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Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2010, Article ID 393025, 18 pages
doi:10.1155/2010/393025
Research Article
A New Hilbert-Type Linear Operator with
a Composite Kernel and Its Applications
Wuyi Zhong
Department of Mathematics, Guangdong Institute of Education, Guangzhou, Guangdong 510303, China
Correspondence should be addressed to Wuyi Zhong,
Received 20 April 2010; Accepted 31 October 2010
Academic Editor: Ond
ˇ
rej Do
ˇ
sl
´
y
Copyright q 2010 Wuyi Zhong. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any m edium, provided the original work is properly cited.
A new Hilbert-type linear operator with a composite kernel function is built. As the applications,
two new more accurate operator inequalities and their equivalent forms are deduced. The constant
factors in these inequalities are proved to be the best possible.
1. Introduction
In 1908, Weyl 1 published the well-known Hilbert’s inequality as follows:
if a
n
, b
n
≥ 0 are real sequences, 0 <
∞
n1
a
2
n
< ∞ and 0 <
∞
n1
b
2
n
< ∞,then
∞
n1
∞
m1
a
m
b
n
m n
<π
∞
n1
a
2
n
∞
n1
b
2
n
1/2
,
1.1
where the constant factor π is the best possible.
Under the same conditions, there are the classical inequalities 2
∞
n0
∞
m0
a
m
b
n
m n 1
<π
∞
n1
a
2
n
∞
n1
b
2
n
1/2
,
1.2
∞
n1
∞
m1
ln
m/n
a
m
b
n
m − n
<π
2
∞
n1
a
2
n
∞
n1
b
2
n
1/2
,
1.3
2 Journal of Inequalities and Applications
where the constant factors π and π
2
are the best possible also. Expression 1.2 is called a
more accurate form of 1.1. Some more accurate inequalities were considered by 3–5. In 2009,
Zhong 5 gave a more accurate form of 1.3.
Set p, q, s, r as two pairs of conjugate exponents, and p>1, s>1, α ≥ 1/2, and a
n
,
b
n
≥ 0, such that 0 <
∞
n0
n α
p1−λ/r−1
a
p
n
< ∞ and 0 <
∞
n0
n α
q1−λ/s−1
b
q
n
< ∞,thenit
has
∞
n0
∞
m0
ln
m α
/
n α
a
m
b
n
m α
λ
−
n α
λ
<
∞
n0
n α
p1−λ/r−1
a
p
n
1/p
∞
n0
n α
q1−λ/s−1
b
q
n
1/q
.
1.4
Letting φx :x α
p1−λ/r−1
, ϕx :x α
q1−λ/s−1
,
p
φ
: {a; a {a
n
}
∞
n0
, a
p,φ
:
{
∞
n0
φn|a
n
|
p
}
1/p
< ∞},
q
ϕ
: {b; b {b
n
}
∞
n0
,and b
q,ϕ
: {
∞
n0
ϕn|b
n
|
q
}
1/q
< ∞},the
expression 1.4 canberewrittenas
Ta,b
:
∞
n0
∞
m0
ln
m α
/
n α
a
m
b
n
m α
λ
−
n α
λ
<k
λ
s
a
p,φ
b
q,ϕ
,
1.5
where T :
p
φ
→
p
ψ
is a linear operator, k
λ
sT. a
p,φ
is the norm of the sequence a with
a weight function φ. Ta,b is a formal inner product of the sequences Tan:
∞
m0
lnm
α/n αa
m
/m α
λ
− n α
λ
and b.
By setting two monotonic incr easing functions ux and vx, a new Hilbert-type
inequality, which is with a composite kernel function Kux,vy, and its equivalent are
built in this paper. As the applications, two new more accurate Hilbert-type inequalities
incorporating the linear operator and the norm are deduced.
Firstly, the improved Euler-Maclaurin’s summation formula 6 is introduced.
Set f ∈ C
4
m, ∞m ∈ N
0
.If−1
i
f
i
x > 0, f
i
∞0 i 0 , 1, 2, 3,4,thenithas
∞
nm
f
n
<
∞
m
f
x
dx
1
2
f
m
−
1
12
f
m
,
∞
nm
f
n
>
∞
m
f
x
dx
1
2
f
m
.
1.6
2. Lemmas
Lemma 2.1. Set r, s as a pair of conjugate exponents, s>1, β ≥ α ≥ e
7/12
, 0 <λ≤ 1,anddefine
f
y
:
1
ln
λ
αm ln
λ
βy
ln
λ/r
αm
ln
1−λ/s
βy
y
,y∈
1, ∞
,m∈ N,
2.1
g
y
:
1
ln
λ
αy ln
λ
βn
ln
λ/s
βn
ln
1−λ/r
αy
y
,y∈
1, ∞
,n∈ N, 2.2
Journal of Inequalities and Applications 3
R
λ
m, s
:
ln β/ln αm
0
u
λ/s−1
1 u
λ
du −
1
2
f
1
1
12
f
1
, 2.3
R
λ
n, r
:
ln α/ ln βn
0
u
λ/r−1
1 u
λ
du −
1
2
g
1
1
12
g
1
, 2.4
η
λ
m, s
:
ln β/ln αm
0
u
λ/s−1
1 u
λ
du −
1
2
f
1
. 2.5
Then, it has the following.
