# Criterion onL

Authors: Zuoshunhua Shi; Dunyan Yan
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We investigate the bilinear Hilbert transform with oscillatory factors and the truncated bilinear Hilbert transform. The main result is that theL
p
1
×
L
p
2
→
L
q-boundedness of the two operators is equivalent with 1
≤
p
1
,
p
2
<
∞, and 1
/
q
=
1
/
p
1
+
1
/
p
2. In addition, we also discuss the boundedness of a variant operator of bilinear Hilbert transform with a nontrivial polynomial phase.


---

## Body

## 1. Introduction

In this paper, we mainly discuss the following operator defined by(1)
𝒯
(
f
,
g
)
(
x
)
=
p.v.
∫
ℝ
‍
e
i
P
(
x
,
t
)
f
(
x
-
t
)
g
(
x
+
t
)
d
t
t
,
where P
(
·
,
·
) is a real-valued polynomial defined on ℝ
×
ℝ and f
,
g are smooth functions with compact support.Clearly, whenP
=
0, the operator 𝒯 becomes the normal bilinear Hilbert transform defined as
(2)
T
(
f
,
g
)
(
x
)
=
p.v.
∫
ℝ
‍
f
(
x
-
t
)
g
(
x
+
t
)
d
t
t
.IfP
(
x
,
t
)
=
P
1
(
x
-
t
)
+
P
2
(
x
+
t
)
+
P
3
(
x
) for some one-variable polynomials P
1, P
2, and P
3, then clearly the boundedness of 𝒯 and T is equivalent. For this case, we introduce the following definitions.Definition 1.
For a polynomialP
(
·
,
·
) defined on ℝ
×
ℝ, one calls P degenerate, if
(3)
P
(
x
,
t
)
=
P
1
(
x
-
t
)
+
P
2
(
x
+
t
)
+
P
3
(
x
)
holds with three one-variable polynomials P
1, P
2, and P
3.In fact, a simple verification shows that if the degree ofP
(
·
,
·
) is less than 3, then P is degenerate. However, when d
≥
3, there exist nondegenerate polynomials with the degree d. We will prove this in Section 2. It should be pointed out that Definition 1 may be extended to such polynomials defined on ℝ
n
×
ℝ
n (see [1]). We will give a characterization of nondegenerate polynomials by Ind
β in Section 2. Now we first give the definition of Ind
β.Definition 2.
LetP
(
·
,
·
) be a polynomial with the degree d
≥
3. For 1
≤
β
≤
d
-
2, denote Ind
β by
(4)
Ind
β
(
P
)
(
x
,
t
)
=
(
∂
∂
x
+
∂
∂
t
)
β
(
∂
∂
x
-
∂
∂
t
)
d
-
1
-
β
∂
∂
t
P
(
x
,
t
)
.
For the operator𝒯 defined by (1), the following result has been obtained in [1].Theorem A.
LetP
(
·
,
·
) be a real-valued polynomial. Then the operator 𝒯 extends as a bounded operator from L
p
1
(
ℝ
)
×
L
p
2
(
ℝ
) to L
q
(
ℝ
) with 1
<
p
1, p
2
<
∞, and q
>
2
/
3 such that 1
/
q
=
1
/
p
1
+
1
/
p
2; furthermore, the operator norm ∥
𝒯
∥ depends only on the degree of P.In the study of oscillatory singular integrals, the boundedness of the original operator without oscillatory factors may imply the boundedness of the corresponding truncated operator. For the Calderón-Zygmund singular integral operators of convolution type, one can see Ricci and Stein [2]. For the multilinear singular operators, one may refer to [1]. In [3] Lu and Zhang established a criterion for the boundedness of the following operator:
(5)
𝒦
f
(
x
)
=
p.v.
∫
ℝ
n
‍
e
i
P
(
x
,
t
)
K
(
x
-
y
)
f
(
y
)
d
y
,
where P
(
·
,
·
) is a real-valued polynomial defined on ℝ
n
×
ℝ
n, f
∈
C
0
∞
(
ℝ
n
), and K
(
x
)
=
Ω
(
x
′
)
/
|
x
|
n satisfies(1)
Ω is homogeneous of the degree 0 on S
n
-
1;
(2)
Ω has mean value zero on S
n
-
1;
(3)
Ω
∈
L
q
(
S
n
-
1
) for some 1
<
q
≤
∞.For the sake of clarity, we also present the theorem obtained by Lu and Zhang as follows.Theorem B.
Suppose that𝒦 is defined by (5) and 1
<
p
<
∞. Each of the following statements implies the other two.(i)
IfP
(
·
,
·
) is not of the form P
(
x
,
y
)
=
P
1
(
x
)
+
P
2
(
y
) for some polynomials P
1 and P
2 in ℝ
n, then the operator
(6)
𝒦
f
(
x
)
=
p
.
v
.
∫
ℝ
n
‍
e
i
P
(
x
,
y
)
K
(
x
-
y
)
f
(
y
)
d
y
is bounded onL
p
(
ℝ
n
).
(ii)
IfQ
(
·
,
·
) satisfies Q
(
x
+
h
,
y
+
h
)
=
Q
(
x
,
y
)
+
R
1
(
x
,
h
)
+
R
2
(
y
,
h
) for some polynomials R
1, R
2, and h
∈
ℝ
n, then the operator
(7)
𝒢
f
(
x
)
=
p
.
v
.
∫
ℝ
n
‍
e
i
Q
(
x
,
y
)
K
(
x
-
y
)
f
(
y
)
d
y
extends as a bounded operator fromL
p
(
ℝ
n
) to itself.
(iii)
The following truncated operator(8)
𝒮
f
(
x
)
=
p
.
v
.
∫
|
x
-
y
|
<
1
‍
K
(
x
-
y
)
f
(
y
)
d
y
is bounded onL
p
(
ℝ
n
).For the bilinear Hilbert transform and the corresponding truncated operator, we can formulate some analogous results as Theorem B.Theorem 3.
Suppose that0
<
p
1
,
p
2
,
q
<
∞ satisfy 1
/
q
=
1
/
p
1
+
1
/
p
2. If the truncated operator
(9)
S
(
f
,
g
)
(
x
)
=
p
.
v
.
∫
|
t
|
<
1
‍
f
(
x
-
t
)
g
(
x
+
t
)
d
t
t
is bounded from L
p
1
(
ℝ
)
×
L
p
2
(
ℝ
) to L
q
(
ℝ
), then the bilinear Hilbert transform
(10)
T
(
f
,
g
)
(
x
)
=
p
.
v
.
∫
ℝ
‍
f
(
x
-
t
)
g
(
x
+
t
)
d
t
t
is bounded from L
p
1
(
ℝ
)
×
L
p
2
(
ℝ
) to L
q
(
ℝ
). Moreover, when 1
≤
p
1, p
2
<
∞, the converse also holds; that is, if T extends to a bounded operator from L
p
1
(
ℝ
)
×
L
p
2
(
ℝ
) to L
q
(
ℝ
), then so does S.Theorem 4.
LetP
(
·
,
·
) be a real-valued polynomial and let 1
≤
p
1, p
2
<
∞, and q
≥
1
/
2 satisfy 1
/
q
=
1
/
p
1
+
1
/
p
2. If P
(
x
+
h
,
t
)
-
P
(
x
,
t
) is degenerate with respect to x, t for all h
∈
ℝ in the sense of Definition 1 and the following operator
(11)
𝒯
(
f
,
g
)
(
x
)
=
p
.
v
.
∫
ℝ
‍
e
i
P
(
x
,
t
)
f
(
x
-
t
)
g
(
x
+
t
)
d
t
t
is bounded from L
p
1
(
ℝ
)
×
L
p
2
(
ℝ
) to L
q
(
ℝ
), then the truncated operator
(12)
S
(
f
,
g
)
(
x
)
=
p
.
v
.
