# Hilbert Transforms along Convex Curves forl

Authors: Honghai Liu
Journal: ISRN Mathematical Analysis
 (2014)

---

## Abstract

We show that Hilbert transforms along a large class of convex curves are bounded onL
p
(
ℝ
2
,
l
q
), where 5
/
3
<
p, q
<
5
/
2.


---

## Body

## 1. Introduction

LetΓ
:
ℝ
→
ℝ
n be a curve in ℝ
n with Γ
(
0
)
=
0, n
≥
2. For f
∈
C
0
∞
(
ℝ
n
), the Hilbert transform along curve Γ is defined by the following principal-valued integral:
(1)
ℋ
f
(
x
)
=
p
.
v
.
∫
ℝ
‍
f
(
x
-
Γ
(
t
)
)
d
t
t
.There has been considerable interest to determine for which curvesΓ, and which indices p, one has
(2)
∥
ℋ
f
∥
L
p
≤
A
p
∥
f
∥
L
p
,
for a constant A
p depending only on Γ and p. This problem has been extensively studied by a large number of authors. More results are found in [1–6].The question of whether these results could be extended to the Lebesgue-Bôchner spacesL
p
(
ℝ
n
;
X
) of vector-valued functions was taken up by several authors recently, where X is some suitable Banach space. But the Banach space of most interest to us will be l
q for 1
<
q
<
∞. Let us state some previous theorems which establish the background for our current work. The first is the work done in 1986 by Rubio de Francia et al. [7]. They dealt with well-curved curves and obtained the following result.Theorem 1.
LetΓ
:
ℝ
→
ℝ
n be a well-curved curve in ℝ
n with Γ
(
0
)
=
0. Then the l
q-valued inequality
(3)
∥
(
∑
j
∈
ℤ
‍
|
ℋ
f
j
|
q
)
1
/
q
∥
L
p
≤
C
p
,
q
∥
(
∑
j
∈
ℤ
‍
|
f
j
|
q
)
1
/
q
∥
L
p
holds for all p, q with 1
<
p, q
<
∞, and all f
j
∈
L
p
(
ℝ
n
), n
≥
2.Recently, the author considered the submanifold of finite type which means that the image of the mappingt
→
Γ
(
t
) is of finite type at 0. We first restrict our attention to Γ
(
t
) that are given as polynomial functions, because it is a model problem in this situation. We obtained the following.Theorem 2 (see [8, 9]).
(1) LetΓ
:
ℝ
→
ℝ
n be a polynomial function. Then the l
q-valued inequality
(4)
∥
(
∑
j
∈
ℤ
‍
|
ℋ
f
j
|
q
)
1
/
q
∥
L
p
≤
C
p
,
q
∥
(
∑
j
∈
ℤ
‍
|
f
j
|
q
)
1
/
q
∥
L
p
holds for all p, q with 1
<
p, q
<
∞, and all f
j
∈
L
p
(
ℝ
n
), n
≥
2.
(2) Let the image ofΓ
:
ℝ
→
ℝ
n be of finite type at 0. Then the l
q-valued inequality
(5)
∥
(
∑
j
∈
ℤ
‍
|
ℋ
f
j
|
q
)
1
/
q
∥
L
p
≤
C
p
,
q
∥
(
∑
j
∈
ℤ
‍
|
f
j
|
q
)
1
/
q
∥
L
p
holds for all p, q with 1
<
p, q
<
∞, and all f
j
∈
L
p
(
ℝ
n
), n
≥
2.An important feature of the well-curved curve is that it is in a sense approximated by the model curves; the models are homogeneous with respect to some nonisotropic dilations. This means that corresponding scaling arguments are essential for well-curved curves.“The image ofΓ
:
ℝ
→
ℝ
n is of finite type at 0” is a crucial condition imposed on Γ, this condition bears on finitely many derivatives of Γ at the origin, this property is captured by a Taylor polynomial of Γ at origin of sufficiently high degree and should be modeled by polynomial behavior. For polynomial function Γ, in order to utilize the scaling arguments, the situation need be simplified further. This is accomplished by using the lifting technique.All these results have common roots in that both depend in a key way on the use of scaling arguments or nonisotropic dilations, although these dilations are given implicitly for above curves in different way. More precisely, an analytic family of convolution operatorsℋ
z are defined such that ℋ
0
=
ℋ. The L
2
(
ℝ
n
,
l
2
) result of ℋ
z is obtained with the help of the Fourier transform and Plancherel theorem for all z with Re
(
z
)
<
a, where a is some positive constant. ℋ
z is closer to a more traditional singular integral when Re
(
z
)
<
0, because there is some slack in the estimate of the kernel. Therefore, by the vector-valued Calderón-Zygmund theory of Benedek et al. in [10], one can obtain the L
p
(
ℝ
n
,
l
q
) result for all p, q with 1
<
p, q
<
∞, Re
(
z
)
<
0. By generalized analytic interpolation of operators (see Lemma 12), a combination of above two results gives the desired L
p
(
ℝ
n
,
l
q
) estimate for ℋ.What about the situation when the curvesΓ do not have appropriate homogeneity as well-curved curves or finite type curves? This kind of curves has been extensively considered in [1–5]. The Calderón-Zygmund type argument cannot be used to show that ℋ
z is bounded in L
p
(
ℝ
n
,
l
q
) for 1
<
p, q
<
∞ and Re
(
z
)
<
0. New techniques should be utilized. In this paper, we deal with plane curves of the form Γ
(
t
)
=
(
t
,
γ
(
t
)
), where γ
(
t
) is a convex function for t
≥
0. We have the following main result.Theorem 3.
Suppose thatγ
(
t
) is a continuous odd function, twice continuously differentiable, increasing and convex for t
≥
0. Suppose also that γ
′
′ is monotone for t
>
0 and that there exists C
>
0 so that γ
′
(
t
)
≤
C
t
γ
′
′
(
t
) for t
>
0. Then ℋ is bounded on L
p
(
ℝ
2
,
l
q
) for all p, q with 5
/
3
<
p, q
<
5
/
2.The proof of Theorem3 is given in several steps, which are similar to that in [4]. An analytic family of operators ℋ
z are introduced so that ℋ
0
=
ℋ. The L
2
(
ℝ
2
,
l
2
) result is equivalent to the boundedness of the multiplier associated to ℋ
z with Re
(
z
)
<
a, where a is some positive constant. Van der Corput lemma and integration by part are used to obtain the boundedness. The kernels of ℋ
z do not have appropriate homogeneity, we have to use the vector-valued Fourier multiplier theorem (See Lemma 11) to show that ℋ
z is bounded on L
p
(
ℝ
2
,
l
q
) for p, q with 1
<
p, q
<
∞, and Re
(
z
)
<
b, where b is some negative constant. Finally, the theorem follows from above two facts by using generalized analytic interpolation of operators.Remark 4.
(1) Theorem3 covers a large class of functions γ
(
t
) such as
(6)
γ
(
t
)
=
sgn
⁡
(
t
)
|
t
|
α
,
(
α
≥
2
)
,
γ
(
t
)
=
t
e
-
1
/
|
t
|
.
The first one is homogeneous, while the other one is without homogeneity. We extend the class of curves for which the L
p
(
ℝ
2
,
l
q
)-result is known for 5
/
3
<
p, q
<
5
/
2. On the other hand, we also generalized Theorem  3.1 of Nagel and Wainger in [4] to the l
q-valued setting for 5
/
3
<
q
<
5
/
2.
(2) In the present work, the kernels must be smoothed a definite amount, we are unable to takea
≥
1
/
4 or b
≥
-
1. Hence we obtain bounds for ℋ on L
p
(
ℝ
2
,
l
q
) only when 5
/
3
<
p, q
<
5
/
2.


## 2. Proof of the Theorem

Forz
∈
ℂ, we define an analytic family of operators ℋ
z by
(7)
ℋ
z
f
^
(
ξ
,
η
)
=
m
z
(
ξ
,
η
)
f
^
(
ξ
,
η
)
,
where m
z are given by
(8)
m
z
(
ξ
,
η
)
=
p
.
v
.
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
.
Obviously, ℋ
0 is our original operator ℋ.


