A multiple Hilbert-type integral inequality with a non-homogeneous kernel

Abstract

By using the way of weight functions and the technic of real analysis, a multiple Hilbert-type integral inequality with a non-homogeneous kernel is given. The operator expression with the norm, the reverses and some examples with the particular kernels are also considered.

MSC:47A07, 26D15.

1 Introduction

If $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $f(\ge 0)\in L^p(0,\mathrm{\infty })$, $g(\ge 0)\in L^q(0,\mathrm{\infty })$, ${\parallel f\parallel }_p,{\parallel g\parallel }_q>0$, then we have the following equivalent inequalities (cf. [1]):

(1)

(2)

where the constant factor $\frac{\pi }{sin(\pi /p)}$ is the best possible. (1) is the well-known Hardy-Hilbert integral inequality. Define the Hardy-Hilbert integral operator $T:L^p(0,\mathrm{\infty })\to L^p(0,\mathrm{\infty })$ as follows: for $f\in L^p(0,\mathrm{\infty })$, $Tf(y):={\int }_0^{\mathrm{\infty }}\frac{1}{x+y}f(x)dx$ ($y\in (0,\mathrm{\infty })$).

Then in view of (2), it follows ${\parallel Tf\parallel }_p<\frac{\pi }{sin(\pi /p)}{\parallel f\parallel }_p$ and $\parallel T\parallel \le \frac{\pi }{sin(\pi /p)}$. Since the constant is the best possible, we find $\parallel T\parallel =\frac{\pi }{sin(\pi /p)}$.

Inequalities (1) and (2) and the operator are important in analysis and its applications (cf. [2, 3]). In 2002, [4] considered the property of the Hardy-Hilbert integral operator and gave an improvement of (1) (for $p=q=2$). In 2004, by introducing another pair of conjugate exponents $(r,s)$ ($r>1$, $\frac{1}{r}+\frac{1}{s}=1$) and an independent parameter $\lambda >0$, [5] gave the best extensions of (1) as follows:

$${\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{f(x)g(y)}{x^{\lambda }+y^{\lambda }}dxdy<\frac{\pi }{\lambda sin(\pi /r)}{\parallel f\parallel }_{p,\varphi }{\parallel g\parallel }_{q,\psi },$$

(3)

where $\varphi (x)=x^{p(1-\frac{\lambda }{r})-1}$, $\psi (x)=x^{q(1-\frac{\lambda }{s})-1}$, ${\parallel f\parallel }_{p,\varphi }={\{{\int }_0^{\mathrm{\infty }}\varphi (x)f^p(x)dx\}}^{\frac{1}{p}}>0$, ${\parallel g\parallel }_{q,\psi }>0$. In 2007, [6] gave the following inequality with the best constant $B(\frac{\lambda }{2},\frac{\lambda }{2})$ ($\lambda >0$; $B(u,v)$ is the beta function):

$${\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{f(x)g(y)}{{(1+xy)}^{\lambda }}dxdy<B(\frac{\lambda }{2},\frac{\lambda }{2}){\{{\int }_0^{\mathrm{\infty }}{x}^{1-\lambda }{f}^2(x)dx{\int }_0^{\mathrm{\infty }}{x}^{1-\lambda }{g}^2(x)dx\}}^{\frac{1}{2}}.$$

(4)

In 2009, [7] gave an extension of (4) in ${\mathbf{R}}^2$ with the kernel $\frac{1}{{|1+xy|}^{\lambda }}$ ($0<\lambda <1$); [8] gave another extension of (4) to the general kernel $k_{\lambda }(1,xy)$ ($\lambda >0$) with a pair of conjugate exponents $(p,q)$ and obtained the following multiple Hilbert-type integral inequality. Suppose that $n\in \mathbf{N}\mathrm{\setminus }\{1\}$, $p_i>1$, ${\sum }_{i=1}^n\frac{1}{p_i}=1$, $\lambda >0$, $k_{\lambda }(x_1,\dots ,x_n)\ge 0$ is a measurable function of −λ-degree in ${\mathbf{R}}_{+}^n$, and for any $(r_1,\dots ,r_n)$ ($r_i>1$), satisfies ${\sum }_{i=1}^n\frac{1}{r_i}=1$ and

$$k_{\lambda }={\int }_{{\mathbf{R}}_{+}^{n-1}}{k}_{\lambda }(u_1,\dots ,u_{n-1},1)\prod _{j=1}^{n-1}{u}_j^{\frac{\lambda }{r_j}-1}d{u}_1\cdots d{u}_{n-1}>0.$$

If ${\varphi }_i(x)=x^{p_i(1-\frac{\lambda }{r_i})-1}$, $f_i(\ge 0)\in L_{{\varphi }_i}^{p_i}(0,\mathrm{\infty })$, ${\parallel f\parallel }_{p_{i,{\varphi }_i}}>0$ ($i=1,\dots ,n$), then we have the following inequality:

$${\int }_{{\mathbf{R}}_{+}^n}{k}_{\lambda }(x_1,\dots ,x_n)\prod _{i=1}^n{f}_i(x_i)d{x}_1\cdots d{x}_n<k_{\lambda }\prod _{i=1}^n{\parallel f_i\parallel }_{p_i,{\varphi }_i},$$

(5)

where the constant factor $k_{\lambda }$ is the best possible. For $n=2$, $k_{\lambda }(x,y)=\frac{1}{x^{\lambda }+y^{\lambda }}$ in (5), we obtain (3). Inequality (5) is an extension of the results in [9–12] and [13]. In recent years, [14] and [15] considered some Hilbert-type operators relating (1)-(3); [16] also considered a multiple Hilbert-type integral operator with the homogeneous kernel of $-n+1$-degree and the relating particular case of (5) (for $\lambda =n-1$, $\frac{1}{r_i}=\frac{1}{n-1}(1-\frac{1}{p_i})$).

In this paper, by using the way of weight functions and the technic of real analysis, a multiple Hilbert-type integral inequality with a non-homogeneous kernel is given. The operator expression with the norm, the reverses and some examples with the particular kernels are considered.

