MAPPING PROPERTIES OF THE MARCINKIEWICZ INTEGRALS ON HOMOGENEOUS GROUPS
Youngwoo Choi and Kyung Soo Rim
Abstract. Under the cancellation property and the Lipschitz con- dition on kernels, we prove that the Marcinkiewicz integrals de- fined on a homogeneous group H are bounded from H
1(H) to L
1(H), from L
∞c(H) to BM O (H), and from L
p(H) to L
p(H) for 1 < p < ∞ assuming the L
q-boundedness for some q > 1.
1. Introduction
Stein [8] defined a higher dimensional analogue of the Marcinkiewicz integral by
(1.1) µ Ω f (x) =
Z ∞ 0
|F t (x)| 2 dt t 3
12, where
F t (x) = Z
|y−x|<t
Ω (x − y)
|x − y| n−1 f (y) dy,
Ω is a homogeneous function of degree zero whose restriction to S n−1 belongs to Λ α S n−1 and satisfies the cancellation property,
Z
S
n−1Ω(x ′ ) dσ x ′
= 0.
Here, S n−1 denotes the unit sphere in R n and Λ α S n−1 denotes the Lipschitz space of order α on S n−1 . The continuity of Marcinkiewicz
Received June 26, 2000. Revised April 19, 2001.
2000 Mathematics Subject Classification: Primary 42B20; Secondary 47B38.
Key words and phrases: Marcinkiewicz integrals, homogeneous groups.
The first author was supported by grant No. 1999-2-112-004-3 from the Interdis-
ciplinary Research program of the KOSEF.
integrals is very useful in harmonic analysis [9, 10, 15]. Stein [8] proved that if Ω is in Λ α S n−1 with 0 < α ≤ 1, then
(1.2)
{x ∈ R n : µ Ω f (x) > λ}
≤ C
λ kf k L1
and
(1.3) kµ Ω f k Lp ≤ C p kf k Lp,
,
where 1 < p ≤ 2, and if Ω is an integrable odd function, then (1.4) kµ Ω f k Lp ≤ C p kf k Lp
for 2 < p < ∞.
The problem most immediately suggested by Marcinkiewicz [7], who conjectured the L p -boundedness of (1.1) for n = 1 and for Ω(t) = sign t until Zygmund [14] proved that the conjecture holds for 1 < p < ∞, has been extensively studied beginning with the 1958’s article of Stein [8].
Benedek, Calderon and Panzone [1] proved that if Ω ∈ C 1 (R n \ {0}) is a homogeneous function of degree zero satisfying the cancellation property, then µ Ω is bounded on L p (H) for 1 < p < ∞. Torchinsky and Wang considered the weighted L p -boundedness of µ Ω , and showed that if Ω is in Λ α S Q−1 and µ Ω is bounded on L p (R n ) for 1 < p < ∞, then for ω satisfying an A p condition, µ Ω is bounded on L p (ω) [13]. Further results on (1.1) were obtained when Ω satisfies some smoothness conditions [1], [8] and [13].
In this paper, we prove the H 1 -L 1 , L ∞ c -BM O and L p -L p (1 < p < ∞) boundedness of Marcinkiewicz integrals defined on homogeneous groups under the cancellation property and the Lipschitz condition on Ω and under the L q -boundedness of µ Ω .
This paper is organized as follows: in the next section, some prelim- inary materials are introduced. The main theorem is stated in Section 3. End-point results appear in Sections 4 and 5. Combined with an in- terpolation argument, the L p boundedness for 1 < p < ∞ will be shown in Section 6.
2. Preliminaries and notations
In this section, we introduce notations related to homogeneous groups
along with some preliminary materials. Mainly, we follow [6].
2.1. Homogeneous groups
A nilpotent Lie group H with a dilation group {δ r } r>0 is said to be a homogeneous group. The dilation group is given by
δ r = exp (A ln r)
with a suitable matrix A having positive eigenvalues.
H has a natural homogeneous norm | · |, and the homogeneous di- mension Q. Abusing the notation, the bi-invariant measure on H will be denoted by | · |.
Remark 2.1. Let H, {δ r }, | · | and Q be as above. The eigenvalues of the matrix defining {δ r } are listed as 1 = d 1 ≤ d 2 ≤ · · · ≤ d n and we let ¯ d = max {d i : i = 1, · · · , n}.
