We expand $\Phi_{5}$ as

$\displaystyle\Phi_{5}$
$\displaystyle=:I+\varepsilon\mathcal{A}_{1}=:I+\varepsilon
A_{1}+\varepsilon^{2}\mathcal{A}_{2}=:I+\varepsilon
A_{1}+\frac{\varepsilon^{2}A_{1}^{2}}{2}+\varepsilon^{3}\mathcal{A}_{3},$
(10.158) $\displaystyle\Phi_{5}^{-1}$
$\displaystyle=:I+\varepsilon\tilde{\mathcal{A}}_{1}=:I-\varepsilon
A_{1}+\varepsilon^{2}\tilde{\mathcal{A}}_{2}=:I-\varepsilon
A_{1}+\frac{\varepsilon^{2}A_{1}^{2}}{2}+\varepsilon^{3}\tilde{\mathcal{A}}_{3}.$

We first specify the operators of size $O(\varepsilon)$:

######  Lemma 10.52.

We have that

$\displaystyle\Phi_{5}^{-1}\mathcal{L}^{4}\Phi_{5}=\mathcal{D}_{\omega}-\mathcal{B}_{0}+\varepsilon(\mathcal{D}_{\omega}(A_{1})-[\mathcal{B}_{0,1},A_{1}]-\mathcal{B}_{1})+\mathcal{Q}_{1}+\mathcal{W}.$
(10.159)

where

$\displaystyle\mathcal{Q}_{1}$
$\displaystyle:=\varepsilon(\mathcal{D}_{\omega}-\mathcal{D}_{\overline{\omega}})(A_{1})-\varepsilon[\mathcal{B}_{0,2},A_{1}]+\left(\varepsilon^{2}\mathcal{D}_{\omega}(\mathcal{A}_{2})+(\Phi_{5}^{-1}-I)\left(\varepsilon\mathcal{D}_{\omega}(A_{1})+\varepsilon^{2}\mathcal{D}_{\omega}(\mathcal{A}_{2})\right)\right)$
(10.160) $\displaystyle\
+\varepsilon^{2}\left(A_{1}\mathcal{B}_{0}A_{1}+\tilde{\mathcal{A}_{2}}(\mathcal{B}_{0}+\varepsilon\mathcal{B}_{0}A_{1})-\Phi_{5}^{-1}\mathcal{B}_{0}\mathcal{A}_{2}\right)$
$\displaystyle\
-\left(\varepsilon^{2}\mathcal{B}_{1}\mathcal{A}_{1}+\varepsilon^{2}\tilde{\mathcal{A}}_{1}\mathcal{B}_{1}\Phi_{5}\right)-\varepsilon^{2}\Phi_{5}^{-1}\mathcal{B}_{2}\Phi_{5}+-\Phi_{5}^{-1}\Pi_{S^{\perp}}\partial_{x}Op^{W}(\mathfrak{r}_{-2,\leq
3})\Phi_{5}+\Phi_{5}^{-1}R_{4}\Phi_{5}$ $\displaystyle\mathcal{W}$
$\displaystyle:=\Phi_{5}^{-1}\Pi_{S^{\perp}}\partial_{x}W_{1}\Phi_{5}.$

###### Proof.

Using (10.156), we have that

$\displaystyle\Phi_{5}^{-1}\mathcal{L}^{4}\Phi_{5}=\Phi_{5}^{-1}\mathcal{D}_{\omega}\Phi_{5}-\left(\Phi_{5}^{-1}\mathcal{B}_{0}\Phi_{5}+\varepsilon\Phi_{5}^{-1}\mathcal{B}_{1}\Phi_{5}+\varepsilon^{2}\Phi_{5}^{-1}\mathcal{B}_{2}\Phi_{5}\right)+\Phi_{5}^{-1}\mathcal{Q}\Phi_{5}.$

Conjugation with $\mathcal{D}_{\omega}.$ Using (10.158), we have

$\displaystyle\Phi_{5}^{-1}\mathcal{D}_{\omega}\Phi_{5}$
$\displaystyle=\Phi_{5}^{-1}\left(\mathcal{D}_{\omega}(\Phi_{5})+\Phi_{5}\mathcal{D}_{\omega}\right)$
$\displaystyle=\mathcal{D}_{\omega}+\Phi_{5}^{-1}\left(\varepsilon\mathcal{D}_{\omega}(A_{1})+\varepsilon^{2}\mathcal{D}_{\omega}(\mathcal{A}_{2})\right)$
$\displaystyle=\mathcal{D}_{\omega}+\varepsilon\mathcal{D}_{\omega}(A_{1})+\left(\varepsilon^{2}\mathcal{D}_{\omega}(\mathcal{A}_{2})+(\Phi_{5}^{-1}-I)\left(\varepsilon\mathcal{D}_{\omega}(A_{1})+\varepsilon^{2}\mathcal{D}_{\omega}(\mathcal{A}_{2})\right)\right)$
$\displaystyle=\mathcal{D}_{\omega}+\varepsilon\mathcal{D}_{\overline{\omega}}(A_{1})+\varepsilon(\mathcal{D}_{\omega}-\mathcal{D}_{\overline{\omega}})(A_{1})+\left(\varepsilon^{2}\mathcal{D}_{\omega}(\mathcal{A}_{2})+(\Phi_{5}^{-1}-I)\left(\varepsilon\mathcal{D}_{\omega}(A_{1})+\varepsilon^{2}\mathcal{D}_{\omega}(\mathcal{A}_{2})\right)\right).$
(10.161)

Conjugation with $\mathcal{B}_{0}.$ Using (10.158) again, we have that

$\displaystyle\Phi_{5}^{-1}\mathcal{B}_{0}\Phi_{5}$
$\displaystyle=\Phi_{5}^{-1}\mathcal{B}_{0}(I+\varepsilon
A_{1}+\varepsilon^{2}\mathcal{A}_{2})$
$\displaystyle=\Phi_{5}^{-1}\mathcal{B}_{0}(I+\varepsilon
A_{1})+\varepsilon^{2}\Phi_{5}^{-1}\mathcal{B}_{0}\mathcal{A}_{2}$
$\displaystyle=\mathcal{B}_{0}+\varepsilon[\mathcal{B}_{0},A_{1}]-\varepsilon^{2}\left(A_{1}\mathcal{B}_{0}A_{1}-\tilde{\mathcal{A}_{2}}(\mathcal{B}_{0}+\varepsilon\mathcal{B}_{0}A_{1})-\Phi_{5}^{-1}\mathcal{B}_{0}\mathcal{A}_{2}\right).$

Conjugation with $\mathcal{B}_{1}.$ Similarly,

$\displaystyle\varepsilon\Phi_{5}^{-1}\mathcal{B}_{1}\Phi_{5}=\varepsilon\mathcal{B}_{1}+\varepsilon^{2}\mathcal{B}_{1}\mathcal{A}_{1}+\varepsilon^{2}\tilde{\mathcal{A}}_{1}\mathcal{B}_{1}\Phi_{5}.$

Conjugation with $\mathcal{B}_{2}.$ We do not rewrite it since it is already
$O(\varepsilon^{2})$.

Conjugation with $\mathcal{Q}$. Recalling $\mathcal{Q}$ from (10.155), we have

$\displaystyle\Phi_{5}^{-1}\mathcal{Q}\Phi_{5}$
$\displaystyle=-\Phi_{5}^{-1}\Pi_{S^{\perp}}\partial_{x}Op^{W}(\mathfrak{r}_{-2,\leq
3})\Phi_{5}+\Phi_{5}^{-1}\Pi_{S^{\perp}}\partial_{x}W_{1}\Phi_{5}+\Phi_{5}^{-1}R_{4}\Phi_{5}.$

Summing everything up, we obtain

$\displaystyle\Phi_{5}^{-1}\mathcal{L}^{4}\Phi_{5}=\mathcal{D}_{\omega}-\mathcal{B}_{0}+\varepsilon\left(\mathcal{D}_{\overline{\omega}}(A_{1})-[\mathcal{B}_{0},A_{1}]-\mathcal{B}_{1}\right)+\varepsilon(\mathcal{D}_{\omega}-\mathcal{D}_{\overline{\omega}})(A_{1})+Q_{1}+\mathcal{W},$
(10.162)

where $Q_{1}$ and $\mathcal{W}$ are

$\displaystyle{Q}_{1}$
$\displaystyle:=\left(\varepsilon^{2}\mathcal{D}_{\omega}(\mathcal{A}_{2})+(\Phi_{5}^{-1}-I)\left(\varepsilon\mathcal{D}_{\omega}(A_{1})+\varepsilon^{2}\mathcal{D}_{\omega}(\mathcal{A}_{2})\right)\right)$
(10.163) $\displaystyle\
+\varepsilon^{2}\left(A_{1}\mathcal{B}_{0}A_{1}-\tilde{\mathcal{A}_{2}}(\mathcal{B}_{0}+\varepsilon\mathcal{B}_{0}A_{1})-\Phi_{5}^{-1}\mathcal{B}_{0}\mathcal{A}_{2}\right)$
$\displaystyle\
-\left(\varepsilon^{2}\mathcal{B}_{1}\mathcal{A}_{1}+\varepsilon^{2}\tilde{\mathcal{A}}_{1}\mathcal{B}_{1}\Phi_{5}\right)-\varepsilon^{2}\Phi_{5}^{-1}\mathcal{B}_{2}\Phi_{5}+-\Phi_{5}^{-1}\Pi_{S^{\perp}}\partial_{x}Op^{W}(\mathfrak{r}_{-2,\leq
3})\Phi_{5}+\Phi_{5}^{-1}R_{4}\Phi_{5}$ $\displaystyle\mathcal{W}$
$\displaystyle:=\Phi_{5}^{-1}\Pi_{S^{\perp}}\partial_{x}W_{1}\Phi_{5}.$

Recalling $\mathcal{B}_{0,1}$ and $\mathcal{B}_{0,2}$ from (10.4.1), we have

$[\mathcal{B}_{0},A_{1}]=[\mathcal{B}_{0,1},A_{1}]+[\mathcal{B}_{0,2},A_{1}].$

Plugging this into (10.162) and defining
$\mathcal{Q}_{1}:=Q_{1}+\varepsilon(\mathcal{D}_{\omega}-\mathcal{D}_{\overline{\omega}})(A_{1})-\varepsilon[\mathcal{B}_{0,2},A_{1}]$,
we obtain (10.159) with (10.160). ∎

#####  10.4.1.2 Choice of $\rho_{1}$

We aim to choose $A_{1}$ by choosing the symbol $\rho_{1}$ (see (10.150)), to
eliminate the terms of size $O(\varepsilon)$ in (10.162), that is,

$\displaystyle\mathcal{D}_{\overline{\omega}}(A_{1})-[\mathcal{B}_{0,1},A_{1}]-\mathcal{B}_{1}=0.$
(10.164)

######  Proposition 10.53.

There exists
$\rho_{1}=\rho_{1}(\omega,\varphi,x,\xi)\in\mathfrak{B}^{-1-\alpha}_{1}$ such
that

  1. (1)

With $A_{1}$ defined as in (10.150), we have that

$\displaystyle\mathcal{D}_{\overline{\omega}}(A_{1})-[\mathcal{B}_{0,1},A_{1}]-\mathcal{B}_{1}=0.$
(10.165)

  2. (2)

$\rho_{1}$ is a real-valued and reversibility preserving symbol. Furthermore,
$\partial_{x}Op^{W}(\rho_{1})$ is a real operator.

###### Proof.

Proof of (1). Using (10.150) and (10.4.1), it is easy to see that (10.164) is
equivalent to (see (2.31) for the notation $[\cdot,\cdot]_{x}$),

$\displaystyle\Pi_{S^{\perp}}\left(Op^{W}(\mathcal{D}_{\overline{\omega}}\rho_{1})-\left[Op^{W}\left(-\frac{1}{2}m_{1,\alpha}(\xi)+\frac{T_{\alpha}}{4}\right),Op^{W}(\rho_{1})\right]_{x}-Op^{W}(\mathfrak{b}_{1})\right)=0.$
(10.166)

Recalling $\mathfrak{b}_{1}\in\mathfrak{B}^{-2}_{1}$, which follows from Lemma
10.45, and the definition of $\mathfrak{B}^{-2}_{1}$ from Definition 8.8, we
can write $\mathfrak{b}_{1}$ as

$\displaystyle\mathfrak{b}_{1}(\omega,\varphi,x,\xi)=\sum_{j_{k}\in
S}C_{\mathfrak{b},j_{k}}(\xi)\sqrt{j_{k}\zeta_{k}}e^{\text{i}(\mathtt{l}(j_{k})\cdot\varphi+j_{k}x)},\text{
for some $C_{\mathfrak{b},j_{k}}\in\mathcal{S}^{-2}$.}$ (10.167)

Using (2.14), let us denote by $\mathfrak{b}_{1,s}$, a symbol such that
$Op^{W}(\mathfrak{b}_{1})=Op(\mathfrak{b}_{1,s})$, that is,

$\displaystyle\widehat{\mathfrak{b}}_{1,s}(\omega,\varphi,x,\xi):=\sum_{j_{k}\in
S}\underbrace{C_{\mathfrak{b},j_{k}}(\xi+\frac{j_{k}}{2})}_{=:C_{\mathfrak{b}_{1,s},j_{k}}(\xi)}\sqrt{j_{k}\zeta_{k}}e^{\text{i}(\mathtt{l}(j_{k})\cdot\varphi+j_{k}x)}.$
(10.168)

Since $S$ is a finite set (see (4.6)), it is clear that
$C_{\mathfrak{b}_{1,s},j_{k}}\in\mathcal{S}^{-2}$, therefore,
$\mathfrak{b}_{1,s}\in\mathfrak{B}^{-2}_{1}$. We look for a symbol
$\rho_{1,s}\in\mathfrak{B}^{-1-\alpha}_{1}$ such that

$\displaystyle\Pi_{S^{\perp}}\left(Op(\mathcal{D}_{\overline{\omega}}\rho_{1,s})-\left[Op\left(-\frac{1}{2}m_{1,\alpha}(\xi)+\frac{T_{\alpha}}{4}\right),Op(\rho_{1,s})\right]_{x}-Op(\mathfrak{b}_{1,s})\right)=0.$
(10.169)

Once we find $\rho_{1,s}$, then we can define
$\widehat{\rho}_{1}(\omega,\varphi,k,\xi):=\widehat{\rho}_{1,s}(\omega,\varphi,k,\xi-k)$
for $k\in S$, so that (2.14) tells us $Op^{W}(\rho_{1})=Op(\rho_{1,s})$,
therefore, $\rho_{1}$ is a solution to (10.166).

To find $\rho_{1,s}$ solving (10.169), we use the notations in (2.36) and
(2.98) to see that (10.169) is equivalent to

$\displaystyle\text{i}\left(\overline{\omega}\cdot
l-\left(\left(-\frac{1}{2}\lambda_{\alpha}(k+j)-(k+j)\frac{T_{\alpha}}{4}\right)-\left(-\frac{1}{2}\lambda_{\alpha}(j)-j\frac{T_{\alpha}}{4}\right)\right)\right)\widehat{\rho}_{1,s}^{\varphi,x}(\omega,l,k,j)=\widehat{\mathfrak{b}}^{\varphi,x}_{1,s}(\omega,l,k,j),$
(10.170)

for all $j\in S^{\perp}$. We set

$\displaystyle{\rho}_{1,s}(\omega,\varphi,x,\xi):=\sum_{j_{k}\in
S}2\text{i}C_{\mathfrak{b}_{1,s},j_{k}}(\xi)\kappa_{-j_{k},1-\alpha}(\xi)\sqrt{j_{k}\zeta_{k}}e^{\text{i}(\mathtt{l}(j_{k})\cdot\varphi+j_{k}x)},$
(10.171)

where $\xi\mapsto\kappa_{-j_{k},1-\alpha}(\xi)$ is the multiplier obtained in
Lemma 2.35. Since $C_{\mathfrak{b}_{1,s,j_{k}}}\in\mathcal{S}^{-2}$ and
$\kappa_{-j_{k},1-\alpha}\in\mathcal{S}^{1-\alpha}$, we have that
$\rho_{1,s}\in\mathfrak{B}^{-1-\alpha}_{1}$. Now, it suffices to show that
$\rho_{1,s}$ solves (10.170). Since $\rho_{1,s}$ and $\mathfrak{b}_{1,s}$ are
supported on a finite number of Fourier modes as seen in (10.171), and
(10.167), we have that both sides in (10.170) are equal to $0$, if
$(l,k)\neq(\mathtt{l}(j_{k}),j_{k})$ for some $j_{k}\in S$. On the other hand,
if $(l,k)=(\mathtt{l}(j_{k}),j_{k})$ for some $j_{k}\in S$, then
$\overline{\omega}$ in (5.9) tells us that (also see $\lambda_{\alpha}$ in
(2.98))

$\displaystyle\overline{\omega}\cdot l$
$\displaystyle-\left(\left(-\frac{1}{2}\lambda_{\alpha}(k+j)-(k+j)\frac{T_{\alpha}}{4}\right)-\left(-\frac{1}{2}\lambda_{\alpha}(j)-j\frac{T_{\alpha}}{4}\right)\right)$
$\displaystyle=\frac{1}{2}\left(\lambda_{\alpha}(j+j_{k})-\lambda_{\alpha}(j)-\lambda_{\alpha}(j_{k})\right),$
(10.172)

hence, the left-hand side of (10.170) is given by

$LHS=\frac{\text{i}}{2}\left(\lambda_{\alpha}(j+j_{k})-\lambda_{\alpha}(j)-\lambda_{\alpha}(j_{k})\right)\times
2\text{i}C_{\mathfrak{b}_{1,s}j_{k}}(j)\kappa_{-j_{k},1-\alpha}(j)=C_{\mathfrak{b}_{1,s},j_{k}}(j)=\widehat{\mathfrak{b}}_{1,s}^{\varphi,x}(\omega,\mathtt{l}(j_{k}),j_{k},j),$

for all $j\in S^{\perp}$, where the second equality follows from (2.100).
Therefore, (10.171) solves (10.170) and thus, (10.169).

Proof of (2). We will prove that $\rho_{1}$ can be chosen to be a real-valued
symbol only. This property follows from the fact that $\rho_{1}$ is a solution
to (10.166), while,
$Op^{W}\left(-\frac{1}{2}m_{1,\alpha}(\xi)+\frac{T_{\alpha}}{4}\right)$ and
$Op^{W}(\mathfrak{b}_{1})$ are symmetric operators, which follows from (2.15)
with the fact that the symbols
$-\frac{1}{2}m_{1,\alpha}(\xi)+\frac{T_{\alpha}}{4}$ and $\mathfrak{b}_{1}$
are real-valued, which follows from Proposition 2.34 for $m_{1,\alpha}(\xi)$
and Lemma 10.47 with (10.146) for $\mathfrak{b}_{1}$). Indeed, Proposition
2.34 and Lemma 10.47 with (10.146) tell us that
$Op^{W}\left(-\frac{1}{2}m_{1,\alpha}(\xi)+\frac{T_{\alpha}}{4}\right)$ and
$Op^{W}(\mathfrak{b}_{1})$ are real and reversibility preserving operators,
the other desired properties of $\rho_{1}$ follow in the same way.

To show that $\rho_{1}$ can be chosen to be a real-valued symbol, we see that
$\rho_{1}$ is a solution to (10.166), therefore, (2.15) tells us that (using
that $Op^{W}\left(-\frac{1}{2}m_{1,\alpha}(\xi)+\frac{T_{\alpha}}{4}\right)$
and $Op^{W}(\mathfrak{b}_{1})$ are symmetric operators),
$Op^{W}(\overline{\rho_{1}}$) solve (10.166) as well. Since the homological
equation (10.166) is linear, we see that
$\frac{1}{2}(\rho_{1}+\overline{\rho_{1}})$ solves (10.166). Hence, if
necesary, we can replace $\rho_{1}$ by
$\frac{1}{2}(\rho_{1}+\overline{\rho_{1}})$, which gives a real-valued symbol.
Clearly, if $\rho_{1}\in\mathfrak{B}^{-1-\alpha}_{1}$, then
$\frac{1}{2}(\rho_{1}+\overline{\rho_{1}})\in\mathfrak{B}^{-1-\alpha}_{1}$ as
well, which follows immediately from its definition in Definition 8.8.

In view of (2.135), we can replace $\rho_{1}$ by
$\frac{1}{2}(\rho_{1}(\varphi,x,\xi)+\overline{\rho_{1}}(\varphi,x,-\xi))$ and
after than, again replace it by
$\frac{1}{2}(\rho_{1}(\varphi,x,\xi)-\rho_{1}(-\varphi,-x,-\xi))$, which
eventually gives us a symbol $\rho_{1}$ with all the desired properties. ∎

#####  10.4.1.3 Analysis of $\Phi_{5}$

######  Lemma 10.54.

For $k=1,2,3$, there exist $\rho_{k,*}(\tau)\in\mathcal{S}^{-k\alpha}$ such
that

$\Psi_{5}(\tau)-I=\varepsilon
Op(\rho_{1,*}(\tau))+\frac{1}{2}\varepsilon^{2}Op(\rho_{2,*}(\tau))+\varepsilon^{3}Op(\rho_{3,*}(\tau)),$

and $\rho_{k,*}$ satisfy

$\displaystyle\sup_{\tau\in[0,1]}|\rho_{k,*}(\tau)|^{\text{Lip}(\gamma,\Omega_{1})}_{-k\alpha,s,\eta_{0}}$
$\displaystyle\leq_{s,\mathtt{pe},\eta_{0}}1,$ (10.173)
$\displaystyle\sup_{\tau\in[0,1]}|d_{i}\rho_{k,*}(\tau)(i_{0})[\hat{\textbf{\i}}]|_{-k\alpha,s,\eta_{0}}$
$\displaystyle=0.$ (10.174)

###### Proof.

Let us pick $\rho_{1,1}\in\mathcal{S}^{-\alpha}$ such that

$\displaystyle Op(\rho_{1,1})=\partial_{x}Op^{W}(\rho_{1}),$ (10.175)

so that (10.152) gives us that

$\displaystyle\Psi_{5}(\tau)-I=\varepsilon\tau
Op(\rho_{1,1})+\frac{(\varepsilon\tau)^{2}}{2}Op(\rho_{1,1})\circ
Op(\rho_{1,1})+(\varepsilon\tau)^{3}\sum_{n=3}^{\infty}(\varepsilon\tau)^{n-3}\frac{1}{n!}\left(Op(\rho_{1,1})\right)^{n}.$
(10.176)

From (2.14) and Proposition 10.53, which tells us that
$\rho_{1}\in\mathfrak{B}^{-1-\alpha}_{1}$, hence
$\rho_{1,1}\in\mathfrak{B}^{-\alpha}_{1}$. This gives us that

$\displaystyle|\rho_{1,1}|^{\text{Lip}(\gamma,\Omega_{1})}_{-\alpha,s,\eta_{0}}\leq
C_{1}(\mathtt{pe},s,\eta_{0}),\quad|d_{i}\rho_{1,1}(i_{0})[\hat{\textbf{\i}}]|_{-\alpha,s,\eta_{0}}=0,$
(10.177)

for some constant $C_{1}(\mathtt{pe},s,\eta_{0})>0$, where the estimate for
$d_{i}\rho_{1,1}$ follows from the fact that $\rho_{1,1}$ does not depend on
the embedding $i_{0}$, since $\rho_{1,1}\in\mathfrak{B}^{-\alpha}_{1}$ (see
Definition 8.8). Also, this property tells us that

$\widehat{\rho}_{1,1}^{\varphi,x}(l,k,\xi)=0,\text{ if $|l|+|k|\geq N_{S}$,}$

for some ${N}_{S}>0$, depending on the choice of the tangential sites in
(4.6). Note that choosing $\rho_{1,*}$ and $\rho_{2,*}$ so that

$\displaystyle\rho_{1,*}(\tau):=\tau\rho_{1,1},\quad
Op(\rho_{2,*}(\tau))=\tau^{2}Op(\rho_{1,1})\circ Op(\rho_{1,1}),$ (10.178)

(10.177) gives us (10.173) and (10.174) for $k=1,2$, using (2.15) and Lemma
2.6.

We focus on $\rho_{3,*}(\tau)$ from now on. For $k=3$, we will prove (10.173)
only, since (10.174) is trivial ($\Phi_{5}$ is independent of the embedding,
since so is $\rho_{1}$.)

