# Decay estimates for bi-Schrödinger operators in dimension one

Avy Soffer, Zhao Wu and Xiaohua Yao † Avy Soffer, Mathematics Department,
Rutgers University, New Brunswick, NJ, 08903, USA<EMAIL_ADDRESS>Zhao Wu, School of Mathematics and Statistics, Central China Normal
University, Wuhan, 430079, P.R. China<EMAIL_ADDRESS>Xiaohua Yao,
Department of Mathematics and Hubei Province Key Laboratory of Mathematical
Physics, Central China Normal University, Wuhan, 430079, P.R. China
<EMAIL_ADDRESS>
###### Abstract.

This paper is devoted to study the time decay estimates for bi-Schrödinger
operators $H=\Delta^{2}+V(x)$ in dimension one with decaying potentials
$V(x)$. We first deduce the asymptotic expansions of resolvent of $H$ at zero
energy threshold without/with the presence of resonances, and then
characterize these resonance spaces corresponding to different types of zero
resonance in suitable weighted spaces $L_{s}^{2}({\mathbf{R}})$. Next we use
them to establish the sharp $L^{1}-L^{\infty}$ decay estimates of Schrödinger
groups $e^{-itH}$ generated by bi-Schrödinger operators also with zero
resonances. As a consequence, Strichartz estimates are obtained for the
solution of fourth-order Schrödinger equations with potentials for initial
data in $L^{2}({\mathbf{R}})$. In particular, it should be emphasized that the
presence of zero resonances does not change the optimal time decay rate of
$e^{-itH}$ in dimension one, except at requiring faster decay rate of the
potential.

###### Key words and phrases:

Decay estimates, Bi-Schrödinger operators, Zero resonances, Resolvent
expansions

† Corresponding author

##  1\. Introduction

###  1.1. Backgrounds and problems

The higher-order elliptic operators $P(D)+V$ have been extensively considered
as general Hamiltonian operator by many people in different contexts. We just
mention some references, not all. For instance, see Schechter [Sch71] for
spectral theory, Kuroda [Kur78], Agmon [Agm75], Hörmander [Hö05] for
scattering theory, Davies [Dav97], Davies and Hinz [DH98b], Deng et al [DDY14]
for the Gaussian estimates of semigroup, and as well [HS15, HS17, BS91, Mou81,
SYY18] for other interesting topics.

In this paper, we will consider the time decay estimates for the bi-
Schrödinger operator:

$H=\Delta^{2}+V(x),\,\,H_{0}=\Delta^{2},$ (1.1)

where $V(x)$ is a real-valued function satisfying
$|V(x)|\lesssim(1+|x|)^{-\beta}$ for some $\beta>0$. Such investigations are
motivated in large part by classical Schrödinger operator $-\Delta+V$ and
their fundamental applications. In the last thirty years, the dispersive
estimates of Schrödinger operator have been one of the key topics, which were
applied broadly to nonlinear Schrödinger equations, see e.g. [Bou91, Caz03,
Tao06, JSS91, KT98, NaSc11, Sch07, Sch20, Sim18a, Sim18b] and references
therein.

For the free propagators $e^{-itH_{0}}$ and $H_{0}=\Delta^{2}$, Ben-Artzi,
Koch and Saut in [BKS] have proved the following pointwise estimates for the
kernel of $e^{-itH_{0}}$ on $\mathbf{R}^{n}$:

$|\partial_{x}^{\alpha}I_{0}(x,t)|\lesssim
t^{-\frac{n+|\alpha|}{4}}\Big{(}1+\frac{|x|}{t^{1/4}}\Big{)}^{-\frac{(n-|\alpha|)}{3}},\
\ x\in\mathbf{R}^{n},$ (1.2)

where $I_{0}(x,t)$ be the kernel of $e^{-itH_{0}}$ and
$\alpha\in\mathbf{N}^{n},$ which immediately implies the following
$L^{1}-L^{\infty}$ estimates:

$\big{\|}\partial^{\alpha}e^{-itH_{0}}\big{\|}_{L^{1}(\mathbf{R}^{n})\rightarrow
L^{\infty}(\mathbf{R}^{n})}\lesssim|t|^{-\frac{n+|\alpha|}{4}},\ \
|\alpha|\leq n.$ (1.3)

Moreover, we notice that the regular term $\partial^{\alpha}$ of the
inequality (1.3) can be replaced by fractional power $H_{0}^{\frac{s}{4}}$ for
any $0\leq s\leq n$ ( see e.g. [CMY12] ).

For the fourth order Schrödinger operators $H=\Delta^{2}+V(x)$, such time
decay estimates were much less understood than Schrödinger operators.
Recently, Feng, Soffer and Yao [FSY18] deduced the asymptotic expansion of the
resolvent $R_{V}(z)=(H-z)^{-1}$ around zero threshold, assuming that zero is a
regular point of $H$ for $n\geq 5$ and $n=3$, and showed that in the regular
case, the Kato-Jensen time decay estimate of $e^{-itH}$ is $(1+|t|)^{-n/4}$
for $n\geq 5$ and the $L^{1}-L^{\infty}$ decay estimates is $O(|t|^{-1/2})$
for $n=3$ (not optimal).

Based on the higher energy estimates of resolvent in [FSY18],
Erdog$\breve{a}$n, Green and Toprak in [EGT21] for $n=3$ and Green-Toprak in
[GT19] for $n=4$ further derived the asymptotic expansion of $R_{V}(z)$ near
zero with the presence of resonance or eigenvalue, and obtained
$L^{1}-L^{\infty}$ estimates for each kind of zero resonances. They proved the
$L^{1}-L^{\infty}$ time decay estimate is $|t|^{-n/4}$ for $n=3,4$ if zero is
a regular point, and zero energy resonance will change the time decay rate.

Moreover, we mention that Kato-Jensen estimates and asymptotic expansion of
resolvent have been established by [FSWY20] for poly-Schrödinger operators
$(-\Delta)^{m}+V$ with $n\geq 2m\geq 4$. In previous works, Murata in [Mur82]
have generalized Kato and Jensen’s work [JK79] to a certain class of $P(D)+V$
assume that $P(D)$ satisfies

$\big{(}\nabla P\big{)}(\xi_{0})=0,\ \ \ \
\det\big{[}\partial_{i}\partial_{j}P(\xi)\big{]}\Big{|}_{\xi_{0}}\neq 0.$
(1.4)

However, the poly-harmonic operators $H_{0}=(-\Delta)^{m}$ do not satisfy the
nondegenerate condition (1.4) at zero but the case $m=1$. Hence the results in
[FSWY20] were not covered by [Mur82]. In particular, the degeneracy of
$(-\Delta)^{m}$ at zero energy certainly leads to more complicated zero
resonances classifications and asymptotic expansions of the resolvent for
$(-\Delta)^{m}+V$ as $m\geq 2$. This is actually one of the main difficult
points for the higher order estimates.

In this paper, we are devoted to establishing the same $L^{1}-L^{\infty}$
estimates of $H=\Delta^{2}+V$ (with regularity terms) as the free estimates
(1.3) in dimension one. We showed that the presence of zero resonance of $H$
does not affect the time decay rate of $e^{-itH}$ but requires higher decay
rate of potential $V$. This seems to be distinct phenomena than other
situations ( see e.g. [FSWY20, EGT21, GT19, Sch20]).

As a consequence, Strichartz estimates were obtained for the fourth-order
Schrodinger equations with potentials for initial data in
$L^{2}({\mathbf{R}})$. Furthermore, by using similar methods, Li, Soffer and
Yao [LSY21] have derived the $L^{1}-L^{\infty}$ estimates for $H$ with zero
resonances in two dimensional case. In particular, we remark that arguments
used here can be applied to the higher-order elliptic operators
$(-\Delta)^{m}+V$.

For Schrödinger operator $-\Delta+V$, the $L^{1}-L^{\infty}$ decay estimates
have been widely investigated by many people. For $n\geq 3$, Journé, Soffer
and Sogge in [JSS91] first established the $L^{1}-L^{\infty}$ dispersive
estimate in the regular case. For $n\leq 3$, one can see Weder [Wed00],
Rodnianski and Schlag [RS04], Schlag and Goldberg [GS04], Schlag [Sch05] and
so on. Yajima in [Yaj95] has also established the $L^{1}-L^{\infty}$ estimates
for $-\Delta+V$ by wave operator method. For many further studies, one can
refer to [Wed99, ES04, ES06, GG15, GG17, Sch07, Sch20] and therein references.

###  1.2. Some notations

In this subsection, we collect some notations used throughout the paper.

  * •

For $a\in\mathbf{R}$, we write $a\pm$ to denote $a\pm\epsilon$ for any small
$\epsilon>0$. $[a]$ denotes the largest integer less than or equal to $a$. For
$a,b\in\mathbf{R^{+}},$ we denote $A\lesssim B$ (or $B\gtrsim A$) which means
that there exists some constant $c>0$ such that $A\leq cB$ (or $B\geq cA$).
Moreover, if $A\lesssim B$ and $B\lesssim A,$ then we write $A\sim B.$

  * •

For $s,s^{\prime}\in\mathbf{R}$, $B(s,s^{\prime})$ denote the space of the
bounded operators from $L^{2}_{s}(\mathbf{R})$ to
$L^{2}_{s^{\prime}}(\mathbf{R})$, where $L^{2}_{s}(\mathbf{R})$ is the
weighted $L^{2}$ space:

$L^{2}_{s}(\mathbf{R})=\Big{\\{}f\in
L^{2}_{loc}(\mathbf{R}):(1+|\cdot|)^{s}f\in L^{2}(\mathbf{R})\Big{\\}}.$

###  1.3. Main Results

In the subsection, we will state our main results. Recall that, if there
exists a nonzero solution $\phi(x)\in L^{2}(\mathbf{R})$ satisfying the
equation $H\phi(x)=0$ in distributional sense, then we say that zero is an
eigenvalue of $H=\Delta^{2}+V(x).$ In general, there may exist a nonzero
resonant solution $\phi(x)$ in $L^{2}_{-s}(\mathbf{R})$ with some $s>0$ such
that $H\phi(x)=0$, which usually results in the obstructions to dispersive
estimates at low energy. In the present case, we first introduce the zero
resonance definitions of $\Delta^{2}+V(x)$ in dimension one. For
$\sigma\in\mathbf{R}$, let $W_{\sigma}(\mathbf{R})$ denotes the intersection
space
$W_{\sigma}(\mathbf{R})=\bigcap\limits_{s>\sigma}L^{2}_{-s}(\mathbf{R}).$ Note
that, $W_{\sigma_{1}}(\mathbf{R})\supset W_{\sigma_{2}}(\mathbf{R})$ if
$\sigma_{1}>\sigma_{2}$ and $W_{0}(\mathbf{R})\supset L^{2}(\mathbf{R})$.

######  Definition 1.1.

Let $H=\Delta^{2}+V$ on $L^{2}(\mathbf{R})$ and
$|V(x)|\lesssim(1+|x|)^{-\beta}$ for some $\beta>0$. Then we say that

(i) zero is a first kind resonance of $H$ if there exists some nonzero
$\phi(x)\in W_{\frac{3}{2}}(\mathbf{R})$ but no any nonzero $\phi\in
W_{\frac{1}{2}}(\mathbf{R})$ such that $H\phi(x)=0$ in the distributional
sense;

(ii) zero is a second kind resonance of $H$ if there exists some nonzero
$\phi(x)\in W_{\frac{1}{2}}(\mathbf{R})$ but no any nonzero $\phi\in
L^{2}(\mathbf{R})$ such that $H\phi(x)=0$ in the distributional sense;

(iii) zero is an eigenvalue of $H$ if there exists some nonzero $\phi(x)\in
L^{2}(\mathbf{R})$ such that $H\phi(x)=0$ in the distributional sense;

(iv) zero is a regular point of $H$ if zero is neither a resonance nor an
eigenvalue of $H$.

######  Remark 1.2.

Some examples of zero resonance and zero eigenvalue:

(i) (Zero resonance) Assume that $\phi\in C^{\infty}(\mathbf{R})$ is a
positive function which is equal to $c|x|+d$ for $|x|>1.$ Then $H\phi(x)=0$
when $\displaystyle V(x)=-(\Delta^{2}\phi)/\phi(x).$ One can easily check that
$V(x)\in C_{0}^{\infty}(\mathbf{R})$ and $\phi(x)\in
W_{\frac{3}{2}}(\mathbf{R}).$ Hence by Definition 1.1, we know that the zero
is a resonance point of $H$ even with a compactly supported smooth potential.

(ii) (Zero eigenvalue) Unlike zero resonance, the zero eigenvalue problem of
$H$ is subtler. For instance, let $\phi(x)=(1+x^{2})^{-s}$ with
$s>\frac{1}{4}$ and $\displaystyle V(x)=-(\Delta^{2}\phi)/\phi(x).$ Then one
has that

$V(x)=O((1+|x|)^{-4}),\ \ |x|\rightarrow\infty,$

and $\phi\in L^{2}(\mathbf{R})$ satisfying $H\phi=0.$ This means that zero is
an eigenvalue of $H$ with such slowly decaying potential.

On the other hand, it was pointed out to us by Prof. Goldberg [Gold21] that if
a potential $V(x)$ has faster decay, then zero eigenvalue of the one-
dimensional bi-Schrödinger operator may not occur, see e.g.[DT79] for
Schrödinger operator case. In fact, if the potential $V$ satisfies the decay
condition of Theorem 1 below, then a proof on the absence of zero eigenvalue
has been stated in Lemma 3.5 of Section 3.

Now we state the first main result under the decay conditions depending on the
types of zero energy defined above. In this paper due to the faster decay rate
of $V$ and the absence of zero eigenvalue, we do not necessarily discuss zero
eigenvalue case in Theorem 1 below. Even so, it should be noticed that
compared with the Schrödinger operator $-\Delta+V$ on $L^{2}(\mathbf{R})$ (
see e.g. Jensen and Nenciu [JN01, JN04] ), the classification of the zero
resonance for $H=\Delta^{2}+V$ in one dimensional case is more complicated due
to the low dimensionality ( i.e. $n=1$ ) and the degeneracy of $\Delta^{2}$ at
zero energy ( i.e. zero is the degenerate critical point of the symbol
$\xi^{4}$ ); it leads to the more singular terms in the zero asymptotic
expansion of resolvent $R_{V}^{\pm}(\mu)$ ( see Theorem 1.8 below or Section 3
).

######  Theorem 1.3.

111After the submission of our paper, we learned that by a different method,
Thomas Hill has proved that the estimates (1.6) and (1.7) hold, assuming that
zero is a regular point for $H=\Delta^{2}+V$ with a compactly supported
potential $V(x)\in C_{c}^{1}(\mathbf{R}),$ see his dissertation [Hill20].

Let $|V(x)|\lesssim(1+|x|)^{-\beta}$ with some $\beta>0$ depending on the
following types:

$\beta>\begin{cases}13,&\text{zero is the regular point},\\\ 17,&\text{zero is
the first kind resonance},\\\ 25,&\text{zero is the second kind
resonance}.\end{cases}$ (1.5)

Assume that $H=\Delta^{2}+V$ has no positive embedded eigenvalues and
$P_{ac}(H)$ denotes the projection onto the absolutely continuous spectrum
space of $H$. Then we have the following decay estimate:

$\big{\|}e^{-itH}P_{ac}(H)\big{\|}_{L^{1}(\mathbf{R})\rightarrow
L^{\infty}(\mathbf{R})}\lesssim|t|^{-\frac{1}{4}},\ \ t\neq 0.$ (1.6)

Furthermore, for any $0\leq\alpha<1,$ the following improved estimate holds:

$\big{\|}H^{\frac{\alpha}{4}}e^{-itH}P_{ac}(H)\big{\|}_{L^{1}(\mathbf{R})\rightarrow
L^{\infty}(\mathbf{R})}\lesssim|t|^{-\frac{1+\alpha}{4}},\ \ t\neq 0.$ (1.7)

As we will see later, the needed decay rates of the potential $V$ mainly
depend on the lower energy expansions of resolvent $R_{V}(z)$ for different
zero resonance types (see Theorem 1.8 below). According to Remark 1.10 after
Theorem 1.8, we still can obtain the decay estimates (1.6) and (1.7) by using
the weak expansions of $R_{V}(z)$ under the assumption of the following less
decay rates:

$\beta>\begin{cases}7,&\text{zero is the regular point},\\\ 9,&\text{zero is
the first kind resonance},\\\ 13,&\text{zero is the second kind
resonance}.\end{cases}$

In the paper, we do not follow the way along with these interesting
improvements. Instead, we focus on the simplicity of the proof based on the
expansions of resolvent $R_{V}(z)$ in Theorem 1.8. Noting that the estimates
(1.6) and (1.7) above are optimal in view of the free estimate (1.3) with
$V=0$.

In particular, the optimal time decay rate ( e.g. $|t|^{-\frac{1}{4}}$ in the
(1.6)) does not change as the zero resonance arises, but only requires the
faster decay rate $\beta$ of potential $V$ as the compensations of zero
singularities. Moreover, note that for Schrödinger operator $-\Delta+V$,
Egorova, Kopylova, Marchenko and Teschl in [EKMT16] have proved that the
resonant case does not need additional decay of the potential, hence it is an
interesting question whether we can show that the sharp decay estimates (1.6)
still hold on the same decay condition on $V(x)$ for $H=\Delta^{2}+V$.

Moreover, in the above Theorem 1, we also assume that $H=\Delta^{2}+V$ has no
positive eigenvalues embedded in the absolutely continuous spectrum, which has
been the indispensable condition in dispersive estiamtes. For Schrödinger
operator $-\Delta+V$, Kato in [Kat59] has shown the absence of positive
eigenvalues for $H=-\Delta+V$ with the decay potentials $V=o(|x|^{-1})$ as
$|x|\rightarrow\infty$. For the bi-Schrödinger operator $H=\Delta^{2}+V$, the
situation seems to be more complicated than the second order operator, since
there exist examples even with compactly supported smooth potentials such that
positive eigenvalues appear ( see e.g. [FSWY20] ).

However, interestingly, a simple criterion has been proved in [FSWY20] that
$H=\Delta^{2}+V$ has no any positive eigenvalues under the assumption that the
potential $V$ is bounded and satisfies the repulsive condition (i.e.
$(x\cdot\nabla)V\leq 0$). In particular, such criterion also works for the
general higher order elliptic operator $P(D)+V$. We also notice that for a
general selfadjoint operator $\mathcal{H}$ on $L^{2}(\mathbf{R}^{n})$, even if
$\mathcal{H}$ has a simple embedded eigenvalue, Costin and Soffer in [CS01]
have proved that $\mathcal{H}+\epsilon W$ can kick off the eigenvalue located
in a small interval under certain small perturbation of potential.

As an interesting consequence of $L^{1}-L^{\infty}$ estimates above, by Keel-
Tao [KT98] method, we immediately obtain Strichartz estimate to the fourth-
order Schrödinger equation on $\mathbf{R}$:

$\left\\{\begin{gathered}i\partial_{t}\phi=\big{(}\Delta^{2}+V\big{)}\phi+F(t,x),\
\ (t,x)\in\mathbf{R}^{1+1},\hfill\\\ \phi(0,\cdot)=\phi_{0}(x).\hfill\\\
\end{gathered}\right.\ \ \ $ (1.8)

Recall that, a pair $(q,r)$ is said to be $\sigma-$admissible if

$\frac{1}{q}+\frac{\sigma}{r}=\frac{\sigma}{2},\ \ 2\leq q\leq\infty,\ \
(q,r,\sigma)\neq(2,\infty,1).$ (1.9)

######  Corollary 1.4.

Let $V$ and $H=\Delta^{2}+V$ satisfy the same conditions as Theorem 1 above.
Then for any $\displaystyle\frac{1}{4}-$admissible pairs $(q,r)$ and
$(\tilde{q},\tilde{r}),$ we have the following estimates:

(i) Homogeneous Strichartz Estimate

$\big{\|}e^{-itH}P_{ac}(H)\phi_{0}\big{\|}_{L_{t}^{q}L_{x}^{r}(\mathbf{R}\times\mathbf{R})}\lesssim\|\phi_{0}\|_{L^{2}(\mathbf{R})}.$
(1.10)

(ii) Dual Homogeneous Strichartz Estimate

$\Big{\|}\int_{\mathbf{R}}e^{-isH}P_{ac}(H)F(s,\cdot)ds\Big{\|}_{L_{x}^{2}(\mathbf{R})}\lesssim\|F\|_{L_{t}^{\tilde{q}^{\prime}}L_{x}^{\tilde{r}^{\prime}}(\mathbf{R}\times\mathbf{R})}.$
(1.11)

(iii) The solution $\phi(t,x)$ of the equation (1.8) satisfies

$\big{\|}P_{ac}(H)\phi\big{\|}_{L_{t}^{q}L_{x}^{r}(\mathbf{R}\times\mathbf{R})}\lesssim\|\phi_{0}\|_{L^{2}(\mathbf{R})}+\|F\|_{L_{t}^{\tilde{q}^{\prime}}L_{x}^{\tilde{r}^{\prime}}(\mathbf{R}\times\mathbf{R})}.$
(1.12)

In order to derive the decay estimates (1.6), it follows from Stone’s formula
that

$e^{-itH}P_{ac}(H)f=\frac{2}{\pi
i}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}[R_{V}^{+}(\mu^{4})-R_{V}^{-}(\mu^{4})]f(x)d\mu.$
(1.13)

Hence we need to study the detailed zero expansions of the resolvent
$R_{V}(z)=(H-z)^{-1}$ by the perturbations of the free resolvent $R_{0}(z)$,
for which the following identity holds:

$R_{0}(z):=\big{(}\Delta^{2}-z\big{)}^{-1}=\frac{1}{2z^{\frac{1}{2}}}\Big{(}R(-\Delta;z^{\frac{1}{2}})-R(-\Delta;-z^{\frac{1}{2}})\Big{)},\
\ \ z\in\mathbf{C}\setminus[0,\infty).$ (1.14)

Here the resolvent
$R(-\Delta;z^{\frac{1}{2}}):=(-\Delta-z^{\frac{1}{2}})^{-1}.$ For
$\lambda\in\mathbf{R}^{+},$ we define the limiting resolvent operators by

$R_{0}^{\pm}(\lambda):=R_{0}^{\pm}(\lambda\pm i0)=\lim_{\epsilon\rightarrow
0^{+}}(\Delta^{2}-(\lambda\pm i\epsilon))^{-1},$ (1.15)
$R_{V}^{\pm}(\lambda):=R_{V}^{\pm}(\lambda\pm i0)=\lim_{\epsilon\rightarrow
0^{+}}(H-(\lambda\pm i\epsilon))^{-1},$ (1.16)

By using the representation (1.14) for $R_{0}(z)$ with $z=\omega^{4}$ for
$\omega$ in the first quadrant of the complex plane, and taking limits as
$\omega\rightarrow\mu$ and $\omega\rightarrow i\mu$, we obtain

$R_{0}^{\pm}(\mu^{4})=\frac{1}{2\mu^{2}}\Big{(}R^{\pm}(-\Delta;\mu^{2})-R(-\Delta;-\mu^{2})\Big{)},\
\ \mu>0.$ (1.17)

It is well-known that by the limiting absorption principle ( see e.g. [Agm75]
), $R^{\pm}(-\Delta;\mu^{2})$ are well-defined as the bounded operators of
$B(L_{s}^{2},L^{2}_{-s})$ for any $s>1/2$, and therefore
$R_{0}^{\pm}(\mu^{4})$ are also well-defined between these weighted spaces.
This limiting absorption property can be extended to $R_{V}^{\pm}(\mu^{4})$
for $\mu>0$ for certain decaying and bounded potentials ( see e.g. [FSY18] ).

The main challenge is to derive the asymptotic expansions of
$R_{V}^{\pm}(\mu^{4})$ as $\mu$ near zero. Recall that the Green’s kernel of
the free Laplace resolvent in dimension one( see [ES04, JN01] for example)

$R^{\pm}(-\Delta;\mu^{2})(x,y)=\frac{\pm i}{2\mu}e^{\pm i\mu|x-y|}.$ (1.18)

Therefore, by (1.17), we have for $\mu>0$,

$\begin{split}R_{0}^{\pm}(\mu^{4})(x,y)=&\frac{1}{4\mu^{3}}\Big{(}\pm ie^{\pm
i\mu|x-y|}-e^{-\mu|x-y|}\Big{)}.\end{split}$ (1.19)

Denote $M^{\pm}(\mu)=U+vR_{0}^{\pm}(\mu^{4})v,$ $v(x)=|V(x)|^{1/2}$ and
$U=\text{sgn}\big{(}V(x)\big{)},$ by the following symmetric identity:

$R_{V}^{\pm}(\mu^{4})=R_{0}^{\pm}(\mu^{4})-R_{0}^{\pm}(\mu^{4})v\Big{(}M^{\pm}(\mu)\Big{)}^{-1}vR_{0}^{\pm}(\mu^{4}).$
(1.20)

It suffices to establish the asymptotic expansions of
$\Big{(}M^{\pm}(\mu)\Big{)}^{-1}$ with the presence of zero resonance. For
this end, let’s introduce some spaces to derive the asymptotic expansions of
$\Big{(}M^{\pm}(\mu)\Big{)}^{-1}$.

