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Volume 2008, Article ID 546829,14pages doi:10.1155/2008/546829

Research Article

On Some Extensions of Hardy-Hilbert’s Inequality and Applications

Laith Emil Azar

Department of Mathematics, Al Al-Bayt University, P.O. Box. 130095, Mafraq 25113, Jordan

Correspondence should be addressed to Laith Emil Azar,azar [email protected] Received 23 October 2007; Accepted 2 January 2008

Recommended by Shusen Ding

By introducing some parameters we establish an extension of Hardy-Hilbert’s integral inequality and the corresponding inequality for series. As an application, the reverses, some particular results and their equivalent forms are considered.

Copyrightq2008 Laith Emil Azar. 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 Iffx,gx≥0, 0<

0 f2xdx <∞, and 0<

0 g2xdx <∞, thensee1

0

fxgy

xy dx dy < π

0

f2xdx

1/2 0

g2xdx 1/2

, 1.1

0

fxgy

max{x, y}dx dy <4

0

f2xdx1/2 0

g2xdx1/2

, 1.2

where the constant factorsπ and 4 are the best possible in1.1and1.2, respectively. In- equality 1.1 is called Hilbert’s integral inequality and1.2 is called Hilbert’s type which have been extended by Hardysee2as follows: ifp > 1, 1/p1/q 1,fx,gx > 0, 0<

0 fpxdx <∞, and 0<

0 gqxdx <∞, then

0

fxgy

xy dx dy < π sinπ/p

0

fpxdx

1/p 0

gqxdx 1/q

, 1.3

0

fxgy

max{x, y}dx dy < pq

0

fpxdx

1/p 0

gqxdx 1/q

, 1.4

(2)

where the constant factorsπ/sinπ/pandpqare the best possible in1.3and1.4, respec- tively. Hardy-Hilbert’s inequality and its applications are important in analysissee3. Re- cently, Yang4gave some generalizations and the reverse form of1.3as follows: ifp > 1, 1/p1/q 1,r >1, 1/r1/s1,λ > 0,fx, gx ≥0, 0<

0 xp1−λ/r−1fpxdx <∞, and 0<

0 xq1−λ/s−1gqxdx <∞, then

0

fxgy

xλyλ dx dy < π λsinπ/r

0

xp1−λ/r−1fpxdx

1/p 0

xq1−λ/s−1gqxdx 1/q

, 1.5

where the constant factorπ/λsinπ/ris the best possible.

The corresponding inequalities for series1.3and1.4are

n1

m1

ambn

mn < π sinπ/p

n0

apn

1/p

n0

bqn 1/q

,

n1

m1

ambn

max{m, n} < pq

n0

apn

1/p n0

bqn 1/q

,

1.6

where the sequences{an}and{bn}are such that 0 <

n0apn <∞, 0 <

n0aqn <∞,and the constant factorπ/sinπ/pandpqare the best possible. By introducing a parameter 0< λ≤2, some extensions of1.6 pq2were given by Yang5,6as

n1

m1

ambn

mλnλ λ

n0

n1−λa2n

1/2 n0

n1−λbn2 1/2

,

n1

m1

ambn max

mλ, nλ < 4 λ

n0

n1−λa2n

1/2

n0

n1−λb2n 1/2

.

1.7

Very recently, in7the following extensions were given:

0

fxgx

Amin{x, y}Bmax{x, y}dx dy < DA, B

0

f2xdx

1/2 0

g2xdx 1/2

,

n1

m1

ambn

Amin{m, n}Bmax{m, n} < DA, B n1

a2n

1/2 n1

bn2 1/2

,

1.8

where the constant factorDA, B see7, Lemma 2.1is the best possible in both inequalities.

For more information related to this subject see, for example,8,9.

In this paper by introducing some parameters, we generalize1.8and we obtain the reverse form for each of them. Some particular results and the equivalent form are also consid- ered.

