Necessary and Sufficient Conditions for the Boundedness of Multiple Integral Operators with Super-Homogeneous Kernels in Weighted Lebesgue Space

Abstract

Super-homogeneous functions including homogeneous functions, quasi-homogeneous functions, and several non-homogeneous functions are considered. Using the weight function method, the construction conditions of Hilbert-type multiple integral inequalities with super-homogeneous kernels are first discussed. Then, using the obtained results, the construction problem of bounded multiple integral operators with super-homogeneous kernels in weighted Lebesgue space is discussed, and the necessary and sufficient conditions for operator boundedness and the operator norm formula are obtained.

1. Introduction

Suppose that ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$$(p>1)$, Hardy obtained the following Hilbert-type integral inequality containing a $-1$ order homogeneous kernel ${\displaystyle \frac{1}{x+y}}$ in [1]

$${\int }_0^{+\infty }{\int }_0^{+\infty }{\displaystyle \frac{f\left(x\right)g\left(y\right)}{x+y}}\mathrm{d}x\mathrm{d}y\le {\displaystyle \frac{\pi }{sin(\pi /p)}}{\left|\right|f\left|\right|}_p{\left|\right|g\left|\right|}_q,$$

(1)

where $f\in L_p(0,+\infty ),$$g\in L_q(0,+\infty ),$ and $\pi /sin(\pi /p)$ is the optimal constant factor. Using (1), it can be concluded that operator $T\left(f\right)\left(y\right)={\int }_0^{+\infty }{\displaystyle \frac{f\left(x\right)}{x+y}}$$\mathrm{d}x$ is a bounded operator in Lebesgue space $L_p(0,+\infty )$, and the operator norm $\left|\right|T\left|\right|=\pi /sin(\pi /p)$. Due to the fact that operator research methods have become fundamental in modern harmonic analysis, it is of great significance to explore Hilbert inequality and related operators. In order to further promote research, scholars introduced the weighted Lebesgue space and studied optimal Hilbert inequalities with homogeneous kernels ${\displaystyle \frac{1}{x^{\lambda }+y^{\lambda }}}$, ${\displaystyle \frac{1}{{(x+y)}^{\lambda }}}$, ${\displaystyle \frac{1}{max\{x^{\lambda },y^{\lambda }\}}}|ln{\displaystyle \frac{x}{y}}|$, etc. [2,3,4,5,6], and then discussed the cases of several quasi-homogeneous and non-homogeneous kernels such as ${\displaystyle \frac{1}{{\left(x^{{\lambda }_1}+y^{{\lambda }_2}\right)}^{\lambda }}}$, ${\displaystyle \frac{1}{(1+x^{{\lambda }_1}{y}^{{\lambda }_2}{)}^{\lambda }}}$, ${\displaystyle \frac{1}{max\{1,x^{{\lambda }_1}{y}^{{\lambda }_2}\}}}|ln{x}^{{\lambda }_1}{y}^{{\lambda }_2}|$ (see [7,8,9,10]), and extended the relevant results to high-dimensional spaces.

The results of a large number of literature indicate that whether an integral operator is bounded is not only related to the integral kernel, but also to the properties of the space and the numerous parameters involved. Hong and Wen [11] discussed the problem of constructing Hilbert-type inequalities and related integral operators for the abstract $\lambda$ order homogeneous kernel $K(x,y)$, and obtained the necessary and sufficient conditions for the optimal matching parameters. Hong [12] investigated the construction parameter conditions for Hilbert-type inequalities with homogeneous kernels, and the necessary and sufficient conditions, achieving a theoretical breakthrough. Afterwards, this result was extended to high-dimensional situations (see [13]). More relevant studies can be found in [14,15,16,17,18].

In this article, we will consider the super-homogeneous function proposed in [19] and use super-homogeneous kernels to unify homogeneous kernels, quasi-homogeneous kernels, and several non-homogeneous kernels, exploring the construction of bounded operators from a broader perspective.

Definition 1

([19]). Let ${\sigma }_1,{\sigma }_2,{\tau }_1,{\tau }_2\in \mathbb{R}$. If the function $K(u,v)$ satisfies for any $t>0$:

$$K(tu,v)=t^{{\sigma }_1}K(u,t^{{\tau }_1}v),K(u,tv)=t^{{\sigma }_2}K(t^{{\tau }_2}u,v),$$

then $K(u,v)$ is called a super-homogeneous function with parameters $\{{\sigma }_1,{\sigma }_2,$${\tau }_1,{\tau }_2\}$. Obviously, $K(t,1)=t^{{\sigma }_1}K(1,t^{{\tau }_1})$, $K(1,t)=t^{{\sigma }_2}K(t^{{\tau }_2},1)$.

If $K_1(u,v)$ is a homogeneous function of $\lambda$ order, then $K_1(u,v)$ is a super-homogeneous function with parameters $\{\lambda ,\lambda ,-1,-1\}$, and $K_2(u,v)=K_1(u^{{\lambda }_1},v^{{\lambda }_2})$ is a super-homogeneous function with parameters $\{\lambda {\lambda }_1$, $\lambda {\lambda }_2$, $-{\displaystyle \frac{{\lambda }_1}{{\lambda }_2}}$, $-{\displaystyle \frac{{\lambda }_2}{{\lambda }_1}}\}$. If $G\left(t\right)$ is a function of one variable, then $K_3(u,v)=$$G\left(u^{{\lambda }_1}{v}^{{\lambda }_2}\right)$ is a super-homogeneous function with parameters $\{0$,0,${\displaystyle \frac{{\lambda }_1}{{\lambda }_2}}$,${\displaystyle \frac{{\lambda }_2}{{\lambda }_1}}\}$. It should be pointed out that the kernel $e^{-ix\xi }$ of Fourier transform $F\left(\xi \right)={\int }_{-\infty }^{+\infty }{e}^{-ix\xi }f\left(x\right)\mathrm{d}x$ is a super-homogeneous function with parameters $\{0$,0,1,$1\}$.

Let ${\mathbb{N}}_{+}:=\{1,2,3,\dots \},$$k\in {\mathbb{N}}_{+},x=(x_1,x_2,\dots ,x_k)\in$${\mathbb{R}}_{+}^k$, $\rho >0,$${\parallel x\parallel }_{\rho ,k}=$${\left(x_1^{\rho }+x_2^{\rho }+\dots +x_k^{\rho }\right)}^{1/\rho },$$r>1,\alpha \in \mathbb{R}.$ We call

$$L_p^{\alpha }\left({\mathbb{R}}_{+}^k\right)=\left\{{f\left(x\right):\parallel f\parallel }_{p,\alpha }={\left({\int }_{{\mathbb{R}}_{+}^k}{\parallel x\parallel }_{\rho ,k}^{\alpha }{\left|f\left(x\right)\right|}^p\mathrm{d}x\right)}^{1/p}<+\infty \right\}$$

a weighted Lebesgue space.

For the sake of simplicity, denote

$$W_1\left(s\right)={\int }_0^{+\infty }K(1,t)t^s\mathrm{d}t,W_2\left(s\right)={\int }_0^{+\infty }K(t,1)t^s\mathrm{d}t,$$

$$\Phi (\rho ,k)={\displaystyle \frac{{\Gamma }^k(1/\rho )}{{\rho }^{k-1}\Gamma (k/\rho )}}(\Gamma \left(t\right)\mathrm{is}\mathrm{the}\mathrm{Gamma}\mathrm{function}).$$

2. Preliminary Lemmas

Lemma 1. 

Let parameters $\{{\sigma }_1,{\sigma }_2,{\tau }_1,{\tau }_2\}$ satisfy ${\tau }_1{\tau }_2=1,{\sigma }_1+{\tau }_1{\sigma }_2=0,$ m, $n\in {\mathbb{N}}_{+},$ $p\ne 0,$ $q\ne 0.$ Then, ${\tau }_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)-\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)={\sigma }_1$ is equivalent to ${\tau }_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)-\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)=$${\sigma }_2.$

Proof. 

It follows from ${\tau }_1{\tau }_2=1,{\sigma }_1+{\tau }_1{\sigma }_2=0$ that ${\sigma }_2+{\tau }_2{\sigma }_1=0.$ If ${\tau }_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)-\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)={\sigma }_1,$ then ${\tau }_1{\tau }_2\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)-{\tau }_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)={\tau }_2{\sigma }_1=-{\sigma }_2$. Hence, ${\tau }_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)-\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)=$${\sigma }_2.$

Conversely, if ${\tau }_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)-\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)=$${\sigma }_2,$ then

$${\tau }_1{\tau }_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)-{\tau }_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)={\tau }_1{\sigma }_2=-{\sigma }_1.$$

Thus, ${\tau }_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)-\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)={\sigma }_1.$ □

Lemma 2. 

