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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 [19]). 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 >1are real numbers such that 1 p 1 + 1 p 2 1and0<λ=2 1 p 1 1 p 2 = 1 q 1 + 1 q 2 1, where, as usual, q 1 and q 2 are the conjugate exponents of p 1 and p 2 respectively, then

m = 1 n = 1 a m b n ( m + n ) λ K ( m = 1 a m p ) 1 / p 1 ( n = 1 b n q ) 1 / p 2 ,
(1.1)

whereK=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 qa(s)b(t)a(0)b(0)a(s)andb(t)be as in [11], then

s = 1 m t = 1 n | a ( s ) | | b ( t ) | q s p 1 + p t q 1 1 p q m ( p 1 ) / p n ( q 1 ) / q ( s = 1 m ( m s + 1 ) | a ( s ) | p ) 1 / p × ( t = 1 n ( n t + 1 ) | b ( t ) | q ) 1 / q .
(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. Iff L p (0,)andg L q (0,), then

0 0 f ( x ) g ( x ) ( x + y ) λ dxdyK ( 0 f p ( x ) d x ) 1 / p ( 0 g q ( y ) d y ) 1 / q ,
(1.3)

whereK=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 qf(s)g(t)f(0)g(0) f (s)and g (t)be as in [11], then

0 x 0 y | f ( s ) | | g ( t ) | q s p 1 + p t q 1 d t d s 1 p q x ( p 1 ) / p y ( q 1 ) / q ( 0 x ( x s ) | f ( s ) | p d s ) 1 / p ( 0 y ( y t ) | g ( t ) | q d t ) 1 / q .
(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 >1be constants and 1 p i + 1 q i =1. Let a i ( s 1 i ,, s n i )be real-valued functions defined for s j i =1,2,, m j i , where m j i (i,j=1,2,,n) are natural numbers. For convenience, we write a i (0,,0)=0and a i (0, s 2 i ,, s n i )= a i ( s 1 i ,0, s 3 i ,, s n i )== a i ( s 1 i ,, s n 1 , i ,0)=0. Define the operators i by i a i ( s 1 i ,, s n i )= a i ( s 1 i ,, s n i ) a i ( s 1 i ,, s i 1 , i , s i i 1, s i + 1 , i ,, s n i )for any function a i ( s 1 i ,, s n i ). Then

s 11 = 1 m 11 s n 1 = 1 m n 1 s 12 = 1 m 12 s n 2 = 1 m n 2 s 1 n = 1 m 1 n s n n = 1 m n n i = 1 n | a i ( s 1 i , , s n i ) | ( i = 1 n ( s 1 i s n i ) / q i ) i = 1 n 1 / q i M i = 1 n ( s n i = 1 m n i s 1 i = 1 m 1 i j = 1 n ( m j i s j i + 1 ) | n 1 a i ( s 1 i , , s n i ) | p i ) 1 / p i ,
(2.1)

where

M=M( m 1 i ,, m n i )= ( n i = 1 n 1 / p i ) i = 1 n 1 / p i n i = 1 n ( m 1 i m n i ) 1 / q i .

Remark 2.1 Let a i ( s 1 i ,, s n i ) change to a i ( s i ) in Theorem 2.1 and in view of a i (0)=0 and a i ( s i )= a i ( s i ) a i ( s i 1) for any function a i ( s i ), i=1,2,,n, then

s 1 = 1 m 1 s 2 = 1 m 2 s n = 1 m n i = 1 n | a i ( s i ) | ( i = 1 n s i / q i ) i = 1 n 1 / q i M ¯ i = 1 n ( s i = 1 m i ( m i s i + 1 ) | a i ( s i ) | p i ) 1 / p i ,
(2.2)

where

M ¯ = M ¯ ( m 1 ,, m n )= ( n i = 1 n 1 p i ) i = 1 n 1 / p i n 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 1 p 1 + 1 p 2 1 and 0<λ=2 1 p 1 1 p 2 = 1 q 1 + 1 q 2 1, then inequality (2.2) reduces to

s 1 = 1 m 1 s 2 = 1 m 2 | a 1 ( s 1 ) | | a 2 ( s 2 ) | ( q 2 s 1 + q 1 s 2 ) λ 1 ( λ q 1 q 2 ) λ m 1 1 / q 1 m 2 1 / q 2 ( s 1 = 1 m 1 ( m 1 s 1 + 1 ) | a 1 ( s 1 ) | p 1 ) 1 / p 1 × ( s 2 = 1 m 2 ( m 2 s 2 + 1 ) | a 2 ( s 2 ) | p 2 ) 1 / p 2 ,
(2.3)

which is an interesting variation of inequality (1.1).