∫
|
t
|
<
1
‍
f
(
x
-
t
)
g
(
x
+
t
)
d
t
t
is bounded from L
p
1
(
ℝ
)
×
L
p
2
(
ℝ
) to L
q
(
ℝ
).We summarize the above two theorems as follows.Theorem 5.
Suppose that1
≤
p
1,  p
2
<
∞, and q
≥
1
/
2 satisfy 1
/
q
=
1
/
p
1
+
1
/
p
2. Each of the following three statements implies the other two.(i)
The bilinear Hilbert transform(13)
T
(
f
,
g
)
(
x
)
=
p
.
v
.
∫
ℝ
‍
f
(
x
-
t
)
g
(
x
+
t
)
d
t
t
is bounded fromL
p
1
(
ℝ
)
×
L
p
2
(
ℝ
) to L
q
(
ℝ
).
(ii)
The oscillatory bilinear Hilbert transform(14)
𝒯
(
f
,
g
)
(
x
)
=
p
.
v
.
∫
ℝ
‍
e
i
P
(
x
,
t
)
f
(
x
-
t
)
g
(
x
+
t
)
d
t
t
is bounded fromL
p
1
(
ℝ
)
×
L
p
2
(
ℝ
) to L
q
(
ℝ
) for any real-valued polynomial P.
(iii)
The truncated operator(15)
S
(
f
,
g
)
(
x
)
=
p
.
v
.
∫
|
t
|
<
1
‍
f
(
x
-
t
)
g
(
x
+
t
)
d
t
t
is bounded fromL
p
1
(
ℝ
)
×
L
p
2
(
ℝ
) to L
q
(
ℝ
).Next, we will consider a variant operator of bilinear Hilbert transform defined by(16)
T
σ
(
f
,
g
)
(
x
)
=
∫
ℝ
‍
e
i
P
(
x
,
t
)
f
(
x
-
t
)
g
(
x
+
t
)
d
t
(
1
+
|
t
|
)
σ
for σ
≥
1. Clearly, the operator T
σ
(
f
,
g
) has no singular property at origin, so it is the key at ∞.We will use the power decay property of a bilinear functional to obtain the following theorem.Theorem 6.
Suppose thatP
(
·
,
·
) is a real-valued nondegenerate polynomial with the degree d
≥
3 and |
Ind
β
(
P
)
|
≥
1 for some 1
≤
β
≤
d
-
2. If 1
≤
p
1, p
2
<
∞, and q
≥
1
/
2 satisfy 1
/
q
=
1
/
p
1
+
1
/
p
2, then the operator T
σ defined by (16) is bounded from L
p
1
(
ℝ
)
×
L
p
2
(
ℝ
) to L
q
(
ℝ
); that is,
(17)
∥
T
σ
(
f
,
g
)
∥
q
≤
C
∥
f
∥
p
1
∥
g
∥
p
2
,
where C is a positive constant.


## 2. Some Lemmas

Throughout the paper, we always assume that1
≤
p
1, p
2
<
∞, and 1
/
q
=
1
/
p
1
+
1
/
p
2, unless the contrary is explicitly stated. We first introduce some lemmas which are useful in the proof of the main theorems.Lemma 7.
Suppose thatP defined on ℝ
×
ℝ is a homogeneous polynomial of the degree d with d
≥
3. Then P is degenerate if and only if Ind
β
(
P
)
=
0 for all 1
≤
β
≤
d
-
2.Proof.
By Definition1, if P is degenerate, then
(18)
P
(
x
,
t
)
=
P
1
(
x
-
t
)
+
P
2
(
x
+
t
)
+
Q
(
x
)
,
where P
1, P
2, and Q are homogeneous single-variable polynomials of the degree d. Observe that
(19)
∂
∂
t
Q
(
x
)
=
0
,
(
∂
∂
x
-
∂
∂
t
)
P
2
(
x
+
t
)
=
0
,
(
∂
∂
x
+
∂
∂
t
)
P
1
(
x
-
t
)
=
0
.
This implies that
(20)
Ind
β
(
P
)
=
0
for all 1
≤
β
≤
d
-
2.
Now we assume thatInd
β
(
P
)
=
0 for all 1
≤
β
≤
d
-
2. Make changes of variables
(21)
u
=
x
+
t
v
=
x
-
t
.
By the chain rule, we have that
(22)
∂
β
∂
u
β
=
1
2
β
(
∂
∂
x
+
∂
∂
t
)
β
,
∂
d
-
1
-
β
∂
v
d
-
1
-
β
=
1
2
d
-
1
-
β
(
∂
∂
x
-
∂
∂
t
)
d
-
1
-
β
.
SinceP is a homogeneous polynomial with the degree d, (
∂
/
∂
t
)
P is a homogeneous polynomial of the degree d
-
1. There, therefore, exists a homogeneous polynomial R of the degree d
-
1 such that
(23)
R
(
u
,
v
)
=
∂
∂
t
P
(
x
,
t
)
.
It follows from (23) that
(24)
Ind
β
(
P
)
=
2
d
-
1
∂
u
β
∂
v
d
-
1
-
β
R
.
The assumption Ind
β
(
P
)
=
0, for all 1
≤
β
≤
d
-
2, implies that
(25)
R
(
u
,
v
)
=
c
1
u
d
-
1
+
c
2
v
d
-
1
for some constants c
1 and c
2. Integrating both sides of the equality (25) with respect to t gives that
(26)
P
(
x
,
t
)
=
c
1
d
(
x
+
t
)
d
-
c
2
d
(
x
-
t
)
d
+
c
3
x
d
,
for some constant c
3. Hence P is degenerate.Remark 8.
Ford
≥
3 and 1
≤
k
≤
d
-
1, (
x
-
t
)
k
(
x
+
t
)
d
-
k and t
d are nondegenerate polynomials. Furthermore, when P is a general polynomial of the degree d, we can write P as sums of homogeneous polynomials,
(27)
P
(
x
,
t
)
=
∑
k
=
0
d
‍
P
k
(
x
,
t
)
,
where P
k is a homogeneous polynomial with the degree k. If P is degenerate, then each P
k is degenerate for 3
≤
k
≤
d. In fact, if P is degenerate, then P
(
x
,
t
)
=
P
1
(
x
-
t
)
+
P
2
(
x
+
t
)
+
Q
(
x
). It follows that P
k
(
x
,
t
)
=
P
1
,
k
(
x
-
t
)
+
P
2
,
k
(
x
+
t
)
+
Q
k
(
x
) for each 0
≤
k
≤
d. Consequently, each P
k is degenerate.Lemma 9.
LetP be a polynomial with the degree d
≥
3. For any τ
∈
ℝ, set
(28)
Q
τ
(
u
,
v
)
=
P
(
u
+
v
2
,
u
-
v
2
+
τ
)
-
P
(
u
+
v
2
,
u
-
v
2
)
.
Then, one has(29)
∂
u
β
∂
v
d
-
1
-
β
Q
τ
(
u
,
v
)
=
2
1
-
d
τ
Ind
β
(
P
)
for each 1
≤
β
≤
d
-
2.Proof.
By the same change of variables as in the proof of Lemma7, we obtain
(30)
Q
τ
(
u
,
v
)
=
P
(
x
,
t
+
τ
)
-
P
(
x
,
t
)
=
τ
∫
0
1
‍
∂
t
P
(
x
,
t
+
s
τ
)
d
s
.
We can use the chain rule as in (22) to differentiate both sides of (28) by the differential operator ∂
u
β
∂
v
d
-
1
-
β and easily yield the desired result.Lemma 10.
