Skip to content

Advertisement

Open Access

On a half-discrete Hilbert-type inequality similar to Mulholland’s inequality

Journal of Inequalities and Applications20132013:290

https://doi.org/10.1186/1029-242X-2013-290

Received: 23 January 2013

Accepted: 30 April 2013

Published: 7 June 2013

Abstract

By using the way of weight functions and Hadamard’s inequality, a half-discrete Hilbert-type inequality similar to Mulholland’s inequality with a best constant factor is given. The extension with multi-parameters, the equivalent forms as well as the operator expressions are also considered.

MSC:26D15.

Keywords

Hilbert-type inequalityweight functionequivalent form

1 Introduction

Assuming that f , g L 2 ( R + ) , f = { 0 f 2 ( x ) d x } 1 2 > 0 , g > 0 , we have the following Hilbert integral inequality (cf. [1]):
0 0 f ( x ) g ( y ) x + y d x d y < π f g ,
(1)
where the constant factor π is the best possible. If a = { a n } n = 1 , b = { b n } n = 1 l 2 , a = { n = 1 a n 2 } 1 2 > 0 , b > 0 , then we still have the following discrete Hilbert inequality:
m = 1 n = 1 a m b n m + n < π a b ,
(2)
with the same best constant factor π. Inequalities (1) and (2) are important in analysis and its applications (cf. [24]). Also we have the following Mulholland inequality with the same best constant factor (cf. [1, 5]):
m = 2 n = 2 a m b n ln m n < π { m = 2 m a m 2 n = 2 n b n 2 } 1 2 .
(3)
In 1998, by introducing an independent parameter λ ( 0 , 1 ] , Yang [6] gave an extension of (1). By generalizing the results from [6], Yang [7] gave some best extensions of (1) and (2) as follows: If p > 1 , 1 p + 1 q = 1 , λ 1 + λ 2 = λ , k λ ( x , y ) is a non-negative homogeneous function of degree −λ with k ( λ 1 ) = 0 k λ ( t , 1 ) t λ 1 1 d t R + , ϕ ( x ) = x p ( 1 λ 1 ) 1 , ψ ( x ) = x q ( 1 λ 2 ) 1 , f ( 0 ) L p , ϕ ( R + ) = { f | f p , ϕ : = { 0 ϕ ( x ) | f ( x ) | p d x } 1 p < } , g ( 0 ) L q , ψ ( R + ) , f p , ϕ , g q , ψ > 0 , then
0 0 k λ ( x , y ) f ( x ) g ( y ) d x d y < k ( λ 1 ) f p , ϕ g q , ψ ,
(4)
where the constant factor k ( λ 1 ) is the best possible. Moreover, if k λ ( x , y ) is finite and k λ ( x , y ) x λ 1 1 ( k λ ( x , y ) y λ 2 1 ) is decreasing for x > 0 ( y > 0 ), then for a m , b n 0 , a = { a m } m = 1 l p , ϕ = { a | a p , ϕ : = { n = 1 ϕ ( n ) | a n | p } 1 p < } , b = { b n } n = 1 l q , ψ , a p , ϕ , b q , ψ > 0 , we have
m = 1 n = 1 k λ ( m , n ) a m b n < k ( λ 1 ) a p , ϕ b q , ψ ,
(5)

with the same best constant factor k ( λ 1 ) . Clearly, for p = q = 2 , λ = 1 , k 1 ( x , y ) = 1 x + y , λ 1 = λ 2 = 1 2 , (4) reduces to (1), while (5) reduces to (2). Some other results about Hilbert-type inequalities are provided by [5, 816].

