Key words and phrases: Hardy-Hilbert’s inequality, Weight coefficient, Hölder’s inequality

http://jipam.vu.edu.au/

Volume 7, Issue 5, Article 180, 2006

ON A NEW STRENGTHENED VERSION OF A HARDY-HILBERT TYPE INEQUALITY AND APPLICATIONS

WEIHONG WANG AND DONGMEI XIN DEPARTMENT OFMATHEMATICS,

GUANGDONGEDUCATIONCOLLEGE, GUANGZHOU, GUANGDONG510303, PEOPLE’SREPUBLIC OFCHINA.

xdm77108@gdei.edu.cn

Received 16 April, 2006; accepted 20 June, 2006 Communicated by B. Yang

ABSTRACT. By improving an inequality of the weight coefficient, we give a new strengthened version of Hardy-Hilbert’s type inequality. As its applications, we build some strengthened versions of the equivalent form and some particular results.

Key words and phrases: Hardy-Hilbert’s inequality, Weight coefficient, Hölder’s inequality.

2000 Mathematics Subject Classification. 26D15.

1. INTRODUCTION

Ifp > 1, ¹_p + ¹_q = 1, a_n, b_n ≥ 0 (n ∈ N0 = N∪ {0}),such that 0 < P∞

n=0a^p_n < ∞and 0<P∞

n=0b^q_n <∞, then we have the famous Hardy-Hilbert inequality as follows [1]:

(1.1)

∞

X

n=0

∞

X

m=0

a_mb_n

m+n+ 1 < π sin

π p

∞

X

n=0

a^p_n

!¹_{p ∞} X

n=0

b^q_n

!¹_q ,

where the constant factor _sin(^ππ

p) is the best possible.

Inequality (1.1) is important in analysis and its applications. In recent years, [2] – [5] consid- ered the strengthened version, generalizations and improvements of inequality (1.1) and Pach- patte [6] built some inequalities similar to inequality (1.1).

Under the same condition with (1.1), we still have Mulholand’s inequality (cf. [7]):

(1.2)

∞

X

n=2

∞

X

n=2

a_mb_n

mnlnmn < π sin

π p

( _∞

X

n=2

1 na^p_n

)¹_p( _∞ X

n=2

1 nb^q_n

)¹_q ,

where the constant factor _sin(^ππ

p) is the best possible.

ISSN (electronic): 1443-5756

c 2006 Victoria University. All rights reserved.

116-06

For the double series:

∞

X

n=1

∞

X

m=1

a_mb_n

m^sn^t(lnm+ lnn+ lnα) =

∞

X

n=1

∞

X

m=1

a_mb_n

m^sn^tlnαm (s, t ∈R, α ≥e^7/6), in 2003, Yang [8] built an inequality of the weight coefficient as follows:

∞

X

m=1

1 mlnαmn

ln√ αn ln√

αm ¹_r

< π

sinπ(1−1/r) (r >1, α ≥e^7/6), then he gave

(1.3)

∞

X

n=1

∞

X

m=1

a_mb_n

m^sn^t(lnαmn) < π sin

π p

∞

X

n=1

n^1q−sa_np!¹_{q ∞} X

n=1

n^1p−tb_nq!¹_p ,

(1.4)

∞

X

n=1

1 n

∞

X

m=1

a_m m^slnαmn

!p

<



 π sin

π p





p ∞

X

n=1

n^1q−sa_np

,

where the constant factors ^π

sin(^π_p) and

π sin(^π_p)

p

are the best possible.

In this paper, by using the refined Euler-Maclaurin formula, we have some strengthened versions of inequalities (1.3) and (1.4).

2. SOME LEMMAS

Iff⁽⁴⁾ ∈C[1,∞),R∞

1 f(x)dx <∞,and(−1)ⁿf⁽ⁿ⁾(x)>0, f⁽ⁿ⁾(∞) = 0 (n = 0,1,2,3,4), then we have the following inequality (see [9]):

(2.1)

∞

X

m=1

f(m)<

Z ∞ 1

f(x)dx+1

2f(1)− 1 12f⁰(1).

Lemma 2.1. Settingr >1, n∈Nandα≥e^7/6, define the functionω(r, n, α)as:

ω(r, n, α) =

∞

X

m=1

1 mlnαmn

ln√ αn ln√

αm ¹_r

.

The we have

(2.2) ω(r, n, α)< π

sinπ(1−1/r) − 3

8(r−1)(2 lnn+ 1)^1−1/r. Proof. For fixedx∈[1,∞), settingf(x) = _xln¹_αnx ·

ln√ αn ln√

αx

¹_r

,we have

f(1) = 1 lnαn·

ln√ αn ln√

α ¹_r

= ln√ αn ln√

α·lnαn ·

ln√ α ln√

αn ¹⁻¹_r

,

f⁰(1) =− 1

lnαn+ 1

ln²αn+ 1 rln√

α·lnαn

·

ln√ αn ln√

α 1−¹_r

=−

ln√ αn ln√

αn·lnαn + ln√ αn rln²√

α·lnαn + ln√ αn ln√

α·ln²αn

·

ln√ α ln√

αn 1−¹_r

,

Z ∞ 1

f(x)dx= Z ∞

1

1 xlnαnx·

ln√ αn ln√

αx ¹_r

dx

= Z ∞

ln√ α ln√

αn

1 1 +u·

1 u

¹_r du

= π

sin(π/r) − Z ^ln

√α ln√

αn

0

1 1 +u ·

1 u

¹_r du.

