Volume 8 (2007), Issue 4, Article 108, 8 pp.
ON THE Lp BOUNDEDNESS OF ROUGH PARAMETRIC MARCINKIEWICZ
FUNCTIONS
AHMAD AL-SALMAN AND HUSSAIN AL-QASSEM
D EPARTMENT OF M ATHEMATICS AND S TATISTICS
S ULTAN Q ABOOS U NIVERSITY
P.O. B OX 36, A L -K HOD 123 M USCAT
S ULTANATE OF O MAN .
alsalman@squ.edu.om
D EPARTMENT OF M ATHEMATICS AND P HYSICS
Q ATAR U NIVERSITY, Q ATAR
husseink@qu.edu.qa
Received 07 November, 2006; accepted 25 November, 2007
Communicated by L. Debnath
A BSTRACT. In this paper, we study the Lp boundedness of a class of parametric Marcinkiewicz
integral operators with rough kernels in L(log+ L)(Sn−1 ). Our result in this paper solves an
open problem left by the authors of ([6]).
Key words and phrases: Parametric Marcinkiewicz operators, rough kernels, Fourier transforms, Parametric maximal functions.
2000 Mathematics Subject Classification. Primary 42B20; Secondary 42B15, 42B25.
1. I NTRODUCTION
Let n ≥ 2 and Sn−1 be the unit sphere in Rn equipped with the normalized Lebesgue measure
dσ. Suppose that Ω is a homogeneous function of degree zero on Rn that satisfies Ω ∈ L1 (Sn−1 )
and
Z
(1.1)
Ω(x)dσ(x) = 0.
Sn−1
ρ
In 1960, Hörmander ([9]) defined the parametric Marcinkiewicz function µΩ of higher dimension by
2 ! 21
Z ∞
Z
−ρt
ρ
−n+ρ
2
dt
(1.2)
µΩ f (x) =
f
(x
−
y)
|y|
Ω(y)dy
,
−∞
|y|≤2t
ρ
where ρ > 0. It is clear that if ρ = 1, then µΩ is the classical Marcinkiewicz integral operator introduced by Stein ([11]) which will be denoted by µΩ . When Ω ∈ Lip α (Sn−1 ),
This paper was written during the authors’ time in Yarmouk University.
285-06
2
A HMAD A L -S ALMAN AND H USSAIN A L -Q ASSEM
(0 < α ≤ 1), Stein proved that µΩ is bounded on Lp for all 1 < p ≤ 2. Subsequently,
Benedek-Calderón-Panzone proved the Lp boundedness of µΩ for all 1 < p < ∞ under the
condition Ω ∈ C 1 (Sn−1 ) ([4]). Recently, under various conditions on Ω, the Lp boundedness
of µΩ and a more general class of operators of Marcinkiewicz type has been investigated (see
[1] – [2], [5], among others).
ρ
In ([9]), Hörmander proved that µΩ is bounded on Lp for all 1 < p < ∞, provided that
Ω ∈ Lipα (Sn−1 ), (0 < α ≤ 1) and ρ > 0.
ρ
A long standing open problem concerning the operator µΩ is whether there are some Lp
ρ
results on µΩ similar to those on µΩ when Ω satisfies only some size conditions. In a recent
paper, Ding, Lu, and Yabuta ([6]) studied the operator
2 ! 21
Z ∞
Z
−ρt
ρ
−n+ρ
dt
2
(1.3)
µΩ,h f (x) =
,
f
(x
−
y)
|y|
h(|y|)Ω(y)dy
−∞
|y|≤2t
where ρ is a complex number, Re(ρ) = α > 0, and h is a radial function on Rn satisfying
h(|x|) ∈ l∞ (Lq )(R+ ), 1 ≤ q ≤ ∞, where l∞ (Lq )(R+ ) is defined as follows: For 1 ≤ q < ∞,


! 1q
Z 2j


dr
l∞ (Lq )(R+ ) = h : khkl∞ (Lq )(R+ ) = sup
|h(r)|q
<C


r
j∈Z
2j−1
and for q = ∞, l∞ (L∞ )(R+ ) = L∞ (R+ ).
Ding, Lu, and Yabuta ([6]) proved the following:
Theorem 1.1. Suppose that Ω ∈ L(log+ L)(Sn−1 ) is a homogeneous function of degree zero
n
on
(1.1) and h(|x|) ∈ l∞ (Lq )(R+ ) for some 1 < q ≤ ∞. If Re(ρ) = α > 0, then
ρR satisfying
µ f ≤ C/√α kf k , where C is independent of ρ and f .
