|y\|_{\beta}^{(n-\lambda_{2})/p}}f(x) \biggr] \biggl[ \frac {\|y\|_{\beta }^{(n-\lambda_{2})/p}}{\|x\|_{\alpha}^{(m-\delta\lambda _{1})/q}} \biggr]\,dx \\ &\quad\leq \biggl[ \int_{\mathbf{R}_{+}^{m}}\frac{|\ln(\|y\|_{\beta }/\|x\|_{\alpha}^{\delta})|^{\eta}}{(\max\{\|x\|_{\alpha}^{\delta },\|y\|_{\beta}\})^{\lambda}} \frac{\|x\|_{\alpha}^{(m-\delta\lambda _{1})(p-1)}}{\|y\|_{\beta}^{n-\lambda_{2}}}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}} \\ &\qquad{}\times \biggl[ \int_{\mathbf{R}_{+}^{m}}\frac{|\ln(\|y\|_{\beta }/\|x\|_{\alpha}^{\delta})|^{\eta}}{(\max\{\|x\|_{\alpha}^{\delta },\|y\|_{\beta}\})^{\lambda}} \frac{\|y\|_{\beta}^{(n-\lambda _{2})(q-1)}}{\|x\|_{\alpha}^{m-\delta\lambda_{1}}}\,dx \biggr] ^{\frac{1}{q}} \\ &\quad=\bigl(\omega(\lambda_{1},y)\bigr)^{\frac{1}{q}}\|y \|_{\beta}^{\frac{n}{p}-\lambda_{2}} \\ &\qquad{}\times \biggl[ \int_{\mathbf{R}_{+}^{m}}\frac{|\ln(\|y\|_{\beta }/\|x\|_{\alpha}^{\delta})|^{\eta}}{(\max\{\|x\|_{\alpha}^{\delta },\|y\|_{\beta}\})^{\lambda}} \frac{\|x\|_{\alpha}^{(m-\delta\lambda _{1})(p-1)}}{\|y\|_{\beta}^{n-\lambda_{2}}}f^{p}(x)\,dx \biggr] ^{\frac {1}{p}}. \end{aligned}$$
(20)
$$\begin{aligned} J_{1} \leq& \biggl\{ \int_{\mathbf{R}_{+}^{m}} \biggl[ \int _{\mathbf{R}_{+}^{i_{0}}}\frac{|\ln(\|y\|_{\beta}/\|x\|_{\alpha}^{\delta })|^{\eta}}{(\max\{\|x\|_{\alpha}^{\delta},\|y\|_{\beta}\})^{\lambda}}\frac{\|x\|_{\alpha}^{(m-\delta\lambda_{1})(p-1)}}{\|y\|_{\beta }^{n-\lambda _{2}}}f^{p}(x)\,dx \biggr]\,dy \biggr\} ^{\frac{1}{p}} \\ =& \biggl\{ \int_{\mathbf{R}_{+}^{m}} \biggl[ \int_{\mathbf {R}_{+}^{n}} \frac{|\ln(\|y\|_{\beta}/\|x\|_{\alpha}^{\delta})|^{\eta}}{(\max \{\|x\|_{\alpha}^{\delta},\|y\|_{\beta}\})^{\lambda}}\frac {\|x\|_{\alpha }^{(m-\delta\lambda_{1})(p-1)}}{\|y\|_{\beta}^{n-\lambda _{2}}}\,dy \biggr] f^{p}(x)\,dx \biggr\} ^{\frac{1}{p}} \\ =& \biggl[ \int_{\mathbf{R}_{+}^{m}}\varpi(\lambda_{2},x)\|x \|_{\alpha }^{p(m-\delta\lambda_{1})-m}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$
(21)
Lemma 5
As the assumptions of Lemma
4, then (i) for
\(p>1\), we have the following inequality equivalent to (19):
(ii) for
\(0< p<1\), or
\(p<0\), we have the reverse of (22) equivalent to the reverse of (19).
