\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 48, pp. 1--24.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE-2004/48\hfil Complex Monge-Amp\`{e}re equation]
{Dirichlet problem for degenerate
elliptic complex Monge-Amp\`{e}re equation}

\author[Saoussen Kallel-Jallouli\hfil EJDE-2004/48\hfilneg]
{Saoussen Kallel-Jallouli}

\address{Saoussen Kallel-Jallouli\hfill\break
Facult\'{e} des Sciences, Campus Universitaire, 1060 Tunis, Tunisie}
\email{Saoussen.Kallel@fst.rnu.tn}

\date{}
\thanks{Submitted May 15, 2003. Published April 6, 2004.}
\subjclass[2000]{35J70, 32W20, 32W05}
\keywords{Degenerate elliptic, omplex Monge-Amp\`ere, \hfill\break\indent
Plurisubharmonic function}

\begin{abstract}
 We consider the Dirichlet problem
 $$
 \det \big({\frac{\partial^2u}{\partial z_i\partial \overline{z_j}}}
 \big)=g(z,u)\quad\mbox{in }\Omega\,,  \quad
 u\big|_{ \partial \Omega }=\varphi\,,
 $$
 where $\Omega $ is a bounded open set of $\mathbb{C}^{n}$ with regular
 boundary, $g$ and $\varphi$ are sufficiently smooth functions, and $g$ is
 non-negative. We prove that, under additional hypotheses on $g$ and
 $\varphi $, if $|\det \varphi _{i\overline{j}}-g|_{C^{s_{\ast }}}$ is
 sufficiently small the problem has a plurisubharmonic solution.
\end{abstract}

\maketitle
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Propositon}
\numberwithin{equation}{section}
\allowdisplaybreaks

\section{Introduction}

Let $\Omega $ be a bounded domain in $\mathbb{R}^{2n}$ with smooth boundary
and let $z_i=x_i+ix_{i+n}(1\leq i\leq n)$. We shall also
denote by $\Omega $ the set of $z=(z_1,z_{2},\dots ,z_n)$
satisfying $(\mathop{\rm Re}z,\mathop{\rm Im}z)\in \Omega $.\ We study the
problem of finding a sufficiently smooth plurisubharmonic solution to the
degenerate problem
\begin{equation}
\begin{gathered}
\det \big({\frac{\partial ^2\phi }{\partial z_i\partial \overline{z_j}
}}\big)=g(z,\phi )\quad\mbox{in }\Omega \,, \\
\phi \big|_{ \partial \Omega }=\varphi \,.
\end{gathered} \label{I.1}
\end{equation}
In \cite{k1,k2}, the author studies local solutions, while, here we consider
global solutions.

This problem has received considerable attention both in the non-degenerate
case ($g>0$) and in the degenerate case ($g\geq 0$). In particular,
Caffarelli, Kohn, Nirenberg and Spruck \cite{c1} established
some existence results in strongly pseudoconvex domains based on the
construction of a subsolution. The recent work of Guan \cite{g2},
extends some of these results to arbitrary smooth bounded domains. Guan
proved for the nondegenerate case that a sufficient condition for the
classical solvability is the existence of a subsolution.
Here we are concerned with degenerate problems in  an arbitrary
smooth bounded domain, which need not be Pseudoconvex.

Counterexamples due to Bedford and Fornaes \cite{b1} show that the
Dirichlet problem, in general, does not have a regular solution. This
implies that we should place some restrictions on $g$ and $\varphi $.

Let us assume that $\varphi $ is a real function defined in
$\overline{\Omega }$, $\Sigma $ is a finite set of points in $\Omega $,
and $g(z,\phi )=K(z)f(\mathop{\rm Re}z,\mathop{\rm Im}z,\phi )$.
 We further assume the following hypotheses.
\begin{itemize}
\item[(A1)] $K\geq 0$ in $\overline{\Omega }$, and  $K^{-1}(0)=\Sigma $
% \label{I.2}

\item[(A2)] $f(x,u)>0$ in $\overline{\Omega }\times \mathbb{R}$, and
 $\frac{\partial f}{\partial u}\geq -\varrho $ in
$\overline{\Omega }\times \mathbb{R}$, with $0\leq \varrho <<1$ %  \label{I.3}

\item[(A3)] $\varphi \big|_{ \overline{\Omega }\backslash \Sigma }$
is strictly plurisubharmonic, $(\varphi _{i\overline{j}})\big|_{\Sigma }$
is of rank $(n-1)$, and
the eigenvalues of $(\varphi _{i\overline{j}})$ on $\Sigma $ are distinct. %\label{I.4}
\end{itemize}

Our main results are the following theorems:

\begin{theorem} \label{thmA}
 Let $s_{\ast }\geq 7+2n$ be an integer, $\alpha \in ] 0,1[ $, and $\Gamma >1$.
If $\varphi \in C^{s_{\ast}+2,\alpha }(\overline{\Omega })$ satisfies the condition
 (A3), {\it  then one can find a constant $\varepsilon _0>0$
(depending on $s_{\ast }$, $\alpha $, $\Gamma $, $\Omega $ and $\varphi $)
such that for any $g=Kf\in C^{s_{\ast }}$ satisfying} (A1), (A2),
\begin{equation}
|\det \varphi _{i\overline{j}}-g(\varphi )|
_{C^{s_{\ast }}}\leq \varepsilon _0  \label{I.5}
\end{equation}
and
$|\frac{\partial g}{\partial u}|_{C^{s_{\ast }}}\leq \Gamma $,
then problem  \eqref{I.1}  has a plurisubharmonic (real
valued) solution $\phi \in C^{s_{\ast }-3-n}(\overline{\Omega })$,
which is unique when $\rho =0$.
\end{theorem}

Let $l_{\alpha }(x)$ denote $\alpha $-th row
the matrix of cofactors of $(\varphi _{i\overline{j}})$, and
$$
D^{k}K(x)(l_{\alpha }(x),l_{\beta }(x))^{(k)}
=D^{k}K(x)\big(l_{\alpha}(x),l_{\beta }(x);\dots ;l_{\alpha }(x)
,l_{\beta }(x)\big).
$$

\begin{theorem} \label{thmB}
 Under the assumptions in Theorem \ref{thmA}, suppose
that $\varphi \in C^{\infty }(\overline{\Omega })$ and for any
point $x_0\in \Sigma $ one can find an integer $k$ such that
$D^{j}K(x_0)=0$ for all $j\leq k-1$ and there exists
$\alpha \neq \beta \in \{1,\dots ,n\}$ such that
$D^{k}K(x_0)(l_{\alpha }(x_0),l_{\beta }(x_0))^{(k)}\neq 0$.
Then there exists an integer $s_{\ast }>0$ and a constant
$\varepsilon_0>0 $ such that for any function $g\in C^{\infty }$ satisfying
 (A2), (A3) and \eqref{I.5}, the plurisubharmonic solution $\phi $ to the
 problem \eqref{I.1} is in  $C^{\infty }(\overline{\Omega })$.
\end{theorem}

In Theorem \ref{thmA}, the assumption concerning $\Sigma $ leads to a-priori
estimates and the assumption on $g$ and $\varphi $ ensures the convergence
of an iteration scheme of Nash-Moser type. It is to be noted that we do not
require demonstrating that a subsolution exists as in \cite{c1} and
[6].

Under some additional conditions on $g$, we can prove the smoothness of the
solution, using the works of Xu \cite{x1} and Xu and Zuily \cite{x2}.

This paper is organized as follows. In Section 2 we state some
preliminary results.
In Section 3, we state fundamental global a-priori estimates for
degenerate linearized operators that are crucial to establish an iteration
scheme of Nash-Moser type. We then prove Theorem \ref{thmA} in Section 4. We prove
Theorem \ref{thmB} in Section 5. Finally, we prove the a-priori estimates stated in
Section 3.

\section{Preliminary results}

We shall use the norms
\begin{equation*}
|\cdot|_{k}=\|\cdot\|_{C^{k}(\overline{\Omega })}, \quad
\|\cdot \|_{k}=\|\cdot\|_{H^{k}(\Omega )}, \quad
|\cdot|_{k,\tau }=\|\cdot \|_{C^{k,\tau }(\overline{\Omega })}\,
\end{equation*}
where $k\in \mathbb{N}$ and $\tau \in ] 0,\alpha [ $.

In this work, we need some technical lemmas which play important
roles in the proof of convergence of our iteration scheme.

\begin{lemma} \label{lmII.1}
Let $s_{\ast }$ be an integer, $s_{\ast }\geq 7+2n$.
We can find a constant $\beta \geq 2$ such that for any
$0\leq i,j,k\leq s_{\ast }+2$, $n_{\ast }=n+\tau $
and $u\in C^{s_{\ast }+2,\alpha }(\overline{\Omega })$ we have:
The Sobolev inequality
\begin{equation}
|u|_{i,\tau }\leq \beta \| u\|_{i+n_{\ast }}\label{II.1}
\end{equation}
The Gagliardo-Nirenberg inequality
\begin{equation}
\| u\|_j\leq \beta \| u\|_i^{\frac{k-j}{k-i}
}\| u\|_{k}^{\frac{j-i}{k-i}},\quad i<j<k  \label{II.2}
\end{equation}
The inequality
\begin{equation}
\| u\|_{s_{\ast }}\leq \beta |u|_{s_{\ast }} \label{II.3}
\end{equation}
For any $\lambda \geq 1$, there exists a family of smoothing linear operators
$S_{\lambda }:\cup_{i\geq 0}H^{i}(\Omega) \to \cap_{j\geq 0}H^{j}(\Omega )$,
satisfying
\begin{gather}
\| S_{\lambda }u\|_i\leq \beta \| u\|_{_j}, \quad\mbox{if }i\leq j
\label{II.4} \\
\| S_{\lambda }u\|_i\leq \beta \lambda ^{i-j}\| u\|_j,\quad \text{if }i\geq j
\label{II.5} \\
\| S_{\lambda }u-u\|_i\leq \beta \lambda ^{i-j}\|u\|_j,\quad \text{if }i\leq j
 \label{II.6}
\end{gather}
\end{lemma}

\begin{lemma}[\cite{a1,h1}] \label{lmII.2}
(1)  For $t>0$; if $u,v\in L^{\infty }\cap H^{t}$, then
$uv\in L^{\infty }\cap H^{t}$ and
\begin{equation}
\| uv\|_{t}\leq K_1(|u|_0\|v\|_{t}+\| u\|_{t}|v|_0),\label{II.7}
\end{equation}
where, $K_1$ is a constant $\geq 1$ independent of $u$ and $v$.