### 2.1. The Boundedness ofℋ
z on L
2
(
ℝ
2
,
l
2
)

In this subsection, we prove that(9)
∥
(
∑
j
∈
ℤ
‍
|
ℋ
z
f
j
|
2
)
1
/
2
∥
L
2

≤

C
δ
[
1
+
|
Im
⁡
(
z
)
|
]
∥
(
∑
j
∈
ℤ
‍
|
f
j
|
2
)
1
/
2
∥
L
2
,
where Re
(
z
)
=
(
1
/
4
)
-
δ for some δ
>
0.Clearly, the boundedness ofℋ
z on L
2
(
ℝ
2
,
l
2
) is equivalent to the uniform boundedness of the corresponding multiplier m
z
(
ξ
,
η
). Thus, we just need to show that
(10)
|
m
z
(
ξ
,
η
)
|
≤
C
δ
[
1
+
|
Im
⁡
(
z
)
|
]
,
where the constant C
δ is independent of Im
⁡
(
z
).In proving (10), some of our estimates will require the following lemmas. The first one is detailed in [4, Lemma  2.1].Lemma 5.
LetA
<
B, let h
∈
L
∞
(
(
A
,
B
)
)
∩
C
1
(
(
A
,
B
)
), and suppose that (
h
(
t
)
/
t
)
′
=
0 has at most k roots in (
A
,
B
)
∖
(
-
1,1
). Then
(11)
|
∫
A
B
‍
sin
⁡
(
t
)
h
(
t
)
d
t
t
|
≤
(
2
+
4
k
)
∥
h
∥
∞
.The second one is Van der Corput lemma which plays an important role in estimating related multipliers. This lemma appears in several books or papers, compare, for example, Stein [11, P.332].Lemma 6.
Supposeϕ is real-valued and smooth in (
a
,
b
), and that |
ϕ
(
k
)
(
x
)
|
≥
1 for all x
∈
(
a
,
b
). Then
(12)
|
∫
a
b
‍
e
i
λ
ϕ
(
x
)
d
x
|
≤
C
k
λ
-
1
/
k
holds when k
≥
2; the bound C
k is independent of ϕ and λ.To estimate|
m
z
(
ξ
,
η
)
|, for fixed η, we choose t
0
>
0 so that |
η
|
γ
(
t
0
)
=
1 and decompose m
z as
(13)
|
m
z
(
ξ
,
η
)
|
≤
|
p
.
v
.
∫
|
t
|
<
t
0
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
|
+
|
∫
|
t
|
≥
t
0
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
|
.
First of all, we consider the first integral. For any 0
<
ε
<
t
0,
(14)
|
∫
ε
<
|
t
|
<
t
0
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
|
≤
|
∫
ε
<
|
t
|
<
t
0
‍
e
-
2
π
i
ξ
t
(
e
-
2
π
i
η
γ
(
t
)
-
1
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
|
+
|
∫
ε
<
|
t
|
<
t
0
‍
e
-
2
π
i
ξ
t
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
|
:
=
I
+
II
.
Notice that 1
≤
1
+
η
2
γ
2
(
t
)
≤
2 for |
t
|
≤
t
0. Because of the convexity of γ
(
t
), we have γ
(
t
)
≤
t
γ
′
(
t
) for t
>
0. Thus,
(15)
I
≤
∫
ε
<
|
t
|
<
t
0
‍
|
e
-
2
π
i
η
γ
(
t
)
-
1
|
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
d
t
|
t
|
≤
C
∫
ε
<
|
t
|
<
t
0
‍
|
η
γ
(
t
)
|
[
1
+
η
2
γ
2
(
t
)
]
1
/
4
-
δ
d
t
|
t
|
≤
C
2
5
/
4
-
δ
|
η
|
∫
ε
t
0
‍
γ
(
t
)
d
t
t
≤
C
2
5
/
4
-
δ
|
η
|
∫
ε
t
0
‍
γ
′
(
t
)
d
t
≤
C
δ
|
η
|
γ
(
t
0
)
≤
C
δ
.
For the integral II, note that γ
(
t
) is odd. An easy calculation shows that
(16)
II
=
2
|
∫
ε
t
0
‍
sin
(
2
π
ξ
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
|
≤
2
|
∫
ε
t
0
‍
sin
(
2
π
ξ
t
)
[
1
+
η
2
γ
2
(
t
)
]
1
/
4
-
δ
×
cos
⁡
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
]
d
t
t
|
+
2
|
∫
ε
t
0
‍
sin
(
2
π
ξ
t
)
[
1
+
η
2
γ
2
(
t
)
]
1
/
4
-
δ
×
sin
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
]
d
t
t
|
.
It is trivial that [
1
+
η
2
γ
2
(
t
)
]
1
/
4
-
δ
cos
⁡
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
] is piecewise monotone in [
ε
,
t
0
]. Further, it has at most C
[
1
+
|
Im
⁡
(
z
)
|
] monotone intervals in [
ε
,
t
0
]. Indeed, the derivative of [
1
+
η
2
γ
2
(
t
)
]
1
/
4
-
δ
cos
⁡
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
] is
(17)
2
η
2
γ
(
t
)
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
3
/
4
+
δ
×
{
(
1
4
-
δ
)
cos
⁡
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
]
-
Im
⁡
(
z
)
sin
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
]
1
4
-
δ
}
.
γ
(
t
)
γ
′
(
t
) is nonzero for t
∈
[
ε
,
t
0
]. So, we consider the following equation:
(18)
(
1
4
-
δ
)
cos
⁡
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
]
-
Im
⁡
(
z
)
sin
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
]
=
0
.
We suppose Im
⁡
(
z
)
≠
0; otherwise, it is obvious. Then, above equation is
(19)
tan
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
]
=
(
1
/
4
-
δ
)
Im
⁡
(
z
)
.
For Im
⁡
(
z
)
>
0 and t
∈
[
ε
,
t
0
], we have Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
∈
(
0
,
Im
⁡
(
z
)
ln
⁡
2
). There are about (
Im
⁡
(
z
)
ln
⁡
2
/
π
)
+
1 periods in interval (
0
,
Im
⁡
(
z
)
ln
⁡
2
), only one root exists in one period. So, (19) has at most (
Im
⁡
(
z
)
ln
⁡
2
/
π
)
+
1 roots. That mean that [
1
+
η
2
γ
2
(
t
)
]
1
/
4
-
δ
cos
⁡
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
] has at most C
[
1
+
|
Im
⁡
(
z
)
|
] stationary points in [
ε
,
t
0
], that is there are at most C
[
1
+
|
Im
⁡
(
z
)
|
] monotone intervals. The case Im
⁡
(
z
)
<
0 can be treated in the same way. We suppose it is monotone, otherwise, we deal with the integral on every monotone interval. By the second mean value theorem and Lemma 5, there exists t
1
∈
(
ε
,
t
0
) such that
(20)
|
∫
ε
t
0
‍
sin
⁡
(
2
π
ξ
t
)
[
1
+
η
2
γ
2
(
t
)
]
1
/
4
-
δ
×
cos
⁡
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
]
d
t
t
|
≤
[
1
+
η
2
γ
2
(
ε
)
]
1
/
4
-
δ
|
cos
⁡
⁡
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
ε
)
)
]
|
×
|
∫
ε
t
1
‍
sin
(
2
π
ξ
t
)
d
t
t
|
+
[
1
+
η
2
γ
2
(
t
0
)
]
1
/
4
-
δ
|
cos
⁡
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
0
)
)
]
|
×
|
∫
t
1
t
0
‍
sin
(
2
π
ξ
t
)
d
t
t
|
≤
C
δ
[
1
+
|
Im
⁡
(
z
)
|
]
.
In the same way, we can prove that
(21)
|
∫
ε
t
0
‍
sin
(
2
π
ξ
t
)
[
1
+
η
2
γ
2
(
t
)
]
1
/
4
-
δ
×
sin
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
]
d
t
t
|
≤
C
δ
[
1
+
|
Im
⁡
(
z
)
|
]
.We next deal with the the part of the integral where|
t
|
≥
t
0. If γ
′
′
(
t
) is monotone increasing, we set
(22)
φ
(
t
)
=
-
∫
t
∞
‍
e
-
2
π
i
[
ξ
s
+
η
γ
(
s
)
]
d
s
,
while if γ
′
′
(
t
) is monotone decreasing, put
(23)
φ
(
t
)
=
∫
t
0
t
‍
e
-
2
π
i
[
ξ
s
+
η
γ
(
s
)
]
d
s
.
Obviously, φ
′
(
t
)
=
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]. By integration by part, we have
(24)
∫
t
0
∞
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
=
∫
t
0
∞
‍
φ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
=
φ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
t
|
t
0
∞
-
∫
t
0
∞
‍
φ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
×
2
z
η
2
t
γ
(
t
)
γ
′
(
t
)
-
1
-
η
2
γ
2
(
t
)
t
2
d
t
.
Notice that |
[
ξ
s
+
η
γ
(
s
)
]
′
′
|
=
|
η
|
γ
′
′
(
s
)
≥
|
η
|
γ
′
′
(
t
) for s
∈
[
t
,
∞
] or s
∈
[
t
0
,
t
]. Van der Corput lemma shows that |
φ
(
t
)
|
≤
C
[
|
η
|
γ
′
′
(
t
)
]
-
1
/
2 in either case. A combination of γ
(
t
)
≤
t
γ
′
(
t
) with the hypothesis γ
′
(
t
)
≤
C
t
γ
′
′
(
t
) implies that γ
(
t
)
≤
C
t
2
γ
′
′
(
t
). So, for the boundary terms, t
∈
[
t
0
,
∞
), we have
(25)
|
φ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
t
|
≤
C
[
|
η
|
γ
′
′
(
t
)
]
-
1
/
2
[
2
η
2
γ
2
(
t
)
]
1
/
4
-
δ
|
t
|
≤
C
[
|
η
|
γ
′
′
(
t
)
t
2
]
-
1
/
2
[
η
γ
(
t
)
]
1
/
2
-
2
δ
≤
C
[
η
γ
(
t
)
]
-
2
δ
≤
C
.
Finally, we consider the integrated terms; it is bounded by
(26)
|
∫
t
0
∞
‍
φ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
×
2
z
η
2
t
γ
(
t
)
γ
′
(
t
)
-
1
-
η
2
γ
2
(
t
)
t
2
d
t
|
≤
C
∫
t
0
∞
‍
[
|
η
|
γ
′
′
(
t
)
]
-
1
/
2
[
|
η
|
γ
(
t
)
]
-
3
/
2
-
2
δ
×
(
2
[
|
Re
(
z
)
|
+
|
Im
⁡
(
z
)
|
]
η
2
t
γ
(
t
)
×
γ
′
(
t
)
+
2
η
2
γ
2
(
t
)
)
×
(
t
2
)
-
1
d
t
≤
C
[
|
Re
(
z
)
|
+
|
Im
⁡
(
z
)
|
]
|
η
|
-
2
δ
×
∫
t
0
∞
‍
[
t
2
γ
′
′
(
t
)
]
-
1
/
2
[
γ
(
t
)
]
-
1
/
2
-
2
δ
γ
′
(
t
)
d
t
+
C
|
η
|
-
2
δ
∫
t
0
∞
‍
[
t
2
γ
′
′
(
t
)
]
-
1
/
2
[
γ
(
t
)
]
-
1
/
2
-
2
δ
γ
(
t
)
t
d
t
≤
C
[
1
+
|
Im
⁡
(
z
)
|
]
|
η
|
-
2
δ
×
∫
t
0
∞
‍
[
γ
(
t
)
]
-
1
-
2
δ
γ
′
(
t
)
d
t
≤
C
δ
[
1
+
|
Im
⁡
(
z
)
|
]
.
The integral
(27)
∫
-
∞
-
t
0
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
can be handled similarly. This completes the proof of (10).


