# Logarithmic Bounds for Oscillatory Singular Integrals on Hardy Spaces

Authors: Hussain Al-Qassem; Leslie Cheng; Yibiao Pan
Journal: Journal of Function Spaces
 (2016)

---

## Abstract

We establish a logarithmic bound for oscillatory singular integrals with quadratic phases on the Hardy spaceH
1
(
R
n
). The logarithmic rate of growth is the best possible.


---

## Body

## 1. Introduction

Forn
∈
N, let K
(
x
) be a Calderón-Zygmund kernel on R
n and let P
(
x
) be a polynomial of n variables with real coefficients. Consider the following oscillatory singular integral operator:(1)
T
P
:
f
⟶
p.v.
∫
R
n
e
i
P
x
-
y
K
x
-
y
f
y
d
y
.It is well known that T
P is bounded from L
p
(
R
n
) to L
p
(
R
n
) when 1
<
p
<
∞ and also from L
1
(
R
n
) to L
1
,
∞
(
R
n
). Additionally, L
p
→
L
p and L
1
→
L
1
,
∞ bounds are dependent on the degree of the phase polynomial P only, not its coefficients (see [1, 2]).However, forH
1
(
R
n
)
→
H
1
(
R
n
) boundedness of T
P, the answers are not nearly as clear-cut. First, it was shown in [3] that, in general, T
P may fail to be bounded on H
1
(
R
n
) and when the coefficients of the first-order terms of P vanish, T
P is bounded from H
1
(
R
n
) to itself with a bound independent of the higher order coefficients of P.More recent work can be found in [4, 5], including the following.Theorem 1 (see [5]).
Letn
∈
N, m
≥
2, and P
(
x
)
=
∑
0
≤
|
α
|
≤
m
a
α
x
α be a polynomial of degree m in R
n with real coefficients. Let K be a Calderón-Zygmund kernel and let T
P be given as in (1). Then, there exists a positive constant C such that(2)
T
P
f
H
1
R
n
≤
C
1
+
∑
α
=
1
a
α
∑
2
≤
α
≤
m
a
α
1
/
α
f
H
1
R
nfor all f
∈
H
1
(
R
n
). The constant C may depend on n
,
m, and K but is independent of the coefficients {
a
α
} of P.In order to determine the optimal bound onT
P
H
1
→
H
1, an example was given in [5] to show that, as ∑
α
=
1
a
α
/
∑
2
≤
|
α
|
≤
m
|
a
α
|
1
/
|
α
|
→
∞, any bound on T
P
H
1
→
H
1 must increase at least at the rate of log
⁡
(
∑
|
α
|
=
1
a
α
/
∑
2
≤
|
α
|
≤
m
|
a
α
|
1
/
|
α
|
). This naturally leads to the following question.Does(3)
T
P
f
H
1
R
n
≤
C
n
,
m
1
+
log
+
⁡
∑
α
=
1
a
α
∑
2
≤
α
≤
m
a
α
1
/
α
f
H
1
R
nhold for all f
∈
H
1
(
R
n
)?In this paper, we will prove that the answer to the above question is affirmative for all quadratic polynomials. Namely, we have the following.Theorem 2.
Letn
∈
N and P
(
x
)
=
∑
0
≤
|
α
|
≤
2
a
α
x
α be a quadratic polynomial in R
n with real coefficients. Let K be a Calderón-Zygmund kernel and let T
P be given as in (1). Then, there exists a positive constant C such that(4)
T
P
f
H
1
R
n
≤
C
1
+
log
+
⁡
∑
α
=
1
a
α
∑
α
=
2
a
α
1
/
2
f
H
1
R
nfor all f
∈
H
1
(
R
n
). The constant C may depend on n and K but is independent of the coefficients {
a
α
} of P.We point out thatC denotes an absolute constant whose value may change from line to line.


