A Hilbert Type Linear Operator with the Norm and Its Applications

Volume 2009, Article ID 494257,18pages doi:10.1155/2009/494257

Research Article

A Hilbert-Type Linear Operator with the Norm 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 9 February 2009; Accepted 9 March 2009

Recommended by Nikolaos Papageorgiou

A Hilbert-type linear operatorT:_φp → ψpis defined. As for applications, a more precise operator

inequality with the norm and its equivalent forms are deduced. Moreover, three equivalent reverses from them are given as well. The constant factors in these inequalities are proved to be the best possible.

Copyrightq2009 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.

1. Introduction

In 1925, Hardy1extended Hilbert inequality as follows. Ifp >1, 1/p1/q1,an, bn≥0,0<

_∞ n1a

p

n<∞, and 0< _∞

n1b q

n<∞, then

∞

n1

∞

m1

ambn

mn< π

sinπ/p

_∞

n1

apn

1/p_∞

n1

bqn 1/q

, 1.1

∞

n1 _∞

m1

am

mn

p

<

π

sinπ/p

p∞

n1

apn, 1.2

Under the same conditions, there are the classic inequalities2:

∞

n1

∞

m1

lnm/nambn

m−n <

π

sinπ/p

2_∞

n1

apn

1/p_∞

n1

bqn 1/q

, 1.3

∞

n1 _∞

m1

lnm/nam

m−n

p

<

π

sinπ/p

2p_∞

n1

apn, 1.4

where the constant factorsπ/sinπ/p2andπ/sinπ/p2pare also the best possible. The expression1.3is well known as a Hilbert-type inequality.

By setting a real space of sequences:p_:{a;a{a

n}∞n0,ap{ _∞

n1|an|p}1/p<∞} and defining a linear operator T : p → p, Tan Cn ∞n1lnm/nam/m−n n∈N0, the expressions1.3and1.4can be rewritten as

Ta, b<Ta_pb_q, 1.5

Tap<Tap, 1.6

respectively, whereT π/sinπ/p2,b ∈ q_{.Ta, bis the formal inner product ofTa} andb.

The inequalities 1.1–1.4 play important roles in theoretical analysis and appli-cations 3. These inequalities and their integral forms have been recently extended

or strengthened in 4–8. Zhao and Debnath 9 obtained a Hilbert-Pachpatte’s reverse inequality. Zhong and Yang10,11have given some reverses concerning some extensions of 1.1. Papers in 12–15 studied some multiple Hardy-Hilbert-type or Hilbert-type inequalities. Articles in16,17got some Hilbert-type linear operator inequalities. In 2006, Yang18deduced a new Hilbert-type inequality as follows.

Setp, qas a pair of conjugate exponents, andp >1,1/2≤α≤1, an, bn≥0, such that 0<∞_n1apn<∞, 0<

_∞ n1b

q

n<∞, then one has

∞

n0

∞

m0

lnmα/nαambn

m−n <

π

sinπ/p

2_∞

n0

apn

1/p_∞

n0

bqn 1/q

, 1.7

∞

n0 _∞

m0

lnmα/nαam

m−n

p

<

π

sinπ/p

2p_∞

n0

apn. 1.8

It has been proved that1.7 and 1.8are two equivalent inequalities and their constant factorsπ/sinπ/p2andπ/sinπ/p2pare the best possible. Whenα1, the expressions 1.7and1.8can be reduced to1.3and1.4, respectively.

This paper reports the studies on a Hilbert-type linear operator T : _φp → ψp. As for the applications, a more precise linear operator’s general form of Hilbert-type inequality 1.3incorporating the norm and its equivalent form are deduced. Moreover, three equivalent reverses of the new general forms are deduced as well. The constant factors in these inequalities are all the best possible.

