A New Hilbert Type Linear Operator with a Composite Kernel and Its Applications

Journal of Inequalities and Applications Volume 2010, Article ID 393025,18pages 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,[email protected]

Received 20 April 2010; Accepted 31 October 2010

Academic Editor: Ondˇrej Doˇsl ´y

Copyrightq2010 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, Weyl1published the well-known Hilbert’s inequality as follows: ifan,bn≥0 are real sequences, 0<

_∞

n1a2n<∞and 0<

_∞

n1b2n<∞, then

∞

n1 ∞

m1

ambn

mn < π

_∞

n1

a2_n

∞

n1

b_n2

1/2

, 1.1

where the constant factorπis the best possible.

Under the same conditions, there are the classical inequalities2

∞

n0 ∞

m0

ambn

mn1 < π

_∞

n1

a2_n

∞

n1

b2_n

1/2

, 1.2

∞

n1 ∞

m1

lnm/nambn

m−n < π

2 _∞

n1

a2_n

∞

n1

b_n2

1/2

where the constant factorsπ andπ2 _{are the best possible also. Expression1.2is called a}

more accurate form of1.1. Some more accurate inequalities were considered by3–5. In 2009, Zhong5gave a more accurate form of1.3.

Setp, q,s, ras two pairs of conjugate exponents, andp >1,s >1,α≥1/2, andan,

bn ≥0, such that 0<

_∞

n0nαp1−λ/r−1a

p

n<∞and 0<

_∞

n0nαq1−λ/s−1b

q

n <∞, then it

has

∞

n0 ∞

m0

lnmα/nαambn

mαλ−nαλ <

_∞

n0

nαp1−λ/r−1apn

1/p_∞

n0

nαq1−λ/s−1bqn

1/q

.

1.4

Lettingφx: xαp1−λ/r−1,ϕx: xαq1−λ/s−1,_φp:{a;a{an}∞n0,ap,φ:

{∞

n0φn|an|p}1/p <∞},ϕq : {b;b{bn}n∞0, and bq,ϕ:{

_∞

n0ϕn|bn|q}1/q < ∞}, the

expression1.4can be rewritten as

Ta, b:

∞

n0 ∞

m0

lnmα/nαambn

mαλ−nαλ < kλsap,φbq,ϕ, 1.5

whereT :_φp → ψpis a linear operator,kλs T.ap,φis the norm of the sequenceawith

a weight functionφ.Ta, bis a formal inner product of the sequencesTan:∞_m0lnm

α/nαam/mαλ−nαλandb.

By setting two monotonic increasing functions ux and vx, a new Hilbert-type inequality, which is with a composite kernel functionKux, 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 formula6is introduced. Setf ∈C4_m,∞m∈N

0. If−1ifix>0,fi∞ 0i0,1,2,3,4, then it has ∞

nm

fn<

_∞

m

fxdx 1

2fm− 1 12f

_m,

∞

nm

fn>

_∞

m

fxdx 1

2fm.

1.6

2. Lemmas

Lemma 2.1. Setr, sas a pair of conjugate exponents,s >1,β≥α≥e7/12_{,0< λ≤1, and define}

f y: 1

lnλ_αmlnλ_βy

lnλ/r_αm

ln1−λ/s_βyy, y∈1,∞, m∈N, 2.1

g y: 1

lnλ_αylnλ_βn

lnλ/s_βn

Rλm, s:

_lnβ/lnαm

0

uλ/s−1

1uλdu−

1 2f1

1 12f

_{1, 2.3}

Rλn, r:

lnα/lnβn

0

uλ/r−1

1uλdu−

1 2g1

1 12g

_{1, 2.4}

ηλm, s:

_lnβ/lnαm

0

uλ/s−1

1uλdu−

1

2f1. 2.5

Then, it has the following.

