\begin{equation}\label{Xi2:estimate:final:1alpha:eps}
\abs{\p_{x}\Xi_{2,\epsilon}(x)- \p_{x}\Xi_{2,\epsilon}(0)} =\abs{M_{0,\epsilon}+M_{1,\epsilon}+M_{2,\epsilon}+M_{3,\epsilon}}\leq C \abs{x}^{\alpha} \norm{H}_{C^{1,\alpha}(\Omega)}, \mbox{ for } x\in \SS^1.
\end{equation}

Recalling that

and bounds (<ref>), (<ref>) we find that

\begin{equation}\label{Xi:final:estimate:1alpha:eps}
\abs{\p_{x}\Xi_{\epsilon}(x)- \p_{x}\Xi_{\epsilon}(0)} \leq C\abs{x}^{\alpha}\norm{H}_{C^{1,\alpha}(\Omega)}, \mbox{ for } x\in \SS^1
\end{equation}

which shows the desired $C^{1,\alpha}$ semi-norm bound in (<ref>).

To conclude the proof of the proposition, we claim that the $C^{1,\alpha}$ semi-norm bound (<ref>) and the $L^{\infty}$ pointwise bound (<ref>) implies the  $C^{1,\alpha}$ Hölder norm (<ref>). This is, we have to show that we can control the $L^\infty$ norm of $\p_{x}\Xi_{\epsilon}$. Indeed, by means of the identity

\begin{equation}
\Xi_{\epsilon}(x_{1})= \Xi_{\epsilon}(x_2)+\p_{x}\Xi_{\epsilon}(x_{2})(x_{1}-x_{2}) -\int_{x_1}^{x_2} \left(\p_{x}\Xi_{\epsilon}(x)-\p_{x}\Xi_{\epsilon}(x_{2})\right) dx, \mbox{ for } x_1,x_2\in \SS^1
\end{equation}

and using the $C^{1,\alpha}$ semi-norm bound (<ref>) and the $L^{\infty}$ pointwise bound (<ref>)

\begin{align}
\abs{\p_{x}\Xi_{\epsilon}(x_{2})} &\leq \frac{1}{2\pi}2\norm{\Xi}_{L^\infty}+\abs{\p_{x}\Xi_{\epsilon}(x)-\p_{x}\Xi_{\epsilon}(x_{2})}  \leq \frac{C}{\pi} \norm{H}_{C^{\alpha}(\Omega)}
+C(2\pi)^{\alpha}\norm{H}_{C^{1,\alpha}(\Omega)} \label{new:der:Linfty:Xi}
\end{align}

since $\abs{x_{1}-x_{2}}\leq 2\pi$. Therefore, estimate (<ref>) and the previous $C^{1,\alpha}$ semi-norm bound (<ref>) implies the desired $C^{1,\alpha}$ Hölder norm estimate (<ref>).

§.§ Lg and Lg estimates for simplified singular integral operators

We provide a similar type of estimate for a simplified operator that do not have the dependence on the function $\Lambda$ (and hence is of convolution type) which reads

Let $H(\eta, y)\in C^{1,\alpha}(\Omega)$ and let Assumption <ref> hold. Then for any $x\in \mathbb{S}^{1}$ the following limit exists

\begin{equation} \label{int:sin:2}
\displaystyle\lim_{\epsilon\to 0^{+}} \widetilde{\Xi}_{\epsilon}(x)=\widetilde{\Xi}(x),
\end{equation}

\begin{equation} \label{int:sin:2:eps}
\widetilde{\Xi}_{\epsilon}(x)= \int_{\epsilon}^{L}dy\int_{\mathbb{S}^{1}}d\eta \  \p_{x} \left( \frac{e^{i(x-\eta)-y}}{1-e^{i(x-\eta)-y}} \right)
H( \eta, y), \ \epsilon>0.
\end{equation}

Moreover, we have that

\begin{align}
\norm{\widetilde{\Xi}}_{C^{\alpha}(\mathbb{S}^{1})}&\leq C \norm{H}_{C^{\alpha}(\Omega)}, \label{cota:singular:3} \\
\norm{\widetilde{\Xi}}_{C^{1,\alpha}(\mathbb{S}^{1})}&\leq C \norm{H}_{C^{1,\alpha}(\Omega)}.\label{cota:singular:4}
\end{align}

with $C>0$.

The proof follows closely the ideas of Proposition <ref> and Proposition <ref> and the estimates can be shown mimicking the arguments used there. For the sake of completeness, we include the computations to derive bound (<ref>), being the $C^{1,\alpha}$ bound (<ref>) analogous. To show that the left hand side of (<ref>) exists, we first notice that

\begin{equation}
\p_{x} \left( \frac{e^{ix-y}}{1-e^{i(x-\eta)-y}} \right)=-\p_{\eta} \left( \frac{e^{ix-y}}{1-e^{i(x-\eta)-y}} \right).
\end{equation}

Using the fact that $\int_{\mathbb{S}^{1}} d\eta \ \p_{\eta}(\ldots)=0$, we rewrite (<ref>) as

\begin{equation}\label{int:sin:2:eps:2}
\widetilde{\Xi}_{\epsilon}(x)=- \int_{\epsilon}^{L}dy\int_{\mathbb{S}^{1}}d\eta \  \p_{\eta} \left( \frac{e^{ix-y}}{1-e^{i(x-\eta)-y}} \right) \left( H( \eta, y)-H(x,0) \right).
\end{equation}

Expanding the derivative and using Lemma <ref> (for the trivial case of $\Lambda\equiv 0$), where for $x\in [-\pi,\pi]$ we identify $\eta\in \SS^1$ with the symmetric interval $I_{x}=[x-\pi,x+\pi]$ we find that

\begin{align}
\abs{\widetilde{\Xi}_{\epsilon}(x)} &\leq C \norm{H}_{C^{\alpha}(\Omega)} \displaystyle\lim_{\epsilon\to 0^{+}} \int_{\epsilon}^{L}dy\int_{I_{x}}d\eta \frac{(\abs{x-\eta}^{\alpha}+y^{\alpha})}{(x-\eta)^{2}+y^{2}}\leq C\norm{H}_{C^{\alpha}(\Omega)}.
\end{align}

Since we are estimating the integrands of (<ref>) by integrable functions that are independent on $\epsilon$ it follows from Lebesgue Dominated Convergence Theorem that
$\lim_{\epsilon\to 0^{+}}\widetilde{\Xi}_{\epsilon}(x)$ exists. Therefore, the limit on the left hand side of (<ref>) exists and the function $\widetilde{\Xi}$ is well-defined.

To obtain the Hölder estimate, we compute the difference $\widetilde{\Xi}_{\epsilon}(x)-\widetilde{\Xi}_{\epsilon}(0)$ and divide the region of integration in the integrals $\int_{\epsilon}^{L} dy \int_{\mathbb{S}^{1}} d\eta (\ldots)$ into sets of the form

\begin{align*}
R^{\epsilon}_{\Omega,\leq}&=\{(\eta,y)\in \Omega:\displaystyle\max\{\abs{y},\abs{\eta}\}\leq 2\abs{x} \mbox{ and } \epsilon\leq y\leq L\}, \\
R^{\epsilon}_{\Omega,>}&=\{(\eta,y)\in \Omega:\displaystyle\max\{\abs{y},\abs{\eta}\}> 2\abs{x} \mbox{ and } \epsilon\leq y\leq L\},
\end{align*}

\begin{align*}
\widetilde{\Xi}_{\epsilon}(x)-\widetilde{\Xi}_{\epsilon}(0)&= \int_{R^{\epsilon}_{\Omega,\leq}}dy\ d\eta \bigg[\left(\frac{ie^{i(x-\eta)-y}}{(1-e^{i(x-\eta)-y})^2}\right) \left(H( \eta, y)-H(x,0)\right)\\
&\hspace{1cm} -\left(\frac{ie^{-i\eta-y}}{(1-e^{-i\eta-y})^2}\right) \left(H( \eta, y)-H(0,0)\right) \bigg] \\
&+  \int_{R^{\epsilon}_{\Omega,\leq}}dy\ d\eta  \bigg[\left(\frac{ie^{i(x-\eta)-y}}{(1-e^{i(x-\eta)-y})^2}\right) \left(H( \eta, y)-H(x,0)\right)\\
&\hspace{1cm} -\left(\frac{ie^{-i\eta-y}}{(1-e^{-i\eta-y})^2}\right) \left(H( \eta, y)-H(0,0)\right)=D_{1,\epsilon}+D_{2,\epsilon}.
\end{align*}

In the inner region, we have that using Lemma <ref> (again in the trivial case of $\Lambda\equiv 0$),

\begin{align}
\abs{D_{1,\epsilon}} &\leq C \norm{H}_{C^{\alpha}(\Omega)}\int_{R_{\Omega,\leq}}dy\ d\eta \bigg[ \frac{(\abs{x-\eta}^{\alpha}+y^{\alpha})}{(x-\eta)^{2}+y^{2}}+\frac{(\abs{\eta}^{\alpha}+y^{\alpha})}{\eta^{2}+y^{2}}\bigg] \nonumber  \\
&\leq C \norm{H}_{C^{\alpha}(\Omega)}\abs{x}^{\alpha} \label{estimate:M_1}.
\end{align}

In the outer region we rewrite the term $D_{2,\epsilon}$ as follows

\begin{align*}
D_{2,\epsilon}&=\int_{R^{\epsilon}_{\Omega,>}} dy \ d\eta \bigg[\left(\frac{ie^{i(x-\eta)-y}}{(1-e^{i(x-\eta)-y})^2}\right)-\left(\frac{ie^{-i\eta-y}}{(1-e^{-i\eta-y})^2}\right)\bigg] \left(H( \eta, y)-H(0,0)\right) \\
& \quad -\left(H(x, 0)-H(0,0)\right)\int_{R{\epsilon}_{\Omega,>}} dy \ d\eta \left(\frac{ie^{i(x-\eta)-y}}{(1-e^{i(x-\eta)-y})^2}\right)=D_{21,\epsilon}+D_{22,\epsilon}.
\end{align*}

The first integral $D_{21,\epsilon}$ can be bounded using the mean value theorem as

\begin{align}
\abs{D_{21,\epsilon}} &\leq C\abs{x} \norm{H}_{C^{\alpha}(\Omega)}\int_{R_{\Omega,>}} dy \ d\eta \bigg[  \frac{(\abs{\eta}^{\alpha}+\abs{y}^{\alpha})}{\left((x-\eta)^{2}+y^{2}\right)^{3/2}}\bigg] \leq C\abs{x}^{\alpha} \norm{H}_{C^{\alpha}(\Omega)}. \label{estimate:M_21}
\end{align}

To deal with $D_{22,\epsilon}$, we make the following change of variables $\widetilde{\eta}=x-\eta$ to find that

\begin{align*}
D_{22,\epsilon}&=-i\left(H(x, 0)-H(0,0)\right)\displaystyle\lim_{\epsilon\to 0^{+}}\bigg[\int_{-\frac{1}{2}}^{\frac{1}{2}} d\widetilde{\eta} \int_{2\abs{x}}^{L} dy \frac{e^{i\widetilde{\eta}-y}}{(1-e^{i\widetilde{\eta}-y})^2}+\int_{-\frac{1}{2}}^{-2\abs{x}}d\widetilde{\eta}  \int_{\epsilon}^{L} dy \frac{e^{i\widetilde{\eta}-y}}{(1-e^{i\widetilde{\eta}-y})^2} \\
&\hspace{6cm} +\int_{2\abs{x}}^{\frac{1}{2}}d\widetilde{\eta}  \int_{\epsilon}^{L} dy\frac{e^{i\widetilde{\eta}-y}}{(1-e^{i\widetilde{\eta}-y})^2}   \bigg] \\
&=-i\left(H(x, 0)-H(0,0)\right)\displaystyle\lim_{\epsilon\to 0^{+}}\bigg[ \int_{-\frac{1}{2}}^{\frac{1}{2}}d\widetilde{\eta} \left( \frac{e^{i\widetilde{\eta}}}{e^{i\widetilde{\eta}}-e^{L}}-\frac{e^{i\widetilde{\eta}}}{e^{i\widetilde{\eta}}-e^{2\abs{x}}} \right) +\int_{-\frac{1}{2}}^{-2\abs{x}}d\widetilde{\eta} \left( \frac{e^{i\widetilde{\eta}}}{e^{i\widetilde{\eta}}-e^{L}}-\frac{e^{i\widetilde{\eta}}}{e^{i\widetilde{\eta}}-e^{\epsilon}} \right) \\
&\hspace{6cm}+ \int_{2\abs{x}}^{\frac{1}{2}}d\widetilde{\eta} \left( \frac{e^{i\widetilde{\eta}}}{e^{i\widetilde{\eta}}-e^{L}}-\frac{e^{i\widetilde{\eta}}}{e^{i\widetilde{\eta}}-e^{\epsilon}} \right)  \bigg] \\
&=\left(H(x, 0)-H(0,0)\right) \displaystyle\lim_{\epsilon\to 0^{+}} \left( \bigg[\displaystyle\log(\frac{e^{2\abs{x}}-e^{i\widetilde{\eta}}}{e^{L}-e^{i\widetilde{\eta}}})\bigg]^{\widetilde{\eta}=\frac{1}{2}}_{\widetilde{\eta}=-\frac{1}{2}}+\bigg[\displaystyle\log(\frac{e^{\epsilon}-e^{i\widetilde{\eta}}}{e^{L}-e^{i\widetilde{\eta}}})\bigg]^{\widetilde{\eta}=-2\abs{x}}_{\widetilde{\eta}=-\frac{1}{2}}+\bigg[\displaystyle\log(\frac{e^{\epsilon}-e^{i\widetilde{\eta}}}{e^{L}-e^{i\widetilde{\eta}}})\bigg]^{\widetilde{\eta}=2\abs{x}}_{\widetilde{\eta}=\frac{1}{2}} \right)
\end{align*}

Therefore we have that

\begin{equation}\label{estimate:M_22}
\abs{D_{22,\epsilon}}\leq C  \abs{x}^{\alpha}\norm{H}_{C^{\alpha}(\Omega)}
\end{equation}

which together with (<ref>) yields

\begin{equation}\label{estimate:M_2:final}
\abs{D_{2,\epsilon}}\leq C  \abs{x}^{\alpha}\norm{H}_{C^{\alpha}(\Omega)}
\end{equation}

Combining estimates (<ref>)-(<ref>) we infer that

\begin{equation}
\abs{\widetilde{\Xi}(x)-\widetilde{\Xi}(0)} \leq \displaystyle\lim_{\epsilon\to 0^{+}} \abs{D_{1,\epsilon}+D_{2,\epsilon}} \leq C \abs{x}^{\alpha}\norm{H}_{C^{\alpha}(\Omega)}
\end{equation}

which shows the desired $C^\alpha$ semi-norm bound (<ref>).

To conclude this subsection, let us state the following result which will be needed later to show the contracting estimates in Subsection <ref>.

Let $H(\eta, y)\in C^{\alpha}(\Omega)$ and let $\Lambda^1,\Lambda^2$ satisfy Assumption <ref>. Then for any $x\in \mathbb{S}^{1}$ the following limit exists

\begin{equation} \label{int:sin:1:diff}
\displaystyle\lim_{\epsilon\to 0^{+}} \Xi^{d}_{\epsilon}(x)=\Xi^{d}(x),
\end{equation}

\begin{equation} \label{int:sin:diff:eps}
\Xi^{d}_{\epsilon}(x)=\displaystyle\lim_{\epsilon\to 0^{+}} \int_{\epsilon}^{L}dy\int_{\mathbb{S}^{1}}d\eta \  \p_{x} \left(\frac{e^{i(x-\eta)-y}y}{(1-e^{i(x-\eta)-y-i\Lambda^1(\eta,y)})(1-e^{i(x-\eta)-y-i\Lambda^2(\eta,y)})}\right) H( \eta, y), \ \epsilon>0.
\end{equation}

Moreover, we have that

\begin{align}\label{cota:singular:diff}
\norm{\Xi^{d}}_{C^{\alpha}(\mathbb{S}^{1})}&\leq C \norm{H}_{C^{\alpha}(\Omega)}
\end{align}

with $C>0$.

This result can be proved by a means an elementary adaption of the Proposition <ref>. Notice that the only difference between this result and Proposition <ref> is that instead of a single function $\Lambda$ appearing as a perturbation in the denominator we have
two different functions $\Lambda^1,\Lambda^2$ affecting the denominator in (<ref>).
Actually, Proposition <ref> is a particular case in which we take $\Lambda^1=0 $ and $\Lambda^2=\Lambda$. The proof of this Proposition <ref> can be showed around similar lines of Proposition <ref> just estimating the corrective terms due to $\Lambda^1,\Lambda^2$  as it was made in the proof of Proposition <ref>. We will not provide the details here.

