Open Access

On a strengthened Hardy-Hilbert type inequality

Journal of Inequalities and Applications20132013:511

https://doi.org/10.1186/1029-242X-2013-511

Received: 1 March 2013

Accepted: 10 October 2013

Published: 8 November 2013

Abstract

We derive a strengthenment of a Hardy-Hilbert type inequality by using the Euler-Maclaurin expansion for the zeta function and estimating the weight function effectively. As applications, some particular results are presented.

MSC:26D15.

Keywords

Hardy-Hilbert type inequalityweight coefficientHölder inequality

1 Introduction

Let p , q > 1 , 1 p + 1 q = 1 , a n , b n 0 , 0 < n = 1 a n p < and 0 < n = 1 b n q < . Then one [1] has
n = 1 m = 1 a m b n m + n < π sin ( π / p ) [ n = 1 a n p ] 1 p [ n = 1 b n q ] 1 q ,
(1.1)
n = 1 m = 1 a m b n max { m , n } < p q [ n = 1 a n p ] 1 p [ n = 1 b n q ] 1 q ,
(1.2)

where the constant factor π sin ( π / p ) and pq are best possible. Inequality (1.1) is well known as Hardy-Hilbert’s inequality, and inequality (1.2) is named a Hardy-Hilbert type inequality. Both of them are important in analysis and applications [2]. In recent years, many results about generalizations of this type of inequality were established (see [3]). Under the same conditions as (1.1) and (1.2), some Hardy-Hilbert type inequalities, which are similar to (1.1) and (1.2), have been studied and generalized by some mathematicians.

By introducing a parameter, Yang gave a generalization of inequality (1.2) with the best constant factor as follows:

If p , q > 1 , 1 p + 1 q = 1 , 2 min { p , q } < λ 2 , a n , b n 0 , such that 0 < n = 1 n 1 λ a n p < and 0 < n = 1 n 1 λ b n q < , then
n = 1 m = 1 a m b n max { m λ , n λ } < k λ ( p ) { n = 1 n 1 λ a n p } 1 p { n = 1 n 1 λ b n q } 1 q ,
(1.3)

where the constant factor k λ ( p ) = λ p q ( p + λ 2 ) ( q + λ 2 ) is best possible.

Furthermore, by introducing a parameter and two pairs of conjugate exponents, Zhong gave a generalization of inequality (1.3) with the best constant factor as follows:

If p > 1 , 1 p + 1 q = 1 , r > 1 , 1 r + 1 s = 1 , 0 < λ min { r , s } , a n , b n 0 , such that 0 < n = 1 n p ( 1 λ r ) 1 a n p < and 0 < n = 1 n q ( 1 λ s ) 1 b n q < , then
n = 1 m = 1 a m b n max { m λ , n λ } < k λ ( r ) { n = 1 n p ( 1 λ r ) 1 a n p } 1 p { n = 1 n q ( 1 λ s ) 1 b n q } 1 q ,
(1.4)

where the constant factor k λ ( r ) = r s λ is best possible.

Recently, in [4], Jiang and Hua established an improvement of inequality (1.3) as follows:

If p , q > 1 , 1 p + 1 q = 1 , 2 min { p , q } < λ 2 , a n 0 , b n 0 , for n 1 , n N and 0 < n = 1 n 1 λ a n p < , 0 < n = 1 n 1 λ b n q < , then
n = 1 m = 1 a m b n max { m λ , n λ } < { n = 1 [ k ( λ ) q 3 ( q + λ 2 ) n q + λ 2 q ] n 1 λ a n p } 1 p × { n = 1 [ k ( λ ) p 3 ( p + λ 2 ) n p + λ 2 p ] n 1 λ b n q } ,
(1.5)

where k ( λ ) = p q λ ( p + λ 2 ) ( q + λ 2 ) > 0 .

In this paper, by introducing a parameter and estimating the weight coefficient, we obtain a strengthenment of inequality (1.4) and generalize inequality (1.5). As applications, some particular results are presented.

2 Some preliminary results

First, we need the following formula of the Riemann-ζ function (see [5]):
ζ ( ρ ) = n = 1 m 1 n ρ m 1 ρ 1 ρ 1 2 m ρ n = 1 l 1 B 2 n 2 n ( ρ 2 n 1 ) 1 m ρ + 2 n 1 B 2 l 2 l ( ρ 2 l 1 ) ε m ρ + 2 l 1 ,
(2.1)

where ρ > 0 , ρ 1 , m , l 1 , m , l N , 0 < ε = ε ( ρ , l , m ) < 1 . The numbers B 1 = 1 / 2 , B 2 = 1 / 6 , B 3 = 0 , B 4 = 1 / 30 , … are Bernoulli numbers. In particular, ζ ( ρ ) = n = 1 1 n ρ ( ρ > 1 ).