\noindent (2) Let $H:\mathbb{R}^{m}\to \mathbb{C}$ be a function
$C^{\infty }$ of its arguments.\\
For $s>0$, if $\omega \in (L^{\infty }\cap H^{s})^{m}$ and
$|\omega |_0\leq M$, then
\begin{equation}
\| H(\omega )\|_s\leq K_{2}(s,H,M)
(\| \omega \|_s+1),  \label{II.8}
\end{equation}
where $K_{2}\geq 1$ and is a constant independent of $\omega $.
\\
If $\omega $ $\in (C^{i,\mu })^{m}$, $\mu \in ] 0,1[ $ and $i\in \mathbb{N}$,
then $H(\omega )\in C^{i,\mu }$.

If we suppose that $|\omega |_0\leq M$, then we can
find a constant $K_{3}=K_{3}(i,\mu ,H,M)\geq 1$ such that
\begin{equation}
|H(\omega )|_{i,\mu }\leq K_{3}(|\omega |_{i,\mu }+1).  \label{II.9}
\end{equation}
\end{lemma}

We shall also need the following technical lemma.


\begin{lemma}[{\cite[Lemma]{k1}}] \label{lmII.3}
Let $F(u_{z_i\overline{z_j}})=\det (u_{z_i\overline{z_j}})$.
For $1\leq i,j,a,b\leq n$, we have
\begin{equation}
F\frac{\partial ^2F}{\partial u_{z_{a}\overline{z_{b}}}\partial u_{z_i
\overline{z_j}}}=\frac{\partial F}{\partial u_{z_{a}\overline{z_{b}}}}
\frac{\partial F}{\partial u_{z_i\overline{z_j}}}-\frac{\partial F}{
\partial u_{z_i\overline{z_{b}}}}\frac{\partial F}{\partial u_{z_{a}
\overline{z_j}}}.  \label{II.10}
\end{equation}
\end{lemma}

\section{A priori estimates for the linearized operator}

Defining $\phi =\varphi +\varepsilon w$, \eqref{I.1} becomes
\begin{equation}
\det (\phi _{z_i\overline{z_j}})=\det (\varphi _{z_i
\overline{z_j}}+\varepsilon w_{z_i\overline{z_j}})=g.
\label{III.1}
\end{equation}
Let
\begin{equation}
G(w)=\frac{1}{\varepsilon }[\det \Phi -g] .\label{III.2}
\end{equation}
Then the  linearization of $G$ at $w$ is
\begin{equation}
L_{G}(w)=\sum_{i,j=1}^n \phi ^{ij}\partial
_{z_i}\partial _{\overline{z_j}}+b,   \label{III.3}
\end{equation}
where $\widetilde{\Phi }=(\phi ^{ij})$ is the matrix of
cofactors of $\Phi =(\phi _{z_i\overline{z_j}}(z,\varepsilon
,w))$ and $b={\frac{\partial g}{\partial u}}$.

Now we construct linear elliptic operators, maybe degenerate,  related
to linearized operators.
For any smooth real valued function $w$, the matrix
$(\phi _{i\overline{j}})$ is Hermitian and we can find a unitary matrix
$T(z,\varepsilon )$ satisfying
\begin{equation}
T(z,\varepsilon )(\phi _{z_i\overline{z_j}})
^{t}\overline{T}(z,\varepsilon )=\mathop{\rm diag}(\lambda_1,\dots ,\lambda _n).  \label{III.4}
\end{equation}
Without loss of generality we may assume that $\Sigma $ is
reduced to one  point,
the origin. By means of change of variables we may assume, using (A3), that
\begin{equation}
\varphi _{z_i\overline{z_j}}(0)=\sigma _i\delta _i^{j}
\quad i,j=1,\dots ,n,  \label{III.5}
\end{equation}
where
$\sigma _i>0$  for $i=1,\dots ,n-1$, $\sigma _n=0$  and
$\sigma_i\neq \sigma _j$  for $i\neq j$.  %\label{III.6}
Let $0<\tau \leq \frac{\alpha }{4}$.

\begin{lemma} \label{lmIII.1}
There exist constants $\varepsilon _1>0$, $\delta _1>0$ and $M>0$
depending only on $\varphi $, $n$, $\Omega $ such that when
$$
V_0=\{(z,\varepsilon ,w)/|z|\leq \delta_1,\,
0\leq \varepsilon \leq \varepsilon _1,\,
w\in C^{3,\tau}(\overline{\Omega }),\,|w|_{3,\tau }\leq 1\},
$$
we have:
(i) The eigenvalues $\lambda _i$, $i=1,\dots ,n$ of $\Phi $ are distinct on
$V_0$ and of class $C^{1}$ in \r{V}$_0$. Moreover, $\lambda _i>0$ in
$V_0$, for $i=1,\dots ,n-1$.\\
(ii) For $(z,\varepsilon ,w)\in $V$_0$,
\begin{equation}
\sum_{i=1}^n |\sigma _i-\lambda _i(z,\varepsilon ,w)|
+|\Phi ^{nn}(z,\varepsilon,w)-\prod_{i=1}^{n-1} \sigma _i|
\leq M(\varepsilon +|z|).  \label{III.7}
\end{equation}
(iii) For $(z,\varepsilon ,w)\in V_0$ and $i=1,\dots ,n-1$,
\begin{equation}
\lambda _i\geq \inf_{1\leq i\leq n-1} \sigma _i
-M\delta_1-(M+1)\varepsilon _1>0\text{ and }\Phi ^{nn}\geq
\prod_{i=1}^{n-1}\sigma _i-M\delta _1-M\varepsilon _1>0.
\label{III.8}
\end{equation}
\end{lemma}

\begin{proof} Let us consider the function
$H(z,\varepsilon ,w,\lambda )=\det (\varphi _{z_i\overline{z_j}}
+\varepsilon w_{z_i\overline{z_j}}-\lambda \delta_i^{j})$.
Then $H\in C^{1}$ and by \eqref{III.5}, we have
$$
H(0,0,0,\sigma _i)=0\mbox{ and }
\frac{\partial H}{\partial\lambda }(0,0,0,\sigma _i)\neq 0, \quad
\forall i\in \{1,\dots ,n\}.
$$
By the implicit function theorem, one can find two constants
$\varepsilon _1>0$ and $\delta _1>0$ such that (i) holds. Moreover by
\eqref{III.5} we have
$$
{\frac{\partial F}{\partial u_{n\overline{n}}}}(\varphi _{i\overline{j}
})(0)=\Phi ^{nn}(0,0,w)
=\prod_{i=1}^{n-1} \sigma _i>0,
$$
which gives (ii) and (iii).
\end{proof}

\begin{lemma} \label{lmIII.2}
There exists  a positive constant $\varepsilon _{2}$
such that for any  $0<\varepsilon <\varepsilon _{2}$, any real valued
function $w\in C^{3,\tau }(\overline{\Omega })$ satisfying
$|w|_{3,\tau }\leq 1$ and $\theta =\max_{z\in \overline{\Omega }}|G(w)|$,
the operator
\begin{equation}
L=-L_{G}(w)-\theta \triangle  \label{III.9}
\end{equation}
is  elliptic, maybe degenerate.
(Here $\triangle =\sum_{i=1}^n (\frac{\partial^2}{\partial x_i^2}
+\frac{\partial ^2}{\partial y_i^2})$)
\end{lemma}

\begin{proof} Let
\begin{equation}
A=\theta |\xi |^2+\sum_{i,j=1}^n \phi^{ij}\xi _i\overline{\xi _j}\geq 0,\quad
 \forall (z,\xi ) \in \overline{\Omega }\times \mathbb{C}^{n}.  \label{III.10}
\end{equation}
If $z\in \overline{\Omega }\backslash \{0\}$, as $\varphi $
is strictly plurisubharmonic, then
$A>0$ for all $\xi \in \mathbb{C}^{n}\backslash \{0\}$.

If $z=0$, for $\xi \in \mathbb{C}^{n}$, we let
$\xi =^{t}T(\tau ,\varepsilon )\widetilde{\xi }$. Then we have
$$
A=\theta |\xi |^2+^{t}\xi \widetilde{\Phi }\overline{\xi }
=\theta |\xi |^2+^{t}\widetilde{\xi }T\widetilde{\Phi }^{t}
\overline{T}\overline{\widetilde{\xi }}.
$$
Since $\Phi \widetilde{\Phi }=\det \Phi \mathop{\rm \mathop{\rm Id}}$,
by \eqref{III.4},
\begin{gather*}
\det \Phi \mathop{\rm \mathop{\rm Id}}=T\Phi ^{t}\overline{T}T\widetilde{\Phi }^{t}\overline{T}
=\mathop{\rm diag}(\lambda _i)T\widetilde{\Phi }^{t}\overline{T}, \\
T\widetilde{\Phi }^{t}\overline{T}=\det \Phi \mathop{\rm diag}
(\frac{1}{\lambda _i})=\prod_{i=1}^n \lambda_i \mathop{\rm diag}
(\frac{1}{\lambda _i})
=(\varepsilon G+g)\mathop{\rm diag}(\frac{1}{\lambda _i}).
\end{gather*}
Thus,
\begin{align*}
A&=\theta |\widetilde{\xi }|^2+\det \Phi
\sum^n_{i=1} \frac{|\widetilde{\xi }_i|^2}{\lambda_i} \\
& =\theta |\widetilde{\xi }|^2+\sum^{n-1}_{i=1} \det
\Phi \frac{|\widetilde{\xi }_i|^2}{\lambda _i}
+\prod_{i=1}^{n-1} \lambda _i|\widetilde{\xi }_n|^2  \\
&=(\theta +\prod_{i=1}^{n-1} \lambda _i)|\widetilde{\xi }_n|^2
+\sum^{n-1}_{i=1} \frac{\varepsilon G+g+\theta \lambda _i}{\lambda _i}|
\widetilde{\xi }_i|^2.
\end{align*}
By \eqref{III.8}, for $i=1,\dots ,n-1$, $\varepsilon \leq
\varepsilon _1$ and $|w|_{3,\tau }\leq 1$, we have
$$
\varepsilon G+\theta \lambda _i\geq \theta (\sigma _i-M\delta
_1-(M+1)\varepsilon _1)\geq 0.
$$
Therefore, $A\geq 0$, which proves the lemma.
\end{proof}

