\mathrm{i}\,x_{22}F_{21}&x_{22}F_{22}+x_{12}F_{12}\end{pmatrix}=\begin{pmatrix}r(\mu\epsilon^{2},\mu^{3}\epsilon,\mu^{5})&\mathrm{i}\,r(\mu^{3}\epsilon,\mu^{2}\epsilon^{2},\mu^{5})\\\
\mathrm{i}\,r(\mu^{4}\epsilon,\mu^{2}\epsilon^{2},\mu^{6})&r(\mu^{4}\epsilon,\mu^{2}\epsilon^{2},\mu^{6})\end{pmatrix}$

and the expansion of $\tilde{G}$ in (5.35) follows by symmetrizing. ∎

We now show that the last term in (5.32) is very small.

######  Lemma 5.7.

The $4\times 4$ Hamiltonian and reversibility matrix

$\frac{1}{2}\int_{0}^{1}(1-\tau^{2})\,\exp(\tau
S)\,\textup{ad}_{S}^{2}(R^{(1)})\,\exp(-\tau
S)\,\mathrm{d}\tau=\begin{pmatrix}\mathtt{J}_{2}\widehat{E}&\mathtt{J}_{2}F^{(2)}\\\
\mathtt{J}_{2}[F^{(2)}]^{*}&\mathtt{J}_{2}\widehat{G}\end{pmatrix}$ (5.36)

where the $2\times 2$ self-adjoint and reversible matrices
$\widehat{E}=\footnotesize\begin{pmatrix}\widehat{E}_{11}&\mathrm{i}\,\widehat{E}_{12}\\\
-\mathrm{i}\,\widehat{E}_{12}&\widehat{E}_{22}\end{pmatrix}$,
$\widehat{G}=\footnotesize\begin{pmatrix}\widehat{G}_{11}&\mathrm{i}\,\widehat{G}_{12}\\\
-\mathrm{i}\,\widehat{G}_{12}&\widehat{G}_{22}\end{pmatrix}$ have entries

$\widehat{E}_{ij}\
,\widehat{G}_{ij}=\mu^{2}r(\epsilon^{3},\mu\epsilon^{2},\mu^{3}\epsilon,\mu^{5})\,,\quad
i,j=1,2\,,$ (5.37)

and the $2\times 2$ reversible matrix $F^{(2)}$ admits an expansion as in
(5.12).

###### Proof.

Since $S$ and $R^{(1)}$ are Hamiltonian and reversibility-preserving then
$\textup{ad}_{S}R^{(1)}=[S,R^{(1)}]$ is Hamiltonian and reversibility-
preserving as well. Thus each $\exp(\tau
S)\,\textup{ad}_{S}^{2}(R^{(1)})\,\exp(-\tau S)$ is Hamiltonian and
reversibility-preserving, and formula (5.36) holds. In order to estimate its
entries we first compute $\textup{ad}_{S}^{2}(R^{(1)})$. Using the form of $S$
in (5.10) and $[S,R^{(1)}]$ in (5.33) one gets

$\textup{ad}_{S}^{2}(R)=\begin{pmatrix}0&\mathtt{J}_{2}\tilde{F}\\\
\mathtt{J}_{2}\tilde{F}^{*}&0\end{pmatrix}\qquad\text{where}\qquad\tilde{F}:=2\left(\Sigma\mathtt{J}_{2}\tilde{G}-\tilde{E}\mathtt{J}_{2}\Sigma\right)$
(5.38)

and $\tilde{E}$, $\tilde{G}$ are defined in (5.34). In order to estimate
$\tilde{F}$, we write
$\footnotesize\tilde{G}=\begin{pmatrix}\tilde{G}_{11}&\mathrm{i}\,\tilde{G}_{12}\\\
-\mathrm{i}\,\tilde{G}_{12}&\tilde{G}_{22}\end{pmatrix}$,
$\footnotesize\tilde{E}=\begin{pmatrix}\tilde{E}_{11}&\mathrm{i}\,\tilde{E}_{12}\\\
-\mathrm{i}\,\tilde{E}_{12}&\tilde{E}_{22}\end{pmatrix}$ and, by (5.35), (5.9)
and $\Sigma=\mathtt{J}_{2}X$, we obtain

$\Sigma\mathtt{J}_{2}\tilde{G}=\begin{pmatrix}x_{21}\tilde{G}_{12}-x_{22}\tilde{G}_{11}&\\!\\!\\!\\!\mathrm{i}\,(x_{21}\tilde{G}_{22}-x_{22}\tilde{G}_{12})\\\
\mathrm{i}\,(x_{11}\tilde{G}_{12}+x_{12}\tilde{G}_{11})&\\!\\!\\!\\!-x_{11}\tilde{G}_{22}-x_{12}\tilde{G}_{12}\end{pmatrix}=\begin{pmatrix}r(\mu^{2}\epsilon^{3},\mu^{3}\epsilon^{2},\mu^{5}\epsilon,\mu^{7})&\\!\\!\\!\\!\mathrm{i}\,r(\mu^{2}\epsilon^{3},\mu^{4}\epsilon^{2},\mu^{5}\epsilon,\mu^{7})\\\
\mathrm{i}\,r(\mu^{2}\epsilon^{3},\mu^{4}\epsilon^{2},\mu^{5}\epsilon,\mu^{7})&\\!\\!\\!\\!r(\mu^{3}\epsilon^{3},\mu^{4}\epsilon^{2},\mu^{6}\epsilon,\mu^{8})\end{pmatrix}\,,$
$\tilde{E}\mathtt{J}_{2}\Sigma=\begin{pmatrix}\tilde{E}_{12}x_{21}-\tilde{E}_{11}x_{11}&\\!\\!\\!\\!-\mathrm{i}\,(\tilde{E}_{11}x_{12}+\tilde{E}_{12}x_{22})\\\
\mathrm{i}\,(\tilde{E}_{12}x_{11}-\tilde{E}_{22}x_{21})&\\!\\!\\!\\!-\tilde{E}_{12}x_{12}-\tilde{E}_{22}x_{22}\end{pmatrix}=\begin{pmatrix}r(\mu^{2}\epsilon^{3},\mu^{3}\epsilon^{2},\mu^{5}\epsilon,\mu^{7})&\\!\\!\\!\\!\mathrm{i}\,r(\mu^{2}\epsilon^{3},\mu^{4}\epsilon^{2},\mu^{6}\epsilon,\mu^{8})\\\
\mathrm{i}\,r(\mu^{2}\epsilon^{3},\mu^{4}\epsilon^{2},\mu^{5}\epsilon,\mu^{7})&\\!\\!\\!\\!r(\mu^{3}\epsilon^{3},\mu^{4}\epsilon^{2},\mu^{6}\epsilon,\mu^{8})\end{pmatrix}\,.$

