&-\Big\langle[L_b, t_{b,*}](\check{g}_1, \check{A}_1)+\frac{1}{2}[[L_b, t_{b,*}], t_{b,*}](\dg_1,\dA_1), (\dg^*, \dA^*)\Big\rangle\\
\label{eq:condition1}&\quad\quad+\Big\langle[L_b, t_{b,*}](\dg_2, \dA_2), (\dg^*, \dA^*)\Big\rangle=\langle f, (\dg^*, \dA^*)\rangle\quad\mbox{for all}\quad(\dg^*,\dA^*)\in\mathcal{K}_b^*,\\\notag
&\Big\langle[L_b, t_{b,*}](\check{g}'_1, \check{A}'_1)+\frac{1}{2}[[L_b, t_{b,*}], t_{b,*}](\dg'_1,\dA'_1), (\dg_1^*, \dA_1^*)\Big\rangle=-\langle f, (\check{g}_1^*, \check{A}_1^*)\rangle\\\notag
&\quad -\Big\langle\frac{1}{2}[[L_b, t_{b,*}], t_{b,*}](\dg_1,\dA_1)+[L_b, t_{b,*}](\dg_2-\check{g}_1, \dA_2-\check{A}_1), (\check{g}_1^*, \check{A}_1^*)\Big\rangle\\
\label{eq:condition2}
&\quad\quad+\Big\langle\frac{1}{2}[[L_b, t_{b,*}], t_{b,*}](\check{g}_1-\dg_2,\check{A}_1-\dA_2), (\dg_1^*, \dA_1^*)\Big\rangle\quad\mbox{for all}\quad (\dg_1^*, \dA_1^*)\in\mathcal{K}_{b,1}^*.
\end{align}

For the simplicity of notation, we define

\begin{gather*} \wt{\mathcal{K}}_0:=\wt{\mathcal{K}}_b^\perp,\quad \wt{\mathcal{K}}_1:=\wt{\mathcal{K}}_{b,1},\quad \wt{\mathcal{K}}_2:=\wt{\mathcal{K}}_{b,2},\\
\mathcal{R}_0:=\mathcal{R}_b,\quad \mathcal{R}_1:=\mathcal{R}^\perp_{b,1},\quad \mathcal{R}_2:=\mathcal{R}^\perp_{b,2}.
\end{gather*}

First, we prove that if an operator $A(\sigma):\wt{\mathcal{K}}_i\to\mathcal{R}_j$ is invertible, and continuous in $\sigma$ (in norm operator topology) in a small neighborhood of $\sigma=0$, then so is $A(\sigma)^{-1}$. We write

\[
\]

\[ A^e(\sigma)=A(\sigma)-A(0)=o(1)\quad\mbox{in}\quad \mathcal{L}_{\mathrm{op}}(\wt{\mathcal{K}}_i,\mathcal{R}_j)\quad \mbox{as}\quad \sigma\to 0.
\]

Since $\norm{A(0)^{-1}A^e(\sigma)}_{\wt{\mathcal{K}}_i\to\wt{\mathcal{K}}_i}<1/2$ when $\sigma$ is near $0$, the inverse of $A(\sigma)$ is given by the Neumann series

\[
\]

from which we find that

\begin{align*}
\norm{A(\sigma)^{-1}-A(0)^{-1}}_{\mathcal{R}_j\to\wt{\mathcal{K}}_i}&\leq\Big(\sum_{k=1}^\infty\norm{A(0)^{-1}A^e(\sigma)}^k_{\wt{\mathcal{K}}_i\to\wt{\mathcal{K}}_i}\Big)\norm{A(0)^{-1}}_{\mathcal{R}_j\to\wt{\mathcal{K}}_i}\\
&\leq2\norm{A(0)^{-1}}^2_{\mathcal{R}_j\to\wt{\mathcal{K}}_i}\norm{A^e(\sigma)}_{\wt{\mathcal{K}}_i\to\mathcal{R}_j}=o(1)\quad \mbox{in}\quad \mathcal{L}_{\mathrm{op}}(\mathcal{R}_j,\wt{\mathcal{K}}_i)\quad \mbox{as}\quad \sigma\to 0.
\end{align*}

This proves the continuity of $A(\sigma)^{-1}$ in the norm operator topology in a small neighborhood of $\sigma=0$. Applying this result to $L_{00}$ and $\wt{L}_{22}^\flat$ (defined in (<ref>)) yields the continuity (in the norm operator topology) of $L_{00}^{-1}$ and $(\wt{L}_{22}^\flat)^{-1}$. Then the continuity of $\wt{R}_{ij}$ follows from the explicit expressions in (<ref>). Furthermore, we find that

\begin{equation}\label{eq:expansionofcomofR}
\begin{split}
\wt{R}_{21}(b,\sigma)&=\wt{R}^0_{21}(b)+\sigma\wt{R}^e_{21}(b,\sigma)\quad \mbox{with}\quad \wt{R}^0_{21}(b)=-(\wt{L}^0_{22}(b))^{-1}\wt{L}^0_{21}(b)(\wt{L}^0_{11}(b))^{-1},\\
\wt{R}_{22}(b,\sigma)&=\wt{R}^0_{22}(b)+\sigma\wt{R}^e_{22}(b,\sigma)\quad \mbox{with}\quad \wt{R}^0_{22}(b)=(\wt{L}^0_{22}(b))^{-1},\\
\wt{R}_{11}(b,\sigma)&=\wt{R}^0_{11}(b)+\sigma\wt{R}^e_{11}(b,\sigma)\quad \mbox{with}\quad \wt{R}^0_{11}(b)=(\wt{L}^0_{11}(b))^{-1}
\end{split}
\end{equation}

where $\wt{R}^e_{21}, \wt{R}^e_{22}, \wt{R}^e_{11}$ are continuous in $(b,\sigma)$ for $(b,\sigma)$ near $(b_0, 0)$ in the norm operator topology. Therefore, we rewrite

\begin{equation}\label{eq:simplifiedR}
\begin{pmatrix}
\frac{\wt{R}_{10}(b,\sigma)}{\sigma}&	\frac{\wt{R}^0_{11}(b)}{\sigma^2}+	\frac{\wt{R}^e_{11}(b,\sigma)}{\sigma}&	\frac{\wt{R}_{12}(b,\sigma)}{\sigma}\\
0&	\frac{\wt{R}^0_{21}(b)}{\sigma}&	\frac{\wt{R}^0_{22}(b)}{\sigma}
\end{pmatrix}+R_{\mathrm{reg}}(b,\sigma)
\end{equation}

where all the entries in $R_{\mathrm{reg}}(b,\sigma)$ are continuous in $(b,\sigma)$ in the norm operator topology.

For $(\tilde{g}, \tilde{A})=\check{L}_b(0)(\dg, \dA)\in\wt{\mathcal{K}}_{b,1}$ with $(\dg,\dA)\in\mathcal{K}_{b,1}$, we have

\begin{equation}\label{eq:exofcheckLon1}
\begin{split}
\check{L}_b(\sigma)^{-1}(\tilde{g}, \tilde{A})&=\check{L}_b(\sigma)^{-1}\check{L}_b(0)(\dg, \dA)=(\dg,\dA)-i\sigma\check{L}_b(\sigma)^{-1}\widehat{L_{b}}(0)(\check{g}, \check{A})-\frac{\sigma^2}{2}\check{L}_b(\sigma)^{-1}\pa_\sigma^2\widehat{L_{b}}(0)(\dg,\dA)\\
&=(\dg,\dA)-i\sigma\check{L}_b(\sigma)^{-1}\big(\check{L}_b(\sigma)+\widehat{L_{b}}(0)-\widehat{L_{b}}(\sigma)-V_b\big)(\check{g}, \check{A})-\frac{\sigma^2}{2}\check{L}_b(\sigma)^{-1}\pa_\sigma^2\widehat{L_{b}}(0)(\dg,\dA)\\
&=(\dg,\dA)-i\sigma(\check{g}, \check{A})+\sigma^2\check{L}_b(\sigma)^{-1}\Big(i(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa_\sigma^2\widehat{L_{b}}(0))(\check{g},\check{A})-\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg,\dA)\Big)
\end{split}
\end{equation}

where the coefficient of $\sigma^2$ is continuous in $(b,\sigma)$ in the norm operator topology.

For $(\tilde{g}, \tilde{A})=\check{L}_b(0)(\dg, \dA)\in\wt{\mathcal{K}}_{b,2}$ with $(\dg,\dA)\in\mathcal{K}_{b,2}$, we have

\begin{equation}\label{eq:exofcheckLon2}
\begin{split}
\check{L}_b(\sigma)^{-1}(\tilde{g}, \tilde{A})&=\check{L}_b(\sigma)^{-1}\check{L}_b(0)(\dg, \dA)=(\dg,\dA)-\sigma\check{L}_b(\sigma)^{-1}\Big(\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa_\sigma^2\widehat{L_{b}}(0)\big)(\dg,\dA)\Big)
\end{split}
\end{equation}

where the coefficient of $\sigma$ is continuous in $(b,\sigma)$ in the norm operator topology.

Since $R_{\mathrm{reg}}(b,\sigma)$ is continuous in $\sigma$ in norm operator topology and $\check{L}_b(\sigma)^{-1}$ is continuous in $\sigma$ with values in $\mathcal{L}_{\mathrm{weak}}(\eHb^{s-1,\ell+2}(\CX),\eHb^{s,\ell}(\CX))\cap\mathcal{L}_{\mathrm{op}}(\eHb^{s-1+\epsilon,\ell+2+\epsilon}(\CX),\eHb^{s-\epsilon,\ell-\epsilon}(\CX))$ for $\epsilon>0$, it follows that $\check{L}_b(\sigma)^{-1}R_{\mathrm{reg}}(b,\sigma)$ is uniformly bounded and has the same continuity as $\check{L}_b(\sigma)^{-1}$. According to (<ref>) and (<ref>), we conclude that $\check{L}_b(\sigma)^{-1}R_{\mathrm{sing}}(b,\sigma)$ is equal to a singular part $P(b,\sigma)$ plus a uniformly bounded and continuous operator with value in $\mathcal{L}_{\mathrm{op}}(\eHb^{s-1,\ell+2}(\CX),\eHb^{s,\ell}(\CX))$. Moreover, $P(b,\sigma)$ satisfies

\begin{equation}\label{eq:exofPbsigma}
P(b,\sigma)f=(\sigma^{-2}(\dg_1,\dA_1)-i\sigma^{-1}(\check{g}_1, \check{A}_1))+i\sigma^{-1}\big((\dg_2, \dA_2)
+(\dg'_1, \dA'_1)\big)+i\sigma^{-1}(\dg_1''(\sigma), \dA_1''(\sigma))
\end{equation}

where $(\dg_1,\dA_1), (\dg_1',\dA_1'),(\dg_1''(\sigma),\dA_1''(\sigma))\in\mathcal{K}_{b,1}$ with $(\dg_1''(\sigma),\dA_1''(\sigma))=o(1)$ in $\mathcal{K}_{b,1}$ as $\sigma\to 0$, and $(\dg_2,\dA_2)\in\mathcal{K}_{b,2}$.

It remains to prove that $(\dg_1,\dA_1), (\dg_1',\dA_1'),(\dg_2,\dA_2)$ are determined by conditions (<ref>) and (<ref>). Let $f\in\eHb^{s-1,\ell+2}(\CX)$ and $(\dg(\sigma), \dA(\sigma)):=\widehat{L_{b}}(\sigma)^{-1}f$, and then we have

\begin{align*}
(\dg(\sigma), \dA(\sigma))&=(\sigma^{-2}(\dg_1,\dA_1)-i\sigma^{-1}(\check{g}_1, \check{A}_1))+i\sigma^{-1}\big((\dg_2, \dA_2)
+(\dg'_1, \dA'_1)\big)+i\sigma^{-1}(\dg_1''(\sigma), \dA_1''(\sigma))\\
&\qquad+(\tilde{\dg},\tilde{\dA})+(\tilde{\dg}(\sigma), \tilde{\dA}(\sigma))
\end{align*}

where $(\tilde{\dg},\tilde{\dA})\in\eHb^{s,\ell}(\CX)$ and $(\tilde{\dg}(\sigma), \tilde{\dA}(\sigma))=o(1)$ in $\eHb^{s-\epsilon, \ell-\epsilon}(\CX)$ as $\sigma\to 0$. Applying $\widehat{L_{b}}(\sigma)$ on both sides gives

\begin{equation}\label{eq:exoff}
\begin{split}
f&=\widehat{L_{b}}(\sigma)(\dg(\sigma), \dA(\sigma))=\widehat{L_{b}}(0)(\dg(\sigma), \dA(\sigma))+\big(\sigma\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma^2}{2}\pa_\sigma^2\widehat{L_{b}}(0)\big)(\dg(\sigma), \dA(\sigma))\\
&=\Big(\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1,\dA_1)-i\pa_\sigma\widehat{L_{b}}(0)(\check{g}_1, \check{A}_1)-\widehat{L_{b}}(0)(\check{g}_1',\check{A}_1')+i\pa_\sigma\widehat{L_{b}}(0)(\dg_2, \dA_2)+\widehat{L_{b}}(0)(\tilde{\dg},\tilde{\dA})\Big)\\
&\quad+i\pa_\sigma\widehat{L_{b}}(0)(\dg_{1}''(\sigma),\dA_1''(\sigma))+\widehat{L_{b}}(0)(\tilde{\dg}(\sigma), \tilde{\dA}(\sigma))\\
&\quad\quad+\frac{i\sigma}{2}\Big(-\pa_\sigma^2\widehat{L_{b}}(0)(\check{g}_1, \check{A}_1)+\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1', \dA_1')+\pa_\sigma^2\widehat{L_{b}}(0)(\dg_2, \dA_2))\Big)\\
&\quad\quad\quad+\frac{i\sigma}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1''(\sigma), \dA_1''(\sigma)+\big(\sigma\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma^2}{2}\pa_\sigma^2\widehat{L_{b}}(0)\big)(\tilde{\dg}+\tilde{\dg}(\sigma), \tilde{\dA}+\tilde{\dA}(\sigma)).
\end{split}
\end{equation}

Letting $\sigma\to0$, we arrive at

\begin{equation}\label{eq:limitofexoff}
f=\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1,\dA_1)-i\pa_\sigma\widehat{L_{b}}(0)(\check{g}_1, \check{A}_1)-\widehat{L_{b}}(0)(\check{g}_1',\check{A}_1')+i\pa_\sigma\widehat{L_{b}}(0)(\dg_2, \dA_2)+\widehat{L_{b}}(0)(\tilde{\dg},\tilde{\dA}).
\end{equation}

Pairing both sides of (<ref>) with $(\dg^*, \dA^*)\in\mathcal{K}_b^*$, we obtain

\begin{equation}\label{eq:combinedpairing}
\begin{split}
\langle f, (\dg^*, \dA^*)\rangle&=\Big\langle\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1,\dA_1)-i\pa_\sigma\widehat{L_{b}}(0)(\check{g}_1, \check{A}_1)+i\pa_\sigma\widehat{L_{b}}(0)(\dg_2, \dA_2), \ (\dg^*, \dA^*)\Big\rangle\\
&=-\Big\langle[L_b, t_{b,*}](\check{g}_1, \check{A}_1)+\frac{1}{2}[[L_b, t_{b,*}], t_{b,*}](\dg_1,\dA_1)-[L_b, t_{b,*}](\dg_2, \dA_2), (\dg^*, \dA^*)\Big\rangle
\end{split}
\end{equation}

Since the pairing on $\mathcal{K}_b\times\mathcal{K}_b^*$ on the right-hand side of (<ref>) is non-degenerate, $(\dg_1,\dA_1)$ (thus $(\check{g}_1, \check{A}_1)$) and $(\dg_2, \dA_2)$ are uniquely determined by (<ref>). This proves the condition (<ref>).

Rewriting (<ref>) gives rise to the following PDE

\begin{equation}\label{eq:PDEfoftwoterms}
\widehat{L_{b}}(0)\big((\check{g}_1',\check{A}_1')-(\tilde{\dg},\tilde{\dA})\big)=-f+\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1,\dA_1)-i\pa_\sigma\widehat{L_{b}}(0)(\check{g}_1, \check{A}_1)+i\pa_\sigma\widehat{L_{b}}(0)(\dg_2, \dA_2)
\end{equation}

whose right-hand side is uniquely determined by $f$. Due to (<ref>), the pairings of the right-hand side of (<ref>) with any $(\dg^*, \dA^*)\in\mathcal{K}_b^*$ is $0$, which implies that (<ref>) can be solved for $(\check{g}_1',\check{A}_1')-(\tilde{\dg},\tilde{\dA})$ and $(\check{g}_1',\check{A}_1')-(\tilde{\dg},\tilde{\dA})$ is uniquely determined by $f$ modulo $\mathcal{K}_b$. Namely,

\begin{equation}\label{eq:exoftwoterms}
(\check{g}_1',\check{A}_1')-(\tilde{\dg},\tilde{\dA})=(\bar{\dg}, \bar{\dA})+(\dg''', \dA''')
\end{equation}

where $(\bar{\dg}, \bar{\dA})$ is uniquely determined by $f$ and $(\dg''', \dA''')\in\mathcal{K}_b$. Pairing both sides of (<ref>) with $(\dg_1^*, \dA_1^*)\in\mathcal{K}^*_{b,1}$ and using the identity (<ref>), we obtain, upon multiplying $\sigma^{-1}$ and letting $\sigma\to0$,

\begin{align*}
0&=\frac{i}{2}\Big\langle-\pa_\sigma^2\widehat{L_{b}}(0)(\check{g}_1, \check{A}_1)+\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1', \dA_1')+\pa_\sigma^2\widehat{L_{b}}(0)(\dg_2, \dA_2)-2i\pa_\sigma\widehat{L_{b}}(0)(\tilde{\dg},\tilde{\dA}),\  (\dg_1^*, \dA_1^*)\Big\rangle\\
&=\frac{i}{2}\Big\langle-\pa_\sigma^2\widehat{L_{b}}(0)(\check{g}_1, \check{A}_1)+\pa_\sigma^2\widehat{L_{b}}(0)(\dg_2, \dA_2)+2i\pa_\sigma\widehat{L_{b}}(0)(\bar{\dg},\bar{\dA}),\  (\dg_1^*, \dA_1^*)\Big\rangle\\
&\quad+\frac{i}{2}\Big\langle\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1', \dA_1')-2i\pa_\sigma\widehat{L_{b}}(0)(\check{g}_1', \check{A}_1'),\  (\dg_1^*, \dA_1^*)\Big\rangle+i\langle(\dg''', \dA'''),\widehat{L_{b}}(0)^*(\check{g}^*_1, \check{A}_1^*)\rangle
\end{align*}

where we use (<ref>) and $-i(\pa_\sigma\widehat{L_{b}}(0))^*(\dg^*_1, \dA_1^*)=\widehat{L_{b}}(0)^*(\check{g}^*_1, \check{A}_1^*)$ in the second step. Since
\[\langle(\dg''', \dA'''),\widehat{L_{b}}(0)^*(\check{g}^*_1, \check{A}_1^*)\rangle=\langle\widehat{L_{b}}(0)(\dg''', \dA'''),(\check{g}^*_1, \check{A}_1^*)\rangle=0,
\]

we arrive at

\begin{equation}\label{eq:determinationofh1'}
\begin{split}
&\Big\langle[[L_b, t_{b,*}], t_{b,*}](\dg_1', \dA_1')+2[L_b, t_{b,*}](\check{g}_1', \check{A}_1'),\  (\dg_1^*, \dA_1^*)\Big\rangle\\
&\quad=\Big\langle[[L_b, t_{b,*}], t_{b,*}](\check{g}_1, \check{A}_1)-[[L_b, t_{b,*}], t_{b,*}](\dg_2, \dA_2)+2[L_b, t_{b,*}](\bar{\dg},\bar{\dA}),\  (\dg_1^*, \dA_1^*)\Big\rangle.
\end{split}
\end{equation}

Since the pairing on $\mathcal{K}_{b,1}\times\mathcal{K}_{b,1}^*$ on the left-hand side of (<ref>) is non-degenerate and the right-hand side is uniquely determined by $f$, so is $(\dg_1', \dA_1')$. Actually, we can further calculate the right-hand side

\begin{align*}
\Big\langle[L_b, t_{b,*}](\bar{\dg},\bar{\dA}),\  (\dg_1^*, \dA_1^*)\Big\rangle&=\Big\langle(\bar{\dg},\bar{\dA}),\  \widehat{L_{b}}(0)^*(\check{g}_1^*, \check{A}_1^*)\Big\rangle=\Big\langle\widehat{L_{b}}(0)(\bar{\dg},\bar{\dA}),\  (\check{g}_1^*, \check{A}_1^*)\Big\rangle\\
&=-\langle f+\frac{1}{2}[[L_b, t_{b,*}], t_{b,*}](\dg_1, \dA_1)+[L_b, t_{b,*}](\dg_2-\check{g}_1, \dA_2-\check{A}_1), (\check{g}_1^*, \check{A}_1^*)\rangle
\end{align*}

where we use (<ref>) and (<ref>) in the third step. Plugging this into (<ref>) proves the condition (<ref>).

We leave $V_{b,1}$ as it is; so in particular $\widehat{L_b}(0) + V_{b,1}$ is invertible (as in Lemma 11.1). We modify our requirements on $V_{b,2}$ in two ways:
(1) we shall only require $\check{\mathcal{K}}_b \subset \ker V_{b,2}$; and
(2) we shall choose $V_{b,2} = \delta W_{b,2}$ for some fixed $W_{b,2}$ and some small $\delta$.

Arranging (1) ensures the validity of (11.38) as before. As for point (2), if $\delta$ is sufficiently small, then $\widehat{L_b}(0) + V_{b,1} + V_{b,2} = \widehat{L_b}(0) + V_{b,1} + \delta W_{b,2}$ will still be invertible. (NB: This smallness condition is what was not arranged in the stated proof, and it seems rather awkward to do so in the context of the stated proof.) So the main task is to arrange (11.39), which we do via arranging

\[
\widehat{L_b}(0)^*\check{\mathcal{K}}_b^* \cap ( \widehat{L_b}(0)^* + V_{b,2}^* ) ( \ker V_{b,1}^* ) = 0,    (111)
\]

which is the displayed equation after eq. (11.42). (We discard the part of the proof starting with `We arrange the stronger condition [...]'.) — To arrange this condition, note three things:

(i) $\widehat{L_b}(0)^*$ is invertible on $\ker V_{b,1}^*$ (by construction of $V_{b,1}$) with range given by the annihilator of $\ker\widehat{L_b}(0) = \mathcal{K}_b$.
(ii) The space $\widehat{L_b}(0)^* ( \check{\mathcal{K}}_b^* )$ is a 4-dimensional subspace of that range space $\ann \mathcal{K}_b$ (which itself is a 7-codimensional subspace of $\dot H_b^{-s+1,-\ell-2}$).
(iii) Requirement (1) above is equivalent to $\Ran W_{b,2}^*$ being contained in $\ann \check{\mathcal{K}}_b$, which by virtue of eq. (11.42) is a subspace transversal to $\ann \mathcal{K}_b$.

The idea for arranging (111) is then as follows: pick (continuously in b, but dropping the b-dependence) four vectors $w_j \in \ker V_{b,1}^*$ so that $\widehat{L_b}(0)^* w_j$ spans $\widehat{L_b}(0)^*\check{\mathcal{K}}_b^*$. Pick moreover 4 linearly independent $q_j \in \ann \check{\mathcal{K}}_b \setminus \ann \mathcal{K}_b$. Fix a subspace $W^\perp$ of $\dot H_{\mathrm b}^{-s+1,-\ell+2}$ complementary to span $ \{ w_j \}$. Put $W_{b,2}w_j=q_j$ and $W_{b,2}=0$ on $W^\perp$.

By construction $(\widehat{L_b}(0)+\delta W_{b,2})^*$ is independent of $\delta$ on $W^\perp$, and maps $W^\perp$ into a subspace of $\ann \mathcal{K}_b$ transversal to $\widehat{L_b}(0)^*\check{\mathcal{K}}_b^*$. On the other hand, for all nonzero $\delta$, and for any nonzero linear combination $w$ of the $w_j$, the vector $(\widehat{L_b}(0)+\delta W_{b,2})^* w$ does not lie in $\ann \mathcal{K}_b^*$. So (111) is satisfied for all nonzero $\delta$. Finally, for small enough $\delta$, we get the required invertibility of $\widehat{L_b}(0) + V_{b,1} + \delta W_{b,2}$, as mentioned before.

§ REGULARITY OF THE RESOLVENT OF THE LINEARIZED GAUGE-FIXED EINSTEIN-MAXWELL OPERATOR

In this section, we continue using the notations from <ref>. In <ref>, we give a refined description of the singular part $P(b,\sigma)$ and the regular part $L^-_b(\sigma)$ of the operator $\widehat{L_{b}}(0)^{-1}$ defined in Corollary <ref>. 	Concretely, we prove that $(\dg_1''(\sigma), \dA_1''(\sigma))$ defined in Corollary <ref> is Hölder regular at $\sigma=0$, while $L^-_b(\sigma)$ is Hölder regular at $\sigma=0$ only when one relaxes the target space. In <ref>, we provide estimates of the bounds of any derivatives (in $\sigma$) of $L^-_b(\sigma)$ for $\sigma$ in the region $0\leq\IM\sigma\leq C, \abs{\sigma}\geq c_0$ for some $c_0,C>0$. Finally, in <ref>, we discuss the conormal regularity of $L^-_b(\sigma)$ at $\sigma=0$. Specifically, after applying any number of derivatives $\sigma\pa_\sigma$ to $L^-_b(\sigma)$, it satisfies the same properties as $L^-_b(\sigma)$.

§.§ Regularity at low frequencies

In <ref>, we study in detail the structure of $\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}$, i.e., prove the $\epsilon$-regularity of $\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}$. In <ref>, we establish the $\epsilon$-regularity of $R(b,\sigma)$, which allows us to give a refined description of the singular part $P(b,\sigma)$ of $\widehat{L_{b}}(\sigma)^{-1}$ near $\sigma=0$. We also discuss a certain Hölder regularity of the regular part $L^-_b(\sigma)$ at $\sigma=0$.