§ THE A PRIORI ESTIMATES FOR THE OPERATORS LG.

In this section, we will provide the $C^\alpha$ and $C^{1,\alpha}$ Hölder estimates for the operators $\mathsf{T}_{1},\ldots,\mathsf{T}_{4}$ and the function $\mathsf{G}$. The key point towards the estimates relies on the $C^\alpha$ and $C^{1,\alpha}$ Hölder bounds showed in the previous Section <ref>.

In order to do so, let us introduce the following new assumption:

We assume that the function $\vartheta:\Omega \to \mathbb{S}^{1}$ belongs to $C^{1,\alpha}(\Omega)$ and also that there exists $\delta_1\in(0,\frac{1}{2})$ such that

\begin{equation*}
\norm{\vartheta}_{C^{1,\alpha}(\Omega)}\leq \delta_1.
\end{equation*}

The new function $\vartheta(\eta,y)$ will play the role of the function $\partial_{\xi} \Theta(X(\eta,y),y))$ in the operators $\mathsf{T}_{1},\ldots,\mathsf{T}_{4}$. However, using this notation reduces the length of the formulas.

In the following we will define the operators $\mathsf{T}_{1},\ldots,\mathsf{T}_{4}$ by means of certain integral expressions which represent operators from $C^\alpha$ to $C^\alpha$ and from $C^{1,\alpha}$ to $C^{1,\alpha}$. For the sake of simplicity, we will use the same symbol to denote these operators, in spite of the fact that they act in different spaces. The space on which they act will be clear in each particular case from the context.

Let Assumption <ref> and Assumption <ref> hold. Then for $j_0\in C^{1,\alpha}(\SS^1)$ we define the operator $\mathsf{T}_{1}$ as follows

\begin{equation}\label{def:T1:second}
\mathsf{T}_{1}[\Lambda,\vartheta]j_{0}(x)=-\frac{1}{2\pi} \displaystyle \lim_{\epsilon\to 0^{+}}\int_{\mathbb{S}^{1}} \mathfrak{G}_{1,\epsilon}(x-\eta,\eta)j_{0}(\eta) \ d\eta, \mbox{ for } x\in \SS^1,
\end{equation}

\begin{equation}\label{def:G1:second}
\mathfrak{G}_{1,\epsilon}(x,\eta)=\displaystyle\sum_{n=-\infty}^{n=\infty}\frac{n\sinh(nL)}{(\cosh(nL)-1)}e^{inx}\int_{\epsilon}^{L} e^{-\abs{n}y}  \left[ \frac{e^{-in\Lambda(\eta,y)}-1}{1+\vartheta(\eta,y)} \right]  dy.
\end{equation}

The limit in (<ref>) exists, moreover we have that

\begin{align}
\norm{\mathsf{T}_{1}[\Lambda,\vartheta]}_{\mathcal{L}(C^{\alpha}(\mathbb{S}^1)) }&\leq C\delta_{0}  \label{estimate:T1:Calpha}, \\
\norm{\mathsf{T}_{1}[\Lambda,\vartheta]}_{\mathcal{L}(C^{1,\alpha}(\mathbb{S}^1))}&\leq C\delta_{0} \label{estimate:T1:C1alpha}.
\end{align}

To show that the right hand side of (<ref>) exists, let us split the function $\mathfrak{G}_{1,\epsilon}(x,\eta)$ for $n>0$ and $n<0$. We will use the notation $\mathfrak{G}^{+}_{1,\epsilon}$ for $n>0$ and $\mathfrak{G}^{-}_{1,\epsilon}$ otherwise. Therefore, for $n>0$ we have that

\begin{equation}\label{operadorG1+:n:m}
\mathfrak{G}^{+}_{1,\epsilon}(x,\eta)=\displaystyle\sum_{n=1}^{n=\infty}n e^{inx}\int_{\epsilon}^{L} e^{-ny}  \left[ \frac{e^{-in\Lambda(\eta,y)}-1}{1+\vartheta(\eta,y)} \right]  dy+ \mathfrak{G}_{12,\epsilon}^{+}(x,\eta)
\end{equation}

\begin{equation}\label{operadorG1+:n:s}
\mathfrak{G}_{12,\epsilon}^{+}(x,\eta)=\displaystyle\sum_{n=1}^{n=\infty}\left(\frac{n\sinh(nL)}{(\cosh(nL)-1)}-n\right)e^{inx}\int_{\epsilon}^{L} e^{-ny}  \left[ \frac{e^{-in\Lambda(\eta,y)}-1}{1+\vartheta(\eta,y)} \right]  dy.
\end{equation}

On the one hand, calculating explicitly the summation via geometric series, we have that the first term in (<ref>) denoted by $ \mathfrak{G}^{+}_{11,\epsilon}$ is given by

\begin{align}
\mathfrak{G}^{+}_{11,\epsilon}(x,\eta)& = \frac{1}{i}\int_{\epsilon}^{L}\p_{x}\left(\frac{e^{ix-y-i\Lambda(\eta,y)}}{1-e^{ix-y-i\Lambda(\eta,y)}}-\frac{e^{ix-y}}{1-e^{ix-y}}\right)\frac{1}{1+\vartheta(\eta,y)}  \  dy \nonumber \\
&=\frac{1}{i}\int_{\epsilon}^{L} \frac{(e^{-i\Lambda(\eta,y)}-1)}{y} \p_{x} \left(\frac{e^{ix-y}y}{(1-e^{ix-y})(1-e^{ix-y-i\Lambda(\eta,y)})}\right)\frac{1}{1+\vartheta(\eta,y)} \ dy  \nonumber \\
&=\frac{1}{i}\int_{\epsilon}^{L} \mathsf{F}(x,\eta,y) \frac{(e^{-i\Lambda(\eta,y)}-1)}{y} \frac{1}{1+\vartheta(\eta,y)} \ dy  \label{G11+:formula}
\end{align}

\begin{equation*}
\mathsf{F}(x,\eta,y)=\p_{x} \left(\frac{e^{ix-y}y}{(1-e^{ix-y})(1-e^{ix-y-i\Lambda(\eta,y)})}\right).
\end{equation*}

Therefore, by means of (<ref>), we have that the operator $\mathsf{T}_{1}$ for $n>0$ given in (<ref>) can be written after changing the order of integration as

\begin{align}
\mathsf{T}^{+}_{1}[\Lambda,\vartheta]j_{0}(x)&=-\frac{1}{2\pi i} \displaystyle \lim_{\epsilon\to 0} \int_{\epsilon}^{L}dy \int_{\mathbb{S}^{1}} d\eta \ \mathsf{F}(x-\eta,\eta,y) \frac{(e^{-i\Lambda(\eta,y)}-1)}{y} \frac{1}{1+\vartheta(\eta,y)}j_{0}(\eta) \nonumber \\
& - \frac{1}{2\pi}\displaystyle \lim_{\epsilon\to 0}\int_{\mathbb{S}^{1}}\mathfrak{G}^{+}_{12,\epsilon}(x-\eta,\eta)j_{0}(\eta) \ d\eta= \mathsf{T}^{+}_{11}[\Lambda,\vartheta]j_{0}(x)+ \mathsf{T}^{+}_{12}[\Lambda,\vartheta]j_{0}(x) \label{T11:T12}
\end{align}

assuming that the limits exist. To show that the limit of the second integral $\mathsf{T}^{+}_{12}j_{0}(x)$ exists, we notice that the function $\mathfrak{G}^{+}_{12}(x,\eta)$ given in (<ref>) is a smooth function in $x$ and decays exponentially in $n$. Therefore, using the bound

\begin{equation*}
\abs{e^{-in\Lambda(\eta,y)}-1}\leq C \abs{n}\norm{\Lambda}_{L^\infty(\Omega)},
\end{equation*}

we find that

\begin{equation*}
\abs{\mathfrak{G}^{+}_{12,\epsilon}(x,\eta)}\leq C\norm{\frac{1}{1+\vartheta}}_{L^{\infty}(\Omega)}  \norm{\Lambda}_{L^{\infty}(\Omega)}\leq C, \mbox{ for } x\in \SS^1,\eta\in \SS^1.
\end{equation*}

Since the estimate is independent of $\epsilon$ using the Lebesgue Dominated Convergence Theorem, we can ensure that the limit exists and that the operator $\mathsf{T}^{+}_{12}[\Lambda,\vartheta]j_{0}(x)$ is well defined. Moreover, we also have the pointwise bound

\begin{equation}
\abs{\mathsf{T}^{+}_{12}[\Lambda,\vartheta]j_{0}(x)}\leq C\norm{\frac{1}{1+\vartheta}}_{L^{\infty}(\Omega)}  \norm{\Lambda}_{L^{\infty}(\Omega)}\norm{j_0}_{L^{\infty}(\SS^1)} \leq C\delta_{0}\norm{j_0}_{L^{\infty}(\SS^1)}, \mbox{ for } x\in \SS^1.
\end{equation}

On the other hand, to ensure that $\mathsf{T}^{+}_{11}j_{0}(x)$ in (<ref>) is well defined we make use of Proposition <ref> where

\begin{equation}\label{selection:H1}
H(\eta,y)= \frac{(e^{-i\Lambda(\eta,y)}-1)}{y} \frac{1}{1+\vartheta(\eta,y)} j_{0}(\eta).
\end{equation}

It is straightforward to check that choosing $H(\eta,y)$ as in (<ref>), we have that $H(\eta,y)\in C^{\alpha}(\Omega)$ since

\begin{align*}
\norm{H}_{C^{\alpha}(\Omega)} &\leq C \norm{\frac{1}{1+\vartheta}}_{C^{\alpha}(\Omega)}\norm{\Lambda}_{C^{1,\alpha}(\Omega)}\norm{j_0}_{C^{\alpha}(\mathbb{S}^1)} \leq C\delta_{0} \norm{j_0}_{C^{\alpha}(\mathbb{S}^1)}
\end{align*}

where in the last inequality we have used Assumption <ref>.
To show the $C^{\alpha}$ and $C^{1,\alpha}$ semi-norm estimate (<ref>) and (<ref>), we check that using estimate

\begin{align}
\abs{e^{i nx_{1}}-e^{in x_2}} &\leq C \abs{n}^\alpha \abs{x_{1}-x_{2}}^{\alpha}, \mbox{ for } \alpha\in(0,1) \mbox{ and } x_{1},x_{2}\in \mathbb{S}^{1} \label{formula:holder:exp}
\end{align}

we find that

\begin{align*}
\abs{\mathfrak{G}^{+}_{12,\epsilon}(x_{1},\eta)-\mathfrak{G}^{+}_{12,\epsilon}(x_{2},\eta)}&\leq C\norm{\frac{1}{1+\vartheta}}_{L^{\infty}(\Omega)}  \norm{\Lambda}_{L^{\infty}(\Omega)} \abs{x_{1}-x_{2}}^{\alpha} \\
\abs{\p_{x}\mathfrak{G}^{+}_{12,\epsilon}(x_{1},\eta)-\p_{x}\mathfrak{G}^{+}_{12,\epsilon}(x_{2},\eta)}&\leq C\norm{\frac{1}{1+\vartheta}}_{L^{\infty}(\Omega)}  \norm{\Lambda}_{L^{\infty}(\Omega)} \abs{x_{1}-x_{2}}^{\alpha}.
\end{align*}

\begin{align}
\norm{\mathsf{T}^{+}_{12}[\Lambda,\vartheta]}_{\mathcal{L}(L^{\infty}(\mathbb{S}^{1}),C^{\alpha}(\mathbb{S}^1))} &\leq C\norm{\frac{1}{1+\vartheta}}_{L^{\infty}(\Omega)}\norm{\Lambda}_{L^{\infty}(\Omega)}\leq C\delta_{0},\label{bounds:T1+sa} \\
\norm{\mathsf{T}^{+}_{12}[\Lambda,\vartheta]}_{\mathcal{L}(L^{\infty}(\mathbb{S}^{1}),C^{1,\alpha}(\mathbb{S}^1))} &\leq C\norm{\frac{1}{1+\vartheta}}_{L^{\infty}(\Omega)}\norm{\Lambda}_{L^{\infty}(\Omega)}\leq C\delta_{0}.\label{bounds:T1+s1a}
\end{align}

Similarly, as before a direct application of Proposition <ref> and Proposition <ref> with $H(\eta,y)$ as in (<ref>) yields

\begin{align}
\norm{\mathsf{T}^{+}_{11}[\Lambda,\vartheta]j_{0}}_{\mathcal{L}(C^{\alpha}(\mathbb{S}^1))}  &\leq C \norm{\frac{1}{1+\vartheta}}_{C^{\alpha}(\Omega)}\norm{\Lambda}_{C^{1,\alpha}(\Omega)} \leq C\delta_{0}, \label{estimate:T11+m:a} \\
\norm{\mathsf{T}^{+}_{11}[\Lambda,\vartheta]j_{0}}_{\mathcal{L}(C^{1,\alpha}(\mathbb{S}^1))}  &\leq C \norm{\frac{1}{1+\vartheta}}_{C^{1,\alpha}(\Omega)}\norm{\Lambda}_{C^{2,\alpha}(\Omega)} \leq C\delta_{0}, \label{estimate:T11+m:1a}
\end{align}

\begin{align*}
\norm{H}_{C^{\alpha}(\Omega)} &\leq C \norm{\frac{1}{1+\vartheta}}_{C^{\alpha}(\Omega)}\norm{\Lambda}_{C^{1,\alpha}(\Omega)}\norm{j_0}_{C^{\alpha}(\mathbb{S}^1)} \leq C \delta_{0}\norm{j_0}_{C^{\alpha}(\mathbb{S}^1)}, \\
\norm{H}_{C^{1,\alpha}(\Omega)} &\leq C \norm{\frac{1}{1+\vartheta}}_{C^{1,\alpha}(\Omega)}\norm{\Lambda}_{C^{2,\alpha}(\Omega)}\norm{j_0}_{C^{1,\alpha}(\mathbb{S}^1)} \leq C\delta_{0} \norm{j_0}_{C^{1,\alpha}(\mathbb{S}^1)}.
\end{align*}

Since the estimates for $n<0$ follow identically, we omit a detailed proof here.

