In this paper, we build a new inequality similar to Hilbert’s inequality with a best constant factor

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

Volume 3, Issue 5, Article 75, 2002

A NEW INEQUALITY SIMILAR TO HILBERT’S INEQUALITY

BICHENG YANG DEPARTMENT OFMATHEMATICS

GUANGDONGEDUCATIONCOLLEGE, GUANGZHOU, GUANGDONG510303,

PEOPLE’SREPUBLIC OFCHINA. [email protected]

Received 17 May, 2001; accepted 17 June, 2002 Communicated by J.E. Peˇcari´c

ABSTRACT. In this paper, we build a new inequality similar to Hilbert’s inequality with a best constant factor. As an application, we consider its equivalent form.

Key words and phrases: Hilbert’s inequality, Weight coefficient, Cauchy’s inequality.

2000 Mathematics Subject Classification. 26D15.

1. INTRODUCTION

If 0 < P∞

n=0a²_n < ∞ and 0 < P∞

n=0b²_n < ∞, then the famous Hilbert’s inequality (see Hardy et al. [1]) is given by

(1.1)

∞

X

n=0

∞

X

m=0

a_mb_n

m+n+ 1 < π

∞

X

n=0

a²_n

∞

X

n=0

b²_n

!¹₂ ,

where the constant factorπ is the best possible. Recently, Yang and Debnath [2, 3] and Yang [4, 5] gave (1.1) some extensions and improvements, and Kuang and Debnath [6] considered its strengthened versions and generalizations.

The major objective of this paper is to build a new inequality similar to (1.1), which relates to the double series form as

(1.2)

∞

X

n=1

∞

X

m=1

ambn

lnm+ lnn+ 1 =

∞

X

n=1

∞

X

m=1

ambn

lnemn. For this, we must estimate the following weight coefficient

(1.3) ω(n) =

∞

X

m=1

1 mlnemn

ln√ en ln√

em ¹₂

(n ∈N), and do some preparatory works.

ISSN (electronic): 1443-5756

c 2002 Victoria University. All rights reserved.

044-01

2. SOME LEMMAS

Let f have its first four derivatives on[1,∞) and (−1)ⁿf⁽ⁿ⁾(x) > 0 (n = 0, . . . ,4), and f(x), f⁰(x)−→0 (x→ ∞),then (see [6, (2.1)])

(2.1)

∞

X

k=1

f(k)<

Z ∞ 1

f(x)dx+ 1

2f(1)− 1 12f⁰(1).

Lemma 2.1. Forn ∈N,defineR(n)as

(2.2) R(n) = 1

(2 ln√ en)¹²

Z _{2 ln}^1√_en

0

1

(1 +u)u¹²du− 2

3 lnen− 1 12(lnen)². Then we haveR(n)>0(n∈N).

Proof. Integrating by parts, we have Z ¹

2 ln√ en

0

1

(1 +u)u¹²du= 2 Z ¹

2 ln√ en

0

1 (1 +u)du¹²

= (2 ln√

en)¹² 1 lnen+ 2

Z ¹

2 ln√ en

0

u¹² 1 (1 +u)²du

= (2 ln√

en)¹² 1 lnen+4

3

Z _{2 ln}^1√_en

0

1

(1 +u)²du^3/2

= (2 ln√

en)¹² 1 lnen+1

3(2 ln√

en)¹² 1 (lnen)² +8

3

Z _{2 ln}^1√_en

0

u^3/2 1 (1 +u)³du

>(2 ln√

en)¹² 1 lnen+1

3(2 ln√

en)¹² 1 (lnen)². Hence by (2.2), we have

R(n)> 1

lnen + 1

3(lnen)² − 2

3 lnen− 1

12(lnen)² = 1

3 lnen+ 1

4(lnen)² >0.

The lemma is thus proved.

Lemma 2.2. Ifω(n)is defined by (1.3), thenω(n)< π,forn∈N. Proof. For fixedn∈N, setting

fn(x) = 1 xlnenx

ln√ en ln√

ex ¹₂

, x∈[1,∞), we findf_n(1) = _lnen¹ (2 ln√

en)¹²,and f_n⁰(x) = − 1

x²lnenx

ln√ en ln√

ex ¹₂

− 1 x²ln²enx

ln√ en ln√

ex ¹₂

− 1

2x²lnenx· (ln√ en)¹² (ln√

ex)³², f_n⁰(1) =−

2

lnen + 1 ln²en

(2 ln√ en)¹².