1The functionsfy,gysatisfy the conditions of 1.6. It means that

−1iFi y>0 Ff, g,y∈1,∞,

Fi∞ 0 F f, g,i0,1,2,3,4,

2.6

2

Rλm, s>0, Rλn, r>0, 2.7

3

0< ηλm, s O

1 lnαm

ρ

ρ >0, m−→ ∞. 2.8

Proof. 1Forβ≥α≥e7/12,y≥1,m∈N, 0< λ≤1, ands >1, set

f1 y

: 1

lnλ_αmlnλ_βy, f2 y

: 1

ln1−λ/s_βy, f3 y

: 1

y. 2.9

It has

f ylnλ/rαmf1 y

f2 y

f3 y

2.10

wheny≥1. It is easy to find that

−1if_ji y>0, f_ji∞ 0 y∈1,∞, j 1,2,3, i0,1,2,3,4,

−1ifi y>0, fi∞ 0 y∈1,∞, i0,1,2,3,4.

Similarly, it can be shown that−1igi_y>0,gi_{∞ 0y≥1,i0,1,2,3,4. These}

tell us that2.6holds and the functionsfy,gysatisfy the conditions of1.6.

2Settuλ_{. With the partial integration, it has}

_lnβ/lnαm

0

uλ/s−1

1uλdu

1

λ

_lnβ/lnαmλ

0

t1/s−1

1tdt s λ

_lnβ/lnαmλ

0

dt1/s

1t

s

λ

lnβ/lnαmλ/s

1 lnβ/lnαmλ s λ

lnβ/lnαmλ

0

t1/s

1t2dt

s

λ

lnλ_β

lnλ_αmlnλ_β

lnαm

lnβ

λ/r

s2

λ1s

_lnβ/lnαmλ

0

dt1/s1 1t2

≥ s

λ

lnλ_β

lnλ_αmlnλ_β

lnαm

lnβ

λ/r

s2

λ1s

ln2λ_β

lnλ_αmlnλ_β2

lnαm lnβ λ/r . 2.12

By2.1, it has

f1

lnαm

lnβ

λ/r _lnλ−1_β

lnλ_αmlnλ_β, 2.13

f1

lnαm

lnβ

λ/r

− λln2λ−2β

lnλ_αmlnλ_β2 −

1−λ/slnλ−2_β

lnλ_αmlnλ_β −

lnλ−1_β

lnλ_αmlnλ_β

. 2.14

In view of2.12∼2.14, it has

Rλm, s≥

lnαm

lnβ

λ/r _lnλ−1_β

lnλ_αmlnλ_β

−1−λ/s

12 lnβ −

7 12

s λlnβ

ln2λβ

lnλ_αmlnλ_β2

s2

λ1s−

λ

12ln2_β

.

2.15

As lnβ≥7/12,s >1r >1, and 0< λ≤1, it has

−1−λ/s

12 lnβ −

7 12

s λlnβ≥

7s

12λ

1− λ

s

− 1

12 lnβ

1− λ

s

1−λ

s

7s

12λ−

1 12 lnβ

>

1−λ

s

7 12−

1 12 lnβ

>0,

s2

λ1s−

λ

12ln2_β >

s

4 −

s

12ln2_β

s

4

1− 1 3ln2_β

>0.

2.16

It means thatRλm, s> 0. Similarly, it can be shown thatRλn, r>0. The expression2.7

3By2.5,2.12,2.13, and 0< λ < s,β≥e7/12_{, it has}

ηλm, s≥ s

λ

lnλ_β

lnλ_αmlnλ_β

lnαm

lnβ

λ/r

−1

2

lnλ−1_β

lnλ_αmlnλ_β

lnαm

lnβ

λ/r

lnαm

lnβ

λ/r _lnλ−1_β

lnλ_αmlnλ_β

_slnβ

λ −

1 2

>

lnαm

lnβ

λ/r _lnλ−1_β

lnλ_αmlnλ_β

lnβ−1

2

>0,

ηλm, s< 1

λ

lnβ/lnαmλ

0

u1/s−1du s λ

lnβ

lnαm

λ/s

.

2.17

The expression2.8holds, andLemma 2.1is proved.