1. |δ r x| = r|x| for each x ∈ H, r > 0.
2. There exist C 1 , C 2 > 0 such that
C 1 kxk ≤ |x| ≤ C 2 kxk d ¯ whenever |x| ≤ 1.
Here, k·k denotes the euclidean norm.
3. There exists a constant γ > 0 such that for all x, y ∈ H,
|x ◦ y| ≤ γ(|x| + |y|) for all x ∈ H (2.1)
|x ◦ y| − |x|
≤ γ|y| for all x, y ∈ H such that |y| ≤ |x|/2.
(2.2)
4. |δ r E| = r Q |E|.
5. We let
S = {x ∈ H : |x| = 1} .
There is a unique Radon measure σ on S such that for all f ∈ L 1 (H),
Z
H
f (x) dx = Z ∞
0
Z
S
f (δ r y)r Q−1 dσ(y) dr.
2.2. The Hardy space H 1 (H)
For the definition of the Hardy space H 1 (H), we refer the interested
readers to [6].
2.2.1. H q,0 1 -atoms. Let q ∈ (1, ∞]. A function a(x) on H is said to be an H q,0 1 -atom (associated to a ball B) if it satisfies the following conditions:
(a) a(x) is supported in ¯ B;
(b) |a(x)| ≤ |B|
1q−1 almost everywhere; and (c) R
H a(x) dx = 0.
Remark 2.2. Let a(x) be an H ∞,0 1 -atom. Then we have kak L1(H) ≤ 1.
(2.3)
2.3. Atomic decomposition
An equivalent way of looking at H 1 (H) is the decomposition of ele- ments in H 1 (H) into H q,0 1 -atoms.
Theorem 2.1 (Decomposition Theorem). Let q ∈ (1, ∞]. For f ∈ H 1 (H) there exist a collection of H q,0 1 atoms {a k } k∈N and a sequence of nonnegative real numbers {λ k } k∈N with P ∞
k=1 λ k < ∞ so that f =
∞
X
k=1
λ k a k
in the sense of distributions, and we have kf k H1 ≈
∞
X
k=1
λ k .
2.4. BMO
Definition 2.3. A locally integrable function f : H → C is said to be in BM O if there exists a constant C such that for each ball B
1
|B|
Z
B
|f (x) − f B | dx ≤ C, holds, where
f B = 1
|B|
Z
B
f (x) dx.
3. Marcinkiewicz integrals
Let Ω be a measurable function on a homogeneous group H, which is homogeneous of degree 0 in the sense that
Ω (δ r x) = Ω(x)
holds for a.e. x ∈ H \ {0} and r > 0. We define the Marcinkiewicz integral µ Ω f as follows:
(3.1) µ Ω f (x) =
Z ∞
0
|F t (x)| 2 dt t 3
12, where
F t (x) = Z
B
t(x)
Ω x ◦ y −1
|x ◦ y −1 | Q−1 f (y) dy.
We will study the mapping properties of µ Ω . To be more specific, we will prove:
Theorem 3.1. Let H, Ω and µ Ω be as above. We assume the follow- ing:
• Ω
S ∈ Λ α (S);
• R
S Ω (x ′ ) dσ (x ′ ) = 0; and
• µ Ω is bounded in L q (H) for some q > 1.
Then the following inequalities hold:
kµ Ω f k L1 ≤ C 1 kf k H1, f ∈ H 1 (H) (3.2)
, f ∈ H 1 (H) (3.2)
kµ Ω f k BM O ≤ C ∞ kf k L∞, f ∈ L ∞ c (H) (3.3)
and
kµ Ω f k Lp ≤ C p kf k Lp, f ∈ L p (H) (3.4)
, f ∈ L p (H) (3.4)
for 1 < p < ∞.
4. H 1 -L 1 boundedness
In this section, we establish H 1 -L 1 boundedness of the Marcinkiewicz
integral. In view of Theorem 2.1 and the sublinearity of µ Ω , it suffices to
verify the inequality (3.2) when f is an arbitrary H ∞,0 1 -atom. Let a(x)
be an H ∞,0 1 -atom supported in B r (x 0 ). We split the integral into two parts,
Z
H
µ Ω a(x) dx = Z
B
2λr(x
0)
µ Ω a(x) dx + Z
H \B
2λr(x
0)
µ Ω a(x) dx
≡ I + II.