2 Some lemmas

Lemma 1 If $n\in \mathbf{N}\mathrm{\setminus }\{1\}$, ${\lambda }_i\in \mathbf{R}$ ($i=1,\dots ,n$), ${\sum }_{i=1}^n\frac{1}{p_i}=1$, then we have

$$A:=\prod _{i=1}^n{\left[x_i^{({\lambda }_i-1)(1-p_i)}\prod _{j=1(j\ne i)}^n{x}_j^{{\lambda }_j-1}\right]}^{\frac{1}{p_i}}=1.$$

(6)

Proof

We find

$$\begin{array}{rcl}A& :=& \prod _{i=1}^n{\left[x_i^{({\lambda }_i-1)(1-p_i)+1-{\lambda }_i}\prod _{j=1}^n{x}_j^{{\lambda }_j-1}\right]}^{\frac{1}{p_i}}\\ =& \prod _{i=1}^n{\left[x_i^{(1-{\lambda }_i)p_i}\prod _{j=1}^n{x}_j^{{\lambda }_j-1}\right]}^{\frac{1}{p_i}}=\prod _{i=1}^n{x}_i^{1-{\lambda }_i}{\left(\prod _{j=1}^n{x}_j^{{\lambda }_j-1}\right)}^{{\sum }_{i=1}^n\frac{1}{p_i}},\end{array}$$

and then (6) is valid. □

Definition 1 If $n\in \mathbf{N}$, ${\mathbf{R}}_{+}^n:=\{(x_1,\dots ,x_n)|x_i>0(i=1,\dots ,n)\}$, $\lambda \in \mathbf{R}$, $k_{\lambda }(x_1,\dots ,x_n)$ is a measurable function in ${\mathbf{R}}_{+}^n$ such that for any $u>0$ and $(x_1,\dots ,x_n)\in {\mathbf{R}}_{+}^n$, $k_{\lambda }(u{x}_1,\dots ,u{x}_n)=u^{-\lambda }{k}_{\lambda }(x_1,\dots ,x_n)$, then we call $k_{\lambda }(x_1,\dots ,x_n)$ the homogeneous function of −λ-degree in ${\mathbf{R}}_{+}^n$.

Lemma 2 Suppose $n\in \mathbf{N}\mathrm{\setminus }\{1\}$, ${\lambda }_i\in \mathbf{R}$ ($i=1,\dots ,n$), ${\lambda }_n={\sum }_{i=1}^{n-1}{\lambda }_i=\frac{\lambda }{2}$, $k_{\lambda }(x_1,\dots ,x_n)\ge 0$ is a homogeneous function of −λ-degree. If

$$\begin{array}{rcl}H(i)& :=& {\int }_{{\mathbf{R}}_{+}^{n-1}}{k}_{\lambda }(u_1,\dots ,u_{i-1},1,u_{i+1},\dots ,u_n)\\ \times \prod _{j=1(j\ne i)}^n{u}_j^{{\lambda }_j-1}d{u}_1\cdots d{u}_{i-1}d{u}_{i+1}\cdots d{u}_n(i=1,\dots ,n)\end{array}$$

satisfying $k_{\lambda }:=H(n)\in \mathbf{R}$, then each $H(i)=H(n)=k_{\lambda }$ and for any $i=1,\dots ,n$,

$$\begin{array}{rcl}{\omega }_i(x_i)& :=& x_i^{{\lambda }_i}{\int }_{{\mathbf{R}}_{+}^{n-1}}{k}_{\lambda }(x_1{x}_n,\dots ,x_{n-1}{x}_n,1)\\ \times \prod _{j=1(j\ne i)}^n{x}_j^{{\lambda }_j-1}d{x}_1\cdots d{x}_{i-1}d{x}_{i+1}\cdots d{x}_n=k_{\lambda }(\in \mathbf{R}).\end{array}$$

(7)

Proof Setting $u_j=u_n{v}_j$ ($j\ne i,n$) in the integral $H(i)$, we find

$$\begin{array}{rcl}H(i)& =& {\int }_{{\mathbf{R}}_{+}^{n-1}}{k}_{\lambda }(v_1,\dots ,v_{i-1},u_n^{-1},v_{i+1},\dots ,v_{n-1},1)\prod _{j=1(j\ne i)}^{n-1}{v}_j^{{\lambda }_j-1}\\ \times u_n^{-1-{\lambda }_i}d{v}_1\cdots d{v}_{i-1}d{v}_{i+1}\cdots d{v}_{n-1}d{u}_n.\end{array}$$

Setting $v_i=u_n^{-1}$ in the above integral, we obtain $H(i)=H(n)$. Setting $x_n^{\mathrm{\prime }}=x_n^{-1}$ in (7), since $\lambda -{\lambda }_n={\lambda }_n$, we find

$$\begin{array}{rcl}{\omega }_i(x_i)& =& x_i^{{\lambda }_i}{\int }_{{\mathbf{R}}_{+}^{n-1}}{k}_{\lambda }(x_1,\dots ,x_{n-1},x_n^{-1})x_n^{-{\lambda }_n-1}\\ \times \prod _{j=1(j\ne i)}^{n-1}{x}_j^{{\lambda }_j-1}d{x}_1\cdots d{x}_{i-1}d{x}_{i+1}\cdots d{x}_n\\ =& x_i^{{\lambda }_i}{\int }_{{\mathbf{R}}_{+}^{n-1}}{k}_{\lambda }(x_1,\dots ,x_{n-1},x_n^{\mathrm{\prime }}){\left(x_n^{\mathrm{\prime }}\right)}^{{\lambda }_n+1}\\ \times \prod _{j=1(j\ne i)}^{n-1}{x}_j^{{\lambda }_j-1}d{x}_1\cdots d{x}_{i-1}d{x}_{i+1}\cdots d{x}_{n-1}{\left(x_n^{\mathrm{\prime }}\right)}^{-2}d{x}_n^{\mathrm{\prime }}\\ =& x_i^{{\lambda }_i}{\int }_{{\mathbf{R}}_{+}^{n-1}}{k}_{\lambda }(x_1,\dots ,x_{n-1},x_n^{\mathrm{\prime }}){\left(x_n^{\mathrm{\prime }}\right)}^{{\lambda }_n-1}\\ \times \prod _{j=1(j\ne i)}^{n-1}{x}_j^{{\lambda }_j-1}d{x}_1\cdots d{x}_{i-1}d{x}_{i+1}\cdots d{x}_{n-1}d{x}_n^{\mathrm{\prime }}.\end{array}$$

