Journal of Inequalities in Pure and
Applied Mathematics
ON A STRENGTHENED HARDY-HILBERT INEQUALITY
BICHENG YANG
Department of Mathematics
Guangdong Education College
Guangzhou
Guangdong 510303
PEOPLE’S REPUBLIC OF CHINA
EMail: bcyang@163.net
volume 1, issue 2, article 22,
2000.
Received 8 May, 2000;
accepted 10 June 2000.
Communicated by: L. Debnath
Abstract
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Abstract
In this paper, a new inequality for the weight coefficient W (n, r) of the form
∞
X
n + 12
1
W (n, r) =
m + n + 1 m + 12
m=0
<
!1
r
π
1
1 (r > 1, n ∈ N0 = N ∪ {0})
π −
sin r
13 (n + 1) (2n + 1)1− r
is proved. This is followed by a strengthened version of the more accurate
Hardy-Hilbert inequality.
On a Strengthened
Hardy-Hilbert Inequality
Bicheng Yang
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2000 Mathematics Subject Classification: 26D15
Key words: Hardy-Hilbert inequality, Weight Coefficient, Hölder’s inequality.
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
Some Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3
Theorem and Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
References
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1.
Introduction
P
P∞
q
p
If p > 1, p1 + 1q = 1, an , bn ≥ 0, and 0 < ∞
n=1−λ an < ∞, 0 <
n=1−λ bn < ∞
(λ = 0, 1), then the Hardy-Hilbert inequality is
(1.1)
∞
X
∞
X
am b n
π
<
π
m
+
n
+
λ
sin
m=1−λ n=1−λ
p
∞
X
n=1−λ
! p1
apn
∞
X
! 1q
bqn
,
n=1−λ
where the constant π/ sin(π/p) is best possible (see [3, Chapter 9]). Inequality (1.1) is important in analysis and it’s applications (see [4, Chapter 5]).
In recent years, Yang and Gao [2, 7], have given a strengthened version of (1.1)
for λ = 0 as
(1.2)


  p1
∞
∞
∞ X

X
X
π
am b n
1 − γ  p

−
an
<
1

m + n  n=1 sin π
p
n
m=1 n=1
p


  1q
∞
X

 π − 1 −1 γ  bqn
×
,
π


q
n
sin
n=1
p
where γ = 0.5772+ , is Euler’s constant. Later, Yang and Debnath [6] proved a
On a Strengthened
Hardy-Hilbert Inequality
Bicheng Yang
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distinctly strengthened version of (1.1) for λ = 0 as
(1.3)


  p1
∞ X
∞
∞


X
X
1
am b n
π
p


−
<
an
1
−1

m + n  n=1 sin π
p + n q
2n
m=1 n=1
p

  1q

∞

X
q

 π − 1 1
b
,
×
1
n

 n=1 sin π
2n q + n− p
p
which is not comparable with (1.2).
Inequality (1.1) for λ = 1 is called the more accurate Hardy-Hilbert’s inequality, which has been strengthened as
On a Strengthened
Hardy-Hilbert Inequality
Bicheng Yang
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(1.4)
∞ X
∞
X
m=0 n=0
am b n
m+n+1


