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, zwy@gdei.edu.cn
Received 20 April 2010; Accepted 31 October 2010
Academic Editor: Ondřej Došlý
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 medium, 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:
∞ 2
2
if an , bn ≥ 0 are real sequences, 0 < ∞
n1 an < ∞ and 0 <
n1 bn < ∞, then
1/2
∞ ∞
∞
∞
am bn
2
2
<π
an bn
,
mn
n1 m1
n1
n1
1.1
where the constant factor π is the best possible.
Under the same conditions, there are the classical inequalities 2
1/2
∞
∞
am bn
2
2
an bn
,
<π
mn1
n0 m0
n1
n1
∞ ∞
∞ ∞
lnm/nam bn
n1 m1
m−n
<π
2
∞
∞
a2n bn2
n1
n1
1.2
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 an ,
p1−λ/r−1 p
q1−λ/s−1 q
bn ≥ 0, such that 0 < ∞
an < ∞ and 0 < ∞
bn < ∞, then it
n0 n α
n0 n α
has
∞ ∞
lnm α/n αam bn
n0 m0
m αλ − n αλ
<
∞
1/p n p
αp1−λ/r−1 an
n0
∞
1/q
n q
αq1−λ/s−1 bn
.
n0
1.4
Letting φx : x αp1−λ/r−1 , ϕx : x αq1−λ/s−1 , φ : {a; a {an }∞
n0 , ap,φ :
∞
∞
q
∞
p 1/p
q 1/q
< ∞}, ϕ : {b; b {bn }n0 , and bq,ϕ : { n0 ϕn|bn | }
< ∞}, the
{ n0 φn|an | }
expression 1.4 can be rewritten as
p
Ta, b :
∞ ∞
lnm α/n αam bn
n0 m0
p
m αλ − n αλ
< kλ sap,φ bq,ϕ ,
1.5
p
where T : φ → ψ is a linear operator, kλ s T. ap,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 λ
λ
α/n αam /m α − n α and b.
By setting two monotonic increasing 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 ∈ C4 m, ∞m ∈ N0 . If −1i f i x > 0, f i ∞ 0 i 0, 1, 2, 3, 4, then it has
∞
nm
∞
1
1
fxdx fm − f m,
2
12
m
∞
∞
1
fn >
fxdx fm.
2
m
nm
fn <
1.6
2. Lemmas
Lemma 2.1. Set r, s as a pair of conjugate exponents, s > 1, β ≥ α ≥ e7/12 , 0 < λ ≤ 1, and define
f y :
lnλ/r αm
1
,
lnλ αm lnλ βy ln1−λ/s βy y
g y :
lnλ αy
lnλ/s βn
1
1−λ/r ,
λ
ln βn ln
αy y
y ∈ 1, ∞, m ∈ N,
2.1
y ∈ 1, ∞, n ∈ N,
2.2
Journal of Inequalities and Applications
Rλ m, s :
3
ln β/ ln αm
0
λ n, r :
R
ln α/ ln βn
0
ηλ m, s :
1
uλ/s−1
1
du − f1 f 1,
2
12
1 uλ
2.3
1
uλ/r−1
1
du − g1 g 1,
2
12
1 uλ
2.4
ln β/ ln αm
0
uλ/s−1
1
du − f1.
2
1 uλ
2.5
Then, it has the following.
1 The functions fy, gy satisfy the conditions of 1.6. It means that
−1i F i y > 0 F f, g, y ∈ 1, ∞ ,
F i ∞ 0 F f, g, i 0, 1, 2, 3, 4 ,
2.6
2
λ n, r > 0,
R
Rλ m, s > 0,
2.7
3
0 < ηλ m, s O
1
ln αm
ρ ρ > 0, m −→ ∞ .