Setting $u_j=x_j/x_i$ ($j\ne i,n$) and $u_n=x_n^{\mathrm{\prime }}/x_i$ in the above integral, we find ${\omega }_i(x_i)=H(i)=H(n)=k_{\lambda }$. □

Lemma 3 With the assumptions given in Lemma 2, then

$$k({\tilde{\lambda }}_1,\dots ,{\tilde{\lambda }}_{n-1}):={\int }_{{\mathbf{R}}_{+}^{n-1}}{k}_{\lambda }(u_1,\dots ,u_{n-1},1)\prod _{j=1}^{n-1}{u}_j^{{\tilde{\lambda }}_j-1}d{u}_1\cdots d{u}_{n-1}$$

is finite in a neighborhood of $({\lambda }_1,\dots ,{\lambda }_{n-1})$ if any only if $k({\tilde{\lambda }}_1,\dots ,{\tilde{\lambda }}_{n-1})$ is continuous at $({\lambda }_1,\dots ,{\lambda }_{n-1})$.

Proof The sufficiency property is obvious. We prove the necessary property of the condition by mathematical induction in the following. For $n=2$, there exists $I:=\{{\tilde{\lambda }}_1|{\tilde{\lambda }}_1={\lambda }_1+{\delta }_1,|{\delta }_1|\le {\delta }_0,{\delta }_0>0\}$ such that for any ${\tilde{\lambda }}_1\in I$, $k({\tilde{\lambda }}_1)\in \mathbf{R}$. Since for ${\tilde{\lambda }}_1={\lambda }_1+{\delta }_1\in I$ (${\delta }_1\ne 0$),

and $k({\lambda }_1-{\delta }_0)+k({\lambda }_1+{\delta }_0)<\mathrm{\infty }$, then by the Lebesgue control convergence theorem (cf. [17]), it follows $k({\lambda }_1+{\delta }_1)=k({\lambda }_1)+o(1)$ (${\delta }_1\to 0$). Assuming that for n (≥2), $k({\tilde{\lambda }}_1,\dots ,{\tilde{\lambda }}_{n-1})$ is continuous at $({\lambda }_1,\dots ,{\lambda }_{n-1})$, then for $n+1$, by the result of $n=2$, since $k({\lambda }_1+{\delta }_1,\dots ,{\lambda }_n+{\delta }_n)$ is finite in a neighborhood of $({\lambda }_1,\dots ,{\lambda }_n)$, we find

then by the assumption for n, it follows

$$\underset{{\delta }_n\to 0}{lim}k({\lambda }_1+{\delta }_1,\dots ,{\lambda }_n+{\delta }_n)=k({\lambda }_1,\dots ,{\lambda }_n)+o(1)({\delta }_i\to 0,i=1,\dots ,n-1).$$

By mathematical induction, we prove that for $n\in \mathbf{N}\setminus \{1\}$, $k({\tilde{\lambda }}_1,\dots ,{\tilde{\lambda }}_{n-1})$ is continuous at $({\lambda }_1,\dots ,{\lambda }_{n-1})$. □

Lemma 4 With the assumptions given in Lemma 2, if there exists $\delta >0$ such that for ${max}_{1\le i\le n-1}\{|{\delta }_i|\}<\delta$, $k({\lambda }_1+{\delta }_1,\dots ,{\lambda }_{n-1}+{\delta }_{n-1})\in \mathbf{R}$, $p_i\in \mathbf{R}\mathrm{\setminus }\{0,1\}$ ($i=1,\dots ,n$), $0<\epsilon <{min}_{1\le i\le n}\{|p_i|\}\delta$, then we have

$$\begin{array}{rcl}{I}_{\epsilon }& :=& \epsilon {\int }_1^{\mathrm{\infty }}\cdots {\int }_1^{\mathrm{\infty }}[{\int }_0^1{x}_n^{{\lambda }_n+\frac{\epsilon }{p_n}-1}{k}_{\lambda }(x_1{x}_n,\dots ,x_{n-1}{x}_n,1)d{x}_n]\\ \times \prod _{j=1}^{n-1}{x}_j^{{\lambda }_j-\frac{\epsilon }{p_j}-1}d{x}_1\cdots d{x}_{n-1}=k_{\lambda }+o(1)(\epsilon \to 0^{+}).\end{array}$$

(8)

Proof Setting $x_n^{\mathrm{\prime }}=x_n^{-1}$ in (8), we find

$$\begin{array}{rcl}{I}_{\epsilon }& :=& \epsilon {\int }_1^{\mathrm{\infty }}\cdots {\int }_1^{\mathrm{\infty }}\left[{\int }_1^{\mathrm{\infty }}{\left(x_n^{\mathrm{\prime }}\right)}^{-{\lambda }_n-\frac{\epsilon }{p_n}-1}{k}_{\lambda }(\frac{x_1}{x_n^{\mathrm{\prime }}},\dots ,\frac{x_1}{x_n^{\mathrm{\prime }}},1)d{x}_n^{\mathrm{\prime }}\right]\\ \times \prod _{j=1}^{n-1}{x}_j^{{\lambda }_j-\frac{\epsilon }{p_j}-1}d{x}_1\cdots d{x}_{n-1}=k_{\lambda }+o(1).\end{array}$$

Setting $u_j=x_j/x_n^{\mathrm{\prime }}$ ($j=1,\dots ,n-1$) in the above integral, since $\lambda -{\lambda }_n={\lambda }_n$, we find (replacing $x_n^{\mathrm{\prime }}$ by $x_n$)

$$I_{\epsilon }=\epsilon {\int }_1^{\mathrm{\infty }}{x}_n^{-1-\epsilon }[{\int }_{x_n^{-1}}^{\mathrm{\infty }}\cdots {\int }_{x_n^{-1}}^{\mathrm{\infty }}{k}_{\lambda }(u_1,\dots ,u_{n-1},1)\prod _{j=1}^{n-1}{u}_j^{{\lambda }_j-\frac{\epsilon }{p_j}-1}d{u}_1\cdots d{u}_{n-1}]d{x}_n.$$

(9)