Lemma 2.2. Setr, sas a pair of conjugate exponents,s > 1,β ≥ α≥ 1/2, and0 < λ≤ 1, and define

f1 y

: 1

λmα

lnyβ/mαλ

yβ/mαλ−1

_yβ

mα

λ/s−1

, y∈0,∞, m∈N0,

g1 y

: 1

λ nβ

lnyα/nβλ

yα/nβλ−1

_yα

nβ

λ/r−1

, y∈0,∞, n∈N0,

Rλm, s: 1

λ2

_β/mαλ

0

lnu u−1u

1/s−1_du−1

2f10 1 12f

10,

Rλn, r: 1

λ2

_α/nβλ

0

lnu u−1u

1/r−1_du−1

2g10 1 12g

10,

ηλm, s:

1

λ2

β/mαλ

0

lnu u−1u

1/s−1_du−1

2f10.

2.18

Then, it has

1The functionsf1y,g1ysatisfy the conditions of 1.6. It means that

−1iFi y>0 Ff1, g1,y∈0,∞

,

Fi∞ 0 Ff1, g1,i0,1,2,3,4

,

2.19

2

3

0< ηλm, s O

1

mα

ρ

ρ >0, m−→ ∞. 2.21

Proof. 1Lettinghu : lnu/u−1,u yβ/mαλ, it can be proved thatf1y

1/λmαhuu1/s−1/λ_{satisfy2.19as in5. Similarly, it can be shown thatg}

1ysatisfy 2.19also.

2Setting u0 : β/mαλ, by u λyβ/mαλ−11/mα λ/y

βyβ/mαλ,u0 λ/βu0, andhu>0, it has

1

λ2

_β/mαλ

0

lnu u−1u

1/s−1_du 1

λ2 _u0

0

huu1/s−1du

s

λ2 _u0

0

hudu1/s s λ2

hu0u1₀/s− _u0

0

u1/shudu

≥ s

λ2

hu0u1₀/s−hu0 u0

0

u1/sdu

s

λ2

hu0u1₀/s− s

s1h

_u 0u1₀/s1

,

2.22

f10

1

λmα

lnβ/mαλ

β/mαλ−1

_β

mα

λ/s−1 1

λβhu0u

1/s

0 ,

2.23

f₁0 1

β2u 1/s

0

hu0u0

1

s−

1

λ

hu0

. 2.24

With2.22∼2.24, it has

Rλm, s≥hu0u1₀/s

s λ2 −

1 2λβ

1 12β2

1

s−

1

λ

−hu0u1₀/s1

s

1s−

1 12β2

. 2.25

Byhu0>0,hu0<0, andβ≥1/2,s >1, 0< λ≤1, it has

s λ2 −

1 2λβ

1 12β2

1

s−

1

λ

6βs 2βs−λ

−λs−λ

12β2_sλ2 >0,

s

1s −

1 12β2

12β2_s−1s

12β2_1s

4β2_{s−s 8β}2_s−1

12βλ1s >0.

2.26

3In view of2.22,2.23, byhu>0,hu<0, it has

ηλm, s> hu0u1₀/s

s λ2 −

1 2λβ

hu0u1₀/s

2βs−λ

2λ2_β >0, 2.27

and by limu→0lnu/u−1u1/2s 0, so there exists a constantL >0, such that|lnu/u−

1u1/2s_{|< Lu∈0,β/mα}λ_{. Then it has}

ηλm, s<

1

λ2

_β/mαλ

0

lnu u−1u

1/s−1_{du <} L

λ2

_β/mαλ

0

u1/2s−1_du 2sL

λ2

β

mα

λ/2s

. 2.28

It means that2.21holds. The proof forLemma 2.2is finished.