On the other hand, if λ=1, then 1 p 1 + 1 p 2 = 1 q 1 + 1 q 2 =1 and so p 1 = q 2 , p 2 = q 1 . In this case inequality (2.3) reduces to

s 1 = 1 m 1 s 2 = 1 m 2 | a 1 ( s 1 ) | | a 2 ( s 2 ) | p 1 s 1 + q 1 s 2 1 p 1 q 1 m 1 ( p 1 1 ) / p 1 m 2 ( q 1 1 ) / q 1 ( s 1 = 1 m 1 ( m 1 s 1 + 1 ) | a 1 ( s 1 ) | p 1 ) 1 / p 1 × ( s 2 = 1 m 2 ( m 2 s 2 + 1 ) | a 2 ( s 2 ) | q 1 ) 1 / q 1 .

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

Theorem 2.2 Let p i >1be constants and 1 p i + 1 q i =1. Let f i ( τ 1 i ,, τ n i )be real-valued nth differentiable functions defined on[0, x 1 i )××[0, x n i ), where0 x j i t j i , t j i (0,)andi,j=1,2,,n. Suppose

f i ( x 1 i ,, x n i )= 0 x 1 i 0 x n i n τ 1 i τ n i f i ( τ 1 i ,, τ n i )d τ 1 i d τ n i ,

then

0 t 11 0 t n 1 0 t 12 0 t n 2 0 t 1 n 0 t n n i = 1 n ( 0 x 1 i 0 x n i | n τ 1 i τ n i f i ( τ 1 i , , τ n i ) | p i d τ 1 i d τ n i ) 1 / p i ( i = 1 n ( x 1 i x n i ) / q i ) i = 1 n 1 / q i d x 11 d x n 1 d x 12 d x n 2 d x 1 n d x n n N i = 1 n ( 0 t 1 i 0 t n i j = 1 n ( t j i x j i ) × | n x 1 i x n i f i ( x 1 i , , x n i ) | p i d x 1 i d x n i ) 1 / p i ,
(2.4)

where

N=N( t 1 i ,, t n i )= ( n i = 1 n 1 p i ) i = 1 n 1 / p i n i = 1 n ( t 1 i t n i ) 1 / q i .

Remark 2.3 Let f i ( x 1 i ,, x n i ) change to f i ( s i ) in Theorem 2.2 and in view of f i (0)=0, i=1,2,,n, then

0 x 1 0 x n i = 1 n | f ( s i ) | ( i = 1 n s i / q i ) i = 1 n 1 / q i d s n d s 1 N ¯ i = 1 n ( 0 x i ( x i s i ) | f i ( s i ) | p i d s i ) 1 / p i ,
(2.5)

where

N ¯ = N ¯ ( x 1 ,, x n )= ( n i = 1 n 1 p i ) i = 1 n 1 / p i n 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 1 p 1 + 1 p 2 1 and 0<λ=2 1 p 1 1 p 2 = 1 q 1 + 1 q 2 1, inequality (2.5) reduces to

0 x 1 0 x 2 | f 1 ( s 1 ) | | f 2 ( s 2 ) | ( q 2 s 1 + q 1 s 2 ) λ d s 2 d s 1 1 ( λ q 1 q 2 ) λ x 1 1 / q 1 x 2 1 / q 2 ( 0 x 1 ( x 1 s 1 ) | f 1 ( s 1 ) | p 1 d s 1 ) 1 / p 1 × ( 0 x 2 ( x 2 s 2 ) | f 2 ( s 2 ) | p 2 d s 2 ) 1 / p 2 ,
(2.6)

which is an interesting variation of inequality (1.3).

On the other hand, if λ=1, then 1 p 1 + 1 p 2 = 1 q 1 + 1 q 2 =1 and so p 1 = q 2 , p 2 = q 1 . In this case inequality (2.6) reduces to

0 x 1 0 x 2 | f 1 ( s 1 ) | | f 2 ( s 2 ) | p 1 s 1 + q 1 s 2 h 1 h 2 p 1 q 1 x 1 ( p 1 1 ) / p 1 x 2 ( q 1 1 ) / q 1 ( 0 x 1 ( x 1 s 1 ) | f 1 ( s 1 ) | p 1 d s 1 ) 1 / p 1 × ( 0 x 2 ( x 2 s 2 ) | f 2 ( s 2 ) | q 1 d s 2 ) 1 / q 1 .

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,,0)= a i (0, s 2 i ,, s n i )= a i ( s 1 i ,0, s 3 i ,, s n i )== a i ( s 1 i ,, s n 1 , i ,0)=0, we have

| a i ( s 1 i ,, s n i )| τ n i = 1 s n i τ 1 i = 1 s 1 i | n 1 a i ( τ 1 i ,, 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

i = 1 n | a i ( s 1 i , , s n i ) | i = 1 n τ n i = 1 s n i τ 1 i = 1 s 1 i | n 1 a i ( τ 1 i , , τ n i ) | i = 1 n ( s 1 i s n i ) 1 / q i ( τ n i = 1 s n i τ 1 i = 1 s 1 i | n 1 a i ( τ 1 i , , τ n i ) | p i ) 1 / p i ( i = 1 n ( s 1 i s n i ) / q i ) i = 1 n 1 / q i ( n i = 1 n 1 / p i ) n i = 1 n 1 / p i × i = 1 n ( τ n i = 1 s n i τ 1 i = 1 s 1 i | n 1 a i ( τ 1 i , , τ n i ) | p i ) 1 / p i .
(3.2)

