Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2007, Article ID 79758, 7 pages
doi:10.1155/2007/79758
Research Article
A Reverse Hardy-Hilbert-Type Inequality
Gaowen Xi
Received 12 December 2006; Accepted 12 March 2007
Recommended by Lars-Erik Persson
By estimating the weight coefficient, a reverse Hardy-Hilbert-type inequality is proved.
As applications, some equivalent forms and a number of particular cases are obtained.
Copyright © 2007 Gaowen Xi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
1. Introduction
Let p > 1, 1/ p + 1/q = 1, an ≥ 0, bn ≥ 0, and 0 <
Hardy-Hilbert inequality is as follows:
∞
p
n =0 a n
∞ ∞
∞
am bn
π
p
an
<
m
+
n
+
1
sin(π/
p)
n=0 m=0
n =0
< ∞, 0 <
1/ p ∞
q
n=0 bn
< ∞. Then the
1/q
∞
q
bn
,
(1.1)
n =0
where the constant factor π/ sin(π p) is the best possible [1].
For (1.1), Yang et al. [2–6] gave some strengthened versions and extensions as follows:
∞ ∞
am bn
<
m
+n+1
n=0 m=0
∞ ∞
am bn
<
m+n+1
n=0 m=0
∞
n =0
∞
n =0
×
∞
7
7
π−
a2n
π− √
b2
5( n + 3)
5(
n
+ 3) n
n =0
√
π
ln 2 − C
p
−
an
sin(π/ p) (2n + 1)1+1/ p
∞
n =0
1/2
(1.2)
1/ p
π
ln2 − C
q
−
bn
sin(π/ p) (2n + 1)1+1/q
(1.3)
1/q
,
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Journal of Inequalities and Applications
where ln 2 − C = 0.1159315+ (C is the Euler constant),
∞ ∞
p+λ−2 q+λ−2
am bn
<B
,
λ
(m + n + 1)
p
q
n=0 m=0
×
∞ n =0
1
n+
2
1−λ
∞ 1
n+
2
n =0
1−λ
1/ p
p
an
(1.4)
1/q
q
bn
,
where the constant B((p + λ − 2)/ p, (q + λ − 2)/q) is the best possible (2 − min{ p, q} <
λ ≤ 2),
∞ ∞
am bn
λ λ
<B ,
λ
(m
+
n
+
1)
p q
n=0 m=0
×
∞ n =0
∞ n =0
1
n+
2
1
n+
2
q−1−λ
p−1−λ
1/ p
p
an
(1.5)
1/q
q
bn
,
where the constant B(λ/ p, λ/q) is the best possible (0 < λ ≤ min{ p, q}),
∞ ∞
am bn
(r − 2)t + λ (s − 2)t + λ
<B
,
λ
(m
+
n
+
1)
r
s
n=0 m=0
×
∞ 1
n+
2
n =0
∞ q(1−t+(2t−λ)/s)−1
n+
n =0
1
2
p(1−t+(2t−λ)/r)−1
1/ p
p
an
1/q
q
bn
,
(1.6)
where the constant B(((r − 2)t + λ)/r,((s − 2)t + λ)/s) is the best possible (r > 1, 1/r +
1/s = 1, t ∈ [0, 1], 2 − min{r, s}t < λ ≤ 2 − min{r, s}t + min{r, s}).
For the reverse Hardy-Hilbert inequality, recently, Yang [7] gave a reverse form of inequalities (1.4), (1.5), and (1.6) for λ = 2. The main objective of this paper is to establish
an extension of the above Yang’s work for 1.5 < λ < 3, by estimating the weight coefficient.
For this, we need the following expression of the β function B(p, q) (see [8]):
B(p, q) = B(q, p) =
∞
0
1
u p−1 du,
(1 + u) p+q
(p, q > 0)
(1.7)
and the following inequality [3]:
∞
0
∞
f (x)dx +
1
f (0) <
f (m) <
2
m=0
where f (x) ∈ C 3 [0, ∞), and
∞
0
∞
0
f (x)dx +
1
1 f (0) −
f (0),
2
12
(1.8)
f (x)dx < ∞, (−1)n f (n) (x) > 0, f (n) (∞) = 0 (n = 0,1,2,3).
