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Canad. Math. Bull. Vol. 52 (4), 2009 pp. 521–534

The Parabolic Littlewood–Paley Operator
with Hardy Space Kernels
Yanping Chen and Yong Ding
Abstract. In this paper, we give the L p boundedness for a class of parabolic Littlewood–Paley g-function with its kernel function Ω is in the Hardy space H 1 (Sn−1 ).

1 Introduction
Let Rn be the Euclidean space with the routine norm |x| for each x ∈ Rn . Denote
by Sn−1 = {x ∈ Rn : |x| = 1} the unit sphere on Rn equipped with the Lebesgue
measure σ(x ′ ). Let α1 , . . . , αn be fixed real numbers with αi ≥ 1. It is easy to see
that for fixed x ∈ Rn , the function
F(x, ρ) =

n
X
xi 2
ρ2αi
i=1

is a strictly decreasing function of ρ > 0. Therefore, there exists a unique ρ(x) such
that F(x, ρ) = 1. It was proved in [7] that ρ(x) is a metric on Rn . For x ∈ Rn , set
x1 = ρα1 cos ϕ1 · · · cos ϕn−2 cos ϕn−1
x2 = ρα2 cos ϕ1 · · · cos ϕn−2 sin ϕn−1
..
.
xn−1 = ραn−1 cos ϕ1 sin ϕ2
xn = ραn sin ϕ1 .
′
′
α−1
Then dx = ρα−1 J(x
J(x ′ ) is the Jacobian of the above transPn)dρdσ(x ), and′ ρ
form, where α = i=1 αi and J(x ) = α1 x1′2 + · · · + αn xn′2 . It is easy to see that
J(x ′ ) ∈ C ∞ (Sn−1 ) with 1 ≤ J(x ′ ) ≤ M for some M ≥ 1. Without loss of generality,
we may assume αn ≥ αn−1 ≥ · · · ≥ α1 ≥ 1.
For t > 0, let At = diag[t α1 , . . . , t αn ]. Suppose that Ω(x) is a real valued and
measurable function defined on Rn . We say Ω(x) is homogeneous of degree zero
with respect to At , if for any t > 0 and x ∈ Rn

(1.1)

Ω(At x) = Ω(x).

Received by the editors October 5, 2006; revised February 19, 2009.
The research was supported by NSF of China (Grants: 10571015, 10826046) and RFDP of China
(Grant: 20050027025).
AMS subject classification: Primary: 42B20; secondary: 42B25.
Keywords: parabolic Littlewood-Paley operator, Hardy space, rough kernel.
c Canadian Mathematical Society 2009.

521

522

Y. Chen and Y. Ding
Moreover, we also assume that Ω(x) satisfies the following cancellation condition:
(1.2)

Z

Ω(x ′ ) J(x ′ )dσ(x ′ ) = 0.

Sn−1

In 1966, Fabes and Rivière [7; ] proved that if Ω ∈ C 1 (Sn−1 ) satisfies (1.1) and (1.2),
then the parabolic singular integral operator TΩ is bounded on L p (Rn ) for 1 <
p < ∞, where TΩ is defined by
TΩ f (x) = p.v.

Z

Rn

Ω(y)
f (x − y) dy.
ρ(y)α

In 1976, Nagel, Rivière and Wainger [9] improved the above result. They showed TΩ
is still bounded on L p (Rn ) for 1 < p < ∞ if replacing Ω ∈ C 1 (Sn−1 ) by a weaker
condition Ω ∈ L log+L(Sn−1 ).
On the other hand, in 1974, Madych considered the L p boundedness with respect
to the transform At of the Littlewood–Paley operator. Let ψ ∈ S(Rn ) satisfy ψ̂(0) =
0, where and below, ψ̂ denotes the Fourier transform of ψ. Let ψt (x) = t −α ψ(At −1 x)
for t > 0. Then the Littlewood–Paley operator related to At is defined by
gψ ( f )(x) =

Z

0

Theorem A

∞

|ψt ∗ f (x)|2

dt  1/2
.
t

[8] The Littlewood–Paley operator gψ is of type (p, p) for 1 < p < ∞.