Let Assumption <ref> hold. Then for $j_0\in C^{1,\alpha}(\SS^1)$ we define the operator $\mathsf{T}_{2}$ as follows

\begin{equation}\label{def:T2:second}
\mathsf{T}_{2}[\vartheta]j_{0}(x)=-\frac{1}{2\pi}\displaystyle\lim_{\epsilon\to 0^{+}}\int_{\mathbb{S}^{1}} \mathfrak{G}_{2,\epsilon}(x-\eta,\eta)j_{0}(\eta) \ d\eta,
\end{equation}

\begin{equation}\label{operadorG2:frak}
\mathfrak{G}_{2,\epsilon}(x,\eta)=\displaystyle\sum_{n=-\infty}^{n=\infty}\frac{n\sinh(nL)}{(\cosh(nL)-1)}e^{inx}\int_{\epsilon}^{L} e^{-\abs{n}y} \left[ \frac{\vartheta(\eta,y)}{1+\vartheta(\eta,y)} \right]  dy.
\end{equation}

The limit (<ref>) exists and in addition the following estimates hold

\begin{align}
\norm{\mathsf{T}_{2}[\vartheta]}_{\mathcal{L}(C^{\alpha}(\mathbb{S}^1)) }&\leq C\delta_{1}, \label{estimate:T2:Calpha}  \\
\norm{\mathsf{T}_{2}[\vartheta]}_{\mathcal{L}(C^{1,\alpha}(\mathbb{S}^1))}&\leq C\delta_{1}. \label{estimate:T2:C1alpha}
\end{align}

The proof follows the same lines of Proposition <ref>. We first, show that the right hand side of (<ref>) exists, and afterwards we provide the Hölder bounds (<ref>) and (<ref>). To that purpose, we split $\mathfrak{G}_{2,\epsilon}(x,\eta)$ for $n>0$ and $n<0$ and just show the estimates for $n>0$, being the case for $n<0$ identical. Therefore, for $n>0$ we have that

\begin{equation}\label{operadorG2+:n:m}
\mathfrak{G}^{+}_{2,\epsilon}(x,\eta)=\sum_{n=1}^{n=\infty}n e^{inx}\int_{\epsilon}^{L} e^{-ny} \left[ \frac{\vartheta(\eta,y)}{1+\vartheta(\eta,y)} \right]  dy+ \mathfrak{G}_{22,\epsilon}^{+}(x,\eta)
\end{equation}

\begin{equation}\label{operadorG2+:n:s}
\mathfrak{G}_{22,\epsilon}^{+}(x,\eta)=\sum_{n=1}^{n=\infty}\left(\frac{n\sinh(nL)}{(\cosh(nL)-1)}-n\right)e^{inx}\int_{\epsilon}^{L} e^{-ny} \left[ \frac{\vartheta(\eta,y)}{1+\vartheta(\eta,y)} \right]  dy.
\end{equation}

Computing the sum in the first term in (<ref>) we obtain

\begin{equation}
\mathfrak{G}^{+}_{21,\epsilon}(x,\eta)= \frac{1}{i}\int_{\epsilon}^{L}\p_{x}\left(\frac{1}{1-e^{ix-y}}\right)\left[ \frac{\vartheta(\eta,y)}{1+\vartheta(\eta,y)} \right]  dy,   \label{finalG+2}
\end{equation}

and changing the order of integration the operator $\mathsf{T}_{2}[\vartheta]j_{0}$ in (<ref>) can be written for $n>0$ as

\begin{align*}
\mathsf{T}^{+}_{2}[\vartheta]j_{0}(x)&=-\frac{1}{2\pi i}\displaystyle\lim_{\epsilon\to 0} \int_{\epsilon}^{L}dy \int_{\mathbb{S}^{1}}d\eta \  \p_{x}\left(\frac{e^{i(x-\eta)-y}}{1-e^{i(x-\eta)-y}}\right) \left[ \frac{\vartheta(\eta,y)}{1+\vartheta(\eta,y)} \right]  j_{0}(\eta) \\
& \quad - \frac{1}{2\pi}\displaystyle\lim_{\epsilon\to 0}\int_{\mathbb{S}^{1}}\mathfrak{G}^{+}_{22,\epsilon}(x-\eta,\eta)j_{0}(\eta) \ d\eta:=\mathsf{T}^{+}_{21}[\vartheta]j_0+\mathsf{T}^{+}_{22}[\vartheta]j_0.
\end{align*}

As before, the remainder smoothing term $\mathfrak{G}^{+}_{22}(x,\eta)$ (<ref>) is a smooth function in $x$ and in particular

\begin{equation}\label{L:inftyG22}
\abs{\mathfrak{G}^{+}_{22,\epsilon}(x,\eta)}\leq C\norm{\frac{\vartheta}{1+\vartheta} }_{L^{\infty}(\Omega)}\leq C\delta_{1}, \mbox{ for } x\in\SS^1, \eta\in \SS^1.
\end{equation}

Hence, since estimate  (<ref>) is independent of $\epsilon$, Lebesgue Dominated Convergence Theorem shows that the limit exists and that the associated operator $\mathsf{T}^{+}_{22}[\vartheta]j_0$ is well-defined. Choosing

\begin{equation}\label{selection:H2}
\end{equation}

and noticing that

\begin{equation}
\norm{H}_{C^{\alpha}(\Omega)} \leq C \norm{\frac{\vartheta}{1+\vartheta}}_{C^{\alpha}(\Omega)}\norm{j_0}_{C^{\alpha}(\mathbb{S}^1)} \leq C\delta_{1} \norm{j_0}_{C^{\alpha}(\mathbb{S}^1)}.
\end{equation}

we can apply Lemma <ref> to obtain that  $\mathsf{T}^{+}_{21}[\vartheta]j_0$ is well defined. We are left to show the $C^\alpha$ and $C^{1,\alpha}$ semi-norm estimates (<ref>)-(<ref>). Making use of the bound

\begin{align*}
\abs{e^{i nx_{1}}-e^{in x_2}} &\leq C \abs{n}^\alpha \abs{x_{1}-x_{2}}^{\alpha}, \mbox{ for } \alpha\in(0,1) \mbox{ and } x_{1},x_{2}\in \mathbb{S}^{1}.
\end{align*}

we infer that for $\mathfrak{G}^{+}_{22}$ defined in (<ref>)

\begin{align*}
\abs{\mathfrak{G}^{+}_{22,\epsilon}(x_{1},\eta)-\mathfrak{G}^{+}_{22,\epsilon}(x_{2},\eta)}&\leq C\norm{\frac{\vartheta}{1+\vartheta} }_{L^{\infty}(\Omega)} \abs{x_{1}-x_{2}}^{\alpha}, \\
\abs{\p_{x}\mathfrak{G}^{+}_{22,\epsilon}(x_{1},\eta)-\p_{x}\mathfrak{G}^{+}_{22,\epsilon}(x_{2},\eta)}&\leq C\norm{\frac{\vartheta}{1+\vartheta} }_{L^{\infty}(\Omega)} \abs{x_{1}-x_{2}}^{\alpha}
\end{align*}

holds. Thus, due to Assumption <ref> we obtain

\begin{align}
\norm{\mathsf{T}^{+}_{22}[\vartheta]}_{\mathcal{L}(L^{\infty}(\mathbb{S}^{1}),C^{\alpha}(\mathbb{S}^1))} &\leq C\norm{\frac{\vartheta}{1+\vartheta}}_{L^{\infty}(\Omega)}\leq C\delta_{1},\label{bounds:T2+s:a} \\
\norm{\mathsf{T}^{+}_{22}[\vartheta]}_{\mathcal{L}(L^{\infty}(\mathbb{S}^{1}),C^{1,\alpha}(\mathbb{S}^1))} &\leq C\norm{\frac{\vartheta}{1+\vartheta}}_{L^{\infty}(\Omega)}\leq C\delta_{1}. \label{bounds:T2+s:1:a}
\end{align}

To deal with the most singular operator $\mathsf{T}_{21}^{+}[\vartheta]$ we make use of Proposition <ref>. Indeed, applying estimates (<ref>) and (<ref>) to the function $H(\eta,y)$ as in (<ref>) we find that

\begin{align}
\norm{\mathsf{T}^{+}_{21}[\vartheta]}_{\mathcal{L}(C^{\alpha}(\mathbb{S}^1))}  &\leq C \norm{\frac{\vartheta}{1+\vartheta}}_{C^{\alpha}(\Omega)} \leq C\delta_{1}, \label{estimate:T21+:a} \\
\norm{\mathsf{T}^{+}_{21}[\vartheta]}_{\mathcal{L}(C^{1,\alpha}(\mathbb{S}^1))}  &\leq C \norm{\frac{\vartheta}{1+\vartheta}}_{C^{1,\alpha}(\Omega)}\leq C\delta_{1},  \label{estimate:T21+:1a}
\end{align}

\begin{align*}
\norm{H}_{C^{\alpha}(\Omega)} &\leq C \norm{\frac{\vartheta}{1+\vartheta}}_{C^{\alpha}(\Omega)}\norm{j_0}_{C^{\alpha}(\mathbb{S}^1)} \leq C\delta_{1} \norm{j_0}_{C^{\alpha}(\mathbb{S}^1)}, \\
\norm{H}_{C^{1,\alpha}(\Omega)} &\leq C \norm{\frac{\vartheta}{1+\vartheta}}_{C^{1,\alpha}(\Omega)}\norm{j_0}_{C^{1,\alpha}(\mathbb{S}^1)} \leq C\delta_{1} \norm{j_0}_{C^{1,\alpha}(\mathbb{S}^1)}.
\end{align*}

Therefore, using (<ref>) and (<ref>) we conclude that

\begin{equation}\label{estimate:T2:alpha}
\norm{\mathsf{T}^{+}_{2}[\vartheta}_{\mathcal{L}(C^{\alpha}(\mathbb{S}^1))} \leq C\delta_{1},
\end{equation}

and similarly invoking (<ref>) and (<ref>) we find that

\begin{equation}\label{estimate:T2:1alpha}
\norm{\mathsf{T}^{+}_{2}[\vartheta]}_{\mathcal{L}(C^{1,\alpha}(\mathbb{S}^1))}  \leq C\delta_{1}.
\end{equation}

Let Assumption <ref> and Assumption <ref> hold. Then for $j_0\in C^{1,\alpha}(\SS^1)$ we define the operators $\mathsf{T}_{3}$ and $\mathsf{T}_{4}$ as follows

\begin{align}
\mathsf{T}_{3}[\Lambda,\vartheta]j_{0}(x)=-\frac{1}{2\pi}\displaystyle\int_{\mathbb{S}^{1}} \mathfrak{G}_{3}(x-\eta,\eta)j_{0}(\eta) \ d\eta,  \label{def:T3:sec}\\
\mathsf{T}_{4}[\Lambda,\vartheta]j_{0}(x)=-\frac{1}{2\pi}
\int_{\mathbb{S}^{1}} \mathfrak{G}_{4}(x-\eta,\eta)j_{0}(\eta) \ d\eta.\label{def:T4:sec}
\end{align}

\begin{align}
\mathfrak{G}_{3}(x,\eta)&=\displaystyle\sum_{n=-\infty}^{n=\infty}\frac{n\sinh(nL)}{(\cosh(nL)-1)}e^{inx}\int_{0}^{L} M(n,y) \frac{(e^{-in\Lambda(\eta,y)}-1)}{1+\vartheta(\eta,y)}  dy,  \\
\mathfrak{G}_{4}(x,\eta)&=\displaystyle\sum_{n=-\infty}^{n=\infty} \frac{n\sinh(nL)}{(\cosh(nL)-1)} e^{inx} \int_{0}^{L} M(n,y)  \frac{\vartheta(\eta,y)}{1+\vartheta(\eta,y)} dy,
\end{align}

\begin{equation}\label{function:M}
M(n,y)=\frac{e^{-2nL}\left(e^{ny}-e^{-ny} \right)}{(1-e^{-2nL})}.
\end{equation}

Then we have that

\begin{align}
\norm{\mathsf{T}_{3}[\Lambda,\vartheta]}_{\mathcal{L}(C^{\alpha}(\mathbb{S}^1),L^\infty(\SS^1))}\leq C\delta_{0},  \label{estimate:prop:T3:Calpha} \\
\norm{\mathsf{T}_{4}[\Lambda,\vartheta]}_{\mathcal{L}(C^{\alpha}(\mathbb{S}^1),L^\infty(\SS^1))}\leq C\delta_{1},\label{estimate:prop:T4:Calpha}
\end{align}

\begin{align}
\norm{\mathsf{T}_{3}[\Lambda,\vartheta]}_{\mathcal{L}(C^{1,\alpha}(\mathbb{S}^1),L^\infty(\SS^1))}\leq C, \label{estimate:prop:T3:C1alpha}  \\
\norm{\mathsf{T}_{4}[\Lambda,\vartheta]}_{\mathcal{L}(C^{1,\alpha}(\mathbb{S}^1),L^\infty(\SS^1))}\leq C, \label{estimate:prop:T4:C1alpha}
\end{align}

As we can see from the estimates the operators, $\mathsf{T}_{3}$ and $\mathsf{T}_{4}$ are smoothing operators that transform functions from $L^\infty(\SS^1)$ to functions in $C^{1,\alpha}(\SS^1)$. Moreover, the series in $\mathfrak{G}_{3}$ and $\mathfrak{G}_{4}$ are uniformly convergent and therefore we do not need to define the operators $\mathsf{T}_{3}$ and $\mathsf{T}_{4}$ in (<ref>), (<ref>) as a limit since the integrals are well defined.

Notice that $M(n,y)$ defined in (<ref>) is a smooth function in $y$ and decays exponential in $n$. Therefore, the simple bounds

\begin{align*}
\abs{M(n,y)}\leq C e^{-nL}, \quad \abs{e^{-in\Lambda(\eta,y)}-1}\leq C \abs{n} \norm{\Lambda}_{L^{\infty}(\Omega)},
\end{align*}

\begin{align*}
\abs{\mathfrak{G}_{3}(x,\eta)}&\leq C \norm{\Lambda}_{L^{\infty}(\Omega)} \norm{\frac{1}{1+\vartheta} }_{L^{\infty}(\Omega)}  \mbox{ and }
\abs{\mathfrak{G}_{4}(x,\eta)}\leq C \norm{\frac{\vartheta}{1+\vartheta}}_{L^{\infty}}, \mbox{ for } x\in \SS^1,\eta\in \SS^1.
\end{align*}

Moreover using that $\abs{e^{i nx_{1}}-e^{in x_2}} \leq C \abs{n}^\alpha \abs{x_{1}-x_{2}}^{\alpha}$ for $\alpha\in(0,1)$ and $x_{1},x_{2}\in \mathbb{S}^{1}$ we have that for $\eta\in \SS^1$

\begin{align*}
\abs{\mathfrak{G}_{3}(x_{1},\eta)-\mathfrak{G}_{3}(x_{2},\eta)}&\leq C \norm{\Lambda}_{L^{\infty}(\Omega)}\norm{\frac{1}{1+\vartheta}}_{L^{\infty}(\Omega)}\abs{x_{1}-x_{2}}^{\alpha}\leq C\delta_{0}\abs{x_{1}-x_{2}}^{\alpha}, \\
\abs{\mathfrak{G}_{4}(x_{1},\eta)-\mathfrak{G}_{4}(x_{2},\eta)}&\leq C \norm{\frac{\vartheta}{1+\vartheta}}_{L^{\infty}(\Omega)}\abs{x_{1}-x_{2}}^{\alpha}\leq C\delta_{1}\abs{x_{1}-x_{2}}^{\alpha}, \\
\abs{\partial_{x}\mathfrak{G}_{3}(x_{1},\eta)-\partial_{x}\mathfrak{G}_{3}(x_{2},\eta)}&\leq C \norm{\Lambda}_{L^{\infty}(\Omega)} \norm{\frac{1}{1+\vartheta}}_{L^{\infty}(\Omega)}\abs{x_{1}-x_{2}}^{\alpha}\leq C\delta_{0}\abs{x_{1}-x_{2}}^{\alpha}, \\
\abs{\partial_{x}\mathfrak{G}_{4}(x_{1},\eta)-\partial_{x}\mathfrak{G}_{4}(x_{2},\eta)}&\leq C \norm{\frac{\vartheta}{1+\vartheta}}_{L^{\infty}(\Omega)}\abs{x_{1}-x_{2}}^{\alpha}\leq C\delta_{1}\abs{x_{1}-x_{2}}^{\alpha}.
\end{align*}

Therefore using the above pointwise estimates we conclude that the Hölder semi-norm of $\mathsf{T}_{3}$, $\mathsf{T}_{4}$ defined in (<ref>)-(<ref>) is bounded as

\begin{align*}
\norm{\mathsf{T}_{3}[\Lambda,\vartheta]}_{\mathcal{L}(L^\infty(\SS^1),C^{\alpha}(\mathbb{S}^1))}\leq  C\delta_{0},  \\
\norm{\mathsf{T}_{4}[\Lambda,\vartheta]}_{\mathcal{L}(L^\infty(\SS^1),C^{\alpha}(\mathbb{S}^1))}\leq C\delta_{1}, \\
\norm{\mathsf{T}_{3}[\Lambda,\vartheta]}_{\mathcal{L}(L^\infty(\SS^1),C^{1,\alpha}(\mathbb{S}^1))}\leq C\delta_{0},  \\
\norm{\mathsf{T}_{4}[\Lambda,\vartheta]}_{\mathcal{L}(L^\infty(\SS^1),C^{1,\alpha}(\mathbb{S}^1))}\leq C\delta_{1},
\end{align*}

concluding the proof.