### 2.2. The Boundedness ofℋ
z on L
p
(
ℝ
2
,
l
q
)

In this subsection, we show that(28)
∥
(
∑
j
∈
ℤ
‍
|
ℋ
z
f
j
|
q
)
1
/
q
∥
L
p

≤
C
[
1
+
|
Im
⁡
(
z
)
|
]
2
∥
(
∑
j
∈
ℤ
‍
|
f
j
|
q
)
1
/
q
∥
L
p
,
where Re
(
z
)
<
-
1, 1
<
p, q
<
∞, the constant C depends on Re
(
z
), and is independent of Im
⁡
(
z
).To prove (28), we need a vector-valued Fourier multiplier theorem. We have to recall some definitions before we present the theorem.Definition 7.
LetX be a Banach space; X is said to be an UMD-space, if the Hilbert transform is bounded on L
p
(
ℝ
,
X
) for some (and then all) p
∈
(
1
,
∞
).Definition 8.
A basis{
e
j
}
j in a Banach space X is called unconditional, if there is a constant C such that for every x
=
∑
j
‍
t
j
e
j
∈
X one has ∥
∑
j
‍
ϵ
j
t
j
e
j
∥
≤
C
∥
x
∥, for all ϵ
j
=
±
1 for j
∈
ℕ.Definition 9.
A BanachX is said to have local unconditional structure if there exists a constant C such that for any finite-dimensional subspace X
0 of X there exists a Banach space F with an unconditional basis and operators u
0
:
X
0
→
F and w
0
:
F
→
X such that the natural embedding j
:
X
0
→
X admits a factorization j
=
w
0
u
0 and ∥
u
0
∥
∥
w
0
∥
≤
C.Remark 10.
Well-known examples ofU
M
D-spaces are L
p
(
ℝ
n
), l
q and L
p
,
q
(
ℝ
n
), 1
<
p, q
<
∞, three examples are also Banach lattices for their usual norm and the pointwise order. Further, all Banach lattices have local unconditional structure.Letm
:
ℝ
n
→
ℂ be a bounded function, we associate operators T
m defined on the test functions f
∈
𝒮
(
ℝ
n
,
X
) by
(29)
T
m
f
(
x
)
=
(
m
f
^
)
∨
(
x
)
.
As a vector-valued Fourier multiplier theorem, we state the following vector-valued Mikhlin theorem which already was proved in [12].Lemma 11.
LetX be an UMD-space with local unconditional structure. Then for any p
∈
(
1
,
∞
) there is a constant C
<
∞ such that
(30)
∥
T
m
∥
L
p
(
ℝ
n
,
X
)
≤
C
sup
⁡
⁡
{
|
x
α
∂
α
m
(
x
)
|
:
x
∈
ℝ
n
∖
{
0
}
,
α
≤
(
1,1
,
…
,
1
)
}
.Forn
=
2, in view of Lemma 11 and Remark 10, it suffices to show that the following functions
(31)
m
z
(
ξ
,
η
)
,
ξ
∂
m
z
∂
ξ
(
ξ
,
η
)
,
η
∂
m
z
∂
η
(
ξ
,
η
)
,
ξ
η
∂
2
m
z
∂
ξ
∂
η
(
ξ
,
η
)
are uniformly bounded on ℝ
2 for Re
(
z
)
<
-
1.The uniform boundedness ofm
z
(
ξ
,
η
) is trivial, it can be established by minor modification of the proof of (10). Without repetition, we omit the proof.


#### 2.2.1. The Boundedness ofξ
(
∂
m
z
/
∂
ξ
)
(
ξ
,
η
)

Integration by part implies that(32)
ξ
∂
m
z
∂
ξ
(
ξ
,
η
)
=
-
2
π
i
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
ξ
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
=
∫
ℝ
‍
d
d
t
(
e
-
2
π
i
ξ
t
)
e
-
2
π
i
η
γ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
=
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
|
-
∞
∞
+
2
π
i
η
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
-
2
z
η
2
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
γ
(
t
)
γ
′
(
t
)
d
t
.Note thatRe
(
z
)
<
-
1, for t
∈
ℝ, we have |
[
1
+
η
2
γ
2
(
t
)
]
z
|
=
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
≤
1. The boundary terms is bounded by 1
/
π.ForRe
(
z
)
<
-
1, making the change of variables u
=
|
η
|
γ
(
t
), we obtain
(33)
|
η
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
|
≤
∫
ℝ
‍
γ
′
(
t
)
|
η
|
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
d
t
≤
∫
ℝ
‍
(
1
+
u
2
)
Re
(
z
)
d
u
≤
π
.In the similar way, the second integrated term can be dominated by(34)
|
z
η
2
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
γ
(
t
)
γ
′
(
t
)
d
t
|
≤
2
|
z
|
∫
0
∞
‍
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
-
1
η
2
γ
(
t
)
γ
′
(
t
)
d
t
≤
|
z
|
∫
0
∞
‍
(
1
+
u
)
Re
(
z
)
-
1
d
u
≤
|
z
|
|
Re
(
z
)
|
≤
1
+
|
Im
⁡
(
z
)
|
.
Therefore, for Re
(
z
)
<
-
1,
(35)
|
ξ
∂
m
z
∂
ξ
(
ξ
,
η
)
|
≤
C
[
1
+
|
Im
⁡
(
z
)
|
]
.


#### 2.2.2. The Boundedness ofη
(
∂
m
z
/
∂
η
)
(
ξ
,
η
)

Integrating by parts, we obtain(36)
η
∂
m
z
∂
η
(
ξ
,
η
)
=
-
2
π
i
p
.
v
.
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
+
2
z
p
.
v
.
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
t
.To estimate above two integrals, we follow the argument similar to that in the proof of (10). For the first integral, for any ε
>
0, it suffices to bound the following two parts
(37)
∫
ε
<
|
t
|
<
t
0
‍
|
η
|
|
γ
(
t
)
|
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
d
t
|
t
|
,
∫
|
t
|
≥
t
0
‍
|
η
|
|
γ
(
t
)
|
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
d
t
|
t
|
.
Recall that t
0
>
0 was chosen so that |
η
|
γ
(
t
0
)
=
1, and γ
(
t
)
≤
t
γ
′
(
t
) because of the convexity. Thus,
(38)
∫
ε
<
|
t
|
<
t
0
‍
|
η
|
|
γ
(
t
)
|
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
d
t
|
t
|
≤
2
|
η
|
∫
0
t
0
‍
γ
(
t
)
t
d
t
≤
2
|
η
|
∫
0
t
0
‍
γ
′
(
t
)
d
t
≤
2
.
For Re
(
z
)
<
-
1, an elementary calculation shows that
(39)
∫
|
t
|
≥
t
0
‍
|
η
|
|
γ
(
t
)
|
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
d
t
|
t
|
≤
2
|
η
|
2
Re
(
z
)
+
1
∫
t
0
∞
‍
γ
2
Re
(
z
)
(
t
)
γ
(
t
)
t
d
t
≤
2
|
2
Re
(
z
)
+
1
|
≤
2
.Similarly, the second integral can be controlled by(40)
|
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
t
|
≤
2
|
z
|
∫
0
t
0
‍
η
2
γ
2
(
t
)
d
t
t
+
2
|
z
|
∫
t
0
∞
‍
η
2
γ
2
(
t
)
[
η
2
γ
2
(
t
)
]
Re
(
z
)
-
1
d
t
t
≤
2
|
z
|
η
2
∫
0
t
0
‍
γ
(
t
)
γ
′
(
t
)
d
t
+
2
|
z
|
η
2
Re
(
z
)
∫
t
0
∞
‍
γ
2
Re
(
z
)
-
1
(
t
)
γ
′
(
t
)
d
t
≤
|
z
|
+
|
z
|
|
Re
(
z
)
|
≤
2
|
Re
(
z
)
|
[
1
+
|
Im
⁡
(
z
)
|
]
.
Therefore, for Re
(
z
)
<
-
1,
(41)
|
ξ
∂
m
z
∂
ξ
(
ξ
,
η
)
|
≤
C
[
1
+
|
Im
⁡
(
z
)
|
]
.


