On Hilbert type inequalities

Abstract

In the present paper we establish new inequalities similar to the extensions of Hilbert’s double-series inequality and also give their integral analogues. Our results provide some new estimates to these types of inequalities.

MSC:26D15.

1 Introduction

In recent years several authors have given considerable attention to Hilbert’s double-series inequality together with its integral version, inverse version, and various generalizations (see [1–9]). In this paper, we establish multivariable sum inequalities for the extensions of Hilbert’s inequality and also obtain their integral forms. Our results provide some new estimates to these types of inequalities.

The well-known classical extension of Hilbert’s double-series theorem can be stated as follows [10], p.253].

Theorem A If$p_1,p_2>1$are real numbers such that$\frac{1}{p_1}+\frac{1}{p_2}\ge 1$and$0<\lambda =2-\frac{1}{p_1}-\frac{1}{p_2}=\frac{1}{q_1}+\frac{1}{q_2}\le 1$, where, as usual, $q_1$and$q_2$are the conjugate exponents of$p_1$and$p_2$respectively, then

$$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{{(m+n)}^{\lambda }}\le K{\left(\sum _{m=1}^{\mathrm{\infty }}{a}_m^p\right)}^{1/p_1}{\left(\sum _{n=1}^{\mathrm{\infty }}{b}_n^q\right)}^{1/p_2},$$

(1.1)

where$K=K(p_1,p_2)$depends on$p_1$and$p_2$only.

In 2000, Pachpatte [11] established a new inequality similar to inequality (1.1) as follows:

Theorem A′ Let p q$a(s)$$b(t)$$a(0)$$b(0)$$\mathrm{\nabla }a(s)$and$\mathrm{\nabla }b(t)$be as in [11], then

$$\begin{array}{rl}\sum _{s=1}^m\sum _{t=1}^n\frac{|a(s)||b(t)|}{q{s}^{p-1}+p{t}^{q-1}}\le & \frac{1}{pq}{m}^{(p-1)/p}{n}^{(q-1)/q}{(\sum _{s=1}^m(m-s+1)|\mathrm{\nabla }a(s){|}^p)}^{1/p}\\ \times {(\sum _{t=1}^n(n-t+1)|\mathrm{\nabla }b(t){|}^q)}^{1/q}.\end{array}$$

(1.2)

The integral analogue of inequality (1.1) is as follows [10], p.254].

Theorem B Let p, q, $p^{\prime }$, $q^{\prime }$and λ be as in Theorem A. If$f\in L^p(0,\mathrm{\infty })$and$g\in L^q(0,\mathrm{\infty })$, then

$${\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{f(x)g(x)}{{(x+y)}^{\lambda }}dxdy\le K{({\int }_0^{\mathrm{\infty }}{f}^p(x)dx)}^{1/p}{({\int }_0^{\mathrm{\infty }}{g}^q(y)dy)}^{1/q},$$

(1.3)

where$K=K(p,q)$depends on p and q only.

In [11], Pachpatte also established a similar version of inequality (1.3) as follows.

Theorem B′ Let p q$f(s)$$g(t)$$f(0)$$g(0)$$f^{\prime }(s)$and$g^{\prime }(t)$be as in [11], then

$$\begin{array}{r}{\int }_0^x{\int }_0^y\frac{|f(s)||g(t)|}{q{s}^{p-1}+p{t}^{q-1}}dtds\\ \le \frac{1}{pq}{x}^{(p-1)/p}{y}^{(q-1)/q}{({\int }_0^x(x-s)|f^{\prime }(s){|}^{p}ds)}^{1/p}{({\int }_0^y(y-t)|g^{\prime }(t){|}^{q}dt)}^{1/q}.\end{array}$$

(1.4)

In the present paper we establish some new inequalities similar to Theorems A, A^′, B and B^′. Our results provide some new estimates to these types of inequalities.

2 Statement of results

Our main results are given in the following theorems.