§.§ Estimates for the differences of Lg

In this subsection, we will derive estimates for the difference operators. This will be needed in order to show the contraction estimate in the general fixed point argument (see Section <ref>) in the lower order Hölder space $C^{\alpha}$. The proof follows the same lines as in the previous subsection  but some extra computations are needed in order to get the desired contraction type estimate. More precisely, our first result reads

Let $\Lambda^1,\Lambda^2$ satisfy Assumption <ref> and $\vartheta^1,\vartheta^2$ satisfy Assumption <ref>. Let $j_{0}\in C^{\alpha}(\SS^1)$ and define

\begin{align}
\mathsf{T}^{d}_{1}j_{0}(x)=\left(\mathsf{T}_{1}[\Lambda^1,\vartheta^1]-\mathsf{T}_{1}[\Lambda^2,\vartheta^2]\right)j_{0}(x), \label{difference:operator:T1}
\end{align}

where $\mathsf{T}_{1}[\cdot,\cdot]$ is given in (<ref>). Then we have that

\begin{align}
\norm{\mathsf{T}_{1}^{d}}_{\mathcal{L}(C^{\alpha}(\mathbb{S}^1))}\leq C\left( \norm{\Lambda^1-\Lambda^2}_{C^{1,\alpha}(\Omega)}+ \norm{\vartheta^2-\vartheta^1}_{C^{\alpha}(\Omega)}\right). \label{estimate:difference:T1:Calpha}
\end{align}

Let us first introduce some notation. We define by $\mathfrak{G}^{d}_{1,\epsilon}$ the difference function

\begin{align}
\mathfrak{G}^{d}_{1,\epsilon}(x,\eta)=\mathfrak{G}_{1,\epsilon}[\Lambda^1,\vartheta^1](x,\eta)-\mathfrak{G}_{1,\epsilon}[\Lambda^2,\vartheta^2](x,\eta)  \label{difference:fucntion:G1}
\end{align}

where $\mathfrak{G}_{1,\epsilon}[\cdot,\cdot]$ is given in (<ref>). Following the lines of Proposition <ref>, we split the function (<ref>) for $n>0$ and $n<0$. Similarly as in the previous propositions, we will use the notation $\mathfrak{G}^{d,+}_{1,\epsilon}$ for $n>0$ and $\mathfrak{G}^{d,-}_{1,\epsilon}$ otherwise.Then, for $n>0$ we find that

\begin{align}
\mathfrak{G}^{d,+}_{1,\epsilon}(x,\eta)&=\displaystyle\sum_{n=1}^{n=\infty}n e^{inx}\int_{\epsilon}^{L} e^{-ny}  \mathsf{L}^{n}(\eta,y) dy \nonumber + \displaystyle\sum_{n=1}^{n=\infty}\left(\frac{n\sinh(nL)}{(\cosh(nL)-1)}-n\right)e^{inx}\int_{\epsilon}^{L} e^{-ny}   \mathsf{L}^{n}(\eta,y)  dy \nonumber \\
&=\mathfrak{G}^{d,+}_{11,\epsilon}(x,\eta)+\mathfrak{G}^{d,+}_{12,\epsilon}(x,\eta) \label{splitting:difference:T1}
\end{align}

\begin{equation}
\mathsf{L}^{n}(\eta,y) =\left[ \frac{e^{-in\Lambda^{1}(\eta,y)}-1}{1+\vartheta^1(\eta,y)} - \frac{e^{-in\Lambda^2(\eta,y)}-1}{1+\vartheta^2(\eta,y)} \right].
\end{equation}

It will be convenient to decompose $\mathsf{L}^{n}(\eta,y)$ as follows

\begin{align}
\mathsf{L}^{n}(\eta,y) &=\frac{1}{1+\vartheta^1(\eta,y)}\left(e^{-in\Lambda^1(\eta,y)}-e^{-in\Lambda^2(\eta,y)}\right) + \left(e^{-in\Lambda^2(\eta,y)}-1\right) \left(\frac{1}{1+\vartheta^1(\eta,y)}-\frac{1}{1+\vartheta^2(\eta,y)}\right) \nonumber \\
&= \mathsf{L}^{n}_{1}+\mathsf{L}^{n}_{2}. \label{definition:L1:L2}
\end{align}

Plugging the decomposition $\mathsf{L}^{n}=\mathsf{L}^{n}_1+\mathsf{L}^{n}_2$  into (<ref>) and calculating the sum in $n$ for the first term $\mathfrak{G}^{d}_{11,\epsilon}(x,\eta)$ we obtain

\begin{align*}
\mathfrak{G}^{d,+}_{11,\epsilon}(x,\eta)&= \frac{1}{i}\int_{\epsilon}^{L}\p_{x}\left(\frac{e^{ix-y-i\Lambda^1(\eta,y)}}{1-e^{ix-y-i\Lambda^1(\eta,y)}}-\frac{e^{ix-y-i\Lambda^2(\eta,y)}}{1-e^{ix-y-i\Lambda^2(\eta,y)}}\right)\frac{1}{1+\vartheta^1(\eta,y)}  \  dy \nonumber \\
&\quad + \frac{1}{i}\int_{\epsilon}^{L}\p_{x}\left(\frac{e^{ix-y-i\Lambda^2(\eta,y)}}{1-e^{ix-y-i\Lambda^2(\eta,y)}}-\frac{e^{ix-y}}{1-e^{ix-y}}\right)\times \left(\frac{1}{1+\vartheta^1(\eta,y)}-\frac{1}{1+\vartheta^2(\eta,y)}\right)  dy \\
&= \mathfrak{G}^{d,+}_{111,\epsilon}(x,\eta)+\mathfrak{G}^{d,+}_{112,\epsilon}(x,\eta).
\end{align*}

Denoting by

\begin{align*}
\mathsf{F}_{1}(x,\eta,y)&= \p_{x}\left(\frac{ye^{ix-y}}{\left(1-e^{ix-y-i\Lambda^1(\eta,y)}\right)\left(1-e^{ix-y-i\Lambda^2(\eta,y)}\right)} \right), \\
\mathsf{F}_{2}(x,\eta,y)&=\p_{x}\left(\frac{ye^{ix-y}}{\left(1-e^{ix-y-i\Lambda^2(\eta,y)}\right)\left(1-e^{ix-y}\right)} \right),
\end{align*}

we obtain that

\begin{align*}
\mathfrak{G}^{d,+}_{111,\epsilon}(x,\eta)= \frac{1}{i}\int_{\epsilon}^{L} \frac{\mathsf{L}^{1}_{1}(\eta,y)}{y}\mathsf{F}_{1}(x,\eta,y) dy \mbox{ and } \mathfrak{G}^{d}_{112,\epsilon}(x,\eta)= \frac{1}{i}\int_{\epsilon}^{L} \frac{\mathsf{L}^{1}_{2}(\eta,y)}{y}\mathsf{F}_{2}(x,\eta,y)dy.
\end{align*}

Therefore, recalling the definition of the operator $\mathsf{T}^{d}_{1}$ in (<ref>) and the previous computations we find, after changing the order of integration that

\begin{align*}
\mathsf{T}^{d,+}_{1}j_{0}(x)&= \mathsf{T}^{d,+}_{11}j_{0}(x)+ \mathsf{T}^{d,+}_{12}j_{0}(x)+ \mathsf{T}^{d,+}_{13}j_{0}(x)
\end{align*}

\begin{align}
\mathsf{T}^{d,+}_{11}j_{0}(x)&=  -\frac{1}{2\pi i} \displaystyle \lim_{\epsilon\to 0} \int_{\epsilon}^{L}dy \int_{\mathbb{S}^{1}} d\eta \frac{\mathsf{L}^{1}_{1}(\eta,y)}{y} \mathsf{F}_{1}(x-\eta,\eta,y) j_{0}(\eta), \label{def:Td:11} \\
\mathsf{T}^{d,+}_{12}j_{0}(x)&=  -\frac{1}{2\pi i} \displaystyle \lim_{\epsilon\to 0} \int_{\epsilon}^{L}dy \int_{\mathbb{S}^{1}} d\eta \frac{\mathsf{L}^{1}_{2}(\eta,y)}{y}\mathsf{F}_{2}(x-\eta,\eta,y) j_{0}(\eta), \label{def:Td:12} \\
\mathsf{T}^{d,+}_{13}j_{0}(x)&= -\frac{1}{2\pi} \displaystyle \lim_{\epsilon\to 0}\int_{\mathbb{S}^{1}} \mathfrak{G}^{d,+}_{12,\epsilon}(x-\eta,\eta)j_{0}(\eta) \ d\eta. \label{def:Td:13}
\end{align}

We start estimating the third term $\mathsf{T}^{d}_{13}j_{0}(x)$. Notice that the function $\mathfrak{G}^{d}_{12,\epsilon}$ in (<ref>) is a smooth function in $x$ due to the exponential decay in $n$ of the terms that define the function (<ref>). Furthermore, we also have that the functions $\mathsf{L}^{n}_{1}(\eta,y),\mathsf{L}^{n}_{2}(\eta,y)$ defined in (<ref>) are bounded by

\begin{align}
\abs{\mathsf{L}^{n}_{1}(\eta,y)}&\leq C \norm{\frac{1}{1+\vartheta^1}}_{L^{\infty}(\Omega)} \abs{n} \norm{\Lambda^1-\Lambda^2}_{L^\infty(\Omega)}, \mbox{ for } (\eta,y)\in\Omega, \label{estimate:L1} \\
\abs{\mathsf{L}^{n}_{2}(\eta,y)}&\leq C \norm{\frac{\vartheta^2-\vartheta^1}{(1+\vartheta^2)(1+\vartheta^1)}}_{L^{\infty}(\Omega)} \abs{n}\norm{\Lambda^2}_{L^\infty(\Omega)}, \mbox{ for } (\eta,y)\in\Omega. \label{estimate:L2}
\end{align}

and thus

\begin{align}
\abs{\mathfrak{G}^{d,+}_{12,\epsilon}(x,\eta)}&\leq C \left(\norm{\frac{1}{1+\vartheta^1}}_{L^{\infty}(\Omega)}\norm{\Lambda^1-\Lambda^2}_{L^\infty(\Omega)}+\norm{\frac{\vartheta^2-\vartheta^1}{(1+\vartheta^2)(1+\vartheta^1)}}_{L^{\infty}(\Omega)}\norm{\Lambda^2}_{L^\infty(\Omega)} \right)\nonumber \\
&\leq C \norm{\Lambda^1-\Lambda^2}_{L^\infty(\Omega)}+ \norm{\vartheta^2-\vartheta^1}_{L^{\infty}(\Omega)}, \mbox{ for } x\in\SS^1,\eta\in\SS^1. \label{estimate:G12d:Linfty}
\end{align}

Since the estimate (<ref>) is independent of $\epsilon$, Lebesgue Dominated Convergence Theorem shows that the limit exists and that the associated operator $\mathsf{T}^{d}_{13}j_0$ in (<ref>) is well-defined. Moreover, we have the pointwise estimate

\begin{equation}
\abs{\mathsf{T}^{d,+}_{13}j_{0}(x)}\leq C  \left(\norm{\Lambda^1-\Lambda^2}_{L^\infty(\Omega)}+ \norm{\vartheta^2-\vartheta^1}_{L^{\infty}(\Omega)}\right)\norm{j_0}_{L^{\infty}(\Omega)}, \quad x\in\SS^1. \label{estimate:Td:13:Linf}
\end{equation}

The $C^{\alpha}$ semi-norm estimate for $\mathsf{T}^{d,+}_{13}j_0$ follows directly by using estimate (<ref>) which combined with (<ref>) shows

\begin{align}
\norm{\mathsf{T}^{d,+}_{13}}_{\mathcal{L}(L^\infty(\SS^1),C^{\alpha}(\mathbb{S}^1))}\leq C\left(\norm{\Lambda^1-\Lambda^2}_{L^\infty(\Omega)}+\norm{\vartheta^2-\vartheta^1}_{L^{\infty}(\Omega)}\right).  \label{bound:T+13:diff}
\end{align}

To deal with $\mathsf{T}^{d,+}_{11}j_0$ in (<ref>) and $\mathsf{T}^{d,+}_{12}j_0$  in (<ref>), we can invoke Proposition <ref> and Proposition <ref>, respectively. Notice that a consequence of the before mentioned lemmas is that the limits of the integrals (<ref>) and (<ref>) are well-defined.

To that purpose, we first notice using the expression (<ref>) that

\begin{align}
\frac{\mathsf{L}^{1}_{1}(\eta,y)}{y}=\frac{e^{-i\Lambda^{1}(\eta,y)}}{1+\vartheta^1(\eta,y)}\frac{\left(1-e^{-(i\Lambda^{2}(\eta,y)-i\Lambda^{1}(\eta,y))} \right)}{y}=\frac{e^{-i\Lambda^{1}(\eta,y)}}{1+\vartheta^1(\eta,y)}i\int_{0}^{1}B(\eta,y)e^{iysB(\eta,y)} ds \label{exp:L1:integral}
\end{align}

since by Assumption <ref> we can write $\Lambda^{2}(\eta,y)-\Lambda^{1}(\eta,y)=yB(\eta,y)$ for $B(\eta,y)\in C^{1,\alpha}(\Omega).$
Thus, choosing

\begin{equation*}
H(\eta,y)=\frac{\mathsf{L}^{1}_{1}(\eta,y)}{y} j_{0}(\eta)\in C^{\alpha}(\Omega)
\end{equation*}

we apply estimate (<ref>) in Proposition <ref> combined with (<ref>) to obtain that

\begin{equation}\label{estimate:Td11}
\norm{\mathsf{T}^{d,+}_{11}}_{\mathcal{L}(C^{\alpha}(\mathbb{S}^1))}\leq  C \norm{\Lambda^1-\Lambda^2}_{C^{1,\alpha}(\Omega)}.
\end{equation}

On the other hand, proceeding in a similar way and choosing

\begin{equation*}
H(\eta,y)=\frac{\mathsf{L}^{1}_{2}(\eta,y)}{y} j_{0}(\eta)\in C^{\alpha}(\Omega)
\end{equation*}

we use Lemma <ref> to show that

\begin{equation}\label{estimate:Td12}
\norm{\mathsf{T}^{d,+}_{12}}_{\mathcal{L}(C^{\alpha}(\mathbb{S}^1))}\leq C \norm{\vartheta^2-\vartheta^1}_{C^{\alpha}(\Omega)}.
\end{equation}

Collecting bounds (<ref>)-(<ref>) we deduce estimate (<ref>) for $n>0$. Repeating the same arguments for $n<0$ concludes the proof.

Let $\vartheta^1,\vartheta^2$ satisfy Assumption <ref>. Let $j_0\in C^{\alpha}(\SS^1)$ and define

\begin{align}
\mathsf{T}^{d}_{2}j_{0}(x)=\left(\mathsf{T}_{2}[\vartheta^1]-\mathsf{T}_{2}[\vartheta^2]\right)j_{0}(x), \label{difference:operator:T2}
\end{align}

where $\mathsf{T}_{2}[\cdot]$ is given in (<ref>). Then we have that

\begin{align}
\norm{\mathsf{T}_{2}^{d}}_{\mathcal{L}(C^{\alpha}(\mathbb{S}^1))}\leq C \norm{\vartheta^2-\vartheta^1}_{C^{\alpha}(\Omega)}. \label{estimate:difference:T2:Calpha}
\end{align}

Following the same notation as in the last proposition and the arguments of the proof of Proposition <ref>, we introduce the difference function $\mathfrak{G}^{d}_{2,\epsilon}$

\begin{align}
\mathfrak{G}^{d}_{2,\epsilon}(x,\eta)=\mathfrak{G}_{2,\epsilon}[\vartheta^1](x,\eta)-\mathfrak{G}_{2,\epsilon}[\vartheta^2](x,\eta).  \label{difference:fucntion:G2}
\end{align}

where $\mathfrak{G}_{2,\epsilon}[\cdot]$ is given in (<ref>). The using the same notation by means of the superscripts $\pm$ as before, we find that for $n>0$ we have that

\begin{align}
\mathfrak{G}^{d,+}_{2,\epsilon}(x,\eta)&=\displaystyle\sum_{n=1}^{n=\infty}n e^{inx}\int_{\epsilon}^{L} e^{-ny}  \mathsf{L}_{3}(\eta,y) dy \nonumber + \displaystyle\sum_{n=1}^{n=\infty}\left(\frac{n\sinh(nL)}{(\cosh(nL)-1)}-n\right)e^{inx}\int_{\epsilon}^{L} e^{-ny}  \mathsf{L}_{3}(\eta,y)  dy \nonumber \\
&=\mathfrak{G}^{d,+}_{21,\epsilon}(x,\eta)+\mathfrak{G}^{d,+}_{22,\epsilon}(x,\eta) \label{splitting:difference:T2}
\end{align}

\begin{equation*}
\mathsf{L}_3(\eta,y) =\frac{1}{1+\vartheta^1(\eta,y)} (\vartheta^1(\eta,y)-\vartheta^2(\eta,y))+\vartheta^{2}(\eta,y)\left( \frac{1}{1+\vartheta^1(\eta,y)}-\frac{1}{1+\vartheta^2(\eta,y)}\right).
\end{equation*}

