ON THREE INEQUALITIES SIMILAR TO HARDY-HILBERT’S INTEGRAL INEQUALITY

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ON THREE INEQUALITIES SIMILAR
TO HARDY-HILBERT’S INTEGRAL INEQUALITY
W. T. SULAIMAN
Abstract. The following three inequalities, which are similar to Hardy-Hilbert integral inequality are proved.
Z∞Z∞ β/q α/p
x
y
F (x)G(y)
λ
p1−1/p q 1−1/q
dxdy ≤
max{xλ , y λ }
(α + 1)1/p (β + 1)1/q (p − 1)(q − 1)
0
0
0∞
11/p 0 ∞
11/q
Z
Z
@ g q (t)dtA
· @ f p (t)dtA
.
0
0
where
Zy
Zx
f (t)dt,
F (x) =
G(y) =
0
Z∞Z∞
0
g(t)dt.
0
x1/p y 1/q F (x)G(y)
pq
dxdy < B 1/p (p, p)B 1/q (q, q)
(x + y)4
(p − 1)(q − 1)
0
0∞
11/p 0 ∞
11/q
Z
Z
p
q
@
A
@
·
f (t)dt
g (t)dtA
.
0
0
Received June 13, 2006.
2000 Mathematics Subject Classification. Primary 26D15.
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Abstract.
Z∞Z∞Z∞ 1/p 1/q 1/r p
x y z
xyzF (x)G(y)H(z)
dxdydz
(x + y + z)8
0
0
0
11/r
11/q 0 ∞
11/p 0 ∞
0∞
Z
Z
Z
r/2
q/2
p/2
@ h (t)dtA
@ g (t)dtA
.
(t)dtA
< Kp Kq Kr @ f
0
0
0
where
Zz
s
h(t)dt,
H(z) =
Kp =
“p p”
p/2
B 1/p
,
B 1/p (p, p).
p/2 − 1
2 2
0
1.
Introduction
Let f, g ≥ 0 satisfy
Z∞
2
f (t)dt < ∞
0<
Z∞
and
0
0<
g 2 (t)dt < ∞,
0
then
Z∞ Z∞
(1)
0
0
∞
1/2
Z
Z∞
f (x)g(y)
dx dy < π  f 2 (t)dt g 2 (t)dt ,
x+y
0
0
where the constant factor π is the best possible (cf. Hardy et al. [3]). Inequality (1) is well known as Hilbert’s
integral inequality. This inequality had been extended by Hardy [1] as follows
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If p > 1,
1
p
+
1
q
= 1, f, g ≥ 0 satisfy
Z∞
0<
Z∞
f p (t)dt < ∞
and
0<
0
g q (t)dt < ∞,
0
then
Z∞ Z∞
(2)
0
π
f (x)g(y)
dx dy <
x+y
sin(π/p)
∞
1/p  ∞
1/q
Z
Z
 f p (t)dt  g q (t)dt ,
0
0
0
π
sin(π/p)
is the best possible. Inequality (2) is called Hardy-Hilbert’s integral inequality
where the constant factor
and is important in analysis and application (cf. Mitrinovic et al. [4]).
B. Yang gave the following extension of (2) as follows:
Theorem ([5]). If λ > 2 − min{p, q}, f, g ≥ 0, satisfy
Z∞
0<
t
Z∞
1−λ p
f (t)dt < ∞
and
0
0<
t1−λ g q (t)dt < ∞,
0
then
Z∞ Z∞
(3)
0
∞
1/p  ∞
1/q
Z
Z
f (x)g(y)
dx dy < kλ (p)  t1−λ f p (t)dt  t1−λ g q (t)dt ,
(x + y)λ
0
where the constant factor kλ (p) = B
0
p+λ−2 q+λ−2
, q
p
0
is the best possible, B is the beta function.
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The object of this paper is that to give some new inequalities similar to that of Hardy-Hilbert’s inequality.
We need the following result for our aim
Theorem A ([2]). Let f be a nonnegative integrable function. Define
Zx
F (x) = f (t)dt.
a
Then
Z∞ F (x)
x
p
dx <
p
p−1
p Z∞
f p (x)dx,
p > 1.
0
0
2.
New results
We state and prove the following:
Ry
Rx
Theorem 1. Let f, g ≥ 0, F (x) = 0 f (t)dt, G(y) = 0 g(t)dt, p > 1, p1 + 1q = 1, p = λ − α − 1 > 1,
q = λ − β − 1 > 1, α, β > −1 then
Z∞ Z∞ β/q α/p
x y F (x)G(y)
λ
p1−1/p q 1−1/q
dx
dy
≤
max{xλ , y λ }
(α + 1)1/p (β + 1)1/q (p − 1)(q − 1)
0
0
∞
1/p  ∞
1/q
Z
Z
·  f p (t)dt  g q (t)dt .
0
0
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Proof. We have
Z∞ Z∞ β/q α/p
x y F (x)G(y)
dx dy
max{xλ , y λ }
0
0
Z∞ Z∞
Z∞ Z∞
y α/q F (x)
·
xβ/q G(y)
dx dy
1/q
(max{xλ , y λ })
0 0
0 0
1/p  ∞ ∞
1/q
∞∞
Z Z
Z Z
β q
α p
x
G
(y)
y
F
(x)
dx dy  
dx dy 
≤
max{xλ , y λ }
max{xλ , y λ }
1/p
(max{xλ , y λ })
0 0
1/p
=M
N
0
1/q
0
.
Observe that
Z∞
M
=
0
=
F p (x)dx
Z∞
yα
dx dy =
max{xλ , y λ }
0
Z∞
p
F (x)
λ
dx =
xλ−α−1
(α + 1)p
0
<
 x