## 2. Some Definitions and Lemmas

Many of the tools we use are known. For readers who wish to see the definitions and some of their properties, the following references are suggested: [6–12].Forx
∈
R
n and r
>
0, let B
(
x
,
r
)
=
{
y
∈
R
n
:
|
y
-
x
|
<
r
} and |
B
(
x
,
r
)
| denote the Euclidean volume of B
(
x
,
r
).Letϕ be a function in the Schwartz space S
(
R
n
) such that ∫
R
n
ϕ
(
x
)
d
x
=
1. For each f
∈
L
l
o
c
1
(
R
n
) and x
∈
R
n, we let(5)
M
ϕ
f
x
=
sup
s
>
0
⁡
f
∗
ϕ
s
x
,where ϕ
s
(
x
)
=
s
-
n
ϕ
(
x
/
s
).Definition 3.
For a nonnegative, locally integrable functionw on R
n, the Hardy space H
1
(
R
n
) is given by(6)
H
1
R
n
=
f
∈
L
l
o
c
1
R
n
:
M
ϕ
f
L
1
R
n
<
∞
,with f
H
1
(
R
n
)
=
M
ϕ
f
L
1
(
R
n
).Definition 4.
A measurable functionf on R
n is called H
1 atom if there exist ζ
∈
R
n and r
>
0 such that(7)
supp
⁡
f
⊆
B
ζ
,
r
;
(8)
f
∞
≤
1
B
ζ
,
r
;
(9)
∫
R
n
f
y
d
y
=
0
.Lemma 5 (see [9, 10]).
For eachf
∈
H
1
(
R
n
), there exist H
1 atoms {
f
ν
} and coefficients {
ω
ν
} such that(10)
f
=
∑
ν
ω
ν
f
ν
,
f
H
1
R
n
≈
inf
⁡
∑
ν
ω
ν
.Definition 6.
AC
1 function K
:
R
n
∖
{
0
}
→
C is called a Calderón-Zygmund kernel if the following are true:(i)
There existsC
>
0 such that(11)
K
x
+
x
∇
K
x
≤
A
x
-
n
holds for allx
∈
R
n
∖
{
0
}.
(ii)
For all0
<
a
<
b,(12)
∫
B
0
,
b
∖
B
0
,
a
K
x
d
x
=
0
.Lemma 7.
LetP
(
x
)
=
∑
0
≤
|
α
|
≤
2
a
α
x
α for x
∈
R
n and λ
≥
0. Define operator U
P
,
λ by(13)
U
P
,
λ
f
x
=
χ
B
0
,
λ
c
x
x
n
∫
B
0,1
e
i
P
x
-
y
f
y
d
y
.Then, there exists C
>
0 independent of P such that(14)
U
P
,
λ
f
L
1
R
n
≤
C
f
L
∞
B
0,1holds for all f
∈
L
∞
(
B
(
0,1
)
) and λ
≥
(
∑
|
α
|
=
2
|
a
α
|
1
/
2
)
-
2.Proof.
We start by treating the more difficult casen
≥
2. The other case, n
=
1, will be briefly considered later.
Write(15)
∑
α
=
2
a
α
x
α
=
∑
j
=
1
n
∑
k
=
1
n
b
j
k
x
j
x
k
,with b
j
k
=
b
k
j for 1
≤
j
,
k
≤
n. Then, there exist l
,
s
∈
{
1
,
…
,
n
} such that(16)
b
l
s
=
max
⁡
b
j
k
:
1
≤
j
,
k
≤
n
.Thus, we have(17)
2
n
4
b
l
s
λ
≥
λ
∑
α
=
2
a
α
1
/
2
2
>
1
.For x
,
y
∈
R
n, let(18)
x
′
=
x
1
,
…
,
x
l
-
1
,
x
l
+
1
