ON A DECOMPOSITION OF HILBERT’S INEQUALITY

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Hilbert’s Inequality Bicheng Yang vol. 10, iss. 1, art. 25, 2009

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ON A DECOMPOSITION OF HILBERT’S INEQUALITY

BICHENG YANG

Department of Mathematics Guangdong Education Institute Guangzhou, Guangdong 510303 P.R. CHINA

EMail:bcyang@pub.guangzhou.gd.cn

Received: 07 October, 2008

Accepted: 20 March, 2009

Communicated by: B. Opic 2000 AMS Sub. Class.: 26D15.

Key words: Hilbert’s inequality; Weight coefficient; Equivalent form; Hilbert-type inequality.

Abstract: By using the Euler-Maclaurin’s summation formula and the weight coefficient, a pair of new inequalities is given, which is a decomposition of Hilbert’s inequality. The equivalent forms and the extended inequalities with a pair of conjugate exponents are built.

Acknowledgement: The author expresses his sincerest thanks to the referee for his thoughtful sug- gestions in improving this work.

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1. Introduction

In 1908, H. Weyl published the following Hilbert inequality: If{a_n},{b_n}are real sequences,0<P∞

n=1a²_n<∞and0<P∞

n=1b²_n<∞,then [1]

(1.1)

∞

X

n=1

∞

X

m=1

ambn

m+n < π

∞

X

n=1

a²_n

∞

X

n=1

b²_n

!¹₂ ,

where the constant factor π is the best possible. In 1925, G. H. Hardy gave an extension of (1.1) by introducing one pair of conjugate exponents(p, q) (¹_p +¹_q = 1) as [2]: Ifp > 1, an, bn ≥0,0<P∞

n=1a^p_n<∞and0<P∞

n=1b^q_n<∞,then (1.2)

∞

X

n=1

∞

X

m=1

a_mb_n

m+n < π sin(^π_p)

∞

X

n=1

a^p_n

!¹_p _∞ X

n=1

b^q_n

!¹_q ,

where the constant factor π/sin(^π_p) is the best possible. We refer to (1.2) as the Hardy-Hilbert inequality. In 1934, Hardy et al. [3] gave some applications of (1.1) and (1.2). By introducing a pair of non-conjugate exponents(p, q)in (1.1), Hardy et al. [3] gave: Ifp, q >1,¹_p +¹_q ≥1,0< λ= 2−(¹_p +¹_q)≤1,then

(1.3)

∞

X

n=1

∞

X

m=1

a_mb_n

(m+n)^λ ≤K(p, q)

∞

X

n=1

a^p_n

!¹_p _∞ X

n=1

b^q_n

!¹_q ,

where the constant factorK(p, q)is the best value only forλ = 1. In 1951, Bonsall [4] considered (1.3) in the case of a general kernel. In 1991, Mitrinovi´c et al. [5]

summarized the above method and results.

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In 1997-1998, by using weight coefficients, Yang and Gao [6], [7] gave a strengthened version of (1.2) as:

(1.4)

∞

X

n=1

∞

X

m=1

a_mb_n m+n

<

( _∞ X

n=1

"

π

sin(^π_p) − 1−γ n^1/p

# a^p_n

)¹_p( _∞ X

n=1

"

π

sin(^π_p) − 1−γ n^1/q

# b^q_n

)¹_q , where, 1−γ = 0.42278433⁺ (γ is the Euler constant). In 2001, Yang [8] gave an extension of (1.1) by introducing an independent parameter0< λ≤4as

(1.5)

∞

X

n=1

∞

X

m=1

ambn

(m+n)^λ < B λ

2,λ 2

^∞ X

n=1

n^1−λa²_n

∞

X

n=1

n^1−λb²_n

!¹₂ ,

where the constant factorB(^λ₂,^λ₂)is the best possible (B(u, v)is the Beta function).

In 2004, Yang [9] published the dual form of (1.2) as follows (1.6)

∞

X

n=1

∞

X

m=1

a_mb_n

m+n < π sin(^π_p)

∞

X

n=1

n^p−2a^p_n

!¹_p _∞ X

n=1

n^q−2b^q_n

!¹_q .