Theorem 2.1 Let$p_i>1$be constants and$\frac{1}{p_i}+\frac{1}{q_i}=1$. Let$a_i(s_{1i},\dots ,s_{ni})$be real-valued functions defined for$s_{ji}=1,2,\dots ,m_{ji}$, where$m_{ji}$ ($i,j=1,2,\dots ,n$) are natural numbers. For convenience, we write$a_i(0,\dots ,0)=0$and$a_i(0,s_{2i},\dots ,s_{ni})=a_i(s_{1i},0,s_{3i},\dots ,s_{ni})=\cdots =a_i(s_{1i},\dots ,s_{n-1,i},0)=0$. Define the operators${\mathrm{\nabla }}_i$by${\mathrm{\nabla }}_i{a}_i(s_{1i},\dots ,s_{ni})=a_i(s_{1i},\dots ,s_{ni})-a_i(s_{1i},\dots ,s_{i-1,i},s_{ii}-1,s_{i+1,i},\dots ,s_{ni})$for any function$a_i(s_{1i},\dots ,s_{ni})$. Then

$$\begin{array}{r}\sum _{s_{11}=1}^{m_{11}}\cdots \sum _{s_{n1}=1}^{m_{n1}}\sum _{s_{12}=1}^{m_{12}}\cdots \sum _{s_{n2}=1}^{m_{n2}}\cdots \sum _{s_{1n}=1}^{m_{1n}}\cdots \sum _{s_{nn}=1}^{m_{nn}}\frac{{\prod }_{i=1}^n|a_i(s_{1i},\dots ,s_{ni})|}{{({\sum }_{i=1}^n(s_{1i}\cdots s_{ni})/q_i)}^{{\sum }_{i=1}^{n}1/q_i}}\\ \le M\prod _{i=1}^n{(\sum _{s_{ni}=1}^{m_{ni}}\cdots \sum _{s_{1i}=1}^{m_{1i}}\prod _{j=1}^n(m_{ji}-s_{ji}+1)|{\mathrm{\nabla }}_n\cdots {\mathrm{\nabla }}_1{a}_i(s_{1i},\dots ,s_{ni}){|}^{p_i})}^{1/p_i},\end{array}$$

(2.1)

where

$$M=M(m_{1i},\dots ,m_{ni})={(n-\sum _{i=1}^{n}1/p_i)}^{{\sum }_{i=1}^{n}1/p_i-n}\prod _{i=1}^n{(m_{1i}\cdots m_{ni})}^{1/q_i}.$$

Remark 2.1 Let $a_i(s_{1i},\dots ,s_{ni})$ change to $a_i(s_i)$ in Theorem 2.1 and in view of $a_i(0)=0$ and $\mathrm{\nabla }{a}_i(s_i)=a_i(s_i)-a_i(s_i-1)$ for any function $a_i(s_i)$, $i=1,2,\dots ,n$, then

$$\sum _{s_1=1}^{m_1}\sum _{s_2=1}^{m_2}\cdots \sum _{s_n=1}^{m_n}\frac{{\prod }_{i=1}^n|a_i(s_i)|}{{({\sum }_{i=1}^n{s}_i/q_i)}^{{\sum }_{i=1}^{n}1/q_i}}\le \overline{M}\prod _{i=1}^n{(\sum _{s_i=1}^{m_i}(m_i-s_i+1)|\mathrm{\nabla }{a}_i(s_i){|}^{p_i})}^{1/p_i},$$

(2.2)

where

$$\overline{M}=\overline{M}(m_1,\dots ,m_n)={(n-\sum _{i=1}^n\frac{1}{p_i})}^{{\sum }_{i=1}^{n}1/p_i-n}\cdot \prod _{i=1}^n{m}_i^{1/q_i}.$$

