Journal of Inequalities in Pure and
Applied Mathematics
A NEW INEQUALITY SIMILAR TO HILBERT’S INEQUALITY
BICHENG YANG
Department of Mathematics
Guangdong Education College
Guangzhou
Guangdong 510303
PEOPLE’S REPUBLIC OF CHINA
EMail: bcyang@pub.guangzhou.gd.cn
volume 3, issue 5, article 75,
2002.
Received 17 May 2001;
accepted 17 June, 2002.
Communicated by: J.E. Pečarić
Abstract
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Abstract
In this paper, we build a new inequality similar to Hilbert’s inequality with a best
constant factor. As an application, we consider its equivalent form.
2000 Mathematics Subject Classification: 26D15
Key words: Hilbert’s inequality, Weight coefficient, Cauchy’s inequality.
A New Inequality Similar to
Hilbert’s Inequality
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Some Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
Main Result and an Application . . . . . . . . . . . . . . . . . . . . . . . .
References
3
4
9
Bicheng Yang
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1.
Introduction
P∞ 2
P∞ 2
If 0 <
n=0 an < ∞ and 0 <
n=0 bn < ∞, then the famous Hilbert’s
inequality (see Hardy et al. [1]) is given by
(1.1)
∞ X
∞
X
∞
∞
X
X
am b n
2
<π
an
b2n
m
+
n
+
1
n=0 m=0
n=0
n=0
! 12
,
where the constant factor π is the best possible. Recently, Yang and Debnath
[2, 3] and Yang [4, 5] gave (1.1) some extensions and improvements, and Kuang
and Debnath [6] considered its strengthened versions and generalizations.
The major objective of this paper is to build a new inequality similar to (1.1),
which relates to the double series form as
(1.2)
∞ X
∞
X
am b n
=
ln m + ln n + 1
n=1 m=1
∞ X
∞
X
am b n
.
ln emn
n=1 m=1
For this, we must estimate the following weight coefficient
(1.3)
∞
X
1
ω(n) =
m ln emn
m=1
√ 1
ln en 2
√
(n ∈ N ),
ln em
A New Inequality Similar to
Hilbert’s Inequality
Bicheng Yang
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and do some preparatory works.
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2.
Some Lemmas
Let f have its first four derivatives on [1, ∞) and (−1)n f (n) (x) > 0 (n =
0, . . . , 4), and f (x), f 0 (x) −→ 0 (x → ∞), then (see [6, (2.1)])
Z ∞
∞
X
1
1
(2.1)
f (k) <
f (x)dx + f (1) − f 0 (1).
2
12
1
k=1
Lemma 2.1. For n ∈ N, define R(n) as
Z 1√
2 ln en
1
2
1
1
−
.
(2.2) R(n) =
√
1 du −
1
3 ln en 12(ln en)2
(2 ln en) 2 0
(1 + u)u 2
A New Inequality Similar to
Hilbert’s Inequality
Bicheng Yang
Then we have R(n) > 0(n ∈ N ).
Proof. Integrating by parts, we have
Z 1√
Z 1√
2 ln en
2 ln en
1
1
1
du 2
1 du = 2
(1 + u)
(1 + u)u 2
0
0
Z 1√
2 ln en
√
1
1
1
1
+2
u2
du
= (2 ln en) 2
ln en
(1 + u)2
0
Z 1
√
1
4 2 ln √en
1
1
du3/2
= (2 ln en) 2
+
(1 + u)2
ln en 3 0
√
√
1
1
1
1
1
+ (2 ln en) 2
= (2 ln en) 2
ln en 3
(ln en)2
Z 1
8 2 ln √en 3/2
1
+
u
du
3 0
(1 + u)3
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> (2 ln
√
1
en) 2
√
1
1
1
1
+ (2 ln en) 2
.
ln en 3
(ln en)2
Hence by (2.2), we have
R(n) >
1
1
2
1
1
1
+
−
−
=
+
> 0.
ln en 3(ln en)2 3 ln en 12(ln en)2
3 ln en 4(ln en)2
The lemma is thus proved.
Lemma 2.2. If ω(n) is defined by (1.3), then ω(n) < π, for n ∈ N .
Proof. For fixed n ∈ N , setting
Bicheng Yang
1
fn (x) =
x ln enx
we find fn (1) =
fn0 (x)
1
(2 ln
ln en
√
√ 1
ln en 2
√
, x ∈ [1, ∞),
ln ex
1
2
en) , and
√ 12
√ 1
ln en 2
1
ln en
√
√
− 2 2
ln ex
x ln enx ln ex
√
1
1
(ln en) 2
− 2
·
,
√
2x ln enx (ln ex) 32
√
1
2
1
0
+ 2
fn (1) = −
(2 ln en) 2 .
ln en ln en
1
=− 2
x ln enx
A New Inequality Similar to
Hilbert’s Inequality
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√
ln ex
√
ln en
Setting u =
∞
Z
in the following integral, we obtain
Z
∞
fn (x)dx =
1
2 ln
1√
en
1
1+u
12
12
Z 1√
2 ln en
1
1
1
du = π −
du.
u
1+u u
0
Hence by (2.1), (2.2) and Lemma 2.1, we have
ω(n) =
∞
X
fn (m)
m=1
Z
∞
1
1
fn (x)dx + fn (1) − fn0 (1)
2
12
1
12
Z 1/(2 ln √en)
√
1
1
1
2
1
=π−
du +
+
(2 ln en) 2
2
1+u u
3 ln en 12 ln en
0
√
1
= π − (2 ln en) 2 R(n) < π.