Setting $D_j:=\{(u_1,\dots ,u_{n-1})|u_j\in (0,x_n^{-1}),u_k\in (0,\mathrm{\infty })(k\ne j)\}$ and

$$A_j(x_n):=\int \cdots {\int }_{D_j}{k}_{\lambda }(u_1,\dots ,u_{n-1},1)\prod _{j=1}^{n-1}{u}_j^{{\lambda }_j-\frac{\epsilon }{p_j}-1}d{u}_1\cdots d{u}_{n-1},$$

by (9), it follows

$$I_{\epsilon }\ge {\int }_{{\mathbf{R}}_{+}^{n-1}}{k}_{\lambda }(u_1,\dots ,u_{n-1},1)\prod _{j=1}^{n-1}{u}_j^{{\lambda }_j-\frac{\epsilon }{p_j}-1}d{u}_1\cdots d{u}_{n-1}-\epsilon \sum _{j=1}^{n-1}{\int }_1^{\mathrm{\infty }}{x}_n^{-1}{A}_j(x_n)d{x}_n.$$

(10)

Without loss of generality, we estimate the case of $j=n$, e.t.

${\int }_1^{\mathrm{\infty }}{x}_n^{-1}{A}_{n-1}(x_n)d{x}_n=O(1)$. In fact, setting $\alpha >0$, such that $|\frac{\epsilon }{p_{n-1}}+\alpha |<\delta$, since $-u_{n-1}^{\alpha }ln{u}_{n-1}\to 0$ ($u_{n-1}\to 0^{+}$), there exists $M>0$ such that $-u_{n-1}^{\alpha }ln{u}_{n-1}\le M$ ($u_{n-1}\in (0,1]$), and then by the Fubini theorem, it follows

$$\begin{array}{rcl}0& \le & {\int }_1^{\mathrm{\infty }}{x}_n^{-1}{A}_{n-1}(x_n)d{x}_n\\ =& {\int }_1^{\mathrm{\infty }}{x}_n^{-1}[{\int }_{{\mathbf{R}}_{+}^{n-2}}{\int }_0^{x_n^{-1}}{k}_{\lambda }(u_1,\dots ,u_{n-1},1)\prod _{j=1}^{n-1}{u}_j^{{\lambda }_j-\frac{\epsilon }{p_j}-1}d{u}_{n-1}d{u}_1\cdots d{u}_{n-2}]d{x}_n\\ =& {\int }_0^1{\int }_{{\mathbf{R}}_{+}^{n-2}}{k}_{\lambda }(u_1,\dots ,u_{n-1},1)\prod _{j=1}^{n-1}{u}_j^{{\lambda }_j-\frac{\epsilon }{p_j}-1}\left({\int }_1^{u_{n-1}^{-1}}{x}_n^{-1}d{x}_n\right)d{u}_1\cdots d{u}_{n-1}\\ =& {\int }_0^1{\int }_{{\mathbf{R}}_{+}^{n-2}}{k}_{\lambda }(u_1,\dots ,u_{n-1},1)\prod _{j=1}^{n-1}{u}_j^{{\lambda }_j-\frac{\epsilon }{p_j}-1}(-ln{u}_{n-1})d{u}_1\cdots d{u}_{n-1}\\ \le & M{\int }_0^1{\int }_{{\mathbf{R}}_{+}^{n-2}}{k}_{\lambda }(u_1,\dots ,u_{n-1},1)\prod _{j=1}^{n-2}{u}_j^{{\lambda }_j-\frac{\epsilon }{p_j}-1}{u}_{n-1}^{{\lambda }_{n-1}-(\frac{\epsilon }{p_{n-1}}+\alpha )-1}d{u}_1\cdots d{u}_{n-1}\\ \le & M{\int }_{{\mathbf{R}}_{+}^{n-1}}{k}_{\lambda }(u_1,\dots ,u_{n-1},1)\prod _{j=1}^{n-2}{u}_j^{{\lambda }_j-\frac{\epsilon }{p_j}-1}{u}_{n-1}^{{\lambda }_{n-1}-(\frac{\epsilon }{p_{n-1}}+\alpha )-1}d{u}_1\cdots d{u}_{n-1}\\ =& M\cdot k({\lambda }_1-\frac{\epsilon }{p_1},\dots ,{\lambda }_{n-2}-\frac{\epsilon }{p_{n-2}},{\lambda }_{n-1}-(\frac{\epsilon }{p_{n-1}}+\alpha ))<\mathrm{\infty }.\end{array}$$

Hence by (10), we have

$$I_{\epsilon }\ge {\int }_{{\mathbf{R}}_{+}^{n-1}}{k}_{\lambda }(u_1,\dots ,u_{n-1},1)\prod _{j=1}^{n-1}{u}_j^{{\lambda }_j-\frac{\epsilon }{p_j}-1}d{u}_1\cdots d{u}_{n-1}-o_1(1).$$

(11)

Since by Lemma 3 we find

$$\begin{array}{rcl}{I}_{\epsilon }& \le & \epsilon {\int }_1^{\mathrm{\infty }}{x}_n^{-1-\epsilon }[{\int }_0^{\mathrm{\infty }}\cdots {\int }_0^{\mathrm{\infty }}{k}_{\lambda }(u_1,\dots ,u_{n-1},1)\prod _{j=1}^{n-1}{u}_j^{{\lambda }_j-\frac{\epsilon }{p_j}-1}d{u}_1\cdots d{u}_{n-1}]d{x}_n\\ =& {\int }_0^{\mathrm{\infty }}\cdots {\int }_0^{\mathrm{\infty }}{k}_{\lambda }(u_1,\dots ,u_{n-1},1)\prod _{j=1}^{n-1}{u}_j^{{\lambda }_j-\frac{\epsilon }{p_j}-1}d{u}_1\cdots d{u}_{n-1}\\ =& k({\lambda }_1-\frac{\epsilon }{p_1},\dots ,{\lambda }_{n-1}-\frac{\epsilon }{p_{n-1}})=k_{\lambda }+o_2(1),\end{array}$$

then combining with (11), we have (8). □

Lemma 5 Suppose that $n\in \mathbf{N}\mathrm{\setminus }\{1\}$, $p_i\in \mathbf{R}\mathrm{\setminus }\{0,1\}$ ($i=1,\dots ,n$), ${\sum }_{i=1}^n\frac{1}{p_i}=1$, $\frac{1}{q_n}=1-\frac{1}{p_n}$, $({\lambda }_1,\dots ,{\lambda }_n)\in {\mathbf{R}}^n$, ${\lambda }_n={\sum }_{i=1}^{n-1}{\lambda }_i=\frac{\lambda }{2}$, $k_{\lambda }(x_1,\dots ,x_n)$ (≥0) is a measurable function of −λ-degree in ${\mathbf{R}}_{+}^n$ such that

$$k_{\lambda }={\int }_{{\mathbf{R}}_{+}^{n-1}}{k}_{\lambda }(u_1,\dots ,u_{n-1},1)\prod _{j=1}^{n-1}{u}_j^{{\lambda }_j-1}d{u}_1\cdots d{u}_{n-1}\in \mathbf{R}.$$