2.8
Proof. 1 For β ≥ α ≥ e7/12 , y ≥ 1, m ∈ N, 0 < λ ≤ 1, and s > 1, set
f1 y :
lnλ αm
1
,
lnλ βy
f2 y :
1
ln1−λ/s βy
,
1
f3 y : .
y
2.9
It has
f y lnλ/r αmf1 y f2 y f3 y
2.10
when y ≥ 1. It is easy to find that
i −1i fj
y > 0,
i
fj ∞ 0
−1i f i y > 0,
y ∈ 1, ∞, j 1, 2, 3, i 0, 1, 2, 3, 4 ,
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 −1i 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 uλ . With the partial integration, it has
ln β/ ln αm
0
uλ/s−1
1
du λ
λ
1u
ln β/ ln αmλ
0
λ/s
s ln β/ ln αm
s
λ
λ 1 ln β/ ln αm
λ
lnλ β
s
λ lnλ αm lnλ β
≥
ln β/ ln αmλ
s
t1/s−1
dt 1t
λ
lnλ β
s
λ lnλ αm lnλ β
ln αm
ln β
ln αm
ln β
0
ln β/ ln αmλ
λ/r
t1/s
1 t2
0
λ/r
dt1/s
1t
s2
λ1 s
dt
ln β/ ln αmλ
dt1/s1
1 t2
ln2λ β
ln αm λ/r
s2
.
λ1 s lnλ αm lnλ β 2 ln β
2.12
0
By 2.1, it has
f1 f 1 ln αm
ln β
λ/r −
ln αm
ln β
λ/r
lnλ−1 β
,
lnλ αm lnλ β
2.13
lnλ−1 β
1 − λ/slnλ−2 β
− λ
.
2 −
lnλ αm lnλ β
ln αm lnλ β
lnλ αm lnλ β
λln2λ−2 β
2.14
In view of 2.12∼2.14, it has
Rλ m, s ≥
ln αm
ln β
λ/r lnλ−1 β
7
s
1 − λ/s
−
ln
β
−
12 ln β
12 λ
lnλ αm lnλ β
ln2λ β
2
lnλ αm lnλ β
λ
s2
−
λ1 s 12ln2 β
2.15
.
As ln β ≥ 7/12, s > 1r > 1, and 0 < λ ≤ 1, it has
7
s
7s
λ
1
λ
1 − λ/s
−
ln β ≥
1−
−
1−
12 ln β
12 λ
12λ
s
12 ln β
s
λ
7s
1
λ
7
1
1−
−
> 1−
−
> 0,
s
12λ 12 ln β
s
12 12 ln β
s2
s
s
λ
s
1
> −
> 0.
−
1−
λ1 s 12ln2 β 4 12ln2 β 4
3ln2 β
−
2.16
λ n, r > 0. The expression 2.7
It means that Rλ m, s > 0. Similarly, it can be shown that R
holds.
Journal of Inequalities and Applications
5
3 By 2.5, 2.12, 2.13, and 0 < λ < s, β ≥ e7/12 , it has
ηλ m, s ≥
lnλ β
s
λ
λ ln αm lnλ β
ηλ m, s <
ln αm
ln β
1
λ
λ/r
ln αm
ln β
λ/r
lnλ−1 β
lnλ αmlnλ β
ln β/ ln αmλ
u1/s−1 du 0
−
lnλ−1 β
1
λ
2 ln αm lnλ β
ln αm
ln β
λ/r
s ln β 1
lnλ−1 β
ln αm λ/r
1
−
>
> 0,
ln
β
−
λ
2
ln β
2
lnλ αm lnλ β
λ/s
s ln β
.
