A NEW EXTENSION OF HILBERT’S INEQUALITY FOR MULTIFUNCTIONS
WITH BEST CONSTANT FACTORS
H. A. AGWO
Abstract. The aim of this paper is to establish a new extension of Hilbert’s inequality and HardyHilbert’s inequality for multifunctions with best constant factors. Also, we present some applications
for Hilbert’s inequality which give new integral inequalities.
1. Introduction
Hilbert’s inequality has a great interest in analysis and its applications (see [10], [11]). The original
Hilbert’s inequality can be stated asR follows
R∞
∞
If f (x), g(x) ≥ 0, such that 0 < 0 f 2 (x)dx < ∞ and 0 < 0 g 2 (x)dx < ∞, then (see [6])
Z∞ Z∞
(1)
JJ J
I II
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0
0
∞
Z
f (x) g(y)
dx dy < π

x+y
0
f 2 (x)dx
Z∞
0
where the constant factor π is the best possible. This inequality was extended by Hardy-Riesz as
(see [5]):
Full Screen
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 21

g 2 (x)dx
,

Received May 25, 2008.
2000 Mathematics Subject Classification. Primary 26D15, 49C99.
Key words and phrases. Hilbert’s inequality; integral inequalities.
If p > 1,
then
1
p
+
1
q
Z
∞
Z
= 1, f (x), g(x) ≥ 0, such that 0 <
(2)
0
0
∞
f (x) g(y)
π
dx dy <
x+y
sin( πp )
where the constant factor
π
sin( π
p)
Z
R∞
0
f p (x)dx < ∞ and 0 <
∞
p
p1 Z
∞
f (x)dx
0
q
g (x)dx
R∞
0
g q (x)dx < ∞,
q1
,
0
is the best possible.
Hardy-Hilbert’s integral inequality is important in analysis and its applications (see[10], [11]).
In recent years, the various improvements and extensions on the inequality (1) and (2) appeared
in some papers (such as [1]–[4], [7], [9], [12]–[14]) and bibliography therein. They focalize on
changing the denominator of the function of the left-hand side of (2). Such as the denominator
(x + y) is replaced by (Ax + By)λ in paper [13], the denominator (x + y) is replaced by (xt + y t )
(t is a parameter which is independent of x and y) in paper [7]. Generally, the denominator (x + y)
is replaced by (xu(x) + yv(y))λ in paper [9].
The main objective of this paper is to build some new Hilbert-type integral inequalities with
best constant factors which are extensions of above results for multi-functions f, g and h. Moreover
the denominator is (m(x) + n(y) + r(z)), where m, n and r are arbitrary functions.
JJ J
2. Main results
I II
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We need the formula of the β function as (see [8]):
Z∞
Full Screen
(3)
β(u, v) =
tu−1
dt = β(v, u)
(1 + t)u+v
0
Close
Quit
Before stating our results we need the following lemmas.
u, v > 0.
Lemma 2.1. Let f (x, y, z) ∈ Lp[0,∞]×[0,∞]×[0,∞] and g(x, y, z) ∈ Lq[0,∞]×[0,∞]×[0,∞] , where
1. Then
1
p
+
1
q
=
Z∞ Z∞ Z∞
|f (x, y, z) g(x, y, z)| dx dy dz
(4)
0
0
0

