#  $L^{p}$-boundedness of wave operators for fourth order Schrödinger
operators with zero resonances on $\mathbb{R}^{3}$

Haruya Mizutani  Department of Mathematics, Graduate School of Science, Osaka
University, Toyonaka, Osaka 560-0043, Japan<EMAIL_ADDRESS>,
Zijun Wan  Department of Mathematics, Central China Normal University, Wuhan,
430079, P.R. China<EMAIL_ADDRESS>and  Xiaohua Yao  School of
Mathematics and Statistics, Key Laboratory of Nonlinear Analysis and
Applications (Ministry of Education), Central China Normal University, Wuhan,
430079, P.R. China<EMAIL_ADDRESS>
###### Abstract.

Let $H=\Delta^{2}+V$ be the fourth-order Schrödinger operator on
$\mathbb{R}^{3}$ with a real-valued fast-decaying potential $V$. If zero is
neither a resonance nor an eigenvalue of $H$, it has recently been shown by
Goldberg and Green [18] that the wave operators $W_{\pm}(H,\Delta^{2})$ are
bounded on $L^{p}(\mathbb{R}^{3})$ for $1<p<\infty$. Additionally, it has been
proved by the authors in [26] that these wave operators are unbounded at the
endpoints, i.e., for $p=1$ and $p=\infty$.

In this paper, our primary focus is to further establish the $L^{p}$-bounds of
$W_{\pm}(H,\Delta^{2})$ that exhibit all types of singularities at the zero
energy threshold. We first prove that $W_{\pm}(H,\Delta^{2})$ are bounded on
$L^{p}(\mathbb{R}^{3})$ for $1<p<\infty$ in the first kind resonance case. We
then proceed to establish for the second kind resonance case that they are
bounded on $L^{p}(\mathbb{R}^{3})$ for $1<p<3$, but not for $3\leq p<\infty$.

In the third kind resonance case, we show that $W_{\pm}(H,\Delta^{2})$ are
bounded on $L^{p}(\mathbb{R}^{3})$ for $1<p<3$. Remarkably, we can also prove
that $W_{\pm}(H,\Delta^{2})$ remain bounded on $L^{p}({\mathbb{R}}^{3})$ for
$3\leq p<\infty$ if in addition $H$ has a zero eigenvalue, but no $p$-wave
zero resonance and all zero eigenfunctions are orthogonal to
$x_{i}x_{j}x_{k}V$ in $L^{2}({\mathbb{R}}^{3})$ for all $i,j,k=1,2,3$ with
$x=(x_{1},x_{2},x_{3})$ and that, without such additional conditions, they are
unbounded on $L^{p}({\mathbb{R}}^{3})$ for $3\leq p<\infty$.

These results precisely delineate all $L^{p}$-bounds of the wave operators
$W_{\pm}(H,\Delta^{2})$ in $\mathbb{R}^{3}$ with the exception of the
endpoints $p=1$ and $p=\infty$. As a result, $L^{p}$-$L^{q}$ decay estimates
are derived for solutions of fourth-order Schrödinger equations and beam
equations with zero resonance singularities.

The proof of these results hinges crucially on the detailed analysis of the
asymptotic expansions of the resolvent of $H$ at low energy for various types
of zero resonance singularities, as well as the incorporation of crucial ideas
by Yajima [35] and Goldberg and Green [18].

###### Key words and phrases:

Wave operator, Fourth-order Schrödinger operator, Zero resonance,
$L^{p}$-boundedness

###### Contents

  1. 1 Introduction
    1. 1.1 Backgrounds
    2. 1.2 The main results and zero resonances
    3. 1.3 Further related results
    4. 1.4 Notations
  2. 2 The stationary formulas and asymptotic expansions of resolvent
    1. 2.1 The stationary formulas of wave operators
    2. 2.2 Integral operators related with wave operators
    3. 2.3 Resolvent asymptotic expansions
  3. 3 The $L^{p}$-boundedness of wave operators with the first kind zero resonances
    1. 3.1 The regular case
    2. 3.2 The first kind of resonance case
  4. 4 The $L^{p}$-boundedness of wave operators for the second and third types
    1. 4.1 The second kind of resonance case
    2. 4.2 The third kind of resonance
  5. 5 The proofs of Lemmas 2.1 and 2.2
    1. 5.1 The proof of Lemma 2.1
    2. 5.2 The proof of Lemma 2.2
    3. 5.3 Two technical lemmas
  6. A Asymptotic expansions of $M^{-1}(\lambda)$ with threshold singularities
    1. A.1  The asymptotic expansion of $M^{\pm}(\lambda)$
    2. A.2  The asymptotic expansions of $M^{\pm}(\lambda)^{-1}$

##  1\. Introduction

###  1.1. Backgrounds

Consider the fourth-order Schrödinger operators $H=\Delta^{2}+V(x)$ on
$\mathbb{R}^{3}$ with real-valued potentials $V(x)$ satisfying
$|V(x)|\lesssim{\langle}x{\rangle}^{-\mu}$ with a specific $\mu>2$ and
${\langle}x{\rangle}=\sqrt{1+|x|^{2}}$. Under this condition, $H$ is self-
adjoint on $L^{2}({\mathbb{R}}^{3})$ and its spectrum consists of negative
discrete eigenvalues and the essential spectrum $[0,\infty)$ (with possible
zero eigenvalue). Throughout the paper we assume that $H$ has no positive
eigenvalue. As $\mu>1$, it has been well-documented (see e.g. Agmon [1],
Kuroda [24]) that the wave operators

$\displaystyle
W_{\pm}=W_{\pm}(H,\Delta^{2}):=\operatorname*{s-lim}_{t\to\pm\infty}e^{itH}e^{-it\Delta^{2}}$
( 1.1)

exist as partial isometries on $L^{2}(\mathbb{R}^{3})$ and are asymptotically
complete, namely the range of $W_{\pm}$ coincides with the absolutely
continuous spectral subspace
${\mathcal{H}}_{\mathop{\mathrm{ac}}\nolimits}(H)$ of $H$. In particular,
$W_{\pm}$ satisfy the identities
$W_{\pm}W_{\pm}^{*}=P_{\mathop{\mathrm{ac}}\nolimits}(H)$,
$W_{\pm}^{*}W_{\pm}=I$ and exhibit the intertwining property
$f(H)W_{\pm}=W_{\pm}f(\Delta^{2})$ for any Borel measurable function $f$ on
$\mathbb{R}$, where $P_{\mathop{\mathrm{ac}}\nolimits}(H)$ denotes the
projection onto ${\mathcal{H}}_{\mathop{\mathrm{ac}}\nolimits}(H)$. These
formulas, in particular, imply

$\displaystyle
f(H)P_{\mathop{\mathrm{ac}}\nolimits}(H)=W_{\pm}f(\Delta^{2})W_{\pm}^{*}.$ (
1.2)

We refer to [27] for a detailed description on the mathematical scattering
theory.

In this paper we are interested in the boundedness of the wave operators
$W_{\pm}$ and $W_{\pm}^{*}$ on $L^{p}({\mathbb{R}}^{3})$. One of our
motivation to study the $L^{p}$-boundedness is that, based on the equation
1.2, the $L^{p}$-boundedness of $W_{\pm}$ and $W_{\pm}^{*}$ can be directly
employed to establish the $L^{p}$-$L^{q}$ estimates for the perturbed operator
$f(H)$ to the same estimates for the free operator $f(\Delta^{2})$. This can
be expressed as follows:

$\displaystyle\|f(H)P_{\mathop{\mathrm{ac}}\nolimits}(H)\|_{L^{p}\to
L^{q}}\leq\|W_{\pm}\|_{L^{q}\to L^{q}}\ \|f(\Delta^{2})\|_{L^{p}\to
L^{q}}\|W_{\pm}^{*}\|_{L^{p}\to L^{p}}.$ ( 1.3)

In many cases, under appropriate conditions on $f$, it is feasible to
establish $L^{p}$-$L^{q}$ bounds for $f(\Delta^{2})$ using Fourier multiplier
methods. Therefore, in order to use the inequality  1.3, it would be
interesting to prove the following $L^{p}$ estimates for $W_{\pm}$ and
$W_{\pm}^{*}$:

$\displaystyle\|W_{\pm}\phi\|_{L^{p}(\mathbb{R}^{3})}\lesssim\|\phi\|_{L^{p}(\mathbb{R}^{3})},\quad\|W_{\pm}^{*}\phi\|_{L^{p}(\mathbb{R}^{3})}\lesssim\|\phi\|_{L^{p}(\mathbb{R}^{3})}.$
( 1.4)

The range of values $p$ for which the wave operators are bounded on $L^{p}$ is
known to heavily depend on the existence/non-existence of zero resonances of
$H$. In the present case, there are three kinds of zero resonances (the third
one coincides with the existence of zero eigenvalues). We refer to Definition
1.1 for the precise definition. For the regular case, the estimates  1.4
have been already established at least for sufficiently fact decaying
potentials as follows: if $\mu>9$ and zero is neither an eigenvalue nor a
resonance of $H$, Goldberg and Green [18] initially established that $W_{\pm}$
are bounded on $L^{p}({\mathbb{R}}^{3})$ for $1<p<\infty$. In the critical
cases $p=1$ and $\infty$, we proved in [26] that $W_{\pm}$ become unbounded on
$L^{1}({\mathbb{R}}^{3})$ and $L^{\infty}({\mathbb{R}}^{3})$, even when
assuming that $V$ has compact support. Furthermore, the weak $(1,1)$-estimates
for $W_{\pm}$ and $W_{\pm}^{*}$ were also established in the regular case if
$\mu>11$. That is,

$\displaystyle\big{|}\\{x\in{\mathbb{R}}^{3};\
|W_{\pm}f(x)|\geq\lambda\\}\big{|}$
$\displaystyle\lesssim\frac{1}{\lambda}\int_{{\mathbb{R}}^{3}}|f(x)|dx,\quad\lambda>0,$
( 1.5)

with a similar estimate for $W_{\pm}^{*}$. Therefore, the main object of the
present paper is establish the case when zero is either an eigenvalue or a
resonance of $H$. We here stress that, due to many intriguing applications, it
would be also crucial to establish the $L^{p}$-bounds  1.4 even when zero is
either an eigenvalue or a resonance of $H$.

Our results are divided into three cases according to the types of the zero
resonances. For the case of the first kind resonance, the same results as in
the regular case has been expected to hold as postulated by Goldberg and Green
[18]. We confirm this conjecture by showing that $W_{\pm}$ are bounded on
$L^{p}({\mathbb{R}}^{3})$ for any $1<p<\infty$. For the second and third kind
resonance cases, the situation is much more complicated. Indeed, for the case
of the second kind resonance, we show that $W_{\pm}$ are bounded on
$L^{p}({\mathbb{R}}^{3})$ for any $1<p<3$, but not for $3\leq p<\infty$. For
the case of th third kind resonance, we first show that $W_{\pm}$ are bounded
on $L^{p}({\mathbb{R}}^{3})$ for $1<p<3$. Remarkably, we can also show in such
a case that $W_{\pm}$ are bounded on $L^{p}({\mathbb{R}}^{3})$ for $3\leq
p<\infty$ if in addition $H$ has no $p$-wave zero resonance and all zero
eigenfunctions $\phi$ satisfy the orthogonality conditions:

$\int_{{\mathbb{R}}^{3}}x_{i}x_{j}x_{k}V(x)\phi(x)dx=0$

for all $i,j,k=1,2,3$ with $x=(x_{1},x_{2},x_{3})$. Finally, without such
additional conditions, they are shown to be unbounded on
$L^{p}({\mathbb{R}}^{3})$ for $3\leq p<\infty$.

The proofs of these results significantly rely on the thorough analysis of the
asymptotic expansions of the resolvent of $H$ at low energy for all types of
zero-resonance singularities, incorporating ideas from Yajima [35] and
Goldberg and Green [18].

###  1.2. The main results and zero resonances

In order to present our main results, we first need to introduce the
definition of zero resonance by Erdogan, Green, and Toprak [10].

Recall that if there is a nontrivial solution $\phi\in
L^{2}({\mathbb{R}}^{3})$ that satisfies the equation $\Delta^{2}\phi+V\phi=0$
in a distributional sense, we classify zero as an eigenvalue of $H$. In
general, it is possible for a nontrivial resonant solution $\phi(x)$ in
$L^{2}_{-s}({\mathbb{R}}^{3})$ with some $s>0$ to exist, such that
$H\phi(x)=0$. This may lead to challenges in achieving dispersive estimates or
establishing $L^{p}$-boundedness for the wave operator of $H$ within the low-
energy part.

To introduce the concept of the zero resonances for $H$, we define
$L^{2}_{-s}(\mathbb{R}^{3})=\langle\cdot\rangle^{s}L^{2}({\mathbb{R}}^{3})$
for $s\in\mathbb{R}$ and set
$W_{-\sigma}(\mathbb{R}^{3})=\bigcap_{s>\sigma}L^{2}_{-s}(\mathbb{R}^{3})$ for
$\sigma\in{\mathbb{R}}$. It is evident that $W_{-\sigma}({\mathbb{R}}^{3})$
increases as $\sigma$ increases, and $L^{2}_{-\sigma}({\mathbb{R}}^{3})\subset
W_{-\sigma}({\mathbb{R}}^{3})$. In particular,
$L^{2}({\mathbb{R}}^{3})\subsetneq
W_{-\frac{1}{2}}({\mathbb{R}}^{3})\subsetneq
W_{-\frac{3}{2}}({\mathbb{R}}^{3})$.

######  Definition 1.1.

Following the approach in [10], zero resonant solutions of $H=\Delta^{2}+V$
can be categorized into the following three types:

  * (i)

Zero is a first kind resonance of $H$ if there exists a $0\neq\phi\in
W_{-{3\over 2}}(\mathbb{R}^{3})$ but no any $0\neq\phi\in W_{-{1\over
2}}(\mathbb{R}^{3})$ such that $H\phi=0$ holds in the distributional sense;

  * (ii)

Zero is a second kind resonance of $H$ if there exists a $0\neq\phi\in
W_{-{1\over 2}}(\mathbb{R}^{3})$ but no any $0\neq\phi\in
L^{2}(\mathbb{R}^{3})$ such that $H\phi=0$ holds in the distributional sense;

  * (iiii)

Zero is a third kind resonance of $H$ if $H$ has a zero eigenvalue, namely
there exists a $0\neq\phi\in L^{2}(\mathbb{R}^{3})$ such that $H\phi=0$ holds
in the distributional sense;

  * (iv)

Zero is a regular point of $H$ if $H$ has neither zero eigenvalue nor zero
resonance.

We also recall the notions of $s$-wave and $p$-wave resonances.

######  Definition 1.2.

We say that zero is a $s$-wave resonance if there exists $\phi\in W_{-{3\over
2}}({\mathbb{R}}^{3})\setminus W_{-{\frac{1}{2}}}({\mathbb{R}}^{3})$ such that
$H\Phi=0$, and is a $p$-wave resonance if there exists $\phi\in W_{-{1\over
2}}({\mathbb{R}}^{3})\setminus L^{2}({\mathbb{R}}^{3})$ such that $H\Phi=0$.
Using these notions, one can rephrase the above definition as follows:

  * •

Zero is a first kind resonance of $H$ if there is only a $s$-wave zero
resonance;

  * •

Zero is a second kind resonance of $H$ if there is a $p$-wave zero resonance,
but no zero eigenvalue. In this case, there may or not may be a $s$-wave
resonance;

######  Remark 1.3.

Some examples of zero resonance or eigenvalue are constructed as follows:

(i) (The $s$-wave and $p$-wave resonances) Assuming that $\phi\in
C^{\infty}(\mathbb{R}^{3})$ is positive, $\phi(x)=c>0$ for $|x|>1$, then
$H\phi(x)=0$ when the potential $V(x)$ is defined as
$V(x)=-(\Delta^{2}\phi)/\phi(x)$. It can be readily verified that $V\in
C_{0}^{\infty}(\mathbb{R}^{3})$ and $\phi\in
W_{-\frac{3}{2}}(\mathbb{R}^{3})\setminus W_{-\frac{1}{2}}(\mathbb{R}^{3})$.
Consequently, zero is a resonance of the operator $H$, even in the presence of
a compactly supported smooth potential.

Moreover, if we consider a positive smoothing function $\phi(x)$ on
$\mathbb{R}^{3}$ satisfying $\phi(x)=|x|^{-1}$ for $|x|>1$, then we still find
$V\in C_{0}^{\infty}(\mathbb{R}^{3})$ such that $H\phi=0$ and $\phi\in
W_{-\frac{1}{2}}(\mathbb{R}^{3})\setminus L^{2}(\mathbb{R}^{3})$.

(ii) (The zero eigenvalue) Unlike the case of zero resonance, the problem of
zero eigenvalues for the operator $H$ is more subtle. For example, consider
$\phi(x)=(1+x^{2})^{-s}$ with $s>\frac{3}{4}$ and define $V(x)$ as
$V(x)=-(\Delta^{2}\phi)/\phi(x)$. In this setup, it can be observed that
$V(x)=O((1+|x|)^{-4})$ as $|x|\rightarrow\infty$, and $\phi\in
L^{2}(\mathbb{R}^{3})$ while satisfying $H\phi=0$. This demonstrates that zero
is an eigenvalue of $H$ in the presence of such slowly decaying potential.

Assume that $\langle x\rangle^{5+}V\in L^{\infty}({\mathbb{R}}^{3})$. Then it
was shown by [10] that any resonance solution $\phi$ of equation
$\Delta^{2}\phi+V\phi=0$ in $W_{-{3\over 2}}({\mathbb{R}}^{3})$ can be
expressed generally as follows

$\phi(x)=c_{0}+\sum_{i=1}^{3}c_{i}\ {x_{i}\over\langle x\rangle}+\sum_{1\leq
i\leq j\leq 3}c_{ij}\ {x_{i}x_{j}\over\langle x\rangle^{3}}+O_{L^{2}}(1),$ (
1.6)

where $c_{0}$, $c_{i}\ (1\leq i\leq 3)$ and $c_{ij}\ (1\leq i\leq j\leq 3)$
are some specific constants. Hence

  * •

If zero is a first kind resonance of $H$, then for any such nonzero solution
$\phi$ in  1.6, there exists some $c_{i}\neq 0$ for $0\leq i\leq 3$.

  * •

If zero is a second kind resonance of $H$, then for any nonzero solution
$\phi$ with $c_{i}=0$ for all $0\leq i\leq 3$ in  1.6, there must exist some
$c_{ij}\neq 0$ for $1\leq i\leq j\leq 3$.

  * •

Finally, if zero is an eigenvalue of $H$, then there exists some nonzero
solution $\phi$ in  1.6 such that $c_{i}=c_{ij}=0$ for all $i,j$.

According to the Fredholm theorem, the eigenspace has finite dimension.
Therefore, when modulating these zero eigenfunctions, the codimension of space
of the zero resonant functions is 10 at most.

We should emphasize that these zero resonances are closely linked to the
asymptotic expansions of the resolvent $R_{V}(z)$ of the operator $H$ near
zero energy. More specifically, the solutions related to zero resonances can
be characterized by the kernel subspace of certain special bounded operators
$T_{1}$ and $T_{2}$ on $L^{2}({\mathbb{R}}^{3})$. These solutions manifest in
the asymptotic expansion through the projection operators $S_{j+1}$ for
$j=0,1,2$ onto the kernel $\mathop{\mathit{ker}}(T_{j})$, respectively (See
Section A below for the details).

Moreover, it is worth noting that these asymptotic expansions play a crucial
role in proving the $L^{p}$-boundedness of $W_{\pm}(H,\Delta^{2})$, especially
in the lower energy part. This will become clear through the stationary
formula  2.1 of $W_{\pm}(H,\Delta^{2})$, which will be presented later in
Section 3 and 4.

Now we state our main results by the following theorem.

Let $\mathbb{B}(X,Y)$ be the space of bounded operators from Banach space $X$
to Banach space $Y$, namely $A\in\mathbb{B}(X,Y)$ if
$\|Af\|_{Y}\lesssim\|f\|_{X}$ for $f\in X$ (or $f$ in some dense set of $X$),
and also set $\mathbb{B}(X)=\mathbb{B}(X,X)$.

######  Theorem 1.4.

Let $H=\Delta^{2}+V$ with $|V(x)|\lesssim(1+|x|)^{-\mu}$ with some $\mu>0$
depending on the type of zero resonance. Assume that $H$ has no positive
embedded eigenvalues, then the following statements hold:

(i) If zero is a first kind resonance of $H$ and $\mu>13$, then the wave
operators $W_{\pm}(H,\Delta^{2})\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for
all $1<p<\infty$.

(ii) If zero is a second kind resonance of $H$ and $\mu>21$, then
$W_{\pm}(H,\Delta^{2})\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<3$
and $W_{\pm}(H,\Delta^{2})\notin\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for any
$3\leq p<\infty$.

(iii) If zero is a third kind resonance of $H$ (i.e. $H$ has a zero
eigenvalue) and $\mu>25$, then
$W_{\pm}(H,\Delta^{2})\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<3$.
Furthermore, we have

  * •

If $S_{2}L^{2}=S_{3}L^{2}=S_{4}L^{2}$, then
$W_{\pm}(H,\Delta^{2})\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $3\leq
p<\infty$;

  * •

If $S_{2}L^{2}\neq S_{3}L^{2}$ or $S_{3}L^{2}\neq S_{4}L^{2}$, then
$W_{\pm}(H,\Delta^{2})\notin\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for any
$3\leq p<\infty$.

Here $S_{2}L^{2}$, $S_{3}L^{2}$, $S_{4}L^{2}$ are three finite dimensional
subspaces of $L^{2}({\mathbb{R}}^{3})$ defined in  2.16- 2.18
respectively, and they satisfy $S_{4}L^{2}\subset S_{3}L^{2}\subset
S_{2}L^{2}$.

######  Remark 1.5.

By virtue of Definition 1.2, we can see that
$S_{2}L^{2}=S_{3}L^{2}=S_{4}L^{2}$ if and only if $H$ has a zero eigenvalue
(and may or may not have an $s$-wave zero resonance), but no $p$-wave zero
resonance, and moreover all zero eigenfunctions $\phi$ of $H$ satisfy

$\displaystyle\displaystyle\int_{{\mathbb{R}}^{3}}x_{i}x_{j}x_{k}V(x)\phi(x)dx=0$
( 1.7)

for all $i,j,k=1,2,3$. Moreover, the condition $S_{2}L^{2}\neq S_{3}L^{2}$
means there exists a $p$-wave resonance, and $S_{3}L^{2}\neq S_{4}L^{2}$ means
 1.7 does not hold for some $i,j,k$. We refer to Appendix A.2 for more
details.

######  Remark 1.6.

Based on the characterizations of the projection subspaces $S_{j}L^{2}$ for
$j=1,2,3$ as presented in Lemma A.4, and the instances of zero resonance
solutions $\phi$ for the operator $H$ including compactly supported potentials
$V$ as detailed in Remark 1.3, it has been demonstrated that the properties
$S_{1}L^{2}\neq S_{2}L^{2}$ or $S_{2}L^{2}\neq S_{3}L^{2}$ may arise, even
dealing with rapidly decaying potentials.

Nevertheless, it remains uncertain for the authors whether there exists a
nontrivial potential for which the relationship
$S_{2}L^{2}=S_{3}L^{2}=S_{4}L^{2}$ holds. Equivalently, this means whether a
potential $V$ can be found such that each nontrivial solution of $H\phi=0$ in
$W_{-1/2}({\mathbb{R}}^{3})$ must be bounded and satisfy that
$\phi(x)=O({\langle}x{\rangle}^{-3})$ as $|x|$ goes to infinity.

