A class of Hilbert-type multiple integral inequalities with the kernel of generalized homogeneous function and its applications

Abstract

Let $x=(x_{1},x_{2},\ldots,x_{n})$, and let $K(u(x),v(y))$ satisfy $u(rx)=ru(x)$, $v(ry)=rv(y)$, $K(ru,v)=r^{\lambda\lambda_{1}}K(u, r^{-\frac{\lambda_{1}}{\lambda_{2}}}v)$, and $K(u,rv)=r^{\lambda\lambda_{2}}K(r^{-\frac{\lambda_{2}}{\lambda_{1}}}u, v)$. In this paper, we obtain a necessary and sufficient condition and the best constant factor for the Hilbert-type multiple integral inequality with kernel $K(u(x),v(y))$ and discuss its applications in the theory of operators.

1 Preliminary

Let $n\ge1$, $x=(x_{1},x_{2},\ldots, x_{n})$, $\|x\|_{\rho}=(x_{1}^{\rho}+\cdots +x_{n}^{\rho})^{1/\rho}$, and $\mathbf {R}_{+}^{n}=\{x=(x_{1},\ldots, x_{n}): x_{1}>0, \ldots, x_{n}>0\}$.

Define the function space

$$L^{p}_{\omega(x)} \bigl(\mathbf {R}^{n}_{+} \bigr)= \biggl\{ f(x)\ge0: \Vert f \Vert _{p,\omega(x)}= \biggl( \int _{\mathbf {R}^{n}_{+}}f^{p}(x)\omega(x)\,dx \biggr)^{\frac{1}{p}}< +\infty \biggr\} . $$

Definition 1

Let λ, $\lambda_{1}$, and $\lambda_{2}$ be constants, and let $u(x)$, $v(y)$ and $K(u,v)$ satisfy: for all $r>0$, $u(rx)=ru(x)$, $v(ry)=rv(y)$, and

$$K(ru,v)=r^{\lambda\lambda_{1}}K \bigl(u, r^{-\frac{\lambda_{1}}{\lambda_{2}}}v \bigr),\qquad K(u,rv)=r^{\lambda\lambda_{2}}K \bigl(r^{-\frac{\lambda_{2}}{\lambda_{1}}}u, v \bigr). $$

Then we call $K(u(x), v(y))$ a generalized homogeneous function with parameters $(\lambda,\lambda_{1},\lambda_{2})$. Obviously, $K(u(x), v(y))$ is a homogeneous function of order $\lambda\lambda_{1}$ when $\lambda _{1}=\lambda_{2}$.

If $p>1$ and $\frac{1}{p}+\frac{1}{q}=1$, then we call the inequality

$$ \int_{\mathbf {R}^{n}_{+}} \int_{\mathbf {R}^{n}_{+}}K \bigl(u(x),v(y) \bigr)f(x)g(y)\,dx \,dy \le M \Vert f \Vert _{p,u^{\alpha}(x)} \Vert g \Vert _{q,v^{\beta}(y)} $$

(1.1)

the Hilbert-type multiple integral inequality with $f\in L^{p}_{u^{\alpha}(x)}(\mathbf {R}^{n}_{+})$ and $g\in L^{q}_{v^{\beta}(y)}(\mathbf {R}^{n}_{+})$.

Define the integral operator T with kernel $K(u(x),v(y))$ as follows:

$$ T(f) (y)= \int_{\mathbf {R}^{n}_{+}}K \bigl(u(x),v(y) \bigr)f(x)\,dx,\quad y\in \mathbf {R}^{n}_{+}. $$

(1.2)

If there exists a constant M such that

$$\bigl\Vert T(f) \bigr\Vert _{p,\omega_{2}(y)}\le M \Vert f \Vert _{p,\omega_{1}(x)},\quad f\in L^{p}_{\omega _{1}(x)} \bigl(\mathbf {R}^{n}_{+} \bigr), $$

then T is called a bounded operator from $L^{p}_{\omega_{1}}(\mathbf {R}^{n}_{+})$ to $L^{p}_{\omega_{2}}(\mathbf {R}^{n}_{+})$. If T is a bounded operator from $L^{p}_{\omega_{1}}(\mathbf {R}^{n}_{+})$ to itself, then we call T a bounded operator in $L^{p}_{\omega_{1}}(\mathbf {R}^{n}_{+})$. The operator norm of T is defined as

$$\Vert T \Vert = \inf M=\sup_{f\in L^{p}_{\omega_{1}}(\mathbf {R}^{n}_{+}) } \frac{ \Vert T(f) \Vert _{p,\omega_{2}}}{ \Vert f \Vert _{p,\omega_{1}}}. $$

By (1.2) inequality (1.1) can be rewritten as

$$\int_{\mathbf {R}^{n}_{+}}T(f) (y)g(y)\,dy \le M \Vert f \Vert _{p,u^{\alpha}(x)} \Vert g \Vert _{q,v^{\beta}(y)}. $$

It is not hard to prove that this inequality is equivalent to

$$ \bigl\Vert T(f) \bigr\Vert _{p,v^{\beta(1-p)}(y)}\le M \Vert f \Vert _{p,u^{\alpha}(x)}. $$

(1.3)

In this paper, we discuss a necessary and sufficient condition and the best constant factor for the Hilbert-type multiple integral inequality with the integral kernel of the generalized homogeneous function $K(u(x),v(y))$. Our research is of some theoretical and application value for the research of Hilbert-type inequalities. Further, these results are used to study the boundedness and norm of the operator. Related studies can be found in [1–16].

Lemma 1

Let$p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $n \ge1$, $\lambda>0$, $\lambda _{1}\lambda_{2}>0$, and let a nonnegative measurable function$K(u(x), v(y))$be a generalized homogeneous function with parameters$(\lambda, \lambda_{1}, \lambda_{2})$. Denote

$$\begin{gathered} W_{1}= \int_{\mathbf {R}^{n}_{+}} \bigl[v(t) \bigr]^{-\frac{\beta+n}{q}}K \bigl(1,v(t) \bigr)\,dt,\\ W_{2}= \int_{\mathbf {R}^{n}_{+}} \bigl[u(t) \bigr]^{-\frac{\alpha+n}{p}}K \bigl(u(t),1 \bigr)\,dt.\end{gathered} $$

Then

$$\begin{aligned}& \omega_{1}(x)= \int_{\mathbf {R}^{n}_{+}} \bigl[v(y) \bigr]^{-\frac{\beta +n}{q}}K \bigl(u(x),v(y) \bigr)\,dy= \bigl[u(x) \bigr]^{\lambda\lambda_{1}-\frac{\lambda _{1}}{\lambda_{2}}(\frac{\beta+n}{q}-n)}W_{1}, \\& \omega_{2}(y)= \int_{\mathbf {R}^{n}_{+}} \bigl[u(x) \bigr]^{-\frac{\alpha +n}{p}}K \bigl(u(x),v(y) \bigr)\,dx= \bigl[v(y) \bigr]^{\lambda\lambda_{2}-\frac{\lambda _{2}}{\lambda_{1}}(\frac{\alpha+n}{p}-n)}W_{2}. \end{aligned}$$

Proof

Since $K(u(x), v(y))$ is a generalized homogeneous function with parameters $(\lambda, \lambda_{1}, \lambda_{2})$, we have

$$\begin{aligned} \omega_{1}(x) =& \int_{\mathbf {R}^{n}_{+}}u^{\lambda\lambda_{1}}(x) \bigl[v(y) \bigr]^{-\frac {\beta+n}{q}}K \bigl(1,u^{-\frac{\lambda_{1}}{\lambda_{2}}}(x)v(y) \bigr)\,dy \\ =& \int_{\mathbf {R}^{n}_{+}}u^{\lambda\lambda_{1}}(x) \bigl[v(y) \bigr]^{-\frac{\beta +n}{q}}K \bigl(1,v \bigl(u^{-\frac{\lambda_{1}}{\lambda_{2}}}(x)y \bigr) \bigr)\,dy \\ =&u^{\lambda\lambda_{1}}(x) \int_{\mathbf {R}^{n}_{+}} \bigl[u^{\frac{\lambda _{1}}{\lambda_{2}}}(x)v(t) \bigr]^{-\frac{\beta+n}{q}}K \bigl(1,v(t) \bigr)u^{\frac{n\lambda _{1}}{\lambda_{2}}} (x)\,dt \\ =& \bigl[u(x) \bigr]^{\lambda\lambda_{1}-\frac{\lambda_{1}}{\lambda_{2}}(\frac{\beta +n}{q}-n)} \int_{\mathbf {R}^{n}_{+}} \bigl[v(t) \bigr]^{-\frac{\beta+n}{q}}K \bigl(1,v(t) \bigr)\,dt \\ =& \bigl[u(x) \bigr]^{\lambda\lambda_{1}-\frac{\lambda_{1}}{\lambda_{2}}(\frac{\beta +n}{q}-n)}W_{1}. \end{aligned}$$