Now we study a boundary-value problem for the degenrate elliptic
operator
$$
L=-L_{G}(w)-\theta \triangle =\sum^n_{i,j=1} b^{ij}\partial _{z_i}
\partial _{\overline{z_j}}+b,
$$
where
$$
b^{ij}=-{\frac{\partial F}{\partial u_{i\overline{j}}}}(\varphi _{i
\overline{j}}+\varepsilon w_{i\overline{j}})-\theta \delta
_i^{j}=-\Phi ^{ij}-\theta \delta _i^{j}
$$
and $b=K{\frac{\partial f}{\partial u}}$.
For $k,s\in \mathbb{N}$ we let
\begin{equation}
\begin{gathered}
A(k)=\max (1,\max_{1\leq i,j\leq n} |b^{ij}|_{k},|b|_{k})  \\
\Lambda _s=\{(i,j): 0\leq i,j\leq s,\;
i+j\leq s,\mbox{ and } i+2\leq \max (s,2)\}
\end{gathered}\label{III.11}
\end{equation}
Now from Lemma \ref{lmIII.2} we have the following statement.

\begin{theorem} \label{thmIII.3}
Suppose that $\theta \leq 1$ and $A(2)\leq M_0$, for some constant $M_0>0$.
 One can find $\varepsilon _{3}>0$ such that for any
$\varepsilon \in ] 0,\varepsilon _{3}] $, any real valued function
$w\in C^{s_{\ast }+2,\tau }(\overline{\Omega })$ satisfying the inequality
$|w|_{3,\tau }\leq 1$ and any real valued function $h\in H^{s_{\ast }}$,
 the problem
\begin{equation}
\begin{gathered}
Lu=h\quad\mbox{in }\Omega  \\
u\big|_{\partial \Omega }=0
\end{gathered} \label{III.12}
\end{equation}
has a unique solution $u\in H^{s_{\ast }}$. Moreover for $0\leq s\leq s_{\ast }$,
\begin{gather}
\| u\|_0\leq C_0\| h\|_0  \label{III.13}\\
\| u\|_1\leq C_1(\| h\|_1+\|u\|_0) \label{III.14}\\
\| u\|_s\leq C_s\{\| h\|_s+\sum_{j\leq s-1,\, (i,j)\in \Lambda _s}
(1+|\varphi +\varepsilon w|_{i+4,\tau })\| u\|_j\},\quad s\geq 2  \label{III.15}
\end{gather}
for some constant $C_s=C_s(\varphi ,s,\Omega ,M_0,\varepsilon
_{3})$ independent of $w$ and $\varepsilon $.
\end{theorem}

For $\nu \in ] 0,1[ $, we denote $L_{\nu }=L-\nu \triangle $.
To solve the Dirichlet problem \eqref{III.12}, we first
establish the following proposition.


\begin{proposition} \label{propIII.4}
Let $\theta \leq 1$ and, for some constant $M_0>0$,  $A(2)\leq M_0$.
Then there exists $\varepsilon _{3}>0$ such that for any
$\varepsilon \in ] 0,\varepsilon_{3}] $, any real valued function
$w\in C^{s_{\ast }+2,\tau }(\overline{\Omega })$ satisfying the inequality
$|w|_{3,\tau }\leq 1$ and any real valued function $h\in H^{s_{\ast}}(\Omega )$,
 the regularized problem
\begin{equation}
\begin{gathered}
L_{\nu }u=h \quad\mbox{in }\Omega , \\
u\big|_{ \partial \Omega }=0,
\end{gathered} \label{III.12'}
\end{equation}
has a unique (real valued) solution $u\in H^{s_{\ast }+1}(\Omega)$.
\end{proposition}

\begin{proof}
Since $L_{G}(w)$ is a second order operator with real coefficients, from
Lemma \ref{lmIII.2}, $L_{\nu }$ is uniformly elliptic with coefficients in
$C^{s_{\ast },\tau }(\overline{\Omega })$. Thus by
\cite[Theorems 6.14 and 8.13]{b2}
we see that \eqref{III.12'} has a real valued solution.

If  \eqref{III.13}--\eqref{III.15} hold for the regularized problem
\eqref{III.12'} with an uniform constant $C_s$ independent of
$\nu \in ] 0,1]$, then by letting $\nu $ tend to zero we  get a solution
$u\in H^{s_{\ast }}(\Omega )$ to the original problem which of course
satisfies \eqref{III.13}--\eqref{III.15}.
\end{proof}

Using Theorem \ref{thmIII.3}, we prove Theorem \ref{thmA} by constructing
a sequence of approximating solutions and  a priori
estimates for linearized operators. The hypothesis \eqref{I.5}
 will play an important role in the proof of the convergence of our iteration
scheme of Nash-Moser type.

\section{Proof of Theorem \ref{thmA}}

\subsection*{Part 1: An iteration scheme of Nash-Moser type}
In this section, we  use the Nash-Moser procedure  \cite{h1,m1} and the results
of Section 3 to prove Theorem \ref{thmA}.
We construct a sequence which converges to a solution to our problem.
We define
\begin{equation}
M_0=1+\underset{H\in \mathcal{F}}{\max }K_{3}(2,\tau ,H,(
1+|\varphi |_{2}))(1+|\varphi |
_{4,\tau }),  \label{IV.1}
\end{equation}
where $\mathcal{F=}\{\frac{\partial F}{\partial u_{i\overline{j}}},
\frac{\partial g}{\partial u}/1\leq i,j\leq n\}$ and $K_{3}$ is the
constant introduced in \eqref{II.9}.
(i.e: $|H(u)|_{j,\mu }\leq K_{3}(j,\mu ,H,M)|u|_{j,\mu }$).
We also define
\begin{equation}
D=\max \big(\max_{0\leq s\leq s_{\ast }} C_s,\,1\big).
\label{IV.2}
\end{equation}
Here $C_s$ is the constant (depending only on $s,\varphi ,\Omega ,M_0$)
given by Theorem \ref{thmIII.3}.
We let
\begin{equation}
\mu =\max (\beta ,3Ds_{\ast }^2(1+|\varphi |
_{s_{\ast }+2,\tau }),n,2^{\frac{1}{\tau }})\quad\text{and}\quad
\widetilde{\mu }=\beta ^2\mu ^{s_{\ast }},  \label{IV.3}
\end{equation}
\begin{equation}
a_1=9K_0\mu ^{5}, \quad
a_{2}=5a_1\mu ^{s_{\ast }+1}, \quad
a_{3}=7K_0\mu ^{5}, \label{IV.4}
\end{equation}
were $K_0$ is the constant given by Proposition \ref{propVI.1}.
Also, we fix $\widetilde{\varepsilon }$ satisfying
\begin{equation}
\widetilde{\varepsilon }\leq \min [1,(\varepsilon _i)
_{1\leq i\leq 4},(3D^2a_{2}+6\widetilde{\mu }D^2)^{-2}],   \label{IV.5}
\end{equation}
were $\varepsilon _i$ are given in Lemma \ref{lmIII.2}, Theorem \ref{thmIII.3}, the proof
of Theorem \ref{thmIII.3} and the proof of \eqref{III.14}.

As a consequence of these inequalities, we have
$6\widetilde{\varepsilon }\mu ^{s_{\ast }}\leq 1/4$.
Let $g\in C^{s_{\ast }}$ satisfy
$$
|\det \varphi _{i\overline{j}}-g(\varphi )|
_{s_{\ast }}\leq \widetilde{\varepsilon }^2,
$$
with $\varepsilon _0$ in Theorem \ref{thmA} equal to $\widetilde{\varepsilon }^2$.
Let $S_n=S_{\mu _n}$ the family of operators given by Lemma \ref{lmII.1},
with $\mu _n=\mu ^{n}$ ($\mu $ is given by \eqref{IV.3}).