Let $T_{0}:=U+vG_{0}v$ with $G_{0}=\big{(}\Delta^{2}\big{)}^{-1}$ and
$P=\|V\|_{L^{1}(\mathbf{R})}^{-1}v\langle v,\cdot\rangle$ be the orthogonal
projection onto the span of $v,$ i.e. $PL^{2}(\mathbf{R})=\text{span}\\{v\\}.$

######  Definition 1.5.

Let $Q=I-P.$ Then we define that

(i) $S_{0}L^{2}(\mathbf{R})=\\{f\in L^{2}~{}|~{}\langle f,x^{k}v\rangle=0,\
k=0,1\\}.$

(ii) $S_{1}L^{2}(\mathbf{R})=\\{f\in L^{2}~{}|~{}\langle f,x^{k}v\rangle=0,\
k=0,1,\ \text{and}\ S_{0}T_{0}f=0\\}.$

(iii) $S_{2}L^{2}(\mathbf{R})=\\{f\in L^{2}~{}|~{}\langle f,x^{k}v\rangle=0,\
k=0,1,2,\ \text{and}\ QT_{0}f=0\\}.$

(iv) $S_{3}L^{2}(\mathbf{R})=\\{f\in L^{2}~{}|~{}\langle f,x^{k}v\rangle=0,\
k=0,1,2,3,\ \text{and}\ T_{0}f=0\\}.$

######  Remark 1.6.

(i) The projection spaces clearly satisfy the following equivalent
relationships:

$\displaystyle f\in S_{0}L^{2}(\mathbf{R})$ $\displaystyle\Longleftrightarrow
f\in\text{span}\\{v,xv\\}^{\perp};$ $\displaystyle f\in
S_{1}L^{2}(\mathbf{R})$ $\displaystyle\Longleftrightarrow
f\in\text{span}\\{v,xv\\}^{\perp}\ \text{and}\
T_{0}f\in\text{span}\\{v,xv\\};$ $\displaystyle f\in S_{2}L^{2}(\mathbf{R})$
$\displaystyle\Longleftrightarrow f\in\text{span}\\{v,xv,x^{2}v\\}^{\perp}\
\text{and}\ T_{0}f\in\text{span}\\{v\\};$ $\displaystyle f\in
S_{3}L^{2}(\mathbf{R})$ $\displaystyle\Longleftrightarrow
f\in\text{span}\\{v,xv,x^{2}v,x^{3}v\\}^{\perp}\ \text{and}\ T_{0}f=0.$

Hence, $S_{0}L^{2}(\mathbf{R})\supseteq S_{1}L^{2}(\mathbf{R})\supseteq
S_{2}L^{2}(\mathbf{R})\supseteq S_{3}L^{2}(\mathbf{R}).$

(ii) Denote by $S_{j}\ (j=0,1,2,3)$ the orthogonal projections onto the
related to projection spaces $S_{j}L^{2}(\mathbf{R})$, then their orthogonal
relationships can be listed in the following Table 1.

Projections on $L^{2}(\mathbf{R})$ |  Orthogonal Relationships
---|---
$S_{0}$ |  $S_{0}(v(x))=S_{0}(xv(x))=0$
$S_{1}$ |  $S_{1}T_{0}S_{0}=S_{0}T_{0}S_{1}=0$ $S_{1}(v(x))=S_{1}(xv(x))=0$
$S_{2}$ |  $S_{2}T_{0}Q=QT_{0}S_{2}=0$ $S_{2}(v(x))=S_{2}(xv(x))=S_{2}(x^{2}v(x))=0$
$S_{3}$ |  $S_{3}T_{0}Q=QT_{0}S_{3}=0$ $S_{3}T_{0}P=PT_{0}S_{3}=0$ $S_{3}(v(x))=S_{3}(xv(x))=0$ $S_{3}(x^{2}v(x))=S_{3}(x^{3}v(x))=0$
Table 1. Orthogonal Relationships of Projections

(iii) From Lemma 3.5 later, we will further prove that
$S_{3}L^{2}(\mathbf{R})=\\{0\\}$, which is important in the proof of Theorem
1.8 (iii), see Section 3 below.

These projection spaces in Definition 1.5 will be used to characterize the
zero energy resonance types of $H$ ( see Theorem 4.1 for the proof).

######  Theorem 1.7.

Assume that $H=\Delta^{2}+V$ with $|V(x)|\lesssim(1+|x|)^{-\beta}$ with some
$\beta$ satisfying the condition (1.5). Then the following statements hold:

(i) zero is a regular point of $H$ if and only if
$S_{1}L^{2}(\mathbf{R})=\\{0\\};$

(ii) zero is a first kind resonance of $H$ if and only if
$S_{1}L^{2}(\mathbf{R})\neq\\{0\\}$ and $S_{2}L^{2}(\mathbf{R})=\\{0\\};$

(iii) zero is a second kind resonance of $H$ if and only if
$S_{2}L^{2}(\mathbf{R})\neq\\{0\\}.$

By the characterizations of zero resonances in Theorem 1.7 and using the
orthogonal relationships in Table 1, we can derive the asymptotic expansions
of $\Big{(}M^{\pm}(\mu)\Big{)}^{-1}$, which are fundamental to study
dispersive estimates and other problems related to $H$.

######  Theorem 1.8.

Assume that $|V(x)|\lesssim(1+|x|)^{-\beta}$ with some $\beta>0$. Then for
$0<|\mu|\ll 1$, we have the following expansions of
$\Big{(}M^{\pm}(\mu)\Big{)}^{-1}$ in $L^{2}(\mathbf{R})$:

(i) If zero is a regular point of $H$ and $\beta>13,$ then

$\begin{split}\Big{(}M^{\pm}(\mu)\Big{)}^{-1}=&S_{0}A_{01}^{0}S_{0}+\mu
QA_{11}^{0}Q+\mu^{2}(QA_{21}^{0}Q+S_{0}A_{22}^{0}+A_{23}^{0}S_{0})+\Gamma_{3}^{0}(\mu),\end{split}$
(1.21)

where $A_{kj}^{0}$ are $\mu$-independent bounded operators in
$L^{2}(\mathbf{R})$ and $\Gamma_{3}^{0}(\mu)$ is $\mu$-dependent operator
which satisfies

$\big{\|}\Gamma_{3}^{0}(\mu)\big{\|}_{L^{2}\rightarrow
L^{2}}+\mu\big{\|}\partial_{\mu}\big{(}\Gamma_{3}^{0}(\mu)\big{)}\big{\|}_{L^{2}\rightarrow
L^{2}}\lesssim\mu^{3}.$

(ii) If zero is a first kind resonance of $H$ and $\beta>17,$ then

$\begin{split}\Big{(}M^{\pm}(\mu)\Big{)}^{-1}=&\frac{S_{1}A_{-1,1}^{1}S_{1}}{\mu}+S_{0}A_{01}^{1}Q+QA_{02}^{1}S_{0}+\mu\big{(}QA_{11}^{1}Q+S_{0}A_{12}^{1}+A_{13}^{1}S_{0}\big{)}\\\
&+\mu^{2}(QA_{21}^{1}+A_{22}^{1}Q)+\Gamma_{3}^{1}(\mu),\end{split}$ (1.22)

where $A_{kj}^{1}$ are $\mu$-independent bounded operators in
$L^{2}(\mathbf{R})$ and $\Gamma_{3}^{1}(\mu)$ is $\mu$-dependent operator
which satisfies

$\big{\|}\Gamma_{3}^{1}(\mu)\big{\|}_{L^{2}\rightarrow
L^{2}}+\mu\big{\|}\partial_{\mu}\big{(}\Gamma_{3}^{1}(\mu)\big{)}\big{\|}_{L^{2}\rightarrow
L^{2}}\lesssim\mu^{3}.$

(iii) If zero is a second kind resonance of $H$ and $\beta>25,$ then

$\begin{split}\Big{(}M^{\pm}(\mu)\Big{)}^{-1}=&\frac{S_{2}A_{-3,1}^{2}S_{2}}{\mu^{3}}+\frac{S_{2}A_{-2,1}^{2}S_{0}+S_{0}A_{-2,2}^{2}S_{2}}{\mu^{2}}+\frac{S_{0}A_{-1,1}^{2}S_{0}+QA_{-1,2}^{2}S_{2}+S_{2}A_{-1,3}^{2}Q}{\mu}\\\
&\ \
+\big{(}QA_{01}^{2}S_{0}+S_{0}A_{02}^{2}Q+S_{2}A_{03}^{2}+A_{04}^{2}S_{2}\big{)}+\mu\big{(}QA_{11}^{2}Q+S_{0}A_{12}^{2}+A_{13}^{2}S_{0}\big{)}\\\
&\ \ +\mu^{2}(QA_{21}^{2}+A_{21}^{2}Q)+\Gamma_{3}^{2}(\mu),\end{split}$ (1.23)

where $A_{kj}^{2}$ are $\mu$-independent bounded operators in
$L^{2}(\mathbf{R})$ and $\Gamma_{3}^{2}(\mu)$ is $\mu$-dependent operator
which satisfies

$\big{\|}\Gamma_{3}^{2}(\mu)\big{\|}_{L^{2}\rightarrow
L^{2}}+\mu\big{\|}\partial_{\mu}\big{(}\Gamma_{3}^{2}(\mu)\big{)}\big{\|}_{L^{2}\rightarrow
L^{2}}\lesssim\mu^{3}.$

######  Remark 1.9.

If we assume the more decay rate of $V(x)$, then we will have more information
about the error term $\Gamma_{3}^{i}(\mu)$ ( here the index $i$ respects to
the zero energy type). More specifically, one actually can obtain that

$\Gamma_{3}^{i}(\mu)=\frac{\mu^{3}}{\tilde{a}^{\pm}}P+\mu^{3}(QA_{31}^{i}+A_{32}^{i}Q)+\Gamma_{4}^{i}(\mu),$

where $A_{31}^{i}$ are $\mu$-independent bounded operators in
$L^{2}(\mathbf{R})$ and $\Gamma_{4}^{i}(\mu)$ are $\mu$-dependent operators
which satisfy

$\big{\|}\Gamma_{4}^{i}(\mu)\big{\|}_{L^{2}\rightarrow
L^{2}}+\mu\big{\|}\partial_{\mu}\big{(}\Gamma_{4}^{i}(\mu)\big{)}\big{\|}_{L^{2}\rightarrow
L^{2}}\lesssim\mu^{4}.$

######  Remark 1.10.

In addtion, assuming the less decay rate of $V(x)$, we can also obtain some
weak expansions of $\big{(}M^{\pm}(\mu)\big{)}^{-1}$ by checking the proof
process of Theorem 1.8 (see Section 3) along with Remark 3.2. Specifically, in
the Regular Case, it is enough for $\beta>7$ to obtain the following
expansions:

$\Big{(}M^{\pm}(\mu)\Big{)}^{-1}=S_{0}\Gamma_{01}^{0}(\mu)S_{0}+Q\Gamma_{11}^{0}(\mu)Q+(Q\Gamma_{21}^{0}(\mu)Q+S_{0}\Gamma_{22}^{0}(\mu)+\Gamma_{23}^{0}(\mu)S_{0})+\Gamma_{31}^{0}(\mu),$

where $\Gamma_{kj}^{0}(\mu)(k=0,1,2,3)$ are $\mu$-dependent operators which
satisfy

$\big{\|}\Gamma_{kj}^{0}(\mu)\big{\|}_{L^{2}\rightarrow
L^{2}}+\mu\big{\|}\partial_{\mu}\big{(}\Gamma_{kj}^{0}(\mu)\big{)}\big{\|}_{L^{2}\rightarrow
L^{2}}\lesssim\mu^{k}.$

Furthermore, similar weak expansions of $\big{(}M^{\pm}(\mu)\big{)}^{-1}$ can
be established for the different resonance cases if $V$ satisfies the
following decay rate:

$\beta>\begin{cases}9,&\text{zero is the first kind resonance},\\\
13,&\text{zero is the second kind resonance}.\end{cases}$

According to the proof of Theorem 1, we remark that even if we use the above
weak form expansion of $\big{(}M^{\pm}(\mu)\big{)}^{-1},$ the
$L^{1}-L^{\infty}$ decay estimates (1.6) of $e^{-itH}$ also hold. But the
expansions of Theorem 1.8 will be helpful to other problems such as the
$L^{p}$ boundedness of the wave operators.

From Theorem 1.8 above, it is obvious that all types of zero energy resonances
lead to the higher singularity than the regular case in the expansions of
$\Big{(}M^{\pm}(\mu)\Big{)}^{-1}$. On the other hand, the zero energy
resonances also offer more orthogonal relationships, which in turn provide the
more cancellations to balance the singularities at the cost of the decay rate
of potentials, as shown in the proof of Theorem 1 below.

We also remark that our analysis for the proof of Theorem 1.8 is longer and
more complicated processes in some ways. First, some singular terms in the
expansions of the free resolvent $R_{0}^{\pm}(\mu^{4})$ will bring more
complicated interacting terms in the expansions of
$\Big{(}M^{\pm}(\mu)\Big{)}^{-1}$, see Section 3. Then these interacting terms
will lead to the difficulties of the inversion processes, especially for the
second kind resonance case. For instance, in the case of the second kind
resonance , we need to analyze the kernel of the non-positive operator $T_{2}$
given by

$\displaystyle T_{2}$
$\displaystyle=S_{2}vG_{2}vS_{2}-\frac{8\cdot(6!)}{\|V\|_{L^{1}}}S_{2}T_{0}^{2}S_{2}+\frac{5}{2}S_{2}vG_{1}vD_{0}vG_{1}vS_{2}$
$\displaystyle+\frac{48\cdot(6!)}{\|V\|_{L^{1}}^{2}}\Big{(}S_{2}T_{0}vG_{-1}v-\frac{\|V\|_{L^{1}}}{6}S_{2}vG_{1}vD_{0}T_{0}\Big{)}D_{2}\Big{(}vG_{-1}vT_{0}S_{2}-\frac{\|V\|_{L^{1}}}{6}T_{0}D_{0}vG_{1}vS_{2}\Big{)},$

which is actually a sum of the four complicated operators with changing signs.
We will identify the $\ker T_{2}$ as the projection space
$S_{3}L^{2}(\mathbf{R})$, see Lemma 3.4 for more details.

Finally, let us comment about the proof of the decay estimates in Theorem 1.

Since the derivative of $\mu$ for the free resolvent kernel
$R_{0}^{\pm}(\mu^{4})(x,y)$ can not be dominated by $\mu^{\alpha}$ in $x,y$
uniformly, in this paper we have used Littlewood-Paley decompositions and
oscillatory integrals theory to solve the uniformity problem, which seems to
be general interesting technical point. Using some similar methods but with
important technical modifications, Li, Soffer and Yao [LSY21] have established
the $L^{1}-L^{\infty}$ estimates for $e^{-itH}$ in dimension two. We also
think that these arguments used here could be extended to general higher-order
elliptic operators $P(D)+V$.

The paper is organized as follows. In Section 2, we give the proof of Theorem
1. In Section 3, we give the process how to obtain the asymptotic expansion of
$\Big{(}M^{\pm}(\mu)\Big{)}^{-1}$ case by case. In Section 4, we characterize
the resonance spaces for each kind of zero resonance.

##  2\. Proof of Theorem 1

In this section, we will give the proof of Theorem 1. Choose a fixed even
function $\varphi\in C_{c}^{\infty}(\mathbf{R})$ such that

$\varphi(s)=\begin{cases}1,\ &\ \text{if}\ |s|\leq\frac{1}{2},\\\ 0,\ &\
\text{if}\ |s|\geq 1.\end{cases}$ (2.1)

For each $N\in\mathbf{Z},$ let
$\varphi_{N}(s)=\varphi(2^{-N}s)-\varphi(2^{-N+1}s),$ then $\text{supp}\
\varphi_{N}\subset[2^{N-2},2^{N}]$ and $\varphi_{N}(s)=\varphi_{0}(2^{-N}s).$
Furthermore, for any $s\neq 0$ we have
$\displaystyle\sum_{N=-\infty}^{+\infty}\varphi_{N}(s)=1$. Using functional
calculus and Stone’s formula, we have

$\displaystyle e^{-itH}P_{ac}(H)f=$ $\displaystyle\frac{2}{\pi
i}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}[R_{V}^{+}(\mu^{4})-R_{V}^{-}(\mu^{4})]fd\mu$
(2.2) $\displaystyle=$ $\displaystyle\frac{2}{\pi
i}\int_{0}^{\infty}e^{-it\mu^{4}}\sum_{N=-\infty}^{+\infty}\varphi_{N}(\mu)\mu^{3}[R_{V}^{+}(\mu^{4})-R_{V}^{-}(\mu^{4})]fd\mu$
$\displaystyle:=$ $\displaystyle\frac{2}{\pi
i}\int_{0}^{\infty}e^{-it\mu^{4}}\chi(\mu)\mu^{3}[R_{V}^{+}(\mu^{4})-R_{V}^{-}(\mu^{4})]fd\mu$
$\displaystyle\ +\frac{2}{\pi
i}\int_{0}^{\infty}e^{-it\mu^{4}}\widetilde{\chi}(\mu)\mu^{3}[R_{V}^{+}(\mu^{4})-R_{V}^{-}(\mu^{4})]fd\mu,$

where $\chi(\mu)=\sum\limits_{N=-\infty}^{N_{0}}\varphi_{N}(\mu)$ and
$\widetilde{\chi}(\mu)=\sum\limits_{N=N_{0}+1}^{+\infty}\varphi_{N}(\mu)$ for
some $N_{0}<0$ ( fixed later but depending on specific zero energy types).
Hence we reduce the proof of Theorem 1 into the estimates of the following two
parts: low and high energy parts.

######  Theorem 2.1.

(the low energy dispersive estimate)
Let $|V(x)|\lesssim(1+|x|)^{-\beta}$ with some $\beta>0$ depending on the
following zero energy types:

$\beta>\begin{cases}13,&\text{zero is the regular point},\\\ 17,&\text{zero is
the first kind resonance},\\\ 25,&\text{zero is the second kind
resonance}.\end{cases}$

Assume that $H=\Delta^{2}+V$ has no positive embedded eigenvalue and
$P_{ac}(H)$ denotes the projection onto the absolutely continuous spectrum
space of $H,$ then the following estimate holds:

$\|e^{-itH}P_{ac}(H)\chi(H)\|_{L^{1}(\mathbf{R})\rightarrow
L^{\infty}(\mathbf{R})}\lesssim|t|^{-\frac{1}{4}}.$ (2.3)

Moreover, for any $0\leq\alpha<1,$ we have the following estimate:

$\big{\|}H^{\frac{\alpha}{4}}e^{-itH}P_{ac}(H)\chi(H)\big{\|}_{L^{1}(\mathbf{R})\rightarrow
L^{\infty}(\mathbf{R})}\lesssim|t|^{-\frac{1+\alpha}{4}}.$ (2.4)

######  Theorem 2.2.

(the high energy dispersive estimate)
Let $|V(x)|\lesssim(1+|x|)^{-2-}$. Assume that $H=\Delta^{2}+V$ has no
positive embedded eigenvalue and $P_{ac}(H)$ denotes the projection onto the
absolutely continuous spectrum space of $H$, then we have

$\|e^{-itH}P_{ac}(H)\widetilde{\chi}(H)\|_{L^{1}(\mathbf{R})\rightarrow
L^{\infty}(\mathbf{R})}\lesssim|t|^{-\frac{1}{4}}.$ (2.5)

Moreover, for any $0<\alpha<1,$ we have the following estimate:

$\big{\|}H^{\frac{\alpha}{4}}e^{-itH}P_{ac}(H)\widetilde{\chi}(H)\big{\|}_{L^{1}(\mathbf{R})\rightarrow
L^{\infty}(\mathbf{R})}\lesssim|t|^{-\frac{1+\alpha}{4}}.$ (2.6)

The proofs of the low and high energy parts will be given in the following
subsections. For the end, let us first prove the decay estimates of the free
group $e^{-it\Delta^{2}}$ in dimension one by the resolvent method, which
display some basic technics how to prove the potential case.

###  2.1. The estimates for the free case

First, we give a useful lemma to estimates a kind of oscillatory integral
appeared repeatedly in this paper.

######  Lemma 2.3.

Let $m\geq 1$, $\Psi(z)$ be a nonnegative real valued function of
$\mathbf{R}^{m}$ and $\Phi(s;z)$ be a function on
$\mathbf{R}\times\mathbf{R}^{m}$, which is smooth for the first variable $s$.
Assume that for any $s\in\text{supp}\ \varphi_{0}(s)$ (here
$\varphi_{0}(s)=\varphi(s)-\varphi(\frac{1}{2}s),$ see (2.1)) and any
$N\in\mathbf{Z},$ $\Phi(s;z)$ satisfies the following uniform bounded
condition:

$\big{|}\partial_{s}^{k}\Phi(2^{N}s;z)\big{|}\lesssim 1,\ \ \ k=0,1,$

then we have

$\sup_{z\in\mathbf{R}^{m}}\Big{|}\int_{0}^{\infty}e^{-it2^{4N}s^{4}}\varphi_{0}(s)e^{\pm
i2^{N}s\Psi(z)}\Phi(2^{N}s;z)ds\Big{|}\lesssim(1+|t|\cdot
2^{4N})^{-\frac{1}{2}}.$ (2.7)

###### Proof.

Denote

$\displaystyle K_{N}^{\pm}(t,z):=$
$\displaystyle\int_{0}^{\infty}e^{-it2^{4N}s^{4}}\varphi_{0}(s)e^{\pm
i2^{N}s\Psi(z)}\Phi(2^{N}s;z)ds.$

Since for any $s\in\text{supp}\ \varphi_{0}(s)\subset\big{[}\frac{1}{4},1],$
$\big{|}\Phi(2^{N}s;z)\big{|}\lesssim 1$, then we have

$\big{|}K_{N}^{\pm}(t,z)\big{|}\lesssim\int_{\frac{1}{4}}^{1}\big{|}\varphi_{0}(s)\big{|}ds\lesssim
1.$ (2.8)

Case 1: $2^{N}\Psi(z)\leq 1.$ Using integrate by parts once, we have

$\begin{split}K_{N}^{\pm}(t,z)=&\frac{e^{-it2^{4N}s^{4}}\big{(}s^{-3}\varphi_{0}(s)e^{\pm
i2^{N}s\Psi(z)}\Phi(2^{N}s;z)\big{)}}{-4it\cdot
2^{4N}}\bigg{|}_{0}^{\infty}\\\ &\ \ \ +\frac{1}{4it\cdot
2^{4N}}\int_{0}^{\infty}e^{-it2^{4N}s^{4}}\partial_{s}\big{(}s^{-3}\varphi_{0}(s)e^{\pm
i2^{N}s\Psi(z)}\Phi(2^{N}s;z)\big{)}ds.\end{split}$

Noting that for any $s\in\text{supp}\ \varphi_{0}(s)$ and $k=0,1,$

$\big{|}\partial_{s}^{k}\big{(}e^{\pm i2^{N}s\Psi(z)}\big{)}\big{|}\lesssim
1,\ \ \ \big{|}\partial_{s}^{k}\big{(}\Phi(2^{N}s;z\big{)}\big{|}\lesssim 1,$

thus we can obtain that

$\big{|}K_{N}^{\pm}(t,z)\big{|}\lesssim\big{(}|t|\cdot 2^{4N}\big{)}^{-1}.$
(2.9)

Combining (2.8) with (2.9), we obtain that

$\sup_{z\in\mathbf{R}^{m}}\big{|}K_{N}^{\pm}(t,z)\big{|}\lesssim(1+|t|\cdot
2^{4N})^{-\frac{1}{2}}.$

Case 2: $2^{N}\Psi(z)\geq 1.$ Set $r=2^{N}\Psi(z).$ In this case, we rewrite
$K_{N}^{\pm}(t,z)$ as follows:

$K_{N}^{\pm}(t,z)=\int_{0}^{\infty}e^{-it2^{4N}s^{4}\pm
isr}\varphi_{0}(s)\Phi(2^{N}s;z)ds.$

Assume that $t>0,$ we consider $K_{N}^{-}(t,z)$ and $K_{N}^{+}(z)$ case by
case.