Inspired by the works in [7–9], recently Ding, Xue and Yabuta [5] improved the
above result. More precisely, the authors in [5] proved that the parabolic Littlewood–
Paley operator is still bounded on L p if ψ(x)is replaced by a kernel function φ(x) =
Ω(x)ρ(x)−α+1 χ{ρ(x)≤1} (x) with Ω ∈ Lq (Sn−1 ) (q > 1) satisfying (1.1) and (1.2).
Theorem B [5] If Ω ∈ Lq (Sn−1 )(q > 1) satisfies (1.1) and (1.2), then gφ is of type
(p, p) for 1 < p < ∞.
Notice that on the unit sphere Sn−1 , there are the following containing relationships:
C ∞ $ Lq (q > 1) $ L log+ L $ H 1 $ L1 ,
where H 1 denotes the Hardy space on Sn−1 (see §2 for its definition). Hence, a natural
question is whether the size condition assumed on Ω can be weakened further. The
purpose of this paper is to give a positive answer to this question.
Theorem 1.1
1 < p < ∞.

If Ω ∈ H 1 (Sn−1 ) satisfies (1.1) and (1.2), then gφ is of type (p, p) for

Remark. If α1 = · · · = αn = 1, then ρ(x) = |x| and α = n. In this case, gφ = µΩ
and the latter is just the classical Marcinkiewicz integral, which was studied by many
authors. (See [1, 4, 10], for example.) Moreover, note also that the Ω in Theorem 1.1
(also Theorem B) has no any smoothness on Sn−1 .

The Parabolic Littlewood–Paley Operator with Hardy Space Kernels

523

2 Definitions and Lemmas
Let us begin with the definition of Hardy space H 1 (Sn−1 ). For f ∈ L1 (Sn−1 ) and
x ′ ∈ Sn−1 , we denote
Z
+
′
P f (x ) = sup
f (y ′ )Ptx ′ (y ′ ) dσ(y ′ ) ,
0<t<1

where Ptx ′ (y ′ ) =

1−t 2
|y ′ −tx ′ |n

Sn−1

for y ′ ∈ Sn−1 . Then

H 1 (Sn−1 ) = { f ∈ L1 (Sn−1 ) : kP+ f kL1 (Sn−1 ) < ∞},
and we define k f kH1 (Sn−1 ) = kP+ f kL1 (Sn−1 ) if f ∈ H 1 (Sn−1 ).
A very useful characterization of the space H 1 (Sn−1 ) is its atomic decomposition.
Let us first recall the definition of atoms. A regular H 1 (Sn−1 ) atom is a function a(x ′ )
on L∞ (Sn−1 ) satisfying the following conditions:
(2.1)

supp(a) ⊂ Sn−1
∩ {y ∈ Rn : |y − ξ ′ | < r for some ξ ′ ∈ Sn−1 and r ∈ (0, 2]};

(2.2)

Z

a(x ′ )Y m (x ′ ) dσ(x ′ ) = 0

Sn−1

for any spherical harmonic polynomial Y m with degree m ≤ N, where N is any fixed
integer;
kakL∞ (Sn−1 ) ≤ r1−n .

(2.3)

An exceptional H 1 (Sn−1 ) atom u(x ′ ) is an L∞ (Sn−1 ) function bounded by 1.
From [3], we find that any Ω ∈ H 1 (Sn−1 ) has an atomic decomposition
Ω=

∞
X
j=1

λjaj +

∞
X

δi u i ,

i=1

where each a j is a regular H 1 (Sn−1 ) atom and each ui is an exceptional atom. Moreover,
∞
∞
X
X
|δi | ≤ CkΩkH1 (Sn−1 ) .
|λ j | +
i=1

j=1

We note that for any x ′ ∈ Sn−1 ,
∞
X
i=1

δi ui (x ′ ) ≤

∞
X
i=1

|δi |.

524

Y. Chen and Y. Ding
Without loss of generality, we can assume
∞
X

δi ui (x ′ ) ≤ kΩkH1 (Sn−1 ) .

i=1

Thus we write

′

with ω(x ) =
(2.4)

P∞

∞
X
i=1

i=1 δi ui (x

Ω(x ′ ) =

δi ui (x ′ ) = kΩkH1 (Sn−1 ) ω(x ′ ),

∞
X

′

)/kΩkH1 (Sn−1 ) . In this new definition, for x ′ ∈ Sn−1 ,

λ j a j (x ′ ) + kΩkH1 (Sn−1 ) ω(x ′ )

and

kωkL∞ (Sn−1 ) ≤ 1.

j=1

The following Lemmas 2.1 and 2.2 can be found in [6].
Lemma 2.1
(2.5)

[6] Suppose that n ≥ 3 and b satisfies (2.1), (2.3), and
Z
b(y ′ ) dσ(y ′ ) = 0.
Sn−1

Let
2 (n−3)/2

Fb (s) = (1 − s )

χ(−1,1) (s)

Z

Sn−2

Gb (s) = (1 − s2 )(n−3)/2 χ(−1,1) (s)

Z

Sn−2

b(s, (1 − s2 )1/2 ye)dσ(e
y ),

|b(s, (1 − s2 )1/2 ye)|dσ(e
y ).