Using Theorem \ref{thmIII.3}, we construct $w_n$, $n=0,1,\dots $, by induction on
$n$ as follows. We let  $u_0$, $w_0=0$, and assume
$w_0,w_1,\dots,w_n$ have been chosen and define $w_{n+1}$ by
\begin{equation}
w_{n+1}=w_n+u_{n+1},  \label{IV.6}
\end{equation}
where $u_{n+1}$ is the solution to the Dirichlet problem
\begin{equation}
\begin{gathered}
L_{G}(\widetilde{w}_n)u_{n+1}+\theta _n\triangle u_{n+1}=g_n,
\quad \mbox{in }\Omega \\
u_{n+1}\big|_{\partial \Omega }=0,
\end{gathered}\label{IV.7}
\end{equation}
given by Theorem \ref{thmIII.3}. Here
\begin{gather}
\widetilde{w}_n=S_nw_n,  \label{IV.8} \\
\theta _n=|G(\widetilde{w}_n)|_0,  \label{IV.9} \\
g_0=-S_0G(0),g_n=S_{n-1}R_{n-1}-S_nR_n+S_{n-1}G(
0)-S_nG(0),  \label{IV.10} \\
R_0=0,\quad R_n=\sum^n_{j=1} r_j,  \label{IV.11} \\
r_0=0,\quad r_j=[L_{G}(w_{j-1})-L_{G}(\widetilde{w}_{j-1})] u_j+Q_j
-\theta _{j-1}\triangle u_j,\quad 1\leq j\leq n, \label{IV.12} \\
Q_j=G(w_j)-G(w_{j-1})-L_{G}(w_{j-1})u_j,\quad 1\leq j\leq n.  \label{IV.13}
\end{gather}
To ensure that the $w_n$'s  are well defined, we prove the following
proposition.

\begin{proposition} \label{propIV.1}
Let $s\in \mathbb{N}$. If $s_{\ast }\geq 7+2n$ and
$4+2n+2\tau \leq \sigma <s_{\ast }-2$, we have
\begin{gather}
\| u_j\|_s\leq \sqrt{\widetilde{\varepsilon }}[\max
(\mu ,\mu _{j-1})] ^{s-\sigma },\quad j\in \mathbb{N}^{\ast },\;
0\leq s\leq s_{\ast },  \label{IV.14} \\
\| w_j\|_s\leq \begin{cases}
2\sqrt{\widetilde{\varepsilon }}, &\mbox{for } s\leq \sigma -\tau  \\
\sqrt{\widetilde{\varepsilon }}\mu _j^{s-\sigma }, &\mbox{for }
\sigma -\tau \leq s\leq s_{\ast }
\end{cases} \quad j\in \mathbb{N}^{\ast },   \label{IV.15} \\
|\widetilde{w}_j|_{4,\tau }\leq 1,\quad j\in \mathbb{N}^{\ast }, \label{IV.16}\\
\| w_j-\widetilde{w}_j\|_s\leq 2\beta \sqrt{\widetilde{
\varepsilon }}\mu _j^{s-\sigma },\quad 0\leq s\leq s_{\ast },\;
j\in \mathbb{N}^{\ast },   \label{IV.17} \\
\| r_j\|_s\leq \widetilde{\varepsilon }a_1[\max
(\mu ,\mu _{j-1})] ^{s-\sigma },\quad 0\leq s\leq s_{\ast }-2,
\; j\in \mathbb{N}^{\ast },   \label{IV.18} \\
\| g_j\|_s\leq \widetilde{\varepsilon }a_{2}\mu
_j^{s-\sigma },\quad 0\leq s\leq s_{\ast },\; j\in \mathbb{N},
\label{IV.19} \\
\theta _j\leq a_{3}\sqrt{\widetilde{\varepsilon }}\mu _j^{-2}\leq 1,
\quad j\in \mathbb{N},   \label{IV.20} \\
A_j(2)\leq M_0,\quad j\in \mathbb{N}. \label{IV.21}
\end{gather}
Here, $A_j(k)$ is defined by using the definition of $A(k)$
in \eqref{III.11}, where the coefficients correspond to
$\widetilde{w}_j$.
\end{proposition}

Let us first show how that Proposition  \ref{IV.1} implies Theorem \ref{thmA}.
The proof of this proposition will be given later in Appendix 1.

\subsection*{Part 2: Proof of Theorem \ref{thmA}}
We prove the convergence of the sequence $(w_n)$ using Proposition  \ref{IV.1}.
Set $\sigma =s_{\ast }-2-\tau $ and $s=\sigma -\tau $. By
\eqref{IV.6} and \eqref{IV.14}, for any $i,k\in \mathbb{N}^{\ast }$, $i>k$,
$$
\| w_i-w_{k}\|_s\leq \sum^i_{j=k+1}
\| u_j\|_s\leq \beta \sqrt{\widetilde{\varepsilon }}
\sum^i_{j=k+1} \mu _{j-1}^{-\tau }=\beta \sqrt{\widetilde{\varepsilon }}
\sum^i_{j=k+1} (\mu ^{-\tau})^{j-1}.
$$
Since $\mu \geq 2$ and $\tau >0$, then $\| w_i-w_{k}\|_s\to 0$ as
$i,k\to \infty $.
Hence, there is a function $w\in H^{s_{\ast }-2-2\tau }(\Omega )$
satisfying $w_n\to w$ in $H^{s_{\ast }-2-2\tau }(\Omega)$.

Since $H^{s_{\ast }-2-2\tau }(\Omega )\subset C^{s_{\ast
}-2-n-3\tau }(\overline{\Omega })$, it follows that
 $w\in C^{s_{\ast}-3-n}(\overline{\Omega })$.
On the other hand, combining \eqref{IV.7}, \eqref{IV.12} and \eqref{IV.13},
 we obtain
\begin{equation*}
r_j=G(w_j)-G(w_{j-1})-g_{j-1}
\end{equation*}
Taking the sum between $j=1$ and $j=n$, using \eqref{IV.10} and \eqref{IV.11},
we get
\begin{equation}
G(w_n)=(I-S_{n-1})R_{n-1}+(I-S_{n-1})G(0)+r_n.   \label{IV.22}
\end{equation}
For $n\geq 2$, using \eqref{II.2} and \eqref{IV.18}, we have
\begin{equation*}
\| r_n\|_{s_{\ast }-2-2\tau }\leq a_1\beta \widetilde{
\varepsilon }\mu _{n-1}^{s_{\ast }-2-2\tau -\sigma }=a_1\beta \widetilde{
\varepsilon }\mu _{n-1}^{-\tau }.
\end{equation*}
Combining \eqref{II.3} with \eqref{II.6} and
\eqref{I.5}, we get
\begin{equation*}
\| (I-S_{n-1})G(0)\|_{s_{\ast
}-2-2\tau }\leq \beta \mu _{n-1}^{-2-2\tau }\| G(0)
\|_{s_{\ast }}\leq \beta ^2\mu _{n-1}^{-2-2\tau }\widetilde{
\varepsilon }.
\end{equation*}
Combining \eqref{II.6}, \eqref{IV.11} and \eqref{IV.18}, we can write
\begin{align*}
\| (I-S_{n-1})R_{n-1}\|_{s_{\ast }-2-2\tau }
&\leq \beta \mu _{n-1}^{-2\tau }\| R_{n-1}\|_{s_{\ast}-2}
 \leq \beta \mu _{n-1}^{-2\tau }\sum^{n-1}_{j=1}\| r_j\|_{s_{\ast }-2} \\
&\leq \beta \mu _{n-1}^{-2\tau }\widetilde{\varepsilon }a_1
Big(\mu^{s_{\ast }-2-\sigma }+\sum^{n-1}_{j=2} \mu_{j-1}^{s_{\ast }-2-\sigma }
\big) \\
&\leq \widetilde{\varepsilon }\beta a_1\mu _{n-1}^{-2\tau }\mu
_{n-1}^{s_{\ast }-2-\sigma }
\leq \beta a_1\widetilde{\varepsilon }\mu _{n-1}^{-\tau }.
\end{align*}
These inequalities imply $G(w_n)\to 0$ in $H^{s_{\ast}-2-2\tau }(\Omega )$ as
$n\to \infty $.

Since $H^{s_{\ast }-2-2\tau }(\Omega )\subset C^2(\overline{\Omega })$ and 
$w_{n\mid \partial \Omega }=0$, we conclude that 
$G(w)=0$ and $w\big|_{ \partial \Omega }=0$. That is 
$u=\varphi+\varepsilon w$ is a solution to the original Monge-Amp\`{e}re 
equation
which is by Lemma \ref{lmIII.1} plurisubharmonic since $g$ is nonnegative. If we
suppose that $\rho =0$, in (A2), then\ the uniqueness of the solution
follows immediately from \cite{c1}.


\section{Proof of Theorem \ref{thmB}}

We shall use the result of  Xu and Zuily \cite{x1,x2} that we recall briefly.
Let us consider a non linear partial
differential equation
$$
F(x,y,u,\nabla u,D^2u)=0\,,
$$
where $F$ is $C^{\infty }$. To any solution $u$\ we can associate the vector
fields $X_j=\sum_{k} \frac{\partial F}{\partial u_{jk}}\partial_{k}$. Then

\begin{theorem}[{[12]}] \label{thmV.1}
Suppose $u\in C_{\rm loc}^{\rho }(\Omega )$ with $\rho >Max(4,r+2)$
for some constant $r\geq 0$ and that the brackets of the $X_j$, up to the
order $r$, span the tangent space at each point of $\Omega $, then $u$
belongs to $C^{\infty }(\Omega )$.
\end{theorem}

To prove this theorem, it is sufficient to prove that the
solution of Theorem \ref{thmA} satisfies Theorem \ref{thmV.1} at any point in $\Sigma $.
Suppose $\Sigma =\{0\}$. For $i=1\dots n$;
\begin{gather}
X_i=\phi ^{ii}\frac{\partial }{\partial x_i}
+\sum_{j\neq i,\, j=1} ^n \frac{\phi ^{ij}+\overline{\phi ^{ij}}}{2}
\frac{\partial }{\partial x_j}
+\sum_{j\neq i,\, j=1}^n \frac{i\phi ^{ij}-i\overline{\phi ^{ij}}}{2}
\frac{\partial }{\partial x_{j+n}},  \label{V.1} \\
X_{i+n}=\phi ^{ii}\frac{\partial }{\partial x_{i+n}}
+\sum_{j\neq i,\, j=1} ^n \frac{\phi ^{ij}+\overline{\phi ^{ij}}}{2}
\frac{\partial }{\partial x_{j+n}}
-\sum_{j\neq i,\, j=1} ^n \frac{i\phi ^{ij}-i\overline{\phi ^{ij}}}{2}
\frac{\partial }{\partial x_j}.  \label{V.2}
\end{gather}

For computing the Lie algebra generated by the $X_i$, we need the
following result.


