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Equivalent conditions and applications of a class of Hilbert-type integral inequalities involving multiple functions with quasi-homogeneous kernels

Journal of Inequalities and Applications20182018:206

https://doi.org/10.1186/s13660-018-1797-5

  • Received: 1 April 2018
  • Accepted: 31 July 2018
  • Published:

Abstract

Let \(K(x_{1},\ldots,x_{n})\) satisfy
$$K(x_{1},\ldots,tx_{i},\ldots,x_{n})=t^{\lambda\lambda _{i}}K \bigl(t^{-\frac{\lambda_{i}}{\lambda_{1}}}x_{1},\ldots,x_{i}, \ldots,t^{-\frac {\lambda _{i}}{\lambda_{n}}}x_{n} \bigr) $$
for \(t>0\). With this integral kernel, by using the method and technique of weight coefficients, the equivalent conditions and the best constant factors for the validity of Hilbert-type integral inequalities involving multiple functions are discussed. Finally, the applications of the integral inequalities are considered.

Keywords

  • Hilbert-type integral inequality
  • Quasi-homogeneous kernel
  • Equivalent conditions
  • Best constant factor

MSC

  • 26D15
  • 47A07

1 Introduction

Let \(x=(x_{1},\ldots,x_{n}), \mathbf{R}_{+}^{n}=\{x=(x_{1},\ldots ,x_{n}):x_{i}>0\ (i=1,\ldots,n)\},r>1,f(t)\geq0\), and α be a constant. Set
$$L_{\alpha}^{r}(0,+\infty)= \biggl\{ f(t): \Vert f \Vert _{r,\alpha}= \biggl( \int_{0}^{+\infty}t^{\alpha}f^{r}(t)\,dt \biggr) ^{1/r}< +\infty \biggr\} . $$
If \(\sum_{i=1}^{n}\frac{1}{p_{i}}=1\ (p_{i}>1,i=1,\ldots,n),\alpha _{i}\in \mathbf{R},f_{i}(x_{i})\in L_{\alpha_{i}}^{p_{i}}(0,+\infty)\) \((i=1,\ldots ,n),K(x_{1},\ldots,x_{n})\geq0\), M is a constant, then we name the following inequality a Hilbert-type integral inequality:
$$\int_{\mathbf{R}_{+}^{n}}K(x_{1},\ldots,x_{n})\prod _{i=1}^{n}f_{i}(x_{i})\,dx_{1} \cdots \,dx_{n}\leq M\prod_{i=1}^{n} \Vert f_{i} \Vert _{p_{i},\alpha_{i}}. $$
An integral kernel \(K(x_{1},\ldots,x_{n})\) is said to be a quasi-homogeneous function with parameters \((\lambda,\lambda _{1},\ldots ,\lambda_{n})\) if, for \(t>0\),
$$K(x_{1},\ldots,tx_{i},\ldots,x_{n})=t^{\lambda\lambda _{i}}K \bigl(t^{-\frac{\lambda_{i}}{\lambda_{1}}}x_{1},\ldots,x_{i}, \ldots,t^{-\frac {\lambda _{i}}{\lambda_{n}}}x_{n} \bigr) \quad (i=1,\ldots,n). $$
Obviously, \(K(x_{1},\ldots,x_{n})\) becomes a homogeneous function of order \(\lambda\lambda_{0}\) when \(\lambda_{1}=\lambda_{2}=\cdots=\lambda _{n}=\lambda_{0}\).

So far, many good results have been obtained in the study of Hilbert-type inequalities (cf. [124]). What are the necessary and sufficient conditions for the validity of a Hilbert-type inequality? What is the best constant factor when the inequality holds? The research on such problems is undoubtedly of great significance to the study and applications of Hilbert-type inequality theory, but unfortunately, the research on this type of problems is rarely seen.

In this paper, we focus on the quasi-homogeneous integral kernels, discuss the equivalent conditions for the validity of Hilbert-type integral inequalities involving multiple functions, and obtain the expressions of the best constant factors when the inequalities are established. Finally, we discuss their applications.

2 Some lemmas

Lemma 1

Let integer \(n\geq2,\sum_{i=1}^{n}\frac{1}{p_{i}}=1\ (p_{i}>1,i=1,\ldots,n),\lambda\in\mathbf{R},\alpha_{i}\in\mathbf {R},\lambda_{i}>0\) (or \(\lambda_{i}<0\)) \((i=1,\ldots,n)\), and \(K(x_{1},\ldots,x_{n})\) be a nonnegative measurable function with parameters \((\lambda,\lambda_{1},\ldots,\lambda_{n}),\sum_{i=1}^{n}\frac{\alpha _{i}+1}{\lambda_{i}p_{i}}=\lambda+\sum_{i=1}^{n}\frac{1}{\lambda_{i}}\). Set
$$\begin{aligned} W_{j} ={}& \int_{\mathbf{R}_{+}^{n-1}}K(u_{1},\ldots,u_{j-1},1,u_{j+1}, \ldots,u_{n}) \\ &{}\times\prod_{i=1(i\neq j)}^{n}u_{i}^{-\frac{\alpha_{i}+1}{p_{i}}}\,du_{1} \cdots \,du_{j-1}\,du_{j+1}\cdots \,du_{n}. \end{aligned}$$
Then \(\frac{1}{\lambda_{1}}W_{1}=\frac{1}{\lambda_{2}}W_{2}=\cdots =\frac{1}{\lambda n}W_{n}\), and
$$\begin{aligned} \omega_{j}(x_{j})& = \int_{\mathbf{R}_{+}^{n-1}}K(x_{1},\ldots,x_{n})\prod _{i=1(i\neq j)}^{n}x_{i}^{-\frac{\alpha_{i}+1}{p_{i}}}\,dx_{1} \cdots \,dx_{j-1}\,dx_{j+1}\cdots \,dx_{n} \\ &= x_{j}^{\lambda_{j} ( \lambda+\sum_{i=1(i\neq j)}^{n}\frac {\alpha _{i}+1}{\lambda_{i}p_{i}}+\sum_{i=1(i\neq j)}^{n}\frac{1}{\lambda_{i}} ) }W_{j}, \end{aligned}$$
where \(j=1,\ldots,n\).