4.1. Estimation on I By hypothesis, we have (4.1)
Z
B
2λr(x
0)
[µ Ω a(x)] q dx . Z
B
r(x
0)
|a(x)| q dx . |B r (x 0 )| −q+1 . By H¨ older’s inequality and (4.1), we obtain
I ≤ Z
B
2λr(x
0)
[µ Ω a(x)] q dx
!
1q
|B 2λr (x 0 )|
q1′. 1.
4.2. Estimation on II
Before we proceed, we introduce some simple facts on balls in H.
Definition 4.1. For E ⊂ H and x 6∈ E, we will use the following notation.
d(x, E) = inf
x ◦ y −1
: y ∈ E . Lemma 4.2. Let x 6∈ B 2λr (x 0 ). Then we have
d (x, B r (x 0 )) ≥ r.
Proof. Suppose
d (x, B r (x 0 )) < r.
Then there exists y ∈ B r (x 0 ) such that
y ◦ x −1
< r. So we get
x ◦ x −1 0
≤ λ
x ◦ y −1 +
y ◦ x −1 0
< 2λr.
A contradiction to x 6∈ B 2λr (x 0 ).
Lemma 4.3. Let x 6∈ B 2λr (x 0 ) and y ∈ B r (x 0 ). Then we have
x ◦ y −1
≤ 2λ
x ◦ x −1 0
≤ 4λ 2
x ◦ y −1
.
Proof. Observe that x ◦ y −1
≤ λ
x ◦ x −1 0 +
x 0 ◦ y −1
≤ 2λ
x ◦ x −1 0
≤ 2λ 2
x ◦ y −1 +
y ◦ x −1 0
≤ 4λ 2
x ◦ y −1 , from
x ◦ y −1
≥ d (x, B r (x 0 )) ≥ r ≥
y ◦ x −1 0 .
Lemma 4.4. Let x 6∈ B 2λr (x 0 ) and t < d (x, B r (x 0 )). Then we have B r (x 0 ) ∩ B t (x) = ∅.
Proof. If y ∈ B r (x 0 ), then we have
y ◦ x −1
≥ d (x, B r (x 0 )) ≥ t.
Also, observe the following fact.
Fact 4.5. There exist constants C > 0, ε ∈ (0, 1) and ρ > 0 such that
δ s x ◦ x −1
≤ C|1 − s| ρ |x|
uniformly in x ∈ H and |1 − s| < ε.
Fix x ∈ H \ B 2λr (x 0 ). Then, by Lemma 4.2 we have d (x, B r (x 0 )) ≤
x ◦ x −1 0
≤ 2λd (x, B r (x 0 )) . We have
µ Ω a(x) 2 = Z ∞
d(x,B
r(x
0))
Z
B
r(x
0)∩B
t(x)
Ω x ◦ y −1
|x ◦ y −1 | Q−1 a(y) dy
2 dt t 3
= Z ∞
d(x,B
r(x
0))
Z
B
r(x
0)∩B
t(x)
Ω x ◦ y −1
t |x ◦ y −1 | Q−1 a(y) dy
· J t a(x) dt t 2 , where
J t a(x) = Z
B
r(x
0)∩B
t(x)
Ω x ◦ y −1
|x ◦ y −1 | Q−1 a(y) dy
.
Lemma 4.6. Let Ω, a, and B r (x 0 ) be as above. Then we have
Z
B
r(x
0)∩B
t(x)
Ω x ◦ y −1
t |x ◦ y −1 | Q−1 a(y) dy
. Ma(x), whenever t > d (x, B r (x 0 )), x ∈ H \ B 2λr (x 0 ).
Proof. There are two cases.
Case 1. t ≤ 2d (x, B r (x 0 )).
Z
B
r(x
0)∩B
t(x)
Ω x ◦ y −1
t |x ◦ y −1 | Q−1 a(y) dy
. kΩk ∞
t · d (x, B r (x 0 )) Q−1 Z
B
t(x)
|a(y)| dy . kΩk ∞
t Q Z
B
t(x)
|a(y)| dy . Ma(x).