Since ζ ( 0 ) = 1 / 2 , the formula of the Riemann-ζ function (2.1) also holds for ρ = 0 .

Lemma 2.1 Let r > 1 , 1 r + 1 s = 1 , 0 < λ min { r , s } , define the weight coefficients ω ( m , λ , s ) and ω ( n , λ , r ) as
ω ( m , λ , s ) = n = 1 1 max { m λ , n λ } ( m n ) 1 λ s ,
(2.2)
ω ( n , λ , r ) = m = 1 1 max { m λ , n λ } ( n m ) 1 λ r .
(2.3)
Then we have
ω ( m , λ , s ) < m 1 λ [ k λ s 3 λ m λ s ]
(2.4)
and
ω ( n , λ , r ) < n 1 λ [ k λ r 3 λ m λ r ] ,
(2.5)

where k λ = r s λ .

Proof For 0 < λ min { r , s } , taking ρ = 1 λ s 0 , l = 1 in (2.1), we get
ζ ( 1 λ s ) = n = 1 m 1 n 1 λ s s m λ s λ 1 2 m 1 λ s + 1 λ s 12 m 2 λ s ε 1 ,
(2.6)

where 0 < ε 1 < 1 .

Set ρ = 1 + λ r , l = 1 , and we can derive
ζ ( 1 + λ r ) = n = 1 m 1 1 n 1 + λ r + r m λ r λ + 1 2 m 1 + λ r + 1 + λ r 12 m 2 + λ r ε 2 ,
(2.7)

where 0 < ε 2 < 1 .

Thus we get
ω ( m , λ , s ) = n = 1 1 max { m λ , n λ } ( m n ) 1 λ s = n = 1 m 1 max { m λ , n λ } ( m n ) 1 λ s 1 m λ + n = m 1 max { m λ , n λ } ( m n ) 1 λ s = n = 1 m 1 m λ ( m n ) 1 λ s 1 m λ + n = m 1 n λ ( m n ) 1 λ s = 1 m λ + λ s 1 n = 1 m 1 n 1 λ s 1 m λ + m 1 λ s n = m 1 n 1 + λ r .
Combining (2.6) and (2.7), we have
ω ( m , λ , s ) < 1 m λ + λ s 1 [ ζ ( 1 λ s ) + s m λ s λ + 1 2 m 1 λ s ] 1 m λ + m 1 λ s [ r m λ r λ + 1 2 m 1 + λ r + 1 + λ r 12 m 2 + λ r ] = 1 m λ + λ s 1 ζ ( 1 λ s ) + s m 1 λ λ + 1 2 m λ 1 m λ + r m 1 λ λ + 1 2 m λ + 1 + λ r 12 m 1 + λ = 1 m λ + λ s 1 ζ ( 1 λ s ) + r s m 1 λ λ + 1 + λ r 12 m 1 + λ = m 1 λ { r s λ 1 m λ s [ ζ ( 1 λ s ) 1 + λ r 12 m 2 λ s ] } .
In (2.6), let m = 1 , by 0 < λ min { r , s } , we obtain
ζ ( 1 λ s ) = 1 s λ 1 2 + ( 1 λ s ) ε 1 12 < 1 2 s λ + 1 λ s 12 = 6 λ 12 s λ ( 1 λ s ) 12 λ < 6 λ 12 s ( λ s ) 12 λ = 5 λ 11 s 12 λ = 11 s 5 λ 12 λ < 0 .
Therefore, for m 1 , m N , 0 < λ min { r , s } , we obtain
ζ ( 1 λ s ) 1 + λ r 12 m 2 λ s > 11 s 5 λ 12 λ 1 + λ r 12 = 11 s 5 λ λ ( 1 + λ r ) 12 λ 11 s 5 λ 2 λ 12 λ = 4 s + 7 ( s λ ) 12 λ 4 s 12 λ = s 3 λ .