Z∞ Z∞ Z∞
≤
0
0
 p1  ∞ ∞ ∞
 q1
Z Z Z
p
q
|f (x, y, z)| dx dy dz  
|g(x, y, z)| dx dy dz  .
0
0
0
0
Lemma 2.2. Let f (x, y, z) ∈ Lp[0,∞]×[0,∞]×[0,∞] , g(x, y, z) ∈ Lq[0,∞]×[0,∞]×[0,∞] and h(x, y, z) ∈
1 1 1
Lk[0,∞]×[0,∞]×[0,∞] where
+ + = 1. Then
p q
k
Z∞ Z∞ Z∞
|f (x, y, z) g(x, y, z)h(x, y, z)| dx dy dz
0
JJ J
I II
(5)
Go back
0
0
 p1  ∞ ∞ ∞
∞∞∞
 q1
Z Z Z
Z Z Z
p
q
≤ 
|f (x, y, z)| dx dy dz  × 
|g(x, y, z)| dx dy dz 
0
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0
0
0
∞∞∞
 k1
Z Z Z
r
×
|h(x, y, z)| dx dy dz  .
0
0
0
0
0
1
1
1
1 1 1
, we have
+
= 1, p > 1, + + = 1 then for 0 < ε <
p1
q1
p q
k
qk
Z∞ p 1qk − pε −1
1
1
v 1
1
dv
=
β
ε → 0+ .
,
+ 0(1)
1
p
qk
q
qk
qk
1
1
(1 + v)
Lemma 2.3. If p1 > 1,
(6)
0
Proof. Since
∞
Z
ε
1
p1 qk − p1 −1
1
1
v
dv − β
,
1
p
qk
q
qk
qk
1
1
(1 + v)
0
∞
Z
ε
1
1
v p1 qk − p1 −1 − v p1 qk −1 dv
= 1
(1 + v) qk
0
1
1
ε
Z∞ v p11qk − pε1 −1 − v p11qk −1 Z1 v p1 qk − p1 −1 − v p1 qk −1 dv +
dv
≤
1
1
(1 + v) qk
(1 + v) qk
0
JJ J
I II
≤
0
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Z1 =
1
v
1
ε
p1 qk − p1
−1
−v
1
p1 qk −1
Z∞
dv +
1
1
1
1
ε
v p1 qk −1 − v p1 qk − p1 −1
dv
v qk
1
1
−1
1
−
+
+
→0
1
ε
1
1
1
ε
−
−
p1 qk p1
p1 qk
q1 qk
q1 qk q1
for
ε → 0.
1
1
1
1 1 1
, setting
+
= 1, p > 1, + + = 1 and 0 < ε <
p1
q1
p q
k
qk
1
qk
Z∞ Z∞ 1
ε
n(y)
J1 :=
(m(x)) p1 qk − p1 −1
m(x) + n(y)
Lemma 2.4. If p1 > 1,
1
1
1
ε
×(n(y))− p1 qk − q1 −1
dm(x) dn(y)
dx dy,
dx
dy
then we have
1
ε
(7)
1
1
β
,
+ o(1) − O(1)
p1 qk q1 qk
1
1
1
β
,
+ o(1) ,
≤ J1 ≤
ε
p1 qk q1 qk
ε → 0+ .
Proof. For fixed y, setting m(x) = n(y)v, then by (6), we obtain
J1
JJ J
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Close
Z∞
1
ε
dn(y)
(n(y))− p1 qk − q1 −1
dy
1
∞

1
qk
Z 1
ε
n(y)
dm(x) 
×
(m(x)) p1 qk − p1 −1
dx dy
m(x) + n(y)
dx
1


∞
∞
1
ε
Z
Z
dn(y) 
v p1 qk − p1 −1 

=
(n(y))−ε−1
dv 
1
 dy
dy 
qk
(1
+
v)
1
=
1
Quit
n(y)