Additionally, we may mention that the absence of embedded positive eigenvalues
is a fundamental requirement for establishing $L^{p}$-bounds of wave
operators. In fact, for any dimension $d\geq 1$, one can readily construct a
potential function $V\in C_{0}^{\infty}(\mathbb{R}^{d})$ for which the
operator $H=\Delta^{2}+V$ has some positive eigenvalues. This is discussed,
for example, in [11, Section 7.1].

However, it is essential to observe that Feng et al. in [11] have demonstrated
that, under the condition that the potential $V$ is bounded and satisfies the
repulsive condition (i.e., $(x\cdot\nabla)V\leq 0$), $H=\Delta^{2}+V$ does not
contain any positive eigenvalues. Additionally, it is well-established by Kato
[23], that Schrödinger operator $-\Delta+V$ lacks positive eigenvalues only
when a bounded potential $V(x)$ approaches zero faster than $|x|^{-1}$ as
$|x|\to\infty$.

As a consequence of Theorem 1.4 and the inequality  1.3, we immediately
obtain the following corollary.

######  Corollary 1.7.

Let $H$ and $V$ satisfy the same conditions as in Theorem 1.4 above. Then

$\big{\|}e^{itH}P_{ac}(H)\big{\|}_{L^{p}({\mathbb{R}}^{3})\rightarrow
L^{p^{\prime}}({\mathbb{R}}^{3})}\lesssim|t|^{-\frac{3}{4}(\frac{1}{2}-\frac{1}{p})},$
( 1.8)
$\big{\|}\cos(t\sqrt{H})P_{ac}(H)\big{\|}_{L^{p}({\mathbb{R}}^{3})\rightarrow
L^{p^{\prime}}({\mathbb{R}}^{3})}+\Big{\|}\frac{\sin(t\sqrt{H})}{t\sqrt{H}}P_{ac}(H)\Big{\|}_{L^{p}({\mathbb{R}}^{3})\rightarrow
L^{p^{\prime}}({\mathbb{R}}^{3})}\lesssim|t|^{-\frac{3}{2}(\frac{1}{2}-\frac{1}{p})},$
( 1.9)

where $1<p\leq 2$ in the regular or first kind resonance cases and
$\frac{3}{2}<p\leq 2$ in the second or third kind resonance cases.

When zero is a regular point or first kind of resonance, the decay estimates
 1.8 and  1.9 for all $1\leq p\leq 2$, were established by Erdogan et al.
[10] and Chen et al [4], respectively. However, the $L^{p}$-estimates  1.8
and  1.9 are new for $3/2<p\leq 2$ if zero is the second or third kind of
resonance.

###  1.3. Further related results

For the classical Schrödinger operator $H=-\Delta+V(x)$, since the seminal
work by Yajima in [29], there has been a significant body of works focused on
the $L^{p}$-boundedness of the wave operators $W_{\pm}$. This research has
gained great importance due to its applications in linear and nonlinear
dispersive equations. In particular, in the one-dimensional space ($d=1$), the
wave operators $W_{\pm}$ are bounded on $L^{p}(\mathbb{R})$ for $1<p<\infty$
in both regular and zero resonance cases. However, they are generally
unbounded for $p=1$ and $p=\infty$ (see, for example, [14, 5, 28]).

In the regular case, for dimension $d=2$, the wave operators $W_{\pm}$ are
bounded on $L^{p}(\mathbb{R}^{2})$ for $1<p<\infty$, but the results at the
endpoints are yet unknown. For dimensions $d\geq 3$, the wave operators
$W_{\pm}$ are bounded on $L^{p}(\mathbb{R}^{d})$ for $1\leq p\leq\infty$ (see,
for example, [3, 21, 30, 29, 31]). However, the presence of threshold
resonances narrows down the range of values for $p$, depending on the
dimension $d$ and the decay characteristics of zero-energy eigenfunctions (as
discussed in [6, 13, 16, 17, 22, 32, 33, 34, 35, 36]).

More recently, there has been some research on the $L^{p}$-boundedness of wave
operators $W_{\pm}$ for higher-order Schrödinger operators
$H=(-\Delta)^{m}+V(x)$, particularly for $m=2$. First, Goldberg and Green in
[18] established that for dimension $d=3$ and $m=2$, the wave operators
$W_{\pm}$ become bounded operators on $L^{p}(\mathbb{R}^{3})$ for $1<p<\infty$
when zero is a regular point (though the endpoint case is not discussed in
[18]). Then, Erdoğan and Green in [7, 8] further demonstrated that for $m>1$
and $d>2m$, $W_{\pm}$ are bounded on $L^{p}(\mathbb{R}^{d})$ for $1\leq
p\leq\infty$ when the potentials $V(x)$ are smooth enough in the regular case.
Additionally, in [9], Erdoğan, Goldberg, and Green found that for dimension
$d>4m-1$ and $\frac{2d}{d-4m+1}<p\leq\infty$, the $L^{p}$ boundedness of the
wave operators may not hold, even for compactly supported continuous
potentials, if the potential is not sufficiently smooth.

In our previous work in [25], we examined the case of $d=1$ and $m=2$. We
found that whether zero is a regular point or a resonance of $H$, the wave
operators $W_{\pm}$ are bounded on $L^{p}(\mathbb{R})$ for $1<p<\infty$.
Furthermore, if the potential $V$ is compactly supported, then $W_{\pm}$ are
also bounded from $L^{1}(\mathbb{R})$ to $L^{1,\infty}(\mathbb{R})$. However,
for the regular case, it has been shown that $W_{\pm}$ are unbounded on both
$L^{1}(\mathbb{R})$ and $L^{\infty}(\mathbb{R})$ in the one-dimensional
setting. More recently, Galtbayar and Yajima in [15] have provided $L^{p}$
estimates for the wave operators $W_{\pm}$ with zero resonances in the case of
$m=2$ and $d=4$.

###  1.4. Notations

Throughout the paper we use the following notations:

  * •

$A\lesssim B$ (resp. $A\gtrsim B$) means $A\leq CB$ (resp. $A\geq CB$) with
some constant $C>0$.

  * •

$L^{p}=L^{p}({\mathbb{R}}^{3}),L^{1,\infty}=L^{1,\infty}({\mathbb{R}}^{3})$
denote the Lebesgue and weak $L^{1}$ spaces on ${\mathbb{R}}^{3}$,
respectively.

  * •

${\langle}f,g{\rangle}_{L^{2}}=\int_{{\mathbb{R}}^{3}}f\overline{g}dx$ denotes
the inner product in $L^{2}({\mathbb{R}}^{3})$.

  * •

$\mathbb{S}({\mathbb{R}}^{3})$ denotes the Schwartz class.

  * •

The Fourier transform of $f$ is defined as

$\hat{f}(\xi)=\mathcal{F}(f)(\xi)=\frac{1}{(2\pi)^{3}}\int_{{\mathbb{R}}^{3}}e^{-ix\xi}f(x)dx,\
\xi\in{\mathbb{R}}^{3},$ ( 1.10)

and denote the inverse Fourier transform of $f$ by $\check{f}(\xi)$ or
$\mathcal{F}^{-1}(f)(\xi)$.

  * •

Denote by $m(D)$ the Fourier multiplier operator defined as

$m(D)f=\mathcal{F}^{-1}(m(\xi)\mathcal{F}(f)),\
f\in\mathbb{S}({\mathbb{R}}^{3}),$ ( 1.11)

where $m(\xi)$ is the corresponding symbol function to $m(D)$.

  * •

Let $\chi(\lambda)$ be a smoothing function such that $\chi(\lambda)=1$ for
$|\lambda|\leq 1/2$ and $\chi(\lambda)=0$ for $|\lambda|\geq 1$. For any
$a>0$, set $\chi_{<a}(\lambda)=\chi(\lambda/a)$ and $\chi_{\geq
a}(\lambda)=1-\chi_{<a}(\lambda)$.

  * •

Let $\chi_{<a}(|D|)$ and $\chi_{\geq a}(|D|)$ be the Fourier multipliers
defined by the symbols $\chi_{<a}(|\xi|)$ and $\chi_{\geq a}(|\xi|)$ on
${\mathbb{R}}^{3}$, respectively.

  * •

$K$ ( or $T_{K}$) denotes an integral operator defined by the kernel $K(x,y)$:

$Kf(x)=\int_{{\mathbb{R}}^{3}}K(x,y)f(y)dy.$

##  2\. The stationary formulas and asymptotic expansions of resolvent

###  2.1. The stationary formulas of wave operators

First of all, we observe that it suffices to deal with $W_{-}$ only since
1.1 implies $W_{+}f=\overline{W_{-}\overline{f}}$. The starting point is the
following well-known stationary representation of $W_{-}$ (see e.g. Kuroda
[24]):

$\displaystyle W_{-}=\mathop{\mathrm{Id}}\nolimits-\frac{2}{\pi
i}\int_{0}^{\infty}\lambda^{3}R_{V}^{+}(\lambda^{4})V\left(R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\right)d\lambda.$
( 2.1)

To explain the formula  2.1, we need to introduce some notations. Let

$\displaystyle R_{0}(z)=(\Delta^{2}-z)^{-1},\quad R_{V}(z)=(H-z)^{-1},\quad
z\in{\mathbb{C}}\setminus[0,\infty),$

be the resolvents of $\Delta^{2}$ and $H=\Delta^{2}+V(x)$, respectively. Then
$R^{\pm}_{0}(\lambda),R^{\pm}_{V}(\lambda)$ are defined as their proper
boundary values (limiting resolvents) on $(0,\infty)$, namely

$R^{\pm}_{0}(\lambda)=\lim_{{\varepsilon}\searrow 0}R_{0}(\lambda\pm
i{\varepsilon}),\quad R_{V}^{\pm}(\lambda)=\lim_{{\varepsilon}\searrow
0}R_{V}(\lambda\pm i{\varepsilon}),\quad\lambda>0.$

More specifically, the existence of $R^{\pm}_{0}(\lambda)$ as bounded
operators from $L^{2}_{s}({\mathbb{R}}^{3})$ to $L^{2}_{-s}({\mathbb{R}}^{3})$
with $s>1/2$ follows from the limiting absorption principle for the resolvent
$(-\Delta-z)^{-1}$ of the free Schrödinger operator $-\Delta$ (see e.g. Agmon
[1]) and the following equality:

$R_{0}(z)=\frac{1}{2\sqrt{z}}\left[(-\Delta-\sqrt{z})^{-1}-(-\Delta+\sqrt{z})^{-1}\right],\quad
z\in{\mathbb{C}}\setminus[0,\infty).$ ( 2.2)

The existence of $R^{\pm}_{V}(\lambda)$ for $\lambda>0$ under our assumption
of Theorem 1.4 has been also already shown (see e.g. [1, 12, 24]).

This formula  2.2 above also leads to the explicit kernels of free resolvent
$R_{0}^{\pm}(\lambda^{4})$ of $\Delta^{2}$:

$\displaystyle
R_{0}^{\pm}(\lambda^{4},x,y)=\frac{1}{8\pi\lambda^{2}|x-y|}\Big{(}e^{\pm
i\lambda|x-y|}-e^{-\lambda|x-y|}\Big{)}=\frac{F_{\pm}(\lambda|x-y|)}{8\pi\lambda},$
( 2.3)

where $x,y\in{\mathbb{R}}^{3}$ and $F_{\pm}(s)=s^{-1}(e^{\pm is}-e^{-s})$. As
a consequence, it follows that for each $\lambda>0$,

$\displaystyle(R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4}))f(x)=\frac{1}{4\pi\lambda^{2}}\int_{{\mathbb{R}}^{3}}\frac{\sin\lambda|x-y|}{|x-y|}\
f(y)dx.$

By the Fourier transform, this convolution above can be transferred into the
following Fourier presentations:

$\displaystyle(R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4}))f(x)=\frac{\pi
i}{2\lambda}\int_{S^{2}}e^{i\lambda\omega x}\hat{f}(\lambda\omega)d\omega,$ (
2.4)

where $S^{2}$ is the unit sphere of ${\mathbb{R}}^{3}$. The equality  2.4
immediately gives the following commuting property:

$\displaystyle(R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4}))m(|D|)=m(\lambda)(R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})),$
( 2.5)

where $m(\lambda)$ is a function on ${\mathbb{R}}_{+}=(0,\infty)$ and $m(|D|)$
is a radial Fourier multiplier with the symbol $m(|\xi|)$ on
${\mathbb{R}}^{3}$.

###  2.2. Integral operators related with wave operators

In this subsection, we give some lemmas on the $L^{p}$ boundedness of some
integral operators related with the wave operators $W_{\pm}$. We first recall
the classical Schur’s test, which say that an operator
$K\in\mathbb{B}\big{(}L^{p}({\mathbb{R}}^{3})\big{)}$ for all $1\leq
p\leq\infty$ if its kernel $K(x,y)$ satisfies:

$\displaystyle\sup_{x\in{\mathbb{R}}^{3}}\int_{{\mathbb{R}}^{3}}|K(x,y)|dy+\sup_{y\in{\mathbb{R}}^{3}}\int_{{\mathbb{R}}^{3}}|K(x,y)|dx<\infty.$
( 2.6)

In many situations faced here, the kernels $K(x,y)$ have critical decaying in
$x$ or $y$ such that $K(x,y)$ may not satisfy the above integral conditions of
Schur’s test. Hence we need to study these integral operators case by case in
order to obtain the $L^{p}$ estimates.

######  Lemma 2.1.

Let $z,w\in{\mathbb{R}}^{3}$ and $A_{z,w}$ be the integral operators given by
the following kernels:

$\displaystyle
A_{z,w}(x,y)=\int_{0}^{\infty}\lambda^{4}\chi_{<a}(\lambda)R_{0}^{+}(\lambda^{4})(x,z)\Big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\Big{)}(w,y)d\lambda,\
\ x,y\in{\mathbb{R}}^{3},$ ( 2.7)

where $R^{\pm}(\lambda^{4})(\cdot,\cdot)$ are the kernels  2.3 of boundary
resolvents $R^{\pm}(\lambda^{4})$. Then for any $z,w\in{\mathbb{R}}^{3}$, we
have $A_{z,w}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<\infty$ and

$\sup_{z,w}\|A_{z,w}\|_{\mathbb{B}(L^{p})}\leq\|A_{0,0}\|_{\mathbb{B}(L^{p})}<\infty.$

By  2.3 and using translation invariance of $L^{p}$-norm, we can first
reduce the boundedness of $A_{z,w}$ into the bounds of $A_{0,0}$ with $z=w=0$,
and then obtain the following type integral:

$\displaystyle
A(x,y)=\frac{1}{64\pi^{2}}\int_{0}^{\infty}\chi_{<a}(\lambda)\Big{(}\frac{e^{i\lambda|x|}-e^{-\lambda|x|}}{|x|}\Big{)}\Big{(}\frac{e^{i\lambda|y|}-e^{-i\lambda|y|}}{|y|}\Big{)}d\lambda,$
( 2.8)

which satisfies the following estimates:

$A(x,y)=-4i\frac{|x|\ \chi_{\\{||x|-|y||\geq
1\\}}}{|x|^{4}-|y|^{4}}+\Psi(x,y),\ \ x,y\in{\mathbb{R}}^{3},$

with $\Psi(x,y)$ satisfying Schur test condition  2.6. Note that this type
of function $\frac{|x|}{|x|^{4}-|y|^{4}}$ do not satisfy Schur test condition
 2.6, we need to use Calderon-Zygmund singular theory to prove the $L^{p}$
boundedness of $A_{0,0}$ for all $1<p<\infty$. Here we left all these proof
details to Section 5 below.

Besides, we also need to use the following lemma, which is related to Fourier
integrals of the following symbols $m(\xi,\eta)$ on ${\mathbb{R}}^{6}$:

$\displaystyle K(x,y)=\frac{1}{(2\pi)^{3}}\int_{{\mathbb{R}}^{6}}e^{ix\xi-
iy\eta}\
\frac{\chi_{<4a}(|\xi|)\chi_{<a}(|\eta|)}{(|\xi|^{2}+|\eta|^{2})(|\xi|+|\eta|)}d\xi
d\eta,$ ( 2.9) $\displaystyle
E(x,y)=\frac{1}{(2\pi)^{3}}\int_{{\mathbb{R}}^{6}}e^{ix\xi-iy\eta}\
\frac{\chi_{<4a}(|\xi|)\chi_{<a}(|\eta|)}{(|\xi|^{2}+|\eta|^{2})\ |\eta|}d\xi
d\eta.$ ( 2.10)

######  Lemma 2.2.

$K(x,y)$ and $E(x,y)$ satisfy the following estimates:

$\displaystyle|K(x,y)|\lesssim{\langle}|x|+|y|{\rangle}^{-3},$ ( 2.11)
$\displaystyle|E(x,y)|\lesssim{\langle}x{\rangle}^{-1}{\langle}|x|+|y|{\rangle}^{-2}.$
( 2.12)

As a consequence, it follows that $K\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$
for all $1<p<\infty$ and $E\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all
$1<p<3$.

We observe that the symbols in  2.9 and  2.10 do not exhibit smoothing
properties on the hyperplane set $\\{(\xi,\eta)\in{\mathbb{R}}^{6};\
\eta=0\\}$. This, in turn, indicates potential technical challenges when
applying conventional methods, such as Littlewood-Paley theory. However, the
multiplication property of Fourier translation enables us to transform the
target integrals into convolution integrals, facilitating the establishment of
the desired estimates.

Remarkably, the estimates  2.11 and  2.12 are essentially optimal, because
the $L^{p}$-boundedness of $K$ and $E$ provides the precise range for the
$L^{p}$-boundedness of wave operators, as asserted in Theorem 1.4. The proof
of Lemma 2.2 is presented in Section 5 below.

Finally, we collect some facts on Fourier multiplier $m(D)$, see e.g. Grafakos
[19].

######  Lemma 2.3 (Mikhlin’s theorem).

Let $k>d/2$ and $m(\xi)$ be a $C^{k}({\mathbb{R}}^{d}\setminus 0)$ function
satisfying the following estimates:

$|\partial^{\alpha}_{\xi}m(\xi)|\leq C_{\alpha}|\xi|^{-|\alpha|},\ \xi\neq 0,\
|\alpha|\leq k.$

Then $m(D)\in\mathbb{B}(L^{p}({\mathbb{R}}^{d}))$ for all $1<p<\infty$.

In particular, Riesz transforms
$\mathcal{R}=(\mathcal{R}_{1},\mathcal{R}_{2},\cdots,\mathcal{R}_{d})$ are
bounded on $L^{p}({\mathbb{R}}^{d})$ for all $1<p<\infty$, where

$\widehat{\mathcal{R}_{j}f}(\xi)=\frac{\xi_{j}}{|\xi|}\hat{f}(\xi),\ \ 1\leq
j\leq d.$

###  2.3. Resolvent asymptotic expansions

This subsection is devoted to the study of asymptotic behaviors of the
resolvent $R_{V}^{\pm}(\lambda^{4})$ at low energy $\lambda\to+0$ when zero is
a resonance of $H$. We begin with recalling the symmetric resolvent formula
for $R_{V}^{\pm}(\lambda^{4})$. Let $v(x)=|V(x)|^{1/2}$ and
$U(x)=\mathop{\mathrm{sgn}}\nolimits V(x)$, that is $U(x)=1$ if $V(x)\geq 0$
and $U(x)=-1$ if $V(x)<0$. Let $M^{\pm}(\lambda)=U+vR_{0}^{\pm}(\lambda^{4})v$
and $(M^{\pm})^{-1}(\lambda):=[M^{\pm}(\lambda)]^{-1}$.

######  Lemma 2.4.

For $\lambda>0$, $M^{\pm}(\lambda)$ is invertible on $L^{2}({\mathbb{R}}^{3})$
and $R_{V}^{\pm}(\lambda^{4})V$ has the form

$\displaystyle
R_{V}^{\pm}(\lambda^{4})V=R_{0}^{\pm}(\lambda^{4})v(M^{\pm})^{-1}(\lambda)v,$
( 2.13)

###### Proof.

Thanks to the absence of embedded positive eigenvalues and the Birman-
Schwinger principle, $M^{\pm}(\lambda)$ is invertible for all $\lambda>0$.
Since $V=vUv$ and $1=U^{2}$, we have

$\displaystyle R_{V}^{\pm}(\lambda^{4})v$
$\displaystyle=R_{0}^{\pm}(\lambda^{4})v-R_{V}^{\pm}(\lambda^{4})vUvR_{0}^{\pm}(\lambda^{4})v=R_{0}^{\pm}(\lambda^{4})v\Big{(}1+UvR_{0}^{\pm}(\lambda^{4})v\Big{)}^{-1}$
$\displaystyle=R_{0}^{\pm}(\lambda^{4})v\Big{(}U+vR_{0}^{\pm}(\lambda^{4})v\Big{)}^{-1}U^{-1}.$

Multiplying $Uv$ from the right, we obtain the desired formula for
$R_{V}^{\pm}(\lambda^{4})V$. ∎

We will now provide asymptotic expansions of
$\big{(}M^{\pm}(\lambda)\big{)}^{-1}$ in the presence of various zero
resonances. Throughout this paper, we exclusively utilize $M^{+}(\lambda)$, so
for simplicity, we denote it as $M(\lambda)=M^{+}(\lambda)$. In the case at
hand (d=3), the asymptotic expansion of $(M(\lambda))^{-1}$ for the operator
$H=\Delta^{2}+V$ was initially explored by Feng, Soffer, and Yao [12] in the
regular case. It was further extended to encompass all zero resonance cases by
Erdogan, Green, and Toprak [10], where the decay estimates for $e^{itH}$ were
derived using these asymptotic expansions.

However, in order to establish the precise $L^{p}$ boundedness and
unboundedness of the wave operator $W_{\pm}$ in the presence of zero resonance
singularities, a more intricate understanding of the asymptotic expansions of
$M^{-1}(\lambda)$ at $\lambda=0$ is essential in the resonance cases.
Achieving this requires an amount of additional calculations, as detailed in
Appendix A.

To state the asymptotic expansion of $(M(\lambda))^{-1}$, we need to introduce
the following several crucial projection operators $P,Q$ and $S_{j}$,
$j=1,2,3$, on $L^{2}({\mathbb{R}}^{3})$:

$\displaystyle QL^{2}=\\{f\in L^{2}\ |\ \langle v,f\rangle=0\\},\ \
P:={\|V\|}_{L^{1}}^{-1}{\langle}\cdot,v{\rangle}v,\quad Q=I-P;$ ( 2.14)
$\displaystyle S_{1}L^{2}=\\{f\in QL^{2}\ |\ QTf=0\\},\quad T:=U+v\
(\Delta^{2})^{-1}v;$ ( 2.15) $\displaystyle S_{2}L^{2}=\\{f\in S_{1}L^{2}\ |\
PTf=0,\ \langle x_{i}v,f\rangle=0,\ \ i=1,2,3\\};$ ( 2.16) $\displaystyle
S_{3}L^{2}=\\{f\in S_{2}L^{2}\ |\ \langle x_{i}x_{j}v,f\rangle=0,\
i,j=1,2,3\\}.$ ( 2.17)

Moreover, due to the need to establish the unboundedness of wave operators
$W_{\pm}$ on $L^{p}$ for $3\leq p<\infty$ in the case of the third kind
resonance, we define the fourth projection operator $S_{4}$ of
$L^{2}({\mathbb{R}}^{3})$:

$\displaystyle S_{4}L^{2}=\\{f\in S_{3}L^{2}\ |\ \langle
x_{i}x_{j}x_{k}v,f\rangle=0,\ i,j,k=1,2,3\\}.$ ( 2.18)

It is easy to see that $S_{j}L^{2}$ are closed subspaces of $L^{2}$ for all
$1\leq j\leq 4$, satisfying

$S_{4}L^{2}\subset S_{3}L^{2}\subset S_{2}L^{2}\subset S_{1}L^{2}\subset
QL^{2}.$

Precisely speaking, $S_{j}$ is defined as the projection onto $S_{j}L^{2}$.

