Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 5, Article 183, 2006
HARDY’S INTEGRAL INEQUALITY FOR COMMUTATORS OF HARDY
OPERATORS
QING-YU ZHENG AND ZUN-WEI FU
D EPARTMENT OF M ATHEMATICS
L INYI N ORMAL U NIVERSITY
L INYI S HANDONG , 276005
P EOPLE ’ S R EPUBLIC OF C HINA
zqy7336@163.com
lyfzw@tom.com
Received 26 March, 2006; accepted 09 October, 2006
Communicated by B. Yang
A BSTRACT. The authors establish the Hardy integral inequality for commutators generated by
Hardy operators and Lipschitz functions.
Key words and phrases: Hardy’s integral inequality, Commutator, Hardy operator, Lipschitz function.
2000 Mathematics Subject Classification. 26D15, 42B25.
1. I NTRODUCTION AND M AIN R ESULTS
Let f be a non-negative and integral function on R+ , Hardy operators are defined by
Z
1 x
(Hf )(x) =
f (t) dt, x > 0,
x 0
and
Z ∞
(Vf )(x) =
f (t) dt, x > 0.
x
The Hardy integral inequality results are well known [9, 10, 11]; in particular
Z ∞
p1
Z ∞
p1
p
p
p
≤
(f (x)) dx ,
(1.1)
(Hf (x)) dx
p−1
0
0
and
Z
∞
p
(Vf (x)) dx
(1.2)
0
p1
Z
≤p
∞
p
(xf (x)) dx
p1
.
0
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
This paper was supported by the National Natural Science Foundation (No.10371080) of People’s Republic of China.
083-06
2
Q ING -Y U Z HENG AND Z UN -W EI F U
p
In (1.1), the constant p−1
is the best possible. In (1.2), the constant p is 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].
Let f be a non-negative and integral function on R+ , and the fractional Hardy operator be
defined by
Z x
1
α
(H f )(x) = 1−α
f (t) dt, x > 0,
x
0
for p1 − 1q = α, 0 < α < 1. There are fractional order Hardy integral inequalities which
correspond to (1.1) and (1.2):
Z ∞
1q
Z ∞
p1
p
q
α
(1.3)
(H f (x)) dx
≤C
(f (x)) dx ,
0
0
and
Z
1q
∞
q
(Vf (x)) dx
(1.4)
0
Z
≤C
∞
(x
1−α
p
f (x)) dx
p1
.
0
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 developments 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 and b(x) be a measurable, locally integrable function. Then the
commutator of the Hardy operator Ubα is defined by
Z x
1
α
Hb f (x) = 1−α
f (t)(b(x) − b(t))dt, x > 0.
x
0
Fu [7] obtained the following results.
Theorem 1.1. For b ∈ ∧˙ β (R+ ), 0 < β < 1, Hbα is a bounded operator from Lp (R+ ) to Lq (R+ ),
1 < p < q < ∞, 0 ≤ α < 1, 0 < α + β < 1, p1 − 1q = α + β.
Remark 1.2. In Theorem 1.1, If we let α = 0, Then the result corresponds to Hardy inequality
(1.3).
Definition 1.2. Let b(x) be a measurable, locally integrable function. Then the commutator of
the Hardy operator Vb is defined by
Z ∞
Vb f (x) =
f (t)(b(x) − b(t))dt, x > 0.
x
In Definition 1.1, if we let α = 0, then we denote Hbα by Hb .
It can be seen that if b ∈ ∧˙ β (Rn ), 0 < β < 1, then Hb has a similar boundedness property
to Hβ . A natural question regarding the boundedness property of Vb , can be answered in the
affirmative by the following inequality (1.5).
1
q
Theorem 1.3. If b ∈ ∧˙ β (R+ ),
1
p
(1.5)
kVb f kq ≤ Ckbk∧˙ β (R+ ) k(·)f (·)kp .
−
J. Inequal. Pure and Appl. Math., 7(5) Art. 183, 2006
= β, 0 < β < 1, p > 1. Then
http://jipam.vu.edu.au/
H ARDY ’ S I NTEGRAL I NEQUALITY
1
p
Theorem 1.4. If b ∈ ∧˙ β (R+ ), 0 < β < 1,
Then
−
1
q
3
= α + β, 0 < α + β < 1, 0 ≤ α < 1, p > 1.
kVb f kq ≤ Ckbk∧˙ β (R+ ) k(·)1−α f (·)kp .
(1.6)
If we let α = 0 in Theorem 1.4, then Theorem 1.3 can be obtained without difficulty. Thus we
just need to prove Theorem 1.4. Before we prove the main theorem, let us state some lemmas
and notations.
The Besov-Lipschitz space ∧˙ β (R+ ) is the space of functions f satisfying
kf k∧˙ β (R+ ) =
|f (x + h) − f (x)|
< ∞.
|h|β
x,h∈R+ ,h6=0
sup
Lemma 1.5 ([8, 15]). For any x, y ∈ R+ , if f ∈ ∧˙ β (R+ ), 0 < β < 1, then
|f (x) − f (y)| ≤ |x − y|β kf k∧˙ β (R+ ) ,
(1.7)
and given any interval I in R+
sup |f (x) − fI | ≤ C|I|β kf k∧˙ β (R+ ) ,
x∈I
where
1
fI =
|I|
Z
f.
I
Lemma 1.6 ([7, 14]). Let s > 0, q ≥ p > 1 , then
! p1 q
Z ∞
pq
Z 2k+1
∞ X
∞
X
(i−k)/s
p
p
2
|f (t)| dt ≤ C
|f (t)| dt ,
k
2
0
i=−∞ k=i
where
C=
2p/2s
2p/2s − 1
0
2q /2s
2q0 /2s − 1
qq0
,
1
1
+ = 1.
0
q
q
There are two different methods to prove Theorem 1.4.
2. P ROOF OF T HEOREM 1.4
First Proof.
Z ∞
∞ Z
X
q
|Vb f (x)| dx =
0
≤
q/q 0
≤2
x
∞ Z
X
2i
+2
2i+1
i=−∞
|(b(x) − b(t))f (t)| dt
∞ Z
X
2i
k=i
∞ Z
X
2i+1
2i
!q
2k+1
dx
2k
k=i
∞ Z
X
q/q 0
q
|(b(x) − b(t))f (t)| dt dx
2i
2i+1
i=−∞
q
(b(x) − b(t))f (t)dt dx
∞
Z
2i
∞ Z
X
i=−∞
∞
Z
i
i=−∞ 2
∞ Z 2i+1
X
i=−∞
=
2i+1
2k+1
!q
(b(x) − b(2i ,2i+1 ] )f (t) dt dx
2k
∞ Z
X
k=i
2k+1
!q
(b(t) − b(2i ,2i+1 ] )f (t) dt dx
2k
:= I + J.
J. Inequal. Pure and Appl. Math., 7(5) Art. 183, 2006
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4
Q ING -Y U Z HENG AND Z UN -W EI F U
By Lemma 1.5 and the Hölder inequality,
∞ Z 2i+1
∞ Z
X
X
q
q/q 0
I=2
b(x) − b(2i ,2i+1 ] dx
2i
i=−∞
∞
X
q/q 0
≤2
sup
x∈(2i ,2i+1 ]
i=−∞
×
≤ C2q/q
∞
X
0
2i(qβ+1)
∞
X
≤ C2
! p1
2k+1
|f (t)|p dt
2k
k=i
2k
2i(qβ+1)
β
i=−∞
Notice that
1
p
−
1
q
2k+1
Z
! p1 0 q

