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volume 7, issue 3, article 115, 2006.

Received 18 January, 2006;

accepted 25 May, 2006.

Communicated by:L.-E. Persson

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Journal of Inequalities in Pure and Applied Mathematics

ON A REVERSE OF A HARDY-HILBERT TYPE INEQUALITY

BICHENG YANG

Department Of Mathematics

Guangdong Education College, Guangzhou Guangdong 510303, People’s Republic of China EMail:bcyang@pub.guangzhou.gd.cn

c

2000Victoria University ISSN (electronic): 1443-5756 016-06

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On a Reverse of a Hardy-Hilbert Type Inequality

Bicheng Yang

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J. Ineq. Pure and Appl. Math. 7(3) Art. 115, 2006

http://jipam.vu.edu.au

Abstract

This paper deals with a reverse of the Hardy-Hilbert’s type inequality with a best constant factor. The other reverse of the form is considered.

2000 Mathematics Subject Classification:26D15.

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

Research supported by Natural Science Foundation of Guangdong Institutions of Higher Learning, College and University (China, No. 0177).

1. Introduction

If p > 1,¹_p + ¹_q = 1, a_n, b_n ≥ 0, such that 0 < P∞

n=0a^p_n < ∞ and 0 <

P∞

n=0b^q_n < ∞, then we have the well known Hardy-Hilbert inequality (Hardy et al. [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 factorπ/sin(π/p)is the best possible. The equivalent form is (see Yang et al. [8]):

(1.2)

∞

X

n=0

∞

X

m=0

am

m+n+ 1

!p

<



 π sin

π p





p ∞

X

n=0

a^p_n,

where the constant factor[π/sin(π/p)]^p is still the best possible.

Inequalities (1.1) and (1.2) are important in analysis and its applications (see Mitrinovi´c, et al. [3]). In recent years, inequality (1.1) had been strengthened by Yang [5] as

(1.3)

∞

X

n=0

∞

X

m=0

a_mb_n m+n+ 1 <







∞

X

n=0



 π sin

π p

− ln 2−γ (2n+ 1)¹⁺¹^p



a^p_n







1 p

×







∞

X

n=0



 π sin

π p

− ln 2−γ (2n+ 1)¹⁺¹^q



b^q_n







1 q

,

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whereln 2−γ = 0.11593⁺(γ is Euler’s constant). Another strengthened version of (1.1) was given in [6]. By introducing a parameterλ, two extensions of (1.1) were proved in [8,7] as:

(1.4)

∞

X

n=0

∞

X

m=0

ambn

(m+n+ 1)^λ

< k_λ(p) ( _∞

X

n=0

n+1

2 1−λ

a^p_n

)¹_p( _∞ X

n=0

n+1

2 1−λ

b^q_n )¹_q

;

(1.5)

∞

X

n=0

∞

X

m=0

a_mb_n

(m+n+ 1)^λ < B λ

p,λ q

× ( _∞

X

n=0

n+1

2

(p−1)(1−λ)

a^p_n

)_p¹ ( _∞ X

n=0

n+1

2

(q−1)(1−λ)

b^q_n )¹_q

,

where, the constant factorskλ(p) = B

p+λ−2

p ,^q+λ−2_q

(2−min{p, q}< λ≤ 2), and B

λ p,^λ_q

(0 < λ ≤ min{p, q}) are the best possible (B(u, v) is the β function). For λ = 1, both (1.4) and (1.5) reduce to (1.1). We call (1.4) and (1.5) Hardy-Hilbert type inequalities. Yang et al. [9] summarized the use of weight coefficients in research for Hardy-Hilbert type inequalities. But the problem on how to build the reverse of this type inequalities is unsolved.

The main objective of this paper is to deal with a reverse of inequality (1.4) forλ= 2. Another related reverse of the form is considered.

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2. Some Lemmas

We need the formula of theβ functionB(p, q)as follows (see [4]):

(2.1) B(p, q) =B(q, p) = Z ∞

0

t^p−1 (1 +t)^p+qdt, and the following inequality (see [5,2]): Iff⁴ ∈C[0,∞),0<R∞

0 f(t)dt <∞ and(−1)ⁿf⁽ⁿ⁾(x)>0, f⁽ⁿ⁾(∞) = 0 (n = 0,1,2,3,4), then

(2.2)

Z ∞ 0

f(t)dt+ 1

2f(0)<

∞

X

k=0

f(k)<

Z ∞ 0

f(t)dt+1

2f(0)− 1 12f⁰(0).

