Volume 9 (2008), Issue 2, Article 49, 9 pp.
A REVERSE HARDY-HILBERT-TYPE INTEGRAL INEQUALITY
GAOWEN XI
D EPARTMENT OF M ATHEMATICS
L UOYANG T EACHERS ’ C OLLEGE
L UOYANG 471022, P. R. C HINA
xigaowen@163.com
Received 07 December, 2007; accepted 25 May, 2008
Communicated by B. Opić
A BSTRACT. By estimating a weight function, a reverse Hardy-Hilbert-type integral inequality
with a best constant factor is obtained. As an application, some equivalent forms and some
particular results have been established.
Key words and phrases: Hardy-Hilbert-type integral inequality; weight function; β function; Hölder’s inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
R∞
R∞
Let p > 1, p1 + 1q = 1, f (x) ≥ 0, g(x) ≥ 0, and 0 < 0 f p (x)dx < ∞, 0 < 0 g q (x)dx <
∞. Then the Hardy-Hilbert’s integral inequality is as follows:
Z ∞
p1 Z ∞
1q
Z ∞Z ∞
π
f (x)g(y)
p
q
(1.1)
dxdy <
g (x)dx ,
f (x)dx
x+y
sin( πp ) 0
0
0
0
where the constant factor sin(π π ) is the best possible (see [1]). For (1.1), Yang et al. [2], [3], [4],
p
[8] and [9] gave some strengthened versions and extensions. In particular, in [7], Yang obtained
Z ∞Z ∞
f (x)g(y)
p+λ−2 q+λ−2
(1.2)
dxdy < B
,
(x + y − 2α)λ
p
q
α
α
Z ∞
p1 Z ∞
1q
1−λ p
1−λ q
(x − α) g (x)dx ,
×
(x − α) f (x)dx
α
α
where α ∈ R, λ > 2 − min{p, q},
The author wishes to thank the referee for several helpful comments. This research is supported by the Natural Science Foundation of
Henan Province (Grant No. 2008B110011).
360-07
2
G AOWEN X I
Recently, in [5], Xi gave a reverse Hardy-Hilbert-type inequality
(1.3)
∞ X
∞
X
am b n
(m + n + 1)λ
n=0 m=0
) 1q
(∞
) p1 (X
∞
2−λ
X (n + 1)2−λ (λ − 1)2
(n
+
1)
2
,
bqn
>
1−
apn
2
2n + 3 − λ
λ − 1 n=0 2n + 3 − λ
4(n + 1)
n=0
P
(n+1)2−λ apn
where 0 < p < 1, p1 + 1q = 1, 1.5 ≤ λ < 3 and an ≥ 0, bn > 0, such that 0 < ∞
n=0 2n+3−λ <
P
(n+1)2−λ bqn
∞, 0 < ∞
n=0 2n+3−λ < ∞.
In this paper, by estimating the weight function, a reverse Hardy-Hilbert-type integral inequality with a best constant factor is obtained. As an application, some equivalent forms and
some particular results have been established.
2. S OME L EMMAS
Lemma 2.1.
Z
∞
B(p, q) = B(q, p) =
(2.1)
0
1
up−1 du
(1 + u)p+q
(p, q > 0),
where B(p, q) is the β function .
Proof. See [6].
Lemma 2.2. Let p < 0 or 0 < p < 1, and p1 + 1q = 1, f (x), g(x) ≥ 0, f ∈ Lp (E), g ∈ Lq (E),
x ∈ Rk and the set E is Borel measurable in Rk , where k is a positive integer. Then
Z
Z
f (x)g(x)dx ≥
(2.2)
p1 Z
p
f (x)dx
E
E
1q
q
g (x)dx
,
E
where the equality holds if and only if there exist non-negative real numbers a and b, such that
they are not all zero and af p (x) = bg q (x), a. e. in E.
Proof. See [1, Section 6.9, Theorem 189].