Gaowen Xi 3
2. Main results
Lemma 2.1. Let N0 be the set of nonnegative integers, N the set of positive integers, and R
the set of real numbers. The weight coefficient ωλ (n) is defined by
ωλ (n) =
∞
1
,
(m + n + 1)λ
m=0
n ∈ N0 , 1.5 ≤ λ < 3.
(2.1)
Then
2(n + 1)2−λ
(λ − 1)2
2(n + 1)2−λ
1−
<
ω
(n)
<
.
λ
(λ − 1)(2n + 3 − λ)
4(n + 1)2
(λ − 1)(2n + 3 − λ)
(2.2)
Proof. If n ∈ N0 , let f (x) = 1/(m + n + 1)λ , x ∈ [0, ∞). By (1.8), we obtain
∞
ωλ (n) >
0
1
1
dx
1
+
=
+
,
(x + n + 1)λ 2(n + 1)λ (λ − 1)(n + 1)λ−1 2(n + 1)λ
0
1
λ
1
dx +
+
λ
λ
(x + n + 1)
2(n + 1)
12(n + 1)λ+1
∞
ωλ (n) <
=
(2.3)
1
λ
1
+
+
.
(λ − 1)(n + 1)λ−1 2(n + 1)λ 12(n + 1)λ+1
Since we find
1
1
2
λ−1
+
2(n + 1)λ−1 − (λ − 1)(n + 1)λ−2 =
−
,
(λ − 1)(n + 1)λ−1 2(n + 1)λ
λ − 1 2(n + 1)2
1
λ
1
+
+
2(n + 1)λ−1 − (λ − 1)(n + 1)λ−2
λ
−
1
λ
λ+1
(λ − 1)(n + 1)
2(n + 1)
12(n + 1)
=
2
2λ − 3
λ(λ − 1)
−
−
.
λ − 1 6(n + 1)2 12(n + 1)3
(2.4)
Then we obtain
1
1
2(n + 1)2−λ
(λ − 1)2
+
=
1−
,
λ
−
1
λ
(λ − 1)(n + 1)
2(n + 1)
(λ − 1)(2n + 3 − λ)
4(n + 1)2
2(n + 1)2−λ
1
λ
1
+
+
=
(λ − 1)(n + 1)λ−1 2(n + 1)λ 12(n + 1)λ+1 (λ − 1)(2n + 3 − λ)
(2.5)
(2λ − 3)(λ − 1) λ(λ − 1)2
× 1−
−
.
12(n + 1)2
24(n + 1)3
Since for 1.5 ≤ λ < 3, 2(n + 1)2−λ /(λ − 1)(2n + 3 − λ) > 0, (2λ − 3)(λ − 1)/12(n + 1)2 ≥ 0,
λ(λ − 1)2 /24(n + 1)3 > 0, then we have (2.2). The lemma is proved.
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Journal of Inequalities and Applications
Theorem 2.2. Let 0 < p < 1, 1/ p + 1/q = 1, 1.5 ≤ λ < 3, and an ≥ 0, bn > 0, such that 0 <
∞
2−λ a p /(2n + 3 − λ)) < ∞, 0 < ∞ ((n + 1)2−λ bq /(2n + 3 − λ)) < ∞. Then
n
n
n=0 ((n + 1)
n =0
∞
2
am bn
(n + 1)2−λ
(λ − 1)2 p
>
1
−
an
(m + n + 1)λ λ − 1 n=0 2n + 3 − λ
4(n + 1)2
n=0 m=0
∞ ∞
×
∞
(n + 1)2−λ q
bn
2n + 3 − λ
n =0
1/ p
(2.6)
1/q
.
