asymptotic expansions of $\big{(}M^{\pm}(\lambda)\big{)}^{-1}$ for
$\lambda\rightarrow 0$.

Let $\displaystyle
M^{\pm}(\lambda)=\frac{a_{\pm}}{\lambda^{2}}\widetilde{M}^{\pm}(\lambda)$.
Then we just need to compute the asymptotic expansion of
$\big{(}\widetilde{M}^{\pm}(\lambda)\big{)}^{-1}$ for $\lambda\rightarrow 0$.
Recall that from the (iii) of Lemma 2.3

$\begin{split}\widetilde{M}^{\pm}(\lambda)=&P+\frac{g_{0}^{\pm}(\lambda)}{a_{\pm}}\lambda^{2}vG_{-1}v+\frac{1}{a_{\pm}}\lambda^{2}T_{0}+\frac{c_{\pm}}{a_{\pm}}\lambda^{4}vG_{1}v\\\
&+\frac{\tilde{g}_{2}^{\pm}(\lambda)}{a_{\pm}}\lambda^{6}vG_{2}v+\frac{\lambda^{6}}{a_{\pm}}vG_{3}v+O\big{(}\lambda^{8}\big{)}.\end{split}$
(4.3)

Note that projection $P$ is not invertible on $L^{2}$, but let $Q=I-P$, then
$\widetilde{M}^{\pm}(\lambda)+Q$ is invertible on $L^{2}$ for small
$\lambda>0$. Hence by Lemma 4.1 we know that $\widetilde{M}^{\pm}(\lambda)$ is
invertible on $L^{2}$ if and only if

$M_{1}^{\pm}(\lambda):=Q-Q\big{(}\widetilde{M}^{\pm}(\lambda)+Q\big{)}^{-1}Q,$

is invertible on $QL^{2}$. In order to get the invertibility of
$M_{1}^{\pm}(\lambda)$, we first need to compute
$\big{(}\widetilde{M}^{\pm}(\lambda)+Q\big{)}^{-1}$ on $L^{2}$, which can be
given by Neumann series

$\begin{split}&\big{(}\widetilde{M}^{\pm}(\lambda)+Q\big{)}^{-1}\\\
=&I-g^{\pm}_{0}(\lambda)\lambda^{2}B_{1}^{\pm}-\lambda^{2}B_{2}^{\pm}-g^{\pm}_{0}(\lambda)^{2}\lambda^{4}B_{3}^{\pm}-g^{\pm}_{0}(\lambda)\lambda^{4}B_{4}^{\pm}-\lambda^{4}B_{5}^{\pm}-g_{0}^{\pm}(\lambda)^{3}\lambda^{6}B_{6}^{\pm}\\\
&-g_{0}^{\pm}(\lambda)^{2}\lambda^{6}B_{7}^{\pm}-g^{\pm}_{0}(\lambda)\lambda^{6}B_{8}^{\pm}-\tilde{g}^{\pm}_{2}(\lambda)\lambda^{6}B_{9}^{\pm}-\lambda^{6}B_{10}^{\pm}+O\big{(}\lambda^{8-\epsilon}\big{)},\end{split}$
(4.4)

where $B_{j}^{\pm}(1\leq j\leq 10)$ are bounded operators in $L^{2}$ as
follows:
$\displaystyle B_{1}^{\pm}=\frac{1}{a_{\pm}}vG_{-1}v$, $\displaystyle
B_{2}^{\pm}=\frac{1}{a_{\pm}}T_{0}$, $\displaystyle
B_{3}^{\pm}=-\frac{1}{a_{\pm}^{2}}(vG_{-1}v)^{2}$, $\displaystyle
B_{4}^{\pm}=-\frac{1}{a_{\pm}^{2}}(vG_{-1}vT_{0}+T_{0}vG_{-1}v)$,
$\displaystyle
B_{5}^{\pm}=\frac{c_{\pm}}{a_{\pm}}vG_{1}v-\frac{1}{a_{\pm}^{2}}T_{0}^{2}$ ,
$\displaystyle B_{6}^{\pm}=\frac{1}{a_{\pm}^{3}}(vG_{-1}v)^{3}$,
$\displaystyle
B_{7}^{\pm}=\frac{1}{a_{\pm}^{3}}\big{[}(vG_{-1}v)^{2}T_{0}+T_{0}(vG_{-1}v)^{2}\big{]}$,
$\displaystyle
B_{8}^{\pm}=\frac{1}{a_{\pm}^{3}}(vG_{-1}vT_{0}^{2}+T_{0}^{2}vG_{-1}v)-\frac{c_{\pm}}{a_{\pm}^{2}}(vG_{-1}v\cdot
vG_{1}v+vG_{1}v\cdot vG_{-1}v)$,
$\displaystyle B_{9}^{\pm}=\frac{1}{a_{\pm}}vG_{2}v$, $\displaystyle
B_{10}^{\pm}=\frac{1}{a_{\pm}}vG_{3}v+\frac{1}{a_{\pm}^{3}}T_{0}^{3}-\frac{c_{\pm}}{a_{\pm}^{2}}(T_{0}vG_{1}v+vG_{1}vT_{0})$.

Before we study the invertibility of $M_{1}^{\pm}(\lambda)$, let’s list the
following orthogonality relations of various operators and projections from
Definition 2.4, which are used frequently later.

$\begin{split}QD_{0}&=D_{0}Q=D_{0};\ \ Q\geq S_{i}\geq S_{j},\ 0\leq i\leq
j\leq 5;\\\ S_{i}D_{j}&=D_{j}S_{i}=S_{i},i\geq j,\
S_{i}D_{j}=D_{j}S_{i}=D_{i},i<j,\ D_{i}D_{j}=D_{j}D_{i}=D_{i},i>j;\end{split}$
(4.5) $\begin{split}&QvG_{-1}vS_{0}=S_{0}QvG_{-1}vS_{0}=0,\
S_{1}T_{0}S_{0}=S_{0}T_{0}S_{1}=0,\ S_{2}T_{0}Q=QT_{0}S_{2}=0;\\\
&S_{3}vG_{-1}v=vG_{-1}vS_{3}=0,\ S_{4}T_{0}P=PT_{0}S_{4}=0,\
S_{4}T_{0}=T_{0}S_{4}=0;\\\ &S_{4}vG_{1}vS_{0}=S_{0}vG_{1}vS_{4}=0,\
S_{5}vG_{1}vQ=QvG_{1}vS_{5}=0.\end{split}$ (4.6)

Moreover, we can obtain

$\begin{split}QB_{1}^{\pm}S_{0}=S_{0}B_{1}^{\pm}Q=S_{3}B_{1}^{\pm}=B_{1}^{\pm}S_{3}=S_{3}B_{3}^{\pm}=B_{3}^{\pm}S_{3}=S_{3}B_{4}^{\pm}S_{3}=0,\,S_{4}B_{8}^{\pm}S_{4}=0;\\\
S_{4}B_{i}^{\pm}=B_{i}^{\pm}S_{4}=0(i=1,2,3,4,6,7),\
S_{4}B_{5}^{\pm}S_{0}=S_{0}B^{\pm}_{5}S_{4}=S_{5}B_{5}^{\pm}Q=QB^{\pm}_{5}S_{5}=0.\end{split}$
(4.7)

Now we turn to the proof of the invertibility of $M_{1}^{\pm}(\lambda)$, We
only prove the assertion for the case $+$ sign, the case $-$ sign proceeds
identically. By Lemma 4.1, we know that $\widetilde{M}^{+}(\lambda)$ is
invertible on $L^{2}$ if and only if

$M_{1}^{+}(\lambda):=Q-Q\big{(}\widetilde{M}^{+}(\lambda)+Q\big{)}^{-1}Q$

is invertible on $QL^{2}$. Moreover, by (4.4) we get

$\begin{split}M_{1}^{+}(\lambda):=\frac{\lambda^{2}}{a_{+}}\widetilde{M}_{1}^{+}(\lambda)=\frac{\lambda^{2}}{a_{+}}\Big{(}A^{+}(\lambda)+a_{+}W^{+}(\lambda)\Big{)},\end{split}$
(4.8)

where $A^{+}(\lambda)=g^{+}_{0}(\lambda)QvG_{-1}vQ+QT_{0}Q$, and

$\begin{split}W^{+}(\lambda)&=g^{+}_{0}(\lambda)^{2}\lambda^{2}QB_{3}^{+}Q+g^{+}_{0}(\lambda)\lambda^{2}QB_{4}^{+}Q+\lambda^{2}QB_{5}^{+}Q+g_{0}^{+}(\lambda)^{3}\lambda^{4}QB_{6}^{+}Q\\\
&+g_{0}^{+}(\lambda)^{2}\lambda^{4}QB_{7}^{+}Q+g^{+}_{0}(\lambda)\lambda^{4}QB_{8}^{+}Q+\tilde{g}^{+}_{2}(\lambda)\lambda^{4}QB_{9}^{+}Q+\lambda^{4}QB_{10}^{+}Q+O\big{(}\lambda^{6-\epsilon}\big{)}.\end{split}$
(4.9)

In order to compute $\big{(}M_{1}^{+}(\lambda)\big{)}^{-1}$, by (4.8 ) we need
to obtain the expansions of
$\big{(}\widetilde{M_{1}^{+}}(\lambda)\big{)}^{-1}$ on $QL^{2}$ when $\lambda$
is near zero. Note that $A^{+}(\lambda)=g^{+}_{0}(\lambda)QvG_{-1}vQ+QT_{0}Q$
has the logarithm factor $g^{+}_{0}(\lambda)$ which becomes $\infty$ as
$\lambda\downarrow 0$, and $QvG_{-1}vQ$ is not invertible on $QL^{2}$ by Lemma
2.7(i), hence we will use Lemma 4.2 to consider the inverse of
$A^{+}(\lambda)$.

To prove the invertibility of $A^{+}(\lambda)$ by Lemma 4.2, we first give a
lemma as follows.

######  Lemma 4.3.

Suppose that $|V(x)|\lesssim(1+|x|)^{-\beta}$ with some $\beta>10$. Then the
inverse operator

$\begin{split}d:=\Big{(}g^{+}_{0}(\lambda)(Q-S_{0})vG_{-1}v(Q-S_{0})+(Q-S_{0})\big{(}T_{0}-T_{0}D_{1}T_{0}\big{)}(Q-S_{0})\Big{)}^{-1}\end{split}$
(4.10)

exists on $(Q-S_{0})L^{2}(\mathbb{R}^{2})$ for $0<\lambda\ll 1$. Moreover, we
have

$\displaystyle d=g^{+}_{0}(\lambda)^{-1}(Q-S_{0})F(\lambda)(Q-S_{0}),$

where $F(\lambda)=\big{(}A+g^{+}_{0}(\lambda)^{-1}B\big{)}^{-1}$ with
$A=(Q-S_{0})vG_{-1}v(Q-S_{0})$ and
$B=(Q-S_{0})\big{(}T_{0}-T_{0}D_{1}T_{0}\big{)}(Q-S_{0})$.

In particular, $A$ and $B$ are bounded operators on $(Q-S_{0})L^{2}$ with
$\hbox{rank}\ A=2$ and $\hbox{rank}\ B\leq 2$, and
$\|F(\lambda)\|_{(Q-S_{0})L^{2}}=O_{1}(1)$ for small enough $\lambda>0$.

###### Proof.

Since $QL^{2}(\mathbb{R}^{2})=\hbox{span}\\{v\\}^{\bot}$ and
$S_{0}L^{2}(\mathbb{R}^{2})=\hbox{span}\\{v,x_{1}v(x),x_{2}v(x)\\}^{\bot}$,
then

$(Q-S_{0})L^{2}(\mathbb{R}^{2})=\hbox{span}\\{Q(x_{1}v),Q(x_{2}v)\\}.$

Note that $\hbox{rank}(Q-S_{0})=2$, and

$\displaystyle d=$ $\displaystyle
g^{+}_{0}(\lambda)^{-1}(Q-S_{0})\Big{[}(Q-S_{0})vG_{-1}v(Q-S_{0})$
$\displaystyle+g^{+}_{0}(\lambda)^{-1}(Q-S_{0})\big{(}T_{0}-T_{0}D_{1}T_{0}\big{)}(Q-S_{0})\Big{]}^{-1}(Q-S_{0}).$

Hence, in order to prove the operator $d$ exists, it suffices to prove that
$(Q-S_{0})vG_{-1}v(Q-S_{0})$ is invertible on $(Q-S_{0})L^{2}$ by Neumamm
series if $\lambda$ is small enough.

Without loss of generality, we may assume that $Qx_{1}v$ and $Qx_{2}v$ are
orthogonal on $(Q-S_{0})L^{2}$. Otherwise, we can do the same arguments after
Schmidt orthogonalization. If $f\in(Q-S_{0})L^{2}$, then
$f=a_{1}Q(x_{1}v)+a_{2}Q(x_{2}v)$. Since
$(Q-S_{0})(Q(x_{i}v))=Q(x_{i}v)(i=1,2)$, then we have

$\displaystyle(Q-S_{0})vG_{-1}v(Q-S_{0})(Q(x_{1}v))(x)=(Q-S_{0})vG_{-1}v(Q(x_{1}v))(x)$
$\displaystyle=$
$\displaystyle(Q-S_{0})v(x)\int_{\mathbb{R}^{2}}|x-y|^{2}v(y)Q(x_{1}v)(y)dy$
$\displaystyle=$ $\displaystyle-2(Q-S_{0})(x_{1}v)\langle
x_{1}v,Qx_{1}v\rangle-2(Q-S_{0})(x_{2}v)\langle x_{2}v,Qx_{1}v\rangle.$

Since $\langle x_{2}v,Qx_{1}v\rangle=\langle Q(x_{2}v),Q(x_{1}v)\rangle=0$ and
$(Q-S_{0})(x_{i}v)=Q(x_{i}v)(i=1,2)$, then we get

$(Q-S_{0})vG_{-1}v(Q-S_{0})(Qx_{1}v)=-2Q(x_{1}v)\|Qx_{1}v\|_{L^{2}}^{2}.$

Similarly, we have

$(Q-S_{0})vG_{-1}v(Q-S_{0})(Qx_{2}v)=-2Q(x_{2}v)\|Qx_{2}v\|_{L^{2}}^{2},$

which leads to

$\begin{split}&(Q-S_{0})vG_{-1}v(Q-S_{0})f=-2a_{1}Q(x_{1}v)\|Qx_{1}v\|_{L^{2}}^{2}-2a_{2}Q(x_{2}v)\|Qx_{2}v\|_{L^{2}}^{2}\\\
=&\begin{pmatrix}f_{1}&f_{2}\end{pmatrix}\begin{pmatrix}-2\|Q(x_{1}v)\|_{L^{2}}^{2}&0\\\
0&-2\|Q(x_{2}v)\|_{L^{2}}^{2}\end{pmatrix}\begin{pmatrix}a_{1}\\\
a_{2}\end{pmatrix}:=\begin{pmatrix}f_{1}&f_{2}\end{pmatrix}A\begin{pmatrix}a_{1}\\\
a_{2}\end{pmatrix}.\end{split}$ (4.11)

Since $\det A=4\|Q(x_{1}v)\|_{L^{2}}^{2}\|Q(x_{2}v)\|_{L^{2}}^{2}\geq 0$, so
we need only to prove $\det A\neq 0$. Indeed, if $\det A=0$, then
$Q(x_{1}v)=0$ or $Q(x_{2}v)=0$, it is contradictory with $rank(Q-S_{0})=2$.
Thus, $(Q-S_{0})vG_{-1}v(Q-S_{0})$ is invertible and strictly negative on the
space $(Q-S_{0})L^{2}$, and then $d$ exists on $(Q-S_{0})L^{2}$ for small
enough $\lambda$.

Let $A=(Q-S_{0})vG_{-1}v(Q-S_{0})$,
$B=(Q-S_{0})\big{(}T_{0}-T_{0}D_{1}T_{0}\big{)}(Q-S_{0})$ and
$F(\lambda)=\big{[}A+g^{+}_{0}(\lambda)^{-1}B\big{]}^{-1}$, we obtain that

$\displaystyle d=g^{+}_{0}(\lambda)^{-1}(Q-S_{0})F(\lambda)(Q-S_{0}).$

Finally, we need to prove that $\|F(\lambda)\|_{(Q-S_{0})L^{2}}=O_{1}(1)$ for
small enough $\lambda$. Indeed, let $G(\lambda)=A+g_{0}^{+}(\lambda)^{-1}B$,
then $F(\lambda)=G(\lambda)^{-1}$. Note that
$G^{\prime}(\lambda)=-\lambda^{-1}g_{0}^{+}(\lambda)^{-2}$ , we have

$\frac{d}{d\lambda}F(\lambda)=\frac{d}{d\lambda}\big{(}G(\lambda)^{-1}\big{)}=-\Big{(}A+g_{0}^{+}(\lambda)^{-1}B\Big{)}^{-1}\cdot\lambda^{-1}g_{0}^{+}(\lambda)^{-2}\cdot\Big{(}A+g_{0}^{+}(\lambda)^{-1}B\Big{)}^{-1}.$

Observe that

$\|F(\lambda)\|_{(Q-S_{0})L^{2}\rightarrow(Q-S_{0})L^{2}}\leq C,\ \lambda\ll
1,$

then

$\|\frac{d}{d\lambda}F(\lambda)\|_{(Q-S_{0})L^{2}\rightarrow(Q-S_{0})L^{2}}\lesssim\lambda^{-1}|\ln\lambda|^{-2}\lesssim\lambda^{-1},\
\lambda\ll 1,$

Hence, we have
$\|F(\lambda)\|_{(Q-S_{0})L^{2}\rightarrow(Q-S_{0})L^{2}}=O_{1}(1)$ for small
enough $\lambda$. ∎

Now we begin to compute the inverse $\big{(}A^{+}(\lambda)\big{)}^{-1}$ as
follows.

######  Lemma 4.4.

Suppose that $|V(x)|\lesssim(1+|x|)^{-\beta}$ with some $\beta>10$. If zero is
a regular point of the spectrum of $H$, then for $0<|\lambda|\ll 1$,

$\begin{split}\big{(}A^{+}(\lambda)\big{)}^{-1}=S_{0}D_{1}S_{0}+g_{0}^{+}(\lambda)^{-1}S(\lambda),\end{split}$
(4.12)

where

$\begin{split}S:=S(\lambda)=\begin{pmatrix}(Q-S_{0})F(\lambda)(Q-S_{0})&-(Q-S_{0})F(\lambda)(Q-S_{0})T_{0}D_{1}\\\
-D_{1}T_{0}(Q-S_{0})F(\lambda)(Q-S_{0})&D_{1}T_{0}(Q-S_{0})F(\lambda)(Q-S_{0})T_{0}D_{1}\end{pmatrix}\end{split}$

with $F(\lambda)=\big{(}A+g^{+}_{0}(\lambda)^{-1}B\big{)}^{-1}$ and
$D_{1}=(S_{0}T_{0}S_{0})^{-1}$ on $S_{0}L^{2}$. Moreover,
$\|S(\lambda)\|_{L^{2}\rightarrow L^{2}}=O_{1}(1)$ for enough small $\lambda$
.

