nach oben

Journal of Inequalities and Applications

Erschienen in:

Open Access 01.12.2012 | Research

On Hilbert type inequalities

verfasst von: Chang-Jian Zhao, Wing-Sum Cheung

Erschienen in: Journal of Inequalities and Applications | Ausgabe 1/2012

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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.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

C-JZ and W-SC jointly contributed to the main results Theorems 2.1 and 2.2. All authors read and approved the final manuscript.

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}} \geq 1

and

0 < λ = 2 - \frac{1}{p_{1}} - \frac{1}{p_{2}} = \frac{1}{q_{1}} + \frac{1}{q_{2}} \leq 1

, where, as usual,

q_{1}

and

q_{2}

are the conjugate exponents of

p_{1}

and

p_{2}

respectively, then

\sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} \frac{a_{m} b_{n}}{{(m + n)}^{λ}} \leq K {(\sum_{m = 1}^{\infty} a_{m}^{p})}^{1 / p_{1}} {(\sum_{n = 1}^{\infty} b_{n}^{q})}^{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)

\nabla a (s)

and

\nabla b (t)

be as in [11], then

\begin{aligned} \sum_{s = 1}^{m} \sum_{t = 1}^{n} \frac{| a (s) | | b (t) |}{q s^{p - 1} + p t^{q - 1}} \leq & \frac{1}{p q} m^{(p - 1) / p} n^{(q - 1) / q} {(\sum_{s = 1}^{m} (m - s + 1) | \nabla a (s) |^{p})}^{1 / p} \\ \times {(\sum_{t = 1}^{n} (n - t + 1) | \nabla b (t) |^{q})}^{1 / q} . \end{aligned}

(1.2)

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

Theorem B Let p, q,

p^{'}

q^{'}

and λ be as in Theorem A. If

f \in L^{p} (0, \infty)

and

g \in L^{q} (0, \infty)

, then

\int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (x)}{{(x + y)}^{λ}} d x d y \leq K {(\int_{0}^{\infty} f^{p} (x) d x)}^{1 / p} {(\int_{0}^{\infty} g^{q} (y) d y)}^{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^{'} (s)

and

g^{'} (t)

be as in [11], then

\begin{aligned} \int_{0}^{x} \int_{0}^{y} \frac{| f (s) | | g (t) |}{q s^{p - 1} + p t^{q - 1}} d t d s \\ \leq \frac{1}{p q} x^{(p - 1) / p} y^{(q - 1) / q} {(\int_{0}^{x} (x - s) | f^{'} (s) |^{p} d s)}^{1 / p} {(\int_{0}^{y} (y - t) | g^{'} (t) |^{q} d t)}^{1 / q} . \end{aligned}

(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_{1 i}, \dots, s_{n i})

be real-valued functions defined for

s_{j i} = 1, 2, \dots, m_{j i}

, where

m_{j i}

(

i, j = 1, 2, \dots, n

) are natural numbers. For convenience, we write

a_{i} (0, \dots, 0) = 0

and

a_{i} (0, s_{2 i}, \dots, s_{n i}) = a_{i} (s_{1 i}, 0, s_{3 i}, \dots, s_{n i}) = \dots = a_{i} (s_{1 i}, \dots, s_{n - 1, i}, 0) = 0

. Define the operators

\nabla_{i}

\nabla_{i} a_{i} (s_{1 i}, \dots, s_{n i}) = a_{i} (s_{1 i}, \dots, s_{n i}) - a_{i} (s_{1 i}, \dots, s_{i - 1, i}, s_{i i} - 1, s_{i + 1, i}, \dots, s_{n i})

for any function

a_{i} (s_{1 i}, \dots, s_{n i})

. Then

\begin{aligned} \sum_{s_{11} = 1}^{m_{11}} \dots \sum_{s_{n 1} = 1}^{m_{n 1}} \sum_{s_{12} = 1}^{m_{12}} \dots \sum_{s_{n 2} = 1}^{m_{n 2}} \dots \sum_{s_{1 n} = 1}^{m_{1 n}} \dots \sum_{s_{n n} = 1}^{m_{n n}} \frac{\prod_{i = 1}^{n} | a_{i} (s_{1 i}, \dots, s_{n i}) |}{{(\sum_{i = 1}^{n} (s_{1 i} \dots s_{n i}) / q_{i})}^{\sum_{i = 1}^{n} 1 / q_{i}}} \\ \leq M \prod_{i = 1}^{n} {(\sum_{s_{n i} = 1}^{m_{n i}} \dots \sum_{s_{1 i} = 1}^{m_{1 i}} \prod_{j = 1}^{n} (m_{j i} - s_{j i} + 1) | \nabla_{n} \dots \nabla_{1} a_{i} (s_{1 i}, \dots, s_{n i}) |^{p_{i}})}^{1 / p_{i}}, \end{aligned}

(2.1)

where

M = M (m_{1 i}, \dots, m_{n i}) = {(n - \sum_{i = 1}^{n} 1 / p_{i})}^{\sum_{i = 1}^{n} 1 / p_{i} - n} \prod_{i = 1}^{n} {(m_{1 i} \dots m_{n i})}^{1 / q_{i}} .