Thus the matrix $\tilde{F}$ in (5.38) has an expansion as in (5.12). Then, for
any $\tau\in[0,1]$, the matrix $\exp(\tau
S)\,\textup{ad}_{S}^{2}(R^{(1)})\,\exp(-\tau
S)=\textup{ad}_{S}^{2}(R^{(1)})(1+\mathcal{O}(\mu,\epsilon))$. In particular
the matrix $F^{(2)}$ in (5.36) has the same expansion of $\tilde{F}$, whereas
the matrices $\widehat{E}$, $\widehat{G}$ has entries at least as in (5.37). ∎

###### Proof of Lemma 5.2..

It follows by Lemmata 5.6 and 5.7. The matrix
$E^{(2)}:=E^{(1)}+\tilde{E}+\widehat{E}$ has the same expansion of $E^{(1)}$
in (4.13). The same holds for $G^{(2)}$. ∎

###  5.3 Complete block-decoupling and proof of the main results

We now block-diagonalize the $4\times 4$ Hamiltonian and reversible matrix
$\mathtt{L}_{\mu,\epsilon}^{(2)}$ in (5.11). First we split it into its
$2\times 2$-diagonal and off-diagonal Hamiltonian and reversible matrices

$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\mathtt{L}_{\mu,\epsilon}^{(2)}=D^{(2)}+R^{(2)}\,,$
$\displaystyle D^{(2)}:=\begin{pmatrix}D_{1}^{(2)}&0\\\
0&D_{0}^{(2)}\end{pmatrix}=\begin{pmatrix}\mathtt{J}_{2}E^{(2)}&0\\\
0&\mathtt{J}_{2}G^{(2)}\end{pmatrix},\quad
R^{(2)}:=\begin{pmatrix}0&\mathtt{J}_{2}F^{(2)}\\\
\mathtt{J}_{2}[F^{(2)}]^{*}&0\end{pmatrix}.$ (5.39)

######  Lemma 5.8.

There exist a $4\times 4$ reversibility-preserving Hamiltonian matrix
$S^{(2)}:=S^{(2)}(\mu,\epsilon)$ of the form (5.10), analytic in
$(\mu,\epsilon)$, of size
$\mathcal{O}(\epsilon^{3},\mu\epsilon^{2},\mu^{3}\epsilon,\mu^{5})$, and a
$4\times 4$ block-diagonal reversible Hamiltonian matrix $P:=P(\mu,\epsilon)$,
analytic in $(\mu,\epsilon)$, of size
$\mu^{2}\mathcal{O}(\epsilon^{4},\mu^{4}\epsilon^{3},\mu^{6}\epsilon^{2},\mu^{8}\epsilon,\mu^{10})$
such that

$\exp(\mu S^{(2)})(D^{(2)}+R^{(2)})\exp(-\mu S^{(2)})=D^{(2)}+P\,.$ (5.40)

###### Proof.

We set for brevity $S=S^{(2)}$. The equation (5.40) is equivalent to the
system

$\begin{cases}\Pi_{D}\big{(}e^{\mu S}\big{(}D^{(2)}+R^{(2)}\big{)}e^{-\mu
S}\big{)}-D^{(2)}=P\\\ \Pi_{\varnothing}\big{(}e^{\mu
S}\big{(}D^{(2)}+R^{(2)}\big{)}e^{-\mu S}\big{)}=0\,,\end{cases}$ (5.41)

where $\Pi_{D}$ is the projector onto the block-diagonal matrices and
$\Pi_{\varnothing}$ onto block-off-diagonal ones. The second equation in
(5.41) is equivalent, by a Lie expansion, and since $[S,R^{(2)}]$ is block-
diagonal, to

$R^{(2)}+\mu\left[S\,,\,D^{(2)}\right]+\mu^{2}\underbrace{\Pi_{\varnothing}\int_{0}^{1}(1-\tau)e^{\mu\tau
S}\text{ad}_{S}^{2}\big{(}D^{(2)}+R^{(2)}\big{)}e^{-\mu\tau
S}\mathrm{d}\tau}_{=:\mathcal{R}(S)}=0\,.$ (5.42)

The “nonlinear homological equation” (5.42), i.e.
$[S,D^{(2)}]=-\frac{1}{\mu}R^{(2)}-\mu\mathcal{R}(S)$, is equivalent to solve
the $4\times 4$ real linear system

${\mathcal{A}}\vec{x}=\vec{f}(\mu,\epsilon,\vec{x})\,,\quad\vec{f}(\mu,\epsilon,\vec{x})=\mu\vec{v}(\mu,\epsilon)+\mu^{2}\vec{g}(\mu,\epsilon,\vec{x})$
(5.43)

associated, as in (5.17), to (5.42). The vector $\mu\vec{v}(\mu,\epsilon)$ is
associated with $-\frac{1}{\mu}R^{(2)}$ with $R^{(2)}$ in (5.39). The vector
$\mu^{2}\vec{g}(\mu,\epsilon,\vec{x})$ is associated with the matrix
$-\mu\mathcal{R}(S)$, which is a Hamiltonian and reversible block-off-diagonal
matrix (i.e of the form (5.13)), of size $\mathcal{R}(S)=\mathcal{O}(\mu)$
since $\Pi_{\varnothing}\text{ad}^{2}_{S}(D^{(2)})=0$. The function
$\vec{g}(\mu,\epsilon,\vec{x})$ is quadratic in $\vec{x}$. In view of (5.12)
one has

$\mu^{2}\vec{v}(\mu,\epsilon):=(-F^{(2)}_{21},F^{(2)}_{22},-F^{(2)}_{11},F^{(2)}_{12})^{\top},\quad
F^{(2)}_{ij}=\mu^{2}r(\epsilon^{3},\mu\epsilon^{2},\mu^{3}\epsilon,\mu^{5})\,.$
(5.44)

System (5.43) is equivalent to
$\vec{x}={\mathcal{A}}^{-1}\vec{f}(\mu,\epsilon,\vec{x})$ and, writing
${\mathcal{A}}^{-1}=\frac{1}{\mu}{\mathcal{B}}(\mu,\epsilon)$ (cfr. (5.31)),
to

$\vec{x}={\mathcal{B}}(\mu,\epsilon)\vec{v}(\mu,\epsilon)+\mu{\mathcal{B}}(\mu,\epsilon)\vec{g}(\mu,\epsilon,\vec{x})\,.$

By the implicit function theorem this equation admits a unique small solution
$\vec{x}=\vec{x}(\mu,\epsilon)$, analytic in $(\mu,\epsilon)$, with size
$\mathcal{O}(\epsilon^{3},\mu\epsilon^{2},\mu^{3}\epsilon,\mu^{5})$ as
$\vec{v}$ in (5.44). The claimed estimate of $P$ follows by the the first
equation of (5.41) and the estimate for $S$ and of $R^{(2)}$ obtained by
(5.12). ∎

Proof of Theorems 2.2 and 1.1.  By Lemma 5.8 and recalling (3.1) the operator
$\mathcal{L}_{\mu,\epsilon}:\mathcal{V}_{\mu,\epsilon}\to\mathcal{V}_{\mu,\epsilon}$
is represented by the $4\times 4$ Hamiltonian and reversible matrix

