Volume 2010, Article ID 393025, 18 pagesdoi:10.1155/2010/393025 Research Article A New Hilbert-Type Linear Operator with a Composite Kernel and Its Applications Wuyi Zhong Department of
Trang 1Volume 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: Ondˇrej Doˇsl ´y
Copyrightq 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, Weyl1 published the well-known Hilbert’s inequality as follows:
if a n , b n ≥ 0 are real sequences, 0 <∞n1 a2
n < ∞ and 0 <∞n1 b2
n <∞, then
∞
n1
∞
m1
a m b n
m n < π
∞
n1
a2n
∞
n1
b n2
1/2
where the constant factor π is the best possible.
Under the same conditions, there are the classical inequalities2
∞
n0
∞
m0
a m b n
m n 1 < π
∞
n1
a2n
∞
n1
b2n
1/2
∞
n1
∞
m1
lnm/nam b n
m − n < π2
n1
a2n
∞
n1
b n2
1/2
Trang 2where the constant factors π and π2 are the best possible also Expression1.2 is called a
more accurate form of1.1 Some more accurate inequalities were considered by 3 5 In 2009, Zhong5 gave a more accurate form of 1.3
Setp, 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 <∞, then it has
∞
n0
∞
m0
lnm α/n αam 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 , φ p: {a; a {an}∞n0,a p,φ: {∞
n0 φ n|a n|p}1/p
< ∞}, q
ϕ : {b; b {bn}∞n0, and b q,ϕ: {∞
n0 ϕ n|b n|q}1/q
< ∞}, the expression1.4 can be rewritten as
Ta, b :∞
n0
∞
m0
lnm α/n αam b n
m α λ − n α λ < k λ sa p,φ b q,ϕ , 1.5
where T : φ p → p
ψ is a linear operator, k λ s T 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 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 formula6 is introduced
Set f ∈ C4m, ∞m ∈ N0 If −1i f i x > 0, f i ∞ 0 i 0, 1, 2, 3, 4, then it has
∞
nm
f n <
∞
m
f xdx 1
2f m − 1
12f
m,
∞
nm
f n >
∞
m
f xdx 1
2f 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, and define
f
y : 1
ln λ αm ln λ βy
ln λ/r αm
ln1−λ/sβy y , y ∈ 1, ∞, m ∈ N, 2.1
g
y : 1
ln λ αy ln λ βn
ln λ/s βn
ln1−λ/rαy y , y ∈ 1, ∞, n ∈ N, 2.2
Trang 3R λ m, s :
ln β/ ln αm
0
u λ/s−1
1 u λ du−1
2f1 1
12f
R λ n, r :
ln α/ ln βn
0
u λ/r−1
1 u λ du−1
2g1 1
12g
η λ m, s :
ln β/ ln αm
0
u λ/s−1
1 u λ du−1
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
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
f1
y : 1
lnλ αm lnλ βy , f2
y : 1
ln1−λ/sβy , f3
y : 1
It has
f
y lnλ/r αmf1
y f2
y f3
when y≥ 1 It is easy to find that
−1i f j i
y > 0, f j i∞ 0 y ∈ 1, ∞, j 1, 2, 3, i 0, 1, 2, 3, 4 ,
−1i f i
y > 0, f i∞ 0 y ∈ 1, ∞, i 0, 1, 2, 3, 4
2.11
Trang 4Similarly, it can be shown that−1i g i y > 0, g i ∞ 0 y ≥ 1, i 0, 1, 2, 3, 4 These
tell us that2.6 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 t2dt
s
λ
lnλ β
lnλ αm lnλ β
ln αm
ln β
λ/r
λ 1 s
ln β/ ln αm λ
0
dt 1/s1
1 t2
≥ s
λ
lnλ β
lnλ αm lnλ β
ln αm
ln β
λ/r
λ 1 s
ln2λ β
lnλ αm lnλ β 2
ln αm
ln β
λ/r
.
