\langl\lt|\int_{\Boule(x)} \phiC - \int_{\Boule(x')} \phiC\rt|^p \rangl^{\frac{1}{p}}
\lesssim
\int_{\mathcal{A}} \langl
\lt(\int_{\Boule(z)} |\nabla \phiC|^2\rt)^{\frac{p}{2}}\rangl^{\frac{1}{p}} |\dd z|.
\end{align*}

Last, inserting (<ref>) into the above estimate, we easily get (<ref>).

Step 3 : Conclusion.
By the Poincaré-Wirtinger estimate, invoking (<ref>) and (<ref>), we have

\begin{equation}\label{key7}
\langl
\lt( \fint_{\Dom(x)} |\phitilde_n|^2 \rt)^{\frac{p}{2}}
\rangl^{\frac{1}{p}}
\lesssim_{\Xi,p,n} (|x|+1)^{\rhobar_n-\nu} \ln^{\tilde{\nu}'}(|x|+2).
\end{equation}

Next, notice that the function $u(\cdot) :=\taubar_n(\cdot)+\phiC_{n}(\cdot) -\taubar_n(x)$
is $a$-harmonic in $\Dom(x)$.
Hence, we may apply the De Giorgi-Nash-Moser theorem <cit.> to $u$ in the form $|\phiC_{n}(x)|=|u(x)|\lesssim_{\lambda,\omega} (\fint_{\Dom(x)}|u|^2)^\frac12$.
Appealing to the triangle inequality on the decomposition (<ref>),
into which we insert (<ref>), (<ref>), and (<ref>), we finally obtain (<ref>).

§ ERROR ESTIMATE FOR THE NONSTANDARD $2$-SCALE EXPANSION: ARGUMENT FOR THEOREM <REF>

§.§ Construction of the corner flux-correctors: Proof of Lemma <ref>

In this proof, the constant $\tilde{\nu}'\lesssim_{\tilde{\nu}} 1$ may change from line to line.
Let $i = 1,2$. This proof proceeds in three steps :
In Step 1, we construct $\tisigmaD_i$; this relies on the extension of a given divergence-free field, that we establish in Step 2.
In Step 3, we establish

\begin{equation}\label{E:nablasigD-2}
\langl  \Big| \fint_{\Dom(x)} |\nabla \tisigmaD_i |^2 \Big|^{\frac{p}{2}} \rangl^{\frac{1}{p}}
\lesssim  (|x|+1)^{-\nu} \ln^{\tilde{\nu}'}(|x|+2) \qquad \pourtout p \in [1,\infty) \et x \in \Dom.
\end{equation}

Last, in Step 4, we prove (<ref>).

Step 1 : Construction of $\tisigmaD_i$.
We begin by constructing the skew-symmetric correction $\tisigmaD_i$ in (<ref>).
From (<ref>) we see that $\tisigmaD_i$ must satisfy

\begin{align}
\label{equation_sigma_tilde_1}
\nabla \cdot \tisigmaD_i = a (\nabla \phiD_i + e_i) - \abar e_i - \nabla \cdot \lt( (1 -\eta_{\Boule,1} ) \lt( \etaup \sigup_i + \etabulk \sigma_i + \etadown \sigdown_i\rt) \rt) \dans \Dom.
\end{align}

Using the ansatz for $\phiD_i$ given by (<ref>) with $\bar{\tau}_n$ replaced by $x_i$, i.e.

\begin{align*}
\phiD_i=(1-\eta_{\Boule,1})\lt( \etaup \phiup_i + \etabulk \phi_i + \etadown \phidown_i \rt) + \phiDtilde_i,
\end{align*}

the equations (<ref>) satisfied by $\sigup_i$ and $\sigdown_i$, and, similarly to (<ref>), the notation

\begin{equation}
\label{defn_hi}
\begin{aligned}
h_i := \eta_{\Boule,1} (a-\abar) e_i
&+\lt(a\phiup_i- \sigup_i\rt)
\nabla \lt( (1-\eta_{\Boule,1}) \etaup   \rt)
+ \lt(a\phi_i - \sigma_i \rt) \nabla \lt( (1-\eta_{\Boule,1}) \etabulk \rt)
\\
&+\lt(a\phidown_i- \sigdown_i\rt)
\nabla \lt( (1-\eta_{\Boule,1})\etadown \rt),
\end{aligned}
\end{equation}

the of (<ref>) becomes

\begin{align}
\label{rewrite_rhs}
\begin{split}
&a \nabla \phiDtilde_i + a \phiup_i \nabla( (1 -\eta_{\Boule,1} ) \etaup) + a \phi_i \nabla( (1 -\eta_{\Boule,1} ) \etabulk) + a \phidown_i \nabla( (1 -\eta_{\Boule,1} ) \etadown)
\\
&+ (1 -\eta_{\Boule,1} )\lt( \etaup a \nabla \phiup_i + \etabulk a \nabla \phi_i + \etadown a \nabla \phidown_i \rt) + (a - \abar) e_i
\\
&- \sigup_i \nabla( (1 -\eta_{\Boule,1} ) \etaup)
- \sigma_i \nabla( (1 -\eta_{\Boule,1} ) \etabulk)
- \sigdown_i \nabla( (1 -\eta_{\Boule,1} )\etadown)
\\
&-(1-\eta_{\Boule,1}) \lt(\etaup \lt(a\nabla\phiup_i + (a-\abar)e_i\rt)
+\etabulk \lt(a\nabla\phi_i + (a-\abar)e_i\rt)
+\etadown \lt(a\nabla\phidown_i + (a-\abar)e_i\rt)\rt)
\\
&= a \nabla \phiDtilde_i + h_i.
\end{split}
\end{align}

Combining (<ref>) and (<ref>), we obtain that $\tisigmaD_i$ is required to solve

\begin{align}
\label{equation_sigma_tilde}
\nabla \cdot \tisigmaD_i =a \nabla \phiDtilde_i + h_i \dans \Dom.
\end{align}

By adapting (<ref>) to $\phiDtilde_i$, we immediately obtain that the of (<ref>) is divergence-free.
To construct a solution of (<ref>) we emulate previous constructions of adapted vector potentials, e.g. in [23] or [32], and for now take for granted the existence of an extension $g_i$ of $a\nabla \phiDtilde_i + h_i$ to $\R^2$ that is divergence-free and satisfies

\begin{align}
\label{growth_g}
\langl \lt( \fint_{B_1(x)} \lt| g_i \rt|^2 \rt)^{\frac{p}{2}} \rangl^{\frac{1}{p}}
\lesssim (|x|+1)^{-\nu} \ln^{\tilde{\nu}'}( |x| +2) \qquad \pourtout x \in \R^2 \et p \in [1, \infty).
\end{align}

Then, for $j=1,2$, we seek strictly subquadratic solutions to

\begin{align}
\label{def_Nji}
-\Delta N_{ij}  = g_i \cdot e_j \dans \R^2
\end{align}

and take the ansatz

\begin{align}
\label{sigma_correction_ansatz}
\tisigmaD_{ijk} = \partial_j N_{ik} - \partial_k N_{ij}.
\end{align}

Taking the divergence of the ansatz for $\tisigmaD_{ijk}$, we see that

\begin{align*}
\partial_k \tisigmaD_{ijk}
= \partial_k\partial_j N_{ik}  - \partial_k\partial_k N_{ij} = g_i \cdot e_j \dans \R^2,
\end{align*}

where we have used (<ref>), and the Liouville principle for harmonic functions and that $g$ is divergence-free to deduce that $\partial_k N_{ik}$ is a constant.