If $f_i\ge 0$ are measurable functions in ${\mathbf{R}}_{+}$ ($i=1,\dots ,n-1$), ${\tilde{k}}_{\lambda }(x_1,\dots ,x_n):=k_{\lambda }(x_1{x}_n,\dots ,x_{n-1}{x}_n,1)$, then (1) for $p_i>1$ ($i=1,\dots ,n$), we have

$$\begin{array}{rcl}J& :=& {\left\{{\int }_0^{\mathrm{\infty }}{x}_n^{\frac{\lambda q_n}{2}-1}{[{\int }_{{\mathbf{R}}_{+}^{n-1}}{\tilde{k}}_{\lambda }(x_1,\dots ,x_n)\prod _{i=1}^{n-1}{f}_i(x_i)d{x}_1\cdots d{x}_{n-1}]}^{q_n}d{x}_n\right\}}^{\frac{1}{q_n}}\\ \le & k_{\lambda }\prod _{i=1}^{n-1}{\{{\int }_0^{\mathrm{\infty }}{x}^{p_i(1-{\lambda }_i)-1}{f}^{p_i}(x)dx\}}^{\frac{1}{p_i}};\end{array}$$

(12)

(2) for $0<p_1<1$, $p_i<0$ ($i=2,\dots ,n$), we have the reverse of (12).

Proof (1) For $p_i>1$ ($i=1,\dots ,n$), by the Hölder inequality (cf. [18]) and (7), it follows

(13)

For $n\ge 3$, by the Hölder inequality again, it follows

$$\begin{array}{rcl}J& \le & {(k_{\lambda })}^{\frac{1}{p_n}}\left\{\prod _{i=1}^{n-1}\right[{\int }_{{\mathbf{R}}_{+}^{n-1}}({\int }_0^{\mathrm{\infty }}{\tilde{k}}_{\lambda }(x_1,\dots ,x_n)x_n^{{\lambda }_n-1}d{x}_n)\\ \times {{x_i^{({\lambda }_i-1)(1-p_i)}\prod _{j=1(j\ne i)}^{n-1}{x}_j^{{\lambda }_j-1}{f}_i^{p_i}(x_i)d{x}_1\cdots d{x}_{n-1}]}^{\frac{q_n}{p_i}}\}}^{\frac{1}{q_n}}\\ =& {(k_{\lambda })}^{\frac{1}{p_n}}\prod _{i=1}^{n-1}\left\{{\int }_0^{\mathrm{\infty }}\right[{\int }_{{\mathbf{R}}_{+}^{n-1}}{\tilde{k}}_{\lambda }(x_1,\dots ,x_n)\\ \times {x_i^{{\lambda }_i}\prod _{j=1(j\ne i)}^n{x}_j^{{\lambda }_j-1}d{x}_1\cdots d{x}_{i-1}d{x}_{i+1}\cdots d{x}_n]x_i^{p_i(1-{\lambda }_i)-1}{f}_i^{p_i}(x_i)d{x}_i\}}^{\frac{1}{p_i}}\\ =& {(k_{\lambda })}^{\frac{1}{p_n}}\prod _{i=1}^{n-1}{\{{\int }_0^{\mathrm{\infty }}{\omega }_i(x_i)x_i^{p_i(1-{\lambda }_i)-1}{f}_i^{p_i}(x_i)d{x}_i\}}^{\frac{1}{p_i}}.\end{array}$$

(14)

Then by (7), we have (12). (Note: for $n=2$, we do not use the Hölder inequality again in the above.) (2) For $0<p_1<1$, $p_i<0$ ($i=2,\dots ,n$), by the reverse Hölder inequality and in the same way, we obtain the reverses of (12). □

3 Main results and applications

With the assumptions given in Lemma 5, setting ${\varphi }_i(x):=x^{p_{{}_i(1-{\lambda }_i)-1}}$ ($x\in (0,\mathrm{\infty })$; $i=1,\dots ,n$), then we find ${\varphi }_n^{1/(1-p_n)}(x)=x^{q_{{}_n\lambda _n-1}}$. If $p_i>1$ ($i=1,\dots ,n$), define the following real function spaces:

and a multiple Hilbert-type integral operator $T:{\prod }_{i=1}^{n-1}{L}_{{\varphi }_i}^{p_i}(0,\mathrm{\infty })\to L_{{\varphi }_n^{1/(1-p_n)}}^{q_n}$ as follows: for $f=(f_1,\dots ,f_{n-1})\in {\prod }_{i=1}^{n-1}{L}_{{\varphi }_i}^{p_i}(0,\mathrm{\infty })$,

$$(Tf)(x_n):={\int }_{{\mathbf{R}}_{+}^{n-1}}{\tilde{k}}_{\lambda }(x_1,\dots ,x_n)\prod _{i=1}^{n-1}{f}_i(x_i)d{x}_1\cdots d{x}_{n-1},x_n\in (0,\mathrm{\infty }).$$

(15)