$\mathrm{i}\,\mu+\exp(\mu S^{(2)})\mathtt{L}_{\mu,\epsilon}^{(2)}\exp(-\mu
S^{(2)})=\mathrm{i}\,\mu+\begin{pmatrix}\mathtt{J}_{2}E^{(3)}&0\\\
0&\mathtt{J}_{2}G^{(3)}\end{pmatrix}=:\begin{pmatrix}\mathtt{U}&0\\\
0&\mathtt{S}\end{pmatrix}\,,$

where the matrices $E^{(3)}$ and $G^{(3)}$ expand as in (4.13)-(4.14).
Consequently the matrices $\mathtt{U}$ and $\mathtt{S}$ expand as in (2.33),
(2.34). Theorem 2.2 is proved. The unstable eigenvalues in Theorem 1.1 arise
from the block $\mathtt{U}$. Its bottom-left entry vanishes for
$\frac{\mu^{2}}{8}(1+r^{\prime}_{1}(\mu,\epsilon))=\epsilon^{2}(1+r^{\prime\prime}_{1}(\mu,\epsilon))$,
which, by taking square roots, amounts to solve
$\mu=2\sqrt{2}\epsilon(1+r(\mu,\epsilon))$. By the implicit function theorem,
it admits a unique analytic solution
$\underline{\mu}(\epsilon)=2\sqrt{2}\epsilon(1+r(\epsilon)).$ The proof of
Theorem 1.1 is complete.

##  Appendix A Proof of Lemma 4.2

We provide the expansion of the basis
$f_{k}^{\pm}(\mu,\epsilon)=U_{\mu,\epsilon}f_{k}^{\pm}$, $k=0,1$, in (4.1),
where $f_{k}^{\pm}$ defined in (4.2) belong to the subspace
$\mathcal{V}_{0,0}:=\text{Rg}(P_{0,0})$. We first Taylor-expand the
transformation operators $U_{\mu,\epsilon}$ defined in (3.10). We denote
$\partial_{\epsilon}$ with an apex and $\partial_{\mu}$ with a dot.

######  Lemma A.1.

The first jets of $U_{\mu,\epsilon}P_{0,0}$ are

$\displaystyle U_{0,0}P_{0,0}$ $\displaystyle=P_{0,0}\,,\quad
U_{0,0}^{\prime}P_{0,0}=P_{0,0}^{\prime}P_{0,0}\,,\quad\dot{U}_{0,0}P_{0,0}=\dot{P}_{0,0}P_{0,0}\,,$
(A.1) $\displaystyle\dot{U}_{0,0}^{\prime}P_{0,0}$
$\displaystyle=\big{(}\dot{P}_{0,0}^{\prime}-\frac{1}{2}P_{0,0}\dot{P}_{0,0}^{\prime}\big{)}P_{0,0}\,,$
(A.2)

where

$\displaystyle P_{0,0}^{\prime}$
$\displaystyle=\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}({\mathcal{L}}_{0,0}-\lambda)^{-1}{\mathcal{L}}_{0,0}^{\prime}({\mathcal{L}}_{0,0}-\lambda)^{-1}\mathrm{d}\lambda\,,$
(A.3) $\displaystyle\dot{P}_{0,0}$
$\displaystyle=\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}({\mathcal{L}}_{0,0}-\lambda)^{-1}\dot{\mathcal{L}}_{0,0}({\mathcal{L}}_{0,0}-\lambda)^{-1}\mathrm{d}\lambda\,,$
(A.4)

and

$\displaystyle\dot{P}_{0,0}^{\prime}$
$\displaystyle=-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}({\mathcal{L}}_{0,0}-\lambda)^{-1}\dot{\mathcal{L}}_{0,0}({\mathcal{L}}_{0,0}-\lambda)^{-1}{\mathcal{L}}_{0,0}^{\prime}({\mathcal{L}}_{0,0}-\lambda)^{-1}\mathrm{d}\lambda$
(A.5a)
$\displaystyle\qquad-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}({\mathcal{L}}_{0,0}-\lambda)^{-1}{\mathcal{L}}_{0,0}^{\prime}({\mathcal{L}}_{0,0}-\lambda)^{-1}\dot{\mathcal{L}}_{0,0}({\mathcal{L}}_{0,0}-\lambda)^{-1}\mathrm{d}\lambda$
(A.5b)
$\displaystyle\qquad+\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}({\mathcal{L}}_{0,0}-\lambda)^{-1}\dot{\mathcal{L}}_{0,0}^{\prime}({\mathcal{L}}_{0,0}-\lambda)^{-1}\mathrm{d}\lambda\,.$
(A.5c)

The operators ${\mathcal{L}}_{0,0}^{\prime}$ and $\dot{\mathcal{L}}_{0,0}$ are

${\mathcal{L}}_{0,0}^{\prime}=\begin{bmatrix}\partial_{x}\circ p_{1}(x)&0\\\
-a_{1}(x)&p_{1}(x)\circ\partial_{x}\end{bmatrix},\quad\dot{\mathcal{L}}_{0,0}=\begin{bmatrix}0&\operatorname*{sgn}(D)+\Pi_{0}\\\
0&0\end{bmatrix},$ (A.6)

with $a_{1}(x)=p_{1}(x)=-2\cos(x)$, cfr. (2.15)-(2.16). The operator
$\dot{\mathcal{L}}_{0,0}^{\prime}$ is

$\dot{\mathcal{L}}_{0,0}^{\prime}=\begin{bmatrix}\mathrm{i}\,p_{1}(x)&0\\\
0&\mathrm{i}\,p_{1}(x)\end{bmatrix}\,.$ (A.7)

###### Proof.

By (3.10) and (3.9) one has the Taylor expansion in $\mathcal{L}(Y)$

$U_{\mu,\epsilon}P_{0,0}=P_{\mu,\epsilon}P_{0,0}+\frac{1}{2}(P_{\mu,\epsilon}-P_{0,0})^{2}P_{\mu,\epsilon}P_{0,0}+\mathcal{O}(P_{\mu,\epsilon}-P_{0,0})^{4}\,,$

where
$\mathcal{O}(P_{\mu,\epsilon}-P_{0,0})^{4}=\mathcal{O}(\epsilon^{4},\epsilon^{3}\mu,\epsilon^{2}\mu^{2},\epsilon\mu^{3},\mu^{4})\in\mathcal{L}(Y)$.
Consequently one derives (A.1), (A.2), using also the identity
$\dot{P}_{0,0}P_{0,0}^{\prime}P_{0,0}+P_{0,0}^{\prime}\dot{P}_{0,0}P_{0,0}=-P_{0,0}\dot{P}_{0,0}^{\prime}P_{0,0}$,
which follows differentiating $P_{\mu,\epsilon}^{2}=P_{\mu,\epsilon}$.
Differentiating (3.5) one gets (A.3)–(A.5c). Formulas (A.6)-(A.7) follow by
(3.2). ∎