(3)

2. Main results

Lemma 2.1. Suppose that λ > 0, A0, B > 0. Define the weight coefficients ωλA, B, x and ωλA, B, yby

ωλA, B, x:

0

xλ/2y−1λ/2 Amin

xλ, yλ Bmax

xλ, yλ dy, 2.1

ωλA, B, y:

0

x−1λ/2yλ/2 Amin

xλ, yλ Bmax

xλ, yλ dx, 2.2

thenωλA, B, x ωλA, B, y CλA, Bis a constant defined by

CλA, B

⎧⎪

⎪⎪

⎪⎪

⎪⎩ 4 λ

ABarctan A

B forA >0, B >0, 4

λB forA0, B >0.

2.3

Proof. Settingt y/xλ, we get

ωλA, B, x

0

xλ/2y−1λ/2 Amin

xλ, yλ Bmax

xλ, yλ dy, 1

λ

0

t−1/2

Amin{1, t}Bmax{1, t}dt:I,

2.4

iforA,B >0, we obtain

I 1 λ

1 0

t−1/2 AtBdt

1

t−1/2 ABtdt

1 λ

2

AB

A/B 0

dt t21 2

AB

A/B 0

dt t21

4 λ

ABarctan

A B;

2.5

ii forA0,B >0, we find

I 1 λ

1 0

t−1/2 B dt

1

t−1/2 Bt dt

4

λB. 2.6

Hence,ωλA, B, x CλA, B. By the symmetry we still haveωλA, B, y CλA, B.

The lemma is proved.

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Lemma 2.2. Forp >1or 0< p <1, 1/p1/q1,λ >0,A0,B >0 and 0< ε < pλ/2, setting

1

xλ/2−1−ε/pyλ/2−1−ε/q Amin

xλ, yλ Bmax

xλ, yλ dx dy, 2.7 then forε→0,

1 ε

CλA, B o1

O1< Jε<1 ε

CλA, B o1

. 2.8

Proof. Settingt x/yλ, we find

1

xλ/2−1−ε/pyλ/2−1−ε/q Amin

xλ, yλ Bmax

xλ, yλ dx dy 1

λ

1

y−1−ε

y−λ

t−1/2−ε/λp

Amin{t,1}Bmax{t,1}dt dy 1

λε

0

t−1/2−ε/λp

Amin{t,1}Bmax{t,1}dt dy

−1 λ

1

y−1−ε y−λ

0

t−1/2−ε/λp

Amin{t,1}Bmax{t,1}dt dy

1 ε

CλA, B o1

−1 λ

1

y−1−ε y−λ

0

t−1/2−ε/λp AtB dt dy

≥ 1 ε

CλA, B o1

−1 λ

1

y−1 y−λ

0

t−1/2−ε/λp B dt dy 1

ε

CλA, B o1

O1.

2.9

On the other hand,

1

xλ/2−1−ε/pyλ/2−1−ε/q Amin

xλ, yλ Bmax

xλ, yλ dx dy

<

1

0

xλ/2−1−ε/p Amin

xλ, yλ Bmax

xλ, yλ dx

yλ/2−1−ε/qdy

1 ε

CλA, B o1 .

2.10

Hence,2.8is valid. The lemma is proved.

(5)

Theorem 2.3. If p > 1, 1/p1/q 1,λ > 0, A0,B > 0, fx,gx0 such that 0 <

0 xp1−λ/2−1fpxdx <∞, 0<

0 xq1−λ/2−1gqxdx <∞, then

S:

0

fxgx Amin

xλ, yλ Bmax

xλ, yλ dx dy

< CλA, B

0

xp1−λ/2−1fpxdx1/p 0

xq1−λ/2−1gqxdx1/q

,

2.11

where the constant factorCλA, Bdefined in2.3is the best possible. In particular,

iforλAB1,C11,1 π, and inequality2.11reduces to Hardy-Hilbert’s inequality

0

fxgx

xy dx dy < π

0

xp/2−1fpxdx

1/p 0

xq/2−1gqxdx 1/q

; 2.12

ii forA0,λB1,C10,1 4 and2.11reduces to Hardy-Hilbert’s-type inequality

0

fxgx

max{x, y}dx dy <4

0

xp/2−1fpxdx

1/p 0

xq/2−1gqxdx 1/q

. 2.13

Proof. By the Holder inequality, taking into account2.1, we get

S

0

1 Amin

xλ, yλ Bmax xλ, yλ

1/p

x1−λ/2/q y1−λ/2/pfx

×

1 Amin

xλ, yλ}Bmax xλ, yλ}

1/qy1−λ/2/p x1−λ/2/qgy

dx dy

0

x1−λ/2p−1yλ/2−1 Amin

xλ, yλ}Bmax

xλ, yλ}fpxdx1/p

×

0

y1−λ/2q−1xλ/2−1 Amin

xλ, yλ Bmax

xλ, yλ gqydy1/q

0

ωλA, B, xxp1−λ/2−1fpxdx

1/p 0

ωλA, B, yyq1−λ/2−1gqydy 1/q

CλA, B

0

xp1−λ/2−1fpxdx1/p 0

yq1−λ/2−1gqydy1/q

.

2.14

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If2.14takes the form of equality, then there exist constantsMandNwhich are not all zero such that

M x1−λ/2p−1yλ/2−1

Amin

xλ, yλ Bmax

xλ, yλ fpx N y1−λ/2q−1xλ/2−1 Amin

xλ, yλ Bmax

xλ, yλ gqy, Mxp1−λ/2fpx Nyq1−λ/2gqy, a.e. in0,∞×0,∞

2.15

Hence, there exists a constantcsuch that

Mxp1−λ/2fpx Nyq1−λ/2gqy c a.e. in0,∞. 2.16

We claim thatM0. In fact, ifM /0, then

xp1−λ/2−1fpx c

Mx a.e. in0,∞ 2.17

which contradicts the fact that 0<

0 xp1−λ/2−1fpxdx <∞. Hence, by2.14we get2.11.

If the constant factorCλA, Bis not the best possible, then there exists a positive constant KwithK < CλA, B, thus2.11is still valid if we replaceCλA, BbyK. For 0< ε < pλ/2, setting fand g asfx gx 0 forx ∈ 0,1,fx xλ/2−1−ε/p; gx xλ/2−1−ε/q for x∈1,∞, then we have

K

0

xp1−λ/2−1fpxdx1/p 0

xq1−λ/2−1gqxdx1/q

K

0

x−1−εdx

1/p 0

x−1−εdx 1/q

K ε.

2.18

By using2.8, we find

0

fx gxdx dy Amin

xλ, yλ Bmax

xλ, yλ

1

1

xλ/2−1−ε/pdx Amin

xλ, yλ Bmax xλ, yλ

yλ/2−1−ε/qdy

> 1 ε

CλA, B o1

O1.

2.19

Therefore, we get

1 ε

CλA, B o1

O1< K

ε 2.20

(7)

or

1 λ

CλA, B o1

εO1< K. 2.21

Forε→ 0, it follows thatCλA, B ≤ Kwhich contradicts the fact thatK < CλA, B.

Hence, the constant factorCλA, Bin2.11is the best possible. The theorem is proved.