Z∞ α
Z α
y
y
F p (x)dx 
dy +
dy 
xλ
yλ
0
λ
(α + 1)(λ − α − 1)
pp−1
λ
(α + 1) (p − 1)p
Z∞
Z∞
x
0
Z∞ F (x)
x
p
dx
0
f p (x)dx,
0
in view of Theorem A.
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Similarly, It can be shown that
λ
q q−1
N<
(β + 1) (q − 1)q
Z∞
g q (x)dx.
0
Combining these inequalities to obtain
Z∞ Z∞
0
λ
p1−1/p q 1−1/q
xβ/q y α/p F (x)G(y)
dx dy ≤
λ
λ
1/p
1/q
max{x , y }
(α + 1) (β + 1) (p − 1)(q − 1)
0
1/q
1/p  ∞
∞
Z
Z
·  f p (t)dt  g q (t)dt .
0
0
Theorem 2. Let f, g ≥ 0, F (x) =
Z∞ Z∞
0
Rx
0
f (t)dt, G(y) =
Ry
0
g(t)dt, p > 1,
1
p
+
1
q
= 1, then, we have
x1/p y 1/q F (x)G(y)
pq
dxdy < B 1/p (p, p)B 1/q (q, q)
(x + y)4
(p − 1)(q − 1)
0

Z∞
· 
0
1/p  ∞
1/q
Z
f p (t)dt  g q (t)dt .
0
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Proof.
Z∞ Z∞
0
x1/p y 1/q F (x)G(y)
dxdy
(x + y)4
0
Z∞ Z∞
y 1/q F (x) x1/p G(y)
·
dx dy
(x + y)2 (x + y)2
=
0
0
∞∞
1/p  ∞ ∞
1/q
Z Z p/q p
Z Z q/p q
y
F
(x)
x
G
(y)
≤ 
dx dy  
dx dy 
(x + y)2p
(x + y)2q
= P
0 0
1/p 1/q
Q
0
0
.
Now, we consider
Z∞ P =
F (x)
x
Z∞
p
dx
0
0
< B(p, p)
p
p−1
y p−1 1
x
x
2p dy
1 + xy
p Z∞
Z∞ =
0
F (x)
x
Z∞
p
dx
up−1
du
(1 + u)2p
0
f p (x)dx,
0
by Theorem A.
Similarly,
Q < B(q, q)
q
q−1
q R∞
g q (y)dy.
0
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Therefore, we have
Z∞ Z∞
0
x1/p y 1/q F (x)G(y)
pq
dxdy < B 1/p (p, p)B 1/q (q, q)
4
(x + y)
(p − 1)(q − 1)
0
∞
1/p  ∞
1/q
Z
Z
·  f p (t)dt  g q (t)dt .
0
0
This completes the proof of the theorem.
Theorem 3. Let f, g, h ≥ 0, p, q, r > 2
Z x
F (x) =
f (t)dt,
1
p
+
1
q
+
1
r
= 1,
Z y
G(y) =
g(t)dt,
0
Z
H(z) =
1
z
h(t)dt.
0
Then
Z∞ Z∞ Z∞ 1/p 1/q 1/r p
x y z
xyzF (x)G(y)H(z)
dxdydz
(x + y + z)8
0
0
0