Settingu= ^ln

√ex ln√

en in the following integral, we obtain Z ∞

1

f_n(x)dx= Z ∞

1 2 ln√

en

1 1 +u

1 u

¹₂

du =π− Z ¹

2 ln√ en

0

1 1 +u

1 u

¹₂ du.

Hence by (2.1), (2.2) and Lemma 2.1, we have ω(n) =

∞

X

m=1

f_n(m)<

Z ∞ 1

f_n(x)dx+1

2f_n(1)− 1 12f_n⁰(1)

=π−

Z 1/(2 ln√ en) 0

1 1 +u

1 u

¹₂ du+

2

3 lnen + 1 12 ln²en

(2 ln√ en)¹²

=π−(2 ln√

en)¹²R(n)< π.

The lemma is proved.

Lemma 2.3. For0< <1,we have (2.3)

∞

X

n=1

∞

X

m=1

1 mnlnemn

1 ln√

emln√ en

¹⁺₂

> 1

(π+o(1)) (→0⁺).

Proof. Settingu= ^ln

√ex ln√

ey in the following integral, we find Z ∞

√e

1 xlnexy

1 ln√

ex ¹⁺₂

dx

= 1

ln√ ey

¹⁺₂ Z ∞

1 ln√

ey

1 1 +u

1 u

¹⁺₂ du

= 1

ln√ ey

¹⁺₂ Z ∞ 0

1 1 +u

1 u

¹⁺₂ du−

1 ln√

ey

¹⁺₂ Z ¹

ln√ ey

0

1 1 +u

1 u

¹⁺₂ du

>

1 ln√

ey

¹⁺₂ Z ∞ 0

1 1 +u

1 u

¹⁺₂ du−

1 ln√

ey

¹⁺₂ Z _ln^√1_ey

0

1 u

¹⁺₂ du

= 1

ln√ ey

¹⁺₂

(π+o(1))− 2 1−

1 ln√

ey

(−→0⁺).

Hence we have

∞

X

n=1

∞

X

m=1

1 mnlnemn

1 ln√

emln√ en

¹⁺₂

>

Z ∞

√e

Z ∞

√e

1 xylnexy

1 ln√

exln√ ey

¹⁺₂ dxdy

= Z ∞

√e

1 y

1 ln√

ey

¹⁺₂ "

Z ∞

√e

1 xlnexy

1 ln√

ex ¹⁺₂

dx

# dy

>(π+o(1)) Z ∞

√e

1 y

1 ln√

ey 1+

dy− 2 1−

Z ∞

√e

1 y

1 ln√

ey

¹⁺₂ +1

dy

= (π+o(1))1

− 4

1−² = 1

(π+o(1)) (→0⁺).

The lemma is proved.

3. MAIN RESULT AND AN APPLICATION

Theorem 3.1. If0<P∞

n=1na²_n <∞and0<P∞

n=1nb²_n <∞, then (3.1)

∞

X

n=1

∞

X

m=1

a_mb_n lnemn < π

∞

X

n=1

na²_n

∞

X

n=1

nb²_n

!¹₂ , where the constant factorπis the best possible.

Proof. By Cauchy’s inequality and (1.3), we have

∞

X

n=1

∞

X

m=1

ambn

lnemn

=

∞

X

n=1

∞

X

m=1

"

a_m (lnemn)¹²

ln√ em ln√

en ¹₄

m n

¹₂

# "

b_n (lnemn)¹²

ln√ en ln√

em ¹₄

n m

¹₂

#

≤

" _∞ X

m=1

∞

X

n=1

a²_m lnemn

ln√ em ln√

en ¹₂

m n

X^∞

n=1

∞

X

m=1

b²_n lnemn

ln√ en ln√

em ¹₂

n m

#¹₂

=

∞

X

m=1

ω(m)ma²_m

∞

X

n=1

ω(n)nb²_n

!¹₂ . By Lemma 2.2, we have (3.1).