Let us define a sequence of symbols $\rho_{1,n+1}(\tau)$ by

$Op(\rho_{1,n+1}(\tau)):=\tau Op(\rho_{1,n}(\tau))\circ Op(\rho_{1,1}),\text{
for $n\geq 1$},$

so that we have

$\displaystyle(\varepsilon\tau)^{3}\sum_{n=3}^{\infty}(\varepsilon\tau)^{n-3}\frac{1}{n!}\left(Op(\rho_{1,1})\right)^{n}=\varepsilon^{3}\sum_{n=3}^{\infty}\varepsilon^{n-3}\frac{1}{n!}Op(\rho_{1,n}(\tau)).$
(10.179)

Note that using Lemma 2.16, there exists $C_{2}(\mathtt{pe},s,\eta_{0})>1$
such that

$\displaystyle|\rho_{1,n+1}(\tau)|^{\text{Lip}(\gamma,\Omega_{1})}_{-(n+1)\alpha,s,\eta_{0}}\leq
C_{2}(\mathtt{pe},s,\eta_{0})\left(|{\rho_{1,n}}(\tau)|^{\text{Lip}(\gamma,\Omega_{1})}_{-n\alpha,s_{0},\eta_{0}}|\rho_{1,1}|^{\text{Lip}(\gamma,\Omega_{1})}_{-\alpha,s,\eta_{0}}+|{\rho_{1,n}}(\tau)|^{\text{Lip}(\gamma,\Omega_{1})}_{-n\alpha,s,\eta_{0}}|\rho_{1,1}|^{\text{Lip}(\gamma,\Omega_{1})}_{-\alpha,s_{0},\eta_{0}}\right).$
(10.180)

Without loss of generality, we can assume that $s\mapsto
C_{1}(\mathtt{pe},s,\eta_{0}),C_{2}(\mathtt{pe},s,\eta_{0})\text{ are
increasing.}$

We claim that

$\displaystyle\begin{cases}|\rho_{1,n}(\tau)|^{\text{Lip}(\gamma,\Omega_{1})}_{-n\alpha,s_{0},\eta_{0}}\leq\left(2C_{1}(\mathtt{pe},s_{0},\eta_{0})C_{2}(\mathtt{pe},s_{0},\eta_{0})\right)^{n},\\\
|\rho_{1,n}(\tau)|^{\text{Lip}(\gamma,\Omega_{1})}_{-n,\alpha,s,\eta_{0}}\leq\left(4C_{1}(\mathtt{pe},s,\eta_{0})C_{2}(\mathtt{pe},s,\eta_{0})\right)^{n},\text{
for $s\geq s_{0}$.}\end{cases}$ (10.181)

We prove $s=s_{0}$ first. For $n=1$, the desired estimate follows immediately
from (10.177). Assuming the estimate holds for $n$, let us prove it for $n+1$.
Using (10.180), we have

$\displaystyle|\rho_{1,n+1}(\tau)|^{\text{Lip}(\gamma,\Omega_{1})}_{-(n+1)\alpha,s_{0},\eta_{0}}$
$\displaystyle\leq
2C_{2}(\mathtt{pe},s_{0},\eta_{0})|{\rho_{1,n}}(\tau)|^{\text{Lip}(\gamma,\Omega_{1})}_{-n\alpha,s_{0},\eta_{0}}|\rho_{1,1}|_{-\alpha,s_{0},\eta_{0}}$
$\displaystyle\leq
2C_{1}(\mathtt{pe},s_{0},\eta_{0})C_{2}(\mathtt{pe},s_{0},\eta_{0})|\rho_{1,n}(\tau)|^{\text{Lip}(\gamma,\Omega_{1})}_{-n\alpha,s_{0},\eta_{0}}$
$\displaystyle\leq\left(2C_{1}(\mathtt{pe},s_{0},\eta_{0})C_{2}(\mathtt{pe},s_{0},\eta_{0})\right)^{n+1},$

where the second inequality follows from (10.177) and the last inequality
follows from the induction hypothesis. Therefore, we prove the estimate
(10.181) for $s=s_{0}$.

Now, we prove (10.181) for $s\geq s_{0}$. Again the estimate for $n=1$ holds,
thanks to (10.177). For $n+1$, we again use (10.180) and (10.181) for
$s=s_{0}$ and obtain

$\displaystyle|\rho_{1,n+1}(\tau)|^{\text{Lip}(\gamma,\Omega_{1})}_{-(n+1)\alpha,s,\eta_{0}}$
$\displaystyle\leq\left(2C_{1}(\mathtt{pe},s,\eta_{0})C_{2}(\mathtt{pe},s,\eta_{0})\right)^{n+1}+C_{1}(\mathtt{pe},s,\eta_{0})C_{2}(\mathtt{pe},s,\eta_{0})|{\rho_{1,n}}(\tau)|^{\text{Lip}(\gamma,\Omega_{1})}_{-n\alpha,s,\eta_{0}}$
$\displaystyle\leq(2^{n+1}+4^{n})\left(C_{1}(\mathtt{pe},s,\eta_{0})C_{2}(\mathtt{pe},s,\eta_{0})\right)^{n+1}$
$\displaystyle\leq\left(4C_{1}(\mathtt{pe},s,\eta_{0})C_{2}(\mathtt{pe},s,\eta_{0})\right)^{n+1},$

where the second inequality follows from the induction hypothesis. Therefore,
we have proved (10.181) holds for $n+1$.

Hence, using Stirling’s formula it follows that there exist a constant
$C_{3}(\mathtt{pe},s,\eta_{0})$ and a symbol
$\rho_{3,*}(\tau)\in\mathcal{S}^{-3\alpha}$ such that

$\sum_{n=1}^{\infty}\varepsilon^{n-3}\frac{1}{n!}Op(\rho_{1,n}(\tau))=Op(\rho_{3,*}(\tau)),\quad|\rho_{3,*}(\tau)|^{\text{Lip}(\gamma,\Omega_{1})}_{-\alpha,s,\eta_{0}}\leq_{\mathtt{pe},s,\eta_{0}}1,$

which is the desired result (10.173) for $k=3$, in view of (10.179) and
(10.176). ∎

######  Lemma 10.55.

$\Phi_{5}\Pi_{S^{\perp}}-\Pi_{S^{\perp}}\Psi_{5}\Pi_{S^{\perp}}\in\mathfrak{R}(i_{0})$.

###### Proof.

The result follows from Lemma 9.10, the construction of $\Phi_{5},\Psi_{5}$ in
(10.151) and (10.152) and $\rho_{1}\in\mathfrak{B}^{-1-\alpha}_{1}$, which
follows from Proposition 10.53. ∎

######  Lemma 10.56.

$\Phi^{\pm}=\Phi_{5},\Phi_{5}^{-1}$ satisfies

$\displaystyle\rVert(\Phi^{\pm}-I)h\rVert_{s}^{\text{Lip}(\gamma,\Omega_{1})}\leq_{\mathtt{pe},s}\varepsilon\rVert
h\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s}$ (10.182) $\displaystyle\rVert
d_{i}\Phi^{\pm}(i_{0})h[\hat{\textbf{\i}}]\rVert_{s}=0.$

###### Proof.

It follows straightforwardly from Lemma 10.54 that $\Psi_{5},\Psi_{5}^{-1}$
satisfy the estimates in (10.182). Thanks to Lemma 10.55, $\Phi_{5}^{\pm}$
satisfies (10.182) for all $h$ such that $h(\varphi,\cdot)\in H_{S^{\perp}}$.
Also, using that $\rho_{1}$ is independent of the embedding $i_{0}$, we have
that $d_{i}\Phi^{\pm}(i_{0})=0$. ∎

######  Lemma 10.57.

$\Phi_{5}$ is a symplectic, reversibility preserving and real operator.

###### Proof.

Using that $\rho_{1}$ is real-valued, reversibility preserving and
$Op^{W}(\rho_{1})$ is a real operator, which follows from Proposition 10.53,
the same proof as in Lemma 10.48 gives the desired result. ∎

#####  10.4.1.4 Analysis of $\mathcal{W}$

######  Lemma 10.58.

$\mathcal{W}$ can be written as
$\mathcal{W}=\Pi_{S^{\perp}}\partial_{x}\Psi_{5}^{T}W_{1}\Psi_{5}+R_{5,1}$ for
some finite dimensional operator $R_{5,1}$ of the form in (8.32).

###### Proof.

We write $\mathcal{W}$ in (10.160) as

$\displaystyle\mathcal{W}$
$\displaystyle=\Phi_{5}^{-1}\Pi_{S^{\perp}}\partial_{x}W_{1}\Phi_{5}=\Phi_{5}^{-1}\Pi_{S^{\perp}}\partial_{x}W_{1}\Pi_{S^{\perp}}\Psi_{5}+\Phi_{5}^{-1}\Pi_{S^{\perp}}\partial_{x}W_{1}\left(\Phi_{5}-\Pi_{S^{\perp}}\Psi_{5}\right)$
$\displaystyle=\Phi_{5}^{-1}\Pi_{S^{\perp}}\partial_{x}W_{1}\Psi_{5}-\Phi_{5}^{-1}\Pi_{S^{\perp}}\partial_{x}W_{1}\Pi_{S}\Psi_{5}+\Phi_{5}^{-1}\Pi_{S^{\perp}}\partial_{x}W_{1}\left(\Phi_{5}-\Pi_{S^{\perp}}\Psi_{5}\right)$
$\displaystyle=\Pi_{S^{\perp}}\Psi_{5}^{-1}\Pi_{S^{\perp}}\partial_{x}W_{1}\Psi_{5}+(\Phi_{5}^{-1}\Pi_{S^{\perp}}-\Pi_{S^{\perp}}\Psi_{5}^{-1}\Pi_{S^{\perp}})\partial_{x}W_{1}\Psi_{5}$
$\displaystyle\
-\Phi_{5}\Pi_{S^{\perp}}\partial_{x}W_{1}\Pi_{S}\Psi_{5}+\Phi_{5}\Pi_{S^{\perp}}\partial_{x}W_{1}\left(\Phi_{5}-\Pi_{S^{\perp}}\Psi_{5}\right)$
$\displaystyle=\Pi_{S^{\perp}}\Psi_{5}^{-1}\partial_{x}W_{1}\Psi_{5}-\Pi_{S^{\perp}}\Psi_{5}^{-1}\Pi_{S}\partial_{x}W_{1}\Psi_{5}+(\Phi_{5}\Pi_{S^{\perp}}-\Pi_{S^{\perp}}\Psi_{5}^{-1}\Pi_{S^{\perp}})\partial_{x}W_{1}\Psi_{5}$
$\displaystyle\
-\Phi_{5}^{-1}\Pi_{S^{\perp}}\partial_{x}W_{1}\Pi_{S}\Psi_{5}+\Phi_{5}^{-1}\Pi_{S^{\perp}}\partial_{x}W_{1}\left(\Phi_{5}-\Pi_{S^{\perp}}\Psi_{5}\right)$
$\displaystyle=\Pi_{S^{\perp}}\Psi_{5}^{-1}\partial_{x}W_{1}\Psi_{5}+R_{5,1},$
(10.183)

where

$\displaystyle R_{5,1}$
$\displaystyle:=-\Pi_{S}\Psi_{5}^{-1}\partial_{x}W\Psi_{5}-\Pi_{S^{\perp}}\Psi_{5}^{-1}\Pi_{S}\partial_{x}W_{1}\Psi_{5}+(\Phi_{5}^{-1}\Pi_{S^{\perp}}-\Pi_{S^{\perp}}\Psi_{5}^{-1}\Pi_{S^{\perp}})\partial_{x}W_{1}\Psi_{5}$
(10.184) $\displaystyle\
-\Phi_{5}^{-1}\Pi_{S^{\perp}}\partial_{x}W_{1}\Pi_{S}\Psi_{5}+\Phi_{5}^{-1}\Pi_{S^{\perp}}\partial_{x}W_{1}\left(\Phi_{5}-\Pi_{S^{\perp}}\Psi_{5}\right).$

It follows from (2.128) and (10.4.1.4) that
$\mathcal{W}=\Pi_{S^{\perp}}\partial_{x}\Psi_{5}^{T}W_{1}\Psi_{5}+R_{5,1}$.

Now, we show that $R_{5,1}$ is a finite dimensional operator of the form
(8.32). Thanks to Lemma 10.55, we can see that each term in $R_{5,1}$ has a
finite dimensional operator of the form
$\Phi_{5}\Pi_{S^{\perp}}-\Pi_{S^{\perp}}\Psi_{5}\Pi_{S^{\perp}}$ or $\Pi_{S}$.
Among all the terms in $R_{5,1}$, we will prove the lemma only for the third
term in (10.184), that is,
$(\Phi_{5}^{-1}\Pi_{S^{\perp}}-\Pi_{S^{\perp}}\Psi_{5}^{-1}\Pi_{S^{\perp}})\partial_{x}W_{1}\Psi_{5}$,
since the other terms can be treated in a similar way.

Note that
$\Phi_{5}\Pi_{S^{\perp}}-\Pi_{S^{\perp}}\Psi_{5}\Pi_{S^{\perp}}\in\mathfrak{R}(i_{0})$,
which follows from Lemma 10.55. Since the equations (10.151) and (10.152) that
determine $\Phi_{5}$ and $\Psi_{5}$ are autonomous, we see that
$\Phi_{5}^{-1}$ and $\Psi_{5}^{-1}$ are the time-$1$ flow maps of their
reversed PDEs, therefore
$\Phi_{5}^{-1}\Pi_{S^{\perp}}-\Pi_{S^{\perp}}\Psi_{5}^{-1}\Pi_{S^{\perp}}\in\mathfrak{R}(i_{0})$.
Recalling the definition of $\mathfrak{R}(i_{0})$ from Definition 8.11, we can
decompose it into

$\Phi_{5}^{-1}\Pi_{S^{\perp}}-\Pi_{S^{\perp}}\Psi_{5}^{-1}\Pi_{S^{\perp}}=\varepsilon^{2}R_{5,1,1}+R_{5,1,2},$

where $R_{5,1,2}$ is already of the form (8.32) and there exist
$\mathfrak{a}_{5,1},\mathfrak{b}_{5,1}\in\mathfrak{B}^{m}_{1}$ such that

$\displaystyle\varepsilon^{2}R_{5,1,1}[h]$
$\displaystyle=\Pi_{S^{\perp}}\varepsilon^{2}Op^{W}(\mathfrak{a}_{5,1})\Pi_{S}\left(Op^{W}(\mathfrak{b}_{5,1}h\right)$
$\displaystyle=\sum_{j\in
S}\Pi_{S^{\perp}}\varepsilon^{2}Op^{W}(\mathfrak{a}_{5,1})\left(\frac{1}{2\pi}(h,\left(Op^{W}(\mathfrak{b}_{5,1})\right)^{T}[e^{\text{i}jx}])_{L^{2}}e^{\text{i}jx}\right)$
$\displaystyle=\sum_{j\in S}(h,g_{j})_{L^{2}}\chi_{j},$

where

$g_{j}:=Op^{W}(\mathfrak{b}_{5,1})^{T}[e^{\text{i}jx}],\quad\chi_{j}:=\varepsilon^{2}Op^{W}(\mathfrak{a}_{5,1})[e^{\text{i}jx}].$

Clearly, using $\mathfrak{a}_{5,1},\mathfrak{b}_{5,1}\in\mathfrak{B}^{m}_{1}$
for some $m\geq 0$, we have that

$\displaystyle\rVert
g_{j}\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s}\rVert\chi_{j}\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s_{0}}+\rVert
g\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s_{0}}\rVert\chi\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s}$
$\displaystyle\leq\varepsilon^{2},\quad\rVert
d_{i}g_{j}(i_{0})[\hat{\textbf{\i}}]\rVert_{s}=\rVert
d_{i}\chi_{j}(i_{0})[\hat{\textbf{\i}}]\rVert_{s}$ $\displaystyle=0.$ (10.185)

Hence, we get

$(\Phi_{5}^{-1}\Pi_{S^{\perp}}-\Pi_{S^{\perp}}\Psi_{5}^{-1}\Pi_{S^{\perp}})\partial_{x}W_{1}\Psi_{5}=\sum_{j\in
S}(h,\underbrace{(\partial_{x}W_{1}\Psi_{5})^{T}[g_{j}])}_{=:\tilde{g}_{j}}\chi_{j}.$

Then, using Lemma 10.54, recalling $W_{1}$ from (10.124) and (10.114), it
follows from the estimates for $\Psi_{3},\Phi_{4}$ in Lemma 10.36, Lemma
10.49, and the estimates for
$\mathfrak{q}_{\omega,\mathtt{N}_{\alpha}\star\mathfrak{a}},\mathfrak{q}_{M_{2},\mathtt{N}_{\alpha}}\star\mathfrak{a}$
in Lemma 10.39 that $\tilde{g}_{j}$ and $\chi_{j}$ satisfy the estimates in
(8.33). This finishes the proof. ∎

#####  10.4.1.5 Analysis of $\mathcal{Q}_{1}$

Now, we analyze the structure of the operator $\mathcal{Q}_{1}$ in (10.160).

######  Lemma 10.59.

$\mathcal{Q}_{1}$ is of the form:

$\displaystyle\mathcal{Q}_{1}=\Pi_{S^{\perp}}Op(\tilde{\mathfrak{b}}_{2})+\Pi_{S^{\perp}}Op(\mathfrak{r}_{-2,\leq
3,*})+{R}_{5,2}$ (10.186)

for some $\tilde{\mathfrak{b}}_{2},\mathfrak{r}_{-2,\leq
3,*}\in\mathcal{S}^{-2}$ and a finite dimensional operator $R_{5,2}$ such that

  1. (1)

$\tilde{\mathfrak{b}}_{2}\in\mathfrak{B}^{-2}_{2}$.

  2. (2)

$\mathfrak{r}_{-2,\leq 3,*}$ satisfies

$\displaystyle|\mathfrak{r}_{-2,\leq
3,*}|^{\text{Lip}(\gamma,\Omega_{1})}_{\alpha-3,s,\eta_{0}}\leq_{\mathtt{pe},s,\eta_{0}}\gamma^{-1}\left(\varepsilon^{5}+\varepsilon^{3}\rVert\mathfrak{I}_{\delta}\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s+\mu_{0}}\right),$
(10.187) $\displaystyle|d_{i}\mathfrak{r}_{-2,\leq
3,*}(i_{0})[\hat{\textbf{\i}}]|_{\alpha-3,s,\eta_{0}}\leq_{\mathtt{pe},s,\eta_{0}}\varepsilon^{3}\gamma^{-1}\left(\rVert\hat{\textbf{\i}}\rVert_{s+\mu_{0}}+\rVert\mathfrak{I}_{\delta}\rVert_{s+\mu_{0}}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{0}}\right).$
(10.188)

  3. (3)

$R_{5,2}\in\mathfrak{R}(i_{0})$.

###### Proof.

Among all the terms of $\mathcal{Q}_{1}$ in (10.160), we prove the lemma for
$\varepsilon^{2}\mathcal{D}_{\omega}(\mathcal{A}_{2})$ only, since all the
other terms can be treated in the same way.

Clearly, it suffices to show that $\varepsilon^{2}\mathcal{A}_{2}$ has the
decomposition in (10.186), since all the properties in (1), (2) and (3) still
hold after taking $\mathcal{D}_{\omega}$. Recalling $\mathcal{A}_{2}$ from
(10.158) and $\Phi_{5}$ from (10.150), we have that

$\displaystyle\varepsilon^{2}\mathcal{A}_{2}\Pi_{S^{\perp}}$
$\displaystyle=\int_{0}^{1}(\varepsilon
A_{1})^{2}\Phi_{5}(\tau)(1-\tau)d\tau\Pi_{S^{\perp}}$
$\displaystyle=(\varepsilon
A_{1})^{2}\int_{0}^{1}\Psi_{5}(\tau)(1-\tau)d\tau\Pi_{S^{\perp}}+(\varepsilon
A_{1})^{2}\int_{0}^{1}\Pi_{S}\Psi_{5}(\tau)(1-\tau)d\tau\Pi_{S^{\perp}}$
$\displaystyle\
+(\varepsilon^{2}A_{1})^{2}\int_{0}^{1}(\Phi_{5}(\tau)\Pi_{S^{\perp}}-\Pi_{S^{\perp}}\Psi_{5}(\tau)\Pi_{S^{\perp}})(1-\tau)d\tau.$

From Lemma 10.54 and Lemma 10.55, we see that the last two operators are
finite dimensional operators in class $\mathfrak{R}(i_{0})$ (Note that Lemma
10.55 for the time-$1$ flow, but we can always reparametrize the time variable
$\tau$ so that
$\Phi_{5}(\tau)\Pi_{S^{\perp}}-\Pi_{S^{\perp}}\Psi_{5}(\tau)\Pi_{S^{\perp}}\in\mathfrak{R}(i_{0})$).
Therefore,

$\displaystyle\varepsilon^{2}\mathcal{A}_{2}\Pi_{S^{\perp}}=(\varepsilon
A_{1})^{2}\int_{0}^{1}\Psi_{5}(\tau)(1-\tau)d\tau\Pi_{S^{\perp}},$ (10.189)

up to a finite dimensional operator in $\mathfrak{R}(i_{0})$. Similarly, using
Lemma 10.54, we can find $\rho_{*}(\tau)\in\mathcal{S}^{-\alpha}$ such that

$\Psi_{5}(\tau)=I+\varepsilon
Op(\rho_{*}(\tau)),\quad\sup_{\tau\in[0,1]}|\rho_{*}(\tau)|^{\text{Lip}(\gamma,\Omega_{1})}_{-\alpha,s,\eta_{0}}\leq_{\mathtt{pe},s,\eta_{0}}1,\quad\sup_{\tau\in[0,1]}|d_{i}\rho_{*}(\tau)(i_{0})[\hat{\textbf{\i}}]|_{-\alpha,s,\eta_{0}}=0.$

Therefore, denoting all the finite dimensional operators in the right-hand
side of (10.189) by $R_{5,2}$, we obtain

$\displaystyle\varepsilon^{2}\mathcal{A}_{2}=\varepsilon^{2}\partial_{x}Op(\rho_{1})\partial_{x}Op(\rho_{1})+\varepsilon^{3}\partial_{x}Op(\rho_{1})\partial_{x}Op(\rho_{1})\int_{0}^{1}(1-\tau)Op(\rho_{*}(\tau))d\tau+{R}_{5,2},$
(10.190)

for some $R_{5,2}\in\mathfrak{R}(i_{0})$. Therefore, using
$\rho_{1}\in\mathfrak{B}^{-1-\alpha}_{1}$ and the composition formulae of
symbols in (2.23), it follows straightforwardly that

$\varepsilon^{2}\mathcal{A}_{2}=Op(\tilde{\mathfrak{b}}_{2})+Op(\mathfrak{r}_{-2,\leq
3,*})+R_{5,2},$

for some
$\tilde{\mathfrak{b}}_{2}\in\mathfrak{B}^{-2\alpha}_{2}\subset\mathfrak{B}^{-2}_{2}$
and $\mathfrak{r}_{-2,\leq 3,*}$ satisfying the estimates (10.187) and
(10.188) (indeed with a better estimates since $\rho_{*}$ does not depend on
the embedding $i_{0}$ and the factor $\varepsilon^{3}$ in (10.190)). ∎

####  10.4.2 Structure of the operator of size $O(\varepsilon^{2})$

Before eliminating the operators of size $O(\varepsilon^{2})$ in
$\mathcal{L}^{5}$, we investigate the structure of such terms in (10.153).

Recalling the definition of $\mathfrak{R}(i_{0})$ from Definition 8.11, (3) in
Proposition 10.51 tells us that we can pick
$\mathfrak{c}_{1},\mathfrak{c}_{2}\in\mathfrak{B}^{m}_{1}$ for some $m\geq 0$
such that

$R_{5}=\Pi_{S^{\perp}}\partial_{x}Op^{W}(\mathfrak{c}_{1})\Pi_{S}Op^{W}(\mathfrak{c}_{2})+\mathcal{R}_{5},$

where $\mathcal{R}_{5}$ is of the form in (8.32). Therefore, we write
$\mathcal{L}^{5}$ as

$\displaystyle\mathcal{L}^{5}=\mathcal{D}_{\omega}-\Pi_{S^{\perp}}\partial_{x}Op^{W}\left(\mathtt{m}_{\alpha}m_{1,\alpha}(\xi)+\left(\frac{T_{\alpha}}{4}+\mathfrak{m}_{\leq
0}(\omega,\xi)\right)\right)-\varepsilon^{2}\partial_{x}\mathcal{C}_{2}+\mathcal{R}_{\leq
3},$ (10.191)

where

$\displaystyle\mathcal{C}_{2}$
$\displaystyle=\Pi_{S^{\perp}}\left(Op^{W}(\tilde{\mathfrak{b}}_{2})-Op^{W}(\mathfrak{c}_{1})\Pi_{S}Op^{W}(\mathfrak{c}_{2})\right),\text{
for some $\mathfrak{c}_{1},\mathfrak{c}_{2}\in\mathfrak{B}^{m}_{1}$ for some
$m\geq 0$,}$ (10.192) $\displaystyle\mathcal{R}_{\leq 3}$
$\displaystyle=\Pi_{S^{\perp}}\partial_{x}Op^{W}(\mathfrak{r}_{-2,\leq
3,*})+\partial_{x}\Pi_{S^{\perp}}W_{2}+\mathcal{R}_{5},\text{ for some
$\mathcal{R}_{5}$ of the form in \eqref{j2j2j2j2j2jssds}.}$

The main result of this subsection is summarized in the following proposition:

######  Proposition 10.60.

$\mathcal{L}^{5}$ has the form:

$\displaystyle\mathcal{L}^{5}$
$\displaystyle=\mathcal{D}_{\omega}-\Pi_{S^{\perp}}\partial_{x}Op^{W}\left(\mathtt{m}_{\alpha}m_{1,\alpha}(\xi)+\left(\frac{T_{\alpha}}{4}+\mathfrak{m}_{\leq
0}(\omega,\xi)+\varepsilon^{2}\mathfrak{m}_{\mathfrak{b}}(\omega,\xi)\right)\right)$
(10.193) $\displaystyle\
-\varepsilon^{2}\Pi_{S^{\perp}}\partial_{x}Op^{W}(\mathfrak{b}_{0})+\mathcal{R}_{\leq
3},$

and satisfies the following:

  1. (1)

$\mathfrak{m}_{\mathfrak{b}}(\omega,\xi):=M_{x}(\tilde{\mathfrak{b}})$ is an
$\omega$-dependent Fourier multiplier and it is a reversible symbol.
Furthermore, we have

$\displaystyle\mathfrak{m}_{\mathfrak{b}}(\xi)+\mathtt{m}_{\alpha,1}(\omega)m_{1,\alpha}(\xi)+\mathfrak{m}_{\leq
0,1}(\omega,\xi)=\mathfrak{F}_{\alpha-1}(\omega,\xi),$ (10.194)

where $\mathfrak{F}_{\alpha-1}$ is as given in (10.209)
($\mathtt{m}_{\alpha,1}$ and $\mathfrak{m}_{\leq 0,1}$ are given in
Proposition 10.10 and Proposition 10.41).