In particular, if zero is not regular point, then the same formula (4.12)
holds for $\big{(}A^{+}(\lambda)+S_{1}\big{)}^{-1}$ with
$D_{1}=(S_{0}T_{0}S_{0}+S_{1})^{-1}$.

###### Proof.

We prove only the statement when $S_{1}\neq 0$ ( i.e.
$S_{1}L^{2}=Ker(S_{0}T_{0}S_{0})\neq\\{0\\}$ ) since the proof of the regular
case is similar by setting $S_{1}=0$. Observe that

$\begin{split}A^{+}(\lambda)+S_{1}=&g^{+}_{0}(\lambda)(Q-S_{0})vG_{-1}v(Q-S_{0})+(Q-S_{0})T_{0}(Q-S_{0})+(Q-S_{0})T_{0}S_{0}\\\
&+S_{0}T_{0}(Q-S_{0})+S_{0}T_{0}S_{0}+S_{1}.\end{split}$

Equivalently, by block format, we write $A^{+}(\lambda)+S_{1}$ as follows:

$\begin{split}A^{+}(\lambda)+S_{1}=&\begin{pmatrix}g^{+}_{0}(\lambda)(Q-S_{0})vG_{-1}v(Q-S_{0})+(Q-S_{0})T_{0}(Q-S_{0})&(Q-S_{0})T_{0}S_{0}\\\
S_{0}T_{0}(Q-S_{0})&S_{0}T_{0}S_{0}+S_{1}\end{pmatrix}.\end{split}$

Note that $QL^{2}=(Q-S_{0})L^{2}\bigoplus S_{0}L^{2}$, and
$S_{0}T_{0}S_{0}+S_{1}$ is invertible on $S_{0}L^{2}$, then by Lemma 4.2
$A^{+}(\lambda)+S_{1}$ has a bounded inverse on $S_{0}L^{2}$ if and only if
the inverse operator $d$ defined in (4.10) exists on $(Q-S_{0})L^{2}$ and is
bounded for enough small $\lambda$.

Let $D_{1}=(S_{0}T_{0}S_{0}+S_{1})^{-1}:S_{0}L^{2}\rightarrow S_{0}L^{2}$.
Then by Lemma 4.3, we know that the operator $d$ exists on $(Q-S_{0})L^{2}$
and

$\displaystyle d=g^{+}_{0}(\lambda)^{-1}(Q-S_{0})F(\lambda)(Q-S_{0}).$

Hence by using Lemma 4.2 again, we have $A^{+}(\lambda)+S_{1}$ is invertible
on $S_{0}L^{2}$, and

$\displaystyle\big{(}A^{+}(\lambda)+S_{1}\big{)}^{-1}$ $\displaystyle=$
$\displaystyle
g^{+}_{0}(\lambda)^{-1}\begin{pmatrix}(Q-S_{0})F(\lambda)(Q-S_{0})&-(Q-S_{0})F(\lambda)(Q-S_{0})T_{0}D_{1}\\\
-D_{1}T_{0}(Q-S_{0})F(\lambda)(Q-S_{0})&D_{1}T_{0}(Q-S_{0})F(\lambda)(Q-S_{0})T_{0}D_{1}+g^{+}_{0}(\lambda)D_{1}\end{pmatrix}$
$\displaystyle:=$ $\displaystyle
S_{0}D_{1}S_{0}+g^{+}_{0}(\lambda)^{-1}S(\lambda),$

where

$S(\lambda)=S=\begin{pmatrix}(Q-S_{0})F(\lambda)(Q-S_{0})&-(Q-S_{0})F(\lambda)(Q-S_{0})T_{0}D_{1}\\\
-D_{1}T_{0}(Q-S_{0})F(\lambda)(Q-S_{0})&D_{1}T_{0}(Q-S_{0})F(\lambda)(Q-S_{0})T_{0}D_{1}\end{pmatrix}.$

It is obvious that $\|S(\lambda)\|_{L^{2}\rightarrow L^{2}}=O_{1}(1)$ for
$\lambda\ll 1$ by Lemma 4.3. ∎

######  Remark 4.5.

We can deduce some orthogonal relations from Lemma 4.4 as follows:

$\begin{split}S_{1}S(\lambda)=&-S_{1}T_{0}(Q-S_{0})F(\lambda)(Q-S_{0})+S_{1}T_{0}(Q-S_{0})F(\lambda)(Q-S_{0})T_{0}D_{1},\end{split}$
(4.13)
$\begin{split}S(\lambda)S_{1}=-(Q-S_{0})F(\lambda)(Q-S_{0})T_{0}S_{1}+D_{1}T_{0}(Q-S_{0})F(\lambda)(Q-S_{0})T_{0}S_{1},\end{split}$
(4.14) $S_{1}SS_{1}=S_{1}T_{0}(Q-S_{0})F(\lambda)(Q-S_{0})T_{0}S_{1},$ (4.15)
$S_{2}T_{0}(Q-S_{0})=(Q-S_{0})T_{0}S_{2}=0,$ (4.16)
$S_{2}S(\lambda)=S(\lambda)S_{2}=0,\quad
S_{2}\big{(}A^{+}(\lambda)+S_{1}\big{)}^{-1}=\big{(}A^{+}(\lambda)+S_{1}\big{)}^{-1}S_{2}=S_{2}.$
(4.17)

Now we use Lemma 4.4 to deduce the expansion of
$\big{(}M^{+}(\lambda)\big{)}^{-1}$ in the regular case corresponding to the
existence of $D_{1}=(S_{0}T_{0}S_{0})^{-1}$ on $S_{0}L^{2}$ ( i.e. the
projection $S_{1}=0$ ).

(I) Regular point case. If zero is a regular point of the spectrum of $H$,
then $S_{0}T_{0}S_{0}$ is invertible on $S_{0}L^{2}$. Note that the (4.8)
gives the identity

$\big{(}\widetilde{M}^{+}_{1}(\lambda)\big{)}^{-1}=\big{(}A^{+}(\lambda)\big{)}^{-1}\Big{(}I+a_{+}W^{+}(\lambda)\big{(}A^{+}(\lambda)\big{)}^{-1}\Big{)}^{-1}.$

By using Lemma 4.4, the (4.9) and Neumann series, for small enough $\lambda$,
we have

$\begin{split}\big{(}\widetilde{M}^{+}_{1}(\lambda)\big{)}^{-1}=&D_{1}+g_{0}^{+}(\lambda)^{-1}S-a_{+}g_{0}^{+}(\lambda)^{2}\lambda^{2}D_{1}B_{3}^{+}D_{1}\\\
&-a_{+}g_{0}^{+}(\lambda)\lambda^{2}\big{(}D_{1}B_{4}^{+}D_{1}+D_{1}B_{3}^{+}QS+SQB_{3}^{+}D_{1}\big{)}+O_{1}(\lambda^{2}).\end{split}$

Note that the equality (4.8) also gives
$\big{(}M^{+}_{1}(\lambda)\big{)}^{-1}=\frac{a_{+}}{\lambda^{2}}\big{(}\widetilde{M}^{+}_{1}(\lambda)\big{)}^{-1}$.
Hence we have

$\displaystyle\big{(}M^{+}_{1}(\lambda)\big{)}^{-1}=$
$\displaystyle\frac{a_{+}}{\lambda^{2}}D_{1}+\frac{a_{+}g_{0}^{+}(\lambda)^{-1}}{\lambda^{2}}S-a_{+}^{2}g_{0}^{+}(\lambda)^{2}D_{1}B_{3}^{+}D_{1}-a_{+}^{2}g_{0}^{+}(\lambda)\big{(}D_{1}B_{4}^{+}D_{1}$
$\displaystyle+D_{1}B_{3}^{+}QS+SQB_{3}^{+}D_{1}\big{)}+O_{1}(1).$

Using the orthogonality $QD_{1}=D_{1}Q=D_{1}$, by (4.4) and Lemma 4.1, we get

$\displaystyle\big{(}\widetilde{M}^{+}(\lambda)\big{)}^{-1}=$
$\displaystyle\frac{a_{+}}{\lambda^{2}}D_{1}+\frac{a_{+}g_{0}^{+}(\lambda)^{-1}}{\lambda^{2}}QSQ-
a_{+}^{2}g_{0}^{+}(\lambda)^{2}D_{1}B_{3}^{+}D_{1}$ $\displaystyle-
a_{+}^{2}g_{0}^{+}(\lambda)\big{(}D_{1}B_{4}^{+}D_{1}+D_{1}B_{3}^{+}QSQ+QSQB_{3}^{+}D_{1}\big{)}+O_{1}(1).$

Since
$\displaystyle\big{(}M^{+}(\lambda)\big{)}^{-1}=\frac{\lambda^{2}}{a_{+}}\big{(}\widetilde{M}^{+}(\lambda)\big{)}^{-1}$,
using the orthogonality $S_{0}D_{1}=D_{1}S_{0}=D_{1}$, we have

$\displaystyle\big{(}M^{+}(\lambda)\big{)}^{-1}=$ $\displaystyle
S_{0}D_{1}S_{0}+g_{0}^{+}(\lambda)^{-1}Q\Gamma^{0}_{0,1}Q+g_{0}^{+}(\lambda)^{2}\lambda^{2}S_{0}\Gamma^{0}_{2,1}S_{0}$
$\displaystyle+g_{0}^{+}(\lambda)\lambda^{2}\Big{(}S_{0}\Gamma^{0}_{2,2}Q+Q\Gamma^{0}_{2,3}S_{0}\Big{)}+O_{1}(\lambda^{2}).$
(4.18)

Thus the expansions of $\big{(}M^{+}(\lambda)\big{)}^{-1}$ in regular case is
completed. $\Box$

######  Remark 4.6.

We know that $S_{0}D_{1}S_{0}$ is absolutely bounded by Proposition 2.6. It is
clear that $B^{+}_{i}(i=1,\cdots,9)$ are Hilbert-Schmidt operator by (4.4),
and $S$ is finite-rank operator by $\dim(Q-S_{0})L^{2}=2$ and Lemma 4.3, thus
$\Gamma^{0}_{i,j}(\lambda)$ are also Hilbert-Schmidt operator, which leads to
each operator in the expansion of $\big{(}M^{+}(\lambda)\big{)}^{-1}$ above
are absolutely bounded uniformly for sufficiently small $\lambda>0$.

Next, let’s turn to consider the resonance cases, that is, zero is not regular
point of the spectrum of $H$. Let $S_{1}$ be the Riesz projection onto the
kernel of $S_{0}T_{0}S_{0}$. Then $S_{1}\neq 0$ and $S_{0}T_{0}S_{0}+S_{1}$ is
invertible on $S_{0}L^{2}$. In this case, we define
$D_{1}=(S_{0}T_{0}S_{0}+S_{1})^{-1}$ as a bounded operator on $S_{0}L^{2}$,
then from the (4.8), Lemma 4.4 and Neumann series one has

$\begin{split}&\Big{(}\widetilde{M}^{+}_{1}(\lambda)+S_{1}\Big{)}^{-1}\\\
=&D_{1}+g_{0}^{+}(\lambda)^{-1}S-g_{0}^{+}(\lambda)^{2}\lambda^{2}\widetilde{B}_{1}-g_{0}^{+}(\lambda)\lambda^{2}\widetilde{B}_{2}-\lambda^{2}\widetilde{B}_{3}-g_{0}^{+}(\lambda)^{-1}\lambda^{2}\widetilde{B}_{4}\\\
&-g_{0}^{+}(\lambda)^{-2}\lambda^{2}\widetilde{B}_{5}-g_{0}^{+}(\lambda)^{4}\lambda^{4}\widetilde{B}_{6}-g_{0}^{+}(\lambda)^{3}\lambda^{4}\widetilde{B}_{7}-g_{0}^{+}(\lambda)^{2}\lambda^{4}\widetilde{B}_{8}-g_{0}^{+}(\lambda)\lambda^{4}\widetilde{B}_{9}\\\
&-\tilde{g}_{2}^{+}(\lambda)\lambda^{4}\widetilde{B}_{10}-\lambda^{4}\widetilde{B}_{11}-\tilde{g}_{2}^{+}(\lambda)g_{0}^{+}(\lambda)^{-1}\lambda^{4}\widetilde{B}_{12}-g_{0}^{+}(\lambda)^{-1}\lambda^{4}\widetilde{B}_{13}\\\
&-\tilde{g}_{2}^{+}(\lambda)g_{0}^{+}(\lambda)^{-2}\lambda^{4}\widetilde{B}_{14}-g_{0}^{+}(\lambda)^{-2}\lambda^{4}\widetilde{B}_{15}-g_{0}^{+}(\lambda)^{-3}\lambda^{4}\widetilde{B}_{16}+O(\lambda^{6-\epsilon}),\end{split}$
(4.19)

where $\widetilde{B}_{j}(1\leq j\leq 15)$ are bounded operators on $L^{2}$ as
follows:

$\displaystyle\widetilde{B}_{1}=a_{+}D_{1}B_{3}D_{1},\widetilde{B}_{2}=a_{+}\big{(}D_{1}B_{4}D_{1}+D_{1}B_{3}QS+SQB_{3}D_{1}\big{)},$
$\displaystyle\widetilde{B}_{3}=a_{+}\big{[}D_{1}B_{5}D_{1}+D_{1}B_{4}QS+SQB_{4}D_{1}+SQB_{3}QS\big{]},$
$\displaystyle\widetilde{B}_{4}=a_{+}\big{[}D_{1}B_{5}QS+SQB_{5}D_{1}+SQB_{4}QS\big{]},\,\widetilde{B}_{5}=a_{+}SQB_{5}QS,\,\widetilde{B}_{6}=-a_{+}^{2}D_{1}B_{3}D_{1}B_{3}D_{1},$
$\begin{split}\widetilde{B}_{7}=&a_{+}D_{1}B_{6}D_{1}-a_{+}^{2}\big{[}D_{1}B_{3}D_{1}B_{4}D_{1}+D_{1}B_{4}D_{1}B_{3}D_{1}+D_{1}B_{3}D_{1}B_{3}QS+SQB_{3}D_{1}B_{3}D_{1}\\\
&+D_{1}B_{3}QSQB_{3}D_{1}\big{]},\\\
\widetilde{B}_{8}=&a_{+}\big{[}D_{1}B_{7}D_{1}+D_{1}B_{6}QS+SQB_{6}D_{1}\big{]}-a_{+}^{2}\big{[}D_{1}B_{4}D_{1}B_{4}D_{1}\\\
&+D_{1}B_{3}D_{1}B_{5}D_{1}+D_{1}B_{5}D_{1}B_{3}D_{1}+D_{1}B_{3}D_{1}B_{4}QS+SQB_{4}D_{1}B_{3}D_{1}\\\
&+D_{1}B_{4}D_{1}B_{3}QS+SQB_{3}D_{1}B_{4}D_{1}+D_{1}B_{3}QSQB_{4}D_{1}+D_{1}B_{4}QSQB_{3}D_{1}\\\
&+D_{1}B_{3}QSQB_{3}QS+SQB_{3}QSQB_{3}D_{1}\big{]},\\\
\widetilde{B}_{9}=&a_{+}\big{[}D_{1}B_{8}D_{1}+D_{1}B_{7}QS+SQB_{7}D_{1}+SQB_{6}QS\big{]}-a_{+}^{2}\big{[}D_{1}B_{4}D_{1}B_{5}D_{1}+D_{1}B_{5}D_{1}B_{4}D_{1}\\\
&+D_{1}B_{3}D_{1}B_{5}QS+SQB_{5}D_{1}B_{3}D_{1}+SQB_{3}D_{1}B_{4}QS+SQB_{4}D_{1}B_{3}QS\\\
&+SQB_{3}D_{1}B_{5}D_{1}+D_{1}B_{5}D_{1}B_{3}QS+D_{1}B_{3}QSQB_{5}D_{1}+D_{1}B_{5}QSQB_{3}D_{1}\\\
&+D_{1}B_{3}QSQB_{4}QS+SQB_{4}QSQB_{3}D_{1}+D_{1}B_{4}QSQB_{3}QS+SQB_{3}QSQB_{4}D_{1}\\\
&+D_{1}B_{4}QSQB_{4}D_{1}+SQB_{3}QSQB_{3}QS\big{]},\,\widetilde{B}_{10}=a_{+}D_{1}B_{9}D_{1},\end{split}$
$\begin{split}\widetilde{B}_{11}=&a_{+}\big{[}D_{1}B_{10}D_{1}+D_{1}B_{8}QS+QSB_{8}D_{1}+SQB_{7}QS\big{]}-a_{+}^{2}\big{[}D_{1}B_{5}D_{1}B_{5}D_{1}\\\
&+D_{1}B_{5}D_{1}B_{4}QS+SQB_{4}D_{1}B_{5}D_{1}+D_{1}B_{4}D_{1}B_{5}QS+SQB_{5}D_{1}B_{4}D_{1}\\\
&+D_{1}B_{4}QSQB_{5}D_{1}+D_{1}B_{5}QSQB_{4}D_{1}+D_{1}B_{3}QSQB_{5}QS+SQB_{5}QSQB_{3}D_{1}\\\
&+D_{1}B_{4}QSQB_{4}QS+QSB_{4}QSQB_{4}D_{1}+D_{1}B_{5}QSQB_{3}QS+SQB_{3}QSQB_{5}D_{1}\\\
&+SQB_{3}D_{1}B_{5}QS+SQB_{5}D_{1}B_{3}QS+SQB_{3}QSQB_{4}QS+SQB_{4}QSQB_{3}QS\\\
&+SQB_{4}D_{1}B_{4}QS\big{]},\,\widetilde{B}_{12}=a_{+}\big{[}D_{1}B_{9}QS+SQB_{9}D_{1}\big{]},\end{split}$
$\begin{split}\widetilde{B}_{13}=&a_{+}\big{[}D_{1}B_{10}QS+SQB_{10}D_{1}+SQB_{8}QS\big{]}-a_{+}^{2}\big{[}D_{1}B_{5}QSQB_{5}D_{1}+SQB_{5}D_{1}B_{5}D_{1}\\\
&+D_{1}B_{5}D_{1}B_{5}QS+D_{1}B_{4}QSQB_{5}QS+SQB_{5}QSQB_{4}D_{1}+SQB_{4}QSQB_{5}D_{1}\\\
&+D_{1}B_{5}QSQB_{4}QS+SQB_{5}D_{1}B_{4}QS+SQB_{4}D_{1}B_{5}QS+SQB_{3}QSQB_{5}QS\\\
&+SQB_{5}QSQB_{3}QS+SQB_{4}QSQB_{4}QS\big{]},\,\widetilde{B}_{14}=a_{+}SQB_{9}QS,\\\
\widetilde{B}_{15}=&a_{+}SQB_{10}QS-
a_{+}^{2}\big{[}SQB_{5}D_{1}B_{5}QS+D_{1}B_{5}QSQB_{5}QS+SQB_{5}QSQB_{5}D_{1}\\\
&+SQB_{4}QSQB_{5}QS+SQB_{5}QSQB_{4}QS\big{]},\widetilde{B}_{16}=-a_{+}^{2}SQB_{5}QSQB_{5}QS.\end{split}$