$$\begin{aligned} I :=&\int_{\mathbf{R}_{+}^{n}}\int_{\mathbf{R}_{+}^{m}} \frac{|\ln (\|y\|_{\beta}/\|x\|_{\alpha}^{\delta})|^{\eta}}{(\max\{ \|x\|_{\alpha }^{\delta},\|y\|_{\beta}\})^{\lambda}}f(x)g(y)\,dx\,dy \\ \leq& \biggl[ \int_{\mathbf{R}_{+}^{m}}\varpi(\lambda _{2},x) \|x\|_{\alpha }^{p(m-\delta\lambda_{1})-m}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}} \\ &{}\times \biggl[ \int_{\mathbf{R}_{+}^{n}}\omega(\lambda _{1},y) \|y\|_{\beta }^{q(n-\lambda_{2})-n}g^{q}(y)\,dy \biggr] ^{\frac{1}{q}}; \end{aligned}$$
(22)
Proof
(i) For \(p>1\), by Hölder’s inequality (cf. [23]), it follows that
Then by (19) we have (22).
$$\begin{aligned} I =&\int_{\mathbf{R}_{+}^{n}}\frac{\|y\|_{\beta}^{\frac {n}{q}-(n-\lambda _{2})}}{(\omega(\lambda_{1},y))^{\frac{1}{q}}} \biggl[ \int _{\mathbf{R}_{+}^{m}}\frac{|\ln(\|y\|_{\beta}/\|x\|_{\alpha}^{\delta})|^{\eta }}{(\max\{\|x\|_{\alpha}^{\delta},\|y\|_{\beta}\})^{\lambda }}f(x)\,dx \biggr] \\ &{}\times\bigl[ \bigl(\omega(\lambda_{1},y)\bigr)^{\frac{1}{q}}\|y \|_{\beta }^{(n-\lambda_{2})-\frac{n}{q}}g(y) \bigr]\,dy \\ \leq&J_{1} \biggl[ \int_{\mathbf{R}_{+}^{n}}\omega(\lambda _{1},y)\|y\|_{\beta}^{q(n-\lambda_{2})-n}g^{q}(y)\,dy \biggr] ^{\frac{1}{q}}. \end{aligned}$$
(23)
On the other hand, assuming that (22) is valid, we set
Then it follows that
If \(J_{1}=0\), then (19) is trivially valid; if \(J_{1}=\infty\), then by (21), (19) keeps the form of equality (=∞). Suppose that \(0< J_{1}<\infty\). By (22), we have
Dividing out \(J_{1}^{p-1}\) in the above inequality, it follows that
and then (19) follows. Hence, (19) and (22) are equivalent.
$$ g(y):=\frac{\|y\|_{\beta}^{p\lambda_{2}-n}}{(\omega(\lambda _{1},y))^{p-1}} \biggl[ \int_{\mathbf{R}_{+}^{m}} \frac{|\ln(\|y\|_{\beta }/\|x\|_{\alpha }^{\delta})|^{\eta}f(x)}{(\max\{\|x\|_{\alpha}^{\delta },\|y\|_{\beta }\})^{\lambda}}\,dx \biggr] ^{p-1},\quad y\in\mathbf{R}_{+}^{n}. $$
$$ J_{1}^{p}=\int_{\mathbf{R}_{+}^{n}}\omega( \lambda_{1},y)\|y\|_{\beta }^{q(n-\lambda_{2})-n}g^{q}(y)\,dy. $$
$$\begin{aligned} 0 < &\int_{\mathbf{R}_{+}^{n}}\omega(\lambda_{1},y)\|y \|_{\beta }^{q(n-\lambda_{2})-n}g^{q}(y)\,dy=J_{1}^{p}=I \\ \leq& \biggl[ \int_{\mathbf{R}_{+}^{m}}\varpi(\lambda _{2},x) \|x\|_{\alpha }^{p(m-\delta\lambda_{1})-m}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}} \\ &{}\times \biggl[ \int_{\mathbf{R}_{+}^{n}}\omega(\lambda _{1},y) \|y\|_{\beta }^{q(n-\lambda_{2})-n}g^{q}(y)\,dy \biggr] ^{\frac{1}{q}}<\infty. \end{aligned}$$
$$\begin{aligned} J_{1} =& \biggl[ \int_{\mathbf{R}_{+}^{n}}\omega(\lambda _{1},y)\|y\|_{\beta }^{q(n-\lambda_{2})-n}g^{q}(y)\,dy \biggr] ^{\frac{1}{p}} \\ \leq& \biggl[ \int_{\mathbf{R}_{+}^{m}}\varpi(\lambda _{2},x) \|x\|_{\alpha }^{p(m-\delta\lambda_{1})-m}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}}, \end{aligned}$$
3 Main results and operator expressions
Setting functions
we have the following.