Firstly, we study $K_{N}^{-}(t,z)$. Let $u_{-}(s,r)=t\cdot 2^{4N}s^{4}+sr,$
then we have

$K_{N}^{-}(t,z)=\int_{0}^{\infty}e^{-iu_{-}(s,r)}\varphi_{0}(s)\Phi(2^{N}s;z)ds.$

It is easy to know that

$\big{|}\partial_{s}u_{-}(s,r)\big{|}=\big{|}4t\cdot
2^{4N}s^{3}+r\big{|}\gtrsim(1+|t|\cdot 2^{4N}).$

Thus $u_{-}(s,r)$ does not exist critical point, by (2.8) and integrating by
parts once, we have

$\big{|}K_{N}^{-}(t,z)\big{|}\lesssim(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$

Next to consider $K_{N}^{+}(t,z)$. Let $u_{+}(s,r)=t\cdot 2^{4N}s^{4}-sr,$
then we have

$K_{N}^{+}(t,z)=\int_{0}^{\infty}e^{-iu_{+}(s,r)}\varphi_{0}(s)\Phi(2^{N}s;z)ds.$

It is obvious that

$\partial_{s}u_{+}(s,r)=4t\cdot 2^{4N}s^{3}-r$

has a critical point $s_{0}$ such $r=4t\cdot 2^{4N}s_{0}^{3}.$

If $r\geq 8t\cdot 2^{4N}$ or $r\leq\frac{1}{32}t\cdot 2^{4N},$ then the such
critical point $s_{0}$ does not exist in the range $[1/4,1]$, so by (2.8) and
using integration by parts once, we have

$\big{|}K_{N}^{+}(t,z)\big{|}\lesssim(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$

If $\frac{1}{32}t\cdot 2^{4N}\leq r\leq 8t\cdot 2^{4N},$ then the critical
point exists. But since
$\big{|}\partial_{s}^{2}u_{+}(s,r)\big{|}=\big{|}12t\cdot
2^{4N}s^{2}\big{|}\gtrsim|t|\cdot 2^{4N},$ thus by Van der Corput lemma ( see
e.g. [Ste93]), we have

$\big{|}K_{N}^{+}(t,z)\big{|}\lesssim(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$

Hence, by summing all cases above we have that

$\sup_{z\in\mathbf{R}^{m}}\big{|}K_{N}^{\pm}(t,z)\big{|}\lesssim(1+|t|\cdot
2^{4N})^{-\frac{1}{2}},$

which concludes the proof of this lemma. ∎

Using Lemma 2.3, we can give the following decay estimates of
$e^{-it\Delta^{2}}$ in one dimensional case by the free resolvent
$R_{0}^{\pm}(\mu^{4}).$ Note that we can write down

$\displaystyle e^{-it\Delta^{2}}f(x)=\int_{\mathbf{R}}\Big{(}\frac{2}{\pi
i}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}[R_{0}^{+}(\mu^{4})-R_{0}^{-}(\mu^{4})](x,y)d\mu\Big{)}f(y)dy,$
(2.10)

where

$\begin{split}R_{0}^{\pm}(\mu^{4})(x,y)=&\frac{1}{4\mu^{3}}\Big{(}\pm ie^{\pm
i\mu|x-y|}-e^{-\mu|x-y|}\Big{)}=\frac{e^{\pm i\mu|x-y|}}{4\mu^{3}}\big{(}\pm
i-e^{-\mu|x-y|\mp i\mu|x-y|}\big{)}.\end{split}$

######  Proposition 2.4.

For any $N\in\mathbf{Z},$ we have

$\sup\limits_{x,y\in\mathbf{R}}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{N}(\mu)R_{0}^{\pm}(\mu^{4})(x,y)d\mu\Big{|}\lesssim
2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$ (2.11)

Furthermore, we have

$\sup\limits_{x,y\in\mathbf{R}}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}R_{0}^{\pm}(\mu^{4})(x,y)d\mu\Big{|}\lesssim|t|^{-\frac{1}{4}}.$
(2.12)

As a consequence, it follows that

$\big{\|}e^{-it\Delta^{2}}\big{\|}_{L^{1}(\mathbf{R})\rightarrow
L^{\infty}(\mathbf{R})}\lesssim|t|^{-\frac{1}{4}},\ \ t\neq 0.$ (2.13)

###### Proof.

Since $\varphi_{N}(\mu)=\varphi_{0}(2^{-N}\mu),$ thus set

$\begin{split}K_{0,N}^{\pm}(t;x,y):=\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{0}(2^{-N}\mu)[R_{0}^{\pm}(\mu^{4})](x,y)d\mu.\end{split}$
(2.14)

Let $\mu=2^{N}s,$ then we have

$\displaystyle K_{0,N}^{\pm}(t;x,y)=$ $\displaystyle
2^{N}\int_{0}^{\infty}e^{-it2^{4N}s^{4}}\varphi_{0}(s)e^{\pm
i2^{N}s|x-y|}\big{(}\pm i-e^{-2^{N}s|x-y|\mp i2^{N}s|x-y|}\big{)}ds.$

Let $z=(x,y),$ and

$\Psi(z)=\Psi(x,y)=|x-y|,\ \ \ \Phi^{\pm}(s;z)=\Phi^{\pm}(s;x,y)=\pm
i-e^{-s|x-y|\mp is|x-y|}.$

It is obvious that for any $s\in\text{supp}\ \varphi_{0}(s)$ and $k=0,1,$
$\big{|}\partial_{s}^{k}\big{(}\Phi^{\pm}(2^{N}s;x,y)\big{)}\big{|}\lesssim
1,$ then by Lemma 2.3, we obtain that

$\sup_{x,y\in\mathbf{R}}\big{|}K_{0,N}^{\pm}(t;x,y)\big{|}\lesssim
2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$

Denote

$K_{0}^{\pm}(t;x,y)=\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}R_{0}^{\pm}(\mu^{4})(x,y)d\mu.$

Thus

$\displaystyle K_{0}^{\pm}(t;x,y)=$
$\displaystyle\sum_{N=-\infty}^{\infty}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{N}(\mu)[R_{0}^{\pm}(\mu^{4})](x,y)d\mu$
$\displaystyle=$
$\displaystyle\displaystyle\sum_{N=-\infty}^{+\infty}K_{0,N}^{\pm}(t;x,y).$

It is obvious that there exists $N^{\prime}\in\mathbf{Z}$ such that $|t|\cdot
2^{4N^{\prime}}\sim 1,$ so we have

$\displaystyle\big{|}K_{0}^{\pm}(t;x,y)\big{|}\lesssim$
$\displaystyle\sum_{N=-\infty}^{N^{\prime}}\big{|}K_{0,N}^{\pm}(t;x,y)\big{|}+\sum_{N=N^{\prime}+1}^{+\infty}\big{|}K_{0,N}^{\pm}(t;x,y)\big{|}$
$\displaystyle\lesssim$
$\displaystyle\sum_{N=-\infty}^{N^{\prime}}2^{N}(1+|t|\cdot
2^{4N})^{-\frac{1}{2}}+\sum_{N=N^{\prime}+1}^{+\infty}2^{N}(1+|t|\cdot
2^{4N})^{-\frac{1}{2}}$ $\displaystyle\lesssim$
$\displaystyle\sum_{N=-\infty}^{N^{\prime}}2^{N}+|t|^{-\frac{1}{2}}\sum_{N=N^{\prime}+1}^{+\infty}2^{-N}$
$\displaystyle\lesssim$ $\displaystyle|t|^{-\frac{1}{4}},$

which immediately gives the (2.12). ∎

###  2.2. Low Energy Dispersive Estimate with $\alpha=0$

In this subsection, we show the low energy dispersive estimate (2.3) case by
case for zero is a regular point or zero energy resonance.

The following lemma play an important role to make use of cancellations from
projections operator $S_{j}$ in the zero asymptotical expansions of resolvent
$R_{V}(z)$ ( see Theorem 1.8 above ), and will be used repeatedly to obtain
low energy dispersive estimates for all cases.

######  Lemma 2.5.

Let $\mu>0$ and ${\rm sgn}(x)$ be the sign function of $x$ on $\mathbf{R}$.
Then

(i) If $F(p)\in C^{1}(\mathbf{R})$, then for any $x,y\in\mathbf{R}$, we have

$\begin{split}F(\mu|x-y|)=F(\mu|x|)-\mu y\int_{0}^{1}{\rm sgn}(x-\theta
y)F^{\prime}(\mu|x-\theta y|)d\theta.\end{split}$ (2.15)

(ii) If $F(p)\in C^{2}(\mathbf{R})$ and $F^{\prime}(0)=0$ , then for any
$x,y\in\mathbf{R}$, we have

$\begin{split}F(\mu|x-y|)=F(\mu|x|)-\mu y\ {\rm
sgn}(x)F^{\prime}(\mu|x|)+\mu^{2}y^{2}\int_{0}^{1}(1-\theta)F^{\prime\prime}(\mu|x-\theta
y|)d\theta.\end{split}$ (2.16)

(iii) If $F(p)\in C^{3}(\mathbf{R})$ and
$F^{\prime}(0)=F^{\prime\prime}(0)=0,$ then for any $x,y\in\mathbf{R}$, we
have

$\begin{split}F(\mu|x-y|)=&F(\mu|x|)-\mu y\ {\rm
sgn}(x)F^{\prime}(\mu|x|)+\frac{\mu^{2}y^{2}}{2!}F^{\prime\prime}(\mu|x|)\\\
&\ -\frac{\mu^{3}y^{3}}{2!}\int_{0}^{1}(1-\theta)^{2}\big{(}{\rm sgn}(x-\theta
y)\big{)}^{3}F^{(3)}(\mu|x-\theta y|)d\theta.\end{split}$ (2.17)

###### Proof.

Denote

$G_{\varepsilon}(y)=F\big{(}\mu\sqrt{\varepsilon^{2}+(x-y)^{2}}\big{)},\ \ \
\varepsilon\neq 0.$

If $F(p)\in C^{n+1}(\mathbf{R}),$ then $G_{\varepsilon}(y)\in
C^{n+1}(\mathbf{R})$ for $\varepsilon\neq 0.$ By Taylor expansions, we have

$G_{\varepsilon}(y)=\sum_{k=0}^{n}\frac{G_{\varepsilon}^{(k)}(0)}{k!}y^{k}+\frac{y^{n+1}}{n!}\int_{0}^{1}(1-\theta)^{n}G_{\varepsilon}^{(n+1)}(\theta
y)d\theta.$ (2.18)

(i) Since $F(p)\in C^{1}(\mathbf{R}),$ then
$\displaystyle\lim_{\varepsilon\rightarrow
0}G_{\varepsilon}(y)=\lim_{\varepsilon\rightarrow
0}F(\mu\sqrt{\varepsilon^{2}+(x-y)^{2}})=F(\mu|x-y|).$ Note that for
$\varepsilon\neq 0,$

$G^{\prime}_{\varepsilon}(y)=\frac{-\mu(x-y)}{\sqrt{(\varepsilon^{2}+(x-y)^{2})}}F^{\prime}\big{(}\mu\sqrt{\varepsilon^{2}+(x-y)^{2}}\big{)},$

then for each $0\leq\theta\leq 1$ and $x,y\in\mathbf{R}$,

$\displaystyle\lim_{\varepsilon\rightarrow 0}G^{\prime}_{\varepsilon}(\theta
y)=-\mu{\rm sgn}(x-\theta y)F^{\prime}\big{(}\mu|x-\theta y|\big{)}.$ (2.19)

Since there exists a constant $C=C(\mu,x,y)>0$ such that
$|G^{\prime}_{\varepsilon}(\theta y)|\leq C$ for all $0\leq\theta\leq 1$ and
$0<\varepsilon\leq 1$, then by taking $\varepsilon\downarrow 0$ in the two
sides of (2.18) with $n=0$ and Lebesgue’s dominated convergence theorem, it
immediately follows that

$F(\mu|x-y|)=F(\mu|x|)-\mu y\int_{0}^{1}{\rm sgn}(x-\theta
y)F^{\prime}(\mu|x-\theta y|)d\theta.$

(ii) If $F(p)\in C^{2}(\mathbf{R}),$ then for $\varepsilon\neq 0,$

$\displaystyle G^{\prime\prime}_{\varepsilon}(y)=$
$\displaystyle\mu^{2}\Big{(}\frac{x-y}{\sqrt{\varepsilon^{2}+(x-y)^{2}}}\Big{)}^{2}F^{\prime\prime}(\mu\sqrt{\varepsilon^{2}+(x-y)^{2}})$
$\displaystyle+\mu\frac{\varepsilon^{2}}{\big{(}\sqrt{\varepsilon^{2}+(x-y)^{2}}\big{)}^{2}}\cdot\frac{F^{\prime}(\mu\sqrt{\varepsilon^{2}+(x-y)^{2}})}{\sqrt{\varepsilon^{2}+(x-y)^{2}}}.$

Since $\displaystyle F^{\prime}(0)=0,$ thus we can obtain that

$\displaystyle\lim_{\varepsilon\rightarrow
0}G^{\prime\prime}_{\varepsilon}(y)=\mu^{2}F^{\prime\prime}\big{(}\mu|x-y|\big{)},\
x,y\in\mathbf{R}.$ (2.20)

Then by using (2.18) with $n=1$, (2.19), (2.20) and Lebesgue’s dominated
convergence theorem, we can conclude that

$\displaystyle F(\mu|x-y|)=$
$\displaystyle\sum_{k=0}^{1}\frac{\displaystyle\lim_{\varepsilon\rightarrow
0}G_{\varepsilon}^{(k)}(0)}{k!}y^{k}+y^{2}\cdot\lim_{\varepsilon\rightarrow
0}\int_{0}^{1}(1-\theta)G_{\varepsilon}^{{}^{\prime\prime}}(\theta y)d\theta.$
$\displaystyle=$ $\displaystyle F(\mu|x|)-\mu y{\rm
sgn}(x)F^{\prime}(\mu|x|)+\mu^{2}y^{2}\int_{0}^{1}(1-\theta)F^{\prime\prime}(\mu|x-\theta
y|)d\theta.$

(iii) If $F(p)\in C^{3}(\mathbf{R}),$ then for $\varepsilon\neq 0,$

$\displaystyle G^{(3)}_{\varepsilon}(y)=$
$\displaystyle-\mu^{3}\Big{(}\frac{x-y}{\sqrt{\varepsilon^{2}+(x-y)^{2}}}\Big{)}^{3}F^{(3)}(\mu\sqrt{\varepsilon^{2}+(x-y)^{2}})$
$\displaystyle-3\mu^{2}\frac{(x-y)\varepsilon^{2}}{\big{(}\sqrt{\varepsilon^{2}+(x-y)^{2}}\big{)}^{3}}\cdot\frac{F^{\prime\prime}(\mu\sqrt{\varepsilon^{2}+(x-y)^{2}})}{\sqrt{\varepsilon^{2}+(x-y)^{2}}}$
$\displaystyle+3\mu\frac{(x-y)\varepsilon^{2}}{\big{(}\sqrt{\varepsilon^{2}+(x-y)^{2}}\big{)}^{3}}\cdot\frac{F^{\prime}(\mu\sqrt{\varepsilon^{2}+(x-y)^{2}})}{\big{(}\sqrt{\varepsilon^{2}+(x-y)^{2}}\big{)}^{2}}.$

Then for each $x,y\in\mathbf{R},$ we have

$\displaystyle\lim_{\varepsilon\rightarrow
0}G^{(3)}_{\varepsilon}(y)=-\mu^{3}({\rm
sgn}(x-y))^{3}F^{(3)}\big{(}\mu|x-y|\big{)}.$ (2.21)

Since $F^{\prime}(0)=F^{\prime\prime}(0)=0,$ so by using the expansion (2.18)
with $n=2$, the equalities (2.19)-(2.21) and Lebesgue’s dominated convergence
theorem, we can similarly obtain that

$\displaystyle F(\mu|x-y|)=$
$\displaystyle\sum_{k=0}^{2}\frac{\displaystyle\lim_{\varepsilon\rightarrow
0}G_{\varepsilon}^{(k)}(0)}{k!}y^{k}+\frac{y^{3}}{2!}\cdot\lim_{\varepsilon\rightarrow
0}\int_{0}^{1}(1-\theta)^{2}G_{\varepsilon}^{(3)}(\theta y)d\theta$
$\displaystyle=$ $\displaystyle F(\mu|x|)-\mu y{\rm
sgn}(x)F^{\prime}(\mu|x|)+\frac{\mu^{2}y^{2}}{2!}F^{\prime\prime}(\mu|x|)$
$\displaystyle\ -\frac{\mu^{3}y^{3}}{2!}\int_{0}^{1}(1-\theta)^{2}\big{(}{\rm
sgn}(x-\theta y)\big{)}^{3}F^{(3)}(\mu|x-\theta y|)d\theta.$

∎

######  Lemma 2.6.

For any $0\leq\alpha<1,$ we have

$\sum_{N=-\infty}^{+\infty}2^{(1+\alpha)N}(1+|t|\cdot
2^{4N})^{-\frac{1}{2}}\lesssim|t|^{-\frac{1+\alpha}{4}}.$ (2.22)

###### Proof.

Since for any $t\neq 0,$ there exists a $N_{0}^{\prime}\in\mathbf{Z}$ such
that $|t|\cdot 2^{4N_{0}^{\prime}}\sim 1,$ thus we have

$\displaystyle\sum_{N=-\infty}^{+\infty}2^{(1+\alpha)N}(1+|t|\cdot
2^{4N})^{-\frac{1}{2}}=$
$\displaystyle\sum_{N=-\infty}^{N_{0}^{\prime}}2^{(1+\alpha)N}(1+|t|\cdot
2^{4N})^{-\frac{1}{2}}+\sum_{N=N_{0}^{\prime}+1}^{+\infty}2^{(1+\alpha)N}(1+|t|\cdot
2^{4N})^{-\frac{1}{2}}$ $\displaystyle\lesssim$
$\displaystyle\sum_{N=-\infty}^{N_{0}^{\prime}}2^{(1+\alpha)N}+|t|^{-\frac{1}{2}}\sum_{N=N_{0}^{\prime}+1}^{+\infty}2^{(\alpha-1)N}$
$\displaystyle\lesssim$ $\displaystyle|t|^{-\frac{1+\alpha}{4}}.$

∎

####  2.2.1. Regular Case

In order to establish the lower energy estimate (2.3) with $\alpha=0$ in
regular case, recall that by Stone’s formula ( also see the equality (2.2))

$\displaystyle e^{-itH}P_{ac}(H)\chi(H)f=$ $\displaystyle\frac{2}{\pi
i}\int_{0}^{\infty}e^{-it\mu^{4}}\chi(\mu)\mu^{3}[R_{V}^{+}(\mu^{4})-R_{V}^{-}(\mu^{4})]fd\mu$
(2.23) $\displaystyle=$
$\displaystyle\sum_{N=-\infty}^{N_{0}}\sum_{\pm}\frac{\pm 1}{2\pi
i}\int_{0}^{\infty}e^{-it\mu^{4}}\varphi_{N}(\mu)\mu^{3}R_{V}^{\pm}(\mu^{4})fd\mu.$
(2.24)

When zero is a regular point of the spectrum of $H$, by (1.20) and (1.21), we
have

$\begin{split}R_{V}^{\pm}(\mu^{4})=&R_{0}^{\pm}(\mu^{4})-R_{0}^{\pm}(\mu^{4})v\Big{(}S_{0}A_{01}^{0}S_{0}\Big{)}vR_{0}^{\pm}(\mu^{4})-R_{0}^{\pm}(\mu^{4})v\Big{(}\mu
QA_{11}^{0}Q\Big{)}vR_{0}^{\pm}(\mu^{4})\\\ &\ \
-R_{0}^{\pm}(\mu^{4})v\Big{(}\mu^{2}QA_{21}^{0}Q\Big{)}vR_{0}^{\pm}(\mu^{4})-R_{0}^{\pm}(\mu^{4})v\Big{(}\mu^{2}S_{0}A_{22}^{0}+\mu^{2}A_{23}^{0}S_{0}\Big{)}vR_{0}^{\pm}(\mu^{4})\\\
&\ \
-R_{0}^{\pm}(\mu^{4})v\Gamma_{3}^{0}(\mu)vR_{0}^{\pm}(\mu^{4}).\end{split}$
(2.25)

If we substitute the (2.25) into the (2.23), and combine Proposition 2.4 and
Lemma 2.6, then it suffices to show the following Propositions 2.7, 2.8, 2.9
and 2.10 corresponding to each different terms of the resolvent formula (2.25)
above.

######  Proposition 2.7.

For any $N\in\mathbf{Z}_{-}$ and $N\leq N_{0},$ we have

$\sup\limits_{x,y\in\mathbf{R}}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{N}(\mu)\Big{[}R_{0}^{\pm}(\mu^{4})v\big{(}S_{0}A_{01}^{0}S_{0}\big{)}vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu\Big{|}\lesssim
2^{2N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$ (2.26)

###### Proof.

Since $\varphi_{N}(\mu)=\varphi_{0}(2^{-N}\mu),$ thus

$\begin{split}K_{1,N}^{0,\pm}(t;x,y):=&\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{N}(\mu)\Big{[}R_{0}^{\pm}(\mu^{4})v\big{(}S_{0}A_{01}^{0}S_{0}\big{)}vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu\\\
=&\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{0}(2^{-N}\mu)\Big{[}R_{0}^{\pm}(\mu^{4})v\big{(}S_{0}A_{01}^{0}S_{0}\big{)}vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu.\end{split}$

Indeed, since

$R_{0}^{\pm}(x,y)=\frac{\pm ie^{\pm
i\mu|x-y|}-e^{-\mu|x-y|}}{4\mu^{3}}:=\frac{1}{4\mu^{3}}F^{\pm}(\mu|x-y|),$

and $(F^{\pm})^{\prime}(0)=0$ , then by Lemma (2.5)(ii) and using the fact
that $S_{0}(v)=S_{0}(xv(x))=0,$ we have

$\begin{split}\
\big{[}R_{0}^{\pm}(\mu^{4})v&\big{(}S_{0}A_{01}^{0}S_{0}\big{)}vR_{0}^{\pm}(\mu^{4})\big{]}(x,y)\\\
=&\frac{1}{16\mu^{6}}\int_{\mathbf{R}^{2}}F^{\pm}(\mu|x-y_{2}|)\big{[}vS_{0}A_{01}^{0}S_{0}v\big{]}(y_{2},y_{1})F^{\pm}(\mu|y-y_{1}|)dy_{1}dy_{2}\\\
=&\frac{1}{16\mu^{2}}\int_{\mathbf{R}^{2}}\Big{(}\int_{0}^{1}\int_{0}^{1}(1-\theta_{2})(1-\theta_{1})(F^{\pm})^{\prime\prime}(\mu|x-\theta_{2}y_{2}|)\times\\\
&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
(F^{\pm})^{\prime\prime}(\mu|y-\theta_{1}y_{1}|)d\theta_{1}d\theta_{2}\Big{)}y_{1}^{2}y_{2}^{2}\big{[}vS_{0}A_{01}^{0}S_{0}v\big{]}(y_{2},y_{1})dy_{1}dy_{2}.\end{split}$

Then we obtain that

$\begin{split}&\ K_{1,N}^{0,\pm}(t;x,y)\\\
=&\frac{1}{16}\int_{0}^{\infty}e^{-it\mu^{4}}\mu\varphi_{0}(2^{-N}\mu)\Bigg{[}\int_{\mathbf{R}^{2}}\Big{(}\int_{0}^{1}\int_{0}^{1}(1-\theta_{2})(1-\theta_{1})(F^{\pm})^{\prime\prime}(\mu|x-\theta_{2}y_{2}|)\\\
&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
(F^{\pm})^{\prime\prime}(\mu|y-\theta_{1}y_{1}|)d\theta_{1}d\theta_{2}\Big{)}y_{1}^{2}y_{2}^{2}\big{[}vS_{0}A_{01}^{0}S_{0}v\big{]}(y_{2},y_{1})dy_{1}dy_{2}\Bigg{]}d\mu\\\
=&\frac{1}{16}\int_{\mathbf{R}^{2}}\Bigg{[}\int_{0}^{1}\int_{0}^{1}\bigg{(}\int_{0}^{\infty}e^{-it\mu^{4}}\mu\varphi_{0}(2^{-N}\mu)(F^{\pm})^{\prime\prime}(\mu|x-\theta_{2}y_{2}|)(F^{\pm})^{\prime\prime}(\mu|y-\theta_{1}y_{1}|)d\mu\bigg{)}\\\
&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
(1-\theta_{1})(1-\theta_{2})d\theta_{1}d\theta_{2}\Bigg{]}y_{1}^{2}y_{2}^{2}\big{[}vS_{0}A_{01}^{0}S_{0}v\big{]}(y_{2},y_{1})dy_{1}dy_{2}.\\\
\end{split}$ (2.27)

Denote

$\displaystyle\ E_{1,N}^{0,\pm}(t;x,y,\theta_{1},\theta_{2},y_{1},y_{2})$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}e^{-it\mu^{4}}\mu\varphi_{0}(2^{-N}\mu)(F^{\pm})^{\prime\prime}(\mu|x-\theta_{2}y_{2}|)(F^{\pm})^{\prime\prime}(\mu|y-\theta_{1}y_{1}|)d\mu.$

It is obvious that

$\Big{|}K_{1,N}^{0,\pm}(t;x,y)\Big{|}\lesssim\int_{\mathbf{R}^{2}}\Big{(}\int_{0}^{1}\int_{0}^{1}\big{|}E_{1,N}^{0,\pm}(t;x,y,\theta_{1},\theta_{2},y_{1},y_{2})\big{|}d\theta_{1}d\theta_{2}\Big{)}\big{|}[vS_{0}A_{0}^{0}S_{0}v](y_{2},y_{1})\big{|}y_{1}^{2}y_{2}^{2}dy_{1}dy_{2}.$

Now we estimate $E_{1,N}^{0,\pm}(t;x,y,\theta_{1},\theta_{2},y_{1},y_{2}).$
Since

$(F^{\pm})^{\prime\prime}(\mu|x-\theta_{2}y_{2}|)=\mp ie^{\pm
i\mu|x-\theta_{2}y_{2}|}-e^{-\mu|x-\theta_{2}y_{2}|}:=e^{\pm
i\mu|x-\theta_{2}y_{2}|}\mathcal{F}^{\pm}(\mu|x-\theta_{2}y_{2}|),$
$(F^{\pm})^{\prime\prime}(\mu|y-\theta_{1}y_{1}|)=\mp ie^{\pm
i\mu|y-\theta_{1}y_{1}|}-e^{-\mu|y-\theta_{1}y_{1}|}:=e^{\pm
i\mu|y-\theta_{1}y_{1}|}\mathcal{F}^{\pm}(\mu|y-\theta_{1}y_{1}|),$

and let $\mu=2^{N}s,$ we can rewrite
$E_{1,N}^{0,\pm}(t;x,y,\theta_{1},\theta_{2},y_{1},y_{2})$ as follows:

$\begin{split}E_{1,N}^{0,\pm}(t;x,y,\theta_{1},\theta_{2},y_{1},y_{2})=&2^{2N}\int_{0}^{\infty}e^{-it2^{4N}s^{4}}s\varphi_{0}(s)e^{\pm
i2^{N}s|x-\theta_{2}y_{2}|}\mathcal{F}^{\pm}(2^{N}s|x-\theta_{2}y_{2}|)\\\ &\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \times e^{\pm
i2^{N}s|y-\theta_{1}y_{1}|}\mathcal{F}^{\pm}(2^{N}s|y-\theta_{1}y_{1}|)ds.\end{split}$

Noting that for any $s\in\text{supp}\ \varphi_{0}(s)$ and $k=0,1,$ we have

$\Big{|}\partial_{s}^{k}\mathcal{F}^{\pm}(2^{N}s|x-\theta_{2}y_{2}|)\Big{|}\lesssim
1,\ \ \ \ \ \
\Big{|}\partial_{s}^{k}\mathcal{F}^{\pm}(2^{N}s|y-\theta_{1}y_{1}|)\Big{|}\lesssim
1,$

Using Lemma 2.3 with $z=(x,y,y_{1},y_{2},\theta_{1},\theta_{2})$ and

$\Psi(z)=|x-\theta_{2}y_{2}|+|y-\theta_{1}y_{1}|,\ \ \ \
\Phi^{\pm}(2^{N}s;z)=\mathcal{F}^{\pm}(2^{N}s|x-\theta_{2}y_{2}|)\cdot\mathcal{F}^{\pm}(2^{N}s|y-\theta_{1}y_{1}|),$

we obtain that

$\sup_{x,y\in\mathbf{R}}\big{|}E_{1,N}^{0,\pm}(t;x,y,\theta_{1},\theta_{2},y_{1},y_{2})\big{|}\lesssim
2^{2N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$