λ ln αm
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, and 0 < λ ≤ 1, and
define
f1 y :
ln y β/m αλ y β λ/s−1
1
,
λm α y β/m αλ − 1 m α
ln y α/n βλ y α λ/r−1
1
,
g1 y : λ n β y α/n βλ − 1 n β
Rλ m, s :
1
λ2
λ n, r : 1
R
λ2
β/mαλ
0
α/nβλ
0
1
ηλ m, s : 2
λ
y ∈ 0, ∞, n ∈ N0 ,
ln u 1/s−1
1
1
u
du − f1 0 f1 0,
u−1
2
12
2.18
ln u 1/r−1
1
1
u
du − g1 0 g1 0,
u−1
2
12
β/mαλ
0
y ∈ 0, ∞, m ∈ N0 ,
1
ln u 1/s−1
du − f1 0.
u
u−1
2
Then, it has
1 The functions f1 y, g1 y satisfy the conditions of 1.6. It means that
−1i F i y > 0 F f1 , g1 , y ∈ 0, ∞ ,
F i ∞ 0 F f1 , g1 , i 0, 1, 2, 3, 4 ,
2.19
2
Rλ m, s > 0,
λ n, r > 0,
R
2.20
6
Journal of Inequalities and Applications
3
0 < ηλ m, s O
1
mα
ρ ρ > 0, m −→ ∞ .
2.21
Proof. 1 Letting hu : ln u/u − 1, u y β/m αλ , it can be proved that f1 y 1/λm αhuu1/s−1/λ satisfy 2.19 as in 5. Similarly, it can be shown that g1 y satisfy
2.19 also.
2 Setting u0 : β/m αλ , by u λy β/m αλ−1 1/m α λ/y βy β/m αλ ,u 0 λ/βu0 , and h u > 0, it has
1
λ2
β/mαλ
0
ln u 1/s−1
1
u
du 2
u−1
λ
≥
f1 0 f1 0 u0
huu1/s−1 du
0
u0
s
1/s
1/s hu
−
u
h
u
udu
0 0
λ2
0
0
2.22
u0
s
1/s
1/s
hu0 u0 − h u0 u du
λ2
0
s s
1/s
1/s1
h
hu
,
−
u0 u0
0 u0
s1
λ2
λ/s−1
ln β/m αλ
β
1
1
hu0 u1/s
0 ,
λ
λm α β/m α − 1 m α
λβ
2.23
1 1
1 1/s h u0 u0 2.24
−
hu0 .
u
s λ
β2 0
s
λ2
u0
hudu1/s With 2.22∼2.24, it has
Rλ m, s ≥ hu0 u1/s
0
1
s
1
−
2
2λβ 12β2
λ
1 1
−
s λ
− h u0 u1/s1
0
1
s
−
.
1 s 12β2
2.25
By hu0 > 0, h u0 < 0, and β ≥ 1/2, s > 1, 0 < λ ≤ 1, it has
6βs 2βs − λ − λs − λ
> 0,
12β2 sλ2
2
4β s − s 8β2 s − 1
12β2 s − 1 s
1
s
−
> 0.
1 s 12β2
12βλ1 s
12β2 1 s
s
1
1
−
λ2 2λβ 12β2
1 1
−
s λ
λ n, r > 0.
So Rλ m, s > 0 holds. Similarly, it can be shown that R
2.26
Journal of Inequalities and Applications
7
3 In view of 2.22, 2.23, by hu > 0, h u < 0, it has
ηλ m, s >
hu0 u1/s
0
2βs − λ
s
1
−
> 0,
hu0 u1/s
0
λ2 2λβ
2λ2 β
2.27
and by limu → 0 ln u/u − 1u1/2s 0, so there exists a constant L > 0, such that |ln u/u −
1u1/2s | < L u ∈ 0, β/m αλ . Then it has
1
ηλ m, s < 2
λ
β/mαλ
0
ln u 1/s−1
L
u
du < 2
u−1
λ
β/mαλ
u1/2s−1 du 0
λ/2s
β
2sL
. 2.28
λ2 m α
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, and r, 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 n0 , ∞ such that Un0 > 0,U∞ ∞U u, v. Give