Remark 2.2 Taking for $n=2$ in Remark 2.1. If $p_1,p_2>1$ satisfy $\frac{1}{p_1}+\frac{1}{p_2}\ge 1$ and $0<\lambda =2-\frac{1}{p_1}-\frac{1}{p_2}=\frac{1}{q_1}+\frac{1}{q_2}\le 1$, then inequality (2.2) reduces to

$$\begin{array}{r}\sum _{s_1=1}^{m_1}\sum _{s_2=1}^{m_2}\frac{|a_1(s_1)||a_2(s_2)|}{{(q_2{s}_1+q_1{s}_2)}^{\lambda }}\\ \le \frac{1}{{(\lambda q_1{q}_2)}^{\lambda }}{m}_1^{1/q_1}{m}_2^{1/q_2}{(\sum _{s_1=1}^{m_1}(m_1-s_1+1)|\mathrm{\nabla }{a}_1(s_1){|}^{p_1})}^{1/p_1}\\ \times {(\sum _{s_2=1}^{m_2}(m_2-s_2+1)|\mathrm{\nabla }{a}_2(s_2){|}^{p_2})}^{1/p_2},\end{array}$$

(2.3)

which is an interesting variation of inequality (1.1).

On the other hand, if $\lambda =1$, then $\frac{1}{p_1}+\frac{1}{p_2}=\frac{1}{q_1}+\frac{1}{q_2}=1$ and so $p_1=q_2$, $p_2=q_1$. In this case inequality (2.3) reduces to

$$\begin{array}{r}\sum _{s_1=1}^{m_1}\sum _{s_2=1}^{m_2}\frac{|a_1(s_1)||a_2(s_2)|}{p_1{s}_1+q_1{s}_2}\\ \le \frac{1}{p_1{q}_1}{m}_1^{(p_1-1)/p_1}{m}_2^{(q_1-1)/q_1}{(\sum _{s_1=1}^{m_1}(m_1-s_1+1)|\mathrm{\nabla }{a}_1(s_1){|}^{p_1})}^{1/p_1}\\ \times {(\sum _{s_2=1}^{m_2}(m_2-s_2+1)|\mathrm{\nabla }{a}_2(s_2){|}^{q_1})}^{1/q_1}.\end{array}$$

This is just a similar version of inequality (1.2) in Theorem A^′.

Theorem 2.2 Let$p_i>1$be constants and$\frac{1}{p_i}+\frac{1}{q_i}=1$. Let$f_i({\tau }_{1i},\dots ,{\tau }_{ni})$be real-valued nth differentiable functions defined on$[0,x_{1i})\times \cdots \times [0,x_{ni})$, where$0\le x_{ji}\le t_{ji}$, $t_{ji}\in (0,\mathrm{\infty })$and$i,j=1,2,\dots ,n$. Suppose

$$f_i(x_{1i},\dots ,x_{ni})={\int }_0^{x_{1i}}\cdots {\int }_0^{x_{ni}}\frac{{\partial }^n}{\partial {\tau }_{1i}\cdots \partial {\tau }_{ni}}{f}_i({\tau }_{1i},\dots ,{\tau }_{ni})d{\tau }_{1i}\cdots d{\tau }_{ni},$$

then

$$\begin{array}{r}{\int }_0^{t_{11}}\cdots {\int }_0^{t_{n1}}{\int }_0^{t_{12}}\cdots {\int }_0^{t_{n2}}\cdots {\int }_0^{t_{1n}}\cdots {\int }_0^{t_{nn}}\\ \frac{{\prod }_{i=1}^n{({\int }_0^{x_{1i}}\cdots {\int }_0^{x_{ni}}|\frac{{\partial }^n}{\partial {\tau }_{1i}\cdots \partial {\tau }_{ni}}{f}_i({\tau }_{1i},\dots ,{\tau }_{ni}){|}^{p_i}d{\tau }_{1i}\cdots d{\tau }_{ni})}^{1/p_i}}{{({\sum }_{i=1}^n(x_{1i}\cdots x_{ni})/q_i)}^{{\sum }_{i=1}^{n}1/q_i}}\\ d{x}_{11}\cdots d{x}_{n1}d{x}_{12}\cdots d{x}_{n2}\cdots d{x}_{1n}\cdots d{x}_{nn}\\ \le N\prod _{i=1}^n({\int }_0^{t_{1i}}\cdots {\int }_0^{t_{ni}}\prod _{j=1}^n(t_{ji}-x_{ji})\\ \times {|\frac{{\partial }^n}{\partial x_{1i}\cdots \partial x_{ni}}{f}_i(x_{1i},\dots ,x_{ni})|}^{p_i}d{x}_{1i}\cdots d{x}_{ni}{)}^{1/p_i},\end{array}$$