  2. (2)

$\mathfrak{b}_{0}\in\mathfrak{B}^{-2}_{2}$ is a real-valued and reversible
symbol such that $M_{x}(\mathfrak{b}_{0})=0$.

###### Proof.

Denoting

$\displaystyle\mathfrak{m}_{\mathfrak{b}}:=M_{x}(\tilde{\mathfrak{b}}_{2}),\quad\mathfrak{b}_{0}=\tilde{\mathfrak{b}}_{2}-M_{x}(\tilde{\mathfrak{b}}_{2}),$
(10.195)

the result follows from (10.191) and Lemma 10.65. Note that
$\mathfrak{m}_{\mathfrak{b}}$ is a reversible symbol since
$\tilde{\mathfrak{b}}_{2}$ is reversible symbol (see Proposition 10.51).
$\mathfrak{b}_{0}$ being real-valued and reversible in (2) also follows from
(2) of Proposition 10.51. $M_{x}(\mathfrak{b}_{0})=0$ by its definition in
(10.195). ∎

In the rest of this subsection, we aim to prove the necessary lemmas to obtain
the above proposition.

#####  10.4.2.1 Structure of $\mathcal{L}^{5}$.

In the next lemma, we recall the notation in (8.58) to compare two linear
operators.

######  Lemma 10.61.

We have that

  1. (1)

$\mathcal{R}_{\leq 3}=0$, up to $O(\varepsilon^{9-6b})$.

  2. (2)

$Op^{W}\left(\mathtt{m}_{\alpha}(\omega)m_{1,\alpha}(\xi)+\left(\frac{T_{\alpha}}{4}+\mathfrak{m}_{\leq
0}(\omega,\xi)\right)\right)=Op^{W}(-\frac{1}{2}m_{1,\alpha}(\xi)+\frac{T_{\alpha}}{4})+\varepsilon^{2}Op^{W}(\mathtt{m}_{\alpha,1}(\omega)m_{1,\alpha}(\xi)+\mathfrak{m}_{\leq
0,1}(\omega,\xi))$, up to $O(\varepsilon^{9-6b})$.

###### Proof.

For (1), (10.192) tells us that we need to show that
$Op^{W}(\mathfrak{r}_{-2,\leq 3,*})=0$, $W_{2}=0$ and $\mathcal{R}_{5}=0$ up
to $O(\varepsilon^{9-6b})$. For $\mathfrak{r}_{-2,\leq 3,*}$, it follows from
(10.154), (7.16) and (6.3) that $Op^{W}(\mathfrak{r}_{-2,\leq 3,*})=0$, up to
$O(\gamma^{-1}(\varepsilon^{5}+\varepsilon^{3}\varepsilon^{6-2b}\gamma^{-1}))=O({\varepsilon^{9-6b}})$.
For $\mathcal{R}_{5}$, we already know from (10.192) that $\mathcal{R}_{5}$ is
of the form in (8.32). Thanks to the estimates in (8.33) and Lemma 8.10, we
have that $\mathcal{R}_{5}=0$ up to
$O(\varepsilon^{3}+\varepsilon^{2}|\mathfrak{I}_{\delta}|)\overset{\eqref{size_assumption_3}}{=}O(\varepsilon^{8-4b})=O(\varepsilon^{9-6b})$.
where the last equality follows from $b\in(1,1+1/12)$, hence $9-6b\leq 8-4b$.
For $W_{2}$, we recall from (3) of Proposition 10.51 and (10.124) and (10.114)
that

$\displaystyle W_{2}$
$\displaystyle=\frac{1}{(\mathtt{N}_{\alpha}-2)!}\int_{0}^{1}\int_{0}^{\tau}(\Psi_{3}(\tau-t)\Phi_{4}\Psi_{5})^{T}Op^{W}(\mathfrak{q}_{\omega,\mathtt{N}-2}\star\mathfrak{a})\Psi_{3}(\tau-t)\Phi_{4}\Psi_{5}t^{\mathtt{N}_{\alpha}-2}dtd\tau$
$\displaystyle\
-\frac{1}{\mathtt{N}_{\alpha}!}\int_{0}^{1}(\Psi_{3}(1-t)\Phi_{4}\Psi_{5})^{T}Op^{W}(\mathfrak{q}_{M_{2},\mathtt{N}_{\alpha}}\star\mathfrak{a})\Psi_{3}(1-t)\Phi_{4}\Psi_{5}t^{\mathtt{N}_{\alpha}}dt.$
(10.196)

From Lemma 10.36, Lemma 10.49, and Lemma 10.56, we have that
$(\Psi_{3}(\tau-t)\Phi_{4}\Psi_{5})=I$ up to $O(\varepsilon)$, and so is
$(\Psi_{3}(\tau-t)\Phi_{4}\Psi_{5})^{T}$, thanks to the fact that all the
transformations are symplectic and (2.128). While, the symbols
$\mathfrak{q}_{\omega,\mathtt{N}-2}\star\mathfrak{a}$ and
$\mathfrak{q}_{M_{2},\mathtt{N}_{\alpha}}\star\mathfrak{a}$ satisfy (10.115),
which tells us that $W_{2}=O(\varepsilon^{\mathtt{N}_{\alpha}})$. Recalling
$\mathtt{N}_{\alpha}$ from (10.70), we have $W_{2}=0$ up to
$O(\varepsilon^{9-6b})$.

(2) follows from the estimates for $\mathtt{m}_{\alpha}$ and
$\mathfrak{m}_{\leq 0}$ in (10.3) and (10.121) and
$\varepsilon^{7-4b}\leq\varepsilon^{9-6b}$ for sufficiently small
$\varepsilon>0$. ∎

Lemma 10.61 tells us that denoting

$\displaystyle
D:=Op^{W}\left(-\frac{1}{2}m_{1,\alpha}(\xi)+\frac{T_{\alpha}}{4}\right),\quad\mathcal{T}:=-\mathcal{C}_{2}-Op^{W}(\mathtt{m}_{\alpha,1}m_{1,\alpha}(\xi)+\mathfrak{m}_{\leq
0,1}(\omega,\xi)),$ (10.197)

(10.191) can be written as

$\displaystyle\mathcal{L}^{5}=\left(\mathcal{D}_{\omega}-\partial_{x}D\right)+\varepsilon^{2}\partial_{x}\mathcal{T},\text{
up to $O(\varepsilon^{9-6b})$.}$ (10.198)

#####  10.4.2.2 Structure of $\mathcal{L}^{5}$ from the evolution of
$\mathcal{L}_{\omega}$.

Now, we analyze the structure of $\mathcal{L}^{5}$ from the expression in
(8.59) and compare it with (10.198) to specify an explicit formula for
$\mathcal{T}$.

We recall from (8.59) that

$\displaystyle\mathcal{L}_{\omega}=\mathcal{D}_{\omega}-\partial_{x}\nabla_{z}^{2}H_{2,2}(\bar{v})-\varepsilon\partial_{x}\underbrace{\nabla_{z}^{2}H_{3,2}(\bar{v})}_{=:S_{1}}-\varepsilon^{2}\partial_{x}\underbrace{\nabla_{z}^{2}H_{4,2}(\bar{v})}_{=:S_{2}},\text{
up to $O(\varepsilon^{7-4b})$.}$ (10.199)

Recalling $H_{2,2}$ from (4.14) and $m_{1,\alpha}$ from (2.68), we see that
$\nabla_{z}^{2}H_{2,2}=D$, where $D$ is as in (10.197), therefore,

$\displaystyle\mathcal{L}_{\omega}=\left(\mathcal{D}_{\omega}-\partial_{x}D\right)-\varepsilon\partial_{x}S_{1}-\varepsilon^{2}\partial_{x}S_{2},\text{
up to $O(\varepsilon^{9-6b})$}.$ (10.200)

From Propositions 10.1, 10.10, 10.30, 10.41 and 10.51, we have that

$\mathcal{L}^{5}=\Phi_{5}^{-1}\Phi_{4}^{-1}\Phi_{3}^{-1}\frac{1}{\rho}\Phi_{2}^{-1}\Phi_{1}^{-1}\mathcal{L}_{\omega}\Phi_{1}\Phi_{2}\Phi_{3}\Phi_{4}\Phi_{5}.$

From Lemma 10.15 and (7.16) that $h\mapsto\frac{1}{\rho}h=I$, up to
$O(\varepsilon^{8-4b})=O(\varepsilon^{9-6b})$, therefore,

$\displaystyle\mathcal{L}^{5}=\left(\Phi_{1}\Phi_{2}\Phi_{3}\Phi_{4}\Phi_{5}\right)^{-1}\mathcal{L}_{\omega}\Phi_{1}\Phi_{2}\Phi_{3}\Phi_{4}\Phi_{5},\text{
up to $O(\varepsilon^{9-6b}$).}$ (10.201)

######  Lemma 10.62.

$\mathcal{L}^{5}$ is of the form:

$\displaystyle\mathcal{L}^{5}$
$\displaystyle=\mathcal{D}_{\omega}-\partial_{x}D-\varepsilon(\partial_{x}S_{1}-[\mathcal{D}_{\omega}-\partial_{x}D,\mathfrak{A}_{1}])$
(10.202) $\displaystyle\
-\varepsilon^{2}\left(\partial_{x}S_{2}-[\mathfrak{A}_{1},\partial_{x}S_{1}]-[\mathcal{D}_{\omega}-\partial_{x}D,\mathfrak{A}_{2}]-\frac{1}{2}[[\mathcal{D}_{\omega}-\partial_{x}D,\mathfrak{A}_{1}],\mathfrak{A}_{1}]\right),$

up to $O(\varepsilon^{9-6b})$, where
$\mathfrak{A}_{i}=\Pi_{S^{\perp}}Op(\mathfrak{t}_{i})$ for some
$\mathfrak{t}_{i}\in\mathfrak{B}^{m}_{i}$ for some $m\in\mathbb{R}$.

###### Proof.

First, we consider an operator $L$ is of the form:

$L=\mathcal{D}_{\omega}-\partial_{x}D+\varepsilon T_{1}+\varepsilon^{2}T_{2},$

for some bounded operators $T_{1},T_{2}:H^{s_{0}}_{\varphi,x}\mapsto
H^{s_{0}+\mu_{0}}_{\varphi,x}$ for sufficiently large $\mu_{0}\geq 0$. we
claim that the conjugations with $\Phi_{2},\Phi_{4}$ do not change its
structure, that is,

$\Phi_{i}^{-1}L\Phi_{i}=L,\text{ up to $O(\varepsilon^{9-6b})$.}$

For Lemma 10.13 and Lemma 10.43, we see that (using (7.16))

$\Phi_{i}=I,\text{ up to
$O(\gamma^{-1}\varepsilon^{2}\rVert\mathfrak{I}_{\delta}\rVert_{s_{0}+\mu_{0}})=O(\varepsilon^{8-6b})$,
for $i=2,4$.}$

Therefore, the conjugation of $\Phi_{i}^{-1}\varepsilon
T_{1}\Phi_{i}-\varepsilon
T_{1}=\Phi_{i}^{-1}\varepsilon^{2}T_{2}\Phi_{i}-\varepsilon^{2}T_{2}=0$ up to
$O(\varepsilon^{9-6b})$. Furthermore, since $\Phi_{2},\Phi_{4}$ commute with
$\partial_{x}D$ (see (10.29) and (10.116)), the conjugation of $\partial_{x}D$
does not change. Furthermore, the estimates $\mathcal{D}_{\omega}p_{1}$ and
$\mathcal{D}_{\omega}\mathfrak{p}_{2}$ in (10.42) and (10.133) tell us that

$\Phi_{i}^{-1}\mathcal{D}_{\omega}\Phi_{i}=\mathcal{D}_{\omega},\text{ up to
$O(\varepsilon^{8-4b})=O(\varepsilon^{9-6b})$ for $i=2,4$},$

therefore the conjugation of $\mathcal{D}_{\omega}-\partial_{x}D$ with
$\Phi_{2},\Phi_{4}$ does not change $L$.

Therefore, in view of (10.201), we have

$\displaystyle\mathcal{L}^{5}=\left(\Phi_{1}\Phi_{3}\Phi_{5}\right)^{-1}\mathcal{L}_{\omega}\Phi_{1}\Phi_{3}\Phi_{5},\text{
up to $O(\varepsilon^{9-6b})$.}$ (10.203)

We observe that each of $\Phi_{i}$ for $i=1,3,5$ satisfies (see (10.6),
(10.73) and (10.151))

$\displaystyle\frac{d}{d\tau}\Phi_{i}(\tau)=\mathfrak{A}(\tau)\Phi_{i}(\tau),\quad\Phi_{i}(0)=I,$
(10.204)

for some operator $\mathfrak{A}(\tau)$ of the form

$\displaystyle\mathfrak{A}(\tau)=\varepsilon\underbrace{\Pi_{S^{\perp}}Op(\mathfrak{t}_{1})}_{=:\mathfrak{A}_{1}}+\varepsilon^{2}\underbrace{\Pi_{S^{\perp}}Op(\mathfrak{t}_{2})}_{=:\mathfrak{A}_{2}},\text{
up to $O(\varepsilon^{9-6b})$,}$ (10.205)

for some $\mathfrak{t}_{1},\in\mathfrak{B}^{1}_{1}$ and
$\mathfrak{t}_{2}\in\mathfrak{B}^{1}_{2}$. Indeed, for $\Phi_{1}$, we expand
$b(\tau)$ in (10.6) using the Taylor expansion in the variable $\tau$ at
$\tau=0$, and use the fact that $\beta\in\mathfrak{C}_{1}(i_{0})$ to see that
$\Phi_{1}$ satisfies the above property. For $\Phi_{3},\Phi_{5}$, we can see
it from (10.73) and (10.151), while the corresponding operator
$\mathfrak{A}(\tau)$ does not depend on $\tau$, since the equations in (10.73)
and (10.151) are autonomous. Since we already observed that (10.204) is well-
posed (Subsection 9.2), the usual Taylor expansion gives us that for possibly
different symbols $\mathfrak{t}_{1},\mathfrak{t}_{2}$ (thus, different
$\mathfrak{A}_{1},\mathfrak{A}_{2}$) from the ones in (10.205) but still in
$\mathfrak{B}^{m}_{1},\mathfrak{B}^{m}_{2}$ for some $m\in\mathbb{R}$,
respectively,

$\displaystyle\Phi_{1}\Phi_{3}\Phi_{5}$
$\displaystyle=I+\varepsilon\mathfrak{A}_{1}+\varepsilon^{2}\left(\frac{1}{2}(\mathfrak{A}_{1})^{2}+\mathfrak{A}_{2}\right),$
$\displaystyle(\Phi_{1}\Phi_{3}\Phi_{5})^{-1}$
$\displaystyle=I-\varepsilon\mathfrak{A}_{1}+\varepsilon^{2}\left(\frac{1}{2}(\mathfrak{A}_{1})^{2}-\mathfrak{A}_{2}\right),$

up to $O(\varepsilon^{9-6b})$. Therefore, using (10.200) and (10.203), we see
that

$\displaystyle\mathcal{L}^{5}$
$\displaystyle=(\mathcal{D}_{\omega}-\partial_{x}D)-\varepsilon(\partial_{x}S_{1}-[\mathcal{D}_{\omega}-\partial_{x}D,\mathfrak{A}_{1}])$
$\displaystyle\
-\varepsilon^{2}\left(\partial_{x}S_{2}-[\mathfrak{A}_{1},\partial_{x}S_{1}]-[\mathcal{D}_{\omega}-\partial_{x}D,\mathfrak{A}_{2}]-\frac{1}{2}[[\mathcal{D}_{\omega}-\partial_{x}D,\mathfrak{A}_{1}],\mathfrak{A}_{1}]\right),$

which is in the form (10.202). ∎

######  Lemma 10.63.

We have that

$\displaystyle\mathcal{L}^{5}=\mathcal{D}_{\omega}-\partial_{x}D-\varepsilon^{2}\left(\partial_{x}S_{2}-\frac{1}{2}[\mathfrak{A}_{1},\partial_{x}S_{1}]-[\mathcal{D}_{\overline{\omega}}-\partial_{x}D,\mathfrak{A}_{2}]\right),\text{up
to $O(\varepsilon^{9-6b})$,}$ (10.206)

where $\mathfrak{A}_{2}=\Pi_{S^{\perp}}Op(\mathfrak{t}_{2})$, for some
$\mathfrak{t}_{2}\in\mathfrak{B}^{m}_{2}$ and
$\mathfrak{A}_{1}=\Pi_{S^{\perp}}Op(\mathfrak{t}_{1})$ with
$\mathfrak{t}_{1}\in\mathfrak{B}^{m}_{1}$ for some $m\in\mathbb{R}$.
$\mathfrak{t}_{2}$ is determined uniquely by

$\displaystyle\partial_{x}S_{1}=[\mathcal{D}_{\overline{\omega}}-\partial_{x}D,\mathfrak{A}_{1}]$
(10.207)

###### Proof.

Comparing $O(\varepsilon)$ terms in (10.198) and (10.202), and using that
$\mathcal{D}_{\omega}-\mathcal{D}_{\overline{\omega}}=0$ up to
$O(\varepsilon^{2})$, which follows from (6.2) and (5.25), we have (10.207).
Therefore,

$\frac{1}{2}[[\mathcal{D}_{\omega}-\partial_{x}D,\mathfrak{A}_{1}],\mathfrak{A}_{1}]=\frac{1}{2}[\partial_{x}S_{1},\mathfrak{A}_{1}].$

Plugging this into (10.202), we obtain (10.206). ∎

In what follows, we will find explicit solution to (10.207).

######  Lemma 10.64.

Let $\mathfrak{A}_{1}$ be a unique solution to (10.207). Then there exist a
Fourier multiplier $\mathfrak{F}_{\alpha-1}(\xi)$ and a symbol
$\mathfrak{D}_{0}\in\mathfrak{B}^{\alpha-1}_{2}(\varphi,x,\xi)$ such that

$\displaystyle\partial_{x}S_{2}-\frac{1}{2}[\mathfrak{A}_{1},\partial_{x}S_{1}]-[\mathcal{D}_{\overline{\omega}}-\partial_{x}D,\mathfrak{A}_{2}]=\Pi_{S^{\perp}}\partial_{x}Op^{W}(\mathfrak{F}_{\alpha-1}+\mathfrak{D}_{0}),$
(10.208)

such that $M_{x}(\mathfrak{D}_{0})=0$ and
$\mathfrak{F}_{\alpha-1}(\xi)=\mathfrak{F}_{\alpha-1}(\omega,\xi)$ can be
explicitly computed

$\displaystyle\mathfrak{F}_{\alpha-1}(\omega,\xi)$
$\displaystyle=\sum_{k=1}^{\nu}\left(\frac{12}{\pi}H_{4,j_{k},-j_{k},\xi,-\xi}+\frac{18}{\pi^{2}}\left(\frac{(\xi-
j_{k})H_{3,-j_{k},\xi,-(\xi-j_{k})}H_{3,j_{k},\xi-
j_{k},-\xi}}{\lambda_{\alpha}(j_{k})+\lambda_{\alpha}(\xi-
j_{k})-\lambda_{\alpha}(\xi)}\right.\right.$ (10.209)
$\displaystyle\left.\left.+\frac{(\xi+j_{k})H_{3,j_{k},\xi,-(\xi+j_{k})}H_{3,-j_{k},\xi+j_{k},-\xi}}{-\lambda_{\alpha}(j_{k})+\lambda_{\alpha}(\xi+j_{k})-\lambda_{\alpha}(\xi)}\right)\right)j_{k}\zeta_{k}.$

###### Proof.

We first find a symbol $\mathfrak{t}_{1}$, that solves (10.207).

Solution to (10.207). Let us find a symbol
$\mathfrak{t}_{1}\in\mathfrak{B}^{m}_{1}$ such that
$\mathfrak{A}_{1}=\Pi_{S^{\perp}}Op(\mathfrak{t}_{1})$ satisfies (10.207). By
Definition 8.8 and recalling $\overline{v}$ from (8.4), we have

$\displaystyle\mathfrak{t}_{1}=\sum_{j_{k}\in
S}C_{j_{k}}(\xi)\sqrt{j_{k}\zeta_{k}}e^{\text{i}\mathtt{l}(j_{k})\cdot\varphi+\text{i}j_{k}x}.$
(10.210)

Let us first compute $\partial_{x}S_{1}$ using symbolic notations. In view of
the definition of $S_{1}$ in (10.199), we see from (4.67) that

$\displaystyle H_{3,2}(f)=\sum_{\begin{subarray}{c}j_{1}+j_{2}+j_{3}=0,\\\
j_{1}\in S,\ j_{2},j_{3}\in
S^{\perp}\end{subarray}}3H_{3,j_{1},j_{2},j_{3}}f_{j_{1}}f_{j_{2}}f_{j_{3}}.$
(10.211)

Using our convention for a Fourier expansion in (2.3) and the $L^{2}$-gradient
in (2.125), we determine $\nabla_{z}^{2}H(\overline{v})$ as a unique linear
map such that

$\displaystyle(\nabla_{z}^{2}H_{3,2}(\overline{v})[h],g)_{L^{2}}=\frac{d^{2}}{dtds}H_{3,2}(\overline{v}+th+sg)\bigg{|}_{t=s=0},\text{
for all $h,g\in H_{S^{\perp}}$. }$ (10.212)

The left-hand side of (10.212) can be computed as

$\displaystyle(\nabla_{z}^{2}H_{3,2}(\overline{v})[h],g)_{L^{2}}=\sum_{j\in\mathbb{Z}}2\pi\left(\nabla_{z}^{2}H_{3,2}(\overline{v})[h]\right)_{j}g_{-j},$
(10.213)

while the right-hand side can be computed from (10.211) as

$\frac{d^{2}}{dtds}H_{3,2}(\overline{v}+th+sg)\bigg{|}_{t=s=0}=\sum_{j\in\mathbb{Z}}\left(\sum_{j_{1}+j_{2}=j}6H_{3,j_{1},j_{2},-j}\overline{v}_{j_{1}}h_{j_{2}}\right)g_{-j}.$

Combining this with (10.213), we get, using (2.3),

$\nabla_{z}^{2}H_{3,2}(\overline{v})[h]=\frac{1}{2\pi}\sum_{k\in
S^{\perp}}\sum_{j_{1}+j_{2}=k}6H_{3,j_{1},j_{2},-k}\overline{v}_{j_{1}}h_{j_{2}}e^{\text{i}kx}.$

Therefore, we have

$\displaystyle\partial_{x}S_{1}[h]$
$\displaystyle=\partial_{x}\nabla_{z}^{2}H_{3,2}(\overline{v})[h]=\frac{1}{2\pi}\sum_{k\in
S^{\perp}}\sum_{\begin{subarray}{c}j_{1}+j_{2}=k,\\\ j_{1}\in S,j_{2}\in
S^{\perp}\end{subarray}}6\text{i}kH_{3,j_{1},j_{2},-k}\overline{v}_{j_{1}}h_{j_{2}}e^{\text{i}kx}$
$\displaystyle=\Pi_{S^{\perp}}\sum_{k\in\mathbb{Z}}\sum_{\begin{subarray}{c}j_{1}+j_{2}=k,\\\
j_{1}\in S,j_{2}\in
S^{\perp}\end{subarray}}\frac{3\text{i}}{\pi}kH_{3,j_{1},j_{2},-k}\overline{v}_{j_{1}}h_{j_{2}}e^{\text{i}kx}$
$\displaystyle=:\Pi_{S^{\perp}}Op(\mathfrak{s}_{1})[h],$ (10.214)

where (using $\overline{v}$ in (8.4)),

$\displaystyle{\mathfrak{s}_{1}}(x,\xi):=\sum_{j_{k}\in
S}\frac{3\text{i}}{\pi}(j_{k}+\xi)H_{3,j_{k},\xi,-(\xi+j_{k})}\sqrt{j_{k}\zeta_{k}}e^{\text{i}\mathtt{l}(j_{k})\cdot\varphi+\text{i}j_{k}x}.$
(10.215)

Now, we compute the right-hand side of (10.207). Using (10.210), we have

$\displaystyle[\mathcal{D}_{\overline{\omega}}-\partial_{x}D,Op(\mathfrak{t}_{1})]$
$\displaystyle=\mathcal{D}_{\overline{\omega}}(Op(\mathfrak{t}_{1}))-[\partial_{x}D,Op(\mathfrak{t}_{1})]$
$\displaystyle=Op(\mathcal{D}_{\overline{\omega}}\mathfrak{t}_{1})-[\partial_{x}D,Op(\mathfrak{t}_{1})]$
$\displaystyle=Op(\mathfrak{t}),$ (10.216)

where (using $\overline{\omega}$, and $D$ in (5.9) and (10.197)),

$\displaystyle\mathfrak{t}$ $\displaystyle=\sum_{j_{k}\in
S}C_{j_{k}}(\xi)\sqrt{j_{k}\zeta_{k}}\left(\text{i}\overline{\omega}_{k}-\left(\text{i}(\xi+j_{k})\left(-\frac{1}{2}m_{1,\alpha}(\xi+j_{k})+\frac{T_{\alpha}}{4}\right)-\text{i}\xi\left(-\frac{1}{2}m_{1,\alpha}(\xi)+\frac{T_{\alpha}}{4}\right)\right)\right)e^{\text{i}\mathtt{l}(j_{k})\cdot\varphi+\text{i}j_{k}x}$
$\displaystyle=\sum_{j_{k}\in
S}\frac{1}{2}\text{i}C_{j_{k}}(\xi)\sqrt{j_{k}\zeta_{k}}\left(\lambda_{\alpha}(\xi+j_{k})-\lambda_{\alpha}(\xi)-\lambda_{\alpha}(j_{k})\right)e^{\text{i}\mathtt{l}(j_{k})\cdot\varphi+\text{i}j_{k}x},$
(10.217)

where the second equality can be obtained from the definition of
$\lambda_{\alpha}$ in (2.98). Therefore, plugging (10.4.2.2) and (10.4.2.2)
into (10.207), we have