Therefore, calculating the sum in $n$ in $\mathfrak{G}^{d}_{21,\epsilon}(x,\eta)$ we find that

\begin{align*}
\mathfrak{G}^{d,+}_{21,\epsilon}(x,\eta)= \frac{1}{i}\int_{\epsilon}^{L} \p_{x}\left(\frac{e^{ix-y}}{1-e^{ix-y}}\right)\mathsf{L}_3(\eta,y) dy
\end{align*}

and thus

\begin{align*}
\mathsf{T}^{d,+}_{2}j_{0}(x)&= \mathsf{T}^{d,+}_{21}j_{0}(x)+ \mathsf{T}^{d,+}_{22}j_{0}(x)
\end{align*}

\begin{align}
\mathsf{T}^{d,+}_{21}j_{0}(x)&=  -\frac{1}{2\pi} \displaystyle \lim_{\epsilon\to 0^{+}} \int_{\epsilon}^{L}dy \int_{\mathbb{S}^{1}} d\eta \p_{x}\left(\frac{e^{ix-y}}{1-e^{i(x-\eta)-y}}\right)\mathsf{L}_3(\eta,y) j_{0}(\eta), \label{def:Td:21} \\
\mathsf{T}^{d,+}_{22}j_{0}(x)&= -\frac{1}{2\pi} \displaystyle \lim_{\epsilon\to 0^{+}}\int_{\mathbb{S}^{1}} \mathfrak{G}^{d,+}_{22}(x-\eta,\eta)j_{0}(\eta) \ d\eta. \label{def:Td:22}
\end{align}

Since the function $\mathfrak{G}^{d,+}_{22}$ is smooth in $x$ since the series that defines it decays exponentially in $n$, we infer using the expression on $\mathsf{L}_3(\eta,y)$  that

\begin{equation}
\abs{\mathfrak{G}^{d,+}_{22,\epsilon}(x,\eta)}\leq C \norm{\mathsf{L}_{3}}_{L^\infty(\Omega)}\leq C \norm{\vartheta^1-\vartheta^2}_{L^{\infty}(\Omega)}, \mbox{ for } x\in\SS^1, \eta\in \SS^1.
\end{equation}

Since estimate is independent of $\epsilon$, Lebesgue Dominated Convergence Theorem shows that the limit exists and that the associated operator $\mathsf{T}^{d,+}_{22}j_0$ in (<ref>) is well-defined. Moreover, we have the pointwise estimate

\begin{equation}
\abs{\mathsf{T}^{d,+}_{22}j_{0}(x)}\leq C   \norm{\vartheta^1-\vartheta^2}_{L^{\infty}(\Omega)}\norm{j_0}_{L^{\infty}(\Omega)}, \quad x\in\SS^1. \label{estimate:Td:22:Linf}
\end{equation}

The Hölder semi-norm estimate follows in a similar way as in (<ref>). Therefore,

\begin{equation}\label{bound:T21:diff}
\norm{\mathsf{T}^{d,+}_{22}}_{\mathcal{L}(L^\infty(\SS^1),C^{\alpha}(\mathbb{S}^1))}\leq  C \norm{\vartheta^2-\vartheta^1}_{C^\alpha(\Omega)}.
\end{equation}

To estimate the remainder operator $\mathsf{T}^{d}_{21}j_0$ we apply Proposition <ref> with $H(\eta,y)=\mathsf{L}_{3}j_{0}(\eta)\in C^{\alpha}(\Omega)$ and hence

\begin{equation}\label{bound:T22:diff}
\norm{\mathsf{T}^{d,+}_{21}}_{\mathcal{L}(C^{\alpha}(\mathbb{S}^1))}\leq  C \norm{\vartheta^2-\vartheta^1}_{C^\alpha(\Omega)}.
\end{equation}

By means of (<ref>)-(<ref>) we deduce that (<ref>) concluding the proof.

Let $\Lambda^1,\Lambda^2$ satisfy Assumption <ref> and let $\vartheta^1,\vartheta^2$ satisfy Assumption <ref>. Let $j_0\in C^{\alpha}(\SS^1)$ and define

\begin{align}
\mathsf{T}^{d}_{3}j_{0}(x)&=\left(\mathsf{T}_{3}[\Lambda^1,\vartheta^1]-\mathsf{T}_{2}[\Lambda^2,\vartheta^2]\right)j_{0}(x), \label{difference:operator:T3}  \\
\mathsf{T}^{d}_{4}j_{0}(x)&=\left(\mathsf{T}_{4}[\vartheta^1]-\mathsf{T}_{4}[\vartheta^2]\right)j_{0}(x),
\end{align}

where $\mathsf{T}_{3}[\cdot,\cdot]$ and $\mathsf{T}_{4}[\cdot]$ are given in (<ref>)-(<ref>). Then we have that

\begin{align}
\norm{\mathsf{T}_{3}^{d}}_{\mathcal{L}(L^\infty(\Omega),C^{\alpha}(\mathbb{S}^1))}&\leq C\left(\norm{\Lambda^1-\Lambda^2}_{C^{1,\alpha}(\Omega)}+ \norm{\vartheta^2-\vartheta^1}_{C^{\alpha}(\Omega)}\right), \label{estimate:difference:T3:Calpha}  \\
\norm{\mathsf{T}_{4}^{d}}_{\mathcal{L}(L^\infty(\Omega),C^{\alpha}(\mathbb{S}^1))}&\leq C  \norm{\vartheta^2-\vartheta^1}_{C^{\alpha}(\Omega)} \label{estimate:difference:T4:Calpha}  \end{align}

The proof follows simply by combining the bounds derived in Proposition <ref>, Proposition <ref>  and Proposition <ref>.

§.§ Estimates for Lg

In this subsection, we provide the $C^{1,\alpha}$ Hölder estimate for the term $\mathsf{G}$ defined in (<ref>). To that purpose, let us start with the following lemma:

Let $f\in C^{2,\alpha}(\partial\Omega)$ and define the functions

\begin{align*}
h^{+}(x)=-J+\int_{0}^{x} (f(\xi,L)-A) \ d\xi, \quad
h^{-}(x)=\int_{0}^{x} (f(\xi,0)-A) \ d\xi, \mbox{ for }  x\in \SS^1.
\end{align*}

Denote by $\widehat{h^{+}}(n),\widehat{h^{-}}(n)$ the Fourier coefficients of $h^{+},h^{-}$ respectively. We define

\begin{equation}\label{definition:widetilde:f}
\mathcal{Z}(x)=\frac{1}{2\pi} \displaystyle\sum_{n=-\infty}^{n=\infty}\left( \widehat{h^{+}}(n) \frac{\abs{n}}{\sinh (\abs{n}L)}-
\widehat{h^{-}}(n)\frac{\abs{n}}{\tanh (\abs{n}L)}\right)  e^{inx}.
\end{equation}

Then the function $\mathcal{Z}\in C^{2,\alpha}(\SS^1)$. Moreover, we have that

\begin{equation}
\norm{\mathcal{Z}}_{C^{2,\alpha}(\SS^1)}\leq C \norm{f}_{C^{2,\alpha}(\SS^1)}.
\end{equation}

Adding and subtracting $\abs{n}\widehat{h^{-}}(n)e^{inx}$ in (<ref>) we obtain that

\begin{equation}
\mathcal{Z}(x)=\frac{1}{2\pi} \displaystyle\sum_{n=-\infty}^{n=\infty}\left( \abs{n}\widehat{h^{-}}(n)e^{inx}+W_{1}(n)\widehat{h^{+}}(n)e^{inx}+W_{2}(n)\widehat{h^{-}}(n)e^{inx}\right)
\end{equation}

where $\abs{W_{i}(n)}\leq C e^{-Ln}$ for $i=1,2$. Therefore, recalling that

$$\partial_{x}\mathcal{H}h^{-}(x)=\mathcal{H}\partial_{x}h^{-}(x)=\frac{1}{2\pi} \displaystyle\sum_{n=-\infty}^{n=\infty} \abs{n}\widehat{h^{-}}(n)e^{inx}$$

where $\mathcal{H}$ is the periodic Hilbert transform we arrive at

\begin{align*}
\norm{\mathcal{Z}}_{C^{2,\alpha}(\SS^1)} &\leq C \left(\norm{ \mathcal{H}\partial_{x}h^{-}}_{C^{2,\alpha}(\SS^1)} + \norm{h^-}_{L^{\infty}(\SS^1)}+\norm{h^+}_{L^{\infty}(\SS^1)} \right)\\
&\leq C \left( \norm{ \mathcal{H}h^{-}}_{C^{3,\alpha}(\SS^1)}+ \norm{h^+}_{L^{\infty}(\SS^1)} \right).
\end{align*}

Moreover, using the fact that the Hilbert transform is a bounded operator in the class of Hölder spaces (cf. [21]) namely

$$\norm{\mathcal{H}}_{\mathcal{L}(C^{3,\alpha}(\SS^1))}\leq C,$$

and that $\partial_{x}h^{-}(x)=(f(x,L)-A)\in C^{2,\alpha}(\SS^1)$ we obtain

\begin{align*}
\norm{\mathcal{Z}}_{C^{2,\alpha}(\SS^1)} &\leq C \left( \norm{ \mathcal{H}h^{-}}_{C^{3,\alpha}(\SS^1)}+ \norm{h^+}_{L^{\infty}(\SS^1)} \right)\leq C  \norm{f}_{C^{2,\alpha}(\SS^1)}.
\end{align*}

Let $g\in C^{2,\alpha}(\partial\Omega_{-})$ and define $\widetilde{g}(x)=g(x)-\mathcal{Z}(x)$ where $\mathcal{Z}(x)$ is given in (<ref>). Then for $x\in\SS^1$ we define
\begin{equation}\label{definition:G:boundary}
\mathsf{G}(x)=-\frac{1}{2\pi}\int_{\mathbb{S}^{1}} \displaystyle\sum_{n=-\infty}^{n=\infty}\left[\frac{n \sinh(nL)}{\cosh(nL)-1}\right]e^{in(x-\eta)}\widetilde{g}(\eta) d\eta.
\end{equation}

Moreover, we have that

\begin{equation}\label{bound:G:C1alpha}
\norm{\mathsf{G}}_{C^{1,\alpha}(\SS^1)}\leq C \left( \norm{f}_{C^{2,\alpha}(\SS^1)}+ \norm{g}_{C^{2,\alpha}(\SS^1)}\right).
\end{equation}

Recalling (<ref>), we have that

\begin{align*}
\mathsf{G}(x)&= \mathcal{H}\p_{x}\widetilde{g}(x)- \int_{\SS^1}\widetilde{Q}(x-\eta)\widetilde{g}(\eta) \ d\eta,
\end{align*}

$$\widetilde{Q}_{n}=\frac{1}{n}\left(n\frac{\sinh(nL)}{\cosh(nL)-1}-\abs{n}\right), \mbox{ for } n\neq 0, \quad \widetilde{Q}_{0}=\frac{2}{L} \mbox{ for } n=0.$$

Therefore, using the fact

$$\norm{\mathcal{H}}_{\mathcal{L}(C^{1,\alpha}(\SS^1))}\leq C, \quad \abs{\widetilde{Q}_{n}}\leq e^{-nL}$$

and  (<ref>) we infer that

\begin{equation}
\norm{\mathsf{G}}_{C^{1,\alpha}(\SS^1)}\leq C \left(\norm{\mathcal{H}\p_{x}\widetilde{g}}_{C^{1,\alpha}(\SS^1)}+ \norm{\widetilde{g}}_{L^{\infty}(\SS^1)}\right)
\leq C \norm{\widetilde{g}}_{C^{2,\alpha}(\SS^1)}.
\end{equation}

To conclude, we recall that $\widetilde{g}(x)=g(x)-\mathcal{Z}(x)$ and thus by means of Lemma <ref> we obtain that

\begin{equation}
\norm{\mathsf{G}}_{C^{1,\alpha}(\SS^1)}\leq C \norm{\widetilde{g}}_{C^{2,\alpha}(\SS^1)}\leq C \left( \norm{f}_{C^{2,\alpha}(\SS^1)}+ \norm{g}_{C^{2,\alpha}(\SS^1)}\right).
\end{equation}

§ EXISTENCE OF SOLUTIONS TO THE INTEGRAL EQUATION FOR LG

In this section we are interested in studying the existence of a solution $j_0\in C^{1,\alpha}(\partial\Omega_{-})$ to the integral equation (<ref>) given by

\begin{equation}\label{integral:eq:fixed:2}
j_0(x)=- \displaystyle\sum_{i=1}^{4}(\mathsf{T}_{i}j_{0}(x)-\langle \mathsf{T}_{i}j_{0} \rangle)-\langle j_0 f^{-}\rangle+\mathsf{G}(x)-\langle \mathsf{G} \rangle
\end{equation}

To that purpose, let us first introduce the following notation. Given $j_{0}\in C^{1,\alpha}(\partial\Omega_{-})$ we define the operator

\begin{equation*}
\Upsilon:B_{\delta_{0}}(C^{2,\alpha}(\Omega))\times B_{\delta_{1}}(C^{1,\alpha}(\Omega)) \to \mathcal{L}\left(C^{1,\alpha}(\partial\Omega_{-})\right)
\end{equation*}

such that for $(\Lambda,\vartheta)\in B_{\delta_{0}}(C^{2,\alpha}(\Omega))\times B_{\delta_{1}}(C^{1,\alpha}(\Omega))$ we have that

\begin{equation}\label{eq:Gamma}
\Upsilon[\Lambda,\vartheta](j_{0})= -\displaystyle\sum_{i=1}^{4}\left(\mathsf{T}_{i}[\Lambda,\vartheta]j_{0}-\langle \mathsf{T}_{i}[\Lambda,\vartheta]j_{0}  \rangle\right)-\langle f^{-}j_{0} \rangle
\end{equation}

where $\mathsf{T}_{i}$ are defined in (<ref>),(<ref>),(<ref>) and (<ref>).

More precisely, the operators
$\mathsf{T}_{i}[\Lambda,\vartheta]$ as stated in (<ref>),(<ref>),(<ref>) and (<ref>) are written for the particular case where $\vartheta(\eta,y)=\partial_{\xi} \Theta(X(\eta,y),y))$. However, we will show the existence of solutions for a more general class of integral equations, namely for general functions $\Lambda$ and $\vartheta$ which satisfied certain regularity and smallness assumptions. We will later check that for the particular case where $\vartheta(\eta,y)=\partial_{\xi} \Theta(X(\eta,y),y))$ the required assumptions are  satisfied (cf. Section <ref>).

In a similar manner as we did for the operators $\mathsf{T}_{1}, \ldots, \mathsf{T}_{4}$, we now define an operator $\Upsilon$ acting either in $C^\alpha$ or in   $C^{1,\alpha}$. We will not use different symbols for operators acting in different spaces for the sake of simplicity, (cf. Remark <ref>).

In the first place, we have the following two lemmas

Let $M_{0}\leq \displaystyle\min\{\delta_0,\delta_1\}$ where $\delta_0, \delta_1$ are defined in Assumption <ref> and  Assumption <ref>, respectively. Let $\widetilde{M}\leq M_{0}$ and

\begin{equation}\label{smallness:lambda:vartheta}
\norm{\Lambda}_{C^{2,\alpha}(\Omega)}+    \norm{\vartheta}_{C^{1,\alpha}(\Omega)}+\norm{f^{-}}_{C^{2,\alpha}(\partial\Omega_{-})} \leq \widetilde{M}.
\end{equation}

Then we have that

\begin{equation}\label{estimate:upsilon:1}
\norm{\Upsilon[\Lambda,\vartheta]}_{\mathcal{L}\left(C^{1,\alpha}(\partial\Omega_{-})\right)}\leq C \widetilde{M}.
\end{equation}

Furthermore, the operator $\Upsilon[\Lambda,\vartheta]$ is Lipschitz in $C^{\alpha}(\partial\Omega_{-})$, i.e. for any  $\Lambda^{1},\Lambda^{2}$ and $\vartheta^{1},\vartheta^{2}$ satisfying (<ref>) we have that

\begin{equation}\label{estimate:upsilon:1:contraction}
\norm{\Upsilon[\Lambda^1,\vartheta^1]- \Upsilon[\Lambda^2,\vartheta^2]}_{\mathcal{L}\left(C^{\alpha}(\partial\Omega_{-})\right)}\leq C\left( \norm{\Lambda^1-\Lambda^2}_{C^{1,\alpha}(\Omega)}+ \norm{\vartheta^2-\vartheta^1}_{C^{\alpha}(\Omega)}\right).
\end{equation}

The proof of (<ref>) is a consequence of the estimates  (<ref>),   (<ref>), (<ref>) and (<ref>), as well as the fact that those estimates are preserved for the averaging operators $\langle \cdot \rangle$. Notice that the derivatives of the averaging operators are zero since they are just constant functions.