Then there exists a constant C, independent of b, such that
(2.6)

supp(Fb ) ⊂ (ξ1′ − 2r(ξ ′ ), ξ1′ + 2r(ξ ′ )),

(2.7)

supp(Gb ) ⊂ (ξ1′ − 2r(ξ ′ ), ξ1′ + 2r(ξ ′ )),

(2.8)

kFb k∞ ≤ C/r(ξ ′ ), kGb k∞ ≤ C/r(ξ ′ ),
Z
Fb (s) ds = 0,

(2.9)

R

where r(ξ ′ ) = |ξ|−1 |Lr ξ| and Lr ξ = (r2 ξ1 , rξ2 , . . . , rξn ).
[6] Suppose that n = 2 and b satisfies (2.1), (2.3) and (2.5). Let

Fb (s) = (1 − s2 )−1/2 χ(−1,1) (s) b(s, (1 − s2 )1/2 ) + b(s, −(1 − s2 )1/2 ) ,

Gb (s) = (1 − s2 )−1/2 χ(−1,1) (s) |b(s, (1 − s2 )1/2 )| + |b(s, −(1 − s2 )1/2 )| .

Lemma 2.2

Then Fb (s) satisfies (2.6) and (2.9), and kFb kq ≤ C|Lr (ξ ′ )|−1+1/q . And Gb (s) satisfies
(2.7) and kGb kq ≤ C|Lr (ξ ′ )|−1+1/q for some q ∈ (1, 2).

The Parabolic Littlewood–Paley Operator with Hardy Space Kernels
Lemma 2.3

525

[5] For Ω ∈ L1 (Sn−1 ), denote
σ2t (x) = 2−t Ω(x)ρ(x)−α+1 χ{ρ(x)≤2t } (x),

and σ ∗ ( f )(x) = supt∈R ||σ2t | ∗ f (x)|. Then kσ2t k1 ≤ C and kσ ∗ ( f )k p ≤ Ck f k p for
1 < p < ∞, where the constant C is independent of f and t.
Lemma 2.4 [5] Suppose that m denotes the distinct numbers of {α j }. Then for any
x, y ∈ Rn , 0 ≤ β ≤ 1
Z 2
β
dλ
e−iAλ x·y
≤ C|x · y|− m ,
λ
1
where C > 0 is independent of x and y.

3 Proof of Theorem 1.1
Since Ω ∈ H 1 (Sn−1 ) satisfies the cancellation condition (1.2), by (2.4) we can
write
∞
X
λ j a j (x ′ ) + kΩkH1 (Sn−1 ) ω(x ′ ),
Ω(x ′ ) =
j=1

1

where each a j is a regular H (Sn−1 ) atom and kωkL∞ (Sn−1 ) ≤ 1. Moreover,
∞
X

|λ j | ≤ CkΩkH1 (Sn−1 ) .

j=1

For y ∈ Rn (y 6= 0), we write
Ω(y) =

∞
X

λ j ã j (y) + kΩkH1 (Sn−1 ) ω̃(y),

j=1

where ã j (y) = a j (Aρ(y)−1 y) and ω̃(y) = ω(Aρ(y)−1 y). It is easy to check that ω̃(y ′ ) =
ω(y ′ ), ã j (y ′ ) = a j (y ′ ) for y ′ ∈ Sn−1 and ω̃ and ã j satisfy (1.1) for any t > 0 and
y ∈ Rn .
x
Noticing that J( |x|
)|x|2 is a homogeneous polynomial of degree 2 on Rn by [11,
Theorem 2.1], we can write
 x 
J
|x|2 = P2 (x) + |x|2 P0 (x),
|x|
where Pk (x) is a harmonic polynomial of degree k (k = 0, 2). Then J(x ′ ) = P2 (x ′ ) +
P0 (x ′ ), where Pk (x ′ ) is a spherical harmonic polynomial of degree k (k = 0, 2). So by
(2.2), we have
(3.1)

Z

a j (y ′ ) J(y ′ ) dσ(y ′ )

Sn−1

=

Z

Sn−1

′

′

′

a j (y )P2 (y ) dσ(y ) +

Z

Sn−1

a j (y ′ )P0 (y ′ ) dσ(y ′ ) = 0,

526

Y. Chen and Y. Ding
for all j = 1, 2, . . . . Thus by (2.4) and (3.1), we know
Z

(3.2)

ω(y ′ ) J(y ′ ) dσ(y ′ ) = 0.
Sn−1

Therefore,
(3.3)

kgφ ( f )k p ≤

∞
X

|λ j |kga j ( f )k p + kΩkH1 (Sn−1 ) kgω ( f )k p ,

j=1

where
ga j ( f )(x) =

Z

∞

ρ(y)≤t

0

gω ( f )(x) =

Z

Z

∞

0

Z

ρ(y)≤t

ã j (y)
f (x − y) dy
ρ(y)α−1
ω̃(y)
f (x − y) dy
ρ(y)α−1

2 dt  1/2

t3

2 dt  1/2

t3

,
.