Proof

When \(j\geq2\), by \(\sum_{i=1}^{n}\frac{\alpha_{i}+1}{\lambda_{i}p_{i}}=\lambda+\sum_{i=1}^{n}\frac{1}{\lambda_{i}}\),
$$\begin{aligned} W_{j} ={}& \int_{\mathbf{R}_{+}^{n-1}}u_{1}^{\lambda\lambda_{1}}K \biggl( 1,\frac{u_{2}}{u_{1}^{\frac{\lambda_{1}}{\lambda_{2}}}},\ldots,\frac {u_{j-1}}{u_{1}^{\frac{\lambda_{1}}{\lambda_{j-1}}}},\frac{1}{u_{1}^{\frac {\lambda _{1}}{\lambda_{j}}}}, \frac{u_{j+1}}{u_{1}^{\frac{\lambda_{1}}{\lambda _{j+1}}}},\ldots,\frac{u_{n}}{u_{1}^{\frac{\lambda_{1}}{\lambda _{n}}}} \biggr) \\ &{}\times\prod_{i=1(i\neq j)}^{n}u_{i}^{-\frac{\alpha_{i}+1}{p_{i}}}\,du_{1} \cdots \,du_{j-1}\,du_{j+1}\cdots \,du_{n} \\ ={}& \frac{\lambda_{j}}{\lambda_{1}} \int_{\mathbf{R}_{+}^{n-1}}K ( 1,t_{2},\ldots,t_{n} ) t_{j}^{\lambda_{j} ( -\lambda -\sum_{i=1}^{n}\frac{1}{\lambda_{i}}+\sum_{i=1(i\neq j)}^{n}\frac {\alpha _{i}+1}{\lambda_{i}p_{i}} ) } \\ &{}\times\prod_{i=2(i\neq j)}^{n}t_{i}^{-\frac{\alpha_{i}+1}{p_{i}}}\,dt_{2} \cdots \,dt_{n} \\ ={}& \frac{\lambda_{j}}{\lambda_{1}} \int_{\mathbf{R}_{+}^{n-1}}K ( 1,t_{2},\ldots,t_{n} ) t_{j}^{\lambda_{j} ( -\frac{\alpha _{j}+1}{\lambda_{j}p_{j}} ) }\prod_{i=2(i\neq j)}^{n}t_{i}^{-\frac{\alpha _{i}+1}{p_{i}}}\,dt_{2} \cdots \,dt_{n} \\ ={}& \frac{\lambda_{j}}{\lambda_{1}} \int_{\mathbf{R}_{+}^{n-1}}K ( 1,t_{2},\ldots,t_{n} ) \prod_{i=2}^{n}t_{i}^{-\frac{\alpha _{i}+1}{p_{i}}}\,dt_{2} \cdots \,dt_{n}=\frac{\lambda_{j}}{\lambda_{1}}W_{1}. \end{aligned}$$
Therefore \(\frac{1}{\lambda_{j}}W_{j}=\frac{1}{\lambda _{1}}W_{1}\ (j\geq2)\). When \(j=1,\ldots,n\), we also get
$$\begin{aligned} \omega_{j}(x_{j}) ={}& \int_{\mathbf{R}_{+}^{n-1}}x_{j}^{\lambda\lambda _{j}}K \biggl( \frac{x_{1}}{x_{j}^{\frac{\lambda_{j}}{\lambda 1}}},\ldots,\frac{x_{j-1}}{x_{j}^{\frac{\lambda_{j}}{\lambda_{j-1}}}},1,\frac {x_{j+1}}{x_{j}^{\frac{\lambda_{j}}{\lambda_{j+1}}}}, \ldots,\frac {x_{n}}{x_{j}^{\frac{\lambda_{j}}{\lambda_{n}}}} \biggr) \\ &{}\times\prod_{i=1(i\neq j)}^{n}x_{i}^{-\frac{\alpha_{i}+1}{p_{i}}}\,dx_{1} \cdots \,dx_{j-1}\,dx_{j+1}\cdots \,dx_{n} \\ ={}& x_{j}^{\lambda\lambda_{j}-\lambda_{j}\sum_{i=1(i\neq j)}^{n}\frac{ \alpha_{i}+1}{\lambda_{i}p_{i}}+\lambda_{j}\sum_{i=1(i\neq j)}^{n}\frac{1}{\lambda_{i}}} \\ &{}\times \int_{\mathbf{R}_{+}^{n-1}}K ( u_{1},\ldots,u_{j-1},1,u_{j+1} ,\ldots,u_{n} ) \\ &{}\times\prod_{i=1(i\neq j)}^{n}u_{i}^{-\frac{\alpha_{i}+1}{p_{i}}}\,du_{1} \cdots \,du_{j-1}\,du_{j+1}\cdots \,du_{n} \\ ={}& x_{j}^{\lambda_{j} ( \lambda+\sum_{i=1(i\neq j)}^{n}\frac {\alpha _{i}+1}{\lambda_{i}p_{i}}+\sum_{i=1(i\neq j)}^{n}\frac{1}{\lambda_{i}} ) }W_{j}. \end{aligned}$$
 □

Lemma 2

([25])

Let \(p_{i}>0,a_{i}>0,\alpha_{i}>0\) \((i=1,\ldots,n),\psi(u)\) be measurable. Then
$$\begin{aligned} &\int\cdots \int_{x_{i}>0, ( \frac{x_{1}}{a_{1}} ) ^{\alpha _{1}}+\cdots+( \frac{x_{n}}{a_{n}} ) ^{\alpha_{n}}\leq1}\psi \biggl( \biggl( \frac{x_{1}}{a_{1}} \biggr) ^{\alpha_{1}}+\cdots+ \biggl( \frac{x_{n}}{a_{n}} \biggr) ^{\alpha_{n}} \biggr) \\ &\qquad{}\times x_{1}^{p_{1}-1}\cdots x_{n}^{p_{n}-1}\,dx_{1} \cdots \,dx_{n} \\ &\quad = \frac{a_{1}^{p_{1}}\cdots a_{n}^{p_{n}}\Gamma( \frac {p_{1}}{\alpha _{1}} ) \cdots\Gamma( \frac{p_{n}}{\alpha n} ) }{\alpha _{1}\cdots\alpha_{n}\Gamma( \frac{p_{1}}{\alpha_{1}}+\cdots +\frac{p_{n}}{\alpha n} ) } \int_{0}^{1}\psi(t)t^{\frac{p_{1}}{\alpha _{1}}+\cdots+\frac{p_{n}}{\alpha n}-1}\,dt, \end{aligned}$$
where Γ represents the gamma function.

3 Main results and their proofs

Theorem 1

Suppose that \(n\geq2,\sum_{i=1}^{n}\frac{1}{p_{i}}=1\ (p_{i}>1),\lambda\in\mathbf{R},\lambda_{i}>0\) (or \(\lambda _{i}<0\)), \(\alpha_{i}\in\mathbf{R}\) \((i=1,\ldots,n)\). If \(K(x_{1},\ldots ,x_{n})\) is a quasi-homogeneous positive function with parameters (\(\lambda ,\lambda_{1},\ldots,\lambda_{n}\)), and
$$W_{1}= \int_{\mathbf{R}_{+}^{n-1}}K(1,u_{2},\ldots,u_{n})\prod _{i=2}^{n}u_{i}^{-\frac{\alpha_{i}+1}{p_{i}}}\,du_{2} \cdots \,du_{n} $$
is convergent, then
  1. (i)
    the inequality
    $$ \int_{\mathbf{R}_{+}^{n}}K(x_{1},\ldots,x_{n})\prod _{i=1}^{n}f_{i}(x_{i})\,dx_{1} \cdots \,dx_{n}\leq M\prod_{i=1}^{n} \Vert f_{i} \Vert _{p_{i},\alpha_{i}} $$
    (1)
    holds for some constant \(M>0\) if and only if \(\sum_{i=1}^{n}\frac{\alpha _{i}+1}{\lambda_{i}p_{i}}=\lambda+\sum_{i=1}^{n}\frac{1}{\lambda_{i}}\), where \(f_{i}(x_{i})\in L_{\alpha_{i}}^{p_{i}}(0,+\infty)\) \((i=1,\ldots ,n)\);
     
  2. (ii)

    if (1) holds, then its best constant factor is \(\inf M=\frac {W_{1}}{|\lambda_{1}|}\prod_{i=1}^{n}|\lambda_{i}|^{1/p_{i}}\).