Lemma 2.3. Let p < 0 or 0 < p < 1, and p1 + 1q = 1, 2 − max{p, q} < λ < 2 − min{p, q}, y >
α, The weight function ωλ (y, p) is defined by
Z
(2.3)
∞
ωλ (y, p) =
α
Then we have
(2.4)
p+λ−2
p
> 0,
1
(x + y − 2α)λ
q+λ−2
q
y−α
x−α
2−λ
p
dx,
y ∈ (0, ∞).
> 0, and
1−λ
ωλ (y, p) = (y − α)
B
p+λ−2 q+λ−2
,
p
q
Proof. If p < 0, then 0 < q < 1. Since 1 < λ < 2 − p, p+λ−2
> 0 and
p
q+λ−2
p+λ−2
we have q < 0, 1 < λ < 2 − q, p > 0, q > 0.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 49, 9 pp.
.
q+λ−2
q
> 0. If 0 < p < 1,
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R EVERSE H ARDY-H ILBERT-T YPE I NTEGRAL I NEQUALITY
Setting t =
Z
∞
α
x−α
,
y−α
by
p+λ−2
p
1
(x + y − 2α)λ
+
q+λ−2
q
y−α
x−α
3
= λ, we have
2−λ
p
Z
∞
2−λ
1
t− p dt
λ
(1 + t)
Z0 ∞
p+λ−2
1
= (y − α)1−λ
t p −1 dt
λ
(1 + t)
0
p+λ−2 q+λ−2
1−λ
= (y − α) B
,
.
p
q
1−λ
dx = (y − α)
The lemma is thus proved.
Lemma 2.4. Let p < 0,
Z
(2.5)
1
p
∞
+
1
q
Z
∞
= 1, 2 − q < λ < 2 − p, 0 < ε <
q+λ−2
,
2
then we have
λ−2−ε
λ−2−ε
1
(x − α) p (y − α) q dxdy
λ
α+1 α+1 (x + y − 2α)
1
q+λ−2 ε p+λ−2 ε
= B
− ,
+
− Oε (1).
ε
q
q
p
q
I=
Proof. Setting x − α = s, y − α = t, then
Z
∞
Z
∞
λ−2−ε
λ−2−ε
1
p
q
(x
−
α)
(y
−
α)
dxdy
λ
α+1 α+1 (x + y − 2α)
Z ∞Z ∞
λ−2−ε λ−2−ε
1
=
s p t q dsdt
λ
(s + t)
Z ∞
Z1 ∞ 1
λ−2−ε
λ−2−ε
1
p
q
=
s
t
dt ds.
(s + t)λ
1
1
I=
Let u = st , we have
#
q+λ−2−ε
1
I=
s−1−ε
u q −1 du ds
λ
1
(1
+
u)
1
s
Z ∞
Z ∞
q+λ−2−ε
1
−1
−1−ε
du ds
=
s
u q
(1 + u)λ
1
0
#
"Z 1
Z ∞
s
q+λ−2−ε
1
u q −1 du ds
−
s−1−ε
λ
0 (1 + u)
1
1
q+λ−2 ε p+λ−2 ε
= B
− ,
+
ε
q
q
p
q
#
"Z 1
Z ∞
s
q+λ−2−ε
1
u q −1 du ds.
−
s−1−ε
λ
(1
+
u)
0
1
Z
∞
"Z
∞
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 49, 9 pp.
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4
G AOWEN X I
By 1 < 2 − q < λ, 0 < ε < q+λ−2
. Hence
2
"Z 1
Z ∞
s
0<
s−1−ε
1
0
∞
Z
<
s
−1−ε
"Z
∞
<
s
−1
u
"Z
=
q+λ−2−ε
−1
q
du ds
#
du ds
1
s
u
q+λ−2
−1
q
#
du ds
0
1
1
s
#
0
1
Z
1
u
(1 + u)λ
q+λ−2−ε
−1
q
q
q+λ−2
2
.
The lemma is proved.