Remark 2.1 Let

a_{i} (s_{1 i}, \dots, s_{n i})

change to

a_{i} (s_{i})

in Theorem 2.1 and in view of

a_{i} (0) = 0

and

\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}} \dots \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}}} \leq \bar{M} \prod_{i = 1}^{n} {(\sum_{s_{i} = 1}^{m_{i}} (m_{i} - s_{i} + 1) | \nabla a_{i} (s_{i}) |^{p_{i}})}^{1 / p_{i}},

(2.2)

where

\bar{M} = \bar{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}} \geq 1

and

0 < λ = 2 - \frac{1}{p_{1}} - \frac{1}{p_{2}} = \frac{1}{q_{1}} + \frac{1}{q_{2}} \leq 1

, then inequality (2.2) reduces to

\begin{array}{rcl} \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})}^{λ}} \\ \leq \frac{1}{{(λ q_{1} q_{2})}^{λ}} m_{1}^{1 / q_{1}} m_{2}^{1 / q_{2}} {(\sum_{s_{1} = 1}^{m_{1}} (m_{1} - s_{1} + 1) | \nabla a_{1} (s_{1}) |^{p_{1}})}^{1 / p_{1}} \\ \times {(\sum_{s_{2} = 1}^{m_{2}} (m_{2} - s_{2} + 1) | \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

λ = 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}{rcl} \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}} \\ \leq \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) | \nabla a_{1} (s_{1}) |^{p_{1}})}^{1 / p_{1}} \\ \times {(\sum_{s_{2} = 1}^{m_{2}} (m_{2} - s_{2} + 1) | \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} (τ_{1 i}, \dots, τ_{n i})

be real-valued nth differentiable functions defined on

[0, x_{1 i}) \times \dots \times [0, x_{n i})

, where

0 \leq x_{j i} \leq t_{j i}

t_{j i} \in (0, \infty)

and

i, j = 1, 2, \dots, n

. Suppose

f_{i} (x_{1 i}, \dots, x_{n i}) = \int_{0}^{x_{1 i}} \dots \int_{0}^{x_{n i}} \frac{\partial^{n}}{\partial τ_{1 i} \dots \partial τ_{n i}} f_{i} (τ_{1 i}, \dots, τ_{n i}) d τ_{1 i} \dots d τ_{n i},

then

\begin{array}{rcl} \int_{0}^{t_{11}} \dots \int_{0}^{t_{n 1}} \int_{0}^{t_{12}} \dots \int_{0}^{t_{n 2}} \dots \int_{0}^{t_{1 n}} \dots \int_{0}^{t_{n n}} \\ \frac{\prod_{i = 1}^{n} {(\int_{0}^{x_{1 i}} \dots \int_{0}^{x_{n i}} | \frac{\partial^{n}}{\partial τ_{1 i} \dots \partial τ_{n i}} f_{i} (τ_{1 i}, \dots, τ_{n i}) |^{p_{i}} d τ_{1 i} \dots d τ_{n i})}^{1 / p_{i}}}{{(\sum_{i = 1}^{n} (x_{1 i} \dots x_{n i}) / q_{i})}^{\sum_{i = 1}^{n} 1 / q_{i}}} \\ d x_{11} \dots d x_{n 1} d x_{12} \dots d x_{n 2} \dots d x_{1 n} \dots d x_{n n} \\ \leq N \prod_{i = 1}^{n} (\int_{0}^{t_{1 i}} \dots \int_{0}^{t_{n i}} \prod_{j = 1}^{n} (t_{j i} - x_{j i}) \\ \times {| \frac{\partial^{n}}{\partial x_{1 i} \dots \partial x_{n i}} f_{i} (x_{1 i}, \dots, x_{n i}) |}^{p_{i}} d x_{1 i} \dots d x_{n i})^{1 / p_{i}}, \end{array}

(2.4)

where

N = N (t_{1 i}, \dots, t_{n i}) = {(n - \sum_{i = 1}^{n} \frac{1}{p_{i}})}^{\sum_{i = 1}^{n} 1 / p_{i} - n} \cdot \prod_{i = 1}^{n} {(t_{1 i} \dots t_{n i})}^{1 / q_{i}} .