1 The functions fy, gy satisfy the conditions of 1.6.Itmeansthat
−1
i
F
i
y
> 0
F f, g, y ∈
1, ∞
,
F
i
∞
0
F f, g, i 0, 1, 2,3, 4
,
2.6
2
R
λ
m, s
> 0,
R
λ
n, r
> 0,
2.7
3
0 <η
λ
m, s
O
1
ln αm
ρ
ρ>0,m−→ ∞
.
2.8
Proof. 1 For β ≥ α ≥ e
7/12
, y ≥ 1, m ∈ N,0<λ≤ 1, and s>1, set
f
1
y
:
1
ln
λ
αm ln
λ
βy
,f
2
y
:
1
ln
1−λ/s
βy
,f
3
y
:
1
y
.
2.9
It has
f
y
ln
λ/r
αmf
1
y
f
2
y
f
3
y
2.10
when y ≥ 1. It is easy to find that
−1
i
f
i
j
y
> 0,f
i
j
∞
0
y ∈
1, ∞
,j 1, 2, 3,i 0, 1, 2, 3, 4
,
−1
i
f
i
y
> 0,f
i
∞
0
y ∈
1, ∞
,i 0, 1, 2, 3, 4
.
2.11
4 Journal of Inequalities and Applications
Similarly, it can be shown that −1
i
g
i
y > 0, g
i
∞0 y ≥ 1, i 0, 1, 2, 3, 4.These
tell us that 2.6 holds and the functions fy, gy satisfy the conditions of 1.6.
2 Set t u
λ
. With the partial integration, it has
ln β/ln αm
0
u
λ/s−1
1 u
λ
du
1
λ
ln β/ ln αm
λ
0
t
1/s−1
1 t
dt
s
λ
ln β/ ln αm
λ
0
dt
1/s
1 t
s
λ
ln β/ ln αm
λ/s
1 ln β/ ln αm
λ
s
λ
ln β/ ln αm
λ
0
t
1/s
1 t
2
dt
s
λ
ln
λ
β
ln
λ
αm ln
λ
β
ln αm
ln β
λ/r
s
2
λ
1 s
ln β/ ln αm
λ
0
dt
1/s1
1 t
2
≥
s
λ
ln
λ
β
ln
λ
αm ln
λ
β
ln αm
ln β
λ/r
s
2
λ
1 s
ln
2λ
β
ln
λ
αm ln
λ
β
2
ln αm
ln β
λ/r
.
2.12
By 2.1,ithas
f
1
ln αm
ln β
λ/r
ln
λ−1
β
ln
λ
αm ln
λ
β
,
2.13
f
1
ln αm
ln β
λ/r
−
λln
2λ−2
β
ln
λ
αm ln
λ
β
2
−
1 − λ/s
ln
λ−2
β
ln
λ
αm ln
λ
β
−
ln
λ−1
β
ln
λ
αm ln
λ
β
.
2.14
In view of 2.12∼2.14,ithas
R
λ
m, s
≥
ln αm
ln β
λ/r
ln
λ−1
β
ln
λ
αm ln
λ
β
−
1 − λ/s
12 ln β
−
7
12
s
λ
ln β
ln
2λ
β
ln
λ
αm ln
λ
β
2
s
2
λ
1 s
−
λ
12ln
2
β
.
2.15
As ln β ≥ 7/12, s>1r>1,and0<λ≤ 1, it has
−
1 − λ/s
12 ln β
−
7
12
s
λ
ln β ≥
7s
12λ
1 −
λ
s
−
1
12 ln β
1 −
λ
s
1 −
λ
s
7s
12λ
−
1
12 ln β
>
1 −
λ
s
7
12
−
1
12 ln β
> 0,
s
2
λ
1 s
−
λ
12ln
2
β
>
s
4
−
s
12ln
2
β
s
4
1 −
1
3ln
2
β
> 0.
2.16
It means that R
λ
m, s > 0. Similarly, it can be shown that
R
λ
n, r > 0. The expression 2.7
holds.
Journal of Inequalities and Applications 5
3 By 2.5, 2.12, 2.13,and0<λ<s, β ≥ e
7/12
,ithas
η
λ
m, s
≥
s
λ
ln
λ
β
ln
λ
αm ln
λ
β
ln αm
ln β
λ/r
−
1
2
ln
λ−1
β
ln
λ
αm ln
λ
β
ln αm
ln β
λ/r
ln αm
ln β
λ/r
ln
λ−1
β
ln
λ
αmln
λ
β
s ln β
λ
−
1
2
>
ln αm
ln β
λ/r
ln
λ−1
β
ln
λ
αm ln
λ
β
ln β −
1
2
> 0,
η
λ
m, s
<
1
λ
ln β/ ln αm
λ
0
u
1/s−1
du
s
λ
ln β
ln αm
λ/s
.