Noting that $v(x)\lesssim(1+|x|)^{-\frac{13}{2}-}$ in this case, thus we have

$\displaystyle\big{|}K_{1,N}^{0,\pm}(t;x,y)\big{|}\lesssim$ $\displaystyle
2^{2N}(1+|t|\cdot
2^{4N})^{-\frac{1}{2}}\int_{\mathbf{R}^{2}}y_{2}^{2}|v(y_{2})|\big{|}[S_{0}A_{0}^{0}S_{0}](y_{2},y_{1})\big{|}|v(y_{1})|y_{1}^{2}dy_{1}dy_{2}$
$\displaystyle\lesssim$ $\displaystyle 2^{2N}(1+|t|\cdot
2^{4N})^{-\frac{1}{2}}\|y_{2}^{2}v(y_{2})\|_{L^{2}_{y_{2}}}\cdot\|S_{0}A_{01}^{0}S_{0}\|_{L^{2}\rightarrow
L_{2}}\cdot\|y_{1}^{2}v(y_{1})\|_{L^{2}_{y_{1}}}$ $\displaystyle\lesssim$
$\displaystyle 2^{2N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}$

uniformly in $x,y.$ ∎

######  Proposition 2.8.

For any $N\in\mathbf{Z}_{-}$ and $N\leq N_{0}$, we have

$\sup\limits_{x,y\in\mathbf{R}}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{N}(\mu)\Big{[}R_{0}^{\pm}(\mu^{4})v\big{(}\mu
QA_{11}^{0}Q\big{)}vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu\Big{|}\lesssim
2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$ (2.28)

###### Proof.

Since $\varphi_{N}(\mu)=\varphi_{0}(2^{-N}\mu)$, thus

$\begin{split}K_{2,N}^{0,\pm}(t;x,y):=&\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{0}(2^{-N}\mu)\Big{[}R_{0}^{\pm}(\mu^{4})v\big{(}\mu
QA_{11}^{0}Q\big{)}vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu.\end{split}$

It suffices to prove that

$\sup_{x,y}\Big{|}K_{2,N}^{0,\pm}(t;x,y)\Big{|}\lesssim 2^{N}(1+|t|\cdot
2^{4N})^{-\frac{1}{2}}.$

Indeed, since

$R_{0}^{\pm}(x,y)=\frac{\pm ie^{\pm
i\mu|x-y|}-e^{-\mu|x-y|}}{4\mu^{3}}:=\frac{1}{4\mu^{3}}F^{\pm}(\mu|x-y|),$

by Lemma (2.5)(i) and using the fact that $Q(v)=0,$ we have

$\begin{split}&\ \Big{[}R_{0}^{\pm}(\mu^{4})v\big{(}\mu
QA_{11}^{0}Q\big{)}vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)\\\
=&\frac{1}{16\mu^{5}}\int_{\mathbf{R}^{2}}F(\mu|x-y_{2}|)\big{[}vQA_{11}^{0}Qv\big{]}(y_{2},y_{1})F(\mu|y-y_{1}|)dy_{1}dy_{2}\\\
=&\frac{1}{16\mu^{3}}\int_{\mathbf{R}^{2}}\Big{(}\int_{0}^{1}\int_{0}^{1}(F^{\pm})^{\prime}(\mu|x-\theta_{2}y_{2}|){\rm
sgn}(x-\theta_{2}y_{2})\big{[}vQA_{11}^{0}Qv\big{]}(y_{2},y_{1})\\\ &\ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
(F^{\pm})^{\prime}(\mu|y-\theta_{1}y_{1}|){\rm
sgn}(y-\theta_{1}y_{1})d\theta_{1}d\theta_{2}\Big{)}y_{1}y_{2}dy_{1}dy_{2}.\end{split}$

Then we obtain that

$\begin{split}&\ K_{2,N}^{0,\pm}(t;x,y)\\\
=&\frac{1}{16}\int_{\mathbf{R}^{2}}\Bigg{[}\int_{0}^{1}\int_{0}^{1}{\rm
sgn}(y-\theta_{1}y_{1}){\rm
sgn}(x-\theta_{2}y_{2})\Big{(}\int_{0}^{\infty}e^{-it\mu^{4}}\varphi_{0}(2^{-N}\mu)(F^{\pm})^{\prime}(\mu|x-\theta_{2}y_{2}|)\\\
&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
(F^{\pm})^{\prime}(\mu|y-\theta_{1}y_{1}|)d\mu\Big{)}d\theta_{1}d\theta_{2}\Bigg{]}\big{[}vQA_{11}^{0}Qv\big{]}(y_{2},y_{1})y_{1}y_{2}dy_{1}dy_{2}.\\\
\end{split}$ (2.29)

Denote

$\displaystyle\
E_{2,N}^{0,\pm}(t;x,y,\theta_{1},\theta_{2},y_{1},y_{2})=\int_{0}^{\infty}e^{-it\mu^{4}}\varphi_{0}(2^{-N}\mu)(F^{\pm})^{\prime}(\mu|x-\theta_{2}y_{2}|)(F^{\pm})^{\prime}(\mu|y-\theta_{1}y_{1}|)d\mu.$

Then

$\Big{|}K_{2,N}^{0,\pm}(t;x,y)\Big{|}\lesssim\int_{\mathbf{R}^{2}}\Big{(}\int_{0}^{1}\int_{0}^{1}\big{|}E_{2,N}^{0,\pm}(t;x,y,\theta_{1},\theta_{2},y_{1},y_{2})\big{|}d\theta_{1}d\theta_{2}\Big{)}\big{|}[vQA_{11}^{0}Qv]\big{|}(y_{2},y_{1})\
|y_{1}y_{2}|dy_{1}dy_{2}.$

Next to estimate $E_{2,N}^{0,\pm}(t;x,y,\theta_{1},\theta_{2},y_{1},y_{2}).$
Since

$(F^{\pm})^{\prime}(\mu|x-\theta_{2}y_{2}|)=-e^{\pm
i\mu|x-\theta_{2}y_{2}|}+e^{-\mu|x-\theta_{2}y_{2}|}:=e^{\pm
i\mu|x-\theta_{2}y_{2}|}\mathcal{F}^{\pm}(\mu|x-\theta_{2}y_{2}|),$
$(F^{\pm})^{\prime}(\mu|y-\theta_{1}y_{1}|)=-e^{\pm
i\mu|y-\theta_{1}y_{1}|}+e^{-\mu|y-\theta_{1}y_{1}|}:=e^{\pm
i\mu|y-\theta_{1}y_{1}|}\mathcal{F}^{\pm}(\mu|y-\theta_{1}y_{1}|),$

and let $\mu=2^{N}s,$ we can rewrite
$E_{2,N}^{0,\pm}(t;x,y,\theta_{1},\theta_{2},y_{1},y_{2})$ as follows:

$\begin{split}E_{2,N}^{0,\pm}(t;x,y,\theta_{1},\theta_{2},y_{1},y_{2})=&2^{N}\int_{0}^{\infty}e^{-it2^{4N}s^{4}}\varphi_{0}(s)e^{\pm
i2^{N}s|x-\theta_{2}y_{2}|}\mathcal{F}^{\pm}(2^{N}s|x-\theta_{2}y_{2}|)\\\ &\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e^{\pm
i2^{N}s|y-\theta_{1}y_{1}|}\mathcal{F}^{\pm}(2^{N}s|y-\theta_{1}y_{1}|)ds.\end{split}$

Noting that for any $s\in\text{supp}\ \varphi_{0}(s)$ and $k=0,1,$ we have

$\big{|}\partial_{s}^{k}\mathcal{F}^{\pm}(2^{N}s|x-\theta_{2}y_{2}|)\big{|}\lesssim
1,\ \ \ \ \ \
\big{|}\partial_{s}^{k}\mathcal{F}^{\pm}(2^{N}s|y-\theta_{1}y_{1}|)\big{|}\lesssim
1,$

then using Lemma 2.3 with $z=(x,y,y_{1},y_{2},\theta_{1},\theta_{2})$ and

$\Psi(z)=|x-\theta_{2}y_{2}|+|y-\theta_{1}y_{1}|,\ \ \ \
\Phi^{\pm}(2^{N}s;z)=\mathcal{F}^{\pm}(2^{N}s|x-\theta_{2}y_{2}|)\cdot\mathcal{F}^{\pm}(2^{N}s|y-\theta_{1}y_{1}|),$

we have

$\sup_{x,y\in\mathbf{R}}\big{|}E_{2,N}^{0,\pm}(t;x,y,\theta_{1},\theta_{2},y_{1},y_{2})\big{|}\lesssim
2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$

Furthermore, since $v(x)\lesssim(1+|x|)^{-\frac{13}{2}-},$ we have

$\displaystyle\big{|}K_{2,N}^{0,\pm}(t;x,y)\big{|}\lesssim$ $\displaystyle
2^{N}(1+|t|\cdot
2^{4N})^{-\frac{1}{2}}\int_{\mathbf{R}^{2}}|y_{2}v(y_{2})|\big{|}[QA_{11}^{0}Q](y_{2},y_{1})\big{|}|y_{1}v(y_{1})|dy_{1}dy_{2}$
$\displaystyle\lesssim$ $\displaystyle 2^{N}(1+|t|\cdot
2^{4N})^{-\frac{1}{2}}\|y_{2}v(y_{2})\|_{L^{2}_{y_{2}}}\cdot\|QA_{11}^{0}Q\|_{L^{2}\rightarrow
L_{2}}\cdot\|y_{1}v(y_{1})\|_{L^{2}_{y_{1}}}$ $\displaystyle\lesssim$
$\displaystyle 2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}$

uniformly in $x,y.$ ∎

By using the method in the proof of Proposition 2.7 and Proposition 2.8, it is
easy similarly to obtain the following estimates ( we omit the details of
calculations here ).

######  Proposition 2.9.

For any $N\in\mathbf{Z}_{-}$ and $N\leq N_{0}$, we have

$\sup\limits_{x,y}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{N}(\mu)\Big{[}R_{0}^{\pm}(\mu^{4})v\big{(}\mu^{2}QA_{21}^{0}Q\big{)}vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu\Big{|}\lesssim
2^{2N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$
$\sup\limits_{x,y}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{N}(\mu)\Big{[}R_{0}^{\pm}(\mu^{4})v\big{(}\mu^{2}S_{0}A_{22}^{0}\big{)}vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu\Big{|}\lesssim
2^{2N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$
$\sup\limits_{x,y}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{N}(\mu)\Big{[}R_{0}^{\pm}(\mu^{4})v\big{(}\mu^{2}A_{23}^{0}S_{0}\big{)}vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu\Big{|}\lesssim
2^{2N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$

Finally, for the last remained term
$R_{0}^{\pm}(\mu^{4})v\Gamma_{3}^{0}(\mu)vR_{0}^{\pm}(\mu^{4})$ in (2.25), we
have the following estimates.

######  Proposition 2.10.

For any $N\in\mathbf{Z}_{-}$ and $N\leq N_{0}$, we have

$\sup\limits_{x,y}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{N}(\mu)\Big{[}R_{0}^{\pm}(\mu^{4})v\Gamma_{3}^{0}(\mu)vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu\Big{|}\lesssim
2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$ (2.30)

###### Proof.

Indeed, it suffices to prove that for any $f,g\in L^{1}(\mathbf{R}),$

$\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{N}(\mu)\big{\langle}[v\Gamma_{3}^{0}(\mu)v]R_{0}^{\pm}(\mu^{4})f,(R_{0}^{\pm})^{*}(\mu^{4})g\big{\rangle}d\mu\Big{|}\lesssim
2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}\|f\|_{L^{1}}\cdot\|g\|_{L^{1}}.$

Equivalently, it is enough to bound the following kernel uniformly by
$O(2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}})$ :

$\displaystyle K_{4,N}^{0,\pm}(t;x,y)=$
$\displaystyle\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{N}(\mu)\Big{\langle}[v\Gamma_{3}^{0}(\mu)v]\big{(}R_{0}^{\pm}(\mu^{4})(\ast,y)\big{)}(\cdot),R_{0}^{\mp}(\mu^{4})(x,\cdot)\Big{\rangle}d\mu$

Noting that

$\displaystyle R_{0}^{\pm}(\mu^{4})(x,y)=$
$\displaystyle\frac{1}{4\mu^{3}}\big{(}\pm ie^{\pm
i\mu|x-y|}-e^{-\mu|x-y|}\big{)}:=\frac{e^{\pm
i\mu|x-y|}}{4\mu^{3}}\mathcal{R}_{0}^{\pm}(\mu)(x,y),$

thus, we have

$\begin{split}&\ \
\Big{\langle}[v\Gamma_{3}^{0}(\mu)v]\big{(}R_{0}^{\pm}(\mu^{4})(\ast,y)\big{)}(\cdot),R_{0}^{\mp}(\mu^{4})(x,\cdot)\Big{\rangle}\\\
=&\frac{1}{16\mu^{6}}\Big{\langle}[v\Gamma_{3}^{0}(\mu)v]\big{(}e^{\pm
i\mu|\ast-y|}\mathcal{R}_{0}^{\pm}(\mu)(\ast,y)\big{)}(\cdot),\big{(}e^{\mp
i\mu|x-\cdot|}\mathcal{R}_{0}^{\mp}(\mu)(x,\cdot)\big{)}\Big{\rangle}\\\
=&\frac{1}{16\mu^{6}}e^{\pm i\mu|x|}e^{\pm
i\mu|y|}\Big{\langle}[v\Gamma_{3}^{0}(\mu)v]\big{(}e^{\pm
i\mu(|\ast-y|-|y|)}\mathcal{R}_{0}^{\pm}(\mu)(\ast,y)\big{)}(\cdot),\big{(}e^{\mp
i\mu(|x-\cdot|-|x|)}\mathcal{R}_{0}^{\mp}(\mu)(x,\cdot)\big{)}\Big{\rangle}\\\
:=&\frac{1}{16\mu^{6}}e^{\pm
i\mu(|x|+|y|)}E_{3,N}^{0,\pm}(\mu;x,y).\end{split}$

Let $\mu=2^{N}s$, we have

$\begin{split}K_{4,N}^{0,\pm}(t;x,y)=&2^{-2N}\int_{0}^{\infty}e^{-it2^{4N}s^{4}}\varphi_{0}(s)s^{-3}e^{\pm
i2^{N}s(|x|+|y|)}E_{3,N}^{0,\pm}(2^{N}s;x,y)ds.\end{split}$

Noting that for any $s\in\text{supp}\ \varphi_{0}(s)$ and any
$N\in\mathbf{Z}_{-}$, we have

$\Big{|}\partial_{s}^{k}\Big{(}e^{\pm
i2^{N}s(\ast-y|-|y|)}\mathcal{R}_{0}^{\pm}(2^{N}s)(\ast,y)\Big{)}\Big{|}\lesssim\langle\ast\rangle^{k},\
\ k=0,1,$ $\Big{|}\partial_{s}^{k}\Big{(}e^{\pm
i2^{N}s(|x-\cdot|-|x|)}\mathcal{R}_{0}^{\pm}(2^{N}s)(x,\cdot)\Big{)}\Big{|}\lesssim\langle\cdot\rangle^{k},\
\ k=0,1,$

and since $\|\Gamma_{3}^{0}(\mu)\|_{L^{2}\rightarrow L^{2}}=O_{1}(\mu^{3})$
for $\mu<<1$ from Theorem 1.8 ( e.g. choosing $\mu<2^{N_{0}}$ for some large
$N_{0}\in\mathbf{Z_{-}}$ ), then we have

$\|\partial_{s}^{k}(\Gamma_{3}^{0}(2^{N}s))\|_{L^{2}\rightarrow L^{2}}\lesssim
2^{3N}s^{3-k},\ \ \ k=0,1.$

Furthermore, since

$\displaystyle\ \partial_{s}E_{3,N}^{0,\pm}(2^{N}s;x,y)$ $\displaystyle=$
$\displaystyle\Big{\langle}[v\partial_{s}\Gamma_{3}^{0}(2^{N}s)v]\big{(}e^{\pm
i2^{N}s(|\ast-y|-|y|)}\mathcal{R}_{0}^{\pm}(2^{N}s)(\ast,y)\big{)}(\cdot),\big{(}e^{\mp
i2^{N}s(|x-\cdot|-|x|)}\mathcal{R}_{0}^{\mp}(2^{N}s)(x,\cdot)\big{)}\Big{\rangle}$
$\displaystyle\
+\Big{\langle}[v\Gamma_{3}^{0}(2^{N}s)v]\partial_{s}\big{(}e^{\pm
i2^{N}s(|\ast-y|-|y|)}\mathcal{R}_{0}^{\pm}(2^{N}s)(\ast,y)\big{)}(\cdot),\big{(}e^{\mp
i2^{N}s(|x-\cdot|-|x|)}\mathcal{R}_{0}^{\mp}(2^{N}s)(x,\cdot)\big{)}\Big{\rangle}$
$\displaystyle\ +\Big{\langle}[v\Gamma_{3}^{0}(2^{N}s)v]\big{(}e^{\pm
i2^{N}s(|\ast-y|-|y|)}\mathcal{R}_{0}^{\pm}(2^{N}s)(\ast,y)\big{)}(\cdot),\partial_{s}\big{(}e^{\mp
i2^{N}s(|x-\cdot|-|x|)}\mathcal{R}_{0}^{\mp}(2^{N}s)(x,\cdot)\big{)}\Big{\rangle},$

and $v(x)\lesssim(1+|x|)^{-\frac{13}{2}-},$ then by Hölder inequality, for any
$s\in\text{supp}\ \varphi_{0}(s),$ we have

$\big{|}\partial_{s}^{k}E_{3,N}^{0,\pm}(2^{N}s;x,y)\big{|}\lesssim
2^{3N}s^{3-k},\ \ \ \ k=0,1.$

Using again Lemma 2.3 with $z=(x,y)$, and

$\Psi(z)=|x|+|y|,\ \ \
\Phi^{\pm}(2^{N}s;z)=2^{-3N}E_{3,N}^{0,\pm}(2^{N}s;x,y),$

we have

$\sup_{x,y\in\mathbf{R}}\big{|}K_{4,N}^{0,\pm}(t;x,y)\big{|}\lesssim
2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$

∎

####  2.2.2. First Kind of Resonance

If there is first kind of resonance at zero, using (1.20) and (1.22), we have

$\begin{split}R_{V}^{\pm}(\mu^{4})=&R_{0}^{\pm}(\mu^{4})-R_{0}^{\pm}(\mu^{4})v\Big{(}\mu^{-1}S_{1}A_{-1,1}^{1}S_{1}\Big{)}vR_{0}^{\pm}(\mu^{4})-R_{0}^{\pm}(\mu^{4})v\Big{(}S_{0}A_{01}^{1}Q+QA_{02}^{1}S_{0}\Big{)}vR_{0}^{\pm}(\mu^{4})\\\
&\ \ -R_{0}^{\pm}(\mu^{4})v\Big{(}\mu
QA_{11}^{1}Q\Big{)}vR_{0}^{\pm}(\mu^{4})-R_{0}^{\pm}(\mu^{4})v\Big{(}\mu
S_{0}A_{12}^{1}+\mu A_{13}^{1}S_{0}\Big{)}vR_{0}^{\pm}(\mu^{4})\\\ &\ \
-R_{0}^{\pm}(\mu^{4})v\Big{(}\mu^{2}QA_{21}^{1}+\mu^{2}A_{22}^{1}Q\Big{)}vR_{0}^{\pm}(\mu^{4})-R_{0}^{\pm}(\mu^{4})v\Gamma_{3}^{1}(\mu)vR_{0}^{\pm}(\mu^{4}).\end{split}$
(2.31)

By checking the analysis of Regular Case, we only need to establish the
following two propositions.

######  Proposition 2.11.

For any $N\in\mathbf{Z}_{-}$ and $N\leq N_{0},$ we have

$\sup\limits_{x,y}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{N}(\mu)\Big{[}R_{0}^{\pm}(\mu^{4})v\big{(}\mu^{-1}S_{1}A_{-1}^{1}S_{1}\big{)}vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu\Big{|}\lesssim
2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$ (2.32)

###### Proof.

Since $S_{1}$ has the same orthogonality relationships with $S_{0},$ thus the
proof is the same as the proof of Proposition 2.7. ∎

According to the orthogonality of $S_{1}$ and $Q,$ using Lemma 2.5, it is easy
to prove the following estimates.

######  Proposition 2.12.

For any $N\in\mathbf{Z}_{-}$ and $N\leq N_{0}$, we have

$\sup\limits_{x,y}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{N}(\mu)\Big{[}R_{0}^{\pm}(\mu^{4})v\big{(}S_{0}A_{01}^{1}Q\big{)}vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu\Big{|}\lesssim
2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$ (2.33)
$\sup\limits_{x,y}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{N}(\mu)\Big{[}R_{0}^{\pm}(\mu^{4})v\big{(}QA_{01}^{1}S_{0}\big{)}vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu\Big{|}\lesssim
2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$ (2.34)

####  2.2.3. Second Kind of Resonance

If there is second kind of resonance at zero, using (1.20) and (1.23), we have

$\begin{split}&R_{V}^{\pm}(\mu^{4})=\\\
&R_{0}^{\pm}(\mu^{4})-R_{0}^{\pm}(\mu^{4})v\Big{(}\mu^{-3}S_{2}A_{-3,1}^{2}S_{2}\Big{)}vR_{0}^{\pm}(\mu^{4})-R_{0}^{\pm}(\mu^{4})v\mu^{-2}\Big{(}S_{2}A_{-2,1}^{2}S_{0}+S_{0}A_{-2,2}^{2}S_{2}\Big{)}vR_{0}^{\pm}(\mu^{4})\\\
&\ \
-R_{0}^{\pm}(\mu^{4})v\Big{(}\mu^{-1}S_{0}A_{-1,1}^{2}S_{0}\Big{)}vR_{0}^{\pm}(\mu^{4})-R_{0}^{\pm}(\mu^{4})v\Big{(}\mu^{-1}S_{2}A_{-1,2}^{2}Q+\mu^{-1}QA_{-1,2}^{2}S_{2}\Big{)}vR_{0}^{\pm}(\mu^{4})\\\
&\ \
-R_{0}^{\pm}(\mu^{4})v\Big{(}QA_{01}^{2}S_{0}+S_{0}A_{02}^{2}Q\Big{)}vR_{0}^{\pm}(\mu^{4})-R_{0}^{\pm}(\mu^{4})v\Big{(}S_{2}A_{03}^{2}+A_{04}^{2}S_{2}\Big{)}vR_{0}^{\pm}(\mu^{4})\\\
&\ \ -R_{0}^{\pm}(\mu^{4})v(\mu
QA_{11}^{2}Q)vR_{0}^{\pm}(\mu^{4})-R_{0}^{\pm}(\mu^{4})v(\mu
S_{0}A_{12}^{2}+\mu A_{13}^{2}S_{0})vR_{0}^{\pm}(\mu^{4})\\\ &\ \
-R_{0}^{\pm}(\mu^{4})v(\mu^{2}QA_{21}^{2}+\mu^{2}A_{22}^{2}Q)vR_{0}^{\pm}(\mu^{4})-R_{0}^{\pm}(\mu^{4})v\Gamma_{3}^{2}(\mu)vR_{0}^{\pm}(\mu^{4}).\end{split}$
(2.35)

By combing the analysis of Regular Case and First Kind of Resonance, it is
suffices to establish the following estimates.