For0< <1,settinga⁰_nas:

a⁰_n= 1 n(ln√

en)¹⁺² , n∈N, then we have

∞

X

n=1

na⁰_n² = 1 (ln√

e)¹⁺ + 1

2(ln 2√

e)¹⁺ +

∞

X

n=3

1 n(ln√

en)¹⁺

< 1 (ln√

e)¹⁺ + 1

2(ln 2√

e)¹⁺ + Z ∞

√e

1 x(ln√

ex)¹⁺dx

= 1

(ln√

e)¹⁺ + 1

2(ln 2√

e)¹⁺ +1 = 1

(1 +o(1)) ( →0⁺).

(3.2)

If the constant factorπ in (3.1) is not the best possible, then there exists a positive number K < π, such that (3.1) is valid if we changeπ toK. In particular, we have

∞

X

n=1

∞

X

m=1

a⁰_ma⁰_n lnemn < K

∞

X

n=1

na⁰_n².

By (2.3) and (3.2), we have (π+o(1)) <

∞

X

n=1

∞

X

m=1

a⁰_ma⁰_n

lnemn < K(1 +o(1)) (→0⁺),

and π ≤ K. This contradicts that K < π. Hence the constant factor π in (3.1) is the best

possible. The theorem is proved.

Remark 3.2. Inequality (3.1) is more similar to the following Mulholland’s inequality forp= q= 2(see [7]):

(3.3)

∞

X

n=2

∞

X

m=2

a_mb_n

mnlnemn < π sin(^π_p)

∞

X

n=2

n⁻¹a^p_n

!_p¹ _∞ X

n=2

n⁻¹b^q_n

!¹_q . Theorem 3.3. If0<P∞

n=1na²_n <∞, then we have (3.4)

∞

X

n=1

1 n

∞

X

m=1

a_m lnemn

!2

< π²

∞

X

n=1

na²_n,

where the constant factorπ² is the best possible. Inequalities (3.1) and (3.4) are equivalent.

Proof. SinceP∞

n=1na²_n >0,there existsk₀ ≥1, such that for anyk > k₀, we havePk

n=1na²_n>

0,andb_n(k) = _n¹ Pk m=1

|am|

lnemn >0 (n ∈N).By (3.1), we have 0<

" _k X

n=1

nb²_n(k)

#²

=





k

X

n=1

1 n

k

X

m=1

|a_m| lnemn

!2



2

=

" _k X

n=1 k

X

m=1

|a_m|b_n(k) lnemn

#2

< π²

k

X

n=1

na²_n

k

X

n=1

nb²_n(k).

(3.5)

Thus we find

(3.6) 0<

k

X

n=1

1 n

k

X

m=1

|a_m| lnemn

!²

=

k

X

n=1

nb²_n(k)< π²

k

X

n=1

na²_n. It follows that0 <P∞

n=1nb²_n(∞) ≤ π²P∞

n=1na²_n < ∞.Hence by (3.1), fork → ∞,neither (3.5) nor (3.6) takes equality, and we have

∞

X

n=1

1 n

∞

X

m=1

a_m lnemn

!2

≤

∞

X

n=1

1 n

∞

X

m=1

|a_m| lnemn

!2

< π²

∞

X

n=1

na²_n. Inequality (3.4) is valid.

On the other hand, if (3.4) holds, by Cauchy’s inequality, we have

∞

X

n=1

∞

X

m=1

a_mb_n lnemn =

∞

X

n=1

1 n¹²

∞

X

m=1

a_m lnemn

!

n¹²b_n

≤





∞

X

n=1

1 n

∞

X

m=1

a_m lnemn

!2 ∞

X

n=1

nb²_n





1 2

. (3.7)

By (3.4), we have (3.1).

Hence inequalities (3.1) and (3.4) are equivalent. If the constant factorπ² in (3.4) is not the best possible, we may show that the constant factorπin (3.1) is not the best possible, by using

(3.7). This is a contradiction. The theorem is proved.

Remark 3.4. Inequality (3.4) is similar to the following equivalent form of (1.1) (see [2]):

(3.8)

∞

X

n=0

∞

X

m=0

a_m m+n+ 1

!2

< π²

∞

X

n=0

a²_n.

Since inequalities (3.1) and (3.4) are similar to (1.1) and its equivalent form with the best constant factors, we have provided some new results.

REFERENCES

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[3] B. YANGANDL. DEBNATH, Some inequalities involvingπand an application to Hilbert’s inequal- ity, Applied Math. Letters, 129 (1999), 101–105.

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