To complete our construction of $\tisigmaD_{i}$ it remains to obtain strictly subquadratic solutions of the equations (<ref>) and to show the existence of the extension $g$. To accomplish the former task we notice that because $g$ is divergence-free, (<ref>) can be re-written in divergence-form as

\begin{align}
\label{def_Nji_2}
-\Delta N_{ij}  = \nabla \cdot \lt(x \cdot e_j g_i\rt) \dans \R^2.
\end{align}

Using a covering argument and (<ref>) we see that the relation

\begin{align}
\label{growth_g_2}
\langl \lt( \fint_{\Boule_{r}} \lt| x \cdot e_j g_i \rt|^2 \rt)^{\frac{p}{2}} \rangl^{\frac{1}{p}}
\lesssim
\sup_{x' \in \Boule_r}
\langl \lt( \fint_{\Boule_{1}(x')} \lt| x \cdot e_j g_i \rt|^2 \rt)^{\frac{p}{2}} \rangl^{\frac{1}{p}}
\overset{\eqref{growth_g}}{\lesssim} r^{1-\nu} \ln^{\tilde{\nu}'}(r+1)
\end{align}

holds for $r\geq 1$ and $p\in [1, \infty)$.

By <cit.>, there exists almost-surely a distributional solution $N_{ij} \in H^{1}_{\loc}(\R^2)$ of (<ref>) satisfying

\begin{align}
\label{subquadratic}
\limsup_{r \uparrow \infty} r^{-1} \lt( \fint_{\Boule_{r}} \lt|\nabla N_{ij} - \fint_{\Boule_r} \nabla N_{ij}  \rt|^2 \, \dd x \rt)^{\frac{1}{2}} =0.
\end{align}

Furthermore, it also follows from <cit.>, using (<ref>), that this solution satisfies

\begin{align}
\label{estimate_Nji_1}
\langl \lt( \fint_{\Boule_r} \lt|\nabla N_{ij} - \fint_{\Boule_r} \nabla N_{ij}  \rt|^2 \, \dd x \rt)^{\frac{p}{2}} \rangl^{\frac{1}{p}} \lesssim_{\nu} r^{1-\nu} \ln^{\tilde{\nu}'}(r+1) \pourtout r\geq 1.
\end{align}

Step 2 : Divergence-free extension of $a\nabla \phiDtilde_i + h_i$.
By Lemma <ref> of the Appendix,
we may construct the extension $g_i$ of $a\nabla \phiDtilde_i + h_i$  to $\R^2$ that has been used in Step 1.

We then check that (<ref>) is satisfied.
Let $x \in \Dom$ and $p \in [1,\infty)$.
By (<ref>), using Proposition <ref>, we see that the analogue of (<ref>) is

\begin{align}
\label{analogue_1}
\langl \lt(\fint_{\Dom(x)} |h_i|^2\rt)^{\frac{p}{2}} \rangl^{\frac{1}{p}}
\lesssim
(|x|+1)^{-\nu} \ln^{\tilde{\nu}'}(|x|+2).
\end{align}

Furthermore, using the argument in Step 1 of the proof of Theorem <ref> and (<ref>), we see that the analogue of (<ref>) is

\begin{align}
\label{analogue_2}
\langl \lt(\fint_{\Dom(x)} |\nabla \phiDtilde_i |^2\rt)^{\frac{p}{2}} \rangl^{\frac{1}{p}}
\lesssim
(|x|+1)^{-\nu} \ln^{\tilde{\nu}'}(|x|+2).
\end{align}

Thus, obtaining (<ref>) is a simple matter of combining (<ref>) and (<ref>) with the definition of $\bar{H}$ in Lemma <ref>.

Step 3: Argument for (<ref>).

W. l. o. g., we assume that $|x|\geq 8$.
By the ansatz (<ref>), the Caccioppoli estimate, and then
(<ref>) and (<ref>) combined with a covering argument in which we use the Minkowski inequality, we obtain that

\begin{align*}
\langl \Bigg( \fint_{\Boule_{\frac{|x|}{4}}(x)} |\nabla \tisigmaD_i|^2 \Bigg)^{\frac{p}{2}} \rangl^{\frac{1}{p}} & \lesssim \langl \Bigg( \fint_{\Boule_{\frac{|x|}{4}}(x)} |\nabla^2 N_i |^2 \Bigg)^{\frac{p}{2}} \rangl^{\frac{1}{p}}\\
&  \lesssim \frac{1}{|x|}  \langl \Bigg( \fint_{\Boule_{\frac{|x|}{2}}(x)} |\nabla N_i - \fint_{\Boule_{\frac{|x|}{2}}(x)} \nabla N_i |^2 \Bigg)^{\frac{p}{2}} \rangl^{\frac{1}{p}}  +  \langl \Bigg( \fint_{\Boule_{\frac{|x|}{2}}(x)} |g_i |^2 \Bigg)^{\frac{p}{2}} \rangl^{\frac{1}{p}}\\
&\lesssim |x|^{-\nu} \ln^{\tilde{\nu}'}|x|.
\end{align*}

To finish the argument for (<ref>) we resort to the following estimate:

\begin{align}
\label{MVP_sigma}
\begin{split}
\langl \Bigg( \fint_{\Boule(x)} |\nabla  \tisigmaD_i |^2 \Bigg)^{\frac{p}{2}} \rangl^{\frac{1}{p}} & \lesssim \langl \Bigg(\fint_{\Boule_{\frac{|x|}{4}}(x)} |\nabla  \tisigmaD_i |^2  \Bigg)^{\frac{p'}{2}} \rangl^{\frac{1}{p'}} +  |x|^{-\nu} \ln^{\tilde{\nu}'}|x| \qquad \pourtout p' > p,
\end{split}
\end{align}

which is proved below.
Indeed, the conjunction of the two above estimates gives (<ref>).

Here comes the proof of (<ref>).
By applying the Laplacian to (<ref>) and using (<ref>), we obtain

\begin{align*}
-\Delta \tisigmaD_{ijk} = - \partial_j \Delta N_{ik} + \partial_k \Delta N_{ij} = \partial_j (g_i e_k) - \partial_k (g_i e_j).
\end{align*}

Thus, we may apply Corollary <ref>,
in the easier case where $a=\Id$ and where the sets $\Dom_r$ are replaced by balls $\Boule_r$ (furthermore, no boundary condition needs to be considered).
This yields (<ref>) through

\begin{align}
\label{M:0040}
\langl \lt( \fint_{\Boule(x)} |\nabla  \tisigmaD_i |^2 \rt)^{\frac{p}{2}} \rangl^{\frac{1}{p}}
& \lesssim \langl \lt(\fint_{\Boule_{\frac{|x|}{4}}(x)} |\nabla  \tisigmaD_i |^2  \rt)^{\frac{p'}{2}} \rangl^{\frac{1}{p'}}
+\ln\left( |x| \right)
\sup_{x' \in \Boule_{\frac{|x|}{2}}(x)}
\langl \lt( \fint_{\Boule(x')} |g_i|^2 \rt)^{\frac{p'}{2}} \rangl^{\frac{1}{p'}},
\end{align}

into which we insert (<ref>).