,
…
,
x
n
,
y
~
=
y
1
,
…
,
y
s
-
1
,
y
s
+
1
,
…
,
y
n
.Then, there are polynomials Q
1
(
·
), Q
2
(
·
) on R
n, Q
3
(
·
), Q
4
(
·
) on R
n
-
1, and Q
5
(
·
) on R
n
-
1
×
R
n
-
1 such that(19)
∑
α
=
2
a
α
x
-
y
α
=
-
2
b
l
s
x
l
y
s
+
Q
1
x
+
Q
2
y
+
x
l
Q
3
y
~
+
y
s
Q
4
x
′
+
Q
5
x
′
,
y
~
.Let g
(
x
)
=
f
(
x
) for x
∈
B
(
0,1
) and g
(
x
)
=
0 if x
∈
B
(
0,1
)
c. Then, (20)
U
P
,
λ
f
L
1
R
n
=
∫
R
n
∖
B
0
,
λ
∫
R
n
e
i
P
x
-
y
g
y
d
y
d
x
x
n
=
∫
R
n
∖
B
0
,
λ
∫
y
~
∈
R
n
-
1
e
i
P
0
+
∑
α
=
1
a
α
x
α
+
Q
1
x
+
x
l
Q
3
y
~
+
Q
5
x
′
,
y
~
∫
y
s
∈
R
e
i
-
2
b
l
s
x
l
y
s
-
∑
α
=
1
a
α
y
α
+
Q
2
y
+
y
s
Q
4
x
′
g
y
d
y
s
d
y
~
d
x
x
n
≤
C
∫
x
′
∈
R
n
-
1
∫
y
~
∈
R
n
-
1
∫
x
l
∈
R
h
x
′
x
l
∫
y
s
∈
R
e
-
i
2
b
l
s
x
l
y
s
g
x
′
,
y
~
y
s
d
y
s
d
x
l
d
y
~
d
x
′
,where(21)
g
x
′
,
y
~
y
s
=
e
i
-
∑
α
=
1
a
α
y
α
+
Q
2
y
+
y
s
Q
4
x
′
g
y
,
h
x
′
x
l
=
χ
λ
2
,
∞
x
l
2
+
x
′
2
x
l
2
+
x
′
2
n
/
2
.Since |
g
x
′
,
y
~
(
y
s
)
|
=
|
g
(
y
)
| and supp
⁡
(
g
x
′
,
y
~
)
⊆
[
-
1,1
], we have (22)
U
P
,
λ
f
L
1
R
n
≤
C
∫
x
′
∈
R
n
-
1
∫
y
~
∈
R
n
-
1
∫
x
l
∈
R
h
x
′
x
l
g
x
′
,
y
~
^
2
b
l
s
x
l
d
x
l
d
y
~
d
x
′
≤
C
∫
x
′
∈
R
n
-
1
∫
y
~
∈
R
n
-
1
∫
R
h
x
′
x
l
2
d
x
l
1
/
2
∫
R
g
x
′
,
y
~
^
2
b
l
s
x
l
2
d
x
l
1
/
2
d
y
~
d
x
′
=
C
b
l
s
-
1
/
2
∫
x
′
∈
R
n
-
1
∫
y
~
∈
R
n
-
1
∫
R
h
x
′
x
l
2
d
x
l
1
/
2
∫
R
g
x
′
,
y
~
y
s
2
d
y
s
1
/
2
d
y
~
d
x
′
≤
C
b
l
s
-
1
/
2
f
L
∞
B
0,1
∫
x
′
≥
λ
∫
R
d
x
l
x
l
2
+
x
′
2
n
1
/
2
d
x
′
+
∫
x
′
<
λ
∫
x
l
≥
λ
2
-
x
′
2
d
x
l
x
l
2
+
x
′
2
n
1
/
2
d
x
′
≤
C
b
l
s
-
1
/
2
f
L
∞
B
0,1
∫
x
′
≥
λ
d
x
′
x
′
n
-
1
/
2
+
λ
1
-
2
n
/
2
∫
x
′
<
λ
∫
1
∞
d
t
t
n
t
-
x
′
/
λ
2
1
/
2
d
x
′
≤
C
b
l
s
-
1
/
2
f
L
∞
B
0,1
λ
-
1
/
2
+
λ
-
n
+
1
/
2
∫
x
′
<
λ
1
-
x
′
λ
2
-
1
/
4
d
x
′
≤
C
λ
b
l
s
-
1
/
2
f
L
∞
B
0,1
≤
C
f
L
∞
B
0,1
.The treatment of the case n
=
1 only involves the Fourier transform step of the preceding argument. Details are omitted.Lemma 8.
Letn
∈
N and P
(
x
)
=
∑
0
≤
|
α
|
≤
2
a
α
x
α be a quadratic polynomial in R
n with real coefficients. Let K be a Calderón-Zygmund kernel satisfying (11)-(12) and let T
