Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 3, Article 115, 2006
ON A REVERSE OF A HARDY-HILBERT TYPE INEQUALITY
BICHENG YANG
D EPARTMENT O F M ATHEMATICS
G UANGDONG E DUCATION C OLLEGE , G UANGZHOU
G UANGDONG 510303, P EOPLE ’ S R EPUBLIC OF C HINA
bcyang@pub.guangzhou.gd.cn
Received 18 January, 2006; accepted 25 May, 2006
Communicated by L.-E. Persson
A BSTRACT. This paper deals with a reverse of the Hardy-Hilbert’s type inequality with a best
constant factor. The other reverse of the form is considered.
Key words and phrases: Hardy-Hilbert’s inequality, Weight coefficient, Hölder’s inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
P
P∞ q
p
If p > 1, p1 + 1q = 1, an , bn ≥ 0, such that 0 < ∞
n=0 an < ∞ and 0 <
n=0 bn < ∞, then
we have the well known Hardy-Hilbert inequality (Hardy et al. [1]):
(1.1)
∞ X
∞
X
am bn
π
<
m
+
n
+
1
sin πp
n=0 m=0
(∞
X
apn
) p1 ( ∞
X
n=0
) 1q
bqn
,
n=0
where the constant factor π/ sin(π/p) is the best possible. The equivalent form is (see Yang et
al. [8]):
p
!p 
∞
∞
∞
X
X
am
π  X p

an ,
(1.2)
<
m+n+1
sin π
n=0
m=0
n=0
p
where the constant factor [π/ sin(π/p)]p is still the best possible.
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
Research supported by Natural Science Foundation of Guangdong Institutions of Higher Learning, College and University (China, No.
0177).
016-06
2
B ICHENG YANG
Inequalities (1.1) and (1.2) are important in analysis and its applications (see Mitrinović, et
al. [3]). In recent years, inequality (1.1) had been strengthened by Yang [5] as