1Ifs >1,r, sis a pair of conjugate exponents, then the Beta function is defined as followscf.2, Theorem 342,

_∞

0 lnu u−1u

1/s−1_du

π

sinπ/s

2

B

1

s,

1

r

2

B

1

r,

1

s

2

. 1.9

2 Euler-Maclaurin’s summation formula. Set f ∈ C3_{0,∞, if −1}i_fi_{x >} 0, fi_{∞ 0 i0,1,2,3, thencf.19, Lemma 1}

∞

n0

fn<

_∞

0

fxdx1

2f0− 1 12f

_{0, 1.10}

∞

n0

fn>

_∞

0

fxdx1

2f0. 1.11

2. Lemmas

Lemma 2.1. Setr, sas a pair of conjugate exponents,s >1,α >0,0< λ≤1, and define

gu: ⎧ ⎪ ⎨ ⎪ ⎩

lnu

u−1, u∈0,1∪1,∞, 1, u1,

2.1

fsx:hm,λ

x,1 s

:g

_xα

mα

λ_xα

mα

λ 1/s−1/λ

, x∈−α,∞, m∈N0. 2.2

Then, one has the following:

1the functionfsxsatisfies the conditions of1.10and1.11. This means

−1i_fi

s x>0 x >−α, fsi∞ 0 i0,1,2,3, 2.3

2

kλs: 1 λmα

_∞

−α

fsxdx

B1/s,1/r λ

2

π λsinπ/s

2

. 2.4

Proof. 1Forα > 0,x >−α,m∈ N0, 0< λ ≤1 ands >1, setzx gxα/mαλ,

−1i_gi_{u > 0, g}i_{∞ 0u > 0, i 0,1,2,3 cf. 16, Lemma 2.2, one has zx > 0,}

tx>0,

zx gu λ mα

xα mα

λ−1

<0,

zx gu

λ mα

xα mα

λ−1 2

guλλ−1

mα2

xα mα

λ−2

>0,

zx gu

λ mα

xα mα

λ−1 3

3guλ

2_λ−1 mα3

xα mα

2λ−3

guλλ−1λ−2

mα3

xα mα

λ−3

<0,

tx λ/s−1 mα

xα mα

λ/s−2

<0,

tx λ/s−1λ/s−2

mα2

xα mα

λ/s−3

>0,

tx λ/s−1λ/s−2λ/s−3

mα3

xα mα

λ/s−4

<0.

2.5

These are followed by

fsx zxtx>0, fs∞ 0,

f_sx zxtx zxtx<0, f_s∞ 0, 2.6 f_sx zxtx 2zxtx zxtx>0, f_s∞ 0,

f_sx zxtx 3zxtx 3zxtx zxtx<0, f_s∞ 0. 2.7

Then inequality2.3holds.

2Forx >−α,m∈N0andλ >0,s >1, setu xα/mαλ,then one has

1

λmα

_∞

−α

fsxdx 1 λ

_∞

−α

lnxα/mαλ

xα/mαλ−1

_xα

mα

λ 1/s−1/λ

d

xα mα

1

λ2 _∞

0 lnu u−1u

1/s−1_du.

2.8

Lemma 2.2. Setr, sas a pair of conjugate exponents,s >1,α≥1/2,0< λ≤1and define

ωλm, s:

∞

n0

lnnα/mα

nαλ−mαλ ·

mαλ/r

nα1−λ/s m∈N0, 2.9

then, one has

0< ωλm, s< kλs, 2.10

0< ωλn, r< kλr kλs n∈N0, 2.11

wherekλsis defined by2.4.

Proof. By2.9and2.2, it is evident that

0< ωλm, s 1 λmα

∞

n0

lnnα/mαλ

nα/mαλ−1

_nα

mα

λ 1/s−1/λ

1

λmα

∞

n0

hm,λ

n,1 s

1

λmα

∞

n0

fsn.