By the previous lemma we have the Taylor expansion

$f_{k}^{\sigma}(\mu,\epsilon)=f_{k}^{\sigma}+\epsilon
P_{0,0}^{\prime}f_{k}^{\sigma}+\mu\dot{P}_{0,0}f_{k}^{\sigma}+\mu\epsilon\big{(}\dot{P}_{0,0}^{\prime}-\frac{1}{2}P_{0,0}\dot{P}_{0,0}^{\prime}\big{)}f_{k}^{\sigma}+\mathcal{O}(\mu^{2},\epsilon^{2})\,.$
(A.8)

In order to compute the vectors $P_{0,0}^{\prime}f_{k}^{\sigma}$ and
$\dot{P}_{0,0}f_{k}^{\sigma}$ using (A.3) and (A.4), it is useful to know the
action of $({\mathcal{L}}_{0,0}-\lambda)^{-1}$ on the vectors

$f_{k}^{+}:=\begin{bmatrix}\cos(kx)\\\ \sin(kx)\end{bmatrix},\quad
f_{k}^{-}:=\begin{bmatrix}-\sin(kx)\\\ \cos(kx)\end{bmatrix},\quad
f_{-k}^{+}:=\begin{bmatrix}\cos(kx)\\\ -\sin(kx)\end{bmatrix},\quad
f_{-k}^{-}:=\begin{bmatrix}\sin(kx)\\\ \cos(kx)\end{bmatrix},\quad
k\in\mathbb{N}\,.$ (A.9)

######  Lemma A.2.

The space $H^{1}(\mathbb{T})$ can be decomposed as
$H^{1}(\mathbb{T})=\mathcal{V}_{0,0}\oplus\mathcal{U}\oplus\mathcal{W}$ where
the subspaces $\mathcal{V}_{0,0},\mathcal{U}$ and $\mathcal{W}$, defined
below, are invariant under ${\mathcal{L}}_{0,0}$ and the following properties
hold:

  * (i)

$\mathcal{V}_{0,0}=\text{span}\\{f^{+}_{1},f^{-}_{1},f^{+}_{0},f^{-}_{0}\\}$
is the generalized kernel of ${\mathcal{L}}_{0,0}$. For any $\lambda\neq 0$
the operator
${\mathcal{L}}_{0,0}-\lambda:\mathcal{V}_{0,0}\to\mathcal{V}_{0,0}$ is
invertible and

$\displaystyle({\mathcal{L}}_{0,0}-\lambda)^{-1}f_{1}^{+}=-\frac{1}{\lambda}f_{1}^{+}\,,\quad({\mathcal{L}}_{0,0}-\lambda)^{-1}f_{1}^{-}=-\frac{1}{\lambda}f_{1}^{-},\quad({\mathcal{L}}_{0,0}-\lambda)^{-1}f_{0}^{-}=-\frac{1}{\lambda}f_{0}^{-}\,,$
(A.10)
$\displaystyle({\mathcal{L}}_{0,0}-\lambda)^{-1}f_{0}^{+}=-\frac{1}{\lambda}f_{0}^{+}+\frac{1}{\lambda^{2}}f_{0}^{-}\,.$
(A.11)

  * (ii)

$\mathcal{U}:=\text{span}\left\\{f_{-1}^{+},f_{-1}^{-}\right\\}$. For any
$\lambda\neq\pm 2\mathrm{i}\,$ the operator
${\mathcal{L}}_{0,0}-\lambda:\mathcal{U}\to\mathcal{U}$ is invertible and

$({\mathcal{L}}_{0,0}-\lambda)^{-1}f_{-1}^{+}=\frac{1}{\lambda^{2}+4}\left(-\lambda
f_{-1}^{+}+2f_{-1}^{-}\right),\quad({\mathcal{L}}_{0,0}-\lambda)^{-1}f_{-1}^{-}=\frac{1}{\lambda^{2}+4}\left(-2f_{-1}^{+}-\lambda
f_{-1}^{-}\right)\,.$ (A.12)

  * (iii)

$\mathcal{W}=\overline{\bigoplus\limits_{k=2}^{\infty}\mathcal{W}_{k}}^{H^{1}}$
where each subspace $\mathcal{W}_{k}:=\text{span}\left\\{f_{k}^{+},\
f_{k}^{-},f_{-k}^{+},\ f_{-k}^{-}\right\\}$ is invariant under
${\mathcal{L}}_{0,0}$. For any $|\lambda|<\frac{1}{2}$, the operator
${\mathcal{L}}_{0,0}-\lambda:\mathcal{W}\to\mathcal{W}$ is invertible and, for
any $f\in\mathcal{W}$,

$({\mathcal{L}}_{0,0}-\lambda)^{-1}f=(\partial_{x}^{2}+|D|)^{-1}\begin{bmatrix}\partial_{x}&-|D|\\\
1&\partial_{x}\end{bmatrix}f+\lambda\varphi_{f}(\lambda,x)\,,$ (A.13)

for some analytic function $\lambda\mapsto\varphi_{f}(\lambda,\cdot)\in
H^{1}(\mathbb{T},\mathbb{C}^{2})$.

###### Proof.

By inspection the spaces $\mathcal{V}_{0,0}$, $\mathcal{U}$ and $\mathcal{W}$
are invariant under ${\mathcal{L}}_{0,0}$ and, by Fourier series, they
decompose $H^{1}(\mathbb{T},\mathbb{C}^{2})$.
$(i)$ Formulas (A.10)-(A.11) follow using that $f_{1}^{+},f_{1}^{-},f_{0}^{-}$
are in the kernel of ${\mathcal{L}}_{0,0}$, and
${\mathcal{L}}_{0,0}f_{0}^{+}=-f_{0}^{-}$.
$(ii)$ Formula (A.12) follows using that
${\mathcal{L}}_{0,0}f^{+}_{-1}=-2f^{-}_{-1}$ and
${\mathcal{L}}_{0,0}f^{-}_{-1}=2f^{+}_{-1}$.
$(iii)$ The operator $({\mathcal{L}}_{0,0}-\lambda\mathrm{Id})_{|\mathcal{W}}$
is invertible for any
$\lambda\notin\\{\pm\mathrm{i}\,\sqrt{|k|}\pm\mathrm{i}\,k,k\geq
2,k\in{\mathbb{N}}\\}$ and
$\footnotesize({\left.\kern-1.2pt{\mathcal{L}}_{0,0}\vphantom{\big{|}}\right|_{\mathcal{W}}})^{-1}=\left(\partial_{x}^{2}+|D|\right)^{-1}\begin{bmatrix}\partial_{x}&-|D|\\\
1&\partial_{x}\end{bmatrix}_{|\mathcal{W}}$. In particular, by Neumann series,
for any $\lambda$ such that
$|\lambda|\|({\left.\kern-1.2pt{\mathcal{L}}_{0,0}\vphantom{\big{|}}\right|_{\mathcal{W}}})^{-1}\|_{\mathcal{L}(Y)}<1$,
e.g. for any $|\lambda|<1/2$,