2.12
By2.1, it has
f1
ln αm
ln β
λ/r
lnλ−1 β
f1
ln αm
ln β
λ/r
− λln 2λ−2 β
lnλ αm lnλ β 2 − 1 − λ/sln λ−2 β
lnλ αm lnλ β − lnλ−1 β
lnλ αm lnλ β
In view of2.12∼2.14, it has
R λ m, s ≥
ln αm
ln β
λ/r
lnλ−1 β
lnλ αm lnλ β −1− λ/s
12 ln β − 7
12 s
λ ln β
ln2λ β
lnλ αm lnλ β 2
s2
λ 1 s −
λ
12ln2β
.
2.15
As ln β ≥ 7/12, s > 1r > 1, and 0 < λ ≤ 1, it has
−1− λ/s
12 ln β − 7
12 s
λ ln β≥ 7s
12λ
1− λ
s
12 ln β
1− λ
s
1−λ
s
7s
12 ln β
>
1−λ
s
7
12 ln β
> 0,
s2
λ 1 s−
λ
12ln2β >
s
12ln2β s
4
3ln2β
> 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
Trang 53 By 2.5, 2.12, 2.13, and 0 < λ < s, β ≥ e 7/12, it has
η λ m, s ≥ s
λ
lnλ β
lnλ αm lnλ β
ln αm
ln β
λ/r
−1 2
lnλ−1 β
lnλ αm lnλ β
ln αm
ln β
λ/r
ln αm
ln β
λ/r lnλ−1 β
lnλ αmlnλ β
s ln β
λ −1 2
>
ln αm
ln β
λ/r lnλ−1 β
lnλ αm lnλ 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 expression2.8 holds, andLemma 2.1is 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 : 1
λ m α
lny β/m α λ
y β/m α λ− 1
y β
m α
λ/s−1
, y ∈ 0, ∞, m ∈ N0,
g1
y : 1
λ
n β
lny α/n β λ
y α/n β λ− 1
y α
n β
λ/r−1
, y ∈ 0, ∞, n ∈ N0,
R λ m, s : 1
λ2
β/mα λ
0
ln u
u− 1u 1/s−1 du−
1
2f10 1
12f
10,
R λ n, r : 1
λ2
α/nβ λ
0
ln u
u− 1u 1/r−1 du−
1
2g10 1
12g
10,
η λ m, s : 1
λ2
β/mα λ
0
ln u
u− 1u 1/s−1 du−
1
2f10.
2.18
Then, it has
1 The functions f1y, g1y 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
Trang 60 < 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 f1y
1/λm αhuu 1/s−1/λsatisfy2.19 as in 5 Similarly, it can be shown that g1y satisfy
2.19 also
2 Setting u0 : β/m αλ , by u λy β/m α λ−1 1/m α λ/y
β y β/m α λ ,u0 λ/βu0, and hu > 0, it has
1
λ2
β/mα λ
0
ln u
u− 1u 1/s−1 du
1
λ2
u0
0
h uu 1/s−1 du
s
λ2
u0
0
h udu 1/s s
λ2 h u0u 1/s
0 −
u0
0
u 1/s hudu
≥ s
λ2 h u0u 1/s
0 − hu0
u0
0
u 1/s du
s
λ2 h u0u 1/s
0 − s
s 1hu0u 1/s10
,
2.22
f10 1
λ m α
lnβ/m α λ
β/m α λ− 1
m α
λ/s−1
1
λβ h u0u 1/s
0 ,
2.23
f10 1
β2u 1/s0 hu0u0
1
s− 1
λ
h u0
With2.22∼2.24, it has
R λ m, s ≥ hu0u 1/s
0
s
λ2 − 1
12β2
1
s− 1
λ
− hu0u 1/s1
0
s
1 s−
1
12β2
By hu0 > 0, hu0 < 0, and β ≥ 1/2, s > 1, 0 < λ ≤ 1, it has
s
λ2 − 1
12β2
1
s− 1
λ
6βs
2βs − λ − λs − λ
12β2sλ2 > 0, s
1 s−
1
12β2 12β2s − 1 s
12β21 s
4β2s − s 8β2s− 1
12βλ1 s > 0.
2.26
So R λ m, s > 0 holds Similarly, it can be shown that R λ n, r > 0.