     

Proof

(i) Sufficiency. Assume that \(\sum_{i=1}^{n}\frac{\alpha _{i}+1}{\lambda_{i}p_{i}}=\lambda+\sum_{i=1}^{n}\frac{1}{\lambda_{i}}\). Since
$$\prod_{j=1}^{n}x_{j}^{\frac{\alpha_{j}+1}{p_{j}}} \Biggl( \prod_{i=1}^{n}x_{i}^{-\frac{\alpha_{i}+1}{p_{i}}} \Biggr) ^{1/p_{j}}=\prod_{j=1}^{n}x_{j}^{\frac{\alpha_{j}+1}{p_{j}}} \prod_{i=1}^{n}x_{i}^{-\frac{\alpha_{i}+1}{p_{i}}\sum_{k=1}^{n}\frac {1}{p_{k}}}=1, $$
by Hölder’s inequality and Lemma 1, we obtain
$$\begin{aligned} &\int_{\mathbf{R}_{+}^{n}}K(x_{1},\ldots,x_{n})\prod _{i=1}^{n}f_{i}(x_{i})\,dx_{1} \cdots \,dx_{n} \\ &\quad= \int_{\mathbf{R}_{+}^{n}}K(x_{1},\ldots,x_{n}) \Biggl[ \prod_{j=1}^{n}x_{j}^{\frac{\alpha_{j}+1}{p_{j}}} \Biggl( \prod_{i=1}^{n}x_{i}^{-\frac{\alpha_{i}+1}{p_{i}}} \Biggr) ^{1/p_{j}}f_{j}(x_{j}) \Biggr] \,dx_{1}\cdots \,dx_{n} \\ &\quad \leq\prod_{j=1}^{n} \Biggl[ \int_{\mathbf{R}_{+}^{n}}x_{j}^{\alpha _{j}+1} \Biggl( \prod _{i=1}^{n}x_{i}^{-\frac{\alpha_{i}+1}{p_{i}}} \Biggr) f_{j}^{p_{j}}(x_{j})K(x_{1}, \ldots,x_{n})\,dx_{1}\cdots \,dx_{n} \Biggr] ^{1/p_{j}} \\ &\quad= \prod_{j=1}^{n} \Biggl[ \int_{0}^{+\infty}x_{j}^{\alpha_{j}+1-\frac{\alpha_{j}+1}{p_{j}}}f_{j}^{p_{j}}(x_{j}) \\ &\qquad{}\times \Biggl( \int_{\mathbf{R}_{+}^{n-1}}K(x_{1},\ldots,x_{n})\prod _{i=1(i\neq j)}^{n}x_{i}^{-\frac{\alpha_{i}+1}{p_{i}}}\,dx_{1} \cdots \,dx_{j-1}\,dx_{j+1}\cdots \,dx_{n} \Biggr) \,dx_{j} \Biggr] ^{1/p_{j}} \\ &\quad= \prod_{j=1}^{n} \biggl( \int_{0}^{+\infty}x_{j}^{\alpha_{j}+1-\frac{\alpha_{j}+1}{p_{j}}}f_{j}^{p_{j}}(x_{j}) \omega_{j}(x_{j})\,dx_{j} \biggr) ^{1/p_{j}} \\ &\quad= \prod_{j=1}^{n} \biggl[ \int_{0}^{+\infty}x_{j}^{\alpha_{j}+1-\frac {\alpha _{j}+1}{p_{j}}+\lambda_{j} ( \lambda-\sum_{i=1(i\neq j)}^{n}\frac{ \alpha_{i}+1}{\lambda_{i}p_{i}}+\sum_{i=1(i\neq j)}^{n}\frac {1}{\lambda _{i}} ) }f_{j}^{p_{j}}(x_{j})W_{j}\,dx_{j} \biggr]^{1/p_{j}} \\ &\quad= \prod_{j=1}^{n}W_{j}^{1/p_{j}} \prod_{j=1}^{n} \biggl[ \int_{0}^{+\infty }x_{j}^{\lambda_{j} ( \frac{\alpha_{j}}{\lambda_{j}}-\sum _{i=1}^{n}\frac{\alpha_{i}+1}{\lambda_{i}p_{i}}+\lambda+\sum_{i=1}^{n}\frac{1}{ \lambda_{i}} ) }f_{j}^{p_{j}}(x_{j})\,dx_{j} \biggr]^{1/p_{j}} \\ &\quad= \prod_{i=1}^{n}W_{i}^{1/p_{i}} \prod_{i=1}^{n} \biggl( \int_{0}^{+\infty }x_{i}^{\alpha_{i}}f_{i}^{p_{i}}(x_{i})\,dx_{i} \biggr) ^{1/p_{i}} \\ &\quad = \frac{W_{1}}{ \vert \lambda_{1} \vert } \Biggl( \prod_{i=1}^{n} \vert \lambda_{i} \vert ^{1/p_{i}} \Biggr) \prod _{i=1}^{n} \Vert f_{i} \Vert _{p_{i},\alpha_{i}}, \end{aligned}$$
thus (1) holds when taking any constant \(M\geq\frac {W_{1}}{|\lambda _{1}|}\prod_{i=1}^{n}|\lambda_{i}|^{1/p_{i}}\).

Necessity. Assume that (1) holds. Set \(c=\sum_{i=1}^{n}\frac {\alpha _{i}+1}{\lambda_{i}p_{i}}-\lambda-\sum_{i=1}^{n}\frac{1}{\lambda_{i}}\). Next we will prove \(c=0\).