P be given as in (1). Then, there exists a positive constant C such that(23)
T
P
f
L
1
R
n
≤
C
1
+
log
+
⁡
∑
α
=
1
a
α
∑
α
=
2
a
α
1
/
2for every H
1 atom f which satisfies (7)–(9) with ζ
=
0 and r
=
1. The constant C may depend on n and A but is independent of {
a
α
}, K, and f.Proof.
By the uniform boundedness ofT
P on L
2
(
R
n
) and (7)-(8),(24)
∫
B
0,2
T
P
f
x
d
x
≤
B
0,2
1
/
2
T
P
f
L
2
R
n
≤
C
f
L
2
R
n
≤
C
.
By (11), we have(25)
∫
R
n
∖
B
0,2
T
P
f
x
-
K
x
∫
B
0,1
e
i
P
x
-
y
f
y
d
y
d
x
≤
∫
R
n
∖
B
0,2
∫
B
0,1
K
x
-
y
-
K
x
f
y
d
y
d
x
≤
C
f
L
1
R
n
∫
R
n
∖
B
0,2
x
-
n
-
1
d
x
≤
C
.Let λ
=
(
∑
|
α
|
=
2
|
a
α
|
1
/
2
)
-
2. It follows from (11) and (7)-(8) and Lemma 7 that(26)
∫
R
n
∖
B
0
,
m
a
x
2
,
λ
T
P
f
x
d
x
≤
C
+
∫
R
n
∖
B
0
,
λ
K
x
∫
B
0,1
e
i
P
x
-
y
f
y
d
y
d
x
≤
C
+
C
U
P
,
λ
f
L
1
R
n
≤
C
.
Ifλ
≤
2, then (23) follows from (24) and (26).
Thus, we may assume thatλ
>
2. To finish the proof, it suffices to show that(27)
∫
B
0
,
λ
∖
B
0,2
T
P
f
x
d
x
≤
C
1
+
log
+
⁡
∑
α
=
1
a
α
∑
α
=
2
a
α
1
/
2
.
We will establish (27) by discussing two cases.
Case
1 (
∑
|
α
|
=
1
|
a
α
|
≥
1
/
2
). In this case, we have(28)
∫
B
0
,
λ
∖
B
0,2
T
P
f
x
d
x
≤
C
∫
B
0
,
λ
∖
B
0,2
∫
B
0,1
x
-
y
-
n
f
y
d
y
d
x
≤
C
f
L
1
R
n
∫
B
0
,
λ
∖
B
0,2
x
-
n
d
x
≤
C
ln
⁡
1
2
∑
α
=
2
a
α
1
/
2
2
≤
C
ln
⁡
2
+
2
ln
⁡
∑
α
=
1
a
α
∑
α
=
2
a
α
1
/
2
≤
C
1
+
log
+
⁡
∑
α
=
1
a
α
∑
α
=
2
a
α
1
/
2
.
Case
2 (
∑
α
=
1
a
α
<
1
/
2
). In this case, we let(29)
Q
x
=
P
0
+
∑
α
=
2
a
α
x
α
.It follows from Theorem 1 of [3] that(30)
T
Q
f
L
1
R
n
≤
C
.
Forx
∈
R
n and y
∈
B
(
0,1
), we have(31)
e
i
P
x
-
y
-
e
i
∑
α
=
1
a
α
x
α
+
Q
x
-
y
≤
∑
α
=
1
a
α
.By (30)-(31) and(32)
sup
0
<
t
<
1
/
2
⁡
t
ln
⁡
1
t
=
1
e
,we have(33)
∫
B
0
,
λ
∖
B
0,2
T
P
f
x
d
x
≤
T
Q
f
L
1
R
n
+
∫
B
0
,
λ
∖
B
0,2
T
P
f
x
-
e
i
∑
α
=
1
a
α
x
α
T
Q
f
x
d
x
≤
C
+
C
∑
α
=
1
a
α
f
L
1
R
n
∫
B
0
,
λ
∖
B
0,2
x
-
n
d
x
≤
C
+
C
∑
α
=
1
a
α
ln
⁡
1
∑
α
=
2
a
α
1
/
2
=
C
+
C
∑
α
=
1
a
α
ln
⁡
1
∑
α
=
1
a
α
+
ln
⁡
∑
α
=
1
a
α
∑
α
=
2
a
α
1
/
2
≤
C
1
+
log
+
⁡
∑
α
=
1
a
α
∑
α
=
2
a
α
1
/
2
.Thus, (27) holds in both cases.