Remark 2.3 Let

f_{i} (x_{1 i}, \dots, x_{n i})

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{aligned} \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} \dots d s_{1} \\ \leq \bar{N} \prod_{i = 1}^{n} {(\int_{0}^{x_{i}} (x_{i} - s_{i}) | f_{i}^{'} (s_{i}) |^{p_{i}} d s_{i})}^{1 / p_{i}}, \end{aligned}

(2.5)

where

\bar{N} = \bar{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}} \geq 1

and

0 < λ = 2 - \frac{1}{p_{1}} - \frac{1}{p_{2}} = \frac{1}{q_{1}} + \frac{1}{q_{2}} \leq 1

, inequality (2.5) reduces to

\begin{array}{rcl} \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})}^{λ}} d s_{2} d s_{1} \\ \leq \frac{1}{{(λ q_{1} q_{2})}^{λ}} x_{1}^{1 / q_{1}} x_{2}^{1 / q_{2}} {(\int_{0}^{x_{1}} (x_{1} - s_{1}) | f_{1}^{'} (s_{1}) |^{p_{1}} d s_{1})}^{1 / p_{1}} \\ \times {(\int_{0}^{x_{2}} (x_{2} - s_{2}) | f_{2}^{'} (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

λ = 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}{rcl} \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}} \\ \leq \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}^{'} (s_{1}) |^{p_{1}} d s_{1})}^{1 / p_{1}} \\ \times {(\int_{0}^{x_{2}} (x_{2} - s_{2}) | f_{2}^{'} (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_{2 i}, \dots, s_{n i}) = a_{i} (s_{1 i}, 0, s_{3 i}, \dots, s_{n i}) = \dots = a_{i} (s_{1 i}, \dots, s_{n - 1, i}, 0) = 0

, we have

| a_{i} (s_{1 i}, \dots, s_{n i}) | \leq \sum_{τ_{n i} = 1}^{s_{n i}} \dots \sum_{τ_{1 i} = 1}^{s_{1 i}} | \nabla_{n} \dots \nabla_{1} a_{i} (τ_{1 i}, \dots, τ_{n i}) | .

(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{aligned} \prod_{i = 1}^{n} | a_{i} (s_{1 i}, \dots, s_{n i}) | & \leq \prod_{i = 1}^{n} \sum_{τ_{n i} = 1}^{s_{n i}} \dots \sum_{τ_{1 i} = 1}^{s_{1 i}} | \nabla_{n} \dots \nabla_{1} a_{i} (τ_{1 i}, \dots, τ_{n i}) | \\ \leq \prod_{i = 1}^{n} {(s_{1 i} \dots s_{n i})}^{1 / q_{i}} {(\sum_{τ_{n i} = 1}^{s_{n i}} \dots \sum_{τ_{1 i} = 1}^{s_{1 i}} | \nabla_{n} \dots \nabla_{1} a_{i} (τ_{1 i}, \dots, τ_{n i}) |^{p_{i}})}^{1 / p_{i}} \\ \leq \frac{{(\sum_{i = 1}^{n} (s_{1 i} \dots s_{n i}) / 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_{τ_{n i} = 1}^{s_{n i}} \dots \sum_{τ_{1 i} = 1}^{s_{1 i}} | \nabla_{n} \dots \nabla_{1} a_{i} (τ_{1 i}, \dots, τ_{n i}) |^{p_{i}})}^{1 / p_{i}} . \end{aligned}

(3.2)

Dividing both sides of (3.2) by

{(\sum_{i = 1}^{n} (s_{1 i} \dots s_{n i}) / q_{i})}^{\sum_{i = 1}^{n} 1 / q_{i}}