#### 2.2.3. The Boundedness ofξ
η
(
∂
2
m
z
/
∂
ξ
∂
η
)
(
ξ
,
η
)

To take care ofξ
η
(
∂
2
m
z
/
∂
ξ
∂
η
)
(
ξ
,
η
), we note that it can be written as
(42)
ξ
η
∂
2
m
z
∂
ξ
∂
η
(
ξ
,
η
)
=
-
4
π
2
ξ
η
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
γ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
-
4
π
i
z
ξ
η
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
η
γ
2
(
t
)
d
t
.
For the first term, integrating by parts, we obtain
(43)
4
π
2
ξ
η
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
γ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
=
2
π
i
∫
ℝ
‍
d
d
t
(
e
-
2
π
i
ξ
t
)
e
-
2
π
i
η
γ
(
t
)
[
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
=
2
π
i
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
|
-
∞
∞
-
4
π
2
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
′
(
t
)
[
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
-
2
π
i
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
-
4
π
i
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
η
3
γ
2
(
t
)
γ
′
(
t
)
d
t
.Obviously, forRe
(
z
)
<
-
1, t
∈
ℝ, |
2
π
i
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
η
γ
(
t
)
] [
1
+
η
2
γ
2
(
t
)
]
z
|
≤
2
π
|
η
|
|
γ
(
t
)
| [
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
≤
2
π. So, the boundary terms is bounded by 2
π.For the first integrated terms, making the change of variablesu
=
η
2
γ
2
(
t
), one has
(44)
|
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
′
(
t
)
[
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
|
≤
2
∫
0
∞
‍
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
η
2
γ
(
t
)
γ
′
(
t
)
d
t
≤
∫
0
∞
‍
(
1
+
u
)
Re
(
z
)
d
u
≤
1
|
Re
(
z
)
+
1
|
.The second integrated terms can be treated in the same way, letu
=
η
γ
(
t
),
(45)
|
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
|
≤
∫
ℝ
‍
(
1
+
u
2
)
Re
(
z
)
d
u
≤
π
.Similarly, a trivial calculation shows that(46)
|
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
η
3
γ
2
(
t
)
γ
′
(
t
)
d
t
|
≤
2
|
z
|
∫
0
∞
‍
u
2
(
1
+
u
2
)
Re
(
z
)
-
1
d
u
≤
π
|
z
|
.
The second term can be handled similarly. Integrating by parts, we decomposed it as
(47)
4
π
i
z
ξ
η
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
η
γ
2
(
t
)
d
t
=
2
π
i
z
∫
ℝ
‍
d
d
t
(
e
-
2
π
i
ξ
t
)
e
-
2
π
i
η
γ
(
t
)
×
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
=
2
π
i
z
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
|
-
∞
∞
-
4
π
2
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
′
(
t
)
×
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
-
4
π
i
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
(
t
)
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
-
4
π
i
z
(
z
-
1
)
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
2
×
η
2
γ
(
t
)
γ
′
(
t
)
d
t
.Obviously, forRe
(
z
)
<
-
1, t
∈
ℝ, |
z
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
|
≤
|
z
|. The boundary term is dominated by 4
π
|
z
|.For the first integrated terms, by making the change of variablesu
=
η
γ
(
t
), we have the estimate
(48)
|
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
′
(
t
)
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
|
≤
|
z
|
∫
ℝ
‍
u
2
(
1
+
u
2
)
Re
(
z
)
-
1
d
t
≤
π
|
z
|
.To estimate the second integrated terms, we make the transformationu
=
η
2
γ
2
(
t
) and get
(49)
|
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
(
t
)
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
|
≤
|
z
|
∫
0
∞
‍
(
1
+
u
)
Re
(
z
)
-
1
d
u
≤
|
z
|
|
Re
(
z
)
|
.Similarly, the third integrated terms can be treated as(50)
|
z
(
z
-
1
)
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
2
η
4
γ
3
(
t
)
γ
′
(
t
)
d
t
|
≤
|
z
(
z
-
1
)
|
∫
0
∞
‍
(
1
+
u
)
Re
(
z
)
-
1
d
u
≤
|
z
(
z
-
1
)
|
|
Re
(
z
)
|
.
Note that for Re
(
z
)
<
-
1, we have the following elementary estimates
(51)
|
z
|
≤
|
Re
(
z
)
|
[
1
+
|
Im
⁡
(
z
)
|
]
,
|
z
-
1
|
≤
|
Re
(
z
)
-
1
|
[
1
+
|
Im
⁡
(
z
)
|
]
.
Finally, combining above eight estimates, we obtain
(52)
|
ξ
η
∂
2
m
z
∂
ξ
∂
η
(
ξ
,
η
)
|
≤
C
[
1
+
Im
⁡
(
z
)
]
2
.


### 2.3.  Conclusion

To complete the proof of our main result, we need al
q-valued analytic interpolation theorem, which is a consequence of Theorem  1 in [13], a concrete proof can be found in [14].LetS denote the closed strip {
a
≤
Re
(
z
)
≤
b
}
⊂
C. {
U
z
}
z
∈
S is a family of uniformly bounded linear operators on L
2
(
R
n
), that is, there is an M
>
0 such that
(53)
∥
U
z
∥
L
2
→
L
2
≤
M
,
∀
z
∈
S
.
Moreover, the function
(54)
z
⟼
∫
R
n
‍
U
z
f
(
x
)
g
(
x
)
d
x
is continuous on S and analytic in the interior of S, whenever f
,
g
∈
L
2
(
R
n
).Lemma 12.
With the above assumptions, if for1
<
p
i
,
q
i
<
∞
,
(
i
=
0,1
), {
U
z
} satisfies the following conditions:(i)
∥
(
∑
j
∈
Z
|
U
z
f
j
|
q
0
)
1
/
q
0
∥
L
p
0
≤
M
0
∥
(
∑
j
∈
Z
|
f
j
|
q
0
)
1
/
q
0
∥
L
p
0
w
h
e
n
Re
(
z
)
=
a;
(ii)
∥
(
∑
j
∈
Z
|
U
z
f
j
|
q
1
)
1
/
q
1
∥
L
p
1
≤
M
1
∥
(
∑
j
∈
Z
|
f
j
|
q
1
)
1
/
q
1
∥
L
p
1
w
h
e
n
Re
(
z
)
=
b. Then we have
(55)
∥
(
∑
j
∈
Z
‍
|
U
a
(
1
-
θ
)
+
θ
b
f
j
|
q
)
1
/
q
∥
L
p
≤
M
0
1
-
θ
M
1
θ
∥
(
∑
j
∈
Z
‍
|
f
j
|
q
)
1
/
q
∥
L
p
,
where 0
<
θ
<
1, 1
/
q
=
(
(
1
-
θ
)
/
q
0
)
+
θ
/
q
1 and 1
/
p
=
(
(
1
-
θ
)
/
q
0
)
+
θ
/
p
1.LetU
z
f
(
x
)
=
e
z
2
ℋ
z
f
(
x
). Note that |
e
z
2
|
=
e
Re
(
z
)
2
-
Im
⁡
(
z
)
2, then there exists a constant M
0 which is independent of Im
⁡
(
z
) such that
(56)
∥
(
∑
j
∈
Z
‍
|
U
z
f
j
|
2
)
1
/
2
∥
L
2
≤
C
δ
e
-
Im
⁡
(
z
)
2
[
1
+
|
Im
⁡
(
z
)
|
]
∥
(
∑
j
∈
Z
‍
|
f
j
|
2
)
1
/
2
∥
L
2
≤
M
0
∥
(
∑
j
∈
Z
‍
|
f
j
|
2
)
1
/
2
∥
L
2
when
Re
(
z
)
=
1
/
4
-
δ
.
Also, for 1
<
p
1,  q
1
<
∞, there exists a constant M
1 which is independent of Im
⁡
(
z
) such that
(57)
∥
(
∑
j
∈
Z
‍
|
U
z
f
j
|
q
1
)
1
/
q
1
∥
L
p
1
≤
M
1
∥
(
∑
j
∈
Z
‍
|
f
j
|
q
1
)
1
/
q
1
∥
L
p
1
∥
(
∑
j
∈
Z
‍
|
U
z
f
j
|
q
1
)
1
/
q
1
∥
L
p
1
≤
M
1
≤
when
Re
(
z
)
<
-
1
.
For 2
≤
p, q
<
5
/
2, there exists 1
<
p
1, q
1
<
∞ so that
(58)
1
p
=
θ
2
+
1
-
θ
p
1
,
1
q
=
θ
2
+
1
-
θ
q
1
,
where θ
∈
(
0,1
) and satisfies the equation
(59)
(
1
4
-
δ
)
θ
+
(
-
1
-
ε
)
(
1
-
θ
)
=
0
for some ε
>
0 and 0
<
δ
<
1
/
4. Therefore, by Lemma 12, we obtain
(60)
∥
(
∑
j
∈
Z
‍
|
ℋ
f
j
|
q
)
1
/
q
∥
L
p
=
∥
(
∑
j
∈
Z
‍
|
ℋ
0
f
j
|
q
)
1
/
q
∥
L
p
=
∥
(
∑
j
∈
Z
‍
|
U
0
f
j
|
q
)
1
/
q
∥
L
p
≤
C
∥
(
∑
j
∈
Z
‍
|
f
j
|
q
)
1
/
q
∥
L
p
for 2
≤
p, q
<
5
/
2. The duality implies the result for 5
/
3
<
p, q
≤
2. This completes the proof of Theorem 3.