On the topic of half-discrete Hilbert-type inequalities with the general non-homogeneous kernels, Hardy et al. provided a few results in Theorem 351 of [1]. But they did not prove that the constant factors in the inequalities are the best possible. Moreover, Yang [17] gave an inequality with the particular kernel 1 ( 1 + n x ) λ and an interval variable, and proved that the constant factor is the best possible. Recently, [18] and [19] gave the following half-discrete Hilbert inequality with the best constant factor π:
0 f ( x ) n = 1 a n ( x + n ) λ d x < π f a .
(6)
In this paper, by using the way of weight functions and Hadamard’s inequality, a half-discrete Hilbert-type inequality similar to (3) and (6) with the best constant factor is given as follows:
0 f ( x ) n = 1 a n ln e ( n + 1 2 ) x d x < π f { n = 1 ( n + 1 2 ) a n 2 } 1 2 .
(7)

Moreover, the best extension of (7) with multi-parameters, some equivalent forms as well as the operator expressions are considered.

2 Some lemmas

Lemma 1 If 0 < λ 2 , α 1 2 , setting weight functions ω ( n ) and ϖ ( x ) as follows:
ω ( n ) : = ln λ 2 ( n + α ) 0 x λ 2 1 ln λ e ( n + α ) x d x , n N ,
(8)
ϖ ( x ) : = x λ 2 n = 1 ln λ 2 1 ( n + α ) ( n + α ) ln λ e ( n + α ) x , x ( 0 , ) ,
(9)
we have
ϖ ( x ) < ω ( n ) = B ( λ 2 , λ 2 ) .
(10)
Proof Substitution of t = x ln ( n + α ) in (8), by calculation, yields
ω ( n ) = 0 1 ( 1 + t ) λ t λ 2 1 d t = B ( λ 2 , λ 2 ) .
Since, for fixed x > 0 and in view of the conditions,
h ( x , y ) : = ln λ 2 1 ( y + α ) ( y + α ) ln λ e ( y + α ) x = ln λ 2 1 ( y + α ) ( y + α ) [ 1 + x ln ( y + α ) ] λ
is decreasing and strictly convex for y ( 1 2 , ) , then by Hadamard’s inequality (cf. [20]), we find
ϖ ( x ) < x λ 2 1 2 ln λ 2 1 ( y + α ) ( y + α ) [ 1 + x ln ( y + α ) ] λ d y = t = x ln ( y + α ) x ln ( 1 2 + α ) t λ 2 1 ( 1 + t ) λ d t B ( λ 2 , λ 2 ) ,