$$ \Phi(x):=\|x\|_{\alpha}^{p(m-\delta\lambda_{1})-m},\qquad \Psi (y):=\|y\|_{\beta}^{q(n-\lambda_{2})-n}\quad \bigl(x\in\mathbf{R}_{+}^{m},y\in \mathbf{R}_{+}^{n} \bigr), $$
Theorem 1
Suppose that
\(m,n\in\mathbf{N}\), \(\alpha,\beta ,\lambda _{1},\lambda_{2}>0\), \(\lambda_{1}+\lambda_{2}=\lambda\), \(\eta>-1\), \(\delta \in \{-1,1\}\), \(p\in\mathbf{R}\backslash\{0,1\}\), \(\frac{1}{p}+\frac{1}{q}=1\), \(f(x)=f(x_{1},\ldots,x_{m})\geq0\), \(g(y)=g(y_{1},\ldots,y_{n})\geq 0\),
$$\begin{aligned}& 0 < \|f\|_{p,\Phi}= \biggl[ \int_{\mathbf{R}_{+}^{m}}\|x \|_{\alpha }^{p(m-\delta\lambda_{1})-m}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}}< \infty,\\& 0 <\|g\|_{q,\Psi}= \biggl[ \int_{\mathbf{R}_{+}^{n}}\|y \|_{\beta }^{q(n-\lambda_{2})-n}g^{q}(y)\,dy \biggr] ^{\frac{1}{q}}< \infty. \end{aligned}$$
(i) For
\(p>1\), we have the following equivalent inequalities with the best possible constant factor
\(K(\lambda_{1})\):
where we define the constant factor as follows:
$$\begin{aligned}& I=\int_{\mathbf{R}_{+}^{n}}\int_{\mathbf{R}_{+}^{m}} \frac{|\ln (\|y\|_{\beta }/\|x\|_{\alpha}^{\delta})|^{\eta}}{(\max\{\|x\|_{\alpha}^{\delta },\|y\|_{\beta}\})^{\lambda}}f(x)g(y)\,dx\,dy< K(\lambda_{1})\|f\|_{p,\Phi }\|g \|_{q,\Psi}, \end{aligned}$$
(24)
$$\begin{aligned}& \begin{aligned}[b] J &:= \biggl\{ \int_{\mathbf{R}_{+}^{n}}\|y\|_{\beta}^{p\lambda _{2}-n} \biggl[ \int_{\mathbf{R}_{+}^{m}}\frac{|\ln(\|y\|_{\beta}/\|x\|_{\alpha }^{\delta })|^{\eta}}{(\max\{\|x\|_{\alpha}^{\delta},\|y\|_{\beta}\} )^{\lambda}}f(x)\,dx \biggr] ^{p}\,dy \biggr\} ^{\frac{1}{p}} \\ &< K(\lambda_{1})\|f\|_{p,\Phi}, \end{aligned} \end{aligned}$$
(25)
$$\begin{aligned} K(\lambda_{1}) :=&\bigl(K_{\beta}(\lambda_{1}) \bigr)^{\frac{1}{p}}\bigl(K_{\alpha }(\lambda_{1}) \bigr)^{\frac{1}{q}} \\ =& \biggl( \frac{\Gamma^{n}(\frac{1}{\beta})}{\beta^{n-1}\Gamma(\frac {n}{\beta})} \biggr) ^{\frac{1}{p}} \biggl( \frac{\Gamma^{m}(\frac{1}{\alpha })}{\alpha^{m-1}\Gamma(\frac{m}{\alpha})} \biggr) ^{\frac{1}{q}} \biggl( \frac{1}{\lambda_{1}^{\eta}}+\frac{1}{\lambda_{2}^{\eta}} \biggr) \Gamma (\eta +1). \end{aligned}$$
Proof
(i) For \(p>1\), by the conditions, we can prove that (20) takes the form of strict inequality. Otherwise, if (20) takes the form of equality for \(y\in\mathbf{R}_{+}^{n}\), then there exist constants A and B, which are not all zero, satisfying
If \(A=0\), then \(B=0\), which is impossible; if \(A\neq0\), then (26) reduces to
which contradicts the fact that \(0<\|f\|_{p,\Phi}<\infty\). In fact, by (9), it follows that \(\int_{\mathbf{R}_{+}^{m}}\|x\|_{\alpha }^{-m}\,dx=\infty \). Hence, (20) takes the form of strict inequality. So does (19). By (15) and (16), we have (25).