Indeed, applying those bounds we readily check that

\begin{equation}
\norm{\Upsilon[\Lambda,\vartheta]}_{\mathcal{L}\left(C^{1,\alpha}(\partial\Omega_{-})\right)}\leq C  \widetilde{M}
\end{equation}

for $  \widetilde{M}\leq M_{0}$. To show estimate (<ref>), we invoke bounds (<ref>), (<ref>), (<ref>) and (<ref>) to obtain

\begin{align*}
\norm{\Upsilon[\Lambda^1,\vartheta^1]- \Upsilon[\Lambda^2,\vartheta^2]}_{\mathcal{L}\left(C^{\alpha}(\partial\Omega_{-})\right)} &\leq  \displaystyle \sum_{i=1}^{4}\norm{\mathsf{T}_{i}[\Lambda^1,\vartheta^1]- \mathsf{T}_{i}[\Lambda^2,\vartheta^2]}_{\mathcal{L}\left(C^{\alpha}(\partial\Omega_{-})\right)}  \\
&\quad \quad + \displaystyle \sum_{i=1}^{4} \norm{ \left(\langle \mathsf{T}_{i}[\Lambda_{1},\vartheta_{1}](\cdot) \rangle- \mathsf{T}_{i}[\Lambda_{2},\vartheta_{2}](\cdot) \right)}_{\mathcal{L}\left(C^{\alpha}(\partial\Omega_{-})\right)}\\
&\leq C\left( \norm{\Lambda^1-\Lambda^2}_{C^{1,\alpha}(\Omega)}+ \norm{\vartheta^2-\vartheta^1}_{C^{\alpha}(\Omega)}\right)
\end{align*}

where the action of the operators $\langle \mathsf{T}_{i}[\Lambda_{\ell},\vartheta_{\ell}](\cdot) \rangle$ for $i=1,\ldots,4$ and $\ell=1,2$ acting on the function $j_0$ is given by $\langle \mathsf{T}_{i}[\Lambda_{\ell},\vartheta_{\ell}]j_{0}  \rangle$. We also use the fact that the terms $\langle f^{-}j_0 \rangle$ cancel out.

There exists $M_{0}\leq \displaystyle\min\{\delta_0,\delta_1\}$ such that for any $  \widetilde{M}\leq M_{0}$ and $\Lambda,\vartheta, f^{-}$ satisfying (<ref>) the operator $\left(\mathsf{I}-\Upsilon\right)$ is invertible in $C^{1,\alpha}(\partial\Omega_{-})$. More precisely, there exists an operator $\Pi[\Lambda,\vartheta]= \left(\mathsf{I}-\Upsilon[\Lambda,\vartheta]\right)^{-1}$ such that

\begin{equation*}
\Pi:B_{\widetilde{M}}(C^{2,\alpha}(\Omega))\times B_{\widetilde{M}}(C^{1,\alpha}(\Omega)) \to \mathcal{L}\left(C^{1,\alpha}(\partial\Omega_{-})\right).
\end{equation*}

Moreover, the operator $\Pi[\Lambda,\vartheta]$ is Lipschitz in $C^{\alpha}(\partial\Omega_{-})$, i.e. for any  $\Lambda^{1},\Lambda^{2}$ and $\vartheta^{1},\vartheta^{2}$ satisfying (<ref>) we have that

\begin{equation}\label{difference:PI:lower}
\norm{\Pi[\Lambda^1,\vartheta^1]- \Pi[\Lambda^2,\vartheta^2]}_{\mathcal{L}\left(C^{\alpha}(\partial\Omega_{-})\right)}\leq C\left( \norm{\Lambda^1-\Lambda^2}_{C^{1,\alpha}(\Omega)}+ \norm{\vartheta^2-\vartheta^1}_{C^{\alpha}(\Omega)}\right).
\end{equation}

Under the hypothesis of Lemma <ref> we have shown that  $\norm{\Upsilon[\Lambda,\vartheta]}_{\mathcal{L}\left(C^{1,\alpha}(\partial\Omega_{-})\right)}\leq C \widetilde{M}$ for some $\widetilde{M}$ sufficiently small. Therefore, the existence of  $\left(\mathsf{I}-\Upsilon\right)^{-1}$
follows from the classical Neumann series (cf. [23]). Indeed,  we have that

\begin{equation}\label{Neumann:series}
\Pi[\Lambda,\vartheta]=\left(\mathsf{I}-\Upsilon[\Lambda,\vartheta]\right)^{-1}=\displaystyle \sum_{n=0}^{\infty}(-1)^{n}\Upsilon^{n}[\Lambda,\vartheta]
\end{equation}

where $n=0$, $\Upsilon^{0}[\Lambda,\vartheta]=\mathcal{I}$ is the identity operator. Moreover, for $\Upsilon[\Lambda,\vartheta]\in \mathcal{L}\left(C^{1,\alpha}(\partial\Omega_{-})\right)$ and using estimate (<ref>) we find that

\begin{equation}
\norm{\Pi[\Lambda,\vartheta]}_{\mathcal{L}\left(C^{1,\alpha}(\partial\Omega_{-})\right)} = \norm{\displaystyle \sum_{n=0}^{\infty}(-1)^{n}\Upsilon^{n}[\Lambda,\vartheta]}_{\mathcal{L}\left(C^{1,\alpha}(\partial\Omega_{-})\right)}\leq \frac{1}{1-C\widetilde{M}}.
\end{equation}

Denoting by $A_{1}^{n}=\Upsilon^{n}[\Lambda^1,\vartheta^1]$, $A_{2}^{n}=\Upsilon^{n}[\Lambda^2,\vartheta^2]$ we can find that

\begin{align}
\end{align}

Thus, combining  (<ref>) and (<ref>) we find that

\begin{align*}
\Pi[\Lambda^1,\vartheta^1]-\Pi[\Lambda^2,\vartheta^2]&=\displaystyle \sum_{n=0}^{\infty}(-1)^{n}A^{n}_{1}-\displaystyle \sum_{n=0}^{\infty}(-1)^{n}A^{n}_{2} \\
&= \displaystyle \sum_{n=0}^{\infty}(-1)^{n} \bigg[ A_{1}^{n-1}\left(A_{1}-A_{2}\right)+A_{1}^{n-2}\left(A_{1}-A_{2}\right)A_{2}+\ldots+A\left(A_{1}-A_{2}\right)A_{2}^{n} \bigg]
\end{align*}

and hence by means of bounds (<ref>)-(<ref>)we infer that if $M_0$ is sufficiently small that

\begin{align*}
\norm{\Pi[\Lambda^1,\vartheta^1]- \Pi[\Lambda^2,\vartheta^2]}_{\mathcal{L}\left(C^{\alpha}(\partial\Omega_{-})\right)}&\leq \norm{A_{1}-A_{2}}_{\mathcal{L}\left(C^{\alpha}(\partial\Omega_{-})\right)}  \displaystyle \sum_{n=0}^{\infty} n (C\widetilde{M})^{n-1} \\
&\leq  C \norm{\Upsilon^{n}[\Lambda^1,\vartheta^1]-\Upsilon^{n}[\Lambda^2,\vartheta^2]}_{\mathcal{L}\left(C^{\alpha}(\partial\Omega_{-})\right)} \\
&\leq C\left( \norm{\Lambda^1-\Lambda^2}_{C^{1,\alpha}(\Omega)}+ \norm{\vartheta^2-\vartheta^1}_{C^{\alpha}(\Omega)}\right).
\end{align*}

Combining both lemmas we can provide the existence of solutions to the integral equation (<ref>) which reads

Let the hypothesis of Lemma <ref> hold. Let also $f\in C^{2,\alpha}(\Omega)$, $g\in C^{2,\alpha}(\partial\Omega_{-})$ and $\mathsf{G}$ given as in (<ref>). Then there exists a solution $j_0\in C^{1,\alpha}(\partial\Omega_{-})$ to (<ref>) given by

\begin{equation}\label{solution:explicit:omega_0}
j_{0}=\Pi[\Lambda,\vartheta]\left( \mathsf{G}-\langle \mathsf{G} \rangle \right).
\end{equation}

\begin{equation}\label{bound:Prop5.4}
\norm{j_{0}}_{C^{1,\alpha}(\partial\Omega_{-})}\leq C \left( \norm{f}_{C^{2,\alpha}(\Omega)}+ \norm{g}_{C^{2,\alpha}(\partial\Omega_{-})}\right).
\end{equation}

The fact that $j_0$ as given in (<ref>) solves (<ref>) is a consequence of the definition of the operator $\Pi[\Lambda,\vartheta]$ in Lemma <ref>. On the other hand, we notice by means of Lemma <ref> and estimate (<ref>) yields

\begin{equation}
\norm{\mathsf{G}-\langle\mathsf{G}\rangle}_{C^{1,\alpha}(\SS^1)}\leq C \left( \norm{f}_{C^{2,\alpha}(\SS^1)}+ \norm{g}_{C^{2,\alpha}(\SS^1)}\right)
\end{equation}

where $\mathsf{G}$ is as in (<ref>). This estimate as well as the fact that $\Pi:B_{\widetilde{M}}(C^{2,\alpha}(\Omega))\times B_{\widetilde{M}}(C^{1,\alpha}(\Omega)) \to \mathcal{L}\left(C^{1,\alpha}(\partial\Omega_{-})\right)$

\begin{equation}
\norm{j_{0}}_{C^{1,\alpha}(\partial\Omega_{-})}\leq C \left( \norm{f}_{C^{2,\alpha}(\Omega)}+ \norm{g}_{C^{2,\alpha}(\partial\Omega_{-})}\right).
\end{equation}

§ THE FIXED POINT ARGUMENT

In this section we will provide the fixed point argument, this is, we will define an adequate operator $\Gamma$ on a subspace of $C^{2,\alpha}(\Omega)$ which has a fixed point $b$ such that $B=(0,1)+b$ is a solution to (<ref>) and (<ref>).

We define the operator $\Gamma: B_{M}(C^{2,\alpha}(\Omega)) \to C^{2,\alpha}(\Omega)$ using several intermediate building blocks. Given $b\in B_{M}(C^{2,\alpha}(\Omega))$ we define the flow map associated with the vector field $B=(0,1)+b$ as the mapping $X[b]: B_{M}(C^{2,\alpha}(\Omega))\to C^{2,\alpha}(\Omega)$ which satisfies the ordinary differential equation

\begin{equation}\label{ode:flow:X}
\left\lbrace
\begin{array}{lll}
\partial_{y} X[b](\xi,y)= \frac{b^{1}(X(\xi,y),y)}{1+b^{2}(X(\xi,y),y)} \\
\end{array}\right.
\end{equation}

Moreover, we denote by $X[b]^{-1}(\xi,y)$ the inverse function of $X[b]$ in the first variable, namely $X[b](X[b]^{-1}(\xi,y),y)=\xi$. Then we define $\Lambda:B_{M}(C^{2,\alpha}(\Omega))\to C^{2,\alpha}(\Omega)$ as

\begin{equation}\label{def:Lambda:sec6}
\Lambda[b](\eta,y)= X[b](\eta,y)-\eta
\end{equation}

Actually, in Lemma <ref> we will show the stronger result $Im(\Lambda)\subset B_{CM}(C^{2,\alpha}(\Omega))$ for $C>0$.
We define also the function $\Theta:B_{M}(C^{2,\alpha}(\Omega))\to C^{2,\alpha}(\Omega)$ as

\begin{equation}\label{def:Theta:V}
\Theta[b](\xi,y)=X[b]^{-1}(\xi,y)-\xi.
\end{equation}

Finally $\vartheta[b]:B_{M}(C^{2,\alpha}(\Omega))\to C^{1,\alpha}(\Omega)$ is given by

\begin{equation}\label{vartheta:Theta:V}
\vartheta[b](\eta,y)=\partial_{\xi} \Theta[b](X[b](\eta,y),y).
\end{equation}

Notice that combining (<ref>)-(<ref>) we can write

\begin{equation}\label{def:Comb}
\vartheta[b](\eta,y)= \partial_{\xi} X^{-1}[b]\left(X[b](\eta,y),y)\right)-1.
\end{equation}

Moreover, arguing as in Lemma <ref> we prove that $Im(\vartheta)\subset B_{CM}(C^{1,\alpha}(\Omega))$.

Therefore, we now introduce the following operator
$$\Psi[b]: B_{M}(C^{2,\alpha}(\Omega))\to B_{CM}(C^{2,\alpha}(\Omega))\times B_{CM}(C^{1,\alpha}(\Omega))$$
defined by $\Psi[b]=(\Lambda[b], \vartheta[b])$ and choose $\widetilde{M}=CM\leq M_{0}$.
Next, notice that the function $j_{0}\in C^{1,\alpha}(\partial\Omega_{-})$ given by (<ref>) solving the integral equation (<ref>) can be expressed as the following composition of operators

\begin{equation}\label{omega:cero:fixed}
j_0= \Pi[\Psi[b]](\mathsf{G}-\langle \mathsf{G} \rangle).
\end{equation}

The condition $\widetilde{M}=CM\leq M_{0}$ must be satisfied so that Proposition <ref> can be applied.
To conclude the construction, we use two additional building blocks. First, for $b\in B_{M}(C^{2,\alpha}(\Omega))$  and  $j_0= \Pi[\Psi[b]](\mathsf{G}-\langle \mathsf{G} \rangle)$, we define
$$j=T[b,j_0]:B_{M}(C^{2,\alpha}(\Omega)) \times C^{1,\alpha}(\Omega_{-})\to C^{1,\alpha}(\Omega)$$

where $j$ is the unique solution to the transport type problem

\begin{equation}\label{transport:problem:full:fixed}
\left\lbrace
\begin{array}{lll}
((0,1)+b)\cdot\nabla j =0, \ \mbox{in} \ \Omega \\
j = j_{0}, \ \mbox{on} \ \partial \Omega_{-}.
\end{array} \right.
\end{equation}

To conclude, the new velocity field $W\in C^{2,\alpha}(\Omega)$ is recovered by means of the div-curl problem (also known as Biot-Savart operator) given by

$$W=B_{s}[j,f,J]:C^{1,\alpha}(\Omega)\times C^{2,\alpha}(\partial\Omega)\times\mathbb{R}\to C^{2,\alpha}(\Omega)$$

where $W$ is the unique solution to

\begin{equation}\label{div:curl:problem:full:fixed}
\left\lbrace
\begin{array}{lll}
\nabla\times W= j, \ \mbox{in} \ \Omega \\
\mbox{div } W=0, \ \mbox{in} \ \Omega \\
W\cdot n=f, \ \mbox{on} \ \partial \Omega \\
\int_{0}^{L} W_{1}(0,y) \ dy=J.
\end{array} \right.
\end{equation}

where $J=J[f,g,b]$ is given by

\begin{equation}\label{combination:5}
\frac{2J}{L^2}=-\frac{1}{2\pi}\int_{0}^{2\pi}\mathsf{G}(x)  dx+\frac{1}{2\pi} \displaystyle\sum_{i=1}^{4}\int_{0}^{2\pi} \mathsf{T}_{i}j_{0}(x) \ dx \ dx -\frac{1}{2\pi}\int_{0}^{2\pi} j_0(x)f^{-}(x) \ dx
\end{equation}

(cf. (<ref>)) with $j_0= \Pi[\Psi[b]](\mathsf{G}-\langle \mathsf{G} \rangle)$.
Then,  we define $\Gamma(b)=W$. In particular, the full operator can be expressed as the following composition of operators

\begin{equation}\label{full:composiotion:operator}
W=\Gamma(b)=B_{s}[T[b,\Pi[\Psi[b]](\mathsf{G}-\langle \mathsf{G}\rangle)], f,J[f,g,b]].
\end{equation}

The precise statement of the theorem reads as follows:

Let $f\in C^{2,\alpha}(\Omega)$ satisfying (<ref>) and $g\in C^{2,\alpha}(\partial\Omega_{-})$. There exist $\epsilon_{0}>0, M_{0}=M_{0}(L,\alpha)$ sufficiently small such that if

\begin{equation}\label{small:f:g}
\norm{f}_{C^{2,\alpha}(\Omega)}+\norm{g}_{C^{2,\alpha}(\partial\Omega_{-})}\leq \epsilon_{0}M, \mbox{ for } M\leq M_{0},
\end{equation}

then $\Gamma(B_{M}(C^{2,\alpha}(\Omega)))\subset B_{M}(C^{2,\alpha}(\Omega))$. Furthermore, the operator $\Gamma$ has a unique fixed point in $B_{M}(C^{2,\alpha}(\Omega))$.

