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A half-discrete Hilbert-type inequality with the non-monotone kernel and the best constant factor

Journal of Inequalities and Applications20122012:184

https://doi.org/10.1186/1029-242X-2012-184

Received: 19 December 2011

Accepted: 9 August 2012

Published: 30 August 2012

Abstract

By introducing two pairs of conjugate exponents and using the improved Euler-Maclaurin summation formula, we estimate the weight functions and obtain a half-discrete Hilbert-type inequality with the non-monotone kernel and the best constant factor. We also consider its equivalent forms.

MSC:26D15.

Keywords

Hilbert-type inequalityconjugate exponentHölder’s inequalitybest constant factorequivalent form

1 Introduction

If a n , b n 0 , such that 0 < n = 1 a n 2 < and 0 < n = 1 b n 2 < , then we have the famous Hilbert’s inequality as follows (cf. [1]):
n = 1 m = 1 a m b n m + n < π ( n = 1 a n 2 n = 1 b n 2 ) 1 2 ,
(1)

where the constant factor π is the best possible.

Under the same condition of (1), Xin et al. [2] gave the following inequality:
n = 1 m = 1 | ln ( m / n ) | m + n a m b n < c 0 ( n = 1 a n 2 n = 1 b n 2 ) 1 2 ,
(2)

where the constant factor c 0 = 8 n = 1 ( 1 ) n 1 ( 2 n 1 ) 2 = 7.3277 + is the best possible. And Yang [3] gave the integral analogues of (2).

In 1934, Hardy et al. [1] established a few results on the half-discrete Hilbert-type inequalities with the non-homogeneous kernel (see Theorem 351). But they did not prove that the constant factors are the best possible. However, Yang [4] gave a result by introducing an interval variable and proved that the constant factor is the best possible. Recently, Yang et al. [59] gave some half-discrete Hilbert-type inequalities and their reverses with the monotone kernels and best constant factors.

Recently, Yang [10] gave the following half-discrete Hilbert-type inequality with the non-monotone kernel and the best constant factor 8:
n = 1 1 | ln ( x / n ) | a n max { x , n } f ( x ) d x < 8 ( n = 1 a n 2 1 f 2 ( x ) d x ) 1 2 .
(3)

Obviously, for a half-discrete Hilbert-type inequality with the monotone kernel, it is easy to build the relating inequality by estimating the series form and the integral form of weight functions. However, for a half-discrete Hilbert-type inequality with the non-monotone kernel, it is much more difficult to prove.

In this paper, by using the way of weight functions, we give a new half-discrete Hilbert-type inequality with the non-monotone kernel as follows:
n = 1 a n 1 | ln ( x n ) | f ( x ) x + n d x < 8 k = 0 ( 1 ) k ( 2 k + 1 ) 2 ( n = 1 a n 2 1 f 2 ( x ) ) 1 2 ,
(4)

where the constant factor 8 k = 0 ( 1 ) k ( 2 k + 1 ) 2 is the best possible. The main objective of this paper is to build the best extension of (4) with parameters and equivalent forms.