######  Remark 2.5.

The subspaces $S_{j}L^{2}$ for $j=1,2,3$ can be characterized via the zero
resonant and eigenfunctions $\phi$ in Definition 1.1. We refer to Lemma A.3
below for details.

Finally, we say that an integral operator
$T_{K}\in\mathbb{B}(L^{2}({\mathbb{R}}^{3}))$ with the kernel $K$ is
absolutely bounded if $T_{|K|}\in\mathbb{B}(L^{2}({\mathbb{R}}^{3}))$.

######  Theorem 2.6.

Let $Q$ and $S_{j}$ for $j=1,2,3$ be the projections defined as above. Assume
that $|V(x)|\lesssim(1+|x|)^{-\mu}$ with some $\mu>0$. Then there exists
$\lambda_{0}>0$ such that the following asymptotic expansions of
$M^{-1}(\lambda)$ hold for $0<\lambda\leq\lambda_{0}$ in
$\mathbb{B}(L^{2}(\mathbb{R}^{3}))$ :

  * (i)

If zero is a regular point of $H$ and $\mu>9$, then

$M^{-1}(\lambda)=QA_{0,1}^{0}Q+\lambda A^{0}_{1,1}+\Gamma^{0}_{2}(\lambda);$ (
2.19)

  * (ii)

If zero is a first kind resonance of $H$ and $\mu>13$, then

$M^{-1}(\lambda)=\lambda^{-1}S_{1}A_{-1,1}^{1}S_{1}+\Big{(}S_{1}A^{1}_{0,1}+A^{1}_{0,2}S_{1}+QA^{1}_{0,3}Q\Big{)}+\lambda
A^{1}_{1,1}+\Gamma^{1}_{2}(\lambda);$ ( 2.20)

  * (iii)

If zero is a second kind resonance of $H$ and $\mu>21$, then

$\begin{split}M^{-1}(\lambda)=&\lambda^{-3}S_{2}A^{2}_{-3,1}S_{2}+\lambda^{-2}\Big{(}S_{2}A^{2}_{-2,1}S_{1}+S_{1}A^{2}_{-2,2}S_{2}\Big{)}+\lambda^{-1}\Big{(}S_{2}A^{2}_{-1,1}+A^{2}_{-1,2}S_{2}\\\
&\ \ \ \ \ \ \ \ \ \ \
+S_{1}A^{2}_{-1,3}S_{1}\Big{)}+\Big{(}S_{1}A_{0,1}^{2}+A^{2}_{0,2}S_{1}+QA^{2}_{0,3}Q\Big{)}+\lambda
A^{2}_{1,1}+\Gamma^{2}_{2}(\lambda);\end{split}$ ( 2.21)

  * (iv)

If zero is the third kind resonance of $H$ and $\mu>25$, then

$\begin{split}M^{-1}(\lambda)=&\lambda^{-4}S_{3}A_{-4,1}^{3}S_{3}+\lambda^{-3}S_{2}A^{3}_{-3,1}S_{2}+\lambda^{-2}\Big{(}S_{2}A^{3}_{-2,1}S_{1}+S_{1}A^{3}_{-2,2}S_{2}\Big{)}+\lambda^{-1}\Big{(}S_{2}A^{3}_{-1,1}\\\
&\ \ \ \ \
+A^{3}_{-1,2}S_{2}+S_{1}A^{3}_{-1,3}S_{1}\Big{)}+\Big{(}S_{1}A_{0,1}^{3}+A^{3}_{0,2}S_{1}+QA^{3}_{0,3}Q\Big{)}+\lambda
A^{3}_{1,1}+\Gamma^{3}_{2}(\lambda).\end{split}$ ( 2.22)

Here in the statements (i)-(iv) above,

  * •

$A_{i,j}^{k}$ are $\lambda$-independent and absolutely bounded operators on
$L^{2}(\mathbb{R}^{3}).$ In this context, the index $k$ indicates the types of
resonance, the index $i$ signifies the degree of $\lambda$, and the index $j$
denotes the position of operators with the same power of $\lambda$.

  * •

$\Gamma^{k}_{2}(\lambda)$ are $\lambda$-dependent absolutely bounded operators
on $L^{2}(\mathbb{R}^{3})$ satisfying

$\displaystyle{\|\partial_{\lambda}^{\ell}\Gamma^{k}_{2}(\lambda)\|}_{L^{2}\to
L^{2}}\leq C_{\ell,k}\lambda^{2-\ell},\quad 0<\lambda\leq\lambda_{0},\
\ell=0,1,2,\ k=1,2,3.$ ( 2.23)

######  Remark 2.7.

For the purpose of this paper, we can disregard the specific expressions of
most operators $A_{i,j}^{k}$ as described in Theorem 2.6 above. However, it is
absolutely necessary to focus on the operators $A^{2}_{-3,1}$, $A^{2}_{-2,1}$,
$A^{2}_{-1,1}$, $A^{3}_{-4,1}$, $A^{3}_{-3,1}$, $A^{3}_{-2,1}$, and
$A^{3}_{-1,1}$, which are relevant in the context of the second and third
types of resonances. Indeed, in order to establish the unboundedness of wave
operators $W_{\pm}$ on $L^{p}(\mathbb{R}^{3})$ for $3\leq p<\infty$ in the
second and third types of resonances, we require more detailed information
about these specific operators. These operators are outlined as follows (for
further details, see Section A below):

  * •

The $A^{2}_{-3,1}$, $A^{2}_{-2,1}$, $A^{2}_{-1,1}$ in the  2.21 can be given
as follows (also see  A.30 for more details):

$\begin{split}A^{2}_{-3,1}&=4\pi\cdot 5!(1-i)D_{2},\,\,\,\ \
A^{2}_{-2,1}=20iD_{2}vG_{1}vTD_{1},\\\
A^{2}_{-1,1}&=S_{2}H^{2}_{-1,1}+S_{2}H^{2}_{-1,2}Q;\\\
H^{2}_{-1,1}&=\frac{80\pi(1+i)}{\|V\|_{L^{1}}}(D_{2}vG_{1}v-D_{2}vG_{1}vTD_{1}T),\,\,\,\,H^{2}_{-1,2}\in\mathbb{B}(L^{2}).\end{split}$
( 2.24)

  * •

The $A^{3}_{-4,1}$, $A^{3}_{-3,1}$, $A^{3}_{-2,1}$, $A^{3}_{-1,1}$ in the
2.22 can be given as follows (also see  A.36 for more details):

$\begin{split}A^{3}_{-4,1}&=D_{3},\,\,A^{3}_{-3,1}=S_{2}H^{3}_{-3,1}S_{2}+S_{2}H^{3}_{-3,2}S_{3},\,\,\,\,A^{3}_{-1,1}=S_{2}H^{3}_{-1,1}+S_{2}H^{3}_{-1,2}Q,\\\
A^{3}_{-2,1}&=i\Big{[}\frac{\|V\|_{L^{1}}}{32\pi^{2}\cdot
5!}D_{3}vG_{3}vD_{1}+20(D_{2}vG_{1}vTD_{1}-D_{3}vG_{4}vD_{2}vG_{1}vTD_{1})\\\
&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\frac{1}{32\pi^{2}(3!)^{2}}D_{3}vG_{3}vD_{1}TvG_{1}vD_{2}vG_{1}vTD_{1}\Big{]};\\\
H^{3}_{-3,1}&=4\pi\cdot 5!(1-i)(D_{2}-D_{3}vG_{4}vD_{2})+\frac{1-i}{8\pi\cdot
3!}D_{3}vG_{3}vD_{1}TvG_{1}vD_{2};\ H^{3}_{-3,2}\in\mathbb{B}(L^{2}).\\\
H^{3}_{-1,1}&=\frac{80\pi(1+i)}{\|V\|_{L^{1}}}(D_{2}vG_{1}v-D_{2}vG_{1}vTD_{1}T+D_{3}vG_{4}vD_{2}vG_{1}vTD_{1}T-D_{3}vG_{4}vD_{2}vG_{1}v)\\\
&\,\,\,\,\,\,\,\,+\frac{1+i}{8\pi(3!)^{2}\|V\|_{L^{1}}}(D_{3}vG_{3}vD_{1}TvG_{1}vD_{2}vG_{1}v-D_{3}vG_{3}vD_{1}TvG_{1}vD_{2}vG_{1}vTD_{1}T)\\\
&\,\,\,\,\,\,\,\,-\frac{1+i}{8\pi\cdot 5!}D_{3}vG_{3}vD_{1}T;\\\
H^{3}_{-1,2}&\in\mathbb{B}(L^{2}).\end{split}$ ( 2.25)

In the aforementioned remark, we also omit the detailed descriptions of
operators $H^{2}_{-1,2}$, $H^{3}_{-3,2}$, and $H^{3}_{-1,2}$ for the sake of
simplicity. It suffices to note that they are $\lambda$-independent operators
that are absolutely bounded in $L^{2}(\mathbb{R}^{3})$ for the purposes of
this paper.

However, it is worth mentioning that the specific contents of all operators
$A_{i,j}^{k}$ and $H_{i,j}^{k}$ mentioned above can be found in the proof of
Theorem 2.6.

##  3\. The $L^{p}$-boundedness of wave operators with the first kind zero
resonances

This section is devoted to the proof of Theorem 1.4 (i.e. the
$L^{p}$-boundedness of wave operators with the first kind of zero resonances),
which are divided into the following two subsections. The first one is to
recall the regular case and the second one is to consider the first kind of
resonances case.

###  3.1. The regular case

In this subsection, let us first sketch the proof of $L^{p}$ boundedness of
wave operator $W_{-}$ for all $1<p<\infty$ in the regular case. For a fixed
$a>0$, we first decompose the stationary formula  2.1 of $W_{-}$ into the
low and high energy parts

$\displaystyle\Omega^{<a}$ $\displaystyle=\frac{2}{\pi
i}\int_{0}^{\infty}\lambda^{3}\chi_{<a}(\lambda)R_{V}^{+}(\lambda^{4})V\left(R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\right)d\lambda,$
( 3.1) $\displaystyle\Omega^{\geq a}$ $\displaystyle=\frac{2}{\pi
i}\int_{0}^{\infty}\lambda^{3}\chi_{\geq
a}(\lambda)R_{V}^{+}(\lambda^{4})V\left(R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\right)d\lambda.$
( 3.2)

Then $W_{-}=\mathop{\mathrm{Id}}\nolimits-\left(\Omega^{<a}+\Omega^{\geq
a}\right)$.

In view of the decomposition of $W_{-}$ above, it suffices to estimate
$\Omega^{<a}$ and $\Omega^{\geq a}$ separately. Indeed, in the work by
Goldberg and Green [18, Proposition 4.1], it has been proved that high energy
part $\Omega^{\geq a}$ is bounded on $L^{p}({\mathbb{R}}^{3})$ for all $1\leq
p\leq\infty$ requiring that $V(x)=O({\langle}x{\rangle}^{-\mu})$ with $\mu>5$.

Hence it only remains to show the $L^{p}$ boundedness of the low energy part
$\Omega^{<a}$, which in turns depends on different types of zero resonance. By
Lemma 2.4 we write that

$\displaystyle\Omega^{<a}$ $\displaystyle=\frac{2}{\pi
i}\int_{0}^{\infty}\lambda^{3}\chi_{<a}(\lambda)R_{0}^{+}(\lambda^{4})vM^{-1}(\lambda)v\left(R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\right)d\lambda.$
( 3.3)

If we choose a sufficiently small but fixed $a>0$, then we will make use of
asymptotic expansion of $M^{-1}(\lambda)$, as stated in Theorem 2.6 for
different kinds of zero resonances.

If zero is a regular point of $H$, then we can put
$M^{-1}(\lambda)=QA_{0,1}^{0}Q+\lambda A^{0}_{1,1}+\Gamma^{0}_{2}(\lambda)$
into the formula 3.3, which are split into the sum of the following three
operators:

$\displaystyle T_{B}$ $\displaystyle=\frac{2}{\pi
i}\int_{0}^{\infty}\lambda^{3}\chi_{<a}(\lambda)R_{0}^{+}(\lambda^{4})vBv\left(R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\right)d\lambda,$
( 3.4)

where $B$ is anyone of the three operators $QA_{0,1}^{0}Q,$ $\lambda
A^{0}_{1,1,}$ and $\Gamma^{0}_{2}(\lambda)$.

Certainly, Goldberg and Green [18] have demonstrated that each operator
$T_{B}$ is bounded on $L^{p}$ for all $1<p<\infty$. As a result, it has been
shown that the low-energy operator $\Omega^{<a}$ and the wave operator $W_{-}$
hold the same $L^{p}$ bounds for all $1<p<\infty$.

It is important to note that the proofs of the $L^{p}$-boundedness for
operators like $T_{B}$ in [18] are not contingent on the specific expressions
of the operators $A_{0,1}$, $A^{0}_{1,1}$, and $\Gamma^{0}_{2}(\lambda)$.
Rather, they rely on the absolute boundedness of these operators on $L^{2}$
and the general estimates in  2.23. Therefore, in the subsequent zero
resonance cases, we will omit the proofs for similar operators like $T_{B}$
for simplicity. Additional details can be found in [18, 26].

###  3.2. The first kind of resonance case

In this subsection we will begin with the proof of $L^{p}$ boundedness of wave
operator $W_{-}$ for all $1<p<\infty$ in the first kind of resonance.

As mentioned in the regular case above, it suffices to show deal with the low
energy part $\Omega_{<a}$ for all $1<p<\infty$ by the formula  3.3 with the
following expansion of $M^{-1}(\lambda)$:

$\displaystyle
M^{-1}(\lambda)=\lambda^{-1}S_{1}A_{-1,1}^{1}S_{1}+S_{1}A^{1}_{0,1}+A^{1}_{0,2}S_{1}+QA^{1}_{0,3}Q+\lambda
A^{1}_{1,1}+\Gamma^{1}_{2}(\lambda).$ ( 3.5)

The three terms $B=QA^{1}_{0,3}Q$, $\lambda A^{1}_{1,1}$ and
$\Gamma^{1}_{2}(\lambda)$ in $M^{-1}(\lambda)$ above can be dealt with by the
same process as the corresponding three terms in regular case.

Hence we only consider the remained three terms
$\lambda^{-1}S_{1}A_{-1,1}^{1}S_{1}$, $S_{1}A^{1}_{0,1}$ and
$A^{1}_{0,2}S_{1}$ in the following two Propositions 3.1\- 3.2, respectively.

######  Proposition 3.1.

Let

$\displaystyle K^{1}_{-1,1}:=\frac{2}{\pi
i}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\
R_{0}^{+}(\lambda^{4})vS_{1}A_{-1,1}^{1}S_{1}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}d\lambda.$
( 3.6)

Then $K^{1}_{-1,1}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all
$1<p<\infty$.

###### Proof.

Let $f\in\mathbb{S}({\mathbb{R}}^{3})$. Then

$\displaystyle(K^{1}_{-1,1}f)(x)=\frac{2}{\pi
i}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\Big{[}R_{0}^{+}(\lambda^{4})vS_{1}A_{-1,1}^{1}S_{1}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}f\Big{]}(x)d\lambda.$
( 3.7)

Firstly, by the formula  2.4 and $S_{1}(v)=0$, we have

$S_{1}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}f=\frac{\pi
i}{2\lambda}\int_{S^{2}}S_{1}v\big{(}e^{i\lambda\omega
z}-1\big{)}\hat{f}(\lambda\omega)d\omega.$

Then by using Taylor expansion of $e^{i\lambda\omega z}$ in
$z\in{\mathbb{R}}^{3}$, that is,

$\displaystyle e^{i\lambda\omega z}$ $\displaystyle=1+i\lambda\omega
z-\lambda^{2}(\omega z)^{2}\int_{0}^{1}(1-\theta)e^{i\lambda\omega\theta
z}d\theta$ $\displaystyle=1+i\sum_{k=1}^{3}\lambda
z_{k}\omega_{k}-\sum_{k,l=1}^{3}\lambda^{2}z_{k}z_{l}\int_{0}^{1}(1-\theta)\omega_{k}\omega_{l}e^{i\lambda\omega\theta
z}d\theta,$ ( 3.8)

one can obtain that

$\displaystyle\frac{2}{\pi i}\Big{(}S_{1}v\big{(}R_{0}^{+}(\lambda^{4})-$
$\displaystyle
R_{0}^{-}(\lambda^{4})\big{)}f\Big{)}=i\sum_{k=1}^{3}S_{1}(z_{k}v)\Big{(}\int_{S^{2}}\omega_{k}\hat{f}(\lambda\omega)d\omega\Big{)}$
$\displaystyle-\sum_{k,l=1}^{3}\lambda\int_{0}^{1}(1-\theta)S_{1}\Big{(}z_{k}z_{l}v\int_{S^{2}}\omega_{k}\omega_{l}e^{i\lambda\omega\theta
z}\hat{f}(\lambda\omega)d\omega\Big{)}d\theta.$ ( 3.9)

Hence putting 3.2 into  3.7, we can rewrite $K_{-1,1}^{1}f$ as follows:

$\displaystyle(K^{1}_{-1,1}f)(x)=i\sum_{k=1}^{3}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\Big{[}R_{0}^{+}(\lambda^{4})vS_{1}A_{-1,1}^{1}S_{1}(z_{k}v)\Big{]}(x)\Big{(}\int_{S^{2}}\omega_{k}\hat{f}(\lambda\omega)d\omega\Big{)}d\lambda$
$\displaystyle-\sum_{k,l=1}^{3}\int_{0}^{1}(1-\theta)\int_{0}^{\infty}\lambda^{3}\chi_{<a}(\lambda)\Big{[}R_{0}^{+}(\lambda^{4})vS_{1}A_{-1,1}^{1}S_{1}\Big{(}z_{k}z_{l}v\int_{S^{2}}\omega_{k}\omega_{l}e^{i\lambda\omega\theta
z}\hat{f}(\lambda\omega)d\omega\Big{)}\Big{]}(x)d\lambda d\theta$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \
:=i\sum_{k=1}^{3}(T_{k}f)(x)-\sum_{k,l=1}^{3}(T_{k,l}f)(x).$ ( 3.10)

Thus in order to show $K^{1}_{-1,1}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for
all $1<p<\infty$, it suffices to obtain the same desired results for each
$T_{k}$ and $T_{k,l}$ with $1\leq k,l\leq 3$, which will be given in the
following times (i) and (ii), separately.
(i) The proof of $T_{k,l}\in\mathbb{B}(L^{p})$ for all $1<p<\infty.$

Note that by the formula  2.4,

$\displaystyle\int_{S^{2}}\omega_{k}\omega_{l}e^{i\lambda\omega\theta
z}\hat{f}(\lambda\omega)d\omega=\frac{2\lambda}{\pi
i}\Big{[}\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}\mathcal{R}_{k}\mathcal{R}_{l}f\Big{]}(\theta
z),$ ( 3.11)

where $\mathcal{R}=(\mathcal{R}_{1},\mathcal{R}_{2},\mathcal{R}_{3})$ are
Riesz transforms given by
$\widehat{\mathcal{R}_{j}f}(\xi)=\frac{\xi_{j}}{|\xi|}\hat{f}(\xi)$ for any
$f\in\mathbb{S}({\mathbb{R}}^{3})$. Hence by  3.11, we can rewrite
$T_{k,l}f$ as follows:

$\displaystyle(T_{k,l}f)(x)$ $\displaystyle=\frac{2}{\pi
i}\int_{0}^{1}(1-\theta)\int_{0}^{\infty}\lambda^{4}\chi_{<a}(\lambda)\Big{[}R_{0}^{+}(\lambda^{4})vS_{1}A_{-1,1}^{1}S_{1}(z_{k}z_{l}v)$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\times\Big{(}\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}\mathcal{R}_{k}\mathcal{R}_{l}f\Big{)}(\theta
z)\Big{]}(x)d\lambda d\theta$ $\displaystyle=\frac{2}{\pi
i}\int_{0}^{1}(1-\theta)\int_{{\mathbb{R}}^{6}}v(\rho)(S_{1}A^{1}_{-1,1}S_{1})(\rho,z)\
z_{k}z_{l}v(z)\Big{[}\int_{{\mathbb{R}}^{3}}\Big{(}\int_{0}^{\infty}\lambda^{4}\chi_{<a}(\lambda)R_{0}^{+}(\lambda^{4},x,\rho)$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\times\big{(}R_{0}^{+}-R_{0}^{-}\big{)}(\lambda^{4},\theta
z,y)d\lambda\Big{)}\big{(}\mathcal{R}_{k}\mathcal{R}_{l}f\big{)}(y)dy\Big{]}dzd\rho
d\theta$ $\displaystyle=\frac{2}{\pi
i}\int_{0}^{1}(1-\theta)\int_{{\mathbb{R}}^{6}}v(\rho)(S_{1}A^{1}_{-1,1}S_{1})(\rho,z)\
z_{k}z_{l}v(z)\Big{(}A_{\rho,z\theta}(\mathcal{R}_{k}\mathcal{R}_{l}f)\Big{)}(x)dzd\rho
d\theta,$ ( 3.12)

where $A_{\rho,\theta z}$ is the integral operator defined by the following
kernel:

$A_{\rho,\theta
z}(x,y)=\int_{0}^{\infty}\lambda^{4}\chi_{<a}(\lambda)R_{0}^{+}(\lambda^{4},x,\rho)\big{(}R_{0}^{+}-R_{0}^{-}\big{)}(\lambda^{4},\theta
z,y)d\lambda.$

Hence by Minkowski’s inequality, Lemma 2.1 and $L^{p}$-boundedness of Riesz
transforms $\mathcal{R}$ (see Lemma 2.3), we obtain that for all $1<p<\infty$,

$\displaystyle\|T_{k,l}f\|_{L^{p}}$
$\displaystyle\lesssim\Big{(}\int_{{\mathbb{R}}^{6}}\big{|}v(\rho)(S_{1}A^{1}_{-1,1}S_{1})(\rho,z)\
z_{k}z_{l}v(z)\big{|}d\rho
dz\Big{)}\big{\|}A_{0,0}\big{\|}_{\mathbb{B}(L^{p})}\
\big{\|}\mathcal{R}_{k}\mathcal{R}_{l}\big{\|}_{\mathbb{B}(L^{p})}\
\|f\|_{L^{p}}$
$\displaystyle\lesssim\Big{(}\int_{{\mathbb{R}}^{6}}\big{|}v(\rho)(S_{1}A^{1}_{-1,1}S_{1})(\rho,z)\
z_{k}z_{l}v(z)\big{|}d\rho dz\Big{)}\|f\|_{L^{p}}.$ ( 3.13)

Since $S_{1}A_{-1,1}^{1}S_{1}$ is a finite rank (Hilbert-Schmidt) operator and
$V$ has fast decay, then we have

$\int_{{\mathbb{R}}^{6}}\big{|}v(\rho)(S_{1}A^{1}_{-1,1}S_{1})(\rho,z)\
z_{k}z_{l}v(z)\big{|}d\rho dz\leq\|v\|_{L^{2}}\
\|S_{1}A^{1}_{-1,1}S_{1}\|_{HS}\ \|z_{k}z_{l}v\|_{L^{2}}<\infty,$

which with 3.2 immediately yields the $T_{k,l}\in\mathbb{B}(L^{p})$ for all
$1<p<\infty$.
(ii) The proof of $T_{k}\in\mathbb{B}(L^{p})$ for all $1<p<\infty$.