According to Lemma 4.1, $\widetilde{M}^{+}_{1}(\lambda)$ has bounded inverse
if and only if

$M_{2}^{+}(\lambda):=S_{1}-S_{1}\big{(}\widetilde{M}^{+}_{1}(\lambda)+S_{1}\big{)}^{-1}S_{1}$
(4.20)

has bounded inverse on $S_{1}L^{2}$. Using the orthogonality in Remark 4.5,
one has

$\begin{split}M_{2}^{+}(\lambda)=&g_{0}^{+}(\lambda)^{-1}\Big{(}\widetilde{T}_{1}+g_{0}^{+}(\lambda)^{3}\lambda^{2}S_{1}\widetilde{B}_{1}S_{1}+g_{0}^{+}(\lambda)^{2}\lambda^{2}S_{1}\widetilde{B}_{2}S_{1}+g_{0}^{+}(\lambda)\lambda^{2}S_{1}\widetilde{B}_{3}S_{1}+\lambda^{2}S_{1}\widetilde{B}_{4}S_{1}\\\
&+g_{0}^{+}(\lambda)^{-1}\lambda^{2}S_{1}\widetilde{B}_{5}S_{1}+g_{0}^{+}(\lambda)^{5}\lambda^{4}S_{1}\widetilde{B}_{6}S_{1}+g_{0}^{+}(\lambda)^{4}\lambda^{4}S_{1}\widetilde{B}_{7}S_{1}+g_{0}^{+}(\lambda)^{3}\lambda^{4}S_{1}\widetilde{B}_{8}S_{1}\\\
&+g_{0}^{+}(\lambda)^{2}\lambda^{4}S_{1}\widetilde{B}_{9}S_{1}+\tilde{g}_{2}^{+}(\lambda)g_{0}^{+}(\lambda)\lambda^{4}S_{1}\widetilde{B}_{10}S_{1}+g_{0}^{+}(\lambda)\lambda^{4}S_{1}\widetilde{B}_{11}S_{1}+\tilde{g}_{2}^{+}(\lambda)\lambda^{4}S_{1}\widetilde{B}_{12}S_{1}\\\
&+\lambda^{4}S_{1}\widetilde{B}_{13}S_{1}+\tilde{g}_{2}^{+}(\lambda)g_{0}^{+}(\lambda)^{-1}\lambda^{4}S_{1}\widetilde{B}_{14}S_{1}+g_{0}^{+}(\lambda)^{-1}\lambda^{4}S_{1}\widetilde{B}_{15}S_{1}\\\
&+g_{0}^{+}(\lambda)^{-2}\lambda^{4}S_{1}\widetilde{B}_{16}S_{1}+O(\lambda^{6-\epsilon})\Big{)}\\\
:=&g_{0}^{+}(\lambda)^{-1}\big{(}\widetilde{T}_{1}+W_{1}^{+}(\lambda)\big{)}:=g_{0}^{+}(\lambda)^{-1}\widetilde{M}_{2}^{+}(\lambda),\end{split}$

where

$\begin{split}\widetilde{T}_{1}=-S_{1}T_{0}(Q-S_{0})\big{(}A+g^{+}_{0}(\lambda)^{-1}B\big{)}^{-1}(Q-S_{0})T_{0}S_{1}.\end{split}$
(4.21)

Let

$T_{1}=-S_{1}T_{0}(Q-S_{0})A^{-1}(Q-S_{0})T_{0}S_{1},$

then $\widetilde{T}_{1}$ is invertible on $S_{1}L^{2}$ for enough small
$\lambda$ if and only if $T_{1}$ is invertible on $S_{1}L^{2}$.

(II) The first kind of resonance. If there is a resonance of first kind at
zero, then the operator $T_{1}$ is invertible on $S_{1}L^{2}$. In this case we
define $D_{2}=T_{1}^{-1}$ as an operator on $S_{1}L^{2}$. Then
$\widetilde{T}_{1}$ is invertible on $S_{1}L^{2}$ and
$D_{2}=\widetilde{T}_{1}^{-1}$ for enough small $\lambda$. Note that

$\displaystyle\big{(}\widetilde{M}_{2}^{+}(\lambda)\big{)}^{-1}=D_{2}\big{(}I+W_{1}^{+}(\lambda)D_{2}\big{)}^{-1}$

and
$\big{(}M_{2}^{+}(\lambda)\big{)}^{-1}=g_{0}^{+}(\lambda)\big{(}\widetilde{M}_{2}^{+}(\lambda)\big{)}^{-1}$,
by Neumann series one has

$\begin{split}\big{(}M_{2}^{+}(\lambda)\big{)}^{-1}=&g_{0}^{+}(\lambda)D_{2}+g_{0}^{+}(\lambda)^{4}\lambda^{2}D_{2}\widetilde{B}_{1}D_{2}+g_{0}^{+}(\lambda)^{3}\lambda^{2}D_{2}\widetilde{B}_{2}D_{2}+g_{0}^{+}(\lambda)^{2}\lambda^{2}D_{2}\widetilde{B}_{3}D_{2}\\\
&+g_{0}^{+}(\lambda)\lambda^{2}D_{2}\widetilde{B}_{4}D_{2}+\lambda^{2}D_{2}\widetilde{B}_{5}D_{2}+O_{1}(\lambda^{4-\epsilon}).\end{split}$

Using Lemma 4.1 and (4.19) one has

$\begin{split}\big{(}\widetilde{M}_{1}^{+}(\lambda)\big{)}^{-1}=&g_{0}^{+}(\lambda)S_{1}\Gamma^{1}_{0,1}S_{1}+\Big{(}Q\Gamma^{1}_{0,2}S_{1}+S_{1}\Gamma^{1}_{0,3}Q\Big{)}+g_{0}^{+}(\lambda)^{-1}Q\Gamma^{1}_{0,5}Q\\\
&+g_{0}^{+}(\lambda)^{-2}Q\Gamma^{1}_{0,6}Q+g_{0}^{+}(\lambda)^{4}\lambda^{2}S_{1}\Gamma^{1}_{2,1}S_{1}+g_{0}^{+}(\lambda)^{3}\lambda^{2}Q\Gamma^{1}_{2,1}Q\\\
&+g_{0}^{+}(\lambda)^{2}\lambda^{2}Q\Gamma^{1}_{2,1}Q+g_{0}^{+}(\lambda)\lambda^{2}Q\Gamma^{1}_{2,1}Q+O_{1}(\lambda^{2}).\end{split}$

By the same arguments as in the proof of Theorem 2.9(i) one has

$\displaystyle\big{(}M^{+}(\lambda)\big{)}^{-1}=$ $\displaystyle
g^{+}_{0}(\lambda)S_{1}D_{2}S_{1}+\Big{(}S_{0}\Gamma_{0,1}^{1}S_{0}+S_{1}\Gamma_{0,2}^{1}Q+Q\Gamma_{0,3}^{1}S_{1}\Big{)}+g^{+}_{0}(\lambda)^{-1}Q\Gamma_{0,4}^{1}Q$
$\displaystyle+g^{+}_{0}(\lambda)^{4}\lambda^{2}S_{1}\Gamma_{2,1}^{1}S_{1}+g^{+}_{0}(\lambda)^{3}\lambda^{2}\Big{(}S_{1}\Gamma_{2,2}^{1}Q+Q\Gamma_{2,3}^{1}S_{1}\Big{)}+g^{+}_{0}(\lambda)^{2}\lambda^{2}$
$\displaystyle\times\Big{(}Q\Gamma_{2,4}^{1}Q+S_{1}\Gamma_{2,5}^{1}+\Gamma_{2,6}^{1}S_{1}\Big{)}+g^{+}_{0}(\lambda)\lambda^{2}\Big{(}Q\Gamma_{2,7}^{1}+\Gamma_{2,8}^{1}Q\Big{)}+O_{1}(\lambda^{2}).$

Thus the expansions of $\big{(}M^{+}(\lambda)\big{)}^{-1}$ in the first kind
resonance is completed. $\Box$If zero is not the first kind of resonance, then
$T_{1}$ is not invertible on $S_{1}L^{2}$. Let $S_{2}$ is the Riesz transform
onto the kernel of $T_{1}$, then $T_{1}+S_{2}$ is invertible on $S_{1}L^{2}$,
which implies that $(\widetilde{T}_{1}+S_{2})^{-1}$ exists for sufficiently
small $\lambda>0$. If we define $D_{2}=(\widetilde{T}_{1}+S_{2})^{-1}$, then
$S_{1}D_{2}=D_{2}S_{1}=D_{2}$, $S_{2}D_{2}=D_{2}S_{2}=S_{2}$ and
$D_{2}D_{1}=D_{1}D_{2}=D_{2}$. Note that

$\begin{split}\big{(}\widetilde{M}_{2}^{+}(\lambda)+S_{2}\big{)}^{-1}=D_{2}\Big{(}I+W_{1}^{+}(\lambda)D_{2}\Big{)}^{-1}.\end{split}$

By using Neumann series and (4.21) one has

$\begin{split}&\big{(}\widetilde{M}_{2}^{+}(\lambda)+S_{2}\big{)}^{-1}\\\
=&D_{2}-g_{0}^{+}(\lambda)^{3}\lambda^{2}D_{2}\widetilde{B}_{1}D_{2}-g_{0}^{+}(\lambda)^{2}\lambda^{2}D_{2}\widetilde{B}_{2}D_{2}-g_{0}^{+}(\lambda)\lambda^{2}D_{2}\widetilde{B}_{3}D_{2}-\lambda^{2}D_{2}\widetilde{B}_{4}D_{2}\\\
&-g_{0}^{+}(\lambda)^{-1}\lambda^{2}D_{2}\widetilde{B}_{5}D_{2}-g_{0}^{+}(\lambda)^{6}\lambda^{4}B_{1}^{1}-g_{0}^{+}(\lambda)^{5}\lambda^{4}B_{2}^{1}-g_{0}^{+}(\lambda)^{4}\lambda^{4}B_{3}^{1}-g_{0}^{+}(\lambda)^{3}\lambda^{4}B_{4}^{1}\\\
&-g_{0}^{+}(\lambda)^{2}\lambda^{4}B_{5}^{1}-\tilde{g}_{2}^{+}(\lambda)g_{0}^{+}(\lambda)\lambda^{4}B_{6}^{1}-g_{0}^{+}(\lambda)\lambda^{4}B_{7}^{1}-\tilde{g}_{2}^{+}(\lambda)\lambda^{4}B_{8}^{1}-\lambda^{4}B_{9}^{1}\\\
&-\tilde{g}_{2}^{+}(\lambda)g_{0}^{+}(\lambda)^{-1}\lambda^{4}B_{10}^{1}-g_{0}^{+}(\lambda)^{-1}\lambda^{4}B_{11}^{1}-g_{0}^{+}(\lambda)^{-2}\lambda^{4}B_{12}^{1}+O(\lambda^{6-\epsilon}),\end{split}$
(4.22)

where $B_{j}^{1}(1\leq j\leq 12)$ are bounded operators on on $L^{2}$ as
follows:

$\begin{split}B_{1}^{1}=&-D_{2}\widetilde{B}_{1}D_{2}\widetilde{B}_{1}D_{2},\
B_{2}^{1}=D_{2}\widetilde{B}_{6}D_{2}-\big{[}D_{2}\widetilde{B}_{1}D_{2}\widetilde{B}_{2}D_{2}+D_{2}\widetilde{B}_{2}D_{2}\widetilde{B}_{1}D_{2}\big{]},\\\
B_{3}^{1}=&D_{2}\widetilde{B}_{7}D_{2}-\big{[}D_{2}\widetilde{B}_{2}D_{2}\widetilde{B}_{2}D_{2}+D_{2}\widetilde{B}_{1}D_{2}\widetilde{B}_{3}D_{2}+D_{2}\widetilde{B}_{3}D_{2}\widetilde{B}_{1}D_{2}\big{]},\\\
B_{4}^{1}=&D_{2}\widetilde{B}_{8}D_{2}-\big{[}D_{2}\widetilde{B}_{1}D_{2}\widetilde{B}_{4}D_{2}+D_{2}\widetilde{B}_{4}D_{2}\widetilde{B}_{1}D_{2}+D_{2}\widetilde{B}_{2}D_{2}\widetilde{B}_{3}D_{2}+D_{2}\widetilde{B}_{3}D_{2}\widetilde{B}_{2}D_{2}\big{]},\\\
B_{5}^{1}=&D_{2}\widetilde{B}_{9}D_{2}-\big{[}D_{2}\widetilde{B}_{3}D_{2}\widetilde{B}_{3}D_{2}+D_{2}\widetilde{B}_{2}D_{2}\widetilde{B}_{4}D_{2}+D_{2}\widetilde{B}_{4}D_{2}\widetilde{B}_{2}D_{2}+D_{2}\widetilde{B}_{1}D_{2}\widetilde{B}_{5}D_{2}\\\
&+D_{2}\widetilde{B}_{5}D_{2}\widetilde{B}_{1}D_{2}\big{]},\
B_{6}^{1}=D_{2}\widetilde{B}_{10}D_{2},\\\
B_{7}^{1}=&D_{2}\widetilde{B}_{11}D_{2}-\big{[}D_{2}\widetilde{B}_{2}D_{2}\widetilde{B}_{5}D_{2}+D_{2}\widetilde{B}_{5}D_{2}\widetilde{B}_{2}D_{2}+D_{2}\widetilde{B}_{3}D_{2}\widetilde{B}_{4}D_{2}+D_{2}\widetilde{B}_{4}D_{2}\widetilde{B}_{3}D_{2}\big{]},\\\
B_{8}^{1}=&D_{2}\widetilde{B}_{12}D_{2},B_{9}^{1}=D_{2}\widetilde{B}_{13}D_{2}-\big{[}D_{2}\widetilde{B}_{4}D_{2}\widetilde{B}_{4}D_{2}+D_{2}\widetilde{B}_{3}D_{2}\widetilde{B}_{5}D_{2}+D_{2}\widetilde{B}_{5}D_{2}\widetilde{B}_{3}D_{2}\big{]},\\\
B_{10}^{1}=&D_{2}\widetilde{B}_{14}D_{2},\
B_{11}^{1}=D_{2}\widetilde{B}_{15}D_{2}-\big{[}D_{2}\widetilde{B}_{4}D_{2}\widetilde{B}_{5}D_{2}+D_{2}\widetilde{B}_{5}D_{2}\widetilde{B}_{4}D_{2}\big{]},\\\
B_{12}^{1}=&D_{2}\widetilde{B}_{16}D_{2}-D_{2}\widetilde{B}_{5}D_{2}\widetilde{B}_{5}D_{2}.\end{split}$

According to Lemma 4.1, $\widetilde{M}_{2}^{+}(\lambda)$ has bounded inverse
on $S_{1}L^{2}$ if and only if

$M_{3}^{+}(\lambda):=S_{2}-S_{2}(\widetilde{M}_{2}^{+}(\lambda)+S_{2})^{-1}S_{2}$
(4.23)

has bounded inverse on $S_{2}L^{2}(\mathbb{R}^{2})$. By Neumann series we have

$\begin{split}M^{+}_{3}(\lambda)=&g_{0}^{+}(\lambda)^{3}\lambda^{2}S_{2}\widetilde{B}_{1}S_{2}+g_{0}^{+}(\lambda)^{2}\lambda^{2}S_{2}\widetilde{B}_{2}S_{2}+g_{0}^{+}(\lambda)\lambda^{2}S_{2}\widetilde{B}_{3}S_{2}+\lambda^{2}S_{2}\widetilde{B}_{4}S_{2}\\\
&+g_{0}^{+}(\lambda)^{-1}\lambda^{2}S_{2}\widetilde{B}_{5}S_{2}+g_{0}^{+}(\lambda)^{6}\lambda^{4}S_{2}B_{1}^{1}S_{2}+g_{0}^{+}(\lambda)^{5}\lambda^{4}S_{2}B_{2}^{1}S_{2}+g_{0}^{+}(\lambda)^{4}\lambda^{4}S_{2}B_{3}^{1}S_{2}\\\
&+g_{0}^{+}(\lambda)^{3}\lambda^{4}S_{2}B_{4}^{1}S_{2}+g_{0}^{+}(\lambda)^{2}\lambda^{4}S_{2}B_{5}^{1}S_{2}+\tilde{g}_{2}^{+}(\lambda)g_{0}^{+}(\lambda)\lambda^{4}S_{2}B_{6}^{1}S_{2}+g_{0}^{+}(\lambda)\lambda^{4}S_{2}B_{7}^{1}S_{2}\\\
&+\tilde{g}_{2}^{+}(\lambda)\lambda^{4}S_{2}B_{8}^{1}S_{2}+\lambda^{4}S_{2}B_{9}^{1}S_{2}+\tilde{g}_{2}^{+}(\lambda)g_{0}^{+}(\lambda)^{-1}\lambda^{4}S_{2}B_{10}^{1}S_{2}+g_{0}^{+}(\lambda)^{-1}\lambda^{4}S_{2}B_{11}^{1}S_{2}\\\
&+g_{0}^{+}(\lambda)^{-2}\lambda^{4}S_{2}B_{12}^{1}S_{2}+O(\lambda^{6-\epsilon}).\end{split}$

Note that $S_{2}S=SS_{2}=0$, $QvG_{-1}vS_{0}=0,QT_{0}S_{2}=0$, hence by (4.19)
we have