Dividing both sides of (3.2) by ( i = 1 n ( s 1 i s n i ) / q i ) i = 1 n 1 / q i and then taking sums over s j i from 1 to m j i (i,j=1,2,,n), respectively and then using again Hölder’s inequality, we obtain

s 11 = 1 m 11 s n 1 = 1 m n 1 s 12 = 1 m 12 s n 2 = 1 m n 2 s 1 n = 1 m 1 n s n n = 1 m n n i = 1 n | n 1 a i ( s 1 i , , s n i ) | ( i = 1 n ( s 1 i s n i ) / q i ) i = 1 n 1 / q i ( n i = 1 n 1 / p i ) i = 1 n 1 / p i n × i = 1 n ( s n i = 1 m n i s 1 i = 1 m 1 i ( τ n i = 1 s n i τ 1 i = 1 s 1 i | n 1 a i ( τ 1 i , , τ n i ) | p i ) 1 / p i ) ( n i = 1 n 1 / p i ) i = 1 n 1 / p i n × i = 1 n ( m 1 i m n i ) 1 / q i ( s n i = 1 m n i s 1 i = 1 m 1 i ( τ n i = 1 s n i τ 1 i = 1 s 1 i | n 1 a i ( τ 1 i , , τ n i ) | p i ) ) 1 / p i = M i = 1 n ( τ n i = 1 m n i τ 1 i = 1 m 1 i j = 1 n ( m j i τ j i + 1 ) | n 1 a i ( τ 1 i , , τ n i ) | p i ) 1 / p i = M i = 1 n ( s n i = 1 m n i s 1 i = 1 m 1 i j = 1 n ( m j i s j i + 1 ) | n 1 a i ( s 1 i , , s n i ) | p i ) 1 / p i .

This concludes the proof. □

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

| f i ( x 1 i ,, x n i )| 0 x 1 i 0 x n i | n τ 1 i τ n i f i ( τ 1 i ,, τ n i )|d τ 1 i 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],

( i = 1 n λ i 1 / q i ) 1 / i = 1 n 1 / q i 1 i = 1 n 1 / q i i = 1 n λ i / q i , λ i >0,

we obtain

i = 1 n | f i ( x 1 i , , x n i ) | i = 1 n 0 x 1 i 0 x n i | n τ 1 i τ n i f i ( τ 1 i , , τ n i ) | d τ 1 i d τ n i i = 1 n ( x 1 i x n i ) 1 / q i × ( 0 x 1 i 0 x n i | n τ 1 i τ n i f i ( τ 1 i , , τ n i ) | p i d τ 1 i d τ n i ) 1 / p i ( i = 1 n ( x 1 i x n i ) / q i ) i = 1 n 1 / q i ( n i = 1 n 1 / p i ) n i = 1 n 1 / p i × i = 1 n ( 0 x 1 i 0 x n i | n τ 1 i τ n i f i ( τ 1 i , , τ n i ) | p i d τ 1 i d τ n i ) 1 / p i .
(3.4)

Dividing both sides of (3.4) by ( i = 1 n ( x 1 i x n i ) / q i ) 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,,n), respectively and then using again Hölder’s integral inequality, we obtain

0 t 11 0 t n 1 0 t 12 0 t n 2 0 t 1 n 0 t n n i = 1 n ( 0 x 1 i 0 x n i | n τ 1 i τ n i f i ( τ 1 i , , τ n i ) | p i d τ 1 i d τ n i ) 1 / p i ( i = 1 n ( x 1 i x n i ) / q i ) i = 1 n 1 / q i d x 11 d x n 1 d x 12 d x n 2 d x 1 n d x n n ( n i = 1 n 1 / p i ) i = 1 n 1 / p i n × i = 1 n 0 t 1 i 0 t n i ( 0 x 1 i 0 x n i | n τ 1 i τ n i f i ( τ 1 i , , τ n i ) | p i d τ 1 i d τ n i ) 1 / p i d x 1 i d x n i ( n i = 1 n 1 / p i ) i = 1 n 1 / p i n i = 1 n ( t 1 i t n i ) 1 / q i × ( 0 t 1 i 0 t n i ( 0 x 1 i 0 x n i | n τ 1 i τ n i f i ( τ 1 i , , τ n i ) | p i d τ 1 i d τ n i ) d x 1 i d x n i ) 1 / p i = N i = 1 n ( 0 t 1 i 0 t n i j = 1 n ( t j i x j i ) | n x 1 i x n i f i ( x 1 i , , x n i ) | p i d x 1 i d x n i ) 1 / p i .

This concludes the proof. □

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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.

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Correspondence to Chang-Jian Zhao.

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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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Zhao, CJ., Cheung, WS. On Hilbert type inequalities. J Inequal Appl 2012, 145 (2012). https://doi.org/10.1186/1029-242X-2012-145

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Keywords

  • Hilbert’s inequality
  • Pachpatte’s inequality
  • Hölder’s inequality