Since ω(x ′ ) ∈ L∞ (Sn−1 ) and satisfies the cancellation condition (3.2), by Theorem B
we get
kgω ( f )k p ≤ Ck f k p ,

(3.4)

where C is independent of ω and f . Thus, to prove Theorem 1.1, by (3.3) and (3.4)
it suffices to show that there exists C > 0, independent of the atoms a j and f , such
that for j = 1, 2, . . . ,
kga j ( f )k p ≤ Ck f k p .

(3.5)

We only prove (3.5) for the case n > 2. The case for n = 2 can be dealt with using
the same method and Lemma 2.2. From now we denote simply a j , ã j and ga j by a, ã,
and ga , respectively. Without loss of generality, we may also assume that supp(a) is
contained in B(1, r) ∩ Sn−1 , where B(1, r) = {y : |y − 1| < r} and 1 = (1, 0, . . . , 0).
Choose a C 0∞ (Rn ) function ϕ such that Rϕ(x) = ϕ(ρ(x)), 0 ≤ ϕ ≤ 1 satisfying
∞
supp(ϕ) ⊂ {y : 1/2 ≤ ρ(y) ≤ 2} and 0 ϕ(t)/t dt = 1. Define functions Φ
b
b
and ∆ by Φ(ξ)
= ϕ(ρ(Lr ξ)) and ∆(ξ)
= ϕ(ρ(ξ)), respectively, where Lr ξ is defined in Lemma 2.1. If we denote Φt (x) = t −α Φ(At −1 x) and ∆t (x) = t −α ∆(At −1 x),
ct (ξ) = ϕ(tρ(Lr ξ)), ∆
ct (ξ) = ϕ(tρ(ξ)), and Φt (x) =
then it is easy to check that Φ
1 −α
−1
−1
t
∆(L
A
x),
where
r
t
r n+1
Lr−1 At −1 x = (r−2t −α1 x1 , r−1t −α2 x2 , . . . , r−1t −αn xn ).
For any f ∈ S(Rn ), by taking Fourier transform we have
(3.6)

f (x) =

Z

∞

−∞

Φ2t ∗ f (x)dt ∼

Z

0

∞

Φt ∗ f (x)

dt
.
t

The Parabolic Littlewood–Paley Operator with Hardy Space Kernels

527

Define
gΦ ( f )(x) =

Z

∞

|Φt ∗ f (x)|2

0

dt  1/2 
∼
t

Z

∞

|Φ2t ∗ f (x)|2 dt

−∞

 1/2

.

Now we claim that
kgΦ ( f )k p ≤ Ck f k p ,

(3.7)

with C independent of r > 0. In fact, by the definition of Φt , we have
Z
1
∆(Lr−1 At −1 y) f (x − y) dy
Φt ∗ f (x) = n+1 t −α
r
Rn
Z
= t −α
∆(At −1 y) f (Lr (Lr−1 x − y)) dy
Rn

= ∆t ∗ h(Lr−1 x),
R
b
= ϕ(0) = 0, by Theorem A we get
where h(x) = f (Lr x). Since Rn ∆(x) dx = ∆(0)
Z
 ∞
dt  1/2
kgΦ ( f )k p =
|Φt ∗ f (·)|2
t
p
0
nZ Z ∞
dt  p/2 o 1/p
=
|∆t ∗ h(Lr−1 x)|2
dx
t
Rn
0
Z Z ∞
n
dt  p/2 o 1/p
dx
= rn+1
|∆t ∗ h(x)|2
t
Rn
0
n+1

≤ Cr p khk p
Z

 1/p
n+1
=C r
= Ck f k p .
| f (Lr x)| p dx
Rn

This is (3.7). Now we denote σ2t (y) = 2−t ã(y)ρ(y)−α+1 χ{ρ(y)≤2t } (y). Then
Z ∞ Z
2 dt  1/2
ã(y)
f
(x
−
y)dy
ga ( f )(x) =
α−1
t3
0
ρ(y)≤t ρ(y)
Z ∞
 1/2
∼
|σ2t ∗ f (x)|2 dt
.
−∞

By (3.6) and the Minkowski inequality, we obtain
Z ∞ Z ∞
2  1/2
ga ( f )(x) ∼
Φ2s+t ∗ σ2t ∗ f (x)ds dt
≤

Z

−∞

∞

−∞

=:

Z

Z

−∞

∞

|σ2t ∗ Φ2s+t ∗ f (x)|2 dt

−∞

∞

−∞

Qs ( f )(x) ds.