§.§.§ $\epsilon$-regularity of $\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}$.

First, we introduce several definitions.

Let $X,Y$ be two Banach spaces and $\al, \be\in\BR$. Suppose $A(\sigma):X\to Y$ is a family of operators depending on parameter $\sigma\in\BC\setminus\{0\}$. Then we write

\[
A(\sigma): \abs{\sigma}^\al X\to\abs{\sigma}^\be Y
\]

if and only if there exists a constant $C>0$ such that $\norm{A(\sigma)}_{X\to Y}\leq  C\abs{\sigma}^{\be-\al}$ for all $\sigma\in\BC\setminus\{0\}$.

Let $\wt{\mathcal{K}}_i, \mathcal{R}_i$ ($i=0,1,2$) be defined as in <ref>.

* An operator $L(\sigma):\wt{\mathcal{K}}_i\to\mathcal{R}_j$ is $\epsilon$-regular at $\sigma=0$ if for $\sigma\in \BC, \IM\sigma\geq 0$ and $\abs{\sigma}$ small, there exists a constant $C>0$ such that

\begin{align}
\norm{L(\sigma)}_{\wt{\mathcal{K}}_i\to\mathcal{R}_j}&\leq C,\\
\norm{\pa_\sigma L(\sigma)}_{\wt{\mathcal{K}}_i\to\mathcal{R}_j}&\leq C\abs{\sigma}^{-\epsilon},\\
\norm{\pa^2_\sigma L(\sigma)}_{\wt{\mathcal{K}}_i\to\mathcal{R}_j}&\leq C\abs{\sigma}^{-\epsilon-1}.
\end{align}

* An operator $L(\sigma):\wt{\mathcal{K}}_i\to\mathcal{R}_j$ has an $\epsilon$-regular expansion at $\sigma=0$ in $\sigma$ up to order one if

\begin{equation}
L(\sigma)=L^0+\sigma L^e(\sigma)
\end{equation}

where $L^0:\wt{\mathcal{K}}_i\to\mathcal{R}_j$ is independent of $\sigma$ and $L^e(\sigma):\wt{\mathcal{K}}_i\to\mathcal{R}_j$ is $\epsilon$-regular at $\sigma=0$.

* An operator $L(\sigma):\wt{\mathcal{K}}_i\to\mathcal{R}_j$ has an $\epsilon$-regular expansion at $\sigma=0$ in $\sigma$ up to order two if

\begin{equation}
L(\sigma)=L^0+\sigma L^1+\sigma^2L^e(\sigma)
\end{equation}

where $L^0, L^1:\wt{\mathcal{K}}_i\to\mathcal{R}_j$ are independent of $\sigma$ and $L^e(\sigma):\wt{\mathcal{K}}_i\to\mathcal{R}_j$ is $\epsilon$-regular at $\sigma=0$.

We recall the analysis of the entries $L_{ij}$ of the operator in the proof of Theorem <ref>, and note that each entry is a sum of polynomials in $\sigma$ whose coefficients are either independent of $\sigma$, terms involving $\check{L}_b(\sigma)^{-1}$ acting on an element in $\eHb^{\infty, 3/2-}(\CX)$, or terms involving $(\check{L}_b(\sigma)^{-1})^*$ acting on an element in $\sHb^{-5/2-C(a,\gamma), 3/2-}(\CX)$ where $C(a,\gamma_0)$ is a sufficiently small constant depending on $a,\gamma$. Therefore, we first discuss the properties of the operator $\check{L}_b(\sigma)^{-1}$.

Let $-\frac 32<\ell<-\frac 12$, $0<\epsilon<1$ with $-\frac 12<\ell+\epsilon<\frac 12$, and $s-\epsilon>4$. For $\IM\sigma\geq 0$, $0<\abs{\sigma}\leq c_0$ with $c_0$ small, we have

\begin{align}\label{eq:firstderiofcheckL-1}
&\pa_\sigma\check{L}_b(\sigma)^{-1}: \eHb^{s-1,\ell+2}(\CX)\to \abs{\sigma}^{-\epsilon}\eHb^{s-\epsilon, \ell+\epsilon-1}(\CX),\\\label{eq:secondderiofcheckL-1}
&\pa^2_\sigma\check{L}_b(\sigma)^{-1}: \eHb^{s-1,\ell+2}(\CX)\to \abs{\sigma}^{-\epsilon-1}\eHb^{s-1-\epsilon, \ell+\epsilon-1}(\CX).
\end{align}

Formally, we have

\[
\pa_\sigma\check{L}_b(\sigma)^{-1}=-\check{L}_b(\sigma)^{-1}(\pa_\sigma\check{L}_b(\sigma))\check{L}_b(\sigma)^{-1}
\]

where by Proposition <ref>

\begin{equation}\label{eq:deriofcheckL}
\pa_\sigma\check{L}_b(\sigma)=\pa_\sigma\widehat{L_{b}}(\sigma)\in2i\rho(\rho\pa_\rho-1)+\rho^3\mbox{Diff}_\bop^1+\rho^2C^\infty+\sigma\rho^2C^\infty.
\end{equation}

Motivated by this, we first analyze $\rho(\rho\pa_\rho-1)\check{L}_b(\sigma)^{-1}$ and $\check{L}_b(\sigma)^{-1}\rho(\rho\pa_\rho-1)$. Since $\rho(\rho\pa_\rho-1): \eHb^{s,\ell}(\CX)\to\eHb^{s-1,\ell+1}(\CX)$, it is clear that

\[
\rho(\rho\pa_\rho-1)\check{L}_b(\sigma)^{-1}: \eHb^{s-1,\ell+2}(\CX)\to\eHb^{s-1, \ell+1}(\CX).
\]

Meanwhile, in view of Proposition <ref>

\[
2i\rho(\rho\pa_\rho-1)=\sigma^{-1}\big(\check{L}_b(\sigma)-\check{L}_b(0)+Q\big)\quad\mbox{with}\quad Q\in\sigma\rho^3\mbox{Diff}_\bop^1+\sigma\rho^2C^\infty+\sigma\rho^2C^\infty,
\]

then it follows that

\[
2i\rho(\rho\pa_\rho-1)\check{L}_b(\sigma)^{-1}=\sigma^{-1}\Big(I-(\widehat{L_{b}}(0)+V_b+Q)\check{L}_b(\sigma)^{-1}\Big): \eHb^{s-1,\ell+2}(\CX)\to\abs{\sigma}^{-1}\eHb^{s-2, \ell+2}(\CX).
\]

Then an interpolation gives

\begin{equation}
\rho(\rho\pa_\rho-1)\check{L}_b(\sigma)^{-1}: \eHb^{s-1,\ell+2}(\CX)\to\abs{\sigma}^{-\epsilon}\eHb^{s-1-\epsilon, \ell+1+\epsilon}(\CX),\quad 0\leq \epsilon\leq 1.
\end{equation}

As for $\check{L}_b(\sigma)^{-1}\rho(\rho\pa_\rho-1)$, interpolation between

\[
\check{L}_b(\sigma)^{-1}\rho(\rho\pa_\rho-1):\eHb^{s,\ell+1}(\CX)\to\eHb^{s,\ell}(\CX)
\]

\[
\check{L}_b(\sigma)^{-1}\big(-2i\rho(\rho\pa_\rho-1)\big)=\sigma^{-1}\Big(I-\check{L}_b(\sigma)^{-1}(\widehat{L_{b}}(0)+V_b-Q)\Big):\eHb^{s+1, \ell}(\CX)\to\abs{\sigma}^{-1}\eHb^{s,\ell}(\CX)
\]

\begin{equation}
\check{L}_b(\sigma)^{-1}\rho(\rho\pa_\rho-1):\eHb^{s+\epsilon,\ell+1-\epsilon}(\CX)\to\abs{\sigma}^{-\epsilon}\eHb^{s,\ell}(\CX), \quad 0\leq \epsilon\leq 1.
\end{equation}

Since $s-1-\epsilon>2$ and $1/2<\ell+1+\epsilon<3/2$, we have

\begin{equation}\label{eq:maincal}
\eHb^{s-1, \ell+2}(\CX)\xrightarrow{2i\rho(\rho\pa_\rho-1)\check{L}_b(\sigma)^{-1}}\abs{\sigma}^{-\epsilon}\eHb^{s-1-\epsilon, \ell+\epsilon+1}(\CX)\xrightarrow{\check{L}_b(\sigma)^{-1}}\abs{\sigma}^{-\epsilon}\eHb^{s-\epsilon, \ell+\epsilon-1}(\CX).
\end{equation}

We also have

\begin{equation}\label{eq:easycal}
\eHb^{s-1, \ell+2}(\CX)\xrightarrow{\check{L}_b(\sigma)^{-1}\!\!}\eHb^{s, \ell}(\CX)\xrightarrow{\rho^2\mbox{Diff}^1_\bop+\sigma\rho^2C^\infty\!\!}\eHb^{s-1,\ell+2}(\CX)\xrightarrow{\check{L}_b(\sigma)^{-1}\!\!\!}\eHb^{s,\ell}(\CX)\subset\abs{\sigma}^{-\epsilon}\eHb^{s-\epsilon, \ell+\epsilon-1}(\CX).
\end{equation}

By (<ref>), we finishes the proof of (<ref>).

For $\pa_\sigma^2\check{L}_b(\sigma)^{-1}$, we note that

\begin{align*}
\pa_\sigma^2\check{L}_b(\sigma)^{-1}&=2\check{L}_b(\sigma)^{-1}(\pa_\sigma\check{L}_b(\sigma))\check{L}_b(\sigma)^{-1}(\pa_\sigma\check{L}_b(\sigma))\check{L}_b(\sigma)^{-1}-\check{L}_b(\sigma)^{-1}(\pa^2_\sigma\check{L}_b(\sigma))\check{L}_b(\sigma)^{-1}\\
\end{align*}

Since $\pa_\sigma^2\check{L}_b(\sigma)\in\rho^2C^\infty$, according to (<ref>), we obtain

\[
\Romanupper{2}: \eHb^{s-1,\ell+2}(\CX)\to\eHb^{s,\ell}(\CX)\subset\abs{\sigma}^{-\epsilon-1}\eHb^{s-1-\epsilon, \ell+\epsilon-1}(\CX).
\]

It remains to analyze $\Romanupper{1}$. By the assumption on $s,\ell, \epsilon$, we find that $1/2<(1+\epsilon)/2<1$, and thus using the reasoning in (<ref>) and (<ref>), we obtain

\begin{align*}
\eHb^{s-1,\ell+2}(\CX)&\xrightarrow{\pa_\sigma\check{L}_b(\sigma)\check{L}_b(\sigma)^{-1}}\abs{\sigma}^{-\frac{1+\epsilon}{2}}\eHb^{s-1-\frac{1+\epsilon}{2}, \ell+\frac{1+\epsilon}{2}+1}(\CX)\xrightarrow{\pa_\sigma\check{L}_b(\sigma)\check{L}_b(\sigma)^{-1}}\abs{\sigma}^{-1-\epsilon}\eHb^{s-2-\epsilon, \ell+\epsilon+1}(\CX)\\
&\xrightarrow{\check{L}_b(\sigma)^{-1}}\abs{\sigma}^{-1-\epsilon}\eHb^{s-1-\epsilon, \ell+\epsilon-1}(\CX)
\end{align*}

where we use $s-\frac{3}{2}-\frac{\epsilon}{2}>2, \frac{1}{2}<\ell+\frac{3}{2}+\frac{\epsilon}{2}<\frac{3}{2}$ in the second step and $s-2-\epsilon>2,\frac{1}{2}< \ell+\epsilon+1<\frac{3}{2}$ in the last step. This finishes the proof of (<ref>).

Recall the expression (<ref>) of $\widehat{L_{b}}(\sigma)\check{L}(\sigma)^{-1}$

\[
\begin{split}
L_{00}(b,\sigma)& \sigma^2\wt{L}_{01}(b,\sigma)&\sigma\wt{L}_{02}(b,\sigma)\\
\sigma\wt{L}_{10}(b,\sigma)& \sigma^2\wt{L}_{11}(b,\sigma)&\sigma^2\wt{L}_{12}(b,\sigma)\\
\sigma\wt{L}_{20}(b,\sigma)& \sigma^2\wt{L}_{21}(b,\sigma)&\sigma\wt{L}_{22}(b,\sigma)\\
\end{pmatrix}\\
L_{00}(b,\sigma)& \sigma^2\wt{L}_{01}(b,\sigma)&\sigma\wt{L}_{02}(b,\sigma)\\
\sigma\big(\wt{L}^0_{10}(b)+\sigma\wt{L}^e_{10}(b,\sigma)\big)&\sigma^2\big(\wt{L}^0_{11}(b)+\sigma\wt{L}^1_{11}(b)+\sigma^2\wt{L}^e_{11}(b,\sigma)\big)&\sigma^2\big(\wt{L}^0_{12}(b)+\sigma\wt{L}^e_{12}(b,\sigma)\big)\\
\sigma\wt{L}_{20}(b,\sigma)&\sigma^2\big(\wt{L}^0_{21}(b)+\sigma\wt{L}^e_{21}(b)\big)&\sigma\big(\wt{L}^0_{22}(b)+\sigma\wt{L}^e_{22}(b,\sigma)\big).
\end{pmatrix}
\end{split}
\]

Let $s,\ell,\epsilon$ be defined as in Proposition <ref>. The entries of $\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}$ satisfy the following properties.

* $L_{00}, \wt{L}_{01}, \wt{L}_{02}, \wt{L}_{20}$ are $\epsilon$-regular at $\sigma=0$.

* $\wt{L}_{10}, \wt{L}_{12}, \wt{L}_{21}, \wt{L}_{22}$ have an $\epsilon$-regular expansion up to order one, that is, $\wt{L}^e_{10}, \wt{L}^e_{12}, \wt{L}^e_{21}, \wt{L}^e_{22}$ are $\epsilon$-regular at $\sigma=0$.

* $\wt{L}_{11}$ has an $\epsilon$-regular expansion up to order two, that is, $\wt{L}^e_{11}$ is $\epsilon$-regular at $\sigma=0$.

The key point in the proof was already mentioned before Proposition <ref>.

* Analysis of $\wt{L}^e_{10}$ and $\wt{L}_{20}$.
Here, we only discuss $\wt{L}^e_{10}$ in detail because the proof of $\wt{L}_{02}$ follows in a similar manner. Recall the calculation in (<ref>)

\[
\begin{split}
&\wt{L}^e_{10}(\tilde{g}_0, \tilde{A}_0),\ (\dg^*_1, \dA^*_1)\rangle\\
&\quad=-\Big\langle (\tilde{g}_0, \tilde{A}_0), i(\check{L}_b(\sigma)^{-1})^*(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa_\sigma^2\widehat{L_{b}}(0))^*\big(I-(\check{L}_b(0)^{-1})^*V_b^*\big)(\check{g}^*_2,\check{A}^*_2)\Big\rangle\\
&\quad\quad+\Big\langle(\tilde{g}_0, \tilde{A}_0),\frac{1}{2}(\check{L}_b(\sigma)^{-1})^*(\pa^2_\sigma\widehat{L_{b}}(0))^*(\dg^*_2,\dA^*_2)\Big\rangle
\end{split}
\]

\[
\]

Since $(\tilde{g}_0, \tilde{A}_0)\in\eHb^{s-1,\ell+2}(\CX)$, and by Proposition (<ref>)

\begin{gather*}
(\check{L}_b^{-1}(\sigma))^*\sHb^{-5/2-C(a,\gamma),\, 3/2-}(\CX)\in\sHb^{-3/2-C(a,\gamma), -1/2-}(\CX),\\	(\pa^j_\sigma\check{L}_b^{-1})^*\sHb^{-5/2-C(a,\gamma),\, 3/2-}(\CX)\in\abs{\sigma}^{-\epsilon-j+1}\sHb^{-s+1, -\ell-2}(\CX),\quad j=1,2
\end{gather*}

where we use the facts that $-5/2-C(a,\gamma)>-3>-s+\epsilon+1$ and $-\ell-\epsilon+1<3/2$, the pairings above (and their up to second order derivatives) are well-defined. This proves the $\epsilon$-regularity of $\wt{L}^e_{10}$.

* Analysis of $\wt{L}^e_{11},\wt{L}^e_{12},\wt{L}^e_{22}$ and $\wt{L}^e_{22}$. The statements for these entries can be proved in the same way as in the proof of $\wt{L}^e_{10}$.

* Analysis of $\wt{L}_{01}$ and $\wt{L}_{02}$. 	Here, we only prove the statement for $\wt{L}_{01}$ as the proof of $\wt{L}_{02}$ follows in an analogous manner. Using the calculation in (<ref>), we find that

\begin{align*}
\wt{L}_{01}(\tilde{g}_1, \tilde{A}_1)\!&=\!\!\Big(\!\!-iV_b\check{L}_b(\sigma)^{-1}\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa^2_\sigma\widehat{L_{b}}(0)\big)(\check{g}_1,\check{A}_1)+\frac{1}{2}V_b\check{L}_b(\sigma)^{-1}\pa^2_\sigma\widehat{L_{b}}(0)(\dg_1,\dA_1)\Big)\\
&\quad-(\wt{L}_{11}+\wt{L}_{21})(\tilde{g}_1, \tilde{A}_1).
\end{align*}

\[
\pa_\sigma\widehat{L_{b}}(0)(\check{g}_1,\check{A}_1),\quad \pa^2_\sigma\widehat{L_{b}}(0)(\check{g}_1,\check{A}_1),\quad\pa^2_\sigma\widehat{L_{b}}(0)(\dg_1,\dA_1)\in\eHb^{\infty, 3/2-}(\CX).
\]

Since by Proposition <ref>

\[
\check{L}_b^{-1}(\sigma)\eHb^{\infty, 3/2-}(\CX)\in\eHb^{\infty, -1/2-}(\CX),\  	\pa^j_\sigma\check{L}_b^{-1}(\sigma)\in\abs{\sigma}^{-\epsilon-j+1}\eHb^{\infty, \ell+\epsilon-1}(\CX),\  j=1,2,
\]

and $V_b$ maps $\mathcal{D}'(X^\circ)$ to $C_c^\infty(X)$, it follows that the first term on the right-hand side is $\epsilon$-regular. Meanwhile, it is clear that the second term is $\epsilon$-regular. This proves the $\epsilon$-regularity for $\wt{L}_{01}$.

* Analysis of $L_{00}$. By the calculation in (<ref>), we have

\[
L_{00}(b,\sigma)(\tilde{g}_0, \tilde{A}_0)=(\tilde{g}_0, \tilde{A}_0)-V_b\check{L}_b(\sigma)^{-1}(\tilde{g}_0, \tilde{A}_0)-\big(\sigma\wt{L}^0_{10}(b)+\sigma^2\wt{L}^e_{10}(b,\sigma)+\sigma\wt{L}_{20}(b,\sigma)\big)(\tilde{g}_0, \tilde{A}_0).
\]

Again, it is clear that the third term on the right-hand side is $\epsilon$-regular.	Since by Proposition <ref>

\[
\check{L}_b^{-1}(\sigma)(\tilde{g}_0, \tilde{A}_0)\in\eHb^{s,\ell}(\CX),\  	\pa^j_\sigma\check{L}_b^{-1}(\sigma)\in\abs{\sigma}^{-\epsilon-j+1}\eHb^{s-\epsilon-j+1, \ell+\epsilon-1}(\CX),\  j=1,2,
\]

and $V_b$ maps $\mathcal{D}'(X^\circ)$ to $C_c^\infty(X)$, it follows that the second term on the right-hand side is $\epsilon$-regular. This finishes the proof.

§.§.§ Hölder regularity of $P(b,\sigma)$ and $L^-_b(\sigma)$ at $\sigma=0$

Now we are ready to analyze the $\epsilon$-regularity of $R(b,\sigma)$. To this end, we need the following lemma.

Let $s,\ell,\epsilon$ be defined as in Proposition <ref>. Suppose that the operator $A(\sigma):\wt{\mathcal{K}}_i\to\mathcal{R}_j$ with $i,j=0,1,2$ has an $\epsilon$-regular expansion up to order one, i.e.,

\[
A(\sigma)=A^0+\sigma A^e(\sigma)
\]

where $A^0:\wt{\mathcal{K}}_i\to\mathcal{R}_j$ is invertible and independent of $\sigma$, and $A^e(\sigma):\wt{\mathcal{K}}_i\to\mathcal{R}_j$ is $\epsilon$-regular. Then for $\abs{\sigma}$ sufficiently small, $A(\sigma)$ is invertible, and its inverse $B(\sigma)=A(\sigma)^{-1}$ also has an $\epsilon$-regular expansion up to order one, that is,

\[
B(\sigma)=B^0+\sigma B^e(\sigma)
\]

where $B^0:\mathcal{R}_j\to\wt{\mathcal{K}}_i$ is independent of $\sigma$, and $B^e(\sigma):\mathcal{R}_j\to\wt{\mathcal{K}}_i$ is $\epsilon$-regular.

Similarly, if $A(\sigma)$ is $\epsilon$-regular and has uniformly bounded inverse $B(\sigma)$, then $B(\sigma)$ is $\epsilon$-regular as well.

Since $A^e(\sigma)$ is $\epsilon$-regular (and thus uniformly bounded), $\norm{\sigma (A^0)^{-1}A^e(\sigma)}_{\wt{\mathcal{K}}_i\to\wt{\mathcal{K}}_i}<1/2$ when $\abs{\sigma}$ is sufficiently small, the inverse of $A(\sigma)$ is given by the Neumann series

\begin{align*}
A(\sigma)^{-1}\!\!&=\!\!\Big(I+\sum_{k=1}^\infty(-1)^k\big(\sigma (A^0)^{-1}A^e(\sigma)\big)^k\Big)(A^0)^{-1}\\
&=(A^0)^{-1}-\sigma\Big((A^0)^{-1}A^e(\sigma)\sum_{k=0}^\infty(-1)^k\big(\sigma (A^0)^{-1}A^e(\sigma)\big)^k(A^0)^{-1}\Big):=B^0+\sigma B^e(\sigma).
\end{align*}

First, we have

\begin{align*}
\norm{B^e(\sigma)}_{\mathcal{R}_j\to\wt{\mathcal{K}}_i}&\leq\Big(\sum_{k=0}^\infty\norm{\sigma (A^0)^{-1}A^e(\sigma)}^k_{\wt{\mathcal{K}}_i\to\wt{\mathcal{K}}_i}\Big)\norm{(A^0)^{-1}}^2_{\mathcal{R}_j\to\wt{\mathcal{K}}_i}\norm{A^e(\sigma)}_{\wt{\mathcal{K}}_i\to\mathcal{R}_j}\\
\end{align*}

which implies the uniform boundedness of $B^e(\sigma)$. Secondly, the first derivative of $B^e(\sigma)$ is a sum of the terms of the types

\begin{align*}
&\sigma^{-2}\big(\sigma (A^0)^{-1}A^e(\sigma)\big)^k(A^0)^{-1}\lesssim 2^{-k+2},\quad k\geq 2\\
&\big(\sigma (A^0)^{-1}A^e(\sigma)\big)^{k_1}\circ\big((A^0)^{-1}\pa_\sigma A^e(\sigma)\big)\circ\big(\sigma (A^0)^{-1}A^e(\sigma)\big)^{k_2}(A^0)^{-1}\lesssim2^{-k_1-k_2}\abs{\sigma}^{-\epsilon},
\end{align*}

where we use $\norm{\pa_\sigma A^e(\sigma)}_{\wt{\mathcal{K}}_i\to\mathcal{R}_j}\lesssim \abs{\sigma}^{-\epsilon}$. As a consequence, $\norm{\pa_\sigma B^e(\sigma)}_{\mathcal{R}_j\to\wt{\mathcal{K}}_i}\lesssim \abs{\sigma}^{-\epsilon}$. Finally, the second derivative of $B^e(\sigma)$ is a sum of the terms of the types

\begin{align*}
&\sigma^{-3}\big(\sigma (A^0)^{-1}A^e(\sigma)\big)^k(A^0)^{-1}\lesssim 2^{-k+3},\quad k\geq 3\\
&\sigma^{-1}\big(\sigma (A^0)^{-1}A^e(\sigma)\big)^{k_1}\circ\big((A^0)^{-1}\pa_\sigma A^e(\sigma)\big)\circ\big(\sigma (A^0)^{-1}A^e(\sigma)\big)^{k_2}(A^0)^{-1}\lesssim2^{-k_1-k_2+1}\abs{\sigma}^{-\epsilon}, \quad k_1+k_2\geq 1,\\
&\sigma\big(\sigma (A^0)^{-1}A^e(\sigma)\big)^{k_1}\circ\big((A^0)^{-1}\pa_\sigma A^e(\sigma)\big)\circ\big(\sigma (A^0)^{-1}A^e(\sigma)\big)^{k_2}\circ\big((A^0)^{-1}\pa_\sigma A^e(\sigma)\big)\circ\big(\sigma (A^0)^{-1}A^e(\sigma)\big)^{k_3}(A^0)^{-1}\\
&\big(\sigma (A^0)^{-1}A^e(\sigma)\big)^{k_1}\circ\big((A^0)^{-1}\pa^2_\sigma A^e(\sigma)\big)\circ\big(\sigma (A^0)^{-1}A^e(\sigma)\big)^{k_2}(A^0)^{-1}\lesssim2^{-k_1-k_2}\abs{\sigma}^{-\epsilon-1},
\end{align*}

where we use $\pa_\sigma A^e(\sigma)\lesssim \abs{\sigma}^{-\epsilon}$ and $\pa^2_\sigma A^e(\sigma)\lesssim \abs{\sigma}^{-\epsilon-1}$. Therefore, $\pa^2_\sigma B^e(\sigma)\lesssim \abs{\sigma}^{-\epsilon-1}$.
This proves that $B^e(\sigma)$ is $\epsilon$-regular.