Forp=q= 2,both (1.6) and (1.2) reduce to (1.1). It means that there are two differ- ent best extensions of (1.1). To generalize (1.2) and (1.6), in 2005, Yang [10] gave an extension of (1.2) and (1.6) with two pairs of conjugate exponents(p, q),(r, s)(p, r >

1)and parametersα, λ > 0 (αλ≤min{r, s})as: If0<P∞

n=1n^p(¹⁻^αλr )⁻¹a^p_n <∞

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and0<P∞

n=1n^q(¹⁻^αλs )⁻¹b^q_n <∞, then (1.7)

∞

X

n=1

∞

X

m=1

a_mb_n (m^α+n^α)^λ

< k_αλ(r) ( _∞

X

n=1

n^p(1−^αλ^r ⁾⁻¹a^p_n

)¹_p( _∞ X

n=1

n^q(1−^αλ^s ⁾⁻¹b^q_n )¹_q

, where the constant factor k_αλ(r) = _α¹B(^λ_r,^λ_s) is the best possible. T. K. Pogány [11] also considered a best extension of (1.2) with the non-homogeneous kernel as

1

(λm+ρn)^µ (µ, λm, ρn>0).

We have a non-negative decomposition of kernel in (1.1):

1

m+n = max{m, n}

(m+n)² + min{m, n}

(m+n)² (m, n∈N)

(Nis the set of positive integer numbers). In this paper, by using the Euler-Maclaurin summation formula and the weight coefficient as in [8], we give a pair of new Hilbert-type inequalities as

∞

X

n=1

∞

X

m=1

max{m, n}

(m+n)² a_mb_n <

π 2 + 1

X^∞

n=1

a²_n

∞

X

n=1

b²_n

!¹₂

; (1.8)

∞

X

n=1

∞

X

m=1

min{m, n}

(m+n)² a_mb_n <π

2 −1 X^∞

n=1

a²_n

∞

X

n=1

b²_n

!¹₂ (1.9) ,

where the sum of two best constant factors isπ. The equivalent forms and extended inequalities with a pair of conjugate exponents are considered.

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2. Some Lemmas

Lemma 2.1 (Euler-Maclaurin’s summation formula, cf. [8, 12, Lemma 1]). If f(x)∈C¹[1,∞),then we have

(2.1)

∞

X

k=1

f(k) = Z ∞

1

f(x)dx+1

2f(1) + Z ∞

1

P₁(x)f⁰(x)dx,

whereP₁(x) =x−[x]−¹₂ is the Bernoulli function of the first order; ifg ∈C³[1,∞), (−1)ⁱg⁽ⁱ⁾(x)>0, g⁽ⁱ⁾(∞) = 0, (i= 0,1,2,3), then

1

12[g(n)−g(1)]<

Z n 1

P₁(x)g(x)dx <0, (2.2)

− 1

12g(n)<

Z ∞ n

P₁(x)g(x)dx <0.

Lemma 2.2. If ¹₂ ≤α <1,setting the weight coefficientω(α, m)as

(2.3) ω(α, m) :=

∞

X

n=1

max{m, n}m^α

(m+n)²n^α (m∈N), then we have

(2.4) k(α) = A_α(m) +ω(α, m); ω 1

2, m

< k 1

2

= π 2 + 1, where

k(α) := 1 α

Z ∞ 0

max u^1/α,1

(u^1/α+ 1)² u^α¹⁻²du andAα(m) = O(m^α−1), (m→ ∞).

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Proof. For fixed ¹₂ ≤ α <1, m ∈N, settingf(x) := _(m+x)^max{m,x}2x^α, x∈(0,∞),then by (2.1), it follows that

ω(α, m) = m^α

∞

X

n=1

f(n) (2.5)

=m^α Z ∞

1

f(x)dx+ 1

2f(1) + Z ∞

1

P₁(x)f⁰(x)dx

=m^α Z ∞

0

f(x)dx−m^αρ(α, m), P₁(x)f⁰(x)dx.

(2.6) ρ(α, m) :=

Z 1 0

f(x)dx−1

2f(1)− Z ∞

1

P₁(x)f⁰(x)dx.