Suppose thatk and n are two positive integers and that l
j, 1
≤
j
≤
n, are linear mappings from ℝ
(
n
-
1
)
k to ℝ
k. Assume that 1
≤
p
j
<
∞ for j
=
1
,
…
,
n satisfy
(31)
1
q
=
∑
j
=
1
n
‍
1
p
j
.
If the linear mapping Λ: ℝ
n
k
→
ℝ
n
k defined by
(32)
Λ
(
x
,
y
)
=
(
x
+
l
1
(
y
)
,
…
,
x
+
l
n
(
y
)
)
is onto for x
∈
ℝ
k, y
∈
ℝ
(
n
-
1
)
k, then one has
(33)
I
(
f
1
,
…
,
f
n
)
=
(
∫
ℝ
k
‍
(
∫
ℝ
(
n
-
1
)
k
‍
∏
j
=
1
n
‍
|
f
j
(
x
+
l
j
(
y
)
)
|
|
η
(
y
)
|
d
y
)
q
d
x
)
1
/
q
≤
C
∏
j
=
1
n
‍
∥
f
j
∥
p
j
,
where η is a bounded measurable function with compact support and f
j
∈
L
p
j
(
ℝ
k
) for 1
≤
j
≤
n.Proof.
For the caseq
≥
1, applying generalized Minkowski’s and Hölder’s inequality gives that
(34)
I
(
f
1
,
…
,
f
n
)
≤
∫
ℝ
(
n
-
1
)
k
(
∫
ℝ
k
‍
(
∏
j
=
1
n
‍
|
f
j
(
x
+
l
j
(
y
)
)
|
|
η
(
y
)
|
)
q
d
x
)
1
/
q
d
y
‍
≤
∫
ℝ
(
n
-
1
)
k
‍
|
η
(
y
)
|
d
y
×
∏
j
=
1
n
‍
(
∫
ℝ
k
‍
|
f
j
(
x
+
l
j
(
y
)
)
|
p
j
d
x
)
1
/
p
j
=
∥
η
∥
1
∏
j
=
1
n
‍
∥
f
j
∥
p
j
.
Now we deal with the case 0
<
q
<
1. We first show that the following estimate
(35)
∫
|
x
-
h
|
≤
1
(
∫
ℝ
(
n
-
1
)
k
‍
∏
j
=
1
n
‍
|
f
j
(
x
+
l
j
(
y
)
)
|
|
η
(
y
)
|
d
y
)
q
d
x
‍
≤
C
∏
j
=
1
n
‍
∥
f
j
∥
L
p
j
(
|
x
-
h
|
≤
N
)
q
holds uniformly for h
∈
ℝ
k, where N will be decided by the supp
⁡
(
η
) and the norm of l
j, 1
≤
j
≤
n.
Since0
<
q
<
1, by Hölder’s inequality, it is clear that we have that
(36)
∫
|
x
-
h
|
≤
1
(
∫
ℝ
(
n
-
1
)
k
‍
∏
j
=
1
n
‍
|
f
j
(
x
+
l
j
(
y
)
)
|
|
η
(
y
)
|
d
y
)
q
‍
d
x
≤
C
(
∫
|
x
-
h
|
≤
1
‍
∫
ℝ
(
n
-
1
)
k
‍
∏
j
=
1
n
|
f
j
(
x
+
l
j
(
y
)
)
|
|
η
(
y
)
|
d
y
d
x
‍
)
q
.
Since the linear mapping Λ: ℝ
n
k
→
ℝ
n
k is onto, we can make change of variables u
j
=
x
+
l
j
(
y
), 1
≤
j
≤
n, and then (
x
,
y
)
=
Λ
-
1
(
u
1
,
…
,
u
n
). We also have
(37)
∫
|
x
-
h
|
≤
1
‍
∫
ℝ
(
n
-
1
)
k
∏
j
=
1
n
‍
|
f
j
(
x
+
l
j
(
y
)
)
|
‍
|
η
(
y
)
|
d
y
d
x
≤
C
∫
|
u
1
-
h
|
≤
N
,
…
,
|
u
n
-
h
|
≤
N
∏
j
=
1
n
‍
‍
|
f
j
(
u
j
)
|
d
u
1
⋯
d
u
n
≤
C
(
N
k
ω
k
)
(
n
-
1
/
q
)
∏
j
=
1
n
‍
∥
f
j
∥
L
p
j
(
|
x
-
h
|
≤
N
)
q
,
where ω
k is the volume of the unit ball in ℝ
k and N is a fixed number such that |
l
j
(
y
)
|
+
1
≤
N for all y
∈
supp
⁡
(
η
), 1
≤
j
≤
n.
Note the following fact. For anyF
∈
L
1
(
ℝ
k
) and r
>
0, it follows that
(38)
∫
ℝ
k
‍
(
∫
|
x
-
h
|
<
r
‍
F
(
x
)
d
x
)
d
h
=
∫
ℝ
k
‍
(
F
*
χ
B
r
)
(
h
)
d
h
=
r
k
ω
k
∫
ℝ
k
‍
F
(
x
)
d
x
,
where χ
B
r is the characteristic function of the ball centered at 0 with radius r.
In view of the equality (38), taking integration on both sides of (35) with respect to h over ℝ
k, we conclude that
(39)
ω
k
∫
ℝ
k
‍
(
∫
ℝ
(
n
-
1
)
k
‍
∏
j
=
1
n
‍
|
f
j
(
x
+
l
j
(
y
)
)
|
|
η
(
y
)
|
d
y
)
q
d
x
≤
C
∫
ℝ
k
‍
∏
j
=
1
n
‍
∥
f
j
∥
L
p
j
(
|
x
-
h
|
≤
N
)
q
d
h
≤
C
∏
j
=
1
n
‍
(
∫
ℝ
k
‍
∫
|
x
-
h
|
≤
N
‍
|
f
j
(
x
)
|
p
j
d
x
d
h
)
q
/
p
j
=
C
N
k
ω
k
∏
j
=
1
n
‍
∥
f
∥
p
j
q
.
The proof is therefore concluded.As a special case of Lemma10, we immediately obtain Corollary 11.Corollary 11.
Letf
∈
L
p
1
(
ℝ
) and g
∈
L
p
2
(
ℝ
). If 1
≤
p
1, p
2
<
∞ satisfy 1
/
q
=
1
/
p
1
+
1
/
p
2, then there exists a constant C
=
C
(
p
1
,
p
2
,
q
) such that
(40)
∫
ℝ
‍
(
∫
|
t
|
<
1
‍
|
f
(
x
-
t
)
g
(
x
+
t
)
|
d
t
)
q
d
x
≤
C
∥
f
∥
p
1
q
∥
g
∥
p
2
q
.Actually, the estimate of Corollary11 may also be found in [4, 5].Generally speaking, the power decay estimates of oscillatory integrals are necessary to investigate the boundedness of oscillatory integral operators. The following lemma to appear in [5] is just an oscillatory integral and will be used to prove Lemma 14.Lemma 12.
Suppose thatP
(
·
,
·
) is a real-valued polynomial with the degree d
≥
2 and that there exists a positive integer number 1
≤
β
≤
d
-
1 such that
(41)
|
∂
x
β
∂
y
d
-
β
P
(
x
,
y
)
|
≥
1
.
Define the functional
(42)
Λ
λ
(
f
,
g
)
=
∬
ℝ
e
i
λ
P
(
x
,
y
)
f
(
x
)
g
(
y
)
η
(
x
,
y
)
d
x
d
y
,
where η is a smooth function with compact support, f
,
g
∈
L
2
(
ℝ
), and λ
∈
ℝ. Then there exists some constant δ
>
0 such that the following power decay estimate
(43)
|
Λ
λ
(
f
,
g
)
|
≤
C
(
1
+
|
λ
|
)
-
δ
∥
f
∥
2
∥
g
∥
2
holds, where the constant C is independent of f, g and λ.Remark 13.