Note that $S_{1}$ is a finite rank operator which means $n_{1}:=\dim
S_{1}L^{2}<\infty$. We may assume that $\\{f_{j}\\}_{j=1}^{n_{1}}$ be the
standard orthogonal bases of $S_{1}L^{2}$. Then

$S_{1}A_{-1,1}^{1}S_{1}(z_{k}v)=\sum_{j=1}^{n_{1}}{\langle}S_{1}A_{-1,1}^{1}S_{1}(z_{k}v),f_{j}{\rangle}f_{j}:=\sum_{j=1}^{n_{1}}c_{j}^{k}f_{j}.$

Hence from 3.2 we can rewrite $T_{k}f(x)$ as follows:

$\displaystyle
T_{k}f(x)=\sum_{j=1}^{n_{1}}c_{j}^{k}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega\Big{)}d\lambda.$
( 3.14)

In the sequel, we decompose $T_{k}=\chi_{\geq
4a}(|D|)T_{k}+\chi_{<4a}(|D|)T_{k}$ and show that both $\chi_{\geq
4a}(|D|)T_{k}\in\mathbb{B}(L^{p})$ and
$\chi_{<4a}(|D|)T_{k}\in\mathbb{B}(L^{p})$ for all $1<p<\infty$.
(A) The proof of $\chi_{\geq 4a}(|D|)T_{k}\in\mathbb{B}(L^{p})$ for all
$1<p<\infty$.

We rewrite by the  3.14 that

$\displaystyle\chi_{\geq 4a}(|D|)T_{k}f(x)$
$\displaystyle=\sum_{j=1}^{n_{1}}c_{j}^{k}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\big{(}\chi_{\geq
4a}(|D|)R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega\Big{)}d\lambda$
$\displaystyle:=\sum_{j=1}^{n_{1}}c_{j}^{k}\ \mathbf{T}_{k}^{j}f(x).$ ( 3.15)

In the following it suffices to show that
$\mathbf{T}_{k}^{j}\in\mathbb{B}(L^{p})$ for all $1<p<\infty$ and each $1\leq
j\leq n_{1}$.

Note that by the Fourier transform

$\chi_{\geq
4a}(|D|)R_{0}^{+}(\lambda^{4})=M(D)+\lambda^{4}M(D)R_{0}^{+}(\lambda^{4}),\ \
0<\lambda\leq a,$

where $M(D)$ is the Fourier multiplier corresponding to $M(\xi)=\chi_{\geq
4a}(|\xi|)|\xi|^{-4}$. Hence for each $1\leq j\leq n_{1}$, one has

$\displaystyle\mathbf{T}_{k}^{j}f(x)=$
$\displaystyle\big{(}M(D)vf_{j}\big{)}(x)\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega\Big{)}d\lambda$
$\displaystyle\ \ \ \
+\int_{0}^{\infty}\lambda^{6}\chi_{<a}(\lambda)\big{(}M(D)R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega\Big{)}d\lambda$
$\displaystyle:=\mathbf{T}_{k}^{j,1}f(x)+\mathbf{T}_{k}^{j,2}f(x).$ ( 3.16)

Let first deal with $\mathbf{T}_{k}^{j,1}f$ with $1\leq k\leq 3$ and $1\leq
j\leq n_{1}$.

In fact, set $\eta=\lambda\omega$, then $d\eta=\lambda^{2}d\lambda\ d\omega$
due to the spherical polar coordinate transform. So one obtain that

$\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega\Big{)}d\lambda=\int_{{\mathbb{R}}^{3}}\chi_{<a}(|\eta|)\widehat{\mathcal{R}_{k}f}(\eta)d\eta=\int_{{\mathbb{R}}^{3}}\widehat{\chi_{<a}(|\cdot|)}(y)\mathcal{R}_{k}f(y)dy.$

Thus it follows that for all $1<p<\infty$,

$\displaystyle\|\mathbf{T}^{j,1}_{k}f\|_{L^{p}}\leq\|M(D)vf_{j}\|_{L^{p}}\
\Big{|}\int_{{\mathbb{R}}^{3}}\widehat{\chi_{<a}(|\cdot|)}(y)\mathcal{R}_{k}f(y)dy\Big{|}.$
( 3.17)

Clearly, by Hölder inequality and the $L^{p}$ bounds of Riesz transform
$\mathcal{R}_{k}$, one has

$\Big{|}\int_{{\mathbb{R}}^{3}}\widehat{\chi_{<a}(|\cdot|)}(y)\mathcal{R}_{k}f(y)dy\Big{|}\leq\|\widehat{\chi_{<a}(|\cdot|)}\|_{L^{p^{\prime}}}\|\mathcal{R}_{k}f\|_{L^{p}}\lesssim\|f\|_{L^{p}}.$

On the other hand, since $M(\xi)=\chi_{\geq 4a}(|\xi|)|\xi|^{-4}$ on
${\mathbb{R}}^{3}$, then the Fourier transform gives that
$|\mathcal{F}^{-1}(M)(x)|\lesssim{\langle}x{\rangle}^{-N}$ for any $N$, which
implies that $\mathcal{F}^{-1}(M)\in L^{p}$ for all $1\leq p\leq\infty$. Hence
Young inequality gives that

$\|M(D)vf_{j}\|_{L^{p}}=\|(2\pi)^{-3}\mathcal{F}^{-1}(M)*(vf_{j})\|_{L^{p}}\lesssim\|\mathcal{F}^{-1}(M)\|_{L^{p}}\|vf_{j}\|_{L^{1}}<\infty.$

Thus by the  3.17 we immediately conclude that
$\mathbf{T}_{k}^{j,1}\in\mathbb{B}(L^{p})$ for all $1<p<\infty$.

Next to consider $\mathbf{T}_{k}^{j,2}$ for each $1\leq k\leq 3$ and $1\leq
j\leq n_{1}$.

From 3.2,  2.4 and  2.5, we have

$\displaystyle\mathbf{T}_{k}^{j,2}f(x)$
$\displaystyle=\int_{0}^{\infty}\lambda^{3}\chi_{<a}(\lambda)\big{(}M(D)R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\mathcal{F}\Big{(}|D|^{3}\chi_{<4a}(|D|)\mathcal{R}_{k}f\Big{)}(\lambda\omega)d\omega\Big{)}d\lambda$
$\displaystyle=\frac{2}{\pi
i}\int_{0}^{\infty}\lambda^{4}\chi_{<a}(\lambda)\big{(}M(D)R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{[}\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}|D|^{3}\chi_{<4a}(|D|)\mathcal{R}_{k}f\Big{]}(0)d\lambda$
$\displaystyle=\frac{2}{\pi
i}\int_{{\mathbb{R}}^{3}}v(\rho)f_{j}(\rho)M(D)\Big{[}\int_{{\mathbb{R}}^{3}}\Big{(}\int_{0}^{\infty}\lambda^{4}\chi_{<a}(\lambda)R_{0}^{+}(\lambda^{4})(x,\rho)\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}(0,y)d\lambda\Big{)}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \
\times\Big{(}|D|^{3}\chi_{<4a}(|D|)\mathcal{R}_{k}f\Big{)}(y)dy\Big{]}d\rho$
$\displaystyle=\frac{2}{\pi
i}\int_{{\mathbb{R}}^{3}}v(\rho)f_{j}(\rho)\Big{(}M(D)A_{\rho,0}|D|^{3}\chi_{<4a}(|D|)\mathcal{R}_{k}f\Big{)}(x)d\rho,$
( 3.18)

where we have used the equality
$\chi_{<a}(\lambda)=\chi_{<a}(\lambda)\chi_{<4a}(\lambda)$ in the 3.2 above.

Note that $M(D)$, $|D|^{3}\chi_{<4a}(|D|)$ and $\mathcal{R}_{k}$ are all
$L^{p}$-Fourier multiplier operators from Mihilin’s theorem ( or see Lemma
2.3), and $A_{\rho,0}$ are uniformly bounded for all $\rho$ on
$L^{p}({\mathbb{R}}^{3})$ by Lemma 2.1. Therefore by 3.2 and Minkowski
inequality, it follows that for all $1<p<\infty$,

$\|\mathbf{T}_{k}^{j,2}\|_{\mathbb{B}(L^{p})}\lesssim\|vf_{j}\|_{L^{1}}\|M(D)\|_{\mathbb{B}(L^{p})}\big{(}\sup_{\rho}\|A_{\rho,0}\|_{\mathbb{B}(L^{p})}\big{)}\big{\|}|D|^{3}\chi_{<4a}(|D|)\mathcal{R}_{k}\big{\|}_{\mathbb{B}(L^{p})}<\infty.$

Thus we obtain that $\mathbf{T}_{k}^{j}\in\mathbb{B}(L^{p})$ for all
$1<p<\infty$ by 3.2, and then conclude that $\chi_{\geq
4a}(|D|)T_{k}\in\mathbb{B}(L^{p})$ for all $1<p<\infty$ by 3.2.
(B) The proof of $\chi_{<4a}(|D|)T_{k}\in\mathbb{B}(L^{p})$ for all
$1<p<\infty$.

Again by the  3.14, we rewrite $\chi_{<4a}(|D|)T_{k}f$ for each $1\leq k\leq
3$ as follows:

$\displaystyle\chi_{<4a}(|D|)T_{k}f(x)=\sum_{j=1}^{n_{1}}c_{j}^{k}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\big{(}\chi_{<4a}(|D|)R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega\Big{)}d\lambda.$
( 3.19)

Note that by the Fourier transform, we have

$\displaystyle\chi_{<4a}(|D|)R_{0}^{\pm}(\lambda^{4})vf_{j}(x)=\frac{1}{2\lambda^{2}}\lim_{\epsilon\to
0^{+}}\int_{{\mathbb{R}}^{3}}e^{ix\xi}\chi_{<4a}(|\xi|)\Big{(}\frac{1}{|\xi|^{2}-(\lambda+i\epsilon)^{2}}-\frac{1}{|\xi|^{2}+\lambda^{2}}\Big{)}\widehat{vf_{j}}(\xi)d\xi.$
( 3.20)

Since $S_{1}(v)=0$ (i.e. $v\perp f_{j}$), then
$\widehat{vf_{j}}(0)=\int_{{\mathbb{R}}^{3}}v(\rho)f_{j}(\rho)d\rho=0$. Hence
one has

$\displaystyle\widehat{vf_{j}}(\xi)=\frac{1}{(2\pi)^{3}}\int_{{\mathbb{R}}^{3}}\big{(}e^{-i\rho\xi}-1\big{)}(vf_{j})(\rho)d\rho=\frac{-i}{(2\pi)^{3}}\sum_{\ell=1}^{3}\int_{0}^{1}\int_{{\mathbb{R}}^{3}}\xi_{\ell}\
e^{-i\theta\rho\xi}(\rho_{\ell}vf_{j})(\rho)d\rho d\theta.$ ( 3.21)

Putting the  3.21 into  3.20, we obtain that

$\displaystyle\chi_{<4a}(|D|)R_{0}^{\pm}(\lambda^{4})vf_{j}(x)$
$\displaystyle=\frac{-i}{(2\pi)^{3}}\sum_{\ell=1}^{3}\lim_{\epsilon\to
0^{+}}\int_{0}^{1}\int_{{\mathbb{R}}^{3}}(\rho_{\ell}vf_{j})(\rho)\Big{[}\frac{1}{2\lambda^{2}}\int_{{\mathbb{R}}^{3}}e^{i(x-\theta\rho)\xi}\chi_{<4a}(|\xi|)\xi_{\ell}$
$\displaystyle\
\times\Big{(}\frac{1}{|\xi|^{2}-(\lambda+i\epsilon)^{2}}-\frac{1}{|\xi|^{2}+\lambda^{2}}\Big{)}d\xi\Big{]}d\rho
d\theta.$

By Lebesgue’s dominated convergence theorem, we will obtain that

$\displaystyle\chi_{<4a}(|D|)R_{0}^{\pm}(\lambda^{4})vf_{j}(x)=-i\sum_{\ell=1}^{3}$
$\displaystyle\int_{0}^{1}\int_{{\mathbb{R}}^{3}}(\rho_{\ell}vf_{j})(\rho)\Big{[}\chi_{<4a}(|D|)\mathcal{R}_{\ell}J_{\theta\rho}\Big{(}\lambda
R^{+}_{0}(\lambda^{4})(x,0)$
$\displaystyle+\frac{1}{(2\pi)^{3}}\int_{{\mathbb{R}}^{3}}e^{ix\xi}\frac{1}{(|\xi|^{2}+\lambda^{2})(|\xi|+\lambda)}d\xi\Big{)}\Big{]}d\rho
d\theta.$ ( 3.22)

In the sequel, we give more details about how to establish 3.2. Firstly, by
Lebesgue’s dominated convergence theorem we will obtain that

$\displaystyle\chi_{<4a}(|D|)R_{0}^{\pm}(\lambda^{4})vf_{j}(x)$
$\displaystyle=\frac{-i}{(2\pi)^{3}}\sum_{\ell=1}^{3}\int_{0}^{1}\int_{{\mathbb{R}}^{3}}(\rho_{\ell}vf_{j})(\rho)\lim_{\epsilon\to
0^{+}}\Big{[}\frac{1}{2\lambda^{2}}\int_{{\mathbb{R}}^{3}}e^{i(x-\theta\rho)\xi}\chi_{<4a}(|\xi|)\xi_{\ell}$
$\displaystyle\
\times\Big{(}\frac{1}{|\xi|^{2}-(\lambda+i\epsilon)^{2}}-\frac{1}{|\xi|^{2}+\lambda^{2}}\Big{)}d\xi\Big{]}d\rho
d\theta.$ ( 3.23)

In fact, in order to get 3.2, it suffices to show that

$\begin{split}B_{\varepsilon}(\lambda,x-\theta\rho):=\Big{|}\int_{{\mathbb{R}}^{3}}e^{i(x-\theta\rho)\xi}\chi_{<4a}(|\xi|)\xi_{\ell}\Big{(}\frac{1}{|\xi|^{2}-(\lambda+i\epsilon)^{2}}-\frac{1}{|\xi|^{2}+\lambda^{2}}\Big{)}d\xi\Big{|}\lesssim
1,\end{split}$ ( 3.24)

uniformly in $\varepsilon$, which actually holds due to the following
estimates:

$\begin{split}B_{\varepsilon}(\lambda,x-\theta\rho)&=\Big{|}\int_{{\mathbb{R}}^{3}}\mathcal{F}^{-1}\big{(}\chi_{<4a}(|\xi|)\xi_{\ell}\big{)}(y)\frac{e^{i(\lambda+i\varepsilon)|x-\theta\rho-y|}-e^{-\lambda|x-\theta\rho-y|}}{4\pi|x-\theta\rho-y|}dy\Big{|}\\\
&\lesssim\int_{{\mathbb{R}}^{3}}\big{|}\mathcal{F}^{-1}\big{(}\chi_{<4a}(|\xi|)\xi_{\ell}\big{)}(y)\big{|}\frac{1}{|x-\theta\rho-y|}dy\\\
&\lesssim\int_{{\mathbb{R}}^{3}}\frac{\langle
y\rangle^{-2-}}{|x-\theta\rho-y|}dy\lesssim 1,\end{split}$ ( 3.25)

uniformly in $\varepsilon,\lambda,x,\theta,\rho$.

Next, we need to establish that

$\displaystyle\lim_{\epsilon\to 0^{+}}$
$\displaystyle\frac{1}{2\lambda^{2}}\int_{{\mathbb{R}}^{3}}e^{i(x-\theta\rho)\xi}\chi_{<4a}(|\xi|)\xi_{\ell}\Big{(}\frac{1}{|\xi|^{2}-(\lambda+i\epsilon)^{2}}-\frac{1}{|\xi|^{2}+\lambda^{2}}\Big{)}d\xi$
$\displaystyle=\chi_{<4a}(|D|)\mathcal{R}_{\ell}J_{\theta\rho}\Big{(}(2\pi)^{3}\lambda
R^{+}_{0}(\lambda^{4})(x,0)+\int_{{\mathbb{R}}^{3}}e^{ix\xi}\frac{1}{(|\xi|^{2}+\lambda^{2})(|\xi|+\lambda)}d\xi\Big{)},$
( 3.26)

where $(J_{\theta\rho}g)(x)=g(x-\theta\rho)$ is a translation operator
satisfying the commutativity $m(D)J_{\theta\rho}=J_{\theta\rho}m(D)$ for any
Fourier multiplier $m(D)$.

In fact, we can write that

$\begin{split}C_{\varepsilon}(\lambda,x-\theta\rho)&:=\frac{1}{2\lambda^{2}}\int_{{\mathbb{R}}^{3}}e^{i(x-\theta\rho)\xi}\chi_{<4a}(|\xi|)\xi_{\ell}\Big{(}\frac{1}{|\xi|^{2}-(\lambda+i\epsilon)^{2}}-\frac{1}{|\xi|^{2}+\lambda^{2}}\Big{)}d\xi\\\
&=\frac{\lambda}{2\lambda^{2}}\int_{{\mathbb{R}}^{3}}e^{i(x-\theta\rho)\xi}\chi_{<4a}(|\xi|)\frac{\xi_{\ell}}{|\xi|}\Big{(}\frac{1}{|\xi|^{2}-(\lambda+i\epsilon)^{2}}-\frac{1}{|\xi|^{2}+\lambda^{2}}\Big{)}d\xi\\\
&\ \ \
+\frac{1}{2\lambda^{2}}\int_{{\mathbb{R}}^{3}}e^{i(x-\theta\rho)\xi}\chi_{<4a}(|\xi|)\frac{\xi_{\ell}}{|\xi|}\big{(}|\xi|-\lambda\big{)}\Big{(}\frac{1}{|\xi|^{2}-(\lambda+i\epsilon)^{2}}-\frac{1}{|\xi|^{2}+\lambda^{2}}\Big{)}d\xi\\\
&:=C_{\varepsilon,1}(\lambda,x-\theta\rho)+C_{\varepsilon,2}(\lambda,x-\theta\rho).\end{split}$

We firstly claim that

$\begin{split}\lim_{\epsilon\to
0^{+}}C_{\varepsilon,2}(\lambda,x-\theta\rho)=\chi_{<4a}(|D|)\mathcal{R}_{\ell}J_{\theta\rho}\int_{{\mathbb{R}}^{3}}e^{ix\xi}\frac{1}{(|\xi|^{2}+\lambda^{2})(|\xi|+\lambda)}d\xi.\end{split}$
( 3.27)

Note that

$\begin{split}C_{\varepsilon,2}(\lambda,x-\theta\rho)&=\frac{1}{2\lambda^{2}}\int_{{\mathbb{R}}^{3}}e^{i(x-\theta\rho)\xi}\chi_{<4a}(|\xi|)\frac{\xi_{\ell}}{|\xi|}\frac{2\lambda^{2}-\varepsilon^{2}+2i\varepsilon\lambda}{(|\xi|+\lambda+i\epsilon)(|\xi|^{2}+\lambda^{2})}d\xi\\\
&\ \ \
+\frac{i\varepsilon}{2\lambda^{2}}\int_{{\mathbb{R}}^{3}}e^{i(x-\theta\rho)\xi}\chi_{<4a}(|\xi|)\frac{\xi_{\ell}}{|\xi|}\Big{(}\frac{1}{|\xi|^{2}-(\lambda+i\epsilon)^{2}}-\frac{1}{|\xi|^{2}+\lambda^{2}}\Big{)}d\xi\\\
&:=C^{1}_{\varepsilon,2}(\lambda,x-\theta\rho)+C^{2}_{\varepsilon,2}(\lambda,x-\theta\rho).\end{split}$

One can easily check that $C^{2}_{\varepsilon,2}(\lambda,x-\theta\rho)$ tends
to zero as $\varepsilon$ tends to zero, since by the same process as in
3.25,

$\begin{split}\Big{|}\int_{{\mathbb{R}}^{3}}e^{i(x-\theta\rho)\xi}\chi_{<4a}(|\xi|)\frac{\xi_{\ell}}{|\xi|}\Big{(}\frac{1}{|\xi|^{2}-(\lambda+i\epsilon)^{2}}-\frac{1}{|\xi|^{2}+\lambda^{2}}\Big{)}d\xi\Big{|}\lesssim
1,\end{split}$ ( 3.28)

uniformly in $\varepsilon,\lambda,x,\theta,\rho$. For
$C^{1}_{\varepsilon,2}(\lambda,x-\theta\rho)$, since

$\begin{split}\Big{|}e^{i(x-\theta\rho)\xi}\chi_{<4a}(|\xi|)\frac{\xi_{\ell}}{|\xi|}\frac{2\lambda^{2}-\varepsilon^{2}+2i\varepsilon\lambda}{(|\xi|+\lambda+i\epsilon)(|\xi|^{2}+\lambda^{2})}\Big{|}\lesssim\chi_{<4a}(|\xi|)\big{(}\frac{\lambda}{|\xi|^{2}}+\frac{1}{\lambda|\xi|^{2}}+\frac{1}{|\xi|^{2}}\big{)}\in
L_{\xi}^{1},\end{split}$

by Lebesgue’s dominated convergence theorem, one has

$\begin{split}\lim_{\epsilon\to
0^{+}}C^{1}_{\varepsilon,2}(\lambda,x-\theta\rho)=\int_{{\mathbb{R}}^{3}}e^{i(x-\theta\rho)\xi}\chi_{<4a}(|\xi|)\frac{\xi_{\ell}}{|\xi|}\frac{1}{(|\xi|^{2}+\lambda^{2})(|\xi|+\lambda)}d\xi,\end{split}$

which implies  3.27. Next, we will claim that

$\begin{split}\lim_{\epsilon\to
0^{+}}C_{\varepsilon,1}(\lambda,x-\theta\rho)=(2\pi)^{3}\lambda\chi_{<4a}(|D|)\mathcal{R}_{\ell}J_{\theta\rho}R^{+}_{0}(\lambda^{4})(x,0).\end{split}$
( 3.29)

Note that $C_{\varepsilon,1}(\lambda,x-\theta\rho)$ can be expressed as

$\begin{split}C_{\varepsilon,1}(\lambda,x-\theta\rho)&=\frac{\lambda}{2\lambda^{2}}\int_{{\mathbb{R}}^{3}}\mathcal{F}^{-1}\big{(}\chi_{<4a}(|\xi|)\frac{\xi_{\ell}}{|\xi|}\big{)}(y)\frac{e^{i(\lambda+i\varepsilon)|x-\theta\rho-y|}-e^{-\lambda|x-\theta\rho-y|}}{4\pi|x-\theta\rho-y|}dy,\end{split}$

and

$\begin{split}\Big{|}\mathcal{F}^{-1}\big{(}\chi_{<4a}(|\xi|)\frac{\xi_{\ell}}{|\xi|}\big{)}(y)\frac{e^{i(\lambda+i\varepsilon)|x-\theta\rho-y|}-e^{-\lambda|x-\theta\rho-y|}}{4\pi|x-\theta\rho-y|}\Big{|}\lesssim\frac{\langle
y\rangle^{-3}}{|x-\theta\rho-y|}\in L_{y}^{1}.\end{split}$

Therefore, by Lebesgue’s dominated convergence theorem, one has

$\begin{split}\lim_{\epsilon\to
0^{+}}C_{\varepsilon,1}(\lambda,x-\theta\rho)&=\frac{\lambda}{2\lambda^{2}}\int_{{\mathbb{R}}^{3}}\mathcal{F}^{-1}\big{(}\chi_{<4a}(|\xi|)\frac{\xi_{\ell}}{|\xi|}\big{)}(y)\frac{e^{i\lambda|x-\theta\rho-y|}-e^{-\lambda|x-\theta\rho-y|}}{4\pi|x-\theta\rho-y|}dy\\\
&=\frac{\lambda}{2\lambda^{2}}\Big{[}\mathcal{F}^{-1}\big{(}\chi_{<4a}(|\xi|)\frac{\xi_{\ell}}{|\xi|}\big{)}\ast\Big{(}R_{-\Delta}^{+}(\lambda^{2})-R_{-\Delta}(-\lambda^{2})\Big{)}\Big{]}(x-\theta\rho)\\\
&=(2\pi)^{3}\lambda\chi_{<4a}(|D|)\mathcal{R}_{\ell}J_{\theta\rho}R^{+}_{0}(\lambda^{4})(x,0),\end{split}$

Thus, we obtain 3.2 by Lebesgue’s dominated convergence theorem.