$\Pi_{S^{\perp}}Op(\mathfrak{s}_{1})=\partial_{x}S_{1}=\Pi_{S^{\perp}}[\mathcal{D}_{\overline{\omega}}-\partial_{x}D,Op(\mathfrak{t}_{1})]=\Pi_{S^{\perp}}Op(\mathfrak{t}),$

which gives (using (10.215) and (10.4.2.2)),

$\displaystyle
C_{j_{k}}(\xi)=\frac{6}{\pi}\frac{(j_{k}+\xi)H_{3,j_{k},\xi,-(\xi+j_{k})}}{\lambda_{\alpha}(\xi+j_{k})-\lambda_{\alpha}(\xi)-\lambda_{\alpha}(j_{k})}.$
(10.218)

Therefore, we obtain from (10.210) that

$\displaystyle\mathfrak{A}_{1}=\Pi_{S^{\perp}}Op(\mathfrak{t}_{1}),\quad\mathfrak{t}_{1}(x,\xi):=\frac{6}{\pi}\sum_{j_{k}\in
S}\frac{(j_{k}+\xi)H_{3,j_{k},\xi,-(\xi+j_{k})}}{\lambda_{\alpha}(\xi+j_{k})-\lambda_{\alpha}(\xi)-\lambda_{\alpha}(j_{k})}\sqrt{j_{k}\zeta_{k}}e^{\text{i}\mathtt{l}(j_{k})\cdot\varphi+\text{i}j_{k}x}.$
(10.219)

Computing $\partial_{x}S_{2}-\frac{1}{2}[\mathfrak{A}_{1},\partial_{x}S_{1}]$.
First, using (10.199) and (4.17), we have that for all $h\in H_{S^{\perp}}$,

$\displaystyle\partial_{x}S_{2}[h]$
$\displaystyle=\partial_{x}\nabla_{z}^{2}\mathcal{H}_{4,2}(\overline{v})[h]+\partial_{x}\nabla_{z}^{2}\mathfrak{H}_{1}(\overline{v})[h]+\partial_{x}\nabla_{z}^{2}\mathfrak{H}_{2}(\overline{v})[h].$

Therefore, we have that

$\displaystyle\partial_{x}S_{2}-\frac{1}{2}[\mathfrak{A}_{1},\partial_{x}S_{1}]=\left(\partial_{x}\nabla_{z}^{2}\mathfrak{H}_{1}(\overline{v})-\frac{1}{2}[\mathfrak{A}_{1},\partial_{x}S_{1}]\right)+\partial_{x}\nabla_{z}^{2}\mathcal{H}_{4,2}(\overline{v})+\partial_{x}\nabla_{z}^{2}\mathfrak{H}_{2}(\overline{v})$
(10.220)

Let us compute
$\partial_{x}\nabla_{z}^{2}\mathcal{H}_{4,2}(\overline{v})+\partial_{x}\nabla_{z}^{2}\mathfrak{H}_{2}(\overline{v})$
first. From (4.68) and (4.70), we have

$\displaystyle\mathcal{H}_{4,2}(f)+\mathfrak{H}_{2}(f)$
$\displaystyle=\sum_{\begin{subarray}{c}j_{1}+j_{2}+j_{3}+j_{4}=0,\\\
j_{1},j_{2}\in S,\ j_{3},j_{4}\in S^{\perp}\\\
j_{1}+j_{2}=0,j_{3}+j_{4}=0\end{subarray}}6H_{4,j_{1},j_{2},j_{3},j_{4}}f_{j_{1}}f_{j_{2}}f_{j_{3}}f_{j_{4}}$
$\displaystyle\ +\sum_{\begin{subarray}{c}j_{1}+j_{2}+j_{3}+j_{4}=0,\
j_{1},j_{2}\in S,\\\ j_{3},j_{4}\in S^{\perp},\ j_{1}+j_{2}\neq
0\end{subarray}}G_{j_{1},j_{2},j_{3},j_{4}}f_{j_{1}}f_{j_{2}}f_{j_{3}}f_{j_{4}}$
$\displaystyle=:A_{1}(f)+A_{2}(f).$ (10.221)

where

$G_{j_{1},j_{2},j_{3},j_{4}}:=6H_{4,j_{1},j_{2},j_{3},j_{4}}-\frac{9}{2}(j_{3}+j_{4})\frac{H_{3,j_{1},j_{2},-(j_{1}+j_{2})}H_{3,j_{3},j_{4},-(j_{3}+j_{4})}}{j_{1}\kappa_{j_{1}}+j_{2}\kappa_{j_{2}}-(j_{1}+j_{2})\kappa_{j_{1}+j_{2}}},$

where we used S.1, which implies that $j_{1}+j_{2}\in S^{\perp}\iff
j_{1}+j_{2}\neq 0$, for $j_{1},j_{2}\in S$. For the first term $A_{1}$, we
have (using the property of our tangential sites $S$ in (4.6)),

$\displaystyle A_{1}(f)=12\sum_{j_{1}\in S^{+},j_{3}\in
S^{\perp}}H_{4,j_{1},-j_{1},j_{3},-j_{3}}f_{j_{1}}f_{-j_{1}}f_{j_{3}}f_{-j_{3}}.$

Therefore, for all $h,g\in H_{S^{\perp}}$,

$\displaystyle(\nabla_{z}^{2}A_{1}(\overline{v})[h],g)_{L^{2}}=\frac{d^{2}}{dtds}A_{1}(\overline{v}+th+sg)\bigg{|}_{t=s=0}=24\sum_{j_{1}\in
S^{+},\ j_{3}\in
S^{\perp}}H_{4,j_{1},-j_{1},j_{3},-j_{3}}|\overline{v}_{j_{1}}|^{2}h_{j_{3}}g_{-j_{3}}.$
(10.222)

Defining a Fourier multiplier $\mathfrak{D}_{1}(\xi)$ by (recalling our
convention (2.3))

$\displaystyle\mathfrak{D}_{1}(\xi):=\frac{12}{\pi}\sum_{j_{1}\in
S^{\perp}}H_{4,j_{1},-j_{1},\xi,-\xi}|\overline{v}_{j_{1}}|^{2},$ (10.223)

we have

$(Op^{W}(\mathfrak{D}_{1})h,g)_{L^{2}}=2\pi\sum_{j_{3}\in
S^{\perp}}(Op^{W}(\mathfrak{D}_{1})h)_{j_{3}}g_{-j_{3}}=(\nabla_{z}^{2}A(\overline{v})[h],g)_{L^{2}},$

hence,

$\displaystyle\nabla_{z}^{2}A_{1}(\overline{v})[h]=\Pi_{S^{\perp}}Op^{W}(\mathfrak{D}_{1})h.$
(10.224)

For $A_{2}$ in (10.4.2.2), a similar argument as above shows that we can find
a symbol $\mathfrak{D}_{2}(\varphi,x,\xi)$ such that

$\nabla_{z}^{2}A_{2}(\overline{v})[h]=\Pi_{S^{\perp}}Op^{W}(\mathfrak{D}_{2})h,\quad\int_{\mathbb{T}}\mathfrak{D}_{2}(\varphi,x,\xi)dx=0.$

Hence, plugging this and (10.224) into (10.4.2.2), we obtain

$\displaystyle\partial_{x}\nabla_{z}^{2}\mathcal{H}_{4,2}(\overline{v})+\partial_{x}\nabla_{z}^{2}\mathfrak{H}_{2}(\overline{v})=\Pi_{S^{\perp}}\partial_{x}Op^{W}(\mathfrak{D}_{1}+\mathfrak{D}_{2}).$
(10.225)

Now, we compute
$\partial_{x}\nabla_{z}^{2}\mathfrak{H}_{1}(\overline{v})-\frac{1}{2}[\mathfrak{A}_{1},\partial_{x}S_{1}]$
in (10.220). Using (4.69), we have that for all $h,g\in H_{S^{\perp}}$,

$\displaystyle(\nabla_{z}^{2}$
$\displaystyle\mathfrak{H}_{1}(\overline{v})[h],g)_{L^{2}}=\frac{d^{2}}{dtds}\mathfrak{H}_{1}(\overline{v}+th+sg)|_{t=s=0}$
$\displaystyle=-9\sum_{\begin{subarray}{c}j_{1}+j_{2}+j_{3}+j_{4}=0,\\\
j_{1},j_{2}\in S,\ j_{3},j_{4}\in S^{\perp},\\\ j_{2}+j_{3}\in
S\end{subarray}}(j_{2}+j_{3})\frac{H_{3,-(j_{2}+j_{3}),j_{2},j_{3}}H_{3,j_{1},j_{4},-(j_{1}+j_{4})}}{j_{1}\kappa_{j_{1}}+j_{4}\kappa_{j_{4}}-(j_{1}+j_{4})\kappa_{j_{1}+j_{4}}}\overline{v}_{j_{1}}\overline{v}_{j_{2}}h_{j_{3}}g_{j_{4}}$
$\displaystyle\ -9\sum_{\begin{subarray}{c}j_{1}+j_{2}+j_{3}+j_{4}=0,\\\
j_{1},j_{2}\in S,\ j_{3},j_{4}\in S^{\perp},\\\ j_{2}+j_{3}\in
S\end{subarray}}(j_{1}+j_{4})\frac{H_{3,-(j_{1}+j_{4}),j_{1},j_{4}}H_{3,j_{2},j_{3},-(j_{2}+j_{3})}}{j_{2}\kappa_{j_{2}}+j_{3}\kappa_{j_{3}}-(j_{2}+j_{3})\kappa_{j_{2}+j_{3}}}\overline{v}_{j_{1}}\overline{v}_{j_{2}}h_{j_{3}}g_{j_{4}},$

where we switched the indices $j_{3}$ with $j_{4}$ and $j_{1}$ with $j_{2}$ so
that the Fourier mode of $h$ is denoted by the subindex $j_{3}$. Hence,

$\displaystyle\partial_{x}$
$\displaystyle\nabla_{z}^{2}\mathfrak{H}_{1}(\overline{v})[h]$
$\displaystyle=-\frac{9}{2\pi}\sum_{\begin{subarray}{c}j_{1}+j_{2}+j_{3}=j,\\\
j_{1},j_{2}\in S,\ j_{3}\in S^{\perp},\\\ j_{2}+j_{3}\in
S\end{subarray}}(j_{2}+j_{3})(j_{1}+j_{2}+j_{3})\text{i}\frac{H_{3,-(j_{2}+j_{3}),j_{2},j_{3}}H_{3,j_{1},-(j_{1}+j_{2}+j_{3}),j_{2}+j_{3}}}{j_{1}\kappa_{j_{1}}+(j_{2}+j_{3})\kappa_{j_{2}+j_{3}}-(j_{1}+j_{2}+j_{3})\kappa_{j_{1}+j_{2}+j_{3}}}\overline{v}_{j_{1}}\overline{v}_{j_{2}}h_{j_{3}}e^{\text{i}jx}$
$\displaystyle\
+\frac{9}{2\pi}\sum_{\begin{subarray}{c}j_{1}+j_{2}+j_{3}=j,\\\ j_{1},j_{2}\in
S,\ j_{3},j_{4}\in S^{\perp},\\\ j_{2}+j_{3}\in
S\end{subarray}}(j_{2}+j_{3})(j_{1}+j_{2}+j_{3})\text{i}\frac{H_{3,j_{2}+j_{3},j_{1},-(j_{1}+j_{2}+j_{3})}H_{3,j_{2},j_{3},-(j_{2}+j_{3})}}{j_{2}\kappa_{j_{2}}+j_{3}\kappa_{j_{3}}-(j_{2}+j_{3})\kappa_{j_{2}+j_{3}}}\overline{v}_{j_{1}}\overline{v}_{j_{2}}h_{j_{3}}e^{\text{i}jx},$
(10.226)

where we used $j_{1}+j_{2}+j_{3}=j$ in the second summation.

Now, we compute
$[\mathfrak{A}_{1},\partial_{x}S_{1}]=\Pi_{S^{\perp}}\left(Op(\mathfrak{t}_{1})\Pi_{S^{\perp}}Op(\mathfrak{s}_{1})-Op(\mathfrak{s}_{1})\Pi_{S^{\perp}}Op(\mathfrak{t}_{1})\right)$.
Using (10.215) and (10.219), let us denote

$\displaystyle\widehat{\mathfrak{s}}_{1}(j_{k},\xi):=\frac{3\text{i}}{\pi}(\xi+j_{k})H_{3,j_{k},\xi,-(\xi+j_{k})}\overline{v}_{j_{k}},\quad\widehat{\mathfrak{t}}_{1}(j_{k},\xi):=\frac{6}{\pi}\frac{(j_{k}+\xi)H_{3,j_{k},\xi,-(\xi+j_{k})}}{\lambda_{\alpha}(\xi+j_{k})-\lambda_{\alpha}(\xi)-\lambda_{\alpha}(j_{k})}\overline{v}_{j_{k}},$
(10.227)

which are the Fourier coefficients of $\mathfrak{s}_{1}$ and
$\mathfrak{t}_{1}$. Therefore, we have that for all $h\in H_{S^{\perp}}$,

$\displaystyle\left(Op(\mathfrak{t}_{1})\Pi_{S^{\perp}}Op(\mathfrak{s}_{1})-Op(\mathfrak{s}_{1})\Pi_{S^{\perp}}Op(\mathfrak{t}_{1})\right)[h]$
(10.228) $\displaystyle\ =\sum_{\begin{subarray}{c}j_{1}+j_{2}+j_{3}=j,\\\
j_{1},j_{2}\in S,j_{3}\in S^{\perp},\\\ j_{1}+j_{3}\in
S^{\perp}\end{subarray}}\left(\widehat{\mathfrak{t}}_{1}(j_{2},j_{1}+j_{3})\widehat{\mathfrak{s}}_{1}(j_{1},j_{3})-\widehat{\mathfrak{s}}_{1}(j_{2},j_{1}+j_{3})\widehat{\mathfrak{t}}_{1}(j_{1},j_{3})\right)h_{j_{3}}e^{\text{i}jx}$
$\displaystyle=\frac{18\text{i}}{\pi^{2}}\sum_{\begin{subarray}{c}j_{1}+j_{2}+j_{3}=j,\\\
j_{1},j_{2}\in S,j_{3}\in S^{\perp},\\\ j_{1}+j_{3}\in
S^{\perp}\end{subarray}}\frac{(j_{1}+j_{3})(j_{1}+j_{2}+j_{3})H_{3,j_{2},j_{1}+j_{3},-(j_{1}+j_{2}+j_{3})}H_{3,j_{1},j_{3},-(j_{1}+j_{3})}}{\lambda_{\alpha}(j_{1}+j_{2}+j_{3})-\lambda_{\alpha}(j_{1}+j_{3})-\lambda_{\alpha}(j_{2})}\overline{v}_{j_{1}}\overline{v}_{j_{2}}h_{j_{3}}e^{\text{i}jx}$
$\displaystyle\
-\frac{18\text{i}}{\pi^{2}}\sum_{\begin{subarray}{c}j_{1}+j_{2}+j_{3}=j,\\\
j_{1},j_{2}\in S,j_{3}\in S^{\perp},\\\ j_{1}+j_{3}\in
S^{\perp}\end{subarray}}\frac{(j_{1}+j_{3})(j_{1}+j_{2}+j_{3})H_{3,j_{2},j_{1}+j_{3},-(j_{1}+j_{2}+j_{3})}H_{3,j_{1},j_{3},-(j_{1}+j_{3})}}{\lambda_{\alpha}(j_{1}+j_{3})-\lambda_{\alpha}(j_{1})-\lambda_{\alpha}(j_{3})}\overline{v}_{j_{1}}\overline{v}_{j_{2}}h_{j_{3}}e^{\text{i}jx}.$

We will combine this with $\partial_{x}\nabla_{z}^{2}\mathfrak{H}_{1}$ in
(10.4.2.2). Recall the definitions of $\kappa_{j}$ in (4.3) and
$\lambda_{\alpha}$ in (2.98) so that we have

$j\kappa_{j}=-\frac{\pi}{2}\lambda_{\alpha}(j)+\frac{\pi T_{\alpha}}{4}j.$

Plugging this into (10.4.2.2) and and switching $j_{1}$ and $j_{2}$ in
(10.228), we see that the summands in each summation are exactly the same, and

$\displaystyle\partial_{x}$
$\displaystyle\nabla_{z}^{2}\mathfrak{H}_{1}(\overline{v})[h]-\frac{1}{2}{[\mathfrak{A}_{1},\partial_{x}S_{1}]}$
$\displaystyle=\frac{9\text{i}}{\pi^{2}}\sum_{\begin{subarray}{c}j_{1}+j_{2}+j_{3}=j,\\\
j_{1},j_{2}\in S,\ j_{3}\in
S^{\perp}\end{subarray}}(j_{2}+j_{3})(j_{1}+j_{2}+j_{3})\frac{H_{3,j_{2},j_{3},-(j_{2}+j_{3})}H_{3,j_{1},j_{2}+j_{3},-(j_{1}+j_{2}+j_{3})}}{\lambda_{\alpha}(j_{1})+\lambda_{\alpha}({j_{2}+j_{3}})-\lambda_{\alpha}(j_{1}+j_{2}+j_{3})}\overline{v}_{j_{1}}\overline{v}_{j_{2}}h_{j_{3}}e^{\text{i}jx}$
$\displaystyle\
-\frac{9\text{i}}{\pi^{2}}\sum_{\begin{subarray}{c}j_{1}+j_{2}+j_{3}=j,\\\
j_{1},j_{2}\in S,\ j_{3}\in
S^{\perp}\end{subarray}}(j_{2}+j_{3})(j_{1}+j_{2}+j_{3})\frac{H_{3,j_{2},j_{3},-(j_{2}+j_{3})}H_{3,j_{1},j_{2}+j_{3},-(j_{1}+j_{2}+j_{3})}}{\lambda_{\alpha}(j_{2})+\lambda_{\alpha}(j_{3})-\lambda_{\alpha}(j_{2}+j_{3})}\overline{v}_{j_{1}}\overline{v}_{j_{2}}h_{j_{3}}e^{\text{i}jx}.$

It is straightforward to see that denoting a symbol $\mathfrak{D}$ as

$\displaystyle\mathfrak{D}(\varphi,x,\xi)$
$\displaystyle:=\sum_{j_{1},j_{2}\in
S}D_{j_{1},j_{2}}(\xi)\overline{v}_{j_{1}}(\varphi,x)\overline{v}_{j_{2}}(\varphi,x),$
(10.229) $\displaystyle D_{j_{1},j_{2}}(\xi)$
$\displaystyle:=\frac{9}{\pi^{2}}\left(\xi+\frac{j_{2}-j_{1}}{2}\right)H_{3,j_{2},\xi-\frac{j_{1}+j_{2}}{2},-(\xi+\frac{j_{2}-j_{1}}{2})}H_{3,j_{1},\xi+\frac{j_{2}-j_{1}}{2},-(\xi+\frac{j_{1}+j_{2}}{2})}$
$\displaystyle\times\left(\frac{1}{\lambda_{\alpha}(j_{1})+\lambda_{\alpha}(\xi+\frac{j_{2}-j_{1}}{2})-\lambda_{\alpha}(\xi+\frac{j_{1}+j_{2}}{2}))}-\frac{1}{\lambda_{\alpha}(j_{2})+\lambda_{\alpha}(\xi-\frac{j_{1}+j_{2}}{2})-\lambda_{\alpha}(\xi+\frac{j_{2}-j_{1}}{2})}\right),$

we have

$\partial_{x}\nabla_{z}^{2}\mathfrak{H}_{1}(\overline{v})-\frac{1}{2}{[\mathfrak{A}_{1},\partial_{x}S_{1}]}=\Pi_{S^{\perp}}\partial_{x}Op^{W}(\mathfrak{D}).$

As in (10.225), we can decompose $\mathfrak{D}$ as
$M_{x}(\mathfrak{D})+(\mathfrak{D}-M_{x}(\mathfrak{D}))$, more precisely,

$\displaystyle\mathfrak{D}(\varphi,x,\xi)$ $\displaystyle=\sum_{j_{1}\in
S^{+}}\left(D_{j_{1},-j_{1}}+D_{-j_{1},j_{1}}\right)(\xi)|\overline{v}_{j_{1}}(\varphi,x)|^{2}+\sum_{j_{1}+j_{2}\neq
0}D_{j_{1},j_{2}}(\xi)\overline{v}_{j_{1}}(\varphi,x)\overline{v}_{j_{2}}(\varphi,x)$
$\displaystyle=\underbrace{\sum_{j_{1}\in
S^{+}}\frac{18}{\pi^{2}}\left(\frac{(\xi-j_{1})H_{3,-j_{1},\xi,-(\xi-
j_{1})}H_{3,j_{1},\xi-
j_{1},-\xi}}{\lambda_{\alpha}(j_{1})+\lambda_{\alpha}(\xi-
j_{1})-\lambda_{\alpha}(\xi)}+\frac{(\xi+j_{1})H_{3,j_{1},\xi,-(\xi+j_{1})}H_{3,-j_{1},\xi+j_{1},-\xi}}{-\lambda_{\alpha}(j_{1})+\lambda_{\alpha}(\xi+j_{1})-\lambda_{\alpha}(\xi)}\right)|\overline{v}_{j_{1}}|^{2}}_{=:\mathfrak{D}_{3}(\xi)}$
$\displaystyle\ +\underbrace{\sum_{j_{1}+j_{2}\neq 0,j_{1},j_{2}\in
S}D_{j_{1},j_{2}}(\xi)\overline{v}_{j_{1}}(\varphi,x)\overline{v}_{j_{2}}(\varphi,x)}_{=:\mathfrak{D}_{4}(\varphi,x\xi)},$

so that we have

$\displaystyle\partial_{x}\nabla_{z}^{2}\mathfrak{H}_{1}(\overline{v})-\frac{1}{2}{[\mathfrak{A}_{1},\partial_{x}S_{1}]}=\partial_{x}Op^{W}(\mathfrak{D}_{3}+\mathfrak{D}_{4}),$
(10.230)

where $\int_{\mathbb{T}}\mathfrak{D}_{4}(\varphi,x,\xi)dx=0$. Plugging this
and (10.225) into (10.220) and using (8.4), we get

$\displaystyle\partial_{x}S_{2}-\frac{1}{2}[\mathfrak{A}_{1},\partial_{x}S_{1}]$
$\displaystyle=\Pi_{S^{\perp}}\partial_{x}Op^{W}(\mathfrak{F}_{\alpha-1}(\xi)+\mathfrak{D}_{5}),$
(10.231)

for some symbol $\mathfrak{D}_{5}\in\mathfrak{B}^{\alpha-1}_{2}$ such that
$M_{x}(\mathfrak{D}_{5})=0$ and
$\mathfrak{F}_{\alpha-1}:=\mathfrak{D}_{1}+\mathfrak{D}_{3}$, which is
(10.209).

Contribution from
$[\mathcal{D}_{\overline{\omega}}-\partial_{x}D,\mathfrak{A}_{2}]$. From Lemma
10.62, we see that there exists a symbol
$\mathfrak{t}_{2}\in\mathfrak{B}^{1}_{2}$ that defines $\mathfrak{A}_{2}$ in
such a way that

$\mathfrak{A}_{2}=\Pi_{S^{\perp}}Op(\mathfrak{t}_{2}).$

Since $\mathcal{D}_{\overline{\omega}}-\partial_{x}D$ commutes with
$\Pi_{S^{\perp}}$ and $Op(M_{x}(\mathfrak{t}_{2}))$, it is straightforward
that there is a symbol $\mathfrak{D}_{6}\in\mathfrak{B}^{m}_{2}$ for some
$m\in\mathbb{R}$ such that

$[\mathcal{D}_{\overline{\omega}}-\partial_{x}D,\mathfrak{A}_{2}]=\Pi_{S^{\perp}}\partial_{x}Op^{W}(\mathfrak{D}_{6}),\quad
M_{x}(\mathfrak{D}_{6})=0.$

Combining this with (10.231), we can denote
$\mathfrak{D}_{0}:=\mathfrak{D}_{5}+\mathfrak{D}_{6}$ so that we obtain
(10.208), with $M_{x}(\mathfrak{D}_{0})=0$. ∎

Now, we specify the structure of $\mathcal{C}_{2}$ in (10.191).

######  Lemma 10.65.

$\mathcal{C}_{2}$ in (10.192) satisfies

$\displaystyle\mathcal{C}_{2}=\Pi_{S^{\perp}}Op^{W}(\tilde{\mathfrak{b}}).$
(10.232)

Furthermore, we have that

$\displaystyle
M_{x}(\tilde{\mathfrak{b}})+\mathtt{m}_{\alpha,1}m_{1,\alpha}(\xi)+\mathfrak{m}_{\leq
0,1}(\omega,\xi)=\mathfrak{F}_{\alpha-1}(\omega,\xi),$ (10.233)

where $\mathfrak{F}_{\alpha-1}$ is as given in (10.209).

