volume 1, issue 2, article 22, 2000.
Received 8 May, 2000;
accepted 10 June 2000.
Communicated by:L. Debnath
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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:[email protected]
c
2000Victoria University ISSN (electronic): 1443-5756 012-00
On a Strengthened Hardy-Hilbert Inequality
Bicheng Yang
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Abstract
In this paper, a new inequality for the weight coefficientW(n, r)of the form
W(n, r) =
∞
X
m=0
1 m+n+ 1
n+12 m+12
!1r
< π
sin πr− 1
13 (n+ 1) (2n+ 1)1−1r
(r >1, n∈N0=N∪ {0})
is proved. This is followed by a strengthened version of the more accurate Hardy-Hilbert inequality.
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
On a Strengthened Hardy-Hilbert Inequality
Bicheng Yang
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1. Introduction
Ifp >1,p1+1q = 1,an, bn≥0, and0<P∞
n=1−λapn <∞,0<P∞
n=1−λbqn <∞ (λ= 0,1), then the Hardy-Hilbert inequality is
(1.1)
∞
X
m=1−λ
∞
X
n=1−λ
ambn
m+n+λ < π sin
π p
∞
X
n=1−λ
apn
!1p ∞ X
n=1−λ
bqn
!1q ,
where the constant π/sin(π/p) is best possible (see [3, Chapter 9]). In- equality (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) forλ= 0as
(1.2)
∞
X
m=1
∞
X
n=1
ambn m+n <
∞
X
n=1
π sin
π p
− 1−γ np1
apn
1 p
×
∞
X
n=1
π sin
π p
−1−γ n1q
bqn
1 q
,
whereγ = 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) forλ = 0as
(1.3)
∞
X
m=1
∞
X
n=1
ambn m+n <
∞
X
n=1
π sin
π p
− 1 2n1p +n−1q
apn
1 p
×
∞
X
n=1
π sin
π p
− 1 2n1q +n−1p
bqn
1 q
,
which is not comparable with (1.2).
Inequality (1.1) for λ = 1 is called the more accurate Hardy-Hilbert’s in- equality, which has been strengthened as
(1.4)
∞
X
m=0
∞
X
n=0
ambn m+n+ 1
<
∞
X
n=0
π sin
π p
− ln 2−γ (2n+ 1)1+1p
apn
1 p
×
∞
X
n=0
π sin
π p
− ln 2−γ (2n+ 1)1+1q
bqn
1 q
,
by introducing an inequality of the weight coefficientW(n, r)in the form (see
On a Strengthened Hardy-Hilbert Inequality
Bicheng Yang
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[5, equation (2.9)]):
W(n, r) =
∞
X
m=0
1 m+n+ 1
n+12 m+ 12
1 r
(1.5)
< π
sin πr − ln 2−γ (2n+ 1)2−1r
(r >1, n∈N0).
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
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2. Some Lemmas
Lemma 2.1. If f ∈ C6[0,∞), f(2q)(x) > 0 (q= 0,1,2,3), f(i−1)(∞) = 0 (i= 1,2, . . . ,6), andP∞
n=0f(n)<∞, then we have
∞
X
k=2
(−1)kf(k)> 1
2f(2) (see [1, equation (4.4)]), (2.1)
∞
X
m=0
f(m) = Z ∞
0
f(t)dt+ 1
2f(0) + Z ∞
0
B¯1f0(t)dt, (2.2)
Z ∞ 0
B¯1f0(t)dt =− 1
12f0(0) +δ2, δ2 = 1 6
Z ∞ 0
B¯3f000(t)dt <0, (2.3)
whereB¯i(t) (i= 1,3)are Bernoulli functions (see [5, equations (1.7)-(1.9)]).
Setting the weight coefficientW(n, r)in the form:
W(n, r) =
∞
X
m=0
1 m+n+ 1
n+12 m+ 12
1r (2.4)
= π
sin πr − θ(n, r) (2n+ 1)2−1r
(r >1, n∈N0), then we find
(2.5)
θ(n, r) = π
sin πr(2n+ 1)2−1r −(2n+ 1)2
∞
X
m=0
1 m+n+ 1
1 2m+ 1
1r .