(2.4)

where

$$N=N(t_{1i},\dots ,t_{ni})={(n-\sum _{i=1}^n\frac{1}{p_i})}^{{\sum }_{i=1}^{n}1/p_i-n}\cdot \prod _{i=1}^n{(t_{1i}\cdots t_{ni})}^{1/q_i}.$$

Remark 2.3 Let $f_i(x_{1i},\dots ,x_{ni})$ change to $f_i(s_i)$ in Theorem 2.2 and in view of $f_i(0)=0$, $i=1,2,\dots ,n$, then

$$\begin{array}{r}{\int }_0^{x_1}\dots {\int }_0^{x_n}\frac{{\prod }_{i=1}^n|f(s_i)|}{{({\sum }_{i=1}^n{s}_i/q_i)}^{{\sum }_{i=1}^{n}1/q_i}}d{s}_n\cdots d{s}_1\\ \le \overline{N}\prod _{i=1}^n{({\int }_0^{x_i}(x_i-s_i)|f_i^{\prime }(s_i){|}^{p_i}d{s}_i)}^{1/p_i},\end{array}$$

(2.5)

where

$$\overline{N}=\overline{N}(x_1,\dots ,x_n)={(n-\sum _{i=1}^n\frac{1}{p_i})}^{{\sum }_{i=1}^{n}1/p_i-n}\cdot \prod _{i=1}^n{x}_i^{1/q_i}.$$

Remark 2.4 Taking for $n=2$ in Remark 2.3, if $p_1,p_2>1$ are such that $\frac{1}{p_1}+\frac{1}{p_2}\ge 1$ and $0<\lambda =2-\frac{1}{p_1}-\frac{1}{p_2}=\frac{1}{q_1}+\frac{1}{q_2}\le 1$, inequality (2.5) reduces to

$$\begin{array}{r}{\int }_0^{x_1}{\int }_0^{x_2}\frac{|f_1(s_1)||f_2(s_2)|}{{(q_2{s}_1+q_1{s}_2)}^{\lambda }}d{s}_{2}d{s}_1\\ \le \frac{1}{{(\lambda q_1{q}_2)}^{\lambda }}{x}_1^{1/q_1}{x}_2^{1/q_2}{({\int }_0^{x_1}(x_1-s_1)|f_1^{\prime }(s_1){|}^{p_1}d{s}_1)}^{1/p_1}\\ \times {({\int }_0^{x_2}(x_2-s_2)|f_2^{\prime }(s_2){|}^{p_2}d{s}_2)}^{1/p_2},\end{array}$$

(2.6)

which is an interesting variation of inequality (1.3).

On the other hand, if $\lambda =1$, then $\frac{1}{p_1}+\frac{1}{p_2}=\frac{1}{q_1}+\frac{1}{q_2}=1$ and so $p_1=q_2$, $p_2=q_1$. In this case inequality (2.6) reduces to

$$\begin{array}{r}{\int }_0^{x_1}{\int }_0^{x_2}\frac{|f_1(s_1)||f_2(s_2)|}{p_1{s}_1+q_1{s}_2}\\ \le \frac{h_1{h}_2}{p_1{q}_1}{x}_1^{(p_1-1)/p_1}{x}_2^{(q_1-1)/q_1}{({\int }_0^{x_1}(x_1-s_1)|f_1^{\prime }(s_1){|}^{p_1}d{s}_1)}^{1/p_1}\\ \times {({\int }_0^{x_2}(x_2-s_2)|f_2^{\prime }(s_2){|}^{q_1}d{s}_2)}^{1/q_1}.\end{array}$$

This is just a similar version of inequality (1.4) in Theorem B^′.