namely, (10) follows. □

Lemma 2 Let the assumptions of Lemma  1 be fulfilled and, additionally, let p > 1 , 1 p + 1 q = 1 , a n 0 , n N , f ( x ) be a non-negative measurable function in ( 0 , ) . Then we have the following inequalities:
J : = { n = 1 ln p λ 2 1 ( n + α ) n + α [ 0 f ( x ) ln λ e ( n + α ) x d x ] p } 1 p J [ B ( λ 2 , λ 2 ) ] 1 q { 0 ϖ ( x ) x p ( 1 λ 2 ) 1 f p ( x ) d x } 1 p ,
(11)
L 1 : = { 0 x q λ 2 1 [ ϖ ( x ) ] q 1 [ n = 1 a n ln λ e ( n + α ) x ] q d x } 1 q L 1 { B ( λ 2 , λ 2 ) n = 1 ( n + α ) q 1 ln q ( 1 λ 2 ) 1 ( n + α ) a n q } 1 q .
(12)
Proof By Hölder’s inequality (cf. [20]) and (10), it follows
[ 0 f ( x ) d x ln λ e ( n + α ) x ] p = { 0 1 ln λ e ( n + α ) x [ x ( 1 λ 2 ) / q ln ( 1 λ 2 ) / p ( n + α ) f ( x ) ( n + α ) 1 p ] [ ln ( 1 λ 2 ) / p ( n + α ) x ( 1 λ 2 ) / q ( n + α ) 1 p ] d x } p 0 ln λ 2 1 ( n + α ) ln λ e ( n + α ) x x ( 1 λ 2 ) ( p 1 ) f p ( x ) d x n + α { 0 ( n + α ) q 1 ln λ e ( n + α ) x ln ( 1 λ 2 ) ( q 1 ) ( n + α ) x 1 λ 2 d x } p 1 = { ω ( n ) ( n + α ) q 1 ln q ( λ 2 1 ) + 1 ( n + α ) } p 1 0 ln λ 2 1 ( n + α ) ln λ e ( n + α ) x x ( 1 λ 2 ) ( p 1 ) f p ( x ) d x n + α = [ B ( λ 2 , λ 2 ) ] p 1 n + α ln p λ 2 1 ( n + α ) 0 ln λ 2 1 ( n + α ) ln λ e ( n + α ) x x ( 1 λ 2 ) ( p 1 ) f p ( x ) d x n + α .
Then by the Lebesgue term-by-term integration theorem (cf. [21]), we have
J [ B ( λ 2 , λ 2 ) ] 1 q { n = 1 0 ln λ 2 1 ( n + α ) ln λ e ( n + α ) x x ( 1 λ 2 ) ( p 1 ) f p ( x ) d x n + α } 1 p = [ B ( λ 2 , λ 2 ) ] 1 q { 0 n = 1 ln λ 2 1 ( n + α ) ln λ e ( ( n + α ) ) x x ( 1 λ 2 ) ( p 1 ) f p ( x ) d x n + α } 1 p = [ B ( λ 2 , λ 2 ) ] 1 q { 0 ϖ ( x ) x p ( 1 λ 2 ) 1 f p ( x ) d x } 1 p ,
and (11) follows. Still by Hölder’s inequality, we have
[ n = 1 a n ln λ e ( n + α ) x ] q = { n = 1 1 ln λ e ( n + α ) x [ x ( 1 λ 2 ) / q ln ( 1 λ 2 ) / p ( n + α ) 1 ( n + α ) 1 p ] [ ln ( 1 λ 2 ) / p ( n + α ) x ( 1 λ 2 ) / q ( n + α ) 1 p a n ] } q { n = 1 ln λ 2 1 ( n + α ) ln λ e ( n + α ) x x ( 1 λ 2 ) ( p 1 ) ( n + α ) } q 1 n = 1 ( n + α ) q 1 ln λ e ( n + α ) x ln ( 1 λ 2 ) ( q 1 ) ( n + α ) x 1 λ 2 a n q = [ ϖ ( x ) ] q 1 x q λ 2 1 n = 1 ( n + α ) q 1 ln λ e ( n + α ) x x λ 2 1 ln ( 1 λ 2 ) ( q 1 ) ( n + α ) a n q .
Then by the Lebesgue term-by-term integration theorem, we have
L 1 { 0 n = 1 ( n + α ) q 1 ln λ e ( n + α ) x x λ 2 1 ln ( 1 λ 2 ) ( q 1 ) ( n + α ) a n q d x } 1 q = { n = 1 [ ln λ 2 ( n + α ) 0 x λ 2 1 d x ln λ e ( n + α ) x ] ( n + α ) q 1 ln q ( 1 λ 2 ) 1 ( n + α ) a n q } 1 q = { n = 1 ω ( n ) ( n + α ) q 1 ln q ( 1 λ 2 ) 1 ( n + α ) a n q } 1 q ,

and then in view of (10), inequality (12) follows. □

3 Main results

We introduce two functions
Φ ( x ) : = x p ( 1 λ 2 ) 1 ( x > 0 ) and Ψ ( n ) : = ( n + α ) q 1 ln q ( 1 λ 2 ) 1 ( n + α ) ( n N ) ,

wherefrom, [ Φ ( x ) ] 1 q = x q λ 2 1 , and [ Ψ ( n ) ] 1 p = ln p λ 2 1 ( n + α ) n + α .

Theorem 1 If  0 < λ 2 , α 1 2 , p > 1 , 1 p + 1 q = 1 , f ( x ) , a n 0 , f L p , Φ ( R + ) , a = { a n } n = 1 l q , Ψ , f p , Φ > 0 and a q , Ψ > 0 , then we have the following equivalent inequalities:
I : = n = 1 0 a n f ( x ) d x ln λ e ( n + α ) x I = 0 n = 1 a n f ( x ) d x ln λ e ( n + α ) x < B ( λ 2 , λ 2 ) f p , Φ a q , Ψ ,
(13)
J = { n = 1 [ Ψ ( n ) ] 1 p [ 0 f ( x ) ln λ e ( n + α ) x d x ] p } 1 p J < B ( λ 2 , λ 2 ) f p , Φ ,
(14)
L : = { 0 [ Φ ( x ) ] 1 q [ n = 1 a n ln λ e ( n + α ) x ] q d x } 1 q L < B ( λ 2 , λ 2 ) a q , Ψ ,
(15)

where the constant B ( λ 2 , λ 2 ) is the best possible in the above inequalities.