It should be pointed out how the constantC depends on the cut-off function η, provided that other conditions are not changed. Indeed, if, for the collection Θ
=
{
η
}, every function η is uniformly supported in a bounded set and ∥
η
∥
C
m has an upper bound independent of η
∈
Θ for each positive integer m, then the constant C can be chosen such that the estimate (43) holds uniformly for η
∈
Θ. The decay estimate (43) and its other variants have been systematically investigated by many scholars. For certain polynomials with some other conditions, the power exponent δ can be explicitly given. One may refer to [6] and other references appearing in this paper.Many oscillatory integral operators have power decay estimates which are indispensable to study mapping properties. The following lemma is significant and will be useful in the next section.Lemma 14.
Suppose thatP
(
·
,
·
) is a real-valued nondegenerate polynomial with the degree d and that there is some β, 1
≤
β
≤
d
-
2, such that |
Ind
β
(
P
)
|
≥
1. Let
(44)
𝒮
λ
(
f
,
g
)
(
x
)
=
∫
ℝ
‍
e
i
λ
P
(
x
,
t
)
f
(
x
-
t
)
g
(
x
+
t
)
η
(
t
)
d
t
,
where η
∈
C
0
∞ has support contained in [
-
1,1
]. If 1
≤
p
1,  p
2
<
∞, and q
≥
1
/
2 satisfy
(45)
1
q
=
1
p
1
+
1
p
2
,
then there exists some ɛ
>
0 such that
(46)
∥
𝒮
λ
(
f
,
g
)
∥
q
≤
C
(
1
+
|
λ
|
)
-
ɛ
∥
f
∥
p
1
∥
g
∥
p
2
,
for f
∈
L
p
1
(
ℝ
), g
∈
L
p
2
(
ℝ
), and λ
∈
ℝ.Proof.
Choose a nonnegative functionφ
∈
C
0
∞ such that
(47)
supp
⁡
φ
⊂
[
-
2,2
]
,
∑
k
∈
ℤ
‍
φ
(
x
-
k
)
=
1
for every x
∈
ℝ.
To obtain the decay estimate (46), it suffices to show that the following estimate
(48)
∫
ℝ
‍
|
𝒮
λ
(
f
,
g
)
(
x
)
φ
(
x
-
k
)
|
q
d
x
≤
C
(
1
+
|
λ
|
)
-
q
ɛ
∥
f
∥
L
p
1
(
B
5
(
k
)
)
q
∥
g
∥
L
p
2
(
B
5
(
k
)
)
q
holds uniformly in k
∈
ℤ, where B
5
(
k
) denotes an interval with the center at k and the radius 5.
If inequality (48) has been established, we may use Hölder’s inequality to obtain (46).
In fact, sincesupp
⁡
φ
⊂
[
-
2,2
], at most five terms are not zero in the summation for any fixed x. Thus, if 0
<
q
<
1, then we have
(49)
(
∑
k
‍
|
φ
(
x
-
k
)
|
)
q
≤
∑
k
‍
|
φ
(
x
-
k
)
|
q
,
and if q
≥
1, then it follows that
(50)
(
∑
k
‍
|
φ
(
x
-
k
)
|
)
q
≤
5
q
-
1
∑
k
‍
|
φ
(
x
-
k
)
|
q
.
Taking the summations of both sides of the inequality (48) with respect to k over ℤ and using the two inequalities (49) and (50), we notice that the summation of the left side of (48) is greater than C
∥
𝒮
λ
(
f
,
g
)
∥
q
q.
For the summation of the right side of (48), using Hölder’s inequality, we obtain that
(51)
∑
k
‍
∥
f
∥
L
p
1
(
B
5
(
k
)
)
q
∥
g
∥
L
p
2
(
B
5
(
k
)
)
q
≤
(
∑
k
‍
∫
|
x
-
k
|
≤
5
‍
|
f
(
x
)
|
p
1
d
x
)
q
/
p
1
×
(
∑
k
‍
∫
|
x
-
k
|
≤
5
‍
|
g
(
x
)
|
p
2
d
x
)
q
/
p
2
≤
10
∥
f
∥
p
1
q
∥
g
∥
p
2
q
.
Next let us turn to the proof of the inequality (48). By Lemma 10 or Corollary 11, we know that 𝒮
λ is bounded from L
p
1
×
L
p
2 to L
q and its norm is independent of λ. By the multilinear interpolation of Riesz-Thörin, it is enough to show that the estimate (46) holds for p
1
=
p
2
=
2 and q
=
1. Thus we merely need to show that the inequality (48) holds with the case p
1
=
p
2
=
2 and q
=
1. Furthermore, since φ has a compact support, it suffices to prove that (48) holds for p
1
=
p
2
=
q
=
2.
This is equivalent to proving the following inequality:(52)
∥
𝒮
λ
(
f
,
g
)
φ
(
·
-
k
)
∥
2
≤
C
(
1
+
|
λ
|
)
-
ɛ
∥
f
∥
L
2
(
B
5
(
k
)
)
∥
g
∥
L
2
(
B
5
(
k
)
)
.
By the change of variables, we can easily deduce that(53)
∫
ℝ
‍
|
𝒮
λ
(
f
,
g
)
(
x
)
φ
(
x
-
k
)
|
2
d
x
=
∭
ℝ
e
i
λ
(
P
(
x
,
t
1
)
-
P
(
x
,
t
2
)
)
f
(
x
-
t
1
)
f
(
x
-
t
2
)
¯
×
g
(
x
+
t
1
)
g
(
x
+
t
2
)
¯
×
η
(
t
1
)
η
(
t
2
)
¯
φ
(
x
-
k
)
φ
(
x
-
k
)
¯
d
t
1
d
t
2
d
x
=
∭
ℝ
e
i
λ
(
P
(
x
,
t
+
τ
)
-
P
(
x
,
t
)
)
×
f
(
x
-
t
-
τ
)
f
(
x
-
t
)
¯
×
g
(
x
+
t
+
τ
)
g
(
x
+
t
)
¯
η
(
t
+
τ
)
η
(
t
)
¯
×
φ
(
x
-
k
)
φ
(
x
-
k
)
¯
d
t
d
τ
d
x
=
1
2
∫
D
τ
‍
∫
D
u
‍
∫
D
v
‍
e
i
λ
Q
τ
(
u
,
v
)
F
τ
(
v
)
G
τ
(
u
)
η
τ
(
u
,
v
)
×
ψ
(
u
,
v
)
d
u
d
v
d
τ
,
where
(54)
Q
τ
(
u
,
v
)
=
P
(
k
+
u
+
v
2
,
u
-
v
2
+
τ
)
-
P
(
k
+
u
+
v
2
,
u
-
v
2
)
,
F
τ
(
v
)
=
f
(
k
+
v
-
τ
)
f
(
k
+
v
)
¯
,
G
τ
(
u
)
=
g
(
k
+
u
+
τ
)
g
(
k
+
u
)
¯
,
η
τ
(
u
,
v
)
=
η
(
u
-
v
2
+
τ
)
η
(
u
-
v
2
)
¯
,
ψ
(
u
,
v
)
=
φ
(
u
+
v
2
)
φ
(
u
+
v
2
)
¯
.
Consequently, for someρ
>
0, we can rewrite the above integral as
(55)
∫
ℝ
‍
|
𝒮
λ
(
f
,
g
)
(
x
)
φ
(
x
-
k
)
|
2
d
x
=
1
2
∫
|
τ
|
<
ρ
‍
∫
D
u
‍
∫
D
v
‍
e
i
λ
Q
τ
(
u
,
v
)
F
τ
(
v
)
G
τ
(
u
)
×
η
τ
(
u
,
v
)
ψ
(
u
,
v
)
d
u
d
v
d
τ
+
1
2
∫
|
τ
|
≥
ρ
‍
∫
D
u
‍
∫
D
v
‍
e
i
λ
Q
τ
(
u
,
v
)
F
τ
(
v
)
G
τ
(
u
)
×
η
τ
(
u
,
v
)
ψ
(
u
,
v
)
d
u
d
v
d
τ
=
I
1
+
I
2
.