some notations as follows:
1
1−p
,
φx : uxp1−λ/r−1 u x
1−q
ϕx : vxq1−λ/s−1 v x
,
1−p
ψx : ϕx
vxpλ/s−1 v x
3.1
x ∈ n0 , ∞,
2 set
p
φ
⎧
⎨
:
⎩
a; a {an }∞
nn0 , ap,φ
:
∞
1/p
φn|an |
nn0
p
⎫
⎬
<∞ ,
⎭
3.2
p
and call φ a real space of sequences, where
ap,φ ∞
1/p
p
φn|an |
3.3
nn0
is called the norm of the sequence with a weight function φ. Similarly, the real spaces of sequences
q
p
ϕ , ψ and the norm bq,ϕ can be defined as well,
8
Journal of Inequalities and Applications
p
p
p
3 define a Hilbert-type linear operator T : φ → ψ , for all a ∈ φ ,
Tan : Cn :
∞
Kum, vnam
n ≥ n0 ,
3.4
mn0
p
q
4 for all a ∈ φ , b ∈ ϕ , define the formal inner product of Ta and b as
Ta, b :
∞
nn0
∞
Kum, vnam bn mn0
∞ ∞
Kum, vnam bn ,
3.5
nn0 mn0
5 define two weight coefficients ωm, s and ϑn, r as
∞
ωλ m, s :
Kum, vn
nn0
∞
ϑλ n, r :
Kum, vn
mn0
umλ/r
1−λ/s
vn
vnλ/s
um1−λ/r
u m,
v n,
3.6
m, n ≥ n0 .
Then it has some results in the following theorems.
Theorem 3.1. Suppose that an ≥ 0, U x/Ux > 0U u, v, and 0 <
∞
1ε
1ε
≤ ∞
< ∞ε > 0. If there exists a positive number
nn0 v n/vn
nn0 u n/un
kλ , such that
0 < ωλ m, s < kλ , 0 < ϑλ n, r < kλ m, n ≥ n0 ,
1
≤ ωλ m, s ρ > 0, m −→ ∞ ,
kλ 1 − O
ρ
um
3.7
3.8
p
then for all a ∈ φ and ap,φ > 0, it has the following:
1
Ta C {Cn }∞
nn0 ∈ ψ ,
p
p
p
It means that T : φ → ψ ,
3.9
Journal of Inequalities and Applications
9
2 T is a bounded linear operator and
Tp,ψ :
Tap,ψ
sup
p
a∈φ a /
θ
kλ ,
ap,φ
3.10
where Cn , T are defined by 3.4, Tap,ψ Cp,ψ is defined as 3.3.
Proof. By using Hölder’s inequality 7 and 3.6, 3.7, it has Cn ≥ 0 and
p
Cn
∞
Kum, vn
um1−λ/r/q v n1/p
am
vn1−λ/s/p u m1/q
∞
ump−11−λ/r v n p
≤
Kum, vn
am
vn1−λ/s u mp−1
mn0
p−1
∞
vnq−11−λ/s u m
×
Kum, vn
um1−λ/r v nq−1
mn0
∞
Kum, vn
mn0
<
p−1
kλ
∞
ump−11−λ/r v n
vn1−λ/s u m
Kum, vn
p
a
p−1 m
mn0
ϑλ n, rϕn
ump−11−λ/r v n
p
um1−λ/r/q v n1/p
mn0
vn1−λ/s/p u m1/q
p
a
p−1 m
vn1−λ/s u m
p−1
ϕp−1 n .
3.11
And by ψn ϕ1−p n, it follows that
p
Tap,ψ
∞
p
ψnCn
nn0
p−1
kλ
∞
<
p−1
kλ
∞ ∞
Kum, vn
mn0 nn0
p
ωλ m, sφmam
mn0
<
p
p
kλ ap,φ
ump−11−λ/r v n
p
a
p−1 m
vn1−λ/s u m
3.12
< ∞.
This means that C {Cn }∞
n0 ∈ ψ , Tap,ψ ≤ kλ ap,φ , and Tp,ψ ≤ kλ . T is a bounded linear
operator.