Since Z ^ln

√α ln√

αn

0

1 1 +u ·

1 u

¹_r du

= r

r−1 Z ^ln

√α ln√

αn

0

1

1 +u ·du^1−1r

= rln√ αn (r−1) lnαn ·

ln√ α ln√

αn 1−¹_r

+ r

r−1 Z ^ln

√α ln√

αn

0

u^1−1r · 1 (1 +u)²du

= rln√ αn (r−1) lnαn ·

ln√ α ln√

αn ¹⁻¹_r

+ r²

(r−1)(2r−1) Z ^ln

√α ln√

αn

0

1

(1 +u)²du^2−1r

> rln√ αn (r−1) lnαn ·

ln√ α ln√

αn ¹⁻¹_r

+ r²

(r−1)(2r−1)· ln√

α lnα n

2

· ln√

α lnα n

²⁻¹_r

=

ln√ α ln√

αn

1−¹_r

rln√ αn

(r−1) lnαn+ r²

(r−1)(2r−1)· ln√

α·ln√ αn ln²αn

, in view of (2.1) and the above result, we have

ω(r, n, α) =

∞

X

m=1

f(m)

<

Z ∞ 1

f(x)dx+1

2f(1)− 1 12f⁰(1)

< π sin

π p

−

ln√ α ln√

αn ¹⁻¹_r

·

rln√ αn

(r−1) lnαn+ r²

(r−1)(2r−1)· ln√

α·ln√ αn ln²αn

+ ln√

αn 2 ln√

α·lnαn ·

ln√ α ln√

αn ¹⁻¹_r

+ 1 12

ln√ αn ln√

α·lnαn

+ ln√

αn rln²√

α·lnαn+ ln√ αn ln√

α·ln²αn ·

ln√ α ln√

αn ¹⁻¹_r

= π

sin

π p

−

ln√ α ln√

αn ¹⁻¹_r

·

r

r−1 − 7

6 lnα − 1 3rln²α

·ln√ αn lnαn

+

r²lnα

2(r−1)(2r−1) − 1 6 lnα

·ln√ αn ln²αn

.

Forn ∈N, r > 1, α≥e^7/6, since r

1−r − 7

6 lnα − 1 3rln²α

· ln√ αn lnαn ≥

1 + 1

r−1− 7

6·⁷₆ − 1 3r·(⁷₆)²

· 1 2

= 3

8(r−1), r²lnα

2(r−1)(2r−1)− 1 6 lnα

·ln√ αn ln²αn

>

lnα 4

1 + 3

2(r−1)

− 1 6 lnα

·ln√ αn ln²αn

>

7 24

1 + 3

2(r−1)

−1 7

· ln√ αn ln²αn >0, and

ln√ α ln√

αn 1−¹_r

> 1

(2 lnn+ 1)^1−1/r, we have

ω(r, n, α)< π

sinπ(1−1/r) − 3

8(r−1)(2 lnn+ 1)^1−1/r.

The lemma is proved.

3. MAIN RESULTS AND APPLICATIONS

Theorem 3.1. Ifp >1, ¹_p + ¹_q = 1, α≥e^7/6, s, t∈R, a_nb_nare two sequences of non-negative real numbers, such that0<P∞

n=1

n^1q−sa_np

<∞and0<P∞ n=1

n^1p−tb_nq

<∞,then we have

(3.1)

∞

X

n=1

∞

X

m=1

a_mb_n

m^sn^tlnαmn <





∞

X

n=1



 π sin

π p



− 3(p−1) 8(2 lnn+ 1)^1p

n^1q−sa_np





1 p

×





∞

X

n=1



 π sin

π p

− 3(q−1) 8(2 lnn+ 1)^1q





n^p1−tb_nq





1 q

.

In particular,

(a) fors= ¹_q, t= ¹_p, we have

(3.2)

∞

X

n=1

∞

X

m=1

ambn

m^1pn^1qlnαmn

<





∞

X

n=1



 π sin

π p

− 3(p−1) 8(2 lnn+ 1)^p1



a^p_n





1 p

×







 π sin

π p

− 3(p−1) 8(2 lnn+ 1)^1q



b^q_n





1 q

;

(b) fors=t= 1, we have

(3.3)

∞

X

n=1

∞

X

m=1

ambn

mnlnαmn <





∞

X

n=1



 π sin

π p

− 3(p−1) 8(2 lnn+ 1)^1p



 a^p_n

n





1 p

×







 π sin

π p

− 3(p−1) 8(2 lnn+ 1)^1q



 b^q_n

n





1 q

(c) fors=t= 0, we have

(3.4)

∞

X

n=1

∞

X

m=1

a_mb_n lnαmn <





∞

X

n=1



 π sin

π p

− 3(p−1) 8(2 lnn+ 1)^1p



n^p−1a^p_n





1 p

×







 π sin

π p

− 3(p−1) 8(2 lnn+ 1)^1q



n^q−1b^q_n





1 q

.