$({\left.\kern-1.2pt{\mathcal{L}}_{0,0}\vphantom{\big{|}}\right|_{\mathcal{W}}}-\lambda)^{-1}=({\left.\kern-1.2pt{\mathcal{L}}_{0,0}\vphantom{\big{|}}\right|_{\mathcal{W}}})^{-1}\big{(}\mathrm{Id}-\lambda({\left.\kern-1.2pt{\mathcal{L}}_{0,0}\vphantom{\big{|}}\right|_{\mathcal{W}}})^{-1}\big{)}^{-1}=({\left.\kern-1.2pt{\mathcal{L}}_{0,0}\vphantom{\big{|}}\right|_{\mathcal{W}}})^{-1}\sum_{k\geq
0}(({\left.\kern-1.2pt{\mathcal{L}}_{0,0}\vphantom{\big{|}}\right|_{\mathcal{W}}})^{-1}\lambda)^{k}\,.$

Formula (A.13) follows with
$\varphi_{f}(\lambda,x):=({\left.\kern-1.2pt{\mathcal{L}}_{0,0}\vphantom{\big{|}}\right|_{\mathcal{W}}})^{-1}\sum_{k\geq
1}\lambda^{k-1}[({\left.\kern-1.2pt{\mathcal{L}}_{0,0}\vphantom{\big{|}}\right|_{\mathcal{W}}})^{-1}]^{k}f$.
∎

We shall also use the following formulas, obtained by (A.6) and (4.2):

$\displaystyle{\mathcal{L}}_{0,0}^{\prime}f_{1}^{+}=2\begin{bmatrix}\sin(2x)\\\
0\end{bmatrix}\,,\quad{\mathcal{L}}_{0,0}^{\prime}f_{1}^{-}=2\begin{bmatrix}\cos(2x)\\\
0\end{bmatrix}\,,\quad{\mathcal{L}}_{0,0}^{\prime}f_{0}^{+}=2\begin{bmatrix}\sin(x)\\\
\cos(x)\end{bmatrix}\,,\quad{\mathcal{L}}_{0,0}^{\prime}f_{0}^{-}=0\,,$ (A.14)
$\displaystyle\dot{\mathcal{L}}_{0,0}f_{1}^{+}=-\mathrm{i}\,\begin{bmatrix}\cos(x)\\\
0\end{bmatrix}\,,\quad\dot{\mathcal{L}}_{0,0}f_{1}^{-}=\mathrm{i}\,\begin{bmatrix}\sin(x)\\\
0\end{bmatrix}\,,\quad\dot{\mathcal{L}}_{0,0}f_{0}^{+}=0,\quad\dot{\mathcal{L}}_{0,0}f_{0}^{-}=f_{0}^{+}\,.$

We finally compute $P_{0,0}^{\prime}f_{k}^{\sigma}$ and
$\dot{P}_{0,0}f_{k}^{\sigma}$.

######  Lemma A.3.

One has

$\displaystyle P_{0,0}^{\prime}f^{+}_{1}=\begin{bmatrix}2\cos(2x)\\\
\sin(2x)\end{bmatrix}\,,\ \ \
P_{0,0}^{\prime}f^{-}_{1}=\begin{bmatrix}-2\sin(2x)\\\
\cos(2x)\end{bmatrix}\,,\ \ \ P_{0,0}^{\prime}f^{+}_{0}=f^{+}_{-1}\,,\ \ \
P_{0,0}^{\prime}f^{-}_{0}=0\,,$ (A.15)
$\displaystyle\dot{P}_{0,0}f_{1}^{+}=\frac{\mathrm{i}\,}{4}f^{-}_{-1}\,,\quad\dot{P}_{0,0}f_{1}^{-}=\frac{\mathrm{i}\,}{4}f^{+}_{-1}\,,\quad\dot{P}_{0,0}f_{0}^{+}=0\,,\quad\dot{P}_{0,0}f_{0}^{-}=0\,.$

###### Proof.

We first compute $P_{0,0}^{\prime}f_{1}^{+}$. By (A.3), (A.10) and (A.14) we
deduce

$P_{0,0}^{\prime}f_{1}^{+}=-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}\frac{1}{\lambda}({\mathcal{L}}_{0,0}-\lambda)^{-1}\begin{bmatrix}2\sin(2x)\\\
0\end{bmatrix}\mathrm{d}\lambda\,.$

We note that $\footnotesize\begin{bmatrix}2\sin(2x)\\\ 0\end{bmatrix}$ belongs
to $\mathcal{W}$, being equal to $f_{-2}^{-}-f_{2}^{-}$ (recall (A.9)). By
(A.13) there is an analytic function $\lambda\mapsto\varphi(\lambda,\cdot)\in
H^{1}(\mathbb{T},\mathbb{C}^{2})$ so that

$P_{0,0}^{\prime}f_{1}^{+}=-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}\frac{1}{\lambda}\Big{(}\begin{bmatrix}-2\cos(2x)\\\
-\sin(2x)\end{bmatrix}+\lambda\varphi(\lambda)\Big{)}\,\mathrm{d}\lambda=\begin{bmatrix}2\cos(2x)\\\
\sin(2x)\end{bmatrix}\,,$

using the residue Theorem. Similarly one computes $P_{0,0}^{\prime}f_{1}^{-}$.
By (A.3), (A.10) and (A.14), one has $P_{0,0}^{\prime}f_{0}^{-}=0$. Next we
compute $P_{0,0}^{\prime}f_{0}^{+}$. By (A.3), (A.10), (A.11) and (A.14) we
get

$P_{0,0}^{\prime}f_{0}^{+}=-\frac{2}{2\pi\mathrm{i}\,}\oint_{\Gamma}\frac{1}{\lambda}({\mathcal{L}}_{0,0}-\lambda)^{-1}f^{-}_{-1}\mathrm{d}\lambda\stackrel{{\scriptstyle\eqref{primainversione3}}}{{=}}-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}\Big{(}-\frac{4}{\lambda(\lambda^{2}+4)}f_{-1}^{+}-\frac{2}{\lambda^{2}+4}f_{-1}^{-}\Big{)}\mathrm{d}\lambda=f^{+}_{-1}\,,$

where in the last step we used the residue theorem. We compute now
$\dot{P}_{0,0}f^{+}_{1}$. First we have $\dot{P}_{0,0}f_{1}^{+}=\
\frac{\mathrm{i}\,}{2\pi\mathrm{i}\,}\oint_{\Gamma}\frac{1}{\lambda}({\mathcal{L}}_{0,0}-\lambda)^{-1}\footnotesize\begin{bmatrix}\cos(x)\\\
0\end{bmatrix}\mathrm{d}\lambda$ and then, writing
$\footnotesize\begin{bmatrix}\cos(x)\\\
0\end{bmatrix}=\frac{1}{2}(f_{1}^{+}+f_{-1}^{+})$ and using (A.12), we
conclude

$\dot{P}_{0,0}f_{1}^{+}=\frac{\mathrm{i}\,}{2}\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}\Big{(}-\frac{1}{\lambda^{2}}f^{+}_{1}-\frac{1}{\lambda^{2}+4}f_{-1}^{+}+\frac{2}{\lambda(\lambda^{2}+4)}f_{-1}^{-}\Big{)}\mathrm{d}\lambda=\frac{\mathrm{i}\,}{4}f_{-1}^{-}$

using again the residue theorem. The computations of $\dot{P}_{0,0}f^{-}_{1}$,
$\dot{P}_{0,0}f_{0}^{+}$, $\dot{P}_{0,0}f_{0}^{-}$ are analogous. ∎

So far we have obtained the linear terms of the expansions (4.4), (4.5),
(4.6), (4.7). We now provide further information about the expansion of the
basis at $\mu=0$.