###### Proof.

From (10.198), (10.197) and (10.192), we have that

$\mathcal{L}^{5}=\mathcal{D}_{\omega}-\partial_{x}D-\varepsilon^{2}\Pi_{S^{\perp}}\partial_{x}\left(Op^{W}(\tilde{\mathfrak{b}}_{2}+\mathtt{m}_{\alpha,1}m_{1,\alpha}(\xi)+\mathfrak{m}_{\leq
0,1}(\omega,\xi))-Op^{W}(\mathfrak{c}_{1})\Pi_{S}Op^{W}(\mathfrak{c}_{2})\right),\text{
up to $O(\varepsilon^{9-6b})$}.$

On the other hand, Lemma 10.63 and Lemma 10.64 tells us that

$\mathcal{L}^{5}=\mathcal{D}_{\omega}-\varepsilon^{2}\partial_{x}D-\Pi_{S^{\perp}}\partial_{x}Op^{W}(\mathfrak{F}_{\alpha-1}(\omega,\xi)+\mathfrak{D}_{0}),\text{
up to $O(\varepsilon^{9-6b})$.}$

Equating the two expressions of $\mathcal{L}^{5}$, we have

$Op^{W}(\tilde{\mathfrak{b}}_{2}+\mathtt{m}_{\alpha,1}m_{1,\alpha}(\xi)+\mathfrak{m}_{\leq
0,1}(\omega,\xi))+Op^{W}(\mathfrak{c}_{1})\Pi_{S}Op^{W}(\mathfrak{c}_{2})=Op^{W}(\mathfrak{F}_{\alpha-1}(\omega,\xi)+\mathfrak{D}_{0}),\text{
up to $O(\varepsilon^{7-6b})$.}$

Note that the operators on the both sides are independent of $\varepsilon>0$,
while, $7-6b>0$, thanks to the range of $b$ in (6.24). Therefore, for
sufficiently small $\varepsilon>0$, we have exact equality and therefore,
$\mathfrak{c}_{1}=\mathfrak{c}_{2}=0$, which gives (10.232), and

$\tilde{\mathfrak{b}}_{2}+\mathtt{m}_{\alpha,1}m_{1,\alpha}(\xi)+\mathfrak{m}_{\leq
0,1}(\omega,\xi)=\mathfrak{F}_{\alpha-1}(\omega,\xi)+\mathfrak{D}_{0}.$

Using Lemma 10.64, which implies $M_{x}(\mathfrak{D}_{0})=0$ and taking the
average in $x$ on both sides in the above equality, we get (10.233).

∎

####  10.4.3 Linear Birkhoff normal form: Step 2

In this subsection, we eliminate the term $\mathfrak{b}_{0}$ in
$\mathcal{L}^{5}$ in (10.60). In order to remove $\mathfrak{b}_{0}$, we use a
transformation $\Phi_{6}$ defined by

$\displaystyle\Phi_{6}:=e^{\varepsilon^{2}A_{1,*}},\quad\text{ where
$A_{1,*}:=\partial_{x}\Pi_{S^{\perp}}Op^{W}({\rho_{2}})$},$ (10.234)

for some symbol
$\rho_{2}=\rho_{2}(\omega,\varphi,\cdot,\cdot)\in\mathcal{S}^{-1-\alpha}$,
which will be chosen later (see Proposition 10.68). Note that $\Phi_{6}$ is a
time-$1$ flow map of a pseudo PDE:

$\displaystyle\frac{d}{d\tau}u=\partial_{x}\Pi_{S^{\perp}}\varepsilon^{2}Op^{W}(\rho_{2})u.$
(10.235)

Since $\partial_{x}Op^{W}(\rho_{2})\in\mathcal{S}^{-\alpha}$,
$e^{\varepsilon^{2}A_{1,*}}:=\sum_{n=0}^{\infty}\frac{1}{n!}(\varepsilon^{2}A_{1,*})^{n}$
is well-defined (see Lemma 10.69).

As in Subsection 9.2 (see Lemma 9.10), we will decompose $\Phi_{6}$ into a
finite dimensional operator and $\Psi_{6}$, which is the time-$1$ flow map of
the pseudo-PDE:

$\displaystyle\partial_{\tau}u=\varepsilon^{2}\partial_{x}Op^{W}(\rho_{2})u,\text{
that is,
}\frac{d}{d\tau}\Psi_{6}(\tau)=\varepsilon^{2}\partial_{x}Op^{W}(\rho_{2})\Psi_{6}(\tau),\quad\Psi_{6}(0)=I.$
(10.236)

The main result is as follows:

######  Proposition 10.66.

There exists a symplectic transformation $\Phi_{6}:H_{S^{\perp}}\mapsto
H_{S^{\perp}}$ such that

$\displaystyle\mathcal{L}^{6}[h]$
$\displaystyle:=(\Phi_{6})^{-1}\mathcal{L}^{5}\Phi_{6}[h]=\mathcal{D}_{\omega}h-\Pi_{S^{\perp}}\partial_{x}M_{6}[h]+\Pi_{S^{\perp}}\partial_{x}Op^{W}(\mathfrak{r}_{-2,\leq
3,\sharp})+\partial_{x}\Pi_{S^{\perp}}W_{3}+R_{6}[h],$ (10.237) $\displaystyle
M_{6}[h]$
$\displaystyle:=Op^{W}\left(\mathtt{m}_{\alpha}m_{1,\alpha}(\xi)+\left(\frac{T_{\alpha}}{4}+\mathfrak{m}_{\leq
0}(\omega,\xi)+\varepsilon^{2}\mathfrak{m}_{\mathfrak{b}}(\omega,\xi)\right)\right),$

satisfies the following:

  1. (1)

The symbol
$\mathtt{m}_{\alpha}m_{1,\alpha}(\xi)+\left(\frac{T_{\alpha}}{4}+\mathfrak{m}_{\leq
0}(\omega,\xi)+\varepsilon^{2}\mathfrak{m}_{\mathfrak{b}}(\omega,\xi)\right)$
is reversible

  2. (2)

$\mathfrak{r}_{-2,\leq 3,\sharp}$ satisfies

$\displaystyle|\mathfrak{r}_{-2,\leq
3,\sharp}|^{\text{Lip}(\gamma,\Omega_{1})}_{-2,s,\eta_{0}}\leq_{\mathtt{pe},s,\eta_{0}}\gamma^{-1}\left(\varepsilon^{5}+\varepsilon^{3}\rVert\mathfrak{I}_{\delta}\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s+\mu_{0}}\right),$
(10.238) $\displaystyle|d_{i}\mathfrak{r}_{-2,\leq
3,\sharp}(i_{0})[\hat{\textbf{\i}}]|_{-2,s,\eta_{0}}\leq_{\mathtt{pe},s,\eta_{0}}\varepsilon^{3}\gamma^{-1}\left(\rVert\hat{\textbf{\i}}\rVert_{s+\mu_{0}}+\rVert\mathfrak{I}_{\delta}\rVert_{s+\mu_{0}}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{0}}\right).$

  3. (3)

$W_{3}:=\Psi_{6}^{T}W_{2}\Psi_{6}$, and $R_{6}$ is of the form in (8.32).

  4. (4)

$\Phi_{6}$ is real, reversibility preserving and symplectic. Therefore,
$\mathcal{L}^{6}$ is real, reversible and Hamiltonian.

  5. (5)

$\mathcal{L}^{6}$ and $\Phi_{6}$ are $\frac{2\pi}{\mathtt{M}}$-translation
invariance preserving.

###### Proof.

The expression for $\mathcal{L}^{6}$ given in (10.237) follows from Lemma
10.67, Proposition 10.68, Lemma 10.73 and Lemma 10.244, denoting
$R_{6}:=R_{6,1}+R_{6,2}$. To see (1), note that $\xi\mapsto m_{1,\alpha}(\xi)$
is even (see (2.69)) Also, $\mathfrak{m}_{\leq_{0}}$ is a reversible symbol,
which follows from Proposition 10.41. From Proposition 10.60, it follows that
$\mathfrak{m}_{\mathfrak{b}}$ is also reversible, therefore (1) follows
immediately. (2) is proved in Lemma 10.74. (3) also follows from Lemma 10.73
and Lemma 10.74. (4) is a result of Lemma 10.72. Lastly, for (5), we note from
(5) of Proposition 10.51 that $\mathcal{L}^{5}$ is
$\frac{2\pi}{\mathtt{M}}$-translation invariance preserving. Therefore, the
symbol $\mathfrak{b}_{0}$ in (10.193) is also
$\frac{2\pi}{\mathtt{M}}$-translation invariant, which implies the Fourier
coefficients of $x\mapsto\mathfrak{b}_{0}(\varphi,x,\xi)$ are supported only
on the modes which are the multiplies of $\mathtt{M}$. This property holds for
$\rho_{2}$, since $\rho_{2}\in\mathfrak{B}^{-1-\alpha}_{2}$ (Proposition
10.68) therefore it is supported on the Fourier modes that are multiples of
$\mathtt{M}$. Hence, $\Phi_{6}$ in (10.234) is
$\frac{2\pi}{\mathtt{M}}$-translation invariance preserving, therefore so is
$\mathcal{L}^{6}$. ∎

From now on, we aim to prove the lemmas used to prove the above proposition.
We see from the expansions of $\mathtt{m}_{\alpha}$ and $\mathfrak{m}_{\leq
0}$ from Proposition 10.10 and Proposition 10.41, and (10.194) that

$\displaystyle\mathtt{m}_{\alpha}m_{1,\alpha}(\xi)+\frac{T_{\alpha}}{4}+\mathfrak{m}_{\leq
0}(\omega,\xi)+\varepsilon^{2}\mathfrak{m}_{\mathfrak{b}}(\omega,\xi):=\left(-\frac{1}{2}m_{1,\alpha}(\xi)+\frac{T_{\alpha}}{4}\right)+\varepsilon^{2}\mathfrak{F}_{\alpha-1}(\omega,\xi)+(\mathtt{m}_{\alpha,2}m_{1,\alpha}(\xi)+\mathfrak{m}_{\leq
0,2}).$

We denote

$\displaystyle\mathcal{B}_{0,*}$
$\displaystyle:=\partial_{x}\Pi_{S^{\perp}}Op^{W}\left(-\frac{1}{2}m_{1,\alpha}(\xi)+\frac{T_{\alpha}}{4}\right),\quad\mathcal{B}_{1,*}:=\partial_{x}\Pi_{S^{\perp}}Op^{W}\left(\mathfrak{F}_{\alpha-1}(\omega,\xi)\right)$
(10.239) $\displaystyle\mathcal{B}_{2,*}$
$\displaystyle:=\partial_{x}\Pi_{S^{\perp}}Op^{W}\left(\mathtt{m}_{\alpha,2}m_{1,\alpha}(\xi)+\mathfrak{m}_{\leq
0,2}\right),\quad\mathcal{B}_{3,*}:=\partial_{x}\Pi_{S^{\perp}}Op^{W}\left(\mathfrak{b}_{0}\right),$

so that

$\displaystyle\partial_{x}\Pi_{S^{\perp}}Op^{W}\left(\mathtt{m}_{\alpha}m_{1,\alpha}(\xi)+\frac{T_{\alpha}}{4}+\mathfrak{m}_{\leq
0}(\omega,\xi)+\varepsilon^{2}\mathfrak{m}_{\mathfrak{b}}(\omega,\xi)\right)=\mathcal{B}_{0,*}+\varepsilon^{2}\mathcal{B}_{1,*}+\mathcal{B}_{2,*}.$
(10.240)

Therefore, (10.193) can be written as

$\displaystyle\mathcal{L}^{5}=\mathcal{D}_{\omega}-\mathcal{B}_{0,*}-\varepsilon^{2}\mathcal{B}_{3,*}-\varepsilon^{2}\mathcal{B}_{1,*}-\mathcal{B}_{2,*}+\mathcal{R}_{\leq
3}.$ (10.241)

#####  10.4.3.1 Computing the conjugation

As in Subsubsection 10.4.1, we expand $\Phi_{6},\Phi_{6}^{-1}$ as

$\displaystyle\Phi_{6}$
$\displaystyle=:I+\varepsilon^{2}\mathcal{A}_{1,*}=:I+\varepsilon^{2}A_{1,*}+\varepsilon^{4}\mathcal{A}_{2,*},\quad\Phi_{6}^{-1}$
$\displaystyle=:I+\varepsilon^{2}\tilde{\mathcal{A}}_{1,*}=:I-\varepsilon^{2}A_{1,*}+\varepsilon^{4}\tilde{\mathcal{A}}_{2,*}.$
(10.242)

######  Lemma 10.67.

We have that

$\displaystyle\Phi_{6}^{-1}\mathcal{L}^{5}\Phi_{6}$
$\displaystyle=\mathcal{D}_{\omega}-\Pi_{S^{\perp}}\partial_{x}Op^{W}\left(\mathtt{m}_{\alpha}m_{1,\alpha}(\xi)+\left(\frac{T_{\alpha}}{4}+\mathfrak{m}_{\leq
0}(\omega,\xi)+\varepsilon^{2}\mathfrak{m}_{\mathfrak{b}}(\omega,\xi)\right)\right)$
(10.243) $\displaystyle\
+\varepsilon^{2}\left(\mathcal{D}_{\overline{\omega}}(A_{1,*})-[\mathcal{B}_{0,*},A_{1,*}]-\mathcal{B}_{3,*}\right)+\mathcal{W}_{*}+\mathcal{Q}_{1,*},$

where

$\displaystyle\mathcal{W}_{*}$
$\displaystyle:=\Phi_{6}^{-1}\partial_{x}\Pi_{S^{\perp}}W_{2}\Phi_{6},$
(10.244) $\displaystyle\mathcal{Q}_{1,*}$
$\displaystyle:=\varepsilon^{2}\left(\mathcal{D}_{{\omega}}(A_{1,*})-\mathcal{D}_{\overline{\omega}}(A_{1,*})\right)+\varepsilon^{4}(\mathcal{D}_{\omega}(\mathcal{A}_{2,*})+\tilde{\mathcal{A}}_{1,*}(\mathcal{D}_{\omega}(A_{1,*})+\varepsilon^{2}\mathcal{D}_{\omega}(\mathcal{A}_{2,*})))$
$\displaystyle\
-\varepsilon^{4}(\tilde{\mathcal{A}}_{2,*}\mathcal{B}_{0,*}+\tilde{\mathcal{A}}_{1,*}\mathcal{B}_{0,*}A_{1,*}+\Phi_{6}^{-1}\mathcal{B}_{0,*}\mathcal{A}_{2,*})$
$\displaystyle\
-\varepsilon^{4}\left(\tilde{\mathcal{A}}_{1,*}\mathcal{B}_{3,*}+\Phi_{6}^{-1}\mathcal{B}_{3,*}\mathcal{A}_{1,*}\right)-\varepsilon^{4}\left(\tilde{\mathcal{A}}_{1,*}\mathcal{B}_{1,*}+\Phi_{6}^{-1}\mathcal{B}_{1,*}\mathcal{A}_{1,*}\right)$
$\displaystyle\
-\varepsilon^{2}\left(\tilde{\mathcal{A}}_{1,*}\mathcal{B}_{2,*}+\Phi_{6}^{-1}\mathcal{B}_{2,*}\mathcal{A}_{1,*}\right)+\Phi_{6}^{-1}\Pi_{S^{\perp}}\partial_{x}Op^{W}(\mathfrak{r}_{-2,\leq
3,*})\Phi_{6}+\Phi_{6}^{-1}\mathcal{R}_{5}\Phi_{6}^{-1}.$

###### Proof.

Using (10.241), we see that

$\displaystyle\Phi_{6}^{-1}\mathcal{L}^{5}\Phi_{6}=\Phi_{6}^{-1}\mathcal{D}_{\omega}\Phi_{6}-\Phi_{6}^{-1}\mathcal{B}_{0,*}\Phi_{6}-\varepsilon^{2}\Phi_{6}^{-1}\mathcal{B}_{3,*}\Phi_{6}-\varepsilon^{2}\Phi_{6}^{-1}\mathcal{B}_{1,*}\Phi_{6}-\Phi_{6}^{-1}\mathcal{B}_{2,*}\Phi_{6}+\Phi_{6}^{-1}\mathcal{R}_{\leq
3}\Phi_{6}.$ (10.245)

Conjugation with $\mathcal{D}_{\omega}.$ Using (10.242), we have that

$\displaystyle\Phi_{6}^{-1}\mathcal{D}_{\omega}\Phi_{6}$
$\displaystyle=\Phi_{6}^{-1}\left(\mathcal{D}_{\omega}(\Phi_{6})+\Phi_{6}\mathcal{D}_{\omega}\right)$
$\displaystyle=\mathcal{D}_{\omega}+\Phi_{6}^{-1}(\varepsilon^{2}\mathcal{D}_{\omega}(A_{1,*})+\varepsilon^{4}\mathcal{D}_{\omega}(\mathcal{A}_{2,*}))$
$\displaystyle=\mathcal{D}_{\omega}+\varepsilon^{2}\mathcal{D}_{\omega}(A_{1,*})+(\varepsilon^{4}\mathcal{D}_{\omega}(\mathcal{A}_{2,*})+\varepsilon^{2}\tilde{\mathcal{A}}_{1,*}(\varepsilon^{2}\mathcal{D}_{\omega}(A_{1,*})+\varepsilon^{4}\mathcal{D}_{\omega}(\mathcal{A}_{2,*})))$
$\displaystyle=\mathcal{D}_{\omega}+\varepsilon^{2}\mathcal{D}_{\overline{\omega}}(A_{1,*})+\varepsilon^{2}\left(\mathcal{D}_{{\omega}}(A_{1,*})-\mathcal{D}_{\overline{\omega}}(A_{1,*})\right)+\varepsilon^{4}(\mathcal{D}_{\omega}(\mathcal{A}_{2,*})+\tilde{\mathcal{A}}_{1,*}(\mathcal{D}_{\omega}(A_{1,*})+\varepsilon^{2}\mathcal{D}_{\omega}(\mathcal{A}_{2,*}))).$
(10.246)

Conjugation with $\mathcal{B}_{0,*}.$ Again, (10.242) gives us that

$\displaystyle\Phi_{6}^{-1}\mathcal{B}_{0,*}\Phi_{6}$
$\displaystyle=\Phi_{6}^{-1}(\mathcal{B}_{0,*}+\varepsilon^{2}\mathcal{B}_{0,*}A_{1,*}+\varepsilon^{4}\mathcal{A}_{2,*})$
$\displaystyle=\mathcal{B}_{0,*}-\varepsilon^{2}A_{1}\mathcal{B}_{0,*}+\varepsilon^{4}\tilde{\mathcal{A}}_{2,*}\mathcal{B}_{0,*}+\varepsilon^{2}\mathcal{B}_{0,*}A_{1}+\varepsilon^{4}\tilde{\mathcal{A}}_{1,*}\mathcal{B}_{0,*}A_{1,*}+\varepsilon^{4}\Phi_{6}^{-1}\mathcal{A}_{2,*}$
$\displaystyle=\mathcal{B}_{0,*}+\varepsilon^{2}[\mathcal{B}_{0,*},A_{1,*}]+\varepsilon^{4}(\tilde{\mathcal{A}}_{2,*}\mathcal{B}_{0,*}+\tilde{\mathcal{A}}_{1,*}\mathcal{B}_{0,*}A_{1,*}+\Phi_{6}^{-1}\mathcal{B}_{0,*}\mathcal{A}_{2,*}).$
(10.247)

Conjugation with $\mathcal{B}_{3,*}.$ Similarly, we have

$\displaystyle\varepsilon^{2}\Phi_{6}^{-1}\mathcal{B}_{3,*}\Phi_{6}$
$\displaystyle=\varepsilon^{2}\Phi_{6}^{-1}(\mathcal{B}_{3,*}+\varepsilon^{2}\mathcal{B}_{3,*}\mathcal{A}_{1,*})$
$\displaystyle=\varepsilon^{2}\mathcal{B}_{3,*}+\varepsilon^{4}\left(\tilde{\mathcal{A}}_{1,*}\mathcal{B}_{3,*}+\Phi_{6}^{-1}\mathcal{B}_{3,*}\mathcal{A}_{1,*}\right).$
(10.248)

Conjugation with $\mathcal{B}_{1,*}.$ Similarly, we have

$\displaystyle\varepsilon^{2}\Phi_{6}^{-1}\mathcal{B}_{1,*}\Phi_{6}=\varepsilon^{2}\mathcal{B}_{1,*}+\varepsilon^{4}\left(\tilde{\mathcal{A}}_{1,*}\mathcal{B}_{1,*}+\Phi_{6}^{-1}\mathcal{B}_{1,*}\mathcal{A}_{1,*}\right).$
(10.249)

Conjugation with $\mathcal{B}_{2,*}.$ Similarly, we have

$\displaystyle\Phi_{6}^{-1}\mathcal{B}_{2,*}\Phi_{6}^{-1}=\mathcal{B}_{2,*}+\varepsilon^{2}\left(\tilde{\mathcal{A}}_{1,*}\mathcal{B}_{2,*}+\Phi_{6}^{-1}\mathcal{B}_{2,*}\mathcal{A}_{1,*}\right)$
(10.250)

Conjugation with $\mathcal{R}_{\leq 3}.$ Recalling $\mathcal{R}_{\leq 3}$ from
(10.192), we have

$\displaystyle\Phi_{6}^{-1}\mathcal{R}_{\leq 3}\Phi_{6}$
$\displaystyle=\Phi_{6}^{-1}\partial_{x}\Pi_{S^{\perp}}W_{2}\Phi_{6}+\Phi_{6}^{-1}\Pi_{S^{\perp}}\partial_{x}Op^{W}(\mathfrak{r}_{-2,\leq
3,*})\Phi_{6}+\Phi_{6}^{-1}\mathcal{R}_{5}\Phi_{6}^{-1}.$ (10.251)

Plugging each conjugation into (10.245), we obtain that

$\displaystyle\mathcal{L}^{5}=\mathcal{D}_{\omega}-(\mathcal{B}_{0,*}+\varepsilon^{2}\mathcal{B}_{1,*}+\mathcal{B}_{2,*})+\varepsilon^{2}\left(\mathcal{D}_{\overline{\omega}}(A_{1,*})-[\mathcal{B}_{0,*},A_{1,*}]-\mathcal{B}_{4,*}\right)+\mathcal{Q}_{1,*}+\mathcal{W}_{*},$

where $\mathcal{Q}_{1,*}$ and $\mathcal{W}_{*}$ are as in (10.244). Recalling
(10.240), we obtain (10.243). ∎

#####  10.4.3.2 Choice of $\rho_{2}$

In view of (10.243), we will choose $\rho_{2}$ in (10.234) so that

$\displaystyle\mathcal{D}_{\overline{\omega}}(A_{1,*})-[\mathcal{B}_{0,*},A_{1,*}]-\mathcal{B}_{3,*}=0,$
(10.252)

######  Proposition 10.68.

There exists $\rho_{2}=\mathfrak{B}_{2}^{-1-\alpha}$ such that

  1. (1)

With $A_{1,*}$ defined as in (10.234), we have that

$\displaystyle\mathcal{D}_{\overline{\omega}}(A_{1,*})-[\mathcal{B}_{0,*},A_{1,*}]-\mathcal{B}_{3,*}=0.$
(10.253)

  2. (2)

$\rho_{2}$ is a real-valued and reversibility preserving symbol. Furthermore,
$\partial_{x}Op^{W}(\rho_{2})$ is a real operator.