Observe that integralsI
1 and I
2 are taken in the domain
(56)
{
(
u
,
v
,
τ
)
:
|
u
|
≤
3
,
|
v
|
≤
3
,
|
τ
|
≤
2
}
.
Recall supp
⁡
(
φ
)
⊂
[
-
2,2
] and supp
⁡
(
η
)
⊂
[
-
1,1
], respectively.
We clearly conclude that(57)
|
I
1
|
≤
C
∫
|
τ
|
≤
min
⁡
⁡
{
ρ
,
2
}
‍
∫
|
u
|
≤
3
‍
∫
|
v
|
≤
3
‍
|
F
τ
(
v
)
G
τ
(
u
)
|
d
u
d
v
d
τ
=
C
∫
|
τ
|
≤
min
⁡
{
ρ
,
2
}
‍
∫
|
u
|
≤
3
∫
|
v
|
≤
3
‍
‍
|
f
(
k
+
v
-
τ
)
f
(
k
+
v
)
¯
g
×
(
k
+
u
+
τ
)
g
(
k
+
u
)
¯
|
d
u
d
v
d
τ
≤
C
∫
|
τ
|
<
min
⁡
{
ρ
,
2
}
‍
(
∫
|
v
|
≤
3
‍
|
f
(
v
+
k
-
τ
)
|
2
d
v
)
1
/
2
×
(
∫
|
v
|
≤
3
‍
|
f
(
v
+
k
)
|
2
d
v
)
1
/
2
×
(
∫
|
u
|
≤
3
‍
|
g
(
u
+
k
+
τ
)
|
2
d
u
)
1
/
2
×
(
∫
|
u
|
≤
3
‍
|
g
(
u
+
k
)
|
2
d
u
)
1
/
2
d
τ
≤
C
min
⁡
{
ρ
,
2
}
∥
f
∥
L
2
(
B
5
(
k
)
)
2
∥
g
∥
L
2
(
B
5
(
k
)
)
2
.
For|
τ
|
≥
ρ, by Lemma 9, it is easily verified that
(58)
∂
u
β
∂
v
d
-
1
-
β
Q
τ
(
u
,
v
)
=
2
1
-
d
τ
Ind
β
(
P
)
holds for each 1
≤
β
≤
d
-
2.
Thus we have that(59)
|
∂
u
β
∂
v
d
-
1
-
β
Q
τ
(
u
,
v
)
|
=
2
1
-
d
|
τ
|
|
Ind
β
(
P
)
|
≥
c
|
ρ
|
,
for some 1
≤
β
≤
d
-
2.
Invoking Lemma12, we obtain, for |
τ
|
≥
ρ, that
(60)
|
∫
u
‍
∫
v
‍
e
i
λ
Q
τ
(
u
,
v
)
F
τ
(
v
)
G
τ
(
u
)
η
τ
(
u
,
v
)
φ
(
u
,
v
)
d
u
d
v
|
≤
C
(
1
+
|
λ
ρ
|
)
-
δ
(
∫
|
v
|
≤
3
‍
|
F
τ
(
v
)
|
2
d
v
)
1
/
2
×
(
∫
|
u
|
≤
3
‍
|
G
τ
(
u
)
|
2
d
u
)
1
/
2
,
since the cut-off function η
τ
ψ is uniformly supported in {
(
u
,
v
)
:
|
u
|
≤
3
,
|
v
|
≤
3
} and its C
m norms have bounds independent of τ for each positive m, and the constant C is independent of τ. Therefore, the Cauchy-Schwartz inequality implies that
(61)
|
I
2
|
≤
C
(
1
+
|
λ
ρ
|
)
-
δ
×
∫
|
τ
|
≤
2
‍
(
∫
|
v
|
≤
3
‍
|
F
τ
(
v
)
|
2
d
v
)
1
/
2
(
∫
|
u
|
≤
3
‍
|
G
τ
(
u
)
|
2
d
u
)
1
/
2
d
τ
≤
C
(
1
+
|
λ
ρ
|
)
-
δ
(
∫
|
τ
|
≤
2
‍
∫
|
v
|
≤
3
‍
|
F
τ
(
v
)
|
2
d
v
d
τ
)
1
/
2
×
(
∫
|
τ
|
≤
2
‍
∫
|
u
|
≤
3
‍
|
G
τ
(
u
)
|
2
d
u
d
τ
)
1
/
2
=
C
(
1
+
|
λ
ρ
|
)
-
δ
×
(
∫
|
τ
|
≤
2
‍
∫
|
v
|
≤
3
‍
|
f
(
v
+
k
-
τ
)
|
2
|
f
(
v
+
k
)
|
2
d
v
d
τ
)
1
/
2
×
(
∫
|
τ
|
≤
2
‍
∫
|
u
|
≤
3
‍
|
g
(
u
+
k
+
τ
)
|
2
|
g
(
u
+
k
)
|
2
d
u
d
τ
)
1
/
2
≤
C
(
1
+
|
λ
ρ
|
)
-
δ
×
(
∫
|
x
-
k
|
≤
5
‍
|
f
(
x
)
|
2
d
x
)
(
∫
|
x
-
k
|
≤
5
‍
|
g
(
x
)
|
2
d
x
)
,
where we use the Fubini theorem in the proof of the above inequality.
Consequently, choosingρ
=
|
λ
|
-
δ
/
(
1
+
δ
)
<
2 for |
λ
|
≥
1 and combining the above estimates for I
1 and I
2, we immediately obtain that
(62)
∥
𝒮
λ
(
f
,
g
)
φ
(
·
-
k
)
∥
2
≤
C
(
1
+
|
λ
|
)
-
ɛ
∥
f
∥
L
2
(
B
5
(
k
)
)
∥
g
∥
L
2
(
B
5
(
k
)
)
,
where ɛ
=
(
-
δ
/
2
(
1
+
δ
)
). This is just the inequality (52).Lemma 15.
Suppose thatP
(
·
,
·
) is a real-valued polynomial and 1
≤
p
1, p
2
<
∞, and q
≥
1
/
2 satisfy
(63)
1
q
=
1
p
1
+
1
p
2
.
If the operator 𝒯 defined by (1) satisfies
(64)
∥
𝒯
(
f
,
g
)
∥
q
≤
C
∥
f
∥
p
1
∥
g
∥
p
2
,
for all f
,
g
∈
C
c
∞
(
ℝ
), then the truncated operator
(65)
𝒮
(
f
,
g
)
(
x
)
=
p
.
v
.
∫
|
t
|
<
1
‍
e
i
P
(
x
,
t
)
f
(
x
-
t
)
g
(
x
+
t
)
d
t
t
is also bounded from L
p
1
×
L
p
2 to L
q.Proof.
To prove Lemma15, we will show that
(66)
∫
|
x
-
h
|
<
1
/
4
‍
|
𝒮
(
f
,
g
)
(
x
)
|
q
d
x
≤
C
∥
f
∥
L
p
1
(
[
h
-
5
/
4
,
h
+
5
/
4
]
)
q
∥
g
∥
L
p
2
(
[
h
-
5
/
4
,
h
+
5
/
4
]
)
q
holds uniformly with respect to h
∈
ℝ.
Fixh
∈
ℝ and set
(67)
f
1
=
f
χ
{
|
·
-
h
|
≤
1
/
2
}
,
f
2
=
f
χ
{
1
/
2
<
|
·
-
h
|
≤
5
/
4
}
,
f
3
=
f
χ
{
|
·
-
h
|
>
5
/
4
}
.