Trang 73 In view of 2.22, 2.23, by hu > 0, hu < 0, it has
η λ m, s > hu0u 1/s
0
s
λ2 − 1
2λβ
hu0u 1/s
0
2βs − λ 2λ2β > 0, 2.27
and by limu→ 0 ln u/u − 1u 1/2s 0, so there exists a constant L > 0, such that |ln u/u −
1u1/2s | < L u ∈ 0, β/m α λ Then it has
η λ m, s < 1
λ2
β/mα λ
0
ln u
u− 1u 1/s−1 du <
L
λ2
β/mα λ
0
u 1/2s−1 du 2sL
λ2
β
m α
λ/2s
2.28
It means that2.21 holds The proof forLemma 2.2is 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
φ x : ux p1−λ/r−1ux1−p,
ϕ x : vx q1−λ/s−1vx1−q,
ψ x :ϕ x1−p vx pλ/s−1 vx x ∈ n0, ∞,
3.1
2 set
φ p:
⎧
⎨
⎩a; a {a n}∞nn0, a p,φ:
nn0
φ n|a n|p
1/p
<∞
⎫
⎬
and call φ p a real space of sequences, where
a p,φ
nn0
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
ϕ q , ψ p and the norm b q,ϕcan be defined as well,
Trang 83 define a Hilbert-type linear operator T : p
φ → p
ψ , for all a ∈ p
φ,
Tan : C n: ∞
mn0
4 for all a ∈ p
φ , b ∈ q
ϕ , define the formal inner product of Ta and b as
Ta, b : ∞
nn0
mn0
K um, vna m
b n ∞
nn0
∞
mn0
K um, vna m b n , 3.5
5 define two weight coefficients ωm, s and ϑn, r as
ω λ m, s : ∞
nn0
K um, vn um λ/r
vn1−λ/sv
n,
ϑ λ n, r : ∞
mn0
K um, vn vn λ/s
um1−λ/ru
m, m, n ≥ n0.
3.6
Then it has some results in the following theorems
Theorem 3.1 Suppose that a n ≥ 0, Ux/Ux > 0U u, v , and 0 <
∞
nn0vn/vn1ε ≤ ∞nn0un/un1ε < ∞ε > 0 If there exists a positive number
k λ , such that
0 < ω λ m, s < k λ , 0 < ϑ λ n, r < k λ m, n ≥ n0, 3.7
k λ
1− O
1
um ρ
then for all a ∈ p
φ and a p,φ > 0, it has the following:
1
Ta C {C n}∞nn0 ∈ p
It means that T : φ p → p
ψ ,
Trang 92 T is a bounded linear operator and
T p,ψ: sup
a∈ p
φ a / θ
Ta p,ψ
where C n , T are defined by3.4, Ta p,ψ C p,ψ is defined as3.3.
Proof By using H ¨older’s inequality7 and 3.6, 3.7, it has C n≥ 0 and
C n p
mn0
K um, vn
um 1−λ/r/q vn 1/p
vn 1−λ/s/p um 1/q a m
vn 1−λ/s/p um 1/q
um 1−λ/r/q vn 1/p
p
≤
mn0
K um, vn um p−11−λ/r vn
vn1−λ/sum p−1 a p m
×
∞
mn0
K um, vn vn q−11−λ/s um
um1−λ/rvn q−1
p−1
∞
mn0
K um, vn um p−11−λ/r vn
vn1−λ/sum p−1 a p m
ϑ λ n, rϕn
p−1
< k p−1 λ
∞
mn0
K um, vn um p−11−λ/r vn
vn1−λ/sum p−1 a p m
ϕ p−1 n .
3.11
And by ψn ϕ1−pn, it follows that
Ta p
p,ψ ∞
nn0
ψ nC p
n < k p−1 λ
∞
mn0
∞
nn0
K um, vn um p−11−λ/r vn
vn1−λ/sum p−1 a
p m
k p−1 λ ∞
mn0
ω λ m, sφma 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, such thatT p,ψ ≤ K, then for ε > 0, setting a m :
um λ/r−ε/p−1 um, b n : vnλ/s−ε/q−1 vn, it has a { a m}∞mn0 ∈ p
φ , b { b n}∞nn0 ∈ q
ϕ, and
T a, b ≤ T p,ψ a p,φ b
q,ϕ ≤ K
∞
mn0
um
um1ε
1/p∞
nn0
vn
vn1ε
1/q
≤ K∞
mn0
um
um1ε.