## 2.1. The Boundedness ofℋ
z on L
2
(
ℝ
2
,
l
2
)

In this subsection, we prove that(9)
∥
(
∑
j
∈
ℤ
‍
|
ℋ
z
f
j
|
2
)
1
/
2
∥
L
2

≤

C
δ
[
1
+
|
Im
⁡
(
z
)
|
]
∥
(
∑
j
∈
ℤ
‍
|
f
j
|
2
)
1
/
2
∥
L
2
,
where Re
(
z
)
=
(
1
/
4
)
-
δ for some δ
>
0.Clearly, the boundedness ofℋ
z on L
2
(
ℝ
2
,
l
2
) is equivalent to the uniform boundedness of the corresponding multiplier m
z
(
ξ
,
η
). Thus, we just need to show that
(10)
|
m
z
(
ξ
,
η
)
|
≤
C
δ
[
1
+
|
Im
⁡
(
z
)
|
]
,
where the constant C
δ is independent of Im
⁡
(
z
).In proving (10), some of our estimates will require the following lemmas. The first one is detailed in [4, Lemma  2.1].Lemma 5.
LetA
<
B, let h
∈
L
∞
(
(
A
,
B
)
)
∩
C
1
(
(
A
,
B
)
), and suppose that (
h
(
t
)
/
t
)
′
=
0 has at most k roots in (
A
,
B
)
∖
(
-
1,1
). Then
(11)
|
∫
A
B
‍
sin
⁡
(
t
)
h
(
t
)
d
t
t
|
≤
(
2
+
4
k
)
∥
h
∥
∞
.The second one is Van der Corput lemma which plays an important role in estimating related multipliers. This lemma appears in several books or papers, compare, for example, Stein [11, P.332].Lemma 6.
Supposeϕ is real-valued and smooth in (
a
,
b
), and that |
ϕ
(
k
)
(
x
)
|
≥
1 for all x
∈
(
a
,
b
). Then
(12)
|
∫
a
b
‍
e
i
λ
ϕ
(
x
)
d
x
|
≤
C
k
λ
-
1
/
k
holds when k
≥
2; the bound C
k is independent of ϕ and λ.To estimate|
m
z
(
ξ
,
η
)
|, for fixed η, we choose t
0
>
0 so that |
η
|
γ
(
t
0
)
=
1 and decompose m
z as
(13)
|
m
z
(
ξ
,
η
)
|
≤
|
p
.
v
.
∫
|
t
|
<
t
0
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
|
+
|
∫
|
t
|
≥
t
0
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
|
.
First of all, we consider the first integral. For any 0
<
ε
<
t
0,
(14)
|
∫
ε
<
|
t
|
<
t
0
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
|
≤
|
∫
ε
<
|
t
|
<
t
0
‍
e
-
2
π
i
ξ
t
(
e
-
2
π
i
η
γ
(
t
)
-
1
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
|
+
|
∫
ε
<
|
t
|
<
t
0
‍
e
-
2
π
i
ξ
t
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
|
:
=
I
+
II
.
Notice that 1
≤
1
+
η
2
γ
2
(
t
)
≤
2 for |
t
|
≤
t
0. Because of the convexity of γ
(
t
), we have γ
(
t
)
≤
t
γ
′
(
t
) for t
>
0. Thus,
(15)
I
≤
∫
ε
<
|
t
|
<
t
0
‍
|
e
-
2
π
i
η
γ
(
t
)
-
1
|
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
d
t
|
t
|
≤
C
∫
ε
<
|
t
|
<
t
0
‍
|
η
γ
(
t
)
|
[
1
+
η
2
γ
2
(
t
)
]
1
/
4
-
δ
d
t
|
t
|
≤
C
2
5
/
4
-
δ
|
η
|
∫
ε
t
0
‍
γ
(
t
)
d
t
t
≤
C
2
5
/
4
-
δ
|
η
|
∫
ε
t
0
‍
γ
′
(
t
)
d
t
≤
C
δ
|
η
|
γ
(
t
0
)
≤
C
δ
.
For the integral II, note that γ
(
t
) is odd. An easy calculation shows that
(16)
II
=
2
|
∫
ε
t
0
‍
sin
(
2
π
ξ
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
|
≤
2
|
∫
ε
t
0
‍
sin
(
2
π
ξ
t
)
[
1
+
η
2
γ
2
(
t
)
]
1
/
4
-
δ
×
cos
⁡
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
]
d
t
t
|
+
2
|
∫
ε
t
0
‍
sin
(
2
π
ξ
t
)
[
1
+
η
2
γ
2
(
t
)
]
1
/
4
-
δ
×
sin
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
]
d
t
t
|
.
It is trivial that [
1
+
η
2
γ
2
(
t
)
]
1
/
4
-
δ
cos
⁡
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
] is piecewise monotone in [
ε
,
t
0
]. Further, it has at most C
[
1
+
|
Im
⁡
(
z
)
|
] monotone intervals in [
ε
,
t
0
]. Indeed, the derivative of [
1
+
η
2
γ
2
(
t
)
]
1
/
4
-
δ
cos
⁡
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
] is
(17)
2
η
2
γ
(
t
)
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
3
/
4
+
δ
×
{
(
1
4
-
δ
)
cos
⁡
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
]
-
Im
⁡
(
z
)
sin
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
]
1
4
-
δ
}
.
γ
(
t
)
γ
′
(
t
) is nonzero for t
∈
[
ε
,
t
0
]. So, we consider the following equation:
(18)
(
1
4
-
δ
)
cos
⁡
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
]
-
Im
⁡
(
z
)
sin
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
]
=
0
.
We suppose Im
⁡
(
z
)
≠
0; otherwise, it is obvious. Then, above equation is
(19)
tan
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
]
=
(
1
/
4
-
δ
)
Im
⁡
(
z
)
.
For Im
⁡
(
z
)
>
0 and t
∈
[
ε
,
t
0
], we have Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
∈
(
0
,
Im
⁡
(
z
)
ln
⁡
2
). There are about (
Im
⁡
(
z
)
ln
⁡
2
/
π
)
+
1 periods in interval (
0
,
Im
⁡
(
z
)
ln
⁡
2
), only one root exists in one period. So, (19) has at most (
Im
⁡
(
z
)
ln
⁡
2
/
π
)
+
1 roots. That mean that [
1
+
η
2
γ
2
(
t
)
]
1
/
4
-
δ
cos
⁡
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
] has at most C
[
1
+
|
Im
⁡
(
z
)
|
] stationary points in [
ε
,
t
0
], that is there are at most C
[
1
+
|
Im
⁡
(
z
)
|
] monotone intervals. The case Im
⁡
(
z
)
<
0 can be treated in the same way. We suppose it is monotone, otherwise, we deal with the integral on every monotone interval. By the second mean value theorem and Lemma 5, there exists t
1
∈
(
ε
,
t
0
) such that
(20)
|
∫
ε
t
0
‍
sin
⁡
(
2
π
ξ
t
)
[
1
+
η
2
γ
2
(
t
)
]
1
/
4
-
δ
×
cos
⁡
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
]
d
t
t
|
≤
[
1
+
η
2
γ
2
(
ε
)
]
1
/
4
-
δ
|
cos
⁡
⁡
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
ε
)
)
]
|
×
|
∫
ε
t
1
‍
sin
(
2
π
ξ
t
)
d
t
t
|
+
[
1
+
η
2
γ
2
(
t
0
)
]
1
/
4
-
δ
|
cos
⁡
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
0
)
)
]
|
×
|
∫
t
1
t
0
‍
sin
(
2
π
ξ
t
)
d
t
t
|
≤
C
δ
[
1
+
|
Im
⁡
(
z
)
|
]
.
In the same way, we can prove that
(21)
|
∫
ε
t
0
‍
sin
(
2
π
ξ
t
)
[
1
+
η
2
γ
2
(
t
)
]
1
/
4
-
δ
×
sin
[
Im
⁡
(
z
)
ln
⁡
(
1
+
η
2
γ
2
(
t
)
)
]
d
t
t
|
≤
C
δ
[
1
+
|
Im
⁡
(
z
)
|
]
.We next deal with the the part of the integral where|
t
|
≥
t
0. If γ
′
′
(
t
) is monotone increasing, we set
(22)
φ
(
t
)
=
-
∫
t
∞
‍
e
-
2
π
i
[
ξ
s
+
η
γ
(
s
)
]
d
s
,
while if γ
′
′
(
t
) is monotone decreasing, put
(23)
φ
(
t
)
=
∫
t
0
t
‍
e
-
2
π
i
[
ξ
s
+
η
γ
(
s
)
]
d
s
.
Obviously, φ
′
(
t
)
=
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]. By integration by part, we have
(24)
∫
t
0
∞
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
=
∫
t
0
∞
‍
φ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
=
φ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
t
|
t
0
∞
-
∫
t
0
∞
‍
φ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
×
2
z
η
2
t
γ
(
t
)
γ
′
(
t
)
-
1
-
η
2
γ
2
(
t
)
t
2
d
t
.
Notice that |
[
ξ
s
+
η
γ
(
s
)
]
′
′
|
=
|
η
|
γ
′
′
(
s
)
≥
|
η
|
γ
′
′
(
t
) for s
∈
[
t
,
∞
] or s
∈
[
t
0
,
t
]. Van der Corput lemma shows that |
φ
(
t
)
|
≤
C
[
|
η
|
γ
′
′
(
t
)
]
-
1
/
2 in either case. A combination of γ
(
t
)
≤
t
γ
′
(
t
) with the hypothesis γ
′
(
t
)
≤
C
t
γ
′
′
(
t
) implies that γ
(
t
)
≤
C
t
2
γ
′
′
(
t
). So, for the boundary terms, t
∈
[
t
0
,
∞
), we have
(25)
|
φ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
t
|
≤
C
[
|
η
|
γ
′
′
(
t
)
]
-
1
/
2
[
2
η
2
γ
2
(
t
)
]
1
/
4
-
δ
|
t
|
≤
C
[
|
η
|
γ
′
′
(
t
)
t
2
]
-
1
/
2
[
η
γ
(
t
)
]
1
/
2
-
2
δ
≤
C
[
η
γ
(
t
)
]
-
2
δ
≤
C
.
Finally, we consider the integrated terms; it is bounded by
(26)
|
∫
t
0
∞
‍
φ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
×
2
z
η
2
t
γ
(
t
)
γ
′
(
t
)
-
1
-
η
2
γ
2
(
t
)
t
2
d
t
|
≤
C
∫
t
0
∞
‍
[
|
η
|
γ
′
′
(
t
)
]
-
1
/
2
[
|
η
|
γ
(
t
)
]
-
3
/
2
-
2
δ
×
(
2
[
|
Re
(
z
)
|
+
|
Im
⁡
(
z
)
|
]
η
2
t
γ
(
t
)
×
γ
′
(
t
)
+
2
η
2
γ
2
(
t
)
)
×
(
t
2
)
-
1
d
t
≤
C
[
|
Re
(
z
)
|
+
|
Im
⁡
(
z
)
|
]
|
η
|
-
2
δ
×
∫
t
0
∞
‍
[
t
2
γ
′
′
(
t
)
]
-
1
/
2
[
γ
(
t
)
]
-
1
/
2
-
2
δ
γ
′
(
t
)
d
t
+
C
|
η
|
-
2
δ
∫
t
0
∞
‍
[
t
2
γ
′
′
(
t
)
]
-
1
/
2
[
γ
(
t
)
]
-
1
/
2
-
2
δ
γ
(
t
)
t
d
t
≤
C
[
1
+
|
Im
⁡
(
z
)
|
]
|
η
|
-
2
δ
×
∫
t
0
∞
‍
[
γ
(
t
)
]
-
1
-
2
δ
γ
′
(
t
)
d
t
≤
C
δ
[
1
+
|
Im
⁡
(
z
)
|
]
.
The integral
(27)
∫
-
∞
-
t
0
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
can be handled similarly. This completes the proof of (10).