Proof By the Lebesgue term-by-term integration theorem, there are two expressions for I in (13). In view of (11), for ϖ ( x ) < B ( λ 2 , λ 2 ) , we have (14). By Hölder’s inequality, we have
I = n = 1 [ Ψ 1 q ( n ) 0 1 ln λ e ( n + α ) x f ( x ) d x ] [ Ψ 1 q ( n ) a n ] J a q , Ψ .
(16)
Then by (14), we have (13). On the other hand, assuming that (13) is valid, setting
a n : = [ Ψ ( n ) ] 1 p [ 0 1 ln λ e ( n + α ) x f ( x ) d x ] p 1 , n N ,
then J p 1 = a q , Ψ . By (11), we find J < . If J = 0 , then (14) is trivially valid; if J > 0 , then by (13), we have
a q , Ψ q = J p = I < B ( λ 2 , λ 2 ) f p , Φ a q , Ψ , i.e. a q , Ψ q 1 = J < B ( λ 2 , λ 2 ) f p , Φ ,
that is, (14) is equivalent to (13). In view of (12), for [ ϖ ( x ) ] 1 q > [ B ( λ 2 , λ 2 ) ] 1 q , we have (15). By Hölder’s inequality, we find
I = 0 [ Φ 1 p ( x ) f ( x ) ] [ Φ 1 p ( x ) n = 1 1 ln λ e ( n + α ) x a n ] d x f p , Φ L .
(17)
Then by (15), we have (13). On the other hand, assuming that (13) is valid, setting
f ( x ) : = [ Φ ( x ) ] 1 q [ n = 1 1 ln λ e ( n + α ) x a n ] q 1 , x ( 0 , ) ,
then L q 1 = f p , Φ . By (12), we find L < . If L = 0 , then (15) is trivially valid; if L > 0 , then by (13), we have
f p , Φ p = L q = I < B ( λ 2 , λ 2 ) f p , Φ a q , Ψ , i.e. , f p , Φ p 1 = L < B ( λ 2 , λ 2 ) a q , Ψ ,

that is, (15) is equivalent to (13). Hence, inequalities (13), (14) and (15) are equivalent.

For 0 < ε < p λ 2 , setting f ˜ ( x ) = x λ 2 + ε p 1 , x ( 0 , 1 ) ; f ˜ ( x ) = 0 , x [ 1 , ) , and a ˜ n = 1 n + α ln λ 2 ε q 1 ( n + α ) , n N , if there exists a positive number k ( B ( λ 2 , λ 2 ) ) such that (13) is valid as we replace B ( λ 2 , λ 2 ) with k, then, in particular, it follows
I ˜ : = n = 1 0 1 ln λ e ( n + α ) x a ˜ n f ˜ ( x ) d x < k f ˜ p , Φ a ˜ q , Ψ I ˜ = k { 0 1 d x x ε + 1 } 1 p { 1 ( 1 + α ) ln ε + 1 ( 1 + α ) + n = 2 1 ( n + α ) ln ε + 1 ( n + α ) } 1 q I ˜ < k ( 1 ε ) 1 p { 1 ( 1 + α ) ln ε + 1 ( 1 + α ) + 1 1 ( x + α ) ln ε + 1 ( x + α ) d x } 1 q I ˜ = k ε { ε ( 1 + α ) ln ε + 1 ( 1 + α ) + 1 ln ε ( 1 + α ) } 1 q ,
(18)
I ˜ = n = 1 1 n + α ln λ 2 ε q 1 ( n + α ) 0 1 1 ln λ e ( n + α ) x x λ 2 + ε p 1 d x = t = x ln ( n + α ) n = 1 1 ( n + α ) ln ε + 1 ( n + α ) 0 ln ( n + α ) 1 ( t + 1 ) λ t λ 2 + ε p 1 d t = B ( λ 2 + ε p , λ 2 ε p ) n = 1 1 ( n + α ) ln ε + 1 ( n + α ) A ( ε ) > B ( λ 2 + ε p , λ 2 ε p ) 1 1 ( y + α ) ln ε + 1 ( y + α ) d y A ( ε ) = 1 ε ln ε ( 1 + α ) B ( λ 2 + ε p , λ 2 ε p ) A ( ε ) ,
A ( ε ) : = n = 1 1 ( n + α ) ln ε + 1 ( n + α ) ln ( n + α ) 1 ( t + 1 ) λ t λ 2 + ε p 1 d t .
(19)
We find
0 < A ( ε ) n = 1 1 ( n + α ) ln ε + 1 ( n + α ) ln ( n + α ) 1 t λ t λ 2 + ε p 1 d t = 1 λ 2 ε p n = 1 1 ( n + α ) ln λ 2 + ε q + 1 ( n + α ) < ,
and then A ( ε ) = O ( 1 ) ( ε 0 + ). Hence by (18) and (19), it follows
1 ln ε ( 1 + α ) B ( λ 2 + ε p , λ 2 ε p ) ε O ( 1 ) < k { ε ( 1 + α ) ln ε + 1 ( 1 + α ) + 1 ln ε ( 1 + α ) } 1 q ,
(20)