Suppose that $p\ne 0,$ $q\ne 0,m,n\in {\mathbb{N}}_{+},$ $\alpha ,\beta \in \mathbb{R}$, $K(u,v)$ is a super-homogeneous function with parameters $\{{\sigma }_1,{\sigma }_2,{\tau }_1,{\tau }_2\},$${\tau }_1{\tau }_2=1,{\sigma }_1+{\tau }_1{\sigma }_2=0$, and ${\tau }_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)$−$\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)={\sigma }_1,$ then

$$\begin{array}{ccc}& & W_1^{1/p}\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right)W_2^{1/q}\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}-1\right)\hfill \\ & =& {\displaystyle \frac{1}{|{\tau }_1{|}^{1/q}}}{W}_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right)={\displaystyle \frac{1}{|{\tau }_2{|}^{1/p}}}{W}_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}-1\right).\hfill \end{array}$$

(2)

Proof. 

Since ${\tau }_1{\tau }_2=1,{\sigma }_1+{\tau }_1{\sigma }_2=0$, ${\tau }_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)-\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)={\sigma }_1$, it follows from Lemma 1 that ${\tau }_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)-\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)={\sigma }_2.$ Therefore,

$$\begin{array}{ccc}\hfill W_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right)& =& {\int }_0^{+\infty }K(1,t)t^{{\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1}\mathrm{d}t={\int }_0^{+\infty }K(t^{{\tau }_2},1)t^{{\sigma }_2+{\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1}\mathrm{d}t\hfill \\ & =& {\displaystyle \frac{1}{|{\tau }_2|}}{\int }_0^{+\infty }K(u,1)u^{{\displaystyle \frac{1}{{\tau }_2}}\left({\sigma }_2+{\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right)+{\displaystyle \frac{1}{{\tau }_2}}-1}\mathrm{d}u\hfill \\ & =& {\displaystyle \frac{1}{|{\tau }_2|}}{\int }_0^{+\infty }K(u,1)u^{{\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}-1}\mathrm{d}u={\displaystyle \frac{1}{|{\tau }_2|}}{W}_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}-1\right).\hfill \end{array}$$

Similarly,

$$W_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}-1\right)={\displaystyle \frac{1}{|{\tau }_1|}}{W}_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right).$$

From $W_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right)={\displaystyle \frac{1}{|{\tau }_2|}}{W}_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}-1\right)$ and

$$W_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}-1\right)={\displaystyle \frac{1}{|{\tau }_1|}}{W}_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right),$$

(2) can be obtained. □

Lemma 3

([20]). Let $k\in {\mathbb{N}}_{+},$$\rho >0$, $r>0$, $x=(x_1,x_2,\dots ,x_k)\in {\mathbb{R}}_{+}^k$, $\phi \left(t\right)$ be a measurable function. Then,

$${\int }_{{0<\parallel x\parallel }_{\rho ,k}\le r}\phi \left({\parallel x\parallel }_{\rho ,k}\right)\mathrm{d}x=\Phi (\rho ,k){\int }_0^r\phi \left(u\right)u^{k-1}\mathrm{d}u,$$

$${\int }_{{\parallel x\parallel }_{\rho ,k}\ge r}\phi \left({\parallel x\parallel }_{\rho ,k}\right)\mathrm{d}x=\Phi (\rho ,k){\int }_r^{+\infty }\phi \left(u\right)u^{k-1}\mathrm{d}u,$$

$${\int }_{{\mathbb{R}}_{+}^k}\phi \left({\parallel x\parallel }_{\rho ,k}\right)\mathrm{d}x=\Phi (\rho ,k){\int }_0^{+\infty }\phi \left(u\right)u^{k-1}\mathrm{d}u.$$

Lemma 4. 

Let $K(u,v)$ be a super-homogeneous function with parameters $\{{\sigma }_1,{\sigma }_2,$${\tau }_1,$ ${\tau }_2\},$$m,$$n\in {\mathbb{N}}_{+}$, ${\rho }_1>0,$ ${\rho }_2>0$, $x=(x_1,x_2,\dots ,x_m)\in$${\mathbb{R}}_{+}^m,$$y=(y_1,y_2,\dots ,$$y_n)$$\in$${\mathbb{R}}_{+}^n$. Then,

$$\begin{array}{ccc}\hfill {\omega }_1(x,\beta )& =& {\int }_{{\mathbb{R}}_{+}^n}{K(\parallel x\parallel }_{{\rho }_1,m}{,\parallel y\parallel }_{{\rho }_2,n}{)\parallel y\parallel }_{{\rho }_2,n}^{-{\displaystyle \frac{\beta +n}{q}}}\mathrm{d}y\hfill \\ & =& \Phi ({\rho }_2,n){\parallel x\parallel }_{{\rho }_1,m}^{{\sigma }_1-{\tau }_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)}{W}_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right),\hfill \end{array}$$

$$\begin{array}{ccc}\hfill {\omega }_2(y,\alpha )& =& {\int }_{{\mathbb{R}}_{+}^m}{K(\parallel x\parallel }_{{\rho }_1,m}{,\parallel y\parallel }_{{\rho }_2,n}{)\parallel x\parallel }_{{\rho }_1,m}^{-{\displaystyle \frac{\alpha +m}{p}}}\mathrm{d}x\hfill \\ & =& \Phi ({\rho }_1,m){\parallel y\parallel }_{{\rho }_2,n}^{{\sigma }_2-{\tau }_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)}{W}_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}-1\right).\hfill \end{array}$$

Proof. 

It follows from Lemma 3 and the properties of super-homogeneous functions that

$$\begin{array}{ccc}\hfill {\omega }_1(x,\beta )& =& \Phi ({\rho }_2,n){\int }_0^{+\infty }{K(\parallel x\parallel }_{{\rho }_1,m},u)u^{-{\displaystyle \frac{\beta +n}{q}}+n-1}\mathrm{d}u\hfill \\ & =& \Phi ({\rho }_2,n){\parallel x\parallel }_{{\rho }_1,m}^{{\sigma }_1}{\int }_0^{+\infty }{K(1,u\parallel x\parallel }_{{\rho }_1,m}^{{\tau }_1})u^{{\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1}\mathrm{d}u\hfill \\ & =& \Phi ({\rho }_2,n){\parallel x\parallel }_{{\rho }_1,m}^{{\sigma }_1-{\tau }_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)}{\int }_0^{+\infty }K(1,t)t^{{\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1}\mathrm{d}t\hfill \\ & =& \Phi ({\rho }_2,n){\parallel x\parallel }_{{\rho }_1,m}^{{\sigma }_1-{\tau }_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)}{W}_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right).\hfill \end{array}$$

Similarly, it can be proven that

$${\omega }_2(y,\alpha )=\Phi ({\rho }_1,m){\parallel y\parallel }_{{\rho }_2,n}^{{\sigma }_2-{\tau }_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)}{W}_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}-1\right).$$

□

3. Construction Conditions for Hilbert-Type Multiple Integral Inequalities with Super-Homogeneous Kernels

Theorem 1. 

Let ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1(p>1),$$m,n\in {\mathbb{N}}_{+},$${\rho }_1>0,$${\rho }_2>0,\alpha ,\beta \in \mathbb{R}$, $x=(x_1,x_2,\dots ,x_m)$∈${\mathbb{R}}_{+}^m,$$y=(y_1,y_2,\dots ,y_n)\in$${\mathbb{R}}_{+}^n$, $K(u,v)>0$ be a super-homogeneous function with parameters $\{{\sigma }_1,{\sigma }_2,{\tau }_1,{\tau }_2\}$, ${\tau }_1{\tau }_2=1,{\sigma }_1+{\tau }_1{\sigma }_2=0,$ $W_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right)$$<+\infty$ or $W_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}-1\right)<+\infty .$

(i) If and only if the parameters satisfy ${\tau }_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)-\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)={\sigma }_1$, there exists a constant $M>0,$ and the following Hilbert-type multiple integral inequality holds

$${\int }_{{\mathbb{R}}_{+}^n}{\int }_{{\mathbb{R}}_{+}^m}{K(\parallel x\parallel }_{{\rho }_1,m}{,\parallel y\parallel }_{{\rho }_2,n}{)f\left(x\right)g\left(y\right)\mathrm{d}x\mathrm{d}y\le M\parallel f\parallel }_{p,\alpha }{\parallel g\parallel }_{q,\beta },$$

(3)

where $f\in L_p^{\alpha }\left({\mathbb{R}}_{+}^m\right),g\in L_q^{\beta }\left({\mathbb{R}}_{+}^n\right).$

(ii) When ${\tau }_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)-\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)={\sigma }_1$, i.e., (3), holds, its optimal constant factor is

$$M_0=inf\left\{M\right\}={\Phi }^{1/q}({\rho }_1,m){\Phi }^{1/p}({\rho }_2,n){\left({\displaystyle \frac{1}{|{\tau }_1|}}\right)}^{1/q}{W}_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right).$$

Proof. 