Case 2. t ≥ 2 (d (x, B r (x 0 ))). From B r (x 0 ) ∩ B t (x) ⊂ B d(x,Br(x
0)) (x), we have
Z
B
r(x
0)∩B
t(x)
Ω x ◦ y −1
t |x ◦ y −1 | Q−1 a(y) dy
. kΩk ∞
d (x, B r (x 0 )) Q Z
B
d(x,Br(x0))(x
0)
|a(y)| dy . Ma(x),
which completes the proof.
For J t a(x) we have the following:
Lemma 4.7. With Ω, a, and B r (x 0 ) as above,
J t a(x) .
tMa(x)
if d (x, B r (x 0 )) ≤ t ≤ λ (d (x, B r (x 0 )) + 2λr) r ν
x ◦ x −1 0
−Q+1−ν
if t ≥ λ (d (x, B r (x 0 )) + 2λr)
for any x ∈ H \ B 2λr (x 0 ), where ν = min {α, ρα, 1}.
Proof. We have two cases.
Case 1. t ≥ λ (d (x, B r (x 0 )) + 2λr).
From B r (x 0 ) ⊂ B t (x), R
B
r(x
0) a(y) dy = 0 and the Lipschitz condition on Ω| S , we get
J t a(x) = Z
B
r(x
0)∩B
t(x)
Ω x ◦ y −1
|x ◦ y −1 | Q−1 a(y) dy
= Z
B
r(x
0)
"
Ω x ◦ y −1
|x ◦ y −1 | Q−1 − Ω x ◦ x −1 0
x ◦ x −1 0
Q−1
#
a(y) dy . Notice the following:
Ω x ◦ y −1
|x ◦ y −1 | Q−1 − Ω x ◦ x −1 0 x ◦ x −1 0
Q−1
≤
Ω x ◦ y −1 − Ω x ◦ x −1 0
|x ◦ y −1 | Q−1 +
Ω x ◦ x −1 0 ·
1
|x ◦ y −1 | Q−1 − 1 x ◦ x −1 0
Q−1
.
δ |x◦y−1|
−1 x ◦ y −1 ◦ δ | x◦x−01|
−1 x ◦ x −1 0 −1
|
−1x ◦ x −1 0 −1
α
x ◦ x −1 0
Q−1 +
y ◦ x −1 0
x ◦ x −1 0
Q . From (2.1), we obtain
δ
1|
x◦y−1| x ◦ y −1 ◦ δ 1
|
x◦x−01| x ◦ x −1 0 −1
.
δ
1|
x◦y−1|
x ◦ y −1 ◦ x ◦ x −1 0 −1 +
δ
1|
x◦y−1| x ◦ x −1 0 ◦ δ 1
|
x◦x−01| x ◦ x −1 0 −1
. r
|x ◦ y −1 | +
1 − | x◦x−01|
|x◦y
−1|
ρ
x ◦ x −1 0
x ◦ x −1 0
. r
x ◦ x −1 0
+ r ρ
x ◦ x −1 0
ρ ,
and so,
Ω x ◦ y −1
|x ◦ y −1 | Q−1 − Ω x ◦ x −1 0 x ◦ x −1 0
Q−1
. r α
x ◦ x −1 0
Q+α−1 + r ρα
x ◦ x −1 0
Q+ρα−1 + r
x ◦ x −1 0
Q
. r ν
x ◦ x −1 0
Q+ν−1 , since
x ◦ x −1 0
≥ 2λr. Therefore we obtain J t a(x) . r ν x ◦ x −1 0
Q−1+ν . Case 2. t ≤ λ (d (x, B r (x 0 )) + 2λr).
Since t ∼ d (x, B r (x 0 )), we get
J t a(x) ≤ kΩk ∞ Z
B
t(x)
1
|x ◦ y −1 | Q−1 |a(y)| dy
. 1
t Q−1 Z
B
t(x)
|a(y)| dy . tMa(x).
The proof of Lemma 4.7 is completed.
Thus, for x ∈ H \ B 2λr (x 0 ),
µ Ω a(x) 2 .
Z ∞ d(x,Br(x
0))
J t a(x) dt t 2
!