and B ( λ 2 , λ 2 ) k ( ε 0 + ). Hence k = B ( λ 2 , λ 2 ) is the best value of (13).

By equivalence, the constant factor B ( λ 2 , λ 2 ) in (14) and (15) is the best possible. Otherwise, we can imply a contradiction by (16) and (17) that the constant factor in (13) is not the best possible. □

Remark 1 (i) Define the first type half-discrete Hilbert-type operator T 1 : L p , Φ ( R + ) l p , Ψ 1 p as follows: For f L p , Φ ( R + ) , we define T 1 f l p , Ψ 1 p , satisfying
T 1 f ( n ) = 0 1 ln λ e ( n + α ) x f ( x ) d x , n N .
Then by (14) it follows T 1 f p . Ψ 1 p B ( λ 2 , λ 2 ) f p , Φ , and then T 1 is a bounded operator with T 1 B ( λ 2 , λ 2 ) . Since by Theorem 1 the constant factor in (14) is the best possible, we have T 1 = B ( λ 2 , λ 2 ) .
  1. (ii)
    Define the second type half-discrete Hilbert-type operator T 2 : l q , Ψ L q , Φ 1 q ( R + ) as follows: For a l q , Ψ , we define T 2 a L q , Φ 1 q ( R + ) , satisfying
    T 2 a ( x ) = n = 1 1 ln λ e ( n + α ) x a n , x ( 0 , ) .
     

Then by (15) it follows T 2 a q , Φ 1 q B ( λ 2 , λ 2 ) a q , Ψ , and then T 2 is a bounded operator with T 2 B ( λ 2 , λ 2 ) . Since by Theorem 1 the constant factor in (15) is the best possible, we have T 2 = B ( λ 2 , λ 2 ) .

Remark 2 For p = q = 2 , λ = 1 , λ 1 = λ 2 = 1 2 , α = 1 2 in (13), (14) and (15), we have (7) and the following equivalent inequalities:
{ n = 1 1 n + 1 2 [ 0 f ( x ) ln e ( n + 1 2 ) x d x ] 2 } 1 2 < π f ,
(21)
{ 0 [ n = 1 a n ln e ( n + 1 2 ) x ] 2 d x } 1 2 < π { n = 1 ( n + 1 2 ) a n 2 } 1 2 .
(22)

Declarations

Acknowledgements

This work is supported by 2012 Knowledge Construction Special Foundation Item of Guangdong Institution of Higher Learning College and University (No. 2012KJCX0079).