 1/2

ds

528

Y. Chen and Y. Ding
Using Minkowski’s inequality again yields
(3.8)

kga ( f )k p ≤ C

Z

∞

kQs ( f )k p ds +

Z

0

−∞

0


kQs ( f )k p ds .

By (3.8), it is easy to see that the proof of (3.5) can be reduced to show the following
estimates
(
C2−sγ k f k p
kQs ( f )k p ≤
C2sτ k f k p

(3.9)

for s > 0,
for s < 0,

where τ and γ are some positive constants, and C is independent s and f .
The proof of (3.9) will be completed in two steps.
Step 1: There exists C > 0, independent of s and f , such that
kQs ( f )k p ≤ Ck f k p

(3.10)

for 1 < p < ∞.

First we consider the case 1 < p < 2. Denote Gs+t (x) = Φ2s+t ∗ f (x). Since a(x ′ ) ∈
L1 (Sn−1 ), by Lemma 2.3, we know kσ2t k1 ≤ C, then
Z

(3.11)

∞

σ ∗ Gs+t (·)dt
2t

−∞

1

≤C

Z

∞

Gt (·)dt

−∞

1

.

On the other hand, for 1 < q < ∞, also by Lemma 2.3, we get
(3.12)

k sup |σ2t ∗ Gs+t |kq ≤ kσ ∗ (sup |Gt |)kq ≤ Ck sup |Gt |kq .
t∈R

t∈R

t∈R

If we define TGs+t (x) = σ2t ∗Gs+t (x), then (3.11) and (3.12) show that T is a bounded
operator on L1 (L1 (R), Rn ) and Lq (L∞ (R), Rn ), respectively. Since 1 < p < 2, we can
take q > 1 such that 1/q = 2/p − 1. Then by using the operator interpolation
theorem between (3.11) and (3.12), we know that the operator T is also bounded on
L p (L2 (R), Rn). That is
Z

∞

|σ2t ∗ Gs+t (·)|2 dt

−∞

 1/2

p

≤C

Z

∞

−∞

|Gt (·)|2 dt

 1/2

p

.

From this and (3.7), we prove (3.10) for 1 < p < 2. Moreover, by (3.7) and the L2
boundedness of σ ∗ , (3.10) holds for the case p = 2. Now let us deal with the case
p > 2. Let q = (p/2) ′ . Then
kQs f k2p

= sup
ν

Z Z
Rn

∞

|σ2t ∗ Φ2s+t ∗ f (x)|2 ν(x) dt dx ,
−∞

The Parabolic Littlewood–Paley Operator with Hardy Space Kernels

529

where the supremum is taken over all ν(x) ∈ Lq (Rn ) with kνkq ≤ 1. Applying
Hölder’s inequality and noting the fact kσ2t k1 ≤ C,
Z Z ∞
|σ2t ∗ Φ2s+t ∗ f (x)|2 ν(x) dt dx
Rn

−∞

≤

Z

∞

−∞

Z Z
Rn

|Φ2s+t ∗ f (y)|2 |σ2t (x − y)| dy

Rn

×

Z

|σ2t (x − y)| dy

Rn

≤ kσ2t k1
≤C

Z

Z

−∞

∞

−∞

=C

∞

Z Z
Rn

Z

Z Z
Rn

 1/2  2

 1/2

|ν(x)| dx dt

|Φ2s+t ∗ f (y)|2 |σ2t (x − y)||ν(x)| dy dx dt

Rn

|Φ2t ∗ f (y)|2 σ ∗ (|ν|)(y) dy dt

Rn

∞

|Φ2t ∗ f (y)|2 dt σ ∗ (|ν|)(y) dy,

−∞

where C is independent of s, f and ν. Using Hölder’s inequality again and (3.7),
Lemma 2.3, we obtain
kQs f k2p ≤ C sup kgΦ ( f )k2p kσ ∗ (|ν|)kq ≤ Ck f k2p .
ν