Ω,h
2
2
ρ
The Lp boundedness of µΩ,h for p 6= 2 was left open by the authors of ([6]). The main
ρ
purpose of this paper is to establish the Lp boundedness of µΩ,h for p 6= 2. Our main result of
this paper is the following:
Theorem 1.2. Suppose that Ω ∈ L(log+ L)(Sn−1 ) is a homogeneous function of degree zero
n
on
satisfying (1.1). If h(|x|) ∈ l∞ (Lq )(R+ ), 1 < q ≤ ∞, and Re(ρ) = α > 0, then
ρR µ f ≤ C/α kf k for all 1 < p < ∞, where C is independent of ρ and f .
Ω,h
p
p
Also, in this paper, we establish the Lp boundedness of the related parametric maximal function. In fact, we have the following result:
Theorem 1.3. Suppose that Ω ∈ L(log+ L)(Sn−1 ) is a homogeneous function of degree zero on
Rn . If h(|x|) ∈ l∞ (Lq )(R+ ), 1 < q ≤ ∞, and α > 0, then
C
kMα f kp ≤ kf kp
α
for all 1 < p < ∞ with a constant C independent of α, where Mα is the operator defined by
Z
−n+ρ
−αt (1.4)
Mα f (x) = sup 2 Ω(y) |y|
h(|y|)f (x − y)dy .
t
t∈R
|y|≤2
The method employed in this paper is based in part on ideas from [1], [2] and [3], among
others. A variation of this method can be applied to deal with more general integral operators of Marcinkiewicz type. An extensive discussion of more general operators will appear in
forthcoming papers.
Throughout the rest of the paper the letter C will stand for a constant but not necessarily the
same one in each occurrence.
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 108, 8 pp.
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Lp B OUNDEDNESS OF ROUGH PARAMETRIC M ARCINKIEWICZ F UNCTIONS
3
2. P REPARATION
Suppose a ≥ 1. For a suitable family of measures τ = {τt : t ∈ R} on Rn and a suitable
family of C ∞ functions Φa = {ϕt : t ∈ R} on Rn , define the family of operators {Λτ,Φa ,s :
t, s ∈ R} by
Z ∞
21
2
(2.1)
Λτ,Φ,s,a (f )(x) =
|τat ∗ ϕt+s ∗ f (x)| dt .
−∞
∗
Also, define the operator τ by
τ ∗ (f )(x) = sup(|τt | ∗ |f |)(x).
(2.2)
t∈R
The proof of our result will be based on the following lemma:
Lemma 2.1. Suppose that for some B > 0, ε > 0, and β > 0, we have
(i) kτt k ≤ β for t ∈ R;
ε
(ii) |τ̂t (ξ)| ≤ β(2t |ξ|)± a for ξ ∈ Rn and t ∈ R;
(iii) kτ ∗ (f )kq ≤ B kf kq for some q > 1;
(iv) The functions ϕt , t ∈ R satisfy
that ϕ̂t is supported in {ξ ∈ Rn :
γ the properties
2−(t+1)a ≤ |ξ| ≤ 2−(t−1)a } and ddξϕ̂γ t (ξ) ≤ Cγ |ξ|−|γ| for any multi-index γ ∈ (N∪(0))n
with constants Cγ depend only on γ and the dimension of the underlying space Rn .
Then for
2q
q+1
<p<
2q
,
q−1
there exists a constant Cp independent of a, β, B, s, and ε such that
θ(p)
1
kΛτ,Φ,s,a (f )kp ≤ Cp (βB) 2 (βB −1 ) 2 2(ε+1)θ(p) 2−εθ(p)|s| kf kp
2q
pq+p−2q
2q
for all f ∈ Lp (Rn ), where θ(p) = 2q−pq+p
if
p
∈
2,
and
θ(p)
=
if
p
∈
,
2
.
p
q−1
p
q+1
(2.3)
Proof. We start with the case p = 2. By Plancherel’s formula and the conditions (i)-(ii), we
obtain
kΛτ,Φ,s,a (f )k2 ≤ β2ε+1 2−ε|s| kf k2
(2.4)
for all f ∈ L2 (Rn ).
Next, set p0 = 2q 0 and choose a non-negative function v ∈ Lq+ (Rn ) with kvkq = 1 such that
Z Z ∞
2
kΛτ,Φ,s,a (f )kp0 =
|τat ∗ ϕt+s ∗ f (x)|2 v(x)dtdx.
Rn
−∞
Now it is easy to see that
(2.5)
p
1
β kga,s (f )kp0 kτ ∗ (v)kq2
kΛτ,Φ,s,a (f )kp0 ≤
where ga,s is the operator
Z
(2.6)
∞
ga,s (f )(x) =
21
|ϕt+s ∗ f (x)| dt .