The proof of the second statement is similar by using

\begin{align*}
\pa_\sigma B(\sigma)&=-B(\sigma)\circ\pa_\sigma A(\sigma) \circ B(\sigma),\\ \pa_\sigma^2B(\sigma)&=-B(\sigma)\circ \pa^2_\sigma A(\sigma) \circ B(\sigma)+2B(\sigma)\circ\pa_\sigma A(\sigma) \circ B(\sigma)\circ \pa_\sigma A(\sigma)\circ B(\sigma).
\end{align*}

Recall the expressions (<ref>) and (<ref>) of $R_{b,\sigma}=(\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1})^{-1}$

\[
\begin{split}
\begin{pmatrix}
\wt{R}_{00}(b,\sigma)&\wt{R}_{01}(b,\sigma)&\wt{R}_{02}(b,\sigma)\\
\sigma^{-1}\wt{R}_{10}(b,\sigma)&\sigma^{-2}\wt{R}_{11}(b,\sigma)&\sigma^{-1}\wt{R}_{12}(b,\sigma)\\
\wt{R}_{20}(b,\sigma)&\sigma^{-1}\wt{R}_{21}(b,\sigma)&\sigma^{-1}\wt{R}_{22}(b,\sigma)
\end{pmatrix}\\
\begin{pmatrix}
\wt{R}_{00}(b,\sigma)&\wt{R}_{01}(b,\sigma)&\wt{R}_{02}(b,\sigma)\\
\sigma^{-1}\wt{R}_{10}(b,\sigma)&\sigma^{-2}\big(\wt{R}^0_{11}(b)+\sigma\wt{R}^e_{11}(b,\sigma)\big)&\sigma^{-1}\wt{R}_{12}(b,\sigma)\\
\wt{R}_{20}(b,\sigma)&\sigma^{-1}\big(\wt{R}^0_{21}(b)+\sigma\wt{R}^e_{21}(b,\sigma)\big)&\sigma^{-1}\big(\wt{R}^0_{22}(b)+\sigma\wt{R}^e_{22}(b,\sigma)\big)
\end{pmatrix}
\end{split}
\]

\begin{equation}
\begin{split}
\wt{R}_{11}=\big(\wt{L}^\sharp_{11}-\sigma\wt{L}^\sharp_{12}(\wt{L}^\flat_{22})^{-1}\wt{L}^\sharp_{21}\big)^{-1},\quad \wt{R}_{10}=\wt{R}_{11}\big(-\wt{L}_{10}+\sigma\wt{L}_{12}^\sharp(\wt{L}^\flat_{22})^{-1}\wt{L}_{20}\big)L_{00}^{-1},\\
\wt{R}_{12}=-\wt{R}_{11}\wt{L}_{12}^\sharp(\wt{L}^\flat_{22})^{-1},\quad \wt{R}_{21}=-(\wt{L}^\flat_{22})^{-1}\wt{L}_{21}^\sharp\wt{R}_{11},\quad\wt{R}_{22}=(\wt{L}^\flat_{22})^{-1}-\sigma(\wt{L}^\flat_{22})^{-1}\wt{L}_{21}^\sharp \wt{R}_{12},\\
\wt{R}_{20}=-(\wt{L}^\flat_{22})^{-1}\big(\wt{L}_{20}L_{00}^{-1}+\wt{L}_{21}^\sharp\wt{R}_{10}\big),\quad \wt{R}_{01}=-L_{00}^{-1}\big(\wt{L}_{01}\wt{R}_{11}+\wt{L}_{02}\wt{R}_{21}\big),\\
\wt{R}_{02}=-L_{00}^{-1}\big(\sigma\wt{L}_{01}\wt{R}_{12}+\wt{L}_{02}\wt{R}_{22}\big),\quad \wt{R}_{00}=L_{00}^{-1}\big(I-\sigma\wt{L}_{01}\wt{R}_{10}-\sigma\wt{L}_{02}\wt{R}_{20}\big)
\end{split}
\end{equation}

\begin{equation}
\begin{split}
&\wt{L}^\sharp_{i1}=\wt{L}_{i1}-\sigma\wt{L}_{i0}L_{00}^{-1}\wt{L}_{01}\quad\mbox{for}\quad i=1,2, \quad \wt{L}^\sharp_{12}=\wt{L}_{12}-\wt{L}_{10}L_{00}^{-1}\wt{L}_{02},\\
&\qquad\qquad\qquad\qquad	\wt{L}_{22}^\flat=\wt{L}_{22}-\sigma\wt{L}_{20}L_{00}^{-1}\wt{L}_{02}.
\end{split}
\end{equation}

Let $s,\ell,\epsilon$ be defined as in Proposition <ref>. The entries of $R(b,\sigma)$ satisfy the following properties.

* $\wt{R}_{00}, \wt{R}_{01}, \wt{R}_{02}, \wt{R}_{10}, \wt{R}_{12}, \wt{R}_{20}$ are $\epsilon$-regular at $\sigma=0$.

* $\wt{R}_{11}, \wt{R}_{21}, \wt{R}_{22}$ have an $\epsilon$-regular expansion up to order one, that is, $\wt{R}^e_{11}, \wt{R}^e_{21}, \wt{R}^e_{22}$ are $\epsilon$-regular at $\sigma=0$.

First, since $L_{00}$ is $\epsilon$-regular, by Lemma <ref>, so is $L_{00}^{-1}$. Then by Proposition <ref>, we find that $\wt{L}^\sharp_{11}, \wt{L}^\sharp_{21}, \wt{L}^\flat_{22}$ (and thus $(\wt{L}^\flat_{22})^{-1}$ by Lemma <ref>) have an $\epsilon$-regular expansion up to order one, while $\wt{L}^\sharp_{12}$ is $\epsilon$-regular. Therefore, we see that

\[
\wt{L}^\sharp_{11}-\sigma\wt{L}^\sharp_{12}(\wt{L}^\flat_{22})^{-1}\wt{L}^\sharp_{21}
\]

has an $\epsilon$-regular expansion up to order one whose coefficient of $\sigma^0$ is $\wt{L}_{11}^0(b)$. Then the $\epsilon$-regular expansion property of $\wt{R}_{11}$ follows from Lemma <ref>. Similarly, the $\epsilon$-regular expansion property of $\wt{R}_{21},\wt{R}_{22}$ follows from that of $\wt{R}_{11}, (\wt{L}^\flat_{22})^{-1},\wt{L}^\sharp_{21}$ and the above expressions of $\wt{R}_{ij}$. Finally, the $\epsilon$-regularity of the remaining components can be obtained by using the above expressions of $\wt{R}_{ij}$ and Proposition <ref>.

Recall the singular part $P(b,\sigma)$ of $\widehat{L_{b}}(\sigma)^{-1}$ as defined in Corollary <ref>, which satisfies

\begin{equation}\label{eq:recallofsingex}
\begin{split}
P(b,\sigma)f&=(\sigma^{-2}(\dg_1,\dA_1)-i\sigma^{-1}(\check{g}_1, \check{A}_1))+i\sigma^{-1}\big((\dg_2, \dA_2)
+(\dg'_1, \dA'_1)\big)+i\sigma^{-1}(\dg_1''(\sigma), \dA_1''(\sigma))\\
\end{split}
\end{equation}

where $(\dg_1''(\sigma),\dA_1''(\sigma))=P^e(b,\sigma)f=o(1)$ in $\mathcal{K}_{b,1}$ as $\sigma\to 0$. With the $\epsilon$-regularity of $R(b,\sigma)$ at our disposal, now we are able to prove that $P^e(b,\sigma)$ is $\epsilon$-regular as well, and thus give a refined description of $\sigma^{-1}(\dg_1''(\sigma),\dA_1''(\sigma))$. Concretely, we shall prove that $(\dg_1''(\sigma),\dA_1''(\sigma))$ is Hölder continuous in a small interval of $0$, and thus $(\dg_1''(\sigma),\dA_1''(\sigma))=\mathcal{O}(\abs{\sigma}^{1-\epsilon})$ in $\mathcal{K}_{b,1}$ for some $0<\epsilon<1$.

We now make a remark that there is a relationship between the Sobolev spave $\sH^k([0,c_0))$, which is the restriction to $[0, c_0)$ of elements in $H^k(\BR)$ with support in $[0, \infty)$, and the $\bop$-Sobolev space $H_{\bop}^k([0, c_0))$, which is a completion of $\{\phi|_{(0, c_0]}\}$ where $\phi\in C_c^\infty((0,\infty))$ with respect the following  norm

\[
\norm{\phi}^2_{H_{\bop}^k([0, C_0))}:=\sum_{j=0}^k\norm{(\sigma\pa_\sigma)^j\phi}^2_{L^2(d\sigma)}.
\]

First, it is clear that $\sigma^kH_{\bop}^k([0, c_0))\subset \sH^k([0, c_0))$. Second, Hardy's inequality tells us that for $\phi\in C_c^\infty(0, \infty)$, we have

\[
\int_0^\infty \frac{\abs{\phi}^2}{r^2}\,dr\leq 4\int_0^\infty \abs{\pa_rf}^2\,dr.
\]

Applying Hardy's inequality $k$ times, we find that

\[
\int_0^\infty \frac{\abs{\phi}^2}{r^{2k}}\,dr\leq C_k\int_0^\infty \abs{\pa^k_rf}^2\,dr.
\]

for $\phi\in C_c^\infty(0, \infty)$. This implies $ \sH^k([0, c_0))\subset\sigma^kH_{\bop}^k([0, c_0))$. Therefore, $\sH^k([0, c_0))=\sigma^kH_{\bop}^k([0, c_0))$. Then an interpolation gives $\sH^\al([0, c_0))=\sigma^\al H_{\bop}^\al([0, c_0))$ for all $\al\geq 0$.

As a preparation, we record the following technical lemma.

Suppose that $u_{\pm}\in H^s(\BR_{\pm})$ with $s\geq 0$ and $s\neq \frac12+\BN_0$ and let $u(x)=u_+(x)$ for $x>0$ and $u(x)=u_-(x)$ for $x<0$. If $\pa_ju_+(x)=\pa_j u_-(x)$ for all $j\in\BN_0, j<s-\frac 12$, then $u(x)\in H^s(\BR)$.

See <cit.>.

Let $\sigma^{-1}(\dg_1''(\sigma),\dA_1''(\sigma))=\sigma^{-1}P^e(b,\sigma)f\in\mathcal{K}_{b,1}$ be defined as in (<ref>) and let $s,\ell,\epsilon$ be defined as in Proposition <ref>. Then

we have

\begin{equation}\label{eq:epsilonregofPe}
P^e(b,\sigma)\in H^{\frac 32-\epsilon-}((-c_0,c_0);\mathcal{L}(\eHb^{s-1,\ell+2}(\CX), \mathcal{K}_{b,1}))
\end{equation}

where $c_0>0$ is a sufficiently small constant. Moreover,

\begin{equation}\label{eq:conormalofPe}
\sigma^{-1}(\dg_1''(\sigma),\dA_1''(\sigma))\in\sum_{\pm}\mathcal{A}^{-\epsilon}\big(\pm[0, c_0);\mathcal{K}_{b,1}\big).
\end{equation}

Using the calculation (<ref>) and (<ref>) in the proof of Corollary <ref>, we find that for $f=(f_0, f_1, f_2)\in\eHb^{s-1,\ell+2}(\CX)=\mathcal{R}_0\oplus\mathcal{R}_{1}\oplus\mathcal{R}_{2}$,

\begin{equation}\label{eq:anotherexofP}
\begin{split}
\end{split}
\end{equation}

According to Proposition <ref>, we have $\norm{\pa_\sigma\wt{R}^e_{11}(b,\sigma)}_{\mathcal{R}_1\to\wt{\mathcal{K}}_1}\lesssim \abs{\sigma}^{-\epsilon}$ and $\norm{\pa^2_\sigma\wt{R}^e_{11}(b,\sigma)}_{\mathcal{R}_1\to\wt{\mathcal{K}}_1}\lesssim \abs{\sigma}^{-\epsilon-1}$ for $0<\abs{\sigma}\leq c_0$ where $c_0$ is a sufficiently small constant, which implies that

\[
\norm{\pa_\sigma\sigma\pa_\sigma\wt{R}^e_{11}(b,\sigma)}_{\mathcal{R}_1\to\wt{\mathcal{K}}_1}\lesssim \abs{\sigma}^{-\epsilon}.\]

Integrating $\pa_\sigma\sigma\pa_\sigma\wt{R}^e_{11}(b,\sigma)$ and $\pa_\sigma\wt{R}^e_{11}(b,\sigma)$ from $\sigma\in(0, c_0)$ to $0$ and using $\sigma\pa_\sigma\wt{R}^e_{11}(\sigma)|_{b,\sigma=0}=0$ yield

\begin{align}\label{eq:firstderoofRe11}
\sigma\pa_\sigma\wt{R}^e_{11}(b,\sigma)\in\sigma^{1-\epsilon}L^\infty([0, c_0);\mathcal{L}(\mathcal{R}_1, \wt{\mathcal{K}}_1))\subset \sigma^{\frac 32-\epsilon-}H^0_{\bop}([0, c_0);\mathcal{L}(\mathcal{R}_1, \wt{\mathcal{K}}_1))
\wt{R}^e_{11}(b,\sigma)-\wt{R}^e_{11}(b,0)\in\sigma^{1-\epsilon}L^\infty([0, c_0);\mathcal{L}(\mathcal{R}_1, \wt{\mathcal{K}}_1))\subset \sigma^{\frac 32-\epsilon-}H^0_{\bop}([0, c_0);\mathcal{L}(\mathcal{R}_1, \wt{\mathcal{K}}_1))
\end{align}

This implies

\[
(\wt{R}^e_{11}(\sigma)-\wt{R}^e_{11}(0))\in\sigma^{\frac32-\epsilon-}H^2_{\bop}([0, c_0);\mathcal{L}(\mathcal{R}_1, \wt{\mathcal{K}}_1))\subset\sH^{\frac 32-\epsilon-}([0, c_0);\mathcal{L}(\mathcal{R}_1, \wt{\mathcal{K}}_1)).
\]

Similarly, we also have

\[
(\wt{R}^e_{11}(\sigma)-\wt{R}^e_{11}(0))\in\sH^{\frac 32-\epsilon-}((-c_0,0];\mathcal{L}(\mathcal{R}_1, \wt{\mathcal{K}}_1)).
\]

Since $	(\wt{R}^e_{11}(\sigma)-\wt{R}^e_{11}(0))\to0$ as $\sigma\to0$ from both sides, by Lemma <ref>, we obtain

\[
(\wt{R}^e_{11}(\sigma)-\wt{R}^e_{11}(0))\in H^{\frac 32-\epsilon-}((-c_0,c_0);\mathcal{L}(\mathcal{R}_1, \wt{\mathcal{K}}_1)).
\]

Similarly, $\wt{R}_{10}(b,\sigma)-\wt{R}_{10}(b,0)$ and $\wt{R}_{12}(b,\sigma)-\wt{R}_{12}(b,0)$ satisfy the same estimates as $\wt{R}^e_{11}(b,\sigma)-\wt{R}^e_{11}(b,0)$. This proves (<ref>).

Moreover, the estimate in (<ref>) implies

\begin{equation}\label{eq:LinfityestimateofR11}
\wt{R}^e_{11}(b,\sigma)-\wt{R}^e_{11}(b,0)\in\sum_{\pm}\sigma^{1-\epsilon}L^\infty(\pm[0, c_0);\mathcal{L}(\mathcal{R}_1, \wt{\mathcal{K}}_1))
\end{equation}

and this also holds for $\wt{R}_{10}(b,\sigma)-\wt{R}_{10}(b,0), \wt{R}_{20}(b,\sigma)-\wt{R}_{20}(b,0)$.
Using Proposition <ref>, the estimate (<ref>) is satisfied after applying any number of derivatives $\sigma\pa_\sigma$. This proves (<ref>).

Now we turn to the analysis of the regular part $L^-_b(\sigma)$.

Let $s,\ell,\epsilon$ be defined as in Proposition <ref>.
For $ \IM\sigma\geq0$ and $0<\abs{\sigma}\leq c_0$ where $c_0>0$ is a small constant, the regular part $L^-_b(\sigma)$ of the resolvent $\widehat{L_{b}}(\sigma)^{-1}$ defined in Corollary <ref> satisfies

\begin{equation}\label{eq:epsilonregofregpart}
\pa_\sigma^jL^-_b(\sigma):\eHb^{s-1,\ell+2}(\CX)\to\abs{\sigma}^{-\epsilon-j+1}\eHb^{s-\epsilon-j+1,\ell+\epsilon+1}(\CX),\quad j=1,2.
\end{equation}

According to Proposition <ref> and the expression (<ref>), we see that $P(b,\sigma)$ has the form $P(b,\sigma)=\sigma^{-2}P^{2}(b)+\sigma^{-1} P^1(b)+\sigma^{-1}P^e(b,\sigma)$

\[
P^2(b), P^1(b):\eHb^{s-1,\ell+2}(\CX)\to\rho C^\infty(\CX)+\eHb^{\infty, \frac12-}(\CX)
\]

and $P^e(b,\sigma):\eHb^{s-1,\ell+2}(\CX)\to\mathcal{K}_{b,1}$ is $\epsilon$-regular as proved in Theorem <ref>. We also write $R(b,\sigma)=R_{\mathrm{sing}}(b,\sigma)+R_{\mathrm{reg}}(b,\sigma)$ where

\[
\sigma^{-1}\wt{R}_{10}(b,\sigma)&\sigma^{-2}\wt{R}^0_{11}(b)+\sigma^{-1}\wt{R}^e_{11}(b,\sigma)&\sigma^{-1}\wt{R}_{12}(b,\sigma)\\
\end{pmatrix}
\]

\[
\wt{R}_{00}(b,\sigma)&\wt{R}_{01}(b,\sigma)&\wt{R}_{02}(b,\sigma)\\
\wt{R}_{20}(b,\sigma)&\wt{R}^e_{21}(b,\sigma)&\wt{R}^e_{22}(b,\sigma)
\end{pmatrix}.
\]

Using the calculation in (<ref>), (<ref>) and (<ref>), we find that

\begin{align*}
\end{align*}

For the term $\Romanupper{1}$, we compute

\begin{align*}
\pa_\sigma\Romanupper{1}&=\pa_\sigma\check{L}_b(\sigma)^{-1}R_{\mathrm{reg}}(b,\sigma)+\check{L}_b(\sigma)^{-1}\pa_\sigma R_{\mathrm{reg}}(b,\sigma),\\
\pa^2_\sigma\Romanupper{1}&=\pa^2_\sigma\check{L}_b(\sigma)^{-1}R_{\mathrm{reg}}(b,\sigma)+2\pa_\sigma\check{L}_b(\sigma)^{-1}\pa_\sigma R_{\mathrm{reg}}(b,\sigma)+\check{L}_b(\sigma)^{-1}\pa^2_\sigma R_{\mathrm{reg}}(b,\sigma).
\end{align*}

Then by proposition <ref> and <ref>, we conclude that

\[
\pa_\sigma^j\Romanupper{1}:\eHb^{s-1,\ell+2}(\CX)\to\abs{\sigma}^{-\epsilon-j+1}\eHb^{s-\epsilon-j+1,\ell+\epsilon+1}(\CX),\quad j=1,2.
\]

As for the term $\Romanupper{2}$, we first note that $P^2(b), P^1(b), P^e(b,\sigma)$ map to $\rho C(\CX)+\eHb^{\infty, 1/2-}(\CX)$, and thus
\[
\pa_\sigma^k\widehat{L_{b}}(0)P^2(b),\  \pa_\sigma^k\widehat{L_{b}}(0)P^1(b),\ \pa_\sigma^k\widehat{L_{b}}(0)P^e(b,\sigma):\eHb^{s-1,\ell+2}(\CX)\to\eHb^{\infty, \frac 32-}(\CX),\quad k=1,2.
\]

Next, we calculate

\begin{align*}
\pa_\sigma\Romanupper{2}&=-\pa_\sigma\check{L}_b(\sigma)^{-1}\Big(\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa_\sigma^2\widehat{L_{b}}(0)\big)(P^1(b)+P^e(b,\sigma))+\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)P^2(b)\Big)\\
&\quad-\check{L}_b(\sigma)^{-1}\Big(\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa_\sigma^2\widehat{L_{b}}(0)\big)\pa_\sigma P^e(b,\sigma)+\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0) P^e(b,\sigma)\Big),\\
\pa_\sigma^2\Romanupper{2}&=-\pa^2_\sigma\check{L}_b(\sigma)^{-1}\Big(\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa_\sigma^2\widehat{L_{b}}(0)\big)(P^1(b)+P^e(b,\sigma))+\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)P^2(b)\Big)\\
&\quad-2\pa_\sigma\check{L}_b(\sigma)^{-1}\Big(\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa_\sigma^2\widehat{L_{b}}(0)\big)\pa_\sigma P^e(b,\sigma)+\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0) P^e(b,\sigma)\Big)\\
&\quad-\check{L}_b(\sigma)^{-1}\Big(\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa_\sigma^2\widehat{L_{b}}(0)\big)\pa^2_\sigma P^e(b,\sigma)+\pa_\sigma^2\widehat{L_{b}}(0) \pa_\sigma P^e(b,\sigma)\Big).
\end{align*}

Therefore, using Proposition <ref> and the fact that $P^e(b,\sigma):\eHb^{s-1,\ell+2}\to\mathcal{K}_{b,1}$ is $\epsilon$-regular, we arrive at

\[
\pa_\sigma^j\Romanupper{2}:\eHb^{s-1,\ell+2}(\CX)\to\abs{\sigma}^{-\epsilon-j+1}\eHb^{s-\epsilon-j+1,\ell+\epsilon+1}(\CX),\quad j=1,2.
\]

This completes the proof of the estimates (<ref>).

Let $s,\ell,\epsilon$ be defined as in Proposition <ref>. Then we have

\begin{equation}\label{eq:holderregularity}
L^-_b(\sigma)\in H^{\frac{3}{2}-\epsilon-}\Big((-c_0, c_0);\mathcal{L}\big(\eHb^{s-1,\ell+2}(\CX), \eHb^{s-\max\{\epsilon, \frac 12\},\ell+\epsilon-1}(\CX)\big)\Big)
\end{equation}

where $c_0>0$ is a sufficiently small constant.

Using the estimates in Proposition <ref> and following the proof at the beginning of Theorem <ref>, we find that

\begin{equation}\label{eq:estimate1}
L_b^-(\sigma)-L_b^-(0)\in\sigma^{\frac32-\epsilon-}H_{\bop}^2\Big([0, c_0);\mathcal{L}\big(\eHb^{s-1,\ell+2}(\CX), \eHb^{s-1-\epsilon,\ell+\epsilon-1}(\CX)\big)\Big).
\end{equation}

Again integrating $\pa_\sigma L^-_b(\sigma)$ from $\sigma\in(0, c_0)$ to $0$ and using the estimate (<ref>) for $k=1$ yield

\begin{equation}\label{eq:estimate2}
L_b^-(\sigma)-L_b^-(0)\in\sigma^{\frac32-\epsilon-}H_{\bop}^1\Big([0, c_0);\mathcal{L}\big(\eHb^{s-1,\ell+2}(\CX), \eHb^{s-\epsilon,\ell+\epsilon-1}(\CX)\big)\Big).
\end{equation}

Interpolating between (<ref>) and (<ref>) gives

\begin{equation}\label{eq:interpo1and2}
L_b^-(\sigma)-L_b^-(0)\in\sigma^{\frac32-\epsilon-}H_{\bop}^{1+\theta}\Big([0, c_0);\mathcal{L}\big(\eHb^{s-1,\ell+2}(\CX), \eHb^{s-\theta-\epsilon,\ell+\epsilon-1}(\CX)\big)\Big), \quad0\leq \theta\leq 1.
\end{equation}

If $\epsilon\in(0,1/2]$, we take $\theta=1/2-\epsilon$, and thus $3/2-\epsilon=1+\theta$. If $\epsilon\in(1/2, 1)$, we take $\theta=0$. This gives

\begin{equation}
\sigma^{\frac32-\epsilon-}H_{\bop}^{\frac32-\epsilon}\Big([0, c_0);\mathcal{L}\big(\eHb^{s-1,\ell+2}(\CX), \eHb^{s-\frac12,\ell+\epsilon-1}(\CX)\big)\Big),\quad &0<\epsilon\leq \frac12,\\
\sigma^{\frac32-\epsilon-}H_{\bop}^{1}\Big([0, c_0);\mathcal{L}\big(\eHb^{s-1,\ell+2}(\CX), \eHb^{s-\epsilon,\ell+\epsilon-1}(\CX)\big)\Big),\quad &\frac 12<\epsilon<1,
\end{cases}
\end{equation}

which implies

\begin{equation}
\sH^{\frac32-\epsilon}\Big([0, c_0);\mathcal{L}\big(\eHb^{s-1,\ell+2}(\CX), \eHb^{s-\frac12,\ell+\epsilon-1}(\CX)\big)\Big),\quad &0<\epsilon\leq \frac12,\\
\sH^{\frac 32-\epsilon-}\Big([0, c_0);\mathcal{L}\big(\eHb^{s-1,\ell+2}(\CX), \eHb^{s-\epsilon,\ell+\epsilon-1}(\CX)\big)\Big),\quad &\frac 12<\epsilon<1.
\end{cases}
\end{equation}

Similarly, we also have the above statement for the other half interval $(-c_0, 0]$. Since $	L_b^-(\sigma)-L_b^-(0)\to0$ as $\sigma\to0$ from both sides, by Lemma <ref>, we obtain

\[
L^-_b(\sigma)-L^-_b(0)\in H^{\frac{3}{2}-\epsilon-}\Big((-c_0, c_0);\mathcal{L}\big(\eHb^{s-1,\ell+2}(\CX), \eHb^{s-\max\{\epsilon, \frac 12\},\ell+\epsilon-1}(\CX)\big)\Big),
\]

and thus the conclusion (<ref>).

§.§ Regularity for frequencies away from $0$.