2.17
The expression 2.8 holds, and Lemma 2.1 is proved.
Lemma 2.2. Set r, s as a pair of conjugate exponents, s>1, β ≥ α ≥ 1/2,and0 <λ≤ 1,and
define
f
1
y
:
1
λ
m α
ln y β/m α
λ
y β/m α
λ
− 1
y β
m α
λ/s−1
,y∈
0, ∞
,m∈ N
0
,
g
1
y
:
1
λ
n β
ln y α/n β
λ
y α/n β
λ
− 1
y α
n β
λ/r−1
,y∈
0, ∞
,n∈ N
0
,
R
λ
m, s
:
1
λ
2
β/mα
λ
0
ln u
u − 1
u
1/s−1
du −
1
2
f
1
0
1
12
f
1
0
,
R
λ
n, r
:
1
λ
2
α/nβ
λ
0
ln u
u − 1
u
1/r−1
du −
1
2
g
1
0
1
12
g
1
0
,
η
λ
m, s
:
1
λ
2
β/mα
λ
0
ln u
u − 1
u
1/s−1
du −
1
2
f
1
0
.
2.18
Then, it has
1 The functions f
1
y, g
1
y satisfy the conditions of 1.6.Itmeansthat
−1
i
F
i
y
> 0
F f
1
,g
1
, y ∈
0, ∞
,
F
i
∞
0
F f
1
,g
1
, i 0, 1, 2, 3, 4
,
2.19
2
R
λ
m, s
> 0,
R
λ
n, r
> 0,
2.20
6 Journal of Inequalities and Applications
3
0 <η
λ
m, s
O
1
m α
ρ
ρ>0,m−→ ∞
.
2.21
Proof. 1 Letting hu : ln u/u − 1, u y β/m α
λ
, it can be proved that f
1
y
1/λm αhuu
1/s−1/λ
satisfy 2.19 as in 5. Similarly, it can be shown that g
1
y satisfy
2.19 also.
2 Setting u
0
:β/m α
λ
,byu
λy β/m α
λ−1
1/m α λ/y
βy β/m α
λ
,u
0λ/βu
0
,andh
u > 0, it has
1
λ
2
β/mα
λ
0
ln u
u − 1
u
1/s−1
du
1
λ
2
u
0
0
h
u
u
1/s−1
du
s
λ
2
u
0
0
h
u
du
1/s
s
λ
2
h
u
0
u
1/s
0
−
u
0
0
u
1/s
h
u
du
≥
s
λ
2
h
u
0
u
1/s
0
− h
u
0
u
0
0
u
1/s
du
s
λ
2
h
u
0
u
1/s
0
−
s
s 1
h
u
0
u
1/s1
0
,
2.22
f
1
0
1
λ
m α
ln β/m α
λ
β/m α
λ
− 1
β
m α
λ/s−1
1
λβ
h
u
0
u
1/s
0
,
2.23
f
1
0
1
β
2
u
1/s
0
h
u
0
u
0
1
s
−
1
λ
h
u
0
. 2.24
With 2.22∼2.24,ithas
R
λ
m, s
≥ h
u
0
u
1/s
0
s
λ
2
−
1
2λβ
1
12β
2
1
s
−
1
λ
− h
u
0
u
1/s1
0
s
1 s
−
1
12β
2
.
2.25
By hu
0
> 0, h
u
0
< 0, and β ≥ 1/2, s>1, 0 <λ≤ 1, it has
s
λ
2
−
1
2λβ
1
12β
2
1
s
−
1
λ
6βs
2βs − λ
− λ
s − λ
12β
2
sλ
2
> 0,
s
1 s
−
1
12β
2
12β
2
s −
1 s
12β
2
1 s
4β
2
s − s
8β
2
s − 1
12βλ
1 s
> 0.
2.26
So R
λ
m, s > 0 holds. Similarly, it can be shown that
R
λ
n, r > 0.
Journal of Inequalities and Applications 7
3 In view of 2.22, 2.23,byhu > 0, h
u < 0, it has
η
λ
m, s
>h
u
0
u
1/s
0
s
λ
2
−
1
2λβ
h
u
0
u
1/s
0
2βs − λ
2λ
2
β
> 0,
2.27
and by lim
u → 0
ln u/u − 1u
1/2s
0, so there exists a constant L>0, such that |ln u/u −
1u
1/2s
| <Lu ∈ 0, β/m α
λ
.Thenithas
η
λ
m, s
<
1
λ
2
β/mα
λ
0
ln u
u − 1
u
1/s−1
du <
L
λ
2
β/mα
λ
0
u
1/2s−1
du
2sL
λ
2
β
m α
λ/2s
.
2.28
It means that 2.21 holds. The proof for Lemma 2.2 is finished.