######  Proposition 2.13.

For any $N\in\mathbf{Z}_{-}$ and $N\leq N_{0},$ we have

$\sup\limits_{x,y}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\varphi_{N}(\mu)\mu^{3}\Big{[}R_{0}^{\pm}(\mu^{4})v\big{(}\mu^{-3}S_{2}A_{-3,1}^{2}S_{2}\big{)}vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu\Big{|}\lesssim
2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}},$ (2.36)
$\sup\limits_{x,y}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\varphi_{N}(\mu)\mu^{3}\Big{[}R_{0}^{\pm}(\mu^{4})v\big{(}\mu^{-2}S_{2}A_{-2,1}^{2}S_{0}\big{)}vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu\Big{|}\lesssim
2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$ (2.37)
$\sup\limits_{x,y}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\varphi_{N}(\mu)\mu^{3}\Big{[}R_{0}^{\pm}(\mu^{4})v\big{(}\mu^{-2}S_{0}A_{-2,2}^{2}S_{2}\big{)}vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu\Big{|}\lesssim
2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$ (2.38)

###### Proof.

We only prove (2.36) here since the other two estimates can be proved by the
similar way. Note that

$\begin{split}K_{1,N}^{2,\pm}(t;x,y):=&\int_{0}^{\infty}e^{-it\mu^{4}}\varphi_{N}(\mu)\mu^{3}\Big{[}R_{0}^{\pm}(\mu^{4})v\big{(}\mu^{-3}S_{2}A_{-3,1}^{2}S_{2}\big{)}vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu.\end{split}$

It suffices to prove that

$\sup_{x,y}\Big{|}K_{1,N}^{2,\pm}(t;x,y)\Big{|}\lesssim 2^{N}(1+|t|\cdot
2^{4N})^{-\frac{1}{2}}.$

Noting that

$R_{0}^{\pm}(x,y)=\frac{\pm ie^{\pm i\mu|x-y|}-e^{-\mu|x-y|}}{4\mu^{3}},$

by Lemma 2.5(iii) and using the fact that
$S_{2}(v)=S_{2}(xv(x))=S_{2}(x^{2}v(x))=0,$ we have

$\begin{split}&\Big{[}R_{0}^{\pm}(\mu^{4})v\big{(}\mu^{-3}S_{2}A_{-3,1}^{2}S_{2}\big{)}vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)\\\
=&\frac{1}{16\mu^{9}}\int_{\mathbf{R}^{2}}\Big{(}\pm ie^{\pm
i\mu|x-y_{2}|}-e^{-\mu|x-y_{2}|}+\frac{1\pm
i}{2}\mu^{2}|x-y_{2}|^{2}\Big{)}[vS_{2}A_{-3,1}^{2}S_{2}v](y_{2},y_{1})\\\ &\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big{(}\pm ie^{\pm
i\mu|y-y_{1}|}-e^{-\mu|y-y_{1}|}+\frac{1\pm
i}{2}\mu^{2}|y-y_{1}|^{2}\Big{)}dy_{1}dy_{2}\\\
:=&\frac{1}{16\mu^{9}}\int_{\mathbf{R}^{2}}F^{\pm}(\mu|x-y_{2}|)[vS_{2}A_{-3,1}^{2}S_{2}v](y_{2},y_{1})F^{\pm}(\mu|y-y_{1}|)dy_{1}dy_{2},\end{split}$

where $\displaystyle F^{\pm}(p)=\pm ie^{\pm ip}-e^{-p}+\frac{1\pm i}{2}p^{2}.$

It is obvious that $F^{\pm}(p)\in C^{\infty}(\mathbf{R})$ and
$\big{(}F^{\pm}\big{)}^{(k)}(0)=0\ (k=1,2),$ then by (2.17) and using the fact
that $S_{2}(v)=S_{2}(xv(x))=S_{2}(x^{2}v(x))=0,$ we obtain that

$\displaystyle\Big{[}R_{0}^{\pm}(\mu^{4})v\big{(}\mu^{-3,1}S_{2}A_{-3}^{2}S_{2}\big{)}vR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)$
$\displaystyle=$
$\displaystyle\frac{1}{16\mu^{9}}\int_{\mathbf{R}^{2}}F^{\pm}(\mu|x-y_{2}|)[vS_{2}A_{-3,1}^{2}S_{2}v](y_{2},y_{1})F^{\pm}(\mu|y-y_{1}|)dy_{1}dy_{2}$
$\displaystyle=$
$\displaystyle\frac{1}{16(2!)^{2}\mu^{3}}\int_{\mathbf{R}^{2}}\Big{(}\int_{0}^{1}\int_{0}^{1}(1-\theta_{1})^{2}(1-\theta_{2})^{2}(F^{\pm})^{(3)}(\mu|x-\theta_{2}y_{2}|)\big{(}{\rm
sgn}(x-\theta_{2}y_{2})\big{)}^{3}$ $\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \
\ (F^{\pm})^{(3)}(\mu|y-\theta_{1}y_{1}|)\big{(}{\rm
sgn}(y-\theta_{1}y_{1})\big{)}^{3}d\theta_{1}d\theta_{2}\Big{)}y_{1}^{3}y_{2}^{3}[vS_{2}A_{-3,1}^{2}S_{2}v](y_{2},y_{1})dy_{1}dy_{2}.$

Then we obtain that

$\begin{split}&\ \ K_{1,N}^{2,\pm}(t;x,y)\\\
=&\frac{1}{16(2!)^{2}}\int_{\mathbf{R}^{2}}\Bigg{[}\int_{0}^{1}\int_{0}^{1}(1-\theta_{1})^{2}(1-\theta_{2})^{2}\big{(}{\rm
sgn}(y-\theta_{1}y_{1})\big{)}^{3}\big{(}{\rm
sgn}(x-\theta_{2}y_{2})\big{)}^{3}\Big{(}\int_{0}^{\infty}e^{-it\mu^{4}}\varphi_{0}(2^{-N}\mu)\\\
&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \
(F^{\pm})^{(3)}(\mu|x-\theta_{2}y_{2}|)(F^{\pm})^{(3)}(\mu|y-\theta_{1}y_{1}|)d\mu\Big{)}d\theta_{1}d\theta_{2}\Bigg{]}y_{1}^{3}y_{2}^{3}[vS_{2}A_{-3,1}^{2}S_{2}v](y_{2},y_{1})dy_{1}dy_{2}.\\\
\end{split}$ (2.39)

Denote

$\displaystyle\
E_{1,N}^{2,\pm}(t;x,y,\theta_{1},\theta_{2},y_{1},y_{2})=\int_{0}^{\infty}e^{-it\mu^{4}}\varphi_{0}(2^{-N}\mu)(F^{\pm})^{(3)}(\mu|x-\theta_{2}y_{2}|)(F^{\pm})^{(3)}(\mu|y-\theta_{1}y_{1}|)d\mu.$

Thus

$\Big{|}K_{1,N}^{2,\pm}(t;x,y)\Big{|}\lesssim\int_{\mathbf{R}^{2}}\Big{(}\int_{0}^{1}\int_{0}^{1}\Big{|}E_{1,N}^{2,\pm}(t;x,y,\theta_{1},\theta_{2},y_{1},y_{2})\Big{|}d\theta_{1}d\theta_{2}\Big{)}\big{|}[vS_{2}A_{-3,1}^{2}S_{2}v](y_{2},y_{1})\big{|}|y_{1}^{3}y_{2}^{3}|dy_{1}dy_{2}.$

Now we consider $E_{1,N}^{2,\pm}(t;x,y,\theta_{1},\theta_{2},y_{1},y_{2})$.
Since

$(F^{\pm})^{(3)}(\mu|x-\theta_{2}y_{2}|)=e^{\pm
i\mu|x-\theta_{2}y_{2}|}+e^{-\mu|x-\theta_{2}y_{2}|}:=e^{\pm
i\mu|x-\theta_{2}y_{2}|}\mathcal{F}^{\pm}(\mu|x-\theta_{2}y_{2}|),$
$(F^{\pm})^{(3)}(\mu|y-\theta_{1}y_{1}|)=e^{\pm
i\mu|y-\theta_{1}y_{1}|}+e^{-\mu|y-\theta_{1}y_{1}|}:=e^{\pm
i\mu|y-\theta_{1}y_{1}|}\mathcal{F}^{\pm}(\mu|y-\theta_{1}y_{1}|),$

let $\mu=2^{N}s,$ we can rewrite
$E_{1,N}^{2,\pm}(t;x,y,\theta_{1},\theta_{2},y_{1},y_{2})$ as follows:

$\begin{split}E_{1,N}^{2,\pm}(t;x,y,\theta_{1},\theta_{2},y_{1},y_{2})=&2^{N}\int_{0}^{\infty}e^{-it2^{4N}s^{4}}\varphi_{0}(s)e^{\pm
i2^{N}s|x-\theta_{2}y_{2}|}\mathcal{F}^{\pm}(2^{N}s|x-\theta_{2}y_{2}|)\\\ &\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e^{\pm
i2^{N}s|y-\theta_{1}y_{1}|}\mathcal{F}^{\pm}(2^{N}s|y-\theta_{1}y_{1}|)ds.\end{split}$

Noting that for any $s\in\text{supp}\ \varphi_{0}(s)$ and $k=0,1,$ we have

$\Big{|}\partial_{s}^{k}\mathcal{F}^{\pm}(2^{N}s|x-\theta_{2}y_{2}|)\Big{|}\lesssim
1,\ \ \ \ \ \
\Big{|}\partial_{s}^{k}\mathcal{F}^{\pm}(2^{N}s|y-\theta_{1}y_{1}|)\Big{|}\lesssim
1,$

then using Lemma 2.3 with $z=(x,y,y_{1},y_{2},\theta_{1},\theta_{2})$ and

$\Psi(z)=|x-\theta_{2}y_{2}|+|y-\theta_{1}y_{1}|,\ \ \ \
\Phi^{\pm}(2^{N}s;z)=\mathcal{F}^{\pm}(2^{N}s|x-\theta_{2}y_{2}|)\cdot\mathcal{F}^{\pm}(2^{N}s|y-\theta_{1}y_{1}|),$

we obtain that

$\sup_{x,y\in\mathbf{R}}\big{|}E_{1,N}^{2,\pm}(t;x,y,\theta_{1},\theta_{2},y_{1},y_{2})\big{|}\lesssim
2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$

Then, since $v(x)\lesssim(1+|x|)^{-\frac{25}{2}-},$ we have

$\displaystyle\big{|}K_{1,N}^{2,\pm}(t;x,y)\big{|}\lesssim$ $\displaystyle
2^{N}(1+|t|\cdot
2^{4N})^{-\frac{1}{2}}\int_{\mathbf{R}^{2}}|y_{2}^{3}v(y_{2})|\big{|}[S_{2}A_{-3,1}^{2}S_{2}](y_{2},y_{1})\big{|}|y_{1}^{3}v(y_{1})|dy_{1}dy_{2}$
$\displaystyle\lesssim$ $\displaystyle 2^{N}(1+|t|\cdot
2^{N})^{-\frac{1}{2}}\|y_{2}^{3}v(y_{2})\|_{L^{2}_{y_{2}}}\|S_{2}\
{-3}^{2}S_{2}\|_{L^{2}\rightarrow L_{2}}\
\|y_{1}^{3}v(y_{1})\|_{L^{2}_{y_{1}}}$ $\displaystyle\lesssim$ $\displaystyle
2^{N}(1+|t|\cdot 2^{N})^{-\frac{1}{2}}$

uniformly in $x,y.$ ∎

###  2.3. High Energy Dispersive Estimates with $\alpha=0$

In this subsection, we will give the proof of the dispersive bound (2.5) with
$\alpha=0$ for the large energy part. To complete the proof, we use the
formula

$\displaystyle
e^{-itH}P_{ac}(H)\widetilde{\chi}(H)f(x)=\sum^{+\infty}_{N=N_{0}+1}\sum_{\pm}\frac{\pm
2}{\pi
i}\int_{\mathbf{R}}\Big{(}\int_{0}^{\infty}e^{-it\mu^{4}}\varphi_{N}(\mu)\mu^{3}R_{V}^{\pm}(\mu^{4})(x,y)d\mu\Big{)}f(y)dy,$
(2.40)

and the following resolvent identity,

$R_{V}^{\pm}(\mu^{4})=R_{0}^{\pm}(\mu^{4})-R_{0}^{\pm}(\mu^{4})VR_{0}^{\pm}(\mu^{4})+R_{0}^{\pm}(\mu^{4})VR_{V}^{\pm}(\mu^{4})VR_{0}^{\pm}(\mu^{4}).$
(2.41)

By combing Proposition 2.4 and Lemma 2.6, it suffices to establish the
following bound is valid for the last two terms in (2.41).

######  Proposition 2.14.

Assume that $|V(x)|\lesssim(1+|x|)^{-1-},$ then for any $N>N_{0},$ we have

$\sup\limits_{x,y\in\mathbf{R}}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{N}(\mu)\Big{[}R_{0}^{\pm}(\mu^{4})VR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu\Big{|}\lesssim
2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$ (2.42)

###### Proof.

Let

$\begin{split}L_{1}^{\pm,N}(t;x,y):=&\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{N}(\mu)\Big{[}R_{0}^{\pm}(\mu^{4})VR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu\\\
\end{split}$ (2.43)

and

$\begin{split}R_{0}^{\pm}(\mu^{4})(x,y)=&\frac{1}{4\mu^{3}}\Big{(}\pm ie^{\pm
i\mu|x-y|}-e^{-\mu|x-y|}\Big{)}:=\frac{e^{\pm
i\mu|x-y|}}{4\mu^{3}}\mathcal{R}_{0}^{\pm}(\mu|x-y|).\end{split}$

Let $\mu=2^{N}s$ we can rewrite $L_{1}^{\pm,N}(x,y)$ as follows:

$\begin{split}&\ L_{1}^{\pm,N}(t;x,y)\\\
=&\frac{2^{-2N}}{16}\int_{0}^{\infty}e^{-it2^{4N}s^{4}}s^{-3}\varphi_{0}(s)\Big{(}\int_{\mathbf{R}}e^{\pm
i2^{N}s|x-y_{1}|}\mathcal{R}_{0}^{\pm}(2^{N}s|x-y_{1}|)V(y_{1})e^{\pm
i2^{N}s|y-y_{1}|}\mathcal{R}_{0}^{\pm}(2^{N}s|y-y_{1}|)dy_{1}\Big{)}ds\\\
=&\frac{2^{-2N}}{16}\int_{\mathbf{R}}\Big{(}\int_{0}^{\infty}e^{-it2^{4N}s^{4}}s^{-3}\varphi_{0}(s)e^{\pm
i2^{N}s|x-y_{1}|}e^{\pm
i2^{N}s|y-y_{1}|}\mathcal{R}_{0}^{\pm}(2^{N}s|x-y_{1}|)V(y_{1})\mathcal{R}_{0}^{\pm}(2^{N}s|y-y_{1}|)ds\Big{)}dy_{1}.\end{split}$

Since for any $s\in\text{supp}\ \varphi_{0}(s)$ and $k=0,1,$ we have

$\sup_{x,y\in\mathbf{R}}\Big{|}\partial_{s}^{k}\big{(}\mathcal{R}_{0}^{\pm}(2^{N}s|x-y_{1}|)V(y_{1})\mathcal{R}_{0}^{\pm}(2^{N}s|y-y_{1}|)\big{)}\Big{|}\lesssim|V(y_{1})|,$

then using Lemma 2.3 with $z=(x,y,y_{1})$ and

$\Psi(z)=|x-y_{1}|+|y-y_{1}|,\ \ \ \
\Phi(2^{N}s;z)=\mathcal{R}_{0}^{\pm}(2^{N}s|x-y_{1}|)\cdot\mathcal{R}_{0}^{\pm}(2^{N}s|y-y_{1}|),$

we can obtain that

$\sup_{x,y\in\mathbf{R}}\big{|}L_{1}^{\pm,N}(t;x,y)\big{|}\lesssim
2^{-2N}(1+|t|\cdot
2^{4N})^{-\frac{1}{2}}\int_{\mathbf{R}}|V(y_{1})|dy_{1}\lesssim
2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$

∎

We next consider the last term
$R_{0}^{\pm}(\mu^{4})VR_{V}^{\pm}(\mu^{4})VR_{0}^{\pm}(\mu^{4})$ in (2.41). To
control this term, we utilize the following estimates.

######  Lemma 2.15.

([FSY18],Theorem 2.23) Let $k\geq 0$ and
$|V(x)|\lesssim\big{(}1+|x|\big{)}^{-k-1-}$ such that $H=(-\Delta)^{2}+V$ has
no embedded positive eigenvalues. Then for any $\sigma>k+\frac{1}{2},\
R^{\pm}_{V}(\lambda)\in\mathcal{B}\big{(}L_{\sigma}^{2}(\mathbf{R}),L_{-\sigma}^{2}(\mathbf{R})\big{)}$
are $C^{k}$-continuous for all $\lambda>0$. Furthermore,

$\Big{\|}\partial^{k}_{\lambda}R^{\pm}_{V}(\lambda)\Big{\|}_{L_{\sigma}^{2}(\mathbf{R})\rightarrow
L_{-\sigma}^{2}(\mathbf{R})}=O\big{(}|\lambda|^{-\frac{3(k+1)}{4}}\big{)},$

as $\lambda\rightarrow+\infty.$

######  Proposition 2.16.

Assume that $|V(x)|\lesssim(1+|x|)^{-2-},$ then for any $N>N_{0},$ we have

$\sup\limits_{x,y}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{N}(\mu)\Big{[}R_{0}^{\pm}(\mu^{4})VR_{V}^{\pm}(\mu^{4})VR_{0}^{\pm}(\mu^{4})\Big{]}(x,y)d\mu\Big{|}\lesssim
2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$ (2.44)

###### Proof.

For the (2.44), it suffices to prove that for any $f,g\in L^{1}(\mathbf{R}),$

$\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{0}(2^{-N}\mu)\big{\langle}VR_{V}^{\pm}(\mu^{4})VR_{0}^{\pm}(\mu^{4})f,\big{(}R_{0}^{\pm}(\mu^{4})\big{)}^{*}g\big{\rangle}d\mu\Big{|}\lesssim
2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}\|f\|_{L^{1}}\cdot\|g\|_{L^{1}}.$
(2.45)

Equivalently, it is enough to bound the following kernel uniformly by
$O(2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}})$ :

$\widetilde{L}_{2}^{\pm,N}(t;x,y)=\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3}\varphi_{0}(2^{-N}\mu)\Big{\langle}VR_{V}^{\pm}(\mu^{4})V\big{(}R_{0}^{\pm}(\mu^{4})(\ast,y)\big{)}(\cdot),(R_{0}^{\pm}(\mu^{4}))^{*}(x,\cdot)\Big{\rangle}d\mu,$

where

$\displaystyle R_{0}^{\pm}(\mu^{4})(x,y)=$
$\displaystyle\frac{1}{4\mu^{3}}\big{(}\pm ie^{\pm
i\mu|x-y|}-e^{-\mu|x-y|}\big{)}:=\frac{e^{\pm
i\mu|x-y|}}{4\mu^{3}}\mathcal{R}_{0}^{\pm}(\mu)(x,y).$

Rewrite that

$\displaystyle\ \
\Big{\langle}VR_{V}^{\pm}(\mu^{4})V\big{(}R_{0}^{\pm}(\mu^{4})(\ast,y)\big{)}(\cdot),R_{0}^{\mp}(\mu^{4})(x,\cdot)\Big{\rangle}$
$\displaystyle=$
$\displaystyle\frac{1}{16\mu^{6}}\Big{\langle}VR_{V}^{\pm}V\big{(}e^{\pm
i\mu|\ast-y|}\mathcal{R}_{0}^{\pm}(\mu)(\ast,y)\big{)}(\cdot),\big{(}e^{\mp
i\mu|x-\cdot|}\mathcal{R}_{0}^{\mp}(\mu)(x,\cdot)\big{)}\Big{\rangle}$
$\displaystyle=$ $\displaystyle\frac{1}{16\mu^{6}}e^{\pm i\mu|x|}e^{\pm
i\mu|y|}\Big{\langle}VR_{V}^{\pm}V\big{(}e^{\pm
i\mu(|\ast-y|-|y|)}\mathcal{R}_{0}^{\pm}(\mu)(\ast,y)\big{)}(\cdot),\big{(}e^{\mp
i\mu(|x-\cdot|-|x|)}\mathcal{R}_{0}^{\mp}(\mu)(x,\cdot)\big{)}\Big{\rangle}$
$\displaystyle:=$ $\displaystyle\frac{1}{16\mu^{6}}e^{\pm
i\mu(|x|+|y|)}E_{2,N}^{L,\pm}(\mu;x,y).$

Let $\mu=2^{N}s,$ then we have

$\begin{split}\widetilde{L}_{2}^{\pm,N}(t;x,y)=&\frac{2^{-2N}}{16}\int_{0}^{\infty}e^{-it2^{4N}s^{4}}e^{\pm
i2^{N}s|x|}e^{\pm
i2^{N}s|y|}s^{-3}\varphi_{0}(s)E_{2,N}^{L,\pm}(2^{N}s;x,y)ds.\end{split}$

Noting that for any $s\in\text{supp}\ \varphi_{0}(s)$ and any $N>N_{0}$, we
have

$\Big{|}\partial_{s}^{k}\Big{(}e^{\pm
i2^{N}s(\ast-y|-|y|)}\mathcal{R}_{0}^{\pm}(2^{N}s)(\ast,y)\Big{)}\Big{|}\lesssim
2^{kN}\langle\ast\rangle^{k},\ \ k=0,1,$ $\Big{|}\partial_{s}^{k}\Big{(}e^{\pm
i2^{N}s(|x-\cdot|-|x|)}\mathcal{R}_{0}^{\pm}(2^{N}s)(x,\cdot)\Big{)}\Big{|}\lesssim
2^{kN}\langle\cdot\rangle^{k},\ \ k=0,1.$

And by Lemma 2.15, for $\sigma>k+\frac{1}{2}$ with $k=0,1,$ we have

$\big{\|}\partial_{s}^{k}R_{V}^{\pm}(2^{4N}s^{4})\big{\|}_{L^{2}_{\sigma}\rightarrow
L^{2}_{-\sigma}}\lesssim 2^{kN}(2^{N}s)^{-3}.$

Furthermore, since $|V(x)|\lesssim(1+|x|)^{-2-},$ then for
$\sigma>k+\frac{1}{2}$ with $k=0,1,$ we can obtain that

$\displaystyle\big{|}\partial_{s}^{k}E_{2,N}^{L,\pm}(2^{N}s;x,y)\big{|}\lesssim$
$\displaystyle\sum_{k=0}^{1}\|V(\cdot)\langle\cdot\rangle^{\sigma+1-k}\|_{L^{2}}^{2}\cdot\big{\|}\partial_{s}^{k}R_{V}^{\pm}(2^{4N}s^{4})\big{\|}_{L^{2}_{\sigma}\rightarrow
L^{2}_{-\sigma}}$ $\displaystyle\lesssim$ $\displaystyle 2^{kN}(2^{N}s)^{-3}.$

Hence, using Lemma 2.3 with $z=(x,y)$ and

$\Psi(z)=|x|+|y|,\ \ \ \Phi(2^{N}s;z)=E_{2,N}^{L,\pm}(2^{N}s;x,y),$

we have

$\big{|}\widetilde{L}_{2}^{\pm,N}(t;x,y)\big{|}\lesssim 2^{(k-5)N}(1+|t|\cdot
2^{4N})^{-\frac{1}{2}}\lesssim 2^{N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}$

uniform in $x,y$ for any $N>N_{0}$. ∎

###  2.4. Proofs of the decay estimates with regular term
$H^{\frac{\alpha}{4}}$ for $0<\alpha<1$

In the final subsection, we will give the proof of the estimates (2.4) and
(2.6). By using functional calculus and the Stone’s formula, we have

$\begin{split}H^{\frac{\alpha}{4}}e^{-itH}P_{ac}(H)f(x)=&\frac{2}{\pi
i}\int_{0}^{\infty}e^{-it\lambda}\lambda^{\frac{\alpha}{4}}[R_{V}^{+}(\lambda)-R_{V}^{-}(\lambda)]fd\lambda\\\
=&\int_{\mathbf{R}}\Big{(}\frac{2}{\pi
i}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3+\alpha}[R_{V}^{+}(\mu^{4})-R_{V}^{-}(\mu^{4})](x,y)d\mu\Big{)}f(y)dy.\end{split}$
(2.46)

Similar to the proof of Theorem 1 for $\alpha=0$ , by using dyadic partition
of unity, we can obtain for any $N\in\mathbf{Z}$ and $\alpha\geq 0$,

$\sup_{x,y\in\mathbf{R}}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3+\alpha}\varphi_{N}(\mu)[R_{V}^{+}(\mu^{4})-R_{V}^{-}(\mu^{4})](x,y)d\mu\Big{|}\lesssim
2^{(\alpha+1)N}(1+|t|\cdot 2^{4N})^{-\frac{1}{2}}.$

Then, by Lemma 2.6, for any $0<\alpha<1,$ we have

$\displaystyle\
\sup_{x,y\in\mathbf{R}}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\mu^{3+\alpha}[R_{V}^{+}(\mu^{4})-R_{V}^{-}(\mu^{4})](x,y)d\mu\Big{|}$
$\displaystyle\lesssim$
$\displaystyle\sum_{N=-\infty}^{+\infty}\sup_{x,y\in\mathbf{R}}\Big{|}\int_{0}^{\infty}e^{-it\mu^{4}}\varphi_{N}(\mu)\mu^{3+\alpha}[R_{V}^{+}(\mu^{4})-R_{V}^{-}(\mu^{4})](x,y)d\mu\Big{|}\lesssim|t|^{-\frac{1+\alpha}{4}}.$

Hence, for any $0<\alpha<1,$ we have

$\big{\|}H^{\frac{\alpha}{4}}e^{-itH}P_{ac}(H)\big{\|}_{L^{1}(\mathbf{R})\rightarrow
L^{\infty}(\mathbf{R})}\lesssim|t|^{-\frac{1+\alpha}{4}},\ t\neq 0.$

##  3\. Proof of Theorem 1.8

In this section, we show the processes how to derive the asymptotic expansions
of $\Big{(}M^{\pm}(\mu)\Big{)}^{-1}.$ Recall that the integral kernel of
$R_{0}^{\pm}(\mu^{4})$ are given by

$R_{0}^{\pm}(\mu^{4})(x,y)=\frac{1}{4\mu^{3}}\Big{(}\pm ie^{\pm
i\mu|x-y|}-e^{-\mu|x-y|}\Big{)}.$

When $\mu|x-y|<1,$ we have the following representation for the
$R_{0}^{\pm}(\mu^{4})(x,y):$

$\begin{split}R_{0}^{\pm}(\mu^{4})(x,y)=&\frac{a^{\pm}}{\mu^{3}}I(x,y)+\frac{a_{-1}^{\pm}}{\mu}G_{-1}(x,y)+G_{0}(x,y)+a_{1}^{\pm}\mu
G_{1}(x,y)+a_{3}^{\pm}\mu^{3}G_{3}(x,y)\\\
&+\mu^{4}G_{4}(x,y)+\sum_{k=5}^{N}a_{k}^{\pm}\mu^{k}G_{k}(x,y)+O(\mu^{N+1}|x-y|^{N+4}).\end{split}$
(3.1)

Here $G_{k}(x,y)$ are integral kernels which defined as follows:

$\begin{split}G_{-1}(x,y)=&|x-y|^{2},\ \ G_{0}(x,y)=\frac{|x-y|^{3}}{12},\ \
G_{1}(x,y)=|x-y|^{4},\\\ G_{3}(x,y)=&|x-y|^{6},\ \
G_{4}(x,y)=\frac{|x-y|^{7}}{2\times(7!)},\ \ G_{k}(x,y)=|x-y|^{3+k}\ (k\geq
5),\end{split}$ (3.2)

and the coefficients $\displaystyle a^{\pm}=\frac{-1\pm i}{4}$,$\displaystyle
a_{k}^{\pm}=\frac{(\pm i)^{k+4}+(-1)^{k+4}}{4\times(k+3)!}.$ Notice that,
$\displaystyle G_{0}(x)=\frac{|x|^{3}}{12}$ is the fundamental solution of
$\Delta^{2}$ in $\mathbf{R}$. In fact, when $\mu|x-y|\gtrsim 1,$ the
expansions remains valid. It is obvious that the first two terms in (3.1) are
singular, which is different from at most one singular term on $\mu$ for
$n\geq 3$, see e.g. [EGT21, FWY18, GT19].

For convenience, we introduce two notations here. For operators
$\varepsilon(\mu)$ with some parameter $\mu$, we write
$\varepsilon(\mu)=O_{1}(\mu^{-\alpha}g(x,y)))$ if it’s kernel
$\varepsilon(\mu)(x,y)$ has the property

$\mu^{\alpha}|\varepsilon(\mu)(x,y)|+\mu^{\alpha+1}|\partial_{\mu}\varepsilon(\mu)(x,y)|\lesssim
g(x,y).$ (3.3)

And we denote $\Gamma_{k}(\mu)$ as a $\mu$ dependent operator which satisfies

$\big{\|}|\Gamma_{k}(\mu)|\big{\|}_{L^{2}\rightarrow
L^{2}}+\mu\big{\|}\partial_{\mu}|\Gamma_{k}(\mu)|\big{\|}_{L^{2}\rightarrow
L^{2}}\lesssim\mu^{k},\ \ \mu>0,$ (3.4)

Noting that the operator $\Gamma_{k}(\mu)$ will vary from line to line.