Now we discuss the regularity of $\widehat{L_{b}}(\sigma)^{-1}$ for $ \abs{\sigma}\geq c_0,\ 0\leq\IM\sigma\leq C$ for some $c_0,\, C>0$.

Let $\ell<-\frac12, s>3+m$ and $s+\ell>-\frac 12+m$ where $m\in\BN_0$. Then for $0\leq \IM\sigma\leq C, \abs{\sigma}\geq c_0$ with $c_0,C>0$, the operator

\begin{equation}\label{eq:regularityaway0}
\pa_\sigma^m \widehat{L_{b}}(\sigma)^{-1}: \bar{H}_{\bop,h}^{s,\ell+1}(\CX)\to h^{-m}\bar{H}_{\bop,h}^{s-m,\ell}(\CX)
\end{equation}

where $h=\abs{\sigma}^{-1}$, is uniformly bounded.

The estimate (<ref>) for the case $m=0$ follows from high energy estimate <ref>. For $m\geq 1$, $\pa_\sigma^m\widehat{L_{b}}(\sigma)^{-1}$ is a linear combination of the operators of the type

\begin{align*}
\Big(	\widehat{L_{b}}(\sigma)^{-1}\circ\pa^{m_1}_\sigma \widehat{L_{b}}(\sigma)\Big)\circ \cdots\circ\Big(\widehat{L_{b}}(\sigma)^{-1}\circ\pa^{m_k}_\sigma \widehat{L_{b}}(\sigma)\Big)\circ \widehat{L_{b}}(\sigma)^{-1}
\end{align*}

where $1\leq m_1, \cdots, m_k\leq 2$ and $m_1+\cdots+m_k=m$. Since

\[
\pa_\sigma \widehat{L_{b}}(\sigma)\in \rho\mbox{Diff}_\bop^1+\rho^2C^\infty\subset h^{-1}\rho\mbox{Diff}_{\bop,h}^1\quad \mbox{and}\quad \pa_\sigma^2\widehat{L_{b}}(\sigma)\in\rho^2C^\infty,
\]

it follows that

\begin{align*}
&\bar{H}_{\bop,h}^{s,\ell+1} \xrightarrow{\widehat{L_{b}}(\sigma)^{-1}}\bar{H}^{s,\ell}_{\bop,h}\xrightarrow{\pa^{m_k}_\sigma\widehat{L_{b}}(\sigma)}h^{m_k-2}\bar{H}^{s+m_k-2,\ell+m_k}_{\bop,h}\subset h^{m_k-2}\bar{H}^{s+m_k-2,\ell+1}_{\bop,h}
\\
&\xrightarrow {\widehat{L_{b}}(\sigma)^{-1}}h^{m_k-2}\bar{H}^{s+m_k-2,\ell}_{\bop,h} \xrightarrow{\cdots}\cdots\xrightarrow{\cdots}h^{m-m_1-2(k-1)}\bar{H}^{s+m-m_1-2(k-1), \ell}_{\bop,h}\\
\end{align*}

Since $1\leq k\leq m$, we have $h^{m-2k}\bar{H}^{s+m-2k,\ell+m-k}_{\bop,h}\subset h^{-m}\bar{H}^{s-m,\ell}_{\bop,h}$. This finishes the proof.

§.§ Conormal regularity

Finally, we consider the conormal regularity of $L^-_b(\sigma)$ at $\sigma=0$. Specifically, after applying any number of derivatives $\sigma\pa_\sigma$ to $L^-_b(\sigma)$, it satisfies the same properties as $L^-_b(\sigma)$.

Recall that

\[
\]

so in order to study the conormal regularity, we need to analyze that of $\check{L}_b(\sigma)^{-1}$, $R_{\mathrm{reg}}(b,\sigma)$ and $P^e(b,\sigma)$.

We first discuss the conormal regularity of $\check{L}_b(\sigma)^{-1}$ and prove a analogous result of Proposition <ref> for $(\sigma\pa_\sigma)^m\check{L}_b(\sigma)^{-1}$.

Let $-\frac 32<\ell<-\frac 12$, $0<\epsilon<1$ with $-\frac 12<\ell+\epsilon<\frac 12$, and $s-\epsilon>4+m$ where $m\in\BN$. Define

\[
\check{\mathcal{L}}^{(m)}_b(\sigma):=(\sigma\pa_\sigma)^m\check{L}_b(\sigma)^{-1}.
\]

For $0<\abs{\sigma}\leq c_0, \IM\sigma\geq 0$ with $c_0$ small, we have

\begin{align}\label{eq:zeroderiofcocheckL-1}
&\check{\mathcal{L}}^{(m)}_b(\sigma):\eHb^{s-1,\ell+2}(\CX)\to \eHb^{s-m, \ell}(\CX),\\\label{eq:firstderiofcocheckL-1}
&\pa_\sigma\check{\mathcal{L}}^{(m)}_b(\sigma): \eHb^{s-1,\ell+2}(\CX)\to \abs{\sigma}^{-\epsilon}\eHb^{s-m-\epsilon, \ell+\epsilon-1}(\CX),\\\label{eq:secondderiofcocheckL-1}
&\pa^2_\sigma\check{\mathcal{L}}^{(m)}_b(\sigma): \eHb^{s-1,\ell+2}(\CX)\to \abs{\sigma}^{-\epsilon-1}\eHb^{s-m-1-\epsilon, \ell+\epsilon-1}(\CX).
\end{align}

\[
\pa_\sigma^2\check{\mathcal{L}}^{(m)}_b(\sigma)=\sigma^{-1}\pa_\sigma(\sigma\pa_\sigma-1)\check{\mathcal{L}}^{(m)}_b(\sigma)=\sigma^{-1}\pa_\sigma\check{\mathcal{L}}^{(m+1)}_b(\sigma)-\sigma^{-1}\pa_\sigma\check{\mathcal{L}}^{(m)}_b(\sigma),\]

it suffices to prove the statements (<ref>) and (<ref>), and then the conclusion (<ref>) follows directly.

Now we prove (<ref>) and (<ref>) by induction on $m$. For $m=1$, we calculate

\[
\check{\mathcal{L}}_b^{(1)}=\sigma\pa_\sigma\check{L}_b(\sigma)^{-1}=-\check{L}_b(\sigma)^{-1}(\sigma\pa_\sigma\check{L}_b(\sigma))\check{L}_b(\sigma)^{-1}
\]

\begin{equation}\label{eq:deriofcocheckL}
\sigma\pa_\sigma\check{L}_b(\sigma)=\check{L}_b(\sigma)-\check{L}_b(0)+\frac{\sigma^2}{2}\pa^2_\sigma\check{L}_b(0)\in\check{L}_b(\sigma)+\rho^2\mbox{Diff}_\bop^2+\sigma^2\rho^2C^\infty.
\end{equation}

As a consequence,

\[
\check{\mathcal{L}}_b^{(1)}=-\check{L}_b(\sigma)^{-1}+\check{L}_b(\sigma)^{-1}\big(\rho^2\mbox{Diff}_\bop^2+\sigma^2\rho^2C^\infty\big)\check{L}_b(\sigma)^{-1}
\]

We then compute

\[
\eHb^{s-1,\ell+2}(\CX)\xrightarrow{\check{L}_b(\sigma)^{-1}}\eHb^{s,\ell}(\CX)\xrightarrow{\rho^2\mbox{Diff}_\bop^2+\sigma^2\rho^2C^\infty}\eHb^{s-2,\ell+2}(\CX)\xrightarrow{\check{L}_b(\sigma)^{-1}}\eHb^{s-1,\ell}(\CX).
\]

This proves (<ref>) for $m=1$. As for the first derivative $\pa_\sigma\check{\mathcal{L}}_b^{(1)}(\sigma)$, we calculate

\begin{align*}
\pa_\sigma\check{\mathcal{L}}_b^{(1)}(\sigma)&=-\pa_\sigma\check{L}_b(\sigma)^{-1}+\pa_\sigma\check{L}_b(\sigma)^{-1}\big(\rho^2\mbox{Diff}_b^2+\sigma^2\rho^2C^\infty\big)\check{L}_b(\sigma)^{-1}+\check{L}_b(\sigma)^{-1}\big(\sigma\rho^2C^\infty\big)\check{L}_b(\sigma)^{-1}\\
\end{align*}

Using Proposition <ref>, we find that

\begin{align*}
\Romanupper{1}:&\eHb^{s-1,\ell+2}(\CX)\to\abs{\sigma}^{-\epsilon}\eHb^{s-\epsilon, \ell+\epsilon+1}(\CX),\\
\Romanupper{2}:&\eHb^{s-1,\ell+2}(\CX)\xrightarrow{\check{L}_b(\sigma)^{-1}}\eHb^{s,\ell}(\CX)\xrightarrow{\rho^2\mbox{Diff}_\bop^2+\sigma^2\rho^2C^\infty}\eHb^{s-2,\ell+2}(\CX)\xrightarrow{\pa_\sigma\check{L}_b(\sigma)^{-1}}\abs{\sigma}^{-\epsilon}\eHb^{s-1-\epsilon, \ell+\epsilon-1}(\CX),\\
\Romanupper{3}:&\eHb^{s-1,\ell+2}(\CX)\xrightarrow{\check{L}_b(\sigma)^{-1}}\eHb^{s,\ell}(\CX)\xrightarrow{\sigma\rho^2C^\infty}\eHb^{s,\ell+2}(\CX)\xrightarrow{\check{L}_b(\sigma)^{-1}}\eHb^{s-1, \ell}(\CX)\subset \abs{\sigma}^{-\epsilon}\eHb^{s-1-\epsilon, \ell+\epsilon-1}(\CX),\\
\Romanupper{4}:&\eHb^{s-1,\ell+2}(\CX)\xrightarrow{\pa_\sigma\check{L}_b(\sigma)^{-1}}\abs{\sigma}^{-\epsilon}\eHb^{s-\epsilon,\ell+\epsilon-1}(\CX)\xrightarrow{\rho^2\mbox{Diff}_\bop^2+\sigma^2\rho^2C^\infty}\abs{\sigma}^{-\epsilon}\eHb^{s-\epsilon-2,\ell+\epsilon+1}(\CX)\\
&\xrightarrow{\check{L}_b(\sigma)^{-1}}\abs{\sigma}^{-\epsilon}\eHb^{s-1-\epsilon, \ell+\epsilon-1}(\CX)
\end{align*}

where we use $1/2<\ell+\epsilon+1<3/2$ in the last step of the analysis of $\Romanupper{4}$. This proves (<ref>) for $m=1$.

Suppose that (<ref>) and (<ref>) hold for $k\leq m$.
Then for $m+1$, by a direct calculation, we obtain that $\check{\mathcal{L}}_b^{(m+1)}(\sigma)$ is a linear combination of the operators of the type

\[
L_1=\check{\mathcal{L}}_b^{(m)}(\sigma),\quad L_2:=\check{\mathcal{L}}_b^{(m_1)}(\sigma)\circ(\rho^2\mbox{Diff}_\bop^2)\circ\check{\mathcal{L}}_b^{(m_2)}(\sigma)
\]

where $m_1, m_2\geq 0$ and $m_1+m_2\leq m$. Then by the induction hypothesis on $k\leq m$, it follows that

\begin{align*}
\end{align*}

This proves (<ref>) for $m+1$. As for the first derivative $\pa_\sigma\check{\mathcal{L}}_b^{(m+1)}$, it is a linear combination of the operators of the following types

\begin{align*}
D_1=\pa_\sigma\check{\mathcal{L}}_b^{(m)}(\sigma),\quad D_2:=\pa_\sigma\check{\mathcal{L}}_b^{(m_1)}(\sigma)\circ(\rho^2\mbox{Diff}_\bop^2)\circ\check{\mathcal{L}}_b^{(m_2)}(\sigma),\quad D_3:=\check{\mathcal{L}}_b^{(m_1)}(\sigma)\circ(\rho^2\mbox{Diff}_\bop^2)\circ\pa_\sigma\check{\mathcal{L}}_b^{(m_2)}(\sigma)
\end{align*}

where $m_1, m_2\geq 0$ and $m_1+m_2\leq m$. Then we have

\begin{align*}
\end{align*}

where we use $1/2<\ell+\epsilon+1<3/2$ in the third step of the analysis of $D_3$. This proves (<ref>) for $m+1$.

This finishes the proof of the proposition.

With the conormal regularity of $\check{L}_b(\sigma)^{-1}$ with our disposal, we are ready to discuss that of the entries $\wt{L}_{ij}$ of the operator $\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}$.

Recall the expression (<ref>) of $\widehat{L_{b}}(\sigma)\check{L}(\sigma)^{-1}$

\[
\begin{split}
L_{00}(b,\sigma)& \sigma^2\wt{L}_{01}(b,\sigma)&\sigma\wt{L}_{02}(b,\sigma)\\
\sigma\wt{L}_{10}(b,\sigma)& \sigma^2\wt{L}_{11}(b,\sigma)&\sigma^2\wt{L}_{12}(b,\sigma)\\
\sigma\wt{L}_{20}(b,\sigma)& \sigma^2\wt{L}_{21}(b,\sigma)&\sigma\wt{L}_{22}(b,\sigma)\\
\end{pmatrix}\\
L_{00}(b,\sigma)& \sigma^2\wt{L}_{01}(b,\sigma)&\sigma\wt{L}_{02}(b,\sigma)\\
\sigma\big(\wt{L}^0_{10}(b)+\sigma\wt{L}^e_{10}(b,\sigma)\big)&\sigma^2\big(\wt{L}^0_{11}(b)+\sigma\wt{L}^1_{11}(b)+\sigma^2\wt{L}^e_{11}(b,\sigma)\big)&\sigma^2\big(\wt{L}^0_{12}(b)+\sigma\wt{L}^e_{12}(b,\sigma)\big)\\
\sigma\wt{L}_{20}(b,\sigma)&\sigma^2\big(\wt{L}^0_{21}(b)+\sigma\wt{L}^e_{21}(b)\big)&\sigma\big(\wt{L}^0_{22}(b)+\sigma\wt{L}^e_{22}(b,\sigma)\big).
\end{pmatrix}
\end{split}
\]

Let $s,\ell,\epsilon,m$ be defined as in Proposition <ref>. Then  $(\sigma\pa_\sigma)^m\wt{L}_{ij}$ satisfy the following properties.

(1) $(\sigma\pa_\sigma)^mL_{00}, (\sigma\pa_\sigma)^m\wt{L}_{01}, (\sigma\pa_\sigma)^m\wt{L}_{02}, (\sigma\pa_\sigma)^m\wt{L}_{20}$ are $\epsilon$-regular at $\sigma=0$.

(2) $(\sigma\pa_\sigma)^m\wt{L}_{10}, (\sigma\pa_\sigma)^m\wt{L}_{12}, (\sigma\pa_\sigma)^m\wt{L}_{21}, (\sigma\pa_\sigma)^m\wt{L}_{22}$ have an $\epsilon$-regular expansion up to order one, that is, $(\sigma\pa_\sigma)^m\wt{L}^e_{10}, (\sigma\pa_\sigma)^m\wt{L}^e_{12}, (\sigma\pa_\sigma)^m\wt{L}^e_{21}, (\sigma\pa_\sigma)^m\wt{L}^e_{22}$ are $\epsilon$-regular at $\sigma=0$.

(3)$(\sigma\pa_\sigma)^m\wt{L}_{11}$ has an $\epsilon$-regular expansion up to order two, that is, $(\sigma\pa_\sigma)^m\wt{L}^e_{11}$ is $\epsilon$-regular at $\sigma=0$.

The proof is analogous to that of Proposition <ref>, now by using Proposition <ref> instead of Proposition <ref>. Here, we only demonstrate the proof of $(\sigma\pa_\sigma)^m\wt{L}_{10}$ in detail because the proof of the other entries follows in a similar manner.

Recall the calculation in (<ref>)

\[
\begin{split}
&(\sigma\pa_\sigma)^m\wt{L}^e_{10}(\tilde{g}_0, \tilde{A}_0),\ (\dg^*_1, \dA^*_1)\rangle\\
&\quad=-\sum_{j+k=m}\Big\langle (\tilde{g}_0, \tilde{A}_0), i\big(\check{\mathcal{L}}_b^{(k)}(\sigma)\big)^*\Big((\sigma\pa_\sigma)^j\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa_\sigma^2\widehat{L_{b}}(0)\big)\Big)^*\big(I-(\check{L}_b(0)^{-1})^*V_b^*\big)(\check{g}^*_2,\check{A}^*_2)\Big\rangle\\
&\quad\quad+\Big\langle(\tilde{g}_0, \tilde{A}_0),\frac{1}{2}(\check{\mathcal{L}}_b^{(m)}(\sigma))^*(\pa^2_\sigma\widehat{L_{b}}(0))^*(\dg^*_2,\dA^*_2)\Big\rangle
\end{split}
\]

\[
\Big((\sigma\pa_\sigma)^j\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa_\sigma^2\widehat{L_{b}}(0)\big)\Big)^*\big(I-(\check{L}_b(0)^{-1})^*V_b^*\big)(\check{g}^*_2,\check{A}^*_2),\quad(\pa^2_\sigma\widehat{L_{b}}(0))^*(\dg^*_2,\dA^*_2)\in\sHb^{-5/2-C(a,\gamma),\,3/2-}(\CX).
\]

Since $(\tilde{g}_0, \tilde{A}_0)\in\eHb^{s-1,\ell+2}(\CX)$ with $s-m-\epsilon>4$, and by Proposition <ref>

\begin{gather*}
(\check{\mathcal{L}}_b^{(k)}(\sigma))^*\sHb^{-5/2-C(a,\gamma), \,3/2-}(\CX)\in\sHb^{-3/2-k-C(a,\gamma), -1/2-}(\CX),\quad 0\leq k\leq m, \ k\in\BN_0,\\	(\pa^j_\sigma\check{\mathcal{L}}_b^{(k)}(\sigma))^*\sHb^{-5/2-C(a,\gamma),\, 3/2-}(\CX)\in\abs{\sigma}^{-\epsilon-j+1}\sHb^{-s+1, -\ell-2}(\CX),\quad j=1,2\quad\mbox{and}\quad 0\leq k\leq m, \ k\in\BN_0
\end{gather*}

where we use the facts that $-5/2-C(a,\gamma)>-3>-s+k+\epsilon+1$ for $0\leq k\leq m$ and $-\ell-\epsilon+1<3/2$, the pairings above (and their up to second order derivatives) are well-defined. This proves the $\epsilon$-regularity of $(\sigma\pa_\sigma)\wt{L}^e_{10}$.

Next, we turn to discussing the conormal regularity of $R(b,\sigma)$. To this end, we need the following lemma, which is an analogy of Lemma <ref>.

Let $s,\ell,\epsilon,m$ be defined as in Proposition <ref>. Suppose that the operator $A(\sigma):\wt{\mathcal{K}}_i\to\mathcal{R}_j$ with $i,j=0,1,2$ has an $\epsilon$-regular expansion up to order one, i.e.,

\[
A(\sigma)=A^0+\sigma A^e(\sigma)
\]

where $A^0:\wt{\mathcal{K}}_i\to\mathcal{R}_j$ is invertible and independent of $\sigma$, and $(\sigma\pa_\sigma)^kA^e(\sigma):\wt{\mathcal{K}}_i\to\mathcal{R}_j$ is $\epsilon$-regular for any $0\leq k\leq m, k\in\BN_0$. Then for $\abs{\sigma}$ sufficiently small, $A(\sigma)$ is invertible with inverse $B(\sigma)=A(\sigma)^{-1}$, and $(\sigma\pa_\sigma)^kB(\sigma)$ has an $\epsilon$-regular expansion up to order one, that is,

\[
(\sigma\pa_\sigma)^kB(\sigma)=B^0_k+\sigma B_k^e(\sigma), \quad 0\leq k\leq m,\  k\in\BN_0
\]

where $B^0_k:\mathcal{R}_j\to\wt{\mathcal{K}}_i$ is independent of $\sigma$, and $B_k^e(\sigma):\mathcal{R}_j\to\wt{\mathcal{K}}_i$ is $\epsilon$-regular.

Similarly, if $A(\sigma)$ has uniformly bounded inverse $B(\sigma)$, and $(\sigma\pa_\sigma)^kA(\sigma)$ is $\epsilon$-regular for any $0\leq k\leq m, k\in\BN_0$, then $(\sigma\pa_\sigma)^kB(\sigma)$ is $\epsilon$-regular as well.

As established in the proof of Lemma <ref>, we have

\begin{align*}
A(\sigma)^{-1}&=\Big(I+\sum_{k=1}^\infty(-1)^k\big(\sigma (A^0)^{-1}A^e(\sigma)\big)^k\Big)(A^0)^{-1}\\
&=(A^0)^{-1}-\sigma\Big((A^0)^{-1}A^e(\sigma)\sum_{k=0}^\infty(-1)^k\big(\sigma (A^0)^{-1}A^e(\sigma)\big)^k(A^0)^{-1}\Big):=B^0+\sigma B^e(\sigma).
\end{align*}

For $1\leq k\leq m$, we have

\[
\]

A direct calculation implies that  $(\sigma\pa_\sigma)^jB^e(\sigma)$ is a sum of the operators of the type $\sigma^{-1}L\circ(A^0)^{-1}$, where $L$ is a composition of the terms of the form $\sigma(A^0)^{-1}(\sigma\pa_\sigma)^lA^e(\sigma)$ with $0\leq l\leq j$.

Then proceeding as in the proof of Lemma <ref>, we conclude that $(\sigma\pa_\sigma)^jB^e(\sigma)$ with $0\leq j\leq k$ is $\epsilon$-regular, and thus $(\sigma\pa_\sigma)^kB^e(\sigma)$ with $1\leq k\leq m$ has an $\epsilon$-regular expansion up to order one.

Since $(\sigma\pa_\sigma)^mB(\sigma)$ is a linear combination of the operators of the form

\[
\]

where $1\leq m_2\leq m, 0\leq m_1, m_2\leq m-1$ and $m_1+m_2+m_3=m$, it follows that the second statement follows by an induction argument on $m$.

Recall the expressions (<ref>) and (<ref>) of $R_{b,\sigma}=(\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1})^{-1}$

\[
\begin{split}
\begin{pmatrix}
\wt{R}_{00}(b,\sigma)&\wt{R}_{01}(b,\sigma)&\wt{R}_{02}(b,\sigma)\\
\sigma^{-1}\wt{R}_{10}(b,\sigma)&\sigma^{-2}\wt{R}_{11}(b,\sigma)&\sigma^{-1}\wt{R}_{12}(b,\sigma)\\
\wt{R}_{20}(b,\sigma)&\sigma^{-1}\wt{R}_{21}(b,\sigma)&\sigma^{-1}\wt{R}_{22}(b,\sigma)
\end{pmatrix}\\
\begin{pmatrix}
\wt{R}_{00}(b,\sigma)&\wt{R}_{01}(b,\sigma)&\wt{R}_{02}(b,\sigma)\\
\sigma^{-1}\wt{R}_{10}(b,\sigma)&\sigma^{-2}\big(\wt{R}^0_{11}(b)+\sigma\wt{R}^e_{11}(b,\sigma)\big)&\sigma^{-1}\wt{R}_{12}(b,\sigma)\\
\wt{R}_{20}(b,\sigma)&\sigma^{-1}\big(\wt{R}^0_{21}(b)+\sigma\wt{R}^e_{21}(b,\sigma)\big)&\sigma^{-1}\big(\wt{R}^0_{22}(b)+\sigma\wt{R}^e_{22}(b,\sigma)\big)
\end{pmatrix}
\end{split}
\]

\begin{equation}
\begin{split}
\wt{R}_{11}=\big(\wt{L}^\sharp_{11}-\sigma\wt{L}^\sharp_{12}(\wt{L}^\flat_{22})^{-1}\wt{L}^\sharp_{21}\big)^{-1},\quad \wt{R}_{10}=\wt{R}_{11}\big(-\wt{L}_{10}+\sigma\wt{L}_{12}^\sharp(\wt{L}^\flat_{22})^{-1}\wt{L}_{20}\big)L_{00}^{-1},\\
\wt{R}_{12}=-\wt{R}_{11}\wt{L}_{12}^\sharp(\wt{L}^\flat_{22})^{-1},\quad \wt{R}_{21}=-(\wt{L}^\flat_{22})^{-1}\wt{L}_{21}^\sharp\wt{R}_{11},\quad\wt{R}_{22}=(\wt{L}^\flat_{22})^{-1}-\sigma(\wt{L}^\flat_{22})^{-1}\wt{L}_{21}^\sharp \wt{R}_{12},\\
\wt{R}_{20}=-(\wt{L}^\flat_{22})^{-1}\big(\wt{L}_{20}L_{00}^{-1}+\wt{L}_{21}^\sharp\wt{R}_{10}\big),\quad \wt{R}_{01}=-L_{00}^{-1}\big(\wt{L}_{01}\wt{R}_{11}+\wt{L}_{02}\wt{R}_{21}\big),\\
\wt{R}_{02}=-L_{00}^{-1}\big(\sigma\wt{L}_{01}\wt{R}_{12}+\wt{L}_{02}\wt{R}_{22}\big),\quad \wt{R}_{00}=L_{00}^{-1}\big(I-\sigma\wt{L}_{01}\wt{R}_{10}-\sigma\wt{L}_{02}\wt{R}_{20}\big)
\end{split}
\end{equation}

\begin{equation}
\begin{split}
&\wt{L}^\sharp_{i1}=\wt{L}_{i1}-\sigma\wt{L}_{i0}L_{00}^{-1}\wt{L}_{01}\quad\mbox{for}\quad i=1,2, \quad \wt{L}^\sharp_{12}=\wt{L}_{12}-\wt{L}_{10}L_{00}^{-1}\wt{L}_{02},\\
&\qquad\qquad\qquad\qquad	\wt{L}_{22}^\flat=\wt{L}_{22}-\sigma\wt{L}_{20}L_{00}^{-1}\wt{L}_{02}.
\end{split}
\end{equation}

Let $s,\ell,\epsilon,m$ be defined as in Proposition <ref>. Then $(\sigma\pa_\sigma)^m\wt{R}_{ij}$ satisfy the following properties.

