Journal of Inequalities in Pure and
Applied Mathematics
ON HARDY-HILBERT’S INTEGRAL INEQUALITY
W.T. SULAIMAN
Mosul University
College of Mathematics and Computer Science
EMail: waadsulaiman@hotmail.com
volume 5, issue 2, article 25,
2004.
Received 17 April, 2003;
accepted 18 January, 2004.
Communicated by: L. Pick
Abstract
Contents
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2000
Victoria University
ISSN (electronic): 1443-5756
053-03
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Abstract
In the present paper, by introducing some parameters, new forms of HardyHilbert’s inequalities are given.
2000 Mathematics Subject Classification: 26D15.
Key words: Hardy-Hilbert’s integral inequality.
On Hardy-Hilbert’s Integral
Inequality
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
New Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
3
5
W.T. Sulaiman
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J. Ineq. Pure and Appl. Math. 5(2) Art. 25, 2004
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1.
Introduction
R∞
R∞
If p > 1, p1 + 1q = 1, f (t), g(t) ≥ 0, 0 < 0 f p (t)dt < ∞, and 0 < 0 g q (t)dt < ∞,
then, the Hardy-Hilbert integral inequality is given by:
∞
Z
∞
Z
(1.1)
0
0
f (x)g(y)
π
dxdy ≤
x+y
sin πp
where the constant
π
sin π
p
Z
p1 Z
f (t)dt
∞
∞
p
0
1q
g (t)dt ,
q
0
is the best possible (see [1]).
On Hardy-Hilbert’s Integral
Inequality
Yang [2] and [3] gave the following generalization of (1.1)
Z
∞
∞
Z
(1.2)
0
0
f (x)g(y)
dxdy
(x + y − 2α)λ
Z
1
1
p
q
≤ Kλ (p)Kλ (q)
W.T. Sulaiman
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p1
f (t)dt
∞
1−λ p
(t − α)
Contents
0
Z
×
∞
1q
g (t)dt ,
1−λ q
(t − α)
0
where
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Z
Kλ (r) =
0
∞
1
u r −1
du = B
(1 + u)λ
1
1
,λ −
r
r
0 < λ ≤ 1, λ >
B is the beta function defined by
Z 1
B(p, q) =
xp−1 (1 − x)q−1 dx, p, q > 0
0
1
> 0,
r
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and, for 0 < T < ∞,
Z
(1.3)
0
T
Z
T
f (x)g(y)
dxdy
(x + y)λ
0
! 12
Z T"
λ #
λ λ
1 t 2 1−λ 2
≤β
,
1−
t f (t)dt
2 2
2 T
α
! 12
λ2 #
Z T"
1 t
1−
t1−λ g 2 (t)dt
.
×
2 T
α
On Hardy-Hilbert’s Integral
Inequality
W.T. Sulaiman
In the present paper, by introducing some parameters, new forms of HardyHilbert’s inequalities are given.
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2.
New Results
We state and prove the following:
Lemma 2.1. Let λ > 0, p, q, r > 1, p1 + 1q + 1r = 1, f (t), g(t), h(t) ≥ 0,
λi (t) ≥ 0, i = p, q, r, and assume that
Z b
0<
λpp (t)f p (t)dt < ∞,
a
Z d
0<
λqq (t)g q (t)dt < ∞
c
and
W.T. Sulaiman
d
Z
On Hardy-Hilbert’s Integral
Inequality
λrr (t)hr (t)dt < ∞.
0<
c
Then the two following inequalities are equivalent
Z bZ dZ k
f (x)g(y)h(z)
(2.1)
dxdydz
hλ (x, y, z)
a
c
e
Z b
p1 Z d
1q Z
≤K
λpp (t)f p (t)dt
λqq (t)g q (t)dt
a
c
Contents
k
r1
λrr (t)hr (t)dt ,
e
where K = K(λ, p, q, r) is a constant, and
qr
Z d Z k
q+r
Z b qr
g(y)h(z)
(2.2)
λpq+r (x)
dydz
dx
λ
a
c
e h (x, y, z)
r Z
Z d
q+r
qr
q+r
q
q
≤K
λq (t)g (t)dt
c
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e
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q
q+r
k
r
r
λr (t)h (t)dt
.