3. M AIN R ESULTS
Let p < 0 or 0 < p < 1, p1 + 1q = 1, 2 − max{p, q} < λ < 2 − min{p, q},
R∞
R∞
f (x), g(x) ≥ 0, and 0 < α (x − α)1−λ f p (x)dx < ∞, 0 < α (x − α)1−λ g q (x)dx < ∞,
α ∈ R. Then
Z ∞Z ∞
f (x)g(y)
p+λ−2 q+λ−2
(3.1)
dxdy > B
,
(x + y − 2α)λ
p
q
α
α
1q
Z ∞
p1 Z ∞
1−λ p
1−λ q
×
(x − α) f (x)dx
(x − α) g (x)dx ,
Theorem 3.1.
α
where the constant factor B
p+λ−2 q+λ−2
, q
p
α
is the best possible.
Proof. By (2.2), we have
2−λ #
Z ∞Z ∞
Z ∞Z ∞"
f (x)g(y)
f (x)
x − α pq
(3.2)
dxdy =
λ
(x + y − 2α)λ
α
α
α
α
(x + y − 2α) p y − α
"
2−λ #
x − α pq
g(y)
dxdy
×
λ
(x + y − 2α) q y − α
2−λ # p1
Z ∞Z ∞"
f p (x)
x−α q
≥
dx
(x + y − 2α)λ y − α
α
α
# 1q
2−λ
Z ∞Z ∞"
q
p
g (y)
y−α
×
dy .
λ
(x + y − 2α)
x−α
α
α
If (3.2) takes the form of equality, then by (2.2), there exist non-negative numbers a and b such
that they are not all zero and
2−λ
2−λ
f p (x)
x−α q
g q (y)
y−α p
=b
, a.e. in (α, ∞) × (α, ∞).
a
(x + y − 2α)λ y − α
(x + y − 2α)λ x − α
It follows that
a(x − α)2−λ f p (x) = b(y − α)2−λ g q (y) = c,
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 49, 9 pp.
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R EVERSE H ARDY-H ILBERT-T YPE I NTEGRAL I NEQUALITY
5
a. e. in (0, ∞) × (0, ∞). Without loss of generality, suppose a 6= 0. One has
c
(x − α)1−λ f p (x) = (x − α)−1 ,
a
R∞
a. e. in (0, ∞) × (0, ∞), which contradicts 0 < α (x − α)1−λ f P (x)dx < ∞. Therefore, by
(2.3), we have
Z
∞
Z
α
∞
α
f (x)g(y)
dxdy >
(x + y − 2α)λ
Z
∞
1−λ p
ωλ (x, q)(x − α)
p1
f (x)dx
α
Z
∞
1−λ q
×
ωλ (y, p)(y − α)
g (y)dy
1q
.
α
By (2.4), we obtain (3.1).
Without loss of generality, suppose p < 0, which implies that 0 < q < 1. For 0 < ε <
set
λ−2−ε
fε (x) = 0, x ∈ (α, α + 1);
fε (x) = (x − α) p , x ∈ [α + 1, ∞),
q+λ−2
2
λ−2−ε
gε (x) = 0, x ∈ (α, α + 1);
gε (x) = (x − α) q , x ∈ [α + 1, ∞).
If the constant factor B p+λ−2
, q+λ−2
in (3.1) is not the best possible, then, there exists a
q
p
q+λ−2
p+λ−2 q+λ−2
positive constant K > B p+λ−2
,
,
such
that
(3.1)
is
still
valid
if
B
,
is
p
q
p
q
replaced by K. By (2.5), we have
q+λ−2 ε p+λ−2 ε
B
− ,
+
− εOε (1)
q
q
p
q
Z ∞Z ∞
fε (x)gε (y)
=ε
dxdy
(x + y − 2α)λ
α
α
1q
Z ∞
p1 Z ∞
1−λ q
1−λ p
> εK
(x − α) fε (x)dx
(y − α) gε (y)dy
= K.
α+1
α+1
Letting ε −→ 0+ , we obtain B p+λ−2
, q+λ−2
≥ K. Hence the constant factor
p
q
B p+λ−2
, q+λ−2
in (3.1) is the best possible when 2 − q < λ < 2 − p, 0 < ε < q+λ−2
.
p
q
2
The theorem is proved.