By the same method we can obtain $\omega_{2}(y)=[v(y)]^{\lambda\lambda _{2}-\frac{\lambda_{2}}{\lambda_{1}}(\frac{\alpha+n}{p}-n)}W_{2}$. □

Lemma 2

([17])

Let$p_{i}>0$, $a_{i}>0$, $\alpha_{i}>0$ ($i=1,2,\ldots,n$), and let$\psi(u)$be measurable. Then

$$\begin{aligned} & \int\cdots \int_{x_{1}>0,\ldots,x_{n}>0;\sum_{i=1}^{n}(\frac {x_{i}}{a_{i}})^{\alpha_{i}}\le1} \psi \Biggl(\sum_{i=1}^{n} \biggl(\frac{x_{i}}{a_{i}} \biggr)^{\alpha_{i}} \Biggr) x_{1}^{p_{1}-1} \cdots x_{n}^{p_{n}-1} \,dx_{1}\cdots dx_{n} \\ &\quad=\frac{a_{1}^{p_{1}}\cdots a_{n}^{p_{n}}\varGamma(\frac{p_{1}}{\alpha_{1}})\cdots \varGamma(\frac{p_{n}}{\alpha_{n}})}{\alpha_{1}\cdots\alpha_{n}\varGamma(\frac {p_{1}}{\alpha_{1}}+\cdots+\frac{p_{n}}{\alpha_{n}})} \int_{0}^{1} \psi(t)t^{\frac{p_{1}}{\alpha_{1}}+\cdots+\frac{p_{n}}{\alpha _{n}}-1}\,dt, \end{aligned}$$

whereΓis the gamma function.

2 Main results

Theorem 1

Let$p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $n\ge1$, $\rho>0$, $\lambda>0$, $\lambda _{1}\lambda_{2}>0$, let there exist positive constants$C_{1}$and$C_{2}$such that$C_{1}\|x\|_{\rho}\le u(x)\le C_{2}\|x\|_{\rho}$, $C_{1}\|y\|_{\rho}\le v(y)\le C_{2}\|y\|_{\rho}$, let a nonnegative measurable function$K(u(x),v(y))$be a generalized homogeneous function with parameters$(\lambda, \lambda_{1}, \lambda_{2})$, and let the convergent integrals$W_{1}$and$W_{2}$be defined as in Lemma 1. Then we have:

• (i) There exists a constantMsuch that the Hilbert-type multiple integral inequality in (1.1) holds if and only if$\frac{\lambda_{2}\alpha-n\lambda_{1}}{p} +\frac{\lambda_{1}\beta-n\lambda_{2}}{q}=\lambda\lambda_{1}\lambda_{2}$.

• (ii) The best constant factor in (1.1) is$\inf M=W_{1}^{\frac{1}{p}}W_{2}^{\frac{1}{q}}$.

Proof

Let $\varOmega(a< b)=\{x=(x_{1},\ldots, x_{n}):a< \|x\|_{\rho}< b \}$.

(i) Suppose there exists a constant M such that (1.1) holds. Denote $l=\frac{\lambda_{2}\alpha-n\lambda_{1}}{p}+\frac{\lambda_{1}\beta -n\lambda_{2}}{q}-\lambda\lambda_{1}\lambda_{2}$. First, we let $\lambda_{1}>0$, $\lambda_{2}>0$. For $l>0$ and $\varepsilon>0$ sufficiently small, we set

$$\begin{aligned}& f(x)=\left \{ \textstyle\begin{array}{l@{\quad}l} [u(x)]^{(-\alpha-n+\lambda_{1}\varepsilon)/p},& 0< \Vert x \Vert _{\rho}< 1, \\ 0,& \Vert x \Vert _{\rho}\geq1. \end{array}\displaystyle \right . \\& g(y)= \left \{ \textstyle\begin{array}{l@{\quad}l} [v(y)]^{(-\beta-n+\lambda_{2}\varepsilon)/q}, &0< \Vert y \Vert _{\rho}< 1, \\ 0, &\|y\|_{\rho}\geq1. \end{array}\displaystyle \right . \end{aligned}$$

Thus we have

$$ \Vert f \Vert _{p,u^{\alpha}(x)} \Vert g \Vert _{q,v^{\beta}(y)}= \biggl( \int _{\varOmega(0< 1)} \bigl[u(x) \bigr]^{-n+\lambda_{1}\varepsilon}\,dx \biggr)^{\frac {1}{p}} \biggl( \int_{\varOmega(0< 1)} \bigl[v(y) \bigr]^{-n+\lambda_{2}\varepsilon}\, dy \biggr)^{\frac{1}{q}}. $$

(2.1)

In view of $\lambda_{1}>0$, $\lambda_{2}>0$, $C_{1}\|x\|_{\rho}\le u(x)\le C_{2}\|x\| _{\rho}$, $C_{1}\|y\|_{\rho}\le v(y)\le C_{2}\|y\|_{\rho}$, the two integrals in (2.1) are all convergent.

Also, since $-\frac{\lambda_{1}}{\lambda_{2}}<0$ and $(C_{2}\|x\|_{\rho})^{-\frac{\lambda_{1}}{\lambda_{2}}}\le u^{-\frac{\lambda_{1}}{\lambda _{2}}}(x)\le(C_{1}\|x\|_{\rho})^{-\frac{\lambda_{1}}{\lambda_{2}}}$, we have

$$\begin{aligned} & \int_{\mathbf {R}^{n}_{+}} \int_{\mathbf {R}^{n}_{+}}K \bigl(u(x),v(y) \bigr)f(x)g(y)\,dx\, dy \\ &\quad= \int_{\varOmega(0< 1)} \bigl[u(x) \bigr]^{(-\alpha-n+\lambda_{1}\varepsilon)/p} \biggl( \int_{\varOmega(0< 1)}K \bigl(u(x),v(y) \bigr) \bigl[v(y) \bigr]^{(-\beta-n+\lambda_{2}\varepsilon )/q}\,dy \biggr)\,dx \\ &\quad= \int_{\varOmega(0< 1)} \bigl[u(x) \bigr]^{\lambda\lambda_{1}+(-\alpha-n+\lambda _{1}\varepsilon)/p} \biggl( \int_{\varOmega(0< 1)}K \bigl(1,v \bigl(u^{-\frac{\lambda _{1}}{\lambda_{2}}}(x)y \bigr) \bigr) \bigl[v(y) \bigr]^{(-\beta-n+\lambda_{2}\varepsilon)/q}\,dy \biggr)\,dx \\ &\quad= \int_{\varOmega(0< 1)} \bigl[u(x) \bigr]^{\lambda\lambda_{1}+(-\alpha-n+\lambda _{1}\varepsilon)/p} \\ &\qquad{}\times \biggl( \int_{\varOmega(0< u^{-\frac {\lambda_{1}}{\lambda_{2}}}(x))}K \bigl(1,v(t) \bigr) \bigl[u^{\frac{\lambda_{1}}{\lambda_{2}}} (x)v(t) \bigr]^{(-\beta-n+\lambda_{2}\varepsilon)/q}u^{\frac{n\lambda _{1}}{\lambda_{2}}}(x)\,dt \biggr)\,dx \\ &\quad= \int_{\varOmega(0< 1)} \bigl[u(x) \bigr]^{-n+\lambda_{1}\varepsilon-\frac {l}{\lambda_{2}}} \biggl( \int_{\varOmega(0< u^{-\frac{\lambda_{1}}{\lambda _{2}}}(x))} K \bigl(1,v(t) \bigr) \bigl[v(t) \bigr]^{-\frac{\beta+n-\lambda_{2}\varepsilon}{q}} \,dt \biggr)\,dx \\ &\quad\ge \int_{\varOmega(0< 1)} \bigl[u(x) \bigr]^{-n+\lambda_{1}\varepsilon-\frac {l}{\lambda_{2}}} \biggl( \int_{\varOmega(0< (C_{2}\|x\|_{\rho})^{-\frac{\lambda _{1}}{\lambda_{2}}})} K \bigl(1,v(t) \bigr) \bigl[v(t) \bigr]^{-\frac{\beta+n-\lambda_{2}\varepsilon }{q}} \,dt \biggr)\,dx \\ &\quad\ge \int_{\varOmega(0< 1)} \bigl[u(x) \bigr]^{-n+\lambda_{1}\varepsilon-\frac{l}{\lambda_{2}}}\,dx \int_{\varOmega (0< C_{2}^{-\frac{\lambda_{1}}{\lambda_{2}}})} K \bigl(1,v(t) \bigr) \bigl[v(t) \bigr]^{-\frac{\beta +n-\lambda_{2}\varepsilon}{q}} \,dt. \end{aligned}$$