3. Main Results
Set λ ∈ R, p>1,r>1,p, q,andr, s as two pairs of conjugate exponents. Kx, y ≥ 0x, y ∈
0, ∞ × 0, ∞ is a measurable kernel function. Both ux and vx are strictly monotonic
increasing differentiable functions in n
0
, ∞ such that Un
0
> 0,U∞∞U u, v.Give
some notations as follows:
1
φ
x
:
u
x
p1−λ/r−1
u
x
1−p
,
ϕ
x
:
v
x
q1−λ/s−1
v
x
1−q
,
ψ
x
:
ϕ
x
1−p
v
x
pλ/s−1
v
x
x ∈
n
0
, ∞
,
3.1
2 set
p
φ
:
⎧
⎨
⎩
a; a
{
a
n
}
∞
nn
0
,
a
p,φ
:
∞
nn
0
φ
n
|
a
n
|
p
1/p
< ∞
⎫
⎬
⎭
, 3.2
and call
p
φ
a real space of sequences,where
a
p,φ
∞
nn
0
φ
n
|
a
n
|
p
1/p
3.3
is called the norm of the sequence with a weight function φ. Similarly, the real spaces of sequences
q
ϕ
,
p
ψ
and the norm b
q,ϕ
can be defined as well,
8 Journal of Inequalities and Applications
3 define a Hilbert-type linear operator T :
p
φ
→
p
ψ
,foralla ∈
p
φ
,
Ta
n
: C
n
:
∞
mn
0
K
u
m
,v
n
a
m
n ≥ n
0
,
3.4
4 for all a ∈
p
φ
, b ∈
q
ϕ
,define the formal inner product of Ta and b as
Ta,b
:
∞
nn
0
∞
mn
0
K
u
m
,v
n
a
m
b
n
∞
nn
0
∞
mn
0
K
u
m
,v
n
a
m
b
n
,
3.5
5 define two weight coefficients ωm, s and ϑn, r as
ω
λ
m, s
:
∞
nn
0
K
u
m
,v
n
u
m
λ/r
v
n
1−λ/s
v
n
,
ϑ
λ
n, r
:
∞
mn
0
K
u
m
,v
n
v
n
λ/s
u
m
1−λ/r
u
m
,m,n≥ n
0
.
3.6
Then it has some results in the following theorems.
Theorem 3.1. Suppose that a
n
≥ 0, U
x/Ux > 0U u, v,and0 <
∞
nn
0
v
n/vn
1ε
≤
∞
nn
0
u
n/un
1ε
< ∞ε>0. If there exists a positive number
k
λ
,suchthat
0 <ω
λ
m, s
<k
λ
, 0 <ϑ
λ
n, r
<k
λ
m, n ≥ n
0
, 3.7
k
λ
1 − O
1
u
m
ρ
≤ ω
λ
m, s
ρ>0,m−→ ∞
,
3.8
then for all a ∈
p
φ
and a
p,φ
> 0, it has the following:
1
Ta C
{
C
n
}
∞
nn
0
∈
p
ψ
,
3.9
It means that T :
p
φ
→
p
ψ
,
Journal of Inequalities and Applications 9
2 T is a bounded linear operator and
T
p,ψ
: sup
a∈
p
φ
a
/
θ
Ta
p,ψ
a
p,φ
k
λ
,
3.10
where C
n
, T are defined by 3.4, Ta
p,ψ
C
p,ψ
is defined as 3.3.
Proof. By using H
¨
older’s inequality 7 and 3.6, 3.7,ithasC
n
≥ 0and
C
p
n
∞
mn
0
K
u
m
,v
n
u
m
1−λ/r/q
v
n
1/p
v
n
1−λ/s/p
u
m
1/q
a
m
v
n
1−λ/s/p
u
m
1/q
u
m
1−λ/r/q
v
n
1/p
p
≤
∞
mn
0
K
u
m
,v
n
u
m
p−11−λ/r
v
n
v
n
1−λ/s
u
m
p−1
a
p
m
×
∞
mn
0
K
u
m
,v
n
v
n
q−11−λ/s
u
m
u
m
1−λ/r
v
n
q−1
p−1
∞
mn
0
K
u
m
,v
n
u
m
p−11−λ/r
v
n
v
n
1−λ/s
u
m
p−1
a
p
m
ϑ
λ
n, r
ϕ
n
p−1
<k
p−1
λ
∞
mn
0
K
u
m
,v
n
u
m
p−11−λ/r
v
n
v
n
1−λ/s
u
m
p−1
a
p
m
ϕ
p−1
n
.
3.11
And by ψnϕ
1−p
n, it follows that
Ta
p
p,ψ
∞
nn
0
ψ
n
C
p
n
<k
p−1
λ
∞
mn
0
∞
nn
0
K
u
m
,v
n
u
m
p−11−λ/r
v
n
v
n
1−λ/s
u
m
p−1
a
p
m
k
p−1
λ
∞
mn
0
ω
λ
m, s
φ
m
a
p
m
<k
p
λ
a
p
p,φ
< ∞.