Theorem 2.4. If 0 < p < 1, 1/p1/q 1,λ > 0, A0, B > 0, fx, gx0 such that 0<

0 xp1−λ/2−1fpxdx <∞, 0<0<

0 xq1−λ/2−1gqxdx <∞, then

0

fxgx Amin

xλ, yλ Bmax

xλ, yλ dx dy

> CλA, B

0

xp1−λ/2−1fpxdx1/p 0

xq1−λ/2−1gqxdx1/q

,

2.22

where the constant factorCλA, Bdefined in2.3is the best possible. In particular,

iforλAB1,C11,1 π, and inequality2.22reduces to Hardy-Hilbert’s inequality

0

fxgx

xy dx dy > π

0

xp/2−1fpxdx

1/p

0

xq/2−1gqxdx 1/q

, 2.23

ii forA0,λB1,C10,1 4 and2.22reduces to Hardy-Hilbert’s-type inequality

0

fxgx

max{x, y}dx dy >4

0

xp/2−1fpxdx

1/p 0

xq/2−1gqxdx 1/q

. 2.24

Proof. By reverse Holder’s inequality, and the same way, we have2.22. If the constant factor CλA, Bin2.22is not the best possible, then there exists a positive constantH withH >

CλA, Bsuch that2.22is still valid if we replaceCλA, BbyH. For 0< ε < pλ/2, settingf andgas inTheorem 2.3, then we have

H

0

xp1−λ/2−1fpxdx1/p 0

xq1−λ/2−1gqxdx1/q

H

0

x−1−εdx

1/p 0

x−1−εdx 1/q

H ε .

2.25

By using2.8, we find

0

fx gxdx dy Amin

xλ, yλ Bmax

xλ, yλ

1

1

xλ/2−1−ε/pdx Amin

xλ, yλ Bmax xλ, yλ

yλ/2−1−ε/qdy

< 1 ε

CλA, B o1 .

2.26

(8)

Therefore, we get

1 ε

CλA, B o1

> H

ε, 2.27

or

CλA, B o1H. 2.28 Forε→0, it follows thatCλA, B≥Hwhich contradicts the fact thatH > CλA, B. Hence, the constant factorCλA, Bin2.22is the best possible. The theorem is proved.

Theorem 2.5. Under the assumption ofTheorem 2.3,

0

yλp/2−1

0

fx Amin

xλ, yλ Bmax

xλ, yλ dxp

dy <

CλA, Bp

0

xp1−λ/2−1fpxdx, 2.29 where the constant factorCλA, Bpis the best possible. Inequalities2.11and2.29are equivalent.

Proof. Setting

gy yλp/2−1

0

fx Amin

xλ, yλ Bmax

xλ, yλ dx p−1

, 2.30

then by2.11we have

0

yq1−λ/2−1gqydy

0

yλp/2−1

0

fx Amin

xλ, yλ Bmax

xλ, yλ dx p

dy

0

0

fx Amin

xλ, yλ Bmax

xλ, yλ dx

×

yλp/2−1

0

fx Amin

xλ, yλ Bmax

xλ, yλ dx p−1

dy

0

fxgy

Amin

xλ, yλ Bmax

xλ, yλ dx dy

CλA, B

0

xp1−λ/2−1fpxdx1/p 0

yq1−λ/2−1gqydy1/q

. 2.31 Hence, we obtain

0

yq1−λ/2−1gqydy≤

CλA, Bp

0

xp1−λ/2−1fpxdx. 2.32

Thus, by2.11, both2.31and2.32keep the form of strict inequalities, then we have2.29.

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By Holder’s inequality, we find

0

fxgy Amin

xλ, yλ Bmax

xλ, yλ dx dy

0

yλ/2−1/p

0

fx Amin

xλ, yλ Bmax

xλ, yλ dx

y1/p−λ/2gy dy

0

ypλ/2−1

0

fx Amin

xλ, yλ Bmax

xλ, yλ dx

p1/p 0

yq1−λ/2−1gqydy 1/q

. 2.33 Therefore, by2.29we have2.11, and inequalities2.29and2.11are equivalent. If the constant factor in2.29is not the best possible, then by2.33we can get a contradiction that the constant factor in2.11is not the best possible. The theorem is proved.