$S_{2}\widetilde{B}_{1}S_{2}=-\frac{1}{a_{+}}S_{2}vG_{-1}vvG_{-1}vS_{2},\
S_{2}\widetilde{B}_{2}S_{2}=-\frac{1}{a_{+}}\big{(}S_{2}vG_{-1}vT_{0}S_{2}+S_{2}T_{0}vG_{-1}vS_{2}\big{)},$
$S_{2}\widetilde{B}_{3}S_{2}=c_{+}S_{2}vG_{1}vS_{2}-\frac{1}{a_{+}}S_{2}T_{0}^{2}S_{2},$
$S_{2}\widetilde{B}_{4}S_{2}=S_{3}\widetilde{B}_{5}S_{2}=S_{2}B_{8}^{1}S_{2}=S_{2}B_{10}^{1}S_{2}=S_{2}B_{12}^{1}S_{2}=0.$

Moreover, we get

$\begin{split}M^{+}_{3}(\lambda)&=-\frac{1}{a_{+}}g_{0}^{+}(\lambda)\lambda^{2}\Big{(}A_{1}^{+}(\lambda)-g_{0}^{+}(\lambda)^{5}\lambda^{2}a_{+}S_{2}B_{1}^{1}S_{2}-g_{0}^{+}(\lambda)^{4}\lambda^{2}a_{+}S_{2}B_{2}^{1}S_{2}\\\
&-g_{0}^{+}(\lambda)^{3}\lambda^{2}a_{+}S_{2}B_{3}^{1}S_{2}-g_{0}^{+}(\lambda)^{2}\lambda^{2}a_{+}S_{2}B_{4}^{1}S_{2}-g_{0}^{+}(\lambda)\lambda^{2}a_{+}S_{2}B_{5}^{1}S_{2}\\\
&-\tilde{g}_{2}^{+}(\lambda)\lambda^{2}a_{+}S_{2}B_{6}^{1}S_{2}-\lambda^{2}a_{+}S_{2}B_{7}^{1}S_{2}-g_{0}^{+}(\lambda)^{-1}\lambda^{2}a_{+}S_{2}B_{9}^{1}S_{2}\\\
&-g_{0}^{+}(\lambda)^{-2}\lambda^{2}a_{+}S_{2}B_{11}^{1}S_{2}+O(\lambda^{4-\epsilon})\Big{)}\\\
&:=-\frac{1}{a_{+}}g_{0}^{+}(\lambda)\lambda^{2}\big{(}A^{+}_{1}(\lambda)+W_{2}^{+}(\lambda)\big{)}:=-\frac{1}{a_{+}}g_{0}^{+}(\lambda)\lambda^{2}\widetilde{M}^{+}_{3}(\lambda),\end{split}$
(4.24)

where

$\begin{split}A_{1}^{+}(\lambda)=&g_{0}^{+}(\lambda)^{2}S_{2}vG_{-1}vvG_{-1}vS_{2}+g_{0}^{+}(\lambda)\big{(}S_{2}vG_{-1}vT_{0}S_{2}+S_{2}T_{0}vG_{-1}vS_{2}\big{)}\\\
&+\big{(}S_{2}T_{0}^{2}S_{2}-c_{+}a_{+}S_{2}vG_{1}vS_{2}\big{)}.\end{split}$

To compute $\big{(}M_{3}^{+}(\lambda)\big{)}^{-1}$, by (4.24) we need to
compute the expansions of $\big{(}\widetilde{M_{3}^{+}}(\lambda)\big{)}^{-1}$
on $S_{2}L^{2}$. Noting that $A_{1}^{+}(\lambda)$ with a logarithm factor, and
$T_{2}$ is not invertible on $S_{2}L^{2}$ by Lemma 2.7(iv). Hence we will use
Lemma 4.2 to study the inverse of $A_{1}^{+}(\lambda)$.

To compute the inverse of $A_{1}^{+}(\lambda)$ by Lemma 4.2, we need the
following lemma.

######  Lemma 4.7.

Suppose that $|V(x)|\lesssim(1+|x|)^{-\beta}$ with some $\beta>18$. Let

$\begin{split}L_{1}=&(S_{2}-S_{3})\Big{(}vG_{-1}vPvG_{-1}v-vG_{-1}vT_{0}D_{4}T_{0}vG_{-1}v\Big{)}(S_{2}-S_{3}),\\\
L_{2}=&(S_{2}-S_{3})\Big{(}vG_{-1}vT_{0}+T_{0}vG_{-1}v-vG_{-1}vT_{0}D_{4}(T_{0}^{2}-c_{+}a_{+}vG_{1}v)\\\
&-(T_{0}^{2}-c_{+}a_{+}vG_{1}v)D_{4}T_{0}vG_{-1}v\Big{)}(S_{2}-S_{3}),\\\
L_{3}=&(S_{2}-S_{3})\Big{(}T_{0}^{2}-c_{+}a_{+}vG_{1}v-(T_{0}^{2}-c_{+}a_{+}vG_{1}v)D_{4}(T_{0}^{2}-c_{+}a_{+}vG_{1}v)\Big{)}(S_{2}-S_{3}).\end{split}$

Then the inverse operator

$d_{1}:=g^{+}_{0}(\lambda)^{-2}\Big{(}L_{1}+g^{+}_{0}(\lambda)^{-1}L_{2}+g^{+}_{0}(\lambda)^{-2}L_{3}\Big{)}^{-1}$

exists on $(S_{2}-S_{3})L^{2}$ for enough small $\lambda>0$. Moreover, we have

$d_{1}=g^{+}_{0}(\lambda)^{-2}h^{+}(\lambda)(S_{2}-S_{3}),$

where
$h^{+}(\lambda)=\Big{(}c_{1}+g^{+}_{0}(\lambda)^{-1}c_{2}+g^{+}_{0}(\lambda)^{-2}c_{3}\Big{)}^{-1}$
with $c_{1}=\hbox{tr}(L_{1})$, $c_{2}=\hbox{tr}(L_{2})$ and
$c_{3}=\hbox{tr}(L_{3})$. In particular, $\displaystyle
h^{+}(\lambda)=O_{1}(1)$.

###### Proof.

We first show that $L_{1}$ is invertible on $(S_{2}-S_{3})L^{2}$. Indeed, by
Lemma 2.7, we know that $(S_{2}-S_{3})L^{2}=\hbox{span}\\{S_{2}(|x|^{2}v)\\}$,
thus it suffices to prove
$\big{\langle}|x|^{2}v,L^{1}(|x|^{2}v)\big{\rangle}>0$. Using
$S_{2}v=S_{2}(x_{i}v)=0(i=1,2)$ and
$S_{3}v=S_{3}(x_{i}v)=S_{3}(|x|^{2}v)=0(i=1,2)$, one has

$\begin{split}PvG_{-1}vS_{2}(|x|^{2}v)(x)=&P\Big{(}v(x)\int_{\mathbb{R}^{2}}\big{(}|x|^{2}-2x\cdot
y+|y|^{2}\big{)}v(y)S_{2}(|x|^{2}v)(y)dy\Big{)}\\\
=&P\big{\langle}|x|^{2}v,S_{2}(|x|^{2}v)\big{\rangle}v(x)=\|S_{2}(|x|^{2}v)\|^{2}_{L^{2}}v(x).\end{split}$
(4.25)

Let $\phi(x)=PvG_{-1}vS_{2}(|x|^{2}v)(x)$, by $S_{3}(|x|^{2}v)=0$ and (4.25),
one has

$\begin{split}&(S_{2}-S_{3})vG_{-1}vPvG_{-1}v(S_{2}-S_{3})(|x|^{2}v)(x)=\big{[}(S_{2}-S_{3})vG_{-1}v\phi\big{]}(x)\\\
=&(S_{2}-S_{3})v(x)\int_{\mathbb{R}^{2}}\big{(}|x|^{2}-2x\cdot
y+|y|^{2}\big{)}v(y)\phi(y)dy\\\
=&\|S_{2}(|x|^{2}v)\|^{2}_{L^{2}}(S_{2}-S_{3})|x|^{2}v(x)\int_{\mathbb{R}^{2}}v(y)v(y)dy\\\
=&\|V\|_{L^{1}}\|S_{2}(|x|^{2}v)\|^{2}_{L^{2}}v(x).\end{split}$

Furthermore,

$\Big{\langle}S_{2}(|x|^{2}v),(S_{2}-S_{3})vG_{-1}vPvG_{-1}v(S_{2}-S_{3})(|x|^{2}v)\Big{\rangle}=\|V\|_{L^{1}}\|S_{2}(|x|^{2}v)\|^{4}_{L^{2}}.$
(4.26)

Using (4.25) and $S_{3}(|x|^{2}v)=0$, we have

$\begin{split}&\big{\langle}S_{2}(|x|^{2}v),(S_{2}-S_{3})vG_{-1}vT_{0}D_{4}T_{0}vG_{-1}vS_{2}(|x|^{2}v)\big{\rangle}\\\
=&\big{\langle}D_{4}T_{0}PvG_{-1}vS_{2}(|x|^{2}v),T_{0}PvG_{-1}vS_{2}(|x|^{2}v)\big{\rangle}=\|S_{2}(|x|^{2}v)\|^{4}_{L^{2}}\big{\langle}D_{4}T_{0}v,T_{0}v\big{\rangle}.\end{split}$
(4.27)

Combing (4.26) and (4.27), one has

$\begin{split}\big{\langle}S_{2}(|x|^{2}v),L_{1}(|x|^{2}v)\big{\rangle}=\|S_{2}(|x|^{2}v)\|^{4}_{L^{2}}\|V\|_{L^{1}}\Big{(}1-\frac{\langle
D_{4}T_{0}v,T_{0}v\rangle}{\|V\|_{L^{1}}}\Big{)}.\end{split}$ (4.28)

Let $\eta=D_{4}S_{3}T_{0}v$ and $\xi=S_{3}T_{0}v$, then
$\xi=D_{4}^{-1}\eta=(T_{3}+S_{4})\eta$. By $S_{3}D_{4}=D_{4}S_{3}=D_{4}$, thus
we have

$1-\frac{\langle D_{4}T_{0}v,T_{0}v\rangle}{\|V\|_{L^{1}}}=1-\frac{\langle
S_{3}D_{4}T_{0}v,S_{3}T_{0}v\rangle}{\|V\|_{L^{1}}}=1-\frac{\langle\eta,\xi\rangle}{\|V\|_{L^{1}}}.$

Noting that

$T_{3}+S_{4}=\big{(}-c_{+}a_{+}S_{3}vG_{1}vS_{3}+S_{4}\big{)}+S_{3}T_{0}^{2}S_{3}:=A_{1}+B_{1},$

by (2.25) we know that for $f\in S_{3}L^{2}$,

$\begin{split}\langle
f,A_{1}f\rangle=&\frac{\|V\|_{L^{1}}}{1024}\Big{(}\langle
x_{1}^{2}v,f\rangle^{2}+\langle x_{2}^{2}v,f\rangle^{2}+2\langle
x_{1}x_{2}v,f\rangle^{2}\Big{)}+\|S_{4}f\|_{L^{2}}^{2}>0,\end{split}$

since $\eta=D_{4}\xi=S_{3}D_{4}\xi\in S_{3}L^{2}$, thus
$\langle\eta,A_{1}\eta\rangle>0$. Furthermore,

$0<\langle\eta,A_{1}\eta\rangle=\langle\eta,(T_{3}+S_{4})\eta-
B_{1}\eta\rangle=\langle\eta,\xi\rangle-\langle\eta,B_{1}\eta\rangle.$ (4.29)

By $S_{3}T_{0}Q=QT_{0}S_{3}=0$, then

$B_{1}\eta=S_{3}T_{0}^{2}S_{3}\eta=S_{3}T_{0}PT_{0}S_{3}\eta=\frac{\langle
T_{0}S_{3}\eta,v\rangle
S_{3}T_{0}v}{\|V\|_{L^{1}}}=\frac{\langle\eta,S_{3}T_{0}v\rangle
S_{3}T_{0}v}{\|V\|_{L^{1}}}=\frac{\langle\eta,\xi\rangle\xi}{\|V\|_{L^{1}}},$

which by (4.29) implies that

$0<\langle\eta,A_{1}\eta\rangle=\langle\eta,\xi\rangle-\frac{\big{|}\langle\eta,\xi\rangle\big{|}^{2}}{\|V\|_{L^{1}}},$

which implies that $\langle\eta,\xi\rangle\neq 0$ and
$1-\frac{\langle\eta,\xi\rangle}{\|V\|_{L^{1}}}\neq 0$. Then we have
$1-\frac{\langle D_{4}T_{0}v,T_{0}v\rangle}{\|V\|_{L^{1}}}\neq 0$ and
$\big{\langle}S_{2}(|x|^{2}v),L_{1}(|x|^{2}v)\big{\rangle}\neq 0$. Thus,
$L_{1}$ is invertible on $(S_{2}-S_{3})L^{2}$. Hence $d_{1}$ exists on
$(S_{2}-S_{3})L^{2}$ for enough small $\lambda$.

Note that $(S_{2}-S_{3})L^{2}=\hbox{span}\\{S_{2}(|x|^{2}v)\\}$ ( i.e.
$rank(S_{2}-S_{3})=1$ ). Let $f\in L^{2}$, then
$L_{1}f=(S_{2}-S_{3})L_{1}f\in(S_{2}-S_{3})L^{2}$. Thus, $L_{1}f=\langle
S_{2}(|x|^{2}v),L_{1}f\rangle(S_{2}|x|^{2}v)$, and
$L_{1}=\hbox{tr}(L_{1}):=c_{1}\neq 0$ by
$\big{\langle}S_{2}(|x|^{2}v),L_{1}(|x|^{2}v)\big{\rangle}\neq 0$. Similarly,
$L_{2}=\hbox{tr}(L_{2}):=c_{2}$ and $L_{3}=\hbox{tr}(L_{3}):=c_{3}$. Thus, we
have

$\begin{split}d_{1}=&\Big{(}g^{+}_{0}(\lambda)^{2}L_{1}+g^{+}_{0}(\lambda)L_{2}+L_{3}\Big{)}^{-1}\\\
=&g^{+}_{0}(\lambda)^{-2}(S_{2}-S_{3})\Big{(}L_{1}+g^{+}_{0}(\lambda)^{-1}L_{2}+g^{+}_{0}(\lambda)^{-2}L_{3}\Big{)}^{-1}(S_{2}-S_{3})\\\
=&g^{+}_{0}(\lambda)^{-2}(S_{2}-S_{3})\Big{(}c_{1}+g^{+}_{0}(\lambda)^{-1}c_{2}+g^{+}_{0}(\lambda)^{-2}c_{3}\Big{)}^{-1}(S_{2}-S_{3})\\\
:=&g^{+}_{0}(\lambda)^{-2}h^{+}(\lambda)(S_{2}-S_{3}),\end{split}$

where
$h^{+}(\lambda)=\Big{(}c_{1}+g^{+}_{0}(\lambda)^{-1}c_{2}+g^{+}_{0}(\lambda)^{-2}c_{3}\Big{)}^{-1}$
and $h^{+}(\lambda)=O_{1}(1)$. ∎

Now we start to compute the inverse of $A_{1}^{+}(\lambda)$ as follows:

######  Lemma 4.8.

Suppose that $|V(x)|\lesssim(1+|x|)^{-\beta}$ with some $\beta>18$. If zero is
the second kind of resonance of the spectrum of $H$, then for $0<|\lambda|\ll
1$,

$\begin{split}\big{(}A_{1}^{+}(\lambda)\big{)}^{-1}=&D_{4}+h^{+}(\lambda)\widetilde{S}(\lambda),\end{split}$

where $D_{4}=T_{3}^{-1}$ and

$\displaystyle\widetilde{S}(\lambda)=\widetilde{S}=\begin{pmatrix}b_{11}&b_{12}\\\
b_{21}&b_{22}\end{pmatrix},$

which $h(\lambda)$ is defined in Lemma 4.7, and $b_{ij}=b_{ij}(\lambda)$ are
absolutely bounded operators for enough small $\lambda$ as follows:

$\begin{split}b_{11}=&g^{+}_{0}(\lambda)^{-2}(S_{2}-S_{3}),\\\
b_{12}=&-\big{(}S_{2}-S_{3}\big{)}\Big{(}g^{+}_{0}(\lambda)^{-1}vG_{-1}vT_{0}+g^{+}_{0}(\lambda)^{-2}(T_{0}^{2}-c_{+}a_{+}vG_{1}v)\Big{)}D_{4},\\\
b_{21}=&-D_{4}\Big{(}g^{+}_{0}(\lambda)^{-1}T_{0}vG_{-1}v+g^{+}_{0}(\lambda)^{-2}(T_{0}^{2}-c_{+}a_{+}vG_{1}v)\Big{)}\big{(}S_{2}-S_{3}\big{)},\\\
b_{22}=&D_{4}\Big{(}T_{0}vG_{-1}v+g^{+}_{0}(\lambda)^{-1}(T_{0}^{2}-c_{+}a_{+}vG_{1}v)\Big{)}\big{(}S_{2}-S_{3}\big{)}\\\
&\times\Big{(}vG_{-1}vT_{0}+g^{+}_{0}(\lambda)^{-1}(T_{0}^{2}-c_{+}a_{+}vG_{1}v)\Big{)}D_{4}.\end{split}$
(4.30)

Moreover, $\|\widetilde{S}(\lambda)\|_{L^{2}\rightarrow L^{2}}=O_{1}(1)$ for
small enough $\lambda$.

In addition, if zero is not the first kind of resonance, the same formula
holds for $\big{(}A^{+}_{1}(\lambda)+S_{4}\big{)}^{-1}$ with
$D_{4}=(T_{3}+S_{4})^{-1}$.

