Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 3, Issue 5, Article 75, 2002
A NEW INEQUALITY SIMILAR TO HILBERT’S INEQUALITY
BICHENG YANG
D EPARTMENT OF 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 17 May, 2001; accepted 17 June, 2002
Communicated by J.E. Pečarić
A BSTRACT. 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.
Key words and phrases: Hilbert’s inequality, Weight coefficient, Cauchy’s inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
P
2
If 0 < n=0 a2n < ∞ and 0 < ∞
n=0 bn < ∞, then the famous Hilbert’s inequality (see
Hardy et al. [1]) is given by
! 12
∞ X
∞
∞
∞
X
X
X
am b n
(1.1)
<π
a2n
b2n
,
m
+
n
+
1
n=0 m=0
n=0
n=0
P∞
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
∞ X
∞
∞ X
∞
X
X
am b n
am b n
(1.2)
=
.
ln m + ln n + 1 n=1 m=1 ln emn
n=1 m=1
For this, we must estimate the following weight coefficient
√ 12
∞
X
1
ln en
√
(n ∈ N ),
(1.3)
ω(n) =
m ln emn ln em
m=1
and do some preparatory works.
ISSN (electronic): 1443-5756
c 2002 Victoria University. All rights reserved.
044-01
2
B ICHENG YANG
2. S OME L EMMAS
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
f (x)dx + f (1) − f 0 (1).
(2.1)
f (k) <
2
12
1
k=1
Lemma 2.1. For n ∈ N, define R(n) as
Z 1√
2 ln en
1
1
2
1
−
.
(2.2)
R(n) =
√
1 du −
1
3 ln en 12(ln en)2
(1 + u)u 2
(2 ln en) 2 0
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
1
4 2 ln √en
1
= (2 ln en) 2
+
du3/2
(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
+
(1 + u)3
3 0
√
√
1
1
1
1
1
+ (2 ln en) 2
.
> (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.
2
2
ln en 3(ln en)
3 ln en 12(ln en)
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
√ 12
1
ln en
√
fn (x) =
, x ∈ [1, ∞),
x ln enx ln ex
√
1
we find fn (1) = ln1en (2 ln en) 2 , and
√ 12
√ 12
√
1
1
ln en
1
ln en
1
(ln en) 2
0
√
√
,
fn (x) = − 2
− 2 2
− 2
·
√
x ln enx ln ex
2x ln enx (ln ex) 32
x ln enx ln ex
√
1
2
1
0
fn (1) = −
+ 2
(2 ln en) 2 .
ln en ln en
J. Inequal. Pure and Appl. Math., 3(5) Art. 75, 2002
http://jipam.vu.edu.au/
A N EW I NEQUALITY S IMILAR TO H ILBERT ’ S I NEQUALITY
√
ln ex
√
ln en
Setting u =
in the following integral, we obtain
∞
Z
3
∞
Z
fn (x)dx =
1
2 ln
1√
en
1
1+u
12
12
Z 1√
2 ln en
1
1
1
du.
du = π −
1+u u
u
0
Hence by (2.1), (2.2) and Lemma 2.1, we have
ω(n) =
∞
X
∞
Z
1
1
fn (x)dx + fn (1) − fn0 (1)
2
12
12
√
1
2
1
1
1
du +
+
(2 ln en) 2
2
3 ln en 12 ln en
1+u u
fn (m) <
1
m=1
Z
1/(2 ln
√
en)
=π−
0
= π − (2 ln
√
1
en) 2 R(n) < π.
The lemma is proved.
Lemma 2.3. For 0 < < 1, we have
(2.3)
∞ X
∞
X
1
mn ln emn
n=1 m=1
Proof. Setting u =
Z
∞
√
e
1
x ln exy
√
ln ex
√
ln ey
1
√
1
√
√
ln em ln en
1+
2
1
> (π + o(1)) ( → 0+ ).
in the following integral, we find
1+
2
ln ex
1+
2
1
√
=
ln ey
1+
2
1
√
=
ln ey
1+
2
1
√
>
ln ey
1+
2
1
√
=
ln ey
dx
1+
1
1 2
du
1
1+u u
√
ln ey
1+
1+
1+
Z √1
Z ∞
2
ln ey
1
1
1 2
1
1 2
√
du −
du
1+u u
1+u u
ln ey
0
0
1+
1+
Z √1 1+
Z ∞
2
ln ey
1 2
1
1 2
1
√
du −
du
1+u u
u
ln ey
0
0
2
1
√
(π + o(1)) −
( −→ 0+ ).
1 − ln ey
Z
∞
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 ex
ln ey
e
e y
∞ X
∞
X
J. Inequal. Pure and Appl. Math., 3(5) Art. 75, 2002
http://jipam.vu.edu.au/
4
B ICHENG YANG
Z
> (π + o(1))
∞
√
1
y
1
√
1+
2
dy −
1−
Z
∞
√
ln ey
e
4
1
1
= (π + o(1)) −
= (π + o(1)) ( → 0+ ).
1 − 2
e
1
y
1
√
1+
+1
2
ln ey
dy
The lemma is proved.
3. M AIN R ESULT AND AN A PPLICATION
P∞
P
2
Theorem 3.1. If 0 < n=1 na2n < ∞ and 0 < ∞
n=1 nbn < ∞, then
! 12
∞ X
∞
∞
∞
X
X
X
am bn
(3.1)
,
<π
na2n
nb2n
ln
emn
n=1 m=1
n=1
n=1
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
#"
√ 14 1 #
√ 1
ln em 4 m 12
ln en
n 2
bn
√
√
=
1
1
n
m
ln en
2
(ln emn) 2 ln em
n=1 m=1 (ln emn)
" ∞ ∞
1
√ 12 # 12
∞ X
∞
2
X X a2 ln √em 2 m X
bn
ln en
n
m
√
√
≤
ln emn ln en
n n=1 m=1 ln emn ln em
m
m=1 n=1
1
!
∞
∞
2
X
X
2
2
=
ω(m)mam
ω(n)nbn
.
∞ X
∞
X
m=1
"
am
n=1
By Lemma 2.2, we have (3.1).
For 0 < < 1, setting a0n as:
a0n =
then we have
∞
X
n=1
(3.2)
1
√
1+ , n ∈ N,
n(ln en) 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+ dx
√ 1+ +
√ 1+ + √
<
(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
J. Inequal. Pure and Appl. Math., 3(5) Art. 75, 2002
http://jipam.vu.edu.au/
A N EW I NEQUALITY S IMILAR TO H ILBERT ’ S I NEQUALITY
5
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.
Remark 3.2. 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
n=2 m=2
P
2
Theorem 3.3. If 0 < ∞
n=1 nan < ∞, then we have
!2
∞
∞
∞
X
X
X
1
am
2
(3.4)
<π
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
P
Proof. Since ∞
na2n > 0, there exists k0 ≥ 1, such that for any k > k0 , we have kn=1 na2n >
n=1
P
|am |
0, and bn (k) = n1 km=1 lnemn
> 0 (n ∈ N ). By (3.1), we have
" k
#2
X
0<
nb2n (k)
n=1