Now return to our purpose. Putting the 3.2 into  3.19, one get that

$\displaystyle\chi_{<4a}(|D|)T_{k}f(x)=-i\sum_{j=1}^{n_{1}}\sum_{\ell=1}^{3}c_{j}^{k}\times\Big{[}$
$\displaystyle\int_{0}^{1}\int_{{\mathbb{R}}^{3}}\big{(}(\rho_{\ell}vf_{j})(\rho)\chi_{<4a}(|D|)\mathcal{R}_{\ell}J_{\theta\rho}\big{)}\Big{(}\int_{0}^{\infty}\lambda^{3}\chi_{<a}(\lambda)R^{+}_{0}(\lambda^{4})(x,0)\int_{S^{2}}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega
d\lambda\Big{)}d\rho d\theta$
$\displaystyle+\frac{1}{(2\pi)^{3}}\int_{0}^{1}\int_{{\mathbb{R}}^{3}}(\rho_{\ell}vf_{j})(\rho)\mathcal{R}_{\ell}J_{\theta\rho}\Big{(}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\int_{{\mathbb{R}}^{3}}\frac{e^{ix\xi}\chi_{<4a}(|\xi|)}{(|\xi|^{2}+\lambda^{2})(|\xi|+\lambda)}d\xi\int_{S^{2}}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega
d\lambda\Big{)}d\rho d\theta\Big{]}$
$\displaystyle:=-i\sum_{j=1}^{n_{1}}\sum_{\ell=1}^{3}c_{j}^{k}\
\Big{(}\mathbf{S}_{k,1}^{j,\ell}f(x)+\frac{1}{(2\pi)^{3}}\mathbf{S}_{k,2}^{j,\ell}f(x)\Big{)}.$
( 3.30)

First deal with $\mathbf{S}_{k,1}^{j,\ell}f$ in 3.2 for each $1\leq j\leq
n_{1}$ and $1\leq\ell\leq 3$. Since

$\int_{S^{2}}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega=\frac{2}{\pi
i}\lambda\big{(}R^{+}_{0}(\lambda^{4})-R^{-}_{0}(\lambda^{4})\big{)}(\mathcal{R}_{k}f)(0),$

then one can write that

$\displaystyle\mathbf{S}_{k,1}^{j,\ell}f(x)=\frac{2}{\pi
i}\int_{0}^{1}\int_{{\mathbb{R}}^{3}}(\rho_{\ell}vf_{j})(\rho)\Big{(}\chi_{<4a}(|D|)\mathcal{R}_{\ell}J_{\theta\rho}A_{0,0}\mathcal{R}_{k}f\Big{)}(x)d\rho
d\theta,$ ( 3.31)

where $A_{0,0}$ is the integral operator defined by the following kernel:

$A_{0,0}(x,y)=\int_{0}^{\infty}\lambda^{4}\chi_{<a}(\lambda)R^{+}_{0}(\lambda^{4})(x,0)\big{(}R^{+}_{0}(\lambda^{4})-R^{-}_{0}(\lambda^{4})\big{)}(0,y)d\lambda.$

Now by using Minkowski’s inequality, the $L^{p}$-boundedness of
$\chi_{<4a}(|D|)$ and Riesz’ transforms $\mathcal{R}$, Lemma 2.1 and the
uniform boundedness of the translation operator $J_{\theta\rho}$ in $L^{p}$,
we can obtain that

$\|\mathbf{S}_{k,1}^{j,\ell}f\|_{L^{p}}\lesssim\Big{(}\int_{{\mathbb{R}}^{3}}|(\rho_{\ell}vf_{j})(\rho)|d\rho\Big{)}\|f\|_{L^{p}}\lesssim\|f\|_{L^{p}},$

which yields that $\mathbf{S}_{k,1}^{j,\ell}\in\mathbb{B}(L^{p})$ for all
$1<p<\infty$.

Next to $\mathbf{S}_{k,2}^{j,\ell}f$ in the 3.2 with $1\leq j\leq n_{1}$ and
$1\leq\ell\leq 3$.

Let $\eta=\lambda\omega$. Then

$\displaystyle\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\int_{{\mathbb{R}}^{3}}\frac{e^{ix\xi}\chi_{<4a}(|\xi|)}{(|\xi|^{2}+\lambda^{2})(|\xi|+\lambda)}$
$\displaystyle
d\xi\int_{S^{2}}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega d\lambda$
$\displaystyle=\int_{{\mathbb{R}}^{6}}\frac{e^{ix\xi}\chi_{<4a}(|\xi|)\chi_{<a}(|\eta|)}{(|\xi|^{2}+|\eta|^{2})(|\xi|+|\eta|)}\widehat{\mathcal{R}_{k}f}(\eta)d\eta
d\xi.$ ( 3.32)

Hence we can obtain that

$\displaystyle\mathbf{S}_{k,2}^{j,\ell}f(x)$
$\displaystyle=\int_{0}^{1}\int_{{\mathbb{R}}^{3}}(\rho_{\ell}vf_{j})(\rho)J_{\theta\rho}\mathcal{R}_{\ell}\Big{(}\int_{{\mathbb{R}}^{6}}\frac{e^{ix\xi}\chi_{<4a}(|\xi|)\chi_{<a}(|\eta|)}{(|\xi|^{2}+|\eta|^{2})(|\xi|+|\eta|)}\widehat{\mathcal{R}_{k}f}(\eta)d\eta
d\xi\Big{)}d\rho d\theta$
$\displaystyle=\int_{0}^{1}\int_{{\mathbb{R}}^{3}}(\rho_{\ell}vf_{j})(\rho)\big{[}J_{\theta\rho}\mathcal{R}_{\ell}K\mathcal{R}_{k}f](x)d\rho
d\theta,$ ( 3.33)

where $K$ is the integral operator defined by the following kernel (also see
 2.9):

$\displaystyle K(x,y)=\frac{1}{(2\pi)^{3}}\int_{{\mathbb{R}}^{6}}e^{ix\xi-
iy\eta}\
\frac{\chi_{<4a}(|\xi|)\chi_{<a}(|\eta|)}{(|\xi|^{2}+|\eta|^{2})(|\xi|+|\eta|)}d\xi
d\eta.$ ( 3.34)

Now by Minkowski’s inequality, the $L^{p}$-boundedness of Riesz’ transforms
$\mathcal{R}$, Lemma 2.2 and the uniform boundedness of translation operators
$J_{\theta\rho}$, it follows that

$\|\mathbf{S}_{k,2}^{j,\ell}f\|_{L^{p}}\lesssim\Big{(}\int_{{\mathbb{R}}^{3}}|(\rho_{\ell}vf_{j})(\rho)|d\rho\Big{)}\|f\|_{L^{p}}\lesssim\|f\|_{L^{p}},$

which leads to $\mathbf{S}_{k,2}^{j,\ell}\in\mathbb{B}(L^{p})$ for all
$1<p<\infty$. Thus, from 3.2 we finally conclude that
$\chi_{<4a}(|D|)T_{k}\in\mathbb{B}(L^{p})$ for all $1<p<\infty$.

Summing up all arguments in the above (A) and (B), we obtain
$T_{k}\in\mathbb{B}(L^{p})$ for $1<p<\infty$. This, together with the above
item (i), completes the whole proof of Proposition 3.1. ∎

In the following, we will consider the left two terms $S_{1}A_{0,1}^{1}$ and
$A_{0,2}^{1}S_{1}$ appearing in the asymptotic expansion  3.5 of
$M^{-1}(\lambda)$ as zero is the first kind resonance.

As in the case of Proposition 3.1, we need to introduce the following two
operators corresponding to $S_{1}A_{0,1}^{1}$ and $A_{0,2}^{1}S_{1}$,
respectively:

$\displaystyle K^{1}_{0,1}f=\frac{2}{\pi
i}\int_{0}^{\infty}\lambda^{3}\chi_{<a}(\lambda)\
R_{0}^{+}(\lambda^{4})vS_{1}A_{0,1}^{1}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}fd\lambda.$
( 3.35) $\displaystyle K^{1}_{0,2}f=\frac{2}{\pi
i}\int_{0}^{\infty}\lambda^{3}\chi_{<a}(\lambda)\
R_{0}^{+}(\lambda^{4})vA_{0,2}^{1}S_{1}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}fd\lambda.$
( 3.36)

######  Proposition 3.2.

One has $K^{1}_{0,1},\ K^{1}_{0,2}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for
all $1<p<\infty$.

###### Proof.

The proof is decomposed into two parts.
(i) The proof of $K^{1}_{0,1}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all
$1<p<\infty.$

Recall the following formula  2.4:

$(R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4}))f(z)=\frac{\pi
i}{2\lambda}\int_{S^{2}}e^{i\lambda\omega z}\hat{f}(\lambda\omega)d\omega,$

and Taylor’s formula:

$e^{i\lambda\omega
z}=1+i\lambda\sum_{k=1}^{3}z_{k}\omega_{k}\int_{0}^{1}e^{i\lambda\omega\theta
z}d\theta.$

By  3.35 we can obtain that

$\displaystyle K^{1}_{0,1}f(x)$
$\displaystyle=\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\big{[}R_{0}^{+}(\lambda^{4})vS_{1}A_{0,1}^{1}v\big{]}(x)\Big{(}\int_{S^{2}}\hat{f}(\lambda\omega)d\omega\Big{)}d\lambda$
$\displaystyle\ \ \ \ \ \ \
+i\sum_{k=1}^{3}\int_{0}^{1}\int_{0}^{\infty}\lambda^{3}\chi_{<a}(\lambda)\Big{[}R_{0}^{+}(\lambda^{4})vS_{1}A_{0,1}^{1}\Big{(}vz_{k}\int_{S^{2}}e^{i\lambda\omega\theta
z}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega\Big{)}\Big{]}(x)d\lambda
d\theta$ $\displaystyle=:Tf(x)+i\sum_{k=1}^{3}T_{k}f(x).$ ( 3.37)

First deal with $T_{k}f$ in 3.2 for each $k=1,2,3$, which is comparably
easy. In fact, by similar processes in 3.2, it can be written as follows:

$\displaystyle T_{k}f(x)=\frac{2}{\pi
i}\int_{0}^{1}\int_{{\mathbb{R}}^{6}}v(\rho)(S_{1}A^{1}_{0,1})(\rho,z)\
z_{k}v(z)\Big{(}A_{\rho,z\theta}\mathcal{R}_{k}f\Big{)}(x)dzd\rho d\theta.$ (
3.38)

Hence as the same as done in 3.2, by Minkowski’s inequality, the
$L^{p}$-boundedness of Riesz’ transform $\mathcal{R}_{k}$ and Lemma 2.1, we
can immediately conclude that $T_{k}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$
for all $1<p<\infty$ and $1\leq k\leq 3$.

Next to consider $Tf(x)$ in 3.2. As before, we decompose $Tf(x)$ into the
sum of $\chi_{\geq 4a}(|D|)Tf$ and $\chi_{<4a}(|D|)Tf$, and set
$\\{f_{j}\\}_{j=1}^{n_{1}}$ is a basis of $S_{1}L^{2}({\mathbb{R}}^{3})$. As
for $\chi_{\geq 4a}(|D|)T$, by similar arguments in Proposition 3.1, we have

$\displaystyle\chi_{\geq 4a}(|D|)Tf(x)$ $\displaystyle=\frac{2}{\pi
i}\sum_{j=1}^{n_{1}}c_{j}\int_{{\mathbb{R}}^{3}}v(\rho)f_{j}(\rho)\Big{(}M(D)A_{\rho,0}|D|^{3}\chi_{<4a}(|D|)f\Big{)}(x)d\rho$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\ \sum_{j=1}^{n_{1}}c_{j}\
\big{(}M(D)vf_{j}\big{)}(x)\Big{(}\int_{{\mathbb{R}}^{3}}\widehat{\chi_{<a}(|\cdot|)}(y)f(y)dy\Big{)}.$
( 3.39)

where $M(\xi)=|\xi|^{-4}\chi_{\geq 4a}(|\xi|)$,
$c_{j}={\langle}S_{1}A_{0,1}^{1}v,f_{j}{\rangle}$ for $j=1,2,\cdots,n_{1}$.
Hence it is easy to show that $\chi_{\geq 4a}(|D|)T\in\mathbb{B}(L^{p})$ for
all $1<p<\infty$ as done for $\chi_{\geq 4a}(|D|)T_{k}$ in Proposition 3.1.

As for $\chi_{<4a}(|D|)Tf$, by using similar arguments in Proposition 3.1
again, i.e., using the Fourier transform, $S_{1}v=0$ and Taylor’s expansion of
$e^{-i\xi\rho}$, we find that

$\displaystyle\chi_{<4a}(|D|)Tf(x)=-i\sum_{\ell=1}^{3}\sum_{j=1}^{n_{1}}$
$\displaystyle c_{j}\Big{[}\frac{2}{\pi
i}\int_{0}^{1}\int_{{\mathbb{R}}^{3}}(\rho_{\ell}vf_{j})(\rho)\Big{(}\chi_{<4a}(|D|)\mathcal{R}_{\ell}J_{\theta\rho}A_{0,0}f\Big{)}(x)d\rho
d\theta$ $\displaystyle\ \ \ \ \ \ \ \ \ \ \
+\frac{1}{(2\pi)^{3}}\int_{0}^{1}\int_{{\mathbb{R}}^{3}}(\rho_{\ell}vf_{j})(\rho)\Big{(}J_{\theta\rho}\mathcal{R}_{\ell}Kf\Big{)}(x)d\rho
d\theta\Big{]},$ ( 3.40)

where $K$ is the integral operator given by the kernel  3.34 and
$J_{\rho\theta}$ is the translation. Thus it follows by the same argument as
in Proposition 3.1 that $\chi_{<4a}(|D|)T\in\mathbb{B}(L^{p})$ for all
$1<p<\infty$.
(ii) The proof of $K^{1}_{0,2}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all
$1<p<\infty.$

Putting the formula  2.4 and the following equality:

$S_{1}(ve^{i\lambda\omega
z})=i\lambda\sum_{k=1}^{3}\omega_{k}S_{1}\Big{(}vz_{k}\int_{0}^{1}e^{i\lambda\omega\theta
z}d\theta\Big{)},$

into the expression  3.36 of $K^{1}_{0,2}f$, we can rewrite
$K^{1}_{0,2}f(x)$ as follows:

$\displaystyle K^{1}_{0,2}f(x)$
$\displaystyle=i\sum_{k=1}^{3}\int_{0}^{1}\int_{0}^{\infty}\lambda^{3}\chi_{<a}(\lambda)\Big{[}R_{0}^{+}(\lambda^{4})vA_{0,2}^{1}S_{1}\Big{(}vz_{k}\int_{S^{2}}e^{i\lambda\omega\theta
z}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega\Big{)}\Big{]}(x)d\lambda
d\theta$
$\displaystyle=\sum_{k=1}^{3}\frac{2}{\pi}\int_{0}^{1}\int_{{\mathbb{R}}^{6}}v(\rho)(A^{1}_{0,2}S_{1})(\rho,z)z_{k}v(z)\Big{(}A_{\rho,z\theta}\mathcal{R}_{k}f\Big{)}(x)dzd\rho
d\theta.$ ( 3.41)

Hence by Minkowski’s inequality, Lemma 2.1 and the $L^{p}$-boundedness of
Riesz transform $\mathcal{R}_{k}$, one can immediately obtain that
$K^{1}_{0,2}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<\infty$ as
done in 3.2.

Finally, summing up the arguments in (i) and (ii) above, we have finished the
whole proof of Proposition 3.2. ∎

As mentioned in the beginning of this subsection, all of the terms in  3.5
except for the three terms treated in Propositions 3.1\- 3.2 can be dealt by
the same argument as in the regular case. Hence, we have finished the proof of
Theorem 1.4 for the case (i).

##  4\. The $L^{p}$-boundedness of wave operators for the second and third
types

In this section, we will finalize the proof of Theorem 1.4 for cases (ii) and
(iii). The section is divided into two subsections to address the second kind
of resonance and the third kind of resonance, respectively.

###  4.1. The second kind of resonance case

In this subsection, we firstly claim the wave operators $W_{-}$ are bounded on
$L^{p}$ for all $1<p<3$, and secondly demonstrate that the wave operators
$W_{-}$ become unbounded on $L^{p}$ for any $3\leq p<\infty$.

To accomplish this, we begin with the expansion of $M^{-1}(\lambda)$ for the
second kind of resonance, as presented in Theorem 2.6:

$\displaystyle M^{-1}(\lambda)=$
$\displaystyle\lambda^{-3}S_{2}A^{2}_{-3,1}S_{2}+\lambda^{-2}\Big{(}S_{2}A^{2}_{-2,1}S_{1}+S_{1}A^{2}_{-2,2}S_{2}\Big{)}+\lambda^{-1}\Big{(}S_{2}A^{2}_{-1,1}+A^{2}_{-1,2}S_{2}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \
+S_{1}A^{2}_{-1,3}S_{1}\Big{)}+\Big{(}S_{1}A_{0,1}^{2}+A^{2}_{0,2}S_{1}+QA^{2}_{0,3}Q\Big{)}+\lambda
A^{2}_{1,1}+\Gamma^{2}_{2}(\lambda).$ ( 4.1)

Taking 4.1 to  3.3 and comparing with the first kind of resonance case, we
can focus our attention on the remaining five terms:
$\lambda^{-3}S_{2}A^{2}_{-3,1}S_{2}$, $\lambda^{-2}S_{2}A^{2}_{-2,1}S_{1}$,
$\lambda^{-1}S_{2}A^{2}_{-1,1}$, $\lambda^{-2}S_{1}A^{2}_{-2,2}S_{2}$ and
$\lambda^{-1}A^{2}_{-1,2}S_{2}$.

Now we start to show that $W_{-}\in\mathbb{B}(L^{p})$ for all $1<p<3$, which
is followed from in the following three propositions (i.e. Proposition
4.1-4.3).

######  Proposition 4.1.

Let

$\displaystyle K^{2}_{-3,1}:=\frac{2}{\pi
i}\int_{0}^{\infty}\chi_{<a}(\lambda)\
R_{0}^{+}(\lambda^{4})vS_{2}A_{-3,1}^{2}S_{2}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}d\lambda.$
( 4.2)

Then $K^{2}_{-3,1}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<3$.

###### Proof.

Let $f\in\mathbb{S}({\mathbb{R}}^{3})$. Then

$\displaystyle(K^{2}_{-3,1}f)(x)=\frac{2}{\pi
i}\int_{0}^{\infty}\chi_{<a}(\lambda)\Big{[}\
R_{0}^{+}(\lambda^{4})vS_{2}A_{-3,1}^{2}S_{2}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}f\Big{]}(x)d\lambda.$
( 4.3)

Note that the Taylor expansion of $e^{i\lambda\omega z}$ can be written as
follows:

$\displaystyle e^{i\lambda\omega z}=$ $\displaystyle 1+i\sum_{k=1}^{3}\lambda
z_{k}\omega_{k}-\frac{1}{2}\sum_{k,l=1}^{3}\lambda^{2}z_{k}z_{l}\omega_{k}\omega_{l}-\frac{i}{6}\sum_{k,l,s=1}^{3}\lambda^{3}z_{kls}\omega_{kls}$
$\displaystyle\ \ \ \ \ \ \ \
+\frac{1}{6}\sum_{k,l,s,t=1}^{3}\lambda^{4}z_{klst}\omega_{klst}\int_{0}^{1}(1-\theta)^{3}e^{i\lambda\omega\theta
z}d\theta,$ ( 4.4)

where $z_{kls}=z_{k}z_{l}z_{s}$, $z_{klst}=z_{k}z_{l}z_{s}z_{t}$,
$\omega_{kls}=\omega_{k}\omega_{l}\omega_{s}$, and
$\omega_{klst}=\omega_{k}\omega_{l}\omega_{s}\omega_{t}$.

Furthermore, combining with 4.1 and $S_{2}(v)=S_{2}(z_{j}v)=0$ for
$j=1,2,3$, we have

$\displaystyle\frac{2}{\pi
i}S_{2}\Big{(}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}f\Big{)}=\frac{1}{\lambda}S_{2}\Big{(}v\int_{S^{2}}\big{(}e^{i\lambda\omega
z}-1-i\sum_{k=1}^{3}\lambda
z_{k}\omega_{k}\big{)}\hat{f}(\lambda\omega)d\omega\Big{)}$
$\displaystyle=-\frac{1}{2}\sum_{k,l=1}^{3}\lambda
S_{2}(vz_{k}z_{l})\int_{S^{2}}\omega_{k}\omega_{l}\hat{f}(\lambda\omega)d\omega-\frac{i}{6}\sum_{k,l,s=1}^{3}\lambda^{2}S_{2}(vz_{kls})\int_{S^{2}}\omega_{kls}\hat{f}(\lambda\omega)d\omega$
$\displaystyle\quad\quad+\frac{1}{6}\sum_{k,l,s,t=1}^{3}\lambda^{3}\int_{0}^{1}(1-\theta)^{3}S_{2}\big{(}vz_{klst}\int_{S^{2}}\omega_{klst}e^{i\lambda\omega\theta
z}\hat{f}(\lambda\omega)d\omega\big{)}d\theta.$

Then, we can obtain that

$\displaystyle\frac{2}{\pi
i}S_{2}\Big{(}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}f\Big{)}=-\frac{1}{2}\sum_{k,l=1}^{3}\lambda
S_{2}(vz_{k}z_{l})\int_{S^{2}}\widehat{\mathcal{R}_{k}\mathcal{R}_{l}f}(\lambda\omega)d\omega$
$\displaystyle-\frac{i}{6}\sum_{k,l,s=1}^{3}\lambda^{2}S_{2}(vz_{kls})\int_{S^{2}}\widehat{\mathcal{R}_{kls}f}(\lambda\omega)d\omega+\frac{1}{6}\sum_{k,l,s,t=1}^{3}\lambda^{3}\int_{0}^{1}(1-\theta)^{3}S_{2}\big{(}vz_{klst}\int_{S^{2}}e^{i\lambda\omega\theta
z}\widehat{\mathcal{R}_{klst}f}(\lambda\omega)d\omega\big{)}d\theta,$ ( 4.5)

where $\mathcal{R}_{kls}=\mathcal{R}_{k}\mathcal{R}_{l}\mathcal{R}_{s}$ and
$\mathcal{R}_{klst}=\mathcal{R}_{k}\mathcal{R}_{l}\mathcal{R}_{s}\mathcal{R}_{t}$.