\begin{lemma} \label{lmV.2}
For any integer $1\leq m\leq k$,
\begin{equation}
\begin{aligned}
&(adX_n)^{m-1}[X_n-iX_{2n},X_i-iX_{i+n}] \\
&=\sum^{2n}_{l=1} \sum_{|\beta |\leq m,\, i\neq j}
\big[(C_{i\beta p})\partial _x^{\beta }g+\varepsilon d_{pij}\big]
\partial _{x_{l}} \\
&\quad + [A_n(\varphi _{i\overline{j}})]^{m-1}A_i(\varphi _{i\overline{j}})
\big[(\partial_{x_n}^{m}g+i\partial _{x_n}^{m-1}\partial _{x_{2n}}g)(
\partial _{x_i}+i\partial _{x_i+n})\big] ,
\end{aligned} \label{V.3}
\end{equation}
where $C_{i\beta p}$ and $d_{pij}$ are $C^{s_{\ast }-m,\tau }(\Omega
)$ (depending on $w$ and $\varphi $ bounded for $\varepsilon $ small
enough) satisfying for $|\beta |=m$, $C_{i\beta p}(
0)=0,p=1,\dots ,n$ if $n\geq 3$ and $C_{i\beta 1}(0)=0$ if $n=2$.
$A_n=\frac{\partial F}{\partial u_{n\overline{n}}}$ and
$A_i=\frac{\partial ^2F}{\partial u_{n\overline{n}}\partial u_{i\overline{i}}}$.
\end{lemma}

\begin{proof}
We use induction on the size of the brackets. First we calculate
$D_{in}=[X_n+iX_{2n},X_i+iX_{i+n}]$,
for $i\leq n-1$.
\begin{align*}
D_{in} & =\Big[\sum_{j=1}^n \Phi ^{nj}\partial
_{x_j}+i\sum_{j=1}^n \Phi ^{nj}\partial _{x_{j+n}},
\sum_{l=1}^n \Phi ^{il}\partial _{x_{l}}+i\sum_{l=1}^n
\Phi ^{il}\partial _{x_{l+n}}\Big]  \\
&  =\sum_{l=1}^n \underset{(1)}{
\underbrace{\sum_{j=1}^n \{\Phi ^{nj}\partial
_{x_j}(\Phi ^{il})-\Phi ^{ij}\partial _{x_j}(\Phi
^{nl})\}}}\partial _{x_{l}}  \\
& \quad +i\underset{(2)}{\sum_{l=1}^n
\underbrace{\sum_{j=1}^n \{\Phi ^{nj}\partial
_{x_{j+n}}(\Phi ^{il})-\Phi ^{ij}\partial _{x_{j+n}}(\Phi
^{nl})\}}}\partial _{x_{l}}  \\
& \quad -\sum_{l=1}^n \underset{(2)}{
\underbrace{\sum_{j=1}^n \{\Phi ^{nj}\partial
_{x_{j+n}}(\Phi ^{il})-\Phi ^{ij}\partial _{x_{j+n}}(\Phi
^{nl})\}}}\partial _{x_{l+n}}  \\
& \quad  +i\underset{(1)}{\sum_{l=1}^n
\underbrace{\sum_{j=1}^n \{\Phi ^{nj}\partial
_{x_j}(\Phi ^{il})-\Phi ^{ij}\partial _{x_j}(\Phi^{nl})\}}}
\partial _{x_{l}+n},
\end{align*}
where
$$
(1)=\sum_{j=1}^n \sum_{p,q=1}^n
\{\frac{\partial F}{\partial u_{n\overline{j}}}\frac{
\partial ^2F}{\partial u_{i\overline{l}}\partial u_{p\overline{q}}}-\frac{
\partial F}{\partial u_{i\overline{j}}}\frac{\partial ^2F}{\partial u_{n
\overline{l}}\partial u_{p\overline{q}}}\}\partial _{x_j}u_{p
\overline{q}}.
$$
Using \eqref{II.10}, we get
\begin{align*}
F.(1) & =\sum_{j=1}^n \sum_{p,q=1}^n {\frac{\partial F}{\partial u_{n\overline{j}}}(
\frac{\partial F}{\partial u_{i\overline{l}}}\frac{\partial F}{\partial u_{p
\overline{q}}}-\frac{\partial F}{\partial u_{i\overline{q}}}\frac{\partial F
}{\partial u_{p\overline{l}}})\partial _{x_j}u_{p\overline{q}}}  \\
&\quad -\sum_{j=1}^n \sum_{p,q=1}^n
\frac{\partial F}{\partial u_{i\overline{j}}}(\frac{\partial F}{
\partial u_{n\overline{l}}}\frac{\partial F}{\partial u_{p\overline{q}}}-
\frac{\partial F}{\partial u_{n\overline{q}}}\frac{\partial F}{\partial u_{p
\overline{l}}})\partial _{x_j}u_{p\overline{q}}  \\
& =\sum_{j=1}^n \underset{(5)}{
\underbrace{\sum_{p,q=1}^n \frac{\partial F}{\partial
u_{p\overline{q}}}\partial _{x_j}u_{p\overline{q}}(\frac{\partial F}{
\partial u_{n\overline{j}}}\frac{\partial F}{\partial u_{i\overline{l}}}-
\frac{\partial F}{\partial u_{i\overline{j}}}\frac{\partial F}{\partial u_{n
\overline{l}}})}} \\
&\quad +\underset{(6)}{\underbrace{\sum_{j,p,q=1}^n
\frac{\partial F}{\partial u_{p\overline{l}}}(\frac{\partial F}{
\partial u_{i\overline{j}}}\frac{\partial F}{\partial u_{n\overline{q}}}-
\frac{\partial F}{\partial u_{n\overline{j}}}\frac{\partial F}{\partial u_{i
\overline{q}}})\partial _{x_j}u_{p\overline{q}}}}.
\end{align*}
Using \eqref{II.10}, we have
$$
(5) =  \partial _{x_j}(F)F\frac{\partial ^2F}
{\partial u_{n\overline{j}}\partial u_{i\overline{l}}}.
$$
Similarly,  we prove that
\begin{align*}
F.(2)& =\sum_{j=1}^n \partial
_{x_{j+n}}(F)F\frac{\partial ^2F}{\partial u_{n\overline{j}
}\partial u_{i\overline{l}}}  \\
&\quad +\underset{(7)}{\underbrace{\sum_{j,p,q=1}^n
\frac{\partial F}{\partial u_{p\overline{l}}}(\frac{\partial F}{
\partial u_{i\overline{j}}}\frac{\partial F}{\partial u_{n\overline{q}}}-
\frac{\partial F}{\partial u_{n\overline{j}}}\frac{\partial F}{\partial u_{i
\overline{q}}})\partial _{x_{j+n}}u_{p\overline{q}}}}.
\end{align*}
We can easily see that $(6)+i(7)=0$,
so,
$$
(1)+i(2)=  \sum_{j=1}^n (\partial _{x_j}(F)+i\partial _{x_{j+n}}(F)
)\frac{\partial ^2F}{\partial u_{n\overline{j}}\partial u_{i\overline{l}}}
$$
and
$$
D_{in}=  \sum_{l=1}^n \sum_{j=1}^n
(\partial _{x_j}(f)+i\partial _{x_{j+n}}(f)
)\frac{\partial ^2F}{\partial u_{n\overline{j}}\partial u_{i
\overline{l}}}[\partial _{x_{l}}+i\partial _{x_{l}+n}] .
$$
Since $F$ is the determinant function,
then, $\frac{\partial F}{\partial u_{i\overline{j}}}$ is independent of $u_{i
\overline{l}}$ and $u_{l\overline{j}}$ for $l=1,\dots ,n$. Therefore $\frac{
\partial ^2F}{\partial u_{i\overline{j}}\partial u_{p\overline{q}}}$
vanishes unless $i\neq p$, $j\neq q$. So,
$$
D_{in}=\sum_{(l,j)\neq (i,n),\, l,j\leq n}(\partial _{x_j}(f)+i\partial
_{x_{j+n}}(f))\frac{\partial ^2F}{\partial u_{n
\overline{j}}\partial u_{i\overline{l}}}[\partial _{x_{l}}+i\partial
_{x_{l}+n}] .
$$
We have
$\varphi _{i\overline{j}}(0)=(1-\delta _i^{n})\sigma _i\delta _i^{j}$;
Therefore, if $n\geq 3$ and $(l,s)\neq (i,n)$,
$$
\frac{\partial ^2F}{\partial u_{n\overline{s}}\partial u_{i\overline{l}}}
(\varphi _{i\overline{j}})(0)=0.
$$
If $n=2$ and $l=1$, then $s=1$ and we also have
$$
\frac{\partial ^2F}{\partial u_{2\overline{1}}\partial u_{1\overline{1}}}
(\varphi _{i\overline{j}})(0)=0.
$$
So, \eqref{V.3} is proved for $m=1$. By a recursion on $m$,
we deduce this lemma.
\end{proof}

On the other hand, we have by \eqref{III.5}
\begin{equation}
\begin{gathered}
\Phi ^{ij}(\varphi _{i\overline{j}})(0)=0, \quad\mbox{for }(i,j)\neq (n,n), \\
A_n(\varphi _{i\overline{j}})(0)=\prod^{n-1}_{i=1} \sigma _i>0, \\
A_i(\varphi _{i\overline{j}})(0)=
\prod^{n-1}_{j\neq i,\, i=1} \sigma _i>0.
\end{gathered} \label{V.4}
\end{equation}
Or by the hypothesis, $\partial _x^{\beta }g(0)=0$ for all
$|\beta |<k$, and by \eqref{V.4}, we can
suppose that $\partial _{x_n}^{k}g(0)\neq 0$ 
($\partial_{x_{2n}}^{k}g(0)\neq 0$ leads  to the same result, just
consider $(adX_{2n})^{m-1}$ instead of $(adX_n)^{m-1}$).