2
−∞
By the condition (iv) and a well-known argument (see [12, p. 26-28]), it is easy to see that
(2.7)
kga,s (f )kp0 ≤ Cp0 kf kp0
for all f ∈ Lp0 (Rn ) with constant Cp0 independent of a and s. Thus, by (2.5) and (2.7), we have
(2.8)
kΛτ,Φ,s,a (f )kp0 ≤ Cp0
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 108, 8 pp.
p
βB kf kp0 .
http://jipam.vu.edu.au/
4
A HMAD A L -S ALMAN AND H USSAIN A L -Q ASSEM
By duality, we get
kΛτ,Φ,s,a (f )k(p0 )0 ≤ C(p0 )0
(2.9)
p
βB kf k(p0 )0 .
Therefore, by interpolation between (2.4), (2.8), and (2.9), we obtain (2.3). This concludes the
proof of the lemma.
Now we establish the following oscillatory estimates:
Lemma 2.2. Suppose that Ω ∈ L∞ (Sn−1 ) is a homogeneous function of degree zero on Rn
satisfying (1.1) and h(|x|) ∈ l∞ (Lq )(R+ ), 1 < q ≤ 2. Then for a complex number ρ with
Re(ρ) = α > 0, we have
Z
−αt
−n+ρ
−iξ·y
(2.10)
2
e
Ω(y)
|y|
h(|y|)dy
|y|≤2t
≤2
0
0
C
t
−ε
khkl∞ (Lq )(R+ ) kΩk11−2/q kΩk2/q
∞ (2 |ξ|) ,
α
and
(2.11)
Z
−αt
2
−iξ·y
e
−n+ρ
Ω(y) |y|
|y|≤2t
C
ε
h(|y|)dy ≤ 2 khkl∞ (Lq )(R+ ) kΩk1 (2t |ξ|)
α
for all 0 < ε < min{1/2, α}. The constant C is independent of Ω, α, and t.
R
0
Proof. For ξ ∈ Rn and r ∈ R+ , let G(ξ, r) = Sn−1 e−irξ·y Ω(y 0 )dσ(y 0 ). Then it is easy to see
that
Z
Z 2t−j
∞
−αt
X
−n+ρ
−iξ·y
−αj
(2.12) 2
e
Ω(y) |y|
h(|y|)dy ≤
2
|h(r)| |G(ξ, r)| r−1 dr.
t
t−j−1
|y|≤2
j=0
2
Using the assumption that 1 < q ≤ 2, it is straightforward to show that the right hand side of
(2.12) is dominated by
! 10
Z 2t−j
∞
q
X
1−2/q 0
2
−αj
−1
(2.13)
2 khkl∞ (Lq )(R+ ) kΩk1
2
|G(ξ, r)| r dr
.
2t−j−1
j=0
Now, for ξ ∈ Rn , y 0 , z 0 ∈ Sn−1 , j ≥ 0, and t ∈ R, set
Z 2t−j
0
0
0 0
Ij,t (ξ, y , z ) =
e−irξ·(y −z ) r−1 dr.
2t−j−1
Then, we have
! 10
Z
Z 2t−j
q
2 −1
2/q 0
≤ kΩk∞
(2.14)
|G(ξ, r)| r dr
10
q
|Ij,t (ξ, y , z )| dσ(y )dσ(z ) .
0
0
0
0
Sn−1 ×Sn−1
2t−j−1
By integration by parts, we have
(2.15)
|Ij,t (ξ, y 0 , z 0 )| ≤ (2t−j−1 |ξ| |ξ 0 · (y 0 − z 0 )|)−1 .
On the other hand, we have
(2.16)
|Ij,t (ξ, y 0 , z 0 )| ≤ ln 2.
Thus, by combining (2.15) and (2.16), we get
(2.17)
|Ij,t (ξ, y 0 , z 0 )| ≤ (2t−j−1 |ξ| |ξ 0 · (y 0 − z 0 )|)−ε
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Lp B OUNDEDNESS OF ROUGH PARAMETRIC M ARCINKIEWICZ F UNCTIONS
5
for 0 < ε < min{1/2, α}. Therefore, by (2.14) and (2.17), we obtain that
! 10
Z t−j
2
q
2
−1
|G(ξ, r)| r dr
(2.18)
2t−j−1
0
t−j−1
≤ kΩk2/q
|ξ|)−ε ,
∞ C(2
where the constant C is independent of Ω, j, and t. Moreover, since ε ≤ 1/2, it can be shown
that C is also independent of α. Hence by (2.12), (2.13), and (2.18), we get (2.10).