Step 4: Argument for (<ref>).
Let $x \in \Dom$.
We integrate

\begin{align*}
\lt| \fint_{\Boule(x)} \tisigmaD_i -  \fint_{\Boule} \tisigmaD_i \rt|
\leq
|x|\lt| \int_{0}^{1} \lt(\fint_{\Boule(tx)} \nabla \tisigmaD_i\rt)  \dd t \rt|.
\end{align*}

Hence, by the Minkowski inequality and (<ref>), we get

\begin{align*}
\langl\lt| \fint_{\Boule(x)} \tisigmaD_i -  \fint_{\Boule} \tisigmaD_i \rt|^{p} \rangl^{\frac{1}{p}}
\leq &
\langl \lt(\fint_{\Boule(tx)} |\nabla \tisigmaD_i|^2\rt)^{\frac{p}{2}} \rangl^{\frac{1}{p}} \dd t
\\
\lesssim&
(|x|+1)^{1-\nu} \ln^{\tilde{\nu}'}(|x|+2) \int_0^1 t^{-\nu} \dd t
\lesssim (|x|+1)^{1-\nu} \ln^{\tilde{\nu}'}(|x|+2).
\end{align*}

Using the Poincaré inequality as well as (<ref>), this implies

\begin{align*}
\langl\lt( \fint_{\Boule(x)} \lt|\tisigmaD_i -  \fint_{\Boule} \tisigmaD_i\rt|^2 \rt)^{\frac{p}{2}} \rangl^{\frac{1}{p}}
\lesssim (|x|+1)^{1-\nu} \ln^{\tilde{\nu}'}(|x| +2).
\end{align*}

Up to substracting a constant in the definition of $\tisigmaD_{i}$, we finally get (<ref>)
by appealing to the decomposition (<ref>),
into which we insert (<ref>) and (<ref>).

§.§ Equation satisfied by the nonstandard 2-scale expansion (<ref>): Proof of Lemma <ref>

For use later on, in order to avoid singularities at the tip of the sector (the small scales),
we show a generalization of (<ref>).
In particular, as is seen in (<ref>) below,
we resort to the use of additional singular functions $\taubar_n$ with a very sharp cut-off near the corner.

Step 1 : Generalizing (<ref>).
We choose $M \in \NN$, $\chi:=\eta_{\Boule,1}$, and the cut-off function $\chi_\epsilon := \eta_{\Boule,\epsilon}$, and we generalize (<ref>) by the following 2-scale expansion :

\begin{equation}\label{2-scale-New-bis}
\utieps^{N,M} := (1+\phiD_{\epsilon,i}\partial_i) \ubar_{\reg}^{N,M} + \sum_{n=1}^N \bar{\gamma}_n (\taubar_n + \phiC_{\epsilon,n})\chi + \sum_{n=1}^N \bar{\gamma}_n \taubar_n \phiD_{\epsilon,i} \partial_i \chi
+ \sum_{n=N+1}^{N+M} \bar{\gamma}_n \taubar_n  \chi_\epsilon,
\end{equation}

\begin{equation}\label{Num:7003-bis}
\ubar_{\reg}^{N,M} := \ubar - \sum_{n=1}^N \bar{\gamma}_n \taubar_n \chi - \sum_{n=N+1}^{N+M} \bar{\gamma}_n \taubar_n  \chi_\epsilon.
\end{equation}

We shall establish that

\begin{equation}\label{E:1}
-\nabla \cdot a_\epsilon (\nabla \utieps^{N,M}  - \nabla \ueps)
\nabla \cdot h^{N,M}_\epsilon \quad \dans \Dom, \qquad \text{with} \quad \utieps^{N,M} = \ueps = 0 \sur \Gamma,
\end{equation}

\begin{equation}\label{E:1-bis}
h^{N,M}_\epsilon :=\lt(\sigmaD_{\epsilon,i} - a \phiD_{\epsilon,i} \rt) \partial_i \nabla \bar{v}^{N,M}
-\sum_{n=1}^N \bar{\gamma}_n a_\epsilon \nabla \lt((\chi-1)(\phiC_{\epsilon,n} - \phiD_{\epsilon,i} \partial_i \taubar_n)\rt)
-\sum_{n=N+1}^{N+M} \bar{\gamma}_n \lt( a_\epsilon- \abar \rt) \nabla \lt(\chi_\epsilon \taubar_n \rt),
\end{equation}

\begin{equation}\label{Def:barv}
\bar{v}^{N,M} :=\ubar - \sum_{n=1}^N \bar{\gamma}_n \taubar_n -\sum_{n=N+1}^{N+M} \bar{\gamma}_n \taubar_n  \chi_\epsilon  =\ubar_{\reg}^{N,M} + \sum_{n=1}^N \bar{\gamma}_n \taubar_n (\chi-1).
\end{equation}

Clearly, taking $M=0$ in the above identities establishes Lemma <ref>.

Step 2 : Proof of (<ref>).
We compute

\begin{equation}\label{Num:9001}
\begin{aligned}
-\nabla \cdot a_\epsilon (\nabla \utieps^{N,M}  - \nabla \ueps)
\overset{\eqref{Num:7001}}{=}~&
-\nabla \cdot \lt( a_\epsilon \nabla \utieps^{N,M}  - \abar \nabla \ubar \rt)
\\
\overset{\eqref{2-scale-New-bis}, \eqref{Num:7003-bis}}{=}~&
- \nabla \cdot \lt( a_\epsilon \nabla  \lt( (1+\phiD_{\epsilon,i}\partial_i) \ubar_{\reg}^{N,M} \rt)  - \abar \nabla \ubar_{\reg}^{N,M} \rt)
\\
&-\sum_{n=1}^N \bar{\gamma}_n  \nabla \cdot \lt( a_\epsilon \nabla \lt( (\taubar_n + \phiC_{\epsilon,n})\chi + \taubar_n \phiD_{\epsilon,i} \partial_i \chi \rt) - \abar \nabla \lt( \taubar_n \chi \rt) \rt)
\\
&- \sum_{n=N+1}^{N+M} \bar{\gamma}_n \nabla \cdot (a_\epsilon - \abar) \nabla (\chi_\epsilon \taubar_n).
\end{aligned}
\end{equation}

