Volume 9 (2008), Issue 2, Article 59, 8 pp.
A RELATION TO HILBERT’S INTEGRAL INEQUALITY AND SOME BASE
HILBERT-TYPE INEQUALITIES
BICHENG YANG
D EPARTMENT OF M ATHEMATICS
G UANGDONG E DUCATION I NSTITUTE
G UANGZHOU , G UANGDONG 510303
P. R. C HINA
bcyang@pub.guangzhou.gd.cn
Received 17 April, 2008; accepted 18 May, 2008
Communicated by J. Pečarić
A BSTRACT. In this paper, by using the way of weight function and real analysis techniques,
a new integral inequality with some parameters and a best constant factor is given, which is a
relation to Hilbert’s integral inequality and some base Hilbert-type integral inequalities. The
equivalent form and the reverse forms are considered.
Key words and phrases: Base Hilbert-type integral inequality; Parameter; Weight function.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
R∞
If f, g ≥ 0, 0 < 0 f 2 (x)dx < ∞ and 0 < 0 g 2 (x)dx < ∞ then we have the following
Hilbert’s integral inequality [1]:
Z ∞
12
Z ∞Z ∞
Z ∞
f (x)g(y)
2
2
(1.1)
dxdy < π
f (x)dx
g (x)dx ,
x+y
0
0
0
0
R∞
where the constant factor π is the best possible. Under the same condition of (1.1), we also have
the following basic Hilbert-type integral inequalities [1, 2, 3]:
Z ∞
12
Z ∞Z ∞
Z ∞
f (x)g(y)
2
2
(1.2)
dxdy < 4
f (x)dx
g (x)dx ;
max{x, y}
0
0
0
0
∞
Z
∞
Z
(1.3)
0
Z
0
∞
Z
(1.4)
0
114-08
0
∞
| ln(x/y)|f (x)g(y)
dxdy < c0
x+y
Z
| ln(x/y)|f (x)g(y)
dxdy < 8
max{x, y}
Z
∞
2
2
f (x)dx
g (x)dx
0
;
0
∞
2
Z
f (x)dx
0
12
∞
Z
∞
2
g (x)dx
0
12
,
2
B ICHENG YANG
P
8(−1)k−1
+
where the constant factors 4, c0 = ∞
=
7.3277
and 8 are the best possible. In
k=1 (2k−1)2
2005, Hardy-Riesz gave a best extension of (1.1) by introducing one pair of conjugate exponents
(p, q) (p > 1, p1 + 1q = 1) as [4]
Z
∞
Z
(1.5)
0
0
∞
f (x)g(y)
π
dxdy <
x+y
sin πp
Z
p1 Z
∞
p
∞
q
f (x)dx
1q
g (x)dx
0
,
0
π
where the constant factor sin(π/p)
is the best possible. Inequality (1.5) is referred to as HardyHilbert’s integral inequality, which is important in analysis and its applications [5]. In 1998,
Yang gave a best extension of (1.1) by introducing an independent parameter λ > 0 as [6, 7]
Z ∞
12
Z ∞Z ∞
Z ∞
f (x)g(y)
λ λ
1−λ 2
1−λ 2
(1.6)
dxdy < B
,
x f (x)dx
x g (x)dx ,
(x + y)λ
2 2
0
0
0
0
where the constant factor B λ2 , λ2 is the best possible and the Beta function B(u, v) is defined
by [8]:
Z ∞
1
(1.7)
B(u, v) :=
tu−1 dt
(u, v > 0).
u+v
(1
+
t)
0
In 2004-2005, by introducing two pairs of conjugate exponents and an independent parameter,
Yang et al. [9, 10] gave two different extensions of (1.1) and (1.5) as: If p, r > 1, p1 + 1q = 1,
1
r
+
1
s
λ
λ
= 1, λ > 0, φ(x) = xp(1− r )−1 , ψ(x) = xq(1− s )−1 , f, g ≥ 0,
Z
∞
0 < ||f ||p,φ :=
x
p(1− λ
)−1 p
r
p1
f (x)dx
<∞
0
and
Z
0 < ||g||q,ψ :=
∞
x
q(1− λs )−1 q
g (x)dx
1q
< ∞,
0
then
Z
∞
Z
∞
(1.8)
0
∞
0