In (3.1), when α = − 12 , we have:
Corollary 3.2. Let p < 0, or 0 < p < 1, p1 + 1q = 1, 2 − max{p, q} < λ < 2 − min{p, q},
1−λ p
1−λ q
R∞
R∞
f (x), g(x) ≥ 0, 0 < − 1 x + 12
f (x)dx < ∞, 0 < − 1 x + 12
g (x)dx < ∞. Then
2
∞Z
∞
2
p+λ−2 q+λ−2
(3.3)
,
p
q
− 12 − 12
# p1 "Z # 1q
1−λ
1−λ
∞
∞
1
1
×
x+
f p (x)dx
x+
g q (x)dx ,
1
1
2
2
−2
−2
where the constant factor B p+λ−2
, q+λ−2
is the best possible.
p
q
Z
f (x)g(y)
dxdy > B
(x + y + 1)λ
"Z In (3.1), when λ = 2, we have:
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 49, 9 pp.
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6
G AOWEN X I
R∞
Corollary 3.3. Let p < 0, or 0 < p < 1, p1 + 1q = 1, f (x), g(x) ≥ 0, 0 < α (x − α)−1 f p (x)dx
R∞
< ∞, 0 < α (x − α)−1 g q (x)dx < ∞, α ∈ R. Then
Z ∞ p
p1 Z ∞ q
1q
Z ∞Z ∞
f (x)
f (x)g(y)
g (x)
dx .
(3.4)
dxdy >
dx
(x + y − 2α)2
x−α
x−α
α
α
α
α
In (3.1), when α = − 12 , λ = 2, we have:
R∞
Corollary 3.4. Let p < 0, or 0 < p < 1, p1 + 1q = 1, f (x), g(x) ≥ 0, 0 < − 1 (x + 12 )−1 f p (x)dx
2
−1 q
R∞
< ∞, 0 < − 1 x + 12
g (x)dx < ∞. Then
2
"Z
# p1 "Z
# 1q
Z ∞Z ∞
∞ q
∞ p
f (x)g(y)
f (x)
g (x)
.
(3.5)
dxdy >
1 dx
1 dx
2
− 12 − 12 (x + y + 1)
− 12 x + 2
− 21 x + 2
4. A N E QUIVALENT F ORM
R∞
Theorem 4.1. Let p < 0, p1 + 1q = 1, 2−q < λ < 2−p, f (x) ≥ 0, and 0 < α (x − α)1−λ f p (x)dx
< ∞, α ∈ R. Then
Z ∞
p
Z ∞
f (x)
(p−1)(λ−1)
dx dy
(4.1)
(y − α)
(x + y − 2α)λ
α
α
p Z ∞
p+λ−2 q+λ−2
< B
,
(x − α)1−λ f p (x)dx.
p
q
α
where the constant factor B p+λ−2
, q+λ−2
is the best possible. Inequalities (4.1) and (3.1)
p
q
are equivalent as p < 0.
Proof. Let
(p−1)(λ−1)
Z
∞
g(y) = (y − α)
α
f (x)
dx
(x + y − 2α)λ
p−1
,
y ∈ (α, ∞).
Then by (3.1) and p < 0, we have
Z ∞
(4.2)
0<
(y − α)1−λ g q (y)dy
Z ∞
p
Zα∞
f (x)
(p−1)(λ−1)
=
(y − α)
dx dy
(x + y − 2α)λ
α
α
Z ∞Z ∞
f (x)g(y)
dxdy
=
(x + y − 2α)λ
α
α
Z ∞
p1
p+λ−2 q+λ−2
1−λ p
≥B
,
(x − α) f (x)dx
p
q
α
Z ∞
1q
1−λ q
×
(y − α) g (y)dy .
α
Since p < 0, we obtain that
Z ∞
(4.3)
(y − α)1−λ g q (y)dy
α
p Z ∞
p+λ−2 q+λ−2
≤ B
,
(x − α)1−λ f p (x)dx < ∞.
p
q
α
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R EVERSE H ARDY-H ILBERT-T YPE I NTEGRAL I NEQUALITY
7
By (4.3) and (4.2), we obtain (4.1).