On a Strengthened Hardy-Hilbert Inequality
Bicheng Yang
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If we define the functionf(x)asf(x) = (x+n+1)1 2x+11 1r
,x∈[0,∞), then we havef(0) = (n+1)1 ,
f0(x) =− 1 (x+n+ 1)2
1 2x+ 1
1r
− 2
r(x+n+ 1) 1
2x+ 1 1+1r
, f0(0) =− 1
(n+ 1)2 − 2 r(n+ 1), and
Z ∞ 0
f(x)dx= 1 (2n+ 1)1r
Z ∞
1 2n+1
1 (y+ 1)
1 y
1r dy
= 1
(2n+ 1)1r
"
π sin πr −
∞
X
ν=0
(−1)ν 1− 1r +ν
(2n+ 1)ν+1−1r
# . By (2.2) and (2.5), we have
θ(n, r) =−(2n+ 1)2 2 (n+ 1) +
∞
X
ν=0
(−1)ν 1− 1r +ν
(2n+ 1)ν−1 (2.6)
+ Z ∞
0
B¯1(t)
"
(2n+ 1)2 (t+n+ 1)2(2t+ 1)1r
+ 2 (2n+ 1)2 r(t+n+ 1) (2t+ 1)1+1r
# dt.
By Lemma 4 of [5, p. 1106] and (2.4), we have
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Bicheng Yang
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Lemma 2.2. Forr >1,n ∈N0, we haveθ(n, r)> θ(n,∞), and (2.7) W(n, r)< π
sin πr − θ(n,∞) (2n+ 1)2−1r
(r >1, n∈N0), where
(2.8)
θ(n,∞) = −(2n+ 1)2 2 (n+ 1)+
∞
X
ν=0
(−1)ν
(1 +ν) (2n+ 1)ν−1+ Z ∞
0
B¯1(t)
"
(2n+ 1)2 (t+n+ 1)2
# dt.
Since by (2.3) and (2.1),we have Z ∞
0
B¯1(t) 1
(t+n+ 1)2dt =− 1
12 (n+ 1)2 − 1 3!
Z ∞ 0
B¯3(t)
1 (t+n+ 1)
000
dt
>− 1 12 (n+ 1)2 and
∞
X
ν=0
(−1)ν
(1 +ν) (2n+ 1)ν−1 = (2n+ 1)− 1 2 +
∞
X
ν=2
(−1)ν
(1 +ν) (2n+ 1)ν−1
>(2n+ 1)− 1
2 + 1
6 (2n+ 1). Then by (2.8), we find
(2.9) θ(n,∞)> 1
6− 1
6 (n+ 1) − 1
12 (n+ 1)2 + 1 6 (2n+ 1).
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Lemma 2.3. Forr >1,n ∈N0, we have (2.10) W(n, r)< π
sin πr − 1
13 (n+ 1) (2n+ 1)1−1r .
Proof. Define the functiong(x)as g(x) = 1
12− 1
6 (2x+ 1) + 1
12 (x+ 1) + 1
12 (2x+ 1)2, x∈[0,∞).
Then by (2.8), we have θ(n,∞) > 2n+1n+1g(n). Sinceg(1) >0.0787> 131, and forx∈[1,∞),
g0(x) = 1
3 (2x+ 1)2− 1
12 (x+ 1)2− 1
3 (2x+ 1)3 = 4x2+ 2x−1
12 (x+ 1)2(2x+ 1)3 >0, then for n ≥ 1, we have θ(n,∞) > (n+1)2n+1g(1) > 13(n+1)2n+1 . Hence by (2.7), inequality (2.10) is valid forn ≥ 1. Sinceln 2−γ = 0.1159+ > 131, then by (1.5), we find
(2.11)
W(0, r)< π
sin πr− ln 2−γ (2×0 + 1)2−1r
< π
sin πr− 1
13 (0 + 1) (2×0 + 1)1−1r . It follows that (2.10) is valid forr >1,andn∈N0. This proves the lemma.