2.12

In view of2.3,1.10, and2.4, one has

ωλm, s< 1 λmα

_∞

0

fsxdx1

2fs0− 1 12f

s0

1

λmα

_∞

−α

fsxdx−

0

−α

fsxdx−1

2fs0 1 12f

s0

kλs− 1

λmαRs, m,

2.13

whereRs, m:₋0_αfsxdx−1/2fs0 1/12f_s0 m∈N0. With2.6, it follows that

fs0 z0t0 g

α mα

λ _α

mα

λ/s−1

, 2.14

f_s0 z0t0 z0t0

λ−s

sα g

_α

mα

λ _α

mα

λ/s−1 λ

αg

α

mα

λ _α

mα

λ/sλ−1

Setu xα/mαλ, with the partial integration, by the strictly monotonic increase of

gugu>0andsr/r−1, it gives

0

−α

fxdx

mα

0

−α

lnxα/mαλ

xα/mαλ−1

_xα

mα

λ 1/s−1/λ

dxα mα

mα

λ

_α/mαλ

0

guu1/s−1du

smα

λ

_α/mαλ

0

gudu1/s

smα

λ g

_α

mα

λ _α

mα

λ/s

−smα λ

_α/mαλ

0

u1/sgudu

> sα λg

_α

mα

λ _α

mα

λ/s−1

−rmα λr−1g

α

mα

λα/mαλ

0

u1−1/rdu

sα

λg

α mα

λ _α

mα

λ/s−1

− r2mα λr−12r−1g

α

mα

λ _α

mα

λ 2−1/r

sα

λg

_α

mα

λ _α

mα

λ/s−1

− r2α λr−12r−1g

α

mα

λ _α

mα

λ/sλ−1

.

2.16

In view of2.13–2.16, one has

Rs, m>

sα λ −

1 2

λ−s

12sα

g

α mα

λ _α

mα

λ/s−1

−

r2_α

λr−12r−1− λ

12α g

α

mα

λ _α

mα

λ/sλ−1

.

2.17

Ifα≥1/2,s >1r >1, 0< λ≤1, gu>0,−gu>0, one has

sα λ −

1 2

λ−s

12sα

12s2_α2_{−6sαλλλ−s} 12sαλ

6sα2sα−λ−λs−λ

12sαλ ≥ 6sαs−λ−λs−λ

12sαλ

6sα−λs−λ

12sαλ >0, r2_α

λr−12r−1− λ

12α

12r2_α2_−λ2_r−12r−1 12λαr−12r−1

2r2

6α2_−λ2_λ2_3r−1 12λαr−12r−1 >0.

This means that Rs, m > 0. By 2.13 and 2.4, the inequalities 2.10 and 2.11 hold.

Lemma 2.2is proved.

Lemma 2.3. Setr, sas a pair of conjugate exponents,s > 1,α ≥1/2,0 < λ ≤1, andωλm, s,

kλsare defined by2.9,2.4, respectively, then,

1 ωλm, s> kλs1−ηλm, 2.19

2 0< ηλm< θλr<1

θλr: 1 kλsλ2

₁

0 lnu u−1u

−1/r_du

, 2.20

3 ηλm O

₁

mα

λ/2s

m−→ ∞, 2.21

whereηλm: 1/kλsλmα0_−αfsxdx−1/2fs0,fsxis defined by2.2. Proof. By2.12,1.11, and2.4,

ωλm, s 1 λmα

∞

n0

fsn> 1 λmα

∞

0

fsxdx1

2fs0

1

λmα

∞

−α

fsxdx−

0

−α

fsxdx1

2fs0

kλs

1− 1

kλsλmα

0

−α

fsxdx−1

2fs0 kλs1−ηλm.

2.22

This implies that2.19holds.

From the monotonic decrease of the functionfsx see2.3,fs0>0 andα≥1/2, one hasηλm>1/kλsλmααfs0−1/2fs0≥0. On the other hand, iffs0>0 and by the computation as in2.16,

ηλm 1 kλsλmα

0

−α

fsxdx−1

2fs0

< 1 kλsλmα

₀

−α

fsxdx 1 kλsλ2

_α/mαλ

0

lnu u−1u

1/s−1_{du≤θλr<1.}

2.23

Since limu→0lnu/u−1u1/2s 0s > 1,there exists a constantL > 0, such that

|lnu/u−1u1/2s_{| ≤Lu∈0,α/mα}λ_{. Then,}

0< ηλm< L kλsλ2

α/mαλ

0

u1/2s−1du 2sL kλsλ2

α mα

λ/2s

. 2.24

This means thatηλm O1/mαλ/2s m → ∞,the proof is finished.