Combining this with (1.1) and (2.1), we get

$$\begin{aligned} & \int_{\varOmega(0< 1)} \bigl[u(x) \bigr]^{-n+\lambda_{1}\varepsilon-\frac{l}{\lambda _{2}}}\,dx \int_{\varOmega(0< C_{2}^{-\frac{\lambda_{1}}{\lambda _{2}}})}K \bigl(1,v(t) \bigr) \bigl[v(t) \bigr]^{-\frac{\beta+n-\lambda_{2}\varepsilon}{q}} \, dt \\ &\quad\le M \biggl( \int_{\varOmega(0< 1)} \bigl[u(x) \bigr]^{-n+\lambda_{1}\varepsilon}\, dx \biggr)^{\frac{1}{p}} \biggl( \int_{\varOmega(0< 1)} \bigl[v(y) \bigr]^{-n+\lambda _{2}\varepsilon}\,dy \biggr)^{\frac{1}{q}}. \end{aligned}$$

(2.2)

Since $l>0$ and ε is sufficiently small, $-n+\lambda _{1}\varepsilon-\frac{l}{\lambda_{2}}<-n$, and additionally $C_{1}\|x\|_{\rho}\le u(x)\le C_{2}\|x\|_{\rho}$, then $\int_{\varOmega(0<1)}[u(x)]^{-n+\lambda_{1}\varepsilon-\frac{l}{\lambda _{2}}}\,dx=+\infty$. So (2.2) is a contradiction to $l>0$.

If $l<0$, let $\varepsilon>0$ be sufficient small. Then we set

$$\begin{aligned}& f(x)=\left \{ \textstyle\begin{array}{l@{\quad}l} [u(x)]^{(-\alpha-n-\lambda_{1}\varepsilon)/p}, & \Vert x \Vert _{\rho}>1,\\ 0 &\mbox{otherwise}, \end{array}\displaystyle \right . \\& g(y)=\left \{ \textstyle\begin{array}{l@{\quad}l} [v(y)]^{(-\beta-n-\lambda_{2}\varepsilon)/q}, &\Vert y \Vert _{\rho}>1,\\ 0 &\mbox{otherwise}. \end{array}\displaystyle \right . \end{aligned}$$

Similarly, we can get

$$\begin{aligned} & \int_{\varOmega(1< +\infty)} \bigl[v(y) \bigr]^{-n-\lambda_{2}\varepsilon-\frac {l}{\lambda_{1}}}\,dy \int_{\varOmega(C_{1}^{-\frac{\lambda_{2}}{\lambda _{1}}}< +\infty)}K \bigl(u(t),1 \bigr) \bigl[u(t) \bigr]^{-\frac{\alpha+\beta+\lambda_{1}\varepsilon }{p}} \,dt \\ &\quad\le M \biggl( \int_{\varOmega(1< +\infty)} \bigl[u(x) \bigr]^{-n-\lambda_{1}\varepsilon }\,dx \biggr)^{\frac{1}{p}} \biggl( \int_{\varOmega(1< +\infty )} \bigl[v(y) \bigr]^{-n-\lambda_{2}\varepsilon}\,dy \biggr)^{\frac{1}{q}}. \end{aligned}$$

(2.3)

Since $C_{1}\|x\|_{\rho}\le u(x)\le C_{2}\|x\|_{\rho}$, $C_{1}\|y\|_{\rho}\le v(y)\le C_{2}\|y\|_{\rho}$, $l<0$, $\lambda_{1}>0$, $\lambda_{2}>0$, and $\varepsilon>0$ is sufficient small, the right-hand side of (2.3) converges; also, $\int_{\varOmega(1<+\infty)}[v(y)]^{-n-\lambda_{2}\varepsilon-\frac {l}{\lambda_{1}}}\,dy$ diverges, and thus (2.3) is a contradiction to $l<0$.

In conclusion, when $\lambda_{1}>0$, $\lambda_{2}>0$, then we have $l=0$, that is, $\frac{\lambda_{2}\alpha-n\lambda_{1}}{p}+\frac{\lambda_{1}\beta -n\lambda_{2}}{q}=\lambda\lambda_{1}\lambda_{2}$.

Again, suppose $\lambda_{1}<0$, $\lambda_{2}<0$. If $l>0$, then taking $\varepsilon>0$ sufficiently small, we set

$$\begin{aligned}& f(x)=\left \{ \textstyle\begin{array}{l@{\quad}l} [u(x)]^{(-\alpha-n+\lambda_{1}\varepsilon)/p}, &\Vert x \Vert _{\rho}>1,\\ 0 &\mbox{otherwise}, \end{array}\displaystyle \right . \\& g(y)=\left \{ \textstyle\begin{array}{l@{\quad}l} [v(y)]^{(-\beta-n+\lambda_{2}\varepsilon)/q}, & \Vert y \Vert _{\rho}>1,\\ 0 &\mbox{otherwise}. \end{array}\displaystyle \right . \end{aligned}$$

We thus have

$$\begin{aligned} \Vert f \Vert _{p,u^{\alpha}(x)} \Vert g \Vert _{q,v^{\beta}(y)}= \biggl( \int_{\varOmega (1< +\infty)} \bigl[u(x) \bigr]^{-n+\lambda_{1}\varepsilon}\,dx \biggr)^{\frac{1}{p}} \biggl( \int_{\varOmega(1< +\infty)} \bigl[v(y) \bigr]^{-n+\lambda_{2}\varepsilon}\,dy \biggr)^{\frac{1}{q}} . \end{aligned}$$

(2.4)

Meanwhile, using $C_{1}\|x\|_{\rho}\le u(x)\le C_{2}\|x\|_{\rho}$, $(C_{2}\|x\| _{\rho})^{-\frac{\lambda_{1}}{\lambda_{2}}}\le u^{-\frac{\lambda_{1}}{\lambda _{2}}}\le(C_{1}\|x\|_{\rho})^{-\frac{\lambda_{1}}{\lambda_{2}}}$, we have

$$\begin{aligned} & \int_{R_{+}^{n}} \int_{R_{+}^{n}}K \bigl(u(x),v(y) \bigr)f(x)g(y)\,dx\,dy \\ &\quad= \int_{\varOmega(1< +\infty)} \bigl[u(x) \bigr]^{(-\alpha-n+\lambda_{1}\varepsilon )/p} \biggl( \int_{\varOmega(1< +\infty)}K \bigl(u(x),v(y) \bigr) \bigl[v(y) \bigr]^{(-\beta -n+\lambda_{2}\varepsilon)/q}\,dy \biggr)\,dx \\ &\quad= \int_{\varOmega(1< +\infty)} \bigl[u(x) \bigr]^{\lambda\lambda_{1}+(-\alpha -n+\lambda_{1}\varepsilon)/p} \\ &\qquad{}\times \biggl( \int_{\varOmega(1< +\infty )}K \bigl(1,v \bigl(u^{-\frac{\lambda_{1}}{\lambda_{2}}}(x)y \bigr) \bigr) \bigl[v(y) \bigr]^{(-\beta-n+\lambda _{2}\varepsilon)/q}\,dy \biggr)\,dx \\ &\quad= \int_{\varOmega(1< +\infty)} \bigl[u(x) \bigr]^{\lambda\lambda_{1}-\frac{\alpha +n-\lambda_{1}\varepsilon}{p}} \\ &\qquad{}\times\biggl( \int_{\varOmega(u^{-\frac{\lambda _{1}}{\lambda_{2}}}(x)< +\infty)}K \bigl(1,v(t) \bigr) \bigl[u^{\frac{\lambda_{1}}{\lambda_{2}}}(x) v(t) \bigr]^{-\frac{\beta+n-\lambda_{2}\varepsilon}{q}}u^{\frac{n\lambda _{1}}{\lambda_{2}}}(x)\,dt \biggr)\,dx \\ &\quad\ge \int_{\varOmega(1< +\infty)} \bigl[u(x) \bigr]^{-n+\lambda_{1}\varepsilon-\frac {l}{\lambda_{2}}} \biggl( \int_{\varOmega((C_{1}\|x\|_{\rho})^{-\frac{\lambda _{1}}{\lambda_{2}}}< +\infty)}K \bigl(1,v(t) \bigr) \bigl[v(t) \bigr]^{-\frac{\beta+n-\lambda _{2}\varepsilon}{q}} \,dt \biggr)\,dx \\ &\quad\ge \int_{\varOmega(1< +\infty)} \bigl[u(x) \bigr]^{-n+\lambda_{1}\varepsilon-\frac {l}{\lambda_{2}}}\,dx \int_{\varOmega(C_{1}^{-\frac{\lambda_{1}}{\lambda _{2}}}< +\infty)}K \bigl(1,v(t) \bigr) \bigl[v(t) \bigr]^{-\frac{\beta+n-\lambda_{2}\varepsilon }{q}} \,dt. \end{aligned}$$