If there exists a constant K < kλ , such that Tp,ψ ≤ K, then for ε > 0, setting am :
p q
∞
umλ/r−ε/p−1 u m, bn : vnλ/s−ε/q−1 v n, it has a {
am }∞
mn0 ∈ φ , b {bn }nn0 ∈ ϕ ,
and
p
ap,φ b
T a, b ≤ Tp,ψ q,ϕ
≤K
∞
u m
1ε
mn0 um
1/p ∞
v n
1ε
nn0 vn
1/q
≤K
∞
u m
mn0 um
1ε
.
3.13
10
Journal of Inequalities and Applications
But on the other side, by 3.8, it has
∞ ∞
umλ/r−ε/p−1 T a, b Kum, vn
u mv n
1−λ/sε/q
vn
mn0 nn0
∞
∞
u m
1ε
mn0 um
nn0
Kum, vn
umλ/rε/q
vn1−λ/sε/q
3.14
v n.
By the strictly monotonic increase of vx and vn0 > 0, v∞ ∞, there exists n1 > n0
such that vn > 1 for all n > n1 . So it has
0<
∞
Kum, vn
nn0
ε/q
um
ε/q
≤ um
∞
umλ/rε/q
vn1−λ/sε/q
Kum, vn
vn1−λ/sε/q
∞
umλ/r v n
Kum, vn
um
umλ/r v n
nn1
nn1
ε/q
v n
ωλ m, s −
n
1 −1
vn1−λ/s
Kum, vn
n
1 −1
Kum, vn
umλ/r v n
nn0
vn1−λ/sε/q
n
1 −1
umλ/r v n
Kum, vn
nn0
vn1−λ/sε/q
umλ/r v n
vn1−λ/s
n
1 −1
umλ/r v n
.
Kum, vn
vn1−λ/sε/q
nn0
nn0
3.15
The series is uniformly convergent for ε ≥ 0, so it has
lim
ε→0
∞
umλ/rε/q
Kum, vn
1−λ/sε/q
vn
nn0
v n ωλ m, s
3.16
and for m > n0 , there exists ε0 > 0, when 0 < ε < ε0 , it has
∞
nn0
Kum, vn
umλ/rε/q
1−λ/sε/q
vn
v n > ωλ m, s −
1
.
um
3.17
Journal of Inequalities and Applications
11
By 3.14 and 3.8, when 0 < ε < ε0 , it has
∞
T a, b ≥
ωλ m, s −
1ε
u m
mn0 um
1
um
"
!
1
1
≥ kλ
1−O
−
1ε
kλ um
umρ
mn0 um
⎧
∞
−1
∞
u m
u m ⎨
kλ
1−
1ε ⎩
1ε
mn um
mn um
∞
u m
0
3.18
0
⎫
⎬
1
1
.
O
×
1ε
⎭
kλ um
umρ
mn0 um
∞
u m
In view of 3.13 and 3.18, letting ε → 0 , it has kλ ≤ K. This means that K kλ ; that is,
Tp,ψ kλ . 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
uλ/r m v y ,
f y : Kλ um, v y
v1−λ/s y
vλ/s n g y : Kλ u y , vn 1−λ/r u y .
u
y
3.19
Here, uy, vy satisfy the conditions as in Theorem 3.1. Set
vn0 /um
Rm, s :
0
r :
Rn,
un0 /vn
0
1
1
Kλ 1, uuλ/s−1 du − fn0 f n0 ,
2
12
3.20
1
1
Kλ μ, 1 μλ/r−1 dμ − gn0 g n0 ,
2
12
3.21
vn0 /um
ηm, s :
0
1
Kλ 1, uuλ/s−1 du − fn0 .