Lemma 2.4. Setp, qandr, sas two pairs of conjugate exponents,p > 1,r > 1,α > 0,λ >0, 0< ε < pλ/2r,am: mαλ/r−ε/p−1,bn: nαλ/s−ε/q−1, andkλsis defined by2.4. Defining

I1:ε _∞

m0

mαp1−λ/r−1apm

1/p_∞

n0

nαq1−λ/s−1bqn 1/q

,

I2:ε

_∞

1−α

xαλ/r−ε/p−1yαλ/s−ε/q−1lnxα/yα

xαλ−yαλ dx dy,

2.25

then

1 0< I1< ε

α1ε 1

αε, 2.26

2 I2≥kλs o1 ε−→0. 2.27

Proof. 1Byα >0 andε >0, one has

0< I1 ε _∞

m0

mα−1−ε

1/p_∞

n0

nα−1−ε

1/q

ε

1

α1ε

∞

n1 1

nα1ε < ε

1

α1ε _∞

0 1 xα1εdx

ε

1

α1ε − 1

εxα

−ε∞ 0

,

2.28

2Byy ≥ 1−α,letting 0 < ε < pλ/2r, one hasyα−1−ε ≤ yα−1. And setting

u xα/yαλ, with limu→0lnu/u−1u1/2r 0r > 1,|lnu/u−1u1/2r| ≤

L1u∈0,1, L1>0,one has

I2

ε λ2

_∞

1−α

yα−1−ε

∞

1/yαλ ln u u−1u

1/r−ε/pλ−1_{du dy}

ε

λ2 _∞

1−α

yα−1−ε

⎡ ⎣∞

0 lnu u−1u

1/r−ε/pλ−1_du−

_1/yαλ

0

lnu u−1u

1/r−ε/pλ−1_du ⎤ ⎦dy

B2

1/r−ε/pλ,1/sε/pλ

λ2 −

ε λ2

_∞

1−α

yα−1−ε

⎡

⎣1/yα λ

0

lnu u−1u

1/r−ε/pλ−1_du ⎤ ⎦dy

≥ B2

1/r−ε/pλ,1/sε/pλ

λ2 −

εL1

λ2 _∞

1−α

yα−1

⎡

⎣1/yα λ

0

u1/2r−ε/pλ−1du

⎤ ⎦dy

B1/r−ε/pλ,1/sε/pλ λ

2

− εL1

λ2_{1/2r−ε/pλ ∞}

1−α

yα−λ1/2r−ε/pλ−1dy

B1/r−ε/pλ,1/sε/pλ λ

2

εL1

λ3_{1/2r−ε/pλ}2.

2.29

Setε → 0, then the inequality2.27holds.Lemma 2.4is proved.

3. Main Results

Firstly, the following notations are given.

1Setp >0, p /1, r >1,p, qandr, sare two pairs of conjugate exponents. Let

φx: xαp1−λ/r−1, ϕx: xαq1−λ/s−1,

ψx:ϕx1−p xαpλ/s−1, x∈0,∞.

3.1

2Setp >1,p, qis a pair of conjugate exponents. Let

_φp:

⎧ ⎨

⎩a;a{an}∞n0,ap,φ: _∞

n0

φn|an|p 1/p

<∞

⎫ ⎬

It is a real space of sequences, where

ap,φ _∞

n0

φn|an|p 1/p

3.3

is a norm of sequenceawith the weight functionφ. Similarly, it can define the real spaces of sequences:qϕ,

p

ψ and the norm of sequencebwith the weight functionϕ:bq,ϕas well. For 0 < p < 1 orq < 0,the marksa_p,φ andb_q,ϕ as two formal norms are still used in

Theorem 3.3.

3 Set p > 1,p, q is a pair of conjugate exponents. Define a Hilbert-type linear operatorT, for alla∈_φp, one has

Tan:Cn:

∞

m0

lnmα/nα

mαλ−nαλam n∈N0. 3.4

4Fora∈_φp,b∈ϕq, define the formal inner product ofTaandbas

Ta, b:

∞

n0

_∞

m0

lnmα/nαam mαλ−nαλ

bn

∞

n0

∞

m0

lnmα/nαambn

mαλ−nαλ , 3.5

Then one will have some results in the following theorem.

Theorem 3.1. Suppose thatp, qandr, sare two pairs of conjugate exponents andp >1,r >1, 1/2≤α≤1,0< λ≤1,an≥0. Then for∀a∈p_φ, one has

1

TaC{Cn}∞n0∈ p

ψ. 3.6

It means thatT :_φp → pψ.