3. Main Results

Setλ∈R,p >1,r >1,p, q, andr, sas two pairs of conjugate exponents.Kx, y≥0x, y∈

0,∞×0,∞is a measurable kernel function. Bothuxandvxare strictly monotonic increasing differentiable functions inn0,∞such thatUn0>0,U∞ ∞Uu, v. Give

some notations as follows:

1

φx: uxp1−λ/r−1ux1−p,

ϕx: vxq1−λ/s−1vx1−q,

ψx:ϕx1−p vxpλ/s−1vx x∈n0,∞,

3.1

2 set

_φp:

⎧ ⎨

⎩a;a{an}n∞n0,ap,φ: _∞

nn0

φn|an|p

1/p

<∞

⎫ ⎬

⎭, 3.2

and call_φpa real space of sequences, where

a_p,φ

_∞

nn0

φn|an|p

1/p

3.3

3define a Hilbert-type linear operatorT :_φp → pψ, for alla∈_φp,

Tan:Cn:

∞

mn0

Kum, vnam n≥n0, 3.4

4for alla∈_φp,b∈ϕq, define the formal inner product of Taandbas

Ta, b:

∞

nn0 _∞

mn0

Kum, vnam

bn

∞

nn0 ∞

mn0

Kum, vnambn, 3.5

5define two weight coefficientsωm, sandϑn, ras

ωλm, s:

∞

nn0

Kum, vnum

λ/r

vn1−λ/sv

_n,

ϑλn, r:

∞

mn0

Kum, vn vn

λ/s

um1−λ/ru

_{m, m, n≥n} 0.

3.6

Then it has some results in the following theorems.

Theorem 3.1. Suppose that an ≥ 0, Ux/Ux > 0U u, v, and 0 <

_∞

nn0v

_n/vn1ε _≤∞ nn0u

_n/un1ε _{< ∞ε > 0. If there exists a positive number}

kλ, such that

0< ωλm, s< kλ, 0< ϑλn, r< kλ m, n≥n0, 3.7

kλ

1−O

1

umρ

≤ωλm, s ρ >0, m−→ ∞

, 3.8

then for alla∈_φpandap,φ>0, it has the following:

1

TaC{Cn}∞nn0 ∈

p

ψ, 3.9

2T is a bounded linear operator and

T_p,ψ: sup

a∈_φpa /θ Ta_p,ψ

a_p,φ kλ, 3.10

whereCn,T are defined by3.4,Tap,ψ Cp,ψ is defined as3.3.

Proof. By using H ¨older’s inequality7and3.6,3.7, it hasCn≥0 and

Cnp

_∞

mn0

Kum, vn

um1−λ/r/qvn1/p

vn1−λ/s/pum1/qam

vn1−λ/s/pum1/q

um1−λ/r/qvn1/p

p

≤

_∞

mn0

Kum, vnum

p−11−λ/r_vn vn1−λ/sump−1a

p m

×

_∞

mn0

Kum, vnvn

q−11−λ/s_um um1−λ/rvnq−1

p−1

_∞

mn0

Kum, vnum

p−11−λ/r_vn vn1−λ/sump−1a

p m

ϑλn, rϕn

p−1

< kp_λ−1

_∞

mn0

Kum, vnum

p−11−λ/r_vn vn1−λ/sump−1a

p m

ϕp−1n.

3.11

And byψn ϕ1−p_{n, it follows that}

Tapp,ψ

∞

nn0

ψnCpn< kp_λ−1

_∞

mn0 ∞

nn0

Kum, vnum

p−11−λ/r_vn vn1−λ/sump−1a

p m

kp_λ−1

∞

mn0

ωλm, sφmamp < kp_λap_p,φ<∞.

3.12

This means thatC{Cn}∞n0∈

p

ψ,Tap,ψ ≤kλap,φ, andTp,ψ ≤kλ.T is a bounded linear

operator.

If there exists a constantK < kλ, such thatTp,ψ ≤ K, then forε > 0, settingam :

umλ/r−ε/p−1um,bn : vnλ/s−ε/q−1vn, it hasa {am}∞mn0 ∈

p

φ,b {bn}∞nn0 ∈

q ϕ,

and

Ta, b≤ T_p,ψa_p,φ b

q,ϕ≤K

_∞

mn0

um

um1ε

1/p_∞

nn0

vn

vn1ε

1/q

≤K

∞

mn0

um

um1ε.