Applying the above inequality, we obtain (2.4). Similarly, we can prove (2.5). The lemma is proved. □

3 Main results

Theorem 3.1 Assume that p , q > 1 , 1 p + 1 q = 1 , r > 1 , 1 r + 1 s = 1 , 0 < λ min { r , s } , a n 0 , b n 0 , such that 0 < n = 1 n p ( 1 λ r ) 1 a n p < and 0 < n = 1 n q ( 1 λ s ) 1 b n q < , then
n = 1 m = 1 a m b n max { m λ , n λ } < { n = 1 [ k λ s 3 λ n λ s ] n p ( 1 λ r ) 1 a n p } 1 p × { n = 1 [ k λ r 3 λ n λ r ] n q ( 1 λ s ) 1 b n q } 1 q ,
(3.1)
n = 1 n p λ s 1 [ k λ r 3 λ n λ r ] p 1 [ m = 1 a m max { m λ , n λ } ] p < n = 1 [ k λ s 3 λ n λ s ] n p ( 1 λ r ) 1 a n p ,
(3.2)
where k λ = r s λ > 0 . Inequality (3.1) is equivalent to (3.2). In particular, we have the following equivalent inequalities:
n = 1 m = 1 a m b n max { m λ , n λ } < k λ { n = 1 [ 1 s k λ 3 λ n λ s ] n p ( 1 λ r ) 1 a n p } 1 p × { n = 1 n q ( 1 λ s ) 1 b n q } 1 q ,
(3.3)
n = 1 n p λ s 1 [ m = 1 a m max { m λ , n λ } ] p < k λ p n = 1 [ 1 s 3 k λ λ n λ s ] n p ( 1 λ r ) 1 a n p .
(3.4)
Proof From Hölder inequality (see [6]), we have
n = 1 m = 1 a m b n max { m λ , n λ } = n = 1 m = 1 a m b n max { m λ , n λ } n ( λ / s 1 ) / p m ( λ / r 1 ) / q m ( λ / r 1 ) / q n ( λ / s 1 ) / p { n = 1 m = 1 a m p m p ( 1 λ / r ) + λ 2 max { m λ , n λ } ( m n ) 1 λ s } 1 p { n = 1 m = 1 b n q n q ( 1 λ / s ) + λ 2 max { m λ , n λ } ( n m ) 1 λ r } 1 q = { m = 1 ω ( m , λ , s ) m p ( 1 λ / r ) + λ 2 a m p } 1 p { n = 1 ω ( n , λ , r ) n q ( 1 λ / s ) + λ 2 b n q } 1 q .

Hence, by (2.4), (2.5), inequality (3.1) is true.

Setting b n as
b n = n p λ / s 1 [ k λ r 3 λ n λ r ] p 1 [ m = 1 a m max { m λ , n λ } ] p 1 ,
by using (3.1), we have
n = 1 [ k λ r 3 λ n λ r ] n q ( 1 λ s ) 1 b n q = n = 1 n p λ / s 1 [ k λ r 3 λ n λ r ] p 1 [ m = 1 a m max { m λ , n λ } ] p = n = 1 m = 1 a m b n max { m λ , n λ } { n = 1 [ k λ s 3 λ n λ s ] n p ( 1 λ r ) 1 a n p } 1 p × { n = 1 [ k λ r 3 λ n λ r ] n q ( 1 λ s ) 1 b n q } 1 q .
(3.5)
Hence, we obtain
0 < n = 1 n p λ s 1 [ k λ r 3 λ n λ r ] p 1 [ m = 1 a m max { m λ , n λ } ] p < n = 1 [ k λ s 3 λ n λ s ] n p ( 1 λ r ) 1 a n p < .
(3.6)

By (3.1), both (3.5) and (3.6) take the form of strict inequality, and we have (3.2).

On the other hand, suppose that (3.2) is valid, from Hölder inequality, we find
n = 1 m = 1 a m b n max { m λ , n λ } = n = 1 n [ q ( λ / s 1 ) + 1 ] / q [ k λ r 3 λ n λ r ] 1 q [ m = 1 a m max { m λ , n λ } ] [ k λ r 3 λ n λ r ] 1 q n [ q ( 1 λ / s ) 1 ] / q b n { n = 1 n p λ / s 1 [ k λ r 3 λ n λ r ] p 1 [ m = 1 a m max { m λ , n λ } ] p } 1 p { n = 1 [ k λ r 3 λ n λ r ] n q ( 1 λ / s ) 1 b n q } 1 q .