2
3.22
If (a) Kλ x, y ≥ 0 is a homogeneous measurable kernel function of “λ” degree in R2 , such that
0 < kλ s :
∞
Kλ 1, uuλ/s−1 du < ∞,
3.23
0
(b) functions fy, gy satisfy the conditions of 1.6; that is,
−1i F i y > 0 y > n0 ,
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, such that
Rm, s > 0,
1
0 < ηm, s O ρ
u m
r > 0,
Rn,
m −→ ∞,
3.25
then it has
p
q
1 if a ∈ φ , b ∈ ϕ , and ap,φ > 0, bq,ϕ > 0, then
Ta, b ∞ ∞
Kλ um, vnam bn < kλ sap,φ bq,ϕ ,
3.26
nn0 mn0
p
2 if a ∈ φ and ap,φ > 0, then
Tap,ψ ∞
pλ/s−1 vn
v n
nn0
p 1/p
∞
Kum, vnam
< kλ sap,φ ,
3.27
mn0
inequality 3.27 is equivalent to 3.26 and the constant factor kλ s kλ r :
#where
∞
K
u,
1uλ/r−1 du is the best possible.
λ
0
Proof. By 3.24, 1.6, it has
∞
n0
∞
1
umλ/r f y dy − fn0 < ωλ m, s Kλ um, vn
v n
2
vn1−λ/s
nn0
∞
∞
fn <
n0
nn0
0 < ϑλ n, r ∞
1
1
f y dy − fn0 f n0 ,
2
12
Kλ um, vn
mn0
∞
mn0
∞
gm <
n0
vnλ/s
um1−λ/r
3.28
u m
1
1
g y dy − gn0 g n0 .
2
12
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, it has
∞
n0
∞
uλ/r m Kλ um, v y
Kλ 1, ννλ/s−1 dν
v y dy v1−λ/s y
n0
vn0 /um
∞
vn0 /um
λ/s−1
Kλ 1, νν
dν −
Kλ 1, ννλ/s−1 dν kλ s
f y dy ∞
0
−
∞
n0
vn0 /um
Kλ 1, ννλ/s−1 dν,
0
∞
vλ/s n Kλ u y , vn 1−λ/r u y dy Kλ μ, 1 μλ/r−1 dμ
u
y
n0
un0 /vn
∞
un0 /vn
Kλ μ, 1 μλ/r−1 dμ −
Kλ μ, 1 μλ/r−1 dμ kλ s
g y dy ∞
0
−
3.30
0
3.31
0
un0 /vn
Kλ μ, 1 μλ/r−1 dμ,
0
#∞
#∞
where, letting t 1/u, it has kλ r 0 Kλ u, 1uλ/r−1 du 0 Kλ 1, ttλ/s−1 dt kλ s. In
view of 3.28, 3.30, 3.20, 3.22, and with 3.25, it has
0 < ωλ m, s < kλ s − Rλ m, s < kλ s,
1
ρ > 0, m −→ ∞ .
ωλ m, s > kλ s − ηλ m, s kλ s 1 − O1
ρ
u m
3.32
Similarly, with 3.29, 3.31, 3.21, and 3.25, it has
0 < ϑλ n, r < kλ s
3.33
Tap,ψ < kλ sap,φ ,
3.34
Ta, b ≤ Tap,ψ bq,ϕ ,
3.35
also. By Theorem 3.1, it has
and 3.27 holds. In view of
3.26 holds also.
14
Journal of Inequalities and Applications
If 3.26 holds, from 3.26 and ap,φ > 0, there exists n1 > n0 , such that
H
p
p−1
φma
> 0 when H > n1 . For
m > 0 and bn H : ψn mn0 Kλ um, vnam mn0
H
b : {bn H}nn0 , it has
H
0<
H
q
ϕnbn H
nn0
H
ψn
nn0
H
p
Kλ um, vnam
mn0
< kλ s
H
H H
Kλ um, vnam bn H
3.36
nn0 mn0
1/p p
φnan
nn0
H
1/q
q
ϕnbn H
< ∞.
nn0
By p > 1 and q > 1, it follows that
0<
H
nn0
q
p
ϕnbn H < kλ s
∞
p
φnan < ∞.
