Ann. Funct. Anal. 3 (2012), no. 1, 142–150
A nnals of F unctional A nalysis
ISSN: 2008-8752 (electronic)
URL: www.emis.de/journals/AFA/
A NEW HALF-DISCRETE MULHOLLAND-TYPE INEQUALITY
WITH PARAMETERS
BICHENG YANG
Communicated by J. Soria
Abstract. By means of weight functions and Hadamard’s inequality, a new
half-discrete Mulholland-type inequality with a best constant factor is given.
A best extension with parameters, some equivalent forms, the operator expressions as well as some particular cases are also considered.
1. Introduction
R∞
1
Assuming that f, g ∈ L2 (R+ ), ||f || = { 0 f 2 (x)dx} 2 > 0, ||g|| > 0, we have the
following Hilbert’s integral inequality (cf. [3]):
Z ∞Z ∞
f (x)g(y)
dxdy < π||f ||||g||,
(1.1)
x+y
0
0
∞
where the constant factor π is best possible. If a = {an }∞
n=1 , b = {bn }n=1 ∈
P
1
2 2
l2 , ||a|| = { ∞
n=1 an } > 0, ||b|| > 0, then we have the following analogous discrete
Hilbert’s inequality
∞ X
∞
X
am b n
< π||a||||b||,
(1.2)
m+n
m=1 n=1
with the same best constant factor π. Inequalities (1.1) and (1.2) are important
in analysis and its applications (cf. [10, 11, 12]). On the other-hand, we have the
following Mulholland’s inequality with the same best constant factor (cf. [3, 20]):
( ∞
) 21
∞
∞ X
∞
X
X
X
am b n
nb2n
<π
ma2m
.
(1.3)
ln
mn
m=2
n=2
m=2 n=2
Date: Received: 6 December 2011; Accepted: 3 March 2012.
2010 Mathematics Subject Classification. Primary 26D15; Secondary 47A07.
Key words and phrases. Mulholland-type inequality, weight function, equivalent form.
142
HALF-DISCRETE MULHOLLAND-TYPE INEQUALITY WITH PARAMETERS
143
In 1998, by introducing an independent parameter λ ∈ (0, 1], Yang [14] gave
an extension of (1.1). Refinement the results from [14], Yang [15] gave some best
extensions of (1.1) and (1.2): If p > 1, p1 + 1q = 1, λ1 + λ2 = λ, kλ (x, y) is a nonR∞
negative homogeneous function of degree −λ satisfying k(λ1 ) = 0 kλ (t, 1)tλ1 −1 dt
∈ R+ , φ(x) = xp(1−λ1 )−1 , ψ(x) = xq(1−λ2 )−1 ,
Z ∞
1
φ(x)|f (x)|p dx} p < ∞},
f (≥ 0) ∈ Lp,φ (R+ ) = {f |||f ||p,φ := {
0
g(≥ 0) ∈ Lq,ψ (R+ ), and ||f ||p,φ , ||g||q,ψ > 0, then
Z ∞Z ∞
kλ (x, y)f (x)g(y)dxdy < k(λ1 )||f ||p,φ ||g||q,ψ ,
0
(1.4)
0
where the constant factor k(λ1 ) is best possible. Moreover if kλ (x, y) is finite and
kλ (x, y)xλ1 −1 (kλ (x, y)y λ2 −1 ) is decreasing for x > 0(y > 0), then for am, bn ≥ 0,
P∞
1
∞
p p
a = {am }∞
m=1 ∈ lp,φ = {a|||a||p,φ := {
n=1 φ(n)|an | } < ∞}, and b = {bn }n=1 ∈
lq,ψ , ||a||p,φ , ||b||q,ψ > 0, we have
∞ X
∞
X
kλ (m, n)am bn < k(λ1 )||a||p,φ ||b||q,ψ ,
(1.5)
m=1 n=1
where the constant k(λ1 ) is still best value. Clearly, for p = q = 2, λ =
1
1, k1 (x, y) = x+y
, λ1 = λ2 = 21 , (1.4) reduces to (1.1), while (1.5) reduces to
(1.2).