It is clear that f
=
f
1
+
f
2
+
f
3.
Now for fixedh
∈
ℝ, set
(68)
x
∈
(
h
-
1
4
,
h
+
1
4
)
.
If|
x
-
h
-
t
|
<
1
/
2, then |
t
|
<
3
/
4. We hence obtain that
(69)
𝒮
(
f
1
,
g
)
(
x
)
=
𝒯
(
f
1
,
g
)
(
x
)
.
For|
x
-
h
|
<
1
/
4, when 1
/
2
≤
|
x
-
h
-
t
|
≤
5
/
4, we have 1
/
4
≤
|
t
|
≤
3
/
2. It immediately follows that
(70)
|
𝒮
(
f
2
,
g
)
(
x
)
|
≤
C
∫
|
t
|
<
1
‍
|
f
2
(
x
-
t
)
g
(
x
+
t
)
|
d
t
.
Finally, both|
x
-
h
|
<
1
/
4 and |
x
-
h
-
t
|
>
5
/
4 imply |
t
|
>
1. We clearly have
(71)
𝒮
(
f
3
,
g
)
(
x
)
=
0
.
Similarly, we also have that(72)
𝒮
(
f
1
,
g
)
=
𝒮
(
f
1
,
g
χ
{
|
·
-
h
|
≤
5
/
4
}
)
=
𝒯
(
f
1
,
g
χ
{
|
·
-
h
|
≤
5
/
4
}
)
,
provided that |
x
-
h
|
<
1
/
4 and |
t
|
<
1.
Therefore, if|
x
-
h
|
<
1
/
4, then the equalities (69), (71), and (72) together with (70) imply that
(73)
|
𝒮
(
f
,
g
)
(
x
)
|
≤
|
𝒮
(
f
1
,
g
)
(
x
)
|
+
C
∫
|
t
|
<
1
‍
|
f
2
(
x
-
t
)
g
(
x
+
t
)
|
d
t
=
|
𝒯
(
f
1
,
g
χ
{
|
·
-
h
|
≤
5
/
4
}
)
(
x
)
|
+
C
∫
|
t
|
<
1
‍
|
f
2
(
x
-
t
)
(
g
χ
{
|
·
-
h
|
≤
5
/
4
}
)
(
x
+
t
)
|
d
t
.
Clearly the inequality (66) is easily obtained by combing (73) together with Corollary 11. We conclude the proof by integrating both sides of (66) with respect to h.Lemma 16.
Suppose that1
≤
p
1, p
2
<
∞, and q
≥
1
/
2 satisfy
(74)
1
q
=
1
p
1
+
1
p
2
.
Define the truncated operator as
(75)
S
r
(
f
,
g
)
(
x
)
=
p
.
v
.
∫
|
t
|
<
r
‍
f
(
x
-
t
)
g
(
x
+
t
)
d
t
t
.
If the operator S defined by (9) satisfies
(76)
∥
S
(
f
,
g
)
∥
q
≤
C
∥
f
∥
p
1
∥
g
∥
p
2
,
for all f
,
g
∈
C
c
∞
(
ℝ
), then the truncated operator S
r is also bounded from L
p
1
×
L
p
2 to L
q for every r
>
0; moreover, the operator norm of S
r is independent of r.Proof.
Clearly it implies from the definitions ofS
r and S that S
1
=
S. By a simple dilation argument, we will prove that ∥
S
r
∥, the norm of the operator S
r defined by (75), is independent of r
>
0 and equals ∥
S
1
∥.
Indeed, a simple computation implies that(77)
S
r
(
f
,
g
)
(
r
x
)
=
p.v.
∫
|
t
|
<
r
‍
f
(
r
x
-
t
)
g
(
r
x
+
t
)
d
t
t
=
p.v.
∫
|
t
|
<
1
‍
f
(
r
x
-
r
t
)
g
(
r
x
+
r
t
)
d
t
t
=
p.v.
∫
|
t
|
<
1
‍
(
δ
r
f
)
(
x
-
t
)
(
δ
r
g
)
(
x
+
t
)
d
t
t
=
S
(
δ
r
f
,
δ
r
g
)
(
x
)
.
It follows that
(78)
∥
S
r
(
f
,
g
)
(
r
·
)
∥
q
∥
f
∥
p
1
∥
g
∥
p
2
=
∥
S
(
δ
r
f
,
δ
r
g
)
∥
q
∥
f
∥
p
1
∥
g
∥
p
2
=
(
1
r
)
1
/
q
∥
S
(
δ
r
f
,
δ
r
g
)
∥
q
∥
δ
r
f
∥
p
1
∥
δ
r
g
∥
p
2
,
where the dilation operator is defined by
(79)
(
δ
τ
f
)
(
x
)
=
f
(
τ
x
)
for all τ
>
0 and x
∈
ℝ.
Denote the norm of the operatorS by ∥
S
∥ as follows:
(80)
∥
S
∥
=
sup
⁡
∥
f
∥
p
1
≠
0
,
∥
g
∥
p
2
≠
0
∥
S
(
f
,
g
)
∥
q
∥
f
∥
p
1
∥
g
∥
p
2
.
Taking supremum over allf
,
g
∈
C
c
∞ for (78), we have
(81)
(
1
r
)
1
/
q
∥
S
r
∥
=
(
1
r
)
1
/
q
∥
S
∥
,
and naturally ∥
S
r
∥
=
∥
S
∥.Lemma 17.
Suppose thatP
(
·
,
·
) is a real-valued polynomial and 1
≤
p
1, p
2
<
∞, and q
≥
1
/
2 satisfy
(82)
1
q
=
1
p
1
+
1
p
2
.
If the operator 𝒯 defined by (1) satisfies
(83)
∥
𝒯
(
f
,
g
)
∥
q
≤
C
∥
f
∥
p
1
∥
g
∥
p
2
,
for all f
,
g
∈
C
c
∞
(
ℝ
), then the following operator
(84)
𝒯
r
(
f
,
g
)
(
x
)
=
∫
ℝ
‍
e
i
P
(
r
x
,
r
t
)
f
(
x
-
t
)
g
(
x
+
t
)
d
t
t
is also bounded from L
p
1
×
L
p
2 to L
q for every r
>
0; moreover, the operator norm of 𝒯
r is independent of r.Proof.
By a simple dilation argument, we will prove that∥
𝒯
r
∥, the norm of the operator 𝒯
r, is independent of r
>
0 and equals ∥
𝒯
1
∥ and naturally equals ∥
𝒯
∥.
Indeed, a simple computation implies that(85)
𝒯
r
(
f
,
g
)
(
x
r
)
=
𝒯
(
δ
r
-
1
(
f
)
,
δ
r
-
1
(
g
)
)
(
x
)
,
(86)
∥
𝒯
r
(
f
,
g
)
(
·
/
r
)
∥
q
∥
f
∥
p
1
∥
g
∥
p
2
=
∥
𝒯
(
δ
r
-
1
(
f
)
,
δ
r
-
1
(
g
)
)
∥
q
∥
f
∥
p
1
∥
g
∥
p
2
=
r
1
/
q
∥
𝒯
(
δ
r
-
1
(
f
)
,
δ
r
-
1
(
g
)
)
∥
q
∥
δ
r
-
1
(
f
)
∥
p
1
∥
δ
r
-
1
(
g
)
∥
p
2
.
Taking supremum over all f
,
g
∈
C
c
∞ for (86), we immediately have
(87)
r
1
/
q
∥
𝒯
r
∥
=
r
1
/
q
∥
𝒯
∥
,
and hence ∥
𝒯
r
∥
=
∥
𝒯
∥.Remark 18.