###### Proof.

We only prove the case of $S_{4}\neq 0$ since the case of $S_{4}=0$ is
identical. In this case, $T_{3}+S_{4}$ is invertible on $S_{3}L^{2}$. Observe
that $vG_{-1}vS_{3}=0$, we have

$\begin{split}A_{1}^{+}(\lambda)+S_{4}=&\Big{(}(S_{2}-S_{3})+S_{3}\Big{)}\Big{(}g_{0}^{+}(\lambda)^{2}vG_{-1}vPvG_{-1}v+g_{0}^{+}(\lambda)\big{(}vG_{-1}vT_{0}+T_{0}vG_{-1}v\big{)}\\\
&+T_{0}^{2}-c_{+}a_{+}vG_{1}v\Big{)}\Big{(}(S_{2}-S_{3})+S_{3}\Big{)}.\end{split}$

By block format we write $A_{1}^{+}(\lambda)+S_{4}$ as follows:

$\begin{split}A_{1}^{+}(\lambda)+S_{4}=&\begin{pmatrix}a_{11}&a_{12}\\\
a_{21}&a_{22}\end{pmatrix},\end{split}$ (4.31)

where $a_{ij}=a_{ij}(\lambda)$ are bounded operators for small enough
$\lambda$ as follows:

$\begin{split}a_{11}=&(S_{2}-S_{3})\Big{(}g_{0}^{+}(\lambda)^{2}vG_{-1}vPvG_{-1}v\\\
&+g_{0}^{+}(\lambda)\big{(}vG_{-1}vT_{0}+T_{0}vG_{-1}v\big{)}+T_{0}^{2}-c_{+}a_{+}vG_{1}v\Big{)}(S_{2}-S_{3}),\\\
a_{12}=&(S_{2}-S_{3})\Big{(}g_{0}^{+}(\lambda)vG_{-1}vT_{0}+T_{0}^{2}-c_{+}a_{+}vG_{1}v\Big{)}S_{3},\\\
a_{21}=&S_{3}\Big{(}g_{0}^{+}(\lambda)T_{0}vG_{-1}v+T_{0}^{2}-c_{+}a_{+}vG_{1}v\Big{)}(S_{2}-S_{3}),\\\
a_{22}=&S_{3}T_{0}^{2}S_{3}-c_{+}a_{+}S_{3}vG_{1}vS_{3}+S_{4}=T_{3}+S_{4}.\end{split}$

By Lemma 4.2, then $A^{+}(\lambda)+S_{4}$ has a bounded inverse if and only if

$d_{1}:=(a_{11}-a_{12}a_{22}^{-1}a_{21})^{-1}$

exist and is bounded. Let $D_{4}=(T_{3}+S_{4})^{-1}:S_{3}L^{2}\rightarrow
S_{3}L^{2}$. Note that $rank(S_{2}-S_{3})=1$, and
$S_{3}D_{4}=D_{4}S_{3}=D_{4}$. By Lemma 4.7, the inverse operator

$d_{1}:=g^{+}_{0}(\lambda)^{-2}\Big{(}L_{1}+g^{+}_{0}(\lambda)^{-1}L_{2}+g^{+}_{0}(\lambda)^{-2}L_{3}\Big{)}^{-1}$

exists on $(S_{2}-S_{3})L^{2}$ for enough small $\lambda>0$, and

$d_{1}=g^{+}_{0}(\lambda)^{-2}h^{+}(\lambda)(S_{2}-S_{3}).$

Using Lemma 4.2 again, $A_{1}^{+}(\lambda)+S_{4}$ is invertible and

$\displaystyle\big{(}A_{1}^{+}(\lambda)+S_{4}\big{)}^{-1}=$ $\displaystyle
D_{4}+h^{+}(\lambda)\widetilde{S}(\lambda),$

where

$\displaystyle\widetilde{S}(\lambda)=\widetilde{S}=\begin{pmatrix}b_{11}&b_{12}\\\
b_{21}&b_{22}\end{pmatrix},$

where $b_{i,j}(\lambda)$ are bounded operators defined in (4.30).

It is easy to check that $\|\widetilde{S}(\lambda)\|_{L^{2}\rightarrow
L^{2}}=O_{1}(1)$ for small enough $\lambda$. ∎

(III) The second kind of resonance. If zero is the second kind of resonance,
then $T_{3}$ is invertible on $S_{3}L^{2}$, let $D_{4}=T_{3}^{-1}$. Since

$\big{(}\widetilde{M}^{+}_{3}(\lambda)\big{)}^{-1}=\big{(}A_{1}^{+}(\lambda)\big{)}^{-1}\Big{(}I+W_{2}^{+}(\lambda)\big{(}A_{1}^{+}(\lambda)\big{)}^{-1}\Big{)}^{-1}.$

Note that $D_{4}B_{1}^{1}=B_{1}^{1}D_{4}=0$, $D_{4}B_{2}^{1}D_{4}=0$, by Lemma
4.8 and (4.24), we have

$\begin{split}\big{(}\widetilde{M}^{+}_{3}(\lambda)\big{)}^{-1}=&h^{+}(\lambda)S_{2}\Gamma^{2}_{0,1}S_{2}+g^{+}_{0}(\lambda)^{5}\lambda^{2}S_{2}\Gamma^{2}_{2,1}S_{2}+\lambda^{2}S_{2}\Gamma^{2}_{2,2}S_{2}\\\
&+g^{+}_{0}(\lambda)^{10}\lambda^{4}S_{2}\Gamma^{2}_{4,1}S_{2}+\lambda^{4}S_{2}\Gamma^{2}_{4,2}S_{2}+O_{1}(\lambda^{4}\ln\lambda).\end{split}$

By (4.24) one has

$\begin{split}\big{(}M^{+}_{3}(\lambda)\big{)}^{-1}=&\frac{g_{0}^{+}(\lambda)^{-1}h^{+}(\lambda)}{\lambda^{2}}S_{2}\Gamma^{2}_{-2,1}S_{2}+g_{0}^{+}(\lambda)^{4}S_{2}\Gamma^{2}_{0,1}S_{2}+S_{2}\Gamma^{2}_{0,2}S_{2}\\\
&+g_{0}^{+}(\lambda)^{9}\lambda^{2}S_{2}\Gamma^{2}_{2,1}S_{2}+\lambda^{2}S_{2}\Gamma^{2}_{2,1}S_{2}+O_{1}\big{(}\lambda^{2}(\ln\lambda)^{-1}\big{)}.\end{split}$

According to Lemma 4.1 and (4.22),

$\begin{split}\big{(}\widetilde{M}_{2}^{+}(\lambda)\big{)}^{-1}=&\frac{h^{+}(\lambda)}{\lambda^{2}}S_{2}\Gamma^{2}_{-2,1}S_{2}+g_{0}^{+}(\lambda)^{5}S_{2}\Gamma^{2}_{0,1}S_{2}+g_{0}^{+}(\lambda)S_{2}\Gamma^{2}_{0,2}S_{2}+g_{0}^{+}(\lambda)^{10}\lambda^{2}S_{2}\Gamma^{2}_{2,1}S_{2}\\\
&+g_{0}^{+}(\lambda)\lambda^{2}S_{2}\Gamma^{2}_{2,1}S_{2}+O_{1}\big{(}\lambda^{2}\big{)}.\end{split}$

By the same argument with the proof of the first kind of resonance, we
immediately get

$\begin{split}\big{(}M^{\pm}(\lambda)\big{)}^{-1}=&\frac{h^{\pm}(\lambda)}{\lambda^{2}}S_{2}\Gamma^{2}_{-2,0}S_{2}+g^{\pm}_{0}(\lambda)^{5}\Big{(}S_{2}\Gamma^{2}_{0,1}+\Gamma^{2}_{0,2}S_{2}+S_{1}\Gamma^{2}_{0,3}Q+Q\Gamma^{2}_{0,4}S_{1}\Big{)}\\\
&+g^{\pm}_{0}(\lambda)^{10}\lambda^{2}\Big{(}S_{2}\Gamma^{2}_{2,1}+\Gamma^{2}_{2,2}S_{2}+S_{1}\Gamma^{2}_{2,3}Q+Q\Gamma^{2}_{2,4}S_{1}\Big{)}\\\
&+g^{\pm}_{0}(\lambda)\lambda^{2}\Big{(}S_{2}\Gamma^{2}_{2,5}+\Gamma^{2}_{2,6}S_{2}+S_{1}\Gamma^{2}_{2,7}Q+Q\Gamma^{2}_{2,8}S_{1}\Big{)}+O_{1}(\lambda^{2}).\end{split}$

Thus the expansions of $\big{(}M^{+}(\lambda)\big{)}^{-1}$ of the second kind
resonance is completed. $\Box$If zero is not the second kind of resonance.
Then $T_{3}$ is not invertible on $S_{3}L^{2}$. Let $S_{4}$ is the Riesz
projection onto the kernel of $T_{3}$. Then $T_{3}+S_{4}$ is invertible on
$S_{3}L^{2}$. Let $D_{4}=(T_{3}+S_{4})^{-1}$. Noting that the identity,

$\big{(}\widetilde{M}^{+}_{3}(\lambda)+S_{4}\big{)}^{-1}=\big{(}A_{1}^{+}(\lambda)+S_{4}\big{)}^{-1}\Big{(}I+W_{2}^{+}(\lambda)\big{(}A_{1}^{+}(\lambda)+S_{4}\big{)}^{-1}\Big{)}^{-1},$

by using Lemma 4.8 and (4.24), we have by Neumann series

$\begin{split}&\big{(}\widetilde{M}^{+}_{3}(\lambda)+S_{4}\big{)}^{-1}\\\
=&D_{4}+\widetilde{S}-g^{+}_{0}(\lambda)^{5}\lambda^{2}B^{2}_{1}-g^{+}_{0}(\lambda)^{4}\lambda^{2}B^{2}_{2}-g^{+}_{0}(\lambda)^{3}\lambda^{2}B^{2}_{3}-g^{+}_{0}(\lambda)^{2}\lambda^{2}B^{2}_{4}\\\
&-g^{+}_{0}(\lambda)\lambda^{2}B^{2}_{5}(\lambda)-\widetilde{g}^{+}_{2}(\lambda)\lambda^{2}B^{2}_{6}-\lambda^{2}B^{2}_{7}-g^{+}_{0}(\lambda)^{-1}\lambda^{2}B^{2}_{9}-g^{+}_{0}(\lambda)^{-2}\lambda^{2}B^{2}_{11}\\\
&+O_{1}(\lambda^{4-\epsilon}),\end{split}$ (4.32)

where
$B^{2}_{j}=-a_{+}\big{(}D_{4}+h^{+}(\lambda)\widetilde{S}\big{)}S_{2}B_{j}^{1}S_{2}\big{(}D_{4}+h^{+}(\lambda)\widetilde{S}\big{)},1\leq
j\leq 11$.

According to Lemma 4.1, $\widetilde{M}_{3}^{+}(\lambda)$ has bounded inverse
if and only if

$M_{4}^{+}(\lambda):=S_{4}-S_{4}\big{(}\widetilde{M}_{3}^{+}(\lambda)+S_{4}\big{)}^{-1}S_{4}$
(4.33)

has bounded inverse on $S_{4}L^{2}$. Since

$S_{4}\widetilde{S}=\widetilde{S}S_{4}=0,\
S_{4}\Big{(}A_{1}^{+}(\lambda)+S_{4}\Big{)}^{-1}=\Big{(}A_{1}^{+}(\lambda)+S_{4}\Big{)}^{-1}S_{4}=S_{4},$

thus $S_{4}B_{j}=B_{j}S_{4}=0(j=1,2,3,4,6,7)$,
$S_{4}\widetilde{B}_{j}=\widetilde{B}_{j}S_{4}=0(j=1,2,3,7)$. Moreover,

$S_{4}B_{j}^{2}S_{4}=0(j=1,\cdots,5,8,10,11),$
$S_{4}B_{6}^{2}S_{4}=-a_{+}S_{4}vG_{2}vS_{4},\
S_{4}B_{7}^{2}S_{4}=-a_{+}S_{4}vG_{3}vS_{4},$
$S_{4}B_{9}^{2}S_{4}=c_{+}^{2}a_{+}\Big{(}S_{4}vG_{1}vQSQvG_{1}vS_{4}+S_{4}vG_{1}vQSD_{2}SQvG_{1}vS_{4}\Big{)}.$

Hence, by (4.32) and (4.33) one has

$\begin{split}M^{+}_{4}(\lambda)=&-a_{+}\widetilde{g}_{2}^{+}(\lambda)\lambda^{2}\Big{(}S_{4}vG_{2}vS_{4}+\widetilde{g}_{2}^{+}(\lambda)^{-1}S_{4}vG_{3}vS_{4}-c_{+}^{2}\widetilde{g}_{2}^{+}(\lambda)^{-1}g_{0}^{+}(\lambda)^{-1}\\\
&\times\big{(}S_{4}vG_{1}vQSQvG_{1}vS_{4}+S_{4}vG_{1}vQSD_{2}SQvG_{1}vS_{4}\big{)}+O(\lambda^{2-\epsilon})\Big{)}\\\
:=&-a_{+}\widetilde{g}_{2}^{+}(\lambda)\lambda^{2}\widetilde{M}^{+}_{4}(\lambda).\end{split}$

Let $T_{4}=S_{4}vG_{2}vS_{4}$. Then

$\begin{split}\widetilde{M}^{+}_{4}(\lambda)=&T_{4}+W_{3}^{+}(\lambda),\end{split}$
(4.34)

where

$\begin{split}W_{3}^{+}(\lambda)=&\widetilde{g}_{2}^{+}(\lambda)^{-1}S_{4}vG_{3}vS_{4}-\widetilde{g}_{2}^{+}(\lambda)^{-1}g_{0}^{+}(\lambda)^{-1}c_{+}^{2}\Big{(}S_{4}vG_{1}vQSQvG_{1}vS_{4}\\\
&+S_{4}vG_{1}vQSD_{2}SQvG_{1}vS_{4}\Big{)}+O(\lambda^{2-\epsilon}).\end{split}$

(IV) The third kind of resonance. If zero is the third kind of resonance of
$H$. Then $T_{4}$ is invertible on $S_{4}L^{2}$, and set $D_{5}=T_{4}^{-1}$.
Since

$\big{(}\widetilde{M}^{+}_{4}(\lambda)\big{)}^{-1}=D_{5}\Big{(}I+W_{3}^{+}(\lambda)D_{5}\Big{)}^{-1},\
\displaystyle\big{(}M^{+}_{4}(\lambda)\big{)}^{-1}=-\frac{\widetilde{g}_{2}^{+}(\lambda)^{-1}}{a_{+}\lambda^{2}}\big{(}\widetilde{M}^{+}_{4}(\lambda)\big{)}^{-1},$

by Neumann series one has

$\begin{split}\big{(}M^{+}_{4}(\lambda)\big{)}^{-1}=&\frac{-\widetilde{g}_{2}^{+}(\lambda)^{-1}D_{5}}{a_{+}\lambda^{2}}+\frac{\widetilde{g}_{2}^{+}(\lambda)^{-2}D_{5}vG_{3}vD_{5}}{a_{+}\lambda^{2}}+O_{1}(\lambda^{-2}(\ln\lambda)^{-3}).\end{split}$

Using Lemma 4.1 and (4.32), we have

$\begin{split}\big{(}\widetilde{M}^{+}_{3}(\lambda)\big{)}^{-1}=&\frac{\widetilde{g}_{2}^{+}(\lambda)^{-1}g_{0}^{+}(\lambda)^{-1}D_{5}}{\lambda^{4}}-\frac{\widetilde{g}_{2}^{+}(\lambda)^{-2}g_{0}^{+}(\lambda)^{-1}D_{5}vG_{3}vD_{5}}{\lambda^{4}}+O_{1}\big{(}\lambda^{-4}(\ln\lambda)^{-4}\big{)}.\end{split}$

Using (4.24) one has

$\begin{split}\big{(}M^{+}_{3}(\lambda)\big{)}^{-1}=&\frac{\widetilde{g}_{2}^{+}(\lambda)^{-1}g_{0}^{+}(\lambda)^{-1}D_{5}}{\lambda^{4}}-\frac{\widetilde{g}_{2}^{+}(\lambda)^{-2}g_{0}^{+}(\lambda)^{-1}D_{5}vG_{3}vD_{5}}{\lambda^{4}}+O_{1}\big{(}\lambda^{-4}(\ln\lambda)^{-4}\big{)}.\end{split}$

By the same arguments as in the proof of the second kind of resonance, we
immediately obtain that

$\begin{split}\big{(}M^{+}(\lambda)\big{)}^{-1}=&\frac{\widetilde{g}_{2}^{+}(\lambda)^{-1}}{\lambda^{4}}D_{5}-\frac{\widetilde{g}_{2}^{+}(\lambda)^{-2}}{\lambda^{4}}D_{5}vG_{3}vD_{5}+O_{1}\big{(}\lambda^{-4}(\ln\lambda)^{-3}\big{)}\\\
:=&\frac{\widetilde{g}_{2}^{+}(\lambda)^{-1}}{\lambda^{4}}S_{4}\Gamma_{-4,1}^{3}S_{4}+\frac{\widetilde{g}_{2}^{+}(\lambda)^{-2}}{\lambda^{4}}S_{4}\Gamma_{-4,2}^{3}S_{4}+O_{1}\big{(}\lambda^{-4}(\ln\lambda)^{-3}\big{)}.\end{split}$

Thus the expansions of $\big{(}M^{+}(\lambda)\big{)}^{-1}$ of the third kind
resonance is completed. $\Box$

Before computing the expansions of $\big{(}M^{+}(\lambda)\big{)}^{-1}$ of the
fourth kind resonance, we first give a lemma as follows.

######  Lemma 4.9.

$\hbox{ker}(S_{5}vG_{3}vS_{5})=\\{0\\}$, thus the operator
$T_{5}=S_{5}vG_{3}vS_{5}$ is invertible on $S_{5}L^{2}$.