  p1
∞ X
∞
∞

X
X
am b n
π
ln 2 − γ  p 

−
an
(1.3)
<
1

m + n + 1  n=0 sin π
(2n + 1)1+ p
n=0 m=0
p


  1q
∞
X

 π − ln 2 − γ 1  bqn
×
,
π


(2n + 1)1+ q
n=0 sin p
where ln 2 − γ = 0.11593+ (γ is Euler’s constant). Another strengthened version of (1.1) was
given in [6]. By introducing a parameter λ, two extensions of (1.1) were proved in [8, 7] as:
(∞ 1−λ ) p1 (X
1−λ ) 1q
∞ X
∞
∞ X
X
am b n
1
1
(1.4)
< kλ (p)
n+
apn
n+
bqn
;
λ
(m + n + 1)
2
2
n=0 m=0
n=0
n=0
∞ X
∞
X
λ λ
am b n
<B
,
(1.5)
(m + n + 1)λ
p q
n=0 m=0
(∞ (p−1)(1−λ) ) p1 (X
(q−1)(1−λ) ) 1q
∞ X
1
1
×
apn
bqn
n+
n+
,
2
2
n=0
n=0
p+λ−2 q+λ−2
λ λ
where, the constant factors kλ (p) = B
, q
(2−min{p, q} < λ ≤ 2), and B p , q
p
(0 < λ ≤ min{p, q}) are the best possible (B(u, v) is the β function). For λ = 1, both (1.4)
and (1.5) reduce to (1.1). We call (1.4) and (1.5) Hardy-Hilbert type inequalities. Yang et al.
[9] summarized the use of weight coefficients in research for Hardy-Hilbert type inequalities.
But the problem on how to build the reverse of this type inequalities is unsolved.
The main objective of this paper is to deal with a reverse of inequality (1.4) for λ = 2.
Another related reverse of the form is considered.
2. S OME L EMMAS
We need the formula of the β function B(p, q) as follows (see [4]):
Z ∞
tp−1
(2.1)
B(p, q) = B(q, p) =
dt,
(1 + t)p+q
0
R∞
and the following inequality (see [5, 2]): If f 4 ∈ C[0, ∞), 0 < 0 f (t)dt < ∞ and (−1)n f (n) (x)
> 0, f (n) (∞) = 0 (n = 0, 1, 2, 3, 4), then
Z ∞
Z ∞
∞
X
1
1
1
(2.2)
f (t)dt + f (0) <
f (k) <
f (t)dt + f (0) − f 0 (0).
2
2
12
0
0
k=0
Lemma 2.1. Define the weight function ω(n) as
∞
1 X
1
(2.3)
ω(n) = n +
,
2 k=0 (k + n + 1)2
n ∈ N0 (= N ∪ {0}),
then we have
(2.4)
1−
1
1
< ω(n) < 1 −
2
4(n + 1)
6(n + 1)(2n + 1)
J. Inequal. Pure and Appl. Math., 7(3) Art. 115, 2006
(n ∈ N0 ).
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R EVERSE OF A H ARDY-H ILBERT T YPE I NEQUALITY
1
,
(n + 1)2
1
(t+n+1)2
(t ∈ [0, ∞)), we obtain
Z ∞
2
1
0
f (0) = −
, and
.
f (t)dt =
3
(n + 1)
n+1
0
Proof. For fixed n, setting f (t) =
f (0) =
3
By (2.2), we find
(2.5)
(2.6)
∞
1
1
1
ω(n) >
dt +
n+
(t + n + 1)2
2(n + 1)2
2
0
1
1
1
=
+
(n
+
1)
−
(n + 1) 2(n + 1)2
2
1
=1−
.
4(n + 1)2
Z ∞
1
1
1
1
ω(n) <
dt +
+
n+
(t + n + 1)2
2(n + 1)2 6(n + 1)3
2
0
1
1
1
1
=
+
+
(n
+
1)
−
n + 1 2(n + 1)2 6(n + 1)3
2
1
1
+
.
=1−
2
12(n + 1)
12(n + 1)3
Z
Since for n ∈ N0 ,
1
1
+
6(n + 1)(2n + 1)
12(n + 1)2 12(n + 1)3
2(n + 1) − 1 2(n + 1) − 1
+
=
2(n + 1)
2(n + 1)2
1
1
=1+
−
≥ 1,
2(n + 1) 2(n + 1)2
then we find
1
1
1
+
≥
,
2
3
12(n + 1)
12(n + 1)
6(n + 1)(2n + 1)
and in view of (2.6), it follows that
1
(2.7)
ω(n) < 1 −
.