3.13
Trang 10But on the other side, by3.8, it has
T a, b ∞
mn0
∞
nn0
K um, vn um λ/r−ε/p−1
vn1−λ/sε/qu
mvn
∞
mn0
um
um1ε
∞
nn0
K um, vn um λ/rε/q
vn1−λ/sε/qv
n.
3.14
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 <
∞
nn0
K um, vn um λ/rε/q
vn1−λ/sε/qv
n
um ε/q
nn1
K um, vn um λ/r vn
vn1−λ/sε/q
n1 −1
nn0
K um, vn um λ/r vn
vn1−λ/sε/q
≤ um ε/q
∞
nn1
K um, vn um λ/r vn
vn1−λ/s
n1 −1
nn0
K um, vn um λ/r vn
vn1−λ/sε/q
um ε/q
ω λ m, s − n1−1
nn0
K um, vn um λ/r vn
vn1−λ/s
n1−1
nn0
K um, vn um λ/r vn
vn1−λ/sε/q
.
3.15
The series is uniformly convergent for ε≥ 0, so it has
lim
ε→ 0
∞
nn0
K um, vn um λ/rε/q
vn1−λ/sε/qvn ω λ m, s 3.16
and for m > n0, there exists ε0> 0, when 0 < ε < ε0, it has
∞
nn
K um, vn um λ/rε/q
vn1−λ/sε/qv
n > ω λ m, s − 1
Trang 11By3.14 and 3.8, when 0 < ε < ε0, it has
T a, b ≥ ∞
mn0
um
um1ε
ω λ m, s − 1
u m
≥ k λ
∞
mn0
um
um1ε 1− O
1
um ρ
k λ u m
"
k λ
∞
mn0
um
um1ε
⎧
⎨
⎩1−
mn0
um
um1ε
−1
×
mn0
um
um1ε O
1
um ρ
k λ u m
⎫⎬
⎭.
3.18
In view of3.13 and 3.18, letting ε → 0, it has k λ ≤ K This means that K k λ; that is,
T p,ψ k λ.Theorem 3.1is 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 : Kλ
u m, v y
λ/r m
v1−λ/s
y
y ,
g
y : Kλ
u
y , v n λ/s n
u1−λ/r
y
y
3.19
Here, u y, vy satisfy the conditions as in Theorem 3.1 Set
R m, s :
vn0/um
0
K λ 1, uu λ/s−1 du− 1
2f n0 1
12f
R n, r :
un0/vn
0
K λ
μ, 1 μ λ/r−1 dμ−1
2g n0 1
12g
η m, s :
vn0/um
0
K λ 1, uu λ/s−1 du−1
If (a) K λ x, y ≥ 0 is a homogeneous measurable kernel function of “λ” degree in R2
, such that
0 < k λ s :
∞
0
(b) functions f y, 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
Trang 12(c) there exists ρ > 0, such that
R m, s > 0, R n, r > 0, 0 < ηm, s O
1
u ρ m
then it has
1 if a ∈ p
φ , b ∈ q
ϕ , and a p,φ > 0, b q,ϕ > 0, then
Ta, b ∞
nn0
∞
mn0
K λ um, vna m b n < k λ sa p,φ b q,ϕ , 3.26
2 if a ∈ p
φ and a p,φ > 0, then
Ta p,ψ
∞
nn0
vn pλ/s−1 vn
∞
mn0
K um, vna m
p1/p
< k λ sa 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 By3.24, 1.6, it has
∞
n0
f
y dy−1
2f n0 < ω λ m, s ∞
nn0
K λ um, vn um λ/r
vn1−λ/sv
n
∞
nn0
f n <
∞
n0
f
y dy−1
2f n0 1
12f
n0,
3.28
0 < ϑ λ n, r ∞
mn0
K λ um, vn vn λ/s
um1−λ/ru
m
∞
mn
g m <
∞
n
g
y dy−1
2g n0 1
12g
n0.
3.29
...ψ ,
Trang 92 T is a bounded linear operator and< /i>
T... p and the norm b q,ϕcan be defined as well,
Trang 83 define a Hilbert-type linear. ..
n > ω λ m, s − 1
Trang 11By3.14 and 3.8,