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Proof. Suppose (2.2) is satisfied, then
Z bZ dZ k
f (x)g(y)h(z)
dxdydz
hλ (x, y, z)
a
c
e
Z b
Z dZ k
g(y)h(z)
−1
=
λp (x)f (x) λp (x)
dydz dx
λ
a
c
e h (x, y, z)
! q+r
qr
q+r
Z b
p1 Z b qr Z d Z k
qr
g(y)h(z)
≤
λpp (x)f p (x)dx
λpq+r
dydz
dx
λ
a
a
c
e h (x, y, z)
Z b
p1 Z d
1q Z k
r1
≤K
λpp (t)f p (t)dt
λqq (t)g q (t)dt
λrr (t)hr (t)dt .
c
a
b
qr
q+r
Z
d
k
Z
λp
a
c
e
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g(y)h(z)
dydz
hλ (x, y, z)
Contents
qr
q+r
dx
qr
Z d Z k
q+r
−1
qr
g(y)h(z) q+r
g(y)h(z)
=
.λp
dydz
dxdydz
λ
λ
a
c
e h (x, y, z)
c
e h (x, y, z)
1q Z k
r1
Z d
≤K
λqq (y)g q (y)dy
λrr (z)hr (z)dz
Z bZ
d
k
Z
c
e
Z
×
a
b
−p qr
λpp (x)λp q+r (x)
W.T. Sulaiman
e
Now, suppose that (2.1) is satisfied, then
Z
On Hardy-Hilbert’s Integral
Inequality
Z
c
d
Z
e
k
g(y)h(z)
dydz
hλ (x, y, z)
! p1
qr
p( q+r
−1)
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dx
J. Ineq. Pure and Appl. Math. 5(2) Art. 25, 2004
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Z
d
λqq (y)g q (y)dy
=K
1q Z
c
k
λrr (z)hr (z)dz
r1
e
b
Z
×
Z
qr
q+r
dZ
k
λp (x)
a
c
e
g(y)h(z)
dydz
hλ (x, y, z)
! p1
qr
q+r
dx
,
therefore
Z
b
qr
q+r
Z
d
Z
λp (x)
a
c
e
k
qr
q+r
! 1q + r1
g(y)h(z)
dydz
dx
hλ (x, y, z)
Z d
1q Z
q
q
≤k
λq (t)g (t)dt
c
On Hardy-Hilbert’s Integral
Inequality
k
r1
r
r
λr (t)h (t)dt ,
e
and the desired equivalence is proved.
W.T. Sulaiman
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Lemma 2.2. (a) Let 0 ≤ y ≤ 1, α > 0, f ≥ 0. If we define the function
Z y
−α
g(y) = y
f (x)dx,
0
then g(y) ≥ g(1).
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(b) Let y ≥ 1, α > 0, f ≥ 0. Defining the function,
Z y
−α
h(y) = y
f (x)dx,
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J. Ineq. Pure and Appl. Math. 5(2) Art. 25, 2004
we have h(y) ≥ h(1).
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Proof. (a) Let x = 1t , then
g(y) = y −α
∞
Z
f
y −1
1
t
t2
dt
.
We observe also that
g 0 (y) = y −α −y 2 f (y) +
Z
∞
f
y −1
1
t
t2
!
dt (−α)y −α−1 ≤ 0,
On Hardy-Hilbert’s Integral
Inequality
therefore g is non-increasing, which implies g(y) ≥ g(1).
W.T. Sulaiman
(b) We obviously have
0
α
Z
h (y) = y f (y) +
y
f (x)dx αy
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α−1
≥ 0,
0
therefore h is non-decreasing, and hence h(y) ≥ h(1).
The following result may be stated as well.
Theorem 2.3. Let f (t), g(t), h(t) ≥ 0, p, q, r > 1, p1 + 1q + 1r = 1, 2 < λ < 3,
γ > µ max {p, q, r}, and
1 1 1
λ−1 λ−1 λ−1
< µ < min
,
,
.