  p1
∞
X

 π − ln 2 − γ 1  apn
<
 n=0 sin π

(2n + 1)1+ p
p


  1q
∞
X

 π − ln 2 − γ 1  bqn
×
,


(2n + 1)1+ q
sin π
n=0
p
by introducing an inequality of the weight coefficient W (n, r) in the form (see
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[5, equation (2.9)]):
(1.5)
∞
X
1
W (n, r) =
m+n+1
m=0
<
n + 12
m + 12
r1
ln 2 − γ
π
−
(r > 1, n ∈ N0 ) .
1
π
sin r
(2n + 1)2− r
In this paper we will give another strengthened version of (1.1) for λ = 1,
which is not comparable with (1.4). We need some preparatory works.
On a Strengthened
Hardy-Hilbert Inequality
Bicheng Yang
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2.
Some Lemmas
6
Lemma 2.1. If f ∈ C
∞), f (2q) (x) > 0 (q = 0, 1, 2, 3), f (i−1) (∞) = 0
P[0,
∞
(i = 1, 2, . . . , 6), and n=0 f (n) < ∞, then we have
(2.1)
(2.2)
(2.3)
∞
X
1
(−1)k f (k) > f (2) (see [1, equation (4.4)]),
2
k=2
Z ∞
Z
∞
∞
X
1
B̄1 f 0 (t) dt,
f (m) =
f (t) dt + f (0) +
2
0
0
m=0
Z ∞
Z
1 0
1 ∞
0
B̄1 f (t) dt = − f (0) + δ2 , δ2 =
B̄3 f 000 (t) dt < 0,
12
6
0
0
On a Strengthened
Hardy-Hilbert Inequality
Bicheng Yang
where B̄i (t) (i = 1, 3) are Bernoulli functions (see [5, equations (1.7)-(1.9)]).
Contents
Setting the weight coefficient W (n, r) in the form:
(2.4)
∞
X
1
W (n, r) =
m+n+1
m=0
n + 12
m + 12
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π
θ (n, r)
−
=
(r > 1, n ∈ N0 ) ,
1
π
sin r
(2n + 1)2− r
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then we find
(2.5)
∞
X
π
1
2− r1
2
θ (n, r) =
(2n + 1)
− (2n + 1)
π
m+n+1
sin r
m=0
Page 6 of 14
1
2m + 1
r1
.
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If we define the function f (x) as f (x) =
1
have f (0) = (n+1)
,
1
(x+n+1)
1
2x+1
r1
, x ∈ [0, ∞), then we
r1
1+ r1
1
1
2
1
f (x) = −
−
,
r (x + n + 1) 2x + 1
(x + n + 1)2 2x + 1
1
2
f 0 (0) = −
,
2 −
r (n + 1)
(n + 1)
0
and
Z
On a Strengthened
Hardy-Hilbert Inequality
∞
Z
1
f (x) dx =
1
(2n + 1) r
0
∞
1
1
(y + 1)
r1
1
dy
y
#
" 2n+1
∞
X
1
π
(−1)ν
−
=
.
1
π
ν+1− r1
1
(2n + 1) r sin r
ν=0 1 − r + ν (2n + 1)
By (2.2) and (2.5), we have
(2.6)
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2
∞
X
ν
(2n + 1)
(−1)
+
1
2 (n + 1) ν=0 1 − r + ν (2n + 1)ν−1
"
#
Z ∞
(2n + 1)2
2 (2n + 1)2
+
B̄1 (t)
dt.
1 +
1
0
(t + n + 1)2 (2t + 1) r
r (t + n + 1) (2t + 1)1+ r
θ (n, r) = −
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By Lemma 4 of [5, p. 1106] and (2.4), we have
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Lemma 2.2. For r > 1, n ∈ N0 , we have θ (n, r) > θ (n, ∞), and
W (n, r) <
(2.7)
π
θ (n, ∞)
−
(r > 1, n ∈ N0 ) ,
1
π
sin r
(2n + 1)2− r
where
(2.8)
"
#
Z ∞
∞
(2n + 1)2 X
(−1)ν
(2n + 1)2
θ (n, ∞) = −
+
+
B̄1 (t)
dt.
2 (n + 1) ν=0 (1 + ν) (2n + 1)ν−1 0
(t + n + 1)2
Since by (2.3) and (2.1),we have
Z
∞
B̄1 (t)
0
1
1
1
2 dt = −
2 −
3!
(t + n + 1)
12 (n + 1)
1
>−
12 (n + 1)2
Z
∞
B̄3 (t)
0
1
(t + n + 1)
ν=0
∞
(−1)ν
1 X
(−1)ν
= (2n + 1) − +
2 ν=2 (1 + ν) (2n + 1)ν−1
(1 + ν) (2n + 1)ν−1
1
1
.
> (2n + 1) − +
2 6 (2n + 1)
Then by (2.8), we find
(2.9)
θ (n, ∞) >
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000
dt
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and
∞
X
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1
1
1
1
−
−
.
2 +
6 6 (n + 1) 12 (n + 1)
6 (2n + 1)
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Lemma 2.3. For r > 1, n ∈ N0 , we have
(2.10)
W (n, r) <
1
π
−
1 .
π
sin r
13 (n + 1) (2n + 1)1− r
Proof. Define the function g(x) as
g(x) =
1
1
1
1
−
+
+
, x ∈ [0, ∞).
12 6 (2x + 1) 12 (x + 1) 12 (2x + 1)2
Then by (2.8), we have θ (n, ∞) >
for x ∈ [1, ∞),
2n+1
g
n+1
(n). Since g(1) > 0.0787 >
1
,
13
and
1
1
1
4x2 + 2x − 1
g (x) =
−
−
=
> 0,
3 (2x + 1)2 12 (x + 1)2 3 (2x + 1)3
12 (x + 1)2 (2x + 1)3
0
2n+1
g
(n+1)
2n+1
.
13(n+1)
then for n ≥ 1, we have θ (n, ∞) >
(1) >
Hence by (2.7),
1
inequality (2.10) is valid for n ≥ 1. Since ln 2 − γ = 0.1159+ > 13
, then by
(1.5), we find
(2.11)
ln 2 − γ
π
π
1
−
−
W (0, r) <
1 <
1 .
π
π
2−
sin r
sin r
(2 × 0 + 1) r
13 (0 + 1) (2 × 0 + 1)1− r
It follows that (2.10) is valid for r > 1,and n ∈ N0 . This proves the lemma.
On a Strengthened
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Bicheng Yang
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3.
Theorem and Remarks
Theorem 3.1. If p > 1, p1 +
P
q
0< ∞
n=0 bn < ∞, then
1
q
= 1, an , bn ≥ 0, 0 <
P∞
n=0
apn < ∞, and
(3.1)
∞ X
∞
X
m=0 n=0