Lemma 2.1. Define the weight functionω(n)as

(2.3) ω(n) =

n+1 2

^∞ X

k=0

1

(k+n+ 1)², n ∈N0(=N∪ {0}), then we have

(2.4) 1− 1

4(n+ 1)² < ω(n)<1− 1

6(n+ 1)(2n+ 1) (n∈N0).

Proof. For fixedn, settingf(t) = _(t+n+1)¹ 2 (t∈[0,∞)), we obtain f(0) = 1

(n+ 1)², f⁰(0) =− 2

(n+ 1)³, and

Z ∞ 0

f(t)dt = 1 n+ 1.

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By (2.2), we find ω(n)>

Z ∞ 0

1

(t+n+ 1)²dt+ 1

2(n+ 1)² n+ 1 2

(2.5)

= 1

(n+ 1) + 1

2(n+ 1)² (n+ 1)−1 2

= 1− 1 4(n+ 1)².

ω(n)<

Z ∞ 0

1

(t+n+ 1)²dt+ 1

2(n+ 1)² + 1

6(n+ 1)³ n+ 1 2

(2.6)

= 1

n+ 1 + 1

2(n+ 1)² + 1

6(n+ 1)³ (n+ 1)− 1 2

= 1−

1

12(n+ 1)² + 1 12(n+ 1)³

.

Since forn ∈N0, 1

12(n+ 1)² + 1 12(n+ 1)³

6(n+ 1)(2n+ 1)

= 2(n+ 1)−1

2(n+ 1) +2(n+ 1)−1 2(n+ 1)²

= 1 + 1

2(n+ 1) − 1

2(n+ 1)² ≥1,

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then we find

1

12(n+ 1)² + 1

12(n+ 1)³ ≥ 1

6(n+ 1)(2n+ 1), and in view of (2.6), it follows that

(2.7) ω(n)<1− 1

6(n+ 1)(2n+ 1).

In virtue of (2.5) and (2.7), we have (2.4). The lemma is proved.

Lemma 2.2. For0< ε < p, we have I :=

∞

X

n=0

∞

X

m=0

1 (m+n+ 1)²

m+1

2

⁻^ε_p n+1

2 ⁻^ε_q (2.8)

<(1 +o(1))

∞

X

n=0

1

n+ ¹₂1+ε (ε→0⁺).

Proof. For fixedn, settingf(t) = (^t+¹2)^−ε/p

(t+n+1)² (t∈(−¹₂,∞)), then we have f(0) = 2^ε/p

(n+ 1)², f⁰(0) =− 2^1+ε/p

(n+ 1)³ − ε2^1+ε/p p(n+ 1)². Settingu= t+¹₂ n+¹₂

in the following integral, we obtain Z ∞

0

f(t)dt <

Z ∞

−1/2

f(t)dt= 1 n+ ¹₂1+^ε_p

Z ∞ 0

u⁻^ε^p (1 +u)²du.

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Hence by (2.2) and (2.1), we find I =

∞

X

n=0

n+1

2

−^ε_q " _∞ X

m=0

1 (m+n+ 1)²

m+ 1

2 −^ε_p#

<

∞

X

n=0

n+1

2 −^ε_q "

1 n+¹₂1+_p^ε

Z ∞ 0

u⁻^ε^p (1 +u)²du + 2^ε/p

2(n+ 1)² + 2^1+ε/p

12(n+ 1)³ + ε2^1+ε/p 12p(n+ 1)²

=

∞

X

n=0

1

n+¹₂1+εB

1− ε

p,1 + ε p

+O(1) (ε→0⁺).

Since B

1− ^ε_p,1 + _p^ε

→ B(1,1) = 1 (ε → 0⁺), then (2.8) is valid. The lemma is proved.