3.37
nn0
∞
q
Letting H → ∞ in 3.37, it has 0 <
nn0 ϕnbn ∞ < ∞, and it means that b q
{bn ∞}∞
nn0 ∈ ϕ and bq,ϕ > 0. Therefore, the inequality 3.36 keeps the form of the strict
q
p
inequality when H → ∞. In view of ∞
nn0 ϕnbn ∞ Tap,ψ , inequality 3.27 holds
and 3.27 is equivalent to 3.26. By Tp,ψ kλ s, it is obvious that the constant factor
kλ s kλ 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 ≥
e7/12 , 0 < λ ≤ 1. Then it has the following.
p1−λ/r−1 p
q1−λ/s−1 q
1 If 0 < ∞
an /np−1 < ∞, and 0 < ∞
bn /nq−1 <
n1 ln αn
n1 ln βn
∞, then
∞ ∞
q1−λ/s−1 q 1/q
ln αnp1−λ/r−1 apn 1/p ln βn
bn
am bn
π
<
.
p−1
q−1
λ αm lnλ βn
λ
sinπ/s
ln
n
n
m1 n1
n1
n1
4.1
∞ ∞
2 If 0 <
∞
p1−λ/r−1 p
an /np−1 n1 ln αn
< ∞, then
p p p
∞
∞ ln βn pλ/s−1
∞
π
am
ln αnp1−λ/r−1 an
<
,
λ
λ
n
λ sinπ/s n1
np−1
n1
m1 ln αm ln βn
4.2
where the constant factors kλ s kλ r : π/λ sinπ/s and π/λ sinπ/sp are both the
best possible. Inequality 4.2 is equivalent to 4.1.
Journal of Inequalities and Applications
15
Proof. Setting Kλ x, y 1/xλ yλ , x, y ∈ R2 , it is a homogeneous measurable kernel
function of “λ” degree. Letting t uλ , it has
0 < kλ s :
∞
Kλ 1, uuλ/s−1 du
0
∞
0
uλ/s−1
1
du λ
1 uλ
∞
0
1
1 1
t1/s−1
dt B
,
kλ r < ∞.
1t
λ
s r
4.3
Setting ux ln αx, vx ln βx, then both ux and vx are strictly monotonic increasing
differentiable functions in 1, ∞ and satisfy
U∞ ∞ U u, v,
U1 > 0,
0<
∞
v n
1ε
n1 vn
∞
n1 n
1
ln βn
1ε ≤
∞
u n
1ε
n1 un
∞
1
1ε
n1 nln αn
<∞
4.4
for ε > 0. As β ≥ α ≥ e7/12 , 0 < λ ≤ 1, s > 1, and n0 1, letting
uλ/r m 1
lnλ/r αm
f y Kλ um, v y
v
y
,
lnλ αm lnλ βy ln1−λ/s βy y
v1−λ/s y
vλ/s n lnλ/s βn
1
g y : Kλ u y , vn 1−λ/r u y λ
,
ln αy lnλ βn ln1−λ/r αy y
y
u
4.5
y ∈ 1, ∞, n, m ∈ N,
with 2.1∼2.8, it has
v1/um
1
1
Kλ 1, uuλ/s−1 du − f1 f 1 > 0,
2
12
0
ρ 1
ρ > 0, m −→ ∞ ,
0 < ηλ m, s O
um
u1/vn
1
1
λ n, r :
Kλ μ, 1 μλ/r−1 dμ − g1 g 1 > 0.
R
2
12
0
Rλ m, s 4.6
p1−λ/r−1 p
q1−λ/s−1 q
an /np−1 < ∞ and 0 < ∞
bn /nq−1 < ∞; that
When 0 < ∞
n1 ln αn
n1 ln βn
p
q
is, a ∈ φ , b ∈ ϕ and ap,φ > 0, bq,ϕ > 0, by Theorem 3.2, inequality 4.1 holds, so does
4.2. And 4.2 is equivalent to 4.1, and the constant factors kλ s kλ r : π/λ sinπ/s
and π/λ sinπ/sp 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 <
∞
p
n0
n αp1−λ/r−1 an < ∞ and 0 <
∞
n0
q
n βq1−λ/s−1 bn < ∞, then
∞ ∞ ln
m α/ n β am bn
λ
m αλ − n β
n0 m0
π
<
λ sinπ/s
2 If 0 <
∞
n0
1/p n p
αp1−λ/r−1 an
n0
∞ q1−λ/s−1 q
nβ
bn
4.7
1/q
.