Some other results about Hilbert-type inequalities can be found in [9]-[16]. On
half-discrete Hilbert-type inequalities with the general non-homogeneous kernels,
Hardy et al. provided a few results in Theorem 351 of [3]. But they did not
prove that the constant factors are best possible. In 2005, Yang [18] gave a result
1
with the kernel (1+nx)
λ by introducing a variable and proved that the constant
factor is best possible. Very recently, Yang [19, 20] gave the following half-discrete
Hilbert’s inequality with best constant factor:
Z ∞
∞
X
an
f (x)
dx < π||f ||||a||.
(1.6)
(x + n)λ
0
n=1
In this paper, by means of weight functions and Hadamard’s inequality, a new
half-discrete Mulholland-type inequality similar to (1.3) and (1.6) with a best
constant factor is given as follows:
Z ∞
∞
X
an
f (x)
dx
3
1 + ln(x − 12 ) ln(n − 12 )
n=2
2
) 12
(Z
∞
∞
X
1
1
< π
(x − )f 2 (x)dx
(n − )a2n
.
(1.7)
3
2
2
n=2
2
Moreover, a best extension of (1.7) with multi-parameters, some equivalent forms,
the operator expressions as well as some particular inequalities are considered.
144
B. YANG
2. Some lemmas
Lemma 2.1. If 0 < λ ≤ 2, α ∈ R, β ≤ 12 , and ω(n) and $(x) are weight
functions given by
λ
λ
Z ∞
ln 2 (n − β) ln 2 −1 (x − α)dx
ω(n) : =
, n ∈ N\{1},
(2.1)
λ
1+α (x − α)[1 + ln(x − α) ln(n − β)]
λ
λ
∞
X
ln 2 (x − α) ln 2 −1 (n − β)
$(x) : =
, x > 1 + α,
(2.2)
λ
(n
−
β)[1
+
ln(x
−
α)
ln(n
−
β)]
n=2
then we have
λ λ
(2.3)
$(x) < ω(n) = B( , ).
2 2
Proof. Substituting t = ln(x − α) ln(n − β) in (2.1), and by simple calculation,
we have
Z ∞
λ
1
λ λ
ω(n) =
t 2 −1 dt = B( , ).
λ
(1 + t)
2 2
0
For fixed x > 1 + α, in view of the conditions, it is easy to see that
λ
ln 2 −1 (y − β)
h(x, y) :=
(y − β)[1 + ln(x − α) ln(y − β)]λ
is decreasing and strictly convex with h0y (x, y) < 0 and h00y2 (x, y) > 0, for y ∈
( 23 , ∞). Hence by (2.2) and Hadamard’s inequality (cf. [7]), we find
λ
Z ∞
λ
ln 2 −1 (y − β)dy
$(x) < ln 2 (x − α)
3
(y − β)[1 + ln(x − α) ln(y − β)]λ
2
Z ∞
λ
λ λ
t 2 −1
t=ln(x−α) ln(y−β)
=
dt
≤
B(
, ),
λ
2 2
ln(x−α) ln( 32 −β) (1 + t)
and (2.3) follows.
Lemma 2.2. Let the assumptions of Lemma 1 be fulfilled and additionally, let
p > 1, p1 + 1q = 1, an ≥ 0, n ∈ N\{1}, f (x) is a non-negative measurable function
in (1 + α, ∞). Then we have the following inequalities:
(∞
p ) p1
X ln pλ2 −1 (n − β) Z ∞
f (x)dx
J :=
λ
n−β
1+α [1 + ln(x − α) ln(n − β)]
n=2
1 Z ∞
p1
λ λ q
p(1− λ
)−1
p−1
p
2
≤ B( , )
$(x)(x − α) ln
(x − α)f (x)dx ,
(2.4)
2 2
1+α
(Z
"∞
#q ) 1q
qλ
∞
(x − α) 2 −1 X
an
L1 : =
dx
q−1
λ
[1
+
ln(x
−
α)
ln(n
−
β)]
1+α [$(x)]
n=2
(
) 1q
∞
λ
λ λ X
≤
B( , )
(n − β)q−1 lnq(1− 2 )−1 (n − β)aqn
.