But, since $\taubar_n$ and $\taubar_n + \phiC_{\epsilon,n}$ are $\abar$- and $a_\epsilon$-harmonic, respectively,
the summands in the second term of the read

\begin{align*}
&\bar{\gamma}_n\nabla \cdot \lt( a_\epsilon \nabla \lt( (\taubar_n + \phiC_{\epsilon,n})\chi + \taubar_n \phiD_{\epsilon,i} \partial_i \chi \rt) - \abar \nabla (\taubar_n \chi) \rt)
\\
&\qquad \qquad
\bar{\gamma}_n \nabla \cdot \lt( a_\epsilon \nabla \lt( (\taubar_n + \phiC_{\epsilon,n})(\chi-1) + \taubar_n \phiD_{\epsilon,i} \partial_i (\chi-1) \rt) - \abar \nabla (\taubar_n (\chi-1)) \rt).
\end{align*}

\begin{align*}
(\taubar_n + \phiC_{\epsilon,n})(\chi-1) + \taubar_n \phiD_{\epsilon,i} \partial_i (\chi-1)
=(1+\phiD_{\epsilon,i}\partial_i)(\taubar_n (\chi-1))
+(\phiC_{\epsilon,n} - \phiD_{\epsilon,i} \partial_i \taubar_n)(\chi-1).
\end{align*}

Thus, recalling (<ref>) we may reformulate (<ref>) as

\begin{equation*}
\begin{aligned}
-\nabla \cdot a_\epsilon (\nabla \utieps^{N,M}  - \nabla \ueps)
- \nabla \cdot \lt( a_\epsilon \nabla  \lt( (1+\phiD_{\epsilon,i}\partial_i) \bar{v}^{N,M} \rt)  - \abar \nabla \bar{v}^{N,M} \rt)
\\
&-\sum_{n=1}^N \bar{\gamma}_n \nabla \cdot a_\epsilon \nabla \lt( (\phiC_{\epsilon,n} - \phiD_{\epsilon,i} \partial_i \taubar_n)(\chi-1) \rt)
\\
&- \sum_{n=N+1}^{N+M} \bar{\gamma}_n \nabla \cdot (a_\epsilon - \abar) \nabla (\chi_\epsilon \taubar_n).
\end{aligned}
\end{equation*}

By (<ref>) (replacing $\phiup \rightsquigarrow \phiD$ and $\sigup \rightsquigarrow\sigmaD$), the first term on the can be expressed as

\begin{align*}
-\nabla \cdot \lt( a_\epsilon \nabla \lt( (1+\phiD_{\epsilon,i}\partial_i) \bar{v}^{N,M} \rt) - \abar \nabla \bar{v}^{N,M} \rt)
=-\nabla \cdot \lt(a_\epsilon \phiD_{\epsilon,i} - \sigmaD_{\epsilon,i} \rt) \partial_i \nabla \bar{v}^{N,M},
\end{align*}

so that we get (<ref>) along with (<ref>).

§.§ Preliminary estimates

Under the assumptions of Theorem <ref>, pick $M \in \mathbb{N}$ sufficiently large so that

\begin{equation}\label{Def:M}
\rhobar_{M+1} \geq 2.
\end{equation}

Then, there exists an exponent $\tilde{\nu}'\lesssim_{\tilde{\nu}} 1$ such that $h^{N,M}_\epsilon$ defined by (<ref>) satisfies

\begin{align}\label{M:0003}
\langl\lt( \fint_{\Dom_\epsilon(x)} |h^{N,M}_\epsilon|^2 \rt)^{\frac{p}{2}}\rangl^{\frac{1}{p}}
\lesssim_{\Xi, N, M, p}
\lt\{
\begin{aligned}
&\epsilon^\nu \ln^{\tilde{\nu}'}(\epsilon^{-1}|x|) |x|^{\rhobar_{N}-1-\nu} && \si x \in \Dom \backslash \Dom_{1},
\\
&\epsilon^\nu \ln^{\tilde{\nu}'}(\epsilon^{-1})  |x|^{\rhobar_{N+1}-1-\nu} && \si x \in \Dom_{1} \backslash \Dom_{4\epsilon},
\\
&\epsilon^{\rhobar_{N+1}-1}  && \si x \in \Dom_{4\epsilon},
\end{aligned}
\rt.
\end{align}

for $p \in [1,\infty)$.

In this proof, the constant $\tilde{\nu}'\lesssim_{\tilde{\nu}} 1$ may change from line to line.

In order to show (<ref>), we split

\begin{align*}
h^{N, M}_\epsilon &~=
\underbrace{\lt(\sigmaD_{\epsilon,i} - a_\epsilon \phiD_{\epsilon,i} \rt) \partial_i \nabla \bar{v}^{N,M}}_{h^{N, M}_{\epsilon,1}}
-\underbrace{\sum_{n=1}^N \bar{\gamma}_n a_\epsilon \nabla \lt((\chi-1)(\phiC_{\epsilon,n} - \phiD_{\epsilon,i} \partial_i \taubar_n)\rt)}_{h^{N, M}_{\epsilon,2}}
-\underbrace{\sum_{n=N+1}^{N+M} \bar{\gamma}_n \lt( a_\epsilon- \abar \rt) \nabla \lt(\chi_\epsilon \taubar_n \rt)}_{h^{N, M}_{\epsilon,3}},
\end{align*}

and we estimate each term separately.
Notice that $|\bar{\gamma}_n| \lesssim 1$ by (<ref>), after applying the Cauchy-Schwarz and Poincaré inequalities, followed by the energy estimate for (<ref>).

Step 1: Estimates on $\nabla^2 \bar{v}$
First, we establish that:

\begin{align}
|\nabla^2 \bar{v}^{N,M}(x)|
\lesssim
\lt\{
\begin{aligned}
& |x|^{\rhobar_{N}-2},
&& \pourtout x \in \Dom \backslash \Dom_1,
\\
&& \pourtout x \in \Dom_1 \backslash \Dom_\epsilon,
\\
& \epsilon^{\rhobar_{N+1}-2},
&& \pourtout x \in \Dom_\epsilon.
\end{aligned}
\rt.
\label{Num:7013b}
\end{align}

Notice that these three estimates coincide near the boundaries of their domains of validity.