Proof. By the reverse Hölder inequality [9], we have
∞ ∞
∞
∞
am bn
am
bn
=
·
λ/ p (m + n + 1)λ/q
λ
(m
+
n
+
1)
(m
+
n
+
1)
n=0 m=0
n=0 m=0
∞ ∞
p
am
≥
(m
+
n + 1)λ
m=0 n=0
=
∞
1/ p 1/ p p
ωλ (m)am
·
m=0
∞ ∞
q
bn
·
(m
+
n + 1)λ
n=0 m=0
∞
1/q
(2.7)
1/q
q
ωλ (n)bn
.
n =0
Since 0 < p < 1 and q < 0, then by (2.2), we obtain (2.6). The theorem is proved.
In Theorem 2.2, for λ = 2, we have the following corollary.
Corollary 2.3. Let
0 < p < 1, 1/ p + 1/q = 1, and an ≥ 0, bn > 0, such that 0 <
q
(2n + 1)) < ∞, 0 < ∞
n=0 (bn /(2n + 1)) < ∞. Then
p
∞ am bn
1
an
>
2
1
−
(m + n + 1)2
4(n + 1)2 2n + 1
n=0 m=0
n =0
∞ ∞
1/ p ∞
q
bn
2n + 1
n =0
∞
p
n=0 (an /
1/q
.
(2.8)
Remark 2.4. Inequality (2.8) is inequality [7, Inquality (8)]. Hence, inequality (2.6) is an
extension of Yang’s inequality [7, Inquality (8)] for 1 < λ < 3.
Theorem 2.5. Let 0< p <1, 1/ p + 1/q = 1, 1.5 ≤ λ < 3, and an ≥ 0, such that 0 <
p
1)2−λ an/(2n + 3 − λ)) < ∞. Then
∞ (n + 1)2−λ 1− p
n =0
2n + 3 − λ
∞
am
(m
+
n + 1)λ
m=0
p
>
2
λ−1
p ∞
∞
n=0 ((n+
(n + 1)2−λ
(λ − 1)2 p
1−
an .
2n + 3 − λ
4(n + 1)2
n =0
(2.9)
Inequalities (2.9) and (2.6) are equivalent.
Proof. Let
bn =
(n + 1)2−λ
2n + 3 − λ
1− p ∞
am
(m
+
n + 1)λ
m=0
p −1
,
n ∈ N0 .
(2.10)
Gaowen Xi 5
By (2.6), we have
q
∞
(n + 1)2−λ bn
2n + 3 − λ
n =0
p
∞ (n + 1)2−λ 1− p
=
2n + 3 − λ
n =0
∞ ∞
am bn
=
(m
+
n + 1)λ
n=0 m=0
≥
am
(m
+
n + 1)λ
m=0
pp
(n + 1)2−λ
(λ − 1)2 p
1−
an
2n + 3 − λ
4(n + 1)2
n =0
λ−1
×
∞
p
p ∞
2
q
∞
(n + 1)2−λ bn
2n + 3 − λ
n =0
(2.11)
p −1
.
Then we obtain
q
∞
∞ (n + 1)2−λ bn (n + 1)2−λ 1− p
=
2n + 3 − λ
2n + 3 − λ
n =0
n =0
≥
If
∞
n=0 ((n + 1)
0<
∞
am
(m + n + 1)λ
m=0
p ∞
2
λ−1
p
(n + 1)2−λ
(λ − 1)2 p
1−
an .
2n + 3 − λ
4(n + 1)2
n =0
2−λ bq /(2n + 3 − λ)) = ∞,
n
(2.12)
then in view of
p
p
∞
∞
(n + 1)2−λ an
(λ − 1)2
(n + 1)2−λ an
1−
≤
<∞
2n + 3 − λ
4(n + 1)2
2n + 3 − λ
n =0
n =0
(2.13)
and (2.12), we have
∞ (n + 1)2−λ 1− p
n =0
2n + 3 − λ
∞
am
(m
+
n + 1)λ
m=0
p
>
2
p
λ−1
(2.14)
∞
(n + 1)λ−2
(λ − 1)2 p
×
1−
an ;
2n + 3 − λ
4(n + 1)2
n =0
if 0 <
∞
n=0 ((n + 1)
λ−2 bq /(2n + 3 − λ)) < ∞,
n
∞ (n + 1)2−λ 1− p
n =0
2n + 3 − λ
∞
then by (2.6), we find
am
(m
+
n + 1)λ
m=0
p
∞
(n + 1)2−λ
(λ − 1)2 p
×
1−
an .