(2.5)
2 2 n=2
HALF-DISCRETE MULHOLLAND-TYPE INEQUALITY WITH PARAMETERS
Proof. Setting k(x, n) :=
(2.3), it follows
Z ∞
1
,
[1+ln(x−α) ln(n−β)]λ
145
by Hölder’s inequality (cf. [7]) and
p
f (x)dx
[1 + ln(x − α) ln(n − β)]λ
#
(Z1+α
"
1
)/q
(1− λ
∞
q
2
(x − α) (x − α) f (x)
ln
=
k(x, n)
1
(1− λ
)/p
1+α
ln 2 (n − β) (n − β) p
"
×
Z
λ
1
λ
1
ln(1− 2 )/p (n − β) (n − β) p
ln(1− 2 )/q (x − α) (x − α) q
k(x, n)
1+α
k(x, n)
1+α
dx
(x − α)p−1 ln(1− 2 )(p−1) (x − α)
1− λ
2
(n − β) ln
(n − β)
f p (x)dx
)p−1
λ
∞
×
)p
λ
∞
≤
(Z
#
(n − β)q−1 ln(1− 2 )(q−1) (n − β)
λ
(x − α) ln1− 2 (x − α)
dx
)p−1
λ
lnq(1− 2 )−1 (n − β)
=
ω(n)
(n − β)1−q
λ
Z ∞
(x − α)p−1 ln(1− 2 )(p−1) (x − α) p
×
k(x, n)
f (x)dx
λ
1+α
(n − β) ln1− 2 (n − β)
λ λ p−1
λ
Z
(n − β) ∞
B( 2 , 2 )
(x − α)p−1 ln(1− 2 )(p−1) (x − α)f p (x)
dx.
=
k(x, n)
pλ
λ
1+α
(n − β) ln1− 2 (n − β)
ln 2 −1 (n − β)
(
Then by the Lebesgue term by term integration theorem (cf. [8]), we have
) p1
1 ( ∞ Z
λ
λ λ q X ∞
(x − α)p−1 ln(1− 2 )(p−1) (x − α) p
J ≤ B( , )
k(x, n)
f (x)dx
1− λ
2 2
2 (n − β)
1+α
(n
−
β)
ln
n=2
) p1
1q (Z ∞ X
λ
∞
λ λ
(x − α)p−1 ln(1− 2 )(p−1) (x − α) p
= B( , )
k(x, n)
f (x)dx
λ
2 2
1+α n=2
(n − β) ln1− 2 (n − β)
1 Z ∞
p1
λ λ q
p(1− λ
)−1
p−1
p
2
= B( , )
$(x)(x − α) ln
(x − α)f (x)dx ,
2 2
1+α
hence, (2.4) follows. By Hölder’s inequality again, we have
"∞
#
#q ( ∞
"
1
λ
X
X
(x − α) q ln(1− 2 )/q (x − α)
k(x, n)an =
k(x, n)
1
λ
(n − β) p ln(1− 2 )/p (n − β)
n=2
n=2
146
B. YANG
"
×
≤
1
λ
ln(1− 2 )/p (n − β) (n − β) p an
1
λ
ln(1− 2 )/q (x − α) (x − α) q
(∞
X
λ
k(x, n)
n=2
×
∞
X
#)q
(x − α)p−1 ln(1− 2 )(p−1) (x − α)
)q−1
λ
(n − β) ln1− 2 (n − β)
λ
k(x, n)
(n − β)q−1 ln(1− 2 )(q−1) (n − β)
1− λ
2
aqn
(x − α) ln
(x − α)
λ
∞
λ
[$(x)]q−1 (x − α) X
ln 2 −1 (x − α)
=
k(x, n)
ln(1− 2 )(q−1) (n − β)aqn .