Firstly, according to Lemma 2, when

$${\tau }_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)-\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)={\sigma }_1,$$

we see that $W_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right)$$<+\infty$ and $W_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}-1\right)<+\infty$ hold simultaneously.

(i) If ${\tau }_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)-\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)={\sigma }_1$, by Lemma 1, we have

$${\tau }_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)-\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)={\sigma }_2.$$

It follows from Hölder’s inequality and Lemma 4 that

$$\begin{gathered} A(K,f,g)={\int }_{{\mathbb{R}}_{+}^n}{\int }_{{\mathbb{R}}_{+}^m}{K(\parallel x\parallel }_{{\rho }_1,m}{,\parallel y\parallel }_{{\rho }_2,n})f\left(x\right)g\left(y\right)\mathrm{d}x\mathrm{d}y \\ \begin{array}{ccc}& =& {\int }_{{\mathbb{R}}_{+}^n}{\int }_{{\mathbb{R}}_{+}^m}\left({\displaystyle \frac{{\parallel x\parallel }_{{\rho }_1,m}^{\left(\alpha +m\right)/\left(pq\right)}}{{\parallel y\parallel }_{{\rho }_2,n}^{\left(\beta +n\right)/\left(pq\right)}}}f\left(x\right)\right)\left({\displaystyle \frac{{\parallel y\parallel }_{{\rho }_2,n}^{\left(\beta +n\right)/\left(pq\right)}}{{\parallel x\parallel }_{{\rho }_1,m}^{\left(\alpha +m\right)/\left(pq\right)}}}g\left(y\right)\right){K(\parallel x\parallel }_{{\rho }_1,m}{,\parallel y\parallel }_{{\rho }_2,n})\mathrm{d}x\mathrm{d}y\hfill \\ & \le & {\left({\int }_{{\mathbb{R}}_{+}^n}{\int }_{{\mathbb{R}}_{+}^m}{\parallel x\parallel }_{{\rho }_1,m}^{\left(\alpha +m\right)/q}{\parallel y\parallel }_{{\rho }_2,n}^{-\left(\beta +n\right)/q}{\left|f\left(x\right)\right|}^p{K(\parallel x\parallel }_{{\rho }_1,m}{,\parallel y\parallel }_{{\rho }_2,n})\mathrm{d}x\mathrm{d}y\right)}^{1/p}\hfill \\ & & \times {\left({\int }_{{\mathbb{R}}_{+}^n}{\int }_{{\mathbb{R}}_{+}^m}{\parallel y\parallel }_{{\rho }_2,n}^{\left(\beta +n\right)/p}{\parallel x\parallel }_{{\rho }_1,m}^{-\left(\alpha +m\right)/p}{\left|g\left(y\right)\right|}^q{K(\parallel x\parallel }_{{\rho }_1,m}{,\parallel y\parallel }_{{\rho }_2,n})\mathrm{d}x\mathrm{d}y\right)}^{1/q}\hfill \\ & =& {\left({\int }_{{\mathbb{R}}_{+}^m}{\parallel x\parallel }_{{\rho }_1,m}^{\left(\alpha +m\right)/q}{\left|f\left(x\right)\right|}^p{\omega }_1(x,\beta )\mathrm{d}x\right)}^{1/p}{\left({\int }_{{\mathbb{R}}_{+}^n}{\parallel y\parallel }_{{\rho }_2,n}^{\left(\beta +n\right)/p}{\left|g\left(y\right)\right|}^q{\omega }_2(y,\alpha )\mathrm{d}y\right)}^{1/q}\hfill \\ & =& {\Phi }^{1/p}({\rho }_2,n){\Phi }^{1/q}({\rho }_1,m)W_1^{1/p}\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right)W_2^{1/q}\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}-1\right)\hfill \\ & & \times {\left({\int }_{{\mathbb{R}}_{+}^m}{\parallel x\parallel }_{{\rho }_1,m}^{{\displaystyle \frac{\alpha +m}{q}}+{\sigma }_1-{\tau }_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)}{\left|f\left(x\right)\right|}^p\mathrm{d}x\right)}^{1/p}{\left({\int }_{{\mathbb{R}}_{+}^n}{\parallel y\parallel }_{{\rho }_2,n}^{{\displaystyle \frac{\beta +n}{p}}+{\sigma }_2-{\tau }_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)}{\left|g\left(y\right)\right|}^q\mathrm{d}y\right)}^{1/q}\hfill \\ & =& {\Phi }^{1/q}({\rho }_1,m){\Phi }^{1/p}({\rho }_2,n)W_1^{1/p}\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right)W_2^{1/q}\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}-1\right)\hfill \\ & & \times {\left({\int }_{{\mathbb{R}}_{+}^m}{\parallel x\parallel }_{{\rho }_1,m}^{\alpha }{\left|f\left(x\right)\right|}^p\mathrm{d}x\right)}^{1/p}{\left({\int }_{{\mathbb{R}}_{+}^n}{\parallel y\parallel }_{{\rho }_2,n}^{\beta }{\left|g\left(y\right)\right|}^q\mathrm{d}y\right)}^{1/q}\hfill \\ & =& {\Phi }^{1/q}({\rho }_1,m){\Phi }^{1/p}({\rho }_2,n)W_1^{1/p}\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right)W_2^{1/q}\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}-1\right){\parallel f\parallel }_{p,\alpha }{\parallel g\parallel }_{q,\beta }.\hfill \end{array} \end{gathered}$$

For any given $M\ge {\Phi }^{1/q}({\rho }_1,m){\Phi }^{1/p}({\rho }_2,n)W_1^{1/p}\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right)W_2^{1/q}\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}-1\right)$, (3) can be obtained.

Conversely, if (3) holds, denote ${\sigma }_1-{\tau }_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)+\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)=c$, and it just needs to be proved that $c=0$ below.

If $c>0$, first discuss the case where ${\tau }_1<0$. At this point, take $0<\epsilon <{\displaystyle \frac{c}{-{\tau }_1}}$ and set

$$f\left(x\right)=\left\{\begin{array}{cc}{\parallel x\parallel }_{{\rho }_1,m}^{-(\alpha +m-{\tau }_1\epsilon )/p},& {\parallel x\parallel }_{{\rho }_1,m}\ge 1,\\ 0,& {0<\parallel x\parallel }_{{\rho }_1,m}<1,\end{array}\right.$$

$$g\left(y\right)=\left\{\begin{array}{cc}{\parallel y\parallel }_{{\rho }_2,n}^{-(\beta +n+\epsilon )/q},& {\parallel y\parallel }_{{\rho }_2,n}\ge 1,\\ 0,& {0<\parallel y\parallel }_{{\rho }_2,n}<1.\end{array}\right.$$

Then,

$$\begin{array}{ccc}& & {\parallel f\parallel }_{p,\alpha }{\parallel g\parallel }_{q,\beta }={\left({\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge 1}{\parallel x\parallel }_{{\rho }_1,m}^{-m+{\tau }_1\epsilon }\mathrm{d}x\right)}^{1/p}{\left({\int }_{{\parallel y\parallel }_{{\rho }_2,n}\ge 1}{\parallel y\parallel }_{{\rho }_2,n}^{-n-\epsilon }\mathrm{d}y\right)}^{1/q}\hfill \\ & =& {\Phi }^{1/p}({\rho }_1,m){\Phi }^{1/q}({\rho }_2,n){\left({\int }_1^{+\infty }{u}^{-1+{\tau }_1\epsilon }\mathrm{d}u\right)}^{1/p}{\left({\int }_1^{+\infty }{u}^{-1-\epsilon }\mathrm{d}u\right)}^{1/q}\hfill \\ & =& {\Phi }^{1/p}({\rho }_1,m){\Phi }^{1/q}({\rho }_2,n){\displaystyle \frac{1}{\epsilon }}{\left({\displaystyle \frac{1}{-{\tau }_1}}\right)}^{1/p},\hfill \end{array}$$