Lemma17 shows that, for the oscillatory integral operator, the operator norm only depends on the degree of the polynomial but is independent of the coefficient of the polynomial.


## 3. Proof of the Main Results

Proof of Theorem3.
Associated with each positiver, denote the operator S
r by
(88)
S
r
(
f
,
g
)
(
x
)
=
p.v.
∫
|
t
|
<
r
‍
f
(
x
-
t
)
g
(
x
+
t
)
d
t
t
,
where f
,
g
∈
C
c
∞
(
ℝ
); hence, we have S
=
S
1 by the definition of S in Theorem 3. By the dilation argument as in the proof of Lemma 16, we can easily conclude that if S can be extended to a bounded operator from L
p
1
×
L
p
2 to L
q, then so is S
r and the operator norm is independent of r
>
0. Observe that C
c
∞
(
ℝ
) is dense in all L
p
(
ℝ
) for 0
<
p
<
∞. For 0
<
p
<
1, we regard L
p
(
ℝ
) as a complete metric space. Since we can approximate f
∈
L
p
(
ℝ
) by finite linear combinations of characteristic function of intervals, we may assume that f
=
χ
I for some finite open interval I. By the Lusin theorem, for any closed interval O
⊂
I there exists a function η
∈
C
c
∞ such that supp
⁡
(
η
)
⊂
I, 0
≤
η
≤
1, and η
(
x
)
=
1 for x
∈
O. Hence we obviously have
(89)
∥
f
-
η
∥
p
≤
|
I
-
O
|
1
/
p
.
Fixf
,
g
∈
C
c
∞
(
ℝ
) temporarily and choose a large N
>
0 such that both supp
⁡
(
f
) and supp
⁡
(
g
) are contained in [
-
N
,
N
]. If |
t
|
>
N, we see that either x
-
t
∉
supp
⁡
(
f
) or x
+
t
∉
supp
⁡
(
g
) must hold for every x. Hence we conclude that
(90)
S
N
(
f
,
g
)
(
x
)
=
T
(
f
,
g
)
(
x
)
,
∥
T
(
f
,
g
)
∥
q
=
∥
S
N
(
f
,
g
)
∥
q
≤
∥
S
N
∥
∥
f
∥
p
1
∥
g
∥
p
2
=
∥
S
∥
∥
f
∥
p
1
∥
g
∥
p
2
,
where ∥
S
N
∥ and ∥
S
∥ are operator norms of S
N and S from L
p
1
×
L
p
2 to L
q, respectively.
Obviously it follows that(91)
∥
T
∥
≤
∥
S
∥
.
The validity of the converse follows from Lemma15.
This finishes the proof of Theorem3.Proof of Theorem4.
We assume that(92)
∥
𝒯
(
f
,
g
)
∥
q
≤
C
∥
f
∥
p
1
∥
g
∥
p
2
holds for some polynomial P
(
x
,
t
) with P
(
x
+
h
,
t
)
-
P
(
x
,
t
) being degenerate for all h
∈
ℝ. We wish to obtain
(93)
∫
|
x
-
h
|
<
1
‍
|
S
(
f
,
g
)
(
x
)
|
q
d
x
≤
C
∥
f
∥
L
p
1
(
B
2
(
h
)
)
q
∥
g
∥
L
p
2
(
B
2
(
h
)
)
q
,
where the constant C is independent of h and B
2
(
h
)
=
(
h
-
2
,
h
+
2
).
SinceP
(
x
+
h
,
t
)
-
P
(
x
,
t
) is degenerate with respect to x
,
t, we can rewrite P as
(94)
P
(
x
,
t
)
=
P
(
x
-
h
,
t
)
+
Q
1
(
x
-
t
,
h
)
+
Q
2
(
x
+
t
,
h
)
+
R
(
x
,
h
)
,
for some polynomials Q
1, Q
2, and R.
Forh
∈
ℝ, it follows that
(95)
S
(
f
,
g
)
(
x
)
=
∫
|
t
|
<
1
‍
f
(
x
-
t
)
g
(
x
+
t
)
d
t
t
=
e
-
i
R
(
x
,
h
)
∫
|
t
|
<
1
‍
e
i
P
(
x
,
t
)
f
h
(
x
-
t
)
g
h
(
x
+
t
)
×
(
e
-
i
P
(
x
-
h
,
t
)
-
e
-
i
P
(
x
-
h
,
0
)
)
d
t
t
+
e
-
i
R
(
x
,
h
)
-
i
P
(
x
-
h
,
0
)
×
∫
|
t
|
<
1
‍
e
i
P
(
x
,
t
)
f
h
(
x
-
t
)
g
h
(
x
+
t
)
d
t
t
=
S
h
1
(
f
,
g
)
(
x
)
+
S
h
2
(
f
,
g
)
(
x
)
,
where
(96)
f
h
(
x
)
=
e
-
i
Q
1
(
x
,
h
)
f
(
x
)
,
g
h
(
x
)
=
e
-
i
Q
2
(
x
,
h
)
g
(
x
)
.
For the polynomialP, when |
x
-
h
|
<
1 and |
t
|
<
1, we immediately have
(97)
|
e
-
i
P
(
x
-
h
,
t
)
-
e
-
i
P
(
x
-
h
,
0
)
|
≤
C
|
t
|
.
It is easy to drive that(98)
|
S
h
1
(
f
,
g
)
(
x
)
|
≤
C
∫
|
t
|
<
1
‍
|
f
(
x
-
t
)
g
(
x
+
t
)
|
d
t
,
for |
x
-
h
|
<
1.
It follows from Corollary11 and the inequality (98) that
(99)
∫
|
x
-
h
|
<
1
‍
|
S
h
1
(
f
,
g
)
(
x
)
|
q
d
x
≤
C
∥
f
∥
L
p
1
(
B
2
(
h
)
)
q
∥
g
∥
L
p
2
(
B
2
(
h
)
)
q
.
If|
x
-
h
|
<
1 and |
t
|
<
1, then we have
(100)
f
(
x
-
t
)
=
(
f
χ
(
h
-
2
,
h
+
2
)
)
(
x
-
t
)
,
g
(
x
+
t
)
=
(
f
χ
(
h
-
2
,
h
+
2
)
)
(
x
+
t
)
.
By the definition of S
h
2, we thus obtain
(101)
S
h
2
(
f
,
g
)
(
x
)
=
S
h
2
(
f
χ
(
h
-
2
,
h
+
2
)
,
g
χ
(
h
-
2
,
h
+
2
)
)
(
x
)
for |
x
-
h
|
<
1.
On the other hand, by Lemma15, S
h
2 is bounded from L
p
1
×
L
p
2 to L
q; that is,
(102)
∥
S
h
2
(
f
,
g
)
∥
L
q
(
B
1
(
h
)
)
q
=
∫
|
x
-
h
|
<
1
‍
|
S
h
2
(
f
χ
(
h
-
2
,
h
+
2
)
,
g
χ
(
h
-
2
,
h
+
2
)
)
(
x
)
|
q
d
x
≤
∫
ℝ
‍
|
S
h
2
(
f
χ
(
h
-
2
,
h
+
2
)
,
g
χ
(
h
-
2
,
h
+
2
)
)
(
x
)
|
q
d
x
≤
C
∥
f
∥
L
p
1
(
B
2
(
h
)
)
q
∥
g
∥
L
p
2
(
B
2
(
h
)
)
q
.
Hence the inequality (93) is also valid for S
h
2. The theorem is a consequence of (93) by taking integration with respect to h over ℝ and an application of Hölder’s inequality.Proof of Theorem5.
The implication(i)
⇒
(ii) is contained in [1]. The implication (ii)
⇒
(iii) is a direct consequence of Theorem 4, and (iii)
⇒
(i) can attribute to Theorem 3.Proof of Theorem6.