max − , − , −
p q r
p
q
r
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T
Z
(t − α)2−λ f p (t)dt < ∞,
0<
α
T
Z
(t − α)2−λ g q (t)dt < ∞
0<
α
and
T
Z
(t − α)2−λ hr (t)dt < ∞,
0<
α
then
Z
(2.3)
On Hardy-Hilbert’s Integral
Inequality
T
α
Z
T
α
Z
T
f (x)g(y)h(z)
dxdydz
λ
α (x + y + z)
p1
Z T
≤
φ(t, µ, λ, p)(t − α)2−λ f p (t)dt
W.T. Sulaiman
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α
Z
T
×
1q
2−λ q
φ(t, µ, λ, q)(t − α) g (t)dt
α
Z
T
×
r1
φ(t, µ, λ, r)(t − α)2−λ hr (t)dt ,
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α
where
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φ(t, µ, λ, j) = B(λ − µj − 1, µj + 1) B(λ − 2, 1 − µj)
γ Z 1
2γ
t−α
uλ−3
t−α
−
du −
λ−µj−1
T −α
T −α
0 (1 + u)
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1
Z
×
0
uλ−µj−2
du
(1 + u)λ
Z
0
1
uγ−µj−2
du,
(1 + u)λ−µj−1
j = p, q, r.
Proof. The proof is as follows. We have
Z TZ TZ T
f (x)g(x)h(z)
dxdydz
λ
α
α
α (x + y + z)
µ
µ
Z T Z T Z T f (x) z−α
y−α µ
h(z)
g(y) x−α
y−α
z−α
x−α
dxdydz
=
λ/p (x + y + z)λ/q (x + y + z)λ/r
(x
+
y
+
z)
α
α
α
µp

 p1
Z T Z T Z T f p (x) z−α
y−α
≤
dxdydz 
(x + y + z)λ
α
α
α
Z
T
Z
T
Z
T
×
α
Z
T
α
Z
T
α
Z
T
×
α
1
p
1
q
α
α
x−α µq
! 1q
g q (y) z−α
dxdydz
(x + y + z)λ
! r1
y−α µr
hr (z) x−α
dxdydz
(x + y + z)λ
1
r
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Inequality
W.T. Sulaiman
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Then we have
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Z
T
F =
α
(x − α)2−λ f p (x)dx
Z
T
α
(1
y−α −µp
x−α
y−α λ−µp−1
+ x−α
)
dy
x−α
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Z
T
×
α
−µp
z−α
x+y−2α
z−α
(1 + x+y−2α
)λ
dz
.
x + y − 2α
Now by Lemma 2.2, we can state that
µp
z−α
Z T
x+y−2α
dz
z−α
λ
0 (1 + x+y−2α ) x + y − 2α
Z T −α
x+y−2α
uµp
du
=
(1 + u)λ
0
Z T −α
y−α
uµp
du
≤
(1 + u)λ
0
Z ∞ λ−µp−2
u
du
=
y−α (1 + u)λ
T −α
γ −γ Z y−α ! λ−µp−2
Z ∞ T −α
y−α
y−α
u
=
−
du
T −α
T −α
(1 + u)λ
0
0
γ Z 1 λ−µp−2 y−α
u
≤ B(λ − µp − 1, µp + 1) −
du .
λ
T −α
0 (1 + u)
T
F ≤
α
(x − α)2−λ f p (x)dx
W.T. Sulaiman
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Therefore
Z
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Inequality
Z
T
α
(1
y−α −µp
x−α
y−α λ−µp−1
+ x−α
)
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dy
x−α
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γ Z 1 λ−µp−2 y−α
u
× B(λ − µp − 1, µp + 1) −
du
λ
T −α
0 (1 + u)
Z T
Z T −α
x−α
u−µp
2−λ p
=
(x − α) f (x)dx
du
(1 + u)λ−µp−1
α
0
γ Z 1 λ−µp−2 x−α
u
× B(λ − µp − 1, µp + 1) −
uγ
du
λ
T −α
0 (1 + u)
"
Z
Z T −α
T
(x − α)
=
f (x)dx× B(λ − µp − 1, µp + 1) −
α
0
T −α
x−α
#
uγ−µp
du
−
(1 + u)λ−µp−1
0
0
Z T
Z ∞
2−λ p
=
(x − α) f (x)dx× B(λ − µp − 1, µp + 1) −
x−α
T −α
γ Z
1
uλ−µp−2
du
(1 + u)λ
"
Z
x−α
T −α
α
−
Z
u−µp
(1 + u)λ−µp−1
x−α
2−λ p
x−α
T −α
γ Z
1
0
uλ−µp−2
du
(1 + u)λ
T −α
x−α
Z
0
uγ−µp
du
(1 + u)λ−µp−1
W.T. Sulaiman
uλ−3
du
(1 + u)λ−µp−1
#
T
=
α
∞
−
x−α
T −α
0
λ−µp−2
u
du
(1 + u)λ
Z
−
0
1
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Z
× B(λ − µp − 1, µp + 1) −
γ Z
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(x − α)2−λ f p (x)dx
"
On Hardy-Hilbert’s Integral
Inequality
Z
0
T −α
x−α
x−α
T −α
!