3.12
This means that C {C
n
}
∞
n0
∈
p
ψ
, Ta
p,ψ
≤ k
λ
a
p,φ
,andT
p,ψ
≤ k
λ
. T is a bounded linear
operator.
If there exists a constant K<k
λ
,suchthatT
p,ψ
≤ K,thenforε>0, setting a
m
:
um
λ/r−ε/p−1
u
m,
b
n
:vn
λ/s−ε/q−1
v
n,ithasa {a
m
}
∞
mn
0
∈
p
φ
,
b {
b
n
}
∞
nn
0
∈
q
ϕ
,
and
T a,
b
≤
T
p,ψ
a
p,φ
b
q,ϕ
≤ K
∞
mn
0
u
m
u
m
1ε
1/p
∞
nn
0
v
n
v
n
1ε
1/q
≤ K
∞
mn
0
u
m
u
m
1ε
.
3.13
10 Journal of Inequalities and Applications
But on the other side, by 3.8,ithas
T a,
b
∞
mn
0
∞
nn
0
K
u
m
,v
n
u
m
λ/r−ε/p−1
v
n
1−λ/sε/q
u
m
v
n
∞
mn
0
u
m
u
m
1ε
∞
nn
0
K
u
m
,v
n
u
m
λ/rε/q
v
n
1−λ/sε/q
v
n
.
3.14
By the s trictly monotonic increase of vx and vn
0
> 0, v∞∞,thereexistsn
1
>n
0
such that vn > 1foralln>n
1
.Soithas
0 <
∞
nn
0
K
u
m
,v
n
u
m
λ/rε/q
v
n
1−λ/sε/q
v
n
u
m
ε/q
∞
nn
1
K
u
m
,v
n
u
m
λ/r
v
n
v
n
1−λ/sε/q
n
1
−1
nn
0
K
u
m
,v
n
u
m
λ/r
v
n
v
n
1−λ/sε/q
≤
u
m
ε/q
∞
nn
1
K
u
m
,v
n
u
m
λ/r
v
n
v
n
1−λ/s
n
1
−1
nn
0
K
u
m
,v
n
u
m
λ/r
v
n
v
n
1−λ/sε/q
u
m
ε/q
ω
λ
m, s
−
n
1
−1
nn
0
K
u
m
,v
n
u
m
λ/r
v
n
v
n
1−λ/s
n
1
−1
nn
0
K
u
m
,v
n
u
m
λ/r
v
n
v
n
1−λ/sε/q
.
3.15
The series is uniformly convergent for ε ≥ 0, so it has
lim
ε → 0
∞
nn
0
K
u
m
,v
n
u
m
λ/rε/q
v
n
1−λ/sε/q
v
n
ω
λ
m, s
3.16
and for m>n
0
,thereexistsε
0
> 0, when 0 <ε<ε
0
,ithas
∞
nn
0
K
u
m
,v
n
u
m
λ/rε/q
v
n
1−λ/sε/q
v
n
>ω
λ
m, s
−
1
u
m
.
3.17
Journal of Inequalities and Applications 11
By 3.14 and 3.8,when0<ε<ε
0
,ithas
T a,
b
≥
∞
mn
0
u
m
u
m
1ε
ω
λ
m, s
−
1
u
m
≥ k
λ
∞
mn
0
u
m
u
m
1ε
1 − O
1
u
m
ρ
−
1
k
λ
u
m
k
λ
∞
mn
0
u
m
u
m
1ε
⎧
⎨
⎩
1 −
∞
mn
0
u
m
u
m
1ε
−1
×
∞
mn
0
u
m
u
m
1ε
O
1
u
m
ρ
1
k
λ
u
m
⎫
⎬
⎭
.
3.18
In view of 3.13 and 3.18, letting ε → 0
,ithask
λ
≤ K. This means that K k
λ
;thatis,
T
p,ψ
k
λ
. Theorem 3.1 is proved.
Theorem 3.2. Suppose that p, q and r, s are two pairs of conjugate exponents, r>1, p>1,
λ ∈ R.Let
f
y
: K
λ
u
m
,v
y
u
λ/r
m
v
1−λ/s
y
v
y
,
g
y
: K
λ
u
y
,v
n
v
λ/s
n
u
1−λ/r
y
u
y
.
3.19
Here, uy, vy satisfy the conditions as in Theorem 3.1.Set
R
m, s
:
vn
0
/um
0
K
λ
1,u
u
λ/s−1
du −
1
2
f
n
0
1
12
f
n
0
,
3.20
R
n, r
:
un
0
/vn
0
K
λ
μ, 1
μ
λ/r−1
dμ −
1
2
g
n
0
1
12
g
n
0
, 3.21
η
m, s
:
vn
0
/um
0
K
λ
1,u
u
λ/s−1
du −
1
2
f
n
0
.