Theorem 2.6. Under the assumption ofTheorem 2.4,

0

yλp/2−1

0

fx Amin

xλ, yλ Bmax

xλ, yλ dx p

dy >

CλA, Bp

0

xp1−λ/2−1fpxdx, 2.34 where the constant factorCλA, Bpis the best possible. Inequalities2.22and2.34are equivalent.

The proof ofTheorem 2.6is similar to that ofTheorem 2.5, so we omit it.

3. Discrete analogous

Lemma 3.1. Suppose that 0 < λ2,A0,B > 0. Then the weight coefficientsλA, B, mand λA, B, n,defined, respectively, by

λA, B, m:

n1

mλ/2n−1λ/2 Amin

mλ, nλ Bmax

mλ, nλ m∈N, 3.1

λA, B, n:

m1

nλ/2m−1λ/2 Amin

mλ, nλ Bmax

mλ, nλ n∈N, 3.2

satisfy the following inequalities:

CλA, B

1−θλA, B, m

< λA, B, m< CλA, B, 3.3 CλA, B

1−θλA, B, n

< λA, B, n< CλA, B, 3.4

whereθλA, B, r: 1/C1A, Br−λ

0 t−1/2/AtBdtO1/rλ/2∈0,1 r ∈N r → ∞,

andCλA, Bis defined by2.3.

(10)

Proof. Since 0< λ≤2,A≥0,B >0, byLemma 2.1we get

λA, B, m<

0

mλ/2y−1λ/2 Amin

mλ, yλ Bmax

mλ, yλ dy, ωλA, B, m CλA, B.

3.5

On the other hand, we have

λA, B, m>

1

mλ/2y−1λ/2 Amin

mλ, yλ Bmax

mλ, yλ dy 1

λ

m−λ

t−1/2

Amin{1, t}Bmax{1, t}dt I−1

λ m−λ

0

t−1/2 AtBdt I

1−θλA, B, m ,

3.6

whereI 1/λC1A, Band

0< θλA, B, m 1 C1A, B

m−λ

0

t−1/2

AtBdt <1. 3.7

Since

m−λ

0

t−1/2 AtBdt

m−λ

0

t−1/2

B dt 2

Bmλ/2, 3.8

thenθλA, B, m O1/mλ/2. Therefore,3.3is valid. By the symmetry,3.4is still valid. The lemma is proved.

Lemma 3.2. Ifp >0p /1, 1/p1/q1, 0< λ2,A0,B >0, and 0< ε < pλ/2,setting

n1

m1

mλ/2−1−ε/pnλ/2−1−ε/q Amin

mλ, nλ Bmax

mλ, nλ , 3.9

then forε→0,

CλA, B−o1

n1

1

n < Lε<

CλA, B o1

n1

1

n. 3.10

(11)

Proof. Settingt x/nλin the following, by3.4, we have

Lε<

n1

0

xλ/2−1−ε/pnλ/2−1−ε/q Amin

xλ, nλ Bmax

xλ, nλ dx

n1

1 n

1 λ

0

t−1/2−ε/λp

Amin{t,1}Bmax{t,1}dt

CλA, B o1

n1

1 n

ε−→0 ,

Lε>

n1

1

xλ/2−1−ε/pnλ/2−1−ε/q Amin

xλ, nλ Bmax

xλ, nλ dx

n1

1 n

1 λ

n−λ

t−1/2−ε/λp

Amin{t,1}Bmax{t,1}dt

n1

1 n

CλA, B o1 −1 λ

n−λ

0

t−1/2−ε/λp AtB dt

>

n1

1 n

CλA, B o1

−1 λ

n1

1 n

n−λ

0

t−1/2−ε/λp

B dt

n1

1 n

CλA, B o1

− 1

λB/2−εB/p n1

1 n1λ/2−ε/p

n1

1 n

CλA, B o1 − 1 λB/2−εB/p

n1

1 n1λ/2−ε/p

n1

1 n

−1

n1

1 n

CλA, B−o1 ε−→0 .