(1) $(\sigma\pa_\sigma)^m\wt{R}_{00}, (\sigma\pa_\sigma)^m\wt{R}_{01}, (\sigma\pa_\sigma)^m\wt{R}_{02}, (\sigma\pa_\sigma)^m\wt{R}_{10}, (\sigma\pa_\sigma)^m\wt{R}_{12}, (\sigma\pa_\sigma)^m\wt{R}_{20}$ are $\epsilon$-regular at $\sigma=0$.

(2) $(\sigma\pa_\sigma)^m\wt{R}_{11}, (\sigma\pa_\sigma)^m\wt{R}_{21}, (\sigma\pa_\sigma)^m\wt{R}_{22}$ have an $\epsilon$-regular expansion up to order one, that is, $(\sigma\pa_\sigma)^m\wt{R}^e_{11}$, $(\sigma\pa_\sigma)^m\wt{R}^e_{21}, (\sigma\pa_\sigma)^m\wt{R}^e_{22}$ are $\epsilon$-regular at $\sigma=0$.

Applying Proposition <ref> and Lemma <ref>, we conclude that for $0\leq k\leq m$, $(\sigma\pa_\sigma)^kL_{00}^{-1}$ is $\epsilon$-regular. Then by Proposition <ref>, we find that $(\sigma\pa_\sigma)^k\wt{L}^\sharp_{11}, (\sigma\pa_\sigma)^k\wt{L}^\sharp_{21}, (\sigma\pa_\sigma)^k\wt{L}^\flat_{22}$ (and thus $(\sigma\pa_\sigma)^k(\wt{L}^\flat_{22})^{-1}$ by Lemma <ref>) have an $\epsilon$-regular expansion up to order one, while $(\sigma\pa_\sigma)^k\wt{L}^\sharp_{12}$ is $\epsilon$-regular. Therefore, we see that

\[
\]

has an $\epsilon$-regular expansion up to order one whose coefficient of $\sigma^0$ is $(\sigma\pa_\sigma)^k\wt{L}_{11}^0(b)$. Then the $\epsilon$-regular expansion property of $(\sigma\pa_\sigma)^m\wt{R}_{11}$ follows from Lemma <ref>. Similarly, the $\epsilon$-regular expansion property of $(\sigma\pa_\sigma)^m\wt{R}_{21},(\sigma\pa_\sigma)^m\wt{R}_{22}$ follows from that of $(\sigma\pa_\sigma)^k\wt{R}_{11}, (\sigma\pa_\sigma)^k(\wt{L}^\flat_{22})^{-1},(\sigma\pa_\sigma)^m\wt{L}^\sharp_{21}$ and the above expressions of $\wt{R}_{ij}$. Finally, the $\epsilon$-regularity of the remaining components can be obtained by using the above expressions of $\wt{R}_{ij}$ and Proposition <ref>.

Let $L^-_b(\sigma)$ be the regular part of $\widehat{L_{b}}(\sigma)^{-1}$ defined in Corollary <ref> and define

\[
\mathcal{L}^{-(m)}_b(\sigma):=(\sigma\pa_\sigma)^mL^-_b(\sigma).
\]

Let $-\frac 32<\ell<-\frac12, 0<\epsilon<1, -\frac12<\ell+\epsilon< \frac12$ and $s-\epsilon>4+m$ where $m\in\BN_0$. For $ \IM\sigma\geq0$ and $0<\abs{\sigma}\leq c_0$ where $c_0>0$ is a small constant, $\mathcal{L}^{-(m)}_b(\sigma)$ satisfies

\begin{equation}\label{eq:conormalbound}
\mathcal{L}^{-(m)}_b(\sigma):\eHb^{s-1,\ell+2}(\CX)\to\eHb^{s-m,\ell+\epsilon-1}(\CX)
\end{equation}

\begin{equation}\label{eq:conormalholderreg}
\pa_\sigma^j\mathcal{L}^{-(m)}_b(\sigma):\eHb^{s-1,\ell+2}(\CX)\to\abs{\sigma}^{-\epsilon-j+1}\eHb^{s-m-\epsilon-j+1,\ell+\epsilon-1}(\CX),\quad j=1,2.
\end{equation}

Recall that

\[
\]

where $
P^2(b), P^1(b):\eHb^{s-1,\ell+2}(\CX)\to\rho C^\infty(\CX)+\eHb^{\infty, \frac12-}(\CX)$,
\[(\sigma\pa_\sigma)^kP^e(b,\sigma):\eHb^{s-1,\ell+2}(\CX)\to\mathcal{K}_{b,1}
\]

is $\epsilon$-regular for $0\leq k\leq m$, and

\[
\wt{R}_{00}(b,\sigma)&\wt{R}_{01}(b,\sigma)&\wt{R}_{02}(b,\sigma)\\
\wt{R}_{20}(b,\sigma)&\wt{R}^e_{21}(b,\sigma)&\wt{R}^e_{22}(b,\sigma)
\end{pmatrix}.
\]

Then a direct calculation implies $
(\sigma\pa_\sigma)^mL^-_b(\sigma)$ is a linear combination of the operators of the forms

\begin{gather*}
\check{\mathcal{L}}_b^{(m_1)}(\sigma)\circ(\sigma\pa_\sigma)^{m_2}R_{\mathrm{reg}}(b,\sigma),\quad\check{\mathcal{L}}_b^{(m_1)}(\sigma)\circ\pa^2_\sigma \widehat{L_{b}}(0)\circ P^2(b),\\
\check{\mathcal{L}}_b^{(m_1)}(\sigma)\circ(\sigma\pa_\sigma)^{m_2}\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa_\sigma^2\widehat{L_{b}}(0)\big)\circ(\sigma\pa_\sigma)^{m_3}\big(P^1(b)+P^e(b,\sigma)\big).
\end{gather*}

Applying Proposition <ref> and <ref>, we obtain (<ref>) and (<ref>).

§ DECAY ESTIMATES

In this section, we continue using the notation from <ref>. We will study the decay of the solution $(\dg, \dA)$ to the equations

\begin{equation}\label{eq:mlgEMeqn}
L_b(\dg, \dA)=f,\quad f\in C_c^\infty((0,\infty)_{t_{b,*}};\eHb^{s,\ell+2}(\CX)).
\end{equation}

We define spacetime Sobolev spaces

\begin{equation}
\wt{H}^{s,\ell,k}_{\bop,\cop},\quad k\in\BN_0,
\end{equation}

which is equal to $L^2(t_{b,*}^{-1}([0, \infty));\abs{dg_b})$ for $s,\ell,k=0$. Here, the index $s$ measures the regularity with respect to the derivatives $\pa_{t_{b,*}}$ and the stationary $b$-vector fields on $\CX$ (which is spanned by $r\pa_r$ and rotation vector fields). The index $\ell$ is the weight in $\rho=r^{-1}$, that is, $	\wt{H}^{s,\ell,k}_{\bop,\cop}=\rho^\ell	\wt{H}^{s,k}_{\bop,\cop}$. The index $k\in\BN_0$ measures the regularity with respect to $\jb{t_{b,*}}\pa_{t_{b,*}}$ (which is related to the conormal regularity of $\widehat{L_{b}}(\sigma)^{-1}$ on the Fourier transform side). Therefore, $u\in	\wt{H}^{s,\ell,k}_{\bop,\cop}$ if and only if $(\jb{t_{b,*}}\pa_{t_{b,*}})^ju\in	\wt{H}^{s-j,\ell}_{\bop,\cop}:=\wt{H}^{s-j,\ell,0}_{\bop,\cop}$ for any $0\leq j\leq k$.

Let $-\frac 32<\ell<-\frac12, 0<\epsilon<1, -\frac12<\ell+\epsilon< \frac12$ and $s-\epsilon>4+m$ where $m\in\BN_0$. Let $(\dg, \dA)$ be the solution to the equation (<ref>) with $f\in C_c^\infty((0,\infty)_{t_{b,*}};\eHb^{s,\ell+2}(\CX))$. Then there exists a generalized zero mode $(\hat{g}, \hat{A})\in\widehat{\mathcal{K}}_b$ (defined in Proposition <ref>) such that

\begin{equation}
(\dg, \dA)=(\hat{g}, \hat{A})+(\bar{g}, \bar{A})+(\tilde{g}, \tilde{A})
\end{equation}

and  $(\hat{g}, \hat{A}), (\bar{g}, \bar{A})$ and $(\tilde{g}, \tilde{A})$ satisfy the following properties.

* The term $(\bar{g}, \bar{A})$ can be written as  $
(\bar{g}, \bar{A})=(\bar{g}_1, \bar{A}_1)+(\bar{g}_2, \bar{A}_2)$ where $(\bar{g}_1, \bar{A}_1)$ is a pure gauge solution. Moreover, for $m\geq 1$ and $j\geq 0$, we have the pointwise estimates

\begin{equation}\label{eq:estimatesforbar1}
\abs{\big(\jb{t_{b,*}}\pa_{t_{b,*}}\big)^{m-1}V^j(\bar{g}_1, \bar{A}_1)}\lesssim
\big(\jb{t_{b,*}+r}^{-2+\epsilon}+\jb{t_{b,*}}^{-1+\epsilon}\jb{r}^{-2}\big)\norm{f}_{\jb{t_{b,*}}^{-\frac 32-}	\wt{H}^{s-1,\ell+2}_{\bop,\cop}}
\end{equation}

\begin{equation}\label{eq:estimatesforgbar2}
\abs{\big(\jb{t_{b,*}}\pa_{t_{b,*}}\big)^{m-1}V^j(\bar{g}_2,\bar{A}_2)}\lesssim
\jb{t_{b,*}+r}^{-2+\epsilon}\norm{f}_{\jb{t_{b,*}}^{-\frac 32-}	\wt{H}^{s-1,\ell+2}_{\bop,\cop}}
%\abs{\big(\jb{t_{b,*}}\pa_{t_{b,*}}\big)^{m-1}V^j \bar{A}_2}&\lesssim
%\Big(\jb{t_{b,*}+r}^{-2+\epsilon}\jb{r}^{-1}+	\jb{t_{b,*}+r}^{-2}\jb{t_{b,*}}^{-1+\epsilon}\Big)\norm{f}_{\jb{t_{b,*}}^{-\frac 32-}	\wt{H}^{s,\ell+2}_{\bop,\cop}}
\end{equation}

where $V^j$ denotes the product of $j$ vector fields $V\in\{\pa_{t_{b,*}}, r\pa_r, \mbox{ rotation fields }\}$.

* $(\tilde{g}, \tilde{A})$ satisfies

\begin{equation}
\norm{(\tilde{g}, \tilde{A})}_{\jb{t_{b,*}}^{-\frac 32+\epsilon}	\wt{H}^{s-2,\ell+\epsilon-1,m}_{\bop,\cop}}\lesssim\norm{f}_{\jb{t_{b,*}}^{-\frac 72+\epsilon}	\wt{H}^{s+m,\ell+2,m}_{\bop,\cop}}.
\end{equation}

For $m\geq 1$, we also have

\begin{equation}
\norm{\big(\jb{t_{b,*}}\pa_{t_{b,*}}\big)^{m-1}(\tilde{g}, \tilde{A})}_{\jb{t_{b,*}}^{-2+\epsilon}	L^\infty(\BR_{t_{b,*}};\eHb^{s-2-m,\ell+\epsilon-1}(\CX))}\lesssim\norm{f}_{\jb{t_{b,*}}^{-\frac 72+\epsilon}	\wt{H}^{s+m,\ell+2,m}_{\bop,\cop}}.
\end{equation}

For $m\geq 1$ and $0\leq j<s-4-m$ with $j\in\BN_0$, we have the pointwise decay estimate

\begin{equation}
\abs{\big(\jb{t_{b,*}}\pa_{t_{b,*}}\big)^{m-1}V^j(\tilde{g}, \tilde{A})}\lesssim\jb{t_{b,*}}^{-2+\epsilon}\jb{r}^{-\frac12-\ell-\epsilon}\norm{f}_{\jb{t_{b,*}}^{-\frac 72+\epsilon}	\wt{H}^{s+m,\ell+2,m}_{\bop,\cop}}.
\end{equation}

* Recall the definition of $\mathcal{K}_{b,1}, \mathcal{K}_{b,2}$ and $\mathcal{K}^*_{b}$ at the beginning of <ref>. The leading term $(\hat{g}, \hat{A})\in\widehat{K}_b$ is uniquely determined by $f$ and can be expressed as follows:

\begin{equation}
(\hat{g}, \hat{A})=-\frac{1}{2}\big(t_{b,*}(\dg_1,\dA_1)+(\check{g}_1, \check{A}_1)\big)
+\frac12\Big((\dg'_1, \dA'_1)+(\dg''_1, \dA''_1)+(\dg_2, \dA_2)\Big)
\end{equation}

where $(\dg_1,\dA_1), (\dg_1',\dA_1'),(\dg''_1, \dA''_1)\in\mathcal{K}_{b,1}$, and $(\dg_2,\dA_2)\in\mathcal{K}_{b,2}$. Moreover, $(\dg_1,\dA_1), (\dg_1',\dA_1')$ and $(\dg_2,\dA_2)$ are determined by the conditions (<ref>) and (<ref>) with $f$ there replaced by $\int_{\BR}f(t_{b,*})\,dt_{b,*}$, and $(\dg''_1, \dA''_1)\in\mathcal{K}_{b,1}$ is uniquely determined by the condition

\begin{align}\notag
&-\Big\langle[L_b, t_{b,*}](\check{g}''_1, \check{A}''_1)+\frac{1}{2}[[L_b, t_{b,*}], t_{b,*}](\dg''_1,\dA''_1), (\dg^*, \dA^*)\Big\rangle+\Big\langle[L_b, t_{b,*}](\dg''_2, \dA''_2), (\dg^*, \dA^*)\Big\rangle\\
&\quad\quad=\langle\int_{\BR} t_{b,*}f(t_{b,*})\,dt_{b,*}, \ (\dg^*, \dA^*)\rangle\quad\mbox{for all}\quad(\dg^*,\dA^*)\in\mathcal{K}_b^*.
\end{align}

for some $(\dg''_2, \dA''_2)\in\mathcal{K}_{b,2}$.

The strategy of the proof is to take Fourier transform of the inhomogeneous equation $L_b(\dg,\dA)=f$ in $t_{b,*}$. We define the Fourier transform of $(\dg, \dA)$ as

\begin{equation}
(\hat{\dg}(\sigma), \hat{\dA}(\sigma)):=\int_{\BR}e^{it_{b,*}\sigma}(\dg, \dA)(t_{b,*})\,dt_{b,*}
\end{equation}

and likewise for $f$ (here we drop the notation for the spatial variables.).

First, according to the energy estimate in Proposition <ref>, we see that the solution $(\dg,\dA)$ satisfies
\[\norm{(\dg, \dA)(t_{b,*},\cdot)}_{\eHb^{1,-1}(\CX)}\lesssim e^{Mt_{b,*}}
\]

for some $M>0$. This implies that the Fourier transform $(\hat{\dg}(\sigma), \hat{\dA}(\sigma))$ of $(\dg, \dA)$
is well defined for $\IM\sigma>M$.
Since $f(t_{b,*})$ has compact support in $t_{b,*}$, $\hat{f}(\sigma)$ is holomorphic in the full plane $\sigma\in\BC$ and decays superpolynomially  as $\abs{\RE\sigma}\to\infty$ with $\IM\sigma>0$ fixed.  Fourier transforming of the equation $L_b(\dg,\dA)=f$, we obtain

\begin{equation}
\widehat{L_{b}}(\sigma)(\hat{\dg}(\sigma), \hat{\dA}(\sigma))=\hat{f}(\sigma),\quad 	\IM\sigma>M.
\end{equation}

In view of Theorem <ref>, we see that $\widehat{L_{b}}(\sigma)$ is invertible in $\IM\sigma>M$ and

\[
(\hat{\dg}(\sigma), \hat{\dA}(\sigma))=\widehat{L_{b}}(\sigma)^{-1}\hat{f}(\sigma),\quad 	\IM\sigma>M.
\]

Therefore, we have the following integral representation

\begin{equation}\label{eq:integralrepsoln}
(\dg(t_{b,*}), \ \dA(t_{b,*}))=
\frac{1}{2\pi}\int_{\IM\sigma=M+1}\!e^{-i\sigma t_{b,*}}\widehat{L_{b}}(\sigma)^{-1}\hat{f}(\sigma)\,d\sigma.
\end{equation}

Owing to Theorem <ref>, we see that the integrand in the above integral representation is holomorphic in $\sigma$ (with values in $\eHb^{s,\ell}(\CX)$) in upper half plane $\IM\sigma>0$, and decay superpolynomially in $\abs{\RE\sigma}$ as with $\IM\sigma$ bounded. Using Cauchy's Theorem, we obtain

\begin{equation}\label{eq:integralrepsoln2}
(\dg(t_{b,*}), \ \dA(t_{b,*}))=
\frac{1}{2\pi}\int_{\IM\sigma=C}\!e^{-i\sigma t_{b,*}}\widehat{L_{b}}(\sigma)^{-1}\hat{f}(\sigma)\,d\sigma.
\end{equation}

for any $C>0$.

Now we let $C\to 0+$.

Fixing a frequency cutoff $\chi(s)\in C_c^\infty(\BR)$ such that $\chi(s)=1$ for $\abs{s}\leq c_0/2$ and $\chi(s)=0$ for $\abs{s}\geq c_0$ where $c_0$ is defined as in Theorem <ref>. We now write

\[
\hat{f}(\sigma)=\chi(\RE\sigma)\hat{f}(\sigma)+(1-\chi(\RE\sigma))\hat{f}(\sigma):=\hat{f}_1(\sigma)+\hat{f}_2(\sigma).
\]

Correspondingly, by Theorem <ref> and Corollary <ref>, we split $\widehat{L_{b}}(\sigma)^{-1}\hat{f}(\sigma)$ into two parts

\[
\widehat{L_{b}}(\sigma)^{-1}\hat{f}(\sigma)=(\hat{\dg}_{\mathrm{reg}}(\sigma), \hat{\dA}_{\mathrm{reg}}(\sigma))+(\hat{\dg}_{\mathrm{sing}}(\sigma), \hat{\dA}_{\mathrm{sing}}(\sigma))
\]

\[
(\hat{\dg}_{\mathrm{reg}}(\sigma), \hat{\dA}_{\mathrm{reg}}(\sigma))=L^-_b(\sigma)\hat{f}_1(\sigma)+\widehat{L_{b}}(\sigma)^{-1}\hat{f}_2(\sigma),\quad (\hat{\dg}_{\mathrm{sing}}(\sigma), \hat{\dA}_{\mathrm{sing}}(\sigma))=P(b,\sigma)\hat{f}_1(\sigma).
\]

As discussed above, we can shift the integration contour to $\IM\sigma=C$ for any $C>0$. Letting $C\to0+$, we obtain

\[
(\dg(t_{b,*}), \dA(t_{b,*}))=(\dg_{\mathrm{reg}}(t_{b,*}),\dA_{\mathrm{reg}}(t_{b,*}))+(\dg_{\mathrm{sing}}(t_{b,*}),\dA_{\mathrm{sing}}(t_{b,*}))
\]

\[
(\dg_{\mathrm{reg}}(t_{b,*}),\dA_{\mathrm{reg}}(t_{b,*}))=\frac{1}{2\pi}\int_{\BR}\!e^{-i\sigma t_{b,*}} (\hat{\dg}_{\mathrm{reg}}(\sigma), \hat{\dA}_{\mathrm{reg}}(\sigma))\,d\sigma
\]

\[
(\dg_{\mathrm{sing}}(t_{b,*}),\dA_{\mathrm{sing}}(t_{b,*}))=\lim_{C\to0+}\frac{1}{2\pi}\int_{\IM\sigma=C}\!e^{-i\sigma t_{b,*}} (\hat{\dg}_{\mathrm{sing}}(\sigma), \hat{\dA}_{\mathrm{sing}}(\sigma))\,d\sigma.
\]

* Analysis of $(\dg_{\mathrm{reg}}(t_{b,*}),\dA_{\mathrm{reg}}(t_{b,*}))$. Since $f\in C_c^\infty((0,\infty)_{t_{b,*}};\eHb^{s,\ell+2}(\CX))$, we see that $\hat{f}_1(\sigma), \hat{f}_2(\sigma)\in H^{s'}(\BR_{\sigma};\eHb^{s,\ell+2}(\CX))$ for any $s'\in\BR$. As $H^{3/2-\epsilon'}(\BR)$ is an algebra for any $0<\epsilon'<1$, by Theorem <ref>, we have

\[
L^-_b(\sigma)\hat{f}_1(\sigma)\in H^{\frac32-\epsilon'}((-c_0,c_0);\eHb^{s+1-\max\{\epsilon, \frac 12\}, \ell+\epsilon-1}(\CX)),\quad 0<\epsilon<\epsilon'<1.
\]

Using Moser estimate, we bound norm of $	L^-_b(\sigma)\hat{f}_1(\sigma)$ as follows

\begin{align*}
&	\norm{L^-_b(\sigma)\hat{f}_1(\sigma)}_{H^{\frac32-\epsilon'}(\BR_\sigma;\eHb^{s+1-\max\{\epsilon, \frac 12\}, \ell+\epsilon-1}(\CX))}\lesssim \norm{\hat{f}_1(\sigma)}_{H^{\frac32-\epsilon'}(\BR_\sigma;\eHb^{s, \ell+2}(\CX))}\\
&\quad \lesssim\norm{\hat{f}(\sigma)}_{H^{\frac32-\epsilon'}(\BR_\sigma;\eHb^{s, \ell+2}(\CX))}=\norm{\jb{t_{b,*}}^{\frac32-\epsilon'}f}_{L^2(\BR_{t_{b,*}}; \eHb^{s,\ell+2}(\CX))}.
\end{align*}

By exploiting the conormal regularity of $L^-_b(\sigma)$ proved in Proposition <ref>, we find that for $0\leq k\leq m$,

\[
(\sigma\pa_\sigma)^k\big(L^-_b(\sigma)\hat{f}_1(\sigma)\big)\in H^{\frac32-\epsilon'}((-c_0,c_0);\eHb^{s+1-k-\max\{\epsilon, \frac 12\}, \ell+\epsilon-1}(\CX)), \quad 0<\epsilon<\epsilon'<1.
\]

In a similar manner, we arrive at

\begin{equation}\label{eq:esofhatreg1}
\begin{split}
&\norm{(\sigma\pa_\sigma)^k\big(L^-_b(\sigma)\hat{f}_1(\sigma)\big)}_{H^{\frac32-\epsilon'}(\BR_\sigma;\eHb^{s+1-k-\max\{\epsilon, \frac 12\}, \ell+\epsilon-1}(\CX))}\\
&\quad\lesssim\sum_{j=0}^k\norm{(\sigma\pa_\sigma)^j\hat{f}_1(\sigma)}_{H^{\frac32-\epsilon'}(\BR_\sigma;\eHb^{s-j, \ell+2}(\CX))}\lesssim\sum_{j=0}^k\norm{(\sigma\pa_\sigma)^j\hat{f}(\sigma)}_{H^{\frac32-\epsilon'}(\BR_\sigma;\eHb^{s-j, \ell+2}(\CX))}\\
&\quad=\sum_{j=0}^k\norm{(t_{b,*}\pa_{t_{b,*}})^j\jb{t_{b,*}}^{\frac32-\epsilon'}f}_{L^2(\BR_{t_{b,*}}; \eHb^{s-j,\ell+2}(\CX))}\lesssim \norm{f}_{\jb{t_{b,*}}^{-\frac 32+\epsilon'}	\wt{H}^{s,\ell+2,k}_{\bop,\cop}}.
\end{split}
\end{equation}

As for the second part $\widehat{L_{b}}(\sigma)^{-1}\hat{f}_2(\sigma)$ in 	 $(\hat{\dg}_{\mathrm{reg}}(\sigma), \hat{\dA}_{\mathrm{reg}}(\sigma))$,
by Theorem <ref>, we have

\begin{equation}\label{eq:improvementforfreawar0}
\widehat{L_{b}}(\sigma)^{-1}\in W^{j,\infty}\Big(\BR_{\sigma}\setminus[-\frac{c_0}{2}, \frac{c_0}{2}];\mathcal{L}\big(\bar{H}^{s,\ell+2}_{\bop,h}, h^{-j}\bar{H}^{s-j,\ell}_{\bop,h}\big)\Big), \quad j\in\BN_0
\end{equation}

where we take $h=\jb{\sigma}^{-1}$. Taking $j=2$ gives

\[
\widehat{L_{b}}(\sigma)^{-1}\hat{f}_2(\sigma)\in  H^{\frac32-\epsilon}(\BR; \jb{\sigma}^{-s+2}\bar{H}_{\bop, \jb{\sigma}^{-1}}^{s-2,\ell}(\CX)).
\]

whose norm is bounded by

\[
\norm{\hat{f}_2(\sigma)}_{H^{\frac32-\epsilon}(\BR; \jb{\sigma}^{-s}\bar{H}_{\bop, \jb{\sigma}^{-1}}^{s,\ell+2}(\CX))}\lesssim \norm{f}_{\jb{t_{b,*}}^{-\frac 32+\epsilon}	\wt{H}^{s,\ell+2}_{\bop,\cop}}.
\]