3.22
If (a) K
λ
x, y ≥ 0 is a homogeneous measurable kernel function of “λ”degreeinR
2
,suchthat
0 <k
λ
s
:
∞
0
K
λ
1,u
u
λ/s−1
du < ∞,
3.23
(b) functions fy, gy satisfy the conditions of 1.6;thatis,
−1
i
F
i
y
> 0
y>n
0
,F
i
∞
0
F f, g, i 0, 1, 2, 3, 4
,
3.24
12 Journal of Inequalities and Applications
(c) there exists ρ>0,suchthat
R
m, s
> 0,
R
n, r
> 0, 0 <η
m, s
O
1
u
ρ
m
m −→ ∞
,
3.25
then it has
1 if a ∈
p
φ
, b ∈
q
ϕ
,anda
p,φ
> 0, b
q,ϕ
> 0,then
Ta,b
∞
nn
0
∞
mn
0
K
λ
u
m
,v
n
a
m
b
n
<k
λ
s
a
p,φ
b
q,ϕ
,
3.26
2 if a ∈
p
φ
and a
p,φ
> 0,then
Ta
p,ψ
∞
nn
0
v
n
pλ/s−1
v
n
∞
mn
0
K
u
m
,v
n
a
m
p
1/p
<k
λ
s
a
p,φ
,
3.27
where inequality 3.27 is equivalent to 3.26 and the constant factor k
λ
s
k
λ
r :
∞
0
K
λ
u, 1u
λ/r−1
du is the best possible.
Proof. By 3.24, 1.6,ithas
∞
n
0
f
y
dy −
1
2
f
n
0
<ω
λ
m, s
∞
nn
0
K
λ
u
m
,v
n
u
m
λ/r
v
n
1−λ/s
v
n
∞
nn
0
f
n
<
∞
n
0
f
y
dy −
1
2
f
n
0
1
12
f
n
0
,
3.28
0 <ϑ
λ
n, r
∞
mn
0
K
λ
u
m
,v
n
vn
λ/s
u
m
1−λ/r
u
m
∞
mn
0
g
m
<
∞
n
0
g
y
dy −
1
2
g
n
0
1
12
g
n
0
.
3.29
Journal of Inequalities and Applications 13
Letting ν vy/um and μ uy/vn in the integral of 3.28 and 3.29, respectively, by
3.23,ithas
∞
n
0
f
y
dy
∞
n
0
K
λ
u
m
,v
y
u
λ/r
m
v
1−λ/s
y
v
y
dy
∞
vn
0
/um
K
λ
1,ν
ν
λ/s−1
dν
∞
0
K
λ
1,ν
ν
λ/s−1
dν −
vn
0
/um
0
K
λ
1,ν
ν
λ/s−1
dν k
λ
s
−
vn
0
/um
0
K
λ
1,ν
ν
λ/s−1
dν,
3.30
∞
n
0
g
y
dy
∞
n
0
K
λ
u
y
,v
n
v
λ/s
n
u
1−λ/r
y
u
y
dy
∞
un
0
/vn
K
λ
μ, 1
μ
λ/r−1
dμ
∞
0
K
λ
μ, 1
μ
λ/r−1
dμ −
un
0
/vn
0
K
λ
μ, 1
μ
λ/r−1
dμ k
λ
s
−
un
0
/vn
0
K
λ
μ, 1
μ
λ/r−1
dμ,
3.31
where, letting t 1/u,ithas
k
λ
r
∞
0
K
λ
u, 1u
λ/r−1
du
∞
0
K
λ
1,tt
λ/s−1
dt k
λ
s.In
view of 3.28, 3.30, 3.20, 3.22,andwith3.25,ithas
0 <ω
λ
m, s
<k
λ
s
− R
λ
m, s
<k
λ
s
,
ω
λ
m, s
>k
λ
s
− η
λ
m, s
k
λ
s
1 − O
1
1
u
ρ
m
ρ>0,m−→ ∞
.
3.32
Similarly, with 3.29, 3.31, 3.21,and3.25,ithas
0 <ϑ
λ
n, r
<k
λ
s
3.33
also. By Theorem 3.1,ithas
Ta
p,ψ
<k
λ
s
a
p,φ
,
3.34
and 3.27 holds. In view of
Ta,b
≤
Ta
p,ψ
b
q,ϕ
,
3.35
3.26 holds also.
14 Journal of Inequalities and Applications
If 3.26 holds, from 3.26 and a
p,φ
> 0, there exists n
1
>n
0
,suchthat
H
mn
0
φma
p
m
> 0andb
n
H : ψn
H
mn
0
K
λ
um,vna
m
p−1
> 0whenH>n
1
.For
b : {b
n
H}
H
nn
0
,ithas
0 <
H
nn
0
ϕ
n
b
q
n
H
H
nn
0
ψ
n
H
mn
0
K
λ
u
m
,v
n
a
m
p
H
nn
0
H
mn
0
K
λ
u
m
,v
n
a
m
b
n
H
<k
λ
s
H
nn
0
φ
n
a
p
n
1/p
H
nn
0
ϕ
n
b
q
n
H
1/q
< ∞.