We find

−1

2f(1) = −m

2(m+ 1)² = −1

2(m+ 1) + 1 2(m+ 1)², and

Z 1 0

f(x)dx= Z 1

0

m

(m+x)²x^αdx≥ Z 1

0

m

(m+x)²dx= 1 m+ 1; Z 1

0

f(x)dx≤ Z 1

0

m

m²x^αdx= 1 (1−α)m.

Forx ∈ (0, m), f(x) = _(m+x)^m2x^α, it follows f⁰(x) = _(m+x)^−2m3x^α − _(m+x)^αm2x^α+1; for

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x∈(m,∞), f(x) = _(m+x)^x^1−α2,we find f⁰(x) = −2x^1−α

(m+x)³ + 1−α (m+x)²x^α

= −2(x+m−m)

(m+x)³x^α + 1−α (m+x)²x^α

= −2

(m+x)²x^α + 2m

(m+x)³x^α + 1−α (m+x)²x^α.

In the following, it is obvious that g₁(x) = _(m+x)¹ 3x^α, g₂(x) = _(m+x)¹2x^α+1 and g₃(x) = _(m+x)¹ 2x^α are suited to apply in (2.2). Then by (2.2), we obtain

− Z m

1

P₁(x)f⁰(x)dx (2.7)

= Z m

1

2mP₁(x)dx (m+x)³x^α +

Z m 1

αmP₁(x)dx (m+x)²x^α+1

> 2m 12

1

8m^3+α − 1 (m+ 1)³

+ αm 12

1

4m^3+α − 1 (m+ 1)²

= α+ 1

48m^2+α − α

12(m+ 1) − 2−α

12(m+ 1)² + 1 6(m+ 1)³;

− Z ∞

m

P₁(x)f⁰(x)dx (2.8)

= Z ∞

m

2P₁(x)dx (m+x)²x^α −

Z ∞ m

2mP₁(x)dx

(m+x)³x^α −(1−α) Z ∞

m

P₁(x)dx (m+x)²x^α

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> −1 24m^2+α −

Z ∞ 1

P₁(x)f⁰(x)dx

=− Z m

1

P₁(x)f⁰(x)dx− Z ∞

m

P₁(x)f⁰(x)dx

> α−1

48m^2+α − α

12(m+ 1) − 2−α

12(m+ 1)² + 1 6(m+ 1)³. Hence by (2.6), forα= ¹₂,it follows that

ρ 1

2, m

> −1

2(m+ 1) + 1

2(m+ 1)² + 1 m+ 1 (2.9)

+

1 2 −1 48m^2+1/2 −

1 2

12(m+ 1) − 2− ¹₂

12(m+ 1)² + 1 6(m+ 1)³

= 11

24(m+ 1) + 9

24(m+ 1)² + −1

96m^2+1/2 + 1 6(m+ 1)³

≥ 11 24(m+ 1) +

9

96m² + −1 96m²

+ 1

6(m+ 1)³ >0.

By (2.7) and (2.8), we obtain

− Z ∞

1

P₁(x)f⁰(x)dx=− Z m

1

P₁(x)f⁰(x)dx− Z ∞

m

P₁(x)f⁰(x)dx

< 1

48m^2+α + 1−α 48m^2+α

= 2−α 48m^2+α.

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Then by (2.6), it follows

0< m^1−α[m^αρ(α, m)]

(2.10)

< −m

2(m+ 1) + m

2(m+ 1)² + 1

1−α + 2−α 48m^1+α

→ 1

1−α − 1

2 (m→ ∞).

Settingu= (x/m)^α,we find m^α

Z ∞ 0

f(x)dx=m^α Z ∞

0

max{m, x}

(m+x)²x^αdx (2.11)

= 1 α

Z ∞ 0

max{u^1/α,1}

(u^1/α+ 1)² u^α¹⁻²du=k(α), k

1 2

= 2 Z ∞

0

max{u²,1}

(u²+ 1)² du= 4 Z 1

0

du (u²+ 1)² (2.12)

= 4 Z ^π₄

0

cos²θdθ= π 2 + 1.

Hence by (2.5), (2.9), (2.10) and (2.11), (2.4) is valid and the lemma is proved.