But on the other side, by3.8, it has

Ta, b

∞

mn0 ∞

nn0

Kum, vnum

λ/r−ε/p−1

vn1−λ/sε/qu

_mvn

∞

mn0

um

um1ε

∞

nn0

Kum, vn um

λ/rε/q

vn1−λ/sε/qv

_n.

3.14

By the strictly monotonic increase ofvxandvn0>0,v∞ ∞, there existsn1> n0

such thatvn>1 for alln > n1. So it has

0<

∞

nn0

Kum, vn um

λ/rε/q

vn1−λ/sε/qv

_n

umε/q

_∞

nn1

Kum, vnum

λ/r_vn vn1−λ/sε/q

n1−1

nn0

Kum, vnum

λ/r_vn vn1−λ/sε/q

≤umε/q

_∞

nn1

Kum, vnum

λ/r_vn vn1−λ/s

n1−1

nn0

Kum, vnum

λ/r_vn vn1−λ/sε/q

umε/q

ωλm, s− n1−1

nn0

Kum, vnum

λ/r_vn vn1−λ/s

n1−1

nn0

Kum, vnum

λ/r_vn vn1−λ/sε/q

.

3.15

The series is uniformly convergent forε≥0, so it has

lim

ε→0 ∞

nn0

Kum, vnum

λ/rε/q

vn1−λ/sε/qv

_{n ω}

λm, s 3.16

and form > n0, there existsε0>0, when 0< ε < ε0, it has

∞

nn0

Kum, vn um

λ/rε/q

vn1−λ/sε/qv

_{n> ω}

λm, s−

1

By3.14and3.8, when 0< ε < ε0, it has

Ta, b≥

∞

mn0

um

um1ε

ωλm, s− 1

um

≥kλ

∞

mn0

um

um1ε

!

1−O

1

umρ

− 1

kλum

"

kλ

∞

mn0

um

um1ε

⎧ ⎨ ⎩1−

_∞

mn0

um

um1ε

₋1

×

_∞

mn0

um

um1ε

O

1

umρ

1

kλum

⎫_⎬ ⎭.

3.18

In view of3.13and3.18, lettingε → 0, it haskλ ≤ K. This means thatK kλ; that is,

Tp,ψ kλ.Theorem 3.1is proved.

Theorem 3.2. Suppose thatp, qand r, sare two pairs of conjugate exponents,r > 1,p > 1, λ∈R. Let

f y:Kλ um, v y

uλ/rm v1−λ/s _yv

_y,

g y:Kλ u y

, vn v

λ/s_n

u1−λ/r _yu

_y.

3.19

Here,uy,vysatisfy the conditions as inTheorem 3.1. Set

Rm, s:

vn0/um

0

Kλ1, uuλ/s−1du− 1

2fn0 1 12f

_n

0, 3.20

Rn, r:

_un0/vn

0

Kλ μ,1

μλ/r−1dμ−1

2gn0 1 12g

_n

0, 3.21

ηm, s:

vn0/um

0

Kλ1, uuλ/s−1du−

1

2fn0. 3.22

If (a)Kλx, y≥0is a homogeneous measurable kernel function of “λ” degree inR2, such that

0< kλs:

_∞

0

Kλ1, uuλ/s−1du <∞, 3.23

(b) functionsfy,gysatisfy the conditions of 1.6; that is,

−1iFi y>0 y > n0

(c) there existsρ >0, such that

Rm, s>0, Rn, r>0, 0< ηm, s O

1

uρ_m

m−→ ∞, 3.25

then it has

1ifa∈_φp,b∈ϕq, andap,φ>0,bq,ϕ>0, then

Ta, b

∞

nn0 ∞

mn0

Kλum, vnambn< kλsap,φbq,ϕ, 3.26

2ifa∈_φpandap,φ>0, then

Ta_p,ψ

_∞

nn0

vnpλ/s−1vn

_∞

mn0

Kum, vnam

p1/p

< kλsap,φ, 3.27

where inequality 3.27 is equivalent to 3.26 and the constant factor kλs kλr :