n0
p
n0
∞ pλ/s−1
nβ
2 ∞
n αp1−λ/r−1 an < ∞, then
∞
lnm α/n βam
λ
λ
m0 m α − n β
p
<
π
λ sinπ/s
2p ∞
p
n αp1−λ/r−1 an ,
n0
4.8
where inequality 4.8 is equivalent to 4.7 and the constant factors kλ s kλ r :
π/λ sinπ/s2 and π/λ sinπ/s2p are both the best possible.
Proof. Setting Kλ x, y lnx/y/xλ − yλ x, y ∈ R2 , it is a homogeneous measurable
kernel function of “λ” degree. Letting t uλ , it has 2
∞
1 ∞ − ln uλ λ/s−1
0 < kλ s :
Kλ 1, uu
du u
du
λ 0 1 − uλ
0
2
1 1 2
1
π
1 ∞ ln t 1/s−1
t
B
,
dt kλ r < ∞.
2
λ
s r
λ sinπ/s
λ 0 t−1
λ/s−1
4.9
Setting ux x α, vx x β, then both ux and vx are strictly monotonic increasing
differentiable functions in 0, ∞ and satisfy
U∞ ∞ U u, v,
U0 > 0,
0<
∞
v n
1ε
n0 vn
∞
n0
1
nβ
1ε ≤
∞
u n
1ε
n0 un
∞
n0 n
1
α1ε
<∞
4.10
Journal of Inequalities and Applications
17
for ε > 0. As β ≥ α ≥ 1/2, 0 < λ ≤ 1, s > 1, and n0 0, letting
uλ/r m ln m α/ y β
m αλ/r
f1 y Kλ um, v y
v
y
1−λ/s
λ v1−λ/s y
yβ
m αλ − y β
ln y β/m α
1
λm α y β/m αλ − 1
λ
yβ
mα
λ/s−1
,
vλ/s n ln y α / n β
n βλ/s
g1 y : Kλ u y , vn 1−λ/r u y λ λ
1−λ/r
y
u
y α − n β y α
ln y α/n βλ y α λ/r−1
1
,
λ n β y α/n βλ − 1 n β
4.11
y ∈ 0, ∞, n, m ∈ N,
with 2.18∼2.21, it has
Rλ m, s :
v0/um
0
1
1
Kλ 1, uuλ/s−1 du − f1 0 f1 0
2
12
λ
1 β/mα ln u 1/s−1
1
1
2
u
du − f1 0 f1 0 > 0,
u−1
2
12
λ 0
u0/vn
1
1
λ n, r :
Kλ μ, 1 μλ/r−1 dμ − g1 0 g1 0 > 0,
R
2
12
0
1 ρ
ρ > 0, m −→ ∞ .
0 < ηλ m, s O
um
a∈
p
φ ,
4.12
p1−λ/r−1 p
q1−λ/s−1 q
When 0 < ∞
an < ∞ and 0 < ∞
bn < ∞; that is,
n0 n α
n0 n β
q
b ∈ ϕ and ap,φ > 0, bq,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 kλ s kλ r : π/λ sinπ/s2 and
π/λ sinπ/s2p are both the best possible.
Remark 4.3. It can be proved similarly that, if the conditions “β ≥ α ≥ e7/12 ” in Lemma 2.1 and
“β ≥ α ≥ 1/2” in Lemma 2.2 are changed into “α ≥ β ≥ e7/12 ” and “α ≥ β ≥ 1/2”, respectively,
Lemmas 2.1 and 2.2 are also valid. So the conditions “β ≥ α ≥ e7/12 ” in Example 4.1 and
“β ≥ α ≥ 1/2” in Example 4.2 can be replaced by “β ≥ e7/12 , α ≥ e7/12 ” 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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