2n + 3 − λ
4(n + 1)2
n =0
Hence we obtain (2.9).
>
2
p
λ−1
(2.15)
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Journal of Inequalities and Applications
On the other hand, by the reverse Hölder inequality [9], we have
∞ ∞
am bn
=
(m
+
n + 1)λ
n=0 m=0
∞
(n + 1)
(λ−2)/q
(2n + 3 − λ)
1/q
n =0
×
≥
bn
(n + 1)(λ−2)/q (2n + 3 − λ)1/q
∞ (n + 1)2−λ 1− p
n =0
×
2n + 3 − λ
q
∞
(n + 1)2−λ bn
2n + 3 − λ
n =0
∞
am
(m
+
n + 1)λ
m=0
∞
am
(m
+
n + 1)λ
m=0
p 1/ p
(2.16)
1/q
.
Hence by (2.9), it follows that
1/ p
p
∞
2
am bn
(n + 1)2−λ an
(λ − 1)2
>
1
−
(m + n + 1)λ λ − 1 n=0 2n + 3 − λ
4(n + 1)2
n=0 m=0
∞ ∞
×
q
∞
(n + 1)2−λ bn
2n + 3 − λ
n =0
(2.17)
1/q
.
Then, (2.9) and (2.6) are equivalent. The theorem is proved.
In (2.9), for λ = 2, we have the following corollary.
Corollary 2.6. Let 0 < p < 1, 1/ p + 1/q = 1, an ≥ 0, 0 <
∞
(2n + 1)
n =0
p −1
∞
am
(m + n + 1)2
m=0
p
> 2p
∞ n =0
∞
p
n=0 (an /(2n + 1)) < ∞,
1−
Then
p
1
an
.
2
4(n + 1) 2n + 1
(2.18)
Inequalities (2.18) and (2.8) are equivalent.
References
[1] G. H. Hardy, J. E. Littlewood, and G. Pólya, Inequalities, Cambridge University Press, Cambridge, UK, 2nd edition, 1952.
[2] B. Yang and L. Debnath, “Some inequalities involving π and an application to Hilbert’s inequality,” Applied Mathematics Letters, vol. 12, no. 8, pp. 101–105, 1999.
[3] B. Yang and L. Debnath, “On a new generalization of Hardy-Hilbert’s inequality and its applications,” Journal of Mathematical Analysis and Applications, vol. 233, no. 2, pp. 484–497, 1999.
[4] B. Yang, “On a strengthened version of the more precise Hardy-Hilbert inequality,” Acta Mathematica Sinica. Chinese Series, vol. 42, no. 6, pp. 1103–1110, 1999.
[5] B. Yang and T. M. Rassias, “On a new extension of Hilbert’s inequality,” Mathematical Inequalities
& Applications, vol. 8, no. 4, pp. 575–582, 2005.
[6] B. Yang, “On a new extension of Hilbert’s inequality with some parameters,” Acta Mathematica
Hungarica, vol. 108, no. 4, pp. 337–350, 2005.
[7] B. Yang, “A reverse of the Hardy-Hilbert’s type inequality,” Journal of Southwest China Normal
University (Natural Science), vol. 30, no. 6, pp. 1012–1015, 2005.
Gaowen Xi 7
[8] Z. Wang and G. Dunren, An Introduction to Special Function, Science Press, Beijing, China, 1979.
[9] J. Kuang, Applied Inequalities, Sandong Science and Technology Press, Jinan, China, 2004.
Gaowen Xi: Department of Mathematics, Luoyang Teachers’ College, Luoyang 471022, China
Email address: xigaowen@163.com