Combining this with (1.1) and (2.4), we obtain

$$\begin{aligned} & \int_{\varOmega(1< +\infty)} \bigl[u(x) \bigr]^{-n+\lambda_{1}\varepsilon-\frac {l}{\lambda_{2}}}\,dx \int_{\varOmega(C_{1}^{-\frac{\lambda_{1}}{\lambda _{2}}}< +\infty)}K \bigl(1,v(t) \bigr) \bigl[v(t) \bigr]^{-\frac{\beta+n-\lambda_{2}\varepsilon }{q}} \,dt \\ &\quad\le M \biggl( \int_{\varOmega(1< +\infty)} \bigl[u(x) \bigr]^{-n+\lambda_{1}\varepsilon} \,dx \biggr)^{\frac{1}{p}} \biggl( \int_{\varOmega(1< +\infty)} \bigl[v(y) \bigr]^{-n+\lambda_{2}\varepsilon}\,dy \biggr)^{\frac{1}{q}}. \end{aligned}$$

(2.5)

Since the two integrals of the right-hand side of (2.5) converge, but the integral

$$\int_{\varOmega(1< +\infty)} \bigl[u(x) \bigr]^{-n+\lambda_{1}\varepsilon-\frac {l}{\lambda_{2}}}\,dx $$

diverges, (2.5) is a contradiction to $l>0$.

If $l<0$ and $\varepsilon>0$ is sufficiently small, then we set

$$\begin{aligned}& f(x)=\left \{ \textstyle\begin{array}{l@{\quad}l} [u(x)]^{(-\alpha-n-\lambda_{1}\varepsilon)/p},& 0< \Vert x \Vert _{\rho}< 1,\\ 0 &\mbox{otherwise}, \end{array}\displaystyle \right . \\& g(y)=\left \{ \textstyle\begin{array}{l@{\quad}l} [v(y)]^{(-\beta-n-\lambda_{2}\varepsilon)/q}, &0< \Vert y \Vert _{\rho}< 1,\\ 0 &\mbox{otherwise}. \end{array}\displaystyle \right . \end{aligned}$$

Similarly, we can get

$$\begin{aligned} & \int_{\varOmega(0< 1)} \bigl[v(y) \bigr]^{-n-\lambda_{2}\varepsilon-\frac{l}{\lambda _{1}}}\,dy \int_{\varOmega(0< C_{2}^{-\frac{\lambda_{2}}{\lambda _{1}}})}K \bigl(u(t),1 \bigr) \bigl[u(t) \bigr]^{-\frac{\alpha+\beta+\lambda_{1}\varepsilon }{p}} \,dt \\ &\quad\le M \biggl( \int_{\varOmega(0< 1)} \bigl[u(x) \bigr]^{-n-\lambda_{1}\varepsilon}\, dx \biggr)^{\frac{1}{p}} \biggl( \int_{\varOmega(0< 1)} \bigl[v(y) \bigr]^{-n-\lambda _{2}\varepsilon}\,dy \biggr)^{\frac{1}{q}}. \end{aligned}$$

(2.6)

We now easily get that both integrals on the right-hand side of (2.6) converge, but

$$\int_{\varOmega(0< 1)} \bigl[v(y) \bigr]^{-n-\lambda_{2}\varepsilon-\frac{l}{\lambda _{1}}}\,dy $$

diverges, and thus (2.6) is a contradiction to $l<0$.

To sum up, when $\lambda_{1}<0$, $\lambda_{2}<0$, we also have $l=0$, that is, $\frac{\lambda_{2}\alpha-n\lambda_{1}}{p}+\frac{\lambda_{1}\beta-n\lambda _{2}}{q}=\lambda\lambda_{1}\lambda_{2}$.

On the contrary, if $\frac{\lambda_{2}\alpha-n\lambda_{1}}{p}+\frac{\lambda _{1}\beta-n\lambda_{2}}{q}=\lambda\lambda_{1}\lambda_{2}$, then let $a=\frac {\alpha}{pq}+\frac{n}{pq}$, $b=\frac{\beta}{pq}+\frac{n}{pq}$. By the Hölder inequality and Lemma 1 we have

$$\begin{aligned} & \int_{R^{n}_{+}} \int_{R^{n}_{+}}K \bigl(u(x),v(y) \bigr)f(x)g(y)\,dx\,dy \\ &\quad= \int_{R^{n}_{+}} \int_{R^{n}_{+}} \biggl[f(x)\frac{u^{a}(x)}{v^{b}(y)} \biggr] \biggl[g(y) \frac {v^{b}(y)}{u^{a}(x)} \biggr]K \bigl(u(x),v(y) \bigr)\,dx\,dy \\ &\quad\le \biggl( \int_{R^{n}_{+}} \int_{R^{n}_{+}}f^{p}(x)\frac {u^{ap}(x)}{v^{bp}(y)}K \bigl(u(x),v(y) \bigr)\,dx\,dy \biggr)^{\frac{1}{p}} \\ &\qquad{}\times \biggl( \int_{R^{n}_{+}} \int_{R^{n}_{+}}g^{q}(y)\frac {v^{bq}(y)}{u^{aq}(x)}K \bigl(u(x),v(y) \bigr)\,dx\,dy \biggr)^{\frac{1}{q}} \\ &\quad= \biggl( \int_{R^{n}_{+}} \bigl[u(x) \bigr]^{\frac{\alpha+n}{q}}f^{p}(x) \omega_{1}(x)\,dx \biggr)^{\frac{1}{p}} \biggl( \int_{R^{n}_{+}} \bigl[v(y) \bigr]^{\frac{\beta+n}{p}}g^{q}(y) \omega _{2}(y)\,dy \biggr)^{\frac{1}{q}} \\ &\quad=W_{1}^{\frac{1}{p}}W_{2}^{\frac{1}{q}} \biggl( \int_{R_{+}^{n}} \bigl[u(x) \bigr]^{\frac {\alpha+n}{q}+\lambda\lambda_{1}-\frac{\lambda_{1}}{\lambda_{2}}(\frac{\beta +n}{q}-n)}f^{p}(x) \,dx \biggr)^{\frac{1}{p}} \\ &\qquad{}\times \biggl( \int_{R_{+}^{n}} \bigl[v(y) \bigr]^{\frac{\beta+n}{p}+\lambda\lambda _{2}-\frac{\lambda_{2}}{\lambda_{1}}(\frac{\alpha+n}{p}-n)}g^{q}(y) \,dy \biggr)^{\frac{1}{q}} \\ &\quad=W_{1}^{\frac{1}{p}}W_{2}^{\frac{1}{q}} \biggl( \int_{R_{+}^{n}}u^{\alpha}(x)f^{p}(x)\,dx \biggr)^{\frac{1}{p}} \biggl( \int_{R_{+}^{n}}v^{\beta}(y)g^{q}(y)\, dy \biggr)^{\frac{1}{q}} \\ &\quad=W_{1}^{\frac{1}{p}}W_{2}^{\frac{1}{q}} \Vert f \Vert _{p,u^{\alpha}(x)} \Vert g \Vert _{q,v^{\beta}(y)}. \end{aligned}$$

Taking arbitrary $M\ge W_{1}^{\frac{1}{p}}W_{2}^{\frac{1}{q}}$, inequality (1.1) holds.