f (x)g(y)
π
dxdy <
||f ||p,φ ||g||q,ψ ;
λ
λ
x +y
λ sin( πr )
∞
f (x)g(y)
λ λ
(1.9)
dxdy < B
,
||f ||p,φ ||g||q,ψ ,
(x + y)λ
r s
0
0
π
λ λ
where the constant factors λ sin(
,
are the best possible. Yang [11] also considered
π and B
)
r s
r
the reverse of (1.8) and (1.9).
In this paper, by using weight functions and real analysis techniques, a new integral inequality
with the homogeneous kernel of −λ degree
Z
kλ (x, y) =
Z
| ln(x/y)|β
(x + y)λ−α (max{x, y})α
(λ > 0, α ∈ R, β > −1)
is given, which is a relation to (1.1) and the above basic Hilbert-type integral inequalities (1.2),
(1.3) and (1.4). The equivalent and reverse forms are considered. All the new inequalities
possess the best constant factors.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 59, 8 pp.
http://jipam.vu.edu.au/
H ILBERT ’ S I NTEGRAL I NEQUALITY
3
2. S OME L EMMAS
We introduce the following Gamma function [8]:
Z ∞
(2.1)
Γ(s) =
e−t ts−1 dt
(s > 0).
0
Lemma 2.1. For a, b > 0, it follows that
Z 1
Z ∞
1
a−1
b−1
(2.2)
x (− ln x) dx = b Γ(b) =
y −a−1 (ln y)b−1 dy.
a
0
1
Proof. Setting x = e−t/a in first integral of (2.2), by (2.1), we find the first equation of (2.2).
Setting y = 1/x in the first integral of (2.2), we obtain the second equation of (2.2). The lemma
is hence proved.
(2.3)
1
r
1
s
= 1, λ > 0, α ∈ R and β > −1, define the weight function as
β
λ
Z ∞
ln xy y s −1
λ
$λ (s, x) := x r
dy
(x ∈ (0, ∞)).
(x + y)λ−α (max{x, y})α
0
Lemma 2.2. If r > 1,
+
Then we have
Z
(2.4)
$λ (s, x) = kλ (r) :=
0
∞
λ
| ln u|β u r −1
du,
(u + 1)λ−α (max{u, 1})α
where kλ (r) is a positive number and
(2.5)
kλ (r) = Γ(β + 1)
∞ X
α−λ
k=0
k
"
1
k+
λ β+1
r
#
1
+
k+
λ β+1
s
.
Proof. Setting u = x/y in (2.3), by simplification, we obtain (2.4). In view of (2.2), we obtain
Z 1
Z ∞
λ
λ
(− ln u)β u r −1
(ln u)β u r −α−1
0 < kλ (r) =
du +
du
(u + 1)λ−α
(u + 1)λ−α
0
1
Z 1
Z ∞
|α−λ|
(β+1)−1 λ
−1
(β+1)−1 −λ
−1
r
s
≤2
(− ln u)
u
du +
(ln u)
u
du
0
1
r β+1 s β+1
|α−λ|
=2
+
Γ(β + 1) < ∞.
λ
λ
Hence kλ (r) is a positive number. Using the property of power series, we find
Z 1
Z ∞
−λ
λ
(− ln u)β u r −1
(ln u)β u s −1
kλ (r) =
du +
du
(u + 1)λ−α
(1 + u−1 )λ−α
0
1
Z 1X
Z ∞X
∞ ∞ −λ
α−λ
α−λ
β λ
+k−1
r
=
(− ln u) u
du +
(ln u)β u s −k−1 du
k
k
0 k=0
1
k=0
Z 1
Z ∞
∞
X
−λ
α−λ
β λ
+k−1
β
−k−1
=
(− ln u) u r
du +
(ln u) u s
du .
k
0
1
k=0
Then in view of (2.2), we have (2.5). The lemma is proved.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 59, 8 pp.
http://jipam.vu.edu.au/
4
B ICHENG YANG
Lemma 2.3. If p > 0 (p 6= 1), r > 1, p1 + 1q = 1, 1r + 1s = 1, λ > 0, α ∈ R, β > −1, n ∈ N,
r
n > |q|λ
, then for n → ∞, we have
β
1
1
−1 λs − nq
−1
Z ∞ Z ∞ ln x x λr − np
y
y
1
(2.6)
In :=
dxdy = kλ (r) + o(1).
n 1
(x + y)λ−α (max{x, y})α
1
Proof. Setting u = y/x, by Fubini’s theorem [12], we obtain