Proof. By Hölder’s inequality, we have

∞

X

n=1

∞

X

m=1

a_mb_n

m^sn^tlnαmn =

∞

X

n=1

∞

X

m=1

"

a_m (lnαmn)^1p

ln√ αm ln√

αn _pq¹

m^1q−s n^1p

!#

×

"

b_n (lnαmn)^1q

ln√ αn ln√

αm _pq¹

n^1p−t m^1q

!#

≤

∞

X

m=1

∞

X

n=1

"

a^p_m lnαmn

ln√ αm ln√

αn ¹_q

m^p(1q−s) n

!#¹_p

×

∞

X

n=1

∞

X

m=1

"

b^q_n lnαmn

ln√ αn ln√

αm ¹_p

n^q(1p−t) m

!#¹_q

=

" _∞ X

m=1

ω(q, m, α)

m^1q−sa_mp#_p¹ " _∞ X

n=1

ω(p, n, α)

n^1p−tb_nq#¹_q .

In view of (2.2), we have (3.1). The theorem is proved.

Theorem 3.2. If p > 1, ¹_p + ¹_q = 1, α ≥ e^7/6, s ∈ R, a_n is sequence of non-negative real numbers, such that0<P∞

n=1

n^1q−sa_np

<∞,then we have

(3.5)

∞

X

n=1

1 n

∞

X

m=1

a_m m^slnαmn

!p

<

"

π sin(^π_p)

#p−1 ∞

X

m=1

π

sin(^π_p) − 3(p−1) 8(2 lnm+ 1)^1p

!

m^1q−sa_mp

.

In particular,

(a) fors= ¹_q, we have

(3.6)

∞

X

n=1

1 n

∞

X

m=1

a_m m^1q lnαmn

!p

<

"

π sin(^π_p)

#p−1 ∞

X

m=1

π

sin(^π_p) − 3(p−1) 8(2 lnm+ 1)^1p

! (a_m)^p;

(b) fors= 1, we have (3.7)

∞

X

n=1

1 n

∞

X

m=1

am

mlnαmn

!p

<

"

π sin(^π_p)

#p−1 ∞

X

m=1

π

sin(^π_p)− 3(p−1) 8(2 lnm+ 1)^1p

!a^p_m m; (c) fors= 0, we have

(3.8)

∞

X

n=1

1 n

∞

X

m=1

a_m lnαmn

!p

<

"

π sin(^π_p)

#p−1 ∞

X

m=1

π

sin(^π_p)− 3(p−1) 8(2 lnm+ 1)^p1

!

(m^p−1a_m)^p.

Proof. It is obvious that for anym∈ N₀, ω(r, m, α)< _{sinπ(1−1/r)}^π .By Cauchy’s inequality, we obtain

∞

X

m=1

a_m m^slnαmn

!p

=

" _∞ X

m=1

1 (lnαmn)^1p

ln√ αm ln√

αn _pq¹

m^1q−s

a_m· 1 (lnαmn)^1q

ln√ αn ln√

αm _pq¹

1 m^1q

#p

≤

∞

X

m=1

1 (lnαmn)

ln√ αm ln√

αn ¹_q

m^p(1q−s)a^p_m·

∞

X

m=1

"

1 (lnαmn)

ln√ αn ln√

αm ¹_p

1 m

#^(p−1)

=

∞

X

m=1

1 (lnαmn)

ln√ αm ln√

αn ¹_q

m^p(1q−s)a^p_m·[ω(p, n, α)]^(p−1)

<

"

π sin^π_p

#(p−1) ∞

X

m=1

1 (lnαmn)

ln√ αm ln√

αn ¹_q

m^p(1q−s)a^p_m.

Hence, by (2.2) we find

∞

X

n=1

1 n

∞

X

m=1

a_m m^slnαmn

!p

<

"

π sin^π_p

#^(p−1) _∞ X

n=1

1 n

∞

X

m=1

1 (lnαmn)

ln√ αm ln√

αn ¹_q

m^p(1q−s)a^p_m

=

"

π sin^π_p

#(p−1) ∞

X

m=1

∞

X

n=1

1 nlnαmn

ln√ αm ln√

αn ¹_q

m^p(1q−s)a^p_m

=

"

π sin^π_p

#(p−1) ∞

X

m=1

ω(q, m, α)

m^1q−sa_mp

<

"

π sin^π_p

#(p−1) ∞

X

m=1

π

sin(^π_p) − 3(p−1) 8(2 lnm+ 1)^1p

!

m^1q−sa_mp

.

The theorem is proved.

Remark 3.3. Obviously, inequalities (3.1) and (3.5) are separately strengthened versions of inequalities (1.3) and (1.4).

REFERENCES

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