Thus we have (3.10) for p > 2. From the proof of (3.10) above, it is easy to check
that the constant C is independent of s and f .
Step 2: There exists C > 0, independent of f and s, such that
(
C2−s k f k2
for s > 0,
(3.13)
kQs ( f )k2 ≤
βs/m
C2
k f k2 for s < 0,
where 0 < β < 2α1 n and m denotes the distinct numbers of {α j }.
By Plancherel’s theorem,
Z ∞Z
(3.14)
kQs f k22 ≤
| fb(ξ)|2 |ϕ(2s+t ρ(Lr ξ))|2 |c
σ2t (ξ)|2 dξdt,
Rn

−∞

where

−t
σc
2t (ξ) = 2
1

n−1

Z

0

2t

Z

′

a(y ′ ) J(y ′ )e−2πiξ·Aρ y dσ(y ′ )dρ

Sn−1

and a is a regular H (S ) atom supported in B(1, r) ∩ Sn−1 ,where 1 = (1, 0, . . . , 0).
We first give the estimate of |c
σ2t (ξ)|. Let η(y ′ ) = a(y ′ ) J(y ′ )/k JkL∞ (Sn−1 ) . By (3.1)
′
∞ n−1
and J(y ) ∈ C 0 (S ), we know η(y ′ ) satisfies (2.3) and (2.5), and supp(η) ⊂
B(1, r) ∩ Sn−1 . Then
Z tZ
′
k JkL∞ (Sn−1 ) 2
η(y ′ )e−2πiξ·Aρ y dσ(y ′ )dρ.
(3.15)
σc
2t (ξ) =
t
2
0
Sn−1

530

Y. Chen and Y. Ding
In the following, we want to prove |c
σ2t (ξ)| ≤ C min{|Lr A2t ξ|, |Lr A2t ξ|−β/m }, where
0 < β < 2α1 n and m denotes the distinct numbers of {α j }. For any ξ 6= 0, denote
Aρ ξ
|Aρ ξ|

=: ζ := (ζ1′ , ζ∗ ) ∈ Sn−1 , where ζ∗ ∈ Rn−1 . We choose a rotation O in Rn such
that O(ζ) = 1. Since O−1 = Ot , where O−1 and Ot denote the inverse and transpose
of O, respectively, it is easy to check that ζ is the first row vector of O. Thus, we have
O2 (ζ) = (ζ1′ , γ∗ ), where γ∗ ∈ Rn−1 . Now, we take a rotation Qn−1 in Rn−1 such
0
that Qn−1 (ζ∗ ) = γ∗ . Set R = 10 Qn−1
; then R is a rotation in Rn , such that for any
′
′
′
n−1
′
y := (ℓ, y 2 , . . . , y n ) in S , h1, Ry i = ℓ. Thus
k JkL∞ (Sn−1 )
σc
2t (ξ) =
2t

t
Z 2Z

0

′

η(O−1 (Ry ′ ))e−2πi|Aρ ξ|h1,Ry i dσ(y ′ )dρ.

Sn−1

Now η(O−1 (Ry ′ )) also satisfies (2.3) and (2.5), and is supported in B(ζ, r) ∩ Sn−1 .
Thus we have
t
Z 2Z

k JkL∞ (Sn−1 )
σc
2t (ξ) =
2t

0

Fη (ℓ)e−2πi|Aρ ξ|ℓ dℓdρ,

R

where Fη (ℓ) is the function defined in Lemma 2.1. By Lemma 2.1, we know that Fη
α1
|L A ξ|
ξ1
. Thus
is supported in (−2r(ζ) + δ1 , 2r(ζ) + δ1 ), where r(ζ) = |Ar ρ ρξ| and δ1 = ρ|Aρ ξ|
δ1
δ1
N(ℓ) = r(ζ)Fη (r(ζ)ℓ) is a function with support in the interval (−2 + r(ζ)
, 2 + r(ζ)
),
R
and kNk∞ < C (C is independent of η and ρ) and R N(ℓ) dℓ = 0. After changing a
variable we have

σc
2t (ξ) =

k JkL∞ (Sn−1 )
2t

t
Z 2Z

0

N(ℓ)e−2πiℓ|Lr Aρ ξ| dℓdρ.

R

So by the cancellation property of N, we obtain that

k JkL∞ (Sn−1 )
(3.16) |c
σ2t (ξ)| =
2t
t
Z 2Z
≤ C2−t
0

t
Z 2Z

0

ξ1

] dℓdρ

R

ζ

1 |
|ℓ− r(ζ)

≤ 2|N(ℓ)||Lr Aρ ξ| ℓ −
≤ C|Lr A2t ξ|.