$$ A\frac{\|x\|_{\alpha}^{(m-\delta\lambda_{1})(p-1)}}{\|y\|_{\beta }^{n-\lambda_{2}}}f^{p}(x)=B\frac{\|y\|_{\beta}^{(n-\lambda _{2})(q-1)}}{\|x\|_{\alpha}^{m-\delta\lambda_{1}}} \quad\mbox{a.e. in }x\in \mathbf {R}_{+}^{m}. $$
(26)
$$ \|x\|_{\alpha}^{p(m-\delta\lambda_{1})-m}f^{p}(x)=\frac{B\|y\|_{\beta }^{q(n-\lambda_{2})}}{A\|x\|_{\alpha}^{m}}\quad \mbox{a.e. in }x\in \mathbf{R}_{+}^{i_{0}}, $$
In view of (23) (putting \(\omega(\lambda_{1},y)=1\)), we still have
Then by (27) and (25), we have (24). It is evident that by Lemma 5 and the assumptions, (24) and (25) are also equivalent.
$$ I\leq J \biggl[ \int_{\mathbf{R}_{+}^{n}}\|y\|_{\beta}^{q(n-\lambda _{2})-n}g^{q}(y)\,dy \biggr] ^{\frac{1}{q}}. $$
(27)
For \(0<\varepsilon<\frac{p\lambda_{1}}{2}\), we set \(\widetilde{f}(x)\), \(\widetilde{g}(y)\) as follows:
Then, for \(\widetilde{\lambda}_{1}=\lambda_{1}-\frac{\varepsilon }{p}\in(\frac{\lambda_{1}}{2},\lambda)\) (\(\subset(0,\lambda)\)), by (10) we find
and then by (17) and (18) it follows that
$$\begin{aligned}& \widetilde{f}(x):=\left \{ \begin{array}{@{}l@{\quad}l} 0,& 0< \|x\|_{\alpha}^{\delta}<1, \\ \|x\|_{\alpha}^{\delta(\lambda_{1}-\frac{\varepsilon}{p})-m}, &\|x\|_{\alpha}^{\delta}\geq1,\end{array} \right . \\& \widetilde{g}(y):=\left \{ \begin{array}{@{}l@{\quad}l} 0,& 0<\|y\|_{\beta}<1, \\ \|y\|_{\beta}^{(\lambda_{2}-\frac{\varepsilon}{q})-n},& \|y\|_{\beta }\geq 1.\end{array} \right . \end{aligned}$$
$$\begin{aligned}& \begin{aligned}[b] \|\widetilde{f}\|_{p,\Phi}\|\widetilde{g}\|_{q,\Psi}&= \biggl( \int_{\{ x\in \mathbf{R}_{+}^{m};\|x\|_{\alpha}^{\delta}\geq1\}}\|x\|_{\alpha }^{-m-\delta\varepsilon}\,dx \biggr) ^{\frac{1}{p}} \biggl( \int_{\{y\in\mathbf{R}_{+}^{n};\|y\|_{\beta}\geq 1\}}\|y\|_{\beta}^{-n-\varepsilon}\,dy \biggr) ^{\frac{1}{q}} \\ &=\frac{1}{\varepsilon} \biggl( \frac{\Gamma^{m}(\frac{1}{\alpha })}{\alpha ^{m-1}\Gamma(\frac{m}{\alpha})} \biggr) ^{\frac{1}{p}} \biggl( \frac {\Gamma ^{n}(\frac{1}{\beta})}{\beta^{n-1}\Gamma(\frac{n}{\beta})} \biggr) ^{ \frac{1}{q}}, \end{aligned}\\& \begin{aligned}[b] 0 &\leq\int_{\{y\in\mathbf{R}_{+}^{n};\|y\|_{\beta}\geq1\} }\|y\|_{\beta }^{-n-\varepsilon}O \bigl(\|y\|_{\beta}^{-\frac{\widetilde{\lambda}_{1}}{2}}\bigr)\,dy \\ &\leq