Plugging 4.1 into  4.3, we obtain that

$\displaystyle K^{2}_{-3,1}f$
$\displaystyle=-\frac{1}{2}\sum_{k,l=1}^{3}\int_{0}^{\infty}\lambda\chi_{<a}(\lambda)\big{[}\
R_{0}^{+}(\lambda^{4})vS_{2}A_{-3,1}^{2}S_{2}(vz_{k}z_{l})\big{]}(x)\int_{S^{2}}\widehat{\mathcal{R}_{k}\mathcal{R}_{l}f}(\lambda\omega)d\omega
d\lambda$
$\displaystyle-\frac{i}{6}\sum_{k,l,s=1}^{3}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\big{[}\
R_{0}^{+}(\lambda^{4})vS_{2}A_{-3,1}^{2}S_{2}(vz_{kls})\big{]}(x)\int_{S^{2}}\widehat{\mathcal{R}_{kls}f}(\lambda\omega)d\omega
d\lambda$
$\displaystyle+\frac{1}{6}\sum_{k,l,s,t=1}^{3}\int_{0}^{1}(1-\theta)^{3}\int_{0}^{\infty}\lambda^{3}\chi_{<a}\big{[}\
R_{0}^{+}(\lambda^{4})vS_{2}A_{-3,1}^{2}S_{2}\big{(}vz_{klst}\int_{S^{2}}e^{i\lambda\omega\theta
z}\widehat{\mathcal{R}_{klst}f}(\lambda\omega)d\omega\big{)}\big{]}(x)d\lambda
d\theta$
$\displaystyle=:-\frac{1}{2}\sum_{k,l=1}^{3}T_{kl}f-\frac{i}{6}\sum_{k,l,s=1}^{3}T_{kls}f+\frac{1}{6}\sum_{k,l,s,t=1}^{3}T_{klst}f.$
( 4.6)

As above, we deal with $T_{kl},T_{kls}$ and $T_{klst}$ separately.
(i) The proof of $T_{klst}\in\mathbb{B}(L^{p})$ for all $1<p<\infty.$

In fact, by similar processes as in 3.2, $T_{klst}$ can be written as
follows:

$\displaystyle T_{klst}f(x)=\frac{2}{\pi
i}\int_{0}^{1}(1-\theta)^{3}\int_{{\mathbb{R}}^{6}}v(\rho)(S_{2}A_{-3,1}^{2}S_{2})(\rho,z)v(z)z_{klst}\Big{(}A_{\rho,z\theta}\mathcal{R}_{klst}f\Big{)}(x)dzd\rho
d\theta.$ ( 4.7)

Hence as the same as done in 3.2, by Minkowski’s inequality, the
$L^{p}$-boundedness of Riesz transform $\mathcal{R}_{k}$ and Lemma 2.1, we can
immediately conclude that $T_{klst}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for
all $1<p<\infty$ and $1\leq k,l,s,t\leq 3$.
(ii) The proof of $T_{kls}\in\mathbb{B}(L^{p})$ for all $1<p<\infty$ and
$T_{kl}\in\mathbb{B}(L^{p})$ for all $1<p<3$.

Noting that $S_{2}L^{2}\subset S_{1}L^{2}$, we have $\dim S_{2}L^{2}=n_{2}\leq
n_{1}$. We may assume that $\\{f_{j}\\}_{j=1}^{n_{2}}$ is a standard
orthogonal basis of $S_{2}L^{2}$. Then, one has

$\begin{split}&S_{2}A_{-3,1}^{2}S_{2}(vz_{kls})=\sum_{j=1}^{n_{2}}\langle
S_{2}A_{-3,1}^{2}S_{2}(vz_{kls}),f_{j}\rangle
f_{j}:=\sum_{j=1}^{n_{2}}c_{j}^{kls}f_{j},\\\
&S_{2}A_{-3,1}^{2}S_{2}(vz_{k}z_{l})=\sum_{j=1}^{n_{2}}\langle
S_{2}A_{-3,1}^{2}S_{2}(vz_{k}z_{l}),f_{j}\rangle
f_{j}:=\sum_{j=1}^{n_{2}}c_{j}^{kl}f_{j}.\end{split}$

Furthermore, $T_{kls}$, $T_{klst}$ can be written as:

$\displaystyle
T_{kls}f(x)=\sum_{j=1}^{n_{2}}c_{j}^{kls}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{kls}f}(\lambda\omega)d\omega\Big{)}d\lambda,$
( 4.8) $\displaystyle
T_{kl}f(x)=\sum_{j=1}^{n_{2}}c_{j}^{kl}\int_{0}^{\infty}\lambda\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}\mathcal{R}_{l}f}(\lambda\omega)d\omega\Big{)}d\lambda.$
( 4.9)

At first, we focus on $T_{kls}f(x)$ in  4.8.

We can decompose $T_{kls}f(x)$ into the sum of $\chi_{\geq
4a}(|D|)T_{kls}f(x)$ and $\chi_{<4a}(|D|)T_{kls}f(x)$. As for $\chi_{\geq
4a}(|D|)T_{kls}f(x)$, by similar arguments in Proposition 3.1, we have

$\displaystyle\chi_{\geq 4a}(|D|)T_{kls}f(x)$ $\displaystyle=\frac{2}{\pi
i}\sum_{j=1}^{n_{2}}c_{j}^{kls}\int_{{\mathbb{R}}^{3}}v(\rho)f_{j}(\rho)\Big{(}M(D)A_{\rho,0}|D|^{3}\chi_{<4a}(|D|)\mathcal{R}_{kls}f\Big{)}(x)d\rho$
$\displaystyle\ \ \ +\ \sum_{j=1}^{n_{2}}c_{j}^{kls}\
\big{(}M(D)vf_{j}\big{)}(x)\Big{(}\int_{{\mathbb{R}}^{3}}\mathcal{F}\big{(}\chi_{<a}(|\cdot|)\big{)}(y)(\mathcal{R}_{kls}f)(y)dy\Big{)}.$

where $M(\xi)=|\xi|^{-4}\chi_{\geq 4a}(|\xi|)$. Hence it is easy to check that
$\chi_{\geq 4a}(|D|)T_{kls}\in\mathbb{B}(L^{p})$ for all $1<p<\infty$ as done
for $\chi_{\geq 4a}(|D|)T_{k}$ in Proposition 3.1.

As for $\chi_{<4a}(|D|)T_{kls}f(x)$. Since $S_{2}(v)=S_{2}(\rho_{j}v)=0$ for
$j=1,2,3$, then we obtain that

$\int_{{\mathbb{R}}^{3}}v(\rho)f_{j}(\rho)d\rho=\int_{{\mathbb{R}}^{3}}\rho_{j}v(\rho)f_{j}(\rho)d\rho=0$

for $j=1,2,3$. Note that the Taylor expansion of $e^{-i\xi\rho}$ can be
written as:

$\begin{split}e^{-i\xi\rho}=1-i\sum_{t=1}^{3}\rho_{t}\xi_{t}-\sum_{t,m=1}^{3}\int_{0}^{1}(1-\theta)e^{-i\xi\theta\rho}d\theta.\end{split}$

Therefore, one has

$\displaystyle\widehat{vf_{j}}(\xi)$
$\displaystyle=\frac{1}{(2\pi)^{3}}\int_{{\mathbb{R}}^{3}}\big{(}e^{-i\rho\xi}-1+i\sum_{t=1}^{3}\rho_{t}\xi_{t}\big{)}(vf_{j})(\rho)d\rho$
$\displaystyle=\frac{-1}{(2\pi)^{3}}\sum_{t,m=1}^{3}\int_{0}^{1}(1-\theta)\int_{R^{3}}\xi_{t}\xi_{m}e^{-i\xi\theta\rho}(\rho_{t}\rho_{m}vf_{j})(\rho)d\rho
d\theta.$ ( 4.10)

Putting 4.1 into  3.20 and by the dominated convergence theorem, we obtain
that

$\displaystyle\chi_{<4a}(|D|)R_{0}^{+}$
$\displaystyle(\lambda^{4})vf_{j}(x)=\frac{-1}{(2\pi)^{3}}\sum_{t,m=1}^{3}\lim_{\epsilon\to
0^{+}}\int_{0}^{1}(1-\theta)\int_{{\mathbb{R}}^{3}}(\rho_{t}\rho_{m}vf_{j})(\rho)\Big{[}\frac{1}{2\lambda^{2}}\int_{{\mathbb{R}}^{3}}e^{i(x-\theta\rho)\xi}$
$\displaystyle\ \ \
\times\chi_{<4a}(|\xi|)\xi_{t}\xi_{m}\Big{(}\frac{1}{|\xi|^{2}-(\lambda+i\epsilon)^{2}}-\frac{1}{|\xi|^{2}+\lambda^{2}}\Big{)}d\xi\Big{]}d\rho
d\theta$
$\displaystyle=-\sum_{t,m=1}^{3}\int_{0}^{1}(1-\theta)\int_{{\mathbb{R}}^{3}}(\rho_{t}\rho_{m}vf_{j})(\rho)\mathcal{R}_{t}\mathcal{R}_{m}J_{\theta\rho}$
$\displaystyle\ \ \
\times\Big{(}\lambda^{2}\chi_{<4a}(|D|)R_{0}^{+}(\lambda^{4})(x,0)+\frac{1}{(2\pi)^{3}}\int_{{\mathbb{R}}^{3}}e^{i\xi
x}\frac{\chi_{<4a}(|\xi|)}{|\xi|^{2}+\lambda^{2}}d\xi\Big{)}d\rho d\theta.$ (
4.11)

By the similar argument in Proposition 3.1, combining with 4.1 and  4.8,
$\chi_{<4a}(|D|)T_{kls}f(x)$ can be written as:

$\begin{split}\chi_{<4a}(|D|)&T_{kls}f(x)=-\sum_{t,m=1}^{3}\sum_{j=1}^{n_{2}}c_{j}^{kls}\int_{0}^{1}(1-\theta)\int_{{\mathbb{R}}^{3}}(\rho_{t}\rho_{m}vf_{j})(\rho)\times\\\
&\Big{[}\frac{2}{\pi
i}\big{(}\mathcal{R}_{t}\mathcal{R}_{m}J_{\theta\rho}A_{0,0}|D|\chi_{<4a}(|D|)\mathcal{R}_{kls}f\big{)}(x)+\frac{1}{(2\pi)^{3}}\big{(}\mathcal{R}_{t}\mathcal{R}_{m}J_{\theta\rho}L_{1}\mathcal{R}_{kls}f\big{)}(x)\Big{]}d\rho
d\theta,\end{split}$

where

$L_{1}(x,y)=\frac{1}{(2\pi)^{3}}\int_{\mathbb{R}^{6}}e^{i\xi x-i\eta
y}\frac{\chi_{<4a}(|\xi|)\chi_{<a}(|\eta|)}{|\xi|^{2}+|\eta|^{2}}d\xi d\eta,$

which is dominated by $\langle|x|+|y|\rangle^{-4}$, seen Proposition 5.2.

Note that $A_{0,0},L_{1}\in\mathbb{B}(L^{p})$ for all $1<p<\infty$, by the
same arguments as done for $\chi_{<4a}(|D|)T_{k}$ in Proposition 3.1, we
obtain that $\chi_{<4a}(|D|)T_{kls}\in\mathbb{B}(L^{p})$ for all $1<p<\infty$.

Next, we deal with $T_{kl}f(x)$ in  4.9.

We similarly decompose $T_{kl}f(x)$ into the sum of $\chi_{\geq
4a}(|D|)T_{kl}f(x)$ and $\chi_{<4a}(|D|)T_{kl}f(x)$.

As for $\chi_{\geq 4a}(|D|)T_{kls}f(x)$, by the analogous arguments in
Proposition 3.1, but noting that it is $\lambda$ rather than $\lambda^{2}$
inside the  4.9, then $\chi_{\geq 4a}(|D|)T_{kls}f(x)$ can be written as:

$\displaystyle\chi_{\geq 4a}(|D|)T_{kl}f(x)$ $\displaystyle=\frac{2}{\pi
i}\sum_{j=1}^{n_{2}}c_{j}^{kl}\int_{{\mathbb{R}}^{3}}v(\rho)f_{j}(\rho)\Big{(}M(D)A_{\rho,0}|D|^{2}\chi_{<4a}(|D|)\mathcal{R}_{k}\mathcal{R}_{l}f\Big{)}(x)d\rho$
$\displaystyle\ \ \ +\ \sum_{j=1}^{n_{2}}c_{j}^{kl}\
\big{(}M(D)vf_{j}\big{)}(x)\Big{(}\int_{{\mathbb{R}}^{3}}\mathcal{F}\big{(}\frac{\chi_{<a}(|\cdot|)}{|\cdot|}\big{)}(y)(\mathcal{R}_{k}\mathcal{R}_{l}f)(y)dy\Big{)}.$

where $M(\xi)=|\xi|^{-4}\chi_{\geq 4a}(|\xi|)$. Then, it is easy to obtain
that the first part is bounded on $L^{p}$ for all $1<p<\infty$. As for the
second part, note that by Proposition 5.2,

$\begin{split}\big{|}\mathcal{F}\Big{(}\frac{\chi_{<a}(|\cdot|)}{|\cdot|}\Big{)}(y)\big{|}\lesssim\frac{1}{\langle
y\rangle^{2}}\in L^{q},\ \ \frac{3}{2}<q<\infty.\end{split}$

Furthermore, by Minkowski’s inequality and Hölder’s inequality, we have

$\begin{split}\Big{\|}\sum_{j=1}^{n_{2}}c_{j}^{kl}&\
\big{(}M(D)vf_{j}\big{)}(x)\Big{(}\int_{{\mathbb{R}}^{3}}\mathcal{F}\big{(}\frac{\chi_{<a}(|\cdot|)}{|\cdot|}\big{)}(y)(\mathcal{R}_{k}\mathcal{R}_{l}f)(y)dy\Big{)}\Big{\|}_{L^{p}}\\\
&\lesssim\sum_{j=1}^{n_{2}}|c_{j}^{kl}|\|\mathcal{F}^{-1}(M)\|_{L^{p}}\|vf_{j}\|_{L^{1}}\int_{{\mathbb{R}}^{3}}\
|\mathcal{F}\big{(}\frac{\chi_{<a}(|\cdot|)}{|\cdot|}\big{)}(y)\big{|}\
|(\mathcal{R}_{k}\mathcal{R}_{l}f)(y)\big{|}dy\\\
&\lesssim\sum_{j=1}^{n_{2}}\big{\|}\mathcal{F}\big{(}\frac{\chi_{<a}(|\cdot|)}{|\cdot|}\big{)}(y)\big{\|}_{L_{y}^{q}}\big{\|}\big{(}\mathcal{R}_{k}\mathcal{R}_{l}f\big{)}(y)\big{\|}_{L_{y}^{p}}\lesssim\|f\|_{L^{p}},\end{split}$

where $1<p<3$ and $\frac{1}{p}+\frac{1}{q}=1$.

Hence $\chi_{\geq 4a}(|D|)T_{kl}\in\mathbb{B}(L^{p})$ for all $1<p<3$. As for
$\chi_{<4a}(|D|)T_{kl}f(x)$, similarly for $\chi_{<4a}(|D|)T_{kls}f(x)$ but
noting that it is $\lambda$ rather than $\lambda^{2}$ inside the  4.9, we
can check that $\chi_{<4a}(|D|)T_{kl}f(x)$ can be written as:

$\begin{split}\chi_{<4a}(|D|)T_{kl}f(x)&=-\sum_{t,m=1}^{3}\sum_{j=1}^{n_{2}}c_{j}^{kl}\int_{0}^{1}(1-\theta)\int_{{\mathbb{R}}^{3}}(\rho_{t}\rho_{m}vf_{j})(\rho)\times\\\
&\Big{[}\frac{2}{\pi
i}\big{(}\mathcal{R}_{t}\mathcal{R}_{m}J_{\theta\rho}A_{0,0}\mathcal{R}_{k}\mathcal{R}_{l}f\big{)}(x)+\frac{1}{(2\pi)^{3}}\big{(}\mathcal{R}_{t}\mathcal{R}_{m}J_{\theta\rho}E\mathcal{R}_{k}\mathcal{R}_{l}f\big{)}(x)\Big{]}d\rho
d\theta,\end{split}$

where

$E(x,y)=\frac{1}{(2\pi)^{3}}\int_{{\mathbb{R}}^{6}}e^{ix\xi-iy\eta}\
\frac{\chi_{<4a}(|\xi|)\chi_{<a}(|\eta|)}{(|\xi|^{2}+|\eta|^{2})\ |\eta|}d\xi
d\eta.$

By Lemma 2.2, we have $E\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all
$1<p<3$. Hence it is easy to obtain that
$\chi_{<4a}(|D|)T_{kl}f\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all
$1<p<3$.

By the above items (i) and (ii), we obtain $K^{2}_{-3,1}\in\mathbb{B}(L^{p})$
for all $1<p<3$. ∎

In the proofs of the following propositions, for simplicity, we may omit lots
of similar and repetitive processes mentioned above . Now set

$\displaystyle z$ $\displaystyle=(z_{1},z_{2},z_{3})\in{\mathbb{R}}^{3},\quad
z_{kls}=z_{k}z_{l}z_{s},\quad z_{klst}=z_{k}z_{l}z_{s}z_{t},$ ( 4.12)
$\displaystyle\omega$ $\displaystyle=(\omega_{1},\omega_{2},\omega_{3})\in
S^{2},\quad\omega_{kls}=\omega_{k}\omega_{l}\omega_{s},\quad\omega_{klst}=\omega_{k}\omega_{l}\omega_{s}\omega_{t}.$

Since $S_{2}L^{2}\subset S_{1}L^{2}$, we may let $\dim S_{2}L^{2}:=n_{2}\leq
n_{1}\,(=\dim S_{1}L^{2})$. In addition, since
$S_{1}L^{2}=\overline{S_{1}L^{2}}$ and $S_{2}L^{2}=\overline{S_{2}L^{2}}$, we
can assume that $\\{f_{j}\\}_{j=1}^{n_{1}}$, $\\{f_{j}\\}_{j=1}^{n_{2}}$ is
real-valued orthogonal bases of $S_{1}L^{2}$, $S_{2}L^{2}$, respectively. We
will always use these notations below.

Next, we will focus on the following two operators $K^{2}_{-2,1}$,
$K^{2}_{-1,1}$ corresponding to $\lambda^{-2}S_{2}A^{2}_{-2,1}S_{1}$ and
$\lambda^{-1}S_{2}A^{2}_{-1,1}$, respectively:

$\displaystyle(K^{2}_{-2,1}f)(x)$ $\displaystyle=\frac{2}{\pi
i}\int_{0}^{\infty}\lambda\chi_{<a}(\lambda)\Big{[}\
R_{0}^{+}(\lambda^{4})vS_{2}A^{2}_{-2,1}S_{1}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}f\Big{]}(x)d\lambda,$
( 4.13) $\displaystyle(K^{2}_{-1,1}f)(x)$ $\displaystyle=\frac{2}{\pi
i}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\Big{[}\
R_{0}^{+}(\lambda^{4})vS_{2}A^{2}_{-1,1}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}f\Big{]}(x)d\lambda.$
( 4.14)

######  Proposition 4.2.

Let $K^{2}_{-2,1}$ and $K^{2}_{-1,1}$ be the operators defined as above. Then
one has $K^{2}_{-2,1},\ K^{2}_{-1,1}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$
for all $1<p<3$.

###### Proof.

(i) The proof of $K^{2}_{-2,1}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all
$1<p<3.$

Note that Taylor’s expansion of $e^{i\lambda\omega z}$ is of the form

$\displaystyle e^{i\lambda\omega z}=$ $\displaystyle 1+i\sum_{k=1}^{3}\lambda
z_{k}\omega_{k}-\frac{1}{2}\sum_{k,l=1}^{3}\lambda^{2}z_{k}z_{l}\omega_{k}\omega_{l}-\frac{i}{2}\sum_{k,l,s=1}^{3}\lambda^{3}z_{kls}\omega_{kls}\int_{0}^{1}(1-\theta)^{2}e^{i\lambda\omega\theta
z}d\theta.$ ( 4.15)

By $S_{1}(v)=0$ and  4.15, one has

$\displaystyle\frac{2}{\pi
i}S_{1}\Big{(}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}f\Big{)}=i\sum_{k=1}^{3}S_{1}(vz_{k})\int_{S^{2}}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega-\frac{1}{2}\sum_{k,l=1}^{3}\lambda
S_{1}(vz_{k}z_{l})\times$
$\displaystyle\int_{S^{2}}\widehat{\mathcal{R}_{k}\mathcal{R}_{l}f}(\lambda\omega)d\omega-\frac{i}{2}\sum_{k,l,s=1}^{3}\lambda^{2}\int_{0}^{1}(1-\theta)^{2}S_{1}\big{(}vz_{kls}\int_{S^{2}}e^{i\lambda\omega\theta
z}\widehat{\mathcal{R}_{kls}f}(\lambda\omega)d\omega\big{)}d\theta.$ ( 4.16)

Putting 4.1 into  4.13, we can check that

$\displaystyle K^{2}_{-2,1}f(x)$
$\displaystyle=i\sum_{k=1}^{3}\sum_{j=1}^{n_{2}}c_{j}^{k}\int_{0}^{\infty}\lambda\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega\Big{)}d\lambda$
$\displaystyle\
-\frac{1}{2}\sum_{k,l=1}^{3}\sum_{j=1}^{n_{2}}c_{j}^{kl}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}\mathcal{R}_{l}f}(\lambda\omega)d\omega\Big{)}d\lambda$
$\displaystyle\ -\frac{i}{2}\frac{2}{\pi
i}\sum_{k,l,s=1}^{3}\int_{0}^{1}(1-\theta)^{2}\int_{{\mathbb{R}}^{6}}v(\rho)(S_{2}A^{2}_{-2,1}S_{1})(\rho,z)\
z_{kls}v(z)\Big{(}A_{\rho,\theta z}(\mathcal{R}_{kls}f)\Big{)}(x)dzd\rho
d\theta$
$\displaystyle=:i\sum_{k=1}^{3}T_{k}f(x)-\frac{1}{2}\sum_{k,l=1}^{3}T_{kl}f(x)-\frac{i}{2}\sum_{k,l,s=1}^{3}T_{kls}f(x),$
( 4.17)

where $c_{j}^{k}=\langle S_{2}A^{2}_{-2,1}S_{1}(vz_{k}),f_{j}\rangle$,
$c_{j}^{kl}=\langle S_{2}A^{2}_{-2,1}S_{1}(vz_{k}z_{l}),f_{j}\rangle$.

As the same as done in 3.2, we have
$T_{kls}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<\infty$. And
compared with  4.9 and  4.8 , we obtain that
$T_{k}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<3$ and
$T_{kl}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<\infty$.