Whereas, assume that (4.1) is true, by (2.2), we have
Z ∞Z ∞
f (x)g(y)
dxdy
(x + y − 2α)λ
α
α
h
Z ∞
Z ∞
i
1−λ
λ−1
f
(x)
q g(y) dy
dx
(y
−
α)
=
(y − α) q
(x + y − 2α)λ
α
α
1q
Z ∞
Z ∞
p p1 Z ∞
f (x)
(p−1)(λ−1)
1−λ q
(y − α) g (y)dy .
≥
(y − α)
dx dy
(x + y − 2α)λ
α
α
α
Since p < 0, by (4.1), then (3.1) is proved. Hence inequalities (4.1) and (3.1) are equivalent as
p < 0.
1−λ p
R∞
Corollary 4.2. Let p < 0, p1 + 1q = 1, 2−q < λ < 2−p, f (x) ≥ 0, 0 < − 1 x + 12
f (x)dx
2
< ∞. Then
Z
∞
(4.4)
− 12
1
y+
2
#p
f (x)
dx dy
λ
− 12 (x + y + 1)
1−λ
p Z ∞ p+λ−2 q+λ−2
1
x+
f p (x)dx,
< B
,
1
2
p
q
−2
(p−1)(λ−1) "Z
where the constant factor B
∞
p+λ−2 q+λ−2
, q
p
is the best possible.
R∞
Corollary 4.3. Let p < 0, p1 + 1q = 1, f (x) ≥ 0, 0 < α (x − α)−1 f p (x)dx < ∞, α ∈ R.
Then
Z ∞
p
Z ∞
Z ∞ p
f (x)
f (x)
(p−1)
(4.5)
(y − α)
dx dy <
dx.
2
(x + y − 2α)
x−α
α
α
α
Corollary 4.4. Let p < 0,
Z
∞
(4.6)
− 21
1
p
1
y+
2
+
1
q
= 1, f (x) ≥ 0, 0 <
(p−1) "Z
∞
− 12
R∞
− 12
f (x)
dx
(x + y + 1)2
x+
1 −1
2
#p
Z
f p (x)dx < ∞. Then
∞
dy <
− 12
f p (x)
dx.
x + 12
Theorem 4.5. Let 0 < p < 1, p1 + 1q = 1, 2 − p < λ < 2 − q, f (x) ≥ 0, and 0 <
R∞
(x − α)1−λ f p (x)dx < ∞, α ∈ R. Then
α
p
f (x)
(4.7)
(y − α)
dx dy
(x + y − 2α)λ
α
α
p Z ∞
p+λ−2 q+λ−2
> B
,
(x − α)1−λ f p (x)dx,
p
q
α
q+λ−2
,
is the best possible. Inequalities (4.7) and (3.1)
where the constant factor B p+λ−2
p
q
are equivalent as 0 < p < 1.
Z
∞
(p−1)(λ−1)
Z
∞
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 49, 9 pp.
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8
G AOWEN X I
Proof. Since 0 < p < 1, by (2.2) and (2.4),we have
2−λ #
p (Z ∞ "
Z ∞
f (x)
x − α pq
f (x)
(4.8)
dx =
λ
(x + y − 2α)λ
α
α
(x + y − 2α) p y − α
"
2−λ # )p
1
y − α pq
dx
×
λ
(x + y − 2α) q x − α
2−λ
Z ∞
f p (x)
x−α q
≥
dx
(x + y − 2α)λ y − α
α
"Z
2−λ #p−1
∞
y−α p
1
dx
×
(x + y − 2α)λ x − α
α
p−1
p+λ−2 q+λ−2
,
(y − α)(p−1)(1−λ)
= B
p
q
2−λ
Z ∞
f p (x)
x−α q
×
dx.
(x + y − 2α)λ y − α
α
If (4.8) takes the form of equality, then by (2.2), there exist non-negative numbers a and b, such
that they are not all zero and
f p (x)
a
(x + y − 2α)λ
x−α
y−α
2−λ
q
1
=b
λ
(x + y − 2α) p
y−α
x−α
2−λ
p
,
a.e. (α, ∞).