2Tis a bounded linear operator and

T_p,ψ : sup a∈p_φa /θ

Tap,ψ

ap,φ

kλs, 3.7

Proof. Ifp > 1, by using H ¨older’s inequalitycf.20and the result2.11, forn ∈N0, it is obvious thatCn≥0 and

Cpn _∞

m0

lnmα/nα

mαλ−nαλ

mα1−λ/r/q

nα1−λ/s/pam

nα1−λ/s/p

mα1−λ/r/q

p

≤

_∞

m0

lnmα/nα

mαλ−nαλ

mαp−11−λ/r

nα1−λ/s a

p m

×

_∞

m0

lnmα/nα

mαλ−nαλ

nαq−11−λ/s

mα1−λ/r

p−1

_∞

m0

lnmα/nα

mαλ−nαλ

mαp−11−λ/r

nα1−λ/s a

p m

!

ωλn, rϕn"p−1

≤kp_λ−1s

_∞

m0

lnmα/nα

mαλ−nαλ

mαp−11−λ/r

nα1−λ/s a

p m

#

ϕp−1n$.

3.8

And ifψn ϕ1−p_{n, by2.9and2.10, it follows that}

Tap p,ψ

∞

n0

ψnCpn

≤kp_λ−1s

_∞

m0

∞

n0

lnmα/nα

mαλ−nαλ

mαp−11−λ/r

nα1−λ/s a

p m

kp_λ−1s

∞

m0

ωλm, sφmapm≤k_λpsap_p,φ<∞.

3.9

This means thatC{Cn}∞n0∈ p

ψandTp,ψ ≤kλs.

If there exists a constantK < kλs, such thatT_p,ψ ≤K, then for 0 < ε < pλ/2r, by the definition3.5, and by using H ¨older’s inequality and the result2.26, one has

ε%Ta, b&≤εT_p,ψa _p,φ'''b'''

q,ϕ≤KI1< K

ε α1ε

1

αε

, 3.10

wherea{am}∞m0∈ p

φ,b{bn}

∞

n0 ∈ q

ϕandam,bnare defined as inLemma 2.4. On the other hand, from the strictly monotonic decrease of the function gu

lnu/u−1 and the exponentsλ/r−ε/p−1 < 0, λ/s−ε/q−1 < 0 and 1−α ≥ 0, and byα >0,λ >0, in view of2.27, one has

ε%Ta, b&≥ε

_∞

1−α

xαλ/r−ε/p−1yαλ/s−ε/q−1lnxα/yα

xαλ−yαλ dx dy

I2≥kλs o1 ε−→0.

In view of3.10and3.11, one haskλs o1< Kε/α1ε_1/αε_{.Settingε → 0,one has}

kλs≤K.This means thatKkλs, that is,T_p,ψkλs.Theorem 3.1is proved.

Theorem 3.2. Suppose thatp, qand r, s are two pairs of conjugate exponents,r > 1,p > 1, 1/2≤α≤1,0< λ≤1,an, bn≥0n∈N0. Then one has the following.

1 Ifa∈p_φ,b∈ϕq, andap,φ>0,bq,ϕ>0, then

Ta, b< kλsa_p,φb_q,ϕ. 3.12

2 Ifa∈p_φanda_p,φ>0, then

Tap,ψ < kλsap,φ, 3.13

where the markTa_p,ψis defind as inTheorem 3.1. The inequality3.13is equivalent to3.12, and the constant factorkλs 1/λB1/s,1/r2kλris the best possible.

Proof. In view of3.9and 0<a_p,φ<∞, one has

Tap

p,ψ< kp_λsap_p,φ. 3.14

And byp >1,3.13holds.

By using H ¨older’s inequality and3.13, one has

Ta, b

∞

n0

∞

m0

lnmα/nα

mαλ−nαλ

mα1−λ/r/q

nα1−λ/s/pam

nα1−λ/s/p

mα1−λ/r/qbn ≤ Tap,ψbq,ϕ< kλsap,φbq,ϕ.

3.15

The inequality3.12is obtained.