∞
Z∞
Z p 1qk − pε −1
1
1
dn(y)
v

=
(n(y))−ε−1
dv  dy
1
dy
(1 + v) qk
1
0

 1
n(y)
1
ε
Z∞
Z
−
−1
dn(y) 
v p1 qk p1

− (n(y))−ε−1
dv  dy

1
dy
(1 + v) qk
1
≥
=
=
JJ J
I II
0
 1

Z∞
Z
1
ε
1
1
1
dn(y)
−
−1
 v p1 qk p1 dv  dy
β
,
+ 0(1) − (n(y))−ε−1
ε
p1 qk q1 qk
dy
1
0
1
1
1
1
1
β
,
+ 0(1) − 1
ε
ε
p1 qk q1 qk
ε
−
p1 qk p1
1
1
1
β
,
+ o(1) − O(1).
ε
p1 qk q1 qk
By the same way, we have
Z∞ Z∞ J1 ≤
1
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=
Close
The lemma is proved.
Quit
n(y)
m(x) + n(y)
1
qk
1
ε
(m(x)) p1 qk − p1 −1
0
ε
1
dm(x) dn(y)
dx dy,
× (n(y))− p1 qk − q1 −1
dx
dy
1
1
1
β
,
+ o(1) .
ε
p1 qk q1 qk
Theorem 2.5. Assume that m, n, r are increasing functions defined on [0, ∞[ such that m(0) =
n(0) = r(0) = 0, lim m(x) = lim n(x) = lim r(x) = ∞, and f , g, h satisfy
x→∞
(8)
x→∞
x→∞
−q1 ( kp + p1 )
Z∞
kq1
q1
1
1
dm(x)
p1 (1+ pq )
(m(x))
|f (x)| 2 dx < ∞
dx
0
(9)
1
−p
Z∞
q1
pp1
p1
1
dm(x)
−
(m(x)) q1 qk
|f (x)| 2 dx < ∞,
dx
0
(10)
−q1 ( kp + p1 )
Z∞
pq1
q1
1
1
dn(y)
p1 (1+ qk )
(n(y))
|g(y)| 2 dy < ∞,
dy
(11)
−p1
Z∞
qp1
p1
1
dn(y) q1
q1 − pk
(n(y))
|g(y)| 2 dy < ∞,
dy
(12)
−q1 ( kp + p1 )
Z∞
qq1
q1
1
1
dr(z)
p1 (1+ pk )
(r(z))
|h(z)| 2 dz < ∞,
dz
0
0
JJ J
0
I II
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(13)
1
−p
Z∞
q1
kp1
p1
1
dr(z)
−
(r(z)) q1 pk
|h(z)| 2 dz < ∞,
dz
0
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1 1 1
1
1
+ + = 1,
+
= 1, then
where
p q
k
p1
q1
! p1
|f (x)g(y)h(z)|
π
1
1
dx dy dz ≤
,
π β
m(x) + n(y) + r(z)
sin qk
q1 qk p1 qk
 pp1
∞
1
1
−p
Z
q1
pp
p1
1
1
dm(x)
×  (m(x)) q1 − qk
|f (x)| 2 dx
dx
0
∞
 pq1
1
−q1 ( kp + p1 )
Z
pq
q1
1
1
1
dn(y)
×  (n(y)) p1 (1+ qk )
|g(y)| 2 dy 
dy
0
∞
 qp1
1
1
! q1
−p
Z
q
qp
p
1
1
1
1
π
1
1
dn(y)
2
q1 − pk


(n(y))
×
,
×
|g(y)|
dy
π β
sin pk
q1 pk p1 pk
dy
0
 qq1
∞
1
−q1 ( kp + p1 )
Z
qq1
q1
1
1
dr(z)
2
p1 (1+ pk )