###### Proof.

Using (10.234) and (10.239), we have that the above equation is equivalent to

$\displaystyle\Pi_{S^{\perp}}\left(Op^{W}(\mathcal{D}_{\overline{\omega}}(\rho_{2}))-[Op^{W}\left(-\frac{1}{2}m_{1,\alpha}(\xi)+\frac{T_{\alpha}}{4}\right),Op^{W}(\rho_{2})]_{x}-Op^{W}(\mathfrak{b}_{0})\right)=0.$
(10.254)

We aim to choose $\rho_{2}$ satisfying (10.254). First, it follows from (2) of
Proposition 10.60 that $\mathfrak{b}_{0}\in\mathfrak{B}^{-2}_{2}$, hence,
Definition 8.8 tells us that there exists a smooth symbol
$C_{j_{k_{1}}j_{k_{2}}}(\xi)\in\mathcal{S}^{-2}$ such that

$\displaystyle\mathfrak{b}_{0}(\varphi,x,\xi)=\sum_{\begin{subarray}{c}j_{k_{1}},j_{k_{2}}\in
S,\\\ j_{k_{1}}+j_{k_{2}}\neq
0\end{subarray}}C_{j_{k_{1}},j_{k_{2}}}(\xi)\overline{v}_{j_{k_{1}}}\overline{v}_{j_{k_{2}}}e^{\text{i}\left(\mathtt{l}(j_{k_{1}})+\mathtt{l}(j_{k_{2}})\right)\cdot\varphi+\text{i}(j_{k_{1}}+j_{k_{2}})x}.$
(10.255)

As in the proof of Proposition 10.53, let us choose, using (2.14),
$\mathfrak{b}_{0,s}\in\mathfrak{B}^{-2}_{2}$ so that
$Op^{W}(\mathfrak{b}_{0})=Op(\mathfrak{b}_{0,s})$. That is,

$\displaystyle\mathfrak{b}_{0,s}(\varphi,x,\xi)=\sum_{\begin{subarray}{c}j_{k_{1}},j_{k_{2}}\in
S,\\\ j_{k_{1}}+j_{k_{2}}\neq
0\end{subarray}}\underbrace{C_{j_{k_{1}},j_{k_{2}}}\left(\xi+\frac{j_{k_{1}}+j_{k_{2}}}{2}\right)}_{=:C_{j_{k_{1}},j_{k_{2}},s}(\xi)}\overline{v}_{j_{k_{1}}}\overline{v}_{j_{k_{2}}}e^{\text{i}\left(\mathtt{l}(j_{k_{1}})+\mathtt{l}(j_{k_{2}})\right)\cdot\varphi+\text{i}(j_{k_{1}}+j_{k_{2}})x}.$
(10.256)

Again, by Definition 8.8, $\xi\mapsto C_{j_{k_{1}},j_{k_{2}},s}(\xi)$ is a
Fourier multiplier in $\mathcal{S}^{-2}$, that does not depend on
$i_{0},\omega,\varphi,x,\varepsilon$. Therefore, in view of (10.254), it
suffices to find $\rho_{2,s}\in\mathfrak{B}^{-1-\alpha}_{2}$ such that

$\displaystyle\Pi_{S^{\perp}}\left(Op(\mathcal{D}_{\overline{\omega}}\rho_{2,s})-[Op\left(-\frac{1}{2}m_{1,\alpha}(\xi)+\frac{T_{\alpha}}{4}\right),Op(\rho_{2,s}))]_{x}-Op(\mathfrak{b}_{0,s})\right)=0.$
(10.257)

Using the notation in (2.36), (10.256) tells us that it suffices to find
$\rho_{2,s}$ such that (see the proof of Proposition 10.53) for all $j\in
S^{\perp}$,

$\displaystyle\widehat{\rho}_{2,s}^{\varphi,x}(l,k,j)=\begin{cases}\frac{\widehat{\mathfrak{b}}_{0,s}^{\varphi,x}(l,k,j)}{\text{i}\delta_{ljk}},&\text{if
$(l,k)\in\mathcal{J}_{S}$},\\\ 0,&\text{otherwise,}\end{cases}$ (10.258)

where

$\displaystyle\mathcal{J}_{S}$
$\displaystyle:=\left\\{(\mathtt{l}(j_{k_{1}}+j_{k_{2}}),j_{k_{1}}+j_{k_{2}}):j_{k_{1}},j_{k_{2}}\in
S,\ j_{k_{1}}+j_{k_{2}}\neq 0,\ j_{k_{1}}+j_{k_{2}}+j\in S^{\perp}\text{ for
all $j\in S^{\perp}$}\right\\},$ (10.259) $\displaystyle\delta_{ljk}$
$\displaystyle:=\left(\overline{\omega}\cdot
l-\left(\left(-\frac{1}{2}\lambda_{\alpha}(k+j)-(k+j)\frac{T_{\alpha}}{4}\right)-\left(-\frac{1}{2}\lambda_{\alpha}(j)-j\frac{T_{\alpha}}{4}\right)\right)\right).$
(10.260)

Plugging $(l,k)=(\mathtt{l}(j_{k_{1}}+j_{k_{2}}),j_{k_{1}}+j_{k_{2}})$ in
(10.258), we see that the denominator in the left-hand side can be written as
(see (5.9) and (8.5) for $\overline{\omega}$ and
$\mathtt{l}(j_{k_{1}}+j_{k_{2}}$)),

$\displaystyle\frac{1}{\delta_{ljk}}=\frac{2}{\left(\lambda_{\alpha}(j+j_{k_{1}}+j_{k_{2}})-\lambda_{\alpha}(j)-\lambda_{\alpha}(j_{k_{1}}+j_{k_{2}})\right)}=-2\kappa_{-(j_{k_{1}}+j_{k_{2}}),1-\alpha}(j),$

where the last equality follows from (2.100). Therefore, in view of (10.256)
and Lemma 2.35, we can choose

$\displaystyle\rho_{2,s}(\varphi,x,\xi):=\sum_{\begin{subarray}{c}j_{k_{1}},j_{k_{2}}\in
S,\\\ j_{k_{1}}+j_{k_{2}}\neq
0\end{subarray}}2\text{i}C_{j_{k_{1}},j_{k_{2}},s}(\xi)\kappa_{-(j_{k_{1}}+j_{k_{2}}),1-\alpha}(\xi)\overline{v}_{j_{k_{2}}}e^{\text{i}\left(\mathtt{l}(j_{k_{1}})+\mathtt{l}(j_{k_{2}})\right)\cdot\varphi+\text{i}(j_{k_{1}}+j_{k_{2}})x}.$
(10.261)

Clearly, $\rho_{2,s}$ satisfies (10.258). Since
$C_{j_{k_{1}},j_{k_{2}},s}(\xi)\in\mathcal{S}^{-2}$ and
$\kappa_{-(j_{k_{1}}+j_{k_{2}}),1-\alpha}\in\mathcal{S}^{1-\alpha}$, we have
$\rho_{2,s}\in\mathfrak{B}^{-1-\alpha}_{2}$. For the proof of (2), the proof
follows in the exactly same way as in the proof of (2) of Proposition 10.53.
Indeed, comparing (10.254) and (10.166), we see that $\mathfrak{b}_{0}$ is
also a real-valued, reversible symbol, as proved in (2) of Proposition 10.60.
Also $Op^{W}(\mathfrak{b}_{0})$ is a real operator, since so is
$\mathcal{L}^{5}$ (see (10.193) and (4) of Proposition 10.51). ∎

#####  10.4.3.3 Analysis of $\Phi_{6}$

######  Lemma 10.69.

For $k=1,2$, there exist $\tilde{\rho}_{k,*}(\tau)\in\mathcal{S}^{-k\alpha}$
such that

$\Psi_{6}(\tau)-I=\varepsilon^{2}Op(\tilde{\rho}_{1,*}(\tau))+\frac{1}{2}\varepsilon^{4}Op(\tilde{\rho}_{2,*}(\tau)),$

and $\tilde{\rho}_{k,*}$ satisfy

$\displaystyle\sup_{\tau\in[0,1]}|\tilde{\rho}_{k,*}(\tau)|^{\text{Lip}(\gamma,\Omega_{1})}_{-k\alpha,s,\eta_{0}}$
$\displaystyle\leq_{\mathtt{pe},s,\eta_{0}}1,$ (10.262)
$\displaystyle\sup_{\tau\in[0,1]}|d_{i}\tilde{\rho}_{k,*}(\tau)(i_{0})[\hat{\textbf{\i}}]|_{-k\alpha,s,\eta_{0}}$
$\displaystyle=0$ (10.263)

###### Proof.

The proof is identical to Lemma 10.54. ∎

######  Lemma 10.70.

$\Phi_{6}\Pi_{S^{\perp}}-\Pi_{S^{\perp}}\Psi_{6}\Pi_{S^{\perp}}\in\mathfrak{R}(i_{0})$.

###### Proof.

The proof is identical to Lemma 10.55. ∎

######  Lemma 10.71.

$\Phi^{\pm 1}=\Phi_{6},\Phi_{6}^{-1}$ satisfies

$\displaystyle\rVert(\Phi^{\pm}-I)h\rVert_{s}^{\text{Lip}(\gamma,\Omega_{1})}\leq_{\mathtt{pe},s}\varepsilon^{2}\rVert
h\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s}$ (10.264) $\displaystyle\rVert
d_{i}\Phi^{\pm}(i_{0})h[\hat{\textbf{\i}}]\rVert_{s}=0$

###### Proof.

The proof can be done following the proof of Lemma 10.56, word by word, using
Lemma 10.69. ∎

######  Lemma 10.72.

$\Phi_{6}$ is a symplectic, reversibility preserving, real operator.

###### Proof.

The proof is identical to Lemma 10.57, using (2) of Proposition 10.68. ∎

#####  10.4.3.4 Analysis of $\mathcal{W}_{*}$

######  Lemma 10.73.

$\mathcal{W}_{*}$ in (10.244) can be written as
$\mathcal{W}_{*}=\Pi_{S^{\perp}}\partial_{x}\Psi_{6}^{T}W_{2}\Psi_{6}+R_{6,1}$
for some finite dimensional operator $R_{6,1}$ of the form in (8.32).

###### Proof.

The proof is identical to Lemma 10.58. ∎

#####  10.4.3.5 Analysis of $\mathcal{Q}_{1,*}$

######  Lemma 10.74.

$\mathcal{Q}_{1,*}$ in (10.244) is of the form:

$\displaystyle Q_{1,*}=\Pi_{S^{\perp}}Op(\mathfrak{r}_{-2,\leq
3,\sharp})+R_{6,2},$

where the symbol $\mathfrak{r}_{-2,\leq 3,\sharp}$ satisfies

$\displaystyle|\mathfrak{r}_{-2,\leq
3,\sharp}|^{\text{Lip}(\gamma,\Omega_{1})}_{\alpha-3,s,\eta_{0}}\leq_{\mathtt{pe},s,\eta_{0}}\gamma^{-1}\left(\varepsilon^{5}+\varepsilon^{3}\rVert\mathfrak{I}_{\delta}\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s+\mu_{0}}\right),$
(10.265) $\displaystyle|d_{i}\mathfrak{r}_{-2,\leq
3,\sharp}(i_{0})[\hat{\textbf{\i}}]|_{\alpha-3,s,\eta_{0}}\leq_{\mathtt{pe},s,\eta_{0}}\varepsilon^{3}\gamma^{-1}\left(\rVert\hat{\textbf{\i}}\rVert_{s+\mu_{0}}+\rVert\mathfrak{I}_{\delta}\rVert_{s+\mu_{0}}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{0}}\right),$
(10.266)

and $R_{6,2}$ is a finite dimensional operator of the form in (8.32).

###### Proof.

The proof can be completed following the proof of Lemma 10.59, using Lemma
10.69 and Lemma 10.71. ∎

###  10.5 Lipschitz tame estimates for the remainders

Before we obtain the full reduction of the linear operator, we first derive
tame estimates of the remainder in $\mathcal{L}^{6}$ obtained in Proposition
10.66. We denote

$\displaystyle\mathcal{R}_{0}:=\partial_{x}Op^{W}(\mathfrak{r}_{-2,\leq
3,\sharp})+\partial_{x}W_{3}+R_{6}[h].$ (10.267)

In the following lemma, we recall $\mathtt{b}_{0}\in\mathbb{N}$ is fixed as in
(10.70).

######  Lemma 10.75.

Given $\mathtt{S}\gg s_{0}$, we have the following:

  1. (1)

For $s\in[s_{0},\mathtt{S}]$, and $\vec{\mathtt{b}}\in\mathbb{Z}^{\nu}$ such
that $|\mathtt{b}|\in[0,s_{0}+\mathtt{b}_{0}]$, the operators
$\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{R}_{0}),[\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{R}_{0}),\partial_{x}]$
are Lip-$0$-tame operators and

$\displaystyle\mathfrak{M}^{\gamma}_{\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{R}_{0})}(0,s),\mathfrak{M}^{\gamma}_{[\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{R}_{0}),\partial_{x}]}(0,s)$
$\displaystyle\leq_{\mathtt{pe},\mathtt{S}}\gamma^{-1}\left(\varepsilon^{5}+\varepsilon^{3}\rVert\mathfrak{I}_{\delta}\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s+\mu_{0}}\right).$
(10.268)

  2. (2)

For $s\in[s_{0},\mathtt{S}]$, and $\vec{\mathtt{b}}\in\mathbb{Z}^{\nu}$ such
that $|\mathtt{b}|\in[0,s_{0}+\mathtt{b}_{0}]$, the operators
$\partial_{\varphi}^{\vec{\mathtt{b}}}(d_{i}(\mathcal{R}_{0})(i_{0})[\hat{\textbf{\i}}]),[\partial_{\varphi}^{\vec{\mathtt{b}}}(d_{i}(\mathcal{R}_{0})(i_{0})[\hat{\textbf{\i}}]),\partial_{x}]$
are $0$-tame operators and

$\displaystyle\sup_{\vec{\mathtt{b}}\in\mathbb{N}_{0}^{\nu},\
|\vec{\mathtt{b}}|\leq
s_{0}+\mathtt{b}_{0}}\mathfrak{M}_{\mathcal{Q}_{\vec{\mathtt{b}}}}(0,s_{0})\leq_{\mathtt{pe},\mathtt{S}}\varepsilon^{3}\gamma^{-1}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{0}},\quad\mathcal{Q}_{\vec{\mathtt{b}}}\in\left\\{\partial_{\varphi}^{\vec{\mathtt{b}}}(d_{i}(\mathcal{R}_{0})(i_{0})[\hat{\textbf{\i}}]),[\partial_{\varphi}^{\vec{\mathtt{b}}}(d_{i}(\mathcal{R}_{0})(i_{0})[\hat{\textbf{\i}}]),\partial_{x}]\right\\}.$
(10.269)

###### Proof.

Proof of (1). We claim that

$\displaystyle\mathfrak{M}^{\gamma}_{\partial_{x}\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{R}_{0})}(0,s)$
$\displaystyle\leq_{\mathtt{pe},\mathtt{S}}\gamma^{-1}\left(\varepsilon^{5}+\varepsilon^{3}\rVert\mathfrak{I}_{\delta}\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s+\mu_{0}}\right),$
(10.270)
$\displaystyle\mathfrak{M}^{\gamma}_{\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{R}_{0})\partial_{x}}(0,s)$
$\displaystyle\leq_{\mathtt{pe},\mathtt{S}}\gamma^{-1}\left(\varepsilon^{5}+\varepsilon^{3}\rVert\mathfrak{I}_{\delta}\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s+\mu_{0}}\right).$
(10.271)

Let us assume for a moment that (10.270) and (10.271) hold. Since
$\partial_{x}^{-1}$ is clearly a Lip-$0$-tame operator, the tame estimates for
$\partial_{\varphi}^{\vec{b}}(\mathcal{R}_{0})=\partial_{x}^{-1}(\partial_{x}\partial_{\varphi}^{\vec{b}}(\mathcal{R}_{0}))$
follow from (10.270) and Lemma 2.23. Furthermore, it follows immediately from
Definition 2.18 that, for two operators $A,B$,
$\mathfrak{M}^{\gamma}_{A+B}(0,s)\leq\mathfrak{M}^{\gamma}_{A}(0,s)+\mathfrak{M}^{\gamma}_{B}(0,s)$.
Hence (10.270) and (10.271) give us the tame estimates (10.268) for
$[\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{R}_{0}),\partial_{x}]$. For
the proof of (10.270) and (10.271), we prove (10.270) only, since the other
estimate can be proved in the same way.

In order to show (10.270), we need to show that, in view of (10.267),

$\displaystyle\mathfrak{M}^{\gamma}_{\partial_{x}^{2}Op^{W}(\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathfrak{r}_{-2,\leq
3,\sharp}))}(0,s),\
\mathfrak{M}^{\gamma}_{\partial_{x}\partial_{\varphi}^{\vec{\mathtt{b}}}(R_{6})}(0,s),\
\mathfrak{M}^{\gamma}_{\partial_{x}^{2}\partial_{\varphi}^{\vec{\mathtt{b}}}(W_{3})}(0,s)\leq_{\mathtt{pe},s}\gamma^{-1}\left(\varepsilon^{5}+\varepsilon^{3}\rVert\mathfrak{I}_{\delta}\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s+\mu_{0}}\right).$
(10.272)

For
$\partial_{x}^{2}Op^{W}(\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathfrak{r}_{-2,\leq
3,\sharp}))$, it follows from (10.238) and (1) of Lemma 2.15 that we can find
a symbol $r_{0}\in\mathcal{S}^{0}$ such that
$\partial_{x}^{2}Op^{W}(\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathfrak{r}_{-2,\leq
3,\sharp}))=Op(r_{0})$ and $r_{0}$ satisfies (Recalling Remark 8.6),

$\displaystyle|r_{0}|^{\text{Lip}(\gamma,\Omega_{1})}_{0,s,0}\leq_{s,\mathtt{pe}}\gamma^{-1}\left(\varepsilon^{5}+\varepsilon^{3}\rVert\mathfrak{I}_{\delta}\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s+\mu_{0}}\right),$
(10.273)
$\displaystyle|d_{i}r_{0}(i_{0})[\hat{\textbf{\i}}]|_{0,s,0}\leq_{s,\mathtt{pe}}\varepsilon^{3}\gamma^{-1}\left(\rVert\hat{\textbf{\i}}\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s+\mu_{0}}+\rVert\mathfrak{I}_{\delta}\rVert_{s+\mu_{0}}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{0}}\right).$
(10.274)

Hence the desired tame estimates follow from (10.273) and Lemma 2.25.

For the operator $\partial_{x}\partial_{\varphi}^{\vec{\mathtt{b}}}(R_{6})$,
it follows from (3) of Proposition 10.66 that we can pick
$g_{j}(\tau),\chi_{j}(\tau)$ such that

$R_{6}[h]=\sum_{|j|\leq
C}\int_{0}^{1}(h,g_{j}(\tau))_{L^{2}}\chi_{j}(\tau)d\tau,$

and satisfy

$\displaystyle\sup_{\tau\in[0,1]}\rVert
g(\tau)\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s}\rVert\chi(\tau)\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s_{0}}+\rVert
g(\tau)\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s_{0}}\rVert\chi(\tau)\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s}\leq_{\mathtt{pe},s}\varepsilon^{3}+\varepsilon^{2}\rVert\mathfrak{I}_{\delta}\rVert_{s+\mu_{0}}^{\text{Lip}(\gamma,\Omega_{1})},$
(10.275) $\displaystyle\sup_{\tau\in[0,1]}\left(\rVert
d_{i}g(i_{0})[\hat{\textbf{\i}}]\rVert_{s_{0}}\rVert\chi(i_{0})\rVert_{s}+\rVert
d_{i}g(i_{0})[\hat{\textbf{\i}}]\rVert_{s}\rVert\chi_{j}(i_{0})\rVert_{s_{0}}+\rVert
d_{i}\chi(i_{0})[\hat{\textbf{\i}}]\rVert_{s_{0}}\rVert
g(i_{0})\rVert_{s}+\rVert d_{i}\chi(i_{0})[\hat{\textbf{\i}}]\rVert_{s}\rVert
g(i_{0})\rVert_{s_{0}}\right)$
$\displaystyle\quad\leq_{\mathtt{pe},s}\varepsilon^{2}\rVert\hat{\textbf{\i}}\rVert_{s+\mu_{0}}+\varepsilon^{2b-1}\rVert\mathfrak{I}_{\delta}\rVert_{s+\mu_{0}}\rVert\hat{\textbf{\i}}\rVert_{s+\mu_{0}}.$
(10.276)

It is clear that

$\partial_{x}\partial_{\varphi}^{\vec{\mathtt{b}}}(R_{6})h=\int_{0}^{1}\sum_{\vec{b}_{1}+\vec{b}_{2}=\vec{\mathtt{b}}}\sum_{|j|\leq
C}(h,\underbrace{\partial_{\varphi}^{\vec{b}_{1}}g_{j}(\tau)}_{=:g_{b_{1},b_{2},j}(\tau)})_{L^{2}_{x}}\underbrace{C_{\vec{b}_{1},\vec{b}_{2}}\partial_{x}\partial_{\varphi}^{\vec{b}_{2}}(\chi_{j}(\tau))}_{=:\chi_{b_{1},b_{2},j}(\tau)}d\tau$

Clearly, $g_{b_{1},b_{2},j}(\tau),\chi_{b_{1},b_{2},j}(\tau)$ satisfy the
estimates in (10.275) with possibly larger $\mu_{0}$. Hence, the tame estimate
(10.272) for $\partial_{x}\partial_{\varphi}^{\vec{\mathtt{b}}}(R_{6})$
follows from Lemma 8.10.

Lastly, we deal with $W_{3}$. It follows from (3) of Proposition 10.66 and
(10.4.2.1) that

$\displaystyle\partial_{x}^{2}W_{3}$
$\displaystyle=\frac{1}{(\mathtt{N}_{\alpha}+1)!}\int_{0}^{1}\int_{0}^{\tau}\partial_{x}^{2}\left(\Psi_{3}(\tau-t)\Phi_{4}\Psi_{5}\Psi_{6}\right)^{T}Op^{W}(\mathfrak{q}_{\omega,\mathtt{N}_{\alpha}-2}\star\mathfrak{a})\Psi_{3}(\tau-t)\Phi_{4}\Psi_{5}\Psi_{6}t^{\mathtt{N}_{\alpha}-2}dtd\tau$
(10.277) $\displaystyle\
-\frac{1}{\mathtt{N}_{\alpha}!}\int_{0}^{1}\partial_{x}^{2}\left(\Psi_{3}(1-t)\Phi_{4}\Psi_{5}\Psi_{6}\right)^{T}Op^{W}(\mathfrak{q}_{M_{2},\mathtt{N}_{\alpha}}\star\mathfrak{a})\Psi_{3}(1-t)\Phi_{4}\Psi_{5}\Psi_{6}t^{\mathtt{N}_{\alpha}}dt.$

Note that it follows from (10.75), (10.117), (10.152) and (10.236) that the
transformations $\Psi_{3},\Phi_{4},\Psi_{5},\Psi_{6}$ are generated by the
symbols $\mathfrak{a}\in\mathcal{S}^{1-\alpha},\
\mathfrak{p}_{2}\in\mathcal{S}^{1-\alpha},\
\varepsilon\rho_{1},\varepsilon^{2}\rho_{2}\in\mathcal{S}^{-1-\alpha}\subset\mathcal{S}^{1-\alpha}$.
Furthermore, since $\mathfrak{a}\in\mathfrak{S}^{1-\alpha}_{p_{1}}$ (see
(10.73)), using Lemma 10.27 and also using Lemma 10.43, Proposition 10.53 and
Proposition 10.68 that for
$\mathfrak{u}\in\left\\{\mathfrak{a},\mathfrak{p}_{2},\varepsilon\rho_{1},\varepsilon^{2}\rho_{2}\right\\},$
we have the estimates

$\displaystyle|\mathfrak{u}|^{\text{Lip}(\gamma,\Omega_{1})}_{1-\alpha,s,\eta_{0}}$
$\displaystyle\leq_{\mathtt{pe},s,\eta_{0}}\varepsilon(1+\rVert\mathfrak{I}_{\delta}\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s+\mu_{0}}),$
(10.278) $\displaystyle\rVert
d_{i}\mathfrak{u}(i_{0})[\hat{\textbf{\i}}]\rVert_{1-\alpha,s,\eta_{0}}$
$\displaystyle\leq_{\mathtt{pe},s,\eta_{0}}\varepsilon^{3}\gamma^{-1}\left(\rVert\hat{\textbf{\i}}\rVert_{s+\mu_{0}}+\rVert\mathfrak{I}_{\delta}\rVert_{s+\mu_{0}}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{0}}\right).$
(10.279)

especially, (7.16) tells us that

$\displaystyle|\mathfrak{u}|^{\text{Lip}(\gamma,\Omega_{1})}_{1-\alpha,s_{0}+\mu_{0},\eta_{0}}\leq_{\mathtt{pe},s,\eta_{0}}\varepsilon.$
(10.280)

Therefore, we can apply Proposition 9.8 for sufficiently small
$\varepsilon>0$, depending on $\mathtt{S}$ (Recalling Remark 8.6, we choose
$\eta_{0}=\eta_{0}(\mathtt{b}_{0})$ in (10.280)). Then, (1) and (2) of
Proposition 9.8 tell us that

$\displaystyle\sup_{|k|\leq\mathtt{b}_{0},|\vec{b}|\leq|\vec{\mathtt{b}}|}\mathfrak{M}^{\gamma}_{|D|^{k}\partial_{\varphi}^{\vec{{b}}}(\psi)|D|^{-k-|\vec{b}|(2-\alpha)}}(0,s),\
\mathfrak{M}^{\gamma}_{|D|^{k-|\vec{b}|(2-\alpha)}\partial_{\varphi}^{\vec{{b}}}(\psi)|D|^{-k}}(0,s)\leq_{\mathtt{pe},s}(1+\rVert\mathfrak{I}_{\delta}\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s+\mu_{0}}),$
(10.281)

for $\psi\in\left\\{\Psi_{3}(\tau),\Phi_{4},\Psi_{5},\Psi_{6}\right\\}$ for
$\tau\in[0,1]$. Furthermore, it follows from Lemma 10.39 and (1) of Lemma 2.15
that denoting a symbol $\mathfrak{q}_{*}$

$\displaystyle
Op^{W}(\mathfrak{q}_{*})\in\left\\{|D|^{(|\vec{b}_{1}|+1)(2-\alpha)+2}Op^{W}(\mathfrak{q}_{\mathfrak{a}})|D|^{(|\vec{b}_{2}|+1)(2-\alpha)}:\mathfrak{q}_{\mathfrak{a}}=\mathfrak{q}_{\omega,\mathtt{N}_{\alpha}-2}\star\mathfrak{a}\text{
or }\mathfrak{q}_{M_{2},\mathtt{N}_{\alpha}}\star\mathfrak{a},\
\vec{b}_{1}+\vec{b}_{2}=\vec{\mathtt{b}}\right\\},$ (10.282)

we have that
$\mathfrak{q}_{*}\in\mathcal{S}^{\mathtt{N}_{\alpha}(1-\alpha)+\left((|\vec{\mathtt{b}}|+2)(2-\alpha)+2\right)}\subset\mathcal{S}^{0}$
(see (10.70) and recall that $|\mathtt{b}|\leq s_{0}+\mathtt{b}_{0}$), and

$\displaystyle|\mathfrak{q}_{*}|^{\text{Lip}(\gamma,\Omega_{1})}_{0,s,0}$
$\displaystyle\leq_{s,\mathtt{pe}}\varepsilon^{\mathtt{N}_{\alpha}}(1+\rVert\mathfrak{I}_{\delta}\rVert_{s+\mu_{0}}^{\text{Lip}(\gamma,\Omega_{1})}),$
(10.283)
$\displaystyle|d_{i}(\mathfrak{q}_{*}(i_{0})[\hat{\textbf{\i}}]|_{0,s,0}$
$\displaystyle\leq_{s,\mathtt{pe}}\varepsilon^{\mathtt{N}_{\alpha}+2}\gamma^{-1}\left(\rVert\hat{\textbf{\i}}\rVert_{s+\mu_{0}}+\rVert\mathfrak{I}_{\delta}\rVert_{s+\mu_{0}}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{0}}\right).$
(10.284)

Therefore, taking $\partial_{\varphi}^{\vec{\mathtt{b}}}$ in (10.277), we see,
using (10.281), (10.283) and Lemma 2.23, that

$\mathfrak{M}^{\gamma}_{\partial_{x}^{2}\partial_{\varphi}^{\vec{\mathtt{b}}}(W_{3})}\leq_{\mathtt{pe},s}\varepsilon^{\mathtt{N}_{\alpha}+1}(1+\rVert\mathfrak{I}_{\delta}\rVert_{s+\mu_{0}}^{\text{Lip}(\gamma,\Omega_{1})}),$

which gives us (10.272), since $\mathtt{N}_{\alpha}\geq 2$, thanks to (10.70).