Z∞
< K p Kq Kr 
where
1/p 
f p/2 (x)dx
0

1/q  ∞
1/r
Z
g q/2 (y)dy   hr/2 (z)dz  .
0
s
Kp =
Z∞
0
p p
p/2
B 1/p
,
B 1/p (p, p).
p/2 − 1
2 2
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Proof.
Z∞ Z∞ Z∞
0
0
p
x1/p y 1/q z 1/r xyzF (x)G(y)H(z)
dxdydz
(x + y + z)8
0
Z∞ Z∞ Z∞
=
0
0
0
p
p
F (x) z (1/2−1/q) x(1/p+1/r) G(y)
·
(x + y + z)2

p
x(1/2−1/r) y (1/p+1/q) H(z) 
·
dx dy dz
(x + y + z)2
y (1/2−1/p) z (1/q+1/r)
(x + y + z)2
∞∞∞
1/p
Z Z Z (p/2−1) p−1 p/2
y
z
F
(x)
≤ 
dx dy dz 
(x + y + z)2p
0
0
0
∞∞∞
1/q
Z Z Z (q/2−1) q−1 q/2
z
x G (y)
·
dx dy dz 
(x + y + z)2q
0
0
0
∞∞∞
1/r
Z Z Z (r/2−1) r−1 r/2
x
y
H
(z)
·
dx dy dz 
(x + y + z)2r
0
0
0
= A1/p B 1/q C 1/r .
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Observe that
Z∞
A=
F
p/2
(x)
xp/2
Z∞
dx
0
0
Z∞ =
y p/2−1 1
x
px
1 + xy
F (x)
x
Z∞
p/2
dx
0
Z∞
dy
z
x+y
0
up/2−1
du
(1 + u)p
0
Z∞
p−1
1+
z
x+y
1
x+y
2p
dz
v p−1
dv
(1 + v)2p
0
p/2
Z∞ Z∞
p p
F (x)
p
=B
,
B(p, p)
dx < Kp f p/2 (x)dx,
2 2
x
0
0
in view of Theorem A.
Similarly,
B<
Kqq
Z∞
g
0
The proof is complete.
p/2
(y)dy,
C<
Krr
Z∞
hr/2 (z)dz.
0
1. Hardy G. H., Note on a theorem of Hilbert concerning series of positive terms, Proc. Math. Soc. 23(2) (1925), Records of Proc.
XLV-XLVI.
2.
, Note on a theorem of Hilbert, Math. Z., 6 (1920) 314–317.
3. Hardy G. H., Littlewood J. E. and Polya G., Inequalities, Cambridge University Press, Cambridge, 1952.
4. Mitrinovic D. S., Pecaric J. E. and Fink A. M., Inequalities involving functions and their integrals and derivatives, Kluwer
Academic Publishers, Boston, 1991.
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5. Yang B., On Hardy-Hilbert’s integral inequality, J. Math. Anal. Appl. 261 (2001) 295–306.
W. T. Sulaiman, Department of Mathematics, College of Computer Sciences & Mathematics, University of Mosul, Mosul, Iraq,
e-mail: waadsulaiman@hotmail.com
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