|h(z)|
(r(z))
dz
×
dz
0

 kp1
1
1
! k1 Z∞
−p
q1
kp1
p1
1
π
1
1
dr(z)
2
q1 − pk


×
,
(r(z))
|h(z)|
dz
π β
sin pq
q1 pq p1 pq
dz
0
∞
 kq1
1
−q1 ( kp + p1 )
Z
kq
q1
1
1
1
dm(x)
×  (m(x)) p1 (1+ pq )
|f (x)| 2 dx
,
dx
Z∞ Z∞ Z∞
0
(14)
JJ J
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0
0
0
where the constant factors
π
1
1
β
,
,
π
sin qk
q1 qk p1 qk
π
π β
sin pk
1
1
,
q1 pk p1 pk
,
π
π β
sin pq
1
1
,
q1 pq p1 pq
are best possible.
Proof. Since
Z∞ Z∞ Z∞
|f (x) g(y)h(z)|
dx dy dz
I=
m(x) + n(y) + r(z)
0
0
0
Z∞ Z∞ Z∞
1
1
1
0
0
0
(m(x) + n(y) + r(z)) p
1
(15) ×
×
I II
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×
Close
m(x) + n(y)
r(z)
1
r(z) + m(x)
n(y)
1
(m(x) + n(y) + r(z)) k
dn(y)
dy
k1 dm(x)
dx
−1
q
dx dy dz.
Applying Hölder’s inequality on (15) we get
1
Quit
(16)
1
1
I ≤ I1p I2q I3k
n(y)
m(x) + n(y)
1 pqk
n(y) + r(z)
m(x)
1
|f (x)| 2 |h(z)| 2
1 pqk
(m(x) + n(y) + r(z)) q
Full Screen
1
|g(y)| 2 |h(z)| 2
1
JJ J
|f (x)| 2 |g(y)| 2
=
1 pqk
r(z)
n(y) + r(z)
1 pqk
dr(z)
dz
1 pqk
m(x)
r(z) + m(x)
1
pqk
p1 dm(x)
dx
dn(y)
dy
q1 −1
k
dr(z)
dz
−1
p
where
Z∞ Z∞ Z∞
I1 =
0
(17)
0
p
Z∞ Z∞ Z∞
0
dn(y)
dy
I3 =
(19)
0
0
k
1
qk
n(y) + r(z)
m(x)
1 pk
r(z)
n(y) + r(z)
1
pk
dr(z)
dz
k
−q
p
dx dy dz,
r(z) + m(x)
n(y)
1 pq
m(x)
r(z) + m(x)
1
pq
0
Go back
dn(y)
dy
dm(x)
dx
−k
q
dx dy dz.
Consider the weight coefficient
Close
Z
(20)
Quit
q
|f (x)| 2 |h(z)| 2
m(x) + n(y) + r(z)
×
Full Screen
n(y)
m(x) + n(y)
0
Z∞ Z∞ Z∞
I II
1 qk
dx dy dz,
q
dm(x)
×
dx
JJ J
m(x) + n(y)
r(z)
−p
k
|g(y)| 2 |h(z)| 2
m(x) + n(y) + r(z)
I2 =
0
0
dr(z)
×
dz
(18)
p
|f (x)| 2 |g(y)| 2
m(x) + n(y) + r(z)
w1 (x, y) =
0
∞
1
m(x) + n(y) + r(z)
m(x) + n(y)
r(z)
1
qk
dr(z)
dz.
dz
Let v =
(21)
r(z)
in (20) then we obtain
m(x) + n(y)
w1 (x, y) =
π
sin
π .
qk
Similarly,
Z∞
(22)
w2 (y, z) =
1
m(x) + n(y) + r(z)
1
m(x) + n(y) + r(z)
1
pk
n(y) + r(z)
m(x)
dx =
π
sin
0
π
pk
and
Z∞
(23)
w3 (z, x) =
r(z) + m(x)
n(y)
1
pq
dx =
sin
0
JJ J
I II
0
(24)
×
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π
π
sin pk

Close
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π .
pq
Combining (21), (22), (23) and (16) we get
 p1

1
qk
−p
Z∞ Z∞ k
p
p
n(y)
dn(y)
π
|f (x)| 2 |g(y)| 2
dx dy 
I≤ 
π
m(x) + n(y)
dy
sin qk

Go back
π
×
π
π
sin pq
0
Z∞ Z∞ 0
q
2
|g(y)| |h(z)|
q
2
 q1
−q
p
dr(z)
dx dy 
dz
0
Z∞ Z∞ 0
r(z)
n(y) + r(z)
1
pk
0
m(x)
r(z) + m(x)
1
pq
|f (x)|
k
2
|h(z)|
k
2
 k1
−k
q
dm(x)
dx dy .
dx
Applying Hölder’s inequality with p1 > 1, p11 +
(24),we have
Z∞ Z∞ 0
n(y)
m(x) + n(y)
1
qk
p
2
|f (x)| |g(y)|
dn(y)
dy
= 1 on the first integral on the right side in
−p
k
dx dy
0
p
Z∞ Z∞ |f (x)| 2
=
0
0
×
(25)
1
− p1
1 q1 qk
1
p
1
1
m(x) q1 − p1 dn(y) p11 dm(x) −1
q1
1
dy
dx
(m(x) + n(y)) p1 qk
n(y)
m(x)
p
1
1

1
|g(y)| 2 (n(y)) qk + p1 − q1
(m(x) + n(y))
Z∞ Z∞
≤
|f (x)|
1
q1 qk
pp1
2
1
0
JJ J
p
2
1
q1
(m(x) + n(y)) qk
0
m(x)
n(y)
n(y)
m(x)
1
− q1
1 q1 qk
1
p
q 1qk −1
1
(m(x))
p1
q1
dn(y)
dy
−1
1 −p
k −p
dm(x)
dx
1
1
−p
q
0
0
Go back
Let
Full Screen
Z∞
(26)
w4 (x) =
1
0
Quit
q1
1
dydx
 p1
1
dn(y)

dydx
dy
∞∞
 q1
pq1
q1
q1
1
p 1qk −1 −q1 ( kp + p1 )
Z Z
2
qk + p1 −1
1
1
|g(y)|
m(x)
(n(y))
dn(y)
dm(x)