Proof of (2). The proof of (2) goes similarly as in (1), hence we briefly
explain the outline of the proof. As above, we claim that

$\displaystyle\mathfrak{M}_{\partial_{x}\partial_{\varphi}^{\vec{\mathtt{b}}}(d_{i}(\mathcal{R}_{0})(i_{0})[\hat{\textbf{\i}}])}(0,s_{0})$
$\displaystyle\leq_{\mathtt{pe}}\varepsilon^{3}\gamma^{-1}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{0}},$
(10.285)
$\displaystyle\mathfrak{M}_{\partial_{\varphi}^{\vec{\mathtt{b}}}(d_{i}(\mathcal{R}_{0})(i_{0})[\hat{\textbf{\i}}])\partial_{x}}(0,s_{0})$
$\displaystyle\leq_{\mathtt{pe}}\varepsilon^{3}\gamma^{-1}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{0}}.$
(10.286)

Once we have the above estimates, then the similar argument described above,
using that $\partial_{x}^{-1}$ is $0$-tame operator, gives us the desired
result (10.269). As mentioned in (1), the proofs for (10.285) and (10.286) are
similar. For (10.285), in view of (10.267), we are led to prove

$\displaystyle\mathfrak{M}_{\partial_{x}^{2}Op^{W}(\partial_{\varphi}^{\vec{\mathtt{b}}}(d_{i}\mathfrak{r}_{-2,\leq
3,\sharp}(i_{0})[\hat{\textbf{\i}}]))}(0,s_{0}),\
\mathfrak{M}_{\partial_{x}\partial_{\varphi}^{\vec{\mathtt{b}}}(d_{i}R_{6}(i_{0})[\hat{\textbf{\i}}])}(0,s_{0}),\
\mathfrak{M}_{\partial_{x}^{2}\partial_{\varphi}^{\vec{\mathtt{b}}}(d_{i}W_{3}(i_{0})[\hat{\textbf{\i}}])}(0,s_{0})\leq_{\mathtt{pe}}\varepsilon^{3}\gamma^{-1}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{0}}.$
(10.287)

For each operator in above, one can follow the same argument as in the proof
of (1), using the estimates (10.274), (10.276), (10.279) and (10.284), with
(7.16). ∎

Recalling Remark 7.4 and Remark 7.5, we see that the loss of derivatives,
$\mu_{0}$, arising in the reduction steps in Section 10 is finite number
depending on only $\mathtt{p}$, fixed as in (6.24), assuming that
$\varepsilon$ is small enough depending on $s,\mathtt{pe}$. We summarize the
results that we obtain through Subsections 10.1\- 10.4 and Lemma 10.75.

######  Proposition 10.76.

There exists a constant

$\displaystyle\mu_{\mathtt{p},1}\geq 0,\text{ depending only on $\mathtt{p}$
in \eqref{parametersets1231}},$ (10.288)

such that if (7.3) is satisfied with $\mu=\mu_{\mathtt{p},1}$ and for some
$\mathtt{C}>0$, then the followings hold: $\mathcal{L}^{6}$ in (10.237) can be
written as

$\displaystyle
L_{0}:=\mathcal{L}^{6}=\mathcal{D}_{\omega}-D_{0}+\mathcal{R}_{0},$ (10.289)

where the operators $D_{0}$ and $\mathcal{R}_{0}$ satisfy

  1. (1)

$D_{0}$ is a diagonal operator between $H_{S^{\perp}}$ and $H_{S^{\perp}}$
such that

$\displaystyle D_{0}$ $\displaystyle=Op^{W}(d_{0}(\omega,\xi)),\quad
d_{0}({\omega,\xi}):=\text{i}\left(\mathtt{m}_{\alpha}(\omega)\lambda_{\alpha}(\xi)+\xi\frac{T_{\alpha}}{4}+\mathfrak{m}_{1}(\omega,\xi)\right),$
(10.290) $\displaystyle\mathfrak{m}_{1}$
$\displaystyle:=\xi\left(\mathfrak{m}_{\leq
0}(\omega,\xi)+\varepsilon^{2}\mathfrak{m}_{\mathfrak{b}}(\xi)\right)\in\mathcal{S}^{1},$
(10.291)

and
$\mathtt{m}_{\alpha}=-\frac{1}{2}+\varepsilon^{2}\mathtt{m}_{\alpha,1}+\mathtt{m}_{\alpha,2}$
satisfies

$\displaystyle|\mathtt{m}_{\alpha,1}|^{\text{Lip}(\gamma,\Omega_{1})}\leq_{\mathtt{pe}}1,\quad|\mathtt{m}_{\alpha,2}|^{\text{Lip}(\gamma,\Omega_{1})}\leq_{\mathtt{pe}}\varepsilon^{7-4b},\quad|d_{i}\mathtt{m}_{\alpha,2}(i_{0})[\hat{\textbf{\i}}]|\leq_{\mathtt{pe}}\varepsilon\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{\mathtt{p},1}},$
(10.292)

and $\mathfrak{m}_{1}(\omega,\xi)$ satisfies

$\displaystyle|\mathfrak{m}_{1}|^{\text{Lip}(\gamma,\Omega_{1})}_{1,0,2}$
$\displaystyle\leq_{\mathtt{pe}}\varepsilon^{2},\quad|d_{i}\mathfrak{m}_{1}(i_{0})[\hat{\textbf{\i}}]|_{1,0,2}\leq_{\mathtt{pe}}\varepsilon\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{\mathtt{p},1}}.$
(10.293)

  2. (2)

Given $\mathtt{S}\gg s_{0}$, there exists
$\varepsilon_{0}=\varepsilon_{0}(\mathtt{pe},\mathtt{S})$ such that if
$\varepsilon\in(0,\varepsilon_{0})$, then $\mathcal{R}_{0}$ is a
Lip-$0$-modulo tame operator up $\mathtt{S}$ and
$d_{i}\mathcal{R}_{0}(i_{0})[\hat{\textbf{\i}}]$ is a $0$-tame operator and
satisfies the following estimates:

$\displaystyle\sup_{\vec{b}\in\mathbb{N}_{0}^{\nu},\
|\vec{b}|\leq\mathtt{b}_{0}}\mathfrak{M}^{\sharp,\gamma}_{\partial_{\varphi}^{\vec{b}}(\mathcal{R}_{0})}(0,s)$
$\displaystyle\leq_{\mathtt{pe},s}\gamma^{-1}\left(\varepsilon^{5}+\varepsilon^{3}\rVert\mathfrak{I}_{0}\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s+\mu_{\mathtt{p},1}}\right),$
(10.294) $\displaystyle\sup_{\vec{{b}}\in\mathbb{N}_{0}^{\nu},\
|\vec{{b}}|\leq\mathtt{b}_{0}}\mathfrak{M}^{\sharp}_{\partial_{\varphi}^{\vec{b}}(d_{i}\mathcal{R}_{0}(i_{0})[\hat{\textbf{\i}}])}(0,s_{0})$
$\displaystyle\leq_{\mathtt{pe}}\varepsilon^{3}\gamma^{-1}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{\mathtt{p},1}},$
(10.295)

for all $s\in[s_{0},\mathtt{S}]$.

  3. (3)

$D_{0}$ is a real, reversible operator. Also, $\mathcal{R}_{0}$ is a real and
reversible operator.

  4. (4)

There exist reversibility preserving real operators
$\Phi_{1-6,1},\Phi_{1-6,2}:H_{S^{\perp}}\mapsto H_{S^{\perp}}$ such that

$\displaystyle L_{0}=\Phi_{1-6,1}\mathcal{L}_{\omega}\Phi_{1-6,2},$ (10.296)

and $\Phi:=\Phi_{1-6,1},\Phi_{1-6,2}$ satisfies (assuming
$\varepsilon\in(0,\varepsilon_{0}(\mathtt{pe},\mathtt{S}$)))

$\displaystyle\rVert(\Phi-I)h\rVert_{s}^{\text{Lip}(\gamma,\Omega_{1})}\leq_{\mathtt{pe},s}\varepsilon\left(\rVert
h\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s+\mu_{\mathtt{p},1}}+\rVert\mathfrak{I}_{0}\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s+\mu_{\mathtt{p},1}}\rVert
h\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{s_{0}+\mu_{\mathtt{p},1}}\right),$
(10.297)

for all $s\in[s_{0},\mathtt{S}]$.

  5. (5)

$\mathcal{R}_{0}$ is $\frac{2\pi}{\mathtt{M}}$-translation invariant and
$\Phi_{1-6,1},\Phi_{1-6,2}$ are $\frac{2\pi}{\mathtt{M}}$-translation
invariance preserving.

###### Proof.

Fix $\mu_{\mathtt{p},1}$ be the largest $\mu_{0}$, arising in Section 10,
which certainly depends on only $\mathtt{p}$. We simply rewrite
$\mathcal{L}^{6}$ in (10.237) using $d_{0}$ and $\mathcal{R}_{0}$, defined in
(10.290) and (10.267) to obtain the expression (10.289).

Proof of (1). (10.292) can be found in (10.3). To see (10.293), notice that
(1) of Proposition 10.60 tells us that the symbol
$\mathfrak{m}_{\mathfrak{b}}(\xi)$ is a Fourier multiplier that does not
depend on $\omega$ or $i_{0}$, and it satisfies
$\mathfrak{m}_{\mathfrak{b}}\in\mathcal{S}^{-2}$. Then, (10.293) immediately
follows from the definition of $\mathfrak{m}_{1}$ in (10.291) with the
estimates for $\mathfrak{m}_{\leq 0}$ in (1) of Proposition 10.41.

Proof of (2). (2) follows from Lemma 10.75 and Lemma 2.27, replacing
$\mathfrak{I}_{\delta}$ by $\mathfrak{I}_{0}$ using (7.12).

Proof of (3). It follows from (4) of Proposition 10.66 that $\mathcal{L}^{6}$
is a real and reversible operator, therefore so are $D_{0}$ and
$\mathcal{R}_{0}$.

Proof of (4). We denote

$\Phi_{1-6,1}:=\Phi_{6}\Phi_{5}\Phi_{4}\Phi_{3}\frac{1}{\rho}\Phi_{2}\Phi_{1},\quad\Phi_{1-6,2}:=\Phi_{1}\Phi_{2}\Phi_{3}\Phi_{4}\Phi_{5}\Phi_{6}.$

Then the expression (10.296) follows from (10.1), (10.30), (10.71), (10.120),
(10.153) and (10.237), with (10.289). The estimates (10.297) follows from
Lemma 10.4, 10.18, 10.36, 10.49, 10.56, 10.71, where we can replace
$\mathfrak{I}_{\delta}$ by $\mathfrak{I}_{0}$, using (7.12).

Proof of (5). We see from (5) of Proposition 10.66 that $\mathcal{L}^{6}$ is
$\frac{2\pi}{\mathtt{M}}$-translation invariance preserving, therefore so is
$\mathcal{R}_{0}$. $\frac{2\pi}{\mathtt{M}}$-translation invariance preserving
property of $\Phi_{1-6,1},\Phi_{1-6,2}$ follows from (5) of Proposition 10.1,
10.10, 10.30, 10.41, 10.51, 10.66. ∎

In view of Proposition 7.14, we need to find the inverse image of
$Y_{\perp}\cap X_{\mathtt{M}}$ in $X_{\perp}\cap X_{\mathtt{M}}$. Especially,
we can restrict the operator $L_{0}$ in (10.289) to the space of functions
whose Fourier modes are supported on only $\mathtt{M}$-multiples. More
precisely, we consider $L_{0}$ as a linear operator between $H_{S^{\perp}}\cap
X_{\mathtt{M}}$. Therefore, without loss of generality, we have that (using
(5.30)),

$\displaystyle(\mathcal{R}_{0})^{j_{1}}_{j_{2}}(l)=0,\text{ if
$j_{1},j_{2}\notin\left\\{\mathtt{M}j\in\mathbb{Z}\cap
S^{\perp}:j\in\mathbb{Z}\right\\}=S_{\mathtt{M}}^{\perp}$.}$ (10.298)

###  10.6 KAM reducibility and inversion of $\mathcal{L}_{\omega}$

#####  10.6.0.1 The homological equation

######  Lemma 10.77.

Fix $N\geq 1$, $\mathtt{S}\gg s_{0}$ and let
$\mu_{\mathtt{p},1},\mathtt{b}_{0}$ be as in (10.288) and (10.70). Let
$\Omega$ be a subset of $\Omega_{1}$. Let an $\omega$-dependent operator
$\mathcal{R}$ and an $\omega$-dependent Fourier multiplier $r(\omega,j)$ be
well-defined for $\omega\in\Omega$. We make the following assumptions:

  1. (A1)

$\mathcal{R}$ is a Lip-$0$-modulo-tame operator up to $\mathtt{S}$ and
satisfies

$\displaystyle(\mathcal{R})^{j}_{k}(l)=0,\text{ of $j,k\notin
S_{\mathtt{M}}^{\perp}$.}$ (10.299)

  2. (A2)

$r(\omega,j)$ satisfies

$\displaystyle|r|^{\text{Lip}(\gamma,\Omega)}_{0,0,0}\leq_{\mathtt{pe}}\varepsilon^{9}\gamma^{-3},\quad|d_{i}r(i_{0})|^{\sup}_{0,0,0}\leq_{\mathtt{pe}}\varepsilon^{3}\gamma^{-1}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{\mathtt{p},1}},$
(10.300)

where

$|r|^{\text{Lip}(\gamma,\Omega)}_{0,0,0}:=\sup_{\omega,\omega_{1},\omega_{2}\in\Omega,\
j\in\mathbb{Z}\backslash\left\\{0\right\\}}|r(\omega,j)|+\gamma\frac{|r(\omega_{1},j)-r(\omega_{2},j)|}{|\omega_{1}-\omega_{2}|}.$

We denote

$\displaystyle d_{r}(\omega,j)$
$\displaystyle:=d_{0}(\omega,j)+r(\omega,j),\text{ where $d_{0}$ is as in
\eqref{d0defsd2}},\quad D_{r}:=Op^{W}(d_{r}),$ (10.301)
$\displaystyle\tilde{\Omega}$
$\displaystyle:=\left\\{\omega\in\Omega:|\text{i}\omega\cdot
l-(d_{r}(\omega,j)-d_{r}(\omega,k))|\geq\frac{\gamma}{2}|l|^{-\tau}|\lambda_{\alpha}(j)-\lambda_{\alpha}(k)|,\
\forall|l|\leq N,\ \forall j,k\in
S_{\mathtt{M}}^{\perp}\cup\left\\{0\right\\}\right\\},$ (10.302)

(we recall $\tau=\frac{1}{\alpha-1}+\nu+2$ from (6.2)). Then, for each
$\omega\in\tilde{\Omega}$, the homological equation for $\Psi$:

$\displaystyle\mathcal{D}_{\omega}(\Psi)-[D_{r},\Psi]+\Pi_{\leq
N}\mathcal{R}=[\mathcal{R}],\text{ where
}([\mathcal{R}])^{j}_{k}(l):=\begin{cases}\mathcal{R}^{j}_{j}(0),&\text{ if
$(l,j,k)=(0,j,j)$, $j\in S_{\mathtt{M}}^{\perp}$}\\\ 0,&\text{
otherwise},\end{cases}$ (10.303)

has a unique solution $\Psi$ such that

  1. (1)

For $\vec{b}\in\mathbb{N}_{0}^{\nu}$ such that $|\vec{b}|\leq\mathtt{b}_{0}$,
it holds that for $\omega\in\tilde{\Omega}$ and $s\in[s_{0},\mathtt{S}]$,

$\displaystyle\mathfrak{M}^{\sharp,\gamma}_{\partial_{\varphi}^{\vec{b}}(\Psi)}(0,s)$
$\displaystyle\leq_{\mathtt{pe}}\varepsilon^{2}\gamma^{-2}N^{2\tau+1}\mathfrak{M}^{\sharp,\gamma}_{\partial_{\varphi}^{\vec{b}}(\mathcal{R})}(0,s),$
(10.304)
$\displaystyle\mathfrak{M}^{\sharp}_{\partial_{\varphi}^{\vec{b}}(d_{i}\Psi(i_{0})[\hat{\textbf{\i}}])}(0,s_{0})$
$\displaystyle\leq_{\mathtt{pe}}\left(\varepsilon^{3}\gamma^{-3}N^{2\tau+1}\mathfrak{M}^{\sharp}_{\partial_{\varphi}^{\vec{b}}(\mathcal{R})}(0,s_{0})\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{\mathtt{p},1}}+\gamma^{-1}N^{\tau}\mathfrak{M}^{\sharp}_{\partial_{\varphi}^{\vec{b}}(d_{i}\mathcal{R}(i_{0})[i\hat{]})}(0,s_{0})\right),$
(10.305)

  2. (2)

If $D_{r}$ and $\mathcal{R}$ are reversible, then $\Psi$ is reversibility
preserving.

  3. (3)

If $D_{r}$ and $\mathcal{R}$ are real operators, then $\Psi$ is a real
opeartor.

  4. (4)

$\Psi$ satisfies

$\displaystyle(\Psi)^{j}_{k}(l)=0,\text{ if $j,k\notin
S^{\perp}_{\mathtt{M}}$}$ (10.306)

and $\Psi$ is $\frac{2\pi}{\mathtt{M}}$-translation invariance preserving.

###### Proof.

We first solve the equation (10.303) and prove the properties of the solution
in (1)-(4).

We write (10.303) as

$\displaystyle\underbrace{\left(\text{i}\omega\cdot
l-(d_{r}(\omega,j)-d_{r}(\omega,k))\right)}_{=:\delta_{ljk}(\omega)}\Psi^{k}_{j}(l)+\mathcal{R}^{k}_{j}(l)=([\mathcal{R}])^{k}_{j}(l).$
(10.307)

Therefore, we define

$\displaystyle\Psi^{k}_{j}(l):=\begin{cases}-\frac{\mathcal{R}^{k}_{j}(l)}{\delta_{ljk}(\omega)},&\text{
if $(l,j,k)\neq(0,j,j)$ for $j\in S_{\mathtt{M}}^{\perp}$},\\\ 0,&\text{
otherwise.}\end{cases}$ (10.308)

Clearly, $\Psi$ defined (10.308) is a unique solution to (10.307).

Estimates for $\delta_{ljk}(\omega)$: Supremum in $\omega$. First, we see from
the definition of $\tilde{\Omega}$ in (10.302) that

$\displaystyle\sup_{\omega\in\Omega}\left|\frac{1}{\delta_{ljk}(\omega)}\right|\leq
2\gamma^{-1}N^{\tau}|\lambda_{\alpha}(j)-\lambda_{\alpha}(k)|^{-1}.$ (10.309)

Estimates for $\delta_{ljk}(\omega)$: Lipschitz dependence on $\omega$. For
$\omega_{1},\omega_{2}\in\tilde{\Omega}$, we will estimate the dependence on
$\omega$ of $\delta_{ljk}$, that is,

$\displaystyle\frac{1}{\delta_{ljk}(\omega_{1})}-\frac{1}{\delta_{ljk}(\omega_{2})}=\frac{\delta_{ljk}(\omega_{2})-\delta_{ljk}(\omega_{1})}{\delta_{ljk}(\omega_{1})\delta_{ljk}(\omega_{2})}.$
(10.310)

Clearly, it follows from the definition of $\delta_{ljk}$ in (10.307) that

$\displaystyle\delta_{ljk}(\omega_{2})-\delta_{ljk}(\omega_{1})$
$\displaystyle\leq|\omega_{1}-\omega_{2}|N+|(d_{r}(\omega_{1},j)-d_{r}(\omega_{2},j))-(d_{r}(\omega_{1},k)-d_{r}(\omega_{2},k))|$
(10.311)

From (10.301) and (10.290), we have that

$\displaystyle|(d_{r}(\omega_{1},j)-d_{r}(\omega_{2},j))$
$\displaystyle-(d_{r}(\omega_{1},k)-d_{r}(\omega_{2},k))|$ (10.312)
$\displaystyle\leq|(\mathtt{m}_{\alpha}(\omega_{1})-\mathtt{m}_{\alpha}(\omega_{2}))(\lambda_{\alpha}(j)-\lambda_{\alpha}(k))|+\left|\int_{j}^{k}\partial_{\xi}\mathfrak{m}_{1}(\omega_{1})(\xi)-\partial_{\xi}\mathfrak{m}_{1}(\omega_{2})(\xi)d\xi\right|$
$\displaystyle\
+|r(\omega_{1},j)-r(\omega_{2},j)|+|r(\omega_{1},k)-r(\omega_{2},k)|.$

Using the estimates for $\mathtt{m}_{\alpha}$ in(10.292) and using (10.293)
and (10.300), we have

$\displaystyle|(\mathtt{m}_{\alpha}(\omega_{1})-\mathtt{m}_{\alpha}(\omega_{2}))(\lambda_{\alpha}(j)-\lambda_{\alpha}(k))|$
$\displaystyle\leq_{\mathtt{pe}}\varepsilon^{2}\gamma^{-1}|\omega_{1}-\omega_{2}||\lambda_{\alpha}(j)-\lambda_{\alpha}(k)|,$
(10.313)
$\displaystyle\left|\int_{j}^{k}\partial_{\xi}\mathfrak{m}_{1}(\omega_{1})(\xi)-\partial_{\xi}\mathfrak{m}_{1}(\omega_{2})(\xi)d\xi\right|$
$\displaystyle\leq_{\mathtt{pe}}\varepsilon^{2}\gamma^{-1}|\omega_{1}-\omega_{2}||j-k|,$
$\displaystyle|r(\omega_{1},j)-r(\omega_{2},j)|+|r(\omega_{1},k)-r(\omega_{2},k)|$
$\displaystyle\leq_{\mathtt{pe}}\varepsilon^{2}\gamma^{-1}|\omega_{1}-\omega_{2}|$

Hence, using (2.99), we have that for $j\neq k$,

$\left|(d_{r}(\omega_{1},j)-d_{r}(\omega_{2},j))-(d_{r}(\omega_{1},k)-d_{r}(\omega_{2},k))\right|\leq_{\mathtt{pe}}\gamma^{-1}\varepsilon^{2}|\omega_{1}-\omega_{2}||\lambda_{\alpha}(j)-\lambda_{\alpha}(k)|.$

Plugging this into (10.311) and (10.310), and using (10.309), we obtain

$\displaystyle\gamma\left|\frac{1}{\delta_{ljk}(\omega_{1})}-\frac{1}{\delta_{ljk}(\omega_{2})}\right|\leq_{\mathtt{pe}}|\omega_{1}-\omega_{2}|\varepsilon^{2}\gamma^{-2}N^{2\tau+1}|\lambda_{\alpha}(j)-\lambda_{\alpha}(k)|^{-1}.$
(10.314)

Estimates for $\delta_{ljk}(\omega)$: Dependence on $i_{0}$. For a fixed
$\omega\in\tilde{\Omega}$ (we will omit its notation), we have that

$\displaystyle\left|d_{i}\left(\frac{1}{\delta_{ljk}}\right)(i_{0})[\hat{\textbf{\i}}]\right|=\left|\frac{d_{i}\delta_{ljk}(i_{0})[\hat{\textbf{\i}}]}{(\delta_{ljk})^{2}}\right|.$
(10.315)

Following the same computations as in (10.312), (one can simply replace the
difference on $\omega$ by $d_{i}$), it is straightforward to see that (using
the estimates in (10.292), (10.293), (10.300) and Lemma 2.32),

$\displaystyle\left|d_{i}\delta_{ljk}(i_{0})[\hat{\textbf{\i}}]\right|$
$\displaystyle\leq\left|d_{i}(\mathfrak{m}_{\alpha}(i_{0})[\hat{\textbf{\i}}])(\lambda_{\alpha}(j)-\lambda_{\alpha}(k))\right|+\left|\int_{j}^{k}\partial_{\xi}(d_{i}\mathfrak{m}_{1}(i_{0})[\hat{\textbf{\i}}])(\xi)d\xi\right|+|d_{i}r(i_{0})[\hat{\textbf{\i}}](j)|+|d_{i}r(i_{0})[\hat{\textbf{\i}}](k)|$
$\displaystyle\leq_{\mathtt{pe}}\varepsilon^{3}\gamma^{-1}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{\mathtt{p},1}}|\lambda_{\alpha}(j)-\lambda_{\alpha}(k)|.$