#_∞

0 Kλu,1uλ/r−1duis the best possible.

Proof. By3.24,1.6, it has

_∞

n0

f ydy−1

2fn0< ωλm, s

∞

nn0

Kλum, vnum λ/r

vn1−λ/sv

_n

∞

nn0

fn<

_∞

n0

f ydy−1

2fn0 1 12f

_n 0,

3.28

0< ϑλn, r

∞

mn0

Kλum, vn vn λ/s

um1−λ/ru

_m

∞

mn0

gm<

_∞

n0

g ydy−1

2gn0 1 12g

_n 0.

Lettingνvy/umandμuy/vnin the integral of3.28and3.29, respectively, by

3.23, it has

_∞

n0

f ydy

_∞

n0

Kλ um, v y

uλ/r_m

v1−λ/s _yv

_ydy∞

vn0/um

Kλ1, ννλ/s−1dν

_∞

0

Kλ1, ννλ/s−1dν−

_vn0/um

0

Kλ1, ννλ/s−1dνkλs

−

_vn0/um

0

Kλ1, ννλ/s−1dν,

3.30

_∞

n0

g ydy

_∞

n0

Kλ u y

, vn v

λ/s_n

u1−λ/r _yu

_ydy∞

un0/vn

Kλ μ,1

μλ/r−1dμ

_∞

0

Kλ μ,1

μλ/r−1dμ−

un0/vn

0

Kλ μ,1

μλ/r−1dμkλs

−

un0/vn

0

Kλ μ,1

μλ/r−1dμ,

3.31

where, lettingt 1/u, it haskλr

#_∞

0 Kλu,1uλ/r−1du #_∞

0 Kλ1, ttλ/s−1dt kλs. In

view of3.28,3.30,3.20,3.22, and with3.25, it has

0< ωλm, s< kλs−Rλm, s< kλs,

ωλm, s> kλs−ηλm, s kλs

1−O1

1

uρ_m

ρ >0, m−→ ∞.

3.32

Similarly, with3.29,3.31,3.21, and3.25, it has

0< ϑλn, r< kλs 3.33

also. ByTheorem 3.1, it has

Ta_p,ψ < kλsap,φ, 3.34

and3.27holds. In view of

Ta, b≤ Ta_p,ψb_q,ϕ, 3.35

If 3.26 holds, from 3.26 and ap,φ > 0, there exists n1 > n0, such that H

mn0φma

p

m > 0 andbnH : ψnHmn0Kλum, vnam

p−1 _{> 0 whenH > n} 1. For

b:{bnH}Hnn0, it has

0<

H

nn0

ϕnbqnH

H

nn0

ψn

_H

mn0

Kλum, vnam

p

H

nn0

H

mn0

Kλum, vnambnH

< kλs

_H

nn0

φnapn

1/p_H

nn0

ϕnbnqH

1/q

<∞.

3.36

Byp >1 andq >1, it follows that

0<

H

nn0

ϕnbqnH< k p λs

∞

nn0

φnapn<∞. 3.37

Letting H → ∞ in 3.37, it has 0 < ∞_nn0ϕnbqn∞ < ∞, and it means that b

{bn∞}∞nn0 ∈

q

ϕandbq,ϕ> 0. Therefore, the inequality3.36keeps the form of the strict

inequality whenH → ∞. In view of∞_nn0ϕnbqn∞ Tapp,ψ, inequality3.27 holds

and 3.27 is equivalent to3.26. ByTp,ψ kλs, it is obvious that the constant factor

kλs kλris the best possible. This completes the proof ofTheorem 3.2.