######  Lemma A.4.

The basis $\\{f_{k}^{\sigma}(0,\epsilon),\,k=0,1,\sigma=\pm\\}$ is real. For
any $\epsilon$ it results $f_{0}^{-}(0,\epsilon)\equiv f_{0}^{-}$. The
property (4.8) holds.

###### Proof.

The reality of the basis $f_{k}^{\sigma}(0,\epsilon)$ is a consequence of
Lemma 3.2-$(iii)$. Since, recalling (3.2),
${\mathcal{L}}_{0,\epsilon}f_{0}^{-}=0$ for any $\epsilon$ (cfr. (2.30)), we
deduce
$({\mathcal{L}}_{0,\epsilon}-\lambda)^{-1}f_{0}^{-}=-\frac{1}{\lambda}f_{0}^{-}$
and then, using also the residue theorem,

$P_{0,\epsilon}f_{0}^{-}=-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}({\mathcal{L}}_{0,\epsilon}-\lambda)^{-1}f_{0}^{-}\mathrm{d}\lambda=f_{0}^{-}\
.$

In particular $P_{0,\epsilon}f_{0}^{-}=P_{0,0}f_{0}^{-}$, for any $\epsilon$
and we get, by (3.10),
$f_{0}^{-}(0,\epsilon)=U_{0,\epsilon}f_{0}^{-}=f_{0}^{-}$, for any $\epsilon$.

Let us prove property (4.8). In view of (3.19) and since the basis is real, we
know that $\footnotesize f_{k}^{+}(0,\epsilon)=\begin{bmatrix}even(x)\\\
odd(x)\end{bmatrix}$, $\footnotesize
f_{k}^{-}(0,\epsilon)=\begin{bmatrix}odd(x)\\\ even(x)\end{bmatrix}$, for any
$k=0,1$. By Lemma 4.1 the basis $\\{f_{k}^{\sigma}(0,\epsilon)\\}$ is
symplectic (cfr. (3.17)) and, since
$\mathcal{J}f_{0}^{-}(0,\epsilon)=\mathcal{J}f_{0}^{-}=\footnotesize\begin{bmatrix}1\\\
0\end{bmatrix}$, for any $\epsilon$, we get

$0=\left(\mathcal{J}f_{0}^{-}(0,\epsilon)\,,\,f_{1}^{+}(0,\epsilon)\right)=\Big{(}\begin{bmatrix}1\\\
0\end{bmatrix},f_{1}^{+}(0,\epsilon)\Big{)}\,,\quad
1=\big{(}\mathcal{J}f_{0}^{-}(0,\epsilon),f_{0}^{+}(0,\epsilon)\big{)}=\Big{(}\begin{bmatrix}1\\\
0\end{bmatrix},f_{0}^{+}(0,\epsilon)\Big{)}\,.$

Thus the first component of both $f_{1}^{+}(0,\epsilon)$ and
$f_{0}^{+}(0,\epsilon)-\footnotesize\begin{bmatrix}1\\\ 0\end{bmatrix}$ has
zero average, proving (4.8). ∎

We now provide further information about the expansion of the basis at
$\epsilon=0$.

######  Lemma A.5.

For any small $\mu$, we have $f_{0}^{+}(\mu,0)\equiv f_{0}^{+}$ and
$f_{0}^{-}(\mu,0)\equiv f_{0}^{-}$. Moreover the vectors $f_{1}^{+}(\mu,0)$
and $f_{1}^{-}(\mu,0)$ have both components with zero space average.

###### Proof.

The operator
${\mathcal{L}}_{\mu,0}=\footnotesize\begin{bmatrix}\partial_{x}&|D+\mu|\\\
-1&\partial_{x}\end{bmatrix}$ leaves invariant the subspace
$\mathcal{Z}:=\text{span}\\{f_{0}^{+},\,f_{0}^{-}\\}$ since
${\mathcal{L}}_{\mu,0}f_{0}^{+}=-f_{0}^{-}$ and
${\mathcal{L}}_{\mu,0}f_{0}^{-}=\mu f_{0}^{+}$. The operator
${\left.\kern-1.2pt{\mathcal{L}}_{\mu,0}\vphantom{\big{|}}\right|_{\mathcal{Z}}}$
has the two eigenvalues $\pm\mathrm{i}\,\sqrt{\mu}$, which, for small $\mu$,
lie inside the loop $\Gamma$ around $0$ in (3.5). Then, by (3.8), we have
$\mathcal{Z}\subseteq{\mathcal{V}}_{\mu,0}=\text{Rg}(P_{\mu,0})$ and

$P_{\mu,0}f_{0}^{\pm}=f_{0}^{\pm},\quad
f_{0}^{\pm}(\mu,0)=U_{\mu,0}f_{0}^{\pm}=f_{0}^{\pm},\text{ for any }\mu\text{
small}\,.$

The basis $\\{f_{k}^{\sigma}(\mu,0)\\}$ is symplectic. Then, since
$\footnotesize\mathcal{J}f_{0}^{+}=\begin{bmatrix}0\\\ -1\end{bmatrix}$ and
$\footnotesize\mathcal{J}f_{0}^{-}=\begin{bmatrix}1\\\ 0\end{bmatrix}$, we
have

$0=\left(\mathcal{J}f_{0}^{+}(\mu,0)\,,\,f_{1}^{\sigma}(\mu,0)\right)=\Big{(}{\footnotesize\begin{bmatrix}0\\\
-1\end{bmatrix}},f_{1}^{\sigma}(\mu,0)\Big{)}\,,\ \
0=\Big{(}\mathcal{J}f_{0}^{-}(\mu,0),f_{1}^{\sigma}(\mu,0)\Big{)}=\Big{(}{\footnotesize\begin{bmatrix}1\\\
0\end{bmatrix}},f_{1}^{\sigma}(\mu,0)\Big{)}\,,$

namely both the components of $f_{1}^{\pm}(\mu,0)$ have zero average. ∎

We finally consider the $\mu\epsilon$ term in the expansion (A.8) of the
vectors $f_{k}^{\sigma}(\mu,\epsilon)$, $k=0,1$, $\sigma=\pm$.