L_{1}\int_{\{y\in\mathbf{R}_{+}^{n};\|y\|_{\beta}\geq 1\}}\|y\|_{\beta}^{-n-(\varepsilon+\frac{\widetilde{\lambda }_{1}}{2})}\,dy=\frac{L_{1}\Gamma^{n}(\frac{1}{\beta})}{(\varepsilon+\frac{\widetilde {\lambda}_{1}}{2})\beta^{n-1}\Gamma(\frac{n}{\beta})} \\ &\leq\frac{L_{1}\Gamma^{n}(\frac{1}{\beta})}{(\varepsilon+\frac {\lambda _{1}}{4})\beta^{n-1}\Gamma(\frac{n}{\beta})}< \infty\quad(L_{1}>0), \end{aligned} \end{aligned}$$
$$\begin{aligned} \widetilde{I} :=&\int_{\mathbf{R}_{+}^{n}}\int_{\mathbf {R}_{+}^{m}} \frac{|\ln(\|y\|_{\beta}/\|x\|_{\alpha}^{\delta})|^{\eta}}{(\max \{\|x\|_{\alpha}^{\delta},\|y\|_{\beta}\})^{\lambda}}\widetilde {f}(x)\widetilde{g}(y)\,dx\,dy \\ =&\int_{\{y\in\mathbf{R}_{+}^{n};\|y\|_{\beta}\geq1\}}\|y\|_{\beta }^{-n-\varepsilon}w( \widetilde{\lambda}_{1},y)\,dy \\ =&K_{\alpha}(\widetilde{\lambda}_{1})\int_{\{y\in\mathbf{R}_{+}^{n};\|y\|_{\beta}\geq1\}} \|y\|_{\beta}^{-n-\varepsilon} \bigl( 1-O\bigl(\|y\|_{\beta}^{-\frac{\widetilde{\lambda}_{1}}{2}} \bigr) \bigr)\,dy \\ =&\frac{1}{\varepsilon}K_{\alpha}(\widetilde{\lambda}_{1}) \biggl( \frac{\Gamma^{n}(\frac{1}{\beta})}{\beta^{n-1}\Gamma(\frac{n}{\beta})}-\varepsilon O_{\lambda_{1}}(1) \biggr) . \end{aligned}$$
If there exists a constant \(K\leq K(\lambda_{1})\), such that (24) is valid when replacing \(K(\lambda_{1})\) by K, then in particular we have
and then \(K(\lambda_{1})\leq K(\varepsilon\rightarrow0^{+})\). Hence \(K=K(\lambda_{1})\) is the best possible constant factor of (24).
$$\begin{aligned} &\frac{\Gamma^{m}(\frac{1}{\alpha})}{\alpha^{m-1}\Gamma(\frac {m}{\alpha })}\Gamma(\eta+1) \biggl( \frac{1}{\widetilde{\lambda}_{1}^{\eta }}+ \frac{1}{\widetilde{\lambda}_{2}^{\eta}} \biggr) \biggl( \frac{\Gamma ^{n}(\frac{1}{\beta})}{\beta^{n-1}\Gamma(\frac{n}{\beta})}-\varepsilon O_{\lambda _{1}}(1) \biggr) \\ &\quad\leq\varepsilon\widetilde{I}< \varepsilon K\|\widetilde{f}\|_{p,\Phi } \|\widetilde{g}\|_{q,\Psi}=K \biggl( \frac{\Gamma^{m}(\frac{1}{\alpha})}{ \alpha^{m-1}\Gamma(\frac{m}{\alpha})} \biggr) ^{\frac{1}{p}} \biggl( \frac{\Gamma^{n}(\frac{1}{\beta})}{\beta^{n-1}\Gamma(\frac{n}{\beta })} \biggr) ^{\frac{1}{q}}, \end{aligned}$$
As the assumptions of Theorem 1, for \(p>1\), in view of \(J< K(\lambda _{1})\|f\|_{p,\Phi}\), we give the following definition.