α1

N(ℓ)[e−2πi|Lr Aρ ξ|ℓ − e−2πiρ

ζ1
dℓ dρ ≤ C
r(ζ)

Z

0

1

|Lr A2t ρ ξ| dρ

The Parabolic Littlewood–Paley Operator with Hardy Space Kernels

531

On the other hand, using Hölder’s inequality and (3.15), we have
t
Z Z
2
k JkL∞ (Sn−1 ) 2
′ −2πiξ·Aρ y ′
′
σ2t (ξ)| =
η(y
)e
dσ(y
)
dρ
(3.17)|c
2t
0 Sn−1
t
Z
Z
2
′
1 2
≤C t
η(y ′ )e−2πiξ·Aρ y dσ(y ′ ) dρ
2 0
Sn−1
Z 2t+ j Z
0
X
′
1
2 j−1 t+ j−1
=C
η(y ′ )e−2πiξ·Aρ y dσ(y ′ )
2
n−1
t+
j−1
S
2

2

2

dρ

j=−∞
0
X

≤C

2 j−1

0
X

2t+ j

2t+ j−1

j=−∞

=C

Z

Z

′

η(y ′ )e−2πiξ·Aρ y dσ(y ′ )

2 dρ

ρ

Sn−1

2 j−1 Bt, j (ξ),

j=−∞

where

Z

Bt, j (ξ) =

2t+ j

2t+ j−1

Then we get
Bt, j (ξ) =

Z

2t+ j

2t+ j−1

ZZ

≤C

ZZ

Z

′

η(y ′ )e−2πiξ·Aρ y dσ(y ′ )

2 dρ

ρ

Sn−1

η(y ′ )η(x ′ )e−2πiAρ (y

′

−x ′ )·ξ

.

dσ(y ′ )dσ(x ′ )

Sn−1 ×Sn−1

|η(y ′ )||η(x ′ )|

Sn−1 ×Sn−1

Z

2t+ j

e−2πiAρ (y

′

−x ′ )·ξ dρ

ρ

2t+ j−1

dρ
ρ

dσ(y ′ )dσ(x ′ ).

By Lemma 2.4, we know
Z

2t+ j

e−2πiAρ (y

′

−x ′ )·ξ dρ

2t+ j−1

ρ

=

Z

2

e−2πiA2t+ j−1 ρ (y

′

−x ′ )·ξ dρ

ρ

−2β/m
≤ C |(y ′ − x ′ ) · A2t+ j−1 ξ|
,
1

where 0 < β < 2α1 n and m denotes the distinct numbers of {α j }. Then by the above
inequality we get
(3.18) Bt, j (ξ) ≤ C

ZZ

|η(y ′ )||η(x ′ )|

Sn−1 ×Sn−1

where
I1 (ξ) =

ZZ

Sn−1 ×Sn−1

 −2β/m
× |(y ′ − x ′ ) · A2t+ j−1 ξ|
dσ(y ′ )dσ(x ′ ) = CI1 (ξ),
 −2β/m
|η(y ′ )||η(x ′ )| |(y ′ − x ′ ) · A2t+ j−1 ξ|
dσ(y ′ )dσ(x ′ ).

532

Y. Chen and Y. Ding
As was done above, for any ξ 6= 0, we choose a rotation O in Rn such that
O(A2t+ j−1 ξ) = |A2t+ j−1 ξ|1 = |A2t+ j−1 ξ|(1, 0, . . . , 0).
Thus, we may take another rotation R in Rn such that for any
y ′ = (y 1′ , y 2′ , . . . , y n′ ) ∈ Sn−1 ,
h1, Ry ′ i = y 1′ = h1, y ′ i. Now, let y ′ = (s, y 2′ , y 3′ , . . . , y n′ ), x ′ = (δ, x2′ , x3′ , . . . , xn′ ).
Then it is easy to see that
ZZ
|η(O−1 (Ry ′ ))||η(O−1 (Rx ′ ))|
I1 (ξ) =
Sn−1 ×Sn−1

 −2β/m
× |(y ′ − x ′ ) · |A2t+ j−1 ξ|1|
dσ(y ′ )dσ(x ′ ),

where O−1 is the inverse of O. Now η(O−1 (Ry ′ )) satisfies (2.3) and (2.5), and is
ξ
A
supported in B(ϑ, r) ∩ Sn−1 where ϑ = |A2t+t+ j−1
ξ| . Thus we have
2 j−1
ZZ
 −2β/m
Gη (s)Gη (δ) |A2t+ j−1 ξ||s − δ|
dsdδ,
I1 (ξ) =
R ×R

where Gη (s) is the function defined in Lemma 2.1. By Lemma 2.1, we know
|L A j−1 ξ|
2(t+ j−1)α1 ξ1
supp(Gη ) ⊂ (−2r(ϑ) + ϑ1 , 2r(ϑ) + ϑ1 ), where r(ϑ) = |Ar t+2t+j−1
ξ| and ϑ1 = |A2t+ j−1 ξ| .
2