$$\begin{array}{ccc}& & A(K,f,g)\hfill \\ & =& {\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge 1}{\parallel x\parallel }_{{\rho }_1,m}^{-{\displaystyle \frac{\alpha +m}{p}}+{\displaystyle \frac{{\tau }_1\epsilon }{p}}}\left({\int }_{{\parallel y\parallel }_{{\rho }_2,n}\ge 1}{K(\parallel x\parallel }_{{\rho }_1,m}{,\parallel y\parallel }_{{\rho }_2,n}{)\parallel y\parallel }_{{\rho }_2,n}^{-{\displaystyle \frac{\beta +n}{q}}-{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}y\right)\mathrm{d}x\hfill \\ & =& {\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge 1}{\parallel x\parallel }_{{\rho }_1,m}^{{\sigma }_1-{\displaystyle \frac{\alpha +m}{p}}+{\displaystyle \frac{{\tau }_1\epsilon }{p}}}\left({\int }_{{\parallel y\parallel }_{{\rho }_2,n}\ge 1}{K(1,\parallel x\parallel }_{{\rho }_1,m}^{{\tau }_1}{\parallel y\parallel }_{{\rho }_2,n}{)\parallel y\parallel }_{{\rho }_2,n}^{-{\displaystyle \frac{\beta +n}{q}}-{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}y\right)\mathrm{d}x\hfill \\ & =& \Phi ({\rho }_2,n){\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge 1}{\parallel x\parallel }_{{\rho }_1,m}^{{\sigma }_1-{\displaystyle \frac{\alpha +m}{p}}+{\displaystyle \frac{{\tau }_1\epsilon }{p}}}\left({\int }_1^{+\infty }{K(1,u\parallel x\parallel }_{{\rho }_1,m}^{{\tau }_1})u^{-{\displaystyle \frac{\beta +n}{q}}-{\displaystyle \frac{\epsilon }{q}}+n-1}\mathrm{d}u\right)\mathrm{d}x\hfill \\ & =& \Phi ({\rho }_2,n){\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge 1}{\parallel x\parallel }_{{\rho }_1,m}^{{\sigma }_1-{\displaystyle \frac{\alpha +m}{p}}+{\displaystyle \frac{{\tau }_1\epsilon }{p}}+{\tau }_1({\displaystyle \frac{\beta +n}{q}}+{\displaystyle \frac{\epsilon }{q}}-n)}\left({\int }_{{\parallel x\parallel }_{{\rho }_1,m}^{{\tau }_1}}^{+\infty }K(1,t)t^{-{\displaystyle \frac{\beta +n}{q}}-{\displaystyle \frac{\epsilon }{q}}+n-1}\mathrm{d}t\right)\mathrm{d}x\hfill \\ & =& \Phi ({\rho }_2,n){\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge 1}{\parallel x\parallel }_{{\rho }_1,m}^{-m+c+{\tau }_1\epsilon }\left({\int }_{{\parallel x\parallel }_{{\rho }_1,m}^{{\tau }_1}}^{+\infty }K(1,t)t^{-{\displaystyle \frac{\beta +n}{q}}+n-1-{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t\right)\mathrm{d}x.\hfill \end{array}$$

Since ${\tau }_1<0,$ for ${\parallel x\parallel }_{{\rho }_1,m}\ge 1,$ we have ${\parallel x\parallel }_{{\rho }_1,m}^{{\tau }_1}\le 1$ and

$$\begin{array}{ccc}\hfill A(K,f,g)& \ge & \Phi ({\rho }_2,n){\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge 1}{\parallel x\parallel }_{{\rho }_1,m}^{-m+c+{\tau }_1\epsilon }\mathrm{d}x{\int }_1^{+\infty }K(1,t)t^{-{\displaystyle \frac{\beta +n}{q}}+n-1-{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t\hfill \\ & =& \Phi ({\rho }_1,m)\Phi ({\rho }_2,n){\int }_1^{+\infty }{u}^{-1+c+{\tau }_1\epsilon }\mathrm{d}u{\int }_1^{+\infty }K(1,t)t^{-{\displaystyle \frac{\beta +n}{q}}+n-1-{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t.\hfill \end{array}$$

Consequently,

$$\begin{array}{ccc}& & {\int }_1^{+\infty }{u}^{-1+c+{\tau }_1\epsilon }\mathrm{d}u{\int }_1^{+\infty }K(1,t)t^{-{\displaystyle \frac{\beta +n}{q}}+n-1-{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t\hfill \\ & \le & {\Phi }^{-1/q}({\rho }_1,m){\Phi }^{-1/p}({\rho }_2,n){\displaystyle \frac{M}{\epsilon }}{\left({\displaystyle \frac{1}{-{\tau }_1}}\right)}^{1/p}<+\infty .\hfill \end{array}$$

(4)

Since $0<\epsilon <{\displaystyle \frac{c}{-{\tau }_1}}$, we obtain $1-c-{\tau }_1\epsilon <1$. Thus, ${\int }_1^{+\infty }{u}^{-1+c+{\tau }_1\epsilon }\mathrm{d}u=+\infty$, which contradicts (4).

Further, we discuss the case where ${\tau }_1>0$. Take $0<\epsilon <{\displaystyle \frac{c}{{\tau }_1}}$ and let

$$f\left(x\right)=\left\{\begin{array}{cc}{\parallel x\parallel }_{{\rho }_1,m}^{-(\alpha +m+{\tau }_1\epsilon )/p},& {\parallel x\parallel }_{{\rho }_1,m}\ge 1,\\ 0,& {0<\parallel x\parallel }_{{\rho }_1,m}<1,\end{array}\right.$$

$$g\left(y\right)=\left\{\begin{array}{cc}{\parallel y\parallel }_{{\rho }_2,n}^{-(\beta +n-\epsilon )/q},& {0<\parallel y\parallel }_{{\rho }_2,n}\le 1,\\ 0,& {\parallel y\parallel }_{{\rho }_2,n}>1.\end{array}\right.$$

A similar calculation yields

$${\parallel f\parallel }_{p,\alpha }{\parallel g\parallel }_{q,\beta }={\Phi }^{1/p}({\rho }_1,m){\Phi }^{1/q}({\rho }_2,n){\displaystyle \frac{1}{\epsilon }}{\left({\displaystyle \frac{1}{{\tau }_1}}\right)}^{1/p},$$

$$\begin{array}{ccc}& & A(K,f,g)\hfill \\ & =& {\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge 1}{\parallel x\parallel }_{{\rho }_1,m}^{-{\displaystyle \frac{\alpha +m}{p}}-{\displaystyle \frac{{\tau }_1\epsilon }{p}}}\left({\int }_{{0<\parallel y\parallel }_{{\rho }_2,n}\le 1}{K(\parallel x\parallel }_{{\rho }_1,m}{,\parallel y\parallel }_{{\rho }_2,n}{)\parallel y\parallel }_{{\rho }_2,n}^{-{\displaystyle \frac{\beta +n}{q}}+{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}y\right)\mathrm{d}x\hfill \\ & =& {\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge 1}{\parallel x\parallel }_{{\rho }_1,m}^{{\sigma }_1-{\displaystyle \frac{\alpha +m}{p}}-{\displaystyle \frac{{\tau }_1\epsilon }{p}}}\left({\int }_{{0<\parallel y\parallel }_{{\rho }_2,n}\le 1}{K(1,\parallel x\parallel }_{{\rho }_1,m}^{\tau }{\parallel y\parallel }_{{\rho }_2,n}{)\parallel y\parallel }_{{\rho }_2,n}^{-{\displaystyle \frac{\beta +n}{q}}+{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}y\right)\mathrm{d}x\hfill \\ & =& \Phi ({\rho }_2,n){\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge 1}{\parallel x\parallel }_{{\rho }_1,m}^{-m+c-{\tau }_1\epsilon }\left({\int }_0^{{\parallel x\parallel }_{{\rho }_1,m}^{\tau }}K(1,t)t^{-{\displaystyle \frac{\beta +n}{q}}+n-1+{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t\right)\mathrm{d}x\hfill \\ & \ge & \Phi ({\rho }_2,n){\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge 1}{\parallel x\parallel }_{{\rho }_1,m}^{-m+c-{\tau }_1\epsilon }\mathrm{d}x{\int }_0^{1}K(1,t)t^{-{\displaystyle \frac{\beta +n}{q}}+n-1+{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t\hfill \\ & =& \Phi ({\rho }_1,m)\Phi ({\rho }_2,n){\int }_1^{+\infty }{u}^{-1+c-{\tau }_1\epsilon }\mathrm{d}u{\int }_0^{1}K(1,t)t^{-{\displaystyle \frac{\beta +n}{q}}+n-1+{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t.\hfill \end{array}$$

Consequently,

$$\begin{array}{ccc}& & {\int }_1^{+\infty }{u}^{-1+c-{\tau }_1\epsilon }\mathrm{d}u{\int }_0^{1}K(1,t)t^{-{\displaystyle \frac{\beta +n}{q}}+n-1+{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t\hfill \\ & \le & {\Phi }^{-1/q}({\rho }_1,m){\Phi }^{-1/p}({\rho }_2,n){\displaystyle \frac{M}{\epsilon }}{\left({\displaystyle \frac{1}{{\tau }_1}}\right)}^{1/p}<+\infty .\hfill \end{array}$$

(5)

Noting $0<\epsilon <{\displaystyle \frac{c}{{\tau }_1}}$, we have $1-c-{\tau }_1\epsilon <1$; hence, ${\int }_1^{+\infty }{u}^{-1+c-{\tau }_1\epsilon }\mathrm{d}u=+\infty$, which contradicts (5).