From 3.12 and a_p,φ > 0, there exists k0 ∈ N, such that Km0φma p

m > 0 and

bnK ψnK_m0lnmα/nαam/mαλ−nαλp−1 >0 whenK > k0. By a combination as in3.15and by 1/2≤α≤1, 0< λ≤1, and with2.10and2.11, then,

0<

K

n0

ϕnbqnK

K

n0

ψn

_K

m0

lnmα/nαam mαλ−nαλ

p

K n0

K

m0

lnmα/nαambnK

mαλ−nαλ < kλs

_K

n0

φnapn

1/p_K

n0

ϕnbnq K

1/q

<∞.

3.16

Byp >1 andq >1, it follows that

0<

K

n0

ϕnbqnK< k_λps

∞

n0

LettingK → ∞in3.17, this means 0< _n∞₀ϕnbnq ∞ <∞, that is,b{bn∞}∞_n0 ∈ϕq andb_q,ϕ > 0. Therefore the inequality3.16keeps the form of the strict inequality when

K → ∞. So does3.17. In view of∞_n0ϕnbqn∞ Tapp,ψ, the inequality3.13holds, and 3.12 is equivalent to3.13. ByT_p,ψ kλs, it is obvious that the constant factor kλs kλris the best possible. This completes the proof ofTheorem 3.2.

Theorem 3.3. Setp, qandr, sas two pairs of conjugate exponents,0 < p < 1q <0,r > 1, 1/2≤α≤1,0< λ≤1,an, bn≥0. Let

Ha, b

∞

n0

∞

m0

lnmα/nαambn

mαλ−nαλ . 3.18

Then the reverse inequalities can be established as follows. 1 If0<a_p,φ<∞and0<b_q,ϕ<∞, then

Ha, b> kλs

_∞

m0

1−ηλmφmapm 1/p

bq,ϕ. 3.19

2 If0<a_p,φ<∞, then

∞

n0

ψn

_∞

m0

lnmα/nαam mαλ−nαλ

p

> kp_λs

∞

m0

1−ηλmφmapm. 3.20

3 If0<b_q,ϕ<∞, then

∞

m0

φ−1m

1−ηλm

q−1_∞

n0

lnmα/nαbn mαλ−nαλ

q

< kq_λsbqq,ϕ, 3.21

where the marksa_p,φ andb_q,ϕ 0 < p < 1as two formal norms are still defined like in3.3

and the factorηλmin3.19–3.21is defined inLemma 2.3. The inequalities3.20and3.21are equivalent to3.19. The constant factorskλs,kp_λsandk_λqsin3.19,3.20, and3.21are all the best possible.

Proof. By 0< p <1q <0, with the reverse H ¨older’s inequality, one has the following.

1By the combination as in3.15for3.18, then one has

Ha, b≥

_∞

m0

ωλm, sφmapm

1/p_∞

n0

ωλn, rϕnbqn 1/q

. 3.22

If 1/2 ≤ α≤ 1, 0< λ ≤ 1, the expressions2.19and2.11are established forωλm, sand

Setting a constant K ≥ kλr, 3.19is still valid if we replace kλr byK, then for 0< ε <−qλ/r, by3.19and2.21, one will have

H%a, b&>K

_∞

m0

1−ηλmφmapm 1/p

b_q,ϕ

K

_∞

m0

1−ηλm

mα1ε

1/p_∞

n0 1 nα1ε

1/q

K

∞

n0 1 nα1ε

⎧ ⎨ ⎩1−

(_∞ m0O

%

1/mα1ελ/2s&)

(_∞

m01/mα1ε )

⎫ ⎬ ⎭

1/p

,

3.23

wherea{am}∞m0,b{bn}

∞

n0,am mαλ/r−ε/p−1,bn nαλ/s−ε/q−1and it is apparent that 0<a _p,φ<∞, 0<b_q,ϕ<∞.

On the other hand, by3.18,2.2,2.12, and2.10,

H%a, b&

∞

m0

∞

n0

lnnα/mαmαλ/r−ε/p−1nαλ/s−ε/q−1

nαλ−mαλ

∞ m0

mα−1−ε

1

λmα

∞

n0

lnnα/mαλnα/mαλ/s−ε/q−1

nα/mαλ−1

∞ m0

mα−1−ε

1

λmα

∞

n0

hm

n,1 s−

ε qλ

< B

2_{1/s−ε/qλ,1/rε/qλ}

λ2

∞

n0 1 nα1ε.