More generally, we have

\[
(\sigma\pa_\sigma)^k\big(\widehat{L_{b}}(\sigma)^{-1}\hat{f}_2(\sigma)\big)\in  H^{\frac32-\epsilon}(\BR; \jb{\sigma}^{-s+2+2k}\bar{H}_{\bop, \jb{\sigma}^{-1}}^{s-2-k,\ell}(\CX)).
\]

\begin{equation}\label{eq:esofhatreg2}
\begin{split}
&\norm{(\sigma\pa_\sigma)^k\big(\widehat{L_{b}}(\sigma)^{-1}\hat{f}_2(\sigma)\big)}_{H^{\frac 32-\epsilon}(\BR; \jb{\sigma}^{-s+2+k}\bar{H}_{\bop, \jb{\sigma}^{-1}}^{s-2-k,\ell}(\CX))}\\
&\quad\lesssim\sum_{j=0}^k\norm{(\sigma\pa_\sigma)^j\hat{f}_2(\sigma)}_{H^{\frac32-\epsilon}(\BR; \jb{\sigma}^{-s-k+2j}\bar{H}_{\bop, \jb{\sigma}^{-1}}^{s+k-2j,\ell+2}(\CX))}\\
&\quad\lesssim\sum_{j=0}^k\norm{(\sigma\pa_\sigma)^j\hat{f}(\sigma)}_{H^{\frac32-\epsilon}(\BR; \jb{\sigma}^{-s-k+2j}\bar{H}_{\bop, \jb{\sigma}^{-1}}^{s+k-2j,\ell+2}(\CX))}\\
&\quad\lesssim\sum_{j=0}^k\norm{(t_{b,*}\pa_{t_{b,*}})^jf}_{\jb{t_{b,*}}^{-\frac 32+\epsilon}	\wt{H}^{s+k-2j,\ell+2}_{\bop,\cop}}
\lesssim \norm{f}_{\jb{t_{b,*}}^{-\frac 32+\epsilon}	\wt{H}^{s+k,\ell+2,k}_{\bop,\cop}}.
\end{split}
\end{equation}

Taking inverse Fourier transform of $(\hat{\dg}_{\mathrm{reg}}(\sigma), \hat{\dA}_{\mathrm{reg}}(\sigma))$ and using the estimates (<ref>) and (<ref>), we obtain for $0\leq k\leq m$ and $0<\epsilon<\epsilon'<1$

\begin{equation}\label{eq:L2esofgAreg}
\begin{split}
&	\norm{(\dg_{\mathrm{reg}}(t_{b,*}),\dA_{\mathrm{reg}}(t_{b,*}))}_{\jb{t_{b,*}}^{-\frac 32+\epsilon'}\wt{H}^{s-2,\ell+\epsilon-1,k}_{\bop,\cop}}\\
&\quad\sim\sum_{j=0}^k\norm{(\sigma\pa_\sigma)^j(\hat{\dg}_{\mathrm{reg}}(\sigma), \hat{\dA}_{\mathrm{reg}}(\sigma))}_{H^{\frac 32-\epsilon'}(\BR; \jb{\sigma}^{-s+2+j}\bar{H}_{\bop, \jb{\sigma}^{-1}}^{s-2-j,\ell+\epsilon-1}(\CX))}\\
&\quad\lesssim\sum_{j=0}^k\norm{(\sigma\pa_\sigma)^j\big(L^-_b(\sigma)\hat{f}_1(\sigma)\big)}_{H^{\frac32-\epsilon'}(\BR_\sigma;\eHb^{s+1-j-\max\{\epsilon, \frac 12\}, \ell+\epsilon-1}(\CX))}\\
&\quad\quad+\sum_{j=0}^k\norm{(\sigma\pa_\sigma)^j\big(\widehat{L_{b}}(\sigma)^{-1}\hat{f}_2(\sigma)\big)}_{H^{\frac 32-\epsilon'}(\BR; \jb{\sigma}^{-s+2+j}\bar{H}_{\bop, \jb{\sigma}^{-1}}^{s-2-j,\ell}(\CX))}\\
\lesssim \norm{f}_{\jb{t_{b,*}}^{-\frac 32+\epsilon'}	\wt{H}^{s+k,\ell+2,k}_{\bop,\cop}}.
\end{split}
\end{equation}

Now we integrate $\pa_{t_{b,*}}(\dg_{\mathrm{reg}}(t_{b,*}),\dA_{\mathrm{reg}}(t_{b,*}))=t_{b,*}^{-1}(t_{b,*}\pa_{t_{b,*}})(\dg_{\mathrm{reg}}(t_{b,*}),\dA_{\mathrm{reg}}(t_{b,*}))$ and exploit the above estimate (<ref>) for $k=1$ to obtain the decay estimate of $(\dg_{\mathrm{reg}}(t_{b,*}),\dA_{\mathrm{reg}}(t_{b,*}))$ in $t_{b,*}$. Concretely, for $t_{b,*}\geq 0$ and $0<\epsilon<\epsilon'<1$, we have

\begin{align*}
&	\norm{(\dg_{\mathrm{reg}}(t_{b,*}),\dA_{\mathrm{reg}}(t_{b,*}))}_{\eHb^{s-3,\ell+\epsilon-1}(\CX)}=	\norm{\int_{t_{b,*}}^\infty\!\pa_s\big(\dg_{\mathrm{reg}}(s),\dA_{\mathrm{reg}}(s)\big)\,ds}_{\eHb^{s-3,\ell+\epsilon-1}(\CX)}\\
&\quad\lesssim\Big(\int_{t_{b,*}}^\infty\!\jb{s}^{-3+2\epsilon'}\jb{s}^{-2}\,ds\Big)^{\frac{1}{2}}	\norm{(\jb{t_{b,*}}\pa_{t_{b,*}})(\dg_{\mathrm{reg}}(t_{b,*}),\dA_{\mathrm{reg}}(t_{b,*}))}_{\jb{t_{b,*}}^{-\frac 32+\epsilon'}	\wt{H}^{s-3,\ell+\epsilon-1}_{\bop,\cop}}\\
&\quad\lesssim\jb{t_{b,*}}^{-2+\epsilon'}\norm{f}_{\jb{t_{b,*}}^{-\frac 32+\epsilon'}	\wt{H}^{s+1,\ell+2,1}_{\bop,\cop}}.
\end{align*}

Since $(\dg_{\mathrm{reg}}(t_{b,*}),\dA_{\mathrm{reg}}(t_{b,*}))$ has support in $t_{b,*}\geq -C$ for some $C>0$, it follows that

\[
\norm{(\dg_{\mathrm{reg}}(t_{b,*}),\dA_{\mathrm{reg}}(t_{b,*}))}_{\jb{t_{b,*}}^{-2+\epsilon}	L^\infty(\BR_{t_{b,*}};\eHb^{s-3,\ell+\epsilon-1}(\CX))}\lesssim\norm{f}_{\jb{t_{b,*}}^{-\frac 32+\epsilon'}	\wt{H}^{s+1,\ell+2,1}_{\bop,\cop}}.
\]

By the same reasoning, the above decay estimate can be generalized for all $0\leq k\leq m$,

\begin{equation}\label{eq:LinftyesofgAreg}
\norm{\big(\jb{t_{b,*}}\pa_{t_{b,*}}\big)^{k-1}(\dg_{\mathrm{reg}}(t_{b,*}),\dA_{\mathrm{reg}}(t_{b,*}))}_{\jb{t_{b,*}}^{-2+\epsilon}	L^\infty(\BR_{t_{b,*}};\eHb^{s-2-k,\ell+\epsilon-1}(\CX))}\lesssim\norm{f}_{\jb{t_{b,*}}^{-\frac 32+\epsilon'}	\wt{H}^{s+k,\ell+2,k}_{\bop,\cop}}.
\end{equation}

The estimates (<ref>) and (<ref>) imply that $(\dg_{\mathrm{reg}}(t_{b,*}),\dA_{\mathrm{reg}}(t_{b,*}))$ occurs as a part of the remainder $(\tilde{g}, \tilde{A})$.

* Analysis of $(\dg_{\mathrm{sing}}(t_{b,*}),\dA_{\mathrm{sing}}(t_{b,*}))$.
We now turn to discussing the singular part

\[
(\dg_{\mathrm{sing}}(t_{b,*}),\dA_{\mathrm{sing}}(t_{b,*}))=\lim_{C\to0+}\frac{1}{2\pi}\int_{\IM\sigma=C}\!e^{-i\sigma t_{b,*}} P(b,\sigma)\hat{f}_1(\sigma)\,d\sigma.
\]

Since $\hat{f}$ is holomorphic in the full plane $\sigma\in\BC$, we write

\[
\hat{f}_1(\sigma)=\chi(\RE\sigma)\hat{f}(\sigma)
=\chi(\RE\sigma)\big(f^{(0)}+i\sigma f^{(1)}+\sigma^2f^{(2)}(\sigma))=\chi(\RE\sigma)\big(f^{(0)}+i\sigma f^{(1)}\big)+\sigma^2\hat{f}_3(\sigma)
\]

\[
f^{(0)}=\hat{f}(0)=\int_{\BR} f(t_{b,*})\,dt_{b,*},\quad f^{(1)}=-i\pa_\sigma\hat{f}(0)=\int_{\BR} t_{b,*}f(t_{b,*})\,dt_{b,*},
\]

and $f^{(2)}(\sigma)$ is holomorphic in the full plane $\sigma\in\BC$. Moreover, by Sobolev embedding theorem, we have for $j=0,1$

\[
\norm{f^{(j)}}_{\eHb^{s,\ell+2}(\CX)}\lesssim \norm{\hat{f}(\sigma)}_{H^{3/2+s'}(\BR_\sigma; \eHb^{s,\ell+2}(\CX))}\lesssim\norm{f}_{\jb{t_{b,*}}^{-\frac32-s'}\wt{H}^{s,\ell}_{\bop,\cop}}\quad\mbox{for any}\quad s'>0.
\]

As for $\hat{f}_3(\sigma)$, since $\hat{f}_3(\sigma)=\sigma^{-2}(\hat{f}_1(\sigma)-\chi(\RE\sigma)f^{(0)}-i\sigma \chi(\RE\sigma)f^{(1)})$, it follows that $\hat{f}_3(\sigma)\in C_c^\infty(\BR_\sigma)$ and

\[
\norm{\hat{f}_3(\sigma)}_{H^{s'}(\BR_\sigma;\eHb^{s,\ell+2}(\CX))}\lesssim \norm{\hat{f}(\sigma)}_{H^{s'+2}(\BR_\sigma;\eHb^{s,\ell+2}(\CX))},\quad s'>-\frac12.
\]

Recall that $P(b,\sigma)=\sigma^{-2}P^2(b)+\sigma^{-1} P^1(b)+\sigma^{-1}P^e(\sigma)$ where $P^2(b), P^1(b):\eHb^{s,\ell+2}\to\eHb^{\infty, -1/2-}(\CX)$ and $P^e(b,\sigma):\eHb^{s,\ell+2}\to\mathcal{K}_{b,1}$ is $\epsilon$-regular. Therefore, following the arguments in the proof of (<ref>), we find that for $0\leq k\leq m$,

\[
(\sigma\pa_\sigma)^k\Big(P(b,\sigma)(\sigma^2\hat{f}_3(\sigma))\Big)\in H^{\frac32-\epsilon}((-c_0, c_0);\eHb^{\infty,-\frac12-}(\CX)),
\]

whose norm is bounded in the following manner

\begin{equation}
\begin{split}
&\quad	\lesssim\sum_{j=0}^k\norm{(\sigma\pa_\sigma)^j\hat{f}_3(\sigma)}_{H^{\frac32-\epsilon}(\BR;\eHb^{s,\ell+2}(\CX))}\lesssim\sum_{j=0}^k\norm{(\sigma\pa_\sigma)^j\hat{f}(\sigma)}_{H^{\frac72-\epsilon}(\BR;\eHb^{s,\ell+2}(\CX))}\\
&\quad\lesssim\norm{f}_{\jb{t_{b,*}}^{-\frac 72+\epsilon}	\wt{H}^{s,\ell+2,m}_{\bop,\cop}}.
\end{split}
\end{equation}

Similarly, we have

\[
(\sigma\pa_\sigma)^k\Big(\big(P^1(b)+P^e(b,\sigma)\big)(\chi(\sigma)f^{(1)})\Big)\in H^{\frac32-\epsilon}((-c_0, c_0);\eHb^{\infty,-\frac12-}(\CX)),
\]

\begin{equation}
\norm{(\sigma\pa_\sigma)^k\big(\big(P^1(b)+P^e(b,\sigma)\big)(\chi(\sigma)f^{(1)})\big)}_{H^{\frac32-\epsilon}(\BR;\eHb^{\infty,-\frac12-}(\CX))}\\
\lesssim\norm{f}_{\jb{t_{b,*}}^{-\frac32-}	\wt{H}^{s,\ell+2}_{\bop,\cop}}.
\end{equation}

As a result, it remains to analyze $P(b,\sigma)(\chi(\RE\sigma)f^{(0)})$ and $P^2(b)(i\sigma^{-1}\chi(\RE\sigma)f^{(1)})$. We write

\begin{align*}
&\Big(\sigma^{-2}P^2(b)f^{(0)}+\sigma^{-1}P^1(b)f^{(0)}+P^2(b)(i\sigma^{-1} f^{(1)})\Big)+\sigma^{-1}P^e(b,\sigma)(\chi(\RE\sigma)f^{(0)})\\
&=\Big((\sigma^{-2}(\dg_1,\dA_1)-i\sigma^{-1}(\check{g}_1, \check{A}_1))+i\sigma^{-1}\big((\dg_2, \dA_2)
+(\dg'_1, \dA'_1)+(\dg''_1, \dA''_1)\big)\Big)\\
&\quad+i\sigma^{-1}(\dg_1'''(\sigma), \dA_1'''(\sigma)):=\Romanupper{1}+\Romanupper{2}
\end{align*}

where $(\dg_1,\dA_1),(\dg_1',\dA_1'), (\dg''_1,\dA''_1),(\dg_1'''(\sigma),\dA_1'''(\sigma))\in\mathcal{K}_{b,1}$ and $(\dg_2,\dA_2)\in\mathcal{K}_{b,2}$. Moreover, $(\dg''_1, \dA''_1)$ is uniquely determined by the condition

\begin{align}\notag
&-\Big\langle[L_b, t_{b,*}](\check{g}''_1, \check{A}''_1)+\frac{1}{2}[[L_b, t_{b,*}], t_{b,*}](\dg''_1,\dA''_1), (\dg^*, \dA^*)\Big\rangle+\Big\langle[L_b, t_{b,*}](\dg''_2, \dA''_2), (\dg^*, \dA^*)\Big\rangle\\
\label{eq:condition3}
&\quad\quad=\langle\int_{\BR} t_{b,*}f(t_{b,*})\,dt_{b,*}, \ (\dg^*, \dA^*)\rangle\quad\mbox{for all}\quad(\dg^*,\dA^*)\in\mathcal{K}_b^*
\end{align}

for some $(\dg''_2, \dA''_2)\in\mathcal{K}_{b,2}$.
Since $\sigma^{-j}(1-\chi(\sigma))\in H^{s'}(\BR_\sigma)$ for $j=1,2$ and any $s'\in\BR$, we have

\[
(\sigma\pa_\sigma)^k\Big(\chi(\RE\sigma)\big(P(b,\sigma)f^{(0)}+P^2(b)(i\sigma^{-1}f^{(1)})\big)-(\Romanupper{1}+\Romanupper{2})\Big)\in H^{\frac32-\epsilon}((-c_0, c_0);\eHb^{\infty,-\frac12-}(\CX)),
\]

\begin{equation}
\begin{split}
&\quad\lesssim\norm{f}_{\jb{t_{b,*}}^{-\frac32-}	\wt{H}^{s,\ell+2}_{\bop,\cop}}.
\end{split}
\end{equation}

As a consequence, it suffices to discuss the inverse Fourier transform of $\Romanupper{1}$ and $\Romanupper{2}$.

* Analysis of the leading order term $(\hat{g}, \hat{A})$.

\[
\mathcal{F}^{-1}(\sigma^{-2})=\lim_{C\to0+}\frac{1}{2\pi}\int_{\IM\sigma=C}\!e^{-i\sigma t_{b,*}}\sigma^{-2}\,d\sigma=-\frac{1}{2}t_{b,*}H(t_{b,*})
\]

\[ \mathcal{F}^{-1}(i \sigma^{-1})=\lim_{C\to0+}\frac{1}{2\pi}\int_{\IM\sigma=C}\!e^{-i\sigma t_{b,*}}i\sigma^{-1}\,d\sigma=\frac{1}{2}H(t_{b,*}),
\]

we obtain

\begin{equation}
\begin{split}
\mathcal{F}^{-1}(\Romanupper{1})&=\lim_{C\to0+}\frac{1}{2\pi}\int_{\IM\sigma=C}\!e^{-i\sigma t_{b,*}}\Romanupper{1}\,d\sigma\\
&=-\frac{1}{2}\big(t_{b,*}(\dg_1,\dA_1)+(\check{g}_1, \check{A}_1)\big)
+\frac{1}{2}\Big((\dg'_1, \dA'_1)+(\dg''_1, \dA''_1)+(\dg_2, \dA_2)\Big):=(\hat{g}, \hat{A}),
\end{split}
\end{equation}

where $(\dg_1,\dA_1), (\dg_1',\dA_1'),(\dg''_1, \dA''_1)\in\mathcal{K}_{b,1}$, and $(\dg_2,\dA_2)\in\mathcal{K}_{b,2}$. Moreover, $(\dg_1,\dA_1), (\dg_1',\dA_1')$ and $(\dg_2,\dA_2)$ are determined by the conditions (<ref>) and (<ref>) with $f$ there replaced by $f^{(0)}=\int_{\BR}f(t_{b,*})\,dt_{b,*}$, and $(\dg''_1, \dA''_1)\in\mathcal{K}_{b,1}$ is uniquely determined by the condition (<ref>) for some $(\dg''_2, \dA''_2)\in\mathcal{K}_{b,2}$.

* Analysis of the pure gauge part $(\bar{g}, \bar{A})$. Finally, we  turn to the analysis of the inverse Fourier transform of $\Romanupper{2}=i\sigma^{-1}(\dg_1'''(\sigma), \dA_1'''(\sigma))$ where $\mathrm{supp }\,(\dg_1'''(\sigma), \dA_1'''(\sigma))\subset (-c_0, c_0)$. By Theorem <ref>,

\[			\Romanupper{2}=\sigma^{-1}(\dg_1'''(\sigma),\dA_1'''(\sigma))=\sum_{\pm}\psi^{\pm}(\sigma)\quad\mbox{where}\quad \psi^{\pm}\in \mathcal{A}^{-\epsilon}\big(\pm[0, c_0);\mathcal{K}_{b,1}\big).
\]

We write

\begin{align*}
\int_{-c_0}^{c_0}\! e^{-it_{b,*}\sigma}	\Romanupper{2}\,d\sigma&=\int_0^{c_0}\!e^{-it_{b,*}\sigma}\chi(t_{b,*}\sigma)\psi^+(\sigma)\,d\sigma+\int_0^{c_0}\!e^{-it_{b,*}\sigma}(1-\chi(t_{b,*}\sigma))\psi^+(\sigma)\,d\sigma\\
\end{align*}

First, we have

\begin{align*}
\abs{\Romanupper{2}^+_1}\leq\int_0^{c_0t_{b,*}^{-1}}\!\abs{\psi_+(\sigma)}\,d\sigma\lesssim\int_0^{c_0t_{b,*}^{-1}}\!\sigma^{-\epsilon}\,d\sigma \lesssim t_{b,*}^{-1+\epsilon}.
\end{align*}

Next, we compute for $k\in\BN$

\begin{align*}
\Romanupper{2}^+_2=\int_0^{c_0}\!e^{-it_{b,*}\sigma}(1-\chi(t_{b,*}\sigma))\psi_+(\sigma)\,d\sigma=(it_{b,*})^{-k}\int_0^{c_0}\!e^{-it_{b,*}\sigma}\pa_{\sigma}^k\Big((1-\chi(t_{b,*}\sigma))\psi_+(\sigma)\Big)\,d\sigma.
\end{align*}

Then we have

\[
\abs{	\Romanupper{2}^+_2}\lesssim t_{b,*}^{-k}\int_{c_0t_{b,*}^{-1}/2}^{c_0}\!\sigma^{-k-\epsilon}\,d\sigma\lesssim t_{b,*}^{-1+\epsilon}.
\]

The estimates for $\Romanupper{2}^+_1,\Romanupper{2}^+_2$ also hold after applying any number of derivatives $t_{b_*}\pa_{t_{b,*}}$. We note that $\Romanupper{2}^-_1,\Romanupper{2}^-_2$ can be handled in a similar manner.

\begin{equation}
\Romanupper{2}=\sigma^{-1}(\dg_1'''(\sigma), \dA_1'''(\sigma))=(2\delta^*_{g_b}\omega_{b ,s_1}(\sigma), \wt{\mathcal{L}}_{A_b}(\omega_{b,s_1}(\sigma))^\sharp+d\phi_{b,s_1}(\sigma))
\end{equation}

\[
(\omega_{b,s_1}(\sigma), \phi_{b,s_1}(\sigma))\in \sum_{\pm}\mathcal{A}^{-\epsilon}(\pm[0, c_0); \eHb^{\infty,-\frac32-}(\CX;\scform)\oplus\eHb^{\infty, -\frac12-}(\CX;\BC)),
\]

we can write

\[
(\bar{g}, \bar{A})=\int_{-c_0}^{c_0}\! e^{-it_{b,*}\sigma}	\Romanupper{2}\,d\sigma=(\bar{g}_1, \bar{A}_1)+(\bar{g}_2, \bar{A}_2)
\]

\begin{align*}
(\bar{g}_1, \bar{A}_1)&=\Big(2\delta^*_{g_b}\big(\chi(\frac{\jb{r}}{\jb{t_{b,*}}})\omega_{b ,s_1}(t_{b,*})\big),\  \wt{\mathcal{L}}_{A_b}\big(\chi(\frac{\jb{r}}{\jb{t_{b,*}}})\omega_{b,s_1}(t_{b,*})\big)^\sharp+d\big(\chi(\frac{\jb{r}}{\jb{t_{b,*}}})\phi_{b,s_1}(t_{b,*})\big)\Big),\\
(\bar{g}_2, \bar{A}_2)&=\Big(2[\chi(\frac{\jb{r}}{\jb{t_{b,*}}}),\delta^*_{g_b}]\omega_{b ,s_1}(t_{b,*}),\  [\chi(\frac{\jb{r}}{\jb{t_{b,*}}}),\wt{\mathcal{L}}_{A_b}]\big(\omega_{b,s_1}(t_{b,*})\big)^\sharp+[\chi(\frac{\jb{r}}{\jb{t_{b,*}}}),d]\phi_{b,s_1}(t_{b,*})\Big)\\
&\quad-\chi(\frac{\jb{r}}{\jb{t_{b,*}}})\Big(2\pa_{t_{b,*}}\omega_{b,s_1}\otimes_s dt_{b,*},\  \pa_{t_{b,*}}\big(A_\alpha(\omega_{b,s_1})^{\sharp,\alpha}+\phi_{b,s_1}\big)dt_{b,*}\Big)+\big(1-\chi(\frac{\jb{r}}{\jb{t_{b,*}}})\big)(\bar{g}, \bar{A}).
\end{align*}

where $\chi$ is a smooth nonnegative cutoff such that $\chi(s)=1$ for $s\leq1/2$ and $\chi(s)=0$ for $s\geq 1$. Then the estimates (<ref>) and (<ref>)  for $(\bar{g}_1, \bar{A}_1)$ and $(\bar{g}_2, \bar{A}_2)$ follow from the following estimates

\begin{align*}
\abs{\big(\jb{t_{b,*}}\pa_{t_{b,*}}\big)^{m-1}V^j\omega_{b,s_1}(t_{b,*})}	&\lesssim
\jb{t_{b,*}}^{-1+\epsilon}\norm{f}_{\jb{t_{b,*}}^{-\frac 32-}	\wt{H}^{s,\ell+2}_{\bop,\cop}},\\
\abs{\big(\jb{t_{b,*}}\pa_{t_{b,*}}\big)^{m-1}V^j\phi_{b,s_1}(t_{b,*})}&\lesssim
\jb{t_{b,*}}^{-1+\epsilon}\jb{r}^{-1}\norm{f}_{\jb{t_{b,*}}^{-\frac 32-}	\wt{H}^{s,\ell+2}_{\bop,\cop}},\\
\abs{\big(\jb{t_{b,*}}\pa_{t_{b,*}}\big)^{m-1}V^j(\bar{g}, \bar{A})}&\lesssim
\jb{t_{b,*}}^{-1+\epsilon}\jb{r}^{-2}\norm{f}_{\jb{t_{b,*}}^{-\frac 32-}	\wt{H}^{s,\ell+2}_{\bop,\cop}}.
\end{align*}

§ PROOF OF THE MAIN THEOREM

In the initial value problem for Einstein-Maxwell system as formulated in <ref>, we impose the initial data at the Cauchy hypersurface which terminates at spatial infinity (i.e., it coincides with the hypersurface $t=0$ when $r$ is large enough). In this section, we will discuss how to reduce this initial value problem to the inhomogeneous problem studied in the previous section <ref>.

We recall the time function $\mathfrak{t}$ (corresponding to $g_{b_0}$) defined in (<ref>), which satisfies that $\mathfrak{t}=t$ for $r\geq 4\Bm_0$, and $d\mathfrak{t}$ is timelike everywhere on $M$ with respect to $g_{b_0}$ with $b_0=(\Bm_0, 0,\BQ_0), \abs{\BQ_0}\ll\Bm_0$, hence for $g_b$ when $b=(\Bm, \Ba,\BQ)$ is close to $b_0$. We define

\[
\Sigma_0:= \mathfrak{t}^{-1}(0),
\]

which is a spacelike (with respect to $g_b$) Cauchy hypersurface. Identifying the region $r\geq 4\Bm_0$ in $\Sigma_0$ with a subset of $\BR^3$, we can compactify $\Sigma_0$ at infinity to the manifold $\bar{\Sigma}_0$ (with two boundary components: one is inside the black hole and the other is the spatial infinity). We note that for large $r$, the spacetime scattering cotangent bundle ${}\widetilde{^{\scop}T^*}\bar{\Sigma}_0={}^{\scop}T^*\bar{\Sigma}_0 \oplus \BR\,d\mathfrak{t}$ is spanned over $C^\infty(\bar{\Sigma}_0)$ by $d t$ and $d x^1,d x^2,d x^3$, where $(x^1,x^2,x^3)$ are the standard coordinates on $\BR^3$.