4. Applications

Example 4.1. Setp, q,r, sbe two pairs of conjugate exponents andp > 1,s > 1,β ≥ α ≥ e7/12_{, 0< λ≤1. Then it has the following.}

1If 0<∞_n1lnαnp1−λ/r−1apn/np−1<∞, and 0<

_∞

n1lnβnq1−λ/s−1 b

q

n/nq−1<

∞, then

∞

m1 ∞

n1

ambn

lnλ_αmlnλ_βn <

π

λsinπ/s

_∞

n1

lnαnp1−λ/r−1apn

np−1

1/p_∞

n1

lnβnq1−λ/s−1bqn

nq−1

1/q

.

4.1

2If 0<∞_n1lnαnp1−λ/r−1apn/np−1<∞, then

∞

n1

lnβnpλ/s−1 n

_∞

m1

am

lnλ_αmlnλ_βn

p

<

π

λsinπ/s

p∞

n1

lnαnp1−λ/r−1apn

np−1 , 4.2

where the constant factorskλs kλr : π/λsinπ/sandπ/λsinπ/spare both the

Proof. SettingKλx, y 1/xλyλ,x, y ∈ R2, it is a homogeneous measurable kernel

function of “λ” degree. Lettingtuλ_{, it has}

0< kλs:

_∞

0

Kλ1, uuλ/s−1du

_∞

0

uλ/s−1

1uλdu

1

λ

_∞

0

t1/s−1

1tdt

1

λB

1

s,

1

r

kλr<∞.

4.3

Settingux lnαx,vx lnβx, then bothuxandvxare strictly monotonic increasing differentiable functions in1,∞and satisfy

U1>0, U∞ ∞ Uu, v,

0<

∞

n1

vn

vn1ε

∞

n1

1

nlnβn1ε ≤

∞

n1

un

un1ε

∞

n1

1

nlnαn1ε <∞

4.4

forε >0. Asβ≥α≥e7/12_{, 0< λ≤1,s >1, andn}

01, letting

f yKλ um, v y

uλ/rm v1−λ/s _yv

_y 1

lnλ_αmlnλ_βy

lnλ/rαm

ln1−λ/s_βyy,

g y:Kλ u y

, vn v

λ/s_n

u1−λ/r _yu

_y 1

lnλ_αylnλ_βn

lnλ/s_βn

ln1−λ/r_αyy,

y∈1,∞, n, m∈N,

4.5

with2.1∼2.8, it has

Rλm, s

_v1/um

0

Kλ1, uuλ/s−1du−

1 2f1

1 12f

_1>0,

0< ηλm, s O

1

um

ρ

ρ >0, m−→ ∞,

Rλn, r:

u1/vn

0

Kλ μ,1

μλ/r−1dμ−1

2g1 1 12g

_1>0.

4.6

When 0<∞_n1lnαnp1−λ/r−1apn/np−1<∞and 0<

_∞

n1lnβnq1−λ/s−1b

q

n/nq−1 <∞; that

is,a∈ _φp,b ∈ϕq andap,φ >0,bq,ϕ > 0, byTheorem 3.2, inequality4.1holds, so does

4.2. And4.2is equivalent to4.1, and the constant factorskλs kλr:π/λsinπ/s

andπ/λsinπ/spare both the best possible.

1If 0<∞_n0nαp1−λ/r−1apn<∞and 0<

_∞

n0nβq1−λ/s−1b

q

n<∞, then

∞

n0 ∞

m0

ln mα/ nβambn

mαλ− nβλ

<

π λsinπ/s

2∞

n0

nαp1−λ/r−1apn

1/p_∞

n0

nβq1−λ/s−1bqn

1/q

.

4.7

2If 0<∞_n0nαp1−λ/r−1apn<∞, then

∞

n0

nβpλ/s−1

_∞

m0

lnmα/nβam

mαλ− nβλ

p

<

π

λsinπ/s

2p∞

n0

nαp1−λ/r−1apn,

4.8

where inequality 4.8 is equivalent to 4.7 and the constant factors kλs kλr :

π/λsinπ/s2andπ/λsinπ/s2pare both the best possible.