qλ
1−q
(x − α)(n − β)
ln 2 −1 (x − α) n=2
n=2
By Lebesgue term by term integration theorem, we have
) 1q
(Z
λ
∞
∞ X
λ
ln 2 −1 (x − α)
(n − β)q−1 ln(1− 2 )(q−1) (n − β)aqn dx
k(x, n)
L1 ≤
(x
−
α)
1+α n=2
=
#
) 1q
λ
λ
λ
ln 2 (n − β) ln 2 −1 (x − α)dx lnq(1− 2 )−1 (n − β) q
k(x, n)
an
x−α
(n − β)1−q
1+α
( ∞ "Z
X
n=2
∞
=
(∞
X
) 1q
q(1− λ
)−1
2
ω(n)(n − β)q−1 ln
(n − β)aqn
,
n=2
and in view of (2.3), inequality (2.5) follows.
3. Main results
We introduce the functions
λ
Φ(x) : = (x − α)p−1 lnp(1− 2 )−1 (x − α)(x ∈ (1 + α, ∞)),
λ
Ψ(n) : = (n − β)q−1 lnq(1− 2 )−1 (n − β)(n ∈ N\{1}),
wherefrom [Φ(x)]1−q =
1
x−α
qλ
ln 2 −1 (x − α), and [Ψ(n)]1−p =
1
n−β
pλ
ln 2 −1 (n − β).
Theorem 3.1. If 0 < λ ≤ 2, α ∈ R, β ≤ 12 , p > 1, p1 + 1q = 1, f (x), an ≥ 0,
f ∈ Lp,Φ (1 + α, ∞), a = {an }∞
n=2 ∈ lq,Ψ , ||f ||p,Φ > 0 and ||a||q,Ψ > 0, then we have
the following equivalent inequalities:
∞ Z ∞
X
an f (x)dx
I : =
[1 + ln(x − α) ln(n − β)]λ
n=2 1+α
Z ∞ X
∞
an f (x)dx
λ λ
=
< B( , )||f ||p,Φ ||a||q,Ψ , (3.1)
λ
2 2
1+α n=2 [1 + ln(x − α) ln(n − β)]
HALF-DISCRETE MULHOLLAND-TYPE INEQUALITY WITH PARAMETERS
J =
(∞
X
[Ψ(n)]1−p
Z
∞
1+α
n=2
f (x)dx
[1 + ln(x − α) ln(n − β)]λ
p ) p1
λ λ
< B( , )||f ||p,Φ ,
2 2
(Z
L
:
=
1+α
(3.2)
"
∞
[Φ(x)]1−q
147
∞
X
n=2
an
[1 + ln(x − α) ln(n − β)]λ
) 1q
#q
dx
λ λ
< B( , )||a||q,Ψ ,
2 2
where the constant B( λ2 , λ2 ) is the best possible in the above inequalities.
(3.3)
Proof. The two expressions for I in (3.1) follow from Lebesgue’s term by term
integration theorem. By (2.4) and (2.3), we have (3.2). By Hölder’s inequality,
we have
Z ∞
∞ X
−1
1
f (x)dx
q
I=
Ψ (n)
[Ψ q (n)an ] ≤ J||a||q,Ψ . (3.4)
λ
1+α [1 + ln(x − α) ln(n − β)]
n=2
Then by (3.2), we have (3.1). On the other-hand, assume that (3.1) is valid.