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3. Theorem and Remarks
Theorem 3.1. If p > 1,1p + 1q = 1, an, bn ≥ 0, 0 < P∞
n=0apn < ∞, and 0<P∞
n=0bqn<∞, then (3.1)
∞
X
m=0
∞
X
n=0
ambn m+n+ 1 <
∞
X
n=0
π sin
π p
− 1
13 (n+ 1) (2n+ 1)1p
apn
1 p
×
∞
X
n=0
π sin
π p
− 1
13 (n+ 1) (2n+ 1)1q
bqn
1 q
,
(3.2)
∞
X
m=0
∞
X
n=0
an
m+n+ 1
!p
<
π sin
π p
p−1 ∞
X
n=0
π sin
π p
− 1
13 (n+ 1) (2n+ 1)1p
apn. Proof. By Hölder’s inequality, we have
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∞
X
m=0
∞
X
n=0
ambn
m+n+ 1
=
∞
X
m=0
∞
X
n=0
"
am (m+n+ 1)1p
m+ 12 n+12
pq1 # "
bn (m+n+ 1)1q
n+ 12 m+12
pq1 #
≤ ( ∞
X
m=0
∞
X
n=0
"
1 (m+n+ 1)
m+12 n+12
1 q#
apm )
1 p
× ( ∞
X
m=0
∞
X
n=0
"
1 (m+n+ 1)
n+ 12 m+12
1 p#
bqn )1q
= ( ∞
X
m=0
∞
X
n=0
"
1 (m+n+ 1)
m+12 n+ 12
1q# apm
)
1 p
× ( ∞
X
n=0
∞
X
m=0
"
1 (m+n+ 1)
n+ 12 m+12
1p# bqn
)
1 q
= ( ∞
X
m=0
W(m, q)apm
)1p( ∞ X
n=0
W(n, p)bqn )1q
.
Sincesin(π/q) = sin(π/p), by (2.10) forr =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
n=0
"
an (m+n+ 1)p1
n+ 12 m+12
1 pq# "
1 (m+n+ 1)1q
m+ 12 n+12
1 pq#
≤ ( ∞
X
n=0
"
1 (m+n+ 1)
n+ 12 m+12
1 q#
apn
)1p ( ∞ X
n=0
1 (m+n+ 1)
m+ 12 n+12
1 p)1q
= ( ∞
X
n=0
"
1 (m+n+ 1)
n+12 m+ 12
1q# apn
)
1 p
{W(m, p)}1q
<
( ∞ X
n=0
"
1 (m+n+ 1)
n+12 m+ 12
1q# apn
)
1
p
π sin
π p
1 q
.
Then we find
∞
X
m=0
∞
X
n=0
an m+n+ 1
!p
<
∞
X
m=0
∞
X
n=0
"
1 (m+n+ 1)
n+12 m+ 12
1 q#
apn
π sin
π p
p q
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=
π sin
π p
p−1 ∞
X
n=0
∞
X
m=0
"
1 (m+n+ 1)
n+ 12 m+12
1q# apn
=
π sin
π p
p−1 ∞
X
n=0
W(n, q)apn.
Hence by (2.10) forr=q, we have (3.2). The theorem is proved.
Remark 3.1. Inequality (3.1) is a definite improvement over (1.1) forλ= 1.
Remark 3.2. Since forn≥2,13 (2−γ)<2− n+11 , then (3.3)
π
sin πr− ln 2−γ (2n+ 1)2−1r
> π
sin πr− 1
13 (n+ 1) (2n+ 1)1−1r
(r >1, n≥2). In view of (2.11) and (3.3), it follows that (3.1) and (1.4) represent two dis- tinct versions of strengthened inequalities. However, they are not comparable.
Remark 3.3. Inequality (3.2) reduces to
(3.4)
∞
X
m=0
∞
X
n=0
an m+n+ 1
!p
<
π sin
π p
p ∞
X
n=0
apn.
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 ANDB. YANG, On the extended Hilbert’s inequality, Proc. Amer.
Math. Soc., 126(3) (1998), 751–759.
[3] G.H. HARDY, J.E. LITTLEWOOD AND G. PÓLYA, Inequalities, Cam- bridge University Press, 1934.
[4] D.S. MITRINOVI ´C, J.E. PE ˇCARI ´CANDA.M. FINK, Inequalities Involv- ing Functions and their Integrals and Derivatives, Kluwer Academic Pub- lishers, 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 in- equality, Int. J. Math. Math. Sci., 21(2) (1998), 403–408.
[7] B. YANGANDM. GAO, On a best value of Hardy-Hilbert’s inequality, Adv.
Math., 26(2) (1997), 156–164.