<
The lemma is proved.
Lemma 2.3. For 0 < < 1, we have
1+
∞ X
∞
X
2
1
1
1
√
√
(2.3)
> (π + o(1)) ( → 0+ ).
mn ln emn ln em ln en
n=1 m=1
Proof. Setting u =
Z
∞
√
e
1
x ln exy
√
ln ex
√
ln ey
1
√
ln ex
in the following integral, we find
A New Inequality Similar to
Hilbert’s Inequality
Bicheng Yang
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1+
2
dx
J. Ineq. Pure and Appl. Math. 3(5) Art. 75, 2002
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=
1+
Z
2
1
√
1+
2
1
√
1+
2
ln ey
=
ln ey
>
ln ey
=
1
√
1
√
ln ey
1+
2
1+
1
1 2
du
1
1+u u
√
ln ey
1+
Z ∞
1
1 2
du
1+u u
0
1+
1+
Z √1
2
ln ey
1
1 2
1
√
−
du
1+u u
ln ey
0
1+
Z ∞
1 2
1
du
1+u u
0
1+
Z √1 1+
2
ln ey
1 2
1
√
du
−
u
ln ey
0
1
2
√
(π + o(1)) −
( −→ 0+ ).
1 − ln ey
∞
Hence we have
1+
2
1
1
√
√
mn ln emn ln em ln en
n=1 m=1
1+
Z ∞Z ∞
2
1
1
√
√
> √ √
dxdy
ln ex ln ey
e
e xy ln exy
"Z
#
1+
1+
Z ∞ ∞
2
2
1
1
1
1
√
√
= √
dx dy
√ x ln exy
ln ey
ln ex
e y
e
∞ X
∞
X
A New Inequality Similar to
Hilbert’s Inequality
Bicheng Yang
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Z
> (π + o(1))
∞
√
e
1
y
1
√
ln ey
1+
2
dy −
1−
Z
∞
√
e
1
y
1
√
ln ey
1+
+1
2
dy
1
4
= (π + o(1)) −
1 − 2
1
= (π + o(1)) ( → 0+ ).
The lemma is proved.
A New Inequality Similar to
Hilbert’s Inequality
Bicheng Yang
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3.
Main Result and an Application
Theorem 3.1. If 0 <
P∞
n=1
na2n < ∞ and 0 <
P∞
n=1
nb2n < ∞, then
∞ X
∞
∞
∞
X
X
X
am b n
<π
na2n
nb2n
ln
emn
n=1 m=1
n=1
n=1
(3.1)
! 12
,
where the constant factor π is the best possible.
Proof. By Cauchy’s inequality and (1.3), we have
∞ X
∞
X
am b n
ln emn
n=1 m=1
=
∞ X
∞
X
Bicheng Yang
"
n=1 m=1
#
√ 1
ln em 4 m 12
√
1
n
(ln emn) 2 ln en
"
√ 14 1 #
bn
ln en
n 2
√
×
1
m
(ln emn) 2 ln em
am
√ 12 # 12
√ 1
∞ ∞
ln em 2 m X X b2n
ln en
n
√
√
≤
n n=1 m=1 ln emn ln em
m
ln en
m=1 n=1
! 12
∞
∞
X
X
=
ω(m)ma2m
ω(n)nb2n
.
"
∞ X
∞
X
m=1
A New Inequality Similar to
Hilbert’s Inequality
a2m
ln emn
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n=1
J. Ineq. Pure and Appl. Math. 3(5) Art. 75, 2002
By Lemma 2.2, we have (3.1).
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For 0 < < 1, setting a0n as:
a0n =
1
√
1+ , n ∈ N,
n(ln en) 2
then we have
∞
X
n=1
(3.2)
∞
na0n 2
X
1
1
1
√ 1+ +
√ 1+ +
√ 1+
=
(ln e)
2(ln 2 e)
n(ln en)
n=3
Z
∞
1
1
1
√ 1+ +
√ 1+ + √
√ 1+ dx
<
(ln e)
2(ln 2 e)
e x(ln ex)
1
1
1
1
√ 1+ +
√ 1+ + = (1 + o(1)) ( → 0+ ).
=
(ln e)
2(ln 2 e)
If the constant factor π in (3.1) is not the best possible, then there exists
a positive number K < π, such that (3.1) is valid if we change π to K. In
particular, we have
∞ X
∞
X
∞
X
a0m a0n
<K
na0n 2 .
ln emn
n=1 m=1
n=1
By (2.3) and (3.2), we have
∞ X
∞
X
a0m a0n
(π + o(1)) < < K(1 + o(1)) ( → 0+ ),
ln
emn
n=1 m=1
and π ≤ K. This contradicts that K < π. Hence the constant factor π in (3.1)
is the best possible. The theorem is proved.