  p1

∞


X
π
1
am b n
p


−
an
<
1

m + n + 1  n=0 sin π
p
13
(n
+
1)
(2n
+
1)
p

  1q

∞

X
1
q

 π −
b
,
×
1
n


13 (n + 1) (2n + 1) q
sin π
n=0
(3.2)
∞
X
m=0
∞
X
!p
p
an
m+n+1
n=0


p−1

∞
X
π
1
 apn .
 π −
<  
1
π
13 (n + 1) (2n + 1) p
sin πp
n=0 sin p
Proof. By Hölder’s inequality, we have
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Bicheng Yang
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∞ X
∞
X
am b n
m+n+1
m=0 n=0
"
1 #"
1 #
∞
∞ X
1 pq
1 pq
X
m
+
n
+
am
bn
2
2
=
1
1
1
1
n+ 2
p
(m + n + 1) q m + 2
m=0 n=0 (m + n + 1)
) p1
( ∞ ∞ "
1#
1 q
XX
m+ 2
1
apm
≤
1
(m
+
n
+
1)
n
+
2
m=0 n=0
( ∞ ∞ "
1 # ) 1q
XX
n + 12 p q
1
×
bn
1
(m
+
n
+
1)
m
+
2
m=0 n=0
( ∞ ∞ "
) p1
1#
1 q
XX
m
+
1
2
=
apm
1
(m + n + 1) n + 2
m=0 n=0
(∞ ∞ "
) 1q
1#
1 p
XX
n+ 2
1
×
bqn
1
(m
+
n
+
1)
m
+
2
n=0 m=0
) 1q
( ∞
) p1 ( ∞
X
X
=
W (m, q) apm
W (n, p) bqn
.
m=0
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n=0
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Since sin(π/q) = sin(π/p), by (2.10) for r = p, q, we have (3.1).
By (2.10), we have W (n, p) < π/ sin(π/p). Then by Hölder’s inequality,
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we obtain
∞
X
n=0
=
an
m+n+1
"
∞
X
an
1
p
n + 12
m + 12
pq1 # "
1
1
m + 12
n + 12
pq1 #
(m + n + 1)
(m + n + 1) q
1
) p1 ( ∞
(∞ "
1 )q
1#
1 q
1 p
X
X
m+ 2
n+ 2
1
1
apn
≤
1
(m + n + 1) n + 12
(m + n + 1) m + 2
n=0
n=0
(∞ "
1 # ) p1
X
1
n + 12 q p
1
{W (m, p)} q
=
an
1
(m + n + 1) m + 2
n=0
1
(∞ "
) p1 
1#
q

1 q
X
n+ 2
1
π 
p
.
<
an
 sin π 
(m + n + 1) m + 12
n=0
n=0
p
Then we find
On a Strengthened
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Bicheng Yang
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∞
X
∞
X
m=0
n=0
an
m+n+1
<
!p
∞ X
∞
X
m=0 n=0
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"
1
(m + n + 1)
n + 12
m + 12
1 #
 pq

q
apn 
sin
π 
π
p
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p−1

= 
sin
π 
π
p
= 
sin
"
n=0 m=0
p−1

∞ X
∞
X
π 
π
p
∞
X
1
(m + n + 1)
n + 12
m + 12
1q #
apn
W (n, q) apn .
n=0
Hence by (2.10) for r = q, we have (3.2). The theorem is proved.
Remark 3.1. Inequality (3.1) is a definite improvement over (1.1) for λ = 1.
1
Remark 3.2. Since for n ≥ 2, 13 (2 − γ) < 2 − n+1
, then
(3.3)
π
π
ln 2 − γ
1
−
−
(r > 1, n ≥ 2) .
1 >
1
π
π
sin r
sin r
(2n + 1)2− r
13 (n + 1) (2n + 1)1− r
In view of (2.11) and (3.3), it follows that (3.1) and (1.4) represent two distinct versions of strengthened inequalities. However, they are not comparable.
Remark 3.3. Inequality (3.2) reduces to
(3.4)
∞
X
∞
X
m=0
n=0
an
m+n+1
!p
On a Strengthened
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Bicheng Yang
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p

<
sin
∞
π  X p
an .
π
p
n=0
This is an equivalent form of the more accurate Hardy-Hilbert’s inequality
(see [3, Chapter 9]).
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References
[1] R.P. BOAS, Estimating remainders, Math. Mag., 51(2) (1978), 83–89.
[2] M. GAO AND B. YANG, On the extended Hilbert’s inequality, Proc. Amer.
Math. Soc., 126(3) (1998), 751–759.
[3] G.H. HARDY, J.E. LITTLEWOOD
bridge University Press, 1934.
AND
G. PÓLYA, Inequalities, Cam-
[4] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Inequalities Involving Functions and their Integrals and Derivatives, Kluwer Academic Publishers, 1991.
[5] B. YANG, On a strengthened version of the more accurate Hardy-Hilbert’s
inequality, Acta Math. Sinica, 42(6) (1999), 1103–1110.
[6] B. YANG AND L. DEBNATH, On new strengthened Hardy-Hilbert’s inequality, Int. J. Math. Math. Sci., 21(2) (1998), 403–408.
[7] B. YANG AND M. GAO, On a best value of Hardy-Hilbert’s inequality, Adv.
Math., 26(2) (1997), 156–164.
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Hardy-Hilbert Inequality
Bicheng Yang
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