Let $T_{0}:=U+vG_{0}v$ and $P=\|V\|_{L^{1}(\mathbf{R})}^{-1}v\langle
v,\cdot\rangle$ denote the orthogonal projection onto the span of $v.$ Since
$M^{\pm}(\mu)=U+vR_{0}^{\pm}(\mu^{4})v$ ( see Theorem 1.8 ), then we obtain
the following expansions of $M^{\pm}(\mu).$

######  Lemma 3.1.

Let $\widetilde{a}^{\pm}=a^{\pm}\|V\|_{L^{1}(\mathbf{R})}$. Assume that
$|V(x)|\lesssim(1+|x|)^{-\beta}$ with some $\beta>0$, the following expansions
for $M^{\pm}(\mu)$ in $L^{2}(\mathbf{R})$ hold:

(i) If $\beta>13,$ then we have

$\begin{split}M^{\pm}(\mu)=&\frac{\widetilde{a}^{\pm}}{\mu^{3}}P+\frac{a_{-1}^{\pm}}{\mu}vG_{-1}v+T_{0}+a_{1}^{\pm}\mu
vG_{1}v+\Gamma_{3}(\mu).\end{split}$ (3.5)

(ii) If $\beta>17,$ then we have

$\begin{split}M^{\pm}(\mu)=&\frac{\widetilde{a}^{\pm}}{\mu^{3}}P+\frac{a_{-1}^{\pm}}{\mu}vG_{-1}v+T_{0}+a_{1}^{\pm}\mu
vG_{1}v+a_{3}^{\pm}\mu^{3}vG_{3}v+\mu^{4}vG_{4}v+\Gamma_{5}(\mu).\end{split}$
(3.6)

(iii) If $\beta>25,$ we have

$\begin{split}M^{\pm}(\mu)=&\frac{\widetilde{a}^{\pm}}{\mu^{3}}P+\frac{a_{-1}^{\pm}}{\mu}vG_{-1}v+T_{0}+a_{1}^{\pm}\mu
vG_{1}v+a_{3}^{\pm}\mu^{3}vG_{3}v+\mu^{4}vG_{4}v\\\
&+\sum_{k=5}^{8}a_{k}^{\pm}\mu^{k}vG_{k}v+\Gamma_{9}(\mu).\end{split}$ (3.7)

###### Proof.

We only give a proof of (3.5), the other cases are similar. If $\mu|x-y|<1,$
using (3.1), we have

$R_{0}^{\pm}(\mu^{4})(x,y)=\frac{a^{\pm}}{\mu^{3}}+\frac{a_{-1}^{\pm}}{\mu}G_{-1}+G_{0}+a_{1}^{\pm}\mu
G_{1}+O_{1}(\mu^{3}|x-y|^{6}).$ (3.8)

If $\mu|x-y|>1,$ using (1.19), we have

$\begin{split}&\ R_{0}^{\pm}(\mu^{4})(x,y)\\\
=&\frac{a^{\pm}}{\mu^{3}}+\frac{a_{-1}^{\pm}}{\mu}G_{-1}+G_{0}+a_{1}^{\pm}\mu
G_{1}+\Big{(}\frac{\pm ie^{\pm
i\mu|x-y|}-e^{-\mu|x-y|}}{4\mu^{3}}-\frac{a^{\pm}}{\mu^{3}}-\frac{a_{-1}^{\pm}}{\mu}G_{-1}-G_{0}-a_{1}^{\pm}\mu
G_{1}\Big{)}\\\
=&\frac{a^{\pm}}{\mu^{3}}+\frac{a_{-1}^{\pm}}{\mu}G_{-1}+G_{0}+a_{1}^{\pm}\mu
G_{1}+O_{1}(\mu^{3}|x-y|^{6}).\end{split}$ (3.9)

Using (3.8) and (3.9) in (3.5), we have

$\Big{|}\Big{(}M^{\pm}(\mu)-\frac{\widetilde{a}^{\pm}}{\mu^{3}}P-\frac{a_{-1}^{\pm}}{\mu}vG_{-1}v-T_{0}-a_{1}^{\pm}\mu
vG_{1}v\Big{)}\Big{|}\lesssim v(x)v(y)|x-y|^{6}\mu^{3},$

and

$\Big{|}\partial_{\mu}\Big{(}M^{\pm}(\mu)-\frac{\widetilde{a}^{\pm}}{\mu^{3}}P-\frac{a_{-1}^{\pm}}{\mu}vG_{-1}v-T_{0}-a_{1}^{\pm}\mu
vG_{1}v\Big{)}\Big{|}\lesssim v(x)v(y)|x-y|^{6}\mu^{2}.$

Since the error term is a Hilbert-Schmidt operator if
$v(x)\lesssim\big{(}1+|x|\big{)}^{-\frac{13}{2}-},$ this implies (3.5) . ∎

######  Remark 3.2.

If we just want to get a weak expansions of $\big{(}M^{\pm}(\mu)\big{)}^{-1}$
(see Remark 1.10 of Introduction), then we only need the following expansions
of $M^{\pm}(\mu)$ in the following Table 2.

$\beta>7+2l$ |  $\displaystyle M^{\pm}(\mu)=\frac{\widetilde{a}^{\pm}}{\mu^{3}}P+\frac{a_{-1}^{\pm}}{\mu}vG_{-1}v+T_{0}+\Gamma_{l}(\mu),\ l\in[0,1]$
---|---
$\beta>9+2l$ |  $\displaystyle M^{\pm}(\mu)=\frac{\widetilde{a}^{\pm}}{\mu^{3}}P+\frac{a_{-1}^{\pm}}{\mu}vG_{-1}v+T_{0}+a_{1}^{\pm}\mu vG_{1}v+\Gamma_{1+l}(\mu),\ l\in[0,2]$
$\beta>13+2l$ |  $\displaystyle M^{\pm}(\mu)=\frac{\widetilde{a}^{\pm}}{\mu^{3}}P+\frac{a_{-1}^{\pm}}{\mu}vG_{-1}v+T_{0}+a_{1}^{\pm}\mu vG_{1}v+a_{3}^{\pm}\mu^{3}vG_{3}v+\Gamma_{3+l}(\mu),\ l\in[0,1]$
Table 2. $M^{\pm}(\mu)$ expansions

Now, we give the following lemma used repeatedly to determine the inverse of
$\Big{(}M^{\pm}(\mu)\Big{)}^{-1}$.

######  Lemma 3.3.

([JN01, Lemma 2.1]) Let $A$ be a closed operator and $S$ be a projection.
Suppose $A+S$ has a bounded inverse. Then $A$ has a bounded inverse if and
only if

$B:=S-S(A+S)^{-1}S$ (3.10)

has a bounded inverse in $S\mathcal{H}$, and in this case

$A^{-1}=(A+S)^{-1}+(A+S)^{-1}SB^{-1}S(A+S)^{-1}.$ (3.11)

Recall from (3.5), let $\displaystyle
M^{\pm}(\mu)=\frac{\widetilde{a}^{\pm}}{\mu^{3}}\widetilde{M^{\pm}}(\mu)$, we
have

$\begin{split}\widetilde{M^{\pm}}(\mu)=&P+\frac{a_{-1}^{\pm}}{\widetilde{a}^{\pm}}\mu^{2}vG_{-1}v+\frac{\mu^{3}}{\widetilde{a}^{\pm}}T_{0}+\frac{a_{1}^{\pm}}{\widetilde{a}^{\pm}}\mu^{4}vG_{1}v+\frac{a_{3}^{\pm}}{\widetilde{a}^{\pm}}\mu^{6}vG_{3}v+\frac{\mu^{7}}{\widetilde{a}^{\pm}}vG_{4}v\\\
&+\sum_{k=5}^{10}\frac{a_{k}^{\pm}}{\widetilde{a}^{\pm}}\mu^{k+3}vG_{k}v+\Gamma_{14}(\mu).\end{split}$
(3.12)

Thus it suffices to derive the asymptotic expansion of
$\Big{(}\widetilde{M^{\pm}}(\mu)\Big{)}^{-1}$ as $\mu\rightarrow 0.$ Denote
$Q=I-P,$ it is obvious that

$\begin{split}\widetilde{M^{\pm}}(\mu)+Q=&I+\frac{a_{-1}^{\pm}}{\widetilde{a}^{\pm}}\mu^{2}vG_{-1}v+\frac{\mu^{3}}{\widetilde{a}^{\pm}}T_{0}+\frac{a_{1}^{\pm}}{\widetilde{a}^{\pm}}\mu^{4}vG_{1}v+\frac{a_{3}^{\pm}}{\widetilde{a}^{\pm}}\mu^{6}vG_{3}v+\frac{\mu^{7}}{\widetilde{a}^{\pm}}vG_{4}v\\\
&+\sum_{k=5}^{10}\frac{a_{k}^{\pm}}{\widetilde{a}^{\pm}}\mu^{k+3}vG_{k}v+\Gamma_{14}(\mu).\end{split}$
(3.13)

Since $vG_{-1}v$ is a bounded operator from $L^{2}(\mathbf{R})$ to
$L^{2}(\mathbf{R}),$ then using Neumann series, we can obtain that

$\begin{split}\Big{(}\widetilde{M^{\pm}}(\mu)+Q\Big{)}^{-1}=&I-\sum_{k=1}^{12}\mu^{k+1}B_{k}^{\pm}+\Gamma_{14}(\mu),\end{split}$
(3.14)

where $B_{k}^{\pm}\ (1\leq k\leq 12)$ are bounded operators in
$L^{2}(\mathbf{R})$ and it is easy to know that

$\begin{split}B_{1}^{\pm}=&\frac{a_{-1}^{\pm}}{\widetilde{a}^{\pm}}vG_{-1}v,\
\ \ \ \ \ B_{2}^{\pm}=\frac{1}{\widetilde{a}^{\pm}}T_{0},\ \ \ \ \ \
B_{3}^{\pm}=\frac{a_{1}^{\pm}}{\widetilde{a}^{\pm}}vG_{1}v-\frac{(a_{-1}^{\pm})^{2}}{(\widetilde{a}^{\pm})^{2}}(vG_{-1}v)^{2},\\\
B_{4}^{\pm}=&-\frac{a_{-1}^{\pm}}{(\widetilde{a}^{\pm})^{2}}\big{(}vG_{-1}vT_{0}+T_{0}vG_{-1}v\big{)},\\\
B_{5}^{\pm}=&\frac{a_{3}^{\pm}}{\widetilde{a}^{\pm}}vG_{3}v+\frac{(a_{-1}^{\pm})^{3}}{(\widetilde{a}^{\pm})^{3}}(vG_{-1}v)^{3}-\frac{1}{(\widetilde{a}^{\pm})^{2}}T_{0}^{2}-\frac{a_{-1}^{\pm}a_{1}^{\pm}}{(\widetilde{a}^{\pm})^{2}}\big{(}vG_{-1}v\cdot
vG_{1}v+vG_{1}v\cdot vG_{-1}v\big{)},\\\
B_{6}^{\pm}=&\frac{1}{\widetilde{a}^{\pm}}vG_{4}v+\frac{(a^{\pm}_{-1})^{2}}{(\widetilde{a}^{\pm})^{3}}\Big{(}(vG_{-1}v)^{2}T_{0}+vG_{-1}vT_{0}vG_{-1}v+T_{0}(vG_{-1}v)^{2}\Big{)}-\frac{a_{1}^{\pm}}{(\widetilde{a}^{\pm})^{2}}\big{(}vG_{1}vT_{0}+T_{0}vG_{1}v\big{)}.\end{split}$

For the convenience, we denote some inverse operators as follows:

$\begin{split}&D_{0}=(QvG_{-1}vQ+S_{0})^{-1}\ \ \text{on}\ \
QL^{2}(\mathbf{R});\ \ D_{1}=(S_{0}T_{0}S_{0}+S_{1})^{-1}\ \ \text{on}\ \
S_{0}L^{2}(\mathbf{R});\\\ &D_{2}=(T_{1}+S_{2})^{-1}\ \ \text{on}\ \
S_{1}L^{2}(\mathbf{R});\ \ \ \ \ \ \ \ \ \ \ D_{3}=(T_{2}+S_{3})^{-1}\ \
\text{on}\ \ S_{2}L^{2}(\mathbf{R}).\end{split}$ (3.15)

Next, we list the orthogonality relations of various operators and projections
which we need in the proof of Theorem 1.8.

$\displaystyle QD_{0}=D_{0}Q=D_{0};$ (3.16) $\displaystyle
S_{i}D_{j}=D_{j}S_{i}=D_{j},\ i<j;\ \ S_{i}D_{j}=D_{j}S_{i}=S_{i},\ i\geq j;\
\ D_{i}D_{j}=D_{j}D_{i}=D_{j},i<j;$ (3.17) $\displaystyle
QB_{1}^{\pm}S_{0}=S_{0}B_{1}^{\pm}Q=S_{2}B_{1}^{\pm}=B_{1}^{\pm}S_{2}=S_{3}B_{1}^{\pm}=B_{1}^{\pm}S_{3}=0;$
(3.18) $\displaystyle
S_{0}B_{2}^{\pm}S_{1}=S_{1}B_{2}^{\pm}S_{0}=QB_{2}^{\pm}S_{2}=S_{2}B_{2}^{\pm}Q=S_{3}B_{2}^{\pm}=B_{2}^{\pm}S_{3}=0;$
(3.19) $\displaystyle
S_{0}B_{3}^{\pm}S_{2}=S_{2}B_{3}^{\pm}S_{0}=QB_{3}^{\pm}S_{3}=S_{3}B_{3}^{\pm}Q=0;$
(3.20) $\displaystyle
S_{3}B_{4}^{\pm}=B_{4}^{\pm}S_{3}=S_{2}B_{5}^{\pm}S_{3}=S_{3}B_{5}^{\pm}S_{2}=0.$
(3.21)

Now, we turn to the proof of Theorem 1.8. The following lemma which will be
used in the proof is an important classification of the projection spaces
$S_{j}L^{2}(\mathbf{R})\ (j=0,1,2,3)$.

######  Lemma 3.4.

We have

(1) $f\in S_{0}L^{2}(\mathbf{R})$ if and only if $f\in\ker QvG_{-1}vQ$ on
$QL^{2}(\mathbf{R});$

(2) $f\in S_{1}L^{2}(\mathbf{R})$ if and only if $f\in\ker S_{0}T_{0}S_{0}$ on
$S_{0}L^{2}(\mathbf{R});$

(3) $f\in S_{2}L^{2}(\mathbf{R})$ if and only if $f\in\ker T_{1}$ on
$S_{1}L^{2}(\mathbf{R})$ with

$T_{1}:=S_{1}vG_{1}vS_{1}+\frac{6}{\|V\|_{L^{1}}}S_{1}vG_{-1}vPvG_{-1}vS_{1}-384S_{1}T_{0}D_{0}T_{0}S_{1}.$
(3.22)

(4) $f\in S_{3}L^{2}(\mathbf{R})$ if and only if $f\in\ker T_{2}$ on
$S_{2}L^{2}(\mathbf{R})$ with

$\begin{split}T_{2}:=&S_{2}vG_{3}vS_{2}-\frac{8\cdot(6!)}{\|V\|_{L^{1}}}S_{2}T_{0}^{2}S_{2}+\frac{5}{2}S_{2}vG_{1}vD_{0}vG_{1}vS_{2}+\frac{48\cdot(6!)}{\|V\|_{L^{1}}^{2}}\Big{(}S_{2}T_{0}vG_{-1}v\\\
&-\frac{\|V\|_{L^{1}}}{6}S_{2}vG_{1}vD_{0}T_{0}\Big{)}D_{2}\Big{(}vG_{-1}vT_{0}S_{2}-\frac{\|V\|_{L^{1}}}{6}T_{0}D_{0}vG_{1}vS_{2}\Big{)}.\end{split}$
(3.23)

###### Proof.

(1) Let $f\in QL^{2}(\mathbf{R})$ and $f\in\ker QvG_{-1}vQ,$ that is,
$QvG_{-1}vQf=0$. Then we have

$0=\langle QvG_{-1}vQf,f\rangle=\langle vG_{-1}vQf,Qf\rangle=\langle
vG_{-1}vf,f\rangle.$

Notice that

$G_{-1}vf(x)=\int_{\mathbf{R}}|x-y|^{2}v(y)f(y)dy,$

and $Q(v)=0,$ i.e. $\displaystyle\langle
f,v\rangle=\int_{\mathbf{R}}v(x)f(x)dx=0.$ So we have

$\begin{split}0=\langle
vG_{-1}vf,f\rangle=&\int_{\mathbf{R}}(vG_{-1}vf)(x)\overline{f(x)}\,dx\\\
=&\int_{\mathbf{R}}\overline{v(x)f(x)}\Big{(}\int_{\mathbf{R}}|x-y|^{2}v(y)f(y)\,dy\Big{)}\,dx\\\
=&\int_{\mathbf{R}}\overline{v(x)f(x)}\Big{(}\int_{\mathbf{R}}(x^{2}-2xy+y^{2})v(y)f(y)\,dy\Big{)}\,dx\\\
=&-2\Big{|}\int_{\mathbf{R}}xv(x)f(x)dx\Big{|}^{2}=-2\big{|}\langle
f,xv\rangle\big{|}^{2},\end{split}$

which implies that $\langle f,xv\rangle=0$. Hence $f\in
S_{0}L^{2}(\mathbf{R}).$

On the other hand, if $f\in S_{0}L^{2}(\mathbf{R}),$ that is,

$\langle f,v\rangle=\int_{\mathbf{R}}v(x)f(x)dx=0,\ \ \ \langle
f,xv\rangle=\int_{\mathbf{R}}xv(x)f(x)dx=0.$

Since $S_{0}L^{2}(\mathbf{R})\subseteq QL^{2}(\mathbf{R}),$ thus we have

$QvG_{-1}vQf=Qv(x)\int_{\mathbf{R}}(x^{2}-2xy+y^{2})v(y)f(y)dy=\langle
f,x^{2}v\rangle Q(v(x))=0,$

which means that $f\in\ker QvG_{-1}vQ.$

(2) It is obvious by the definition of $S_{1},$ see Table 1.

(3) First, we prove the Claim: Denote $\displaystyle
v_{1}:=Q(xv(x))=xv(x)-\frac{\langle v,xv\rangle}{\|V\|_{L^{1}}}v(x),$ then we
have

$\langle D_{0}v_{1},v_{1}\rangle=-\frac{1}{2}.$ (3.24)

In fact, using $Q(v(x))=0,S_{0}(v(x))=S_{0}(xv(x))=0$ and $\langle
v,v_{1}\rangle=0,$ we have

$\displaystyle(QvG_{-1}vQ+S_{0})v_{1}=$ $\displaystyle QvG_{-1}vQv_{1}$
$\displaystyle=$ $\displaystyle
Qv(x)\int_{\mathbf{R}}(x^{2}-2xy+y^{2})v(y)v_{1}(y)dy$ $\displaystyle=$
$\displaystyle-2Q\big{(}xv(x)\big{)}\int_{\mathbf{R}}yv(y)v_{1}(y)dy$
$\displaystyle=$ $\displaystyle-2v_{1}\langle v_{1},v_{1}\rangle.$

Hence, we obtain that

$D_{0}v_{1}=(QvG_{-1}vQ+S_{0})^{-1}v_{1}=\displaystyle-\frac{v_{1}}{2\langle
v_{1},v_{1}\rangle},$ (3.25)

furthermore, we have $\langle
D_{0}v_{1},v_{1}\rangle=\displaystyle-\frac{1}{2}.$

Let $f\in S_{1}L^{2}(\mathbf{R})$ and $f\in\ker T_{1}$. Recall (3.22), we have

$\begin{split}0=\langle
T_{1}f,f\rangle=&\big{\langle}S_{1}vG_{1}vS_{1}f,f\big{\rangle}+\frac{6}{\|V\|_{L^{1}}}\big{\langle}S_{1}vG_{-1}vPvG_{-1}vS_{1}f,f\big{\rangle}\\\
&-384\big{\langle}S_{1}T_{0}D_{0}T_{0}S_{1}f,f\big{\rangle}.\end{split}$
(3.26)

Firstly, for the first part of (3.26), using $S_{1}(v(x))=S_{1}(xv(x))=0,$ we
have

$\displaystyle S_{1}vG_{1}vS_{1}f=$ $\displaystyle
S_{1}v(x)\int_{\mathbf{R}}\big{(}x^{4}-4x^{3}y-4xy^{3}+6x^{2}y^{2}+y^{4}\big{)}v(y)f(y)dy$
$\displaystyle=$ $\displaystyle 6\langle f,x^{2}v\rangle S_{1}(x^{2}v).$

Thus, we have

$\displaystyle\big{\langle}S_{1}vG_{1}vS_{1}f,f\big{\rangle}=6\big{|}\langle
f,x^{2}v\rangle\big{|}^{2}.$ (3.27)

Next, for the second part of (3.26), since $Pf=\displaystyle\frac{\langle
f,v\rangle}{\|V\|_{L^{1}}}v$ and $S_{1}(v(x))=S_{1}(xv(x))=0,$ we have

$PvG_{-1}vS_{1}f=Pv(x)\int_{\mathbf{R}^{2}}|x-y|^{2}v(y)f(y)dy=v(x)\int_{\mathbf{R}}x^{2}v(x)f(x)dx=\langle
f,x^{2}v\rangle v(x),$

and

$S_{1}vG_{-1}vPvG_{-1}vS_{1}f=\langle
f,x^{2}v\rangle\|V\|_{L^{1}}S_{1}(x^{2}v(x)).$

Thus

$\displaystyle\big{\langle}S_{1}vG_{-1}vPvG_{-1}vS_{1}f,f\big{\rangle}=$
$\displaystyle\|V\|_{L^{1}}\cdot\big{|}\langle f,x^{2}v\rangle\big{|}^{2}.$
(3.28)

Finally, for the third part of (3.26), using $QD_{0}=D_{0}Q=D_{0}$ and by the
definition of $v_{1},$ we know that $v_{1}\in(Q-S_{0})L^{2}(\mathbf{R}).$
Indeed, we have $(Q-S_{0})L^{2}(\mathbf{R})=\text{span}\\{v_{1}\\},$ then we
can obtain that

$(Q-S_{0})T_{0}f=\frac{\langle T_{0}f,v_{1}\rangle}{\langle
v_{1},v_{1}\rangle}.$

Thus, using (3.25) and $D_{0}P=PD_{0}=0,S_{1}T_{0}S_{0}=S_{0}T_{0}S_{1}=0$,
for all $f\in S_{1}L^{2}(\mathbf{R}),$ we have

$\displaystyle S_{1}T_{0}D_{0}T_{0}S_{1}f=$ $\displaystyle
S_{1}T_{0}[P+(Q-S_{0})+S_{0}]D_{0}[P+(Q-S_{0})+S_{0}]T_{0}f$ $\displaystyle=$
$\displaystyle S_{1}T_{0}(Q-S_{0})D_{0}(Q-S_{0})T_{0}f$ $\displaystyle=$
$\displaystyle-\frac{\langle T_{0}f,v_{1}\rangle}{2\langle
v_{1},v_{1}\rangle^{2}}S_{1}(T_{0}v_{1}).$

Thus,

$\langle S_{1}T_{0}D_{0}T_{0}S_{1}f,f\rangle=-\frac{\big{|}\langle
T_{0}f,v_{1}\rangle\big{|}^{2}}{2\langle v_{1},v_{1}\rangle^{2}}.$ (3.29)

Hence, substituting (3.27), (3.28) and (3.29) into (3.26), we have

$0=\big{\langle}T_{1}f,f\big{\rangle}=12\big{|}\langle
f,x^{2}v\rangle\big{|}^{2}+192\frac{\big{|}\langle
T_{0}f,v_{1}\rangle\big{|}^{2}}{\langle v_{1},v_{1}\rangle^{2}},$

which implies that

$\langle f,x^{2}v\rangle=0,\ \ \ \langle T_{0}f,v_{1}\rangle=0.$

Furthermore, since $f\in S_{1}L^{2}(\mathbf{R}),$ thus $S_{0}T_{0}f=0,$
combined with $\langle T_{0}f,v_{1}\rangle=0,$ we have $QT_{0}f=0.$ Hence we
have $f\in S_{2}L^{2}(\mathbf{R}).$

On the other hand, if $f\in S_{2}L^{2}(\mathbf{R}),$ then it is easy to know
that

$T_{1}f=12\langle f,x^{2}v\rangle S_{1}(x^{2}v)+192\frac{\langle
T_{0}f,v_{1}\rangle}{\langle v_{1},v_{1}\rangle^{2}}S_{1}(T_{0}v_{1})=0,$

which means that $f\in\ker T_{1}.$

(4) Let $f\in S_{2}L^{2}(\mathbf{R})$ and $f\in\ker T_{2}.$ It follows that
from (3.23)

$\begin{split}0=\langle
T_{2}f,f\rangle=&\big{\langle}S_{2}vG_{3}vS_{2}f,f\big{\rangle}-\frac{8\cdot(6!)}{\|V\|_{L^{1}}}\big{\langle}S_{2}T_{0}^{2}S_{2}f,f\big{\rangle}+\frac{5}{2}\big{\langle}S_{2}vG_{1}vD_{0}vG_{1}vS_{2}f,f\big{\rangle}\\\
+\frac{48\cdot(6!)}{\|V\|_{L^{1}}^{2}}&\Big{\langle}\Big{(}S_{2}T_{0}vG_{-1}v-\frac{\|V\|_{L^{1}}}{6}S_{2}vG_{1}vD_{0}T_{0}\Big{)}D_{2}\Big{(}vG_{-1}vT_{0}S_{2}-\frac{\|V\|_{L^{1}}}{6}T_{0}D_{0}vG_{1}vS_{2}\Big{)}f,f\Big{\rangle}.\end{split}$
(3.30)

For the first part of (3.30), using
$S_{2}(v(x))=S_{2}(xv(x))=S_{2}(x^{2}v(x))=0$ and $QT_{0}f=0,$ we have

$S_{2}vG_{3}vS_{2}f=S_{2}v(x)\int_{\mathbf{R}}\big{|}x-y\big{|}^{6}v(y)f(y)dy=-20\langle
f,x^{3}v\rangle S_{2}(x^{3}v(x)),$

thus

$\langle S_{2}vG_{3}vf,f\rangle=-20\big{|}\langle f,x^{3}v\rangle\big{|}^{2}.$
(3.31)

Secondly, for the second part of (3.30), using $QT_{0}f=0,$ we have

$S_{2}T_{0}^{2}S_{2}f=S_{2}T_{0}PT_{0}S_{2}f=\frac{\langle
T_{0}f,v\rangle}{\|V\|_{L^{1}}}S_{2}(T_{0}v),$

so

$\langle S_{2}T_{0}^{2}S_{2}f,f\rangle=\frac{\langle
T_{0}f,v\rangle^{2}}{\|V\|_{L^{1}}}.$ (3.32)