######  Lemma A.6.

The derivatives
$(\partial_{\mu}\partial_{\epsilon}f_{k}^{\sigma})(0,0)=\left(\dot{P}_{0,0}^{\prime}-\frac{1}{2}P_{0,0}\dot{P}_{0,0}^{\prime}\right)f_{k}^{\sigma}$
satisfy

$\displaystyle(\partial_{\mu}\partial_{\epsilon}f_{1}^{+})(0,0)=\mathrm{i}\,\begin{bmatrix}odd(x)\\\
even(x)\end{bmatrix},\qquad(\partial_{\mu}\partial_{\epsilon}f_{1}^{-})(0,0)-=\mathrm{i}\,\begin{bmatrix}even(x)\\\
odd(x)\end{bmatrix},$ (A.16)
$\displaystyle(\partial_{\mu}\partial_{\epsilon}f_{0}^{+})(0,0)=\mathrm{i}\,\begin{bmatrix}odd(x)\\\
even_{0}(x)\end{bmatrix},\qquad(\partial_{\mu}\partial_{\epsilon}f_{0}^{-})(0,0)=\frac{1}{2}\begin{bmatrix}\sin(x)\\\
\cos(x)\end{bmatrix}+\mathrm{i}\,\begin{bmatrix}even_{0}(x)\\\
odd(x)\end{bmatrix}\ .$

###### Proof.

We decompose the Fourier multiplier operator $\dot{\mathcal{L}}_{0,0}$ in
(A.6) as

$\dot{\mathcal{L}}_{0,0}=\dot{\mathcal{L}}_{0,0}^{(I)}+\dot{\mathcal{L}}_{0,0}^{(II)}\,,\qquad\dot{\mathcal{L}}_{0,0}^{(I)}:=\begin{bmatrix}0&\operatorname*{sgn}(D)\\\
0&0\end{bmatrix}\,,\qquad\dot{\mathcal{L}}_{0,0}^{(II)}:=\begin{bmatrix}0&\Pi_{0}\\\
0&0\end{bmatrix}\,,$

and, accordingly, we write
$\dot{P}_{0,0}^{\prime}=\eqref{Pmisto1}^{(I)}+\eqref{Pmisto1}^{(II)}+\eqref{Pmisto2}^{(I)}+\eqref{Pmisto2}^{(II)}+\eqref{Pmisto3}$
defining

$\displaystyle\eqref{Pmisto1}^{(I)}$
$\displaystyle:=-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}({\mathcal{L}}_{0,0}-\lambda)^{-1}\dot{\mathcal{L}}_{0,0}^{(I)}({\mathcal{L}}_{0,0}-\lambda)^{-1}{\mathcal{L}}_{0,0}^{\prime}({\mathcal{L}}_{0,0}-\lambda)^{-1}\mathrm{d}\lambda\,,$
(A.17) $\displaystyle\eqref{Pmisto1}^{(II)}$
$\displaystyle:=-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}({\mathcal{L}}_{0,0}-\lambda)^{-1}\dot{\mathcal{L}}_{0,0}^{(II)}({\mathcal{L}}_{0,0}-\lambda)^{-1}{\mathcal{L}}_{0,0}^{\prime}({\mathcal{L}}_{0,0}-\lambda)^{-1}\mathrm{d}\lambda\,,$
(A.18) $\displaystyle\eqref{Pmisto2}^{(I)}$
$\displaystyle:=-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}({\mathcal{L}}_{0,0}-\lambda)^{-1}{\mathcal{L}}_{0,0}^{\prime}({\mathcal{L}}_{0,0}-\lambda)^{-1}\dot{\mathcal{L}}_{0,0}^{(I)}({\mathcal{L}}_{0,0}-\lambda)^{-1}\mathrm{d}\lambda\,,$
(A.19) $\displaystyle\eqref{Pmisto2}^{(II)}$
$\displaystyle:=-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}({\mathcal{L}}_{0,0}-\lambda)^{-1}{\mathcal{L}}_{0,0}^{\prime}({\mathcal{L}}_{0,0}-\lambda)^{-1}\dot{\mathcal{L}}_{0,0}^{(II)}({\mathcal{L}}_{0,0}-\lambda)^{-1}\mathrm{d}\lambda\,.$
(A.20)

Note that the operators $\eqref{Pmisto1}^{(I)}$, $\eqref{Pmisto2}^{(I)}$ and
$\eqref{Pmisto3}$ are purely imaginary because $\dot{\mathcal{L}}_{0,0}^{(I)}$
is purely imaginary, ${\mathcal{L}}_{0,0}^{\prime}$ in (A.6) is real and
$\dot{\mathcal{L}}_{0,0}^{\prime}$ in (A.7) is purely imaginary (argue as in
Lemma 3.2-$(iii)$). Then, applied to the real vectors $f^{\sigma}_{k}$,
$k=0,1$, $\sigma=\pm$, give purely imaginary vectors.
We first we compute $(\partial_{\mu}\partial_{\epsilon}f_{1}^{+})(0,0)$. Using
(A.10) and (A.14) we get

$\displaystyle\eqref{Pmisto1}^{(II)}f_{1}^{+}=\frac{2}{2\pi\mathrm{i}\,}\oint_{\Gamma}\frac{1}{\lambda}({\mathcal{L}}_{0,0}-\lambda)^{-1}\dot{\mathcal{L}}_{0,0}^{(II)}({\mathcal{L}}_{0,0}-\lambda)^{-1}\begin{bmatrix}\sin(2x)\\\
0\end{bmatrix}\mathrm{d}\lambda=0$

because, by Lemma A.2,
$({\mathcal{L}}_{0,0}-\lambda)^{-1}\footnotesize\begin{bmatrix}\sin(2x)\\\
0\end{bmatrix}\in\mathcal{W}$ and therefore it is a vector with zero average,
so in the kernel of $\dot{\mathcal{L}}_{0,0}^{(II)}$. In addition
$\eqref{Pmisto2}^{(II)}f_{1}^{+}=0$ since
$\dot{\mathcal{L}}_{0,0}^{(II)}({\mathcal{L}}_{0,0}-\lambda)^{-1}f_{1}^{+}=0$.
All together $\dot{P}_{0,0}^{\prime}f_{1}^{+}$ is a purely imaginary vector.
Since $P_{0,0}$ is a real operator, also
$(\dot{P}_{0,0}^{\prime}-\frac{1}{2}P_{0,0}\dot{P}_{0,0}^{\prime})f_{1}^{+}$
is purely imaginary, and Lemma 3.4 implies that
$(\partial_{\mu}\partial_{\epsilon}f_{1}^{+})(0,0)$ has the claimed structure
in (A.16). In the same way one proves the structure for
$(\partial_{\mu}\partial_{\epsilon}f_{1}^{-})(0,0)$.