From the above, it can be concluded that $c>0$ cannot hold.

If $c<0$, it can still be divided into two cases: ${\tau }_1<0$ and ${\tau }_1>0$ for discussion.

Let’s first discuss the case where ${\tau }_1<0$. At this point, take $0<\epsilon <{\displaystyle \frac{c}{{\tau }_1}}$ and set

$$f\left(x\right)=\left\{\begin{array}{cc}{\parallel x\parallel }_{{\rho }_1,m}^{-(\alpha +m+{\tau }_1\epsilon )/p},& {0<\parallel x\parallel }_{{\rho }_1,m}\le 1,\\ 0,& {\parallel x\parallel }_{{\rho }_1,m}>1,\end{array}\right.$$

$$g\left(y\right)=\left\{\begin{array}{cc}{\parallel y\parallel }_{{\rho }_2,n}^{-(\beta +n+\epsilon )/q},& {0<\parallel y\parallel }_{{\rho }_2,n}\le 1,\\ 0,& {\parallel y\parallel }_{{\rho }_2,n}>1.\end{array}\right.$$

Similar to the previous discussion, it can be concluded that

$$\begin{array}{ccc}& & {\int }_0^1{u}^{-1+c-{\tau }_1\epsilon }\mathrm{d}u{\int }_0^{1}K(1,t)t^{-{\displaystyle \frac{\beta +n}{q}}+n-1+{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t\hfill \\ & \le & {\Phi }^{-1/q}({\rho }_1,m){\Phi }^{-1/p}({\rho }_2,n){\displaystyle \frac{M}{\epsilon }}{\left({\displaystyle \frac{1}{-{\tau }_1}}\right)}^{1/p}<+\infty .\hfill \end{array}$$

(6)

Since $0<\epsilon <{\displaystyle \frac{c}{{\tau }_1}}$, we obtain $1-c+{\tau }_1\epsilon >1$; thus, ${\int }_0^1{u}^{-1+c-{\tau }_1\epsilon }\mathrm{d}u=+\infty ,$ which contradicts (6).

Further, let’s discuss the case where ${\tau }_1>0$. Take $0<\epsilon <{\displaystyle \frac{-c}{{\tau }_1}}$ and let

$$f\left(x\right)=\left\{\begin{array}{cc}{\parallel x\parallel }_{{\rho }_1,m}^{-(\alpha +m-{\tau }_1\epsilon )/p},& {0<\parallel x\parallel }_{{\rho }_1,m}\le 1,\\ 0,& {\parallel x\parallel }_{{\rho }_1,m}>1,\end{array}\right.$$

$$g\left(y\right)=\left\{\begin{array}{cc}{\parallel y\parallel }_{{\rho }_2,n}^{-(\beta +n+\epsilon )/q},& {\parallel y\parallel }_{{\rho }_2,n}\ge 1,\\ 0,& {0<\parallel y\parallel }_{{\rho }_2,n}<1.\end{array}\right.$$

Similarly, there holds

$$\begin{array}{ccc}& & {\int }_0^1{u}^{-1+c+{\tau }_1\epsilon }\mathrm{d}u{\int }_0^{+\infty }K(1,t)t^{-{\displaystyle \frac{\beta +n}{q}}+n-1-{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t\hfill \\ & \le & {\Phi }^{-1/q}({\rho }_1,m){\Phi }^{-1/p}({\rho }_2,n){\displaystyle \frac{M}{\epsilon }}{\left({\displaystyle \frac{1}{{\tau }_1}}\right)}^{1/p}<+\infty .\hfill \end{array}$$

(7)

In view of $0<\epsilon <{\displaystyle \frac{-c}{{\tau }_1}}$, we have $1-c-{\tau }_1\epsilon >1$; therefore,

$${\int }_0^1{u}^{-1+c+{\tau }_1\epsilon }\mathrm{d}u=+\infty ,$$

which contradicts (7).

From the above, it can be concluded that $c<0$ cannot hold.

Since neither $c>0$ nor $c<0$ hold, we obtain $c=0$, i.e., ${\tau }_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)-\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)$$={\sigma }_1.$

(ii) When ${\tau }_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)-\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)={\sigma }_1$, according to Lemma 2 and the proof of (i), it can be concluded that

$$A(K,f,g)\le {\Phi }^{1/q}({\rho }_1,m){\Phi }^{1/p}({\rho }_2,n){\left({\displaystyle \frac{1}{|{\tau }_1|}}\right)}^{1/q}{W}_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right){\parallel f\parallel }_{p,\alpha }{\parallel g\parallel }_{q,\beta }.$$

If the optimal constant factor of (3) is not

$${\Phi }^{1/q}({\rho }_1,m){\Phi }^{1/p}({\rho }_2,n){\left({\displaystyle \frac{1}{|{\tau }_1|}}\right)}^{1/q}{W}_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right),$$

then there exists a constant $\overline{M}>0$ such that $A(K,f,g)\le \overline{M}{\parallel f\parallel }_{p,\alpha }{\parallel g\parallel }_{q,\beta },$ and

$$\overline{M}<{\Phi }^{1/q}({\rho }_1,m){\Phi }^{1/p}({\rho }_2,n){\left({\displaystyle \frac{1}{|{\tau }_1|}}\right)}^{1/q}{W}_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1\right).$$

(8)

For ${\tau }_1<0,$ take a sufficiently small $\epsilon >0$ and a sufficiently large $a>0$. Let

$$f\left(x\right)=\left\{\begin{array}{cc}{\parallel x\parallel }_{{\rho }_1,m}^{-(\alpha +m-{\tau }_1\epsilon )/p},& {\parallel x\parallel }_{{\rho }_1,m}\ge a,\\ 0,& {0<\parallel x\parallel }_{{\rho }_1,m}<a,\end{array}\right.$$

$$g\left(y\right)=\left\{\begin{array}{cc}{\parallel y\parallel }_{{\rho }_2,n}^{-(\beta +n+\epsilon )/q},& {\parallel y\parallel }_{{\rho }_2,n}\ge 1,\\ 0,& {0<\parallel y\parallel }_{{\rho }_2,n}<1.\end{array}\right.$$

Then

$$\begin{array}{ccc}& & {\parallel f\parallel }_{p,\alpha }{\parallel g\parallel }_{q,\beta }={\left({\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge a}{\parallel x\parallel }_{{\rho }_1,m}^{-m+{\tau }_1\epsilon }\mathrm{d}x\right)}^{1/p}{\left({\int }_{{\parallel y\parallel }_{{\rho }_2,n}\ge 1}{\parallel y\parallel }_{{\rho }_2,n}^{-n-\epsilon }\mathrm{d}y\right)}^{1/q}\hfill \\ & =& {\Phi }^{1/p}({\rho }_1,m){\Phi }^{1/q}({\rho }_2,n){\left({\int }_a^{+\infty }{u}^{-1+{\tau }_1\epsilon }\mathrm{d}u\right)}^{1/p}{\left({\int }_1^{+\infty }{u}^{-1-\epsilon }\mathrm{d}u\right)}^{1/q}\hfill \\ & =& {\Phi }^{1/p}({\rho }_1,m){\Phi }^{1/q}({\rho }_2,n){\displaystyle \frac{1}{\epsilon }}{\left({\displaystyle \frac{1}{|{\tau }_1|}}\right)}^{1/p}{a}^{{\tau }_1\epsilon /p},\hfill \end{array}$$