β
1
1
−1 λs − nq
−1
Z ∞ Z ∞ ln x x λr − np
y
y
1


dx dy
In =

λ−α
α
(x + y) (max{x, y})
n 1
1
(2.7)
∞
"Z
y
β
λ
1
+ np
−1
s
#
| ln u| u
du dy
λ−α (max{1, u})α
1
0 (1 + u)
#
"Z
λ
1
λ
1
Z
Z y
1
(ln u)β u s + np −1
1 ∞ − 1 −1
(− ln u)β u s + np −1
=
y n
du +
du dy
n 1
(1 + u)λ−α
(1 + u)λ−α uα
0
1
#
"Z
λ
1
λ
1
Z 1
Z
y
(− ln u)β u s + np −1
1 ∞ − 1 −1
(ln u)β u s + np −1
du +
y n
du dy
=
(1 + u)λ−α
n 1
(1 + u)λ−α uα
1
0
1
1
λ
λ
Z Z ∞
Z 1
1
1 ∞
(− ln u)β u s + np −1
(ln u)β u s + np −1
−n
−1
du +
y
du
=
dy
(1 + u)λ−α
n 1
(1 + u)λ−α uα
u
0
λ
1
λ
1
Z 1
Z ∞
(− ln u)β u s + np −1
(ln u)β u s − nq −1
=
du +
du.
(1 + u)λ−α
(1 + u)λ−α uα
0
1
=
1
n
Z
y
1
−n
−1
(i) If p > 0 (p 6= 1) and q > 0, then by Levi’s theorem [12], we find
λ
1
Z 1
Z 1
λ
(− ln u)β u s −1
(− ln u)β u s + np −1
du =
du + o1 (1),
(1 + u)λ−α
(1 + u)λ−α
0
0
λ
1
Z ∞
Z ∞
λ
(ln u)β u s − nq −1
(ln u)β u s −1
du =
du + o2 (1) (n → ∞);
(1 + u)λ−α uα
(1 + u)λ−α uα
1
1
r
(ii) If q < 0, setting n0 ∈ N, n0 > |q|λ
, s10 = 1s − n01qλ , r10 = 1r + n01qλ , then for n ≥ n0 , we find
λ
1
λ
Z ∞
Z ∞
− 1 −1
(ln u)β u s − nq −1
(ln u)β u s n0 q
du ≤
du ≤ kλ (r0 ),
λ−α uα
λ−α uα
(1
+
u)
(1
+
u)
1
1
and by Lebesgue’s control convergence theorem, we have
λ
1
Z ∞
Z ∞
λ
(ln u)β u s − nq −1
(ln u)β u s −1
du =
du + o3 (1) (n → ∞).
(1 + u)λ−α uα
(1 + u)λ−α uα
1
1
Hence by the above results and (2.7), we obtain (2.6). The lemma is proved.
3. M AIN R ESULTS
Theorem 3.1. Assume that p > 0 (p 6= 1), r > 1,
β > −1, φ(x) = x
p(1− λ
)−1
r
0 < ||f ||p,φ
1
p
+
1
q
= 1,
1
r
+
1
s
= 1, λ > 0, α ∈ R,
q(1− λs )−1
, ψ(x) = x
(x ∈ (0, ∞)), f, g ≥ 0,
p1
Z ∞
p(1− λ
)−1 p
r
< ∞,
0 < ||g||q,ψ < ∞.
=
x
f (x)dx
0
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 59, 8 pp.
http://jipam.vu.edu.au/
H ILBERT ’ S I NTEGRAL I NEQUALITY
5
(i) For p > 1, we have the following inequality:
β
Z ∞ Z ∞ ln x f (x)g(y)
y
(3.1)
I :=
dxdy < kλ (r)||f ||p,φ ||g||q,ψ ;
λ−α
(x + y) (max{x, y})α
0
0
(ii) For 0 < p < 1, we have the reverse of (3.1), where the constant factor kλ (r) expressed
by (2.5) in (3.1) and its reverse is the best possible.
Proof. (i) By Hölder’s inequality with weight [13], in view of (2.3), we find
β
"
#"
#
Z ∞Z ∞
λ
λ
ln xy x(1− r )/q
y (1− s )/p
f (x)
g(y) dxdy
I=
λ
(x + y)λ−α (max{x, y})α y (1− λs )/p
x(1− r )/q
0
0