So, by taking the real and the imaginary parts of \eqref{V.3} at the origin,
we obtain
\begin{align*}
&(adX_n)^{k-1}([X_n,X_i] -[X_{2n},X_{i+n}] )\\
&=\sum^{2n}_{l=1} \sum_{j\neq i}
\varepsilon d_{pij}'(0)\partial_{x_{l}}
+[A_n(\varphi _{i\overline{j}})(0)]
^{k-1}A_i(\varphi _{i\overline{j}})(0)
[\partial _{x_n}^{k}g\partial _{x_i}-\partial _{x_n}^{k-1}
\partial_{x_{2n}}g\partial _{x_i+n}]
\end{align*}
and
\begin{align*}
&(adX_n)^{k-1}([X_{2n},X_i] +[X_n,X_{i+n}] )\\
&=\sum^{2n}_{l=1} \sum_{j\neq i} \varepsilon d_{pij}'' (0)
\partial_{x_{l}}-[A_n(\varphi _{i\overline{j}})(0)]
^{k-1}A_i(\varphi _{i\overline{j}})(0)[
\partial _{x_n}^{k-1}\partial _{x_{2n}}g\partial _{x_i}
+\partial_{x_n}^{k}g \partial _{x_i+n}].
\end{align*}

Suppose now that $|w|_{k+2}\leq 1$. We will get at the origin
for $\varepsilon \leq \widetilde{\varepsilon }$ small enough the determinant
of the vectors
\begin{equation}
\begin{gathered}
(adX_n)^{k-1}([X_n,X_i] -[X_{2n},X_{i+n}]),\\
(adX_n)^{k-1}([X_{2n},X_i] +[X_n,X_{i+n}] )_{i=1,\dots ,n-1},  \\
X_n,X_{2n} \mbox{ is different from zero}.
\end{gathered} \label{V.5}
\end{equation}
Now, choose $s_{\ast }$ so big that $s_{\ast }\geq \max (7+2n,6+k+n)$ by
means of Theorem \ref{thmA} there exists $\varepsilon _0<\widetilde{\varepsilon }^2$
such that for any $g$ satisfying \eqref{I.5}
 there exists a unique solution $u=\varphi +\varepsilon _0^{
\frac{1}{2}}w\in C^{k+3}(\Omega )$ to the problem \eqref{I.1}.
Moreover; by \eqref{II.1}, $|w|_{k+2}\leq \beta \| w\|_{k+2+n+\tau }$.
Since $\sigma =s_{\ast}-2-\tau $, $s_{\ast }\geq 6+k+n$ and
$\tau \leq \frac{\alpha }{4}<\frac{1}{4}$, then
$$
k+2+n+\tau \leq s_{\ast }-4+\tau =\sigma -2+2\tau \leq \sigma -\tau .
$$
We have then, using \eqref{IV.3}, \eqref{IV.5}
and \eqref{IV.15},
$$
|w|_{k+2}\leq 2\beta \sqrt{\widetilde{\varepsilon }}\leq 1.
$$
So, by \eqref{V.5}, we can conclude that for
$\widetilde{\varepsilon }$ sufficiently small, the vector fields at the origin;
$[(adX_n)^{k-1}([X_{\delta n},X_i] )] _{\delta =1,2;i=1,\dots ,2n-1}$,
$X_n$ and $X_{2n}$ span all the tangent space.
Theorem \ref{thmB} follows then from Theorem \ref{thmV.1}.

\section{Appendix 1}

To prove proposition  \ref{IV.1}, we need the following result.

\begin{proposition} \label{propVI.1}
There exists a constant $K_0\geq 1$ such that
for any function $w^{i}\in C^{s_{\ast }+2,\tau }(\overline{\Omega })$,
$|w^{i}|_{2}\leq 1$, $i=1,2,3$ and for any $\varepsilon \leq 1$ we have
\begin{equation}
|G(w^{1})-G(w^2)|_0\leq
K_0|w^{1}-w^2|_{2}\big(\| \varphi \|
_{2+n_{\ast }}+\| w^{1}\|_{2+n_{\ast }}+\| w^2\|
_{2+n_{\ast }}+1\big).  \label{VI.1}
\end{equation}
Also for $t\in [0,1] $, $s\in [0,s_{\ast }] $,
\begin{equation} \label{VI.2}
\begin{aligned}
&\| \frac{d}{dt}[L_{G}(w^{1}+tw^2)w^{3}]\|_s \\
&\leq \varepsilon K_0[(\| \varphi \|_{2+s}+\varepsilon
\| w^{1}\|_{2+s}+\varepsilon \| w^2\|_{2+s}+1)|w^2|_{2}|w^{3}|_{2}\\
&\quad + (\| \varphi \|_{2+n_{\ast }}+\varepsilon \|
w^{1}\|_{2+n_{\ast }}+\varepsilon \| w^2\|_{2+n_{\ast}}+1)(|w^2|_{2}\| w^{3}\|
_{2+s}+|w^{3}|_{2}\| w^2\|_{2+s})].
\end{aligned}
\end{equation}
\end{proposition}


\begin{proof} Just write
\begin{align*}
& G(w^{1}) - G(w^2)\\
&= \frac{1}{\varepsilon }
[\det (\varphi _{i\overline{j}}+\varepsilon w_{i\overline{j}
}^{1})-\det (\varphi _{i\overline{j}}+\varepsilon w_{i\overline{
j}}^2)+g(w^{1})-g(w^2)]  \\
&=\int_0^{1} \sum^n_{i,j=1}
\frac{\partial F}{\partial u_{i\overline{j}}}(\varphi _{i\overline{j}}
+\varepsilon w_{i\overline{j}}^2+t\varepsilon (w_{i\overline{j}}^{1}
-w_{i\overline{j}}^2))(w_{i\overline{j}}^{1}-w_{i\overline{j}}^2)dt \\
&\quad +\int_0^{1}\frac{\partial g}{\partial u}(\varphi+\varepsilon
w^2+t\varepsilon (w^{1}-w^2))(w^{1}-w^2) \\
&\quad +\int_0^{1}\frac{\partial g}{\partial p_i}(\varphi
+\varepsilon w^2+t\varepsilon (w^{1}-w^2))(w_i^{1}-w_i^2),
\end{align*}
and
\begin{align*}
&\frac{d}{dt}[L_{G}(w^{1}+tw^2)w^{3}]\\
& =\frac{d}{dt}[\sum_{i,j=1}^n \frac{\partial F}{
\partial u_{i\overline{j}}}(\varphi _{i\overline{j}}+\varepsilon w_{i
\overline{j}}^{1}+t\varepsilon w_{i\overline{j}}^2)w_{i\overline{j}
}^{3}+\frac{\partial g}{\partial u}(\varphi +\varepsilon
w^{1}+t\varepsilon w^2)w^{3}+\dots ]  \\
&=\varepsilon \overset{n}{\underset{i,j,p,q=1}{\sum }}\frac{\partial ^2F}{
\partial u_{i\overline{j}}\partial u_{p\overline{q}}}(\varphi _{i
\overline{j}}+\varepsilon w_{i\overline{j}}^{1}+t\varepsilon w_{i\overline{j}
}^2)w_{p\overline{q}}^2w_{i\overline{j}}^{3}+\dots .
\end{align*}
Combining \eqref{II.1}, \eqref{II.7}, \eqref{II.8} and
\eqref{II.9} with the inequalities
$$
|\varphi _{i\overline{j}}+\varepsilon w_{i\overline{j}
}^2+t\varepsilon (w_{i\overline{j}}^{1}-w_{i\overline{j}}^2)
|_0\leq |\varphi |_{2}+2|w^2|_{2}+|w^{1}|_{2}\leq 3+|\varphi |_{2}
$$
and
$$
|\varphi _{i\overline{j}}+\varepsilon w_{i\overline{j}
}^{1}+t\varepsilon w_{i\overline{j}}^2|_0\leq |\varphi
|_{2}+\varepsilon |w^{1}|_{2}+t\varepsilon |
w^2|_{2}\leq 2+|\varphi |_{2},
$$
we  deduce \eqref{VI.1} and \eqref{VI.2}.
\end{proof}

\begin{proof}[Proof of the proposition  \ref{IV.1}]
The proposition is proved by induction.
We have $u_0=0$. Let begin by proving \eqref{IV.19}$_0$
to \eqref{IV.21}$_0$. (i.e. \eqref{IV.19}
to \eqref{IV.21} corresponding to $j=0$).
\\
(a) \eqref{IV.19}$_0$: Using \eqref{III.2}
and \eqref{IV.10}, we have
$$
g_0=-S_0G(0)\quad\mbox{ and }\quad
G(0)=\frac{1}{\widetilde{\varepsilon }}(\det (\varphi _{ij})-g(\varphi)).
$$
But $\varphi \in C^{s_{\ast }+2,\alpha }(\overline{\Omega })$,
$g\in C^{s_{\ast }}$ and $S_n$ are smoothing operators, so
$g_0\in H^{s_{\ast }}(\Omega )$. \eqref{II.3}, \eqref{II.4}, \eqref{III.2} and
 \eqref{I.5} show that
$$
\| g_0\|_s\leq \beta \| G(0)\|_s
\leq \frac{\beta }{\widetilde{\varepsilon }}\| \det (\varphi
_{ij})-g(\varphi )\|_{s_{\ast }}
\leq \frac{\beta ^2}{\widetilde{\varepsilon }}|\det (
\varphi _{ij})-g(\varphi )|_{s_{\ast }}\leq \beta
^2\widetilde{\varepsilon }.
$$
Using \eqref{IV.4} and $\beta \leq \mu $, we get
$\| g_0\|_s\leq \mu ^2\widetilde{\varepsilon }\leq a_{2}
\widetilde{\varepsilon }$

\noindent(b) \eqref{IV.20}$_0$: \eqref{III.2}, \eqref{IV.4},
\eqref{IV.5} and \eqref{I.5} give
$$
\theta _0=|G(0)|_0\leq \frac{1}{\widetilde{
\varepsilon }}|\det (\varphi _{i\overline{j}})-g(
\varphi )|_{s_{\ast }}\leq \widetilde{\varepsilon }\leq \sqrt{
\widetilde{\varepsilon }}a_{3}\leq 1.
$$
(c) \eqref{IV.21}$_0$: We have
$$
A_0(2)=  \max (1,|\frac{\partial g}{\partial u}
(\varphi )|_{2},\underset{i,j}{\max }|\frac{
\partial F}{\partial \varphi _{i\overline{j}}}(\varphi _{l\overline{q}
})|_{2}+\theta _0).
$$
Then, by \eqref{II.9}, \eqref{IV.1} and \eqref{IV.20}$_0$,
$A_0(2)\leq M_0$.