Similar to Lemma2.2, we still have

Lemma 2.3. If ¹₂ ≤α <1,setting the weight coefficient$(α, m)as

(2.13) $(α, m) :=

∞

X

n=1

min{m, n}m^α

(m+n)²n^α (m∈N), then we have

(2.14) ek(α) = B_α(m) +$(α, m); $ 1

2, m

<ek 1

2

= π 2 −1,

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where

ek(α) = 1 α

Z ∞ 0

min{u^1/α,1}

(u^1/α+ 1)² u^α¹⁻²du andB_α(m) =O(m^α−2), (m→ ∞).

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3. Main Results and their Equivalent Forms

Theorem 3.1. If p > 1,¹_p + ¹_q = 1, a_n, b_n ≥ 0, 0 < P∞

n=1n^p²⁻¹a^p_n < ∞ and 0<P∞

n=1n^q²⁻¹b^q_n<∞,then we have the following equivalent inequalities I :=

∞

X

n=1

∞

X

m=1

max{m, n}

(m+n)² a_mb_n (3.1)

<

π 2 + 1

X^∞

n=1

n^p²⁻¹a^p_n

!¹_p _∞ X

n=1

n^q²⁻¹b^q_n

!¹_q

;

(3.2) J :=

∞

X

n=1

n^p²⁻¹

" _∞ X

m=1

max{m, n}

(m+n)² a_m

#p

<π

2 + 1p ∞

X

n=1

n^q²⁻¹a^p_n, where the constant factors ^π₂ + 1and ^π₂ + 1p

are the best possible.

Proof. By Hölder’s inequality and (2.3) – (2.4), we find

" _∞ X

m=1

max{m, n}

(m+n)² a_m

#p

(3.3)

= ( _∞

X

m=1

max{m, n}

(m+n)²

m^1/(2q)

n^1/(2p)a_m n^1/(2p) m^1/(2q)

)p

≤

" _∞ X

m=1

max{m, n}

(m+n)²

m^p/(2q) n^1/2 a^p_m

# " _∞ X

m=1

max{m, n}

(m+n)²

n^q/(2p) m^1/2

#p−1

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=ω^p−1 1

2, n

n¹⁻^p²

∞

X

m=1

max{m, n}

(m+n)²

m^p/(2q) n^1/2 a^p_m

≤π 2 + 1

p−1

n¹⁻^p²

∞

X

m=1

max{m, n}

(m+n)²

m^p/(2q) n^1/2 a^p_m;

J ≤π

2 + 1^p−1X^∞

n=1

∞

X

m=1

max{m, n}

(m+n)²

m^p/(2q) n^1/2 a^p_m

=π

2 + 1p−1 ∞

X

m=1

" _∞ X

n=1

max{m, n}

(m+n)²

m^p/(2q) n^1/2

# a^p_m

=π

2 + 1p−1 ∞

X

m=1

ω 1

2, m

m^p²⁻¹a^p_m <π

2 + 1p ∞

X

m=1

m^p²⁻¹a^p_m. Therefore (3.2) is valid. By Hölder’s inequality, we find that

I =

∞

X

n=1

"

n¹²⁻¹^p

∞

X

m=1

max{m, n}

(m+n)² a_m

# h

n⁻¹² ⁺¹^pb_ni (3.4)

≤J¹^p

∞

X

n=1

n^q²⁻¹b^q_n

!¹_q .

Then by (3.2), we have (3.1). On the other hand, suppose that (3.1) is valid. Setting b_n :=n^p²⁻¹

" _∞ X

m=1

max{m, n}

(m+n)² a_m

#p−1

, n∈N,

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then it followsP∞

n=1n^q²⁻¹b^q_n =J.By (3.3), we confirm thatJ <∞.IfJ = 0,then (3.2) is naturally valid; if0< J <∞,then by (3.1), we find

∞

X

n=1

n^q²⁻¹b^q_n =J =I <π

2 + 1 X^∞

n=1

n^p²⁻¹a^p_n

!¹_p _∞ X

n=1

n^q²⁻¹b^q_n

!¹_q

;

∞

X

n=1

n^q²⁻¹b^q_n

!¹_p

=J¹^p <π

2 + 1 X^∞

n=1

n^p²⁻¹a^p_n

!¹_p , and inequality (3.2) is valid, which is equivalent to (3.1).