6(n + 1)(2n + 1)
In virtue of (2.5) and (2.7), we have (2.4). The lemma is proved.
Lemma 2.2. For 0 < ε < p, we have
(2.8)
∞ X
∞
X
1
I :=
(m + n + 1)2
n=0 m=0
< (1 + o(1))
∞
X
1
m+
2
1
n+
n=0
1 1+ε
2
− pε 1
n+
2
− qε
(ε → 0+ ).
−ε/p
Proof. For fixed n, setting f (t) =
f (0) =
(t+ 12 )
(t+n+1)2
2ε/p
,
(n + 1)2
J. Inequal. Pure and Appl. Math., 7(3) Art. 115, 2006
(t ∈ (− 12 , ∞)), then we have
f 0 (0) = −
21+ε/p
ε21+ε/p
−
.
(n + 1)3 p(n + 1)2
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4
B ICHENG YANG
Setting u = t +
1
2
n+
1
2
in the following integral, we obtain
ε
Z ∞
Z ∞
Z ∞
1
u− p
f (t)dt <
f (t)dt =
du.
1+ pε
(1 + u)2
0
−1/2
0
n + 12
Hence by (2.2) and (2.1), we find
− ε #
− ε " ∞
∞ X
1 p
1 q X
1
m+
I=
n+
(m + n + 1)2
2
2
m=0
n=0
"
ε
− ε
Z ∞
∞ X
u− p
1 q
1
du
<
n+
1+ pε
(1 + u)2
2
0
n + 12
n=0
21+ε/p
ε21+ε/p
2ε/p
+
+
+
2(n + 1)2 12(n + 1)3 12p(n + 1)2
∞
X
ε
1
ε
=
+ O(1) (ε → 0+ ).
B 1 − ,1 +
1 1+ε
p
p
n=0 n + 2
Since B 1 − pε , 1 + pε → B(1, 1) = 1 (ε → 0+ ), then (2.8) is valid. The lemma is proved.
3. M AIN R ESULTS
Theorem 3.1. If 0 < p < 1,
P
bqn
0< ∞
n=0 2n+1 < ∞, then
(3.1)
1
p
+
1
q
= 1, an , bn ≥ 0, such that 0 <
apn
n=0 2n+1
P∞
< ∞ and
(∞ ) p1
X
am b n
1
apn
>2
1−
2
(m
+
n
+
1)
4(n + 1)2 2n + 1
n=0 m=0
n=0
(∞ ) 1q
X
1
bqn
×
1−
,
6(n
+
1)(2n
+
1)
2n
+
1
n=0
∞ X
∞
X
where the constant factor 2 is the best possible. In particular, one has
) p1 ( ∞
(∞ ) 1q
∞ X
∞
X
X
X bq
am b n
1
apn
n
(3.2)
>2
1−
.
2
2 2n + 1
(m
+
n
+
1)
4(n
+
1)
2n
+
1
n=0 m=0
n=0
n=0
Proof. By the reverse of Hölder’s inequality and (2.3), one has
∞ X
∞
X
am b n
(m + n + 1)2
n=0 m=0
#
#"
"
∞ X
∞
X
am
bn
=
2
2
p
(m + n + 1) q
n=0 m=0 (m + n + 1)
(∞ ∞
) p1 ( ∞ ∞
) 1q
XX
XX
apm
bqn
≥
(m + n + 1)2
(m + n + 1)2
n=0 m=0
n=0 m=0
J. Inequal. Pure and Appl. Math., 7(3) Art. 115, 2006
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R EVERSE OF A H ARDY-H ILBERT T YPE I NEQUALITY
5
#
) p1 ( ∞ " ∞
#
) 1q
X X
m + 12
n + 12
2apm
2bqn
=
(m + n + 1)2 2m + 1
(m + n + 1)2 2n + 1
m=0 n=0
n=0 m=0
) 1q
( ∞
) p1 ( ∞
X
X
apm
bqn
.
ω(n)
=2
ω(m)
2n + 1
2m + 1
m=0
n=0
(
"∞
∞
X
X
Since 0 < p < 1 and q < 0, by (2.4), it follows that (3.1) is valid.
−ε/p
−ε/q
For 0 < ε < p, setting ãn = n + 12
, b̃n = n + 12
, n ∈ N0 , we find
(3.3)
) p1 ( ∞
) 1q
X b̃q
ãpn
1
n
2 2n + 1
4(n
+
1)
2n
+
1
n=0
n=0
) p1 ( ∞
) 1q
(∞ X
X
1
1
1
=
1−
1 1+ε
4(n + 1)2 2 n + 1 1+ε
n=0 2 n + 2
n=0
2
(∞
) p1 ( ∞
) 1q
∞
X
X
1
1 X
1
1
>
−
1 1+ε
2 n=0 n + 1 1+ε n=0 2(n + 1)2 (2n + 1)
n=0 n + 2
2
∞
1 X
1
1
= {1 − õ(1)} p
(ε → 0+ ).
1 1+ε
2
n=0 n + 2
(∞ X
1−
If the constant factor 2 in (3.1) is not the best possible, then there exists a real number k with