3 Proofs of results

Proof of Theorem 2.1 From the hypotheses $a_i(0,\dots ,0)=a_i(0,s_{2i},\dots ,s_{ni})=a_i(s_{1i},0,s_{3i},\dots ,s_{ni})=\cdots =a_i(s_{1i},\dots ,s_{n-1,i},0)=0$, we have

$$|a_i(s_{1i},\dots ,s_{ni})|\le \sum _{{\tau }_{ni}=1}^{s_{ni}}\cdots \sum _{{\tau }_{1i}=1}^{s_{1i}}|{\mathrm{\nabla }}_n\cdots {\mathrm{\nabla }}_1{a}_i({\tau }_{1i},\dots ,{\tau }_{ni})|.$$

(3.1)

From the hypotheses of Theorem 2.1 and in view of Hölder’s inequality (see [10]) and inequality for mean [10], we obtain

$$\begin{array}{rl}\prod _{i=1}^n|a_i(s_{1i},\dots ,s_{ni})|& \le \prod _{i=1}^n\sum _{{\tau }_{ni}=1}^{s_{ni}}\cdots \sum _{{\tau }_{1i}=1}^{s_{1i}}|{\mathrm{\nabla }}_n\cdots {\mathrm{\nabla }}_1{a}_i({\tau }_{1i},\dots ,{\tau }_{ni})|\\ \le \prod _{i=1}^n{(s_{1i}\cdots s_{ni})}^{1/q_i}{(\sum _{{\tau }_{ni}=1}^{s_{ni}}\cdots \sum _{{\tau }_{1i}=1}^{s_{1i}}|{\mathrm{\nabla }}_n\cdots {\mathrm{\nabla }}_1{a}_i({\tau }_{1i},\dots ,{\tau }_{ni}){|}^{p_i})}^{1/p_i}\\ \le \frac{{({\sum }_{i=1}^n(s_{1i}\cdots s_{ni})/q_i)}^{{\sum }_{i=1}^{n}1/q_i}}{{(n-{\sum }_{i=1}^{n}1/p_i)}^{n-{\sum }_{i=1}^{n}1/p_i}}\\ \times \prod _{i=1}^n{(\sum _{{\tau }_{ni}=1}^{s_{ni}}\cdots \sum _{{\tau }_{1i}=1}^{s_{1i}}|{\mathrm{\nabla }}_n\cdots {\mathrm{\nabla }}_1{a}_i({\tau }_{1i},\dots ,{\tau }_{ni}){|}^{p_i})}^{1/p_i}.\end{array}$$

(3.2)

Dividing both sides of (3.2) by ${({\sum }_{i=1}^n(s_{1i}\cdots s_{ni})/q_i)}^{{\sum }_{i=1}^{n}1/q_i}$ and then taking sums over $s_{ji}$ from 1 to $m_{ji}$ ($i,j=1,2,\dots ,n$), respectively and then using again Hölder’s inequality, we obtain