§.§ Preliminary estimates: ODE, transport problem and div-curl problem

Before showing the proof of Theorem <ref>, let us first show several Lemmas that will be needed to provide the proof of Theorem <ref>. The first result summarizes general Hölder estimates for $\Lambda[b]$ and $\vartheta[b]$.

Let $M_{0}$ be sufficiently small and let
$b\in B_{M}(C^{2,\alpha}(\Omega))$ with $M\leq M_0$. Then, for $\Lambda[b], \vartheta[b]$ given in (<ref>) and (<ref>), we have that $\Lambda[b]\in C^{2,\alpha}(\Omega)$ $\vartheta[b]\in C^{1,\alpha}(\Omega)$ and

\begin{align}
\norm{\Lambda[b]}_{C^{2,\alpha}(\Omega)}\ &\leq C M, \label{bound:Lambda:V} \\
\norm{\vartheta[b]}_{C^{1,\alpha}(\Omega)} & \leq  CM. \label{bound:vartheta:V}
\end{align}

Moreover, the operator $\Lambda[b]$ and $\vartheta[b]$ are Lipschitz in $C^{1,\alpha}(\Omega)$ and $C^{\alpha}(\Omega)$ respectively. This is for any $b^{1},b^{2} \in B_{M}(C^{2,\alpha}(\Omega))$

\begin{align}
\norm{\Lambda[b^1]- \Lambda[b^2]}_{C^{1,\alpha}(\Omega)}\leq C \norm{b^1-b^2}_{C^{1,\alpha}(\Omega)}, \label{estimate:Lambda:contraction} \\
\norm{\vartheta[b^1]- \vartheta[b^2]}_{C^{\alpha}(\Omega)}\leq C \norm{b^1-b^2}_{C^{\alpha}(\Omega)}.\label{estimate:vartheta:contraction}
\end{align}

The proof of Lemma <ref> is the standard argument used to compute the dependence of the solutions for an ODE in their parameters. More precisely, the main idea of the proof is to control incremental quotients of the form $\frac{f(x+h)-f(x)}{h}$ for $h>0$, as well as terms quotients of the form
$\frac{\abs{f(x)-f(y)}}{\abs{x-y}^{\alpha}}$ using Grönwall type arguments. A bound similar to (<ref>), (<ref>) in Lemma <ref> but estimating only the $C^{1,\alpha}$ Hölder norm have been shown in <cit.>. Moreover, the proof of (<ref>), (<ref>) is obtained computing the differences of the solutions of the differential equations which define $\Lambda$, $\vartheta$ (cf. (<ref>), (<ref>) and (<ref>)) with $b=b^1$ and $b=b^2$.

The next results deals with Hölder estimates for solutions to the hyperbolic transport type problem (<ref>). For a proof of this result we refer the reader to <cit.>, where a more general result is shown.

Let $M_{0}$ be sufficiently small. Then for every $M\leq M_{0}$, $b\in B_{M}(C^{2,\alpha}(\Omega))$ and $j_0\in  C^{1,\alpha}(\partial\Omega_{-})$,  there exists a unique $j\in C^{1,\alpha}(\Omega)$ solving

\begin{equation}\label{transport:propo:6}
\left\lbrace
\begin{array}{lll}
(B_{0}+b)\cdot\nabla j =0 \ \mbox{in} \ \Omega, \\
j = j_{0} \ \mbox{on} \ \partial \Omega_{-}.
\end{array} \right.
\end{equation}

Moreover, there exists a constant $C=C(\alpha,L)>0$ such that the following estimate holds

\begin{equation}\label{estimate:TP}
\norm{j}_{C^{1,\alpha}(\Omega)} \leq C  \norm{j_0}_{C^{1,\alpha}(\partial\Omega_{-})}.
\end{equation}

Furthermore, let $j_{1},j_{2}\in C^{\alpha}(\Omega)$ be  two different solutions to  (<ref>) with  $b$ given by $b^{1},b^{2}\in B_{M}(C^{2,\alpha}(\Omega))$  respectively. Then

\begin{equation}\label{estimate:difference:TP}
\norm{j^{1}-j^{2}}_{C^{\alpha}(\Omega)} \leq  C \left(\norm{j^{1}_{0}-j^{2}_{0}}_{C^{\alpha}(\partial\Omega_{-})}+ \norm{j^{1}_{0}}_{C^{1,\alpha}(\partial\Omega_{-})}\norm{V^1-V^2}_{C^{\alpha}(\Omega)}\right)
\end{equation}

where $C=C(\alpha,L)>0$.

To conclude let us also recall the following result regarding Hölder estimate for the div-curl problem, cf.  <cit.> for a detailed proof.

For every $J\in \mathbb{R}$, $j\in C^{1,\alpha}(\Omega)$ and $f\in C^{2,\alpha}(\partial\Omega)$ satisfying (<ref>),  there exists a unique solution $W\in C^{2,\alpha}(\Omega)$ solving

\begin{equation}\label{DCP:problem}
\left\lbrace
\begin{array}{lll}
\nabla \times W = j, \quad \mbox{in } \Omega \\
\mbox{div } W =0,  \quad \mbox{in } \Omega  \\
W \cdot n = f ,  \quad \mbox{on } \partial \Omega \\
\int_{\mathcal{C}}W\cdot n \ dS= J.
\end{array}\right.
\end{equation}

where the curve $\mathcal{C}=\{ (0,y), y\in [0,L]\}.$ Moreover, the solution satisfies the inequality

\begin{equation}\label{estimate:DCP}
\norm{W}_{C^{2,\alpha}(\Omega)} \leq C \left(\norm{j}_{C^{1,\alpha}(\Omega)}+ \norm{f}_{C^{2,\alpha}(\partial\Omega)} + \abs{J}\right),
\end{equation}

where  $C=C(L,\alpha)>0$.

§.§ Proof of Theorem <ref>

First, we show that the operator  $\Gamma$ maps $ b\in B_{M}(C^{2,\alpha}(\Omega))$ into itself and second, that the operator  $\Gamma$ is a contraction mapping in the lower order norm $C^{1,\alpha}(\Omega)$. By combining both ingredients, we can invoke Banach fixed point theorem to infer that the operator $\Gamma$ has a unique fixed point in $B_{M}(C^{2,\alpha}(\Omega)).$ Let us start with the former assertion.  By means of (<ref>) we find that for $b\in B_{M}(C^{2,\alpha}(\Omega))$

\begin{align*}
\norm{\Gamma(b)}_{C^{2,\alpha}(\Omega)}&=\norm{B_{s}[T[b,\Pi[\Psi[b]](\mathsf{G}-\langle \mathsf{G} \rangle)], f,J]}_{C^{2,\alpha}(\Omega)} \\
&\leq C\left(\norm{T[b,\Pi[\Psi[b]](\mathsf{G}-\langle \mathsf{G} \rangle)]}_{C^{1,\alpha}(\Omega)}+\norm{f}_{C^{2,\alpha}(\partial\Omega)} + \abs{J}\right) \\
&\leq C\left(\norm{\Pi[\Psi[b]](\mathsf{G}-\langle \mathsf{G} \rangle)]}_{C^{1,\alpha}(\Omega)}+\norm{f}_{C^{2,\alpha}(\partial\Omega)} + \abs{J}\right).
\end{align*}

where in the first inequality we have used (<ref>) in Proposition <ref> and in the latter we invoked (<ref>) in Proposition <ref>.
On the other hand, recall that by definition $\Psi[b]=(\Lambda[b],\vartheta[b])$. Hence, combining inequalities (<ref>)-(<ref>) in Lemma <ref> and estimate (<ref>) in Proposition <ref> we have that

\begin{equation*}
\norm{\Pi[\Psi[b]](\mathsf{G}-\langle \mathsf{G} \rangle)]}_{C^{1,\alpha}(\Omega)}\leq C \left( \norm{f}_{C^{2,\alpha}(\Omega)}+ \norm{g}_{C^{2,\alpha}(\partial\Omega_{-})}\right).
\end{equation*}

Moreover, we can show using the expression  of $J$ given in (<ref>) and the previous estimates (<ref>),   (<ref>), (<ref>), (<ref>) and  (<ref>) that

\begin{equation}
\abs{J}\leq C \left( \norm{f}_{C^{2,\alpha}(\Omega)}+ \norm{g}_{C^{2,\alpha}(\partial\Omega_{-})}\right).
\end{equation}

Thus, we readily check that

\begin{align}\label{Gamma:estimate:ball}
\norm{\Gamma(b)}_{C^{2,\alpha}(\Omega)}\leq C \left( \norm{f}_{C^{2,\alpha}(\Omega)}+ \norm{g}_{C^{2,\alpha}(\partial\Omega_{-})} \right) \leq C\epsilon_{0}M
\end{align}

where in the second inequality we have used the smallness assumption (<ref>). Choosing $\epsilon_{0}=\frac{1}{4C}$, we obtain that $\Gamma(B_{M}(C^{2,\alpha}(\Omega)))\subset B_{M}(C^{2,\alpha}(\Omega))$.

We now claim that the $B_{M}(C^{2,\alpha}(\Omega))$ endowed with the topology $C^{1,\alpha}$ is a complete metric space which we will denote by $(B_{M}(C^{2,\alpha}(\Omega)), \norm{\cdot}_{C^{1,\alpha}})$.  In order to show this it is sufficient to check that $B_{M}(C^{2,\alpha}(\Omega))$ is a closed subset of $C^{1,\alpha}(\Omega)$, (cf. <cit.>).

Moreover, we also claim that

$$\Gamma: (B_{M}(C^{2,\alpha}(\Omega)), \norm{\cdot}_{C^{1,\alpha}}) \to (B_{M}(C^{2,\alpha}(\Omega)), \norm{\cdot}_{C^{1,\alpha}})$$

is a contraction mapping. To this end, we have to show that for $b^{1},b^{2}\in  B_{M}(C^{2,\alpha}(\Omega))$, we need to estimate the difference $\norm{\Gamma(b^1)-\Gamma(b^2)}_{C^{1,\alpha}(\Omega)}$. To that purpose, using the expression of the full operator given in (<ref>) and noticing that the Biot-Savart operator defined in (<ref>) is a linear operator, we obtain by means of Proposition <ref> that

\begin{align}
\norm{\Gamma(b^1)-\Gamma(b^2)}_{C^{1,\alpha}(\Omega)}&=\norm{B_{s}[T[b^1,\Pi[\Psi[b^1]](\mathsf{G}-\langle \mathsf{G} \rangle)], f,J^{1}]-B_{s}[T[b^2,\Pi[\Psi[b^2]](\mathsf{G}-\langle \mathsf{G} \rangle)], f,J^{2}]}_{C^{1,\alpha}(\Omega)} \nonumber \\
&\leq C\left(\norm{T[b^1,\Pi[\Psi[b^1]](\mathsf{G}-\langle \mathsf{G} \rangle)-T[b^2,\Pi[\Psi[b^2]](\mathsf{G}-\langle \mathsf{G} \rangle)}_{C^{1,\alpha}(\Omega)}\right)+\abs{J^{1}-J^{2}} \label{contraction:gamma:est1}
\end{align}

where $J^{\ell}$ is given by (<ref>) with $j_0= \Pi[\Psi[b^{\ell}]](\mathsf{G}-\langle \mathsf{G} \rangle)$ for $\ell=1,2$.
To deal with the transport type operator $T$  given in (<ref>), we invoke inequality (<ref>) in Proposition <ref> to find that

\begin{align*}
&\norm{T[b^1,\Pi[\Psi[b^1]](\mathsf{G}-\langle \mathsf{G} \rangle)-T[b^2,\Pi[\Psi[b^2]](\mathsf{G}-\langle \mathsf{G} \rangle)}_{C^{1,\alpha}(\Omega)}  \\
& \quad \quad \quad \leq C\bigg[ \norm{\Pi[\Psi[b^1]](\mathsf{G}-\langle \mathsf{G} \rangle)-\Pi[\Psi[b^2]](\mathsf{G}-\langle \mathsf{G} \rangle)}_{C^{\alpha}(\partial\Omega_{-})} \\
&\quad \quad  \quad \quad \quad+ \norm{\Pi[\Psi[b^1]](\mathsf{G}-\langle \mathsf{G} \rangle}_{C^{1,\alpha}(\partial\Omega_{-})}\norm{b_{1}-b_{2}}_{C^{1,\alpha}(\Omega)}\bigg].
\end{align*}

On other hand by means of (<ref>) in Lemma <ref> and recalling that $\Psi[b^1]=(\Lambda[b^1],\vartheta[b^1])$ we arrive at

\begin{align}
&\norm{\Pi[\Psi[b^1]](\mathsf{G}-\langle \mathsf{G} \rangle)-\Pi[\Psi[b^2]](\mathsf{G}-\langle \mathsf{G} \rangle)}_{C^{\alpha}(\partial\Omega_{-})} \nonumber \\
& \hspace{4cm}  \leq C\left(\norm{\Lambda^1-\Lambda^2}_{C^{1,\alpha}(\Omega)}+ \norm{\vartheta^2-\vartheta^1}_{C^{\alpha}(\Omega)}\right)\norm{\mathsf{G}-\langle \mathsf{G} \rangle)}_{C^{\alpha}(\partial\Omega_{-})} \label{cota:diferencia:psi:b1b2}
\end{align}

\begin{align*}
\norm{\Pi[\Psi[b^1]]}_{\mathcal{L}(C^{1,\alpha}(\partial\Omega_{-}))} \leq C.
\end{align*}

Hence, combining both estimates with the fact that

\begin{equation*}
\norm{\mathsf{G}}_{C^{1,\alpha}(\SS^1)}+	\norm{\langle \mathsf{G}\rangle}_{C^{1,\alpha}(\SS^1)}\leq C \left( \norm{f}_{C^{2,\alpha}(\SS^1)}+ \norm{g}_{C^{2,\alpha}(\SS^1)}\right)\leq C\epsilon_{0}M
\end{equation*}

and using (<ref>) we find that

\begin{align}
\norm{T[b^1,\Pi[\Psi[b^1]](\mathsf{G}-\langle \mathsf{G}\rangle)-T[b^2,\Pi[\Psi[b^2]](\mathsf{G}-\langle \mathsf{G}\rangle)}_{C^{1,\alpha}(\Omega)} &\leq C\epsilon_{0}M\bigg[\norm{\Lambda^1-\Lambda^2}_{C^{1,\alpha}(\Omega)}\nonumber \\
&+ \norm{\vartheta^2-\vartheta^1}_{C^{\alpha}(\Omega)} + \norm{b_{1}-b_{2}}_{C^{1,\alpha}(\Omega)}\bigg]. \label{combination:7}
\end{align}

On the other hand using equation (<ref>) and noticing that the first term on the right hand side of (<ref>) cancels out we infer that

\begin{align*}
\abs{J^{1}-J^{2}} &\leq \frac{1}{2\pi} \displaystyle\sum_{i=1}^{4}\int_{0}^{2\pi}  \abs{\mathsf{T}_{i}[\Psi[b^{1}]]j^{1}_{0}(x) -  \mathsf{T}_{i}[\Psi[b^{2}]]j^{2}_{0}(x)} \ dx \\
&  \hspace{3cm}+ \frac{1}{2\pi}\int_{0}^{2\pi} \abs{\left(j^{1}_0(x)-j^{2}_{0}(x)\right)f^{-}(x)} \ dx  \\
& \leq C\epsilon_{0}M\bigg[\norm{\Lambda^1-\Lambda^2}_{C^{1,\alpha}(\Omega)}+ \norm{\vartheta^2-\vartheta^1}_{C^{\alpha}(\Omega)}+ \norm{b_{1}-b_{2}}_{C^{1,\alpha}(\Omega)}\bigg]
\end{align*}

where we have argued as in the derivation of (<ref>) and using that $j^{\ell}_0= \Pi[\Psi[b^{\ell}]](\mathsf{G}-\langle \mathsf{G}\rangle)$ with $\ell=1,2.$

Combining the later estimate with (<ref>), (<ref>) and making use of the estimates (<ref>)-(<ref>) in Lemma <ref> we conclude

\begin{equation}
\norm{\Gamma(b^1)-\Gamma(b^2)}_{C^{1,\alpha}(\Omega)}\leq C\epsilon_{0}M\norm{b_{1}-b_{2}}_{C^{1,\alpha}(\Omega)} \leq \beta \norm{b_{1}-b_{2}}_{C^{1,\alpha}(\Omega)}
\end{equation}

where $\beta$ is strictly less than one for  $\epsilon_{0}= \frac{1}{2CM}$. Therefore,

$$\Gamma: (B_{M}(C^{2,\alpha}(\Omega)), \norm{\cdot}_{C^{1,\alpha}}) \to (B_{M}(C^{2,\alpha}(\Omega)), \norm{\cdot}_{C^{1,\alpha}})$$

is a contraction mapping for $M\leq M_{0}$.
Invoking Banach fixed point theorem we find that $\Gamma$ admits a unique fixed point	$b\in B_{M}(C^{2,\alpha}(\Omega))$ and thus $\Gamma(b)=b$, which concludes the proof.

