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volume 7, issue 5, article 183, 2006.

Received 26 March, 2006;

accepted 09 October, 2006.

Communicated by:B. Yang

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Journal of Inequalities in Pure and Applied Mathematics

HARDY’S INTEGRAL INEQUALITY FOR COMMUTATORS OF HARDY OPERATORS

QING-YU ZHENG AND ZUN-WEI FU

Department of Mathematics Linyi Normal University Linyi Shandong, 276005 People’s Republic of China EMail:zqy7336@163.com EMail:lyfzw@tom.com

c

2000Victoria University ISSN (electronic): 1443-5756 083-06

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Hardy’s Integral Inequality For Commutators Of Hardy

Operators

Qing-Yu Zheng and Zun-Wei Fu

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J. Ineq. Pure and Appl. Math. 7(5) Art. 183, 2006

Abstract

The authors establish the Hardy integral inequality for commutators generated by Hardy operators and Lipschitz functions.

2000 Mathematics Subject Classification:26D15, 42B25.

Key words: Hardy’s integral inequality, Commutator, Hardy operator, Lipschitz func- tion.

This paper was supported by the National Natural Science Foundation (No.10371080) of People’s Republic of China.

1. Introduction and Main Results

Let f be a non-negative and integral function onR⁺, Hardy operators are defined by

(Hf)(x) = 1 x

Z x 0

f(t)dt, x >0, and

(Vf)(x) = Z ∞

x

f(t)dt, x >0.

The Hardy integral inequality results are well known [9,10,11]; in particular

(1.1)

Z ∞ 0

(Hf(x))^pdx ¹_p

≤ p p−1

Z ∞ 0

(f(x))^pdx ¹_p

,

and (1.2)

Z ∞ 0

(Vf(x))^pdx ¹_p

≤p Z ∞

0

(xf(x))^pdx ¹_p

.

In (1.1), the constant _p−1^p is the best possible. In (1.2), the constantpis also the best possible. The inequality (1.1) was first proved by Hardy in an attempt to give a simple proof of Hilbert’s double series theorem [12].

Letfbe a non-negative and integral function onR⁺, and the fractional Hardy operator be defined by

(H^αf)(x) = 1 x^1−α

Z x 0

f(t)dt, x >0,

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for ¹_p−¹_q =α,0< α <1. There are fractional order Hardy integral inequalities which correspond to (1.1) and (1.2):

(1.3)

Z ∞ 0

(H^αf(x))^qdx ¹_q

≤C Z ∞

0

(f(x))^pdx ¹_p

,

and (1.4)

Z ∞ 0

(Vf(x))^qdx ¹_q

≤C Z ∞

0

(x^1−αf(x))^pdx ¹_p

. The Hardy inequality (1.3) can be found in [2], (1.4) in [13].

The Hardy integral inequalities have received considerable attention and a large number of papers have appeared which deal with its alternative proofs, various generalizations, numerous variants and applications. For earlier devel- opments of this kind of inequality and many important applications in analysis, see [11]. Among numerous papers dealing with such inequalities, we choose to refer to the papers [3], [5], [9], [10], [16] – [21] and some of the references cited therein.

Definition 1.1. Let 0 ≤ α < 1 andb(x)be a measurable, locally integrable function. Then the commutator of the Hardy operatorU_b^α is defined by

H^α_bf(x) = 1 x^1−α

Z x 0

f(t)(b(x)−b(t))dt, x >0.

Fu [7] obtained the following results.

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Theorem 1.1. For b ∈ ∧˙_β(R⁺),0 < β < 1, H_b^α is a bounded operator from L^p(R⁺)toL^q(R⁺),1< p < q < ∞,0≤α <1,0< α+β <1,¹_p−¹_q =α+β.

Remark 1. In Theorem 1.1, If we let α = 0, Then the result corresponds to Hardy inequality (1.3).

Definition 1.2. Letb(x)be a measurable, locally integrable function. Then the commutator of the Hardy operatorVb is defined by

V_bf(x) = Z ∞

x

f(t)(b(x)−b(t))dt, x >0.

In Definition1.1, if we letα= 0, then we denoteH^α_b byH_b.

It can be seen that ifb ∈∧˙_β(Rⁿ),0 < β <1, thenH_b has a similar boundedness property toH^β. A natural question regarding the boundedness property ofV_b, can be answered in the affirmative by the following inequality (1.5).