## 2.2. The Boundedness ofℋ
z on L
p
(
ℝ
2
,
l
q
)

In this subsection, we show that(28)
∥
(
∑
j
∈
ℤ
‍
|
ℋ
z
f
j
|
q
)
1
/
q
∥
L
p

≤
C
[
1
+
|
Im
⁡
(
z
)
|
]
2
∥
(
∑
j
∈
ℤ
‍
|
f
j
|
q
)
1
/
q
∥
L
p
,
where Re
(
z
)
<
-
1, 1
<
p, q
<
∞, the constant C depends on Re
(
z
), and is independent of Im
⁡
(
z
).To prove (28), we need a vector-valued Fourier multiplier theorem. We have to recall some definitions before we present the theorem.Definition 7.
LetX be a Banach space; X is said to be an UMD-space, if the Hilbert transform is bounded on L
p
(
ℝ
,
X
) for some (and then all) p
∈
(
1
,
∞
).Definition 8.
A basis{
e
j
}
j in a Banach space X is called unconditional, if there is a constant C such that for every x
=
∑
j
‍
t
j
e
j
∈
X one has ∥
∑
j
‍
ϵ
j
t
j
e
j
∥
≤
C
∥
x
∥, for all ϵ
j
=
±
1 for j
∈
ℕ.Definition 9.
A BanachX is said to have local unconditional structure if there exists a constant C such that for any finite-dimensional subspace X
0 of X there exists a Banach space F with an unconditional basis and operators u
0
:
X
0
→
F and w
0
:
F
→
X such that the natural embedding j
:
X
0
→
X admits a factorization j
=
w
0
u
0 and ∥
u
0
∥
∥
w
0
∥
≤
C.Remark 10.
Well-known examples ofU
M
D-spaces are L
p
(
ℝ
n
), l
q and L
p
,
q
(
ℝ
n
), 1
<
p, q
<
∞, three examples are also Banach lattices for their usual norm and the pointwise order. Further, all Banach lattices have local unconditional structure.Letm
:
ℝ
n
→
ℂ be a bounded function, we associate operators T
m defined on the test functions f
∈
𝒮
(
ℝ
n
,
X
) by
(29)
T
m
f
(
x
)
=
(
m
f
^
)
∨
(
x
)
.
As a vector-valued Fourier multiplier theorem, we state the following vector-valued Mikhlin theorem which already was proved in [12].Lemma 11.
LetX be an UMD-space with local unconditional structure. Then for any p
∈
(
1
,
∞
) there is a constant C
<
∞ such that
(30)
∥
T
m
∥
L
p
(
ℝ
n
,
X
)
≤
C
sup
⁡
⁡
{
|
x
α
∂
α
m
(
x
)
|
:
x
∈
ℝ
n
∖
{
0
}
,
α
≤
(
1,1
,
…
,
1
)
}
.Forn
=
2, in view of Lemma 11 and Remark 10, it suffices to show that the following functions
(31)
m
z
(
ξ
,
η
)
,
ξ
∂
m
z
∂
ξ
(
ξ
,
η
)
,
η
∂
m
z
∂
η
(
ξ
,
η
)
,
ξ
η
∂
2
m
z
∂
ξ
∂
η
(
ξ
,
η
)
are uniformly bounded on ℝ
2 for Re
(
z
)
<
-
1.The uniform boundedness ofm
z
(
ξ
,
η
) is trivial, it can be established by minor modification of the proof of (10). Without repetition, we omit the proof.


### 2.2.1. The Boundedness ofξ
(
∂
m
z
/
∂
ξ
)
(
ξ
,
η
)

Integration by part implies that(32)
ξ
∂
m
z
∂
ξ
(
ξ
,
η
)
=
-
2
π
i
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
ξ
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
=
∫
ℝ
‍
d
d
t
(
e
-
2
π
i
ξ
t
)
e
-
2
π
i
η
γ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
=
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
|
-
∞
∞
+
2
π
i
η
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
-
2
z
η
2
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
γ
(
t
)
γ
′
(
t
)
d
t
.Note thatRe
(
z
)
<
-
1, for t
∈
ℝ, we have |
[
1
+
η
2
γ
2
(
t
)
]
z
|
=
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
≤
1. The boundary terms is bounded by 1
/
π.ForRe
(
z
)
<
-
1, making the change of variables u
=
|
η
|
γ
(
t
), we obtain
(33)
|
η
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
|
≤
∫
ℝ
‍
γ
′
(
t
)
|
η
|
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
d
t
≤
∫
ℝ
‍
(
1
+
u
2
)
Re
(
z
)
d
u
≤
π
.In the similar way, the second integrated term can be dominated by(34)
|
z
η
2
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
γ
(
t
)
γ
′
(
t
)
d
t
|
≤
2
|
z
|
∫
0
∞
‍
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
-
1
η
2
γ
(
t
)
γ
′
(
t
)
d
t
≤
|
z
|
∫
0
∞
‍
(
1
+
u
)
Re
(
z
)
-
1
d
u
≤
|
z
|
|
Re
(
z
)
|
≤
1
+
|
Im
⁡
(
z
)
|
.
Therefore, for Re
(
z
)
<
-
1,
(35)
|
ξ
∂
m
z
∂
ξ
(
ξ
,
η
)
|
≤
C
[
1
+
|
Im
⁡
(
z
)
|
]
.


### 2.2.2. The Boundedness ofη
(
∂
m
z
/
∂
η
)
(
ξ
,
η
)

Integrating by parts, we obtain(36)
η
∂
m
z
∂
η
(
ξ
,
η
)
=
-
2
π
i
p
.
v
.
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
+
2
z
p
.
v
.
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
t
.To estimate above two integrals, we follow the argument similar to that in the proof of (10). For the first integral, for any ε
>
0, it suffices to bound the following two parts
(37)
∫
ε
<
|
t
|
<
t
0
‍
|
η
|
|
γ
(
t
)
|
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
d
t
|
t
|
,
∫
|
t
|
≥
t
0
‍
|
η
|
|
γ
(
t
)
|
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
d
t
|
t
|
.
Recall that t
0
>
0 was chosen so that |
η
|
γ
(
t
0
)
=
1, and γ
(
t
)
≤
t
γ
′
(
t
) because of the convexity. Thus,
(38)
∫
ε
<
|
t
|
<
t
0
‍
|
η
|
|
γ
(
t
)
|
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
d
t
|
t
|
≤
2
|
η
|
∫
0
t
0
‍
γ
(
t
)
t
d
t
≤
2
|
η
|
∫
0
t
0
‍
γ
′
(
t
)
d
t
≤
2
.
For Re
(
z
)
<
-
1, an elementary calculation shows that
(39)
∫
|
t
|
≥
t
0
‍
|
η
|
|
γ
(
t
)
|
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
d
t
|
t
|
≤
2
|
η
|
2
Re
(
z
)
+
1
∫
t
0
∞
‍
γ
2
Re
(
z
)
(
t
)
γ
(
t
)
t
d
t
≤
2
|
2
Re
(
z
)
+
1
|
≤
2
.Similarly, the second integral can be controlled by(40)
|
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
t
|
≤
2
|
z
|
∫
0
t
0
‍
η
2
γ
2
(
t
)
d
t
t
+
2
|
z
|
∫
t
0
∞
‍
η
2
γ
2
(
t
)
[
η
2
γ
2
(
t
)
]
Re
(
z
)
-
1
d
t
t
≤
2
|
z
|
η
2
∫
0
t
0
‍
γ
(
t
)
γ
′
(
t
)
d
t
+
2
|
z
|
η
2
Re
(
z
)
∫
t
0
∞
‍
γ
2
Re
(
z
)
-
1
(
t
)
γ
′
(
t
)
d
t
≤
|
z
|
+
|
z
|
|
Re
(
z
)
|
≤
2
|
Re
(
z
)
|
[
1
+
|
Im
⁡
(
z
)
|
]
.
Therefore, for Re
(
z
)
<
-
1,
(41)
|
ξ
∂
m
z
∂
ξ
(
ξ
,
η
)
|
≤
C
[
1
+
|
Im
⁡
(
z
)
|
]
.