§ PROOF OF THEOREM <REF>

Take $\epsilon_{0}>0$ and $M_{0}=M_{0}(L,\alpha)$ be the constants defined in Theorem <ref>. Let also $M\leq M_{0}$. Then, Theorem <ref> implies that $\Gamma$ has a unique fixed point $b\in B_{M}(C^{2,\alpha}(\Omega))$.
We claim that if $b\in B_{M}(C^{2,\alpha}(\Omega))$ is a fixed point operator of $\Gamma$ then $B=(0,1)+b$ is the velocity field which is a solution
$(B,p)\in C^{{2,\alpha}}(\Omega)\times C^{{2,\alpha}}(\Omega)$ to (<ref>) satisfying the boundary conditions

B\cdot n =1+f \ \mbox{on} \ \partial \Omega, \ B\cdot \tau=g  \ \mbox{on} \ \partial \Omega_{-}.

On the one hand, assuming that $b\in B_{M}(C^{2,\alpha}(\Omega))$ is a fixed point operator of $\Gamma$ it is straightforward to check by construction (see that $b$ solves (<ref>)) that

\begin{equation*}
\nabla \cdot B= \nabla \cdot b=0, \mbox{ in } \Omega, \quad B\cdot n= 1+b\cdot n=1+f,   \mbox{ on } \partial\Omega.
\end{equation*}

On the other hand, since $b$ is a fixed point of of $\Gamma$ we find that

\begin{equation*}
\nabla \times B= \nabla\times b=\nabla\times \Gamma(b)=j
\end{equation*}

where in the last equality we have use the first equation in (<ref>)  where $j$ solves the transport system (<ref>). Thus,

\begin{equation*}
0=(B\cdot\nabla)j= \nabla\times \big[j\times B], \mbox{ in } \Omega
\end{equation*}

and $j_0$ as in (<ref>). Then we can define a uni-valued function $p$ in $\Omega$ given by means of

\begin{equation}\label{construcion:p:non:final}
p(\bm{x})=\int_{\bm{0}}^{\bm{x}} \big[j\times B\big](\bm{y})
\ d\bm{y}
\end{equation}

where the integral on the right hand side is the line integration computed along any curve connecting $\bm{0}=(0,0)$ and $\bm{x}\in\Omega$. In order to check that $p$ is a uni-valued function on $\Omega$, we only need to show that (<ref>) holds or equivalently that (<ref>) is satisfied. However, this follows because $J$ has been chosen as in (<ref>) (cf. (<ref>)).

Finally, since $B\in C^{2,\alpha}(\Omega)$ and $j\in C^{1,\alpha}(\Omega)$ it follows from (<ref>) that $p\in C^{2,\alpha}(\Omega)$ and

$$j\times B=\nabla p, \mbox{ in } \Omega$$
Therefore, $(B,p)\in C^{{2,\alpha}}(\Omega)\times C^{{2,\alpha}}(\Omega)$ solves (<ref>).

§.§ Checking the tangential boundary value condition

To conclude the proof of Theorem <ref>, it is only left to show that $B\cdot \tau=g, \mbox{ on } \partial\Omega_{-}.$  To that purpose, let us first show some consequences of the a priori estimates in Section <ref>.

Let $j_{0}\in C^{1,\alpha}(\partial\Omega_{-})$. Then, for $i=1,\ldots,4$ the operators

\begin{equation}
\mathcal{T}_{i}^{NL}j_{0}(x)=\mathcal{T}_{0}^{NL}\mathsf{T}_{i}j_{0}(x): C^{1,\alpha}(\partial\Omega_{-})\to C^{2,\alpha}(\partial\Omega_{-})
\end{equation}

are well defined operators.  Furthermore, they can be expressed as the perturbation of convolution operators given in (<ref>).

By means of Proposition <ref>, Proposition <ref> and Proposition <ref> we have that for $j_0\in C^{1,\alpha}(\partial\Omega_{-})$ the operators $\mathsf{T}_{i}j_{0}$ are well defined and $\mathsf{T}_{i}j_{0}\in C^{1,\alpha}(\partial\Omega_{-})$, $i=1,\ldots,4.$  On the other hand, for $j_{0}\in C^{1,\alpha}(\partial\Omega_{-})$, the operator $\mathcal{T}_{0}^{NL}j_{0}$ given as in (<ref>) is well defined and $\mathcal{T}_{0}^{NL}j_{0}\in C^{2,\alpha}(\partial\Omega_{-})$. Thus combining both facts yields that

$$\mathcal{T}_{i}^{NL}=\mathcal{T}_{0}^{NL}\mathsf{T}_{i}: C^{1,\alpha}(\partial\Omega_{-})\to C^{2,\alpha}(\partial\Omega_{-}), \quad  i=1,\ldots,4.$$

We now show that we can express $\mathcal{T}_{i}^{NL}j_{0}(x)$ as the convolution operators given in (<ref>), for $i=1,\ldots,4.$ We will just provide the proof for $i=1$, since the cases $i=2,3,4$ are very similar. Recalling that the $\mathsf{T}_{1}$ is understood as the limit operator (<ref>)

\begin{equation}
\mathsf{T}_{1}j_{0}(x)=- \frac{1}{2\pi}\displaystyle \lim_{\epsilon\to 0^{+}}\int_{\mathbb{S}^{1}} \mathfrak{G}_{1,\epsilon}(x-\eta,\eta)j_{0}(\eta) \ d\eta:= \displaystyle \lim_{\epsilon\to 0^{+}}\mathsf{T}_{1,\epsilon}j_{0}(x)
\end{equation}

we have that

$$\mathcal{T}_{0}^{NL}\mathsf{T}_{1}j_{0}=\mathcal{T}_{0}^{NL}\lim_{\epsilon\to 0^{+}}\mathsf{T}_{1,\epsilon}j_{0}(x).$$

By means of Proposition <ref>, we have shown the uniform estimate

\begin{equation}
\norm{\mathsf{T}_{1,\epsilon}j_{0}}_{C^{1,\alpha}(\partial\Omega_{-})}\leq C, \mbox{ and }    \lim_{\epsilon\to 0^{+}}\mathsf{T}_{1,\epsilon}j_{0}(x)=\mathsf{T}_{1}j_{0}(x)
\end{equation}

Therefore, by the Lebesgue Dominated Convergence Theorem and the fact that
$\mathcal{T}_{0}^{NL}$ has an integrable kernel, we conclude that

\begin{equation}
\mathcal{T}_{0}^{NL}\lim_{\epsilon\to 0^{+}}\mathsf{T}_{1,\epsilon}j_{0}(x)=  \lim_{\epsilon\to 0^{+}}(\mathcal{T}_{0}^{NL}\mathsf{T}_{1,\epsilon}j_{0})(x)
\end{equation}

Using the formal Fourier computations in Section <ref>, we have that for $\epsilon>0$

\begin{align*}
\mathcal{T}_{0}^{NL}\mathsf{T}_{1,\epsilon}j_{0}(x)=\int_{\SS^1}\mathcal{G}_{1,\epsilon}^{NL}(x-\eta,\eta) \omega_0(\eta) d\eta
\end{align*}

\begin{equation}
\mathcal{G}_{1,\epsilon}^{NL}(x,\eta)= \displaystyle\sum_{n=-\infty}^{n=\infty}e^{inx}\int_{\epsilon}^{L}e^{-\abs{n}y} \left[ \frac{e^{-in\Lambda(\eta,y)}-1}{1+\partial_{\xi} \Theta(X(\eta,y),y)} \right]  dy.
\end{equation}

Computing the summation in $n$, in a similar fashion in (<ref>) we find that

$$\abs{\mathcal{G}_{1,\epsilon}^{NL}(x,\eta)}\leq C \displaystyle\log(\abs{x}+\epsilon) \leq C \displaystyle\log(\abs{x})$$

and hence by Lebesgue Dominated Convergence Theorem we conclude that

\begin{equation}
\lim_{\epsilon\to 0^{+}}\int_{\SS^1}\mathcal{G}_{1,\epsilon}^{NL}(x-\eta,\eta) \omega_0(\eta) d\eta= \int_{\SS^1} \lim_{\epsilon\to 0^{+}}\mathcal{G}_{1,\epsilon}^{NL}(x-\eta,\eta) \omega_0(\eta) d\eta.
\end{equation}

Moreover $ \lim_{\epsilon\to 0^{+}}  \mathcal{G}_{1,\epsilon}^{NL}(x,\eta)\to \mathcal{G}_{1}^{NL}(x,\eta),  \forall x\neq 0$. Therefore, combining the previous computations we obtain that

\begin{equation}
\mathcal{T}_{1}^{NL}\omega_0(x)= \mathcal{T}_{0}^{NL}\mathsf{T}_{1}\omega_0(x)= \int_{\SS^1}\mathcal{G}_{1}^{NL}(x-\eta,\eta)j_{0}(\eta
) \ d\eta
\end{equation}

which shows the desired asserted expression as in (<ref>).

The following lemma gives the tangential velocity in terms of the Biot-Savart system (<ref>).

Let  $b\in C^{2,\alpha}(\Omega), X^{-1}\in C^{2,\alpha}(\Omega)$,  $f\in C^{2,\alpha}(\partial\Omega)$ and $j_{0}\in C^{1,\alpha}(\partial\Omega_{-})$ satisfy the following system

\begin{equation}\label{div:curl:problem:full:fixed:2}
\left\lbrace
\begin{array}{lll}
\nabla\times b= j_{0}(X^{-1}(x,y)), \ \mbox{in} \ \Omega \\
\mbox{div } b=0, \ \mbox{in} \ \Omega \\
b\cdot n=f, \ \mbox{on} \ \partial \Omega \\
\int_{0}^{L} b_{1}(0,y) \ dy=J.
\end{array} \right.
\end{equation}

with $J$ as in (<ref>) (cf. (<ref>)).

Then, we have that

$$(b\cdot \tau)(x,0)=-\frac{J}{L}+\mathcal{Z}(x) +\frac{1}{2\pi}\mathcal{T}_{0}^{NL}j_{0}(x)+\frac{1}{2\pi}\displaystyle\sum_{i=1}^{4}\mathcal{T}_{i}^{NL}j_{0}(x)$$

$$\mathcal{Z}(x)=\frac{1}{2\pi} \displaystyle\sum_{n=-\infty}^{n=\infty}\left(\widehat{h^{+}}(n) \frac{\abs{n}}{\sinh (\abs{n}L)}-
\widehat{h^{-}}(n)\frac{\abs{n}}{\tanh (\abs{n}L)}\right) e^{inx}. $$

Arguing as in Subsection <ref>, we have that since $b$ solves (<ref>) there exists a stream function $\psi$ such that

\begin{equation}
\left\lbrace
\begin{array}{lll}
\Delta \psi= j_{0}(X^{-1}(x,y)),\ \mbox{in } \Omega \\
\psi(x,L)= -J+\displaystyle\int_{0}^{x} (f(\xi,L)-A) \ d\xi  ,  \ x\in \mathbb{R} \\
\psi(x,0)= \displaystyle\int_{0}^{x} (f(\xi,0)-A) \ d\xi  ,  \ x\in \mathbb{R}
\end{array}\right.
\end{equation}

for $A=\int_{\partial\Omega_{+}} f \ dS=\int_{\partial\Omega_{-}} f\ dS$. Moreover, using the fundamental solution $\Phi(x,y)$ solving the problem

\begin{equation}
\left\lbrace
\begin{array}{lll}
\Delta \Phi (x,y)= \delta(x)\delta(y-y_{0}), \ \mbox{in } \Omega, \\
\Phi=0, \ \mbox{on } \partial \Omega. \\
\end{array}\right.
\end{equation}

we can readily check (cf. Subsection <ref>) that the normal derivative at $y=0$ is given by

\[\partial_{y} \Phi(x,0,y_{0})= -\frac{1}{2\pi} \displaystyle\sum_{n=-\infty}^{n=\infty} \frac{\sinh(n(L-y_{0}))}{\sinh(nL)} e^{inx}. \]

Computing an homogeneous solution and imposing the boundary value conditions using Fourier techniques as in (<ref>) in Section <ref>
we conclude  that

\begin{equation}
\partial_{y} \psi (x,0)= -\frac{J}{L}+
\mathcal{Z}(x)-\frac{1}{2\pi}\int_{0}^{L} dy_{0} \int_{\mathbb{S}^{1}}\displaystyle\sum_{n=-\infty}^{n=\infty} \frac{\sinh(n(L-y_{0}))}{\sinh(nL)} e^{in(x-\xi)}j_{0}(X^{-1}(\xi,y_{0})) \ d \xi \label{psi:invers:full:2}
\end{equation}

where $\mathcal{Z}(x)$ as in (<ref>). Since $f\in C^{2,\alpha}(\partial\Omega)$, it is straightforward to check that boundary condition term $\mathcal{Z}(x)$ is well defined.
On the other hand, invoking Corollary <ref> we have that for $j_0\in C^{1,\alpha}(\partial\Omega_{-})$ the operators

\begin{equation*}
\mathcal{T}_{i}^{NL}j_{0}(x)=\mathcal{T}_{0}^{NL}\mathsf{T}_{i}j_{0}(x): C^{1,\alpha}(\partial\Omega_{-})\to C^{2,\alpha}(\partial\Omega_{-})
\end{equation*}

are well defined operators and can be expressed as the convolution operators given in (<ref>). Hence, by recalling the definition  (<ref>) we infer that

\begin{equation*}
\mathcal{T}^{NL}j_{0}(x)=\mathcal{T}^{NL}_{0}j_{0}(x)+ \displaystyle\sum_{i=1}^{4}\mathcal{T}^{NL}_{i}j_{0}(x)
\end{equation*}

it admit the representation formula

\begin{align*}
\mathcal{T}^{NL}j_{0}(x)=-\frac{1}{2\pi}\int_{\mathbb{S}^1}\mathcal{G}^{NL}(x-\eta,\eta) j_{0}(\eta)  \ d\eta,
\end{align*}

\begin{align*}
\mathcal{G}^{NL}(x,\eta) = \ \displaystyle\sum_{n=-\infty}^{n=\infty}  a_{n}(\eta)e^{inx}, a_{n}(\eta)=  \int_{0}^{L} \frac{\sinh(n(L-y))}{\sinh(nL)} \frac{e^{-in \Lambda (\eta,y)}}{(1+\partial_{\xi} \Theta(X(\eta,y),y))} dy.
\end{align*}

where $\Lambda(\eta,y)=X(\eta,y)-\eta \in C^{2,\alpha}(\Omega)$ and $X^{-1}(\xi,y)=\xi+\Theta(\xi,y)=\eta$. Unraveling notation (cf. computations (<ref>)-(<ref>)), it is easy to check that

\begin{equation}\label{def:TNL:sec7}
\mathcal{T}^{NL}j_{0}(x)=-\frac{1}{2\pi}\int_{0}^{L} dy_{0} \int_{\mathbb{S}^{1}}\displaystyle\sum_{n=-\infty}^{n=\infty} \frac{\sinh(n(L-y_{0}))}{\sinh(nL)} e^{in(x-\xi)}j_{0}(X^{-1}(\xi,y_{0})) \ d \xi
\end{equation}

is a well defined operator. Therefore, combining (<ref>)-(<ref>) and noticing that $-\dfrac{\partial \psi }{\partial y}=b\cdot \tau$ on $\partial\Omega_{-}$ provides our claim.

We have that $B\cdot\tau =g \ \mbox{on} \ \partial \Omega_{-}.$

Applying Lemma <ref> and noticing that by construction $j_0(x)$ solves

\begin{equation}\label{full:integral:eq:2}
\mathcal{T}^{NL}j_{0}(x)=-\frac{1}{2\pi}\int_{\mathbb{S}^1}\mathcal{G}^{NL}(x-\eta,\eta) j_{0}(\eta)  \ d\eta=\widetilde{g}+\frac{J}{L}
\end{equation}

where $\widetilde{g}(x)=-g(x)-\mathcal{Z}(x)$ we conclude that  $B\cdot\tau=b\cdot \tau=g  \ \mbox{on} \ \partial \Omega_{-}$.

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