3.11

Thus, inequality3.10holds. The lemma is proved.

Theorem 3.3. If p > 1, 1/p1/q 1, 0 < λ2,A0, B > 0,an, bn0 such that 0 <

n1np1−λ/2−1apn<∞, 0<

n1nq1−λ/2−1bqn<∞, then

D:

n1

m1

ambn Amin

mλ, nλ Bmax mλ, nλ

< CλA, B

n1

np1−λ/2−1apn

1/p

n1

nq1−λ/2−1bqn 1/q

,

3.12

(12)

where the constant factorCλA, Bdefined in2.3is the best possible. In particular,

iforλAB1,C11,1 π, and inequality3.12reduces to Hardy-Hilbert’s inequality

n1

m1

ambn

mn< π

n1

np/2−1apn

1/p n1

nq/2−1bqn 1/q

; 3.13

ii forA0,λB1,C10,1 4 and3.12reduces to Hardy-Hilbert’s-type inequality

n1

m1

ambn

max{m, n} <4

n1

np/2−1apn

1/p n1

nq/2−1bqn 1/q

. 3.14

Proof. By the Holder inequality, taking into account3.1, we get

D

n1

m1

1 Amin

mλ, nλ Bmax mλ, nλ

1/p

m1−λ/2/q n1−λ/2/pam

×

1 Amin

mλ, nλ Bmax mλ, nλ

1/q

n1−λ/2/p m1−λ/2/qbn

n1

m1

1 Amin

mλ, nλ Bmax mλ, nλ

m1−λ/2p−1 n1−λ/2 apm

1/p

×

n1

m1

1 Amin

mλ, nλ Bmax mλ, nλ

n1−λ/2q−1 m1−λ/2 bnq

1/q

m1

λA, B, mmp1−λ/2−1apm

1/p n1

λA, B, nnq1−λ/2−1bqn 1/q

.

3.15

Then, by3.3and3.4we obtain3.12.

It remains to show that the constant factorCλA, Bis the best possible, to do that we set for 0< ε < pλ/2,ammλ/2−1−ε/p;bnnλ/2−1−ε/q, by3.9we have

n1

m1

ambn Amin

mλ, nλ Bmax

mλ, nλ Lε. 3.16

If there exists a constant 0< KCλA, Bsuch that3.12is still valid if we replaceCλA, B byK, then in particular by3.10we find

CλA, B−o1

n1

1

n < Lε< K

n1

np1−λ/2−1apn

1/p n1

nq1−λ/2−1bnq 1/q

K n1

1 n,

3.17

it follows thatCλA, B−o1< Kand thenCλA, B≤Kε→0. Therefor,KCλA, Bis the best constant factor in3.12. The theorem is proved.

(13)

Theorem 3.4. If 0 < p < 1, 1/p1/q 1, 0 < λ2,A0, B > 0,an,bn0 such that 0<

n1np1−λ/2−1apn<∞, 0<

n1nq1−λ/2−1bnq<∞, then

n1

m1

ambn

Amin

mλ, nλ Bmax mλ, nλ

> CλA, B n1

1−θλA, B, n

np1−λ/2−1apn

1/p n1

nq1−λ/2−1bnq 1/q

,

3.18

where the constant factorCλA, Bdefined in2.4is the best possible. In particular, iforλAB1,C11,1 π, and

n1

m1

ambn mn> π

n1

1− 2

πarctan 1 n1/2

np/2−1apn

1/p n1

nq/2−1bqn 1/q

; 3.19

ii forA0,λB1,C10,1 4, and

n1

m1

ambn max{m, n}>4

n1

1− 1

2n1/2

np/2−1apn

1/p

n1

nq/2−1bnq 1/q

. 3.20

Proof. By reverse Holder’s inequality, we get D

n1

m1

ambn

Amin

mλ, nλ Bmax mλ, nλ

n1

m1

1 Amin

mλ, nλ Bmax mλ, nλ

1/p

m1−λ/2/q n1−λ/2/pam

×

1 Amin

mλ, nλ Bmax mλ, nλ

1/q

n1−λ/2/p m1−λ/2/qbn

m1

λA, B, mmp1−λ/2−1apm

1/p n1

λA, B, nnq1−λ/2−1bqn

1/q

.