### 2.2.3. The Boundedness ofξ
η
(
∂
2
m
z
/
∂
ξ
∂
η
)
(
ξ
,
η
)

To take care ofξ
η
(
∂
2
m
z
/
∂
ξ
∂
η
)
(
ξ
,
η
), we note that it can be written as
(42)
ξ
η
∂
2
m
z
∂
ξ
∂
η
(
ξ
,
η
)
=
-
4
π
2
ξ
η
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
γ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
-
4
π
i
z
ξ
η
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
η
γ
2
(
t
)
d
t
.
For the first term, integrating by parts, we obtain
(43)
4
π
2
ξ
η
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
γ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
=
2
π
i
∫
ℝ
‍
d
d
t
(
e
-
2
π
i
ξ
t
)
e
-
2
π
i
η
γ
(
t
)
[
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
=
2
π
i
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
|
-
∞
∞
-
4
π
2
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
′
(
t
)
[
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
-
2
π
i
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
-
4
π
i
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
η
3
γ
2
(
t
)
γ
′
(
t
)
d
t
.Obviously, forRe
(
z
)
<
-
1, t
∈
ℝ, |
2
π
i
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
η
γ
(
t
)
] [
1
+
η
2
γ
2
(
t
)
]
z
|
≤
2
π
|
η
|
|
γ
(
t
)
| [
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
≤
2
π. So, the boundary terms is bounded by 2
π.For the first integrated terms, making the change of variablesu
=
η
2
γ
2
(
t
), one has
(44)
|
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
′
(
t
)
[
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
|
≤
2
∫
0
∞
‍
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
η
2
γ
(
t
)
γ
′
(
t
)
d
t
≤
∫
0
∞
‍
(
1
+
u
)
Re
(
z
)
d
u
≤
1
|
Re
(
z
)
+
1
|
.The second integrated terms can be treated in the same way, letu
=
η
γ
(
t
),
(45)
|
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
|
≤
∫
ℝ
‍
(
1
+
u
2
)
Re
(
z
)
d
u
≤
π
.Similarly, a trivial calculation shows that(46)
|
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
η
3
γ
2
(
t
)
γ
′
(
t
)
d
t
|
≤
2
|
z
|
∫
0
∞
‍
u
2
(
1
+
u
2
)
Re
(
z
)
-
1
d
u
≤
π
|
z
|
.
The second term can be handled similarly. Integrating by parts, we decomposed it as
(47)
4
π
i
z
ξ
η
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
η
γ
2
(
t
)
d
t
=
2
π
i
z
∫
ℝ
‍
d
d
t
(
e
-
2
π
i
ξ
t
)
e
-
2
π
i
η
γ
(
t
)
×
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
=
2
π
i
z
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
|
-
∞
∞
-
4
π
2
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
′
(
t
)
×
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
-
4
π
i
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
(
t
)
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
-
4
π
i
z
(
z
-
1
)
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
2
×
η
2
γ
(
t
)
γ
′
(
t
)
d
t
.Obviously, forRe
(
z
)
<
-
1, t
∈
ℝ, |
z
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
|
≤
|
z
|. The boundary term is dominated by 4
π
|
z
|.For the first integrated terms, by making the change of variablesu
=
η
γ
(
t
), we have the estimate
(48)
|
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
′
(
t
)
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
|
≤
|
z
|
∫
ℝ
‍
u
2
(
1
+
u
2
)
Re
(
z
)
-
1
d
t
≤
π
|
z
|
.To estimate the second integrated terms, we make the transformationu
=
η
2
γ
2
(
t
) and get
(49)
|
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
(
t
)
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
|
≤
|
z
|
∫
0
∞
‍
(
1
+
u
)
Re
(
z
)
-
1
d
u
≤
|
z
|
|
Re
(
z
)
|
.Similarly, the third integrated terms can be treated as(50)
|
z
(
z
-
1
)
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
2
η
4
γ
3
(
t
)
γ
′
(
t
)
d
t
|
≤
|
z
(
z
-
1
)
|
∫
0
∞
‍
(
1
+
u
)
Re
(
z
)
-
1
d
u
≤
|
z
(
z
-
1
)
|
|
Re
(
z
)
|
.
Note that for Re
(
z
)
<
-
1, we have the following elementary estimates
(51)
|
z
|
≤
|
Re
(
z
)
|
[
1
+
|
Im
⁡
(
z
)
|
]
,
|
z
-
1
|
≤
|
Re
(
z
)
-
1
|
[
1
+
|
Im
⁡
(
z
)
|
]
.
Finally, combining above eight estimates, we obtain
(52)
|
ξ
η
∂
2
m
z
∂
ξ
∂
η
(
ξ
,
η
)
|
≤
C
[
1
+
Im
⁡
(
z
)
]
2
.


## 2.2.1. The Boundedness ofξ
(
∂
m
z
/
∂
ξ
)
(
ξ
,
η
)

Integration by part implies that(32)
ξ
∂
m
z
∂
ξ
(
ξ
,
η
)
=
-
2
π
i
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
ξ
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
=
∫
ℝ
‍
d
d
t
(
e
-
2
π
i
ξ
t
)
e
-
2
π
i
η
γ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
=
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
|
-
∞
∞
+
2
π
i
η
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
-
2
z
η
2
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
γ
(
t
)
γ
′
(
t
)
d
t
.Note thatRe
(
z
)
<
-
1, for t
∈
ℝ, we have |
[
1
+
η
2
γ
2
(
t
)
]
z
|
=
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
≤
1. The boundary terms is bounded by 1
/
π.ForRe
(
z
)
<
-
1, making the change of variables u
=
|
η
|
γ
(
t
), we obtain
(33)
|
η
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
|
≤
∫
ℝ
‍
γ
′
(
t
)
|
η
|
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
d
t
≤
∫
ℝ
‍
(
1
+
u
2
)
Re
(
z
)
d
u
≤
π
.In the similar way, the second integrated term can be dominated by(34)
|
z
η
2
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
γ
(
t
)
γ
′
(
t
)
d
t
|
≤
2
|
z
|
∫
0
∞
‍
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
-
1
η
2
γ
(
t
)
γ
′
(
t
)
d
t
≤
|
z
|
∫
0
∞
‍
(
1
+
u
)
Re
(
z
)
-
1
d
u
≤
|
z
|
|
Re
(
z
)
|
≤
1
+
|
Im
⁡
(
z
)
|
.
Therefore, for Re
(
z
)
<
-
1,
(35)
|
ξ
∂
m
z
∂
ξ
(
ξ
,
η
)
|
≤
C
[
1
+
|
Im
⁡
(
z
)
|
]
.


## 2.2.2. The Boundedness ofη
(
∂
m
z
/
∂
η
)
(
ξ
,
η
)

Integrating by parts, we obtain(36)
η
∂
m
z
∂
η
(
ξ
,
η
)
=
-
2
π
i
p
.
v
.
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
t
+
2
z
p
.
v
.
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
t
.To estimate above two integrals, we follow the argument similar to that in the proof of (10). For the first integral, for any ε
>
0, it suffices to bound the following two parts
(37)
∫
ε
<
|
t
|
<
t
0
‍
|
η
|
|
γ
(
t
)
|
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
d
t
|
t
|
,
∫
|
t
|
≥
t
0
‍
|
η
|
|
γ
(
t
)
|
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
d
t
|
t
|
.
Recall that t
0
>
0 was chosen so that |
η
|
γ
(
t
0
)
=
1, and γ
(
t
)
≤
t
γ
′
(
t
) because of the convexity. Thus,
(38)
∫
ε
<
|
t
|
<
t
0
‍
|
η
|
|
γ
(
t
)
|
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
d
t
|
t
|
≤
2
|
η
|
∫
0
t
0
‍
γ
(
t
)
t
d
t
≤
2
|
η
|
∫
0
t
0
‍
γ
′
(
t
)
d
t
≤
2
.
For Re
(
z
)
<
-
1, an elementary calculation shows that
(39)
∫
|
t
|
≥
t
0
‍
|
η
|
|
γ
(
t
)
|
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
d
t
|
t
|
≤
2
|
η
|
2
Re
(
z
)
+
1
∫
t
0
∞
‍
γ
2
Re
(
z
)
(
t
)
γ
(
t
)
t
d
t
≤
2
|
2
Re
(
z
)
+
1
|
≤
2
.Similarly, the second integral can be controlled by(40)
|
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
t
|
≤
2
|
z
|
∫
0
t
0
‍
η
2
γ
2
(
t
)
d
t
t
+
2
|
z
|
∫
t
0
∞
‍
η
2
γ
2
(
t
)
[
η
2
γ
2
(
t
)
]
Re
(
z
)
-
1
d
t
t
≤
2
|
z
|
η
2
∫
0
t
0
‍
γ
(
t
)
γ
′
(
t
)
d
t
+
2
|
z
|
η
2
Re
(
z
)
∫
t
0
∞
‍
γ
2
Re
(
z
)
-
1
(
t
)
γ
′
(
t
)
d
t
≤
|
z
|
+
|
z
|
|
Re
(
z
)
|
≤
2
|
Re
(
z
)
|
[
1
+
|
Im
⁡
(
z
)
|
]
.
Therefore, for Re
(
z
)
<
-
1,
(41)
|
ξ
∂
m
z
∂
ξ
(
ξ
,
η
)
|
≤
C
[
1
+
|
Im
⁡
(
z
)
|
]
.