Then by (12), it follows $Tf\in L_{{\varphi }_n^{1/(1-p_n)}}^{q_n}$, T is bounded, ${\parallel Tf\parallel }_{q_n,{\varphi }_n^{1/(1-p_n)}}\le k_{\lambda }{\prod }_{i=1}^{n-1}{\parallel f_i\parallel }_{p_i,{\varphi }_i}$ and $\parallel T\parallel \le k_{\lambda }$, where

$$\parallel T\parallel :=\underset{f\in {\prod }_{i=1}^{n-1}{L}_{{\varphi }_i}^{p_i}(0,\mathrm{\infty })(f_i\ne \theta ,i=1,\dots ,n-1)}{sup}\frac{{\parallel Tf\parallel }_{q_n,{\varphi }_n^{1/(1-p_n)}}}{{\prod }_{i=1}^{n-1}{\parallel f_i\parallel }_{p_i,{\varphi }_i}}.$$

(16)

Define the formal inner product of $T(f_1,\dots ,f_{n-1})$ and $f_n$ as

$$(T(f_1,\dots ,f_{n-1}),f_n):={\int }_{{\mathbf{R}}_{+}^n}{\tilde{k}}_{\lambda }(x_1,\dots ,x_n)\prod _{i=1}^n{f}_i(x_i)d{x}_1\cdots d{x}_n.$$

(17)

Theorem 1 With the assumptions given in Lemma 5, suppose that for any $({\lambda }_1,\dots ,{\lambda }_n)\in {\mathbf{R}}^n$, it satisfies ${\lambda }_n={\sum }_{i=1}^{n-1}{\lambda }_i=\frac{\lambda }{2}$, and

$$0<k_{\lambda }={\int }_{{\mathbf{R}}_{+}^{n-1}}{k}_{\lambda }(u_1,\dots ,u_{n-1},1)\prod _{j=1}^{n-1}{u}_j^{{\lambda }_j-1}d{u}_1\cdots d{u}_{n-1}<\mathrm{\infty }.$$

(18)

If $f_i(\ge 0)\in L_{{\varphi }_i}^{p_i}(0,\mathrm{\infty })$, ${\parallel f\parallel }_{p_{i,{\varphi }_i}}>0$ ($i=1,\dots ,n$), then (i) for $p_i>1$ ($i=1,\dots ,n$), we have $\parallel T\parallel =k_{\lambda }$ and the following equivalent inequalities:

(19)

(20)

where the constant factor $k_{\lambda }$ is the best possible; (ii) for $0<p_1<1$, $p_i<0$ ($i=2,\dots ,n$), using the formal symbols in the case of (i), we have the equivalent reverses of (19) and (20) with the same best constant factor.

Proof (i) For all $p_i>1$, if (12) takes the form of equality, then for $n\ge 3$ in (14), there exist $C_i$ and $C_k$ ($i\ne k$) such that they are not all zero and

e.t. $C_i{x}_i^{p_i(1-{\lambda }_i)}{f}_i^{p_i}(x_i)=C_k{x}_k^{p_k(1-{\lambda }_k)}{f}_k^{p_k}(x_k)=C$ a.e. in ${\mathbf{R}}_{+}^n$. Assuming that $C_i>0$, then $x_i^{p_i(1-{\lambda }_i)-1}{f}_i^{p_i}(x_i)=C/(C_i{x}_i)$, which contradicts ${\parallel f\parallel }_{p_{i,{\varphi }_i}}>0$. (Note: for $n=2$, we consider (13) for $f_k^{p_i}(x_k)=1$ in the above.) Hence we have (19). By the Hölder inequality, it follows

$$\begin{array}{rcl}(Tf,f_n)& =& {\int }_0^{\mathrm{\infty }}(x_n^{{\lambda }_n-\frac{1}{q_n}}{\int }_{{\mathbf{R}}_{+}^{n-1}}{\tilde{k}}_{\lambda }(x_1,\dots ,x_n)\prod _{i=1}^{n-1}{f}_i(x_i)d{x}_1\cdots d{x}_{n-1})(x_n^{\frac{1}{q_n}-{\lambda }_n}{f}_n(x_n))d{x}_n\\ \le & {\parallel T(f_1,\dots ,f_{n-1})\parallel }_{q_n,{\varphi }_n^{1/(1-p_n)}}{\parallel f_n\parallel }_{p_n,{\varphi }_n},\end{array}$$

(21)

and then by (19), we have (20). Assuming that (20) is valid, setting

$$f_n(x_n):=x_n^{q_n{\lambda }_n-1}{[{\int }_{{\mathbf{R}}_{+}^{n-1}}{\tilde{k}}_{\lambda }(x_1,\dots ,x_n)\prod _{i=1}^{n-1}{f}_i(x_i)d{x}_1\cdots d{x}_{n-1}]}^{q_n-1},$$

then $J={\{{\int }_0^{\mathrm{\infty }}{x}_n^{p_n(1-{\lambda }_n)-1}{f}_n^{p_n}(x_n)d{x}_n\}}^{\frac{1}{q_n}}$. By (12), it follows $J<\mathrm{\infty }$. If $J=0$, then (19) is naturally valid. Assuming that $0<J<\mathrm{\infty }$, by (20), it follows

and then (19) is valid, which is equivalent to (20).

For $\epsilon >0$ small enough, setting ${\tilde{f}}_i(x)$ as: ${\tilde{f}}_i(x)=0$, $x\in (0,1)$; ${\tilde{f}}_i(x)=x^{{\lambda }_i-\frac{\epsilon }{p_i}-1}$, $x\in [1,\mathrm{\infty })$ ($i=1,\dots ,n-1$), ${\tilde{f}}_n(x)=x^{{\lambda }_n+\frac{\epsilon }{p_n}-1}$, $x\in (0,1)$; ${\tilde{f}}_n(x)=0$, $x\in [1,\mathrm{\infty })$, if there exists $k\le k_{\lambda }$ such that (20) is still valid as we replace $k_{\lambda }$ by k, then in particular, by Lemma 4, we have

$$k_{\lambda }+o(1)=I_{\epsilon }=\epsilon (T({\tilde{f}}_1,\dots ,{\tilde{f}}_{n-1}),{\tilde{f}}_n)<\epsilon k\prod _{i=1}^n{\parallel {\tilde{f}}_i\parallel }_{p_i,{\varphi }_i}=k$$

and $k_{\lambda }\le k(\epsilon \to 0^{+})$. Hence $k=k_{\lambda }$ is the best value of (20). We confirm that the constant factor $k_{\lambda }$ in (19) is the best possible, otherwise we can get a contradiction by (21) that the constant factor in (20) is not the best possible. Therefore $\parallel T\parallel =k_{\lambda }$.