Then, by using (3.2), we have (3.1). Hence, (3.2) and (3.1) are equivalent. The proof of Theorem 3.1 is completed. □

Since 0 < λ min { r , s } , by Theorem 3.1, we have the following.

Corollary 3.2 Assume that p , q > 1 , 1 p + 1 q = 1 , r > 1 , 1 r + 1 s = 1 , 0 < λ min { r , s } , a n 0 , b n 0 , such that 0 < n = 1 n p ( 1 λ r ) 1 a n p < and 0 < n = 1 n q ( 1 λ s ) 1 b n q < , then
n = 1 m = 1 a m b n max { m λ , n λ } < { n = 1 [ k λ 1 3 n λ s ] n p ( 1 λ r ) 1 a n p } 1 p × { n = 1 [ k λ 1 3 n λ r ] n q ( 1 λ s ) 1 b n q } 1 q ,
(3.7)
n = 1 n p λ s 1 [ k λ 1 3 n λ r ] p 1 [ m = 1 a m max { m λ , n λ } ] p < n = 1 [ k λ 1 3 n λ s ] n p ( 1 λ r ) 1 a n p ,
(3.8)

where k λ = r s λ > 0 . Inequality (3.7) is equivalent to (3.8).

For r = s = 2 , by using (3.1) and (3.2), we have the following.

Corollary 3.3 Assume that p , q > 1 , 1 p + 1 q = 1 , 0 < λ 2 , a n 0 , b n 0 , such that 0 < n = 1 n p ( 1 λ 2 ) 1 a n p < and 0 < n = 1 n q ( 1 λ 2 ) 1 b n q < , then
n = 1 m = 1 a m b n max { m λ , n λ } < { n = 1 [ k λ 2 3 λ n λ 2 ] n p ( 1 λ 2 ) 1 a n p } 1 p × { n = 1 [ k λ 2 3 λ n λ 2 ] n q ( 1 λ 2 ) 1 b n q } 1 q ,
(3.9)
n = 1 n p λ 2 1 [ k λ 2 3 λ n λ 2 ] p 1 [ m = 1 a m max { m λ , n λ } ] p < n = 1 [ k λ 2 3 λ n λ 2 ] n p ( 1 λ 2 ) 1 a n p ,
(3.10)
where k λ = 4 λ > 0 . Inequality (3.9) is equivalent to (3.10). In particular, we have the equivalent inequalities as follows.
n = 1 m = 1 a m b n max { m λ , n λ } < k λ { n = 1 [ 1 2 k λ 3 λ n λ 2 ] n p ( 1 λ 2 ) 1 a n p } 1 p × { n = 1 n q ( 1 λ 2 ) 1 b n q } 1 q ,
(3.11)
n = 1 n p λ 2 1 [ m = 1 a m max { m λ , n λ } ] p < k λ p n = 1 [ 1 2 3 k λ λ n λ 2 ] n p ( 1 λ 2 ) 1 a n p .
(3.12)

For r = q , s = p , by using (3.1) and (3.2), we have the following.

Corollary 3.4 Assume that p , q > 1 , 1 p + 1 q = 1 , 0 < λ min { p , q } , a n 0 , b n 0 , such that 0 < n = 1 n ( p 1 ) ( 1 λ ) a n p < and 0 < n = 1 n ( q 1 ) ( 1 λ ) b n q < , then
n = 1 m = 1 a m b n max { m λ , n λ } < { n = 1 [ k λ p 3 λ n λ p ] n ( p 1 ) ( 1 λ ) a n p } 1 p × { n = 1 [ k λ q 3 λ n λ q ] n ( q 1 ) ( 1 λ ) b n q } 1 q ,
(3.13)
n = 1 n λ 1 [ k λ q 3 λ n λ q ] p 1 [ m = 1 a m max { m λ , n λ } ] p < n = 1 [ k λ p 3 λ n λ p ] n ( p 1 ) ( 1 λ ) a n p ,
(3.14)
where k λ = p q λ > 0 . Inequality (3.13) is equivalent to (3.14). In particular, we have the equivalent inequalities as follows.
n = 1 m = 1 a m b n max { m λ , n λ } < k λ { n = 1 [ 1 p k λ 3 λ n λ p ] n ( p 1 ) ( 1 λ ) a n p } 1 p × { n = 1 n ( q 1 ) ( 1 λ ) b n q } 1 q ,
(3.15)
n = 1 n λ 1 [ m = 1 a m max { m λ , n λ } ] p < k λ p n = 1 [ 1 p 3 k λ λ n λ p ] n ( p 1 ) ( 1 λ ) a n p .
(3.16)

For r = p , s = q , by using (3.1) and (3.2), we have the following.