2 Some lemmas

Lemma 2.1 If x 1 R , n 1 Z (Z is the set of non-negative integers), [ x 1 ] = n 1 , ρ ( y ) = y [ y ] 1 2 ( y R ) is the Bernoulli function of first order[11], then we have (cf. [10])
n 1 x 1 ρ ( y ) d y = ε 1 8 ( ε 1 [ 0 , 1 ] ) .
(5)
Lemma 2.2 If r > 1 , 1 r + 1 s = 1 , f ( x , y ) : = | ln ( x y ) | x + y ( x y ) 1 r ( x , y ( 0 , ) ), N is the set of positive integers, define the weight functions as follows:
(6)
(7)
Then we have
ω ( n ) < c r : = k = 0 ( 1 ) k [ 1 ( k + 1 r ) 2 + 1 ( k + 1 s ) 2 ] , ϖ ( x ) < c r .
(8)
Proof Setting u = x n in (6), we have
ω ( n ) = 1 n | ln u | u + 1 u 1 s d u < 0 | ln u | u + 1 u 1 s d u = c r .
Setting u = y x , then it follows
0 1 f ( x , y ) d y = 0 1 x ( ln u ) u + 1 u 1 r d u > s 0 1 x ( ln u ) 1 x + 1 d u 1 r + 1 = x 1 r x + 1 ( s ln x + s 2 ) .
(9)
For 1 y < x , f ( x , y ) = ln ( y x ) x + y ( x y ) 1 r , we have
f y ( x , y ) = x 1 r [ 1 ( y + x ) y 1 + 1 r + ln ( y x ) ( y + x ) 2 y 1 r + ln ( y x ) r ( y + x ) y 1 + 1 r ] = x 1 r [ 1 ( y + x ) y 1 + 1 r + 2 x ln ( y x ) ( y x ) ( y + x ) 2 y 1 r + 2 x ln ( y x ) r ( y x ) ( y + x ) y 1 + 1 r ] + x 1 r [ ln ( y x ) ( y x ) ( y + x ) y 1 r + ln ( y x ) r ( y x ) y 1 + 1 r ] .
For y x , f ( x , y ) = ln ( y x ) x + y ( x y ) 1 r , we find
f y ( x , y ) = x 1 r [ 1 ( y + x ) y 1 + 1 r ln ( y x ) ( y + x ) 2 y 1 r ln ( y x ) r ( y + x ) y 1 + 1 r ] = x 1 r [ 1 ( y + x ) y 1 + 1 r + 2 x ln ( y x ) ( y x ) ( y + x ) 2 y 1 r + 2 x ln ( y x ) r ( y x ) ( y + x ) y 1 + 1 r ] x 1 r [ ln ( y x ) ( y x ) ( y + x ) y 1 r + ln ( y x ) r ( y x ) y 1 + 1 r ] .
Define two functions as follows:
Then we have f y ( x , y ) = g ( y ) h ( y ) . Setting a = 1 2 x 2 , b = 1 2 x 2 , then a b = 1 x 2 . Define two functions as follows:
g ˜ ( y ) = { g 1 ( y ) a , y < x ; g 2 ( y ) , y x , h ˜ ( y ) = { h 1 ( y ) b , y < x ; h 2 ( y ) , y x .
Since g 1 ( x 0 ) a = g 2 ( x ) , h 1 ( x 0 ) b = h 2 ( x ) , then both g ˜ ( y ) and h ˜ ( y ) ( y [ 1 , ) ) are decreasing and continuous. Besides y = x , we have ( 1 ) i g ˜ ( i ) ( y ) 0 , ( 1 ) i h ˜ ( i ) ( y ) 0 ( i = 0 , 1 ), and g ˜ ( ) = h ˜ ( ) = 0 . By the improved Euler-Maclaurin summation formula (cf. [11], Theorem 2.2.2) and (5), for ε 1 [ 0 , 1 ] , ε i ( 0 , 1 ) ( i = 2 , 3 ), it follows
Since g 1 ( 1 ) a g 1 ( x 0 ) a = g 2 ( x ) > 0 , h 1 ( 1 ) b h 1 ( x 0 ) b = h 2 ( x ) > 0 , then we have
1 ρ ( y ) f y ( x , y ) d y > 1 8 ( g 1 ( 1 ) a ) 1 8 ( a b ) = x 1 r 8 ( x + 1 ) x 1 + 1 r ln x 4 ( x + 1 ) 2 ( x 1 ) x 1 + 1 r ln x 4 r ( x + 1 ) ( x 1 ) 1 16 x 2 .
(10)
By the improved Euler-Maclaurin summation formula [11], we have
ϖ ( x ) = k = 1 f ( x , k ) = 1 f ( x , y ) d y + 1 2 f ( x , 1 ) + 1 ρ ( y ) f y ( x , y ) d y = 0 f ( x , y ) d y ( 0 1 f ( x , y ) d y 1 2 f ( x , 1 ) 1 ρ ( y ) f y ( x , y ) d y ) = c r θ ( x ) ,
where
θ ( x ) : = 0 1 f ( x , y ) d y 1 2 f ( x , 1 ) 1 ρ ( y ) f y ( x , y ) d y .
Since 1 2 f ( x , 1 ) = x 1 r ln x 2 ( x + 1 ) , in view of (9), (10), (i) for 1 x < 2 , ln x x 1 1 , we have
θ ( x ) > x 1 r x + 1 ( s ln x + s 2 ) x 1 r ln x 2 ( x + 1 ) x 1 r 8 ( x + 1 ) x 1 + 1 r ln x 4 ( x + 1 ) 2 ( x 1 ) x 1 + 1 r ln x 4 r ( x + 1 ) ( x 1 ) 1 16 x 2 x 1 r ln x ( x + 1 ) ( s 1 2 ) + x 1 r x + 1 [ s 2 1 8 x 4 ( x + 1 ) x 4 r x + 1 16 x 2 + 1 r ] > x 1 r ( x + 1 ) ( 1 1 8 1 4 1 2 1 8 ) = 0 ;
(ii) for x 2 , 1 x 1 2 x , we have
θ ( x ) x 1 r ln x ( x + 1 ) [ s 1 2 1 2 ( x + 1 ) 1 2 r ] + x 1 r x + 1 ( s 2 1 8 x + 1 16 x 2 + 1 r ) > x 1 r ln x ( x + 1 ) ( s 1 2 1 6 1 2 r ) = ( 6 s + 2 r 3 ) x 1 r ln x 6 r ( x + 1 ) > 0 .