## 3. Proof of Main Theorem

To finish the proof, we recall the following result concerning Riesz transforms and Hardy spaces.Lemma 9 (see [10, 13]).
For1
≤
j
≤
n, let R
j denote the jth Riesz transform; that is,(34)
R
j
f
^
ξ
=
i
ξ
j
ξ
f
^
ξ
.Then, there exist C
,
C
1
,
C
2
>
0 such that(35)
R
j
f
H
1
R
n
≤
C
f
H
1
R
nfor 1
≤
j
≤
n, and(36)
C
1
f
H
1
R
n
≤
f
L
1
R
n
+
∑
j
=
1
n
R
j
f
L
1
R
n
≤
C
2
f
H
1
R
nfor all f
∈
H
1
(
R
n
).We will now give the proof of Theorem2.Proof.
Forf
∈
H
1
(
R
n
), let {
ω
ν
} be a sequence of complex numbers and let {
f
ν
} be a sequence of H
1 atoms such that(37)
f
=
∑
ν
ω
ν
f
ν
.For each ν, let ζ
ν
∈
R
n and r
ν
>
0 such that supp
⁡
(
f
ν
)
⊆
B
(
ζ
ν
,
r
ν
) and f
ν
∞
≤
|
B
(
ζ
ν
,
r
ν
)
|
-
1
=
|
B
(
0,1
)
|
-
1
r
ν
-
n. Then,(38)
r
ν
n
T
P
f
ν
r
ν
x
+
ζ
ν
=
p.v.
∫
R
n
e
i
P
ν
x
-
y
K
ν
x
-
y
r
ν
n
f
ν
r
ν
y
+
ζ
ν
d
y
,where P
ν
(
x
)
=
P
(
r
ν
x
) and K
ν
(
x
)
=
r
ν
n
K
(
r
ν
x
). Observe that, for each ν, K
ν satisfies (11)-(12) with the same constant A and r
ν
n
f
ν
(
r
ν
y
+
ζ
ν
) satisfies (7)–(9) with ζ
=
0, r
=
1. Since(39)
P
ν
x
=
∑
0
≤
α
≤
2
r
ν
α
a
α
x
α
,by Lemma 8,(40)
T
P
f
ν
L
1
R
n
=
∫
R
n
r
ν
n
T
P
f
ν
r
ν
x
+
ζ
ν
d
x
=
C
1
+
l
o
g
+
∑
α
=
1
a
α
∑
α
=
2
a
α
1
/
2
,which implies that(41)
T
P
f
L
1
R
n
≤
C
1
+
log
+
⁡
∑
α
=
1
a
α
∑
α
=
2
a
α
1
/
2
∑
ν
ω
ν
.It follows from Lemma 5 that(42)
T
P
f
L
1
R
n
≤
C
1
+
log
+
⁡
∑
α
=
1
a
α
∑
α
=
2
a
α
1
/
2
f
H
1
R
n
.By the translation invariance of T
P and (42) and (35), we have(43)
∑
j
=
1
n
R
j
T
P
f
L
1
R
n
=
∑
j
=
1
n
T
P
R
j
f
L
1
R
n
≤
C
1
+
log
+
⁡
∑
α
=
1
a
α
∑
α
=
2
a
α
1
/
2
∑
j
=
1
n
R
j
f
H
1
R
n
≤
C
1
+
log
+
⁡
∑
α
=
1
a
α
∑
α
=
2
a
α
1
/
2
f
H
1
R
n
.By applying (36), (42), and (43), we obtain (4).



---

*Source: 1570109-2016-02-24.xml*