×
dydx
.
1
n(y)
dy
dx
(m(x) + n(y)) qk
I II
Close
1
dm(x)
dx
1+
n(y)
m(x)
1
qk
n(y)
m(x)
q 1qk −1
1
dn(y)
dy
dy
Putting v1 =
n(y)
in (26) we have
m(x)
(27)
w4 (x) = m(x) β
Similarly
Z∞
w5 (y) =
1
1
1
,
−
q1 qk qk q1 qk
1
1
qk
= m(x) β
m(x)
n(y)
p 1qk −1
1
m(x)
1+
n(y)
1
1
= n(y) β
,
.
q1 qk p1 qk
1
1
,
q1 qk p1 qk
.
dm(x)
dx
dx
0
(28)
From (27), (28) and (25) we get
Z∞ Z∞ JJ J
I II
Go back
0
n(y)
m(x) + y
≤β
1
1
,
q1 qk p1 qk
Quit
|g(y)|
p
2
dn(y)
dy
−p
k
dx dy
 p1
1
1
−p
Z∞
q
pp
p1
1
1
1
2
 (m(x) q1 − qk dm(x)

|f (x)|
dx
dx

0
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Close
|f (x)|
p
2
0
(29)
1
qk
∞
 q1
1
−q1 ( kp + p1 )
Z
pq
q1
1
1
1
dn(y)
|g(y)| 2 dy  .
×  (n(y)) p1 (1+ qk )
dy
0
For the second and third integrals on the right side of (24), following the same steps used for
obtaining (29), we obtain
Z∞ Z∞ 0
r(z)
n(y) + r(z)
|g(y)|
q
2
|h(z)|
q
2
dr(z)
dz
−q
p
dydz
0
≤β
(30)
1
pk
1
1
,
q1 qk p1 qk
∞
 p1
1
1
−p
Z
q1
qp1
p1
1
dn(y)
 (n(y)) q1 − pk
|g(y)| 2 dy 
dy
0
 q1
∞
1
−q1 ( kp + p1 )
Z
qq1
q1
1
1
dr(z)
2
p1 (1+ pk )


(r(z))
|h(z)|
dz
×
,
dz
0
and
Z∞ Z∞ JJ J
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0
m(x)
r(z) + m(x)
≤β
1
1
,
q1 pq p1 pq
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|f (x)|
k
2
dm(x)
dx
−k
q
dzdx
 p1
1
1
−p
Z∞
q
kp
p1
1
1
1
2
 (r(z)) q1 − pk dr(z)

|h(z)|
dz
dz

0
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|h(z)|
k
2
0
(31)
1
pq
∞
 q1
1
−q1 ( kp + p1 )
Z
kq
q1
1
1
1
dm(x)
|f (x)| 2 dx .
×  (m(x)) p1 (1+ pq )
dx
0
Let
γ1 = β
1
1
,
q1 qk p1 qk
,
γ2 = β
1
1
,
q1 qk p1 qk
,
γ3 = β
1
1
,
q1 pq p1 pq
.
To prove the constant factors γ1 , γ2 and γ3 are best possible. Assume that the constant factor
γ1 is not the best possible, then there exists a positive constant λ1 with λ1 < γ1 , such that (29)
is still valid if we replace γ1 by λ1 . Without loss of generality, we assume that m(1) = n(1) = 1.
For 0 < ε < 1, setting fε and gε as fε (x) = gε (x) = 0, for x ∈ (0, 1), and for x ∈ [1, ∞)
“
”
2 p1 +1
dm(x) pp1 q1
1
|fε (x)| = (m(x))
dx
“ “
” ”
pq2 q1 kp + p1 +1
”
“
q1
2
1
1
1
dn(x)
1+
−ε−1
−
|gε (x)| = (n(x)) pq1 p1 ( qk )
,
dx
2
pp1
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p
1
−ε−1
− q1 + qk
”
then we obtain
∞
−p1
Z
p1
pp1
p1
1
dm(x) q1
1
|f (x)| 2 dx
λ1  (m(x)) q1 − qk
dx
0
∞
−q1 ( kp + p1 )
Z
q1
pq1
q1
1
1
dn(y)
1
p1 (1+ qk )