By the energy estimate for (<ref>), there holds

\begin{equation}\label{Num:10002}
\lt(\int_{\Dom}|\nabla \bar{u}|^2\rt)^{\frac{1}{2}}
\lesssim 1.
\end{equation}

We pick $r \geq 1$.
Since $\ubar$ is harmonic in $\Dom \backslash \Dom_1$,
a duality argument in the spirit of Step 3 of the proof of Lemma <ref> yields

\begin{equation}\label{Num:10001}
\lt(\int_{\Dom_{2r} \backslash \Dom_r}|\nabla \bar{u}|^2\rt)^{\frac{1}{2}}
\lesssim r^{-\rhobar_1}.
\end{equation}

Let $x \in \Dom \backslash \Dom_{1/4}$.
The above estimates (<ref>) and (<ref>) combined with classical Schauder regularity theory applied
to the function $\ubar$ in $\Dom_{|x|/2}(x)$ implies that

\begin{align}\label{Num:10003}
|x|^{-1}\lt|\nabla \ubar_{\reg}^{N}(x)\rt| + \lt|\nabla^2 \ubar_{\reg}^{N}(x)\rt| \lesssim |x|^{-\rhobar_{1}-2}
\end{align}

(note that $\ubar_{\reg}^N=\ubar$ in $\Dom\backslash \Dom_2$, cf. (<ref>)).
Recalling (<ref>), in which the third term on the vanishes in $\Dom \backslash \Dom_{\epsilon}$, this yields the first inequality of (<ref>) in form of

\begin{align}\label{M:0004}
|\nabla^2 \bar{v}^{N,M}(x)|
\lesssim~& |x|^{-\rhobar_1-2} + \sum_{n=1}^{N} |\bar{\gamma}_n| |\nabla^2\taubar_n(x)|
\overset{\eqref{taubar_Homog}}{\lesssim}
|x|^{-\rhobar_1-2} + \sum_{n=1}^{N} |x|^{\rhobar_n -2}
\lesssim |x|^{\rhobar_{{N}}-2}.
\end{align}

Next, we apply Theorem <ref> to $\ubar_{\reg}^N$, which is harmonic $\Dom_{1/2}$.
Thus, for $x \in \Dom_{1/4}$, we obtain :

\begin{align*}
\lt(\fint_{\Dom_{2|x|}} \lt|\ubar_{\reg}^{N}\rt|^2 \rt)^\frac{1}{2}\lesssim |x|^{\rhobar_{{N}+1}} \lt( \fint_{\Dom_1} |\ubar|^2 \rt)^{\frac{1}{2}}
\lesssim |x|^{\rhobar_{{N}+1}}.
\end{align*}

Hence, the classical Schauder regularity theory applied to $\ubar_{\reg}^N$
in $\Dom_{|x|/2}(x)$ yields $\lt|\nabla^2 \ubar_{\reg}^{N}(x)\rt| \lesssim |x|^{\rhobar_{{N}+1}-2}$.
Resorting to the triangle inequality as in (<ref>) combined with $\ubar_{\reg}^{N}=\ubar_{\reg}^{N,M}$ on $D\setminus D_\epsilon$,
this establishes the second estimate of (<ref>).

Last, inside $\Dom_{\epsilon}$, we have

\begin{align*}
\bar{v}^{{N},M} = \ubar_\reg^{{N}+M} + \sum_{n={N}+1}^M \bar{\gamma}_n (\chi_\epsilon - 1) \taubar_n.
\end{align*}

Hence, applying the same reasoning as above to $\ubar_\reg^{{N}+M}$, we get, for $x \in \Dom_\epsilon$, that $|\nabla^2 \ubar_\reg^{N+M}(x)| \lesssim |x|^{\rhobar_{{N}+M+1}-2}$, to the effect of

\begin{align*}
\lt|\nabla^2 \bar{v}^{{N},M}(x)\rt|
\lesssim~& |x|^{\rhobar_{{N}+M+1}-2} + \sum_{n={N}+1}^M |\bar{\gamma}_n| \lt| \nabla^2\lt( (\chi_\epsilon - 1) \taubar_n\rt)(x)\rt|
\\
\overset{\eqref{taubar_Homog}}{\lesssim}~& |x|^{\rhobar_{{N}+M+1}-2} + \sum_{n={N}+1}^M \epsilon^{-2+\rhobar_n} \mathds{1}(|x| > \epsilon / 2)
\overset{\eqref{Def:M}}{\lesssim} \epsilon^{\rhobar_{{N}+1}-2}.
\end{align*}

This finally proves the last inequality in (<ref>).

Step 2: Estimate on $h^{N, M}_{\epsilon,1}$
Let $x \in \Dom$.
By Theorem <ref> and Lemma <ref>, we get

\begin{align*}
\langl \lt( \fint_{\Dom_\epsilon(x)} |h^{N, M}_{\epsilon,1}|^2 \rt)^{\frac{p}{2}} \rangl^{\frac{1}{p}}
\langl \lt( \fint_{\Dom_\epsilon(x)} \lt|\lt( \sigmaD_{\epsilon}, \phiD_{\epsilon} \rt)\rt|^2 \rt)^{\frac{p}{2}} \rangl^{\frac{1}{p}}
\lt\|\nabla^2 \bar{v}^{N,M} \rt\|_{\LL^\infty(\Dom_\epsilon(x))}
\\
&\overset{\eqref{Opti_Decay_Dir}, \eqref{M:0005}}{\lesssim}
\epsilon^{\nu}(\epsilon + |x|)^{1-\nu} \ln^{\tilde{\nu}'}\lt(\epsilon^{-1}|x| + 2\rt) \lt\|\nabla^2 \bar{v}^{N,M} \rt\|_{\LL^\infty(\Dom_\epsilon(x))}.
\end{align*}

Inserting (<ref>) entails that $h^{N, M}_{\epsilon,1}$ satisfies (<ref>).

Step 3: Estimate on $h^{N, M}_{\epsilon,2}$
Since $\chi-1$ vanishes inside $\Dom_{1/2}$, we already have that $h^{N, M}_{\epsilon,2}$ satisfies third estimate in (<ref>), and the second estimate in (<ref>) for $x \in \Dom_{1/2}$.
Hence, it remains to establish that the first estimate in (<ref>) is valid for  $x \in \Dom \backslash \Dom_{\frac{1}{2}}$.
By (<ref>), using $\phiDtilde_i$, as defined in Proposition <ref>, and (<ref>), we have

\begin{align*}
|\nabla \lt(\phiC_{n} - \phiD_{i} \partial_i \taubar_n\rt)(x)|
|\nabla\phitilde_n(x)| +
\lt|\nabla\lt(\lt( \etaup \phiup_i  + \etabulk \phi_i + \etadown \phidown_i - \phiD_{i}  \rt) \partial_i \taubar_n\rt)(x)\rt|
\\
|\nabla \phiDtilde(x)| |x|^{\rhobar_n - 1}
|\phiDtilde(x)| |x|^{\rhobar_n - 2},
\end{align*}

By scaling and the triangle inequality, appealing to (<ref>) and (<ref>), we get

\begin{equation}\label{M:0006}
\langl \lt( \fint_{\Dom_\epsilon(x)} |\nabla \lt(\phiC_{\epsilon,n} - \phiD_{\epsilon,i} \partial_i \taubar_n\rt)|^2 \rt)^{\frac{p}{2}} \rangl^{\frac{1}{p}}
\lesssim \epsilon^\nu \ln^{\tilde{\nu}'}(\epsilon^{-1}|x|) |x|^{\rhobar_n - 1 - \nu}.
\end{equation}

As a consequence, this yields that $h^{N, M}_{\epsilon,2}$ satisfies the first estimate in (<ref>).