First consider the case of \(\lambda_{i}>0\ (i=1,\ldots,n)\). If \(c>0\), for \(0<\varepsilon<c\), take
$$f_{i}(x_{i})= \textstyle\begin{cases} x{}_{i}^{(-\alpha_{i}-1+\lambda_{i}\varepsilon)/p_{i}}, & 0< x_{i}\leq1, \\ 0, & x_{i}>1,\end{cases} $$
where \(i=1,\ldots,n\). Then
$$\begin{aligned} &\prod_{i=1}^{n} \Vert f_{i} \Vert _{p_{i},\alpha_{i}}=\prod_{i=1}^{n} \biggl( \int_{0}^{1}x_{i}^{-1+\lambda_{i}\varepsilon} \, dx_{i} \biggr) ^{1/p_{i}}=\frac{1}{\varepsilon{}}\prod _{i=1}^{n} \biggl( \frac {1}{\lambda _{i}{}} \biggr) ^{1/p_{i}}, \end{aligned}$$
(2)
$$\begin{aligned} &\int_{\mathbf{R}_{+}^{n}}K(x_{1},\ldots,x_{n})\prod _{i=1}^{n}f_{i}(x_{i})\,dx_{1}\cdots\,dx_{n} \\ &\quad= \int_{0}^{1}x_{1}^{(-\alpha_{1}-1+\lambda_{1}\varepsilon)/p_{1}} \\ &\qquad{}\times \Biggl( \int_{0}^{1}\cdots \int_{0}^{1}K(x_{1},x_{2}, \ldots,x_{n})\prod_{i=2}^{n}x_{i}^{(-\alpha_{i}-1+\lambda _{i}\varepsilon )/p_{i}}\,dx_{2}\cdots\,dx_{n} \Biggr) \, dx_{1} \\ &\quad= \int_{0}^{1}x_{1}^{(-\alpha_{1}-1+\lambda_{1}\varepsilon )/p_{1}+\lambda\lambda_{1}} \Biggl( \int_{0}^{1}\cdots \int_{0}^{1}K \bigl(1,x_{1}^{-\frac{\lambda_{1}}{\lambda_{2}}}x_{2}, \ldots,x_{1}^{-\frac{\lambda_{1}}{\lambda n}}x_{n} \bigr) \\ &\qquad{}\times\prod_{i=2}^{n}x_{i}^{(-\alpha_{i}-1+\lambda _{i}\varepsilon)/p_{i}} \,dx_{2}\cdots\,dx_{n} \Biggr) \,{d}x_{1} \\ &\quad= \int_{0}^{1}x_{1}^{-\frac{\alpha_{1}+1-\lambda_{1}\varepsilon }{p_{1}}+\lambda\lambda_{1}+\lambda_{1}\sum_{i=2}^{n}\frac{1}{\lambda_{i}}-\lambda_{1}\sum_{i=2}^{n}\frac{\alpha_{i}+1-\lambda_{i}\varepsilon }{\lambda_{i}p_{i}}} \Biggl( \int_{0}^{x_{1}^{-\lambda_{1}/\lambda _{2}}}\cdots \int_{0}^{x_{n}^{-\lambda_{1}/\lambda_{n}}} \\ &\qquad{}\times K(1,u_{2},\ldots,u_{n})\prod _{i=2}^{n}u_{i}^{(-\alpha _{i}-1+\lambda_{i}\varepsilon)/p_{i}} \, du_{2}\cdots\,du_{n} \Biggr) \, dx_{1} \\ &\quad = \int_{0}^{1}x_{1}^{\lambda_{1} ( \lambda+\sum_{i=1}^{n}\frac {1}{\lambda_{i}}-\sum_{i=1}^{n}\frac{\alpha_{i}+1}{\lambda _{i}p_{i}}-\frac{1}{\lambda_{1}}+\varepsilon) } \Biggl( \int_{0}^{x_{1}^{-\lambda _{1}/\lambda_{2}}}\cdots \int_{0}^{x_{n}^{-\lambda_{1}/\lambda _{n}}} \\ &\qquad{}\times K(1,u_{2},\ldots,u_{n})\prod _{i=2}^{n}u_{i}^{(-\alpha _{i}-1+\lambda_{i}\varepsilon)/p_{i}} \, du_{2}\cdots\,du_{n} \Biggr) \, dx_{1} \\ &\quad\geq \int_{0}^{1}x_{1}^{-1-\lambda_{1}c+\lambda_{1}\varepsilon } \,{d}x_{1} \Biggl( \int_{0}^{1}\cdots \int_{0}^{1}K(1,u_{2},\ldots ,u_{n}) \\ &\qquad{}\times\prod_{i=2}^{n}u_{i}^{(-\alpha_{i}-1+\lambda _{i}\varepsilon)/p_{i}} \,du_{2}\cdots\,du_{n} \Biggr). \end{aligned}$$
(3)
It follows from (1), (2), and (3) that
$$\begin{aligned} &\int_{0}^{1}x_{1}^{-1-\lambda_{1}c+\lambda_{1}\varepsilon}\, {d}x_{1} \Biggl( \int_{0}^{1}\cdots \int_{0}^{1}K(1,u_{2},\ldots ,u_{n}) \prod_{i=2}^{n}u_{i}^{(-\alpha_{i}-1+\lambda _{i}\varepsilon)/p_{i}} \,du_{2}\cdots\,du_{n} \Biggr) \\ &\quad \leq\frac{M}{\varepsilon}\prod_{i=1}^{n} \biggl( \frac{1}{\lambda _{i}{}} \biggr) ^{1/p_{i}}. \end{aligned}$$
(4)
Since \(-1-\lambda_{1}c+\lambda_{1}\varepsilon<-1\), \(\int_{0}^{1}x_{1}^{-1-\lambda_{1}c+\lambda_{1}\varepsilon}\, {d}x_{1}\) diverges to +∞. Whence it is a contradiction to (4). In other words, it is not valid for \(c>0\).
If \(c<0\), for \(0<\varepsilon<-c\), take
$$f_{i}(x_{i})= \textstyle\begin{cases} x{}_{i}^{(-\alpha_{i}-1-\lambda_{i}\varepsilon)/p_{i}}, & x_{i}\geq 1, \\ 0, & 0< x_{i}< 1,\end{cases} $$
where \(i=1,\ldots,n\). Similarly, we get
$$\begin{aligned} &\int_{1}^{+\infty}x_{1}^{-1+\lambda_{1}c-\lambda_{1}\varepsilon}\,dx_{1} \\ &\qquad{}\times \Biggl( \int_{1}^{+\infty}\cdots \int_{1}^{+\infty }K(1,u_{2}, \ldots,u_{n})\prod_{i=2}^{n}u_{i}^{(-\alpha_{i}-1-\lambda _{i}\varepsilon)/p_{i}} \,du_{2}\cdots\,du_{n} \Biggr) \\ &\quad \leq\frac{M}{\varepsilon}\prod_{i=1}^{n} \biggl( \frac{1}{\lambda _{i}{}} \biggr) ^{1/p_{i}}. \end{aligned}$$
(5)
Since \(-1-\lambda_{1}c+ \lambda_{1}\varepsilon>-1\) and \(\int_{1}^{+\infty }x_{1}^{-1-\lambda_{1}c-\lambda_{1}\varepsilon} \,dx_{1}\) diverges to +∞, which contradicts the above inequality, hence it does not hold for \(c<0\).

To sum up, we have \(c=0\) for \(\lambda_{i}>0\ ( i=1,\ldots,n ) \).