$$\begin{array}{ccc}& & A(K,f,g)\hfill \\ & =& {\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge a}{\parallel x\parallel }_{{\rho }_1,m}^{-{\displaystyle \frac{\alpha +m}{p}}+{\displaystyle \frac{{\tau }_1\epsilon }{p}}}\left({\int }_{{\parallel y\parallel }_{{\rho }_2,n}\ge 1}{K(\parallel x\parallel }_{{\rho }_1,m}{,\parallel y\parallel }_{{\rho }_2,n}{)\parallel y\parallel }_{{\rho }_2,n}^{-{\displaystyle \frac{\beta +n}{q}}-{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}y\right)\mathrm{d}x\hfill \\ & =& {\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge a}{\parallel x\parallel }_{{\rho }_1,m}^{{\sigma }_1-{\displaystyle \frac{\alpha +m}{p}}+{\displaystyle \frac{{\tau }_1\epsilon }{p}}}\left({\int }_{{\parallel y\parallel }_{{\rho }_2,n}\ge 1}{K(1,\parallel x\parallel }_{{\rho }_1,m}^{{\tau }_1}{\parallel y\parallel }_{{\rho }_2,n}{)\parallel y\parallel }_{{\rho }_2,n}^{-{\displaystyle \frac{\beta +n}{q}}-{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}y\right)\mathrm{d}x\hfill \\ & =& \Phi ({\rho }_2,n){\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge a}{\parallel x\parallel }_{{\rho }_1,m}^{{\sigma }_1-{\displaystyle \frac{\alpha +m}{p}}+{\displaystyle \frac{{\tau }_1\epsilon }{p}}}\left({\int }_1^{+\infty }{K(1,u\parallel x\parallel }_{{\rho }_1,m}^{{\tau }_1})u^{-{\displaystyle \frac{\beta +n}{q}}+n-1-{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}u\right)\mathrm{d}x\hfill \\ & =& \Phi ({\rho }_2,n){\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge a}{\parallel x\parallel }_{{\rho }_1,m}^{{\sigma }_1-{\displaystyle \frac{\alpha +m}{p}}+{\displaystyle \frac{{\tau }_1\epsilon }{p}}+{\tau }_1({\displaystyle \frac{\beta +n}{q}}-n+1+{\displaystyle \frac{\epsilon }{q}})+{\tau }_1}\hfill \\ & & \times \left({\int }_{{\parallel x\parallel }_{{\rho }_1,m}^{{\tau }_1}}^{+\infty }K(1,t)t^{-{\displaystyle \frac{\beta +n}{q}}+n-1-{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t\right)\mathrm{d}x\hfill \\ & =& \Phi ({\rho }_2,n){\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge a}{\parallel x\parallel }_{{\rho }_1,m}^{-m+{\tau }_1\epsilon }\left({\int }_{{\parallel x\parallel }_{{\rho }_1,m}^{{\tau }_1}}^{+\infty }K(1,t)t^{-{\displaystyle \frac{\beta +n}{q}}+n-1-{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t\right)\mathrm{d}x\hfill \\ & \ge & \Phi ({\rho }_2,n){\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge a}{\parallel x\parallel }_{{\rho }_1,m}^{-m+{\tau }_1\epsilon }\mathrm{d}x{\int }_{a^{{\tau }_1}}^{+\infty }K(1,t)t^{-{\displaystyle \frac{\beta +n}{q}}+n-1-{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t\hfill \\ & =& \Phi ({\rho }_1,m)\Phi ({\rho }_2,n){\int }_a^{+\infty }{u}^{-1+{\tau }_1\epsilon }\mathrm{d}u{\int }_{a^{{\tau }_1}}^{+\infty }K(1,t)t^{-{\displaystyle \frac{\beta +n}{q}}+n-1-{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t\hfill \\ & =& \Phi ({\rho }_1,m)\Phi ({\rho }_2,n){\displaystyle \frac{1}{|{\tau }_1|\epsilon }}{a}^{{\tau }_1\epsilon }{\int }_{a^{{\tau }_1}}^{+\infty }K(1,t)t^{{\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1-{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t.\hfill \end{array}$$

Consequently,

$$\begin{array}{ccc}& & \Phi ({\rho }_1,m)\Phi ({\rho }_2,n){\displaystyle \frac{1}{|{\tau }_1|\epsilon }}{a}^{{\tau }_1\epsilon }{\int }_{a^{{\tau }_1}}^{+\infty }K(1,t)t^{{\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1-{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t\hfill \\ & \le & {\Phi }^{1/p}({\rho }_1,m){\Phi }^{1/q}{\displaystyle \frac{1}{\epsilon }}{\left({\displaystyle \frac{1}{|{\tau }_1|}}\right)}^{1/p}{a}^{{\tau }_1\epsilon /p},\hfill \end{array}$$

and it follows that

$${\Phi }^{1/q}({\rho }_1,m){\Phi }^{1/p}({\rho }_2,n){\left({\displaystyle \frac{1}{|{\tau }_1|}}\right)}^{1/q}{a}^{{\tau }_1\epsilon /q}{\int }_{a^{{\tau }_1}}^{+\infty }K(1,t)t^{{\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1-{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t\le \overline{M}.$$

(9)

Without losing scientific nature, consider $\epsilon$ as a decreasing sequence $\left\{c_k\right\}$ that converges to 0. It follows from Fatou’s lemma that

$$\begin{array}{ccc}& & {\int }_{a^{{\tau }_1}}^{+\infty }K(1,t)t^{{\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1}\mathrm{d}t={\int }_{a^{{\tau }_1}}^{+\infty }\underset{k\to \infty }{lim\; inf}K(1,t)t^{{\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1-{\displaystyle \frac{c_k}{q}}}\mathrm{d}t\hfill \\ & \le & \underset{k\to \infty }{lim\; inf}{\int }_{a^{{\tau }_1}}^{+\infty }K(1,t)t^{{\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1-{\displaystyle \frac{c_k}{q}}}\mathrm{d}t.\hfill \end{array}$$

(10)

Let $\epsilon \to 0^{+}$ in (9), that is, $k\to +\infty$, and according to (10), there holds

$${\Phi }^{1/q}({\rho }_1,m){\Phi }^{1/p}({\rho }_2,n){\left({\displaystyle \frac{1}{|{\tau }_1|}}\right)}^{1/q}{\int }_{a^{{\tau }_1}}^{+\infty }K(1,t)t^{{\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1}\mathrm{d}t\le \overline{M}.$$

Then, let $a\to +\infty$, and note that ${\tau }_1<0$, and we obtain

$${\Phi }^{1/q}({\rho }_1,m){\Phi }^{1/p}({\rho }_2,n){\left({\displaystyle \frac{1}{|{\tau }_1|}}\right)}^{1/q}{W}_1({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1)\le \overline{M},$$

(11)

which contradicts (8).

If ${\tau }_1>0,$ take sufficiently small $\epsilon >0$ and sufficiently large $a>0.$ Set

$$f\left(x\right)=\left\{\begin{array}{cc}{\parallel x\parallel }_{{\rho }_1,m}^{-(\alpha +m+{\tau }_1\epsilon )/p},& {\parallel x\parallel }_{{\rho }_1,m}\ge 1,\\ 0,& {0<\parallel x\parallel }_{{\rho }_1,m}<1,\end{array}\right.$$

$$g\left(y\right)=\left\{\begin{array}{cc}{\parallel y\parallel }_{{\rho }_2,n}^{-(\beta +n-\epsilon )/q},& {0<\parallel y\parallel }_{{\rho }_2,n}\le a,\\ 0,& {\parallel y\parallel }_{{\rho }_2,n}>a.\end{array}\right.$$

Then, similar calculations yield

$${\parallel f\parallel }_{p,\alpha }{\parallel g\parallel }_{q,\beta }={\Phi }^{1/p}({\rho }_1,m){\Phi }^{1/q}({\rho }_2,n){\displaystyle \frac{1}{\epsilon }}{\left({\displaystyle \frac{1}{|{\tau }_1|}}\right)}^{1/p}{a}^{\epsilon /q},$$

$$\begin{array}{ccc}& & A(K,f,g)\hfill \\ & =& {\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge 1}{\parallel x\parallel }_{{\rho }_1,m}^{-{\displaystyle \frac{\alpha +m}{p}}-{\displaystyle \frac{{\tau }_1\epsilon }{p}}}\left({\int }_{{0<\parallel y\parallel }_{{\rho }_2,n}\le a}{K(\parallel x\parallel }_{{\rho }_1,m}{,\parallel y\parallel }_{{\rho }_2,n}{)\parallel y\parallel }_{{\rho }_2,n}^{-{\displaystyle \frac{\beta +n}{q}}+{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}y\right)\mathrm{d}x\hfill \\ & =& {\int }_{{\parallel x\parallel }_{{\rho }_1,m}\ge 1}{\parallel x\parallel }_{{\rho }_1,m}^{{\sigma }_1-{\displaystyle \frac{\alpha +m}{p}}-{\displaystyle \frac{{\tau }_1\epsilon }{p}}}\left({\int }_{{0<\parallel y\parallel }_{{\rho }_2,n}\le a}{K(1,\parallel x\parallel }_{{\rho }_1,m}^{\tau }{\parallel y\parallel }_{{\rho }_2,n}{)\parallel y\parallel }_{{\rho }_2,n}^{-{\displaystyle \frac{\beta +n}{q}}+{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}y\right)\mathrm{d}x\hfill \\ & =& \Phi ({\rho }_1,m)\Phi ({\rho }_2,n){\displaystyle \frac{1}{\epsilon |{\tau }_1|}}{\int }_0^{a}K(1,t)t^{{\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1+{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t,\hfill \end{array}$$

and it follows that

$$\Phi ({\rho }_1,m)\Phi ({\rho }_2,n){\displaystyle \frac{1}{|{\tau }_1|\epsilon }}{\int }_0^{a}K(1,t)t^{{\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1+{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t\le {\Phi }^{1/p}({\rho }_1,m){\Phi }^{1/q}({\rho }_2,n){\displaystyle \frac{1}{\epsilon }}{\left({\displaystyle \frac{1}{|{\tau }_1|}}\right)}^{1/q}{a}^{\epsilon /q}.$$

Therefore,

$${\Phi }^{1/q}({\rho }_1,m){\Phi }^{1/p}({\rho }_2,n){\left({\displaystyle \frac{1}{|{\tau }_1|}}\right)}^{1/q}{\int }_0^{a}K(1,t)t^{{\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1+{\displaystyle \frac{\epsilon }{q}}}\mathrm{d}t\le \overline{M}{a}^{\epsilon /q},$$

Similarly, letting $\epsilon \to 0^{+}$, it follows from Fatou’s lemma that

$${\Phi }^{1/q}({\rho }_1,m){\Phi }^{1/p}({\rho }_2,n){\left({\displaystyle \frac{1}{|{\tau }_1|}}\right)}^{1/q}{\int }_0^{a}K(1,t)t^{{\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}-1}\mathrm{d}t\le \overline{M}.$$

Letting $a\to +\infty$, similarly, (11) is obtained, which contradicts (8).