Step 4: Estimate on $h^{N, M}_{\epsilon,3}$
Since $\chi_\epsilon$ is supported in $\Boule_\epsilon$, $h^{N, M}_{\epsilon,3}$ satisfies the first two estimates of (<ref>).
Last, we use the triangle inequality to get for $x \in \Dom_\epsilon$

\begin{align*}
|h^{N, M}_{\epsilon,3}(x)|
&\lesssim \sum_{n=N+1}^{N+M} \lt(| \taubar_n(x)\nabla \chi_\epsilon(x)| +  |\chi_\epsilon(x) \nabla \taubar_n(x)| \rt)
\overset{\eqref{taubar_Homog}}{\lesssim}
\epsilon^{-1} |x|^{\rhobar_{{N}+1}} \mathds{1}(|x|> \epsilon/2)  + |x|^{\rhobar_{{N}+1}-1}
\lesssim |x|^{\rhobar_{{N}+1}-1}.
\end{align*}

Since $\rhobar_{{N}+1}-1 \geq \rhobar_1 -1 > -1$, we have $h^{N, M}_{\epsilon,3} \in \LL^2(\Dom_{4\epsilon})$ and

\begin{align*}
\lt(\fint_{\Dom_{4\epsilon}}\lt|h^{N, M}_{\epsilon,3}\rt|^2 \rt)^{\frac{1}{2}}
\lesssim
\epsilon^{-1} \lt( \int_0^\epsilon t^{2 \rhobar_{N+1} - 1} \dd t \rt)^{\frac{1}{2}} \lesssim \epsilon^{\rhobar_{N+1}-1}.
\end{align*}

In other words, $h^{N, M}_{\epsilon,3}$ satisfies the third estimate in (<ref>).

§.§ Quasi-optimal error estimate: Proof of Theorem <ref>

In this proof, the constant $\tilde{\nu}'\lesssim_{\tilde{\nu}} 1$ may change from line to line.
The core of the proof relies on (<ref>).

In Step 1 we establish

\begin{equation}\label{Num:7004}
\langl\lt( \int_\Dom |\nabla \utieps^{N,M} - \nabla \ueps|^2 \rt)^{\frac{p}{2}}\rangl^{\frac{1}{p}}
\lesssim_{\Xi,q,p, N, M}
\epsilon^{\nu} \ln^{\tilde{\nu}'} (\epsilon^{-1}),
\end{equation}

where $\utieps^{N,M}$ is defined by (<ref>), for $N=N_0$ and $M \in \mathbb{N}$ satisfying (<ref>).
In Step 2, we generalize (<ref>) for general $N>N_0$.
In Step 3 we show that, for any $\epsilon < r \leq 2$, there holds

\begin{equation}\label{M:0013}
\langl \biggl(\fint_{\Dom_{r}}| \nabla \utieps^{N,M}  - \nabla \ueps |^{2}\biggr)^\frac{p}2 \rangl^{\frac{1}{p}}
\lesssim_{\Xi, q, p, N, M}
\epsilon^\nu \ln^{\tilde{\nu}'}(\epsilon^{-1})r^{\rhobar_1-1}.
\end{equation}

In Step 4, we localize (<ref>) near all points $x \in \Dom$
to the effect of (<ref>).

Step 1: Proof of (<ref>) for $N=N_0$.
By definition, we have  $\nabla \ueps \in \LL^2(\Dom)$.
Moreover, we note that $\nabla \utieps^{N_0,M}\in\LL^2(\Dom)$,
cf. (<ref>).
Indeed, Theorem <ref> and (<ref>), and $\rhobar_{N+M+1} \geq 2$ (by (<ref>)), imply that the gradient of the first term on the of (<ref>) is in $\LL^2(\Dom)$.
Moreover, the remaining terms are in $\HH^1_{\rm{loc}}(\Dom)$ with compact support, so that $\nabla \utieps^{N_0,M}\in\LL^2(\Dom)$ follows.
Hence, by the energy estimate applied to (<ref>), we have

\begin{equation*}
\langl\lt( \int_\Dom |\nabla \utieps^{N_0,M} - \nabla \ueps|^2 \rt)^{\frac{p}{2}}\rangl^{\frac{1}{p}}
\lesssim
\langl\lt( \int_\Dom |h^{N_0,M}_\epsilon|^2 \rt)^{\frac{p}{2}}\rangl^{\frac{1}{p}}.
\end{equation*}

Thus, (<ref>) comes as a consequence of

\begin{equation}\label{Num:7024}
\langl\lt( \int_\Dom |h^{N_0,M}_\epsilon|^2 \rt)^{\frac{p}{2}}\rangl^{\frac{1}{p}}
\lesssim \epsilon^{\nu} \ln^{\tilde{\nu}'} (\epsilon^{-1}).
\end{equation}

Here comes the argument for (<ref>).
W. l. o. g. we may assume $p \geq 2$, so that we get from a covering argument and the Minkowski inequality

\begin{align}
\label{last_label}
\langl\lt( \int_\Dom |h^{N_0,M}_\epsilon|^2 \rt)^{\frac{p}{2}}\rangl^{\frac{1}{p}}
\lesssim
\lt(\int_\Dom \langl\lt( \fint_{\Dom_\epsilon(x)} |h^{N_0,M}_\epsilon|^2 \rt)^{\frac{p}{2}}\rangl^{\frac{2}{p}} \dd x\rt)^{\frac{1}{2}}.
\end{align}

Then, we decompose the outer integral onto three domains in order to employ (<ref>) :

\begin{align*}
\int_\Dom \langl\lt( \fint_{\Dom_\epsilon(x)} |h^{N_0,M}_\epsilon|^2 \rt)^{\frac{p}{2}}\rangl^{\frac{2}{p}} \dd x
&=\lt( \int_{\Dom \backslash \Dom_1} + \int_{\Dom_1 \backslash \Dom_{4\epsilon}} + \int_{\Dom_{4\epsilon}}\rt) \lt( \langl\lt( \fint_{\Dom_\epsilon(x)} |h^{N_0,M}_\epsilon|^2 \rt)^{\frac{p}{2}}\rangl^{\frac{2}{p}} \rt) \dd x
\\
\epsilon^{2\nu}
\int_1^\infty \ln^{2\tilde{\nu}'}(\epsilon^{-1}t) t^{2 \rhobar_{N_0} - 2\nu -1} \dd t
\\
&\qquad + \epsilon^{2\nu} \ln^{2\tilde{\nu}'}(\epsilon^{-1})
\int_\epsilon^1 t^{2\rhobar_{{N_0}+1}-2\nu -1} \dd t
\\
&\qquad + \int_0^\epsilon \epsilon^{2\rhobar_{{N_0}+1}-2} t\dd t.
\end{align*}

Yet, recall that $N_0$ satisfies (<ref>), so that $2 \rhobar_{N_0} - 2\nu -1 < -1$ and $2\rhobar_{{N_0}+1}-2\nu -1 \geq -1$.

\begin{align*}
\int_\Dom \langl\lt( \fint_{\Dom_\epsilon(x)} |h^{N_0,M}_\epsilon|^2 \rt)^{\frac{p}{2}}\rangl^{\frac{2}{p}} \dd x
\epsilon^{2\nu} \ln^{2\tilde{\nu}'}(\epsilon^{-1})
+ \epsilon^{2\nu} \ln^{2\tilde{\nu}'+1}(\epsilon^{-1})
+ \epsilon^{2\rhobar_{{N_0}+1}}
\lesssim \epsilon^{2\nu} \ln^{2\tilde{\nu}'+1}(\epsilon^{-1}).
\end{align*}

Up to redefining $\tilde{\nu}'$, this proves (<ref>).