Proof. SettingKλx, y lnx/y/xλ −yλx, y ∈ R2, it is a homogeneous measurable

kernel function of “λ” degree. Lettingtuλ_{, it has2}

0< kλs:

_∞

0

Kλ1, uuλ/s−1du

1

λ

_∞

0

−lnuλ

1−uλu λ/s−1_du

1

λ2 _∞

0

lnt t−1t

1/s−1_dt

1

λB

1

s,

1

r

2

π

λsinπ/s

2

kλr<∞.

4.9

Settingux xα,vx xβ, then bothuxandvxare strictly monotonic increasing differentiable functions in0,∞and satisfy

U0>0, U∞ ∞ Uu, v,

0<

∞

n0

vn

vn1ε

∞

n0

1

nβ1ε ≤

∞

n0

un

un1ε

∞

n0

1

nα1ε <∞

forε >0. Asβ≥α≥1/2, 0< λ≤1,s >1, andn00, letting

f1 y

Kλ um, v y

uλ/rm v1−λ/s _yv

_yln mα/ yβ

mαλ− yβλ

mαλ/r yβ1−λ/s

1

λmα

lnyβ/mαλ

yβ/mαλ−1

yβ

mα

λ/s−1

,

g1 y

:Kλ u y

, vn v

λ/s_n

u1−λ/r _yu

_y ln

yα/ nβ

yαλ− nβλ

nβλ/s

yα1−λ/r

1

λ nβ

lnyα/nβλ

yα/nβλ−1

_yα

nβ

λ/r−1

, y∈0,∞, n, m∈N,

4.11

with2.18∼2.21, it has

Rλm, s:

v0/um

0

Kλ1, uuλ/s−1du−

1 2f10

1 12f

10

1

λ2

_β/mαλ

0

lnu u−1u

1/s−1_du−1

2f10 1 12f

10>0,

Rλn, r:

_u0/vn

0

Kλ μ,1

μλ/r−1dμ−1

2g10 1 12g

10>0,

0< ηλm, s O

1

um

ρ

ρ >0, m−→ ∞.

4.12

When 0 < ∞_n0nαp1−λ/r−1apn < ∞and 0 <

_∞

n0nβq1−λ/s−1b

q

n < ∞; that is,

a∈_φp,b∈ϕqandap,φ>0,bq,ϕ>0, byTheorem 3.2, inequality4.7holds, so does4.8.

And4.8is equivalent to4.7, and the constant factorskλs kλr: π/λsinπ/s2and

π/λsinπ/s2pare both the best possible.

Remark 4.3. It can be proved similarly that, if the conditions “β≥α≥e7/12_{” inLemma 2.1and}

“β≥α≥1/2” inLemma 2.2are changed into “α≥β≥e7/12_{” and “α≥β≥1/2”, respectively,}

Lemmas2.1and 2.2are also valid. So the conditions “β ≥ α ≥ e7/12” in Example 4.1and “β ≥ α ≥ 1/2” in Example 4.2can be replaced by “β ≥ e7/12_{,α ≥ e}7/12_{” and “β ≥ 1/2,}

α≥1/2”, respectively.

Acknowledgment

References

1 H. Weyl,Singulare Integralgleichungen mit besonderer Beriicksichtigung des Fourierschen Integraltheorems, Inaugeral dissertation, University of G ¨ottingen, G ¨ottingen, Germany, 1908.

2 G. Hardy, J. Littiewood, and G. Polya,Inequalities, Cambridge University Press, Cambridge, UK, 1934. 3 B. C. Yang, “On a more accurate Hardy-Hilbert-type inequality and its applications,”Acta Mathematica

Sinica, vol. 49, no. 2, pp. 363–368, 2006Chinese.

4 B. Yang, “On a more accurate Hilbert’s type inequality,”International Mathematical Forum, vol. 2, no. 37–40, pp. 1831–1837, 2007.

5 W. Zhong, “A Hilbert-type linear operator with the norm and its applications,”Journal of Inequalities and Applications, vol. 2009, Article ID 494257, 18 pages, 2009.