Setting
Z ∞
p−1
f (x)dx
1−p
an := [Ψ(n)]
, n ∈ N\{1},
λ
1+α [1 + ln(x − α) ln(n − β)]
where J p−1 = ||a||q,Ψ . By (2.4), we find J < ∞. If J = 0, then (3.2) is trivially
valid; if J > 0, then by (3.1), we have
λ λ
||a||qq,Ψ = J q(p−1) = J p = I < B( , )||f ||p,Φ ||a||q,Ψ ,
2 2
λ λ
therefore ||a||q−1
q,Ψ = J < B( 2 , 2 )||f ||p,Φ , that is, (3.2) is equivalent to (3.1). On
the other-hand, by (2.3) we have [$(x) ]1−q > [B( λ2 , λ2 )]1−q . Then in view of (2.5),
we have (3.3). By Hölder’s inequality, we find
"
#
Z ∞
∞
X
−1
1
a
n
I=
[Φ p (x)f (x)] Φ p (x)
dx ≤ ||f ||p,Φ L.
λ
[1
+
ln(x
−
α)
ln(n
−
β)]
1+α
n=2
(3.5)
Then by (3.3), we have (3.1). On the other-hand, assume that (3.1) is valid.
Setting
"∞
#q−1
X
a
n
f (x) := [Φ(x)]1−q
, x ∈ (1 + α, ∞),
[1 + ln(x − α) ln(n − β)]λ
n=2
then Lq−1 = ||f ||p,Φ . By (2.5), we find L < ∞. If L = 0, then (3.3) is trivially
valid; if L > 0, then by (3.1), we have
λ λ
||f ||pp,Φ = Lp(q−1) = I < B( , )||f ||p,Φ ||a||q,Ψ ,
2 2
148
B. YANG
λ λ
therefore ||f ||p−1
p,Φ = L < B( 2 , 2 )||a||q,Ψ , that is, (3.3) is equivalent to (3.1). Hence,
(3.1), (3.2) and (3.3) are equivalent.
ε
λ
1
, setting fe(x) = x−α
ln 2 + p −1 (x − α), x ∈ (1 + α, e + α); fe(x) =
For 0 < ε < pλ
2
λ
ε
1
0, x ∈ [e + α, ∞), and e
an = n−β
ln 2 − q −1 (n − β), n ∈ N\{1}, if there exists a
positive number k(≤ B( λ2 , λ2 )), such that (3.1) is valid as we replace B( λ2 , λ2 ) with
k, then in particular, it follows that
∞ Z ∞
X
e
an fe(x)dx
Ie :=
< k||fe||p,Φ ||e
a||q,Ψ
λ
[1
+
ln(x
−
α)
ln(n
−
β)]
1+α
n=2
p1
Z e+α
dx
= k
−ε+1
(x − α)
1+α (x − α) ln
) 1q
(
∞
X
1
1
+
×
(2 − β) lnε+1 (2 − β) n=3 (n − β) lnε+1 (n − β)
1q
Z ∞
1 p1
1
dy
< k( )
+
ε
(2 − β) lnε+1 (2 − β)
(y − β) lnε+1 (y − β)
2
1q
k
ε
1
=
+
,
(3.6)
ε (2 − β) lnε+1 (2 − β) lnε (2 − β)
Ie =
λ
ε
Z
∞
X
ln 2 − q −1 (n − β)
n−β
n=2
e+α
1+α
λ
ε
ln 2 + p −1 (x − α)dx
(x − α)[1 + ln(x − α) ln(n − β)]λ
Z ln(n−β) λ2 + pε −1
1
t
=
dt
ε+1
(1 + t)λ
(n − β) ln (n − β) 0
n=2
∞
λ ε λ ε X
1
= B
− A(ε)
+ , −
2 p 2 p n=2 (n − β) lnε+1 (n − β)
Z ∞
λ ε λ ε
dy
> B
+ , −
− A(ε)
2 p 2 p
(y − β) lnε+1 (y − β)
2
1
λ ε λ ε
=
B
+ , −
− A(ε),
ε lnε (2 − β)
2 p 2 p
Z ∞
∞
X
λ
ε
1
1
A(ε) : =
t 2 + p −1 dt.