Now we prove (2.11). Using the cancellation property (1.1), it is clear that
1
Z
q0
−αt
2(ln
2)
−n+ρ
2
(2.19)
e−iξ·y Ω(y) |y|
h(|y|)dy ≤
khkl∞ (Lq )(R+ ) kΩk1 2t |ξ| .
α
|y|≤2t
On the other hand, we have
1
Z
−αt
2(ln 2) q0
−n+ρ
−iξ·y
khkl∞ (Lq )(R+ ) kΩk1 .
(2.20)
e
Ω(y) |y|
h(|y|)dy ≤
2
α
|y|≤2t
Thus, by interpolation between (2.19) and (2.20), we get (2.11). This completes the proof of
Lemma 2.2.
3. ROUGH PARAMETRIC M AXIMAL F UNCTIONS
In this section we shall establish the boundedness of certain maximal functions which will be
needed to prove our main result.
Theorem 3.1. Suppose that Ω ∈ L∞ (Sn−1 ) is a homogeneous function of degree zero on Rn
with kΩk1 ≤ 1 and kΩk∞ ≤ 2a for some a > 1. Suppose also that h(|x|) ∈ l∞ (Lq )(R+ ),
1 < q ≤ ∞ and let Mα be the operator defined as in (1.4). Then
aC
kf kp
(3.1)
kMα f kp ≤
α
for all 1 < p < ∞ with constant C independent of a, f , and α.
Proof. Since l∞ (Lq1 )(R+ ) ⊂ l∞ (Lq2 )(R+ ) whenever q2 ≤ q1 , it suffices to assume that 1 <
q ≤ 2. By a similar argument as in ([2]), choose a collection of C ∞ functions Φa = {ϕt :
n
−(t+1)a
t ∈ R} on Rn that
≤
the following properties: ϕ̂t is supported in {ξ ∈ R : 2
γsatisfies
−|γ|
n
d
ϕ̂
|ξ| ≤ 2−(t−1)a }, dξγ t (ξ) ≤ Cγ |ξ|
for any multi-index γ ∈ (N∪{0}) with constants Cγ
depending only on the underlying dimension and γ, and
X
(3.2)
ϕ̂t+j (ξ) = 1.
j∈Z
For t ∈ R, let {σt : t ∈ R} be the family of measures on Rn defined via the Fourier transform
by
Z
−αt
e−iξ·y |Ω(y)| |y|−n+ρ |h(|y|)| dy
(3.3)
σ̂t (ξ) = 2
|y|≤2t
Then it is easy to see that
(3.4)
Mα f (x) = sup{|σt | ∗ |f (x)|}.
t∈R
Now choose φ ∈ S(Rn ) such that φ̂(η) = 1 for |η| ≤ 12 , and φ̂(η) = 0 for |η| ≥ 1. Let
{τt : t ∈ R} be the family of measures on Rn defined via the Fourier transform by
(3.5)
τ̂t (ξ) = σ̂t (ξ) − φ̂(2t ξ)σ̂t (0).
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 108, 8 pp.
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6
A HMAD A L -S ALMAN AND H USSAIN A L -Q ASSEM
Then by Lemma 2.2, the choice of φ, the definitions of σt , τt , and the assumptions on Ω, we
have
(3.6)
|τ̂t (ξ)| ≤
C2la t
(2 |ξ|)−ε
α
for some l, ε > 0 . Moreover, it is easy to see that
kτt k ≤
(3.7)
C
α
Therefore by interpolation between (3.6) and (3.7), we get
(3.8)
|τ̂t (ξ)| ≤
ε
C t
(2 |ξ|)− a .
α
Now by (3.2), and the definitions of σt , and τt , it is easy to see that
√ X
(3.9)
Mα f (x) ≤ 2 a
Λτ,Φ,j,a (f )(x) + Cα−1 M H(f )(x),
j∈Z
(3.10)
√ X
τ ∗ (f )(x) ≤ 2 a
Λτ,Φ,j,a (f )(x) + Cα−1 M H(f )(x),
j∈Z
where M H stands for the Hardy-Littlewood maximal function on Rn , τ ∗ the maximal function
that corresponds to {τt : t ∈ R}, and Λτ,Φ,s,a is the operator defined by (2.1).
By (3.8), it is easy to see that
(3.11)
kΛτ,Φ,j,a (f )k2 ≤ C2−ε|j| α−1 kf k2
for all f ∈ L2 (Rn ). Therefore, by (3.10) and (3.11) we have
(3.12)
kτ ∗ (f )k2 ≤ Cα−1 a kf k2 .