• (ii) For $0<p_1<1$, $p_i<0$ ($i=2,\dots ,n$), by using the reverse Hölder inequality and in the same way, we have the equivalent reverses of (19) and (20) with the same best constant factor. □

Example 1 For $\lambda >0$, ${\lambda }_i=\frac{\lambda }{r_i}$ ($i=1,\dots ,n$), $r_n=2$, ${\sum }_{i=1}^n\frac{1}{r_i}=1$, $k_{\lambda }(x_1,\dots ,x_n)=\frac{1}{{({\sum }_{i=1}^n{x}_i)}^{\lambda }}$, by mathematical induction, we can show

$$k_{\lambda }={\int }_{R_{+}^{n-1}}\frac{1}{{({\sum }_{i=1}^{n-1}{u}_i+1)}^{\lambda }}\prod _{j=1}^{n-1}{u}_j^{\frac{\lambda }{r_j}-1}d{u}_1\cdots d{u}_{n-1}=\frac{1}{\mathrm{\Gamma }(\lambda )}\prod _{i=1}^n\mathrm{\Gamma }\left(\frac{\lambda }{r_i}\right).$$

(22)

In fact, for $n=2$, we obtain

$$k_{\lambda }={\int }_{R_{+}}\frac{1}{{(u_1+1)}^{\lambda }}{u}_1^{\frac{\lambda }{r_1}-1}d{u}_1=\frac{1}{\mathrm{\Gamma }(\lambda )}\mathrm{\Gamma }\left(\frac{\lambda }{r_1}\right)\mathrm{\Gamma }\left(\frac{\lambda }{r_2}\right).$$

Assuming that for n (≥2) (22) is valid, then for $n+1$, it follows

$$\begin{array}{rcl}{k}_{\lambda }& =& {\int }_{R_{+}^n}\frac{1}{{({\sum }_{i=1}^n{u}_i+1)}^{\lambda }}\prod _{j=1}^n{u}_j^{\frac{\lambda }{r_j}-1}d{u}_1\cdots d{u}_n\\ =& {\int }_{R_{+}^{n-1}}\prod _{j=2}^n{u}_j^{\frac{\lambda }{r_j}-1}\left[{\int }_{R_{+}}\frac{1}{{[u_1+({\sum }_{i=2}^n{u}_i+1)]}^{\lambda }}{u}_1^{\frac{\lambda }{r_1}-1}d{u}_1\right]d{u}_2\cdots d{u}_n\\ =& {\int }_{R_{+}^{n-1}}\frac{1}{{({\sum }_{i=2}^n{u}_i+1)}^{\lambda }}\prod _{j=2}^n{u}_j^{\frac{\lambda }{r_j}-1}\left[{\int }_{R_{+}}\frac{1}{{(v_1+1)}^{\lambda }}{v}_1^{\frac{\lambda }{r_1}-1}d{v}_1\right]d{u}_2\cdots d{u}_n\\ =& \frac{\mathrm{\Gamma }(\frac{\lambda }{r_1})\mathrm{\Gamma }(\lambda -\frac{\lambda }{r_1})}{\mathrm{\Gamma }(\lambda )}{\int }_{R_{+}^{n-1}}\frac{1}{{({\sum }_{i=2}^n{u}_i+1)}^{\lambda (1-\frac{1}{r_1})}}\prod _{j=2}^n{u}_j^{\frac{\lambda }{r_j}-1}d{u}_2\cdots d{u}_n\\ =& \frac{\mathrm{\Gamma }(\frac{\lambda }{r_1})\mathrm{\Gamma }(\lambda -\frac{\lambda }{r_1})}{\mathrm{\Gamma }(\lambda )}\frac{1}{\mathrm{\Gamma }(\lambda -\frac{\lambda }{r_1})}\prod _{i=2}^{n+1}\mathrm{\Gamma }\left(\frac{\lambda }{r_i}\right)=\frac{1}{\mathrm{\Gamma }(\lambda )}\prod _{i=1}^{n+1}\mathrm{\Gamma }\left(\frac{\lambda }{r_i}\right).\end{array}$$

Then by mathematical induction, (22) is valid for $n\in \mathbf{N}\mathrm{\setminus }\{1\}$.

Example 2 For $\lambda >0$, ${\lambda }_i=\frac{\lambda }{r_i}$ ($i=1,\dots ,n$), $r_n=2$, ${\sum }_{i=1}^n\frac{1}{r_i}=1$, $k_{\lambda }(x_1,\dots ,x_n)=\frac{1}{{({max}_{1\le i\le n}\{x_i\})}^{\lambda }}$, by mathematical induction, we can show

$$k_{\lambda }={\int }_{R_{+}^{n-1}}\frac{1}{{({max}_{1\le i\le n-1}\{u_i\}+1)}^{\lambda }}\prod _{j=1}^{n-1}{u}_j^{\frac{\lambda }{r_j}-1}d{u}_1\cdots d{u}_{n-1}=\frac{{\prod }_{i=1}^n{r}_i}{{\lambda }^{n-1}}.$$

(23)

In fact, for $n=2$, we obtain

$$k_{\lambda }={\int }_{R_{+}}\frac{u_1^{\frac{\lambda }{r_1}-1}}{{(max\{u_1,1\})}^{\lambda }}d{u}_1={\int }_0^1{u}_1^{\frac{\lambda }{r_1}-1}d{u}_1+{\int }_1^{\mathrm{\infty }}{u}_1^{\frac{-\lambda }{r_2}-1}d{u}_1=\frac{1}{\lambda }{r}_1{r}_2.$$