· Ma(x)
=
Z λ(d(x,Br(x
0))+2λr)
d(x,B
r(x
0))
J t a(x) dt t 2 +
Z ∞
λ(d(x,B
r(x
0))+2λr)
J t a(x) dt t 2
!
· Ma(x)
.
Z λ(d(x,Br(x
0))+2λr) d(x,B
r(x
0))
dt t
!
Ma(x) 2
+ Z ∞
λ(d(x,B
r(x
0))+2λr)
r ν x ◦ x −1 0
Q−1+ν
dt t 2
!
· Ma(x)
. r
d (x, B r (x 0 )) · Ma(x) 2 + r ν
x ◦ x −1 0
Q+ν · Ma(x), and so we obtain
µ Ω a(x) . r
12· Ma(x)
x ◦ x −1 0
1 2
+ r
ν2· Ma(x)
12x ◦ x −1 0
Q+ν 2
. Pick p 1 , p 2 , q 1 , and q 2 with the following properties:
• 2Q < p 1 < ∞;
• Q+ν 2Q < p 2 < 2;
• p 1
1
+ q 1
1
= 1; and
• p 12 + q 1
2
= 1.
From H¨ older’s inequality, the Maximal theorem, and (2.3), we obtain r
12Z
H \B
2λr(x
0)
Ma(x)
x ◦ x −1 0
1 2
dx
. r
12Z
H \B
2λr(x
0)
x ◦ x −1 0
−
p12dx
!
1p1
kMak q1
. r
12Z ∞
2r
ρ −p12+Q−1 dρ
1
p1
kak q1
. 1
and
r
ν2Z
H \B
2λr(x
0)
Ma(x)
12x ◦ x −1 0
Q+ν 2
dx
. r
ν2Z
H \B
2λr(x
0)
|x − x 0 | −(Q+ν)p22 dx
!
1p2
kMak
12q2 2. r
ν2Z ∞
2r
ρ −(Q+ν)p22 +Q−1 dρ
1
p2
kak
1 2q2
2
. 1.
This shows
II . 1.
Altogether, we obtain Z
H
µ Ω a(x) dx . 1 and the proof is complete.
5. L ∞ -BM O boundedness
In this section, we study the L ∞ -BM O boundedness of the Mar- cinkiewicz integrals. Let f ∈ L ∞ (H) be compactly supported and let B r (x 0 ) be any ball. We write
f = f χ B2λr(x
0) + f χ H \B
2λr(x
0)
≡ f 1 + f 2 .
Then f 1 ∈ L 2 (H). By H¨ older’s inequality and by hypothesis, Z
B
r(x
0)
µ Ω f 1 (x) dx ≤ |B r (x 0 )|
q1′Z
B
r(x
0)
[µ Ω f 1 (x)] q dx
!
1q. |B r (x 0 )|
q1′kf 1 k Lq(H)
. |B r (x 0 )|
q1′|B 2λr (x 0 )|
1qkf 1 k L∞
. |B r (x 0 )| · kf k L∞.
.
Thus we obtain
(5.1) 1
|B r (x 0 )|
Z
B
r(x
0)
µ Ω f 1 (x) dx . kf k L∞.
For x ∈ B r (x 0 ) we have
x ◦ y −1
> r whenever y ∈ H \ B 2λr (x 0 ). Let x ∈ B r (x 0 ) and let
III =
Z ∞
0
|F t (x) − F t (x 0 )| 2 dt t 3
12.
Then we have
III =
Z ∞
0
Z
B
t(x)
"
Ω x ◦ y −1
|x ◦ y −1 | Q−1 − Ω x 0 ◦ y −1
|x 0 ◦ y −1 | Q−1
#
f 2 (y) dy
2 dt t 3
1 2
=
Z ∞
r
Z
B
t(x)
"
Ω x ◦ y −1
|x ◦ y −1 | Q−1 − Ω x 0 ◦ y −1
|x 0 ◦ y −1 | Q−1
#
f 2 (y) dy
2 dt t 3
1 2
.
Z ∞
r
Z
B
2λt(x
0)\B
2λr(x
0)
r ν
|y − x 0 | Q−1+ν dy
! 2
dt t 3
1 2