Thus we have $K^{2}_{-2,1}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all
$1<p<3$.
(ii) The proof of $K^{2}_{-1,1}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all
$1<p<3.$

By Taylor’s expansion of $e^{i\lambda\omega z}$ 3.2, one has

$\displaystyle\frac{2}{\pi
i}\Big{(}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}f\Big{)}=$
$\displaystyle\frac{1}{\lambda}v\int_{S^{2}}\hat{f}(\lambda\omega)d\omega+i\sum_{k=1}^{3}vz_{k}\int_{S^{2}}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega$
$\displaystyle-\sum_{k,l=1}^{3}\lambda
vz_{k}z_{l}\int_{0}^{1}(1-\theta)\int_{S^{2}}e^{i\lambda\omega\theta
z}\widehat{\mathcal{R}_{k}\mathcal{R}_{l}f}(\lambda\omega)d\omega d\theta$ (
4.18)

Put 4.1 into  4.14, then we can obtain that

$\displaystyle K^{2}_{-1,1}f(x)$
$\displaystyle=\sum_{j=1}^{n_{2}}\tilde{c}_{j}\int_{0}^{\infty}\lambda\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\hat{f}(\lambda\omega)d\omega\Big{)}d\lambda$
$\displaystyle\
+i\sum_{k=1}^{3}\sum_{j=1}^{n_{2}}\tilde{c}_{j}^{k}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega\Big{)}d\lambda$
$\displaystyle\ -\frac{2}{\pi
i}\sum_{k,l=1}^{3}\int_{0}^{1}(1-\theta)\int_{{\mathbb{R}}^{6}}v(\rho)(S_{2}A^{2}_{-1,1})(\rho,z)\
z_{k}z_{l}v(z)\Big{(}A_{\rho,\theta
z}(\mathcal{R}_{k}\mathcal{R}_{l}f)\Big{)}(x)dzd\rho d\theta$
$\displaystyle:=\widetilde{T}f(x)+i\sum_{k=1}^{3}\widetilde{T}_{k}f(x)-\sum_{k,l=1}^{3}\widetilde{T}_{kl}f(x),$
( 4.19)

where $\tilde{c}_{j}=\langle S_{2}A^{2}_{-1,1}(v),f_{j}\rangle$,
$\tilde{c}_{j}^{k}=\langle S_{2}A^{2}_{-1,1}(vz_{k}),f_{j}\rangle$.

As the same as done in 3.2, we have
$\widetilde{T}_{kl}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all
$1<p<\infty$. Comparing with  4.9 and  4.8, we obtain that
$\widetilde{T}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<3$ and
$\widetilde{T}_{k}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all
$1<p<\infty$. Hence we have
$K^{2}_{-1,1}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<3$. ∎

Next, we will deal with the terms $\lambda^{-2}S_{1}A^{2}_{-2,2}S_{2}$ and
$\lambda^{-1}A^{2}_{-1,2}S_{2}$.

We define the following two operators:

$\displaystyle K^{2}_{-2,2}f$ $\displaystyle=\frac{2}{\pi
i}\int_{0}^{\infty}\lambda\chi_{<a}(\lambda)\
R_{0}^{+}(\lambda^{4})vS_{1}A^{2}_{-2,2}S_{2}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}fd\lambda,$
( 4.20) $\displaystyle K^{2}_{-1,2}f$ $\displaystyle=\frac{2}{\pi
i}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\
R_{0}^{+}(\lambda^{4})vA^{2}_{-1,2}S_{2}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}fd\lambda.$
( 4.21)

######  Proposition 4.3.

Let $K^{2}_{-2,2}$ and $K^{2}_{-1,2}$ be the operators defined as above. Then
one has $K^{2}_{-2,2},\ K^{2}_{-1,2}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$
for all $1<p<\infty$.

###### Proof.

(i) The proof of $K^{2}_{-2,2}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all
$1<p<\infty.$

With the formulas $S_{2}(v)=S_{2}(z_{j}v)=0$ for $j=1,2,3$ and the expansion
 4.15 of $e^{i\lambda\omega z}$ at hand, we have

$\displaystyle\frac{2}{\pi i}S_{2}\Big{(}v\big{(}R_{0}^{+}$
$\displaystyle(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}f\Big{)}=-\frac{1}{2}\sum_{k,l=1}^{3}\lambda
S_{2}(vz_{k}z_{l})\int_{S^{2}}\widehat{\mathcal{R}_{k}\mathcal{R}_{l}f}(\lambda\omega)d\omega$
$\displaystyle-\frac{i}{2}\sum_{k,l,s=1}^{3}\lambda^{2}\int_{0}^{1}(1-\theta)^{2}S_{2}\big{(}vz_{kls}\int_{S^{2}}e^{i\lambda\omega\theta
z}\widehat{\mathcal{R}_{kls}f}(\lambda\omega)d\omega\big{)}d\theta.$ ( 4.22)

Putting 4.1 into  4.20, we can check that $K^{2}_{-2,2}$ can be written
as:

$\begin{split}K^{2}_{-2,2}f(x)&=-\frac{1}{2}\sum_{k,l=1}^{3}\sum_{j=1}^{n_{1}}c_{j}^{kl}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}\mathcal{R}_{l}f}(\lambda\omega)d\omega\Big{)}d\lambda\\\
&\ -\frac{i}{2}\frac{2}{\pi
i}\sum_{k,l,s=1}^{3}\int_{0}^{1}(1-\theta)^{2}\int_{{\mathbb{R}}^{6}}v(\rho)(S_{1}A^{2}_{-2,2}S_{2})(\rho,z)\
z_{kls}v(z)\Big{(}A_{\rho,\theta z}(\mathcal{R}_{kls}f)\Big{)}(x)dzd\rho
d\theta\\\
&:=-\frac{1}{2}\sum_{k,l=1}^{3}T_{kl}f(x)-\frac{i}{2}\sum_{k,l,s=1}^{3}T_{kls}f(x),\end{split}$

where $c_{j}^{kl}=\langle S_{1}A^{2}_{-2,2}S_{2}(vz_{k}),f_{j}\rangle$.

As the same as done in 3.2, we have
$T_{kls}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<\infty$.
Comparing with  3.14, we obtain that
$T_{kl}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<\infty$. Thus
$K^{2}_{-2,2}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<\infty$.
(ii) The proof of $K^{2}_{-1,2}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all
$1<p<\infty.$

Using $S_{2}(v)=S_{2}(z_{j}v)=0$ for $j=1,2,3$ and Taylor’s expansion of
$e^{i\lambda\omega z}$ 3.2, one has

$\begin{split}&\frac{2}{\pi
i}S_{2}\Big{(}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}f\Big{)}=-\sum_{k,l=1}^{3}\lambda\int_{0}^{1}(1-\theta)S_{2}\big{(}vz_{k}z_{l}\int_{S^{2}}e^{i\lambda\omega\theta
z}\widehat{\mathcal{R}_{k}\mathcal{R}_{l}f}(\lambda\omega)d\omega\big{)}d\theta.\end{split}$
( 4.23)

Putting  4.23 into  4.21, $K^{2}_{-1,2}$ can be written as:

$\begin{split}K^{2}_{-1,2}f(x)&=-\frac{2}{\pi
i}\sum_{k,l=1}^{3}\int_{0}^{1}(1-\theta)\int_{{\mathbb{R}}^{6}}v(\rho)(A^{2}_{-1,2}S_{2})(\rho,z)\
z_{k}z_{l}v(z)\Big{(}A_{\rho,\theta
z}(\mathcal{R}_{k}\mathcal{R}_{l}f)\Big{)}(x)dzd\rho d\theta.\end{split}$

As the same as done in 3.2, it is easy to check that
$K^{2}_{-1,2}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<\infty$. ∎

In the following, we will show that $W_{-}$ is unbounded on
$L^{p}({\mathbb{R}}^{3})$ for any $3\leq p<\infty$ in the case of the second
kind resonance.

Building on our prior analysis and the expansion of $M^{-1}(\lambda)$ as in
equation 4.1 for the second kind of resonance, it becomes clear that most
terms associated with 4.1 are bounded on $L^{p}$ for all $1<p<\infty$.

However, there are three specific terms corresponding to
$\lambda^{-3}S_{2}A^{2}_{-3,1}S_{2}$, $\lambda^{-2}S_{2}A^{2}_{-2,1}S_{1}$,
and $\lambda^{-1}S_{2}A^{2}_{-1,1}$ for which we only establish $L^{p}$ bounds
for all $1<p<3$.

We note that $S_{2}A^{2}_{-1,1}$ can be expressed as
$S_{2}H^{2}_{-1,1}+S_{2}H^{2}_{-1,2}Q$, as derived from  2.24. Therefore,
our focus should primarily be on the term corresponding to
$\lambda^{-1}S_{2}H^{2}_{-1,1}$, rather than the term corresponding to
$\lambda^{-1}S_{2}A^{2}_{-1,1}$ since, given $Q(v)=S_{1}(v)$, the term linked
to $\lambda^{-1}S_{2}H^{2}_{-1,2}Q$ is similar to the one associated with
$\lambda^{-1}S_{1}A^{2}_{-1,3}S_{1}$. The latter is bounded on $L^{p}$ for all
$1<p<\infty$.

As a result, our objective is to establish that the following operator
$\Lambda^{2}$, which encompasses the sum of
$\lambda^{-3}S_{2}A^{2}_{-3,1}S_{2}$, $\lambda^{-2}S_{2}A^{2}_{-2,1}S_{1}$,
and $\lambda^{-1}S_{2}H^{2}_{-1,1}$, is unbounded on $L^{p}$ for any $3\leq
p<\infty$:

$\displaystyle(\Lambda^{2}f)(x)=\frac{2}{\pi
i}\int_{0}^{\infty}\lambda^{3}\chi_{<a}(\lambda)\Big{[}R_{0}^{+}(\lambda^{4})v$
$\displaystyle\Big{(}\lambda^{-3}S_{2}A^{2}_{-3,1}S_{2}+\lambda^{-2}S_{2}A^{2}_{-2,1}S_{1}$
$\displaystyle+\lambda^{-1}S_{2}H^{2}_{-1,1}\Big{)}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}f\Big{]}(x)d\lambda.$
( 4.24)

######  Proposition 4.4.

Let $\Lambda^{2}$ be the operator defined as above. Then the operator
$\Lambda^{2}$ is unbounded on $L^{p}({\mathbb{R}}^{3})$ for any $3\leq
p<\infty$.

###### Proof.

According to 4.1, 4.1 and 4.1, it is easy to check that

$\Lambda^{2}f(x)=\Lambda_{1}^{2}f(x)+\Lambda_{2}^{2}f(x)+\Lambda_{3}^{2}f(x),$

where $\Lambda_{1}^{2}f(x)$, $\Lambda_{2}^{2}f(x)$, $\Lambda_{3}^{2}f(x)$ are
given as follows:

$\begin{split}\Lambda_{1}^{2}f(x)=\sum_{j=1}^{n_{2}}\Big{[}&-\frac{1}{2}\sum_{k,l=1}^{3}c_{-3j}^{kl}\int_{0}^{\infty}\lambda\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}\mathcal{R}_{l}f}(\lambda\omega)d\omega\Big{)}d\lambda\\\
&+i\sum_{k=1}^{3}c_{-2j}^{k}\int_{0}^{\infty}\lambda\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega\Big{)}d\lambda\\\
&+c_{-1j}\int_{0}^{\infty}\lambda\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\hat{f}(\lambda\omega)d\omega\Big{)}d\lambda\Big{]},\end{split}$
$\begin{split}\Lambda_{2}^{2}f(x)=\sum_{j=1}^{n_{2}}\Big{[}&-\frac{i}{6}\sum_{k,l,s=1}^{3}c_{-3j}^{kls}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{kls}f}(\lambda\omega)d\omega\Big{)}d\lambda\\\
&-\frac{1}{2}\sum_{k,l=1}^{3}c_{-2j}^{kl}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}\mathcal{R}_{l}f}(\lambda\omega)d\omega\Big{)}d\lambda\\\
&+i\sum_{k=1}^{3}c_{-1j}^{k}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega\Big{)}d\lambda\Big{]},\end{split}$
$\begin{split}\Lambda_{3}^{2}f(x)=\frac{2}{\pi
i}\Big{[}&\frac{1}{6}\sum_{k,l,s,t=1}^{3}\int_{0}^{1}(1-\theta)^{3}\int_{{\mathbb{R}}^{6}}v(\rho)(S_{2}A_{-3,1}^{2}S_{2})(\rho,z)v(z)z_{klst}\Big{(}A_{\rho,z\theta}(\mathcal{R}_{klst}f)\Big{)}(x)dzd\rho
d\theta\\\
&-\frac{i}{2}\sum_{k,l,s=1}^{3}\int_{0}^{1}(1-\theta)^{2}\int_{{\mathbb{R}}^{6}}v(\rho)(S_{2}A^{2}_{-2,1}S_{1})(\rho,z)\
z_{kls}v(z)\Big{(}A_{\rho,\theta z}(\mathcal{R}_{kls}f)\Big{)}(x)dzd\rho
d\theta\\\
&-\sum_{k,l=1}^{3}\int_{0}^{1}(1-\theta)\int_{{\mathbb{R}}^{6}}v(\rho)(S_{2}H^{2}_{-1,1})(\rho,z)\
z_{k}z_{l}v(z)\Big{(}A_{\rho,\theta
z}(\mathcal{R}_{k}\mathcal{R}_{l}f)\Big{)}(x)dzd\rho
d\theta\Big{]},\end{split}$

and

$\begin{split}&c_{-3j}^{kl}=\langle
S_{2}A_{-3,1}^{2}S_{2}(vz_{k}z_{l}),f_{j}\rangle,\ \ \ c_{-2j}^{k}=\langle
S_{2}A^{2}_{-2,1}S_{1}(vz_{k}),f_{j}\rangle,\ \ \ c_{-1j}=\langle
S_{2}H^{2}_{-1,1}(v),f_{j}\rangle,\\\ &c_{-3j}^{kls}=\langle
S_{2}A_{-3,1}^{2}S_{2}(vz_{kls}),f_{j}\rangle,\ \ \ c_{-2j}^{kl}=\langle
S_{2}A^{2}_{-2,1}S_{1}(vz_{k}z_{l}),f_{j}\rangle,\ \ \ c_{-1j}^{k}=\langle
S_{2}H^{2}_{-1,1}(vz_{k}),f_{j}\rangle.\end{split}$

Based on the analyses and conclusions presented in Propositions 4.1 and 4.2,
it becomes evident that $\Lambda_{2}^{2}$ and $\Lambda_{3}^{2}$ belong to
$\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<\infty$, and
$\Lambda_{1}^{2}$ belongs to $\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all
$1<p<3$.

Therefore, our primary objective is to demonstrate that $\Lambda_{1}^{2}$ is
unbounded on $L^{p}$ for any $3\leq p<\infty$. It is worth noting that
$\chi_{\geq 4a}(|D|)$ is also in $\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all
$1<p<\infty$. Hence, our task now is to establish the unboundedness of
$\chi_{\geq 4a}(|D|)\Lambda_{1}^{2}$ on $L^{p}$ for any $3\leq p<\infty$.

We can check that $\chi_{\geq 4a}(|D|)\Lambda_{1}^{2}f(x)$ can be written as:

$\begin{split}\chi_{\geq
4a}(|D|)\Lambda_{1}^{2}f(x)&=\sum_{j=1}^{n_{2}}\big{(}M(D)vf_{j}\big{)}(x)\int_{{\mathbb{R}}^{3}}\mathcal{F}\Big{(}-\frac{1}{2}\sum_{k,l=1}^{3}c_{-3j}^{kl}\frac{\eta_{k}\eta_{l}}{|\eta|^{3}}\chi_{<a}(|\eta|)\\\
&\ \
+i\sum_{k=1}^{3}c_{-2j}^{k}\frac{\eta_{k}}{|\eta|^{2}}\chi_{<a}(|\eta|)+c_{-1j}\frac{\chi_{<a}(|\eta|)}{|\eta|}\Big{)}(y)f(y)dy+\Lambda_{1,2}^{2}f(x)\\\
&=\Lambda_{1,1}^{2}f(x)+\Lambda_{1,2}^{2}f(x),\end{split}$

where $\Lambda_{1,1}^{2}f(x)$, $\Lambda_{1,2}^{2}f(x)$ are given as follows:

$\begin{split}&\Lambda_{1,1}^{2}f(x)=\sum_{j=1}^{n_{2}}\big{(}M(D)vf_{j}\big{)}(x)\ell_{j}(f),\
\ \ \ell_{j}(f)=\int_{{\mathbb{R}}^{3}}\hat{h}_{j}(y)f(y)dy,\\\
&h_{j}(\eta)=\frac{\chi_{<a}(|\eta|)}{|\eta|}\Big{(}-\frac{1}{2}\sum_{k,l=1}^{3}c_{-3j}^{kl}\frac{\eta_{k}\eta_{l}}{|\eta|^{2}}+i\sum_{k=1}^{3}c_{-2j}^{k}\frac{\eta_{k}}{|\eta|}+c_{-1j}\Big{)},\\\
&\eta=(\eta_{1},\eta_{2},\eta_{3})\in{\mathbb{R}}^{3}\setminus\\{0\\};\end{split}$
$\begin{split}\Lambda_{1,2}^{2}f(x)=\frac{2}{\pi
i}\sum_{j=1}^{n_{2}}\Big{[}&-\frac{1}{2}\sum_{k,l=1}^{3}c_{-3j}^{kl}\int_{{\mathbb{R}}^{3}}v(\rho)f_{j}(\rho)\Big{(}M(D)A_{\rho,0}|D|^{2}\chi_{<4a}(|D|)\mathcal{R}_{k}\mathcal{R}_{l}f\Big{)}(x)d\rho\\\
&+i\sum_{k=1}^{3}c_{-2j}^{k}\int_{{\mathbb{R}}^{3}}v(\rho)f_{j}(\rho)\Big{(}M(D)A_{\rho,0}|D|^{2}\chi_{<4a}(|D|)\mathcal{R}_{k}f\Big{)}(x)d\rho\\\
&+c_{-1j}\int_{{\mathbb{R}}^{3}}v(\rho)f_{j}(\rho)\Big{(}M(D)A_{\rho,0}|D|^{2}\chi_{<4a}(|D|)f\Big{)}(x)d\rho\Big{]}.\end{split}$

By the previous analysis, it is easy to show that
$\Lambda_{1,2}^{2}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all
$1<p<\infty$. Thus, it suffices to show that $\Lambda_{1,1}^{2}$ is unbounded
on $L^{p}$ for any $3\leq p<\infty$.

Next, we suppose $\Lambda_{1,1}^{2}\in\mathbb{B}(L^{p_{0}}({\mathbb{R}}^{3}))$
for some $3\leq p_{0}<\infty$ to lead to a contradiction.

Since $\Lambda_{1,1}^{2}\in\mathbb{B}(L^{p_{0}}({\mathbb{R}}^{3}))$ for some
$3\leq p_{0}<\infty$, then by Lemma 5.6 below and the Hahn-Banach theorem, it
follows that $\ell_{j}(f)$ is a continuous functional on $L^{p_{0}}$ for
$3\leq p_{0}<\infty,$ and each $j=1,\cdots,n_{2}$. Furthermore, by Riesz
representation theorem, for each $j=1,\cdots,n_{2}$, one has

$\hat{h}_{j}(y)\in L^{q_{0}}({\mathbb{R}}^{3}),\ \ \ 1<q_{0}\leq\frac{3}{2},\
\ \ \frac{1}{p_{0}}+\frac{1}{q_{0}}=1.$

By Hausdorff-Young’s inequality, we have

$h_{j}(\eta)\in L^{p_{0}}({\mathbb{R}}^{3}),\ \ \ 3\leq p_{0}<\infty,\ \forall
j=1,\cdots,n_{2}.$

By the  2.24 and noting that $D_{2}vG_{1}vQ=0,$
$D_{2}vG_{1}v^{2}=\|V\|_{L^{1}}D_{2}(|z|^{2}v),$ and

$D_{2}vG_{1}vTD_{1}Tv=\langle D_{1}Tv,Tv\rangle D_{2}(|z|^{2}v),$

we can obtain that

$\begin{split}&c_{-3j}^{kl}=\big{\langle}S_{2}A_{-3,1}^{2}S_{2}(vz_{k}z_{l}),f_{j}\big{\rangle}=4\pi\cdot
5!(1-i)\big{\langle}D_{2}(vz_{k}z_{l}),f_{j}\big{\rangle},\\\
&c_{-2j}^{k}=\big{\langle}S_{2}A^{2}_{-2,1}S_{1}(vz_{k}),f_{j}\big{\rangle}=20i\big{\langle}TD_{1}(vz_{k}),v\big{\rangle}\big{\langle}D_{2}(|z|^{2}v),f_{j}\big{\rangle}\\\
&c_{-1j}=\big{\langle}S_{2}H^{2}_{-1,1}(v),f_{j}\big{\rangle}=80\pi(1+i)\big{(}1-\frac{\langle
D_{1}Tv,Tv\rangle}{\|V\|_{L^{1}}}\big{)}\big{\langle}D_{2}(|z|^{2}v),f_{j}\big{\rangle},\end{split}$

where $1-\langle D_{1}Tv,Tv\rangle\|V\|_{L^{1}}^{-1}\geq 0$ follows from Lemma
4.7 in [10].

Hence, $h_{j}(\eta)$ $(j=1,\cdots,n_{2})$ can be written as

$\begin{split}h_{j}(\eta)=&\frac{\chi_{<a}(|\eta|)}{|\eta|}\Big{[}-\sum_{k,l=1}^{3}2\pi\cdot
5!\big{\langle}D_{2}(vz_{k}z_{l}),f_{j}\big{\rangle}\frac{\eta_{k}\eta_{l}}{|\eta|^{2}}-20\sum_{k=1}^{3}\big{\langle}TD_{1}(vz_{k}),v\big{\rangle}\big{\langle}D_{2}(|z|^{2}v),f_{j}\big{\rangle}\frac{\eta_{k}}{|\eta|}\\\
&+80\pi\big{(}1-\frac{\langle
D_{1}Tv,Tv\rangle}{\|V\|_{L^{1}}}\big{)}\big{\langle}D_{2}(|z|^{2}v),f_{j}\big{\rangle}+i\Big{(}\sum_{k,l=1}^{3}2\pi\cdot
5!\big{\langle}D_{2}(vz_{k}z_{l}),f_{j}\big{\rangle}\frac{\eta_{k}\eta_{l}}{|\eta|^{2}}\\\
&+80\pi\big{(}1-\frac{\langle
D_{1}Tv,Tv\rangle}{\|V\|_{L^{1}}}\big{)}\big{\langle}D_{2}(|z|^{2}v),f_{j}\big{\rangle}\Big{)}\Big{]}:=\frac{\chi_{<a}(|\eta|)}{|\eta|}g_{j}(\eta).\end{split}$

We claim that both $D_{2}g$ and $D_{1}g$ are real-valued, provided that $g$ is
a real-valued function on $L^{2}({\mathbb{R}}^{3})$. In fact, let

$\begin{split}\eta=D_{2}g=(\hbox{Re}D_{2}g)+i(\hbox{Im}D_{2}g):=\varphi+i\psi.\end{split}$

Since $\eta\in S_{2}L^{2}$ and $S_{2}L^{2}=\overline{S_{2}L^{2}}$, so both
$\varphi$ and $\psi$ must be real-valued functions belonging to
$S_{2}L^{2}({\mathbb{R}}^{3})$. Furthermore, by Definition A.2, we have

$\displaystyle g=D_{2}^{-1}\eta=T_{2}\eta=T_{2}\varphi+iT_{2}\psi,$ ( 4.25)

where $T_{2}$ is defined in  A.10.

To demonstrate that $D_{2}g$ is real-valued, it suffices to establish that
$T_{2}\varphi$ and $T_{2}\psi$ are real-valued. Indeed, once this is
confirmed, since $g$ is real-valued,  4.25 implies $T_{2}\psi=0$, leading to
the real-valued property of $D_{2}g$ as $\psi=D_{2}T_{2}\psi=0$. Therefore,
our focus is to confirm that $T_{2}\varphi$ and $T_{2}\psi$ are real-valued.