We first prove Theorem6 for the case σ
>
1. One will see that the oscillatory factor is not necessary in this case.
It follows that(103)
|
T
σ
(
f
,
g
)
(
x
)
|
≤
∫
|
t
|
<
1
‍
|
f
(
x
-
t
)
g
(
x
+
t
)
|
d
t
+
∑
j
=
0
∞
‍
∫
2
j
≤
|
t
|
<
2
j
+
1
‍
|
f
(
x
-
t
)
g
(
x
+
t
)
|
d
t
|
t
|
σ
=
S
(
|
f
|
,
|
g
|
)
(
x
)
+
∑
j
=
0
∞
‍
T
σ
j
(
f
,
g
)
(
x
)
.
If we can show that(104)
∥
T
σ
j
(
f
,
g
)
∥
q
≤
C
2
-
j
ɛ
∥
f
∥
p
1
∥
g
∥
p
2
holds for some ɛ
>
0 and each j
=
1,2
,
…, then our desired result follows immediately.
Now let us consider the operatorT
σ
j
(
f
,
g
).
We conclude that(105)
T
σ
j
(
f
,
g
)
(
x
)
=
∫
2
j
≤
|
t
|
<
2
j
+
1
‍
|
f
(
x
-
t
)
g
(
x
+
t
)
|
d
t
|
t
|
σ
≤
2
-
j
(
σ
-
1
)
+
1
×
∫
1
/
2
≤
|
t
|
<
1
‍
|
f
(
x
-
2
j
+
1
t
)
g
(
x
+
2
j
+
1
t
)
|
d
t
≤
2
-
j
(
σ
-
1
)
+
1
∫
|
t
|
<
1
‍
|
f
(
x
-
2
j
+
1
t
)
g
(
x
+
2
j
+
1
t
)
|
d
t
.
Therefore, it follows that(106)
∥
T
σ
j
(
f
,
g
)
∥
q
q
≤
2
(
-
j
(
σ
-
1
)
+
1
)
q
×
∫
ℝ
‍
(
∫
|
t
|
<
1
‍
|
f
(
x
-
2
j
+
1
t
)
g
(
x
+
2
j
+
1
t
)
|
d
t
)
q
d
x
≤
2
(
-
j
(
σ
-
1
)
+
1
)
q
2
j
+
1
×
∫
ℝ
‍
(
∫
|
t
|
<
1
‍
|
f
(
2
j
+
1
(
x
-
t
)
)
g
(
2
j
+
1
(
x
+
t
)
)
|
d
t
)
q
d
x
=
2
(
-
j
(
σ
-
1
)
+
1
)
q
2
j
+
1
∫
ℝ
‍
(
∫
|
t
|
<
1
‍
|
f
~
(
x
-
t
)
g
~
(
x
+
t
)
|
d
t
)
q
d
x
,
where f
~
(
·
)
=
f
(
2
j
+
1
(
·
)
) and g
~
(
·
)
=
g
(
2
j
+
1
(
·
)
).
Clearly,f
∈
L
p
(
ℝ
) implies f
~
∈
L
p
(
ℝ
).
It implies from Corollary11 that
(107)
∫
ℝ
‍
(
∫
|
t
|
<
1
‍
|
f
~
(
x
-
t
)
g
~
(
x
+
t
)
|
d
t
)
q
d
x
≤
C
(
∫
ℝ
‍
|
f
~
(
x
)
|
p
1
d
x
)
q
/
p
1
(
∫
ℝ
‍
|
g
~
(
x
)
|
p
2
d
x
)
q
/
p
2
=
C
2
-
(
j
+
1
)
∥
f
∥
p
1
q
∥
g
∥
p
2
q
.
Thus we have(108)
∥
T
σ
j
(
f
,
g
)
∥
q
≤
C
2
-
j
(
σ
-
1
)
∥
f
∥
p
1
∥
g
∥
p
2
.
It remains to consider the caseσ
=
1. By Lemma 14, we can reduce the theorem to boundedness of the following operator defined as
(109)
T
φ
(
f
,
g
)
(
x
)
=
∫
ℝ
‍
e
i
P
(
x
,
t
)
f
(
x
-
t
)
g
(
x
+
t
)
×
(
1
-
φ
(
t
)
)
d
t
1
+
|
t
|
,
where φ is a smooth function with compact support contained in [
-
1,1
] and equals 1 near the origin. Hence it suffices to show that T
φ is bounded from L
p
1
×
L
p
2 to L
q.
Sinceφ is smooth with compact support and equals 1 near the origin, it is not hard to have that
(110)
∑
j
=
0
∞
‍
(
φ
(
x
2
j
+
1
)
-
φ
(
x
2
j
)
)
=
1
-
φ
(
x
)
.
We rewrite the integral (109) as
(111)
T
φ
(
f
,
g
)
(
x
)
=
∑
j
=
0
∞
‍
∫
ℝ
‍
e
i
P
(
x
,
t
)
f
(
x
-
t
)
g
(
x
+
t
)
ψ
j
(
t
)
d
t
|
t
|
+
W
(
f
,
g
)
(
x
)
,
=
∑
j
=
0
∞
‍
T
φ
j
(
f
,
g
)
(
x
)
+
W
(
f
,
g
)
(
x
)
,
where
(112)
ψ
j
(
t
)
=
φ
(
x
2
j
+
1
)
-
φ
(
x
2
j
)
,
W
(
f
,
g
)
(
x
)
=
-
∫
ℝ
‍
e
i
P
(
x
,
t
)
f
(
x
-
t
)
g
(
x
+
t
)
(
1
-
φ
(
t
)
)
d
t
(
1
+
|
t
|
)
|
t
|
,
T
φ
j
(
f
,
g
)
(
x
)
=
∫
ℝ
‍
e
i
P
(
x
,
t
)
f
(
x
-
t
)
g
(
x
+
t
)
ψ
j
(
t
)
d
t
|
t
|
,
for j
=
0,1
,
2
,
….
It is clear that we have(113)
|
W
(
f
,
g
)
(
x
)
|
≤
C
∫
ℝ
‍
|
f
(
x
-
t
)
g
(
x
+
t
)
|
d
t
(
1
+
|
t
|
)
2
,
which implies that W is bounded from L
p
1
×
L
p
2 to L
q as shown for σ
>
1. Hence it is enough to prove that
(114)
∥
T
φ
j
(
f
,
g
)
∥
q
≤
C
2
-
j
ɛ
∥
f
∥
p
1
∥
g
∥
p
2
for some ɛ
>
0.
Let(115)
𝒯
φ
j
(
f
,
g
)
(
x
)
=
∫
ℝ
‍
e
i
P
(
2
j
x
,
2
j
t
)
f
(
x
-
t
)
g
(
x
+
t
)
ψ
0
(
t
)
d
t
|
t
|
,
for j
=
0,1
,
2
,
….
By the dilation argument as in the proof of Lemma17, we easily conclude that the operator norms, ∥
T
φ
j
∥ and ∥
𝒯
φ
j
∥, from L
p
1
×
L
p
2 to L
q, are equal.
Observe that(116)
Ind
β
(
P
(
2
j
(
·
)
,
2
j
(
·
)
)
)
=
2
j
d
Ind
β
(
P
)
.
We can now apply Lemma 14 to obtain that the decay estimate (114) holds, since |
Ind
β
(
P
)
|
≥
1 for some 1
≤
β
≤
d
-
2.
This finishes the proof of Theorem6.Remark 19.
It should be pointed out that if the polynomialP
(
·
,
·
) is degenerate, then the operator T
σ defined by (16) may not be bounded from L
p
1
(
ℝ
)
×
L
p
2
(
ℝ
) to L
q
(
ℝ
).



---

*Source: 712051-2014-02-25.xml*