$$\begin{array}{r}\sum _{s_{11}=1}^{m_{11}}\cdots \sum _{s_{n1}=1}^{m_{n1}}\sum _{s_{12}=1}^{m_{12}}\cdots \sum _{s_{n2}=1}^{m_{n2}}\cdots \sum _{s_{1n}=1}^{m_{1n}}\cdots \sum _{s_{nn}=1}^{m_{nn}}\frac{{\prod }_{i=1}^n|{\mathrm{\nabla }}_n\cdots {\mathrm{\nabla }}_1{a}_i(s_{1i},\dots ,s_{ni})|}{{({\sum }_{i=1}^n(s_{1i}\cdots s_{ni})/q_i)}^{{\sum }_{i=1}^{n}1/q_i}}\\ \le {(n-\sum _{i=1}^{n}1/p_i)}^{{\sum }_{i=1}^{n}1/p_i-n}\\ \times \prod _{i=1}^n(\sum _{s_{ni}=1}^{m_{ni}}\cdots \sum _{s_{1i}=1}^{m_{1i}}{(\sum _{{\tau }_{ni}=1}^{s_{ni}}\cdots \sum _{{\tau }_{1i}=1}^{s_{1i}}|{\mathrm{\nabla }}_n\cdots {\mathrm{\nabla }}_1{a}_i({\tau }_{1i},\dots ,{\tau }_{ni}){|}^{p_i})}^{1/p_i})\\ \le {(n-\sum _{i=1}^{n}1/p_i)}^{{\sum }_{i=1}^{n}1/p_i-n}\\ \times \prod _{i=1}^n{(m_{1i}\cdots m_{ni})}^{1/q_i}{(\sum _{s_{ni}=1}^{m_{ni}}\cdots \sum _{s_{1i}=1}^{m_{1i}}(\sum _{{\tau }_{ni}=1}^{s_{ni}}\cdots \sum _{{\tau }_{1i}=1}^{s_{1i}}|{\mathrm{\nabla }}_n\cdots {\mathrm{\nabla }}_1{a}_i({\tau }_{1i},\dots ,{\tau }_{ni}){|}^{p_i}))}^{1/p_i}\\ =M\prod _{i=1}^n{(\sum _{{\tau }_{ni}=1}^{m_{ni}}\cdots \sum _{{\tau }_{1i}=1}^{m_{1i}}\prod _{j=1}^n(m_{ji}-{\tau }_{ji}+1)|{\mathrm{\nabla }}_n\cdots {\mathrm{\nabla }}_1{a}_i({\tau }_{1i},\dots ,{\tau }_{ni}){|}^{p_i})}^{1/p_i}\\ =M\prod _{i=1}^n{(\sum _{s_{ni}=1}^{m_{ni}}\cdots \sum _{s_{1i}=1}^{m_{1i}}\prod _{j=1}^n(m_{ji}-s_{ji}+1)|{\mathrm{\nabla }}_n\cdots {\mathrm{\nabla }}_1{a}_i(s_{1i},\dots ,s_{ni}){|}^{p_i})}^{1/p_i}.\end{array}$$

This concludes the proof. □

Proof of Theorem 2.2 From the hypotheses of Theorem 2.2, we have

$$|f_i(x_{1i},\dots ,x_{ni})|\le {\int }_0^{x_{1i}}\cdots {\int }_0^{x_{ni}}|\frac{{\partial }^n}{\partial {\tau }_{1i}\cdots \partial {\tau }_{ni}}{f}_i({\tau }_{1i},\dots ,{\tau }_{ni})|d{\tau }_{1i}\cdots d{\tau }_{ni}.$$

(3.3)

On the other hand, by using Hölder’s integral inequality (see [10]) and the following inequality for mean [10],

$${\left(\prod _{i=1}^n{\lambda }_i^{1/q_i}\right)}^{1/{\sum }_{i=1}^{n}1/q_i}\le \frac{1}{{\sum }_{i=1}^{n}1/q_i}\sum _{i=1}^n{\lambda }_i/q_i,{\lambda }_i>0,$$

we obtain

$$\begin{array}{r}\prod _{i=1}^n|f_i(x_{1i},\dots ,x_{ni})|\\ \le \prod _{i=1}^n{\int }_0^{x_{1i}}\cdots {\int }_0^{x_{ni}}|\frac{{\partial }^n}{\partial {\tau }_{1i}\cdots \partial {\tau }_{ni}}{f}_i({\tau }_{1i},\dots ,{\tau }_{ni})|d{\tau }_{1i}\cdots d{\tau }_{ni}\\ \le \prod _{i=1}^n{(x_{1i}\cdots x_{ni})}^{1/q_i}\\ \times {({\int }_0^{x_{1i}}\cdots {\int }_0^{x_{ni}}|\frac{{\partial }^n}{\partial {\tau }_{1i}\cdots \partial {\tau }_{ni}}{f}_i({\tau }_{1i},\dots ,{\tau }_{ni}){|}^{p_i}d{\tau }_{1i}\cdots d{\tau }_{ni})}^{1/p_i}\\ \le \frac{{({\sum }_{i=1}^n(x_{1i}\cdots x_{ni})/q_i)}^{{\sum }_{i=1}^{n}1/q_i}}{{(n-{\sum }_{i=1}^{n}1/p_i)}^{n-{\sum }_{i=1}^{n}1/p_i}}\\ \times \prod _{i=1}^n{({\int }_0^{x_{1i}}\cdots {\int }_0^{x_{ni}}|\frac{{\partial }^n}{\partial {\tau }_{1i}\cdots \partial {\tau }_{ni}}{f}_i({\tau }_{1i},\dots ,{\tau }_{ni}){|}^{p_i}d{\tau }_{1i}\cdots d{\tau }_{ni})}^{1/p_i}.\end{array}$$