3.21

Then by3.3and3.4, in view ofq <0, we have3.18. For 0< ε < pλ/2, settingam mλ/2−1−ε/p,bnnλ/2−1−ε/qm, n∈N. If there exists a constantKCλA, Bsuch that3.18is still valid if we replaceCλA, BbyK, then in particular by3.9and3.10we find

CλA, B o1

n1

1

n > Lε

> K

n1

1−θλA, B, n

np1−λ/2−1apn

1/p n1

nq1−λ/2−1bqn 1/q

K

n1

1 n

n1

O

1 nλ/2

1 n

1/p

n1

1 n

1/q

K n1

1 n

1−

n1

1 n

−1 n1

O

1 nλ/2

1 n

1/p ,

3.22

(14)

it follows that

CλA, B o1 > K

1−

n1

1 n

−1 n1

O

1 nλ/2

1 n

1/p

. 3.23

Hence, ifε → 0, we get CλA, B ≥ K. Thus,K CλA, Bis the best constant factor in 3.18.

Theorem 3.5. Under the assumption ofTheorem 3.3,

n1

nλp/2−1

m1

am Amin

mλ, nλ Bmax mλ, nλ

p

<

CλA, Bp m1

mp1−λ/2−1apm, 3.24 where the constant factorCλA, Bpis the best possible. Inequalities3.12and3.24are equivalent.

Proof. Setting

bnnλp/2−1

m1

am

Amin

mλ, nλ Bmax mλ, nλ

p−1

, 3.25

we get

n1

nq1−λ/2−1bnq

n1

m1

ambn

Amin

mλ, nλ Bmax

mλ, nλ . 3.26 By3.12and using the same method ofTheorem 2.5, we obtain3.24. We may show that the constant factor in3.24is the best possible and inequality3.12is equivalent to3.24.

Theorem 3.6. Under the assumption ofTheorem 3.4,

n1

nλp/2−1

m1

am Amin

mλ, nλ Bmax mλ, nλ

p

>

CλA, Bp m1

mp1−λ/2−1apm, 3.27 where the constant factorCλA, Bpis the best possible. Inequalities3.18and3.27are equivalent.

References

1G. H. Hardy, J. E. Littlewood, and G. P ´olya, Inequalities, Cambridge University Press, London, UK, 2nd edition, 1952.

2G. H. Hardy, “Note on a theorem of Hilbert concerning series of positive terms,” Proceedings of the London Mathematical Society, vol. 23, no. 2, pp. 45–46, 1925.

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

4B. Yang, “On an extension of Hilbert’s integral inequality with some parameters,” The Australian Journal of Mathematical Analysis and Applications, vol. 1, no. 1, article 11, pp. 1–8, 2004.

5B. Yang, “On the extended Hilbert’s integral inequality,” Journal of Inequalities in Pure and Applied Math- ematics, vol. 5, no. 4, article 96, pp. 1–8, 2004.

6B. Yang, “Generalization of a Hilbert type inequality and its applications,” Chinese Journal of Engineering Mathematics, vol. 21, no. 5, pp. 821–824, 2004.

7Y. 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.

8B. Yang, “On a Hilbert-type operator with a symmetric homogeneous Kernel of 1-order and applica- tions,” Journal of Inequalities and Applications, vol. 2007, Article ID 47812, 9 pages, 2007.

9G. Xi, “A reverse Hardy-Hilbert-type inequality,” Journal of Inequalities and Applications, vol. 2007, Arti- cle ID 79758, 7 pages, 2007.

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