Assume that $u_0$, $u_1,\dots , u_{n-1}\in H^{s_{\ast }}(\Omega)$ satisfy
 \eqref{III.13}--\eqref{III.15} and
\eqref{IV.14}--\eqref{IV.21} for $j\leq n-1$.
We shall construct $u_n\in H^{s_{\ast }}(\Omega )$ satisfying
\eqref{III.13}--\eqref{III.15} and prove
that \eqref{IV.14}--\eqref{IV.21} are
satisfied for $j=n$.

Combining \eqref{IV.16}$_{n-1}$--\eqref{IV.21}$_{n-1}$, we have
$|\widetilde{w}_{n-1}|_{4,\kappa}\leq 1$, $\theta _{n-1}\leq 1$,
$A_{n-1}(2)\leq M_0$ and $g_{n-1}\in H^{s_{\ast }}(\Omega )$.
We can then apply Theorem \ref{thmIII.3} to get a solution
$u_n\in H^{s_{\ast}}(\Omega )$ to the problem \eqref{IV.7}$_n$
satisfying \eqref{III.13}--\eqref{III.15}. Then:

\noindent (a) \eqref{IV.14}$_n$:
For $n=1$, using \eqref{I.5}, \eqref{II.3}, \eqref{III.2},
\eqref{III.13}, and \eqref{IV.2}, we have
$$
\| u_1\|_0\leq D\| g_0\|_0\leq D\beta
\| G(0)\|_0\leq D\frac{\beta ^2}{\widetilde{
\varepsilon }}|\det (\varphi _{ij})-g(\varphi
)|_{s_{\ast }}\leq D\beta ^2\widetilde{\varepsilon }.
$$
\eqref{IV.3}, \eqref{IV.5}, and $s_{\ast }\geq \sigma $ give
\begin{equation}
\| u_1\|_0\leq \sqrt{\widetilde{\varepsilon }}\mu ^{-\sigma}.  \label{VI.3}
\end{equation}
By \eqref{III.14}, we have
$\| u_1\|_1\leq D(\| g_0\|_1+\|u_1\|_0)$.
Therefore, using \eqref{I.5}, \eqref{II.3}, % (2,3) 
\eqref{VI.3}, and
$s_{\ast }\geq \sigma $, we get
$$
\| u_1\|_1\leq D(\beta ^2\widetilde{\varepsilon }+
\sqrt{\widetilde{\varepsilon }}\mu ^{-\sigma })\leq \sqrt{\widetilde{
\varepsilon }}\mu ^{1-\sigma }.
$$
Suppose that  for $0\leq l\leq s$ and $s\geq 2$ we have
\begin{equation}
\| u_1\|_{l}\leq \sqrt{\widetilde{\varepsilon }}\mu ^{l-\sigma}.  \label{VI.4}
\end{equation}
Using \eqref{III.15}, we have, for $s\geq 2$,
$$
\| u_1\|_s\leq D\big(\| g_0\|_s+\sum_{l\leq s-1,\, (i,l)\in \Lambda _s}
(1+|\varphi |_{i+4,\tau })\| u_1\|_{l}\big).
$$
\eqref{I.5}, \eqref{II.3}, \eqref{II.4}, \eqref{IV.3}, \eqref{IV.10},
 and $s_{\ast }\geq \sigma $ imply
$$
\| g_0\|_s\leq \beta \| G(0)\|
_s\leq \beta ^2|G(0)|_s\leq \beta ^2
\widetilde{\varepsilon }\leq \widetilde{\mu }\widetilde{\varepsilon }\mu
^{s-\sigma },
$$
which by \eqref{VI.3} and \eqref{VI.4} gives
\begin{align*}
\| u_1\|_s&\leq D\big(\widetilde{\mu }\widetilde{
\varepsilon }\mu ^{s-\sigma }+
\sum_{l\leq s-1,\, (i,l)\in \Lambda _s} (1+|\varphi |_{i+4,\tau })
\sqrt{\widetilde{\varepsilon }}\mu ^{l-\sigma }\big) \\
& \leq D\big(\widetilde{\mu }\widetilde{\varepsilon }\mu ^{s-\sigma
}+s_{\ast }^2(1+|\varphi |_{i+4,\tau })\mu ^{-1}
\sqrt{\widetilde{\varepsilon }}\mu ^{s-\sigma }\big),
\end{align*}
which by \eqref{IV.3} and \eqref{IV.5} shows that
$\| u_1\|_s\leq \sqrt{\widetilde{\varepsilon }}\mu
^{s-\sigma }$.

 For $n\geq 2$, \eqref{III.13}, \eqref{IV.2}, \eqref{IV.5},
 and \eqref{IV.19}$_{n-1}$ imply
\begin{equation}
\| u_n\|_0\leq D\| g_{n-1}\|_0\leq D\widetilde{
\varepsilon }a_{2}\mu _{n-1}^{-\sigma }\leq \sqrt{\widetilde{\varepsilon }}
\mu _{n-1}^{-\sigma }.  \label{VI.5}
\end{equation}
In the same way; \eqref{III.14}, \eqref{IV.2}, \eqref{IV.5},
\eqref{IV.19}$_{n-1}$ and \eqref{VI.5} give
$$
\| u_n\|_1\leq \sqrt{\widetilde{\varepsilon }}\mu_{n-1}^{1-\sigma }.
$$
Suppose that, for $0\leq l<s$ and $s\geq 2$,
$\| u_n\|_{l}\leq \sqrt{\widetilde{\varepsilon }}\mu_{n-1}^{l-\sigma }$.
By \eqref{III.15}, we have
$$
\| u_n\|_s\leq D(\| g_{n-1}\|_s+
\sum_{l\leq s-1,\, (i,l)\in \Lambda _s}
(1+|\varphi +\widetilde{\varepsilon }\widetilde{w}_{n-1}|
_{i+4,\tau })\| u_n\|_{l}).
$$
But, \eqref{II.1}, \eqref{II.5}, \eqref{IV.15}$_{n-1}$, and
$4+n_{\ast }\leq \sigma -\tau $ imply that, for $0\leq i\leq s-2$,
$$
|\widetilde{w}_{n-1}|_{i+4,\tau }\leq \beta \| \widetilde{
w}_{n-1}\|_{4+n_{\ast }+i}\leq \beta ^2\mu _{n-1}^{i}\|
\widetilde{w}_{n-1}\|_{4+n_{\ast }}\leq 2\beta ^2\sqrt{\widetilde{
\varepsilon }}\mu _{n-1}^{i}.
$$
Therefore, using \eqref{IV.19}$_{n-1}$, we get
\begin{align*}
\| u_n\|_s&\leq D\big(\widetilde{\varepsilon }
a_{2}\mu _{n-1}^{s-\sigma }+\sum (1+|\varphi |_{s_{\ast
}+2,\tau }+2\beta ^2\sqrt{\widetilde{\varepsilon }}\mu _{n-1}^{i})
\sqrt{\widetilde{\varepsilon }}\mu _{n-1}^{l-\sigma }\big) \\
&\leq D\big(\widetilde{\varepsilon }a_{2}\mu _{n-1}^{s-\sigma }+2\beta
^2s_{\ast }^2\widetilde{\varepsilon }\mu _{n-1}^{s-\sigma }+(
1+|\varphi |_{s_{\ast }+2,\tau })s_{\ast }^2\sqrt{
\widetilde{\varepsilon }}\mu _{n-1}^{s-1-\sigma }\big),
\end{align*}
which combined with \eqref{IV.4} and \eqref{IV.5} gives
$\| u_n\|_s\leq \sqrt{\widetilde{\varepsilon }}\mu
_{n-1}^{s-\sigma }$.

\noindent(b) \eqref{IV.15}$_n$: \eqref{IV.6} shows
that $w_n=\sum_{j=1}^n u_j$. By \eqref{IV.14}$_j$,
$1\leq j\leq n$, we have
$$
\| w_n\|_s\leq \sum_{j=1}^n \|
u_j\|_s\leq \sqrt{\widetilde{\varepsilon }}\mu ^{s-\sigma }+
\overset{n}{\underset{j=2}{\sum }}\sqrt{\widetilde{\varepsilon }}\mu
_{j-1}^{s-\sigma }\leq \sqrt{\widetilde{\varepsilon }}\mu ^{s-\sigma }+
\overset{n-1}{\underset{j=1}{\sum }}\sqrt{\widetilde{\varepsilon }}\mu
_j^{s-\sigma }.
$$
For $s\leq \sigma -\tau $, since $\mu \geq 2^{1/\tau }\geq 2$,
we have $\mu _j^{s-\sigma }\leq \mu _j^{-\tau }\leq \frac{1}{2^{j}}$ and
$$
\| w_n\|_s\leq \sum^{n-1}_{j=0} \sqrt{
\widetilde{\varepsilon }}\mu _j^{s-\sigma }\leq \sqrt{\widetilde{\varepsilon }}
\sum^{n-1}_{j=0} \frac{1}{2^{j}}\leq 2\sqrt{\widetilde{\varepsilon }}.
$$
For $s\geq \sigma -\tau $, we have
$$
\| w_n\|_s\leq \sqrt{\widetilde{\varepsilon }}\mu
^{s-\sigma }+\sqrt{\widetilde{\varepsilon }}\frac{\mu ^{n(s-\sigma
)}-\mu ^{s-\sigma }}{\mu ^{s-\sigma }-1}.
$$
Since $\mu \geq 2^{1/\tau}$, it follows that 
$\mu ^{s-\sigma }\geq \mu ^{\tau}\geq 2$. Therefore,
$\| w_n\|_s\leq \sqrt{\widetilde{\varepsilon }}\mu_n^{s-\sigma }$.