Hence, for x 1 , we have θ ( x ) > 0 , it follows ϖ ( x ) < c r . The lemma is proved. □

Lemma 2.3 As the assumption of Lemma  2.2, if 0 < ε < p r , R = ( 1 r ε p ) 1 , S = ( 1 s + ε p ) 1 , then we have
I ¯ : = n = 1 n 1 S ε 1 1 | ln ( x n ) | x + n x 1 R 1 d x c R ε O ( 1 ) .
(11)
Proof It is obvious that R > 1 , 1 R + 1 S = 1 . Setting u = x n , we have
I ¯ = n = 1 n ε 1 1 n | ln u | u + 1 u 1 R 1 d u = n = 1 n ε 1 [ 0 | ln u | u + 1 u 1 R 1 d u + 0 1 n ln u u + 1 u 1 R 1 d u ] > c R 1 x ε 1 d x + R n = 1 n ε 1 0 1 n ln u d u 1 R = c R ε [ R n = 1 ln n n 1 + ε + 1 R + R 2 n = 1 1 n 1 + ε + 1 R ] = c R ε O ( 1 ) ( ε 0 + ) .

The lemma is proved. □

Lemma 2.4 If p , r > 1 , 1 r + 1 s = 1 p + 1 q = 1 , a n 0 , f ( x ) is a non-negative measurable function, then we have
(12)
(13)
Proof By Hölder’s inequality [12], in view of (6) and (7), we have
Hence we have (12). Still by Hölder’s inequality [12], (6) and (7), we have

Then we have (13). The lemma is proved. □

3 Main results and applications

Theorem 3.1 If p , r > 1 , 1 p + 1 q = 1 r + 1 s = 1 , a n , f ( x ) 0 such that 0 < 1 x p s 1 f p ( x ) d x < , n = 1 n q r 1 a n q < , then we have the following equivalent inequalities:
(14)
(15)
(16)

where the constant factors c r = k = 0 ( 1 ) k [ 1 ( k + 1 r ) 2 + 1 ( k + 1 s ) 2 ] , c r p , c r q are the best possible.