It follows that
a(x − α)2−λ f p (x) = b(y − α)2−λ ,
a.e. in (α, ∞).
Obviously a 6= 0, (otherwise a = b = 0), one has
b
(x − α)1−λ f p (x) = (y − α)2−λ x−1 , a.e. (α, ∞)
a
R∞
which contradicts 0 < α (x − α)1−λ f P (x)dx < ∞. Hence
Z ∞
p
Z ∞
f (x)
(p−1)(λ−1)
(y − α)
dx dy
(x + y − 2α)λ
α
α
p−1 Z ∞ "Z ∞
2−λ #
p+λ−2 q+λ−2
f p (x)
x−α q
,
dx dy
> B
p
q
(x + y − 2α)λ y − α
α
α
p−1
p+λ−2 q+λ−2
= B
,
p
q
2−λ #
Z ∞ "Z ∞
1
x−α q
×
dy f p (x)dx.
λ
(x
+
y
−
2α)
y
−
α
α
α
By (2.3) and (2.4), we obtain (4.7).
Obviously, inequalities
(4.7) and (3.1) are equivalent as 0 < p < 1 and the constant factor
p+λ−2 q+λ−2
B
, q
is the best possible.
p
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 49, 9 pp.
http://jipam.vu.edu.au/
R EVERSE H ARDY-H ILBERT-T YPE I NTEGRAL I NEQUALITY
9
1−λ p
R∞
Corollary 4.6. Let 0 < p < 1, p1 + 1q = 1, 2−p < λ < 2−q, 0 < − 1 x + 12
f (x)dx < ∞,
2
f (x) ≥ 0. Then
#p
(p−1)(λ−1) "Z ∞
Z ∞
f (x)
1
dx dy
y+
λ
2
− 12 (x + y + 1)
− 12
1−λ
p Z ∞ p+λ−2 q+λ−2
1
f p (x)dx,
> B
,
x+
1
2
p
q
−2
where the constant factor B p+λ−2
, q+λ−2
is the best possible.
p
q
R∞
Corollary 4.7. Let 0 < p < 1, p1 + 1q = 1, f (x) ≥ 0, 0 < α (x − α)−1 f p (x)dx < ∞, α ∈ R.
Then
Z ∞
p
Z ∞ p
Z ∞
f (x)
f (x)
(p−1)
dx dy >
dx.
(4.9)
(y − α)
2
(x + y − 2α)
x−α
α
α
α
−1 p
R∞
Corollary 4.8. Let 0 < p < 1, p1 + 1q = 1, f (x) ≥ 0, 0 < − 1 x + 12
f (x)dx < ∞. Then
2
"
#
p
(p−1) Z ∞
Z ∞
Z ∞ p
f (x)
f (x)
1
dx
dy
>
(4.10)
y+
1 dx.
2
2
− 12
− 12 x + 2
− 12 (x + y + 1)
R EFERENCES
[1] G.H. HARDY, J.E. LITTLEWOOD
Cambridge, 1952.
AND
G. POLYA, Inequalities, Cambridge University Press,
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(2005), 29-51.
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[4] A. SHAMANDY, On new extensions of integral inequalities similar to Hilbert’s inequality(II), Proc.
Pakistan Acad. Sci., 42(2) (2005), 99–109.
[5] G. XI, A reverse Hardy-Hilbert-type inequality, J. Inequal. Appl., 2007 (2007), Art. 79758.
[6] B. YANG AND L. DEBNATH, On a new generalization of Hardy-Hilbert’s inequality and its applications, J. Math. Anal. Appl., 233 (1999), 484–497.
[7] B. Yang, On Hardy-Hilbert’s integral inequality, J. Math. Anal. Appl., 261 (2001), 295–306.
[8] B. YANG, On best extensions of Hardy-Hilbert’s inequality with two parameters, J. Inequal. Pure
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sid=554].
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J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 49, 9 pp.
http://jipam.vu.edu.au/