and then taking sums over

s_{j i}

from 1 to

m_{j i}

(

i, j = 1, 2, \dots, n

), respectively and then using again Hölder’s inequality, we obtain

\begin{array}{rcl} \sum_{s_{11} = 1}^{m_{11}} \dots \sum_{s_{n 1} = 1}^{m_{n 1}} \sum_{s_{12} = 1}^{m_{12}} \dots \sum_{s_{n 2} = 1}^{m_{n 2}} \dots \sum_{s_{1 n} = 1}^{m_{1 n}} \dots \sum_{s_{n n} = 1}^{m_{n n}} \frac{\prod_{i = 1}^{n} | \nabla_{n} \dots \nabla_{1} a_{i} (s_{1 i}, \dots, s_{n i}) |}{{(\sum_{i = 1}^{n} (s_{1 i} \dots s_{n i}) / q_{i})}^{\sum_{i = 1}^{n} 1 / q_{i}}} \\ \leq {(n - \sum_{i = 1}^{n} 1 / p_{i})}^{\sum_{i = 1}^{n} 1 / p_{i} - n} \\ \times \prod_{i = 1}^{n} (\sum_{s_{n i} = 1}^{m_{n i}} \dots \sum_{s_{1 i} = 1}^{m_{1 i}} {(\sum_{τ_{n i} = 1}^{s_{n i}} \dots \sum_{τ_{1 i} = 1}^{s_{1 i}} | \nabla_{n} \dots \nabla_{1} a_{i} (τ_{1 i}, \dots, τ_{n i}) |^{p_{i}})}^{1 / p_{i}}) \\ \leq {(n - \sum_{i = 1}^{n} 1 / p_{i})}^{\sum_{i = 1}^{n} 1 / p_{i} - n} \\ \times \prod_{i = 1}^{n} {(m_{1 i} \dots m_{n i})}^{1 / q_{i}} {(\sum_{s_{n i} = 1}^{m_{n i}} \dots \sum_{s_{1 i} = 1}^{m_{1 i}} (\sum_{τ_{n i} = 1}^{s_{n i}} \dots \sum_{τ_{1 i} = 1}^{s_{1 i}} | \nabla_{n} \dots \nabla_{1} a_{i} (τ_{1 i}, \dots, τ_{n i}) |^{p_{i}}))}^{1 / p_{i}} \\ = M \prod_{i = 1}^{n} {(\sum_{τ_{n i} = 1}^{m_{n i}} \dots \sum_{τ_{1 i} = 1}^{m_{1 i}} \prod_{j = 1}^{n} (m_{j i} - τ_{j i} + 1) | \nabla_{n} \dots \nabla_{1} a_{i} (τ_{1 i}, \dots, τ_{n i}) |^{p_{i}})}^{1 / p_{i}} \\ = M \prod_{i = 1}^{n} {(\sum_{s_{n i} = 1}^{m_{n i}} \dots \sum_{s_{1 i} = 1}^{m_{1 i}} \prod_{j = 1}^{n} (m_{j i} - s_{j i} + 1) | \nabla_{n} \dots \nabla_{1} a_{i} (s_{1 i}, \dots, s_{n i}) |^{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_{1 i}, \dots, x_{n i}) | \leq \int_{0}^{x_{1 i}} \dots \int_{0}^{x_{n i}} | \frac{\partial^{n}}{\partial τ_{1 i} \dots \partial τ_{n i}} f_{i} (τ_{1 i}, \dots, τ_{n i}) | d τ_{1 i} \dots d τ_{n i} .

(3.3)

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

{(\prod_{i = 1}^{n} λ_{i}^{1 / q_{i}})}^{1 / \sum_{i = 1}^{n} 1 / q_{i}} \leq \frac{1}{\sum_{i = 1}^{n} 1 / q_{i}} \sum_{i = 1}^{n} λ_{i} / q_{i}, λ_{i} > 0,

we obtain

\begin{aligned} \prod_{i = 1}^{n} | f_{i} (x_{1 i}, \dots, x_{n i}) | \\ \leq \prod_{i = 1}^{n} \int_{0}^{x_{1 i}} \dots \int_{0}^{x_{n i}} | \frac{\partial^{n}}{\partial τ_{1 i} \dots \partial τ_{n i}} f_{i} (τ_{1 i}, \dots, τ_{n i}) | d τ_{1 i} \dots d τ_{n i} \\ \leq \prod_{i = 1}^{n} {(x_{1 i} \dots x_{n i})}^{1 / q_{i}} \\ \times {(\int_{0}^{x_{1 i}} \dots \int_{0}^{x_{n i}} | \frac{\partial^{n}}{\partial τ_{1 i} \dots \partial τ_{n i}} f_{i} (τ_{1 i}, \dots, τ_{n i}) |^{p_{i}} d τ_{1 i} \dots d τ_{n i})}^{1 / p_{i}} \\ \leq \frac{{(\sum_{i = 1}^{n} (x_{1 i} \dots x_{n i}) / 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_{1 i}} \dots \int_{0}^{x_{n i}} | \frac{\partial^{n}}{\partial τ_{1 i} \dots \partial τ_{n i}} f_{i} (τ_{1 i}, \dots, τ_{n i}) |^{p_{i}} d τ_{1 i} \dots d τ_{n i})}^{1 / p_{i}} . \end{aligned}