For0< ε < ^q₂,settingea ={ean}^∞_n=1,eb ={ebn}^∞_n=1asea

−1 2 −^ε

p

n ,eb

−1 2 −^ε

q

n ,forn ∈N,if there exists a constant0< k ≤ ^π₂ + 1,such that (3.1) is still valid when we replace

π

2 + 1byk,then we find Ie:=

∞

X

n=1

∞

X

m=1

max{m, n}ea_meb_n (m+n)²

< k

∞

X

n=1

n^p²⁻¹ea^p_n

!¹_p _∞ X

n=1

n²^q⁻¹eb^q_n

!¹_q

=k

∞

X

n=1

1 n^1+ε;

Ie=

∞

X

n=1

" _∞ X

m=1

max{m, n}

(m+n)² m⁻¹² ⁻^ε^p

# n⁻¹² ⁻^ε^q

=

∞

X

m=1

1 m^1+ε

∞

X

n=1

max{m, n}m¹²⁺^ε^q (m+n)²n¹²⁺^q^ε

=

∞

X

m=1

1 m^1+εω

1 2 +ε

q, m

.

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And then by (2.4) and the above results, it follows that k

∞

X

n=1

1 n^1+ε > k

1 2+ ε

q ^∞

X

m=1

1 m^1+ε



1− 1 k

1

2 +_q^εA¹

2+^ε_q(m)

 (3.5) 

=k 1

2+ ε q





∞

X

m=1

1

m^1+ε − 1 k

1 2 + ^ε_q

∞

X

m=1

1 m^1+εA¹

2+^ε_q(m)





=k 1

2+ ε q

^∞ X

m=1

1 m^1+ε



1− P∞

m=1 1 m^1+εA¹

2+^ε_q(m)]

k

1 2 +_q^ε

P∞ m=1

1 m^1+ε



;

k > k 1

2+ ε q



1− P∞

m=1 1 m^1+εO

1 m

¹₂−^ε_q

k

1 2 +^ε_q

P∞ m=1

1 m^1+ε



. Settingα= ¹₂ +^ε_q,by Fatou’s Lemma, it follows that

ε→0lim⁺k 1

2 +ε q

= lim

α→¹₂⁺

1 α

Z ∞ 0

max{u^1/α,1}

(u^1/α+ 1)² u^α¹⁻²du

≥2 Z ∞

0

lim

α→¹₂⁺

max{u^1/α,1}

(u^1/α+ 1)² u^α¹⁻²du=k 1

2

= π 2 + 1.

Then by (3.5), we have k ≥ ^π₂ + 1 (ε → 0⁺).Hence k = ^π₂ + 1 is the best value of (3.1). We confirm that the constant factor in (3.2) is the best, otherwise we would obtain a contradiction by (3.4) that the constant factor in (3.1) is not the best possible.

The theorem is proved.

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In the same manner, by Lemma2.3, we have:

Theorem 3.2. If p > 1,¹_p + ¹_q = 1, a_n, b_n ≥ 0, 0 < P∞

n=1n^p²⁻¹a^p_n < ∞ and 0<P∞

n=1n^q²⁻¹b^q_n<∞,then we have the following equivalent inequalities (3.6)

∞

X

n=1

∞

X

m=1

min{m, n}

(m+n)² a_mb_n<π

2 −1 X^∞

n=1

n^p²⁻¹a^p_n

!¹_p _∞ X

n=1

n^q²⁻¹b^q_n

!¹_q

;

∞

X

n=1

n^p²⁻¹

" _∞ X

m=1

min{m, n}

(m+n)² a_m

#p

<π

2 −1p ∞

X

n=1

n^q²⁻¹a^p_n, where the constant factors ^π₂ −1and(^π₂ −1)^pare the best possible.

Remark 1. Forp=q= 2,(3.1) reduces to (1.8) and (3.6) reduces to (1.9).

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References

[1] H. WEYL, Singuläre Integralgleichungen mit besonderer Berücksichtigung des Fourierschen Integraltheorems, Göttingen : Inaugural–Dissertation, 1908.

[2] G.H. HARDY, Note on a theorem of Hilbert concerning series of positive term, Proceedings of the London Math. Society, 23 (1925),45–46.

[3] G.H. HARDY, J.E. LITTLEWOOD ANDG. PÓLYA, Inequalities, Cambridge University Press, Cambridge,1934.

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