k > 2, such that (3.1) is still valid if one replaces 2 by k. In particular, one has
(3.4)
∞ X
∞
X
ãm b̃n
(m + n + 1)2
n=0 m=0
(∞ X
>k
1−
n=0
ãpn
1
4(n + 1)2 2n + 1
) p1 ( ∞
X
n=0
b̃qn
) 1q
2n + 1
.
Hence by (2.8) and (3.3), it follows that
(1 + o(1))
∞
X
n=0
1
n+
1 1+ε
2
∞
1 X
k
1
> {1 − õ(1)} p
,
1 1+ε
2
n
+
n=0
2
and then 2 ≥ k (ε → 0+ ). This contradicts the fact that k > 2. Hence the constant factor 2 in
(3.1) is the best possible. The theorem is proved.
Theorem 3.2. If 0 < p < 1,
(3.5)
∞ X
n=0
1
n+
2
1
p
+
1
q
= 1, an ≥ 0, such that 0 <
p−1 " X
∞
am
(m + n + 1)2
m=0
#p
∞ X
>2
1−
n=0
apn
n=0 2n+1
P∞
< ∞, then
1
apn
,
4(n + 1)2 2n + 1
where the constant factor 2 is the best possible.
J. Inequal. Pure and Appl. Math., 7(3) Art. 115, 2006
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6
B ICHENG YANG
Proof. By the reverse of Hölder’s inequality, (2.3) and (2.4), one has ω(n) < 1 and
#)p
#"
" ∞
#p ( ∞ "
X
X
1
am
am
(3.6)
=
2
2
(m + n + 1)2
p
(m + n + 1) q
m=0
m=0 (m + n + 1)
)( ∞
)p−1
( ∞
X
X
1
apm
≥
(m + n + 1)2
(m + n + 1)2
m=0
m=0
)(
( ∞
−1 )p−1
X
1
apm
=
ω(n) n +
2
(m
+
n
+
1)
2
m=0
1−p X
∞
apm
1
> n+
.
2
(m + n + 1)2
m=0
Hence we find
#p
p−1 " X
∞ ∞
∞ X
∞
X
X
1
am
apm
(3.7)
n+
>
2
(m + n + 1)2
(m + n + 1)2
n=0
m=0
n=0 m=0
"
#
∞
∞
X
X
m + 21
2apm
=
(m + n + 1)2 2m + 1
m=0 n=0
=2
∞
X
m=0
By (2.4), we have (3.5).
P
Setting bn ≥ 0 and 0 < ∞
n=0
(3.8)
bqn
2n+1
ω(m)
apm
.
2m + 1
< ∞, by the reverse of Hölder’s inequality, one has
∞ X
∞
X
am b n
(m + n + 1)2
n=0 m=0
# "
"
1 ∞
− 1 #
∞
X
1 qX
am
1 q
n+
bn
=
n+
2
(m + n + 1)2
2
n=0
m=0
(∞ #p ) p1 ( ∞
) 1q
p−1 " X
∞
q
X
X
1
am
2bn
≥
n+
.
2
2
(m + n + 1)
2n + 1
n=0
m=0
n=0
If the constant factor 2 in (3.5) is not the best possible, then by (3.6), we can get a contradiction
that the constant factor 2 in (3.1) is not the best possible. The theorem is proved.
Remark 3.3. If an , bn satisfy the conditions of (1.4) for λ = 2, r > 1, 1r + 1s = 1, and (3.2) for
0 < p < 1, p1 + 1q = 1, then one can get the following two-sided inequality:
( ∞
) p1 ( ∞
) 1q
X bq
3 X apn
n
0<
(3.9)
4 n=0 2n + 1
2n
+
1
n=0
∞
∞
1XX
am b n
<
2 n=0 m=0 (m + n + 1)2
(∞
) r1 ( ∞
) 1s
X ar
X bs
n
n
<
< ∞.
2n
+
1
2n
+
1
n=0
n=0
J. Inequal. Pure and Appl. Math., 7(3) Art. 115, 2006
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R EVERSE OF A H ARDY-H ILBERT T YPE I NEQUALITY
7
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[9] BICHENG YANG AND Th.M. RASSIAS, On the way of weight coefficient and research for the
Hilbert-type inequalities, Math. Inequal. Appl., 6 (2003), 625–658.
J. Inequal. Pure and Appl. Math., 7(3) Art. 115, 2006
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