Then, for the third part of (3.30), using (3.25) and

$QD_{0}Q=D_{0},\ \ Q(xv(x))=v_{1},\ \
S_{2}(v(x))=S_{2}(xv(x))=S_{2}(x^{2}v(x))=0,$

we have

$S_{2}vG_{1}vD_{0}vG_{1}vS_{2}f=S_{2}vG_{1}vQD_{0}QvG_{1}vf=-8\langle
f,x^{3}v\rangle S_{2}(x^{3}v),$

and

$\langle S_{2}vG_{1}vD_{0}vG_{1}vS_{2}f,f\rangle=-8\big{|}\langle
f,x^{3}v\rangle\big{|}^{2}.$ (3.33)

Finally, for the last part of (3.30), using (3.25) and

$S_{1}D_{2}S_{1}=D_{2},\ \ S_{1}(v)=S_{1}(xv)=0,\ \
S_{2}(v(x))=S_{2}(xv(x))=S_{2}(x^{2}v(x))=0,$

we have

$S_{2}T_{0}vG_{-1}vD_{2}vG_{-1}vT_{0}S_{2}f=\langle
T_{0}f,v\rangle\cdot\langle D_{2}S_{1}(x^{2}v),S_{1}(x^{2}v)\rangle
S_{2}(T_{0}v),$ $S_{2}T_{0}vG_{-1}vD_{2}T_{0}D_{0}vG_{1}vS_{2}f=2\frac{\langle
f,x^{3}v\rangle}{\langle v_{1},v_{1}\rangle}\cdot\langle
D_{2}S_{1}(T_{0}v_{1}),S_{1}(x^{2}v)\rangle S_{2}(T_{0}v),$
$S_{2}vG_{1}vD_{0}T_{0}D_{2}vG_{-1}vT_{0}S_{2}f=2\frac{\langle
T_{0}f,v\rangle}{\langle v_{1},v_{1}\rangle}\cdot\langle
D_{2}S_{1}(T_{0}v_{1}),S_{1}(x^{2}v)\rangle S_{2}(x^{3}v),$
$S_{2}vG_{1}vD_{0}T_{0}D_{2}T_{0}D_{0}vG_{1}vS_{2}f=4\frac{\langle
f,x^{3}v\rangle}{\langle v_{1},v_{1}\rangle^{2}}\cdot\langle
D_{2}S_{1}(T_{0}v_{1}),S_{1}(T_{0}v_{1})\rangle S_{2}(x^{3}v).$

Denote $X=\langle f,x^{3}v\rangle,Y=\langle T_{0}f,v\rangle,$ we obtain that

$\begin{split}&\
\Big{\langle}\Big{(}S_{2}T_{0}vG_{-1}v-\frac{\|V\|_{L^{1}}}{6}S_{2}vG_{1}vD_{0}T_{0}\Big{)}D_{2}\Big{(}vG_{-1}vT_{0}S_{2}-\frac{\|V\|_{L^{1}}}{6}T_{0}D_{0}vG_{1}vS_{2}\Big{)}f,f\Big{\rangle}\\\
=&\frac{\|V\|_{L^{1}}^{2}\cdot\langle
D_{2}S_{1}(T_{0}v_{1}),S_{1}(T_{0}v_{1})\rangle}{9\langle
v_{1},v_{1}\rangle^{2}}\cdot\big{|}X\big{|}^{2}+\langle
D_{2}S_{1}(x^{2}v),S_{1}(x^{2}v)\rangle\cdot\big{|}Y\big{|}^{2}\\\ &\
-\frac{2\|V\|_{L^{1}}\cdot\langle
D_{2}S_{1}(T_{0}v_{1}),S_{1}(x^{2}v)\rangle}{3\langle
v_{1},v_{1}\rangle}\cdot{\rm Re}(X\overline{Y}).\end{split}$ (3.34)

Then, substituting (3.31),(3.32),(3.33) and (3.34) into (3.30), we have

$\displaystyle 0=\langle T_{2}f,f\rangle$
$\displaystyle=\Big{(}\frac{32\cdot(5!)\cdot\langle
D_{2}S_{1}(T_{0}v_{1}),S_{1}(T_{0}v_{1})\rangle}{\langle
v_{1},v_{1}\rangle^{2}}-40\Big{)}\cdot\big{|}X\big{|}^{2}$
$\displaystyle+\frac{8\cdot(6!)}{\|V\|^{2}_{L^{1}}}\Big{(}6\langle
D_{2}S_{1}(x^{2}v),S_{1}(x^{2}v)\rangle-1\Big{)}\cdot\big{|}Y\big{|}^{2}-\frac{32\cdot(6!)\langle
D_{2}S_{1}(T_{0}v_{1}),S_{1}(x^{2}v)\rangle}{\langle
v_{1},v_{1}\rangle\cdot\|V\|_{L^{1}}}\cdot{\rm Re}(X\overline{Y})$
$\displaystyle:=a\big{|}X\big{|}^{2}+b\big{|}Y\big{|}^{2}+c~{}{\rm
Re}(X\overline{Y}).$

Let $X=X_{1}+iX_{2}$ and $Y=Y_{1}+iY_{2}$ with $X_{1},X_{2},Y_{1},Y_{2}$ are
real. Then we have

$\displaystyle 0=\langle T_{2}f,f\rangle=$ $\displaystyle
a(X_{1}^{2}+X_{2}^{2})+b(Y_{1}^{2}+Y_{2}^{2})+c(X_{1}Y_{1}+X_{2}Y_{2})=\sum_{k=1}^{2}\Big{(}aX_{k}^{2}+bY_{k}^{2}+cX_{k}Y_{k}\Big{)}.$

Hence, without loss of generality, we only need to consider the case that $X$
and $Y$ are real, that is,

$0=\langle T_{2}f,f\rangle=aX^{2}+bY^{2}+cXY.$

Let $g_{1}=S_{1}(x^{2}v),h_{1}=D_{2}g_{1}\in
S_{1}L^{2}(\mathbf{R}),g_{2}=S_{1}(T_{0}v_{1}),h_{2}=D_{2}g_{2},$ since
$D_{2}=(T_{1}+S_{2})^{-1},$ using $S_{1}(v(x))=S_{1}(xv(x))=0,$ we have

$g_{1}=(T_{1}+S_{2})h_{1}=12\langle g_{1},h_{1}\rangle g_{1}+192\frac{\langle
g_{2},h_{1}\rangle}{\langle v_{1},v_{1}\rangle^{2}}g_{2}+S_{2}h_{1},$

and

$g_{2}=12\langle g_{1},h_{2}\rangle g_{1}+192\frac{\langle
g_{2},h_{2}\rangle}{\langle v_{1},v_{1}\rangle^{2}}g_{2}+S_{2}h_{2}.$

Then we have

$\displaystyle\langle D_{2}S_{1}(x^{2}v),S_{1}(x^{2}v)\rangle=$
$\displaystyle\langle g_{1},h_{1}\rangle=12\langle
g_{1},h_{1}\rangle^{2}+192\frac{\langle g_{2},h_{1}\rangle^{2}}{\langle
v_{1},v_{1}\rangle^{2}}+\langle S_{2}h_{1},h_{1}\rangle,$

since $\langle S_{2}h_{1},h_{1}\rangle\geq 0$ and $\displaystyle\frac{\langle
g_{2},h_{1}\rangle^{2}}{\langle v_{1},v_{1}\rangle^{2}}\geq 0,$ we have

$\langle g_{1},h_{1}\rangle\geq 12\langle g_{1},h_{1}\rangle^{2},$

thus, we obtain that $\displaystyle\langle
g_{1},h_{1}\rangle\leq\frac{1}{12}.$ Furthermore, if $\langle
g_{2},h_{1}\rangle\neq 0,$ then $\displaystyle\langle
g_{1},h_{1}\rangle<\frac{1}{12}.$

Similarly, we have

$\displaystyle\langle D_{2}S_{1}(T_{0}v_{1}),S_{1}(T_{0}v_{1})\rangle=$
$\displaystyle\langle g_{2},h_{2}\rangle=12\langle
g_{1},h_{2}\rangle^{2}+192\frac{\langle g_{2},h_{2}\rangle^{2}}{\langle
v_{1},v_{1}\rangle^{2}}+\langle S_{2}h_{2},h_{2}\rangle,$

thus we also have

$\langle g_{2},h_{2}\rangle\leq\frac{\langle v_{1},v_{1}\rangle^{2}}{192},$

and if $\langle g_{1},h_{2}\rangle\neq 0,$ we have $\displaystyle\langle
g_{2},h_{2}\rangle<\frac{\langle v_{1},v_{1}\rangle^{2}}{192}.$
Hence, we obtain that

$\displaystyle a=$ $\displaystyle\Big{(}\frac{32\cdot(5!)\cdot\langle
D_{2}S_{1}(T_{0}v_{1}),S_{1}(T_{0}v_{1})\rangle}{\langle
v_{1},v_{1}\rangle^{2}}-40\Big{)}\leq\frac{32\cdot(5!)}{192}-40=-20<0,$
$\displaystyle b=$ $\displaystyle\Big{(}6\langle
D_{2}S_{1}(x^{2}v),S_{1}(x^{2}v)\rangle-1\Big{)}\leq\frac{6}{12}-1=-\frac{1}{2}<0.$

Furthermore, by Cauchy-Schwarz inequality, we obtain that

$\langle g_{1},h_{2}\rangle^{2}=\langle g_{1},D_{2}g_{2}\rangle^{2}\leq\langle
g_{1},D_{2}g_{1}\rangle\langle g_{2},D_{2}g_{2}\rangle=\langle
g_{1},h_{1}\rangle\langle g_{2},h_{2}\rangle.$

Thus

$\displaystyle c^{2}-4ab=$ $\displaystyle\ \frac{32^{2}\cdot(6!)^{2}\langle
g_{1},h_{2}\rangle^{2}}{\langle
v_{1},v_{1}\rangle^{2}\cdot\|V\|_{L^{1}}^{2}}-\Big{(}\frac{32\cdot(5!)\cdot\langle
g_{2},h_{2}\rangle}{\langle
v_{1},v_{1}\rangle^{2}}-40\Big{)}\cdot\frac{32\cdot(6!)}{\|V\|^{2}_{L^{1}}}\Big{(}6\langle
g_{1},h_{1}\rangle-1\Big{)}$ $\displaystyle\leq$ $\displaystyle\
\frac{32\cdot(6!)}{\|V\|^{2}_{L^{1}}}\Big{(}\frac{32\cdot(5!)\cdot\langle
g_{2},h_{2}\rangle}{\langle v_{1},v_{1}\rangle^{2}}+240\langle
g_{1},h_{1}\rangle-40\Big{)}$ $\displaystyle\leq$ $\displaystyle\
\frac{32\cdot(6!)}{\|V\|^{2}_{L^{1}}}\Big{(}\frac{32\cdot(5!)}{192}+\frac{240}{12}-40\Big{)}=0.$

Since “=” holds if and only if $\langle g_{1},h_{2}\rangle=\langle
g_{2},h_{1}\rangle=0,$ and in this case, we have

$0=\langle T_{2}f,f\rangle\leq aX^{2}+bY^{2}\leq 0,$

thus it is necessary that

$X=\langle f,x^{3}v\rangle=0,\ \ \ Y=\langle T_{0}f,v\rangle=0.$

Then $f\in S_{3}L^{2}(\mathbf{R}).$ On the other hand, it is obvious that if
$f\in S_{3}L^{2}(\mathbf{R}),$ we can obtain that $T_{2}f=0,$ which means that
$f\in\ker T_{2}.$ ∎

######  Lemma 3.5.

$S_{3}L^{2}(\mathbf{R})=\\{0\\}.$

###### Proof.

The proof can be divided into the following two steps.

Step I: We first show that $f(x)\in S_{3}L^{2}(\mathbf{R})$ if and only if
$f(x)=Uv\phi$ with some $\phi\in L^{2}(\mathbf{R})$ such that $H\phi=0$ in the
distributional sense. Furthermore, $\phi=-G_{0}vf.$

Since $S_{3}\leq S_{2}$, by Theorem 4.1(ii) below, we have known that if $f\in
S_{2}L^{2}(\mathbf{R})$, then $f=Uv\phi$ with $\phi\in
W_{\frac{1}{2}}(\mathbf{R})$ such that $H\phi=0$ in the distributional sense
and

$\phi=-G_{0}vf+\frac{\langle v,T_{0}f\rangle}{\|V\|_{L^{1}}}.$

Thus, we just need to prove that $\displaystyle\frac{\langle
T_{0}f,v\rangle}{\|V\|_{L^{1}(\mathbf{R})}}=0$ and $\phi\in
L^{2}(\mathbf{R})$. Indeed, if $f\in S_{3}L^{2}(\mathbf{R})$, then

$\int_{\mathbf{R}}v(y)f(y)dy=\int_{\mathbf{R}}yv(y)f(y)dy=\int_{\mathbf{R}}y^{2}v(y)f(y)dy=\int_{\mathbf{R}}y^{3}v(y)f(y)dy=QT_{0}f=PT_{0}f=0.$

Thus, we have $\displaystyle\frac{\langle v,T_{0}f\rangle}{\|V\|_{L^{1}}}=0$.

Next we show that $\phi\in L^{2}(\mathbf{R})$. Since $\phi=-G_{0}vf$ , and
using

$S_{3}(v)=S_{3}(xv)=S_{3}(x^{2}v)=S_{3}(x^{3}v)=0,$

we have

$\begin{split}\phi(x)=&-\frac{1}{12}\int_{\mathbf{R}}\big{(}|x-y|^{3}-|x|^{3}+3|x|x\cdot
y-3|x|y^{2}+\frac{x}{|x|}y^{3}\big{)}v(y)f(y)dy\\\
:=&-\frac{1}{12}\int_{\mathbf{R}}K_{3}(x,y)v(y)f(y)dy,\end{split}$

where

$K_{3}(x,y)=|x-y|^{3}-|x|^{3}+3|x|x\cdot
y-3|x|y^{2}+\displaystyle\frac{x}{|x|}y^{3}.$

Indeed, we can divided $K_{3}(x,y)$ into five parts as follows

$\begin{split}K_{3}(x,y)=&\Big{[}y^{2}\big{(}|x-y|-|x|\big{)}+\frac{x}{|x|}y^{3}\Big{]}+\Big{[}\frac{x}{|x|}y^{3}-\frac{2xy^{3}}{|x-y|+|x|}\Big{]}\\\
&+\frac{5}{2}\Big{[}\frac{|x|y^{2}(|x|-|x-y|)}{|x-y|+|x|}-\frac{1}{2}\frac{x}{|x|}y^{3}\Big{]}+\Big{[}\frac{x^{2}y^{2}(|x-y|-|x|)}{(|x-y|+|x|)^{2}}+\frac{1}{4}\frac{x}{|x|}y^{3}\Big{]}\\\
&+\frac{|x|y^{4}}{2(|x-y|+|x|)^{2}}\\\
:=&K_{31}(x,y)+K_{32}(x,y)+K_{33}(x,y)+K_{34}(x,y)+K_{35}(x,y).\end{split}$

For the first part $K_{31}(x,y),$ we have

$\begin{split}|K_{31}(x,y)|=&\Big{|}\frac{|x|y^{2}(y^{2}-2x\cdot
y)}{(|x-y|+|x|)|x|}+\frac{xy^{3}(|x-y|+|x|)}{(|x-y|+|x|)|x|}\Big{|}\\\
\lesssim&\frac{y^{4}}{|x-y|+|x|}+\Big{|}\frac{xy^{3}(|x-y|-|x|)}{(|x-y|+|x|)|x|}\Big{|}\\\
\lesssim&\frac{y^{4}}{|x-y|+|x|}.\end{split}$

Moreover,

$|K_{31}(x,y)|\lesssim\frac{y^{4}}{|x-y|+|x|}\lesssim\begin{cases}|y|^{3},\,\,\,&0\leq|x|\leq
1;\\\ \displaystyle\frac{y^{4}}{|x|},\,\,\,&|x|>1.\end{cases}$

Thus

$\Big{|}\int_{\mathbf{R}}K_{31}(x,y)v(y)f(y)dy\Big{|}\lesssim\int_{\mathbf{R}}\frac{y^{4}}{|x-y|+|x|}|v(y)f(y)|dy\in
L^{2}(\mathbf{R}).$

For the second part $K_{32}(x,y),$ we have

$\begin{split}|K_{32}(x,y)|=&\Big{|}\frac{x|x|y^{3}+xy^{3}|x-y|-2|x|xy^{3}}{|x|(|x-y|+|x)|}\Big{|}\\\
=&\Big{|}\frac{xy^{3}(|x-y|-|x|)}{|x|(|x-y|+|x)|}\Big{|}\lesssim\frac{y^{4}}{|x-y|+|x|}.\end{split}$

Thus

$\Big{|}\int_{\mathbf{R}}K_{32}(x,y)v(y)f(y)dy\Big{|}\lesssim\int_{\mathbf{R}}\frac{y^{4}}{|x-y|+|x|}|v(y)f(y)|dy\in
L^{2}(\mathbf{R}).$

For the third part $K_{33}(x,y),$ we have

$\begin{split}|K_{33}(x,y)|=&\frac{5}{2}\Big{|}\frac{x^{2}y^{2}(2x\cdot
y-y^{2})}{|x|(|x-y|+|x|)^{2}}-\frac{1}{2}\frac{xy^{3}(2|x-y|\cdot|x|+2x^{2}-2x\cdot
y+y^{2})}{|x|(|x-y|+|x|)^{2}}\Big{|}\\\
\lesssim&\frac{|x||y|^{5}}{|x|(|x-y|+|x|)^{2}}+\Big{|}\frac{xy^{3}(|x-y|-|x|)}{(|x-y|+|x|)^{2}}\Big{|}\\\
\lesssim&\frac{|x|y^{4}}{(|x-y|+|x|)^{2}}.\end{split}$

Noting that

$|K_{33}(x,y)|\lesssim\frac{|x|y^{4}}{(|x-y|+|x|)^{2}}\lesssim\begin{cases}|y|^{3},\,\,\,&0\leq|x|\leq
1;\\\ \displaystyle\frac{y^{4}}{|x|},\,\,\,&|x|>1.\end{cases}$ (3.35)

Thus

$\Big{|}\int_{\mathbf{R}}K_{33}(x,y)v(y)f(y)dy\Big{|}\lesssim\int_{\mathbf{R}}\frac{|x|y^{4}}{(|x-y|+|x|)^{2}}|v(y)f(y)|dy\in
L^{2}(\mathbf{R}).$

Then, for the fourth part $K_{34}(x,y),$ we have

$\begin{split}|K_{34}(x,y)|=&\Big{|}\frac{|x|x^{2}y^{2}(y^{2}-2x\cdot
y)}{|x|(|x-y|+|x|)^{3}}+\frac{xy^{3}(|x-y|+|x|)^{3}}{4|x|(|x-y|+|x|)^{3}}\Big{|}\\\
\lesssim&\frac{x^{2}y^{4}}{(|x-y|+|x|)^{3}}+\frac{|x|\cdot|y|^{5}|x-y|}{(|x-y|+|x|)^{3}\cdot|x|}\\\
&+\frac{|x|\cdot|y|^{5}}{(|x-y|+|x|)^{3}}+\Big{|}\frac{x^{3}y^{3}(|x-y|-|x|)}{(|x-y|+|x|)^{3}|x|}\Big{|}\\\
\lesssim&\frac{x^{2}y^{4}}{(|x-y|+|x|)^{3}}+\frac{|y|^{5}}{(|x-y|+|x|)^{2}}+\frac{|x|\cdot|y|^{5}}{(|x-y|+|x|)^{3}},\end{split}$

and it is obvious that

$\frac{x^{2}y^{4}}{(|x-y|+|x|)^{3}}+\frac{|y|^{5}}{(|x-y|+|x|)^{2}}+\frac{|x|\cdot|y|^{5}}{(|x-y|+|x|)^{3}}\lesssim\begin{cases}|y|^{3},\,\,\,&0\leq|x|\leq
1;\\\ \displaystyle\frac{y^{4}}{|x|},\,\,\,&|x|>1.\end{cases}$

Thus again

$\begin{split}&\Big{|}\int_{\mathbf{R}}K_{34}(x,y)v(y)f(y)dy\Big{|}\\\
&\lesssim\int_{\mathbf{R}}\Big{[}\frac{x^{2}y^{4}}{(|x-y|+|x|)^{3}}+\frac{|y|^{5}}{(|x-y|+|x|)^{2}}+\frac{|x|\cdot|y|^{5}}{(|x-y|+|x|)^{3}}\Big{]}|v(y)f(y)|dy\in
L^{2}(\mathbf{R}).\end{split}$

Finally, for the fifth part $K_{35}(x,y),$ using (3.35), it is easy to know
that

$\Big{|}\int_{\mathbf{R}}K_{35}(x,y)v(y)f(y)dy\Big{|}=\int_{\mathbf{R}}\frac{|x|y^{4}}{2(|x-y|+|x|)^{2}}|v(y)f(y)|dy\in
L^{2}(\mathbf{R}).$

Thus, we obtain that $\phi=-G_{0}vf\in L^{2}(\mathbf{R})$, and
$\int_{\mathbf{R}}K_{3}(x,y)v(y)f(y)dy\in L^{2}(\mathbf{R})$.

On the other hand, if $\phi\in L^{2}(\mathbf{R})$ such that $H\phi=0.$ Denote
$f=Uv\phi,$ using the similar method of (4.2) and (4.6), we have

$\begin{split}\int_{\mathbf{R}}v(y)f(y)dy&=0,\
\int_{\mathbf{R}}yv(y)f(y)dy=0,\ \int_{\mathbf{R}}y^{2}v(y)f(y)dy=0,\\\
&\int_{\mathbf{R}}y^{3}v(y)f(y)dy=0,S_{0}T_{0}S_{0}f=0,\end{split}$

then we can obtain that $f\in S_{1}L^{2}(\mathbf{R})$ and $G_{0}vf\in
L^{2}(\mathbf{R}).$ Moreover, by Theorem 4.1, we know that
$\phi=-G_{0}vf+c_{1}x+c_{2}$ since $L^{2}(\mathbf{R})\subset
W_{\frac{3}{2}}(\mathbf{R}).$ Thus we need that $c_{1}x+c_{2}=\phi+G_{0}vf\in
W_{\frac{1}{2}}(\mathbf{R})$ this necessitates that $c_{1}=c_{2}=0,$ which
implies that $T_{0}f=0$ and $\phi=-G_{0}vf$. Hence, we have $f\in
S_{3}L^{2}(\mathbf{R})$.

222The proof was suggested to us by M. Goldberg [Gold21]

Step II: Now we turn to prove that zero is not an eigenvalue of $H$ for $V(x)$
satisfies the condition in Theorem 1.

Let $\phi(x)\in L^{2}(\mathbf{R})$ such that $H\phi(x)=0.$ Denote $f=Uv\phi,$
by the Step I, then it follows that $f\in S_{3}L^{2}(\mathbf{R})$ and

$\displaystyle L^{2}(\mathbf{R})\ni\phi(x)=-G_{0}vf=$
$\displaystyle-\frac{1}{12}\int_{\mathbf{R}}|y-x|^{3}v(y)f(y)dy$
$\displaystyle=$
$\displaystyle-\frac{1}{12}\Big{(}\int_{x}^{\infty}(y-x)^{3}v(y)f(y)dy+\int_{-\infty}^{x}(x-y)^{3}v(y)f(y)dy\Big{)}.$

Then using the following orthogonality relationships from
$S_{3}L^{2}(\mathbf{R})$:

$\int_{\mathbf{R}}v(y)f(y)dy=0,\ \int_{\mathbf{R}}yv(y)f(y)dy=0,\
\int_{\mathbf{R}}y^{2}v(y)f(y)dy=0,\int_{\mathbf{R}}y^{3}v(y)f(y)dy=0,$

and $v(y)f(y)=V(y)\phi(y),$ we have

$\phi(x)=-\frac{1}{6}\int_{x}^{\infty}(y-x)^{3}v(y)f(y)dy=-\frac{1}{6}\int_{x}^{\infty}(y-x)^{3}V(y)\phi(y)dy.$

Based on the decay condition of Theorem 1, that is, $|V(x)|\lesssim\langle
x\rangle^{-\beta}$ for $\beta>25$, and $\langle x\rangle=(1+|x|^{2})^{1/2}$,
then $\displaystyle\lim_{x\rightarrow+\infty}|\phi(x)|=0$ and for any
$x_{0}>x,$

$\displaystyle|\phi(x)|\leq{1\over
6}\int_{x}^{x_{0}}(y-x)^{3}|V(y)|\cdot|\phi(y)|dy+\big{(}\sup_{y\in({x_{0}},+\infty)}|\phi(y)|\big{)}\int_{x_{0}}^{+\infty}{1\over
6}(y-x)^{3}|V(y)|dy.$

Hence, we have

$\displaystyle\langle x\rangle^{-3}|\phi(x)|\lesssim\int_{x}^{x_{0}}\langle
y\rangle^{6}|V(y)|\cdot\langle
y\rangle^{-3}|\phi(y)|dy+\Big{(}\sup_{y\in({x_{0}},+\infty)}|\phi(y)|\Big{)}\int_{x_{0}}^{+\infty}\langle
y\rangle^{3}|V(y)|dy.$

Let

$\displaystyle u(x)=\langle x\rangle^{-3}|\phi(x)|,\ \ B(y)=\langle
y\rangle^{6}|V(y)|$

and

$A(x_{0})=\displaystyle\big{(}\sup_{y\in({x_{0}},+\infty)}|\phi(y)|\big{)}\int_{x_{0}}^{+\infty}\langle
y\rangle^{3}|V(y)|dy.$

Then we have

$u(x)\lesssim\int_{x}^{x_{0}}B(y)u(y)dy+A(x_{0}).$

Hence by Gronwall’s inequality ( see e.g. [Tao06, Theorem 1.10] ), we conclude
that

$u(x)\lesssim A(x_{0})e^{C\int_{x}^{x_{0}}B(y)dy}.$

Note that $\langle y\rangle^{3}|V(y)|\in L^{1}(\mathbf{R})$, hence combining
with $\displaystyle\lim_{y\rightarrow+\infty}\phi(y)=0,$ we can obtain that
$\displaystyle\lim_{{x_{0}}\rightarrow+\infty}A({x_{0}})=0.$ Moreover,
$\int_{x}^{x_{0}}B(y)dy$ can be bounded by a fixed constant. Hence for any
$x\in\mathbf{R},$

$u(x)\lesssim\lim_{x_{0}\rightarrow+\infty}A(x_{0})e^{C\int_{x}^{x_{0}}B(y)dy}=0,$

which implies $\phi(x)\equiv 0.$ That is, zero can not be an eigenvalue of
$H$.