3.36
By p>1andq>1, it follows that
0 <
H
nn
0
ϕ
n
b
q
n
H
<k
p
λ
s
∞
nn
0
φ
n
a
p
n
< ∞.
3.37
Letting H →∞in 3.37,ithas0 <
∞
nn
0
ϕnb
q
n
∞ < ∞, and it means that b
{b
n
∞}
∞
nn
0
∈
q
ϕ
and b
q,ϕ
> 0. Therefore, the inequality 3.36 keeps the form of the strict
inequality when H →∞.Inviewof
∞
nn
0
ϕnb
q
n
∞Ta
p
p,ψ
, inequality 3.27 holds
and 3.27 is equivalent to 3.26.ByT
p,ψ
k
λ
s, it is obvious that the constant factor
k
λ
sk
λ
r is the best possible. This completes the proof of Theorem 3.2.
4. Applications
Example 4.1. Set p, q, r, s be two pairs of conjugate exponents and p>1, s>1, β ≥ α ≥
e
7/12
,0<λ≤ 1. Then it has the following.
1 If 0 <
∞
n1
ln αn
p1−λ/r−1
a
p
n
/n
p−1
< ∞,and0<
∞
n1
ln βn
q1−λ/s−1
b
q
n
/n
q−1
<
∞,then
∞
m1
∞
n1
a
m
b
n
ln
λ
αm ln
λ
βn
<
π
λ sin
π/s
∞
n1
ln αn
p1−λ/r−1
a
p
n
n
p−1
1/p
∞
n1
ln βn
q1−λ/s−1
b
q
n
n
q−1
1/q
.
4.1
2 If 0 <
∞
n1
ln αn
p1−λ/r−1
a
p
n
/n
p−1
< ∞,then
∞
n1
ln βn
pλ/s−1
n
∞
m1
a
m
ln
λ
αm ln
λ
βn
p
<
π
λ sinπ/s
p
∞
n1
ln αn
p1−λ/r−1
a
p
n
n
p−1
,
4.2
where the constant factors k
λ
sk
λ
r : π/λsinπ/s and π/λsinπ/s
p
are both the
best possible. Inequality 4.2 is equivalent to 4.1.
Journal of Inequalities and Applications 15
Proof. Setting K
λ
x, y1/x
λ
y
λ
, x, y ∈ R
2
, it is a homogeneous m easurable kernel
function of “λ” degree. Letting t u
λ
,ithas
0 <k
λ
s
:
∞
0
K
λ
1,u
u
λ/s−1
du
∞
0
u
λ/s−1
1 u
λ
du
1
λ
∞
0
t
1/s−1
1 t
dt
1
λ
B
1
s
,
1
r
k
λ
r
< ∞.
4.3
Setting uxln αx, vxln βx,thenbothux and vx are strictly monotonic increasing
differentiable functions in 1, ∞ and satisfy
U
1
> 0,U
∞
∞
U u, v
,
0 <
∞
n1
v
n
v
n
1ε
∞
n1
1
n
ln βn
1ε
≤
∞
n1
u
n
u
n
1ε
∞
n1
1
n
ln αn
1ε
< ∞
4.4
for ε>0. As β ≥ α ≥ e
7/12
,0<λ≤ 1, s>1, and n
0
1, letting
f
y
K
λ
u
m
,v
y
u
λ/r
m
v
1−λ/s
y
v
y
1
ln
λ
αm ln
λ
βy
ln
λ/r
αm
ln
1−λ/s
βy
y
,
g
y
: K
λ
u
y
,v
n
v
λ/s
n
u
1−λ/r
y
u
y
1
ln
λ
αy ln
λ
βn
ln
λ/s
βn
ln
1−λ/r
αy
y
,
y ∈
1, ∞
,n,m∈ N,
4.5
with 2.1∼2.8,ithas
R
λ
m, s
v1/um
0
K
λ
1,u
u
λ/s−1
du −
1
2
f
1
1
12
f
1
> 0,
0 <η
λ
m, s
O
1
u
m
ρ
ρ>0,m−→ ∞
,
R
λ
n, r
:
u1/vn
0
K
λ
μ, 1
μ
λ/r−1
dμ −
1
2
g
1
1
12
g
1
> 0.
4.6
When 0 <
∞
n1
ln αn
p1−λ/r−1
a
p
n
/n
p−1
< ∞ and 0 <
∞
n1
ln βn
q1−λ/s−1
b
q
n
/n
q−1
< ∞;that
is, a ∈
p
φ
, b ∈
q
ϕ
and a
p,φ
> 0, b
q,ϕ
> 0, by Theorem 3.2, inequality 4.1 holds, so does
4.2.And4.2 is equivalent to 4.1, and the constant factors k
λ
sk
λ
r : π/λsinπ/s
and π/λsinπ/s
p
are both the best possible.
Example 4.2. Set p, q, r, s be two pairs of conjugate exponents and p>1, s>1, β ≥ α ≥ 1 /2,
0 <λ≤ 1. Then it has the following.