(3.4)

Dividing both sides of (3.4) by

{(\sum_{i = 1}^{n} (x_{1 i} \dots x_{n i}) / q_{i})}^{\sum_{i = 1}^{n} 1 / q_{i}}

and then integrating the result inequality over

x_{j i}

from 1 to

t_{j i}

(

i, j = 1, 2, \dots, n

), respectively and then using again Hölder’s integral inequality, we obtain

\begin{array}{rcl} \int_{0}^{t_{11}} \dots \int_{0}^{t_{n 1}} \int_{0}^{t_{12}} \dots \int_{0}^{t_{n 2}} \dots \int_{0}^{t_{1 n}} \dots \int_{0}^{t_{n n}} \\ \frac{\prod_{i = 1}^{n} {(\int_{0}^{x_{1 i}} \dots \int_{0}^{x_{n i}} | \frac{\partial^{n}}{\partial τ_{1 i} \dots \partial τ_{n i}} f_{i} (τ_{1 i}, \dots, τ_{n i}) |^{p_{i}} d τ_{1 i} \dots d τ_{n i})}^{1 / p_{i}}}{{(\sum_{i = 1}^{n} (x_{1 i} \dots x_{n i}) / q_{i})}^{\sum_{i = 1}^{n} 1 / q_{i}}} \\ d x_{11} \dots d x_{n 1} d x_{12} \dots d x_{n 2} \dots d x_{1 n} \dots d x_{n n} \\ \leq {(n - \sum_{i = 1}^{n} 1 / p_{i})}^{\sum_{i = 1}^{n} 1 / p_{i} - n} \\ \times \prod_{i = 1}^{n} \int_{0}^{t_{1 i}} \dots \int_{0}^{t_{n i}} (\int_{0}^{x_{1 i}} \dots \int_{0}^{x_{n i}} \\ | \frac{\partial^{n}}{\partial τ_{1 i} \dots \partial τ_{n i}} f_{i} (τ_{1 i}, \dots, τ_{n i}) |^{p_{i}} d τ_{1 i} \dots d τ_{n i})^{1 / p_{i}} d x_{1 i} \dots d x_{n i} \\ \leq {(n - \sum_{i = 1}^{n} 1 / p_{i})}^{\sum_{i = 1}^{n} 1 / p_{i} - n} \prod_{i = 1}^{n} {(t_{1 i} \dots t_{n i})}^{1 / q_{i}} \\ \times (\int_{0}^{t_{1 i}} \dots \int_{0}^{t_{n i}} (\int_{0}^{x_{1 i}} \dots \int_{0}^{x_{n i}} \\ | \frac{\partial^{n}}{\partial τ_{1 i} \dots \partial τ_{n i}} f_{i} (τ_{1 i}, \dots, τ_{n i}) |^{p_{i}} d τ_{1 i} \dots d τ_{n i}) d x_{1 i} \dots d x_{n i})^{1 / p_{i}} \\ = N \prod_{i = 1}^{n} {(\int_{0}^{t_{1 i}} \dots \int_{0}^{t_{n i}} \prod_{j = 1}^{n} (t_{j i} - x_{j i}) | \frac{\partial^{n}}{\partial x_{1 i} \dots \partial x_{n i}} f_{i} (x_{1 i}, \dots, x_{n i}) |^{p_{i}} d x_{1 i} \dots d x_{n i})}^{1 / p_{i}} . \end{array}

This concludes the proof. □

Acknowledgement

CJZ is supported by National Natural Science Foundation of China (10971205). WSC is partially supported by a HKU URG grant. The authors express their grateful thanks to the referees for their many very valuable suggestions and comments.

Open AccessThis article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

C-JZ and W-SC jointly contributed to the main results Theorems 2.1 and 2.2. All authors read and approved the final manuscript.

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Titel: On Hilbert type inequalities
verfasst von: Chang-Jian Zhao
Wing-Sum Cheung
Publikationsdatum: 01.12.2012
Verlag: Springer International Publishing
Erschienen in: Journal of Inequalities and Applications / Ausgabe 1/2012
Elektronische ISSN: 1029-242X
DOI: https://doi.org/10.1186/1029-242X-2012-145

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On Hilbert type inequalities

Abstract

Competing interests

Authors’ contributions

1 Introduction

2 Statement of results

3 Proofs of results

Acknowledgement

Competing interests

Authors’ contributions

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Springer Professional

Abstract

Competing interests

Authors’ contributions

1 Introduction

2 Statement of results

3 Proofs of results

Acknowledgement

Competing interests

Authors’ contributions

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