(3.7)
ε+1
λ
(t
+
1)
(n
−
β)
ln
(n
−
β)
ln(n−β)
n=2
t=ln(x−α) ln(n−β)
∞
X
We find
0 < A(ε) ≤
=
λ
2
1
−
∞
X
n=2
∞
X
ε
p n=2
1
(n − β) lnε+1 (n − β)
(n − β) ln
Z
1
λ
+ qε +1
2
∞
ln(n−β)
< ∞,
(n − β)
1 λ2 + pε −1
t
dt
tλ
HALF-DISCRETE MULHOLLAND-TYPE INEQUALITY WITH PARAMETERS
149
and so A(ε) = O(1)(ε → 0+ ). Hence by (3.6) and (3.7), it follows that
1
λ ε λ ε
B
+ , −
− εO(1)
lnε (2 − β)
2 p 2 p
1q
ε
1
< k
+
,
(2 − β) lnε+1 (2 − β) lnε (2 − β)
and B( λ2 , λ2 ) ≤ k(ε → 0+ ). Hence k = B( λ2 , λ2 ) is the best value of (3.1).
By the equivalence of the inequalities, in view of (3.4) and (3.5), the constant
factor B( λ2 , λ2 ) in (3.2) and (3.3) is the best possible.
Remark 3.2. (i) Define the first type half-discrete Hilbert-type operator T1 :
Lp,Φ (1 + α, ∞) → lp,Ψ1−p as follows: For f ∈ Lp,Φ (1 + α, ∞), we define T1 f
∈ lp,Ψ1−p by
Z ∞
1
f (x)dx, n ∈ N\{1}.
T1 f (n) =
λ
1+α [1 + ln(x − α) ln(n − β)]
Then by (3.2), ||T1 f ||p.Ψ1−p ≤ B( λ2 , λ2 )||f ||p,Φ and so T1 is a bounded operator
with ||T1 || ≤ B( λ2 , λ2 ). Since by Theorem 3.1, the constant factor in (3.2) is best
possible, we have ||T1 || = B( λ2 , λ2 ).
(ii) Define the second type half-discrete Hilbert-type operator T2 : lq,Ψ →
Lq,Φ1−q (1 + α, ∞) as follows: For a ∈ lq,Ψ , we define T2 a ∈ Lq,Φ1−q (1 + α, ∞) by
T2 a(x) =
∞
X
n=2
1
an , x ∈ (1 + α, ∞).
[1 + ln(x − α) ln(n − β)]λ
Then by (3.3), ||T2 a||q.Φ1−q ≤ B( λ2 , λ2 )||a||q,Ψ and so T2 is a bounded operator
with ||T2 || ≤ B( λ2 , λ2 ). Since by Theorem 3.1, the constant factor in (3.3) is best
possible, we have ||T2 || = B( λ2 , λ2 ).
Remark 3.3. For p = q = 2, λ = 1 in (3.1), (3.2) and (3.3), (i) if α = β = 12 , then
we have (1.7) and the following equivalent inequalities:
"Z
#2
Z ∞
∞
∞
X
1
f (x)dx
1
2
<π
(x − )f 2 (x)dx,
1
1
1
3
3
2
n− 2
1 + ln(x − 2 ) ln(n − 2 )
n=2
2
2
∞
Z
3
2
1
x−
"
1
2
∞
X
n=2
an
1 + ln(x − 21 ) ln(n − 12 )
#2
dx < π
2
∞
X
1
(n − )a2n ;
2
n=2
(ii) if α = β = 0, then we have the following equivalent inequalities
(Z
) 21
Z ∞
∞
∞
∞
X
X
an
f (x)
dx < π
xf 2 (x)dx
na2n
,
1
+
ln
x
ln
n
1
1
n=2
n=2
Z ∞
2
Z ∞
∞
X
f (x)
1
2
dx < π
xf 2 (x)dx,
n
1
+
ln
x
ln
n
1
1
n=2
150
B. YANG
Z
1
∞
"∞
#2
∞
X
1 X
an
2
na2n .
dx < π
x n=2 1 + ln x ln n
n=2
Acknowledgement. This work is supported by Chinese Guangdong Natural
Science Foundation (No. 7004344).