(ii) Suppose inequality (1.1) holds. If $\inf M \neq W_{1}^{\frac {1}{p}}W_{2}^{\frac{1}{q}}$, then there exists a constant $M_{0}< W_{1}^{\frac {1}{p}}W_{2}^{\frac{1}{q}}$ such that

$$\begin{aligned} \int_{R_{+}^{n}} \int_{R_{+}^{n}}K \bigl(u(x),v(y) \bigr)f(x)g(y)\,dx\,dy\le M_{0} \Vert f \Vert _{p,u^{\alpha}(x)} \Vert g \Vert _{q,v^{\beta}(y)} \end{aligned}$$

(2.7)

for all $f\in L^{p}_{u^{\alpha}(x)}(R_{+}^{n})$ and $g\in L^{q}_{v^{\beta}(y)}(R_{+}^{n})$.

Let $\varepsilon>0$ and $\delta>0$ be sufficient small. We take

$$\begin{aligned}& f(x)=\left \{ \textstyle\begin{array}{l@{\quad}l} [u(x)]^{(-\alpha-n- \vert \lambda_{1} \vert \varepsilon)/p}, &\Vert x \Vert _{\rho}>\delta,\\ 0& \mbox{otherwise}, \end{array}\displaystyle \right . \\& g(y)=\left \{ \textstyle\begin{array}{l@{\quad}l} [v(y)]^{(-\beta-n- \vert \lambda_{2} \vert \varepsilon)/q}, &\Vert y \Vert _{\rho}>1,\\ 0& \mbox{otherwise}. \end{array}\displaystyle \right . \end{aligned}$$

Then we have

$$\begin{aligned} \Vert f \Vert _{p,u^{\alpha}(x)} \Vert g \Vert _{q,v^{\beta}(y)}= \biggl( \int_{\varOmega(\delta < +\infty)} \bigl[u(x) \bigr]^{-n- \vert \lambda_{1} \vert \varepsilon}\,dx \biggr)^{\frac{1}{p}} \biggl( \int_{\varOmega(1< +\infty)} \bigl[v(y) \bigr]^{-n- \vert \lambda_{2} \vert \varepsilon}\, dy \biggr)^{\frac{1}{q}}. \end{aligned}$$

(2.8)

Since $\frac{\lambda_{2}\alpha-n\lambda_{1}}{p}+\frac{\lambda_{1}\beta -n\lambda_{2}}{q}=\lambda\lambda_{1}\lambda_{2}$ and $v^{-\frac{\lambda _{2}}{\lambda_{1}}}(y)\le(C_{1}\|y\|_{\rho})^{-\frac{\lambda_{2}}{\lambda_{1}}}$, we have

$$\begin{aligned} & \int_{R_{+}^{n}} \int_{R_{+}^{n}}K \bigl(u(x),v(y) \bigr)f(x)g(y)\,dx\,dy \\ &\quad= \int_{\varOmega(1< +\infty)} \bigl[v(y) \bigr]^{(-\beta-n-|\lambda_{2}|\varepsilon )/q} \biggl( \int_{\varOmega(\delta< +\infty)} K \bigl(u(x),v(y) \bigr) \bigl[u(x) \bigr]^{(-\alpha-n-|\lambda_{1}|\varepsilon)/p}\,dx \biggr)\,dy \\ &\quad= \int_{\varOmega(1< +\infty)} \bigl[v(y) \bigr]^{\lambda\lambda_{2}-\frac{\beta +n+|\lambda_{2}|\varepsilon}{q}} \biggl( \int_{\varOmega(\delta< +\infty)} K \bigl(u \bigl(v^{-\frac{\lambda_{2}}{\lambda_{2}}}(y)x \bigr),1 \bigr) \bigl[u(x) \bigr]^{-\frac{\alpha +n+|\lambda_{1}|\varepsilon}{p}}\,dx \biggr)\,dy \\ &\quad= \int_{\varOmega(1< +\infty)} \bigl[v(y) \bigr]^{\lambda\lambda_{2}-\frac{\beta +n+|\lambda_{2}|\varepsilon}{q}} \\ &\qquad{}\times \biggl( \int_{\varOmega(\delta v^{-\frac{\lambda_{2}}{\lambda _{1}}}(y)< +\infty)} K \bigl(u(t),1 \bigr) \bigl[v^{\frac{\lambda_{2}}{\lambda_{1}}}(y)u(t) \bigr]^{-\frac{\alpha +n+|\lambda_{1}|\varepsilon}{p}}v^{\frac{n\lambda_{2}}{\lambda_{1}}}(y)\, dt \biggr)\,dy \\ &\quad= \int_{\varOmega(1< +\infty)} \bigl[v(y) \bigr]^{-n-|\lambda_{2}|\varepsilon} \biggl( \int_{\varOmega(\delta v^{-\frac{\lambda_{2}}{\lambda_{1}}}(y)< +\infty )}K \bigl(u(t),1 \bigr) \bigl[u(t) \bigr]^{-\frac{\alpha+n+|\lambda_{1}|\varepsilon}{p}} \,dt \biggr)\,dy \\ &\quad\ge \int_{\varOmega(1< +\infty)} \bigl[v(y) \bigr]^{-n-|\lambda_{2}|\varepsilon} \biggl( \int_{\varOmega(\delta(C_{1}\|y\|_{\rho})^{-\frac{\lambda_{2}}{\lambda _{1}}}< +\infty)}K \bigl(u(t),1 \bigr) \bigl[u(t) \bigr]^{-\frac{\alpha+n+|\lambda_{1}|\varepsilon }{p}} \,dt \biggr)\,dy \\ &\quad\ge \int_{\varOmega(1< +\infty)} \bigl[v(y) \bigr]^{-n-|\lambda_{2}|\varepsilon} \int _{\varOmega(\delta C_{1}^{-\frac{\lambda_{2}}{\lambda_{1}}}< +\infty )}K \bigl(u(t),1 \bigr) \bigl[u(t) \bigr]^{-\frac{\alpha+n+|\lambda_{1}|\varepsilon}{p}} \, dt . \end{aligned}$$

Combining this with (2.7) and (2.8), we obtain

$$\begin{aligned} & \int_{\varOmega(1< +\infty)} \bigl[v(y) \bigr]^{-n-|\lambda_{2}|\varepsilon}\,dy \int _{\varOmega(\delta C_{1}^{-\frac{\lambda_{2}}{\lambda_{1}}}< +\infty )}K \bigl(u(t),1 \bigr) \bigl[u(t) \bigr]^{-\frac{\alpha+n+|\lambda_{1}|\varepsilon}{p}} \, dt \\ &\quad\le M_{0} \biggl( \int_{\varOmega(\delta< +\infty)} \bigl[u(x) \bigr]^{-n-|\lambda _{1}|\varepsilon}\,dx \biggr)^{\frac{1}{p}} \biggl( \int_{\varOmega(1< +\infty )} \bigl[v(y) \bigr]^{-n-|\lambda_{2}|\varepsilon}\,dy \biggr)^{\frac{1}{q}}. \end{aligned}$$

Thus

$$\begin{aligned} & \biggl( \int_{\varOmega(1< +\infty)} \bigl[v(y) \bigr]^{-n-|\lambda_{2}|\varepsilon}\, dy \biggr)^{\frac{1}{p}} \int_{\varOmega(\delta C_{1}^{-\frac{\lambda _{2}}{\lambda_{1}}}< +\infty)}K \bigl(u(t),1 \bigr) \bigl[u(t) \bigr]^{-\frac{\alpha+n+|\lambda _{1}|\varepsilon}{p}} \,dt \\ &\quad\le M_{0} \biggl( \int_{\varOmega(\delta< +\infty)} \bigl[u(x) \bigr]^{-n-|\lambda _{1}|\varepsilon}\,dx \biggr)^{\frac{1}{p}}. \end{aligned}$$

(2.9)

We also take

$$\begin{aligned}& f(x)=\left \{ \textstyle\begin{array}{l@{\quad}l} [u(x)]^{(-\alpha-n- \vert \lambda_{1} \vert \varepsilon)/p}, & \Vert x \Vert _{\rho}>1,\\ 0 &\mbox{otherwise}, \end{array}\displaystyle \right . \\& g(y)=\left \{ \textstyle\begin{array}{l@{\quad}l} [v(y)]^{(-\beta-n- \vert \lambda_{2} \vert \varepsilon)/q},& \Vert y \Vert _{\rho}>\delta,\\ 0 &\mbox{otherwise}. \end{array}\displaystyle \right . \end{aligned}$$