Assuming that for n (≥2), (22) is valid, then for $n+1$, it follows

$$\begin{array}{rcl}{k}_{\lambda }& =& {\int }_{R_{+}^{n-1}}\prod _{j=2}^n{u}_j^{\frac{\lambda }{r_j}-1}\left[{\int }_0^{\mathrm{\infty }}\frac{1}{{({max}_{1\le i\le n}\{u_i,1\})}^{\lambda }}{u}_1^{\frac{\lambda }{r_1}-1}d{u}_1\right]d{u}_2\cdots d{u}_n\\ =& {\int }_{R_{+}^{n-1}}\prod _{j=2}^n{u}_j^{\frac{\lambda }{r_j}-1}[{\int }_0^{max\{u_2,\dots ,u_n,1\}}\frac{1}{{({max}_{2\le i\le n}\{u_i,1\})}^{\lambda }}{u}_1^{\frac{\lambda }{r_1}-1}d{u}_1\\ +{\int }_{max\{u_2,\dots ,u_n,1\}}^{\mathrm{\infty }}\frac{1}{u_1^{\lambda }}{u}_1^{\frac{\lambda }{r_1}-1}d{u}_1]d{u}_2\cdots d{u}_n\\ =& \frac{r_1^2}{\lambda (r_1-1)}{\int }_{R_{+}^{n-1}}\frac{1}{{({max}_{2\le i\le n}\{u_i,1\})}^{\lambda (1-\frac{1}{r_1})}}\prod _{j=2}^n{u}_j^{\frac{\lambda }{r_j}-1}d{u}_2\cdots d{u}_n\\ =& \frac{r_1^2}{\lambda (r_1-1)}{\left(\frac{r_1}{r_1-1}\right)}^{n-1}{\int }_{R_{+}^{n-1}}\frac{1}{{({max}_{2\le i\le n}\{v_i,1\})}^{\lambda }}\prod _{j=2}^n{v}_j^{\frac{\lambda }{r_j}\frac{r_1}{r_1-1}-1}d{v}_2\cdots d{v}_n\\ =& \frac{r_1^2}{\lambda (r_1-1)}{\left(\frac{r_1}{r_1-1}\right)}^{n-1}\frac{1}{{\lambda }^{n-1}}\prod _{i=2}^{n+1}\frac{r_1-1}{r_1}{r}_i=\frac{1}{{\lambda }^n}\prod _{i=1}^{n+1}{r}_i.\end{array}$$

Then by mathematical induction, (23) is valid for $n\in \mathbf{N}\mathrm{\setminus }\{1\}$.

Example 3 For $\lambda >0$, ${\lambda }_i=\frac{-\lambda }{r_i}$ ($i=1,\dots ,n$), $r_n=2$, ${\sum }_{i=1}^n\frac{1}{r_i}=1$, $k_{\lambda }(x_1,\dots ,x_n)={({min}_{1\le i\le n}\{x_i\})}^{\lambda }$, by mathematical induction, we can show

$$k_{-\lambda }={\int }_{R_{+}^{n-1}}{(min\{u_1,\dots ,u_{n-1},1\})}^{\lambda }\prod _{j=1}^{n-1}{u}_j^{\frac{-\lambda }{r_j}-1}d{u}_1\cdots d{u}_{n-1}=\frac{{\prod }_{i=1}^n{r}_i}{{\lambda }^{n-1}}.$$

(24)

In fact, for $n=2$, we obtain

$$k_{-\lambda }={\int }_0^1{u}_1^{\frac{\lambda }{r_2}-1}d{u}_1+{\int }_1^{\mathrm{\infty }}{u}_1^{\frac{-\lambda }{r_1}-1}d{u}_1=\frac{1}{\lambda }{r}_1{r}_2.$$

Assuming that for n (≥2), (24) is valid, then for $n+1$, it follows

$$\begin{array}{rcl}{k}_{-\lambda }& =& {\int }_{R_{+}^{n-1}}\prod _{j=2}^n{u}_j^{\frac{-\lambda }{r_j}-1}\left[{\int }_0^{\mathrm{\infty }}{(min\{u_1,\dots ,u_n,1\})}^{\lambda }{u}_1^{\frac{-\lambda }{r_1}-1}d{u}_1\right]d{u}_2\cdots d{u}_n\\ =& {\int }_{R_{+}^{n-1}}\prod _{j=2}^n{u}_j^{\frac{-\lambda }{r_j}-1}[{\int }_0^{min\{u_2,\dots ,u_n,1\}}{u}_1^{\lambda }{u}_1^{\frac{-\lambda }{r_1}-1}d{u}_1\\ +{\int }_{min\{u_2,\dots ,u_n,1\}}^{\mathrm{\infty }}{(min\{u_2,\dots ,u_n,1\})}^{\lambda }{u}_1^{\frac{-\lambda }{r_1}-1}d{u}_1]d{u}_2\cdots d{u}_n\\ =& \frac{r_1^2}{\lambda (r_1-1)}{\int }_{R_{+}^{n-1}}{(min\{u_2,\dots ,u_n,1\})}^{\lambda (1-\frac{1}{r_1})}\prod _{j=2}^n{u}_j^{\frac{-\lambda (1-\frac{1}{r_1})}{(1-\frac{1}{r_1})r_j}-1}d{u}_2\cdots d{u}_n\\ =& \frac{r_1^2}{\lambda (r_1-1)}\frac{1}{{[\lambda (1-\frac{1}{r_1})]}^{n-1}}\prod _{i=2}^{n+1}(1-\frac{1}{r_1})r_i=\frac{1}{{\lambda }^n}\prod _{i=1}^{n+1}{r}_i.\end{array}$$

Then, by mathematical induction, (24) is valid for $n\in \mathbf{N}\mathrm{\setminus }\{1\}$.

Remarks (i) In particular, for $n=2$ in (20), we have

(25)

where $k_{\lambda }={\int }_0^{\mathrm{\infty }}{k}_{\lambda }(u,1)u^{\frac{\lambda }{2}-1}du>0$ ($\lambda \in \mathbf{R}$) is the best possible. Inequality (25) is an extension of (4) and (8.1.7) in [8].

• (ii) In Examples 1 and 2, by Theorem 1, since for any $({\lambda }_1,\dots ,{\lambda }_n)\in {\mathbf{R}}^n({\lambda }_n={\sum }_{i=1}^n{\lambda }_i=\frac{\lambda }{2})$, we obtain $0<k_{\lambda }<\mathrm{\infty }$, then we have $\parallel T\parallel =k_{\lambda }$ and the equivalent inequalities (19) and (20) with the particular kernels and some equivalent reverses. In Example 3, still using Theorem 1, we find $0<\parallel T\parallel =k_{-\lambda }<\mathrm{\infty }$ and the relating particular inequalities.