×
(n(y))
|g(y)| 2 dy
dy
0
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“
∞
 p1  ∞
 q1
1
1
Z
Z
dm(x)  
dn(y) 
λ1
= λ1  (m(x))−ε−1
dx
(n(y))−ε−1
dy
=
.
dx
dy
ε
1
1
But we have
Z∞ Z∞ 0
n(y)
m(x) + y
1
qk
p
p
|f (x)| 2 |g(y)| 2
dn(y)
dy
−p
k
dx dy
0
Z∞ Z∞ =
1
≥
1
1
qk
1
ε
1
ε
n(y)
(m(x)) p1 qk − p1 −1 (n(y))− p1 qk − q1 −1
m(x) + y
dm(x)
dn(y)
dx dy = J1
dx
dy
1
(γ1 + o(1)) − O(1).
ε
Hence we find
(32)
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1
λ1
(γ1 + o(1)) − O(1) <
ε
ε
or
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(33)
γ1 + o(1) − εO(1) < λ1 .
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For ε → 0+ , it follows that γ1 ≤ λ1 . This contradicts the fact that λ1 < γ1 . Hence the constant
factor γ1 in (29) is the best possible. Similarly γ2 and γ3 are the best possible. Substituting from
(29), (30), (31) in (24), the result of the theorem follows.
Remark 1. Putting p = q = k =
1
3
in (14), we get a new inequality in the form
Z∞ Z∞ Z∞
0
(34)
0
|f (x) g(y)h(z)|
dx dy dz
m(x) + n(y) + r(z)
0

 3p1
1
1
Z∞
−p
q
p
3p
1
1
1
1
π
1
1
2
 (m(x)) q1 − 9 dm(x)

≤
β
,
dx
|f
(x)|
sin π9
9q1 9p1
dx
0
 3q1
∞
1
−q1 (1+ p1 )
Z
3q
10q1
1
1
dn(y)
2
9p1


(n(y))
|g(y)|
dy
×
dy
0
 3p1
∞
1
1
−p
Z
q1
3p1
p1
1
dn(y)
2
q1 − 9


(n(y))
|g(y)|
dy
×
dy
0
 3q1
1
−q1 (1+ p1 )
Z∞
3q
10q1
1
1
dr(z)
2
9p1


×
(r(z))
|h(z)|
dz
dz

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0
 3p1
1
1
−p
Z∞
q
kp
p1
1
1
1
dr(z)
2
q1 − 9


×
(r(z))
|h(z)|
dz
dz

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0
 3q1
1
−q1 (1+ p1 )
Z∞
3q
10q1
1
1
dm(x)
.
×  (m(x)) 9p1
|f (x)| 2 dx
dx

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0
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Remark 2. Putting p1 = q1 = 2 in (14) we obtain
Z∞ Z∞ Z∞
|f (x) g(y)h(z)|
dx dy dz
m(x) + n(y) + r(z)
0 0 0
1
 2p

! p1 Z∞
−1
1
1
1
dm(x)
π
q
 (m(x))1− qk
,
|f (x)| dx
≤
π β
sin qk
2qk 2qk
dx
0
(35)
×
1
∞
 2p
−2( kp + 12 )
Z
1
dn(y)
p
|g(y)| dy 
×  (n(y))(1+ qk )
dy
0
 2q1

1
! q1 Z∞
−p
q1
1
1
1
π
dn(y)
p
 (n(y))1− pk
,
|g(y)| dy 
π β
sin pk
2pk 2pk
dy
0
∞
 2q1
−2( kp + 12 )
Z
1
dr(z)
q
×  (r(z))(1+ pk )
|h(z)| dz 
dz
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×
π
π β
sin pq
1
1
,
2pq 2pq
! k1
0
1
 2k
−2( kp + 12 )
Z∞
1
dm(x)
k
|f (x)| dx ,
×  (m(x))(1+ pq )
dx