(3.4)

Dividing both sides of (3.4) by ${({\sum }_{i=1}^n(x_{1i}\cdots x_{ni})/q_i)}^{{\sum }_{i=1}^{n}1/q_i}$ and then integrating the result inequality over $x_{ji}$ from 1 to $t_{ji}$ ($i,j=1,2,\dots ,n$), respectively and then using again Hölder’s integral inequality, we obtain

$$\begin{array}{r}{\int }_0^{t_{11}}\cdots {\int }_0^{t_{n1}}{\int }_0^{t_{12}}\cdots {\int }_0^{t_{n2}}\cdots {\int }_0^{t_{1n}}\cdots {\int }_0^{t_{nn}}\\ \frac{{\prod }_{i=1}^n{({\int }_0^{x_{1i}}\cdots {\int }_0^{x_{ni}}|\frac{{\partial }^n}{\partial {\tau }_{1i}\cdots \partial {\tau }_{ni}}{f}_i({\tau }_{1i},\dots ,{\tau }_{ni}){|}^{p_i}d{\tau }_{1i}\cdots d{\tau }_{ni})}^{1/p_i}}{{({\sum }_{i=1}^n(x_{1i}\cdots x_{ni})/q_i)}^{{\sum }_{i=1}^{n}1/q_i}}\\ d{x}_{11}\cdots d{x}_{n1}d{x}_{12}\cdots d{x}_{n2}\cdots d{x}_{1n}\cdots d{x}_{nn}\\ \le {(n-\sum _{i=1}^{n}1/p_i)}^{{\sum }_{i=1}^{n}1/p_i-n}\\ \times \prod _{i=1}^n{\int }_0^{t_{1i}}\cdots {\int }_0^{t_{ni}}({\int }_0^{x_{1i}}\cdots {\int }_0^{x_{ni}}\\ |\frac{{\partial }^n}{\partial {\tau }_{1i}\cdots \partial {\tau }_{ni}}{f}_i({\tau }_{1i},\dots ,{\tau }_{ni}){|}^{p_i}d{\tau }_{1i}\cdots d{\tau }_{ni}{)}^{1/p_i}d{x}_{1i}\cdots d{x}_{ni}\\ \le {(n-\sum _{i=1}^{n}1/p_i)}^{{\sum }_{i=1}^{n}1/p_i-n}\prod _{i=1}^n{(t_{1i}\cdots t_{ni})}^{1/q_i}\\ \times ({\int }_0^{t_{1i}}\cdots {\int }_0^{t_{ni}}({\int }_0^{x_{1i}}\cdots {\int }_0^{x_{ni}}\\ |\frac{{\partial }^n}{\partial {\tau }_{1i}\cdots \partial {\tau }_{ni}}{f}_i({\tau }_{1i},\dots ,{\tau }_{ni}){|}^{p_i}d{\tau }_{1i}\cdots d{\tau }_{ni})d{x}_{1i}\cdots d{x}_{ni}{)}^{1/p_i}\\ =N\prod _{i=1}^n{({\int }_0^{t_{1i}}\cdots {\int }_0^{t_{ni}}\prod _{j=1}^n(t_{ji}-x_{ji})|\frac{{\partial }^n}{\partial x_{1i}\cdots \partial x_{ni}}{f}_i(x_{1i},\dots ,x_{ni}){|}^{p_i}d{x}_{1i}\cdots d{x}_{ni})}^{1/p_i}.\end{array}$$

This concludes the proof. □