In fact, by  A.10 and $S_{2}(v)=S_{2}(x_{j}v)=0$ for $j=1,2,3$, we can check
that

$\begin{split}T_{2}f=&\sum_{i,j=1}^{3}4\langle f,y_{i}y_{j}v\rangle
S_{2}(x_{i}x_{j}v)+2\langle f,|y|^{2}v\rangle S_{2}(|x|^{2}v)\\\
&+\frac{10}{3}\big{(}1-\frac{\langle
D_{1}Tv,Tv\rangle}{\|V\|_{L^{1}}}\big{)}\langle f,|y|^{2}v\rangle
S_{2}(|x|^{2}v),\,\,\forall f\in S_{2}L^{2}({\mathbb{R}}^{3}).\end{split}$ (
4.26)

Thus, replace $f$ with $\varphi$, $\psi$ in  4.26 respectively and note that
$\varphi$, $\psi$ are real-valued, then we have $T_{2}\varphi,T_{2}\psi$ are
real-valued.

By the similar way and note that

$\begin{split}T_{1}f=\|V\|_{L^{1}}^{-1}\langle Tf,v\rangle
S_{1}(Tv)+\frac{2\|V\|_{L^{1}}}{3\cdot(8\pi)^{2}}\sum_{i=1}^{3}\langle
f,y_{i}v\rangle S_{1}(x_{i}v),\,\,\forall f\in
S_{1}L^{2}({\mathbb{R}}^{3}),\end{split}$

we can conclude that $D_{1}g$ is real-valued, provided that $g\in
L^{2}({\mathbb{R}}^{3})$ is real-valued.

Return to our purpose, by the above arguments, one has $D_{2}(vz_{k}z_{l})$
and $D_{1}(vz_{k})$ are real valued for all $k,l=1,2,3.$ Then we obtain the
imaginary part of $g_{j}(\eta)$ $(j=1,\cdots,n_{2})$ as follows:

$\begin{split}\hbox{Im}g_{j}(\eta)=\sum_{k,l=1}^{3}2\pi\cdot
5!\big{\langle}D_{2}(vz_{k}z_{l}),f_{j}\big{\rangle}\frac{\eta_{k}\eta_{l}}{|\eta|^{2}}+80\pi\big{(}1-\frac{\langle
D_{1}Tv,Tv\rangle}{\|V\|_{L^{1}}}\big{)}\big{\langle}D_{2}(|z|^{2}v),f_{j}\big{\rangle}.\end{split}$

Next we mainly focus on the imaginary part of each $g_{j}(\eta)$, which can be
written as

$\begin{split}\hbox{Im}g_{j}(\eta)=\sum_{k,l=1}^{3}b_{j,kl}\frac{\eta_{k}\eta_{l}}{|\eta|^{2}},\
\ \ \forall j=1,\cdots,n_{2},\end{split}$ ( 4.27)

where $b_{j,kl}(k,l=1,2,3,\ \ j=1,\cdots,n_{2})$ are given as follows:

$\begin{split}&b_{j,kl}=b_{j,lk}=2\pi\cdot
5!\big{\langle}D_{2}(vz_{k}z_{l}),f_{j}\big{\rangle},\ k\neq l;\\\
\end{split}$ ( 4.28) $\begin{split}&b_{j,kk}=b_{j,ll}=2\pi\cdot
5!\big{\langle}D_{2}(vz_{k}^{2}),f_{j}\big{\rangle}+80\pi\big{(}1-\frac{\langle
D_{1}Tv,Tv\rangle}{\|V\|_{L^{1}}}\big{)}\big{\langle}D_{2}(v|z|^{2}),f_{j}\big{\rangle},\
k=l.\end{split}$ ( 4.29)

Since

$h_{j}(\eta)=\frac{\chi_{<a}(|\eta|)}{|\eta|}g_{j}(\eta)\in
L^{p_{0}}({\mathbb{R}}^{3}),\ \ \ 3\leq p_{0}<\infty,\ \ \forall
j=1,\cdots,n_{2},$

and consider the homogeneity and continuity of $g_{j}(\eta)$ for each
$j=1,\cdots,n_{2}$, then

$g_{j}(\eta)\equiv 0,\ \ \ \ \ \ \ \forall j=1,\cdots,n_{2},$

which implies that $\hbox{Im}g_{j}(\eta)\equiv 0$ for each $j=1,\cdots,n_{2}.$
Hence, by  4.27 and Lemma 5.7, one has $b_{j,kl}=0$ for each $k,l=1,2,3$,
$j=1,\cdots,n_{2}$. Furthermore, noting that  4.29, for each $k=1,2,3$ and
$j=1,\cdots,n_{2},$ we have

$2\pi\cdot
5!\big{\langle}D_{2}(vz_{k}^{2}),f_{j}\big{\rangle}=-80\pi\big{(}1-\frac{\langle
D_{1}Tv,Tv\rangle}{\|V\|_{L^{1}}}\big{)}\big{\langle}D_{2}(v|z|^{2}),f_{j}\big{\rangle},$

which implies that $\big{\langle}D_{2}(vz_{k}^{2}),f_{j}\big{\rangle}$ is a
constant independent $k$ when we fix a $1\leq j\leq n_{2}$. Hence for each
$k=1,2,3$ we have

$2\pi\cdot
5!\big{\langle}D_{2}(vz_{k}^{2}),f_{j}\big{\rangle}=-240\pi\big{(}1-\frac{\langle
D_{1}(Tv),Tv\rangle}{\|V\|_{L^{1}}}\big{)}\big{\langle}D_{2}(vz_{k}^{2}),f_{j}\big{\rangle}.$

Note that $2\pi\cdot 5!=240\pi$ and $1-\langle
D_{1}Tv,Tv\rangle\|V\|_{L^{1}}^{-1}\geq 0$, then

$\begin{split}\big{\langle}D_{2}(vz_{k}^{2}),f_{j}\big{\rangle}=0,\ \ \
\forall k=1,2,3.\end{split}$

Therefore, combining with  4.28 and $b_{j,kl}=0$ for each $k,l=1,2,3$,
$j=1,\cdots,n_{2}$, one has

$\big{\langle}D_{2}(vz_{k}z_{l}),f_{j}\big{\rangle}=\big{\langle}vz_{k}z_{l},D_{2}f_{j}\big{\rangle}=0,\
\ \ \forall k,l=1,2,3,\ \ \ j=1,\cdots,n_{2}.$

Since $\\{f_{j}\\}_{j=1}^{n_{2}}$ is a basis of $S_{2}L^{2}({\mathbb{R}}^{3})$
and $D_{2}$ is invertible on $S_{2}L^{2}({\mathbb{R}}^{3})$, hence it follows
that $S_{2}(vz_{k}z_{l})=0$ for all $k,l=1,2,3$, which implies that
$S_{2}(L^{2}({\mathbb{R}}^{3}))=S_{3}(L^{2}({\mathbb{R}}^{3}))$. This clearly
contradicts with the fact that $S_{2}(L^{2}({\mathbb{R}}^{3}))\neq\\{0\\}$ and
$S_{3}(L^{2}({\mathbb{R}}^{3}))=\\{0\\}$ under the assumption that zero is the
second kind resonance of $H$ (also see Definition A.2).

Hence we complete the proof of this proposition. ∎

With Propositions 4.1, 4.2, 4.3 and 4.4 at hand, we have finished the proof of
the second kind resonance case of Theorem 1.4.

###  4.2. The third kind of resonance

In this subsection, we investigate the $L^{p}$ boundedness of the wave
operators $W_{-}$ in the context of the third kind of resonance.

Suppose that $S_{3}L^{2}\neq\\{0\\}$. In the following, we first establish
that $W_{-}$ is bounded on $L^{p}$ for all $1<p<3$. Secondly, we show that
$W_{\pm}(H,\Delta^{2})\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $3\leq
p<\infty$ if and only if $S_{2}L^{2}=S_{3}L^{2}=S_{4}L^{2}$. This also implies
$W_{\pm}(H,\Delta^{2})\notin\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for any
$3\leq p<\infty$ if and only if $S_{2}L^{2}\neq S_{3}L^{2}$ or $S_{3}L^{2}\neq
S_{4}L^{2}$.

We begin with the expansion of $M^{-1}(\lambda)$ in the third kind of
resonance as follows:

$\displaystyle M^{-1}(\lambda)=$
$\displaystyle\lambda^{-4}S_{3}A_{-4,1}^{3}S_{3}+\lambda^{-3}S_{2}A^{3}_{-3,1}S_{2}+\lambda^{-2}\Big{(}S_{2}A^{3}_{-2,1}S_{1}+S_{1}A^{3}_{-2,2}S_{2}\Big{)}+\lambda^{-1}\Big{(}S_{2}A^{3}_{-1,1}$
$\displaystyle+A^{3}_{-1,2}S_{2}+S_{1}A^{3}_{-1,3}S_{1}\Big{)}+\Big{(}S_{1}A_{0,1}^{3}+A^{3}_{0,2}S_{1}+QA^{3}_{0,3}Q\Big{)}+\lambda
A^{3}_{1,1}+\Gamma^{3}_{2}(\lambda).$ ( 4.30)

Taking 4.2 to  3.3 and considering the results in the second kind of
resonance, it suffices to show that the term
$\lambda^{-4}S_{3}A_{-4,1}^{3}S_{3}$ is bounded on $L^{p}$ for all $1<p<3,$
which is expressed as in the following proposition.

In addition to the notations  4.12, we set
$z_{klstm}=z_{k}z_{l}z_{s}z_{t}z_{m}$,
$\omega_{klstm}=\omega_{k}\omega_{l}\omega_{s}\omega_{t}\omega_{m}$ and
$\mathcal{R}_{klstm}=\mathcal{R}_{k}\mathcal{R}_{l}\mathcal{R}_{s}\mathcal{R}_{t}\mathcal{R}_{m}$.
Since $S_{3}L^{2}\subset S_{2}L^{2}\subset S_{1}L^{2}$,

$\dim S_{1}L^{2}=n_{1}\geq\dim S_{2}L^{2}=n_{2}\geq\dim S_{3}L^{2}=n_{3}.$

Moreover, we can choose three families of real-valued standard orthogonal
bases $\\{f_{j}\\}_{j=1}^{n_{1}}$, $\\{f_{j}\\}_{j=1}^{n_{2}}$ and
$\\{f_{j}\\}_{j=1}^{n_{3}}$ of $S_{1}L^{2}$, $S_{2}L^{2}$ and $S_{3}L^{2}$,
respectively.

######  Proposition 4.5.

Let

$\displaystyle K^{3}_{-4,1}:=\frac{2}{\pi
i}\int_{0}^{\infty}\lambda^{-1}\chi_{<a}(\lambda)\
R_{0}^{+}(\lambda^{4})vS_{3}A_{-4,1}^{3}S_{3}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}d\lambda.$
( 4.31)

Then $K^{3}_{-4,1}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<3$.

###### Proof.

Let $f\in\mathbb{S}({\mathbb{R}}^{3})$. Then

$\displaystyle(K^{3}_{-4,1}f)(x):=\frac{2}{\pi
i}\int_{0}^{\infty}\lambda^{-1}\chi_{<a}(\lambda)\Big{[}\
R_{0}^{+}(\lambda^{4})vS_{3}A_{-4,1}^{3}S_{3}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}f\Big{]}(x)d\lambda.$
( 4.32)

Note that Taylor expansion of $e^{i\lambda\omega z}$ can be written as
follows:

$\displaystyle e^{i\lambda\omega z}=$ $\displaystyle 1+i\sum_{k=1}^{3}\lambda
z_{k}\omega_{k}-\frac{1}{2}\sum_{k,l=1}^{3}\lambda^{2}z_{k}z_{l}\omega_{k}\omega_{l}-\frac{i}{6}\sum_{k,l,s=1}^{3}\lambda^{3}z_{kls}\omega_{kls}$
$\displaystyle+\frac{1}{24}\sum_{k,l,s,t=1}^{3}\lambda^{4}z_{klst}\omega_{klst}+\frac{i}{24}\sum_{k,l,s,t,m=1}^{3}\lambda^{5}z_{klstm}\omega_{klstm}\int_{0}^{1}(1-\theta)^{4}e^{i\lambda\omega\theta
z}d\theta.$ ( 4.33)

With 4.2 and $S_{3}(v)=S_{3}(z_{j}v)=S_{3}(z_{j}z_{k}v)=0$ for $j,k=1,2,3$
and hand, one has

$\begin{split}&\frac{2}{\pi
i}S_{3}\Big{(}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}f\Big{)}=-\frac{i}{6}\sum_{k,l,s=1}^{3}\lambda^{2}S_{3}(vz_{kls})\int_{S^{2}}\widehat{\mathcal{R}_{kls}f}(\lambda\omega)d\omega+\frac{1}{24}\sum_{k,l,s,t=1}^{3}\lambda^{3}S_{3}(vz_{klst})\\\
&\times\int_{S^{2}}\widehat{\mathcal{R}_{klst}f}(\lambda\omega)d\omega+\frac{i}{24}\sum_{k,l,s,t=1}^{3}\lambda^{4}\int_{0}^{1}(1-\theta)^{4}S_{3}\big{(}vz_{klstm}\int_{S^{2}}e^{i\lambda\omega\theta
z}\widehat{\mathcal{R}_{klstm}f}(\lambda\omega)d\omega\big{)}d\theta.\end{split}$
( 4.34)

Putting  4.34 into  4.32, we can check that

$\displaystyle($ $\displaystyle
K^{3}_{-4,1}f)(x)=-\frac{i}{6}\sum_{k,l,s=1}^{3}\sum_{j=1}^{n_{3}}c_{j}^{kls}\int_{0}^{\infty}\lambda\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{kls}f}(\lambda\omega)d\omega\Big{)}d\lambda$
$\displaystyle+\frac{1}{24}\sum_{k,l,s,t=1}^{3}\sum_{j=1}^{n_{3}}c_{j}^{klst}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{klst}f}(\lambda\omega)d\omega\Big{)}d\lambda$
$\displaystyle+\frac{i}{24}\frac{2}{\pi
i}\sum_{k,l,s,t,m=1}^{3}\int_{0}^{1}(1-\theta)^{4}\int_{{\mathbb{R}}^{6}}v(\rho)(S_{3}A^{3}_{-4,1}S_{3})(\rho,z)\
z_{klstm}v(z)\Big{(}A_{\rho,\theta z}(\mathcal{R}_{klstm}f)\Big{)}(x)dzd\rho
d\theta$
$\displaystyle=:-\frac{i}{6}\sum_{k,l,s=1}^{3}T_{kls}f(x)+\frac{1}{24}\sum_{k,l,s,t=1}^{3}T_{klst}f(x)+\frac{i}{24}\sum_{k,l,s,t,m=1}^{3}T_{klstm}f(x),$
( 4.35)

where
$c_{j}^{kls}=\big{\langle}S_{3}A^{3}_{-4,1}S_{3}(vz_{kls}),f_{j}\big{\rangle}$,
$c_{j}^{klst}=\big{\langle}S_{3}A^{3}_{-4,1}S_{3}(vz_{klst}),f_{j}\big{\rangle}$.

As the same as done in 3.2, we have
$T_{klstm}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<\infty$.
Moreover, comparing with  4.9 and  4.8 , we obtain that
$T_{kls}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<3$ and
$T_{klst}\in\mathbb{B}(L^{p}({\mathbb{R}}^{3}))$ for all $1<p<\infty$. Hence,
we complete the proof of this proposition. ∎

It is worth noting that we can replace $n_{3}$ with $n_{2}$ in the definition
of $T_{kls}$ and $T_{klst}$ in 4.2. Indeed, since
$\\{f_{j}\\}_{j=1}^{n_{2}}$, $\\{f_{j}\\}_{j=1}^{n_{3}}$ are the standard
orthogonal bases of $S_{2}L^{2}$, $S_{3}L^{2}$, respectively, we have for all
$n_{3}<l\leq n_{2}$,

$S_{3}f_{l}=\sum_{j=1}^{n_{3}}\langle S_{3}f_{l},f_{j}\rangle
f_{j}=\sum_{j=1}^{n_{3}}\langle f_{l},S_{3}f_{j}\rangle
f_{j}=\sum_{j=1}^{n_{3}}\langle f_{l},f_{j}\rangle f_{j}=0.$

Next, we are devoted to investigating whether the wave operators $W_{-}$ are
bounded on $L^{p}$ for $3\leq p<\infty$.

Based on our previous analysis and the expansion of $M^{-1}(\lambda)$ in
4.2, most terms associated with 4.2 are bounded on $L^{p}$ for all
$1<p<\infty$. However, there are four specific terms corresponding to
$\lambda^{-4}S_{3}A^{3}_{-4,1}S_{3}$, $\lambda^{-3}S_{2}A^{3}_{-3,1}S_{2}$,
$\lambda^{-2}S_{2}A^{3}_{-2,1}S_{1}$, and $\lambda^{-1}S_{2}A^{3}_{-1,1}$ for
which we only establish $L^{p}$ bounds for all $1<p<3$.

We can further simplify our focus by noting that
$A^{3}_{-3,1}=S_{2}H^{3}_{-3,1}S_{2}+S_{2}H^{3}_{-3,2}S_{3}$ and
$S_{2}A^{3}_{-1,1}=S_{2}H^{3}_{-1,1}+S_{2}H^{3}_{-1,2}Q$, as derived from
2.25. By utilizing the properties $Q(v)=S_{1}(v)=0$ and
$S_{3}(v)=S_{3}(z_{j}v)=S_{3}(z_{j}z_{k}v)=0$ (where $j$ and $k$ range from 1
to 3), and drawing from previous experience, we can confirm that both the two
terms corresponding to $\lambda^{-1}S_{2}H^{2}_{-1,2}Q$ and
$\lambda^{-3}S_{2}H^{3}_{-3,2}S_{3}$ are bounded on $L^{p}$ for all
$1<p<\infty$. This implies that we only need to address the two terms
corresponding to $S_{2}H^{3}_{-3,1}S_{2}$ and $\lambda^{-1}S_{2}H^{3}_{-1,1}$,
rather than the two terms linked to $\lambda^{-3}S_{2}A^{3}_{-3,1}S_{2}$ and
$\lambda^{-1}S_{2}A^{3}_{-1,1}$, respectively.

Hence, our focus is on determining whether the following operator
$\Lambda^{3}$, which encompasses the sum of
$\lambda^{-4}S_{3}A^{3}_{-4,1}S_{3}$, $\lambda^{-3}S_{2}H^{3}_{-3,1}S_{2}$,
$\lambda^{-2}S_{2}A^{3}_{-2,1}S_{1}$, and $\lambda^{-1}S_{2}H^{3}_{-1,1}$, is
bounded on $L^{p}$ for $3\leq p<\infty$:

$\displaystyle(\Lambda^{3}f)(x)=\frac{2}{\pi
i}\int_{0}^{\infty}\lambda^{3}\chi_{<a}(\lambda)$ $\displaystyle\Big{[}\
R_{0}^{+}(\lambda^{4})v\Big{(}\lambda^{-4}S_{3}A^{3}_{-4,1}S_{3}+\lambda^{-3}S_{2}H^{3}_{-3,1}S_{2}+\lambda^{-2}S_{2}A^{3}_{-2,1}S_{1}$
$\displaystyle+\lambda^{-1}S_{2}H^{3}_{-1,1}\Big{)}v\big{(}R_{0}^{+}(\lambda^{4})-R_{0}^{-}(\lambda^{4})\big{)}f\Big{]}(x)d\lambda.$
( 4.36)

In view of the proofs of Propositions 4.4 and 4.5, we decompose
$\Lambda^{3}=\Lambda_{1}^{3}+\Lambda_{2}^{3}+\Lambda_{3}^{3}$, where

$\displaystyle\Lambda_{1}^{3}f(x)=\sum_{j=1}^{n_{2}}\Big{[}$
$\displaystyle-\frac{i}{6}\sum_{k,l,s=1}^{3}c_{-4j}^{kls}\int_{0}^{\infty}\lambda\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{kls}f}(\lambda\omega)d\omega\Big{)}d\lambda$
$\displaystyle-\frac{1}{2}\sum_{k,l=1}^{3}c_{-3j}^{kl}\int_{0}^{\infty}\lambda\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}\mathcal{R}_{l}f}(\lambda\omega)d\omega\Big{)}d\lambda$
$\displaystyle+i\sum_{k=1}^{3}c_{-2j}^{k}\int_{0}^{\infty}\lambda\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega\Big{)}d\lambda$
$\displaystyle+c_{-1j}\int_{0}^{\infty}\lambda\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\hat{f}(\lambda\omega)d\omega\Big{)}d\lambda\Big{]},$
( 4.37) $\displaystyle\Lambda_{2}^{3}f(x)=\sum_{j=1}^{n_{2}}\Big{[}$
$\displaystyle\frac{1}{24}\sum_{k,l,s,t=1}^{3}c_{-4j}^{klst}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{klst}f}(\lambda\omega)d\omega\Big{)}d\lambda$
$\displaystyle-\frac{i}{6}\sum_{k,l,s=1}^{3}c_{-3j}^{kls}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{kls}f}(\lambda\omega)d\omega\Big{)}d\lambda$
$\displaystyle-\frac{1}{2}\sum_{k,l=1}^{3}c_{-2j}^{kl}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}\mathcal{R}_{l}f}(\lambda\omega)d\omega\Big{)}d\lambda$
$\displaystyle+i\sum_{k=1}^{3}c_{-1j}^{k}\int_{0}^{\infty}\lambda^{2}\chi_{<a}(\lambda)\big{(}R_{0}^{+}(\lambda^{4})vf_{j}\big{)}(x)\Big{(}\int_{S^{2}}\widehat{\mathcal{R}_{k}f}(\lambda\omega)d\omega\Big{)}d\lambda\Big{]},$
$\displaystyle\Lambda_{3}^{3}f(x)=\frac{2}{\pi i}\Big{[}$
$\displaystyle\frac{i}{24}\sum_{k,l,s,t,m=1}^{3}\int_{0}^{1}(1-\theta)^{4}\int_{{\mathbb{R}}^{6}}v(\rho)(S_{3}A^{3}_{-4,1}S_{3})(\rho,z)\
z_{klstm}v(z)\Big{(}A_{\rho,\theta z}(\mathcal{R}_{klstm}f)\Big{)}(x)dzd\rho
d\theta$
$\displaystyle+\frac{1}{6}\sum_{k,l,s,t=1}^{3}\int_{0}^{1}(1-\theta)^{3}\int_{{\mathbb{R}}^{6}}v(\rho)(S_{2}H_{-3,1}^{3}S_{2})(\rho,z)v(z)z_{klst}\Big{(}A_{\rho,z\theta}(\mathcal{R}_{klst}f)\Big{)}(x)dzd\rho
d\theta$
$\displaystyle-\frac{i}{2}\sum_{k,l,s=1}^{3}\int_{0}^{1}(1-\theta)^{2}\int_{{\mathbb{R}}^{6}}v(\rho)(S_{2}A^{3}_{-2,1}S_{1})(\rho,z)\
z_{kls}v(z)\Big{(}A_{\rho,\theta z}(\mathcal{R}_{kls}f)\Big{)}(x)dzd\rho
d\theta$
$\displaystyle-\sum_{k,l=1}^{3}\int_{0}^{1}(1-\theta)\int_{{\mathbb{R}}^{6}}v(\rho)(S_{2}H^{3}_{-1,1})(\rho,z)\
z_{k}z_{l}v(z)\Big{(}A_{\rho,\theta
z}(\mathcal{R}_{k}\mathcal{R}_{l}f)\Big{)}(x)dzd\rho d\theta\Big{]},$

and

$\displaystyle c_{-4j}^{kls}$
$\displaystyle=\big{\langle}S_{3}A_{-4,1}^{3}S_{3}(vz_{kls}),f_{j}\big{\rangle},\quad
c_{-3j}^{kl}=\big{\langle}S_{2}H_{-3,1}^{3}S_{2}(vz_{k}z_{l}),f_{j}\big{\rangle},$
$\displaystyle c_{-2j}^{k}$