Thus by (3.7), (3.8), (3.11), (3.12), and Lemma 2.1 with q = 2, we get
√
(3.13)
kΛτ,Φ,j,a (f )kp ≤ Cα−1 a kf kp
for p ∈ ( 43 , 4). Hence, by interpolation between (3.11) and (3.13), we obtain
√
0
(3.14)
kΛτ,Φ,j,a (f )kp ≤ Cα−1 a2−ε |j| kf kp
for p ∈ ( 34 , 4). Hence by (3.10) and (3.14), we get
(3.15)
kτ ∗ (f )kp ≤ Cα−1 a kf kp
for p ∈ ( 34 , 4). Next, by repeating the above argument with q = 43 + ε (ε → 0+ ), we get that
√
0
(3.16)
kΛτ,Φ,j,a (f )kp ≤ Cα−1 a2−ε |j| kf kp
(3.17)
kτ ∗ (f )kp ≤ Cα−1 a kf kp
for p ∈ ( 87 , 8). Now the result follows by successive applications of the above argument. This
completes the proof.
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 108, 8 pp.
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Lp B OUNDEDNESS OF ROUGH PARAMETRIC M ARCINKIEWICZ F UNCTIONS
7
4. P ROOFS OF T HE M AIN R ESULTS
Proof of Theorem 1.2. Suppose that Ω ∈ L(log+ L)(Sn−1 ) and h(|x|) ∈ l∞ (Lq )(R+ ), 1 < q ≤
∞. A key element in proving our results is decomposing the function Ω as follows (for more
information see [3]): For a natural number w, let Ew be the set of points x0 ∈ Sn−1 which
w+1
satisfy 2
≤ |Ω (x0 )| < 2w+2 . Also, we let E0 be the set of points x0 ∈ Sn−1 which satisfy
0
|Ω (x )| < 22 . Set bw = ΩχEw . Set D = {w : kbw k1 ≥ 2−3w } and define the sequence of
functions {Ωw }w∈D∪{0} by
Z
X
X Z
(4.1)
Ω0 (x) = b0 (x) +
bw (x) −
b0 (x)dσ(x) −
bw (x)dσ(x)v
Sn−1
w∈D
/
w∈D
/
Sn−1
and for w ∈ D,
(4.2)
−1
Ωw (x) = (kbw k1 )
Z
bw (x) −
bw (x)dσ(x) .
Sn−1
Then, it is easy to see that for w ∈ D∪ {0}, Ωw satisfies (1.1),
(4.3)
(4.4)
kΩw k1 ≤ C, kΩw k∞ ≤ C24(w+2) ,
X
Ω(x) =
θw Ωw (x),
w∈D∪{0}
where θ0 = 1, and θw = kbw k1 if w ∈ D.
ρ
For w ∈ D∪ {0}, let µΩw ,h be the operator defined as in (1.3) with Ω replaced by Ωw . Then
by (4.4), we have
X
ρ
ρ
θw µΩw ,h f (x).
(4.5)
µΩ,h f (x) ≤
w∈D∪{0}
Now, for w ∈ D∪ {0}, let τw = {τt,w : t ∈ R} be the family of measures on Rn defined via the
Fourier transform by
Z
−αt
(4.6)
τ̂t,w (ξ) = 2
e−iξ·y Ωw (y) |y|−n+ρ h(|y|)dy
|y|≤2t
and let Φw+2 = {ϕt : t ∈ R} be a collection of C ∞ functions on Rn defined as in the proof
of Theorem 3.1. Let Λτw ,Φw+2 ,j,w+2 , j ∈ Z be the operators given by (2.1). Then by a simple
change of variable we obtain
X
√
ρ
(4.7)
µΩw ,h f (x) ≤ w + 2
Λτw ,Φw+2 ,j,w+2 (f )(x).
j∈Z
Thus by Lemma 2.2, the properties of Ωw , Theorem 3.1, and Lemma 2.1, we get
ρ
µΩ ,h f ≤ (w + 2)C kf k
(4.8)
p
w
p
α
for all 1 < p < ∞.
Therefore, for 1 < p < ∞, by (4.7) and (4.8), we get




X
ρ µΩ,h f ≤ C
(w + 2)θw kf kp
p

α
w∈D∪{0}
≤
C
kΩkL(log L)(Sn−1 ) kf kp .
α
Hence the proof is complete.
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 108, 8 pp.
http://jipam.vu.edu.au/
8
A HMAD A L -S ALMAN AND H USSAIN A L -Q ASSEM
Proof of Theorem 1.3. A proof of Theorem 1.3 can be obtained using the decomposition (4.4)
and Theorem 3.1. We omit the details
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