We first study the initial value problem for linearized gauge-fixed Einstein-Maxwell system $L_b(\dg,\dA)=0$.

Let $0<\alpha<1$ and $s>8+m$, $m\in\BN$. Suppose

\begin{align*}
&(\dg_0, \dg_1,\dA_0,\dA_1)\in \eHb^{s,-1/2+\alpha}(\bar{\Sigma}_0;S^2{}\widetilde{^{\scop}T^*}\bar{\Sigma}_0)\oplus \eHb^{s-1,1/2+\alpha}(\bar{\Sigma}_0;S^2{}\widetilde{^{\scop}T^*}\bar{\Sigma}_0)\\ &\hspace{8em}\oplus\eHb^{s-1,-1/2+\alpha}(\bar{\Sigma}_0;{}\widetilde{^{\scop}T^*}\bar{\Sigma}_0)\oplus \eHb^{s-2,1/2+\alpha}(\bar{\Sigma}_0;{}\widetilde{^{\scop}T^*}\bar{\Sigma}_0).
\end{align*}

Then the solution $(\dg,\dA)$ of the initial value problem

\[
L_{b}(\dg,\dA) = 0, \quad
(\dg,\dA)|_{\bar{\Sigma}_0}=(\dg_0,\dA_0),\quad (\mathcal{L}_{\pa_{\mathfrak{t}}}\dg,\mathcal{L}_{\pa_{\mathfrak{t}}}\dA)|_{\bar{\Sigma}_0}=(\dg_1,\dA_1)
\]

satisfies the following properties. Let

\[
\]

* For $t_{b,*}\geq 0$, the solution $(\dg,\dA)$ can be written as

\begin{equation}
(\dg, \dA)=(\hat{g}, \hat{A})+(\bar{g}, \bar{A})+(\tilde{g}, \tilde{A})
\end{equation}

where $(\hat{g}, \hat{A})\in\widehat{\mathcal{K}}_b$ is a generalized zero mode  (defined in Proposition <ref>), and $(\bar{g}, \bar{A})$, $(\tilde{g}, \tilde{A})$ satisfy the following estimates.

(1) The term $(\bar{g}, \bar{A})$ can be written as  $
(\bar{g}, \bar{A})=(\bar{g}_1, \bar{A}_1)+(\bar{g}_2, \bar{A}_2)$ where $(\bar{g}_1, \bar{A}_1)$ is a pure gauge solution. Moreover, for $m\geq 1$ and $j\geq 0$, we have the pointwise estimates

\begin{equation}\label{EqModifiedGauge}
\abs{\big(\jb{t_{b,*}}\pa_{t_{b,*}}\big)^{m-1}V^j(\bar{g}_1, \bar{A}_1)}\lesssim
\big(\jb{t_{b,*}+r}^{-2+\epsilon}+\jb{t_{b,*}}^{-1+\epsilon}\jb{r}^{-2}\big)	N^{s,\alpha}%\norm{f}_{\jb{t_{b,*}}^{-\frac 32-}	\wt{H}^{s,\ell+2}_{\bop,\cop}}
\end{equation}

\begin{equation}\label{EqNonstaGuage}
\abs{\big(\jb{t_{b,*}}\pa_{t_{b,*}}\big)^{m-1}V^j(\bar{g}_2,\bar{A}_2)}\lesssim
\jb{t_{b,*}+r}^{-2+\epsilon}	N^{s,\alpha}%\norm{f}_{\jb{t_{b,*}}^{-\frac 32-}	\wt{H}^{s,\ell+2}_{\bop,\cop}}
%	\abs{\big(\jb{t_{b,*}}\pa_{t_{b,*}}\big)^{m-1}V^j \bar{A}_2}&\lesssim
%	\Big(\jb{t_{b,*}+r}^{-2+\epsilon}\jb{r}^{-1}+	\jb{t_{b,*}+r}^{-2}\jb{t_{b,*}}^{-1+\epsilon}\Big)\norm{f}_{\jb{t_{b,*}}^{-\frac 32-}	\wt{H}^{s,\ell+2}_{\bop,\cop}}
\end{equation}

where $V^j$ denotes the product of $j$ vector fields $V\in\{\pa_{t_{b,*}}, r\pa_r, \mbox{ rotation fields }\}$.

$(\tilde{g}, \tilde{A})$ satisfies: for $0<\epsilon<1$ with $\alpha+\epsilon>1$ and $0\leq m\leq \frac{s-6}{2}$, we have

\begin{equation}\label{EqPfIVP1Hb}
\norm{(\tilde{g}, \tilde{A})}_{\jb{t_{b,*}}^{-\frac 32+\epsilon}	\wt{H}^{s-6-m,-5/2+\alpha+\epsilon-,m}_{\bop,\cop}}\lesssim N^{s,\alpha}
\end{equation}

For $1\leq m\leq \frac{s-6}{2}$, we also have

\begin{equation}\label{EqPfIVP1Linfty}
\norm{\big(\jb{t_{b,*}}\pa_{t_{b,*}}\big)^{m-1}(\tilde{g}, \tilde{A})}_{\jb{t_{b,*}}^{-2+\epsilon}	L^\infty(\BR_{t_{b,*}};\eHb^{s-6-m,-5/2+\alpha+\epsilon-}(\CX))}\lesssim N^{s,\alpha}.
\end{equation}

For $m\geq 1$ and $0\leq j<s-8-m$ with $j\in\BN_0$, we have the pointwise decay estimate

\begin{equation}\label{EqPfIVP1HighLinfty}
\abs{\big(\jb{t_{b,*}}\pa_{t_{b,*}}\big)^{m-1}V^j(\tilde{g}, \tilde{A})}\lesssim\jb{t_{b,*}}^{-2+\epsilon}\jb{r}^{-\alpha+\epsilon+1+}N^{s,\alpha}.
\end{equation}

*  In $\mathfrak{t}\geq 0$, $t_{b,*}\leq 0$, we have $(\dg,\dA)=r^{-1}H^{s-3}(\BR_{t_{b,*}}\times \BS^2)+(\tilde{g},\tilde{A})$ with $\abs{(\tilde{g},\tilde{A})}\lesssim r^{-1}(1+\abs{t_{b,*}})^{-\alpha}$.

Let $T_1< T_2$ and let $\chi(t_{b,*})$ be a nonnegative smooth cutoff such that $\chi=1$ for $ t_{b,*}\geq T_2$ and $\chi(t_{b,*})=0$ for $t_{b,*}\leq T_1$. Then we obtain the following equation

\[
L_b\Big(\chi(t_{b,*})(\dg,\dA)\Big)=[L_b, \chi(t_{b,*})](\dg,\dA),
\]

of which the right-hand side is supported in $t_{b,*}\in(T_1,T_2)$. In view of Theorem <ref>, it suffices to prove that

\begin{equation}
\label{eq:decayofRHS}
L_b\Big(\chi(t_{b,*})(\dg,\dA)\Big)\in \widetilde{H}_{\bop,\cop}^{s-4-k,1/2+\alpha-,k}, \quad k\in\BN.
\end{equation}

Since the regularity with respect to $\langle t_{b,*}\rangle D_{t_{b,*}}$ and $D_{t_{b,*}}$ is equivalent in the support of 	$L_b\Big(\chi(t_{b,*})(\dg,\dA)\Big)$, it suffices to prove (<ref>) for $k=0$. First, the global wellposedness theory of linear wave equations implies that  $(\dg,\dA)\in H^{s-1}$ for $\mathfrak{t}\geq 0$, $t_*\leq C$, $r\leq C$ for any $C$; this implies that $[L_b,\chi]\in H^{s-2}$ in such compact sets. Therefore, it suffices to work in an arbitrarily small neighborhood $r>R_0\gg 1$ of infinity, which we shall do from now on.

The proof closely follows the argument in  <cit.>. Let $b=(\Bm,\Ba,\BQ)$. Then we define $b_1=(\Bm,0,\BQ)$ and introduce the incoming and outgoing null coordinates

\[
x^0=t+r_{b_1,*},\quad x^1=t-r_{b_1,*},
\]

with respect to which we have

\[
g_{b_1}=-\mu_{b_1}d x^0\,d x^1+r^2\sg
\]

\[
2\pa_0\equiv 2\pa_{x^0}=\pa_t+(1-\frac{2\Bm}{r}+\frac{\BQ^2}{r^2})\pa_r,\quad
2\pa_1\equiv 2\pa_{x^1}=\pa_t-(1-\frac{2\Bm}{r}+\frac{\BQ^2}{r^2})\pa_r.
\]

Let $x^2,x^3$ denote local coordinates on $\BS^2$, and denote spherical indices by $c,d=2,3$; let $\pa_c\equiv\pa_{x^c}$; let $\CX$ be the compactification of $t_{b,*}^{-1}(0) $. Since $g_b-g_{b_1}\in\rho^2 C^\infty(\CX;\scsym)$, we have

\begin{align*}
&g_b(\pa_0,\pa_0),\ (g_b-g_{b_1})(\pa_0,\pa_1),\ g_b(\pa_0,r^{-1}\pa_c), \\
&\quad g_b(\pa_1,\pa_1),\ g_b(\pa_1,r^{-1}\pa_c),\ (g_b-g_{b_1})(r^{-1}\pa_b,r^{-1}\pa_c) \in \rho^2C^\infty.
\end{align*}

We let

\[
\rho_I=(r_{b_1,*}-t)/r,\quad \rho_0=(r_{b_1,*}-t)^{-1},\quad \rho=\rho_I\rho_0=\frac{1}{r}.
\]

Then we have

\[
\rho^{-3}L_b\rho=-\Box_{\rho^2g_b}\otimes\mathrm{Id}_{14\times 14}+R
\]

with $R\in\widetilde{\mbox{Diff}_{\bop}^1}$ acting on sections of $\scsym\oplus\scform$, where $\widetilde{\mbox{Diff}_{\bop}^k}$ consists of up to $k$-fold products of the vector fields $\{\rho_I\pa_{\rho_I},\, \rho_0\pa_{\rho_0},\, \mbox{rotation vector fields}\}$. We note that switching from $L_b$ to $\rho^{-3}L_b\rho$ is related to the Friedlander rescaling for the scalar wave equation [45].

Then one can use the multiplier $W:=\rho_0^{-2a_0}\rho_I^{-2a_I}V$ with $V:=-(1+c_V)\rho_{I}	\pa_{\rho_I}+\rho_0\pa_{\rho_0}$ and $a_0=\alpha,\, a_I<0$ as introduced in <cit.> to derive an energy estimate, which allows us to conclude that near $\mathscr{I}^+$, $(\dg,\dA)$ lies in $\wt H_\bop^{s-1,-1/2-}$ (see <cit.> for detail. Actually, the present setting is much simpler than that in the reference as we do not need to get sharp decay rate for a certain component as did in <cit.>). To further determine the leading behavior of $(\dg,\dA)$ near $\mathscr{I}$, one rewrite the equations $\rho^{-3}L_{b}\rho(\rho^{-1}\dg,\rho^{-1}\dA)=0$ as $2\rho^{-2}\pa_0\pa_1 (\rho^{-1}\dg,\rho^{-1}\dA)=(\sL+\tilde{R})(\rho^{-1}\dg,\rho^{-1}\dA)$ where $\tilde{R}\in \rho\widetilde{\mbox{Diff}^2_{\bop}}+\widetilde{\mbox{Diff}^1_{\bop}}$. Since $-2\rho^{-2}\pa_0\pa_1=\rho_I^{-1}(\rho_0\pa_{\rho_0}-\rho_I\pa_{\rho_I})\rho_I\pa_{\rho_I}+\widetilde{\mbox{Diff}^1_{\bop}}$, it follows that

\[
(\rho_0\pa_{\rho_0}-\rho_I\pa_{\rho_I})\rho_I\pa_{\rho_I}(\rho^{-1}\dg,\rho^{-1}\dA)\in \rho_I\widetilde{\mbox{Diff}^2_{\bop}}(\rho^{-1}\dg,\rho^{-1}\dA)\sim \rho_0^\alpha\rho_I^{a_I+1}.	\]

We choose $a_I<0$ such that $0<1+a_I<\alpha$ (see <cit.>). Integrating first along the vector field $\rho_0\pa_{\rho_0}-\rho_I\pa_{\rho_I}$ from $\rho_I\geq \epsilon$, where $(\rho^{-1}\dg,\rho^{-1}\dA)\sim \rho_0^{\alpha}$, yields
\[\rho_I\pa_{\rho_I}(\rho^{-1}\dg,\rho^{-1}\dA)\sim \rho_0^\alpha\rho_I^{1+a_I}.
\]

Then integrating out the vector field $\rho_I\pa_{\rho_I}$ gives

\[
(\dg,\dA)-\rho H^{s-3}(\BR_{t_{b_1,*}}\times\BS^2)\sim \rho\rho_0^\alpha\rho_I^{1+a_I}.
\]

This proves statement <ref>.
Moreover, in view of Lemma <ref>, we see that

\begin{equation}
\label{EqPfIVPComm}
\end{equation}

Since $(\rho\pa_\rho-1)$ annihilates the leading order term $\rho H^{s-3}(\BR_{t_{b_1,*}}\times\BS^2)$ in $(\dg,\dA)$, it follows that $L_b(\chi (\dg,\dA))=[L_b,\chi](\dg,\dA)\in\widetilde{H}_{\bop,\cop}^{s-4,1/2+\alpha-}$.

Since $s-4>4+m$, the estimates (<ref>)–(<ref>) now follow from Theorem <ref>, with $\ell+2=\frac12+\alpha-$, so $\ell=-\tfrac32+\alpha-$ and $-1/2<\ell+\epsilon<1/2$ with $\alpha+\epsilon>1, 0<\epsilon<1$.

We now state our main theorem.

Let $0<\alpha<1$ and $s>8+m$, $m\in\BN$. Let

\begin{align*}
&(\dot{h}, \dot{k}, \dot{\mathbf{E}}, \dot{\mathbf{H}})
\in \eHb^{s,-1/2+\alpha}(\Sigma_0;S^2\,{}^{\scop}T^*\Sigma_0)\oplus
\in \eHb^{s-1,1/2+\alpha}(\Sigma_0;S^2\,{}^{\scop}T^*\Sigma_0)\\
&\hspace{8em}\eHb^{s-1,1/2+\alpha}(\Sigma_0; {}^{\scop}T^*\Sigma_0)\oplus
\in \eHb^{s,1/2+\alpha}(\Sigma_0;{}^{\scop}T^*\Sigma_0)
\end{align*}

be the initial data for the linearized Einstein-Maxwell equations linearized around the KN solution $(g_b, A_b)$. Suppose that the initial data satisfy the linearized constraint equations, which is the linearization of the constraint equations around the KN initial data $(h_b, k_b,\mathbf{E}_b, \mathbf{H}_b)=\tau(g_b, dA_b)$. Assume that the linearized magnetic charge vanishes, see the discussion in <ref>–<ref>.

Then there exists a solution $(\dg, \dA)$ of the initial value problem

\[
D_{g_b}\Ric(\dg) - 2 D_{(g_b,d A_b)}T(\dg,\dA)=0,\quad
D_{g_b,A_b}(\delta_{(\cdot)}d(\cdot))(\dg,\dA)=0,\quad D_{(g_b, A_b)}\tau(\dg,\dA)=(h_b, k_b,\mathbf{E}_b, \mathbf{H}_b)
\]

satisfying the linearized generalized wave map gauge and Lorenz gauge conditions

\[
D_{g_b}\widetilde{\Upsilon}^E(\dg;g_b)=0,\quad \delta_{g_b}\dA=0.
\]

Moreover, $(\dg,\dA)$ has the asymptotic behavior stated in Theorem <ref>.

In view of Corollary <ref>, there exist sections

\begin{align*}
&(\dot{g}_0,\dot{g}_1,\dot{A}_0,\dot{A}_1)\in \eHb^{s,-1/2+\alpha}(\bar{\Sigma}_0;S^2{}\widetilde{^{\scop} T^*}\bar{\Sigma}_0)\oplus \eHb^{s-1,1/2+\alpha}(\bar{\Sigma}_0;S^2 {}\widetilde{^{\scop}T^*}{\bar{\Sigma}_0}) \\
&\hspace{8em}\oplus \eHb^{s-1,-1/2+\alpha}(\bar{\Sigma}_0;{}\widetilde{^{\scop}T^*}{\bar{\Sigma}_0}) \oplus \eHb^{s-2,1/2+\alpha}(\bar{\Sigma}_0;{}\widetilde{^{\scop}T^*}{\Sigma_0}),
\end{align*}

such that they induce the data $(\dot{h},\dot{k},\dot{\mathbf{E}},\dot{\mathbf{H}})$ on $\Sigma_0=\{\mathfrak{t}=0\}$, and satisfy the linearized gauge conditions $D_{g_b}\widetilde{\Upsilon}^E(\dg)=0$ and $D_{(g_b, A_b)}\widetilde{\Upsilon}^M(\dg,\dA)=-\delta_{g_b}\dot{A}=0$ at $\Sigma_0$.

Then we solve the initial value problem $L_{b}(\dg,\dA)=0$ with initial data $(\dot{g}_0,\dot{g}_1,\dot{A}_0,\dot{A}_1)$. Since $(\dot{h},\dot{k},\dot{\mathbf{E}},\dot{\mathbf{H}})$ satisfies the linearized constraint equations, it follows from Lemma <ref> that

\[
D_{g_b}\widetilde{\Upsilon}^E(\dg;g_b)=0,\quad \delta_{g_b}\dA=0
\]

and thus $(\dg,\dA)$ solves the linearized Einstein-Maxwell equations.

§ THE LINEARIZED EINSTEIN-MAXWELL OPERATOR AROUND REISSNER-NORDSTRÖM SPACETIME

§.§ Detailed calculation of the linearized Einstein-Maxwell system

We now calculate the linearized Einstein-Maxwell system
\[\mathscr{L}(\dg,\dA):=(\mathscr{L}_1(\dg,\dA),\mathscr{L}_2(\dg,\dA)):=(D_g\Ric(\dg, d\dA)-2D_{(g, dA)}T(\dg, d\dA),\ D_{(g, F)}\left(\delta_{g}F\right)(\dg, \dF))=0
\]

in more detail.

First, following <cit.>,<cit.> we see that

\[
D_g\Ric(\dot{g})=-\frac 12\Box_{g}\dot{g}-\delta^*_g\delta_gG_g\dot{g}+\mathscr{R}_g\dot{g}\quad\mbox{where}\quad (\mathscr{R}_g\dg)_{\m\n}=R^{\al\ \ \beta}_{\ \m\n}\dg_{\al\be}+\frac 12\left(\Ric_{\m}^{\ \kappa}\dg_{\n\ka}+\Ric_{\n}^{\ \kappa}\dg_{\m\ka}\right)
\]

Since $T(g, F)=F_{\mu\alpha}F_{\nu}^{~\alpha}-\frac 14 g_{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}$, we find

\begin{equation}\label{eq:DT}
\begin{split}
D_{(g, F)}T(\dg, \dF)&=-\dg^{\alpha\beta}F_{\mu\alpha}F_{\nu\beta}-\frac 14\left(\dg_{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}-g_{\mu\nu}\dg^{\kappa\alpha}g^{\lambda\beta}F_{\alpha\beta}F_{\kappa\lambda}-g_{\mu\nu}g^{\kappa\alpha}\dg^{\lambda\beta}F_{\alpha\beta}F_{\kappa\lambda}\right)\\
&\quad +\left(g^{\alpha\beta}\dF_{\mu\alpha}F_{\nu\beta}+g^{\alpha\beta}F_{\mu\alpha}\dF_{\nu\beta}\right)-\frac 14\left(g_{\mu\nu}\dF_{\alpha\beta}F^{\alpha\beta}+g_{\mu\nu}F_{\alpha\beta}\dF^{\alpha\beta}\right)\\
&=-\dg^{\alpha\beta}F_{\mu\alpha}F_{\nu\beta}-\frac 14\left(\dg_{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\!\!-\!2g_{\mu\nu}\dg^{\kappa\alpha}F_{\alpha\beta}F_{\kappa}^{~\be}\right)\!\!+\!\!\left(g^{\alpha\beta}\dF_{\mu\alpha}F_{\nu\beta}+g^{\alpha\beta}F_{\mu\alpha}\dF_{\nu\beta}\right)\!\!-\!\frac 12g_{\mu\nu}\dF_{\alpha\beta}F^{\alpha\beta}
\end{split}
\end{equation}

and thus the Einstein part $\mathscr{L}_1$ of the linearized Einstein-Maxwell system is given by

\begin{equation}\label{eq:L_1}
\mathscr{L}_1(\dg, d\dA)=D_g\Ric(\dg, d\dA)-2D_{(g, dA)}T(\dg, d\dA)=-\frac12\Box_{g}\dot{g}-\delta^*_g\delta_gG_g\dot{g}+\mathscr{R}_g\dot{g}-2D_{(g, dA)}T(\dg, d\dA)=0.
\end{equation}

Next we analyze the Maxwell part $\mathscr{L}_2$ of the linearized Einstein-Maxwell system. Since

\[
(\delta_g F)_\mu=-\nabla^\nu F_{\nu\mu}=-g^{\nu\alpha}\nabla_\alpha F_{\nu\mu}=-g^{\nu\alpha}\left(\pa_\al F_{\n\m}-\cs^{\ka}_{\al\n}F_{\ka\m}-\cs^{\ka}_{\al\m}F_{\n\ka}\right),
\]

\[
\dot{\cs}_{\al\n}^{\ka}(\dg)=\frac 12 g^{\ka\la}\left(\nabla_\n \dg_{\al\la}+\nabla_\al \dg_{\n\la}-\nabla_\la\dg_{\al\n}\right),
\]

we obtain

\begin{equation}
\begin{split}
&D_{(g, F)}\left(\delta_{g}F\right)(\dg, \dF)=\delta_g\dF+\dg^{\n\al}\nabla_\al F_{\n\m}+g^{\n\al}\left(\dot{\cs}_{\al\n}^{\ka}(\dg)F_{\ka\m}+\dot{\cs}_{\al\m}^{\ka}(\dg)F_{\n\ka}\right)\\
&\quad=\delta_g\dF+\dg^{\n\al}\nabla_\al F_{\n\m}+g^{\ka\la}\nabla^\al(\dg_{\al\la}-\frac 12g^{\n\al}\dg_{\al\n}g_{\al\la})F_{\ka\m}+\frac 12(\nabla^{\n}\dg^{\ka}_{~\m}-\nabla^{\ka}\dg^{\n}_{~\m})F_{\n\ka}\\
&\quad=\delta_g\dF+\dg^{\n\al}\nabla_\al F_{\n\m}-(\delta_gG_g\dg)^\ka F_{\ka\m}+\frac 12(\nabla^{\n}\dg^{\ka}_{~\m}-\nabla^{\ka}\dg^{\n}_{~\m})F_{\n\ka}
\end{split}
\end{equation}

where we use $(\nabla_\m\dg^{\nu\ka})F_{\n\ka}=0$ in the second equality. Therefore, the Maxwell part $\mathscr{L}_2$ of the linearized Einstein-Maxwell system is given by

\begin{equation}\label{eq:L_2}
\mathscr{L}_2(\dg, d\dA)=\delta_gd\dA+\dg^{\n\al}\nabla_\al (dA)_{\n\m}-(\delta_gG_g\dg)^\ka (dA)_{\ka\m}+\frac 12(\nabla^{\n}\dg^{\ka}_{~\m}-\nabla^{\ka}\dg^{\n}_{~\m})(dA)_{\n\ka}=0.
\end{equation}

We then proceed to calculate the linearized Einstein-Maxwell system $\mathscr{L}(\dg,\dA)=0$ around a Reissner-Nordström spacetime $g=\hg+r^2\sg$ under the splitting (<ref>) and (<ref>) .