Now let us consider the case of \(\lambda_{i}<0\ ( i=1,\ldots,n ) \). If \(c>0\), for \(0< \varepsilon<c\), take
$$f_{i}(x_{i})= \textstyle\begin{cases} x{}_{i}^{(-\alpha_{i}-1+\lambda_{i}\varepsilon)/p_{i}}, & x_{i}\geq 1, \\ 0, & 0< x_{i}< 1,\end{cases} $$
where \(i=1,\ldots,n\). Consequently,
$$\begin{aligned} &\prod_{i=1}^{n} \Vert f_{i} \Vert _{p_{i},\alpha_{i}}=\prod_{i=1}^{n} \biggl( \int_{1}^{+\infty}x_{i}^{-1+\lambda_{i}\varepsilon}\, {d}x_{i} \biggr) ^{1/p_{i}}=\frac{1}{\varepsilon}\prod _{i=1}^{n} \biggl( \frac{1}{-\lambda _{i}{}} \biggr) ^{1/p{}_{i}}{}, \end{aligned}$$
(6)
$$\begin{aligned} &\int_{\mathbf{R}_{+}^{n}}K(x_{1},\ldots,x_{n})\prod _{i=1}^{n}f_{i}(x_{i})\,dx_{1}\cdots\,dx_{n} \\ &\quad= \int_{1}^{+\infty}x_{1}^{(-\alpha_{1}-1+\lambda_{1}\varepsilon )/p_{1}} \Biggl( \int_{1}^{+\infty}\cdots \int_{1}^{+\infty }x_{1}^{\lambda \lambda_{1}}K \bigl(1,x_{1}^{-\frac{\lambda_{1}}{\lambda_{2}}}x_{2},\ldots ,x_{1}^{-\frac{\lambda_{1}}{\lambda_{n}}}x_{n} \bigr) \\ &\qquad{}\times\prod_{i=2}^{n}x_{i}^{(-\alpha_{i}-1+\lambda _{i}\varepsilon)/p_{i}} \,dx_{2}\cdots\,dx_{n} \Biggr) \,{d}x_{1} \\ &\quad= \int_{1}^{+\infty}x_{1}^{-\frac{\alpha_{1}+1-\lambda _{1}\varepsilon}{p_{1}}+\lambda\lambda_{1}+\lambda_{1}\sum_{i=2}^{n}\frac{1}{\lambda _{i}}-\lambda_{1}\sum_{i=2}^{n}\frac{\alpha_{i}+1-\lambda_{i}\varepsilon }{\lambda_{i}p_{i}}} \Biggl( \int_{x_{1}^{-\lambda_{1}/\lambda _{2}}}^{+\infty }\cdots \int_{x_{n}^{-\lambda_{1}/\lambda_{n}}}^{+\infty} \\ &\qquad{}\times K(1,u_{2},\ldots,u_{n})\prod _{i=2}^{n}u_{i}^{(-\alpha _{i}-1+\lambda_{i}\varepsilon)/p_{i}} \, du_{2}\cdots\,du_{n} \Biggr) \, dx_{1} \\ &\quad\geq \int_{1}^{+\infty}x_{1}^{-1-\lambda_{1}c+\lambda _{1}\varepsilon}\,dx_{1} \Biggl( \int_{1}^{+\infty}\cdots \int_{1}^{+\infty }K(1,u_{2}, \ldots,u_{n}) \\ &\qquad{}\times\prod_{i=2}^{n}u_{i}^{(-\alpha_{i}-1+\lambda _{i}\varepsilon)/p_{i}} \,du_{2}\cdots\,du_{n} \Biggr). \end{aligned}$$
(7)
It follows from (1), (6), and (7) that
$$\begin{aligned} &\int_{1}^{+\infty}x_{1}^{-1-\lambda_{1}c+\lambda_{1}\varepsilon}\,dx_{1} \\ &\qquad{}\times \Biggl( \int_{1}^{+\infty}\cdots \int_{1}^{+\infty }K(1,u_{2}, \ldots,u_{n})\prod_{i=2}^{n}u_{i}^{(-\alpha_{i}-1+\lambda _{i}\varepsilon)/p_{i}} \,du_{2}\cdots\,du_{n} \Biggr) \\ &\quad \leq\frac{M}{\varepsilon}\prod_{i=1}^{n} \biggl( \frac{1}{-\lambda _{i}{}} \biggr) ^{1/p_{i}}. \end{aligned}$$
(8)
Since \(-1-\lambda_{1}c+\lambda_{1}\varepsilon>-1\), \(\int_{1}^{+\infty }x_{1}^{-1-\lambda_{1}c+\lambda_{1}\varepsilon}\,dx_{1}\) diverges to +∞. Thus it is a contradiction to the above inequality. That is, it does not hold for \(c>0\).
If \(c<0\), for \(0<\varepsilon<-c\), take
$$f_{i}(x_{i})= \textstyle\begin{cases} x{}_{i}^{(-\alpha_{i}-1-\lambda_{i}\varepsilon)/p_{i}}, & 0< x_{i}\leq1, \\ 0, & x_{i}>1,\end{cases} $$
where \(i=1,\ldots,n\). Similarly, one can get
$$\begin{aligned} &\int_{0}^{1}x_{1}^{-1-\lambda_{1}c-\lambda_{1}\varepsilon} \, dx_{1} \Biggl( \int_{0}^{1}\cdots \int_{0}^{1}K(1,u_{2},\ldots ,u_{n}) \\ &\qquad{}\times\prod_{i=2}^{n}u_{i}^{(-\alpha_{i}-1-\lambda _{i}\varepsilon)/p_{i}} \,du_{2}\cdots\,du_{n} \Biggr) \\ &\quad \leq\frac{M}{\varepsilon}\prod_{i=1}^{n} \biggl( \frac{1}{-\lambda _{i}{}} \biggr) ^{1/p_{i}}. \end{aligned}$$
(9)
Since \(-1-\lambda_{1}c- \lambda_{1}\varepsilon<-1\), \(\int_{0}^{1}x_{1}^{-1-\lambda_{1}c-\lambda_{1}\varepsilon}\, {d}x_{1}\) diverges to +∞, which also contradicts the above inequality. It does not hold for \(c<0\).

To sum up, we also get \(c=0\) for \(\lambda_{i}<0\) \((i=1,\ldots,n)\).