 p1
β


x λ
Z ∞ Z ∞

ln y x(1− r )(p−1) p
≤
·
f (x)dxdy
λ


(x + y)λ−α (max{x, y})α
y 1− s
0
 0

×


Z


∞
Z
0
0
(x +
p1 Z
p
=
λ
y)λ−α (max{x, y})α
∞
Z
(3.2)
∞
β
ln xy $λ (s, x)φ(x)f (x)dx
0
·
y (1− s )(q−1)
x
1− λ
r
g q (y)dxdy
 1q




1q
∞
q
$λ (r, y)ψ(y)g (y)dy
.
0
We confirm that the middle of (3.2) keeps the form of strict inequality. Otherwise, there exist
constants A and B, such that they are not all zero and [13]
λ
A
x(1− r )(p−1)
y
1− λs
p(1− λ
)
r
λ
p
f (x) = B
y (1− s )(q−1)
x
1− λ
r
g q (y) a.e. in (0, ∞) × (0, ∞).
q(1− λs )
f p (x) = By
g q (y) a.e.
It follows that Ax
h in (0,λ ∞) × (0,
i ∞). Assuming that A 6= 0,
1
p(1− λ
)−1
q(1−
)
p
q
r
s g (y)
there exists y > 0, such that x
f (x) = By
a.e. in x ∈ (0, ∞). This
Ax
contradicts the fact that 0 < ||f ||p,φ < ∞. Then inequality (3.1) is valid by using (2.4) and
(2.5).
r
, setting fn , gn as
For n ∈ N, n > |q|λ
(
(
0,
0 < x ≤ 1;
0,
0 < x ≤ 1;
g
(x)
:=
fn (x) :=
n
λ
1
λ
1
x r − np −1 , x > 1;
x r − nq −1 , x > 1;
if there exists a constant factor 0 < k ≤ kλ (r), such that (3.1) is still valid if we replace kλ (r)
by k, then by (2.6), we have
β
Z ∞ Z ∞ ln x fn (x)gn (y)
y
1
dxdy
kλ (r) + o(1) = In =
n 0
(x + y)λ−α (max{x, y})α
0
1
< k||fn ||p,φ ||gn ||q,ψ = k,
n
and kλ (r) ≤ k (n → ∞). Hence k = kλ (r) is the best constant factor of (3.1).
(ii) For 0 < p < 1, by the reverse Hölder’s inequality with weight [13], in view of (2.3), we
find the reverse of (3.2), which still keeps the strict form. Then by (2.4) and (2.5), we have the
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 59, 8 pp.
http://jipam.vu.edu.au/
6
B ICHENG YANG
reverse of (3.1). By using (2.6) and the same manner as mentioned above , we can show that
the constant factor in the reverse of (3.1) is still the best possible. The theorem is proved.
Theorem 3.2. Assume that p > 0 (p 6= 1), r > 1,
p(1− λ
)−1
r
1
p
+
1
q
= 1,
1
r
+
1
s
= 1, λ > 0, α ∈ R,
q(1− λs )−1
β > −1, φ(x) = x
, ψ(x) = x
(x ∈ (0, ∞)), f ≥ 0, 0 < ||f ||p,φ < ∞.
(i) For p > 1, we have the following inequality, which is equivalent to (3.1) and with the
best constant factor kλp (r):

p
β
x Z ∞
Z ∞
ln y f (x)
pλ


(3.3)
J :=
y s −1 
dx dy < kλp (r)||f ||pp,φ ;
λ−α (max{x, y})α
(x
+
y)
0
0
(ii) For 0 < p < 1, we have the reverse of (3.3), which is equivalent to the reverse of (3.1),
with the best constant factor kλp (r).
Proof. (i) For p > 1, x > 0, setting a bounded measurable function as
(
f (x), for f (x) < n;
[f (x)]n := min{f (x), n} =
n,
for f (x) ≥ n,
Rn
since ||f ||p,φ > 0, there exists n0 ∈ N, such that 1 φ(x)[f (x)]pn dx > 0 (n ≥ n0 ). Setting gen (y)
n
(y ∈ ( n1 , n); n ≥ n0 ) as
(3.4)