Theorem 1.2. Ifb ∈∧˙_β(R⁺), ¹_p − ¹_q =β,0< β <1, p >1. Then

(1.5) kV_bfk_q≤Ckbk∧˙_β(R⁺)k(·)f(·)k_p.

Theorem 1.3. Ifb ∈ ∧˙_β(R⁺),0 < β < 1, ¹_p − ¹_q = α+β, 0 < α+β < 1, 0≤α <1, p >1. Then

(1.6) kV_bfk_q ≤Ckbk∧˙β(R⁺)k(·)^1−αf(·)k_p.

If we letα = 0 in Theorem1.3, then Theorem1.2 can be obtained without difficulty. Thus we just need to prove Theorem1.3. Before we prove the main theorem, let us state some lemmas and notations.

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The Besov-Lipschitz space∧˙_β(R⁺)is the space of functionsf satisfying kfk∧˙_β(R⁺)= sup

x,h∈R⁺,h6=0

|f(x+h)−f(x)|

|h|^β <∞.

Lemma 1.4 ([8,15]). For anyx, y ∈R⁺, iff ∈∧˙_β(R⁺),0< β <1, then

(1.7) |f(x)−f(y)| ≤ |x−y|^βkfk∧˙β(R⁺), and given any intervalIinR⁺

sup

x∈I

|f(x)−f_I| ≤C|I|^βkfk∧˙_β(R⁺),

where

f_I = 1

|I|

Z

I

f.

Lemma 1.5 ([7,14]). Lets >0, q ≥p >1, then

∞

X

i=−∞

∞

X

k=i

2^(i−k)/s

Z 2^k+1 2^k

|f(t)|^pdt

!¹_p

q

≤C Z ∞

0

|f(t)|^pdt ^q_p

,

where

C=

2^p/2s 2^p/2s−1

2^q⁰^/2s 2^q⁰^/2s−1

q q0

, 1

q⁰ +1 q = 1.

There are two different methods to prove Theorem1.3.

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2. Proof of Theorem 1.3

First Proof.

Z ∞ 0

|V_bf(x)|^qdx

=

∞

X

i=−∞

Z 2ⁱ⁺¹ 2ⁱ

Z ∞ x

(b(x)−b(t))f(t)dt

q

dx

≤

∞

X

i=−∞

Z 2ⁱ⁺¹ 2ⁱ

Z ∞ 2ⁱ

|(b(x)−b(t))f(t)|dt q

dx

=

∞

X

i=−∞

Z 2ⁱ⁺¹ 2ⁱ

∞

X

k=i

Z 2^k+1 2^k

|(b(x)−b(t))f(t)|dt

!q

dx

≤2^q/q⁰

∞

X

i=−∞

Z 2ⁱ⁺¹ 2ⁱ

∞

X

k=i

Z 2^k+1 2^k

(b(x)−b₍₂ⁱ_,2ⁱ⁺¹_])f(t) dt

!q

dx

+ 2^q/q⁰

∞

X

i=−∞

Z 2ⁱ⁺¹ 2ⁱ

∞

X

k=i

Z 2^k+1 2^k

(b(t)−b₍₂ⁱ_,2ⁱ⁺¹_])f(t) dt

!q

dx

:=I+J.

By Lemma1.4and the Hölder inequality, I = 2^q/q⁰

∞

X

i=−∞

Z 2ⁱ⁺¹ 2ⁱ

b(x)−b₍₂ⁱ_,2ⁱ⁺¹_]

qdx

∞

X

k=i

Z 2^k+1 2^k

|f(t)|dt

!q

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≤2^q/q⁰

∞

X

i=−∞

2ⁱ sup

x∈(2ⁱ,2ⁱ⁺¹]

b(x)−b₍₂ⁱ_,2ⁱ⁺¹_]

!q

×







∞

X

k=i

Z 2^k+1 2^k

|f(t)|^pdt

!¹_p

Z 2^k+1 2^k

dt

!¹_p

0





q

≤C2^q/q⁰

∞

X

i=−∞

2^i(qβ+1) kbk∧˙_β(R⁺)

q

×







∞

X

k=i

Z 2^k+1 2^k

|f(t)|^pdt

!¹_p

Z 2^k+1 2^k

dt

!_p¹0





q

≤C2^q/q⁰

∞

X

i=−∞

2^i(qβ+1) kbk∧˙_β(R⁺)

q

×





∞

X

k=i

2^k/p⁰

Z 2^k+1 2^k

2^{−k(1−α)p}|t^1−αf(t)|^pdt

!¹_p



q

.