Proof By the Lebesgue term-by-term integration theorem [13], there are two kinds of representation in (14). By the conditions of Theorem 3.1, (12) takes the form of a strict inequality, and we have (15). By Hölder’s inequality [12], we have
I = n = 1 [ n 1 s 1 p 1 | ln ( x n ) | f ( x ) x + n d x ] [ n 1 s + 1 p a n ] J 1 1 p { n = 1 n q r 1 a n q } 1 q .
(17)
By (15), we have (14). On the other hand, suppose that (14) is valid. Setting a n : = n p s 1 [ 1 | ln ( x n ) | f ( x ) x + n d x ] p 1 , n N , then it follows J 1 = n = 1 n q r 1 a n q . By (12), we have J < . If J = 0 , then (15) is obvious value; if 0 < J < , then by (14), we obtain
n = 1 n q r 1 a n q = J 1 = I < c r { 1 x p s 1 f p ( x ) d x } 1 p { n = 1 n q r 1 a n q } 1 q , J 1 1 p = { n = 1 n q r 1 a n q } 1 p < c r { 1 x p s 1 f p ( x ) d x } 1 p .
(18)

Hence we have (15), which is equivalent to (14).

By Hölder’s inequality [12], we have
I = 1 [ x 1 r 1 q f ( x ) ] [ x 1 r + 1 q n = 1 | ln ( x n ) | a n x + n ] d x { 1 x p s 1 f p ( x ) d x } 1 p J 2 1 q .
(19)
By (16), we have (14). On the other hand, suppose that (14) is valid. Setting f ( x ) : = x q r 1 [ n = 1 | ln ( x n ) | a n x + n ] q 1 , x [ 1 , ) , then it follows J 2 = 1 x p s 1 f p ( x ) d x . By (13), we have J 2 < . If J 2 = 0 , then (16) is obvious value; if 0 < J 2 < , then by (14), we obtain
1 x p s 1 f p ( x ) d x = J 2 = I < c r { 1 x p s 1 f p ( x ) d x } 1 p { n = 1 n q r 1 a n q } 1 q , J 2 1 q = { 1 x p s 1 f p ( x ) d x } 1 q < c r { n = 1 n q r 1 a n q } 1 q .
(20)

Hence we have (16), which is equivalent to (14). Therefore (14), (15) and (16) are equivalent.

If the constant factor c r in (14) is not best possible, then there exists a positive number K, with 0 < K < c r , such that (14) is still valid if we replace c r by K. For 0 < ε < ε 0 , setting f ¯ ( x ) = n 1 r ε p 1 , a ¯ n = n 1 s ε q 1 ( n N ), we have
I ¯ = n = 1 a ¯ n 1 | ln ( x n ) | f ¯ ( x ) x + n d x < K { 1 x p s 1 f ¯ p ( x ) d x } 1 p { n = 1 n q r 1 a ¯ n q } 1 q = K ( 1 x 1 ε d x ) 1 p ( 1 + n = 2 n 1 ε ) 1 q < K ( 1 ε ) 1 p ( 1 + 1 x 1 ε d x ) 1 q = K ε ( ε + 1 ) 1 q .
(21)

By (11) and (21), we have c R ε O ( 1 ) < K ( ε + 1 ) 1 q and for ε 0 + , by Fatou lemma [13], we have c r lim ε 0 + ( c R ε O ( 1 ) ) K . This is a contradiction. Hence we can conclude that the constant c r in (14) is the best possible. If the constant factors in (15) and (16) are not the best possible, then we can imply a contradiction that the constant factor in (14) is not the best possible by (17) and (19). The theorem is proved. □

Remark For p = q = r = s = 2 , (14) reduces to (4). Inequality (4) is a new basic half-discrete Hilbert-type inequality with the non-monotone kernel.

Declarations

Acknowledgements

This work was supported by the Emphases Natural Science Foundation of Guangdong Institutions of Higher Learning, College and University (No. 05Z026) and the Natural Science Foundation of Guangdong (7004344).

Authors’ Affiliations

(1)
Department of Mathematics, Guangdong Education University, Guangzhou, People’s Republic of China

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© Xin and Yang; licensee Springer 2012

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.

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