0
γ−µp
λ−3
u
du
(1 + u)λ−µp−1
#
u
du
(1 + u)λ−µp−1
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Page 12 of 20
J. Ineq. Pure and Appl. Math. 5(2) Art. 25, 2004
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T
Z
(x − α)2−λ f p (x)dx
"
=
α
(
× B(λ − µp − 1, µp + 1) B(λ − 2, 1 − µp) −
x−α
T −α
γ x−α
T −α
−γ
x−α
T −α
2γ Z 1 λ−µp−2
uλ−3
x−α
u
×
du −
du
λ−µp−1
λ
(1 + u)
T −α
0
0 (1 + u)
γ Z T −α
x−α
T −α
uγ−µp
×
(1 + u)λ−µp−1
x−α
0
Z
T
Z
(x − α)2−λ f p (x)dx
α
γ
x−α
× B(λ − µp − 1, µp + 1) B(λ − 2, 1 − µp) −
T −α
2γ Z 1 λ−µp−2 Z 1
Z 1
λ−3
x−α
u
uγ−µp
u
×
du −
du
λ−µp−1
λ
λ−µp−1
T −α
0 (1 + u)
0 (1 + u)
0 (1 + u)
≤
Z
T
=
φ(x, λ, µ, p)(x − α)2−λ f p (x)dx,
α
where
φ(x, λ, µ, p) = B(λ − µp − 1, µp + 1) B(λ − 2, 1 − µp)
γ Z 1
x−α
uλ−3
−
du
λ−µp−1
T −α
0 (1 + u)
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Inequality
W.T. Sulaiman
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−
x−α
T −α
2γ Z
0
1
uλ−µp−2
du
(1 + u)λ
Z
1
0
#
uγ−µp
du .
(1 + u)λ−µp−1
Similarly
T
Z
φ(y, λ, µ, q)(y − α)2−λ g q (y)dy,
G=
α
and
Z
T
φ(z, λ, µ, r)(z − α)2−λ hr (z)dz.
H=
α
This completes the proof.
W.T. Sulaiman
Corollary 2.4. Let f (t), g(t), h(z) ≥ 0, p, q, r > 1, p1 + 1q + 1r = 1, 2 < λ < 3,
and
λ−1 λ−1 λ−1
1 1 1
max − , − , −
< µ < min
,
,
.
p q r
p
q
r
If
∞
Z
(t − α)2−λ f p (t)dt < ∞,
0<
Zα∞
0<
(t − α)2−λ g q (t)dt < ∞
α
and
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Inequality
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∞
0<
α
(t − α)2−λ hr (t)dt < ∞,
J. Ineq. Pure and Appl. Math. 5(2) Art. 25, 2004
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then we have the inequality
Z
(2.4)
∞
Z
∞
Z
∞
f (x)g(y)h(z)
dxdydz
(x + y + z)λ
α
α
α
Z ∞
p1 Z
2−λ p
≤K
(t − α) f (t)dt
α
∞
1q
g (t)dt
2−λ q
(t − α)
α
Z
∞
×
r1
h (t)dt ,
2−λ r
(t − α)
α
where
On Hardy-Hilbert’s Integral
Inequality
W.T. Sulaiman
K=
Y
B 1/j (λ − µj − 1, µj + 1)B 1/j (λ − 2, 1 − µj)
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j=p,q,r
Contents
and
qr
q+r
g(y)h(z)
(2.5)
(x − α)
dxdydz
dx
(x + y + z)λ
α
α
α
q
r Z
Z ∞
q+r
q+r
∞
qr
2−λ
q
2−λ
r
≤ K q+r
(t − α) g (t)dt
(t − α) h (t)dt
.
Z
∞
qr(λ−2)
p(q+r)
Z
∞
Z
∞
α
α
The inequalities (2.4)and (2.5) are equivalent.
Proof. Follows from Theorem 2.3 and Lemma 2.1, on choosing γ = 1, T = ∞,
2−λ
and λj (t) = (t − α) j . We omit the details.