Similarly, we can get

$$\begin{aligned} & \biggl( \int_{\varOmega(1< +\infty)} \bigl[u(x) \bigr]^{-n-|\lambda_{1}|\varepsilon}\, dx \biggr)^{\frac{1}{q}} \int_{\varOmega(\delta C_{1}^{-\frac{\lambda _{2}}{\lambda_{1}}}< +\infty)}K \bigl(1,v(t) \bigr) \bigl[v(t) \bigr]^{-\frac{\beta+n+|\lambda _{2}|\varepsilon}{q}} \,dt \\ &\quad\le M_{0} \biggl( \int_{\varOmega(\delta< +\infty)} \bigl[v(y) \bigr]^{-n-|\lambda _{2}|\varepsilon}\,dy \biggr)^{\frac{1}{q}}. \end{aligned}$$

(2.10)

By (2.9) and (2.10) we have

$$\begin{aligned} & \biggl( \int_{\varOmega(\delta C_{1}^{-\frac{\lambda_{1}}{\lambda_{2}}}< +\infty )}K \bigl(1,v(t) \bigr) \bigl[v(t) \bigr]^{-\frac{\beta+n+|\lambda_{2}|\varepsilon}{q}} \,dt \biggr)^{\frac{1}{p}} \\ &\qquad{}\times \biggl( \int_{\varOmega(\delta C_{1}^{-\frac{\lambda_{2}}{\lambda _{1}}}< +\infty)}K \bigl(u(t),1 \bigr) \bigl[u(t) \bigr]^{-\frac{\alpha+n+|\lambda_{1}|\varepsilon }{p}} \,dt \biggr)^{\frac{1}{q}} \\ &\quad\le M_{0} \biggl(\frac{ \int_{\varOmega(\delta< +\infty)}[u(x)]^{-n-|\lambda _{1}|\varepsilon}\,dx}{ \int_{\varOmega(1< +\infty)}[u(x)]^{-n-|\lambda _{1}|\varepsilon}\,dx} \biggr)^{\frac{1}{pq}} \biggl(\frac{ \int_{\varOmega(\delta< +\infty)}[v(y)]^{-n-|\lambda _{2}|\varepsilon}\,dy}{ \int_{\varOmega(1< +\infty)}[v(y)]^{-n-|\lambda _{2}|\varepsilon}\,dy} \biggr)^{\frac{1}{pq}} \\ &\quad=M_{0} \biggl(1+\frac{ \int_{\varOmega(\delta < 1)}[u(x)]^{-n-|\lambda_{1}|\varepsilon}\,dx}{ \int_{\varOmega(1< +\infty )}[u(x)]^{-n-|\lambda_{1}|\varepsilon}\,dx} \biggr)^{\frac{1}{pq}} \biggl(1+\frac{ \int_{\varOmega(\delta< 1)}[v(y)]^{-n-|\lambda _{2}|\varepsilon}\,dy}{ \int_{\varOmega(1< +\infty)}[v(y)]^{-n-|\lambda _{2}|\varepsilon}\,dy} \biggr)^{\frac{1}{pq}}. \end{aligned}$$

(2.11)

Since $C_{1}\|x\|_{\rho}\le u(x)\le C_{2}\|x\|_{\rho}$, $\int_{\varOmega(\delta <1)}[u(x)]^{-n}\,dx$ is a usual integral, but $\int_{\varOmega(1<+\infty )}[u(x)]^{-n}\,dx$ diverges, and thus

$$\lim_{\varepsilon\to0^{+}}\frac{ \int_{\varOmega(\delta < 1)}[u(x)]^{-n-|\lambda_{1}|\varepsilon}\,dx}{ \int_{\varOmega(1< +\infty )}[u(x)]^{-n-|\lambda_{1}|\varepsilon}\,dx}=0. $$

In the same way, we have

$$\lim_{\varepsilon\to0^{+}}\frac{ \int_{\varOmega(\delta < 1)}[v(y)]^{-n-|\lambda_{2}|\varepsilon}\,dy}{ \int_{\varOmega(1< +\infty )}[v(y)]^{-n-|\lambda_{2}|\varepsilon}\,dy}=0. $$

Letting $\varepsilon\to0^{+}$ in (2.11), we get

$$\begin{gathered} \biggl( \int_{\varOmega(\delta C_{1}^{-\frac{\lambda_{1}}{\lambda_{2}}}< +\infty )}K \bigl(1,v(t) \bigr) \bigl[v(t) \bigr]^{-\frac{\beta+n}{q}} \,dt \biggr)^{\frac{1}{p}}\\\quad{}\times \biggl( \int_{\varOmega(\delta C_{1}^{-\frac{\lambda_{2}}{\lambda_{1}}}< +\infty )}K \bigl(u(t),1 \bigr) \bigl[u(t) \bigr]^{-\frac{\alpha+n}{p}} \,dt \biggr)^{\frac{1}{q}} \le M_{0}.\end{gathered} $$

Letting $\delta\to0^{+}$, we obtain

$$\begin{gathered} W_{1}^{\frac{1}{p}}W_{2}^{\frac{1}{q}}= \biggl( \int_{\varOmega(0< +\infty )}K \bigl(1,v(t) \bigr) \bigl[v(t) \bigr]^{-\frac{\beta+n}{q}} \,dt \biggr)^{\frac{1}{p}}\\\quad{}\times \biggl( \int_{\varOmega(0< +\infty)}K \bigl(u(t),1 \bigr) \bigl[u(t) \bigr]^{-\frac{\alpha+n}{p}} \, dt \biggr)^{\frac{1}{q}}\le M_{0}.\end{gathered} $$

This is a contradiction, and hence $\inf M=W_{1}^{\frac{1}{p}}W_{2}^{\frac{1}{q}}$. □

Theorem 2

Let$p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $n\ge1$, $\lambda>0$, $\lambda _{1}\lambda_{2}>0$, $\gamma=(1-p)\beta$, and let there exist constants$C_{1}$and$C_{2}$such that$C_{1}\|x\|_{\rho}\le u(x)\le C_{2}\|x\|_{\rho}$and$C_{1}\|y\|_{\rho}\le v(y)\le C_{2}\|y\|_{\rho}$. Let a nonnegative measurable function$K(u(x), v(y))$be a generalized homogeneous function for parameters$(\lambda, \lambda_{1}, \lambda_{2})$. Let the operatorTbe defined by (1.2), and let$W_{1}$and$W_{2}$defined by Lemma 1be also convergent. Then

• (i) Tis a bounded operator from$L^{p}_{u^{\alpha}(x)}(R^{n}_{+})$to$L^{p}_{v^{\gamma}(y)}(R^{n}_{+})$if and only if$\frac{1}{p}[\lambda_{2}(\alpha +n)-\lambda_{1}(\gamma+n)]=n\lambda_{2}+\lambda\lambda_{1}\lambda_{2}$.

• (ii) IfTis a bounded operator from$L^{p}_{u^{\alpha}(x)}(R^{n}_{+})$to$L^{p}_{v^{\gamma}(y)}(R^{n}_{+})$, then the operator norm ofTis$\|T\|=W_{1}^{\frac{1}{p}}W_{2}^{\frac{1}{q}}$.

Proof

Since $\frac{1}{p}+\frac{1}{q}=1$, $\beta=\frac{\gamma}{1-p}$, $\frac {\lambda_{2}\alpha-n\lambda_{1}}{p}+\frac{\lambda_{1}\beta-n\lambda _{2}}{q}=\lambda\lambda_{1}\lambda_{2}$ leads to $\frac{1}{p} [\lambda_{2}(\alpha+n)-\lambda_{1}(\gamma+n)]=n\lambda_{2}+\lambda\lambda _{1}\lambda_{2}$, and since equality (1.1) is equivalent to (1.3), Theorem 2 holds by Theorem 1. □

3 Applications

Theorem 3

Let$p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $n\ge1$, $\rho>0$, $\lambda >0$, $\lambda_{1}>0$, $\lambda_{2}>0$, $a_{i}>0$, $b_{i}>0$, $\alpha< n(p-1)$, $\beta< n(q-1)$, $u(x)=(\sum_{i=1}^{n} a_{i}x_{i}^{\rho})^{1/\rho}$, and$v(y)=(\sum_{i=1}^{n} b_{i}y_{i}^{\rho})^{1/\rho}$. Then:

• (i) There exists a constantMsuch that

  $$\begin{aligned} \int_{R_{+}^{n}} \int_{R_{+}^{n}}\frac{1}{(u^{\lambda _{1}}(x)+v^{\lambda_{2}}(y))^{\lambda}}f(x)g(y)\,dx\,dy\le M \Vert f \Vert _{p,u^{\alpha}(x)} \Vert g \Vert _{q,v^{\beta}(y)} \end{aligned}$$

  (3.1)

  if and only if$\frac{n\lambda_{1}-\lambda_{2}\alpha}{p}+\frac{n\lambda _{2}-\lambda_{1}\beta}{q}=\lambda\lambda_{1}\lambda_{2}$, where$f\in L^{p}_{u^{\alpha}(x)}(R_{+}^{n})$and$g\in L^{q}_{v^{\beta}(y)}(R_{+}^{n})$.