Authors’ Affiliations

(1)
Basic Education College of Zhanjiang Normal University, Zhanjiang, Guangdong, P.R. China
(2)
Department of Mathematics, Guangdong University of Education, Guangzhou, Guangdong, P.R. China

References

  1. Hardy GH, Littlewood JE, Pólya G: Inequalities. Cambridge University Press, Cambridge; 1934.MATHGoogle Scholar
  2. Mitrinović DS, Pečarić JE, Fink AM: Inequalities Involving Functions and Their Integrals and Derivatives. Kluwer Acaremic, Boston; 1991.View ArticleMATHGoogle Scholar
  3. Yang B: Hilbert-Type Integral Inequalities. Bentham Science Publishers, Sharjah; 2009.Google Scholar
  4. Yang B: Discrete Hilbert-Type Inequalities. Bentham Science Publishers, Sharjah; 2011.Google Scholar
  5. Yang B: On a new extension of Hilbert’s inequality with some parameters. Acta Math. Hung. 2005, 108(4):337–350. 10.1007/s10474-005-0229-4View ArticleMATHGoogle Scholar
  6. Yang B: On Hilbert’s integral inequality. J. Math. Anal. Appl. 1998, 220: 778–785. 10.1006/jmaa.1997.5877MathSciNetView ArticleMATHGoogle Scholar
  7. Yang B: The Norm of Operator and Hilbert-Type Inequalities. Science Press, Beijin; 2009.Google Scholar
  8. Yang B, Brnetić I, Krnić M, Pečarić J: Generalization of Hilbert and Hardy-Hilbert integral inequalities. Math. Inequal. Appl. 2005, 8(2):259–272.MathSciNetMATHGoogle Scholar
  9. Krnić M, Pečarić J: Hilbert’s inequalities and their reverses. Publ. Math. (Debr.) 2005, 67(3–4):315–331.MATHGoogle Scholar
  10. Jin J, Debnath L: On a Hilbert-type linear series operator and its applications. J. Math. Anal. Appl. 2010, 371: 691–704. 10.1016/j.jmaa.2010.06.002MathSciNetView ArticleMATHGoogle Scholar
  11. Azar L: On some extensions of Hardy-Hilbert’s inequality and applications. J. Inequal. Appl. 2009., 2009: Article ID 546829Google Scholar
  12. Yang B, Rassias TM: On the way of weight coefficient and research for Hilbert-type inequalities. Math. Inequal. Appl. 2003, 6(4):625–658.MathSciNetMATHGoogle Scholar
  13. Arpad B, Choonghong O: Best constant for certain multilinear integral operator. J. Inequal. Appl. 2006., 2006: Article ID 28582Google Scholar
  14. Kuang J, Debnath L: On Hilbert’s type inequalities on the weighted Orlicz spaces. Pac. J. Appl. Math. 2007, 1(1):95–103.MathSciNetMATHGoogle Scholar
  15. Zhong W: The Hilbert-type integral inequality with a homogeneous kernel of − λ -degree. J. Inequal. Appl. 2008., 2008: Article ID 917392Google Scholar
  16. Li Y, He B: On inequalities of Hilbert’s type. Bull. Aust. Math. Soc. 2007, 76(1):1–13. 10.1017/S0004972700039423View ArticleMATHGoogle Scholar
  17. Yang B: A mixed Hilbert-type inequality with a best constant factor. Int. J. Pure Appl. Math. 2005, 20(3):319–328.MathSciNetMATHGoogle Scholar
  18. Yang B: A half-discrete Hilbert’s inequality. J. Guangdong Univ. Educ. 2011, 31(3):1–7.MATHGoogle Scholar
  19. Yang B, Chen Q: A half-discrete Hilbert-type inequality with a homogeneous kernel and an extension. J. Inequal. Appl. 2011., 2011: Article ID 124. doi:10.1186/1029–242X-2011–124Google Scholar
  20. Kuang J: Applied Inequalities. Shangdong Science Technic Press, Jinan; 2004.Google Scholar
  21. Kuang J: Introduction to Real Analysis. Hunan Education Press, Chansha; 1996.Google Scholar

Copyright

© Huang and Yang; licensee Springer 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Advertisement