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3. Main Results

Theorem 3.1. If0< p <1,¹_p+¹_q = 1, a_n, b_n≥0, such that0<P∞ n=0

a^pn

2n+1 <

∞and0<P∞ n=0

b^qn

2n+1 <∞, then

(3.1)

∞

X

n=0

∞

X

m=0

a_mb_n

(m+n+ 1)² >2 ( _∞

X

n=0

1− 1

4(n+ 1)²

a^p_n 2n+ 1

)¹_p

× ( _∞

X

n=0

1− 1

6(n+ 1)(2n+ 1) b^q_n

2n+ 1 )¹_q

,

where the constant factor 2 is the best possible. In particular, one has

(3.2)

∞

X

n=0

∞

X

m=0

a_mb_n (m+n+ 1)²

>2 ( _∞

X

n=0

1− 1

4(n+ 1)² a^p_n

2n+ 1

)_p¹ ( _∞ X

n=0

b^q_n 2n+ 1

)¹_q .

Proof. By the reverse of Hölder’s inequality and (2.3), one has

∞

X

n=0

∞

X

m=0

a_mb_n (m+n+ 1)²

=

∞

X

n=0

∞

X

m=0

"

a_m (m+n+ 1)²^p

# "

b_n (m+n+ 1)²^q

#

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≥ ( _∞

X

n=0

∞

X

m=0

a^p_m (m+n+ 1)²

)¹_p( _∞ X

n=0

∞

X

m=0

b^q_n (m+n+ 1)²

)¹_q

= ( _∞

X

m=0

" _∞ X

n=0

m+ ¹₂ (m+n+ 1)²

# 2a^p_m 2m+ 1

)¹_p ( _∞ X

n=0

" _∞ X

m=0

n+ ¹₂ (m+n+ 1)²

# 2b^q_n 2n+ 1

)¹_q

= 2 ( _∞

X

m=0

ω(m) a^p_m 2m+ 1

)¹_p( _∞ X

n=0

ω(n) b^q_n 2n+ 1

)¹_q .

Since0< p < 1and q <0, by (2.4), it follows that (3.1) is valid.

For0 < ε < p, setting˜a_n = n+¹₂−ε/p

,˜b_n = n+¹₂−ε/q

, n ∈ N0,we find

( _∞ X

n=0

1− 1

4(n+ 1)² ˜a^p_n

2n+ 1

)_p¹ ( _∞ X

n=0

˜b^q_n 2n+ 1

)¹_q (3.3)

= ( _∞

X

n=0

1− 1

4(n+ 1)²

1 2 n+¹₂1+ε

)¹_p( _∞ X

n=0

1 2 n+ ¹₂1+ε

)¹_q

> 1 2

( _∞ X

n=0

1

n+¹₂1+ε −

∞

X

n=0

1

2(n+ 1)²(2n+ 1) )¹_p

× ( _∞

X

n=0

1 n+¹₂1+ε

)¹_q

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= 1

2{1−˜o(1)}¹^p

∞

X

n=0

1

n+¹₂1+ε(ε→0⁺).

If the constant factor 2 in (3.1) is not the best possible, then there exists a real number k with k > 2, such that (3.1) is still valid if one replaces 2 byk. In particular, one has

(3.4)

∞

X

n=0

∞

X

m=0

˜ a_m˜b_n (m+n+ 1)²

> k ( _∞

X

n=0

1− 1

4(n+ 1)² ˜a^p_n

2n+ 1

)_p¹ ( _∞ X

n=0

˜b^q_n 2n+ 1

)¹_q .

Hence by (2.8) and (3.3), it follows that (1 +o(1))

∞

X

n=0

1

n+ ¹₂1+ε > k

2{1−o(1)}˜ ¹^p

∞

X

n=0

1 n+¹₂1+ε,

and then 2 ≥ k(ε → 0⁺). This contradicts the fact that k > 2. Hence the constant factor 2 in (3.1) is the best possible. The theorem is proved.

Theorem 3.2. If0< p <1,¹_p+¹_q = 1, a_n ≥0, such that0<P∞ n=0

a^pn

2n+1 <∞, then

(3.5)

∞

X

n=0

n+ 1

2

p−1" _∞ X

m=0

a_m (m+n+ 1)²

#p

>2

∞

X

n=0

1− 1

4(n+ 1)² a^p_n

2n+ 1,

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where the constant factor2is the best possible.