• (ii) If inequality (3.1) holds, then its best constant factor is

  $$\inf M= \Biggl(\prod_{i=1}^{n} a_{i}^{-\frac{1}{\rho}} \Biggr)^{\frac{1}{q}} \Biggl(\prod _{i=1}^{n} b_{i}^{-\frac{1}{\rho}} \Biggr)^{\frac{1}{p}}\frac{\varGamma^{n}(\frac {1}{\rho})}{\rho^{n-1}\varGamma(\lambda)\varGamma(\frac{n}{\rho})} \varGamma \biggl( \frac{1}{\lambda_{1}} \biggl(\frac{n}{q}-\frac{\alpha}{p} \biggr) \biggr)\varGamma \biggl(\frac{1}{\lambda_{2}} \biggl(\frac{n}{p}- \frac{\beta}{q} \biggr) \biggr). $$

Proof

Set $K(u(x), v(y))=\frac{1}{(u^{\lambda_{1}}(x)+v^{\lambda_{2}}(y))^{\lambda}}$. Then $K(u(x),v(y))$ is a generalized homogeneous function for parameters $(\lambda, -\lambda_{1}, -\lambda_{2})$, and $\frac{n\lambda_{1}-\lambda _{2}\alpha}{p}+\frac{n\lambda_{2}-\lambda_{1}\beta}{q}=\lambda\lambda_{1}\lambda _{2}$ is equivalent to $\frac{(-\lambda_{2})\alpha-n(-\lambda_{1})}{ p}+\frac{(-\lambda_{1})\beta-n(-\lambda_{2})}{q}=\lambda(-\lambda _{1})(-\lambda_{2})$. Further, we have $\lambda-\frac{1}{\lambda_{2}}(\frac {n}{p}-\frac{\beta}{q})=\frac{1}{\lambda_{1}}(\frac{n}{q}-\frac{\alpha}{p})$, and $\frac{n}{p}-\frac{\beta}{q}>0$ and $\frac{n}{q}-\frac{\alpha }{p}>0$ when $\alpha< n(p-1)$ and $\beta< n(q-1)$. By Lemma 1 we have

$$\begin{aligned} W_{1} =& \int_{R_{+}^{n}} \bigl[v(t) \bigr]^{-\frac{\beta+n}{q}}K \bigl(1,v(t) \bigr)\,dt\\ =& \int _{R_{+}^{n}}\frac{1}{[1+(\sum_{i=1}^{n} b_{i}t_{i}^{\rho})^{\lambda_{2}/\rho }]^{\lambda}} \Biggl(\sum _{i=1}^{n} b_{i}t_{i}^{\rho}\Biggr)^{-\frac{\beta+n}{q\rho}}\, dt \\ =&\prod_{i=1}^{n} b_{i}^{-\frac{1}{\rho}} \int_{R_{+}^{n}}\frac{1}{[1+(\sum_{i=1}^{n}x_{i}^{\rho})^{\lambda_{2}/\rho}]^{\lambda}} \Biggl(\sum _{i=1}^{n}x_{i}^{\rho}\Biggr)^{-\frac{\beta+n}{q\rho}}\,dx \\ =&\prod_{i=1}^{n} b_{i}^{-\frac{1}{\rho}}\lim_{r\to+\infty} \int\cdots \int _{x_{i}>0,x_{1}^{\rho}+\cdots+x_{n}^{\rho}\le r^{\rho}}\frac{1}{[1+r^{\lambda _{2}}(\sum_{i=1}^{n}(\frac{x_{i}}{r})^{\rho})^{\lambda_{2}/\rho}]^{\lambda}} \\ & {}\times r^{-\frac{\beta+n}{q}} \Biggl(\prod_{i=1}^{n} \biggl(\frac{x_{i}}{r} \biggr)^{\rho}\Biggr)^{-\frac{\beta+n}{q\rho }}x_{1}^{1-1} \cdots x_{n}^{1-1}\,dx_{1}\cdots dx_{2} \\ =&\prod_{i=1}^{n} b_{i}^{-\frac{1}{\rho}}\lim_{r\to+\infty}r^{-\frac {\beta+n}{q}} \frac{r^{n}\varGamma^{n}(\frac{1}{\rho})}{\rho^{n}\varGamma(\frac {n}{\rho})} \int_{0}^{1}\frac{1}{(1+r^{\lambda_{2}}u^{\lambda_{2}/\rho})^{\lambda}}u^{-\frac {\beta+n}{q\rho}}u^{\frac{n}{\rho}-1}du \\ =&\prod_{i=1}^{n} b_{i}^{-\frac{1}{\rho}}\frac{\varGamma^{n}(\frac{1}{\rho })}{\rho^{n-1}\varGamma(\frac{n}{\rho})\lambda_{2}} \int_{0}^{\infty}\frac {1}{(1+t)^{\lambda}}t^{\frac{1}{\lambda_{2}}(\frac{n}{p}-\frac{\beta }{q})-1} \,dt \\ =&\prod_{i=1}^{n} b_{i}^{-\frac{1}{\rho}}\frac{\varGamma^{n}(\frac{1}{\rho })}{\lambda_{2}\rho^{n-1}\varGamma(\frac{n}{\rho})}B \biggl( \frac{1}{\lambda _{2}} \biggl(\frac{n}{p}-\frac{\beta}{q} \biggr), \lambda-\frac{1}{\lambda_{2}} \biggl(\frac {n}{p}-\frac{\beta}{q} \biggr) \biggr) \\ =&\prod_{i=1}^{n} b_{i}^{-\frac{1}{\rho}}\frac{\varGamma^{n}(\frac{1}{\rho })}{\lambda_{2}\rho^{n-1}\varGamma(\frac{n}{\rho})\varGamma(\lambda )}\varGamma \biggl( \frac{1}{\lambda_{2}} \biggl(\frac{n}{p}-\frac{\beta}{q} \biggr) \biggr)\varGamma \biggl(\frac{1}{\lambda_{1}} \biggl(\frac{n}{q}- \frac{\alpha}{p} \biggr) \biggr). \end{aligned}$$

In the same way, we get

$$\begin{aligned} W_{2}&= \int_{R_{+}^{n}} \bigl[u(x) \bigr]^{-\frac{\alpha+n}{p}}K \bigl(u(t),1 \bigr)\,dt\\&=\prod_{i=1}^{n} a_{i}^{-\frac{1}{\rho}}\frac{\varGamma^{n}(\frac{1}{\rho})}{\lambda _{1}\rho^{n-1}\varGamma(\frac{n}{\rho})\varGamma(\lambda)}\varGamma \biggl( \frac {1}{\lambda_{1}} \biggl(\frac{n}{q}-\frac{\alpha}{p} \biggr) \biggr)\varGamma \biggl(\frac{1}{\lambda _{2}} \biggl(\frac{n}{p}- \frac{\beta}{q} \biggr) \biggr).\end{aligned} $$

Thus

$$\begin{aligned} W_{1}^{\frac{1}{p}}W_{2}^{\frac{1}{q}}&= \Biggl( \prod_{i=1}^{n}a_{i}^{-\frac{1}{\rho }} \Biggr)^{\frac{1}{q}} \Biggl(\prod_{i=1}^{n}b_{i}^{-\frac{1}{\rho}} \Biggr)^{\frac{1}{p}} \frac{\varGamma^{n}(\frac{1}{\rho})}{\lambda_{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}\rho^{n-1}\varGamma(\lambda)\varGamma(\frac{n}{\rho })}\\&\quad\times\varGamma \biggl( \frac{1}{\lambda_{1}} \biggl(\frac{n}{q}-\frac{\alpha}{p} \biggr) \biggr)\varGamma \biggl(\frac{1}{\lambda_{2}} \biggl(\frac{n}{p}- \frac{\beta}{q} \biggr) \biggr).\end{aligned} $$

Hence Theorem 3 holds by Theorem 1. □

Corollary 1

Let$p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $n\ge1$, $\rho>0$, $\lambda>0$, $\lambda_{1}>0$, $\lambda_{2}>0$, $u(x)=(\sum_{i=1}^{n} x_{i}^{\rho})^{\frac{1}{\rho }}$, and$v(y)=(\sum_{i=1}^{n} y_{i}^{\rho})^{\frac{1}{\rho}}$. Then:

• (i) The operatorTdefined by

  $$T(f) (y)= \int_{R_{+}^{n}}\frac{1}{(u^{\lambda_{1}}(x)+v^{\lambda_{2}}(y))^{\lambda}}f(x)\,dx,\quad y\in R_{+}^{n}, $$

  is a bounded operator in$L^{p}(R_{+}^{n})$if and only if$\frac{n\lambda _{1}}{p}+\frac{n\lambda_{2}}{q}=\lambda\lambda_{1}\lambda_{2}$.