dt


! p1 Z k+1 ! 10 q
Z 2k+1
∞
p 
2
q X
p
kbk∧˙ β (R+ )
|f (t)| dt
dt


2k
2k
k=i

! p1 q
Z 2k+1
∞
q X
0
kbk∧˙ (R+ ) 
2k/p
2−k(1−α)p |t1−α f (t)|p dt  .
i=−∞
q/q 0
b(x) − b(2i ,2i+1 ] 
Z
∞
X

|f (t)| dt
2k
k=i
!q
2i
!q
2k+1
2k
k=i
= α + β,
0
I ≤ C2q/q kbk∧˙ β (R+ )

∞
q X

i=−∞
By Lemma 1.6 s =
q
1+qβ
∞
X
2k+1
2k
k=i
! p1 q
|t1−α f (t)|p dt  .
,
I ≤ C kbk∧˙ β (R+ )
(2.1)
1
2(i−k)( q +β)
Z
q
Z
∞
1−α
|t
pq
f (t)| dt .
p
0
Now estimate J,
∞ Z
X
q/q 0
J =2
i=−∞
2q/q 0
≤2
2i
∞ Z
X
i=−∞
2q/q 0
+2
∞ Z
X
2i+1
2k+1
2k
k=i
∞ Z
X
2i+1
2i
!q
(b(t) − b(2k ,2k+1 ] )f (t) dt dx
∞ Z
X
2i+1
2i
i=−∞
2k+1
2k
k=i
∞ Z
X
!q
(b(t) − b(2i ,2i+1 ] )f (t) dt dx
k=i
2k+1
!q
(b(2k ,2k+1 ] − b(2i ,2i+1 ] )f (t) dt dx
2k
:= J1 + J2 .
Notice that
1
p
−
2q/q 0
J1 ≤ 2
1
q
= α + β,
∞ Z
X
i=−∞
2q/q 0
≤ C2
2i+1
2i
kbk∧˙ β (R+ )
1
p
+
1
p0
= 1, by Lemma 1.5, it can be inferred that
! Z k+1
!q
∞
2
X
sup b(t) − b(2k ,2k+1 ] |f (t)|dt dx
k=i
t∈(2k ,2k+1 ]
∞ Z
q X
i=−∞
J. Inequal. Pure and Appl. Math., 7(5) Art. 183, 2006
2i+1
2i
2k
∞
X
k=i
2kβ
Z
2k+1
!q
|f (t)|dt
dx
2k
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H ARDY ’ S I NTEGRAL I NEQUALITY
0
≤ C22q/q kbk∧˙ β (R+ )
∞
q X