3.24

In view of3.23and3.24, one has

K

⎧ ⎨ ⎩1−

(_∞ m0O

%

1/mα1ελ/2s&)

(_∞

m01/mα1ε )

⎫ ⎬ ⎭

1/p

< B

2_{1/s−ε/qλ,1/rε/qλ}

λ2 . 3.25

Settingε → 0, one hasK ≤kλs, which meansK kλs. The constant factorkλsin3.19

2By 0 < a_p,φ < ∞, one has _n∞₀ψn∞_m0lnmα/nαam/mαλ−

nαλp > 0. Setting bn : ψn∞m0lnmα/nαam/mαλ−nαλ p−1

,

making the calculations as in3.8and3.9, one has 0 < bqq,ϕ <∞.By using3.19in the following:

bq q,ϕ

∞

n0

ϕnbqn

∞

n0

ψn

_∞

m0

lnmα/nαam mαλ−nαλ

p

Ha, b> kλs

_∞

m0

1−ηλmφmapm 1/p

bq,ϕ,

3.26

one has

∞

n0

ψn

_∞

m0

lnmα/nαam mαλ−nαλ

p

> kp_λs

∞

m0

1−ηλmφmapm. 3.27

Therefore,3.20holds.

On the other hand, if 3.20is valid, by 0 < p < 1q < 0and by using the reverse H ¨older’s inequality, it has

Ha, b ∞

n0

nαλ/s−1/p

∞

m0

lnmα/nαam mαλ−nαλ

(

nα1/p−λ/sbn )

≥

_∞

n0

ψn

_∞

m0

lnmα/nαam mαλ−nαλ

p1/p_∞

n0

nαq1−λ/s−1bnq 1/q

> kλs

_∞

m0

1−ηλmφmapm 1/p

bq,ϕ.

3.28

Then3.19holds. It means that3.20is equivalent to3.19.

3By 0<b_q,ϕ<∞, it is obvious that there existn0∈N, such that

K

m0

φ−1_m 1−ηλm

q−1⎡ ⎢ ⎣K

n0

ln_% mα/nαbn mαλ−nαλ&

⎤ ⎥ ⎦ q

>0,

_K

n0

ϕnbqn

SettingamK φ−1m/1−ηλmK_n0lnmα/nαbn/mαλ−nαλq−1>

0,one has 0<K_m0φmapmK<∞. By3.19,

K

m0

1−ηλmφmapmK

K m0

φ−1_m 1−ηλm

q−1_K

n0

lnmα/nαbn mαλ−nαλ

q

K m0

K

n0

lnmα/nαbnamK mαλ−nαλ

> kλs

_K

m0

1−ηλmφmapmK

1/p_K

n0

ϕnbqn 1/q

.

3.30

Further one has

_K

m0

1−ηλmφmapmK

1/q

> kλs

_K

n0

ϕnbqn 1/q

>0. 3.31

Byq <0, one has

0<

K

m0

1−ηλmφmapmK< k_λqs

K

n0

ϕnbnq < kq_λs

∞

n0

ϕnbqn<∞. 3.32

SettingK → ∞in3.32, via2.20, one has

0<

∞

m0

φmapm∞< 1

1−θλr

∞

m0

1−ηλmφmapm∞<∞. 3.33

Also, from3.21, byq <0 and by using the reverse H ¨older’s inequality,

Ha, b

∞

m0

_1−ηλm

φ−1_m 1/p

am ⎧_⎨

⎩

φ−1_m 1−ηλm

1/p_∞

n0

lnmα/nαbn mαλ−nαλ

⎫ ⎬ ⎭

≥

_∞

m0

1−ηλmφmapm 1/p⎧_⎨

⎩

∞

m0

φ−1_m 1−ηλm

q−1_∞

n0

lnmα/nαbn mαλ−nαλ

q⎫_⎬

⎭ 1/q

> kλs

_∞

m0

1−ηλmφmapm 1/p

bq,ϕ.

3.34

Therefore,3.19holds. Equation3.21is equivalent to3.19.

If the constant factorkp_λs orkq_λsin3.20 or in3.21is not the best possible, by3.28 or by3.34, then it leads to a contradiction in which the constant factorkλsin 3.19is not the best possible.Theorem 3.3is proved.