Step 2 : Proof of (<ref>) for general $N \geq N_0$.

For $M'=M+N-N_0 \geq M$, we estimate the difference

\begin{align*}
\utieps^{N,M} - \utieps^{N_0,M'} \overset{\eqref{2-scale-New-bis}}{=} \sum_{n=N_0+1}^N \bar{\gamma}_n \lt( \chi \lt(\phiC_{\epsilon,n} - \phiD_{\epsilon,i} \partial_i \taubar_n \rt)  + \phiD_{\epsilon,i}\partial_i \lt( \taubar_n\chi_\epsilon\rt)\rt).
\end{align*}

Towards this end, we decompose

\begin{align*}
\nabla \lt(\utieps^{N,M} - \utieps^{N_0,M'}\rt)
\sum_{n=N_0+1}^N \bar{\gamma}_n \lt( \lt(\phiC_{\epsilon,n} - \phiD_{\epsilon,i} \partial_i \taubar_n \rt) \nabla \chi
+(\chi-\chi_\epsilon) \nabla \lt(\phiC_{\epsilon,n} - \phiD_{\epsilon,i} \partial_i \taubar_n \rt)
\rt)
\\
\sum_{n=N_0+1}^N \bar{\gamma}_n
\lt(
\chi_\epsilon \nabla \phiC_{\epsilon,n}
\phiD_{\epsilon,i}\partial_i \taubar_n \nabla \chi_\epsilon + \nabla \lt( \taubar_n \phiD_{\epsilon,i} \partial_i \chi_\epsilon\rt)
\rt).
\end{align*}

Thus, by the triangle inequality followed by the Poincaré inequality, we get

\begin{align*}
\lt(
\int_{\Dom} \lt| \nabla \lt(\utieps^{N,M} - \utieps^{N_0,M'}\rt)\rt|^2
\rt)^{\frac{1}{2}}
\lesssim~&
\sum_{n=N_0+1}^N
\lt(
\int_{\Dom_1 \backslash \Dom_{\epsilon/2}} \lt|\nabla \lt(\phiC_{\epsilon,n} - \phiD_{\epsilon,i} \partial_i \taubar_n \rt)\rt|^2
\rt)^{\frac{1}{2}}
\\
\lt(
\int_{\Dom_\epsilon} \lt|\lt(\epsilon^{-1}\phiD_{\epsilon}\nabla \taubar_n,\epsilon^{-1} \taubar_n\nabla\phiD_{\epsilon}, \epsilon^{-2} \phiD_{\epsilon}\taubar_n, \nabla \phiC_{\epsilon,n} \rt)\rt|^2
\rt)^{\frac{1}{2}}.
\end{align*}

We estimate the summands of the first above by means of a covering argument combined with the Minkowski estimate (as in (<ref>)), along with (<ref>), to the effect of
\begin{align*}
\langl
\lt(
\int_{\Dom_1 \backslash \Dom_{\epsilon/2}} \lt|\nabla \lt(\phiC_{\epsilon,n} - \phiD_{\epsilon,i} \partial_i \taubar_n \rt)\rt|^2
\rt)^{\frac{p}{2}}\rangl^{\frac{1}{p}}
\lt(
\int_{\Dom_1 \backslash \Dom_{\epsilon/2}}
\langl\lt(\fint_{\Dom_{\epsilon/4}}\lt|\nabla \lt(\phiC_{\epsilon,n} - \phiD_{\epsilon,i} \partial_i \taubar_n \rt)\rt|^2 \rt)^{\frac{p}{2}}\rangl^{\frac{1}{p}}
\rt)^{\frac{1}{2}}
\\
&\overset{\eqref{M:0006}}{\lesssim} \epsilon^\nu \ln^{\tilde{\nu}'}(\epsilon^{-1})
\lt(\int_{\epsilon}^1  t^{2\rhobar_n -  2\nu - 1} \dd t\rt)^{\frac{1}{2}}.
\end{align*}

For $n \geq N_0+1$, by definition (<ref>), the integral on the above is bounded by $\ln(\epsilon^{-1})$.
Hence, also using (<ref>) and (<ref>), we obtain

\begin{align*}
\langl \lt(
\int_{\Dom} \lt| \nabla \lt(\utieps^{N,M} - \utieps^{N_0,M'}\rt)\rt|^2
\rt)^{\frac{p}{2}}\rangl^{\frac{1}{p}}
\lesssim~& \epsilon^{\nu} \ln^{\tilde{\nu}'}(\epsilon^{-1}) + \epsilon^{\rhobar_{{N_0}+1}},
\end{align*}

which, by (<ref>), yields (<ref>).

Step 3: Averaged estimate (<ref>).
W. l. o. g. we may assume that $r=2^{-K}$ for $K \in \NN \backslash\{0\}$.
By Corollary <ref> applied to $\utieps^{N,M}  -\ueps$, which satisfies (<ref>), followed by (<ref>) and (<ref>), we have for $p'>p$ :

\begin{align*}
\langl \lt( \fint_{\Dom_r} |\nabla \utieps^{N,M}  - \nabla \ueps|^2 \rt)^{\frac{p}{2}} \rangl^{\frac{1}{p}}
\lesssim~~~~&
\langl \lt( \fint_{\Dom_{r}} |h^{N,M}_\epsilon|^2 \rt)^{\frac{p'}{2}} \rangl^{\frac{1}{p'}}
\sum_{k=1}^{K-1} 2^{k(1-\rhobar_1)}
\langl \lt( \fint_{\Dom_{2^{(k+1)}r} \backslash \Dom_{2^{k}r}} |h^{N,M}_\epsilon|^2 \rt)^{\frac{p'}{2}} \rangl^{\frac{1}{p'}}
\\
\langl \lt( \fint_{\Dom_{1}} |\nabla \utieps^{N,M}  - \nabla \ueps|^2 \rt)^{\frac{p'}{2}} \rangl^{\frac{1}{p'}}
\\
\overset{\eqref{M:0003}, \eqref{Num:7004}}{\lesssim}&
\epsilon^\nu \ln^{\tilde{\nu}'}(\epsilon^{-1})
\lt(
\sum_{k=0}^{K-1} 2^{k(1-\rhobar_1)}
\\
\lesssim~~~~&
\epsilon^\nu \ln^{\tilde{\nu}'}(\epsilon^{-1})
\lt(
1+\sum_{k=0}^{K-1} (2^{(k+1)}r)^{\rhobar_{N}-\nu}
\rt)r^{\rhobar_1-1}.
\end{align*}

Since we assume that $N > N_0$ and that $N_0$ satisfies (<ref>), and $\rhobar_N-\nu>0$, and since $r = 2^{-K}$,  then we get that the above geometrical sum is bounded, to the effect of (<ref>).