Definition 2
We define a multidimensional Hilbert-type integral operator
as follows:
$$ T:\mathbf{L}_{p,\Phi}\bigl(\mathbf{R}_{+}^{m}\bigr) \rightarrow\mathbf{L}_{p,\Psi ^{1-p}}\bigl(\mathbf{R}_{+}^{n} \bigr) $$
For \(f\in\mathbf{L}_{p,\Phi}(\mathbf{R}_{+}^{m})\), there exists a unique representation \(Tf\in\mathbf{L}_{p,\Psi^{1-p}}(\mathbf{R}_{+}^{n})\), satisfying
For \(g\in\mathbf{L}_{q,\Psi}(\mathbf{R}_{+}^{n})\), we define the following formal inner product of Tf and g as follows:
$$ (Tf) (y):=\int_{\mathbf{R}_{+}^{m}}\frac{|\ln(\|y\|_{\beta }/\|x\|_{\alpha }^{\delta})|^{\eta}}{(\max\{\|x\|_{\alpha}^{\delta},\|y\|_{\beta }\})^{\lambda}}f(x)\,dx\quad\bigl(y\in \mathbf{R}_{+}^{n}\bigr). $$
(28)
$$ (Tf,g):=\int_{\mathbf{R}_{+}^{n}}\int_{\mathbf{R}_{+}^{m}} \frac{|\ln (\|y\|_{\beta}/\|x\|_{\alpha}^{\delta})|^{\eta}}{(\max\{ \|x\|_{\alpha }^{\delta},\|y\|_{\beta}\})^{\lambda}}f(x)g(y)\,dx\,dy. $$
(29)
Then by Theorem 1, for \(p>1\), \(0<\|f\|_{p,\Phi},\|g\|_{q,\Psi}<\infty\), we have the following equivalent inequalities:
It follows that T is bounded with
Since the constant factor \(K(\lambda_{1})\) in (31) is best possible, we have
$$\begin{aligned}& (Tf,g) < K(\lambda_{1})\|f\|_{p,\Phi}\|g\|_{q,\Psi}, \end{aligned}$$
(30)
$$\begin{aligned}& \|Tf\|_{p,\Psi^{1-p}} < K(\lambda_{1})\|f\|_{p,\Phi}. \end{aligned}$$
(31)
$$ \|T\|:=\sup_{f(\neq\theta)\in\mathbf{L}_{p,\Phi}(\mathbf {R}_{+}^{m})}\frac{\|Tf\|_{p,\Psi^{1-p}}}{\|f\|_{p,\Phi}}\leq K( \lambda_{1}). $$
$$\begin{aligned} \|T\| =&K(\lambda_{1})= \biggl( \frac{\Gamma^{n}(\frac{1}{\beta })}{\beta ^{n-1}\Gamma(\frac{n}{\beta})} \biggr) ^{\frac{1}{p}} \biggl( \frac {\Gamma ^{m}(\frac{1}{\alpha})}{\alpha^{m-1}\Gamma(\frac{m}{\alpha})} \biggr) ^{\frac{1}{q}} \biggl( \frac{1}{\lambda_{1}^{\eta}}+\frac{1}{\lambda _{2}^{\eta}} \biggr) \Gamma(\eta+1). \end{aligned}$$
(32)
4 Some corollaries
We also set functions
For \(\delta=-1\) in Theorem 1, setting \(F(x)=\|x\|_{\alpha }^{\lambda }f(x)\), by simplification, we have the following.
$$ \widetilde{\Phi}(x):=\|x\|_{\alpha}^{p(m-\lambda_{2})-m},\qquad\widehat { \Phi}(x):=\|x\|_{\alpha}^{p(m-\lambda_{1})-m}\quad\bigl(x\in \mathbf{R}_{+}^{m}\bigr). $$
Corollary 1
Suppose that
\(m,n\in\mathbf{N}\), \(\alpha,\beta ,\lambda _{1},\lambda_{2}>0\), \(\lambda_{1}+\lambda_{2}=\lambda\), \(\eta>-1\), \(p\in \mathbf{R}\backslash\{0,1\}\), \(\frac{1}{p}+\frac{1}{q}=1\), \(F(x)=F(x_{1},\ldots,x_{m})\geq0\), \(g(y)=g(y_{1},\ldots,y_{n})\geq 0\), \(0<\|F\|_{p,\widetilde{\Phi}},\|g\|_{q,\Psi}<\infty\). (i) For
\(p>1\), we have the following equivalent inequalities with the non-homogeneous kernel and the best possible constant factor
\(K(\lambda_{1})\):
(ii) for
\(0< p<1\), or
\(p<0\), we still have the equivalent reverses of (33) and (34) with the same best constant factor
\(K(\lambda_{1})\).