Therefore, (10.315) and (10.309) give us that

$\displaystyle\left|d_{i}\left(\frac{1}{\delta_{ljk}}\right)(i_{0})[\hat{\textbf{\i}}]\right|\leq
C\varepsilon^{3}\gamma^{-3}N^{2\tau}|\lambda_{\alpha}(j)-\lambda_{\alpha}(k)|^{-1}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{\mathtt{p},1}}.$
(10.316)

Proof of (1). Plugging (10.309), (10.314) and (10.316) into (10.308), it is
straightforward that (using Lemma 2.32, which tells us that
$|\lambda_{\alpha}(j)-\lambda_{\alpha}(k)|^{-1}\leq_{\mathtt{pe}}1$, unless
$j=k$),

$\displaystyle|\Psi^{k}_{j}(l)|$
$\displaystyle\leq_{\mathtt{pe}}\gamma^{-1}N^{\tau}|\mathcal{R}^{k}_{j}(l)|,$
(10.317)
$\displaystyle\gamma\left|\Psi^{k}_{j}(l)(\omega_{1})-\Psi^{k}_{j}(l)(\omega_{2})\right|$
$\displaystyle\leq_{\mathtt{pe}}|\omega_{1}-\omega_{2}|\varepsilon^{2}\gamma^{-2}N^{2\tau+1}|R^{k}_{j}(l)|+N^{\tau}|\mathcal{R}^{k}_{j}(l)(\omega_{1})-\mathcal{R}^{k}_{j}(l)(\omega_{2})|,$
(10.318) $\displaystyle|(d_{i}\Psi(i_{0})[\hat{\textbf{\i}}])^{k}_{j}(l)$
$\displaystyle\leq_{\mathtt{pe}}\varepsilon^{3}\gamma^{-3}N^{2\tau}|\mathcal{R}^{k}_{j}(l)|\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{0}}+\gamma^{-1}N^{\tau}|\left(d_{i}\mathcal{R}(i_{0})[\hat{\textbf{\i}}]\right)^{k}_{j}(l)|$
(10.319)

Towards the estimates (10.304), we pick $h\in C^{\infty}_{\varphi,x}$, and
$\vec{b}\in\mathbb{N}_{0}^{\nu}$ such that $|\vec{b}|\leq\mathtt{b}_{0}$. We
have that (recall the notations (2.9) and (2.55) for a linear operator),

$\displaystyle\rVert\underline{\partial_{\varphi}^{\vec{b}}(\Psi)}h\rVert_{s}^{2}$
$\displaystyle\leq\sum_{(l,j)\in\mathbb{Z}^{\nu+1}}\langle
l,j\rangle^{2s}\left(\sum_{(l^{\prime},k)\in\mathbb{Z}^{\nu+1}}\left|(\partial_{\varphi}^{\vec{b}}(\Psi))^{k}_{j}(l-l^{\prime})\right||h_{k}(l^{\prime})|\right)^{2}$
$\displaystyle=\sum_{(l,j)\in\mathbb{Z}^{\nu+1}}\langle
l,j\rangle^{2s}\left(\sum_{(l^{\prime},k)\in\mathbb{Z}^{\nu+1}}|(l-l^{\prime})^{\vec{b}}|\left|\Psi^{k}_{j}(l-l^{\prime})\right||h_{k}(l^{\prime})|\right)^{2}$
$\displaystyle\overset{\eqref{taem1hanson2}}{\leq}_{\mathtt{pe}}\gamma^{-2}N^{2\tau}\sum_{(l,j)\in\mathbb{Z}^{\nu+1}}\langle
l,j\rangle^{2s}\left(\sum_{(l^{\prime},k)\in\mathbb{Z}^{\nu+1}}|(l-l^{\prime})^{\vec{b}}|R^{k}_{j}(l-l^{\prime})|h_{k}(l^{\prime})|\right)^{2}$
$\displaystyle\overset{\eqref{maojisd2sd1},\eqref{maojisd2sd}}{=}\gamma^{-2}N^{2\tau}\rVert\underline{\partial_{\varphi}^{\vec{b}}(\mathcal{R})}\underline{h}\rVert_{s}^{2}$
$\displaystyle\overset{\text{Definition~{}\ref{def_tame1}}}{\leq}\gamma^{-2}N^{2\tau}\left(\mathfrak{M}^{\sharp}_{\partial_{\varphi}^{\vec{b}}(\mathcal{R})}(0,s)\right)^{2}\rVert\underline{h}\rVert^{2}_{s}$
$\displaystyle\overset{\eqref{endedupspending}}{=}\gamma^{-2}N^{2\tau}\left(\mathfrak{M}^{\sharp}_{\partial_{\varphi}^{\vec{b}}(\mathcal{R})}(0,s)\right)^{2}\rVert{h}\rVert^{2}_{s}.$

Therefore, we have

$\displaystyle\mathfrak{M}^{\sharp}_{\partial_{\varphi}^{\vec{b}}(\Psi)}(0,s)\leq_{\mathtt{pe}}\gamma^{-1}N^{\tau}\mathfrak{M}^{\sharp}_{\partial_{\varphi}^{\vec{b}}(\mathcal{R})}(0,s).$
(10.320)

Similarly, following the same computations but using (10.318) and (10.319),
one can easily see the desired estimates (10.304) and (10.305).

Proof of (2) and (3) To see items (2) and (3), in view of (2.136), let us
denote involutions $P_{rev},P_{real}$ acting on linear operators by

$(P_{rev}\mathcal{A})^{j_{1}}_{j_{2}}(l):=\overline{\mathcal{A}^{-j_{1}}_{-j_{2}}(-l)},\quad\left(P_{real}\mathcal{A}\right)^{j_{1}}_{j_{2}}(l):=\overline{(\mathcal{A})^{j_{2}}_{j_{1}}(l)}.$

Clearly, $P_{rev},P_{real}$ are involutions, since $P_{rev}\circ P_{rev}=I$
and $P_{real}\circ P_{real}=I$. Recalling that $D_{r}$ and $\mathcal{R}$ are
reversible, it follows immediately, taking the involutions in the homological
equation (10.303) that $-P_{rev}\Psi$ and $P_{real}\Psi$ are also solutions to
(10.303). Since the solution is unique, we have that
$\Psi=-P_{rev}\Psi=P_{real}\Psi$. Therefore, $\Psi$ is reversibility
preserving and real. ∎

Proof of (4). It is clear that $\Psi$ is $\frac{2\pi}{\mathtt{M}}$-translation
invariant if and only if

$\displaystyle(\Psi)^{j_{1}}_{j_{2}}(l)=0,\text{ if
$j_{1}-j_{2}\notin\left\\{\mathtt{M}j:j\in\mathbb{Z}\right\\}$},$ (10.321)

Therefore, the result follows immediately from (10.299) and (10.308).

#####  10.6.0.2 Choice of the increasing sequence $N_{n}$

We fix

$\displaystyle N_{n}$ $\displaystyle:=N_{0}^{\chi^{n}},\text{ for $n\geq
0$},\quad\chi:=\frac{3}{2},\quad N_{-1}:=1,\quad
N_{0}:=\left(\varepsilon\gamma^{-1}\right)^{\rho},$ (10.322)
$\displaystyle\sigma_{1}$
$\displaystyle:=3(2\tau+1),\quad\mu_{\mathtt{p},0}:=2\mu_{\mathtt{p},1}+2\tau+1$
$\displaystyle\rho^{-1}$
$\displaystyle:=\max\left\\{22(2b-1)(\sigma_{1}+2\tau+1),\frac{2\tau(2b-1)}{9-8b},\frac{(2b-1)(2\tau+1)}{2b-2},\frac{40(\mu_{\mathtt{p},0}+\mu_{\mathtt{p},2})(2b-1)}{5-4b}\right\\}.$

######  Remark 10.78.

Note that the above sequence is well-defined in the sense that $\tau$ is fixed
in (6.3), $b$ is fixed in (6.24), which gives $\sigma_{1}$ as above.
$\mu_{\mathtt{p},1}$ is fixed in (10.288), which gives us $\mu_{\mathtt{p},0}$
and this gives us $\mu_{\mathtt{p},2}$ as described in Remark 7.16, therefore
$\rho$ is well-defined.

One can easily check that for all sufficiently small $\varepsilon>0$,
depending on $\mathtt{p}$, and for all $n\geq 0$,

$\displaystyle\mathtt{b}_{0}$
$\displaystyle\overset{\eqref{nsjdjwdsdnsde}}{\geq}2\sigma_{1},$ (10.323)
$\displaystyle\varepsilon^{\frac{1}{22}}N_{n}^{2\tau+1}N_{n-1}^{-2\sigma_{1}}$
$\displaystyle\leq N_{n}^{-\sigma_{1}},$ (10.324) $\displaystyle
N_{n}^{-\mathtt{b}_{0}}N_{n-1}$ $\displaystyle\leq
N_{n}^{1-\sigma_{1}}N_{n}^{-\sigma_{1}}\leq
N_{0}^{1-\sigma_{1}}N_{n}^{-\sigma_{1}}$ (10.325)
$\displaystyle\varepsilon^{11-10b}N_{n}^{2\tau+1}N_{n-1}^{-2\sigma_{1}}$
$\displaystyle=\varepsilon^{1-\frac{10}{11}}\underbrace{\varepsilon^{10(\frac{1}{11}-(b-1))}}_{\leq
1,\
\because\eqref{parametersets1231}}N_{n}^{2\tau+1}N_{n-1}^{-2\sigma_{1}}\leq\varepsilon^{\frac{1}{22}}\varepsilon^{\frac{1}{22}}N_{n}^{2\tau+1}N_{n-1}^{-2\sigma_{1}}\overset{\eqref{para1chs12s}}{\leq}\varepsilon^{\frac{1}{22}}N_{n}^{-\sigma_{1}}.$
(10.326)

#####  10.6.0.3 KAM reducibility

We take $L_{0},D_{0},\mathcal{R}_{0},d_{0}$ given in Proposition 10.76. For
$n\geq 0$, we define inductively:

$\displaystyle r_{0}$
$\displaystyle:=0,\quad\Omega^{\gamma}_{0}(i_{0}):=\Omega_{1},$ (10.327)
$\displaystyle\phi_{n}$ $\displaystyle:=I+\psi_{n},\
\phi_{n}^{-1}=:I-\tilde{\psi}_{n},\text{ where $\psi_{n}$ solves
$\mathcal{D}_{\omega}(\psi_{n})-[D_{n},\psi_{n}]+\Pi_{\leq
N_{n}}\mathcal{R}_{n}=[\mathcal{R}_{n}]$ in $H_{S^{\perp}}\cap
X_{\mathtt{M}}$},$ (10.328) $\displaystyle r_{n+1}$
$\displaystyle:=r_{n+1}(\omega,j):=\begin{cases}-[\mathcal{R}_{n}(\omega)]^{j}_{j},&\text{
if $j\in S_{\mathtt{M}}^{\perp}$},\\\ 0,&\text{ if $j\neq
S_{\mathtt{M}}^{\perp}$},\end{cases}\text{ for
$\omega\in\Omega^{\gamma}_{n}(i_{0})$},$ (10.329) $\displaystyle D_{n+1}$
$\displaystyle:=Op^{W}(d_{n+1}),\quad
d_{n+1}:=d_{n}+r_{n+1}=d_{0}+\sum_{k=0}^{n+1}r_{k}=:d_{0}+r_{\leq{n+1}},$
(10.330) $\displaystyle\mathcal{R}_{n+1}$
$\displaystyle:=\tilde{\psi}_{n}\circ
Op^{W}(r_{n})+\phi_{n}^{-1}\circ\left(\Pi_{\geq
N_{n}}\mathcal{R}_{n}+\mathcal{R}_{n}\psi_{n}\right).$ (10.331)

Furthermore, we consider a sequence of sets:

$\displaystyle\Omega^{\gamma}_{n+1}(i_{0})$
$\displaystyle\quad:=\left\\{\omega\in\Omega_{n}^{\gamma}(i_{0}):|\text{i}\omega\cdot
l-(d_{n}(\omega,j)-d_{n}(\omega,k))|\geq\frac{1}{2}\gamma|l|^{-\tau}|\lambda_{\alpha}(j)-\lambda_{\alpha}(k)|,\
\forall|l|\leq N_{n},\ \forall j,k\in
S_{\mathtt{M}}^{\perp}\cup\left\\{0\right\\}\right\\}.$ (10.332)

Trivially, we have that

$\displaystyle\Omega^{2\gamma}_{n+1}(i_{0})\subset\Omega^{\gamma}_{n+1}(i_{0}).$
(10.333)

The motivation is that we aim to conjugate
$L_{n}=\mathcal{D}_{\omega}-D_{n}+\mathcal{R}_{n}$ with a transformation
$\phi_{n}=I+\psi_{n},\phi^{-1}_{n}=I+\tilde{\psi}_{n}$ to reduce the size of
the remainder. One can easily show that for each $n\geq 0$,

$\displaystyle\phi_{n}^{-1}L_{n}\phi_{n}$
$\displaystyle=\mathcal{D}_{\omega}+\underbrace{\left(-D_{n}+\left(\mathcal{D}_{\omega}(\psi_{n})-[D_{n},\psi_{n}]+\Pi_{\leq
N_{n}}\mathcal{R}_{n}\right)\right)}_{=-D_{n+1}}$ $\displaystyle\
+\underbrace{\tilde{\psi}_{n}\circ\left(\mathcal{D}_{\omega}(\psi_{n})-[D_{n},\psi_{n}]+\Pi_{\leq
N_{n}}\mathcal{R}_{n}\right)+\Pi_{\geq
N_{n}}\mathcal{R}_{n}+\mathcal{R}_{n}\psi_{n}}_{=\mathcal{R}_{n+1}}.$ (10.334)

Therefore, we simply choose $\psi_{n}$ to be a solution to the corresponding
homological equation in (10.328), while the diagonal part is updated. Since
our initial remainder in this scheme, $\mathcal{R}_{0}$, is already a $0$-tame
operator, we can repeat this procedure infinitely many times without any loss
of derivatives (since a composition of $0$-tame operators is still a $0$-tame
operator). The parameters in (10.322) are carefully chosen so that we can
close the infinite number of conjugations. Whether they are well-defined or
not will be checked in Proposition 10.79. Note that for all $n\geq 0$ such
that

$\displaystyle|r_{\leq
n}|^{\text{Lip}(\gamma,\Omega^{\gamma}_{n}(i_{0}))}_{0,0,0}\leq_{\mathtt{pe}}\varepsilon^{9}\gamma^{-3},\quad|d_{i}r_{\leq
n}(i_{0})|^{\sup}_{0,0,0}\leq_{\mathtt{pe}}\varepsilon^{3}\gamma^{-1}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{\mathtt{p},1}},$
(10.335)

Lemma 10.77 tells us that $\psi_{n}=\psi_{n}(\omega)$ exists in (10.328) for
$\omega\in\Omega_{n}^{\gamma}(i_{0})$. Once we have $\psi_{n}$, then assuming
$\mathfrak{M}^{\sharp,\gamma}_{\psi_{n}}(0,s_{0})$ is small enough, Lemma 2.28
gives us the existence of $\tilde{\psi}_{n}$ in (10.328), which allows us to
have $\mathcal{R}_{n+1}$ and $D_{n+1}$ in (10.331) and (10.330) well-defined
for each $\omega\in\Omega^{\gamma}_{n}(i_{0})$. Indeed, we have the following:

######  Proposition 10.79.

Given $\mathtt{S}\gg s_{0}$, let us denote

$\displaystyle\mathbb{M}_{0}(s)$
$\displaystyle:=\sup_{\vec{b}\in\mathbb{N}_{0}^{\nu},|\vec{b}|\leq\mathtt{b}_{0}}\mathfrak{M}^{\sharp,\gamma}_{\partial_{\varphi}^{\vec{b}}\mathcal{R}_{0}}(0,s),\text{
for $s\in[s_{0},\mathtt{S}]$.}$ (10.336)

Assuming (7.3) for $\mu=\mu_{\mathtt{p},1}$, there exists
$\varepsilon_{0}(\mathtt{p},\mathtt{S})>0$ such that for all
$\varepsilon\in(0,\varepsilon_{0})$, we have that, for every $n\geq 0$:

  1. (1)

(Estimates for $\mathcal{R}_{n}$) $\mathcal{R}_{n}$ is well defined for
$\omega\in\Omega^{\gamma}_{n}(i_{0})$ as a real, reversible operator such that

$\displaystyle(\mathcal{R}_{n})^{j}_{k}(l)=0,$ if $j,k\notin
S^{\perp}_{\mathtt{M}}$. (10.337)

Furthermore, it satisfies the following estimates: For all
$s\in[s,\mathtt{S}]$,

$\displaystyle\mathfrak{M}^{\sharp,\gamma}_{\mathcal{R}_{n}}(0,s)$
$\displaystyle\leq\mathbb{M}_{0}(s)N_{n-1}^{-\sigma_{1}},$ (10.338)
$\displaystyle\sup_{\vec{b}\in\mathbb{N}^{\nu}_{0},\
|\vec{b}|=\mathtt{b}_{0}}\mathfrak{M}^{\sharp,\gamma}_{\partial_{\varphi}^{\vec{b}}(\mathcal{R}_{n})}(0,s)$
$\displaystyle\leq\mathbb{M}_{0}(s)N_{n-1},$ (10.339)
$\displaystyle\mathfrak{M}^{\sharp}_{d_{i}(\mathcal{R}_{n})(i_{0})[\hat{\textbf{\i}}]}(0,s_{0})$
$\displaystyle\leq_{\mathtt{pe}}\varepsilon^{3}\gamma^{-1}N_{n-1}^{-\sigma_{1}}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{\mathtt{p},1}},$
(10.340) $\displaystyle\sup_{\vec{b}\in\mathbb{N}^{\nu}_{0},\
|\vec{b}|=\mathtt{b}_{0}}\mathfrak{M}^{\sharp}_{\partial_{\varphi}^{\vec{b}}(d_{i}\mathcal{R}_{n}(i_{0})[\hat{\textbf{\i}}])}(0,s_{0})$
$\displaystyle\leq_{\mathtt{pe}}\varepsilon^{3}\gamma^{-1}N_{n-1}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{\mathtt{p},1}},.$
(10.341)

  2. (2)

(Estimates for $D_{n+1}$) $D_{n+1}$ is well defined (that is, $r_{n+1}$ is
well-defined) for $\omega\in\Omega^{\gamma}_{n}(i_{0})$ as a diagonal, real
and reversible operator. Furthermore, it satisfies the following estimates:

$\displaystyle|r_{n+1}|^{\text{Lip}(\gamma,\Omega^{\gamma}_{n}(i_{0}))}_{0,0,0}$
$\displaystyle\leq_{\mathtt{pe}}\mathbb{M}_{0}(s_{0})N_{n-1}^{-{\sigma_{1}}},$
(10.342) $\displaystyle|r_{\leq
n+1}|^{\text{Lip}(\gamma,\Omega^{\gamma}_{n}(i_{0}))}_{0,0,0}$
$\displaystyle\leq_{\mathtt{pe}}\mathbb{M}_{0}(s_{0}),$ (10.343)
$\displaystyle|d_{i}r_{n+1}(i_{0})[\hat{\textbf{\i}}]|^{\sup}_{0,0,0}$
$\displaystyle\leq_{\mathtt{pe}}\varepsilon^{3}\gamma^{-1}N_{n-1}^{-{\sigma_{1}}}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{\mathtt{p}}},$
(10.344)
$\displaystyle|d_{i}r_{\leq{n+1}}(i_{0})[\hat{\textbf{\i}}]|^{\sup}_{0,0,0}$
$\displaystyle\leq_{\mathtt{pe}}\varepsilon^{3}\gamma^{-1}\rVert\hat{\textbf{\i}}\rVert_{s_{0}+\mu_{\mathtt{p}}}.$
(10.345)

  3. (3)

(Estimates for $\psi_{n}$) $\psi_{n}$ is well defined for
$\omega\in\Omega^{\gamma}_{n+1}(i_{0})$ as a real, reversibility preserving
operator such that

$\displaystyle(\psi_{n})^{j}_{k}(l)=0,$ if $j,k\notin S^{\perp}_{\mathtt{M}}$.
(10.346)

Furthermore, it satisfies the following estimate: For all
$s\in[s,\mathtt{S}]$,

$\displaystyle\mathfrak{M}^{\sharp,\gamma}_{\psi_{n}}(0,s)$
$\displaystyle\leq_{\mathtt{pe}}\varepsilon^{2}\gamma^{-2}N_{n-1}^{-\sigma_{1}}N_{n}^{2\tau+1}\mathbb{M}_{0}(s)$
(10.347)

  4. (4)

Let $i_{1},i_{2}$ be two embeddings satisfying (7.3), (7.4) for
$\mu_{\mathtt{p},1}$ and some $\mathtt{C}>0$. There exists
$C=C(\mathtt{pe})>0$ such that if
$C\varepsilon^{3}\gamma^{-1}N_{n-1}^{\tau}\rVert
i_{1}-i_{2}\rVert_{s_{0}+\mu_{\mathtt{p},1}}\leq\delta$, for some
$\delta\in(0,\gamma/2)$, then we have that

$\displaystyle\implies\Omega_{n}^{2\gamma}(i_{1})\subset\Omega^{2(\gamma-\delta)}_{n}(i_{2}),$
(10.348)

for sufficiently small $\varepsilon$ depending on $\mathtt{pe}$.

Lastly for all $n\geq 0$ and $\omega\in\cap_{n\geq
0}\Omega_{n}^{\gamma}(i_{0})$, we have that
$L_{n}:=\mathcal{D}_{\omega}-D_{n}+\mathcal{R}_{n}$ satisfies

$\displaystyle L_{n+1}=\phi_{n}^{-1}L_{n}\phi_{n}.$ (10.349)

Furthermore, $L_{n}$ is real and reversible and
$\frac{2\pi}{\mathtt{M}}$-translation invariant.

###### Proof.

We will prove items (1)-(4) by induction. Once (1)-(4) are verified, then we
will prove (10.349), and the reversibility, reality and
$\frac{2\pi}{\mathtt{M}}$-translation invariance preserving properties of
$L_{n}$.

Proof of (1)-(4) for $n=0$. For item (1), $\mathcal{R}_{0}$ is already well-
defined in Proposition 10.76 for
$\omega\in\Omega_{1}=\Omega_{0}^{\gamma}(i_{0})$. Furthermore, (3) of
Proposition 10.76 tells us that $\mathcal{R}_{0}$ is real and reversible.
(10.337) follows from (10.298).

For the estimates, recalling from (10.322) that $N_{-1}=1$, the estimates
(10.338) and (10.339) hold trivially, thanks to the definition of
$\mathbb{M}_{0}(s)$ in (10.336). The estimates (10.340) and (10.341) follow
from (10.295).

For item (2), recalling $r_{1}$ from (10.329), we see that $r_{1}$ is well-
defined, since so is $\mathcal{R}_{0}$, as proved above. Since
$\mathcal{R}_{0},D_{0}$ is real and reversible (see (3) of Proposition 10.76),
it follows straightforwardly from the definition of $r_{1}$ in (10.329) and
(2.136) that $Op^{W}(r_{1})$ is also real and reversible. For the estimates,
using $N_{-1}=1$ and using the estimates in item (1) for $n=0$ and Lemma 2.29,
we see that the estimates (10.342)-(10.345) follow immediately.

For item (3), we apply Lemma 10.77 to obtain $\psi_{0}$ from (10.328) (since
(A1) follows from (10.337) for $\mathcal{R}_{0}$ and $r_{0}=0$, hence (A2)
holds trivially). Then for $\omega\in\Omega^{\gamma}_{0}(i_{0})$, we see that
$\psi_{0}(\omega)$ is well-defined. From (2), (3) and (4) of Lemma 10.77, we
see that $\Psi$ is real, reversibility preserving and satisfies (10.306).
Also, the estimate (10.347) follows from (10.304) with
$\vec{b}=0,\mathcal{R}=\mathcal{R}_{0}$ and $N=N_{0}$.

For (4), (10.348) holds trivially since $\Omega_{0}^{2\gamma}(i_{0})$ actually
does not depend on the embedding $i_{0}$, see (10.327).

Now, we assume that (1)-(3) hold true for some $n\geq 0$ and aim to prove (1)
for $n+1$.

Inductive step for (1). First, we check the well-definedness and reality,
reversibility and (10.337) for $\mathcal{R}_{n+1}$. We notice that our
induction hypotheses for (10.343) and (10.345) imply that the condition
(10.300) is satisfied for the homological equation for $\psi_{n}$ in (10.328),
which guarantees that $\psi_{n}$ is well-defined for
$\omega\in\Omega^{\gamma}_{n+1}(i_{0})$. We use the induction hypotheses that
$\mathcal{R}_{n}$ and $D_{n}$ are real, reversible and
$\frac{2\pi}{\mathtt{M}}$-translation invariant to see that (2) of Lemma 10.77
implies that $\Psi_{n}$ is a real and reversibility and
$\frac{2\pi}{\mathtt{M}}$-translation invariance preserving operator, and so
are $\phi_{n},\phi_{n}^{-1}$. In view of (10.6.0.3), we have that
$\mathcal{D}_{\omega}-D_{n+1}+\mathcal{R}_{n+1}$ is real, reversible. Thanks
to the induction hypotheses for (2), $D_{n+1}$ is real and reversible.
Therefore, $\mathcal{R}_{n+1}$ is real and reversible. Since $\mathcal{R}_{n}$
and $\Psi_{n}$ satisfy (10.337) and (10.346), it follows from the definition
of $\mathcal{R}_{n+1}$ in (10.331) that $\mathcal{R}_{n+1}$ satisfies (10.337)
as well.