To sum up, $M_0$ is the best factor of (3). □

4. Necessary and Sufficient Conditions for the Boundedness of Multiple Integral Operators with Super-Homogeneous Kernels

Assume that $m,n\in {\mathbb{N}}_{+},$ ${\rho }_1>0,$ ${\rho }_2>0,p>1,\alpha ,\gamma \in \mathbb{R}$, $x=(x_1,x_2,\dots ,x_m)\in$${\mathbb{R}}_{+}^m,$$y=$$(y_1,y_2,\dots ,y_n)\in$${\mathbb{R}}_{+}^n$, $K(u,v)$ is a super-homogeneous function with parameters $\{{\sigma }_1,{\sigma }_2,{\tau }_1,{\tau }_2\}$; the integral operator T is

$$T\left(f\right)\left(y\right)={\int }_{{\mathbb{R}}_{+}^m}{K(\parallel x\parallel }_{{\rho }_1,m}{,\parallel y\parallel }_{{\rho }_2,n})f\left(x\right)\mathrm{d}x.$$

(12)

Discussing whether T is a bounded operator from $L_p^{\alpha }\left({\mathbb{R}}_{+}^m\right)$ to $L_p^{\gamma }\left({\mathbb{R}}_{+}^n\right)$ is obviously related to the integral kernel and the parameters in the corresponding space. According to the basic theory [20] of Hilbert-type integral inequality, for any kernel $K(\parallel x{\parallel }_{{\rho }_1,m},\parallel y{\parallel }_{{\rho }_2,n})\ge 0$, (3) is equivalent to the following inequality of operator T:

$${\left|\right|T\left(f\right)\left|\right|}_{p,\beta (1-p)}\le M{\parallel f\parallel }_{p,\alpha }.$$

Denote $\beta (1-p)=\gamma$ and let ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$. Then, ${\tau }_1\left({\displaystyle \frac{n}{p}}-{\displaystyle \frac{\beta }{q}}\right)-\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)={\sigma }_1$ is transformed into ${\tau }_1{\displaystyle \frac{\gamma +n}{p}}+$$\left({\displaystyle \frac{\alpha }{p}}-{\displaystyle \frac{m}{q}}\right)={\sigma }_1$. According to Theorem 1, the following theorem can be obtained.

Theorem 2. 

Let ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1(p>1),$ $m,n\in {\mathbb{N}}_{+},$ ${\rho }_1>0,$ ${\rho }_2>0,\alpha ,\gamma \in \mathbb{R}$, $x=(x_1,x_2,\dots ,x_m)$∈${\mathbb{R}}_{+}^m,$$y=(y_1,y_2,\dots ,y_n)\in$${\mathbb{R}}_{+}^n$, $K(u,v)>0$ be a super-homogeneous function with parameters $\{{\sigma }_1,{\sigma }_2,{\tau }_1,{\tau }_2\}$, ${\tau }_1{\tau }_2=1,{\sigma }_1+{\tau }_1{\sigma }_2=0,$ $W_1\left({\displaystyle \frac{\gamma +n}{p}}-1\right)$$<0$ or $W_2\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}-1\right)<+\infty$; the multiple integral operator T can be defined as (12).

(i) T is a bounded operator from $L_p^{\alpha }\left({\mathbb{R}}_{+}^m\right)$ to $L_p^{\gamma }\left({\mathbb{R}}_{+}^n\right)$ if and only if

$${\tau }_1{\displaystyle \frac{\gamma +n}{p}}+\left({\displaystyle \frac{\alpha }{p}}-{\displaystyle \frac{m}{q}}\right)={\sigma }_1.$$

(ii) When ${\tau }_1{\displaystyle \frac{\gamma +n}{p}}+\left({\displaystyle \frac{\alpha }{p}}-{\displaystyle \frac{m}{q}}\right)={\sigma }_1$, that is, T is a bounded operator from $L_p^{\alpha }\left({\mathbb{R}}_{+}^m\right)$ to $L_p^{\gamma }\left({\mathbb{R}}_{+}^n\right)$, the operator norm of T is

$$∥T∥={\Phi }^{1/q}({\rho }_1,m){\Phi }^{1/p}({\rho }_2,n){\left({\displaystyle \frac{1}{|{\tau }_1|}}\right)}^{1/q}{W}_1\left({\displaystyle \frac{\gamma +n}{p}}-1\right).$$

Remark 1. 

If $K(u,v)$ is a λ-order homogeneous kernel, then

$${\tau }_1{\displaystyle \frac{\gamma +n}{p}}+\left({\displaystyle \frac{\alpha }{p}}-{\displaystyle \frac{m}{q}}\right)={\sigma }_1$$

is transformed into ${\displaystyle \frac{\alpha }{p}}-{\displaystyle \frac{m}{q}}-{\displaystyle \frac{\gamma +n}{p}}=\lambda$. For other cases such as quasi-homogeneous kernels, corresponding parameter conditions can also be obtained.

In Theorem 2, take $\alpha =\gamma =0.$ Then, the corresponding results in the ordinary Lebesgue space without weight can be obtained.

Corollary 1. 

If the condition of Theorem 2 is satisfied, then

(i) T is a bounded operator from $L_p\left({\mathbb{R}}_{+}^m\right)$ to $L_p\left({\mathbb{R}}_{+}^n\right)$ if and only if

$${\tau }_1{\displaystyle \frac{\gamma }{p}}-{\displaystyle \frac{m}{q}}={\sigma }_1.$$

(ii) When ${\tau }_1{\displaystyle \frac{\gamma }{p}}-{\displaystyle \frac{m}{q}}={\sigma }_1$, the operator norm of T is

$$∥T∥={\Phi }^{1/q}({\rho }_1,m){\Phi }^{1/p}({\rho }_2,n){\left({\displaystyle \frac{1}{|{\tau }_1|}}\right)}^{1/q}{W}_1\left({\displaystyle \frac{n}{p}}-1\right).$$

Corollary 2. 

Assuming ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$$(p>1),m,n\in {\mathbb{N}}_{+},{\rho }_1>0,$${\rho }_2>0$, ${\lambda }_1>0,$${\lambda }_2>0,$$\alpha ,\gamma \in \mathbb{R}$, $x=(x_1,x_2,\dots ,x_m)\in$ ${\mathbb{R}}_{+}^m,$$y=(y_1,y_2,\dots ,y_n)\in$${\mathbb{R}}_{+}^n$, $0<{\displaystyle \frac{\gamma +n}{p}}<{\lambda }_2{\rho }_2$, and the operator T is

$$T\left(f\right)\left(y\right)={\int }_{{\mathbb{R}}_{+}^m}{\displaystyle \frac{ln\left[{\left({\sum }_{k=1}^m{x}_k^{{\rho }_1}\right)}^{{\lambda }_1}/{\left({\sum }_{k=1}^n{y}_k^{{\rho }_2}\right)}^{{\lambda }_2}\right]}{{\left({\sum }_{k=1}^m{x}_k^{{\rho }_1}\right)}^{{\lambda }_1}-{\left({\sum }_{k=1}^n{y}_k^{{\rho }_2}\right)}^{{\lambda }_2}}}f\left(x\right)\mathrm{d}x.$$

(13)

(i) T is a bounded operator from $L_p^{\alpha }\left({\mathbb{R}}_{+}^m\right)$ to $L_p^{\gamma }\left({\mathbb{R}}_{+}^n\right)$ if and only if

$${\displaystyle \frac{1}{{\lambda }_1{\rho }_1}}\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)+{\displaystyle \frac{1}{{\lambda }_2{\rho }_2}}{\displaystyle \frac{\gamma +n}{p}}=1.$$

(ii) When ${\displaystyle \frac{1}{{\lambda }_1{\rho }_1}}\left({\displaystyle \frac{m}{q}}-{\displaystyle \frac{\alpha }{p}}\right)+{\displaystyle \frac{1}{{\lambda }_2{\rho }_2}}{\displaystyle \frac{\gamma +n}{p}}=1$, the operator norm of $T:$$L_p^{\alpha }\left({\mathbb{R}}_{+}^m\right)\to L_p^{\gamma }\left({\mathbb{R}}_{+}^n\right)$ is

$$∥T∥={\Phi }^{1/q}({\rho }_1,m){\Phi }^{1/p}({\rho }_2,n){\displaystyle \frac{1}{{\left({\lambda }_1{\rho }_1\right)}^{1/q}{\left({\lambda }_2{\rho }_2\right)}^{1/p}}}{\left({\displaystyle \frac{\pi }{sin\frac{\pi (\gamma +n)}{p{\lambda }_2{\rho }_2}}}\right)}^2.$$

Proof. 