References
1. L. Azar, On some extensions of Hardy-Hilbert’s inequality and Applications, J. Inequal.
App. 2008, Art. ID 546829, 14 pp.
2. B. Arpad and O. choonghong, Best constant for certain multilinear integral operator, J.
Inequal. Appl. 2006, Art. ID 28582, 12 pp.
3. G.H. Hardy, J.E. Littlewood and G. Pćlya, Inequalities, Cambridge University Press, Cambridge, 1934.
4. J. Jin and L. Debnath, On a Hilbert-type linear series operator and its applications, J.
Math. Anal. Appl. 371 (2010), no. 2, 691–704.
5. M. Krnić and J. Pečarić, Hilbert’s inequalities and their reverses, Publ. Math. Debrecen,
67 (2005), no. 3-4, 315–331.
6. J. Kuang and L. Debnath, On Hilbert’s type inequalities on the weighted Orlicz spaces,
Pacific J. Appl. Math. 1 (2007), no. 1, 95–103.
7. J. Kuang, Applied Inequalities, Shangdong Science Technic Press, Jinan, 2004 (China).
8. J. Kuang, Introduction to Real Analysis, Hunan Education Press, Chansha, 1996 (China).
9. Y. Li and B. He, On inequalities of Hilbert’s type, Bull. Austral. Math. Soc. 76 (2007), no.
1, 1–13.
10. D.S. Mitrinović, J. Pečarić and A.M. Fink, Inequalities Involving Functions and their Integrals and Derivatives, Kluwer Acaremic Publishers, Boston, 1991.
11. B. Yang, Hilbert-type Integral Inequalities, Bentham Science Publishers Ltd., 2009.
12. B. Yang, Discrete Hilbert-type Inequalities, Bentham Science Publishers Ltd., 2011.
13. B. Yang, An extension of Mulholand’s inequality, Jordan J. Math. Stat 3 (2010), no.3,
151–157.
14. B. Yang, On Hilbert’s integral inequality, J. Math. Anal. Appl. 220 (1998), 778–785.
15. B. Yang, The Norm of Operator and Hilbert-type Inequalities, Science Press, Beijin, 2009
(China).
16. B. Yang and I. Brnetić, M. Krnić and J. Pečarić, Generalization of Hilbert and Hardy-Hilbert
integral inequalities, Math. Ineq. and Appl. 8 (2005), no. 2, 259–272.
17. B. Yang and Th.M. Rassias, On the way of weight coefficient and research for Hilbert-type
inequalities, Math. Inequal. Appl., 6 (2003), no. 4, 625–658.
18. B. Yang, A mixed Hilbert-type inequality with a best constant factor, Int. J. Pure Appl.
Math. 20 (2005), no. 3, 319–328.
19. B. Yang, A half-discrete Hilbert’s inequality, J. Guangdong Univ. Edu. 31 (2011), no. 3,
1–7.
20. B. Yang and Q. Chen, A half-discrete Hilbert-type inequality with a homogeneous kernel
and an extension, J. Inequal. Appl. 124 (2011), doi: 10.1186/1029-242X-2011-124.
21. B. Yang and Th.M. Rassias, On a Hilbert-type integral inequality in the subinterval and its
operator expression, Banach J. Math. Anal. 4 (2010), no. 2, 100–110.
22. W. Zhong, The Hilbert-type integral inequality with a homogeneous kernel of Lambda-degree,
J. Inequal. Appl. 2008, no. 917392, 12 pp.
Department of Mathematics, Guangdong University of and Education, Guangzhou,
Guangdong 510303, P. R. China.
E-mail address: bcyang@gdei.edu.cn