(ii) Suppose that (1) holds. If the constant factor \(\inf M\neq\frac{W_{1}}{|\lambda_{1}|} \prod_{i=1}^{n}|\lambda_{i}|^{1/p_{i}}\), then there exists a constant \(M_{0}<\frac{W_{1}}{|\lambda_{1}|}\prod_{i=1}^{n}|\lambda_{i}|^{1/p_{i}}\) such that
$$\int_{\mathbf{R}_{+}^{n}}K(x_{1},\ldots,x_{n})\prod _{i=1}^{n}f_{i}(x_{i})\,dx_{1}\cdots\,dx_{n}\leq M_{0} \prod_{i=1}^{n} \Vert f_{i} \Vert _{p_{i},\alpha_{i}}. $$
For sufficiently small \(\varepsilon>0\) and \(\delta>0\), take
$$f_{1}(x_{1})= \textstyle\begin{cases} x{}_{1}^{(-\alpha_{1}-1-|\lambda_{1}|\varepsilon)/p_{1}}, & x_{1}\geq1, \\ 0, & 0< x_{1}< 1.\end{cases} $$
For \(i=2,3,\ldots,n\), take
$$f_{i}(x_{i})= \textstyle\begin{cases} x{}_{i}^{(-\alpha_{i}-1-|\lambda_{i}|\varepsilon)/p_{i}}, & x_{i}\geq \delta, \\ 0, & 0< x_{i}< \delta.\end{cases} $$
Therefore,
$$\begin{aligned} &\prod_{i=1}^{n} \Vert f_{i} \Vert _{p_{i},\alpha_{i}} \\ &\quad= \biggl( \int_{1}^{+\infty }x{}_{1}^{-1- \vert \lambda_{1} \vert \varepsilon} \, dx_{1} \biggr) ^{1/p_{1}}\prod _{i=2}^{n} \biggl( \int_{\delta}^{+\infty }x_{i}^{1- \vert \lambda _{i} \vert \varepsilon} \, dx_{i} \biggr) ^{1/p_{i}} \\ &\quad= \biggl( \frac{1}{ \vert \lambda_{1} \vert \varepsilon} \biggr) ^{1/p_{1}}\prod _{i=2}^{n} \biggl( \frac{1}{ \vert \lambda_{i} \vert \varepsilon }\cdot \frac{1}{\delta^{ \vert \lambda_{i} \vert \varepsilon}} \biggr) ^{1/p_{i}} \\ &\quad= \frac{1}{\varepsilon}\prod_{i=1}^{n} \biggl( \frac{1}{ \vert \lambda_{i} \vert } \biggr) ^{1/p_{i}}\prod _{i=2}^{n} \biggl( \frac{1}{\delta^{ \vert \lambda _{i} \vert \varepsilon}} \biggr) ^{1/p_{i}}, \end{aligned}$$
(10)
$$\begin{aligned} &\int_{\mathbf{R}_{+}^{n}}K(x_{1},\ldots,x_{n})\prod _{i=1}^{n}f_{i}(x_{i})\,dx_{1}\cdots\,dx_{n} \\ &\quad= \int_{0}^{1}x_{1}^{-\frac{\alpha_{1}+1+|\lambda_{1}|\varepsilon }{p_{1}}} \Biggl( \int_{\delta}^{+\infty}\cdots \int_{\delta}^{+\infty }K(x_{1}, \ldots,x_{n}) \\ &\qquad{}\times\prod_{i=2}^{n}x_{i}^{-\frac{\alpha_{i}+1+|\lambda _{i}|\varepsilon}{p_{i}}} \,dx_{2}\cdots\,dx_{n} \Biggr) \, dx_{1} \\ &\quad= \int_{0}^{1}x{}_{1}^{\lambda\lambda_{1}-\frac{\alpha _{1}+1+|\lambda _{1}|\varepsilon}{p_{1}}} \Biggl( \int_{\delta}^{+\infty}\cdots \int_{\delta}^{+\infty}K \bigl(1,x_{1}^{-\lambda_{1}/\lambda _{2}}x_{2}, \ldots,x_{1}^{-\lambda_{1}/\lambda_{n}}x_{n} \bigr) \\ &\qquad{}\times\prod_{i=2}^{n}x_{i}^{-\frac{\alpha_{i}+1+|\lambda _{i}|\varepsilon}{p_{i}}} \,dx_{2}\cdots\,dx_{n} \Biggr) \, dx_{1} \\ &\quad= \int_{1}^{+\infty}x_{1}^{\lambda\lambda_{1}-\frac{\alpha _{1}+1+|\lambda_{1}|\varepsilon}{p_{1}}-\lambda_{1}\sum _{i=2}^{n}\frac{\alpha_{i}+1+|\lambda_{i}|\varepsilon}{\lambda_{i}p_{i}}+\sum _{i=2}^{n}\frac{\lambda_{1}}{\lambda_{i}}} \Biggl( \int_{\delta x_{1}^{-\lambda _{1}/\lambda_{2}}}^{+\infty}\cdots \int_{\delta x_{n}^{-\lambda _{1}/\lambda_{n}}}^{+\infty} \\ &\qquad{}\times K(1,u_{2},\ldots,u_{n})\prod _{i=2}^{n}u_{i}^{-\frac {\alpha _{i}+1+|\lambda_{i}|\varepsilon}{p_{i}}} \, du_{2}\cdots\,du_{n} \Biggr) \, dx_{1} \\ &\quad\geq \int_{1}^{+\infty}x_{1}^{\lambda_{1} ( \lambda-\sum _{i=1}^{n}\frac{\alpha_{i}+1}{\lambda_{i}p_{i}}+\sum_{i=1}^{n}\frac{1}{\lambda _{i}}-\frac{1}{\lambda_{1}}-\sum_{i=1}^{n}\frac{|\lambda_{i}|\varepsilon}{\lambda_{i}p_{i}} ) } \, dx_{1} \Biggl( \int_{\delta}^{+\infty }\cdots \int_{\delta}^{+\infty} \\ &\qquad{}\times K(1,u_{2},\ldots,u_{n})\prod _{i=2}^{n}u_{i}^{-\frac {\alpha _{i}+1+|\lambda_{i}|\varepsilon}{p_{i}}} \, du_{2}\cdots\,du_{n} \Biggr) \\ &\quad= \int_{1}^{+\infty}x_{1}^{-1- \vert \lambda_{1} \vert \varepsilon} \, dx_{1} \Biggl( \int_{\delta}^{+\infty}\cdots \int_{\delta}^{+\infty }K(1,u_{2}, \ldots,u_{n}) \times\prod_{i=2}^{n}u_{i}^{-\frac{\alpha_{i}+1+ \vert \lambda _{i} \vert \varepsilon}{p_{i}}} \,du_{2}\cdots\,du_{n} \Biggr) \\ &\quad= \frac{1}{ \vert \lambda_{1} \vert \varepsilon} \int_{\delta}^{+\infty}\cdots \int_{\delta}^{+\infty}K(1,u_{2}, \ldots,u_{n})\prod_{i=2}^{n}u_{i}^{- \frac{\alpha_{i}+1+ \vert \lambda_{i} \vert \varepsilon}{p_{i}}} \,du_{2}\cdots\,du_{n}. \end{aligned}$$
(11)
It follows from (1), (10), and (11) that
$$\begin{aligned} &\frac{1}{ \vert \lambda_{1} \vert } \int_{\delta}^{+\infty}\cdots \int_{\delta }^{+\infty}K(1,u_{2}, \ldots,u_{n})\prod_{i=2}^{n}u_{i}^{-\frac{\alpha _{i}+1+ \vert \lambda_{i} \vert \varepsilon}{p_{i}}} \,du_{2}\cdots\,du_{n} \\ &\quad \leq M_{0}\prod_{i=1}^{n} \biggl( \frac{1}{ \vert \lambda_{i} \vert } \biggr) ^{1/p_{i}}\prod _{i=2}^{n} \biggl( \frac{1}{\delta^{ \vert \lambda _{i} \vert \varepsilon}} \biggr) ^{1/p_{i}}. \end{aligned}$$
Consequently,
$$\frac{1}{ \vert \lambda_{1} \vert }\prod_{i=1}^{n} \vert \lambda_{i} \vert ^{1/p_{i}} \int_{\delta }^{+\infty}\cdots \int_{\delta}^{+\infty}K(1,u_{2},\ldots ,u_{n})\prod_{i=2}^{n}u_{i}^{-(\alpha_{i}+1)/p_{i}} \,du_{2}\cdots\,du_{n}\leq M_{0} $$
as \(\varepsilon\rightarrow0^{+}\). And then let \(\delta\rightarrow 0^{+}\), we eventually get
$$\begin{aligned} &\frac{W_{1}}{ \vert \lambda_{1} \vert }\prod_{i=1}^{n} \vert \lambda_{i} \vert ^{1/p_{i}} \\ &\quad= \frac{1}{ \vert \lambda_{1} \vert }\prod_{i=1}^{n} \vert \lambda_{i} \vert ^{1/p_{i}} \int_{ \mathbf{R}_{+}^{n-1}}K(1,u_{2},\ldots,u_{n})\prod _{i=2}^{n}u_{i}^{-(\alpha _{i}+1)/p_{i}} \,du_{2}\cdots\,du_{n} \\ &\quad \leq M_{0}, \end{aligned}$$
this is a contradiction. Hence \(\inf M=\frac{W_{1}}{|\lambda_{1}|}\prod_{i=1}^{n}|\lambda_{i}|^{1/p_{i}}\), i.e., the constant factor \(\frac{W_{1}}{|\lambda_{1}|}\prod_{i=1}^{n}|\lambda_{i}|^{1/p_{i}}\) is the best. □