Denote

$$K(u,v)=ln\left(u^{{\lambda }_1{\rho }_1}/v^{{\lambda }_2{\rho }_2}\right)/\left(u^{{\lambda }_1{\rho }_1}-v^{{\lambda }_2{\rho }_2}\right),u>0,v>0.$$

Then, $K(u,v)>0$ is a super-homogeneous function with parameters $\{-{\lambda }_1{\rho }_1,$$-{\lambda }_2{\rho }_2,$$-{\displaystyle \frac{{\lambda }_1{\rho }_1}{{\lambda }_2{\rho }_2}},$$-{\displaystyle \frac{{\lambda }_2{\rho }_2}{{\lambda }_1{\rho }_1}}\}$, ${\sigma }_1=-{\lambda }_1{\rho }_1,{\sigma }_2=-{\lambda }_2{\rho }_2,{\tau }_1=-{\displaystyle \frac{{\lambda }_1{\rho }_1}{{\lambda }_2{\rho }_2}}$, ${\tau }_2=-{\displaystyle \frac{{\lambda }_2{\rho }_2}{{\lambda }_1{\rho }_1}}$, ${\tau }_1{\tau }_2=1,{\sigma }_1+{\tau }_1{\sigma }_2=0$, and

$${K(\parallel x\parallel }_{{\rho }_1,m}{,\parallel y\parallel }_{{\rho }_2,n})={\displaystyle \frac{ln\left[{\left({\sum }_{k=1}^m{x}_k^{{\rho }_1}\right)}^{{\lambda }_1}/{\left({\sum }_{k=1}^n{y}_k^{{\rho }_2}\right)}^{{\lambda }_2}\right]}{{\left({\sum }_{k=1}^m{x}_k^{{\rho }_1}\right)}^{{\lambda }_1}-{\left({\sum }_{k=1}^n{y}_k^{{\rho }_2}\right)}^{{\lambda }_2}}}.$$

Since $0<{\displaystyle \frac{\gamma +n}{p}}<{\lambda }_2{\rho }_2,$ we have

$$\begin{array}{ccc}\hfill W_1\left({\displaystyle \frac{\gamma +n}{p}}-1\right)& =& {\int }_0^{+\infty }K(1,t)t^{{\displaystyle \frac{\gamma +n}{p}}-1}\mathrm{d}t={\int }_0^{+\infty }{\displaystyle \frac{ln{t}^{{\lambda }_2{\rho }_2}}{t^{{\lambda }_2{\rho }_2}-1}}{t}^{{\displaystyle \frac{\gamma +n}{p}}-1}\mathrm{d}t\hfill \\ & =& {\displaystyle \frac{1}{{\lambda }_2{\rho }_2}}{\int }_0^{+\infty }{\displaystyle \frac{lnu}{u-1}}{u}^{{\displaystyle \frac{\gamma +n}{{\lambda }_2{\rho }_{2}p}}-1}\mathrm{d}u\hfill \\ & =& {\displaystyle \frac{1}{{\lambda }_2{\rho }_2}}{B}^2\left({\displaystyle \frac{\gamma +n}{p{\lambda }_2{\rho }_2}},1-{\displaystyle \frac{\gamma +n}{p{\lambda }_2{\rho }_2}}\right)\hfill \\ & =& {\displaystyle \frac{1}{{\lambda }_2{\rho }_2}}{\left({\displaystyle \frac{\pi }{sin\frac{\pi (\gamma +n)}{p{\lambda }_2{\rho }_2}}}\right)}^2<+\infty .\hfill \end{array}$$

Moreover, it follows from ${\sigma }_1=-{\lambda }_1{\rho }_1$ and ${\tau }_1=-{\displaystyle \frac{{\lambda }_1{\rho }_1}{{\lambda }_2{\rho }_2}}$ that

$${\tau }_1{\displaystyle \frac{\gamma +n}{p}}+\left({\displaystyle \frac{\alpha }{p}}-{\displaystyle \frac{m}{q}}\right)={\sigma }_1\iff {\displaystyle \frac{1}{{\lambda }_1{\rho }_1}}\left({\displaystyle \frac{\alpha }{p}}-{\displaystyle \frac{m}{q}}\right)+{\displaystyle \frac{1}{{\lambda }_2{\rho }_2}}{\displaystyle \frac{\gamma +n}{p}}=1.$$

Based on the above and Theorem 2, it is known that Corollary 2 holds. □

In Corollary 2, choose $\alpha =\gamma =0.$ Then:

Corollary 3. 

Supposing ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$$(p>1)$, $m,n\in {\mathbb{N}}_{+},$${\rho }_1>0,{\rho }_2>0,$${\lambda }_1>0,{\lambda }_2>0$, $x=(x_1,x_2,\dots ,x_m)\in$${\mathbb{R}}_{+}^m,$$y=(y_1,y_2,\dots ,y_n)\in$${\mathbb{R}}_{+}^n$, ${\displaystyle \frac{n}{p}}<{\lambda }_2{\rho }_2$, T is defined as (13).

(i) T is a bounded operator from $L_p\left({\mathbb{R}}_{+}^m\right)$ to $L_p\left({\mathbb{R}}_{+}^n\right)$ if and only if

$${\displaystyle \frac{m}{q{\lambda }_1{\rho }_1}}+{\displaystyle \frac{n}{p{\lambda }_2{\rho }_2}}=1.$$

(ii) When ${\displaystyle \frac{m}{q{\lambda }_1{\rho }_1}}+{\displaystyle \frac{n}{p{\lambda }_2{\rho }_2}}=1$, the operator norm of $T:$$L_p\left({\mathbb{R}}_{+}^m\right)\to L_p\left({\mathbb{R}}_{+}^n\right)$ is

$$∥T∥={\Phi }^{1/q}({\rho }_1,m){\Phi }^{1/p}({\rho }_2,n){\displaystyle \frac{1}{{\left({\lambda }_1{\rho }_1\right)}^{1/q}{\left({\lambda }_2{\rho }_2\right)}^{1/p}}}{\left({\displaystyle \frac{\pi }{sin\frac{\pi }{p{\lambda }_2{\rho }_2}}}\right)}^2.$$

In Corollary 3, select ${\rho }_1={\rho }_2=\rho ,m=n=k.$ Then:

Corollary 4. 

Assuming ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$$(p>1)$, $k\in {\mathbb{N}}_{+}$, $x=(x_1,x_2,\dots ,x_k)$$\in {\mathbb{R}}_{+}^k$, $y=(y_1,y_2,\dots ,y_k)\in {\mathbb{R}}_{+}^k$, ${\displaystyle \frac{k}{p}}<{\lambda }_2\rho$, T is defined by

$$T\left(f\right)\left(y\right)={\int }_{{\mathbb{R}}_{+}^k}{\displaystyle \frac{ln\left[{\left({\sum }_{i=1}^k{x}_i^{\rho }\right)}^{{\lambda }_1}/{\left({\sum }_{i=1}^k{y}_i^{\rho }\right)}^{{\lambda }_2}\right]}{{\left({\sum }_{i=1}^k{x}_i^{\rho }\right)}^{{\lambda }_1}-{\left({\sum }_{i=1}^k{y}_i^{\rho }\right)}^{{\lambda }_2}}}f\left(x\right)\mathrm{d}x.$$

(i) T is a bounded operator in $L_p\left({\mathbb{R}}_{+}^k\right)$ if and only if ${\displaystyle \frac{1}{q{\lambda }_1}}+{\displaystyle \frac{1}{p{\lambda }_2}}={\displaystyle \frac{\rho }{k}}$.

(ii) If ${\displaystyle \frac{1}{q{\lambda }_1}}+{\displaystyle \frac{1}{p{\lambda }_2}}={\displaystyle \frac{\rho }{k}}$, then the operator norm of T is

$$∥T∥={\displaystyle \frac{{\Gamma }^k(1/\rho )}{{\rho }^k\Gamma (k/\rho )}}{\displaystyle \frac{1}{{\lambda }_1^{1/q}{\lambda }_2^{1/p}}}{\left({\displaystyle \frac{\pi }{sin\frac{\pi }{p{\lambda }_2\rho }}}\right)}^2.$$