$$\begin{aligned}& \int_{\mathbf{R}_{+}^{n}}\int_{\mathbf{R}_{+}^{m}} \frac{|\ln (\|x\|_{\alpha }\|y\|_{\beta})|^{\eta}}{(\max\{1,\|x\|_{\alpha}\|y\|_{\beta }\})^{\lambda}}F(x)g(y)\,dx\,dy< K(\lambda_{1})\|F\|_{p,\widetilde{\Phi}}\|g \|_{q,\Psi}, \end{aligned}$$
(33)
$$\begin{aligned}& \biggl[ \int_{\mathbf{R}_{+}^{n}}\|y\|_{\beta}^{p\lambda_{2}-n} \biggl( \int_{\mathbf{R}_{+}^{m}}\frac{|\ln(\|x\|_{\alpha}\|y\|_{\beta})|^{\eta }F(x)}{(\max\{1,\|x\|_{\alpha}\|y\|_{\beta}\})^{\lambda}}\,dx \biggr) ^{p}\,dy \biggr] ^{\frac{1}{p}}< K(\lambda_{1})\|F \|_{p,\widetilde{\Phi}}; \end{aligned}$$
(34)
For \(\delta=1\) in Theorem 1, we have the following.
Corollary 2
Suppose that
\(m,n\in\mathbf{N}\), \(\alpha,\beta ,\lambda _{1},\lambda_{2}>0\), \(\lambda_{1}+\lambda_{2}=\lambda\), \(\eta>-1\), \(p\in \mathbf{R}\backslash\{0,1\}\), \(\frac{1}{p}+\frac{1}{q}=1\), \(f(x)=f(x_{1},\ldots,x_{m})\geq0\), \(g(y)=g(y_{1},\ldots,y_{n})\geq 0\), \(0<\|f\|_{p,\widehat{\Phi}},\|g\|_{q,\Psi}<\infty\). (i) For
\(p>1\), we have the following equivalent inequalities with the homogeneous kernel of degree −λ
and the best possible constant factor
\(K(\lambda _{1})\):
(ii) for
\(0< p<1\), or
\(p<0\), we still have the equivalent reverses of (35) and (36) with the same best constant factor
\(K(\lambda_{1})\).
$$\begin{aligned}& \int_{\mathbf{R}_{+}^{n}}\int_{\mathbf{R}_{+}^{m}} \frac{|\ln (\|y\|_{\beta }/\|x\|_{\alpha})|^{\eta}}{(\max\{\|x\|_{\alpha},\|y\|_{\beta }\})^{\lambda}}f(x)g(y)\,dx\,dy< K(\lambda_{1})\|f\|_{p,\widehat{\Phi}}\|g \|_{q,\Psi}, \end{aligned}$$
(35)
$$\begin{aligned}& \biggl\{ \int_{\mathbf{R}_{+}^{n}}\|y\|_{\beta}^{p\lambda_{2}-n} \biggl[ \int_{\mathbf{R}_{+}^{m}}\frac{|\ln(\|y\|_{\beta}/\|x\|_{\alpha })|^{\eta }f(x)}{(\max\{\|x\|_{\alpha},\|y\|_{\beta}\})^{\lambda}}\,dx \biggr] ^{p}\,dy \biggr\} ^{\frac{1}{p}}< K(\lambda_{1})\|f \|_{p,\widehat{\Phi}}; \end{aligned}$$
(36)
Acknowledgements
This work is supported by the National Natural Science Foundation of China (No. 61370186), and 2012 Knowledge Construction Special Foundation Item of Guangdong Institution of Higher Learning College and University (No. 2012KJCX0079).
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
BY carried out the mathematical studies, participated in the sequence alignment and drafted the manuscript. ZH participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.