4 Applications

Theorem 2

Suppose that \(n\geq2,\sum_{i=1}^{n}\frac {1}{p_{i}}=1\ (p_{i}>1),\lambda_{i}>0\) (or \(\lambda_{i}<0\)), \(f_{i}(x_{i})\in L_{p_{i}-1}^{p_{i}}(0,+\infty), i=1,\ldots,n\). Then
$$\int_{\mathbf{R}_{+}^{n}}\frac{\min\{x_{1}^{\lambda_{1}},\ldots ,x_{n}^{\lambda_{n}}\}}{\max\{x_{1}^{\lambda_{1}},\ldots ,x_{n}^{\lambda _{n}}\}}\prod_{i=1}^{n}f_{i}(x_{i}) \,dx_{1}\cdots\,dx_{n}\leq \Biggl( n!\prod _{i=1}^{n} \vert \lambda_{i} \vert ^{\frac{1}{p_{i}}-1} \Biggr) \prod_{i=1}^{n} \Vert f_{i} \Vert _{p_{i},p_{i}-1}, $$
where the constant factor is the best.

Proof

Set \(\alpha_{i}=p_{i}-1,\lambda=0\), then \(\sum_{i=1}^{n} \frac{\alpha_{i}+1}{\lambda_{i}p_{i}}=\lambda+\sum_{i=1}^{n}\frac{1}{ \lambda_{i}}\). Take
$$K(x_{1},\ldots,x_{n})=\frac{\min\{x_{1}^{\lambda_{1}},\ldots ,x_{n}^{\lambda_{n}}\}}{\max\{x_{1}^{\lambda_{1}},\ldots ,x_{n}^{\lambda _{n}}\}}, $$
then \(K(x_{1},\ldots,x_{n})\) is a quasi-homogeneous positive function with parameters \((\lambda,\lambda_{1},\ldots,\lambda_{n})\), and
$$\begin{aligned} W_{1} &= \int_{\mathbf{R}_{+}^{n-1}}K(1,u_{2},\ldots,u_{n})\prod _{i=2}^{n}u_{i}^{-(\alpha_{i}+1)/p_{i}} \,du_{2}\cdots\,du_{n} \\ &= \int_{\mathbf{R}_{+}^{n-1}}\frac{\min\{1,u_{2}^{\lambda_{2}},\ldots ,u_{n}^{\lambda_{n}}\}}{\max\{1,u_{2}^{\lambda_{2}},\ldots ,u_{n}^{\lambda_{n}}\}}\prod_{i=2}^{n}u_{i}^{-1} \,du_{2}\cdots\,du_{n} \\ &= \prod_{i=2}^{n}\frac{1}{ \vert \lambda_{i} \vert } \int_{\mathbf {R}_{+}^{n-1}}\frac{\min\{1,t_{2},\ldots,t_{n}\}}{\max\{1,t_{2},\ldots,t_{n}\}}\prod _{i=2}^{n}t_{i}^{-1} \, dt_{2}\cdots\,dt_{n}. \end{aligned}$$
In view of [1], we get
$$\int_{\mathbf{R}_{+}^{n-1}}\frac{\min\{1,t_{2},\ldots,t_{n}\}}{\max \{1,t_{2},\ldots,t_{n}\}}\prod_{i=2}^{n}t_{i}^{-1} \,dt_{2}\cdots\,dt_{n}=n!, $$
it follows that
$$\frac{W_{1}}{ \vert \lambda_{1} \vert }\prod_{i=1}^{n} \vert \lambda_{i} \vert ^{\frac {1}{p_{i}}}=n!\frac{1}{ \vert \lambda_{1} \vert } \prod _{i=2}^{n} \vert \lambda _{i} \vert ^{-1}\prod_{i=1}^{n} \vert \lambda_{i} \vert ^{\frac{1}{p_{i}}}=n!\prod _{i=1}^{n} \vert \lambda_{i} \vert ^{\frac{1}{p_{i}}-1}. $$
According to Theorem 1, we know that Theorem 2 holds. □

Theorem 3

Suppose that \(n\geq2,\sum_{i=1}^{n}\frac {1}{p_{i}}=1\ (p_{i}>1),a>0,\lambda_{i}>0\), \(\alpha_{i}\in\mathbf {R},p_{i}>1+\alpha _{i}\). Then
  1. (i)
    the inequality
    $$ \int_{\mathbf{R}_{+}^{n}}\frac{1}{ ( x_{1}^{\lambda _{1}}+x_{2}^{\lambda _{2}}+\cdots+x_{n}^{\lambda_{n}} ) ^{a}}\prod_{i=1}^{n}x_{i}^{-\frac{\alpha_{i}+1}{p_{i}}} \,dx_{1}\cdots\,dx_{n}\leq M\prod _{i=1}^{n} \Vert f_{i} \Vert _{p_{i},\alpha_{i}} $$
    (12)
    holds for some constant \(M>0\) if and only if \(\sum_{i=1}^{n}\frac{\alpha _{i}+1}{\lambda_{i}p_{i}}=-a+\sum_{i=1}^{n}\frac{1}{\lambda_{i}}\).
     