Combined Step I with Step II, we can conclude that
$S_{3}L^{2}(\mathbf{R})=\\{0\\}.$ ∎

######  Remark 3.6.

Since the proof of Step II depends on the orthogonality relationships of
$S_{3}L^{2}(\mathbf{R})$, so we need the decay rate $\beta$ of $V$ satisfies
(1.5). However, we remark that the required decay rate $\beta$ in the proof
above may not be optimal to ensure the absence of zero eigenvalue of $H$. In
fact, in view of Remark 1.2, it is reasonable to guess that $\beta>4$ might be
the best one.

In the following, we come to show the proof of Theorem 1.8 case by case. For
the convenience, we will use the notation $A_{kj}^{i}$ ($i$ means the kind of
zero resonance, $k$ respects to the power of $\mu$) as some bounded operators
in $L^{2}(\mathbf{R})$, and these operators may vary from line to line.

Proof of Theorem 1.8: We prove the assertion for $+$ sign. By Lemma 3.3, we
know that $\widetilde{M}^{+}(\mu)$ is invertible on $L^{2}(\mathbf{R})$ if and
only if

$M_{1}^{+}(\mu):=Q-Q\Big{(}\widetilde{M^{+}}(\mu)+Q\Big{)}^{-1}Q$ (3.36)

is invertible on $QL^{2}(\mathbf{R})$. Moreover, using (3.14) and by the
definition of $B_{1}^{+},$ we get

$\begin{split}M_{1}^{+}(\mu)=&\sum_{k=1}^{12}\mu^{k+1}QB_{k}^{+}Q+Q\Gamma_{14}(\mu)Q\\\
=&\frac{a_{-1}^{+}}{\widetilde{a}^{+}}\mu^{2}\Big{(}QvG_{-1}vQ+\frac{\widetilde{a}^{+}}{a_{1}^{+}}\sum_{k=2}^{12}\mu^{k-1}QB_{k}^{+}Q+Q\Gamma_{12}(\mu)Q\Big{)}\\\
:=&\frac{a_{-1}^{+}}{\widetilde{a}^{+}}\mu^{2}\widetilde{M_{1}}^{+}(\mu).\end{split}$
(3.37)

Thus, in order to compute the expansions of
$\Big{(}M_{1}^{+}(\mu)\Big{)}^{-1}$, we need to obtain the expansions of
$\Big{(}\widetilde{M_{1}^{+}}(\mu)\Big{)}^{-1}$ as $\mu\rightarrow 0$ on
$QL^{2}(\mathbf{R})$.

By Lemma 3.4(1), we know that $QvG_{-1}vQ$ is not invertible on
$QL^{2}(\mathbf{R})$. Since $S_{0}$ be the Riesz projection onto the kernel of
$QvG_{-1}vQ$, then $QvG_{-1}vQ+S_{0}$ is invertible on $QL^{2}(\mathbf{R})$.
By using (3.37), we have

$\begin{split}\widetilde{M_{1}^{+}}(\mu)+S_{0}=&\big{(}QvG_{-1}vQ+S_{0}\big{)}+\frac{\widetilde{a}^{+}}{a_{1}^{+}}\sum_{k=2}^{12}\mu^{k-1}QB_{k}^{+}Q+Q\Gamma_{12}(\mu)Q.\end{split}$
(3.38)

Then, using the Neumann series, we obtain that

$\begin{split}\Big{(}\widetilde{M_{1}^{+}}(\mu)+S_{0}\Big{)}^{-1}=&D_{0}-\sum_{k=1}^{11}\mu^{k}B_{k}^{0}+D_{0}\Gamma_{12}(\mu)D_{0},\end{split}$
(3.39)

where $B_{k}^{0}\ (1\leq k\leq 11)$ are bounded operators in
$QL^{2}(\mathbf{R})$ and we have

$\begin{split}B_{1}^{0}=&\frac{\widetilde{a}^{+}}{a_{-1}^{+}}D_{0}B_{2}^{+}D_{0},\
\ \ \ \ \
B_{2}^{0}=\frac{\widetilde{a}^{+}}{a_{-1}^{+}}D_{0}B_{3}^{+}D_{0}-\frac{(\widetilde{a}^{+})^{2}}{(a_{-1}^{+})^{2}}D_{0}(B_{2}^{+}D_{0})^{2},\\\
B_{3}^{0}=&\frac{\widetilde{a}^{+}}{a_{-1}^{+}}D_{0}B_{4}^{+}D_{0}+\frac{(\widetilde{a}^{+})^{3}}{(a_{-1}^{+})^{3}}D_{0}(B_{2}^{+}D_{0})^{3}-\frac{(\widetilde{a}^{+})^{2}}{(a_{-1}^{+})^{2}}\big{(}D_{0}B_{2}^{+}D_{0}B_{3}^{+}D_{0}+D_{0}B_{3}^{+}D_{0}B_{2}^{+}D_{0}\big{)},\\\
B_{4}^{0}=&\frac{\widetilde{a}^{+}}{a_{-1}^{+}}D_{0}B_{5}^{+}D_{0}-\frac{(\widetilde{a}^{+})^{4}}{(a_{-1}^{+})^{4}}D_{0}(B_{2}^{+}D_{0})^{4}+\frac{(\widetilde{a}^{+})^{3}}{(a^{+}_{-1})^{3}}\Big{(}D_{0}(B_{2}^{+}D_{0})^{2}B_{3}^{+}D_{0}+D_{0}B_{2}^{+}D_{0}B_{5}^{+}D_{0}B_{2}^{+}D_{0}\\\
&+D_{0}B_{3}^{+}D_{0}(B_{2}^{+}D_{2})^{2}\Big{)}-\frac{(\widetilde{a}^{+})^{2}}{(a_{-1}^{+})^{2}}\Big{(}D_{0}B_{2}^{+}D_{0}B_{4}^{+}D_{0}+D_{0}B_{4}^{+}D_{0}B_{2}^{+}D_{0}+D_{0}(B_{3}^{+}D_{0})^{2}\Big{)},\\\
\end{split}$
$\begin{split}B_{5}^{0}=&\frac{\widetilde{a}^{+}}{a_{-1}^{+}}D_{0}B_{6}^{+}D_{0}+\frac{(\widetilde{a}^{+})^{5}}{(a_{-1}^{+})^{5}}D_{0}(B_{2}^{+}D_{0})^{5}\\\
&-\frac{(\widetilde{a}^{+})^{4}}{(a_{-1}^{+})^{4}}\Big{(}D_{0}(B_{2}^{+}D_{0})^{3}B_{3}^{+}D_{0}+D_{0}(B_{2}^{+}D_{0})^{2}B_{3}^{+}D_{0}B_{2}^{+}D_{0}+D_{0}B_{2}^{+}D_{0}B_{3}^{+}D_{0}(B_{2}^{+}D_{0})^{2}\\\
&+D_{0}B_{3}^{+}D_{0}(B_{2}^{+}D_{0})^{3}\Big{)}+\frac{(\widetilde{a}^{+})^{3}}{(a_{-1}^{+})^{3}}\Big{(}D_{0}B_{2}^{+}D_{0}(B_{3}^{+}D_{0})^{2}+D_{0}B_{3}^{+}D_{0}B_{2}^{+}D_{0}B_{3}^{+}D_{0}+D_{0}(B_{3}^{+}D_{0})^{2}B_{2}^{+}D_{0}\\\
&+D_{0}(B_{2}^{+}D_{0})^{2}B_{4}^{+}D_{0}+D_{0}B_{2}^{+}D_{0}B_{4}^{+}D_{0}B_{2}^{+}D_{0}+D_{0}B_{4}^{+}D_{0}(B_{2}^{+}D_{0})^{2}\Big{)}\\\
&-\frac{(\widetilde{a}^{+})^{2}}{(a_{-1}^{+})^{2}}\Big{(}D_{0}B_{3}^{+}D_{0}B_{4}^{+}D_{0}+D_{0}B_{4}^{+}D_{0}B_{3}^{+}D_{0}+D_{0}B_{2}^{+}D_{0}B_{5}^{+}D_{0}+D_{0}B_{5}^{+}D_{0}B_{2}^{+}D_{0}\Big{)}.\end{split}$

Using Lemma 3.3 again, we know that $\widetilde{M_{1}^{+}}(\mu)$ invertible on
$QL^{2}(\mathbf{R})$ if and only if

$M_{2}^{+}(\mu):=S_{0}-S_{0}\big{(}\widetilde{M_{1}^{+}}(\mu)+S_{0}\big{)}^{-1}S_{0}$
(3.40)

invertible on $S_{0}L^{2}(\mathbf{R})$. Notice that
$S_{0}D_{0}=D_{0}S_{0}=S_{0}$, and using (3.39), we obtain that

$\begin{split}M_{2}^{+}(\mu)=&\mu
S_{0}B_{1}^{0}S_{0}+\sum_{k=2}^{11}\mu^{k}B_{k}^{0}+S_{0}\Gamma_{12}(\mu)S_{0}\\\
=&\frac{\mu}{a_{-1}^{+}}\Big{(}S_{0}T_{0}S_{0}+a_{-1}^{+}\sum_{k=2}^{11}\mu^{k-1}B_{k}^{0}+S_{0}\Gamma_{11}(\mu)S_{0}\Big{)}\\\
:=&\frac{\mu}{a_{-1}^{+}}\widetilde{M_{2}^{+}}(\mu).\end{split}$ (3.41)

(1) Regular Case. In this case, we using $\eqref{eq-M-expansions-1}$ which
need $|V(x)|\lesssim(1+|x|)^{-13-},$ see Lemma 3.1. Then by the same process,
we can rewrite $M_{2}^{+}(\mu)$ as follows:

$\begin{split}M_{2}^{+}(\mu)=&\mu
S_{0}B_{1}^{0}S_{0}+\mu^{2}B_{2}^{0}+\mu^{3}B_{3}^{0}+S_{0}\Gamma_{4}(\mu)S_{0}\\\
=&\frac{\mu}{a_{-1}^{+}}\Big{(}S_{0}T_{0}S_{0}+a_{-1}^{+}\mu
B_{2}^{0}+a_{-1}^{+}\mu^{2}B_{3}^{0}+S_{0}\Gamma_{3}(\mu)S_{0}\Big{)}\\\
:=&\frac{\mu}{a_{-1}^{+}}\widetilde{M_{2}^{+}}(\mu).\end{split}$

If zero is a regular point of the spectrum of $H,$ then $S_{0}T_{0}S_{0}$ is
invertible on $S_{0}L^{2}(\mathbf{R}).$ Denote
$\widetilde{D_{0}}=\big{(}S_{0}T_{0}S_{0}\big{)}^{-1},$ by (3.41), using
Neumann series and
$S_{0}\widetilde{D}_{0}=\widetilde{D}_{0}S_{0}=\widetilde{D}_{0},$ we obtain
that

$\begin{split}\Big{(}\widetilde{M_{2}^{+}}(\mu)\Big{)}^{-1}=&\widetilde{D_{0}}-a_{-1}^{+}\mu\widetilde{D}_{0}B_{2}^{0}\widetilde{D}_{0}+\mu^{2}\big{(}(a_{-1}^{+})^{2}\widetilde{D}_{0}(B_{2}^{0}\widetilde{D}_{0})^{2}-a_{-1}^{+}\widetilde{D}_{0}B_{3}^{0}\widetilde{D}_{0}\big{)}+\Gamma_{3}(\mu).\end{split}$

Then, using (3.41), we have

$\begin{split}\Big{(}M_{2}^{+}(\mu)\Big{)}^{-1}=&\frac{a_{-1}^{+}}{\mu}\widetilde{D_{0}}-(a_{-1}^{+})^{2}\widetilde{D}_{0}B_{2}^{0}\widetilde{D}_{0}+\mu\big{(}(a_{-1}^{+})^{3}\widetilde{D}_{0}(B_{2}^{0}\widetilde{D}_{0})^{2}-(a_{-1}^{+})^{2}\widetilde{D}_{0}B_{3}^{0}\widetilde{D}_{0}\big{)}+\Gamma_{2}(\mu).\end{split}$

By Lemma 3.3 with $A=\widetilde{M_{1}^{+}}(\mu)$, $S=S_{0}$ and
$B=M_{2}^{+}(\mu)$, using the orthogonality properties ( see (3.16), (3.17)),
we obtain that

$\begin{split}\Big{(}\widetilde{M_{1}^{+}}(\mu)\Big{)}^{-1}=\
&\Big{(}\widetilde{M_{1}^{+}}(\mu)+S_{0}\Big{)}^{-1}+\Big{(}\widetilde{M_{1}^{+}}(\mu)+S_{0}\Big{)}^{-1}S_{0}\Big{(}M_{2}^{+}(\mu)\Big{)}^{-1}S_{0}\Big{(}\widetilde{M_{1}^{+}}(\mu)+S_{0}\Big{)}^{-1}\\\
=\ &\frac{S_{0}A_{-1,1}^{0}S_{0}}{\mu}+QA_{01}^{0}Q+\mu
QA_{11}^{0}Q+\Gamma_{2}(\mu).\end{split}$

Then, using (3.37), we have

$\begin{split}\Big{(}M_{1}^{+}(\mu)\Big{)}^{-1}=&\frac{S_{0}A_{-3,1}^{0}S_{0}}{\mu^{3}}+\frac{QA_{-2,1}^{0}Q}{\mu^{2}}+\frac{QA_{-1,1}^{0}Q}{\mu}+\Gamma_{0}(\mu).\end{split}$

Then, we use Lemma 3.3 with $A=\widetilde{M^{+}}(\mu),S=Q$ and
$B=M_{1}^{+}(\mu)$, and using the orthogonality properties ( see (3.16),
(3.17)), we have

$\begin{split}\Big{(}\widetilde{M^{+}}(\mu)\Big{)}^{-1}=\
&\Big{(}\widetilde{M^{+}}(\mu)+Q\Big{)}^{-1}+\Big{(}\widetilde{M^{+}}(\mu)+Q\Big{)}^{-1}Q\Big{(}M_{1}^{+}(\mu)\Big{)}^{-1}Q\Big{(}\widetilde{M^{+}}(\mu)+Q\Big{)}^{-1}\\\
=&\frac{S_{0}A_{-3,1}^{0}S_{0}}{\mu^{3}}+\frac{QA_{-2,1}^{0}Q}{\mu^{2}}+\frac{QA_{-1,1}^{0}Q+S_{0}{A_{-1,2}^{0}}+{A_{-1,3}^{0}}S_{0}}{\mu}+\Gamma_{0}(\mu).\end{split}$

Finally, since $\displaystyle
M^{+}(\mu)=\frac{\widetilde{a}^{+}}{\mu^{3}}\widetilde{M^{+}}(\mu),$ then we
can say that

$\begin{split}\Big{(}M^{+}(\mu)\Big{)}^{-1}=&S_{0}A_{01}^{0}S_{0}+\mu
QA_{11}^{0}Q+\mu^{2}(QA_{21}^{0}Q+S_{0}A_{22}^{0}+A_{23}^{0}S_{0})+\Gamma_{3}^{0}(\mu).\end{split}$
(3.42)

(2) First Kind of Resonance. In this case, we using $\eqref{eq-M-
expansions-1-2}$ with $|V(x)|\lesssim(1+|x|)^{-17-},$ see Lemma 3.1. Then by
the same process, we can rewrite $M_{2}^{+}(\mu)$ as follows:

$\begin{split}M_{2}^{+}(\mu)=&\mu
S_{0}B_{1}^{0}S_{0}+\sum_{k=2}^{5}\mu^{k}B_{k}^{0}+S_{0}\Gamma_{6}(\mu)S_{0}\\\
=&\frac{\mu}{a_{-1}^{+}}\Big{(}S_{0}T_{0}S_{0}+a_{-1}^{+}\sum_{k=2}^{5}\mu^{k-1}B_{k}^{0}+S_{0}\Gamma_{5}(\mu)S_{0}\Big{)}\\\
:=&\frac{\mu}{a_{-1}^{+}}\widetilde{M_{2}^{+}}(\mu).\end{split}$

For the first kind of resonance, we aim to derive the expansions of
$\Big{(}\widetilde{M_{2}^{+}}(\mu)\Big{)}^{-1}.$ By the definition of the
first kind of resonance, we know that the operator $S_{0}T_{0}S_{0}$ is not
invertible on $S_{0}L^{2}(\mathbf{R}).$ Since $S_{1}$ is the Riesz projection
onto the kernel of $S_{0}T_{0}S_{0}$ on $S_{0}L^{2}(\mathbf{R}),$ thus
$S_{0}T_{0}S_{0}+S_{1}$ is invertible on $S_{0}L^{2}(\mathbf{R}).$ By Neumann
series, we have

$\begin{split}\Big{(}\widetilde{M_{2}^{+}}(\mu)+S_{1}\Big{)}^{-1}=&D_{1}-\sum_{k=1}^{4}\mu^{k}B_{k}^{1}+\Gamma_{5}(\mu),\end{split}$
(3.43)

where $B_{k}^{1}\ (1\leq k\leq 4)$ are bounded operators in
$S_{0}L^{2}(\mathbf{R})$ with the following definition

$\begin{split}B_{1}^{1}=&a_{-1}^{+}D_{1}B_{2}^{0}D_{1},\ \ \ \ \ \
B_{2}^{1}=a_{-1}^{+}D_{1}B_{3}^{0}D_{1}-(a_{-1}^{+})^{2}D_{1}(B_{2}^{0}D_{1})^{2},\\\
B_{3}^{1}=&a_{-1}^{+}D_{1}B_{4}^{0}D_{1}+(a_{-1}^{+})^{3}D_{1}(B_{1}^{0}D_{1})^{3}-(a_{-1}^{+})^{2}\big{(}D_{1}B_{2}^{0}D_{1}B_{3}^{0}D_{1}+D_{1}B_{3}^{0}D_{1}B_{2}^{0}D_{1}\big{)},\\\
B_{4}^{1}=&(a_{-1}^{+})D_{1}B_{5}^{0}D_{1}-(a_{-1}^{+})^{4}D_{1}(B_{2}^{0}D_{1})^{4}\\\
&+(a_{-1}^{+})^{3}\Big{(}D_{1}(B_{2}^{0}D_{1})^{2}B_{3}^{0}D_{1}+D_{1}B_{2}^{0}D_{1}B_{3}^{0}D_{1}B_{2}^{0}D_{1}+D_{1}B_{3}^{0}D_{1}(B_{2}^{0}D_{1})^{2}\Big{)}\\\
&+(a_{-1}^{+})^{2}\Big{(}D_{1}B_{2}^{0}D_{1}B_{4}^{0}D_{1}+D_{1}B_{4}^{0}D_{1}B_{2}^{0}D_{1}+D_{1}(B_{3}^{0}D_{1})^{2}\Big{)}.\end{split}$
(3.44)

According to Lemma 3.3, we know that $\widetilde{M_{2}^{+}}(\mu)$ is
invertible on $S_{0}L^{2}(\mathbf{R})$ if and only if

$M_{3}^{+}(\mu):=S_{1}-S_{1}\big{(}\widetilde{M_{2}^{+}}(\mu)+S_{1}\big{)}^{-1}S_{1}$
(3.45)

is invertible on $S_{1}L^{2}(\mathbf{R})$. Moreover, using (3.43) and (3.17),
we obtain

$\begin{split}M_{3}^{+}(\mu)=&\mu
S_{1}B_{1}^{1}S_{1}+\mu^{2}S_{1}B_{2}^{1}S_{1}+\mu^{3}S_{1}B_{3}^{1}S_{1}+\mu^{4}S_{1}B_{4}^{1}S_{1}+S_{1}\Gamma_{5}(\mu)S_{1}.\\\
\end{split}$ (3.46)

Now, we shall compute the $S_{1}B_{1}^{1}S_{1}$ in $M_{3}^{+}(\mu)$. Noting
that $S_{1}D_{1}=D_{1}S_{1}=S_{1}$ , we have
$S_{1}B_{1}^{1}S_{1}=a_{-1}^{+}S_{1}B_{2}^{0}S_{1}$. Since

$\begin{split}B_{2}^{0}&=\frac{\widetilde{a}^{+}}{a_{-1}^{+}}D_{0}B_{3}^{+}D_{0}-\frac{(\widetilde{a}^{+})^{2}}{(a_{-1}^{+})^{2}}D_{0}(B_{2}^{+}D_{0})^{2}\\\
&=\frac{a_{1}^{+}}{a_{-1}^{+}}D_{0}vG_{1}vD_{0}-\frac{a_{-1}^{+}}{\widetilde{a}^{+}}D_{0}(vG_{-1}v)^{2}D_{0}-\frac{1}{(a_{-1}^{+})^{2}}D_{0}T_{0}D_{0}T_{0}D_{0}\end{split}$

and $S_{1}D_{0}=D_{0}S_{1}=S_{1}$, we have

$\begin{split}S_{1}B^{1}_{1}S_{1}=a_{-1}^{+}\Big{(}\frac{a_{1}^{+}}{a_{-1}^{+}}S_{1}vG_{1}vS_{1}-\frac{a_{-1}^{+}}{\widetilde{a}^{+}}S_{1}(vG_{-1}v)^{2}S_{1}-\frac{1}{(a_{-1}^{+})^{2}}S_{1}T_{0}D_{0}T_{0}S_{1}\Big{)}.\end{split}$
(3.47)

Notice that $(vG_{-1}v)^{2}=vG_{-1}vvG_{-1}v=vG_{-1}vPvG_{-1}v$, we have

$S_{1}B_{1}^{1}S_{1}=a_{1}^{+}T_{1},$ (3.48)

where

$\begin{split}T_{1}=&S_{1}vG_{1}vS_{1}-\frac{(a_{-1}^{+})^{2}}{\widetilde{a}^{+}a_{1}^{+}}S_{1}vG_{-1}vPvG_{-1}vS_{1}-\frac{1}{a_{1}^{+}a_{-1}^{+}}S_{1}T_{0}D_{0}T_{0}S_{1}\\\
=&S_{1}vG_{1}vS_{1}+\frac{6}{\|V\|_{L^{1}}}S_{1}vG_{-1}vPvG_{-1}vS_{1}-384S_{1}T_{0}D_{0}T_{0}S_{1}.\end{split}$
(3.49)

Recall that (3.46) and (3.48), we can rewrite $M_{3}^{+}(\mu)$ as follows:

$\begin{split}M_{3}^{+}(\mu)=&a_{1}^{+}\mu
T_{1}+\mu^{2}S_{1}B_{2}^{1}S_{1}+\mu^{3}S_{1}B_{3}^{1}S_{1}+\mu^{4}S_{1}B_{4}^{1}S_{1}+S_{1}\Gamma_{5}(\mu)S_{1}\\\
=&a_{1}^{+}\mu\Big{(}T_{1}+\frac{1}{a_{1}^{+}}(\mu
S_{1}B_{2}^{1}S_{1}+\mu^{2}S_{1}B_{3}^{1}S_{1}+\mu^{3}S_{1}B_{4}^{1}S_{1})+S_{1}\Gamma_{4}(\mu)S_{1}\Big{)}\\\
:=&a_{1}^{+}\mu\widetilde{M_{3}^{+}}(\mu).\end{split}$

Since $T_{1}$ is invertible on $S_{1}L^{2}(\mathbf{R}),$ we have
$S_{1}T_{1}^{-1}=T_{1}^{-1}S_{1}=T_{1}^{-1},$ then using Neumann series, we
have the following asymptotic expansions of
$\Big{(}\widetilde{M_{3}^{+}}(\mu)\Big{)}^{-1}$ as $\mu\rightarrow 0:$

$\begin{split}\Big{(}\widetilde{M_{3}^{+}}(\mu)\Big{)}^{-1}=\
&S_{1}A_{01}^{1}S_{1}+\mu
S_{1}A_{11}^{1}S_{1}+\mu^{2}S_{1}A_{21}^{1}S_{1}+\mu^{3}S_{1}A_{31}^{1}S_{1}+\Gamma_{4}(\mu).\end{split}$

Then, we have

$\begin{split}\Big{(}M_{3}^{+}(\mu)\Big{)}^{-1}=&\frac{S_{1}A_{-1,1}^{1}S_{1}}{\mu}+S_{1}A_{01}^{1}S_{1}+\mu
S_{1}A_{11}^{1}S_{1}+\mu^{2}S_{1}A_{21}^{1}S_{1}+\Gamma_{3}(\mu).\end{split}$

Next, by Lemma 3.3 and using (3.17), we have

$\begin{split}\Big{(}\widetilde{M_{2}^{+}}(\mu)\Big{)}^{-1}=&\Big{(}\widetilde{M_{2}^{+}}(\mu)+S_{1}\Big{)}^{-1}+\Big{(}\widetilde{M_{2}^{+}}(\mu)+S_{1}\Big{)}^{-1}S_{1}\Big{(}M_{3}^{+}(\mu)\Big{)}^{-1}S_{1}\Big{(}\widetilde{M_{2}^{+}}(\mu)+S_{1}\Big{)}^{-1}\\\
=&\frac{S_{1}A_{-1,1}^{1}S_{1}}{\mu}+S_{0}A_{01}^{1}S_{0}+\mu
S_{0}A_{11}^{1}S_{0}+\mu^{2}S_{0}A_{21}^{1}S_{0}+\Gamma_{3}(\mu).\end{split}$

Furthermore, we have

$\Big{(}M_{2}^{+}(\mu)\Big{)}^{-1}=\frac{S_{1}A_{-2,1}^{1}S_{1}}{\mu^{2}}+\frac{S_{0}A_{-1,1}^{1}S_{0}}{\mu}+S_{0}A_{01}^{1}S_{0}+\mu
S_{0}A_{11}^{1}S_{0}+\Gamma_{2}(\mu).$

Then using the orthogonality properties (see (3.16),(3.16)), and following the
similar process of Regular Case, we obtain that