Next we prove that $(\partial_{\mu}\partial_{\epsilon}f_{0}^{+})(0,0)$, in
addition to being purely imaginary, has zero average. We have, by (A.11) and
(A.14)

$\displaystyle\eqref{Pmisto1}^{(I)}f_{0}^{+}:=\frac{2}{2\pi\mathrm{i}\,}\oint_{\Gamma}({\mathcal{L}}_{0,0}-\lambda)^{-1}\dot{\mathcal{L}}_{0,0}^{(I)}({\mathcal{L}}_{0,0}-\lambda)^{-1}\frac{1}{\lambda}\begin{bmatrix}\sin(x)\\\
\cos(x)\end{bmatrix}\,\mathrm{d}\lambda$

and since the Fourier multipliers $({\mathcal{L}}_{0,0}-\lambda)^{-1}$ and
$\dot{\mathcal{L}}_{0,0}^{(I)}$ are both Fourier multipliers (so they preserve
the average of the vectors) then $\eqref{Pmisto1}^{(I)}f_{0}^{+}$ has zero
average. In addition $\eqref{Pmisto1}^{(II)}f_{0}^{+}=0$ as
$\dot{\mathcal{L}}_{0,0}^{(II)}({\mathcal{L}}_{0,0}-\lambda)^{-1}\footnotesize\begin{bmatrix}\sin(x)\\\
\cos(x)\end{bmatrix}=0$. Next $\eqref{Pmisto2}^{(I)}f_{0}^{+}=0$ since
$\dot{\mathcal{L}}_{0,0}^{(I)}f_{0}^{\pm}=0$, cfr. (2.24). Using also that
$\dot{\mathcal{L}}_{0,0}^{(II)}f_{0}^{+}=0$ and
$\dot{\mathcal{L}}_{0,0}^{(II)}f_{0}^{-}=f_{0}^{+}$,

$\displaystyle\eqref{Pmisto2}^{(II)}f_{0}^{+}$
$\displaystyle\stackrel{{\scriptstyle\eqref{primainversione2}}}{{=}}-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}({\mathcal{L}}_{0,0}-\lambda)^{-1}{\mathcal{L}}_{0,0}^{\prime}({\mathcal{L}}_{0,0}-\lambda)^{-1}\frac{1}{\lambda^{2}}f_{0}^{+}\mathrm{d}\lambda$
$\displaystyle\stackrel{{\scriptstyle\eqref{primainversione2},\eqref{derivoeps}}}{{=}}\frac{2}{2\pi\mathrm{i}\,}\oint_{\Gamma}\frac{1}{\lambda^{3}}({\mathcal{L}}_{0,0}-\lambda)^{-1}\begin{bmatrix}\sin(x)\\\
\cos(x)\end{bmatrix}\mathrm{d}\lambda=0$

using (A.12) and the residue theorem. Finally, by (A.11) and (A.7) where
$p_{1}(x)=-2\cos(x)$,

$\eqref{Pmisto3}f_{0}^{+}=-\frac{\mathrm{i}\,2}{2\pi\mathrm{i}\,}\oint_{\Gamma}({\mathcal{L}}_{0,0}-\lambda)^{-1}\Big{(}-\frac{1}{\lambda}\begin{bmatrix}\cos(x)\\\
0\end{bmatrix}+\frac{1}{\lambda^{2}}\begin{bmatrix}0\\\
\cos(x)\end{bmatrix}\Big{)}\,\mathrm{d}\lambda$

is a vector with zero average. We conclude that
$\dot{P}_{0,0}^{\prime}f_{0}^{+}$ is an imaginary vector with zero average, as
well as $(\partial_{\mu}\partial_{\epsilon}f_{0}^{+})(0,0)$ since $P_{0,0}$
sends zero average functions in zero average functions. Finally, by Lemma 3.4,
$(\partial_{\mu}\partial_{\epsilon}f_{0}^{+})(0,0)$ has the claimed structure
in (A.16).

We finally consider $(\partial_{\mu}\partial_{\epsilon}f_{0}^{-})(0,0)$. By
(A.10) and ${\mathcal{L}}_{0,0}^{\prime}f_{0}^{-}=0$ (cfr. (A.14)), it results

$\eqref{Pmisto1}^{(M)}f_{0}^{-}=-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}\frac{({\mathcal{L}}_{0,0}-\lambda)^{-1}}{\lambda}\dot{\mathcal{L}}_{0,0}^{(M)}({\mathcal{L}}_{0,0}-\lambda)^{-1}{\mathcal{L}}_{0,0}^{\prime}f_{0}^{-}\mathrm{d}\lambda=0\,,\quad
M=I,II\ .$

Next by (A.10) and $\dot{\mathcal{L}}_{0,0}^{(I)}f_{0}^{-}=0$ we get
$\eqref{Pmisto2}^{(I)}f_{0}^{-}=0$. Then, since
$\dot{\mathcal{L}}_{0,0}^{(II)}f_{0}^{-}=f_{0}^{+}$,

$\displaystyle\eqref{Pmisto2}^{(II)}f_{0}^{-}$
$\displaystyle\stackrel{{\scriptstyle\eqref{primainversione1}-\eqref{primainversione2}}}{{=}}\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}\frac{({\mathcal{L}}_{0,0}-\lambda)^{-1}}{\lambda}\
{\mathcal{L}}_{0,0}^{\prime}\Big{(}-\frac{1}{\lambda}f_{0}^{+}+\frac{1}{\lambda^{2}}f_{0}^{-}\Big{)}\mathrm{d}\lambda$
$\displaystyle\stackrel{{\scriptstyle\eqref{derivoeps},\eqref{primainversione3}}}{{=}}-\frac{2}{2\pi\mathrm{i}\,}\oint_{\Gamma}\frac{1}{\lambda^{2}}\frac{1}{\lambda^{2}+4}(-2f_{-1}^{+}-\lambda
f_{-1}^{-})\mathrm{d}\lambda=\frac{1}{2}f_{-1}^{-}=\frac{1}{2}\begin{bmatrix}\sin(x)\\\
\cos(x)\end{bmatrix}\,,$

which is the only real term of
$(\partial_{\mu}\partial_{\epsilon}f_{0}^{-})(0,0)$ in (A.16). Finally by
(A.10) and (A.7)

$\displaystyle\eqref{Pmisto3}f_{0}^{-}=\frac{2\mathrm{i}\,}{2\pi\mathrm{i}\,}\oint_{\Gamma}({\mathcal{L}}_{0,0}-\lambda)^{-1}\frac{1}{\lambda}\begin{bmatrix}0\\\
\cos(x)\end{bmatrix}\mathrm{d}\lambda\
=-\frac{\mathrm{i}\,}{2}\begin{bmatrix}\cos(x)\\\ -\sin(x)\end{bmatrix}$

by (A.10), (A.12) and the residue theorem. In conclusion
$\dot{P}_{0,0}^{\prime}f_{0}^{-}=\frac{1}{2}\footnotesize\begin{bmatrix}\sin(x)\\\
\cos(x)\end{bmatrix}-\frac{\mathrm{i}\,}{2}\begin{bmatrix}\cos(x)\\\
-\sin(x)\end{bmatrix}\in\mathcal{U}$ and, since $P_{0,0}|_{\mathcal{U}}=0$, we
find that
$\left(\dot{P}_{0,0}^{\prime}-\frac{1}{2}P_{0,0}\dot{P}_{0,0}^{\prime}\right)f_{0}^{-}=\frac{1}{2}\footnotesize\begin{bmatrix}\sin(x)\\\
\cos(x)\end{bmatrix}-\frac{\mathrm{i}\,}{2}\begin{bmatrix}\cos(x)\\\
\sin(x)\end{bmatrix}$. ∎

This completes the proof of Lemma 4.2.

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