§.§ Calculation of $\mathscr{R}_g\dg$

Here and in what follows the components we do not list are $0$. Christoffel symbols of Reissner-Nordström metric $g$ are,

\begin{gather*}
\cs_{ij}^k=\hcs_{ij}^k \quad %\mbox{where} \quad \hcs_{tr}^t=\hcs_{rt}^t=\frac 12\weight^{-1}\weight', \quad \hcs_{tt}^r=\frac 12\weight\weight',\quad \hcs_{rr}^r=-\frac 12 \weight^{-1}\weight',\\
\cs_{ij}^a=0,\quad \cs_{ia}^j=0,\quad\cs_{ab}^k=-r\pa^{k}r\sg_{ab},\\
\cs_{ak}^b= r^{-1}\pa_kr\delta^b_{a},\quad	\cs_{ab}^c=\scs_{ab}^c.
\end{gather*}

The components of Riemann curvature tensor are given by

\begin{gather*}
R_{ijkm}=\hR_{ijkm}=\frac 12\hR(\hg)\left(\hg_{ik}\hg_{jm}-\hg_{im}\hg_{jk}\right)=-\frac 12\weight''\left(\hg_{ik}\hg_{jm}-\hg_{im}\hg_{jk}\right),\\
R_{aijk}=0,\quad R_{abij}=0,\quad R_{abci}=0,\quad R_{aibj}=-r\hnabla_i\hnabla_jr\sg_{ab}=-\frac 12r\weight'\sg_{ab}\hg_{ij}	,	\\
R_{abcd}=\sR_{abcd}-r^2\weight\left(\sg_{ac}\sg_{bd}-\sg_{ad}\sg_{bc}\right)	=r^2(1-\weight)\left(\sg_{ac}\sg_{bd}-\sg_{ad}\sg_{bc}\right).
\end{gather*}

Correspondingly, the Ricci curvature tensor takes the form

\begin{gather*}
\Ric_{ij}=\hRic_{ij}-r^{-1}\weight'\hg_{ij}=-\frac 12\weight''\hg_{ij}-r^{-1}\weight'\hg_{ij},\quad \Ric_{ia}=0,\\
\Ric_{ab}=(1-\weight-r\weight')\sg_{ab}.
\end{gather*}

Now we are ready to calculate $\mathscr{R}_g\dg$ in detail. Since $(\mathscr{R}_g\dg)_{\m\n}=R^{\al\ \ \beta}_{\ \m\n}\dg_{\al\be}+\frac 12\left(\Ric_{\m}^{\ \kappa}\dg_{\n\ka}+\Ric_{\n}^{\ \kappa}\dg_{\m\ka}\right)$, we have

\begin{gather*}
(\mathscr{R}_g\dg)_{ij}=-\weight''\left(\dg_{ij}-\frac 12\hg^{km}\dg_{km}\hg_{ij}\right)-r^{-1}\weight'\dg_{ij}+\frac 12r^{-3}\weight'\sg^{ab}\dg_{ab}\hg_{ij},\\		(\mathscr{R}_g\dg)_{ia}=-\frac 32r^{-1}\weight'\dg_{ia}-\frac 14\weight''\dg_{ia}+\frac 12 r^{-2}(1-\weight)\dg_{ia},\\		(\mathscr{R}_g\dg)_{ab}=2r^{-2}(1-\weight)\left(\dg_{ab}-\frac 12\sg^{cd}\dg_{cd}\sg_{ab}\right)-r^{-1}\weight'\dg_{ab}+\frac 12r\weight'\hg^{ij}\dg_{ij}\sg_{ab}.
\end{gather*}

Therefore, with respect to the splitting of $S^2T^*M$ (<ref>), we can express $\mathscr{R}_g: S^2T^*M\to S^2T^*M$ as follows

\begin{equation}
\mathscr{R}_g=
\begin{pmatrix}
-\weight'' G_{\hg}-r^{-1}\weight'&0&\frac 12 r^{-3}\weight' \hg\str\\
0&-\frac 32 r^{-1}\weight'-\frac 14\weight''+\frac 12r^{-2}(1-\weight)&0\\
\frac 12r\weight'\sg\htr&0&2r^{-2}(1-\weight)G_{\sg}-r^{-1}\weight'
\end{pmatrix}
\end{equation}

where $(G_{\hg} h)_{ij}=h_{ij}-\frac 12\hg_{ij}\htr h$ and $(G_{\sg}h)_{ab}=h_{ab}-\frac 12\sg_{ab}\str h$.

§.§ Calculation of $\delta_g^*\delta_g G_g\dg$

We now calculate the first covariant derivatives on $1$-forms and symmetric $2$-tensors. For $\eta\in T^*M$, we compute

\begin{gather*}
\nabla_i\eta_j=\hnabla_i\eta_j,
\quad \nabla_i\eta_{a}=\pa_i\eta_{a}-(r^{-1}\pa_i r)\eta_a,
\quad \nabla_{a}\eta_i=\pa_{a}\eta_i-(r^{-1}\pa_i r)\eta_a,
\quad \nabla_{a}\eta_b=\slashed{\nabla}_a\eta_b+r(\pa^kr)\eta_k\sg_{ab}.
\end{gather*}

With respect to the splitting of $T^*M$ (<ref>), we write the symmetric gradient $\delta_g^*: T^*M\to S^2T^*M$ with $(\delta^*_g\eta)_{\m\n}=\frac 12(\nabla_\m\eta_\n+\nabla_\n\eta_\m)$ as follows

\begin{equation}\label{eq:symgradient}
\delta_g^*=
\begin{pmatrix}
\hdelta^*&0\\
\frac 12\sd&\frac 12r^2\hd r^{-2}\\
\end{pmatrix}.
\end{equation}

where $\iota_{dr}: T^*\hX\to\BR$ is the contraction with the aspherical vector field $dr^{\sharp}$, i.e. $\iota_{dr}(\cdot)=\hg^{-1}(dr, \cdot)$. We then calculate the first covariant derivatives on symmetric $2$-tensor $h\in S^2T^*M$

\begin{gather*}
\nabla_ih_{jk}=\hnabla_ih_{jk},
\quad \nabla_ih_{ja}=\hnabla_ih_{ja}-r^{-1}(\pa_ir)h_{ja},
\quad \nabla_i h_{ab}=\pa_ih_{ab}-2r^{-1}(\pa_ir)h_{ab}\\
\nabla_ah_{ij}=\pa_ah_{ij}-r^{-1}(\pa_ir)h_{aj}-r^{-1}(\pa_jr)h_{ia},
\quad \nabla_ah_{ib}=\slashed{\nabla}_ah_{ib}+r(\pa^kr)h_{ik}\sg_{ab}-r^{-1}(\pa_ir)h_{ab}, \\ \nabla_ah_{bc}=\slashed{\nabla}_ah_{bc}+r(\pa^kr)(h_{kc}\sg_{ab}+h_{kb}\sg_{ac}).
\end{gather*}

Therefore the negative divergence $(\delta_gh)_\m=-\nabla^\n h_{\n\m}$ which maps symmetric $2$-tensors to $1$-form takes the following form (under the splitting (<ref>) and (<ref>))

\begin{equation}\label{eq:ndivergence}
\delta_g=
\begin{pmatrix}
r^{-2}\hdelta r^2& r^{-2}\sdelta& r^{-3}(dr)\str\\
0&r^{-2}\hdelta r^{2}&r^{-2}\sdelta
\end{pmatrix}
\end{equation}

where $(\sdelta h)_i=-\slashed{\nabla}^ah_{ai}$ and $(\hdelta h)_a=-\hnabla^i h_{ia}$. We can also express $(G_gh)_{\m\n}=h_{\m\n}-\frac 12g_{\m\n}\tr_gh$ as

\begin{equation}\label{eq:G_g}
\begin{pmatrix}
G_{\hg}&0&-\frac 12r^{-2}\hg\str\\
0&1&0 \\
-\frac 12r^2\sg\htr&0&G_{\sg}
\end{pmatrix}.
\end{equation}

Putting (<ref>), (<ref>) and (<ref>) together and using $\sdelta G_{\sg}=\sdelta+\frac 12\sd\str,\  \hdelta G_{\hg}=\hdelta+\frac 12\hd\htr$ yield

\begin{equation}
\delta_g^*\delta_gG_g=%\delta_g^*
%	\begin{pmatrix}
%	r^{-2}\hdelta r^2+\frac 12\hd\htr&r^{-2}\sdelta&\frac 12r^{-2}\hd\str\\
%-\frac 12\sd\htr&r^{-2}\hdelta r^2&	r^{-2}\sdelta +\frac 12 r^{-2}\sd\str
%	\end{pmatrix}
\begin{pmatrix}
\hdelta^*r^{-2}\hdelta r^2+\frac 12\hdelta^*\hd\htr&\hdelta^*r^{-2}\sdelta&\frac 12\hdelta^*r^{-2}\hd\str	\\
\frac 12 r^{-2}\sd\hdelta r^2+\frac 14 \hd\sd\htr+\frac 14 r^2\hd r^{-2}\sd\htr&\frac 12r^{-2}\sd\sdelta+\frac 12 r^2\hd r^{-4}\hdelta r^2&\frac 14 r^{-2}\hd\sd\str+\frac 12r^2\hd r^{-4}\sdelta+\frac14 r^2\hd r^{-4}\sd\str\\
r^{-1}\sg\iota_{dr}\hdelta r^2+\frac12 r\sg\iota_{dr}\hd\htr+\frac 12\sdelta^*\sd\htr&r^{-1}\sg\iota_{dr}\sdelta+r^{-2}\sdelta^*\hdelta r^2&\frac 12 r^{-1}\sg\iota_{dr}\hd\str+r^{-2}\sdelta^*\sdelta+\frac 12r^{-2}\sdelta^*\sd\str
\end{pmatrix}.
\end{equation}

§.§ Calculation of $\Box_g$

Next we compute the second covariant derivatives on symmetric $2$-tensor $h\in S^2T^*M$. Since our goal is to find the form of $\Box_g=g^{\m\n}\nabla_\m\nabla_\n$ applied to $h$, it suffices to compute the following second covariant derivatives

\begin{equation}
\begin{split}
\nabla_i\nabla_j h_{km}&=\hnabla_i\hnabla_j h_{km},\\
\nabla_a\nabla_b h_{km}&=\snabla_a\snabla_bh_{km}-r^{-1}\pa_k r\left(\snabla_a h_{bm}+\snabla_b h_{am}\right)-r^{-1}\pa_m r\left(\snabla_a h_{kb}+\snabla_b h_{ka}\right)\\
&\quad+r\pa^i r\hnabla_i h_{km}\sg_{ab}+2r^{-2}\pa_kr\pa_mr h_{ab}-\sg_{ab}\left(\pa_k r\pa^irh_{im}+\pa_m r\pa^irh_{ik}\right) \\
\nabla_i\nabla_jh_{kc}&=\hnabla_i\hnabla_jh_{kc}-r^{-1}\pa_i r\hnabla_j h_{kc}-r^{-1}\pa_j r\hnabla_i h_{kc}+2r^{-2}\pa_i r\pa_j rh_{kc}-\frac 12r^{-1}\weight'\hg_{ij}h_{kc},\\
\nabla_a\nabla_b h_{kc}&=\snabla_a\snabla_bh_{kc}-r^{-1}\pa_kr\left(\snabla_ah_{bc}+\snabla_bh_{ac}\right)+r\sg_{ac}\pa^ir\snabla_bh_{ik}+r\sg_{bc}\pa^ir\snabla_ah_{ik}+r\sg_{ab}\pa^i r\hnabla_i h_{kc}\\
\nabla_i\nabla_j h_{cd}&=\hnabla_i\hnabla_j h_{cd}-2r^{-1}\left(\pa_ir\hnabla_j h_{cd}+\pa_j r\hnabla_i h_{cd}\right)+6r^{-2}\pa_ir\pa_jrh_{cd}-r^{-1}\weight'\hg_{ij}h_{cd},\\
\nabla_a\nabla_b h_{cd}&=\snabla_{a}\snabla_bh_{cd}+r\pa^ir\left(\sg_{bc}\snabla_{a}h_{di}+\sg_{bd}\snabla_{a}h_{ci}+\sg_{ac}\snabla_{b}h_{di}+\sg_{ad}\snabla_{b}h_{ci}+\sg_{ab}\hnabla_ih_{cd}\right)\\
\end{split}
\end{equation}

where $\hBox$ acting on aspherical symmetric $2$-tensors is the tensor wave operator, on mixed symmetric tensors is the $1$-form wave operator acting on the aspherical part, and on spherical symmetric tensors is the scalar wave operator acting on the aspherical variables; the operator $\sL$ is defined in a similar manner. Then we find

\begin{equation}\label{eq:Box}
\begin{split}
\frac 12\Box_g&=
\begin{pmatrix}
\frac 12\hBox+\frac 12r^{-2}\sL&2r^{-3}(dr)\otimes_s \sdelta&r^{-4}(dr\otimes dr)\str\\
r^{-1}\iota_{dr}\sd&\frac 12\hBox+\frac 12r^{-2}\sL &r^{-3}(dr)\otimes\sdelta\\
\sg\iota_{dr}\iota_{dr}&2r^{-1}\iota_{dr}\sdelta^*&\frac 12	\hBox+\frac 12r^{-2}\sL
\end{pmatrix}\\
&	\quad+	\begin{pmatrix}
r^{-1}\pa^ir\hnabla_i\!-\!2r^{-2}(d r)\otimes_s\iota_{dr}&0&0\\
0& -\frac 12r^{-2}(\weight+r\weight')\!-\!2r^{-2}(dr)\otimes\iota_{dr}&0\\
0&0&-r^{-1}\pa^ir \hnabla_i\!-\!r^{-1}\weight'
\end{pmatrix}.
\end{split}
\end{equation}

On the aspherical part $\hX$, we further calculate the relevant operators on the static region where $\weight>0$ by working in $(t,r)$ coordinates. On the static region, the metric $\hg$ reads

\[
\hg=-\weight dt^2+\weight^{-1}dr^2.
\]

We further split

\begin{equation}\label{eq:splitofX}
T^*\hX=\langle \hd t\rangle\oplus\langle \hd r\rangle,\quad S^2T^*\hX=\langle\hd t^2\rangle\oplus\langle2\hd t\hd r\rangle\oplus\langle\hd r^2\rangle.
\end{equation}

Since $\hcs_{tr}^t=\hcs_{rt}^t=\frac 12\weight^{-1}\weight', \hcs_{tt}^r=\frac 12\weight\weight', \hcs_{rr}^r=-\frac 12 \weight^{-1}\weight'$, with respect to the above split (<ref>), we see that acting on scalar functions, $\pa^ir \hnabla_i=\weight\pa_r$,

\begin{equation}\label{eq:static1form1}
\hd=\begin{pmatrix}
\pa_t\\
\pa_r
\end{pmatrix},\quad \hBox=-\weight^{-1}\pa_t^2+\weight\pa_r^2+\weight'\pa_r=-\weight^{-1}\pa_t^2+\pa_r\weight\pa_r
\end{equation}

On $1$-forms, we have $\iota_{dr}=(0,\ \weight)$,

\begin{equation}\label{eq:static1form2}
\begin{gathered}
\hdelta=(\weight^{-1}\pa_t,\  -\weight\pa_r-\weight')=(\weight^{-1}\pa_t,\  -\pa_r\weight),\quad\hdelta^*=\begin{pmatrix}
\pa_t & -\frac 12\weight\weight'\\
\frac 12 \weight\pa_r\weight^{-1}&\frac 12\pa_t\\
\end{pmatrix},\\
\pa^i r\hnabla_i=
\begin{pmatrix}
\weight\pa_r-\frac12\weight'&0\\
0&	\weight\pa_r+\frac12\weight'
\end{pmatrix}=
\begin{pmatrix}
\weight^{3/2}\pa_r\weight^{-1/2}&0\\
0&	\weight^{1/2}\pa_r\weight^{1/2}
\end{pmatrix},\\
\hstar=\begin{pmatrix}
\end{pmatrix},\quad
\begin{pmatrix}
\end{pmatrix}.
\end{gathered}
\end{equation}

\begin{equation}\label{eq:static1form3}
\hBox=\begin{pmatrix}
-\weight^{-1}\pa_t^2+\weight\pa_r^2-\frac 12\weight''&\weight'\pa_t\\
\weight^{-2}\weight'\pa_t&	-\weight^{-1}\pa_t^2+\weight\pa_r^2+2\weight'\pa_r+\frac 12\weight''
\end{pmatrix}.
\end{equation}

On symmetric $2$-tensors, we find that

\begin{equation}
\begin{gathered}
\hdelta=\begin{pmatrix}
\weight^{-1}\pa_t&-\pa_r\weight&0\\
\end{pmatrix},\quad \iota_{dr}=
\begin{pmatrix}
\end{pmatrix},\\
\pa^ir \hnabla_i=
\begin{pmatrix}
\weight\pa_r-\weight'&0&0\\
0&0&	\weight\pa_r+\weight'
\end{pmatrix}=
\begin{pmatrix}
\weight^2\pa_r\weight^{-1}&0&0\\
\end{pmatrix},\\
\hBox=	-\weight^{-1}\pa_t^2+\weight\pa_r^2+
\begin{pmatrix}
-\weight'\pa_r+\frac 12\weight^{-1}\weight'^2-\weight''&2\weight'\pa_t&-\frac12\weight\weight'^2\\
\weight^{-2}\weight'\pa_t&\weight'\pa_r-\weight^{-1}\weight'^2&\weight'\pa_t\\
-\frac 12 \weight^{-3}\weight'^2&2\weight^{-2}\weight'\pa_t&3\weight'\pa_r+\frac 12\weight^{-1}\weight'^2+\weight''
\end{pmatrix}.
\end{gathered}
\end{equation}

§.§ Calculation of $D_{(g, dA)}T(\dg, d\dA)$
Recall that the Reissner-Nordström electromagnetic $4$-potential is given by

\begin{equation}\label{eq:4potentialt*}
A=\BQ r^{-1}dt_{*},\quad F=\BQ r^{-2}dt_*\wedge dr=\BQ r^{-2}\widehat{\mbox{vol}}
\end{equation}

Since $\hvol_{ij}=\epsilon_{ij}, \hvol_{ia}=\hvol_{ab}=0$ where $\epsilon$ is the Levi-Civita symbol, i.e., $\epsilon_{12}=-\epsilon_{21}=1$, it is clear that $\hvol^{\al\be}=-\hvol_{\al\be}$. A direct computation implies

\[
F^{\al\be}F_{\al\be}=-\frac{2\BQ^2}{r^4},\quad \dg^{\al\be}F_{\m\al}F_{\n\be}=\begin{cases}\frac{\BQ^2}{r^4}\left(\dg_{\m\n}-\hg_{\m\n}\htr\dg\right)\quad \mbox{if } \m, \n\in\{i,j\}\\
\end{cases}.
\]

Then according to (<ref>), we find

\begin{equation}
\begin{split}
\left(D_{(g, dA)}T(\dg, 0)	\right)_{ij}&=-\frac{\BQ^2}{r^4}\left(\dg_{ij}-\hg_{ij}\htr\dg-\frac 12\dg_{ij}+\frac 12\hg_{ij}\htr\dg\right)=-\frac{\BQ^2}{2r^4}\left(\dg_{ij}-\hg_{ij}\htr\dg\right),\\
\left(D_{(g, dA)}T(\dg, 0)	\right)_{ia}&=\frac{\BQ^2}{2r^4}\dg_{ia},\quad \left(	D_{(g, dA)}T(\dg, 0)\right)_{ab}=\frac{\BQ^2}{2r^4}\left(\dg_{ab}-r^2\sg_{ab}\htr\dg\right)
\end{split}
\end{equation}

which means

\begin{equation}\label{eq:DTofg}
D_{(g, dA)}T(\cdot, 0)=\frac{\BQ^2}{2r^4}
\begin{pmatrix}
\hg\htr	\!-\!1&0&0\\
\end{pmatrix}.
\end{equation}

\begin{gather*}
F^{\al\be}\dF_{\al\be}=\BQ r^{-2}\hvol^{ij}\dF_{ij}
%=-2\BQ r^{-2}\dF_{t_* r}=2\BQ r^{-2}\dF_{t_* r}\hstar\hvol%=2\BQ r^{-2}\hstar\dF
g^{\al\be}\dF_{i\al}F_{j\be}+g^{\al\be}F_{i\al}\dF_{j\be}=-2\BQ r^{-2}\dF_{t_* r}\hg_{ij}=2\BQ r^{-2}\dF_{t_* r}(\hstar\hvol)\hg_{ij},\\
g^{\al\be}\dF_{a\al}F_{i\be}=\hg^{kj}\dF_{ak}F_{ij}=-\BQ r^{-2}\hg^{kj}\epsilon_{ij}\dF_{ka}=\BQ r^{-2}\hstar\dF_{\bullet a}
\end{gather*}

where we use the fact $\hstar\hvol=-1$ and the Hodge star operator $\hstar$ only acts on the aspherical part, we have

\begin{equation}\label{eq:DTofA}
D_{(g, dA)}T(0, d(\cdot))=-\BQ r^{-2}
\begin{pmatrix}
\hstar\sd&-\hstar\hd\\
\end{pmatrix}.
\end{equation}

This finishes the calculation of $\mathscr{L}_1(\dg, d\dA)$.

§.§ Calculation of $D_{(g, dA)}(\delta_g dA)(\dg, d\dA)$

It remains to consider $\mathscr{L}_2(\dg, d\dA)$ which is the linearization of $\delta_gdA$. We again use the following splitting

\begin{equation}\label{eq:splitof2form}
T^*M=T^*_{AS}\oplus T^*_{S}, \quad\wedge^2T^*M=\wedge^2T^*_{AS}\oplus(T^*_{AS}\wedge T^*_{S})\oplus\wedge^2T^*_{S}.
\end{equation}

With respect to the above splitting (<ref>), $d:T^*M\to \wedge^2T^*M$ and $\delta_g: \wedge^2T^*M\to T^*M$ take the form

\begin{equation*}
\begin{split}
\hd&0\\
\end{pmatrix}, \quad
\delta_g=
\begin{pmatrix}
r^{-2}\hdelta r^2& -r^{-2}\sdelta& 0\\
0&\hdelta &r^{-2}\sdelta
\end{pmatrix},
\end{split}
\end{equation*}

and thus by (<ref>)

\begin{equation}\label{eq:DMofA}
D_{(g, dA)}(\delta_g dA)(0, d(\cdot))=
\begin{pmatrix}
r^{-2}\hdelta r^2\hd+r^{-2}\sdelta\sd&-r^{-2}\sdelta\hd\\
\end{pmatrix}.
\end{equation}

Finally we are left with dealing with $D_{(g, dA)}(\delta_g dA)(\dg, 0)$. We first calculate the first covariant derivative on the $2$-form $F$. A direct calculation implies $\hnabla_k\hvol_{ij}=0$, and then

\begin{gather*}
\nabla_k F_{ij}=-2r^{-1}(\pa_k r)F_{ij},\quad
\nabla_iF_{ja}=0,\quad \nabla_i F_{ab}=0,\quad \nabla_a F_{ij}=0, \quad \nabla_aF_{ib}=r(\pa^kr)\sg_{ab}F_{ik},\quad \nabla_aF_{bc}=0.
\end{gather*}

Therefore we have

\begin{gather*}
\dg^{\n\al}\nabla_\al F_{\n i}%=\dg^{jk}\nabla_j F_{ki}+\dg^{ab}\nabla_a F_{bi}
=-2r^{-1}(\pa_jr)F_{ki}\dg^{jk}+r^{-3}\sg^{ab}\dg_{ab}(\pa^kr)F_{ki}=-2\BQ r^{-3}\hstar\bigl((\pa^j r)\dg_{jk}\bigr)+\BQ r^{-5}(\hstar dr)_i\sg^{ab}\dg_{ab}\\
\dg^{\n\al}\nabla_\al F_{\n a}=\dg^{bi}\nabla_b F_{ia}=\frac{\BQ}{r^3}(\pa^k r) (\hstar\dg_{\bullet a})_k
\end{gather*}

We also find $(\delta_g G_g\dg)^\kappa F_{\kappa a}=0$ and $-(\delta_g G_g\dg)^\kappa F_{\kappa i}=-\BQ r^{-2}\hstar\left(\delta_g G_g\dg\right)$, thus

\begin{gather*}
-(\delta_g G_g(\cdot))^\kappa F_{\kappa i}=-\frac{\BQ}{r^2}\begin{pmatrix}
r^{-2}\hstar\hdelta r^2+\frac 12\hstar\hd\htr&r^{-2}\hstar\sdelta&\frac 12r^{-2}\hstar\hd\str\\
\end{pmatrix}.
\end{gather*}

Since $\hg^{ij}\epsilon_{ik}\epsilon_{kl}=\hg_{kl}-\hg_{kl}\htr\hg=-\hg_{kl}$ and $\hnabla_k\hvol_{ij}=0$, we find

\begin{align*}
\frac 12\left(\nabla^\n\dg^{\kappa}_{\ i}-\nabla^\kappa\dg^{\n}_{\ i}\right)F_{\n\kappa}&=-\frac{\BQ}{2r^2}(\hnabla^{k}\dg^{lj}-\hnabla^{l}\dg^{kj})\hg^{mn}\epsilon_{jm}\epsilon_{in}\epsilon_{kl}\\
\end{align*}

and using $\hstar\hstar=(-1)^{k(2-k)+1}I$ and $\delta=\hstar\hd\hstar$ on $\wedge^kT^*\hX$ for $1\leq k\leq 2$, we see that

\begin{align*}
\frac 12\left(\nabla^\n\dg^{\kappa}_{\ a}-\nabla^\kappa\dg^{\n}_{\ a}\right)F_{\n\kappa}&=\frac{\BQ}{2r^2}\hg^{ik}\hg^{jl}\left(\pa_k\dg_{la}-\pa_l\dg_{ka}-r^{-1}(\pa_kr)\dg_{la}+r^{-1}(\pa_lr)\dg_{ka}\right)\epsilon_{ij}\\
&=-\frac{\BQ}{r^3}(\pa^k r) (\hstar\dg_{\bullet a})_k+\frac{\BQ}{r^2}\left(2r^{-1}(\pa^k r) (\hstar\dg_{\bullet a})_k+\hstar\hd\dg_{\bullet a}\right)\\
&=-\frac{\BQ}{r^3}(\pa^k r) (\hstar\dg_{\bullet a})_k+\frac{\BQ}{r^2}\left(-r^2(\pa^k r^{-2}) (\hstar\dg_{\bullet a})_k+\hdelta\hstar\dg_{\bullet a}\right)\\
&=-\frac{\BQ}{r^3}(\pa^k r) (\hstar\dg_{\bullet a})_k+\BQ\hdelta r^{-2}\hstar\dg_{\bullet a}.
\end{align*}

Combining the above calculation and (<ref>) yields

\begin{equation}\label{eq:DMofg}
D_{(g, dA)(\delta_g dA)}(\dg, 0)=\frac{\BQ}{r^2}\begin{pmatrix}
\frac 12\hstar\hd\htr&-r^{-2}\hstar\sdelta&-\frac 12\hstar\hd r^{-2}\str\\
0&r^2\hdelta r^{-2}\hstar&0
\end{pmatrix}.
\end{equation}

Lastly, we discuss the calculation of the gauge terms for the electromagnetic field. Since $A=-\BQ r^{-1}dt_*$, we have

\begin{align*}
(\mathcal{L}_{\omega^\sharp}A)_i=\omega^\al\pa_\al A_i+A_\al\pa_i\omega^\al,\quad 	(\mathcal{L}_{\omega^\sharp}A)_a=\omega^\al\pa_\al A_a+A_\al\pa_a\omega^\al=\pa_a(A_\al\omega^\al)
\end{align*}

and as a consequence

\begin{equation}\label{eq:liederiofA}
\mathcal{L}_{(\cdot)^\sharp}A=	\begin{pmatrix}
\hd\iota_A+\iota_{(\cdot)}\hd A&0\\
\sd\iota_{A}&0
\end{pmatrix}.
\end{equation}

§ CALCULATION OF THE SUBPRINCIPAL OPERATOR AT TRAPPING