## 2.2.3. The Boundedness ofξ
η
(
∂
2
m
z
/
∂
ξ
∂
η
)
(
ξ
,
η
)

To take care ofξ
η
(
∂
2
m
z
/
∂
ξ
∂
η
)
(
ξ
,
η
), we note that it can be written as
(42)
ξ
η
∂
2
m
z
∂
ξ
∂
η
(
ξ
,
η
)
=
-
4
π
2
ξ
η
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
γ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
-
4
π
i
z
ξ
η
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
η
γ
2
(
t
)
d
t
.
For the first term, integrating by parts, we obtain
(43)
4
π
2
ξ
η
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
γ
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
=
2
π
i
∫
ℝ
‍
d
d
t
(
e
-
2
π
i
ξ
t
)
e
-
2
π
i
η
γ
(
t
)
[
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
=
2
π
i
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
|
-
∞
∞
-
4
π
2
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
′
(
t
)
[
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
-
2
π
i
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
-
4
π
i
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
η
3
γ
2
(
t
)
γ
′
(
t
)
d
t
.Obviously, forRe
(
z
)
<
-
1, t
∈
ℝ, |
2
π
i
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
η
γ
(
t
)
] [
1
+
η
2
γ
2
(
t
)
]
z
|
≤
2
π
|
η
|
|
γ
(
t
)
| [
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
≤
2
π. So, the boundary terms is bounded by 2
π.For the first integrated terms, making the change of variablesu
=
η
2
γ
2
(
t
), one has
(44)
|
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
′
(
t
)
[
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
|
≤
2
∫
0
∞
‍
[
1
+
η
2
γ
2
(
t
)
]
Re
(
z
)
η
2
γ
(
t
)
γ
′
(
t
)
d
t
≤
∫
0
∞
‍
(
1
+
u
)
Re
(
z
)
d
u
≤
1
|
Re
(
z
)
+
1
|
.The second integrated terms can be treated in the same way, letu
=
η
γ
(
t
),
(45)
|
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
d
t
|
≤
∫
ℝ
‍
(
1
+
u
2
)
Re
(
z
)
d
u
≤
π
.Similarly, a trivial calculation shows that(46)
|
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
η
3
γ
2
(
t
)
γ
′
(
t
)
d
t
|
≤
2
|
z
|
∫
0
∞
‍
u
2
(
1
+
u
2
)
Re
(
z
)
-
1
d
u
≤
π
|
z
|
.
The second term can be handled similarly. Integrating by parts, we decomposed it as
(47)
4
π
i
z
ξ
η
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
η
γ
2
(
t
)
d
t
=
2
π
i
z
∫
ℝ
‍
d
d
t
(
e
-
2
π
i
ξ
t
)
e
-
2
π
i
η
γ
(
t
)
×
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
=
2
π
i
z
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
|
-
∞
∞
-
4
π
2
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
′
(
t
)
×
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
-
4
π
i
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
(
t
)
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
-
4
π
i
z
(
z
-
1
)
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
2
×
η
2
γ
(
t
)
γ
′
(
t
)
d
t
.Obviously, forRe
(
z
)
<
-
1, t
∈
ℝ, |
z
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
|
≤
|
z
|. The boundary term is dominated by 4
π
|
z
|.For the first integrated terms, by making the change of variablesu
=
η
γ
(
t
), we have the estimate
(48)
|
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
γ
′
(
t
)
η
2
γ
2
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
|
≤
|
z
|
∫
ℝ
‍
u
2
(
1
+
u
2
)
Re
(
z
)
-
1
d
t
≤
π
|
z
|
.To estimate the second integrated terms, we make the transformationu
=
η
2
γ
2
(
t
) and get
(49)
|
z
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
η
2
γ
(
t
)
γ
′
(
t
)
[
1
+
η
2
γ
2
(
t
)
]
z
-
1
d
t
|
≤
|
z
|
∫
0
∞
‍
(
1
+
u
)
Re
(
z
)
-
1
d
u
≤
|
z
|
|
Re
(
z
)
|
.Similarly, the third integrated terms can be treated as(50)
|
z
(
z
-
1
)
∫
ℝ
‍
e
-
2
π
i
[
ξ
t
+
η
γ
(
t
)
]
[
1
+
η
2
γ
2
(
t
)
]
z
-
2
η
4
γ
3
(
t
)
γ
′
(
t
)
d
t
|
≤
|
z
(
z
-
1
)
|
∫
0
∞
‍
(
1
+
u
)
Re
(
z
)
-
1
d
u
≤
|
z
(
z
-
1
)
|
|
Re
(
z
)
|
.
Note that for Re
(
z
)
<
-
1, we have the following elementary estimates
(51)
|
z
|
≤
|
Re
(
z
)
|
[
1
+
|
Im
⁡
(
z
)
|
]
,
|
z
-
1
|
≤
|
Re
(
z
)
-
1
|
[
1
+
|
Im
⁡
(
z
)
|
]
.
Finally, combining above eight estimates, we obtain
(52)
|
ξ
η
∂
2
m
z
∂
ξ
∂
η
(
ξ
,
η
)
|
≤
C
[
1
+
Im
⁡
(
z
)
]
2
.


## 2.3.  Conclusion

To complete the proof of our main result, we need al
q-valued analytic interpolation theorem, which is a consequence of Theorem  1 in [13], a concrete proof can be found in [14].LetS denote the closed strip {
a
≤
Re
(
z
)
≤
b
}
⊂
C. {
U
z
}
z
∈
S is a family of uniformly bounded linear operators on L
2
(
R
n
), that is, there is an M
>
0 such that
(53)
∥
U
z
∥
L
2
→
L
2
≤
M
,
∀
z
∈
S
.
Moreover, the function
(54)
z
⟼
∫
R
n
‍
U
z
f
(
x
)
g
(
x
)
d
x
is continuous on S and analytic in the interior of S, whenever f
,
g
∈
L
2
(
R
n
).Lemma 12.
With the above assumptions, if for1
<
p
i
,
q
i
<
∞
,
(
i
=
0,1
), {
U
z
} satisfies the following conditions:(i)
∥
(
∑
j
∈
Z
|
U
z
f
j
|
q
0
)
1
/
q
0
∥
L
p
0
≤
M
0
∥
(
∑
j
∈
Z
|
f
j
|
q
0
)
1
/
q
0
∥
L
p
0
w
h
e
n
Re
(
z
)
=
a;
(ii)
∥
(
∑
j
∈
Z
|
U
z
f
j
|
q
1
)
1
/
q
1
∥
L
p
1
≤
M
1
∥
(
∑
j
∈
Z
|
f
j
|
q
1
)
1
/
q
1
∥
L
p
1
w
h
e
n
Re
(
z
)
=
b. Then we have
(55)
∥
(
∑
j
∈
Z
‍
|
U
a
(
1
-
θ
)
+
θ
b
f
j
|
q
)
1
/
q
∥
L
p
≤
M
0
1
-
θ
M
1
θ
∥
(
∑
j
∈
Z
‍
|
f
j
|
q
)
1
/
q
∥
L
p
,
where 0
<
θ
<
1, 1
/
q
=
(
(
1
-
θ
)
/
q
0
)
+
θ
/
q
1 and 1
/
p
=
(
(
1
-
θ
)
/
q
0
)
+
θ
/
p
1.LetU
z
f
(
x
)
=
e
z
2
ℋ
z
f
(
x
). Note that |
e
z
2
|
=
e
Re
(
z
)
2
-
Im
⁡
(
z
)
2, then there exists a constant M
0 which is independent of Im
⁡
(
z
) such that
(56)
∥
(
∑
j
∈
Z
‍
|
U
z
f
j
|
2
)
1
/
2
∥
L
2
≤
C
δ
e
-
Im
⁡
(
z
)
2
[
1
+
|
Im
⁡
(
z
)
|
]
∥
(
∑
j
∈
Z
‍
|
f
j
|
2
)
1
/
2
∥
L
2
≤
M
0
∥
(
∑
j
∈
Z
‍
|
f
j
|
2
)
1
/
2
∥
L
2
when
Re
(
z
)
=
1
/
4
-
δ
.
Also, for 1
<
p
1,  q
1
<
∞, there exists a constant M
1 which is independent of Im
⁡
(
z
) such that
(57)
∥
(
∑
j
∈
Z
‍
|
U
z
f
j
|
q
1
)
1
/
q
1
∥
L
p
1
≤
M
1
∥
(
∑
j
∈
Z
‍
|
f
j
|
q
1
)
1
/
q
1
∥
L
p
1
∥
(
∑
j
∈
Z
‍
|
U
z
f
j
|
q
1
)
1
/
q
1
∥
L
p
1
≤
M
1
≤
when
Re
(
z
)
<
-
1
.
For 2
≤
p, q
<
5
/
2, there exists 1
<
p
1, q
1
<
∞ so that
(58)
1
p
=
θ
2
+
1
-
θ
p
1
,
1
q
=
θ
2
+
1
-
θ
q
1
,
where θ
∈
(
0,1
) and satisfies the equation
(59)
(
1
4
-
δ
)
θ
+
(
-
1
-
ε
)
(
1
-
θ
)
=
0
for some ε
>
0 and 0
<
δ
<
1
/
4. Therefore, by Lemma 12, we obtain
(60)
∥
(
∑
j
∈
Z
‍
|
ℋ
f
j
|
q
)
1
/
q
∥
L
p
=
∥
(
∑
j
∈
Z
‍
|
ℋ
0
f
j
|
q
)
1
/
q
∥
L
p
=
∥
(
∑
j
∈
Z
‍
|
U
0
f
j
|
q
)
1
/
q
∥
L
p
≤
C
∥
(
∑
j
∈
Z
‍
|
f
j
|
q
)
1
/
q
∥
L
p
for 2
≤
p, q
<
5
/
2. The duality implies the result for 5
/
3
<
p, q
≤
2. This completes the proof of Theorem 3.



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*Source: 827072-2014-02-06.xml*