k
X
1
=
n
n=1
"
(3.5)
k
X
|am |
ln emn
m=1
k X
k
X
|am |bn (k)
=
ln emn
n=1 m=1
!2 2

#2
<π
2
k
X
n=1
na2n
k
X
nb2n (k).
n=1
Thus we find
!2
k
k
k
k
X
X
X
1 X |am |
2
2
(3.6)
0<
=
nbn (k) < π
na2n .
n
ln
emn
n=1
m=1
n=1
n=1
P∞
P∞
2
2
2
It follows that 0 < n=1 nbn (∞) ≤ π
n=1 nan < ∞. Hence by (3.1), for k → ∞, neither
(3.5) nor (3.6) takes equality, and we have
!2
!2
∞
∞
∞
∞
∞
X
X
X
1 X am
1 X |am |
2
≤
<π
na2n .
n
ln
emn
n
ln
emn
n=1
m=1
n=1
m=1
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
m=1
n=1

 21
!2 ∞
∞
∞
X
X
X
1
am
(3.7)
≤
nb2n  .
n
ln
emn
n=1
m=1
n=1
J. Inequal. Pure and Appl. Math., 3(5) Art. 75, 2002
http://jipam.vu.edu.au/
6
B ICHENG YANG
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.
Remark 3.4. Inequality (3.4) is similar to the following equivalent form of (1.1) (see [2]):
!2
∞
∞
∞
X
X
X
am
(3.8)
< π2
a2n .
m
+
n
+
1
m=0
n=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.
R EFERENCES
[1] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cambridge Univ. Press, London,
1952.
[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.
J. Inequal. Pure and Appl. Math., 3(5) Art. 75, 2002
http://jipam.vu.edu.au/