Proof. By the reverse of Hölder’s inequality, (2.3) and (2.4), one hasω(n) <1 and

" _∞ X

m=0

a_m (m+n+ 1)²

#p

(3.6)

= ( _∞

X

m=0

"

a_m (m+n+ 1)²^p

# "

1 (m+n+ 1)²^q

#)p

≥ ( _∞

X

m=0

a^p_m (m+n+ 1)²

) ( _∞ X

m=0

1 (m+n+ 1)²

)p−1

= ( _∞

X

m=0

a^p_m (m+n+ 1)²

) ( ω(n)

n+ 1

2

−1)p−1

>

n+ 1

2

1−p ∞

X

m=0

a^p_m (m+n+ 1)². Hence we find

∞

X

n=0

n+ 1

2

p−1" _∞ X

m=0

a_m (m+n+ 1)²

#p

(3.7)

>

∞

X

n=0

∞

X

m=0

a^p_m (m+n+ 1)²

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=

∞

X

m=0

" _∞ X

n=0

m+ ¹₂ (m+n+ 1)²

# 2a^p_m 2m+ 1

= 2

∞

X

m=0

ω(m) a^p_m 2m+ 1. By (2.4), we have (3.5).

Setting b_n ≥ 0 and 0 < P∞ n=0

b^qn

2n+1 < ∞, by the reverse of Hölder’s inequality, one has

∞

X

n=0

∞

X

m=0

a_mb_n (m+n+ 1)² (3.8)

=

∞

X

n=0

"

n+1

2 ¹_q ^∞

X

m=0

a_m (m+n+ 1)²

# "

n+ 1

2 −¹

q

b_n

#

≥ ( _∞

X

n=0

n+ 1

2

p−1" _∞ X

m=0

a_m (m+n+ 1)²

#p)¹_p( _∞ X

n=0

2b^q_n 2n+ 1

)¹_q .

If the constant factor 2 in (3.5) is not the best possible, then by (3.6), we can get a contradiction that the constant factor 2 in (3.1) is not the best possible. The theorem is proved.

Remark 1. Ifa_n, b_nsatisfy the conditions of (1.4) forλ= 2, r >1, ¹_r+¹_s = 1, and (3.2) for0 < p < 1, ¹_p + ¹_q = 1,then one can get the following two-sided

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inequality:

0<

(3 4

∞

X

n=0

a^p_n 2n+ 1

)¹_p( _∞ X

n=0

b^q_n 2n+ 1

)¹_q (3.9)

< 1 2

∞

X

n=0

∞

X

m=0

a_mb_n (m+n+ 1)²

<

( _∞ X

n=0

a^r_n 2n+ 1

)¹_r ( _∞ X

n=0

b^s_n 2n+ 1

)¹_s

<∞.

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References

[1] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cam- bridge Univ. Press, London, 1952.

[2] JICHANG KUANG, On new generalizations of Hilbert’s inequality and their applications, J. Math. Anal. Appl., 245 (2000), 248–265.

[3] D.S. MITRINOVI ´C, J.E. PE ˇCARI ´C AND A.M. FINK, Inequalities In- volving Functions and their Integrals and Derivatives, Kluwer Academic, Boston, 1991.

[4] ZHUXI WANGANDDUNRIN GUO, An Introduction to Special Functions, Science Press, Beijing, 1979.

[5] BICHENG YANG, On a strengthened version of the more accurate Hardy- Hilbert’s inequality, Acta Math. Sinica, 42 (1999), 1103–1110.

[6] BICHENG YANG, On a strengthened Hardy-Hilbert’s inequality, J. Ineq.

Pure. Appl. Math., 1(2) (2000), Art. 22. [ONLINE:http://jipam.vu.

edu.au/article.php?sid=116]

[7] BICHENG YANG, On a new extension of Hardy-Hilbert’s inequality and its applications, Internat. J. Pure Appl. Math., 5 (2003), 57–66.

[8] BICHENG YANG ANDL. DEBNATH, On a new generalization of Hardy- Hilbert’s inequality and its applications, J. Math. Anal. Appl., 233 (1999), 484–497.

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[9] BICHENG YANG AND Th.M. RASSIAS, On the way of weight coeffi- cient and research for the Hilbert-type inequalities, Math. Inequal. Appl., 6 (2003), 625–658.