Remark 3.4. Setr q, s p, λ 1, the inequalities3.12and3.13can be reduced to1.7

and1.8, respectively. So3.12 or3.13is an extension of1.7 or1.8.

Acknowledgments

The work is supported by the National Natural Science Foundation of Chinano.10871073. The author would like to thank the anonymous referee for his or her suggestions and corrections.

References

1 G. Hardy, “Not on a theorem of Hilbert concering series of positive terms,”Proceedings of the London Mathematical Society, vol. 23, no. 2, pp. 415–416, 1925.

2 G. H. Hardy, J. E. Littlewood, and G. P ´olya,Inequalities, Cambridge University Press, Cambridge, UK, 2nd edition, 1952.

3 D. S. Mitrinovi´c, J. Peˇcari´c, and A. M. Fink,Inequalities Involving Functions and Their Integrals and Derivatives, vol. 53 of Mathematics and Its Applications (East European Series), Kluwer Academic Publishers, Dordrecht, The Netherlands, 1991.

4 M. Gao and B. Yang, “On the extended Hilbert’s inequality,”Proceedings of the American Mathematical Society, vol. 126, no. 3, pp. 751–759, 1998.

5 B. G. Pachpatte, “On some new inequalities similar to Hilbert’s inequality,”Journal of Mathematical Analysis and Applications, vol. 226, no. 1, pp. 166–179, 1998.

6 B. Yang, I. Brneti´c, M. Krni´c, and J. Peˇcari´c, “Generalization of Hilbert and Hardy-Hilbert integral inequalities,”Mathematical Inequalities & Applications, vol. 8, no. 2, pp. 259–272, 2005.

7 W. Zhong and B. Yang, “A best extension of Hilbert inequality involving seveial parameters,”Journal of Jinan University (Natural Science), vol. 28, no. 1, pp. 20–23, 2007Chinese.

8 H. Leping, G. Xuemei, and G. Mingzhe, “On a new weighted Hilbert inequality,”Journal of Inequalities and Applications, vol. 2008, Article ID 637397, 10 pages, 2008.

9 C.-J. Zhao and L. Debnath, “Some new inverse type Hilbert integral inequalities,” Journal of Mathematical Analysis and Applications, vol. 262, no. 1, pp. 411–418, 2001.

11 W. Zhong, “A reverse Hilbert’s type integral inequality,”International Journal of Pure and Applied Mathematics, vol. 36, no. 3, pp. 353–360, 2007.

12 I. Brneti´c and J. Peˇcari´c, “Generalization of Hilbert’s integral inequality,”Mathematical Inequalities & Applications, vol. 7, no. 2, pp. 199–205, 2004.

13 B. Yang and T. M. Rassias, “On the way of weight coefficient and research for the Hilbert-type inequalities,”Mathematical Inequalities & Applications, vol. 6, no. 4, pp. 625–658, 2003.

14 H. Yong, “On multiple Hardy-Hilbert integral inequalities with some parameters,” Journal of Inequalities and Applications, vol. 2006, Article ID 94960, 11 pages, 2006.

15 W. Zhong and B. Yang, “On a multiple Hilbert-type integral inequality with the symmetric kernel,” Journal of Inequalities and Applications, vol. 2007, Article ID 27962, 17 pages, 2007.

16 Y. Li, Z. Wang, and B. He, “Hilbert’s type linear operator and some extensions of Hilbert’s inequality,” Journal of Inequalities and Applications, vol. 2007, Article ID 82138, 10 pages, 2007.

17 W. Zhong, “The Hilbert-type integral inequalities with a homogeneous kernel of−λ-degree,”Journal of Inequalities and Applications, vol. 2008, Article ID 917392, 12 pages, 2008.

18 B. C. Yang, “A more accurate Hardy-Hilbert-type inequality and its applications,”Acta Mathematica Sinica, vol. 49, no. 2, pp. 363–368, 2006Chinese.

19 B. C. Yang and Y. H. Zhu, “Inequalities for the Hurwitz zeta-function on the real axis,” Acta Scientiarum Naturalium Universitatis Sunyatseni, vol. 36, no. 3, pp. 30–35, 1997.