Corollary 3.5 Assume that p , q > 1 , 1 p + 1 q = 1 , 0 < λ min { p , q } , a n 0 , b n 0 , such that 0 < n = 1 n p λ 1 a n p < and 0 < n = 1 n q λ 1 b n q < , then
n = 1 m = 1 a m b n max { m λ , n λ } < { n = 1 [ k λ q 3 λ n λ q ] n p λ 1 a n p } 1 p × { n = 1 [ k λ p 3 λ n λ p ] n q λ 1 b n q } 1 q ,
(3.17)
n = 1 n ( p 1 ) λ 1 [ k λ p 3 λ n λ p ] p 1 [ m = 1 a m max { m λ , n λ } ] p < n = 1 [ k λ q 3 λ n λ q ] n p λ 1 a n p ,
(3.18)
where k λ = p q λ > 0 . Inequality (3.17) is equivalent to (3.18). In particular, we have the equivalent inequalities as follows.
n = 1 m = 1 a m b n max { m λ , n λ } < k λ { n = 1 [ 1 q k λ 3 λ n λ q ] n p λ 1 a n p } 1 p { n = 1 n q λ 1 b n q } 1 q ,
(3.19)
n = 1 n ( p 1 ) λ 1 [ m = 1 a m max { m λ , n λ } ] p < k λ p n = 1 [ 1 q 3 k λ λ n λ q ] n p λ 1 a n p .
(3.20)

Set λ = 1 , combining (3.1) and (3.2), we have the following.

Corollary 3.6 Assume that p , q > 1 , 1 p + 1 q = 1 , r > 1 , 1 r + 1 s = 1 , a n 0 , b n 0 , such that 0 < n = 1 n p s 1 a n p < and 0 < n = 1 n q r 1 b n q < , then
n = 1 m = 1 a m b n max { m , n } < { n = 1 [ r s s 3 n 1 s ] n p s 1 a n p } 1 p × { n = 1 [ r s r 3 n 1 r ] n q r 1 b n q } 1 q ,
(3.21)
n = 1 n p s 1 [ r s r 3 n 1 r ] p 1 [ m = 1 a m max { m , n } ] p < n = 1 [ r s s 3 n 1 s ] n p s 1 a n p .
(3.22)
In particular, we have the equivalent inequalities as follows.
n = 1 m = 1 a m b n max { m , n } < r s { n = 1 [ 1 1 3 r n 1 s ] n p s 1 a n p } 1 p { n = 1 n q r 1 b n q } 1 q ,
(3.23)
n = 1 n p s 1 [ m = 1 a m max { m , n } ] p < ( r s ) p n = 1 [ 1 1 3 r n 1 s ] n p s 1 a n p .
(3.24)
Taking p = q = r = s = 2 , in (3.23) and (3.24), we have
n = 1 m = 1 a m b n max { m , n } < 4 { n = 1 [ 1 1 6 n ] a n 2 } 1 2 { n = 1 [ 1 1 6 n ] b n 2 } 1 2 ,
(3.25)
n = 1 [ m = 1 a m max { m , n } ] 2 < 16 n = 1 [ 1 1 6 n ] a n 2 .
(3.26)

Remark 3.1 For r = λ p λ + p 2 and s = λ q λ + q 2 in Theorem 3.1, we get the results of [4].

Declarations

Acknowledgements

The authors would like to thank the editors and the referees for their valuable suggestions to improve the quality of this paper. The first author was supported by the scientific research foundation of National Natural Science Foundation (51109180), the National Science & Technology Supporting Plan from the Ministry of Science & Technology of P.R. China (2011BAD29B08), the ‘111’ Project from the Ministry of Education of P.R. China and the State Administration of Foreign Experts Affairs of P.R. China (B12007), Fundamental Research Funds for the Central Universities (Z109021310) and the scientific research foundation of National Natural Science Foundation (51279167).

Authors’ Affiliations

(1)
Department of Electrical Engineering, Northwest A&F University
(2)
Department of Construction and Information Engineering, Guangxi Modern Vocational Technology College
(3)
School of Civil Engineering, Hebei University of Technology
(4)
Institute of Information Technology, Guilin University of Electronic Technology

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© Chen et al.; licensee Springer. 2013

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