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Corollary 2.5. Let f (t), g(t), h(t) ≥ 0, 2 < λ < 3, γ > 3µ, and − 31 < µ <
λ−1
. If
3
Z T
0<
(t − α)2−λ f 3 (t)dt < ∞,
α
Z T
0<
(t − α)2−λ g 3 (t)dt < ∞
α
and
T
Z
On Hardy-Hilbert’s Integral
Inequality
(t − α)2−λ h3 (t)dt < ∞,
0<
α
W.T. Sulaiman
then
Z
T
T
Z
(2.6)
α
α
Z
T
f (x)g(x)h(z)
dxdydz
λ
α (x + y + z)
Z T
13
≤
φ(t, λ, µ, 3)(t − α)2−λ f 3 (t)dt
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α
Z
T
×
13
g (t)dt
II
I
2−λ 3
φ(t, λ, µ, 3)(t − α)
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α
Z
1
3
T
h (t)dt ,
2−λ 3
×
φ(t, λ, µ, 3)(t − α)
α
and
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Z
T
− 12
φ
(2.7)
α
(t, λ, µ, 3)(x − α)
λ
−1
2
Z
T
α
Z
T
α
g(y)h(z)
dydz
(x + y + z)λ
12
dx
J. Ineq. Pure and Appl. Math. 5(2) Art. 25, 2004
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Z
T
≤
13
g (t)dt
2−λ 3
φ(t, λ, µ, 3)(t − α)
α
Z
13
h (t)dt .
T
×
2−λ 3
φ(t, λ, µ, 3)(t − α)
α
The inequalities (2.6) and (2.7) are equivalent.
Proof. Follows from Theorem 2.3 and Lemma 2.1, by putting p = q = r = 3,
2−λ
1
and λj (t) = (t − α) 3 φ 3 (t, λ, µ, 3), j = p, q, r.
Note. In Corollary 2.5, we may take as a special case µ = 1 −
λ
3
to obtain
φ(t, λ, 1 − λ/3, 3)
γ 1 t−α
= B(2λ − 4, 4 − λ)B(λ − 2, λ − 2) 1 −
2 T −α
2γ Z 1 2λ−5
Z 1
t−α
u
uγ+λ−3
−
du
du.
λ
2λ−4
T −α
0 (1 + u)
0 (1 + u)
Corollary 2.6. Let f (t), g(t), h(t) ≥ 0, 2 < λ < 3, and − 13 < µ <
Z ∞
0<
(t − α)2−λ f 3 (t)dt < ∞,
Zα∞
0<
(t − α)2−λ g 3 (t)dt < ∞
α
λ−1
.
3
If
On Hardy-Hilbert’s Integral
Inequality
W.T. Sulaiman
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and
Z
∞
(t − α)2−λ h3 (t)dt < ∞,
0<
α
then
Z
(2.8)
∞
Z
∞
Z
∞
f (x)g(y)h(z)
dxdydz
(x + y + z)λ
α
α
α
13 Z
Z ∞
2−λ 3
(t − α) f (t)dt
≤K
α
∞
13
g (t)dt
2−λ 3
(t − α)
α
Z
∞
13
h (t)dt ,
2−λ 3
×
(t − α)
On Hardy-Hilbert’s Integral
Inequality
W.T. Sulaiman
α
Title Page
where
K = (B(λ − 3µ − 1, 3µ + 1)B(λ − 2, 1 − 3µ)
and
Z
∞
− 21
Z
∞
Z
∞
g(y)h(z)
dydz
(x + y + z)λ
(x − α)
(2.9)
α
α
≤ K 3/2
α
Z
Contents
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J
12
dx
12
∞
2−λ 3
(t − α) g (t)dt
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α
Z
∞
×
12
2−λ 3
(t − α) h (t)dt .
Page 18 of 20
α
J. Ineq. Pure and Appl. Math. 5(2) Art. 25, 2004
The inequalities (2.8) and (2.9) are equivalent.
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Proof. Follows from Corollary 2.4 and Lemma 2.1, by putting p = q = r = 3,
T = ∞.
On Hardy-Hilbert’s Integral
Inequality
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References
[1] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cambridge Univ., Cambridge, UK, (1952).
[2] B. YANG, On generalization of Hardy-Hilbert’s integral inequality, Acta.
Math. Sinica, 41 (1998), 839–844.
[3] B. YANG, On Hilbert’s integral inequality, J. Math. Anal. Appl., 220 (2000),
778–785.
On Hardy-Hilbert’s Integral
Inequality
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