ϑ1
) is a function supported in the interval (−2, 2),
Thus ϕ(s) = r(ϑ)Gη r(ϑ)(s − r(ϑ)
and kϕk∞ < C (C is independent of r, t, j and ϑ ). Since 2β/m < 1, we get
Z 2Z 2
 −2β/m
I1 (ξ) =
ϕ(s)ϕ(δ) |Lr A2t+ j−1 ξ||s − δ|
dsdδ
−2

−2

≤ C|Lr A2t+ j−1 ξ|−2β/m

Z

2

−2

Z

2

|s − δ|−2β/m dsdδ

−2

≤ C|Lr A2t+ j ξ|−2β/m .
This together with (3.18) gives
Bt, j (ξ) ≤ C|Lr A2t+ j ξ|−2β/m .

(3.19)
Since 0 < β <
(3.20)

1
2αn

and m ≥ 1, then by (3.17) and (3.19), we get
|c
σ2t (ξ)|2 ≤ C

0
X

2 j−1 |Lr A2t+ j ξ|−2β/m

j=−∞

≤C

0
X

2 j(1−2βαn /m) |Lr A2t ξ|−2β/m

j=−∞

≤C

0
X

2 j(1−2βαn /m) |Lr A2t ξ|−2β/m

j=−∞

≤ C|Lr A2t ξ|−2β/m .

The Parabolic Littlewood–Paley Operator with Hardy Space Kernels

533

By (3.16) and (3.20), we have
|c
σ2t (ξ)| ≤ C min{|Lr A2t ξ|, |Lr A2t ξ|−β/m }.
Now we give the estimates kQs ( f )k2 . For s > 0, by (3.14) and the properties of ϕ,
using the estimate |c
σ2t (ξ)| ≤ C|Lr A2t ξ| and the Plancherel theorem, we get

kQs ( f )k22 ≤ C

=C

≤C

×

1
rn+1
1
rn+1
Z

≤ C2

Z

Z

−∞

∞

−∞

Z

∞

∞

−∞

2−s−1 ≤2t ρ≤2−s+1

Z

Sn−1

ρ
−s−1− log
log 2

−∞

≤ C2

1
rn+1

Z

Rn

Z

Sn−1

| fb(ξ)|2 |Lr A2t ξ|2 dξdt

J(ξ ′ )| fb(Lr−1 Aρ ξ ′ )|2 |A2t Aρ ξ ′ |2 ρα−1 dσ(ξ ′ )dρdt

J(ξ ′ )| fb(Lr−1 Aρ ξ ′ )|2 (2−s+1 )2α1 + · · · + (2−s+1 )2αn




dt ρα−1 dσ(ξ ′ )dρ

Z∞ Z

1
rn+1

−2s

2−s−1 ≤ρ(Lr A2t ξ)≤2−s+1

Z

ρ
−s+1− log
log 2

−2sα1

Z

Sn−1

J(ξ ′ )| fb(Lr−1 Aρ ξ ′ )|2 ρα−1 dσ(ξ ′ )dρ

| fb(Lr−1 ξ)|2 dξ

≤ C2−2s k f k22 .
So we have kQs ( f )k2 ≤ C2−s k f k2 for s > 0. Using the estimate
|c
σ2t (ξ)| ≤ C|Lr A2t ξ|−β/m
and the same idea, we have kQs ( f )k2 ≤ C2βs/m k f k2 for s < 0. Thus we get (3.13),
and obviously, the constant C is independent of s and f .
Applying the Riesz–Thorin interpolation theorem of sub-linear operators [2] between (3.10) and (3.13), we know that there exist two constants γ, τ > 0 such that
kQs ( f )k p ≤ C2−γs k f k p

for s > 0, 1 < p < ∞,

kQs ( f )k p ≤ C2τ s k f k p

for s < 0, 1 < p < ∞.

Thus, we obtain (3.9) and (3.5) follows.

534

Y. Chen and Y. Ding
Acknowledgement The authors would like to express their gratitude to the referee
for valuable comments and suggestions.

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Department of Mathematics and Mechanics Applied Science School, University of Science and Technology
Beijing, Beijing 100083, The People’s Republic of China
e-mail: yanpingch@126.com
School of Mathematical Sciences, Beijing Normal University, Laboratory of Mathematics and Complex Systems (BNU), Ministry of Education, Beijing 100875, The People’s Republic of China
e-mail: dingy@bnu.edu.cn