###### Proof.

If $f\in\hbox{ker}(S_{5}vG_{3}vS_{5})$, then $S_{5}vG_{3}vS_{5}f=0$ for $f\in
S_{5}L^{2}$, and

$0=\langle S_{5}vG_{3}vS_{5}f,f\rangle=\langle G_{3}vf,vf\rangle.$

Since

$R_{0}^{+}(\mu^{4})=\frac{b_{+}}{\lambda^{2}}I+g^{+}_{0}(\lambda)G_{-1}+G_{0}+c_{+}\lambda^{2}G_{1}+\widetilde{g}^{+}_{2}(\lambda)G_{2}+\lambda^{4}G_{3}+O(\lambda^{6-\epsilon}|x-y|^{8-\epsilon}),$

so,

$G_{3}=\frac{R_{0}^{+}(\lambda^{4})-\frac{b_{+}}{\lambda^{2}}I-g^{+}_{0}(\lambda)G_{-1}-G_{0}-c_{+}\lambda^{2}G_{1}-\widetilde{g}^{+}_{2}(\lambda)G_{2}+O(\lambda^{6-\epsilon}|x-y|^{8-\epsilon})}{\lambda^{4}}.$

Noting that for $f\in S_{5}L^{2}$,

$\langle G_{2}vf,vf\rangle=\langle G_{1}vf,vf\rangle=\langle
G_{-1}vf,vf\rangle=0,$ $\langle Ivf,vf\rangle=\langle
v(x)Iv(y)f(y),f(y)\rangle=\langle Pf,f\rangle=0,$

we get that

$\begin{split}0=&\langle S_{5}vG_{3}vS_{5}f,f\rangle=\langle
G_{3}vf,vf\rangle\\\ =&\lim_{\lambda\rightarrow
0}\Big{\langle}\frac{R_{0}^{+}(\lambda^{4})-\frac{b_{+}}{\lambda^{2}}I-g^{+}_{0}(\lambda)G_{-1}-G_{0}-c_{+}\lambda^{2}G_{1}-\widetilde{g}^{+}_{2}(\lambda)G_{2}+O(\lambda^{6-\epsilon}|x-y|^{8-\epsilon})}{\lambda^{4}}vf,vf\Big{\rangle}\\\
=&\lim_{\lambda\rightarrow
0}\frac{1}{\lambda^{4}}\big{\langle}\big{(}R_{0}^{+}(\lambda^{4})-G_{0}\big{)}vf,vf\big{\rangle}.\end{split}$

Notice that $G_{0}=(\Delta^{2})^{-1}$, by using the Fourier transform one has

$\displaystyle 0$ $\displaystyle=\lim_{\lambda\rightarrow
0}\frac{1}{\lambda^{4}}\Big{\langle}\Big{(}\frac{1}{\xi^{4}-\lambda^{4}}-\frac{1}{\xi^{4}}\Big{)}\widehat{vf}(\xi),\widehat{vf}(\xi)\Big{\rangle}$
$\displaystyle=\lim_{\lambda\rightarrow
0}\Big{\langle}\frac{\widehat{vf}(\xi)}{(\xi^{4}-\lambda^{4})\xi^{4}},\widehat{vf}(\xi)\Big{\rangle}=\Big{\langle}\frac{\widehat{vf}(\xi)}{\xi^{8}},\widehat{vf}(\xi)\Big{\rangle}$
$\displaystyle=\int_{\mathbb{R}^{2}}\frac{|\widehat{vf}(\xi)|^{2}}{\xi^{8}}d\xi=\langle
G_{0}vf,G_{0}vf\rangle,$

where we use the monotone convergence theorem as $\lambda\rightarrow 0$ by
choosing $\lambda\in\mathbb{C}$ such that $\lambda^{4}<0$. Thus, $vf=0$ since
$\widehat{vf}=0$. Noting that $f\in S_{5}L^{2}$, hence $f=-UvG_{0}vf=0$. ∎

(V) The fourth kind of resonance. If zero is not the fourth kind of resonance,
then $T_{4}$ is not invertible on $S_{4}L^{2}$. Let $S_{5}$ is the Riesz
projection onto the kernel of $T_{4}$, then $T_{4}+S_{5}$ is invertible on
$S_{4}L^{2}$, and set $D_{5}=(T_{4}+S_{5})^{-1}$. Noting that the identity

$\big{(}\widetilde{M}^{+}_{4}(\lambda)+S_{5}\big{)}^{-1}=D_{5}\Big{(}I+W_{3}^{+}(\lambda)D_{5}\Big{)}^{-1},$

by (4.34) and Neumann series, we have

$\begin{split}\big{(}\widetilde{M}^{+}_{4}(\lambda)+S_{5}\big{)}^{-1}=&D_{5}-\widetilde{g}_{2}^{+}(\lambda)^{-1}D_{5}vG_{3}vD_{5}+\widetilde{g}_{2}^{+}(\lambda)^{-1}g_{0}^{+}(\lambda)^{-1}c_{+}^{2}\Big{(}D_{5}vG_{1}vQSQvG_{1}vD_{5}\\\
&+D_{5}vG_{1}vQSD_{2}SQvG_{1}vD_{5}\Big{)}+O((\ln\lambda)^{-3}).\end{split}$

According to Lemma 4.1, $\widetilde{M}_{4}^{+}(\lambda)$ has bounded inverse
if and only if

$M_{5}^{+}(\lambda):=S_{5}-S_{5}\big{(}\widetilde{M}_{4}^{+}(\lambda)+S_{5}\big{)}^{-1}S_{5}$
(4.35)

has bounded inverse on $S_{5}L^{2}(\mathbb{R}^{2})$. Note that
$S_{5}vG_{1}vQ=QvG_{1}vS_{5}=0$, by Neumann series one has

$\begin{split}M_{5}(\lambda)=\widetilde{g}_{2}^{+}(\lambda)^{-1}S_{5}vG_{3}vS_{5}+O\big{(}(\ln\lambda)^{-2}\big{)}.\end{split}$

By Lemma 4.9, $T_{5}=S_{5}vG_{3}vS_{5}$ is invertible on $S_{5}L^{2}$. Let
$D_{6}=T_{5}^{-1}$. Then we have

$\big{(}M_{5}(\lambda)\big{)}^{-1}=\widetilde{g}_{2}^{+}(\lambda)D_{6}+O\big{(}(\ln\lambda)^{-2}\big{)}.$

By the same arguments as in the proof of the third kind of resonance, we
immediately obtain that

$\begin{split}\big{(}M^{+}(\lambda)\big{)}^{-1}=&\frac{1}{\lambda^{4}}S_{5}D_{6}S_{5}+\frac{\widetilde{g}_{2}^{+}(\lambda)^{-1}}{\lambda^{4}}\Big{(}S_{4}\Gamma_{-4,1}^{4}S_{4}\Big{)}+\frac{\widetilde{g}_{2}^{+}(\lambda)^{-2}}{\lambda^{4}}S_{4}\Gamma_{-4,2}^{4}S_{4}+O_{1}\big{(}\lambda^{-4}(\ln\lambda)^{-3}\big{)}.\end{split}$

Thus, the expansions of $\big{(}M^{+}(\lambda)\big{)}^{-1}$ of the fourth kind
resonance is completed. $\Box$

##  5\. Classification of threshold spectral subspaces

By Definition 2.4 we know that the four projection subspaces
$S_{i}L^{2}(i=1,2,4,5)$ define the four zero resonance types. In this section,
we will identity these four subspace by the distributional solutions of
$H\phi=0$ in the weighted spaces $L^{\infty}_{s}$ with some $s\in\mathbb{R}$.

We define $\ln^{+}t=\ln t$ for $t>1$ and $\ln^{+}t=0$ for $0<t\leq 1$;
$\ln^{-}t=0$ for $t>1$ and $\ln^{-}t=-\ln t$ for $0<t\leq 1$. Then $\ln
t=\ln^{+}t-\ln^{-}t$.

######  Proposition 5.1.

Assume that $|V(x)|\lesssim(1+|x|)^{-\beta}$ with $\beta>14$. Then $f\in
S_{1}L^{2}(\mathbb{R}^{2})\setminus\\{0\\}$ if and only if $f=Uv\phi$ with
$\phi\in L^{\infty}_{-1}(\mathbb{R}^{2})$ such that $H\phi=0$ in the
distributional sense, and

$\phi=-G_{0}vf+c_{3}x_{2}+c_{2}x_{1}+c_{1},$

where $c_{1},c_{2},c_{3}$ are some constants as follows:

$c_{1}=\frac{\langle f_{2},T_{0}f\rangle\langle x_{2}v,f_{1}\rangle\langle
x_{1}v,v\rangle}{\|f_{2}\|^{2}_{L^{2}(\mathbb{R}^{2})}\|f_{1}\|^{2}_{L^{2}(\mathbb{R}^{2})}\|V\|_{L^{1}(\mathbb{R}^{2})}}-\frac{\langle
f_{1},T_{0}f\rangle\langle
x_{1}v,v\rangle}{\|f_{1}\|^{2}_{L^{2}(\mathbb{R}^{2})}\|V\|_{L^{1}(\mathbb{R}^{2})}}+\frac{\langle
v,T_{0}f\rangle}{\|V\|_{L^{1}(\mathbb{R}^{2})}},$ $\displaystyle
c_{2}=-\frac{\langle f_{2},T_{0}f\rangle\langle
x_{2}v,f_{1}\rangle}{\|f_{2}\|^{2}_{L^{2}(\mathbb{R}^{2})}\|f_{1}\|^{2}_{L^{2}(\mathbb{R}^{2})}}+\frac{\langle
f_{1},T_{0}f\rangle}{\|f_{1}\|^{2}_{L^{2}(\mathbb{R}^{2})}},\,\
\,\displaystyle c_{3}=\frac{\langle
f_{2},T_{0}f\rangle}{\|f_{2}\|^{2}_{L^{2}(\mathbb{R}^{2})}}.$

###### Proof.

Let $f\in S_{1}L^{2}$. Then one has $S_{0}(U+vG_{0}v)f=0$. Hence we have

$0=S_{0}(U+vG_{0}v)f=(I-\widetilde{P})(U+vG_{0}v)f=Uf+vG_{0}vf-\widetilde{P}T_{0}f,$

where $\widetilde{P}$ is the projection onto the
span$\\{v,x_{1}v,x_{2}v\\}=(S_{0}L^{2})^{\perp}$. Let $v,f_{1}$ and $f_{2}$
are the orthogonal basis of $(S_{0}L^{2})^{\perp}$ by the Schmidt
orthogonalization, i.e. setting $\displaystyle f_{1}=x_{1}v-\frac{\langle
x_{1}v,v\rangle}{\|V\|_{L^{1}}}v,$ $\displaystyle f_{2}=x_{2}v-\frac{\langle
x_{2}v,f_{1}\rangle}{\|f_{1}\|^{2}_{L^{2}}}f_{1}-\frac{\langle
x_{2}v,v\rangle}{\|V\|_{L^{1}}}v$. Then it follows that

$\begin{split}\widetilde{P}T_{0}f=&\frac{\langle
f_{2},T_{0}f\rangle}{\|f_{2}\|^{2}_{L^{2}}}f_{2}+\frac{\langle
f_{1},T_{0}f\rangle}{\|f_{1}\|^{2}_{L^{2}}}f_{1}+\frac{\langle
v,T_{0}f\rangle}{\|V\|_{L^{1}}}v\\\ =&x_{2}v\frac{\langle
f_{2},T_{0}f\rangle}{\|f_{2}\|^{2}_{L^{2}}}+x_{1}v\Big{(}-\frac{\langle
f_{2},T_{0}f\rangle\langle
x_{2}v,f_{1}\rangle}{\|f_{2}\|^{2}_{L^{2}}\|f_{1}\|^{2}_{L^{2}}}+\frac{\langle
f_{1},T_{0}f\rangle}{\|f_{1}\|^{2}_{L^{2}}}\Big{)}\\\ &+v\bigg{(}\frac{\langle
f_{2},T_{0}f\rangle\langle x_{2}v,f_{1}\rangle\langle
x_{1}v,v\rangle}{\|f_{2}\|^{2}_{L^{2}}\|f_{1}\|^{2}_{L^{2}}\|V\|_{L^{1}}}-\frac{\langle
f_{1},T_{0}f\rangle\langle
x_{1}v,v\rangle}{\|f_{1}\|^{2}_{L^{2}}\|V\|_{L^{1}}}+\frac{\langle
v,T_{0}f\rangle}{\|V\|_{L^{1}}}\bigg{)}\\\
:=&c_{3}x_{2}v+c_{2}x_{1}v+c_{1}v.\end{split}$

Furthermore, we have

$Uf=-vG_{0}vf+\widetilde{P}T_{0}f=-vG_{0}vf+c_{3}x_{2}v+c_{2}x_{1}v+c_{1}v.$

Since $U^{2}=1$, then

$f=U^{2}f=Uv(-G_{0}vf+c_{3}x_{2}+c_{2}x_{1}+c_{1}):=Uv\phi.$

Thus, $vf=vUv\phi=V\phi$. To show that $H\phi=(\Delta^{2}+V)\phi=0$, noting
that $\Delta^{2}(c_{3}x_{2}+c_{2}x_{1}+c_{1})=0$ and
$G_{0}=\big{(}\Delta^{2}\big{)}^{-1}$, one has
$(\Delta^{2}+V)\phi=-vf+V\phi=0.$

Next, we show that $\phi\in L^{\infty}_{-1}:=\\{\langle x\rangle^{-1}\phi\in
L^{\infty}\\}$. Since $c_{3}x_{2}+c_{2}x_{1}+c_{1}\in L^{\infty}_{-1}$, it
suffices to show that

$G_{0}vf(x)=\frac{1}{8\pi}\int_{\mathbb{R}^{2}}|x-y|^{2}\ln|x-y|v(y)f(y)dy\in
L^{\infty}_{-1}.$

In fact, if $f\in S_{0}L^{2}$, by the orthogonality $\langle
x_{1}v,f\rangle=\langle x_{2}v,f\rangle=\langle v,f\rangle=0$, then

$\int_{\mathbb{R}^{2}}|x|^{2}\ln|x|v(y)f(y)dy=\int_{\mathbb{R}^{2}}x\cdot
y\ln|x|v(y)f(y)dy=0,$

which leads to

$\begin{split}G_{0}vf(x)=&\frac{1}{8\pi}\int_{\mathbb{R}^{2}}|x|^{2}(\ln|x-y|-\ln|x|)v(y)f(y)dy\\\
&-\frac{1}{4\pi}\int_{\mathbb{R}^{2}}x\cdot y(\ln|x-y|-\ln|x|)v(y)f(y)dy\\\
&+\frac{1}{8\pi}\int_{\mathbb{R}^{2}}|y|^{2}\ln|x-y|v(y)f(y)dy:=I_{1}+I_{2}+I_{3}.\end{split}$
(5.1)

We now begin to estimate the first integral $I_{1}$. We have

$\begin{split}I_{1}=&\frac{1}{8\pi}\int_{|x|>2|y|}|x|^{2}(\ln|x-y|-\ln|x|)v(y)f(y)dy\\\
&+\frac{1}{8\pi}\int_{|x|\leq
2|y|}|x|^{2}(\ln|x-y|-\ln|x|)v(y)f(y)dy:=I_{11}+I_{12}.\end{split}$ (5.2)

For $I_{11}$. If $|x|>2|y|$, then $|x-y|<2|x|$, hence
$\frac{|x-y|-|x|}{|x|}<1$. By Taylor’s formula we have

$\ln\frac{|x-y|}{|x|}=\ln\Big{(}1+\frac{|x-y|-|x|}{|x|}\Big{)}=O\Big{(}\frac{|y|}{|x|}\Big{)}.$
(5.3)

By Hölder’s inequality and (5.3), one has

$\begin{split}|I_{11}|=&\Big{|}\int_{|x|>2|y|}|x|^{2}(\ln|x-y|-\ln|x|)v(y)f(y)dy\Big{|}\\\
\lesssim&|x|\int_{\mathbb{R}^{2}}|yv(y)|\cdot|f(y)|dy\lesssim|x|\|yv(y)\|_{L^{2}}\|f\|_{L^{2}}\lesssim|x|\in
L_{-1}^{\infty}.\end{split}$

For $I_{12}$. We have

$\begin{split}\int_{|x|\leq
2|y|}|x|^{2}(\ln|x-y|-\ln|x|)v(y)&f(y)dy=\int_{|x|\leq
2|y|}|x|^{2}\big{(}\ln^{+}|x-y|-\ln^{+}|x|\big{)}\cdot v(y)f(y)dy\\\
&-\int_{|x|\leq 2|y|}|x|^{2}\big{(}\ln^{-}|x-y|-\ln^{-}|x|\big{)}\cdot
v(y)f(y)dy:=J_{1}+J_{2}.\end{split}$

For the first term $J_{1}$. If $|x|\leq 2|y|$, then $|x-y|\leq 3(1+|y|)$ and
$|x|\leq 2(1+|y|)$. Since $\log^{+}(\cdot)$ is a nondecreasing function, then
one has

$\big{|}\ln^{+}|x-y|-\ln^{+}|x|\big{|}\leq\ln^{+}|x-y|+\ln^{+}|x|\lesssim
1+\log^{+}(1+|y|)\lesssim\langle y\rangle^{0+},\,\,\hbox{as}\,\,|x|\leq 2|y|.$
(5.4)

Thus by Hölder’s inequality and (5.4), one has

$\begin{split}|J_{1}|&\lesssim\int_{|x|\leq
2|y|}|x|^{2}\big{(}\ln^{+}|x-y|+\ln^{+}|x|\big{)}\cdot|v(y)f(y)|dy\\\
&\lesssim\int_{|x|\leq 2|y|}|x|^{2}\langle y\rangle^{0+}\cdot|v(y)f(y)|dy\\\
&\lesssim\int_{|x|\leq 2|y|}\langle
y\rangle^{2+}|v(y)f(y)|dy\leq\|v(\cdot)\langle\cdot\rangle^{2+}\|_{L^{2}}\|f\|_{L^{2}}<\infty.\end{split}$

For the second term $J_{2}$. By using the definition of function
$\ln^{-}(\cdot)$, we have $|x|^{2}\ln^{-}|x|\lesssim 1$. Then by Hölder’s
inequality again, we have

$\begin{split}|J_{2}|\leq&\int_{|x|\leq
2|y|,\,|x-y|<1}|x|^{2}\big{(}-\ln|x-y|\big{)}\cdot|v(y)f(y)|dy+\int_{|x|\leq
2|y|}|x|^{2}\ln^{-}|x||\cdot|v(y)f(y)|dy\\\
\lesssim&\int_{|x-y|<1}|y|^{2}\frac{1}{|x-y|^{0+}}\cdot|v(y)f(y)|dy+|x|^{2}\ln^{-}|x|\int_{|x|\leq
2|y|}|v(y)f(y)|dy\\\
\lesssim&\int_{|x-y|<1}\frac{1}{|x-y|^{0+}}\cdot|y^{2}v(y)f(y)|dy+\int_{|x|\leq
2|y|}|yv(y)f(y)|dy\\\
\lesssim&\left(\int_{|x-y|<1}\Big{(}\frac{1}{|x-y|^{0+}}\Big{)}^{2}dy\right)^{1/2}\,\|y^{2}v(y)\|_{L^{\infty}}\|f\|_{L^{2}}+\|yv(y)\|_{L^{2}}\|f\|_{L^{2}}\\\
\lesssim&\|y^{2}v(y)\|_{L^{\infty}}\|f\|_{L^{2}}+\|yv(y)\|_{L^{2}}\|f\|_{L^{2}}<\infty.\end{split}$

Combining the estimates for $J_{1}$ and $J_{2}$ above, then
$|I_{12}|\leq|J_{1}|+|J_{2}|<\infty$. Hence combining with the estimates for
$I_{11}$ and $I_{12}$ above, it follows that $I_{1}\in L^{\infty}_{-1}$.