• (ii) WhenTis a bounded operator in$L^{p}(R_{+}^{n})$, the operator norm ofTis

  $$\Vert T \Vert =\frac{\varGamma^{n}(\frac{1}{\rho})}{\rho^{n-1}\lambda_{1}^{\frac {1}{q}}\lambda_{2}^{\frac{1}{p}}\varGamma(\lambda)\varGamma(\frac{n}{\rho })}\varGamma \biggl(\frac{n}{\lambda_{1}q} \biggr)\varGamma \biggl(\frac{n}{\lambda_{2}p} \biggr). $$

Proof

The corollary follows from Theorems 2 and 3. □

Theorem 4

Let$p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $n\ge1$, $\rho>0$, $\lambda >0$, $\lambda_{1}>0$, $\lambda_{2}>0$, $\alpha< n(p-1)$, $\beta< n(q-1)$, $u(x)=(\sum_{i=1}^{n} x_{i}^{\rho})^{1/\rho}$, and$v(y)=(\sum_{i=1}^{n} y_{i}^{\rho})^{1/\rho}$. Then

• (i) There existsMsuch that

  $$\begin{aligned} \int_{R_{+}^{n}} \int_{R_{+}^{n}}\frac{1}{(\max\{u^{\lambda _{1}}(x),v^{\lambda_{2}}(y)\})^{\lambda}}f(x)g(y)\,dx\,dy\le M \Vert f \Vert _{p,u^{\alpha}(x)} \Vert g \Vert _{q,v^{\beta}(y)} \end{aligned}$$

  (3.2)

  if and only if$\frac{n\lambda_{1}-\lambda_{2}\alpha}{p}+\frac{n\lambda _{2}-\lambda_{1}\beta}{q}=\lambda\lambda_{1}\lambda_{2}$, where$f\in L^{p}_{u^{\alpha}(x)}(R_{+}^{n})$and$g\in L^{q}_{v^{\beta}(y)}(R_{+}^{n})$.

• (ii) If inequality (3.2) holds, then its best constant factor is

  $$\inf M=\frac{\varGamma^{n}(\frac{1}{\rho})}{\lambda_{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}\rho^{n-1}\varGamma(\frac{n}{\rho})} \biggl[ \biggl(\frac{1}{\lambda _{1}} \biggl( \frac{n}{q}-\frac{\alpha}{p} \biggr) \biggr)^{-1} + \biggl(\frac{1}{\lambda_{2}} \biggl(\frac{n}{p}-\frac{\beta}{q} \biggr) \biggr)^{-1} \biggr]. $$

Proof

Set $K(u(x), v(y))=\frac{1}{(\max\{u^{\lambda_{1}}(x),v^{\lambda_{2}}(y)\} )^{\lambda}}$. Then $K(u(x),v(y))$ is a generalized homogeneous function for parameters $(\lambda, -\lambda_{1}, -\lambda_{2})$. By Lemma 2 we get

$$\begin{aligned} W_{1} =& \int_{R_{+}^{n}}K \bigl(1,v(t) \bigr) \bigl[v(t) \bigr]^{-\frac{\beta+n}{q}} \,dt \\ =& \int_{v(t)\le1} \bigl[v(t) \bigr]^{-\frac{\beta+n}{q}}\,dt+ \int _{v(t)>1} \bigl[v(t) \bigr]^{-\lambda\lambda_{2}-\frac{\beta+n}{q}}\,dt \\ =&\frac{\varGamma^{n}(\frac{1}{\rho})}{\lambda_{2}\rho^{n-1}\varGamma(\frac {n}{\rho})} \biggl(\frac{1}{\lambda_{2}} \biggl( \frac{n}{p}- \frac{\beta}{q} \biggr) \biggr)^{-1}+ \frac{\varGamma^{n}(\frac{1}{\rho})}{\lambda_{2}\rho^{n-1}\varGamma(\frac {n}{\rho})} \biggl(\frac{1}{\lambda_{1}} \biggl(\frac{n}{q}- \frac{\alpha}{p} \biggr) \biggr)^{-1} \\ =&\frac{\varGamma^{n}(\frac{1}{\rho})}{\lambda_{2}\rho^{n-1}\varGamma(\frac {n}{\rho})} \biggl[ \biggl(\frac{1}{\lambda_{2}} \biggl( \frac{n}{p}-\frac{\beta }{q} \biggr) \biggr)^{-1}+ \biggl(\frac{1}{\lambda_{1}} \biggl(\frac{n}{q}-\frac{\alpha}{p} \biggr) \biggr)^{-1} \biggr]. \end{aligned}$$

Similarly, we obtain

$$\begin{aligned} W_{2} =& \int_{R_{+}^{n}}K \bigl(u(t),1 \bigr) \bigl[u(t) \bigr]^{-\frac{\alpha+n}{p}} \,dt \\ =&\frac{\varGamma^{n}(\frac{1}{\rho})}{\lambda_{1}\rho^{n-1}\varGamma(\frac {n}{\rho})} \biggl[ \biggl(\frac{1}{\lambda_{1}} \biggl( \frac{n}{q}-\frac{\alpha }{p} \biggr) \biggr)^{-1}+ \biggl(\frac{1}{\lambda_{2}} \biggl(\frac{n}{p}-\frac{\beta}{q} \biggr) \biggr)^{-1} \biggr]. \end{aligned}$$

Then we have

$$W_{1}^{\frac{1}{p}}W_{2}^{\frac{1}{q}}= \frac{\varGamma^{n}(\frac{1}{\rho })}{\lambda_{1}^{\frac{1}{q}}\lambda_{2}^{\frac{1}{p}}\rho^{n-1}\varGamma (\frac{n}{\rho})} \biggl[ \biggl(\frac{1}{\lambda_{1}} \biggl( \frac{n}{q}-\frac{\alpha}{p} \biggr) \biggr)^{-1}+ \biggl(\frac{1}{\lambda_{2}} \biggl(\frac {n}{p}-\frac{\beta}{q} \biggr) \biggr)^{-1} \biggr]. $$

In summary, Theorem 4 holds by Theorem 1. □

Corollary 2

Let$p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $n\ge1$, $\rho>0$, $\lambda>0$, $\lambda _{1}>0$, $\lambda_{2}>0$, $u(x)=(\sum_{i=1}^{n} x_{i}^{\rho})^{\frac{1}{\rho}}$, and$v(y)=(\sum_{i=1}^{n} y_{i}^{\rho})^{\frac{1}{\rho}}$. Then

• (i) The operatorTdefined by

  $$T(f) (y)= \int_{R_{+}^{n}}\frac{1}{\max\{u^{\lambda_{1}}(x),v^{\lambda_{2}}(y)\} )^{\lambda}}f(x)\,dx, y\in R_{+}^{n}, $$

  is a bounded operator in$L^{p}(R_{+}^{n})$if and only if$\frac{n\lambda _{1}}{p}+\frac{n\lambda_{2}}{q}=\lambda\lambda_{1}\lambda_{2}$.

• (ii) WhenTis a bounded operator in$L^{p}(R_{+}^{n})$, the operator norm ofTis

  $$\Vert T \Vert =\frac{\varGamma^{n}(\frac{1}{\rho})}{\lambda_{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}\rho^{n-1}\varGamma(\frac{n}{\rho})} \biggl(\frac{\lambda_{1} q}{n}+ \frac{\lambda_{2} p}{n} \biggr). $$

Proof

The corollary follows from Theorems 2 and 4. □