  2. (ii)
    if (12) holds, then its best constant factor is
    $$\inf M=\frac{1}{\Gamma(a)}\prod_{i=1}^{n} \lambda_{i}{}^{\frac {1}{p_{i}}-1}\prod_{i=1}^{n} \Gamma \biggl( \frac{1}{\lambda_{i}} \biggl( 1-\frac {\alpha _{i}+1}{p_{i}} \biggr) \biggr). $$
     

Proof

Set \(K(x_{1},\ldots,x_{n})=1/ ( x_{1}^{\lambda _{1}}+x_{2}^{\lambda_{2}}+\cdots+x_{n}^{\lambda_{n}} ) ^{a}\), then \(K(x_{1},\ldots,x_{n})\) is a quasi-homogeneous positive funct+ion with parameters \((-a,\lambda_{1},\ldots,\lambda_{n})\). By Lemma 2,
$$\begin{aligned} W_{1} ={}& \int_{\mathbf{R}_{+}^{n-1}}\frac{1}{ ( 1+x_{2}^{\lambda _{2}}+\cdots+x_{n}^{\lambda_{n}} ) ^{a}}\prod_{i=2}^{n}x_{i}^{-\frac{\alpha_{i}+1}{p_{i}}} \,dx_{2}\cdots\,dx_{n} \\ = {}&\lim_{r\rightarrow+\infty} \int_{x_{i}>0,x_{2}^{\lambda_{2}}+\cdots +x_{n}^{\lambda_{n}}\leq r}\frac{1}{ \{ 1+r [ ( \frac {x_{1}}{r^{1/\lambda_{1}}} ) ^{\lambda_{1}}+\cdots+( \frac{x_{n}}{r^{1/\lambda_{n}}} ) ^{\lambda_{n}} ] \} ^{n}} \\ &{}\times\prod_{i=2}^{n}x_{i}^{ ( 1-\frac{\alpha _{i}+1}{p_{i}} ) -1} \,dx_{2}\cdots\,dx_{n} \\ ={}& \lim_{r\rightarrow+\infty}\frac{r^{\sum_{i=2}^{n}\frac{1}{\lambda _{i}}-\sum_{i=2}^{n}\frac{1}{\lambda_{i}} ( 1-\frac{\alpha _{i}+1}{p_{i}}) }\prod_{i=2}^{n}\Gamma( \frac{1}{\lambda_{i}} ( 1-\frac{\alpha_{i}+1}{p_{i}} ) ) }{\prod_{i=2}^{n}\lambda_{i}\Gamma ( \sum_{i=2}^{n}\frac{1}{\lambda_{i}} ( 1-\frac{\alpha _{i}+1}{p_{i}} ) ) } \\ &{}\times \int_{0}^{1}\frac{1}{ ( 1+rt ) ^{a}}t^{\sum _{i=2}^{n}\frac{1}{\lambda_{i}} ( 1-\frac{\alpha_{i}+1}{p_{i}} ) -1} \, dt \\ ={}& \lim_{r\rightarrow+\infty}\frac{\prod_{i=2}^{n}\Gamma( \frac {1}{\lambda_{i}} ( 1-\frac{\alpha_{i}+1}{p_{i}} ) ) }{\prod_{i=2}^{n}\lambda_{i}\Gamma( \sum_{i=2}^{n}\frac{1}{\lambda _{i}}( 1-\frac{\alpha_{i}+1}{p_{i}} ) ) } \int_{0}^{r}\frac {1}{( 1+u ) ^{a}}u^{\sum_{i=2}^{n}\frac{1}{\lambda_{i}} ( 1-\frac{\alpha_{i}+1}{p_{i}} ) -1} \, du \\ ={}& \frac{\prod_{i=2}^{n}\Gamma( \frac{1}{\lambda_{i}} ( 1-\frac{\alpha_{i}+1}{p_{i}} ) ) }{\prod_{i=2}^{n}\lambda_{i}\Gamma ( \sum_{i=2}^{n}\frac{1}{\lambda_{i}} ( 1-\frac{\alpha _{i}+1}{p_{i}} ) ) } \int_{0}^{+\infty}\frac{1}{ ( 1+u ) ^{a}}u^{\sum_{i=2}^{n}\frac{1}{\lambda_{i}} ( 1-\frac{\alpha _{i}+1}{p_{i}}) -1} \,du \\ ={}& \frac{\prod_{i=2}^{n}\Gamma( \frac{1}{\lambda_{i}} ( 1-\frac{\alpha_{i}+1}{p_{i}} ) ) }{\prod_{i=2}^{n}\lambda_{i}\Gamma ( \sum_{i=2}^{n}\frac{1}{\lambda_{i}} ( 1-\frac{\alpha _{i}+1}{p_{i}} ) ) } \\ &{}\times\frac{\Gamma( \sum_{i=2}^{n}\frac{1}{\lambda_{i}} ( 1-\frac{\alpha_{i}+1}{p_{i}} ) ) \Gamma( a-\sum_{i=2}^{n}\frac{1}{\lambda_{i}} ( 1-\frac{\alpha_{i}+1}{p_{i}} ) ) }{\Gamma( a ) } \\ = {}&\frac{1}{\Gamma( a ) }\prod_{i=2}^{n} \frac{1}{\lambda _{i}}\prod_{i=2}^{n} \Gamma \biggl( \frac{1}{\lambda_{i}} \biggl( 1-\frac{\alpha _{i}+1}{p_{i}} \biggr) \biggr) \Gamma \biggl( \frac{1}{\lambda_{1}} \biggl( 1-\frac{\alpha_{1}+1}{p_{1}} \biggr) \biggr) \\ ={}& \frac{1}{\Gamma( a ) }\prod_{i=2}^{n} \frac{1}{\lambda _{i}}\prod_{i=1}^{n} \Gamma \biggl( \frac{1}{\lambda_{i}} \biggl( 1-\frac{\alpha _{i}+1}{p_{i}} \biggr) \biggr). \end{aligned}$$
Based on this, we can obtain
$$\frac{W_{1}}{\lambda_{1}}\prod_{i=1}^{n} \lambda_{i}^{\frac{1}{p_{i}}}= \frac{1}{\Gamma( a ) }\prod _{i=1}^{n}\lambda_{i}^{\frac {1}{p_{i}}-1} \prod_{i=1}^{n}\Gamma \biggl( \frac{1}{\lambda_{i}} \biggl( 1-\frac {\alpha _{i}+1}{p_{i}} \biggr) \biggr). $$
According to Theorem 1, we know that Theorem 3 holds. □

Declarations

Authors’ information

Yong Hong (1959-), Professor, the major field of interest is in the area of inequality and harmonic analysis. Junfei Cao, email: jfcaomath@163.com. Bing He, email: hzs314@163.com. Bicheng Yang, email: bcyang@gdei.edu.cn.

Funding

This work is supported by the Guangdong Province Natural Science Foundation (No. 2015A030313896), the Characteristic Innovation Project (Natural Science) of Guangdong Province (Nos. 2015KTSCX097, 2016KTSCX094), the Science and Technology Program Project of Guangzhou (No. 201707010230).

Authors’ contributions

JC carried out the mathematical studies, participated in the sequence alignment, and drafted the manuscript. BH, YH, and BY participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, Guangdong University of Education, Guangzhou, P.R. China
(2)
College of Mathematics and Statistics, Guangdong University of Finance and Economics, Guangzhou, P.R. China

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