By the same arguments with $I_{1}$, and notice that $|x\cdot y|\leq|x||y|$, we
immediately obtain that $I_{2}\in L^{\infty}_{-1}$. Next, we turn to estimate
the last integral $I_{3}$. By Hölder’s inequality again we have

$\begin{split}|I_{3}|=&\Big{|}\int_{\mathbb{R}^{2}}\ln|x-y|\cdot|y|^{2}v(y)f(y)dy\Big{|}\\\
=&\Big{|}\int_{|x-y|<1}\ln|x-y|\cdot|y|^{2}v(y)f(y)dy+\int_{|x-y|\geq
1}\ln|x-y|\cdot|y|^{2}v(y)f(y)dy\Big{|}\\\
\lesssim&\int_{|x-y|<1}\frac{|y|^{2}|v(y)f(y)|}{|x-y|^{0+}}dy+\int_{|x-y|\geq
1}|x-y|^{0+}\cdot|y|^{2}v(y)f(y)dy\\\
\lesssim&\int_{|x-y|<1}\frac{|y|^{2}|v(y)f(y)|}{|x-y|^{0+}}dy+|x|^{0+}\int_{\mathbb{R}^{2}}|y|^{2+}|v(y)f(y)|dy\\\
\lesssim&1+|x|^{0+}\in L^{\infty}_{-1}.\end{split}$

Hence combining with all estimates for $I_{i}(i=1,2,3)$ above and (5.1), we
obtain that $G_{0}vf\in L^{\infty}_{-1}$. That is, we get $\phi\in
L^{\infty}_{-1}$.

Conversely, let $\phi\in L^{\infty}_{-1}$ satisfy $H\phi=0$ and $f=Uv\phi$.
Then we will prove $f\in S_{2}L^{2}$ below. Firstly, we show that $f\in
S_{0}L^{2}$, i.e. $\langle v,f\rangle=\langle x_{1}v,f\rangle=\langle
x_{2}v,f\rangle=0$. Indeed, let $\eta\in C_{0}^{\infty}(\mathbb{R}^{2})$ such
that $\eta(x)=1$ for $|x|\leq 1$ and $\eta(x)=0$ for $|x|>2$. Notice that
$vf=V\phi=-\Delta^{2}\phi$, then we have for any $\delta>0$,

$\begin{split}\Big{|}\int_{\mathbb{R}^{2}}v(y)f(y)\eta(\delta
y)dy\Big{|}=&\Big{|}\int_{\mathbb{R}^{2}}\big{(}\Delta^{2}\phi(y)\big{)}\eta(\delta
y)dy\Big{|}=\Big{|}\int_{\mathbb{R}^{2}}\phi(y)\cdot\Delta^{2}\big{(}\eta(\delta
y)\big{)}dy\Big{|}\\\ =&\delta^{4}\Big{|}\int_{\mathbb{R}^{2}}\langle
y\rangle^{-1}\phi(y)\cdot\langle y\rangle(\Delta^{2}\eta)(\delta
y)dy\Big{|}\\\ \leq&\delta^{4}\|\langle
y\rangle^{-1}\phi(y)\|_{L^{\infty}}\int_{\mathbb{R}^{2}}\big{|}\langle
y\rangle(\Delta^{2}\eta)(\delta y)\big{|}dy\\\ \lesssim&\delta^{2}\|\langle
y\rangle^{-1}\phi(y)\|_{L^{\infty}}\|\Delta^{2}\eta\|_{L^{1}}.\end{split}$
(5.5)

Taking the limit $\delta\rightarrow 0$ for (5.5), by the dominated convergence
theorem we obtain that $\langle v,f\rangle=0$. Similarly, we also have
$\langle x_{1}v,f\rangle=\langle x_{2}v,f\rangle=0.$ Hence we get $f\in
S_{0}L^{2}$.

Next, we turn to prove $\phi=-G_{0}vf+c_{3}x_{2}+c_{2}x_{1}+c_{1}$. In fact,
because of $f\in S_{0}L^{2}$, by using the estimates of $I_{i}(i=1,2,3)$ in
(5.1), we have $G_{0}vf\in L^{\infty}_{-1}$ and $\phi+G_{0}vf\in
L^{\infty}_{-1}$. Noting that

$\Delta^{2}(\phi+G_{0}vf)=\Delta^{2}\big{(}\phi+(\Delta^{2})^{-1}vf\big{)}=\Delta^{2}(\phi+vf)=(\Delta^{2}+V)(\phi)=0,$
(5.6)

then it follows from (5.6) that Fourier transform of the function
$\phi+G_{0}vf$ is supported at the origin, which is the sum of finite
derivatives of Dirac distribution $\delta$ ( See e.g. [Gr14] ). This
immediately implies that $\phi+G_{0}vf$ is a polynomial on $\mathbb{R}^{2}$.
Since $\phi+G_{0}vf\in L^{\infty}_{-1}$, the degree of the polynomial
$\phi+G_{0}vf$ is at most $1$. Hence we may write that

$\phi=-G_{0}vf+\widetilde{c}_{3}x_{2}+\widetilde{c}_{2}x_{1}+\widetilde{c}_{1},$

where $\widetilde{c}_{i}(i=1,2,3)$ are some constants. Since $T_{0}=U+vG_{0}v$
and

$\begin{split}0=H\phi=&\big{(}\Delta^{2}+V\big{)}\phi=-\big{(}\Delta^{2}+V\big{)}\big{(}G_{0}vf\big{)}+\big{(}\Delta^{2}+V\big{)}\big{(}\widetilde{c}_{3}x_{2}+\widetilde{c}_{2}x_{1}+\widetilde{c}_{1}\big{)}\\\
=&-Uv\big{(}U+vG_{0}v\big{)}f+U\big{(}\widetilde{c}_{3}x_{2}v^{2}+\widetilde{c}_{2}x_{1}v^{2}+\widetilde{c}_{1}v^{2}\big{)}\\\
=&U\big{(}-vT_{0}f+\widetilde{c}_{3}x_{2}v^{2}+\widetilde{c}_{2}x_{1}v^{2}+\widetilde{c}_{1}v^{2}\big{)},\end{split}$

we obtain

$\begin{split}\widetilde{c}_{1}v^{2}+\widetilde{c}_{2}x_{1}v^{2}+\widetilde{c}_{3}x_{2}v^{2}=vT_{0}f,\end{split}$
(5.7)

which leads to

$\widetilde{c}_{1}\langle v,v\rangle+\widetilde{c}_{2}\langle
v,x_{1}v\rangle+\widetilde{c}_{3}\langle v,x_{2}v\rangle=\langle
v,T_{0}f\rangle.$ (5.8)

By (5.7), we also have

$\widetilde{c}_{1}x_{1}v^{2}+\widetilde{c}_{2}x_{1}^{2}v^{2}+\widetilde{c}_{3}x_{1}x_{2}v^{2}=x_{1}vT_{0}f,\,\
\,\widetilde{c}_{1}x_{2}v^{2}+\widetilde{c}_{2}x_{1}x_{2}v^{2}+\widetilde{c}_{3}x_{2}^{2}v^{2}=x_{2}vT_{0}f,$

which imply that

$\begin{split}&\widetilde{c}_{1}\langle
x_{1}v,v\rangle+\widetilde{c}_{2}\langle
x_{1}v,x_{1}v\rangle+\widetilde{c}_{3}\langle x_{1}v,x_{2}v\rangle=\langle
x_{1}v,T_{0}f\rangle,\\\ &\widetilde{c}_{1}\langle
x_{2}v,v\rangle+\widetilde{c}_{2}\langle
x_{2}v,x_{1}v\rangle+\widetilde{c}_{3}\langle x_{2}v,x_{2}v\rangle=\langle
x_{2}v,T_{0}f\rangle.\end{split}$ (5.9)

Let
$\widetilde{c}=\big{(}\widetilde{c}_{1},\widetilde{c}_{2},\widetilde{c}_{3}\big{)}^{\intercal}$,
$b=\big{(}\langle v,T_{0}f\rangle,\langle x_{1}v,T_{0}f\rangle,\langle
x_{2}v,T_{0}f\rangle\big{)}^{\intercal}$, and

$A=\begin{pmatrix}\langle v,v\rangle&\langle v,x_{1}v\rangle&\langle
v,x_{2}v\rangle\\\ \langle x_{1}v,v\rangle&\langle
x_{1}v,x_{1}v\rangle&\langle x_{1}v,x_{2}v\rangle\\\ \langle
x_{2}v,v\rangle&\langle x_{2}v,x_{1}v\rangle&\langle
x_{2}v,x_{2}v\rangle\end{pmatrix}.$

Then the identities (5.8) and (5.9) can be written as $A\widetilde{c}=b$. Let
$k:=(k_{1},k_{2},k_{3})\in\mathbb{R}^{3}$. Since $v,x_{1}v,x_{2}v$ are linear
independent, so for each $k\neq 0$, we have
$k_{1}v+k_{2}x_{1}v+k_{3}x_{2}v\neq 0$ and

$0<\big{(}k_{1}v+k_{2}x_{1}v+k_{3}x_{2}v,k_{1}v+k_{2}x_{1}v+k_{3}x_{2}v\big{)}=kAk^{\intercal},$

hence $kAk^{\intercal}$ is a positive definite quadratic form, and the metrix
$A$ is positive definite, which gives that $\det(A)>0$. Thus the linear
equations $A\widetilde{c}=b$ has an unique solution. Notice that

$x_{1}v=f_{1}+\frac{\langle x_{1}v,v\rangle}{\|V\|_{L^{1}}}v,\,\
\,x_{2}v=f_{2}+\frac{\langle
x_{2}v,f_{1}\rangle}{\|f_{1}\|^{2}_{L^{2}}}f_{1}+\frac{\langle
x_{2}v,v\rangle}{\|V\|_{L^{1}}}v,$

it is easy to check that the specific $c=(c_{1},c_{2},c_{2})^{\intercal}$ from
Proposition 5.1 is a solution of the linear equations $A\widetilde{c}=b$.
Hence, we must have $\widetilde{c}_{i}=c_{i}(i=1,2,3)$.

Finally, by $f=Uv\phi$ and $H\phi=0$ one has

$\displaystyle\begin{split}S_{0}T_{0}S_{0}f&=S_{0}T_{0}(Uv\phi)=S_{0}(U+vG_{0}f)(Uv\phi)=S_{0}(vG_{0}v\phi+v\phi)\\\
&=S_{0}vG_{0}\big{(}\Delta^{2}+V\big{)}\phi=S_{0}vG_{0}H\phi=0.\end{split}$
(5.10)

Thus we have concluded that $f\in S_{1}L^{2}$. ∎

######  Proposition 5.2.

Assume that $|V(x)|\lesssim(1+|x|)^{-\beta}$ with $\beta>18$. Then $f\in
S_{2}L^{2}(\mathbb{R}^{2})\setminus\\{0\\}$ if and only if $f=Uv\phi$ with
$\phi\in L^{\infty}(\mathbb{R}^{2})$ such that $H\phi=0$ in the distributional
sense, and

$\phi=-G_{0}vf+\frac{\langle v,T_{0}f\rangle}{\|V\|_{L^{1}(\mathbb{R}^{2})}}.$

###### Proof.

Since $S_{2}\leq S_{1}$, by Proposition 5.1, we know that $f\in S_{1}L^{2}$ if
and only if $f=Uv\phi$ with $\phi\in L^{\infty}_{-1}$ such that $H\phi=0$ in
the distributional sense and

$\phi=-G_{0}vf+c_{3}x_{2}+c_{2}x_{1}+c_{1}.$

Thus, we just need to further prove that $c_{1}=\frac{\langle
v,T_{0}f\rangle}{\|V\|_{L^{1}}}$, $c_{2}=c_{3}=0$, and $\phi\in L^{\infty}$.
Indeed, by the proof of Proposition 5.1, we know that

$f_{1}=x_{1}v-\frac{\langle x_{1}v,v\rangle}{\|V\|_{L^{1}}}v\in QL^{2},\
f_{2}=x_{2}v-\frac{\langle
x_{2}v,f_{1}\rangle}{\|f_{1}\|^{2}_{L^{2}}}f_{1}-\frac{\langle
x_{2}v,v\rangle}{\|V\|_{L^{1}}}v\in QL^{2}.$

If $f\in S_{2}L^{2}$, then $QT_{0}f=0$. Hence

$\langle f_{i},T_{0}f\rangle=\langle Qf_{i},T_{0}f\rangle=\langle
f_{i},QT_{0}f\rangle=0,i=1,2,$

which leads to $c_{1}=\frac{\langle v,T_{0}f\rangle}{\|V\|_{L^{1}}}$, and
$c_{2}=c_{3}=0$.

Next, we show that $\phi\in L^{\infty}$. Since $\frac{\langle
v,T_{0}f\rangle}{\|V\|_{L^{1}}}\in L^{\infty}$, it suffices to show that

$G_{0}vf(x)=\frac{1}{8\pi}\int_{\mathbb{R}^{2}}|x-y|^{2}\ln|x-y|v(y)f(y)dy\in
L^{\infty}.$

If $f\in S_{2}L^{2}\subset S_{1}L^{2}$, then using the orthogonality
$S_{2}v=0,S_{2}(x_{i}v)=0(i=1,2)$, one has

$\int_{\mathbb{R}^{2}}|x|^{2}\ln|x|v(y)f(y)dy=\int_{\mathbb{R}^{2}}x\cdot
y\ln|x|v(y)f(y)dy=\int_{\mathbb{R}^{2}}\frac{x\cdot y}{|x|^{2}}v(y)f(y)dy=0.$

Furthermore, we have

$\begin{split}G_{0}vf(x)=&\frac{1}{16\pi}\int_{\mathbb{R}^{2}}|x|^{2}\Big{[}\ln\Big{(}\frac{|x-y|^{2}}{|x|^{2}}\Big{)}+2\frac{x\cdot
y}{|x|^{2}}\Big{]}v(y)f(y)dy\\\ &-\frac{1}{4\pi}\int_{\mathbb{R}^{2}}x\cdot
y(\ln|x-y|-\ln|x|)v(y)f(y)dy\\\
&+\frac{1}{8\pi}\int_{\mathbb{R}^{2}}|y|^{2}\ln|x-y|v(y)f(y)dy:=I_{1}+I_{2}+I_{3}.\end{split}$
(5.11)

For the first integral $I_{1}$. We have

$\begin{split}I_{1}=&\frac{1}{16\pi}\int_{|x|>4y}|x|^{2}\Big{[}\ln\Big{(}\frac{|x-y|^{2}}{|x|^{2}}\Big{)}+2\frac{x\cdot
y}{|x|^{2}}\Big{]}v(y)f(y)dy\\\ &+\frac{1}{16\pi}\int_{|x|\leq
4|y|}|x|^{2}\Big{[}\ln\Big{(}\frac{|x-y|^{2}}{|x|^{2}}\Big{)}+2\frac{x\cdot
y}{|x|^{2}}\Big{]}v(y)f(y)dy:=I_{11}+I_{12}.\end{split}$ (5.12)

For $I_{11}$. Since $|x|>4|y|$, then $\frac{|x-y|^{2}-|x|^{2}}{|x|^{2}}<1$. By
Taylor’s formula one has

$\begin{split}\ln\Big{(}\frac{|x-y|^{2}}{|x|^{2}}\Big{)}+2\frac{x\cdot
y}{|x|^{2}}=\ln\Big{(}1+\frac{|x-y|^{2}-|x|^{2}}{|x|^{2}}\Big{)}+2\frac{x\cdot
y}{|x|^{2}}=\frac{|y|^{2}}{|x|^{2}}+O\Big{(}\frac{|y|^{2+}}{|x|^{2+}}\Big{)}.\end{split}$
(5.13)

By (5.13) and Hölder’s inequality, one has

$\begin{split}|I_{11}|\leq\Big{|}\int_{|x|>4|y|}|x|^{2}\Big{[}\ln\Big{(}\frac{|x-y|^{2}}{|x|^{2}}\Big{)}+2\frac{x\cdot
y}{|x|^{2}}\Big{]}v(y)f(y)dy\Big{|}\lesssim\|y^{2}v(y)\|_{L^{2}}\|f\|_{L^{2}}<\infty.\end{split}$

For $I_{12}$. Note that the definition of the functions $\ln^{+}(\cdot)$ and
$\ln^{-}(\cdot)$, by Hölder’s inequality one has

$\begin{split}|I_{12}|=&\Big{|}\int_{|x|\leq
4|y|}|x|^{2}\Big{[}\ln\Big{(}\frac{|x-y|^{2}}{|x|^{2}}\Big{)}+2\frac{x\cdot
y}{|x|^{2}}\Big{]}v(y)f(y)dy\Big{|}\\\ \lesssim&\Big{|}\int_{|x|\leq
4|y|}|x|^{2}\big{(}\ln|x-y|-\ln|x|\big{)}v(y)f(y)dy\Big{|}+\int_{|x|\leq
4|y|}|x|^{2}\frac{|y|^{2}}{|x|^{2}}|v(y)f(y)|dy\\\
\lesssim&\big{|}\int_{|x|\leq
4|y|}|x|^{2}\big{(}\ln^{+}|x-y|-\ln^{+}|x|\big{)}\cdot|v(y)f(y)|dy\Big{|}\\\
&+\Big{|}\int_{|x|\leq
4|y|}|x|^{2}\big{(}\ln^{-}|x-y|-\ln^{-}|x|\big{)}\cdot|v(y)f(y)|dy\Big{|}+\int_{\mathbb{R}^{2}}y^{2}|v(y)f(y)|dy\\\
:=&J_{1}+J_{2}+\int_{\mathbb{R}^{2}}y^{2}|v(y)f(y)|dy.\end{split}$

Hence by the same arguments as in the proof of the terms $J_{1}$ and $J_{2}$
of Proposition 5.1, we get that $I_{12}\in L^{\infty}$. Combining with the
estimates $I_{11}$ and $I_{12}$ above, we immediately get $I_{1}\in
L^{\infty}$.

For $I_{2}$ and $I_{3}$, by using the same arguments with the proof of $I_{2}$
and $I_{3}$ in Proposition 5.1, we obtain that both $I_{2}$ and $I_{3}$ in
(5.11) are in $L^{\infty}$.

Combining all computations above, we obtain that $G_{0}vf\in L^{\infty}$ and
$\phi\in L^{\infty}$.