16 Journal of Inequalities and Applications
1 If 0 <
∞
n0
n α
p1−λ/r−1
a
p
n
< ∞ and 0 <
∞
n0
n β
q1−λ/s−1
b
q
n
< ∞,then
∞
n0
∞
m0
ln
m α
/
n β
a
m
b
n
m α
λ
−
n β
λ
<
π
λ sinπ/s
2
∞
n0
n α
p1−λ/r−1
a
p
n
1/p
∞
n0
n β
q1−λ/s−1
b
q
n
1/q
.
4.7
2 If 0 <
∞
n0
n α
p1−λ/r−1
a
p
n
< ∞,then
∞
n0
n β
pλ/s−1
∞
m0
lnm α/n βa
m
m α
λ
−
n β
λ
p
<
π
λ sinπ/s
2p
∞
n0
n α
p1−λ/r−1
a
p
n
,
4.8
where inequality 4.8 is equivalent to 4.7 and the constant factors k
λ
sk
λ
r :
π/λsinπ/s
2
and π/λsinπ/s
2p
are both the best possible.
Proof. Setting K
λ
x, ylnx/y/x
λ
− y
λ
x, y ∈ R
2
, it is a homogeneous measurable
kernel function of “λ” degree. Letting t u
λ
,ithas2
0 <k
λ
s
:
∞
0
K
λ
1,u
u
λ/s−1
du
1
λ
∞
0
− ln u
λ
1 − u
λ
u
λ/s−1
du
1
λ
2
∞
0
ln t
t − 1
t
1/s−1
dt
1
λ
B
1
s
,
1
r
2
π
λ sinπ/s
2
k
λ
r
< ∞.
4.9
Setting uxx α, vxx β,thenbothux and vx are strictly monotonic increasing
differentiable functions in 0, ∞ and satisfy
U
0
> 0,U
∞
∞
U u, v
,
0 <
∞
n0
v
n
v
n
1ε
∞
n0
1
n β
1ε
≤
∞
n0
u
n
u
n
1ε
∞
n0
1
n α
1ε
< ∞
4.10
Journal of Inequalities and Applications 17
for ε>0. As β ≥ α ≥ 1/2, 0 <λ≤ 1, s>1, and n
0
0, letting
f
1
y
K
λ
u
m
,v
y
u
λ/r
m
v
1−λ/s
y
v
y
ln
m α
/
y β
m α
λ
−
y β
λ
m α
λ/r
y β
1−λ/s
1
λ
m α
ln y β/m α
λ
y β/m α
λ
− 1
y β
m α
λ/s−1
,
g
1
y
: K
λ
u
y
,v
n
v
λ/s
n
u
1−λ/r
y
u
y
ln
y α
/
n β
y α
λ
−
n β
λ
n β
λ/s
y α
1−λ/r
1
λ
n β
ln y α/n β
λ
y α/n β
λ
− 1
y α
n β
λ/r−1
,y∈
0, ∞
,n,m∈ N,
4.11
with 2.18∼2.21,ithas
R
λ
m, s
:
v0/um
0
K
λ
1,u
u
λ/s−1
du −
1
2
f
1
0
1
12
f
1
0
1
λ
2
β/mα
λ
0
ln u
u − 1
u
1/s−1
du −
1
2
f
1
0
1
12
f
1
0
> 0,
R
λ
n, r
:
u0/vn
0
K
λ
μ, 1
μ
λ/r−1
dμ −
1
2
g
1
0
1
12
g
1
0
> 0,
0 <η
λ
m, s
O
1
u
m
ρ
ρ>0,m−→ ∞
.
4.12
When 0 <
∞
n0
n α
p1−λ/r−1
a
p
n
< ∞ and 0 <
∞
n0
n β
q1−λ/s−1
b
q
n
< ∞;thatis,
a ∈
p
φ
, b ∈
q
ϕ
and a
p,φ
> 0, b
q,ϕ
> 0, by Theorem 3.2, inequality 4.7 holds, so does 4.8.
And 4.8 is equivalent to 4.7, and the constant factors k
λ
sk
λ
r :π/λsinπ/s
2
and
π/λsinπ/s
2p
are both the best possible.
Remark 4.3. It can be proved similarly that, if the conditions “β ≥ α ≥ e
7/12
”inLemma 2.1 and
“β ≥ α ≥ 1/2” in Lemma 2.2 are changed into “α ≥ β ≥ e
7/12
”and“α ≥ β ≥ 1/2”, respectively,
Lemmas 2.1 and 2.2 are also valid. So the conditions “β ≥ α ≥ e
7/12
”inExample 4.1 and
“β ≥ α ≥ 1/2” in Example 4.2 can be replaced by “β ≥ e
7/12
, α ≥ e
7/12
”and“β ≥ 1/2,
α ≥ 1/2”, respectively.
Acknowledgment
This paper is supported by the National Natural Science Foundation of China (no. 10871073).The
author would like to thank the anonymous referee for his or her suggestions and corrections.
18 Journal of Inequalities and Applications
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ottingen, G
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