Notice that ¹_p − ¹_q =α+β,

I ≤C2^q/q⁰ kbk∧˙_β(R⁺)

q

∞

X

i=−∞





∞

X

k=i

2^(i−k)(¹^q^+β)

Z 2^k+1 2^k

|t^1−αf(t)|^pdt

!¹_p



q

.

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By Lemma1.5

s= _1+qβ^q

,

(2.1) I ≤C kbk∧˙_β(R⁺)

qZ ∞ 0

|t^1−αf(t)|^pdt _p^q

.

Now estimateJ, J = 2^q/q⁰

∞

X

i=−∞

Z 2ⁱ⁺¹ 2ⁱ

∞

X

k=i

Z 2^k+1 2^k

(b(t)−b₍₂ⁱ_,2ⁱ⁺¹_])f(t) dt

!q

dx

≤2^2q/q⁰

∞

X

i=−∞

Z 2ⁱ⁺¹ 2ⁱ

∞

X

k=i

Z 2^k+1 2^k

(b(t)−b₍₂k,2^k+1])f(t) dt

!q

dx

+ 2^2q/q⁰

∞

X

i=−∞

Z 2ⁱ⁺¹ 2ⁱ

∞

X

k=i

Z 2^k+1 2^k

(b₍₂^k_,2^k+1_]−b₍₂ⁱ_,2ⁱ⁺¹_])f(t) dt

!q

dx

:=J₁+J₂.

Notice that ¹_p − ¹_q =α+β, ¹_p + _p¹0 = 1, by Lemma1.4, it can be inferred that

J₁ ≤2^2q/q⁰

∞

X

i=−∞

Z 2ⁱ⁺¹ 2ⁱ

∞

X

k=i

sup

t∈(2^k,2^k+1]

b(t)−b₍₂k,2^k+1]

!Z 2^k+1

2^k

|f(t)|dt

!q

dx

≤C2^2q/q⁰ kbk∧˙_β(R⁺)

q

∞

X

i=−∞

Z 2ⁱ⁺¹ 2ⁱ

∞

X

k=i

2^kβ Z 2^k+1

2^k

|f(t)|dt

!q

dx

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≤C2^2q/q⁰ kbk∧˙_β(R⁺)

q

∞

X

i=−∞

2ⁱ





∞

X

k=i

2^k(β+^p¹⁰⁾

Z 2^k+1 2^k

|f(t)|^pdt

!¹_p



q

≤C2^2q/q⁰ kbk∧˙_β(R⁺)

q

×

∞

X

i=−∞

2ⁱ





∞

X

k=i

2^k(β+^p¹⁰⁾

Z 2^k+1 2^k

2^{−k(1−α)p}|t^1−αf(t)|^pdt

!¹_p



q

=C2^2q/q⁰ kbk∧˙_β(R⁺)

q

∞

X

i=−∞





∞

X

k=i

2^(i−k)/q

Z 2^k+1 2^k

|t^1−αf(t)|^pdt

!¹_p



q

.

In the third inequality, the Hölder inequality is applied.

By Lemma1.5(s=q),

(2.2) J₁ ≤C kbk∧˙_β(R⁺)

qZ ∞ 0

|t^1−αf(t)|^pdt ^q_p

.

To estimateJ₂, fori > k, by Lemma1.4, the following result is obtained.

b₍₂k,2^k+1]−b₍₂ⁱ_,2ⁱ⁺¹_] ≤ 1

2ⁱ Z 2ⁱ⁺¹

2ⁱ

b(y)−b₍₂k,2^k+1]

dy

≤ 1 2ⁱ

1 2^k

Z 2^k+1 2^k

Z 2ⁱ⁺¹ 2ⁱ

|b(y)−b(z)|dydz

≤ kbk∧˙β(R⁺)

1 2ⁱ

1 2^k

Z 2^k+1 2^k

Z 2ⁱ⁺¹ 2ⁱ

|y−z|^βdydz

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≤ kbk∧˙_β(R⁺)

1 2ⁱ

Z 2ⁱ⁺¹ 2ⁱ

y^βdy+ 1 2^k

Z 2^k+1 2^k

z^βdz

!