On the other hand, let $\phi\in L^{\infty}$ satisfy $H\phi=0$ and $f=Uv\phi$.
By the similar methods of (5.5) and (5.10), we have

$\langle v,f\rangle=0,\,\ \,\langle x_{i}v,f\rangle=0,\,i=1,2,\,\
\,S_{0}T_{0}S_{0}f=0.$

Then we can obtain that $f\in S_{1}L^{2}$. By the estimates of
$I_{i}(i=1,2,3)$ in (5.11), we have $G_{0}vf\in L^{\infty}\subset
L^{\infty}_{-1}$. By Proposition 5.1, then
$\phi=-G_{0}vf+c_{3}x_{2}+c_{2}x_{1}+c_{1}$. Thus
$c_{3}x_{2}+c_{2}x_{1}=\phi+G_{0}vf-c_{1}\in L^{\infty}$, hence we must have
$c_{2}=c_{3}=0$, which implies that $\phi=-G_{0}vf+\frac{\langle
v,T_{0}f\rangle}{\|V\|_{L^{1}}}$ and $T_{0}f\in span\\{v\\}$ which means
$QT_{0}f=0$. Hence $f\in S_{2}L^{2}$.

The proof of this proposition is completed. ∎

######  Proposition 5.3.

Assume that $|V(x)|\lesssim(1+|x|)^{-\beta}$ with $\beta>18$. Then $f\in
S_{4}L^{2}(\mathbb{R}^{2})\setminus\\{0\\}$ if and only if $f=Uv\phi$ with
$\phi\in L_{1}^{\infty}(\mathbb{R}^{2})$ such that $H\phi=0$ in the
distributional sense and

$\phi=-G_{0}vf.$

######  Remark 5.4.

By this proposition above, if $f\in S_{4}L^{2}(\mathbb{R}^{2})$, then
$f=Uv\phi$ with $\phi\in L_{1}^{\infty}(\mathbb{R}^{2})$ such that $H\phi=0$
in the distributional sense, and we have $f=-UvG_{0}vf$. Notice that
$UvG_{0}v$ is Hilbert-Schmidt operator, then $S_{4}$ is a finite rank operator
by Fredholm alternative theorem.

###### Proof.

Since $S_{4}\leq S_{3}\leq S_{2}$, by Proposition 5.2, we know that $f\in
S_{2}L^{2}$ if and only if $f=Uv\phi$ with $\phi\in L^{\infty}$ 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 only need to prove that $\langle v,T_{0}f\rangle=0$ and $\phi\in
L^{\infty}_{1}$. Indeed, if $f\in S_{4}L^{2}$, by Lemma 2.7, then
$T_{0}f=PT_{0}f+QT_{0}f=0$, which gives $\langle v,T_{0}f\rangle=0$. Thus
$\phi=-G_{0}vf$.

Next we begin to prove $\phi=-G_{0}vf\in L^{\infty}_{1}$. Indeed, if $f\in
S_{4}L^{2}$, then

$\int_{\mathbb{R}^{2}}x\cdot
y\ln|x|v(y)f(y)dy=\int_{\mathbb{R}^{2}}|x|^{2}\ln|x|v(y)f(y)dy=\int_{\mathbb{R}^{2}}\ln|x||y|^{2}v(y)f(y)dy=0,$

which leads to

$\begin{split}G_{0}vf(x)=&\frac{1}{16\pi}\int_{\mathbb{R}^{2}}|x|^{2}\Big{[}\ln\Big{(}\frac{|x-y|^{2}}{|x|^{2}}\Big{)}+2\frac{x\cdot
y}{|x|^{2}}-\frac{|y|^{2}}{|x|^{2}}\Big{]}v(y)f(y)dy\\\
&-\frac{1}{4\pi}\int_{\mathbb{R}^{2}}x\cdot
y\Big{[}\ln\Big{(}\frac{|x-y|^{2}}{|x|^{2}}\Big{)}+2\frac{x\cdot
y}{|x|^{2}}\Big{]}v(y)f(y)dy\\\
&+\frac{1}{8\pi}\int_{\mathbb{R}^{2}}|y|^{2}(\ln|x-y|-\ln|x|)v(y)f(y)dy:=I_{1}+I_{2}+I_{3}.\end{split}$
(5.14)

Now we estimate the first integral $I_{1}$. We have

$\begin{split}I_{1}=&\frac{1}{16\pi}\int_{|x|>4|y|}|x|^{2}\Big{[}\ln\Big{(}\frac{|x-y|^{2}}{|x|^{2}}\Big{)}+2\frac{x\cdot
y}{|x|^{2}}-\frac{|y|^{2}}{|x|^{2}}\Big{]}v(y)f(y)dy\\\
&+\frac{1}{16\pi}\int_{|x|\leq
4|y|}|x|^{2}\Big{[}\ln\Big{(}\frac{|x-y|^{2}}{|x|^{2}}\Big{)}+2\frac{x\cdot
y}{|x|^{2}}-\frac{|y|^{2}}{|x|^{2}}\Big{]}v(y)f(y)dy:=I_{11}+I_{12}.\end{split}$
(5.15)

For $I_{11}$. Since $|x|>4|y|$, then by Taylor’s formula one has

$\begin{split}\ln\Big{(}\frac{|x-y|^{2}}{|x|^{2}}\Big{)}+2\frac{x\cdot
y}{|x|^{2}}-\frac{|y|^{2}}{|x|^{2}}=O\Big{(}\frac{|y|^{3}}{|x|^{3}}\Big{)}.\end{split}$

By Hölder’s inequality, one has

$\displaystyle|I_{11}|\lesssim\int_{|x|>4|y|}|x|^{2}\frac{|y|^{3}}{|x|^{3}}|v(y)f(y)|dy\lesssim\frac{1}{|x|}\langle|y|^{3}|v(y),f\rangle\lesssim\frac{1}{|x|}.$

For $I_{12}$. By the same argument with the proof of $I_{12}$ in (5.2), one
has

$\begin{split}\left|\int_{|x|\leq
4|y|}|x|^{2}\ln\Big{(}\frac{|x-y|^{2}}{|x|^{2}}\Big{)}v(y)f(y)dy\right|\lesssim\frac{1}{|x|}.\end{split}$
(5.16)

Hence by (5.16) and Hölder’s inequality again, one has

$\begin{split}|I_{12}|\lesssim&\Big{|}\int_{|x|\leq
4|y|}|x|^{2}\Big{[}\ln\Big{(}\frac{|x-y|^{2}}{|x|^{2}}\Big{)}+2\frac{x\cdot
y}{|x|^{2}}-\frac{|y|^{2}}{|x|^{2}}\Big{]}v(y)f(y)dy\Big{|}\\\
\lesssim&\frac{1}{|x|}+\int_{|x|\leq 4|y|}|y|^{2}|v(y)f(y)|dy\\\
\lesssim&\frac{1}{|x|}+\int_{|x|\leq
4|y|}\frac{|y|}{|x|}|y|^{2}|v(y)f(y)|dy\lesssim\frac{1}{|x|}.\end{split}$

Hence, combining the estimates $I_{11}$ and $I_{12}$ above, we get that
$I_{1}$ is bounded by $|x|^{-1}$ which belongs to $L_{1}^{\infty}$. Similarly,
we obtain that both $I_{2}$ and $I_{3}$ above are bounded by $|x|^{-1}$.

On the other hand, let $\phi\in L^{\infty}_{1}$ satisfy $H\phi=0$ and
$f=Uv\phi$, by the similar methods of (5.5) and (5.10), we have

$\langle v,f\rangle=0,\,\ \,\langle x_{i}v,f\rangle=0,i=1,2,\,\ \,\langle
x_{i}x_{j},f\rangle=0,i,j=1,2,\,\ \,S_{0}T_{0}S_{0}f=0.$

By the proof of Proposition 5.2, we get $QT_{0}f=0$, which gives $f\in
S_{2}L^{2}$. By the estimates of $I_{i}(i=1,2,3)$ in (5.14), we have
$G_{0}vf\in L^{\infty}_{1}$. By Proposition 5.2, then
$\phi=-G_{0}vf+\frac{\langle v,T_{0}f\rangle}{\|V\|_{L^{1}}}$. Thus
$\frac{\langle v,T_{0}f\rangle}{\|V\|_{L^{1}}}=\phi+G_{0}vf\in
L^{\infty}_{1}$, so we must have $\langle v,T_{0}f\rangle=0$, which implies
that $\phi=-G_{0}vf$ and $PT_{0}f=0$. Hence $f\in S_{4}L^{2}$.

The proof of this proposition is completed. ∎

######  Proposition 5.5.

Assume that $|V(x)|\lesssim(1+|x|)^{-\beta}$ with $\beta>18$. Then $f\in
S_{5}L^{2}(\mathbb{R}^{2})\setminus\\{0\\}$ if and only if $f=Uv\phi$ with
$\phi\in L_{2}^{\infty}(\mathbb{R}^{2})$ such that $H\phi=0$ in the
distributional sense, and

$\phi=-G_{0}vf.$

######  Remark 5.6.

Noting that $L^{\infty}_{2}(\mathbb{R}^{2})\subset L^{2}(\mathbb{R}^{2})$,
hence the fourth kind zero resonance is actually an eigenvalue of $H$.
Conversely, let $\phi\in L^{2}(\mathbb{R}^{2})$ such that $H\phi=0$, then we
can adapt the proof of Proposition 5.5 to show the eigenfunction $\phi\in
L_{2}^{\infty}(\mathbb{R}^{2})$, which implies that zero eigenvalue is exactly
the fourth kind resonance from Definition 2.4.

###### Proof.

Since $S_{5}\leq S_{4}$, by Proposition 5.3, we know that $f\in S_{4}L^{2}$ if
and only if $f=Uv\phi$ with $\phi\in L_{2}^{\infty}$ such that $H\phi=0$ in
the distributional sense and $\phi=-G_{0}vf.$ Thus, we just need to prove that
$\phi=-G_{0}vf\in L_{2}^{\infty}$.

Now we begin to prove $\phi=-G_{0}vf\in L^{\infty}_{2}$. Indeed, if $f\in
S_{5}L^{2}$, then

$\langle v,f\rangle=\langle x_{j}v,f\rangle=\langle
x_{i}x_{j}v,f\rangle=\langle x_{i}x_{j}x_{k}v,f\rangle=0,\ i,j,k=1,2.$

Moreover, we obtain that

$\begin{split}G_{0}vf(x)=&\frac{1}{32\pi}\int_{\mathbb{R}^{2}}|x|^{2}\Big{[}\ln\Big{(}\frac{|x-y|^{4}}{|x|^{4}}\Big{)}+4\frac{x\cdot
y}{|x|^{2}}-2\frac{|y|^{2}}{|x|^{2}}+4\frac{(x\cdot
y)^{2}}{|x|^{4}}\Big{]}v(y)f(y)dy\\\
&+\frac{1}{16\pi}\int_{\mathbb{R}^{2}}|y|^{2}\Big{[}\ln\Big{(}\frac{|x-y|^{2}}{|x|^{2}}\Big{)}+2\frac{x\cdot
y}{|x|^{2}}\Big{]}v(y)f(y)dy\\\ :=&I_{1}+I_{2}+I_{3}.\end{split}$ (5.17)

For $I_{1}$. We have

$\begin{split}I_{1}=&\frac{1}{32\pi}\int_{|x|>16|y|}|x|^{2}\Big{[}\ln\Big{(}\frac{|x-y|^{4}}{|x|^{4}}\Big{)}+4\frac{x\cdot
y}{|x|^{2}}-2\frac{|y|^{2}}{|x|^{2}}+4\frac{(x\cdot
y)^{2}}{|x|^{4}}\Big{]}v(y)f(y)dy\\\ &+\frac{1}{32\pi}\int_{|x|\leq
16|y|}|x|^{2}\Big{[}\ln\Big{(}\frac{|x-y|^{4}}{|x|^{4}}\Big{)}+4\frac{x\cdot
y}{|x|^{2}}-2\frac{|y|^{2}}{|x|^{2}}+4\frac{(x\cdot
y)^{2}}{|x|^{4}}\Big{]}v(y)f(y)dy:=I_{11}+I_{12}.\end{split}$

For $I_{11}$. Since

$|x-y|^{4}=|x|^{4}+|y|^{4}-4x\cdot y|y|^{2}-4x\cdot
y|x|^{2}+2|x|^{2}|y|^{2}+4(x\cdot y)^{2},$

then for $|x|>16|y|$,

$\Big{|}|x-y|^{4}-|x|^{4}\Big{|}=\Big{|}|y|^{4}-4x\cdot y|y|^{2}-4x\cdot
y|x|^{2}+2|x|^{2}|y|^{2}+4(x\cdot y)^{2}\Big{|}<|x|^{4}.$

By Taylor’s formula we have

$\begin{split}\ln\Big{(}\frac{|x-y|^{4}}{|x|^{4}}\Big{)}=&\ln\Big{(}1+\frac{|x-y|^{4}-|x|^{4}}{|x|^{4}}\Big{)}\\\
=&-4\frac{x\cdot y}{|x|^{2}}+2\frac{|y|^{2}}{|x|^{2}}+4\frac{(x\cdot
y)^{2}}{|x|^{4}}+O\Big{(}\frac{|y|^{4}}{|x|^{4}}\Big{)}.\end{split}$ (5.18)

Thus, by Hölder inequality again, we obtain

$\displaystyle|I_{1}|\lesssim$
$\displaystyle\Big{|}\int_{|x|>16|y|}|x|^{2}\Big{[}\ln\Big{(}\frac{|x-y|^{4}}{|x|^{4}}\Big{)}+4\frac{x\cdot
y}{|x|^{2}}-2\frac{|y|^{2}}{|x|^{2}}+4\frac{(x\cdot
y)^{2}}{|x|^{4}}\Big{]}v(y)f(y)dy\Big{|}$ $\displaystyle\lesssim$
$\displaystyle\int_{|x|>16|y|}|x|^{2}\frac{|y|^{4}}{|x|^{4}}|v(y)f(y)|dy\lesssim\frac{1}{|x|^{2}}.$

For $I_{12}$. By the same argument with the proof of $I_{12}$ in (5.14), we
obtain that $I_{12}$ is bounded by $|x|^{-2}$. Combining with the estimates
$I_{11}$ and $I_{12}$ above, we get that $I_{1}$ is bounded by $|x|^{-2}$.

We also obtain that $I_{2}$ and $I_{3}$ above are bounded by $|x|^{-2}$ by
using similar method with the proof of Proposition 5.3. Hence $G_{0}vf\in
L_{2}^{\infty}$.

On the other hand, let $\phi\in L^{\infty}_{2}$ satisfy $H\phi=0$ and
$f=Uv\phi$, by the similar methods of (5.5) and (5.10), we have

$\langle v,f\rangle=0,\,\ \,\langle x_{i}v,f\rangle=0,i=1,2,\,\ \,\langle
x_{i}x_{j},f\rangle=0,i,j=1,2,\,\,\,\langle
x_{i}x_{j}x_{k},f\rangle=0,\,i,j,k=0.$

By the proof of Proposition 5.3, we also have $T_{0}f=0$, which gives $f\in
S_{5}L^{2}$.

The proof of this proposition is completed. ∎

As the consequence of Definition 2.4 and Propositions 5.1–5.5, we immediately
have the following theorem.

######  Theorem 5.7.

Let $H=\Delta^{2}+V$ on $L^{2}(\mathbb{R}^{2})$ and
$|V(x)|\lesssim(1+|x|)^{-\beta}$ for some $\beta>0$. Then we have following
statements:

(i) If $\beta>10$, then zero is the regular point of $H$ if and only if there
exists only zero solution $\phi(x)\in L^{\infty}_{-1}(\mathbb{R}^{2})$ such
that $H\phi=0$ in the distributional sense,

(ii) If $\beta>14$, then zero is the first kind resonance of $H$ if and if
there exists some nonzero $\phi(x)\in L^{\infty}_{-1}(\mathbb{R}^{2})$ but no
nonzero $\phi\in L^{\infty}(\mathbb{R}^{2})$ such that $H\phi=0$.

(iii) If $\beta>18$, then zero is the second kind resonance of $H$ if and only
if there exists some nonzero $\phi(x)\in L^{\infty}(\mathbb{R}^{2})$ but no
nonzero $\phi\in L^{\infty}_{1}(\mathbb{R}^{2})$ such that $H\phi=0$ .

(iv) If $\beta>18$, then zero is the third kind resonance of $H$ if and only
if there exists some nonzero $\phi(x)\in L^{\infty}_{1}(\mathbb{R}^{2})$ but
no nonzero $\phi\in L^{2}(\mathbb{R}^{2})$ such that $H\phi=0$.

(v) If $\beta>18$, then zero is the fourth kind resonance ( i.e. eigenvalue )
of $H$ if and only if there exists some nonzero $\phi(x)\in
L^{2}(\mathbb{R}^{2})$ such that $H\phi=0$.

Let $\phi\in C^{\infty}(\mathbb{R}^{2})$ be a positive function which is equal
to $c\ln|x|+d$ for $|x|>1$, where $c,d$ are some positive constants. Then we
can easily check that $H\phi(x)=0$ when taking $\displaystyle
V(x)=-(\Delta^{2}\phi)/\phi.$ It is obvious that $V(x)\in
C_{0}^{\infty}(\mathbb{R}^{2})$ and $\phi(x)\in
L_{-1}^{\infty}(\mathbb{R}^{2}),$ hence by Theorem 5.7, it follows that the
zero is a resonance point of $H$. On the other hand, as shown in Lemma 3.5 of
[SWY21], zero cannot be an eigenvalue of $H$ in one dimensional case assuming
that $V$ has fast enough decay ( e.g. $|V(x)|\leq C(1+|x|)^{-\beta}$ for
$\beta>25$ ). However, it would be an interesting problem to find whether
there exist some sufficient decay potentials $V$ such that zero is an
eigenvalue of $H=\Delta^{2}+V$ in $\mathbb{R}^{2}$. Clearly, note that let
$\phi=(1+|x|^{2})^{-s/2}$, then $\phi\in L^{2}(\mathbb{R}^{2})$ for any $s>1$
and $(\Delta^{2}+V)\phi=0$ as we take

$V(x)=-{\Delta^{2}\phi\over\phi}\sim|x|^{-4},\ \ {\text{as}}\
|x|\rightarrow\infty,$

which actually means that the our required decay rate of $V$ in question above
is bigger than $4$ at least.

Acknowledgements: The authors would like to thank the anonymous referees for
careful reading the manuscript and providing valuable suggestions, which
substantially helped improving the quality of this paper. Avy Soffer is
partially supported by the Simon’s Foundation (No. 395767). Ping Li and
Xiaohua Yao are partially supported by NSFC ( No.11771165 and 12171182). We
would like to thank Dr. Zhao Wu for his help and useful discussions. The
authors also thank Professor William Green for his insightful comments that
improves the presentation in the previous version.

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