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1
∞
 2k
−1
Z
1
k
 (r(z))1− pk dr(z)
|h(z)| dz 
dz
0
which is a new inequality.
Remark 3. Let m(x) = x, n(y) = y and r(z) = z in (34) and (35) we get respectively
Z∞ Z∞ Z∞
|f (x) g(y)h(z)|
dx dy dz
x+y+z
0 0 0
 ∞
 ∞

Z p
Z p
3p1
3p1
1 −1
1 −1
π
1
1 
≤
β
,
x q1 9 |f (x)| 2 dx  y q1 9 |g(y)| 2 dy 
sin π9
9q1 9p1
0

Z∞
×
0
 3p1
1
p1
1
z q1 − 9 |h(z)|
3p1
2
dz 
0
 ∞
 ∞

Z 10q
Z 10q
3q1
3q1
1
1
×  x 9p1 |f (x)| 2 dx  y 9p1 |g(y)| 2 dy 
0

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1
z
0
0
 3q1
Z∞
10q1
9p1
|h(z)|
3q1
2
dz 
.
and
Z∞ Z∞ Z∞
|f (x) g(y)h(z)|
dx dy dz
x+y+z
0
0
0
≤
×
π
π β
sin qk
1

 ∞
 2p
! p1 Z∞
Z
1
1
1
1
 x1− qk |f (x)|p dx y (1+ qk ) |g(y)|p dy 
,
2qk 2qk
0
0
×
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0

 ∞
 2q1
! q1 Z∞
Z
1
1
1
1
π
 y 1− pk |g(y)|q dy  z 1+ pk |h(z)|q dz 
,
π β
sin pk
2pk 2pk
0
1

 ∞
 2k
! k1 Z∞
Z
1
1
1
π
1
 z 1− pk |h(z)|k dz  x1+ pq |f (x)|k dx .
,
π β
sin pq
2pq 2pq
0
0
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1. Yang B., On Hardy-Hilbert’s integral inequality, J. Math. Anal. Appl., 261 (2001), 295–306.
2.
, On the extended Hilbert’s integral inequality, JIPAM, 5(4) (2004), Article 96.
3. Bicheng Yang, Bnaetić I., Krnic M. and Pečarić J., Generalization of Hilbert and Hardy- -Hilbert integral
inequalities, Math. Inequal. Appl., 8(2) (2005), 259–272.
4. Gao Mingzhe and Gao Xuemel, On the generalized Hardy-Hilbert inequality and its applications, Math. Inequal.
Appl., 7(1) (2004), 19–26.
5. Hardy G., Note on a theorem of Hilbert concerning series of positive terms, Proc. London Math. Soc., Records
of Proc. XLV–XLIV, 23 (2) (1925).
6. Hardy G., Littlewood J. E. and Pólya G., Inequalities, Cambridge University Press, Cambridge, 1952.
7. Kuang Jichang, On new extensions of Hilbert’s integral inequality, J. Math. Anal. Appl., 235(2) (1999), 608–
614.
8. Larry C. Andrews, Special Functions for Engineers and Applied Mathematicians, Macmillan Publishing Company, New York, 1985.
9. Krnic M., Gao Mingzhe, Pečarić J. E and Gao Xuemel, On the best constant in Hilbert’s inequality, Math.
Inequal. Appl. 8(2) (2005), 317–329.
10. Mitrinović D. S., Pečarić J. E., and Fink A. M., Inequalities Involving Functions and Their Integral and
Derivatives, Boston, Kluwer Academic, 1991.
11.
, Classical and New Inequalities in Analysis, Mathematics and Its Applications (Eastern European
Series), Kluwer Academic Publishers, Dordrecht, Boston and London, 1993.
12. Pečarić J. E., Generalization of inequalities of Hardy-Hilbert type, Math. Inequal. Appl. 7(2) (2004), 217–225.
13. Yang B. and Debnath L., On the extended Hardy-Hilbert’sinequality, J. Math. Anal. Appl., 272(1) (2002),
187–199.
14. Zitian Xie and Yang B., A New Hilbert-type integral inequality with some parameters and its reverse, Kyungpook Math. J. 48 (2008), 93–100.
H. A. Agwo, Department of Mathematics, Faculty of Education, Ain Shams University, Roxy, Cairo, Egypt, e-mail:
Hassanagwa@yahoo.com
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