Z
pλ

−1
gen (y) := y s 
then by (3.1), we find
Z
0<
=
(x + y)λ−α (max{x, y})α
1
n

Z
n
pλ

−1
ys 
n
n
Z
=
n
1
n
β
ln xy [f (x)]n dx
(x + y)λ−α (max{x, y})α
p

 dy
1
n
n
n
< kλ (r)
1
n
Z
(3.6)
,
β
ln xy [f (x)]n gen (y)
dxdy
(x + y)λ−α (max{x, y})α
(Z
) p1 (Z
1
n
(3.5)

[f (x)]n dx
ψ(y)e
gnq (y)dy
1
n
Z
n
n
1
n
Z
p−1
β
ln xy φ(x)[f (x)]pn dx
n
0<
1
n
ψ(y)e
gnq (y)dy
<
kλp (r)
1
n
Z
ψ(y)e
gnq (y)dy
) 1q
< ∞;
∞
φ(x)f p (x)dx < ∞.
0
It follows 0 < ||g||q,ψ < ∞. For n → ∞, by (3.1), both (3.5) and (3.6) still keep the forms of
strict inequality. Hence we have (3.3). On the other-hand, suppose (3.3) is valid. By Hölder’s
inequality, we have


β
x Z ∞
Z ∞
h
i
ln y f (x)dx
1
 −1
 p1 − λs
+λ
y
g(y)
dy ≤ J p ||g||q,ψ .
(3.7)
I=

y p s
λ−α
α
(x + y) (max{x, y})
0
0
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 59, 8 pp.
http://jipam.vu.edu.au/
H ILBERT ’ S I NTEGRAL I NEQUALITY
7
In view of (3.3), we have (3.1), which is equivalent to (3.3). We confirm that the constant factor
in (3.3) is the best possible. Otherwise, we may get a contradiction by (3.7) that the constant
factor in (3.1) is not the best possible.
(ii) For 0 < p < 1, since ||f ||p,φ > 0, we confirm that J > 0. If J = ∞, then the reverse of
(3.3) is naturally valid. Suppose 0 < J < ∞. Setting

p−1
β
x Z ∞
ln
pλ
y 

g(y) := y s −1 
f (x)dx ,
λ−α
α
(x + y) (max{x, y})
0
by the reverse of (3.1), we obtain
∞ > ||g||qq,ψ = J = I > kλ (r)||f ||p,φ ||g||q,ψ > 0;
1
J p = ||g||q−1
q,ψ > kλ (r)||f ||p,φ .
Hence we have the reverse of (3.3). On the other-hand, suppose the reverse of (3.3) is valid. By
the reverse Hölder’s inequality, we can get the reverse of (3.7). Hence in view of the reverse
of (3.3), we obtain the reverse of (3.1), which is equivalent to the reverse of (3.3). We confirm
that the constant factor in the reverse of (3.3) is the best possible. Otherwise, we may get a
contradiction by the reverse of (3.7) that the constant factor in the reverse of (3.1) is not the best
possible. The theorem is proved.
Remark 1. For p = r = 2 in (3.1), setting α = β = 0, λ = 1, we obtain (1.1); setting
α = 0, β = λ = 1, we obtain (1.3). For α = λ > 0, β > −1 in (3.1), we have
β
Z ∞ Z ∞ ln x f (x)g(y)
y
rβ+1 + sβ+1
(3.8)
dxdy
<
Γ(β + 1)||f ||p,φ ||g||q,ψ ,
(max{x, y})λ
λβ+1
0
0
1
where the constant factor λβ+1
(rβ+1 + sβ+1 )Γ(β + 1) is the best possible. For p = r = 2 in
(3.8), setting λ = 1, β = 0, we obtain (1.2); setting λ = 1, β = 1, we obtain (1.4). Hence
inequality (3.1) is a relation to (1.1), (1.2), (1.3) and (1.4).
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[1] G.H. HARDY, J.E. LITTLEWOOD
bridge, 1952.
AND
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