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:

  1. (i)

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

  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

  1. (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\(\frac{1}{p}[\lambda_{2}(\alpha +n)-\lambda_{1}(\gamma+n)]=n\lambda_{2}+\lambda\lambda_{1}\lambda_{2}\).

  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:

  1. (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\(g\in L^{q}_{v^{\beta}(y)}(R_{+}^{n})\).

  2. (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:

  1. (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\(\frac{n\lambda _{1}}{p}+\frac{n\lambda_{2}}{q}=\lambda\lambda_{1}\lambda_{2}\).

  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

  1. (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\(g\in L^{q}_{v^{\beta}(y)}(R_{+}^{n})\).

  2. (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

  1. (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\(\frac{n\lambda _{1}}{p}+\frac{n\lambda_{2}}{q}=\lambda\lambda_{1}\lambda_{2}\).

  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. □