A New Inequality Similar to
Hilbert’s Inequality
Bicheng Yang
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Remark 3.1. Inequality (3.1) is more similar to the following Mulholland’s
inequality for p = q = 2 (see [7]):
! p1 ∞
! 1q
∞
∞ X
∞
X
X
X
π
am b n
n−1 apn
n−1 bqn
.
<
(3.3)
π
mn
ln
emn
sin(
)
p
n=2
n=2 m=2
n=2
P
2
Theorem 3.2. If 0 < ∞
n=1 nan < ∞, then we have
!2
∞
∞
∞
X
X
1 X am
2
<π
(3.4)
na2n ,
n m=1 ln emn
n=1
n=1
where the constant factor π 2 is the best possible. Inequalities (3.1) and (3.4)
are equivalent.
P
2
Proof. Since ∞
k ≥ 1, such that for any k > k0 , we
n=1 nan > 0, there exists
Pk 0 |am |
Pk
1
2
have n=1 nan > 0, and bn (k) = n m=1 lnemn > 0 (n ∈ N ). By (3.1), we
have
" k
#2
X
0<
nb2n (k)
n=1
(3.5)

!2 2
k
k
X
X
1
|am |

=
n
ln
emn
n=1
m=1
" k k
#2
k
k
X X |am |bn (k)
X
X
=
< π2
na2n
nb2n (k).
ln
emn
n=1 m=1
n=1
n=1
A New Inequality Similar to
Hilbert’s Inequality
Bicheng Yang
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Thus we find
(3.6)
k
X
1
0<
n
n=1
k
X
|am |
ln emn
m=1
!2
P∞
=
k
X
nb2n (k) < π 2
n=1
k
X
na2n .
n=1
P∞
2
2
It follows that 0 < n=1 nb2n (∞) ≤ π
n=1 nan < ∞. Hence by (3.1), for
k → ∞, neither (3.5) nor (3.6) takes equality, and we have
∞
X
1
n
n=1
∞
X
am
ln emn
m=1
!2
∞
X
1
≤
n
n=1
∞
X
|am |
ln emn
m=1
!2
<π
2
∞
X
na2n .
n=1
Inequality (3.4) is valid.
On the other hand, if (3.4) holds, by Cauchy’s inequality, we have
!
∞ X
∞
∞
∞
1 X
X
am b n
1 X am
=
n 2 bn
1
ln
emn
ln
emn
2
n
n=1 m=1
n=1
m=1

 12
!2 ∞
∞
∞
X 1 X am
X
(3.7)
≤
nb2n  .
n m=1 ln emn
n=1
n=1
By (3.4), we have (3.1).
Hence inequalities (3.1) and (3.4) are equivalent. If the constant factor π 2 in
(3.4) is not the best possible, we may show that the constant factor π in (3.1)
is not the best possible, by using (3.7). This is a contradiction. The theorem is
proved.
A New Inequality Similar to
Hilbert’s Inequality
Bicheng Yang
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Remark 3.2. Inequality (3.4) is similar to the following equivalent form of (1.1)
(see [2]):
!2
∞
∞
∞
X
X
X
am
2
(3.8)
<π
a2n .
m
+
n
+
1
n=0
m=0
n=0
Since inequalities (3.1) and (3.4) are similar to (1.1) and its equivalent form
with the best constant factors, we have provided some new results.
A New Inequality Similar to
Hilbert’s Inequality
Bicheng Yang
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References
[1] G.H. HARDY, J.E. LITTLEWOOD
bridge Univ. Press, London, 1952.
AND
G. POLYA, Inequalities, Cam-
[2] B. YANG AND L. DEBNATH, on a new generalization of Hardy-Hilbert’s
inequality and its application, J. Math. Anal. Appl., 233 (1999), 484–497.
[3] B. YANG AND L. DEBNATH, Some inequalities involving π and an application to Hilbert’s inequality, Applied Math. Letters, 129 (1999), 101–105.
[4] B. YANG, On a strengthened version of the more accurate Hardy-Hilbert’s
inequality, Acta Math. Sinica, 42(6) (1999), 1103–1110.
[5] B. YANG, On a strengthened Hardy-Hilbert’s inequality, J.
Ineq. Pure. and Appl. Math., 1(2), Art. 22 (2000). [ONLINE:
http://jipam.vu.edu.au/v1n2/012_00.html]
[6] J. KUANG AND L. DEBNATH, On new generalizations of Hilbert’s inequality and their applications, J. Math. Anal. Appl., 245 (2000), 248–265.
[7] H.P. MULHOLLAND, Some theorem on Dirichlet series with coefficients
and related integrals, Proc. London Math. Soc., 29(2) (1999), 281–292.
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Hilbert’s Inequality
Bicheng Yang
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