2i 
i=−∞
0
≤ C22q/q kbk∧˙ β (R+ )
∞
q X
0
= C22q/q kbk∧˙ β (R+ )
k(β+ p10 )
2i 
∞
X
∞
X

i=−∞
|f (t)|p dt
2

2k
k(β+ p10 )
! p1 q
2k+1
Z
2−k(1−α)p |t1−α f (t)|p dt
2

2k
k=i

! p1 q
2k+1
Z
k=i

i=−∞
∞
q X
∞
X
5
2(i−k)/q
! p1 q
2k+1
Z
|t1−α f (t)|p dt
 .
2k
k=i
In the third inequality, the Hölder inequality is applied.
By Lemma 1.6 (s = q),
Z ∞
pq
q
1−α
p
(2.2)
J1 ≤ C kbk∧˙ β (R+ )
|t f (t)| dt .
0
To estimate J2 , for i > k, by Lemma 1.5, the following result is obtained.
Z 2i+1
b(2k ,2k+1 ] − b(2i ,2i+1 ] ≤ 1
b(y) − b(2k ,2k+1 ] dy
i
2 2i
Z k+1 Z 2i+1
1 1 2
≤ i k
|b(y) − b(z)| dydz
2 2 2k
2i
Z k+1 Z 2i+1
1 1 2
≤ kbk∧˙ β (R+ ) i k
|y − z|β dydz
2 2 2k
2i
!
Z 2i+1
Z 2k+1
1
1
≤ kbk∧˙ β (R+ )
y β dy + k
z β dz
2i 2i
2 2k
≤ C2iβ+1 kbk∧˙ β (R+ ) .
Notice that
1
p
−
1
q
0
J2 ≤ C22q/q +q
= α + β,
1
p
+
1
p0
= 1; by Lemma 1.5 and the Hölder inequality, we have
!q
∞ Z 2i+1
∞ Z 2k+1
X
q X
kbk∧˙ β (R+ )
2iβ
|f (t)|dt dx
i=−∞
0
≤ C22q/q +q kbk∧˙ β (R+ )
q
∞
X
2i
1
2i(β+ q )
i=−∞
0
≤ C22q/q +q kbk∧˙ β (R+ )
q
∞
X
0
∞
X
2k/p
1
2i(β+ q )
∞
X
! p1 q
2k+1
|f (t)|p dt

2k
2k/p
0
Z
! p1 q
2k+1
2−k(1−α)p |t1−α f (t)|p dt
2k
k=i

∞
∞
q X X
1

+
kbk∧˙ β (R )
2(i−k)(β+ q )
i=−∞
0
Z
k=i

i=−∞
= C22q/q +q
2k
k=i

k=i
Z

! p1 q
2k+1
|t1−α f (t)|p dt
2k
 .
In the second inequality,
the Hölder inequality is applied.
q
By Lemma 1.6 s = qβ+1 ,
(2.3)
J2 ≤ C kbk∧˙ β (R+ )
q
Z
∞
pq
|t1−α f (t)|p dt .
0
J. Inequal. Pure and Appl. Math., 7(5) Art. 183, 2006
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6
Q ING -Y U Z HENG AND Z UN -W EI F U
Combining (2.1), (2.2) and (2.3), we complete the proof of Theorem 1.4.
Second Proof. By inequality (1.7) and comparing the size of x and t, we have
q
Z ∞
∞ Z 2i+1 Z ∞
X
q
dx
|Vb f (x)| dx =
f
(t)|b(x)
−
b(x)|dt
0
i=−∞
≤
2i
∞ Z
X
i=−∞
x
∞ Z
X
2i+1
2i
k=i
≤ C kbk∧˙ β (R+ )
|t − x|β kbk∧˙ β (R+ ) |f (t)| dt
2k
∞ Z
q X
∞ Z
q X
i=−∞
= C kbk∧˙ β (R+ )
∞
q X
2i+1
2i
i=−∞
= C kbk∧˙ β (R+ )
2i
2i
Z
2k+1
2k
Notice that
Z
0
∞
1
p
−
1
p
+
1
p0
t1−α |f (t)|
dt ≤
t1−α−β
1
q
∞ Z
X
dx
!q
2k+1
tβ |f (t)| dt
dx
2k
k=i
!q
t1−α |f (t)|
dt dx
1−α−β
t
k
2
k=i
!q
Z
∞
X 2k+1 t1−α |f (t)|
dt .
t1−α−β
k
k=i 2
2i+1
i=−∞
By the Hölder inequality,
!q
2k+1
∞ Z
X
2k+1
= 1, the following estimate is obtained.
! p1 Z k+1
Z k+1
2
2
(α+β−1)p0
|t1−α f (t)|p dt
2k
t
! 10
p
dt
.
2k
= α + β,

∞ X
∞
X
q
q
|Vb f (x)| dx ≤ C kbk∧˙ β (R+ )
2(i−k)/q

i=−∞
k=i
Z
2k+1
2k
! p1 q

1−α
p
|t f (t)| dt
.

By Lemma 1.6, the desired result is obtained.
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