Step 4: Local estimate (<ref>).
We now consider $x \in \Dom_1 \backslash \Dom_{2\epsilon}$.
Applying (a rescaled version of) Corollary <ref> to $\utieps^{N,M}  -\ueps$, which satisfies (<ref>), we have, for $p'>p$,

\begin{align*}
\langl \biggl(\fint_{\Dom_\epsilon(x)}| \nabla \utieps^{N,M}  - \nabla \ueps |^{2}\biggr)^\frac{p}2 \rangl^{\frac{1}{p}}
\overset{\eqref{M:0002}}{\lesssim}~&
\langl  \lt(\fint_{\Dom_{|x|/2}(x)} | \nabla \utieps^{N,M}  - \nabla \ueps|^2\rt)^{\frac{p'}{2}} \rangl^{\frac{1}{p'}}
\\
&+ \ln(\epsilon^{-1}|x|) \sup_{x' \in \Dom_{|x|/2}(x)} \langl\lt(\fint_{\Dom_\epsilon(x')}\lt|h^{N,M}_\epsilon\rt|^2\rt)^{\frac{p'}{2}} \rangl^{\frac{1}{p'}}
\\
\overset{\eqref{M:0013},\eqref{M:0003}}{\lesssim}&
\epsilon^\nu \ln^{\tilde{\nu}'}(\epsilon^{-1})|x|^{\rhobar_1-1}
\epsilon^\nu \ln^{\tilde{\nu}'+1}(\epsilon^{-1})  |x|^{\rhobar_{N+1}-1-\nu}
\lesssim \epsilon^\nu \ln^{\tilde{\nu}'}(\epsilon^{-1})|x|^{\rhobar_1-1}.
\end{align*}

Here, the last estimate comes from the fact that $N > N_0$, for $N_0$ satisfying (<ref>).
Last, noticing that $\utieps^{N,M}$ differs from $\utieps^N$ only inside $\Dom_\epsilon$, we obtain (<ref>).

§ ACKNOWLEDGEMENT

We thank Monique Dauge, with whom Marc Josien had an interesting discussion while invited at the Université de Rennes, and who gently pointed out accurate references for homogeneous elliptic operators in corners.
Also, we are indebted to Julian Fischer, who suggested to make use of an ansatz based on Dirichlet correctors instead of the one we first studied (<ref>).
As discussed in Section <ref>, this proved more efficient.

We wish to acknowledge the support of the Max Planck Institute Mathematik in den Naturwissenschaften in Leipzig, at which Marc Josien and Mathias Schäffner held post-doc positions when this work begun, and which supported this project, in particular by funding visits of Claudia Raithel.
Claudia Raithel also gratefully acknowledges partial support from the Austrian Science Fund (FWF), grants P30000, P33010, W1245, and F65.

$'$ $'$ $'$

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§ EXTENSION OF A DIVERGENCE-FREE FIELD

Assume that $\omega \in (0,2\pi)$, and that $H \in \LL^1_\loc(\bar{\Dom})$ is a divergence-free vector field on $\Dom$, written in polar coordinates as

\begin{align*}
H=(H_r,H_\theta) = H_r e_r + H_\theta e_\theta.
\end{align*}

Then it can be extended as a divergence-free vector field on the whole space $\R^2$ by

\begin{align}
\label{extension_2}
\bar{H}(r,\theta)
\lt\{
\begin{aligned}
&H(r,\theta) &&\pour \theta \in \lt[0,\omega \rt],
\\
&\tilde{H}(r,\theta) &&\pour \theta \in \lt( \omega, 2\pi \rt),
\end{aligned}
\rt.
\text{ where }
&\tilde{H}(r,\theta) := \lt( - \alpha H_r, H_\theta \rt)\lt( r, \alpha (2 \pi -\theta) \rt),
\pour \alpha := \frac{\omega}{2 \pi - \omega}.
\end{align}

First, we may compute

\begin{align*}
\nabla \cdot \tilde{H}(r,\theta)
\frac{1}{r} \partial_r (r \tilde{H}_r(r,\theta)) + \frac{1}{r}\partial_\theta \tilde{H}_\theta(r,\theta)
\\
\Big(-\alpha \frac{1}{r} \partial_r (r H_r)  - \alpha \frac{1}{r} \partial_\theta H_\theta \Big) \lt( r, \alpha (2\pi-\theta) \rt)
\\
&= - \alpha \lt(\nabla \cdot H\rt)\lt( r, \alpha (2\pi-\theta) \rt) = 0.
\end{align*}

Moreover, since

\begin{align*}
\tilde{H}(r,\omega) = \lt( - \alpha H_r, H_\theta \rt)(r,\omega) \quad \textrm{and} \quad \tilde{H}(r, 2 \pi) = \lt( - \alpha H_r, H_\theta \rt)(r,0),
\end{align*}

the normal component $\bar{H}_\theta$ of $\bar{H}$ is continuous across $\partial\Dom \backslash\{0\}$.
This proves that $\nabla \cdot \bar{H}=0$ in $\R^2 \backslash \{0\}$.[The argument of continuity may not apply strictly speaking to a function $H$ that is only $\LL^1_\loc(\bar{\Dom})$; however, we may use this argument in combination with a mollifying argument.]

It remains to establish that $\nabla \cdot \bar{H} = 0$ in the sense of distributions in $\R^2$ (that is, $\nabla \cdot \bar{H}$ does not suffer from any singularity in $0$).
By the theory of distributions <cit.>, since $\bar{H} \in \LL^1_\loc(\R^2)$, there holds

\begin{align*}
\nabla \cdot \bar{H} = \delta_0 C, \quad \pour C \in \R^2.
\end{align*}

It remains to prove that $C=0$.
In this perspective, let us define the radial function $\psi= \eta(r)$ where $\eta$ is compactly supported in $[-2,2]$ with $\eta = 1 $ in $[-1,1]$.
By the theory of distributions, we have (abusively using the symbol $\int$) :

\begin{align*}
\int_{\R^2} \bar{H} \cdot \nabla \psi = -	\int_{\R^2}  (\nabla\cdot\bar{H}) \psi =  -\psi(0)C = - C.
\end{align*}

On the other hand, (since $\nabla \psi$ is supported into $\Boule_2 \backslash \Boule_1$), by a change of coordinates, we also have

\begin{align*}
\int_{\R^2} \bar{H} \cdot \nabla \psi
\int_{1}^2 \eta'(r) \lt( \int_0^{2\pi} \bar{H}(r,\theta) \cdot e_r \dd \theta\rt) \dd r.
\end{align*}

By definition (<ref>) followed by a change of variables $\tilde{\theta} = \alpha (2\pi - \theta)$, the above inner integral vanishes :

\begin{align*}
\int_0^{2\pi} \bar{H}(r,\theta) \cdot e_r \dd \theta
\int_0^\omega H_r(r,\theta)\dd \theta -  \alpha \int_\omega^{2\pi} H_r(r,\alpha(2\pi - \theta)) \dd \theta
\end{align*}

Hence, we obtain that $C=0$, which in turn implies that $\nabla \cdot \bar{H} = 0$ in $\R^2$.