≤C2^iβ+1kbk∧˙_β(R⁺).

Notice that ¹_p−¹_q =α+β, ¹_p+_p¹0 = 1; by Lemma1.4and the Hölder inequality, we have

J₂ ≤C2^2q/q⁰^+q kbk∧˙_β(R⁺)

q

∞

X

i=−∞

Z 2ⁱ⁺¹ 2ⁱ

2^iβ

∞

X

k=i

Z 2^k+1 2^k

|f(t)|dt

!q

dx

≤C2^2q/q⁰^+q kbk∧˙_β(R⁺)

q

∞

X

i=−∞



2^i(β+¹^q⁾

∞

X

k=i

2^k/p⁰

Z 2^k+1 2^k

|f(t)|^pdt

!_p¹



q

≤C2^2q/q⁰^+q kbk∧˙β(R⁺)

q

×

∞

X

i=−∞



2^i(β+¹^q⁾

∞

X

k=i

2^k/p⁰

Z 2^k+1 2^k

2^{−k(1−α)p}|t^1−αf(t)|^pdt

!¹_p



q

=C2^2q/q⁰^+q kbk∧˙_β(R⁺)

q

∞

X

i=−∞





∞

X

k=i

2^(i−k)(β+¹^q⁾

Z 2^k+1 2^k

|t^1−αf(t)|^pdt

!¹_p



q

.

In the second inequality, the Hölder inequality is applied.

By Lemma1.5

s= _qβ+1^q , (2.3) J₂ ≤C kbk∧˙_β(R⁺)

qZ ∞ 0

|t^1−αf(t)|^pdt ^q_p

.

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Combining (2.1), (2.2) and (2.3), we complete the proof of Theorem1.3.

Second Proof. By inequality (1.7) and comparing the size ofxandt, we have Z ∞

0

|V_bf(x)|^qdx=

∞

X

i=−∞

Z 2ⁱ⁺¹ 2ⁱ

Z ∞ x

f(t)|b(x)−b(x)|dt

q

dx

≤

∞

X

i=−∞

Z 2ⁱ⁺¹ 2ⁱ

∞

X

k=i

Z 2^k+1 2^k

|t−x|^βkbk∧˙β(R⁺)|f(t)|dt

!q

dx

≤C kbk∧˙_β(R⁺)

q

∞

X

i=−∞

Z 2ⁱ⁺¹ 2ⁱ

∞

X

k=i

Z 2^k+1 2^k

t^β|f(t)|dt

!q

dx

=C kbk∧˙_β(R⁺)

q

∞

X

i=−∞

Z 2ⁱ⁺¹ 2ⁱ

∞

X

k=i

Z 2^k+1 2^k

t^1−α|f(t)|

t^1−α−β dt

!q

dx

=C kbk∧˙_β(R⁺)

q

∞

X

i=−∞

2ⁱ

∞

X

k=i

Z 2^k+1 2^k

t^1−α|f(t)|

t^1−α−β dt

!q

.

By the Hölder inequality, ¹_p +_p¹0 = 1,the following estimate is obtained.

Z 2^k+1 2^k

t^1−α|f(t)|

t^1−α−β dt ≤

Z 2^k+1 2^k

|t^1−αf(t)|^pdt

!_p¹

Z 2^k+1 2^k

t^(α+β−1)p⁰dt

!_p¹0

.

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Notice that ¹_p − ¹_q =α+β, Z ∞

0

|V_bf(x)|^qdx

≤C kbk∧˙_β(R⁺)

q

∞

X

i=−∞







∞

X

k=i

2^(i−k)/q

Z 2^k+1 2^k

|t^1−αf(t)|^pdt

!¹_p





q

.

By Lemma1.5, the desired result is obtained.

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References

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[11] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Lon- don/New York: Cambridge Univ. Press, 1934.

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[21] B.C. YANG, Z.H. ZENG AND L. DEBNATH, Generalizations of Hardy integral inequality, Internat. J. Math. Math. Sci, 3 (1999), 535–542.

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