\noindent(c) \eqref{IV.16}$_n$: Combining \eqref{II.1}, \eqref{II.4},
\eqref{IV.5}, \eqref{IV.15}$_n$ and $4+n_{\ast }\leq \sigma -\tau $, we obtain
$$
|\widetilde{w}_n|_{4,\tau }\leq \beta \| \widetilde{w}
_n\|_{4+n_{\ast }}\leq \beta ^2\| w_n\|_{4+n_{\ast
}}\leq 2\beta ^2\sqrt{\widetilde{\varepsilon }}\leq 1.
$$
(d) \eqref{IV.17}$_n)$:
In the case $s\leq \sigma -\tau $, using \eqref{II.6}
 and \eqref{IV.15}$_n$, we obtain
$$
\| w_n-\widetilde{w}_n\|_s \leq \beta \mu _n^{s-
[\sigma +\tau ] -1}\| w_n\|_{[\sigma +\tau] +1}
\leq \beta \mu _n^{s-[\sigma +\tau ] -1}\sqrt{\widetilde{
\varepsilon }}\mu _n^{[\sigma +\tau ] +1-\sigma }\leq \beta
\sqrt{\widetilde{\varepsilon }}\mu _n^{s-\sigma }.
$$
In the case $s>\sigma -\tau $, \eqref{II.6} \eqref{IV.15}$_n)$ and
$\beta \geq 1$ give
$$
\| w_n-\widetilde{w}_n\|_s\leq \beta \|
w_n\|_s\leq \beta \sqrt{\widetilde{\varepsilon }}\mu
_n^{s-\sigma }.
$$
(e) \eqref{IV.18}$_n$: By \eqref{IV.12}, we have
$$
r_n=\underset{(1)}{\underbrace{[L_{G}(
w_{n-1})-L_{G}(\widetilde{w}_{n-1})] u_n}}-
\underset{(2)}{\underbrace{\theta _{n-1}\triangle u_n}}+\underset{(
3)}{\underbrace{Q_n}}
$$
When $n=1$, $(1)=0$. In the case $n\geq 2$, since
$$
(1)= \int_0^{1}\frac{d}{dt}[L_{G}(
\widetilde{w}_{n-1}+t(w_{n-1}-\widetilde{w}_{n-1}))u_n] dt,
$$
by \eqref{II.1} and \eqref{IV.17}$_{n-1}$, we get
$$
|w_{n-1}-\widetilde{w}_{n-1}|_{2}\leq \beta \| w_{n-1}-
\widetilde{w}_{n-1}\|_{2+n_{\ast }}\leq 2\beta ^2\sqrt{\widetilde{
\varepsilon }}\mu _{n-1}^{3+n_{\ast }-\sigma }.
$$
But $2\beta ^2\sqrt{\widetilde{\varepsilon }}\leq 1$ and $3+n_{\ast }\leq
4+2n_{\ast }\leq \sigma $, so, $|w_{n-1}-\widetilde{w}_{n-1}|
_{2}\leq 1$. In the same way, \eqref{II.1}, \eqref{IV.5} and
\eqref{IV.14}$_n$ give
$$
|u_n|_{2}\leq \beta \| u_n\|_{2+n_{\ast }}\leq
\beta \sqrt{\widetilde{\varepsilon }}\mu _{n-1}^{3+n_{\ast }-\sigma }\leq 1.
$$
By \eqref{IV.16}$_{n-1}$, we also have
$|\widetilde{w}_{n-1}|_{2}\leq 1$.
Hence, we can apply Proposition  \ref{VI.1} to get
\begin{align*}
\| (1)\|_s\leq &\widetilde{\varepsilon }
K_0\{[\| \varphi \|_{s+2}+\| \widetilde{w}
_{n-1}\|_{s+2}+\| w_{n-1}\|_{s+2}+1] |w_{n-1}-
\widetilde{w}_{n-1}|_{2}|u_n|_{2}  \\
&\quad+(\| \varphi \|_{2+n_{\ast }}+\| \widetilde{w}
_{n-1}\|_{2+n_{\ast }}+\| w_{n-1}\|_{2+n_{\ast}}+1)\\
&\quad \times (|w_{n-1}-\widetilde{w}_{n-1}|_{2}\|
u_n\|_{s+2}+\| w_{n-1}-\widetilde{w}_{n-1}\|
_{s+2}|u_n|_{2})\}.
\end{align*}
Using \eqref{II.3} and \eqref{IV.3}, we get
for $0\leq s\leq s_{\ast }$,
$$
\| \varphi \|_{s+2}\leq \beta |\varphi |_{s_{\ast}+2}\leq \beta \mu \leq \mu ^2.
$$
By \eqref{II.2}, it suffices to prove \eqref{IV.18}$_n$ for $s=0$ and
$s=s_{\ast }-2$.

\noindent Case $s=0$: combining \eqref{II.1}, \eqref{IV.14}$_n$,
\eqref{IV.15}$_{n-1}$ and \eqref{IV.17}$_{n-1}$, we have
\begin{align*}
\| (1)\|_0&\leq \widetilde{\varepsilon }
K_0\{(\mu ^2+2\beta \sqrt{\widetilde{\varepsilon }}+2\sqrt{
\widetilde{\varepsilon }}+1)2\beta ^{3}\widetilde{\varepsilon }\mu
_{n-1}^{4+2n_{\ast }-2\sigma } \\
& \quad +(\mu ^2+2\beta \sqrt{\widetilde{\varepsilon }}+2\sqrt{
\widetilde{\varepsilon }}+1)4\beta ^{3}\widetilde{\varepsilon }\mu
_{n-1}^{4+n_{\ast }-2\sigma }\},
\end{align*}
which using \eqref{IV.5} and $\sigma \geq 4+2n_{\ast}\geq 4+n_{\ast }$ gives
$\| (1)\|_0\leq \widetilde{\varepsilon }K_0\mu_{n-1}^{-\sigma }$.

\noindent Case $s=s_{\ast }-2$: \eqref{IV.5} and $s_{\ast }\geq
\sigma +\tau $, as in the previous case, imply
$$
\| (1)\|_{s_{\ast }-2}\leq \widetilde{\varepsilon }
K_0\mu _{n-1}^{s_{\ast }-2-\sigma }.
$$
By \eqref{II.2}, we obtain for $0\leq s\leq s_{\ast }-2$,
$$
\| (1)\|_s\leq \beta \widetilde{\varepsilon }K_0\mu _{n-1}^{s-\sigma }.
$$
Next,
$$\| (2)\|_s\leq \theta _{n-1}\|u_n\|_{s+2}.
$$
If $n=1$ combining \eqref{IV.5}, \eqref{IV.9} and \eqref{IV.14}$_n$, we obtain
$$
\| (2)\|_s\leq |G(0)|_0\| u_1\|_{s+2}\leq \widetilde{\varepsilon }\sqrt{
\widetilde{\varepsilon }}\mu ^{s+2-\sigma }\leq \widetilde{\varepsilon }\mu
^{s-\sigma }.
$$
In the case $n\geq 2$: \eqref{IV.14}$_n$ and \eqref{IV.20}$_{n-1}$ imply
$$
\| (2)\|_s\leq a_{3}\widetilde{\varepsilon }\mu
_{n-1}^{-2}\mu _{n-1}^{s+2-\sigma }=a_{3}\widetilde{\varepsilon }\mu
_{n-1}^{s-\sigma }.
$$
Finally, since by \eqref{IV.13},
\begin{align*}
(3)&=Q_n=G(w_{n-1}+u_n)-G(w_{n-1})-L_{G}(w_{n-1})u_n \\
& =\int_0^{1}(\int_0^{t}\frac{d}{dh}[L_{G}(w_{n-1}+hu_n)u_n] dh)dt\,.
\end{align*}
Then, using \eqref{II.1}, \eqref{IV.5} and \eqref{IV.15}$_{n-1}$, we obtain
$$
|w_{n-1}|_{2}\leq \beta \| w_{n-1}\|_{2+n_{\ast
}}\leq 2\beta \sqrt{\widetilde{\varepsilon }}\leq 1.
$$
Since we proved that $|u_n|_{2}\leq 1$, we can apply
proposition  \ref{VI.1} to have
\begin{align*}
\| (3)\|_s&\leq \widetilde{\varepsilon }
K_0[(\| \varphi \|_{s+2}+\| u_n\|
_{s+2}+\| w_{n-1}\|_{s+2}+1)|u_n|_{2}^2\\
&\quad +2|u_n|_{2}\| u_n\|_{s+2}(
\| \varphi \|_{2+n_{\ast }}+\| u_n\|_{2+n_{\ast}}
+\| w_{n-1}\|_{2+n_{\ast }}+1)].
\end{align*}
Combining \eqref{II.1}, \eqref{IV.14}$_n$ and \eqref{IV.15}$_{n-1}$, we get

\noindent For $s=0$:
\begin{align*}
&\| (3)\|_0\\
&\leq \widetilde{\varepsilon }K_0\{(\mu ^2
+\sqrt{\widetilde{\varepsilon}}[\max (\mu ,\mu _{n-1})] ^{2-\sigma }
+2\sqrt{\widetilde{\varepsilon }}+1)\beta ^2\widetilde{\varepsilon }[
\max (\mu ,\mu _{n-1})] ^{4+2n_{\ast }-2\sigma } \\
&\quad +8(\mu ^2+\sqrt{\widetilde{\varepsilon }}\beta [\max (
\mu ,\mu _{n-1})] ^{2+n_{\ast }-\sigma }+2\sqrt{\widetilde{
\varepsilon }}+1)\widetilde{\varepsilon }\beta [\max (\mu
,\mu _{n-1})] ^{4+n_{\ast }-2\sigma }\},
\end{align*}
which  combined with \eqref{IV.5} and $\sigma \geq 4+2n_{\ast }$ gives
$$
\| (3)\|_0\leq \widetilde{\varepsilon }K_0
[\max (\mu ,\mu _{n-1})] ^{-\sigma }.
$$
For $s=s_{\ast }-2$; since $\sigma \geq 4+2n_{\ast }$, we also get
$$
\| (3)\|_{s_{\ast }-2}\leq \widetilde{\varepsilon }
K_0[\max (\mu ,\mu _{n-1})] ^{s_{\ast }-2-\sigma}.
$$
Then \eqref{II.2} shows that, for $0\leq s\leq s_{\ast }-2$,
$$
\| (3)\|_s\leq \beta \widetilde{\varepsilon }
K_0[\max (\mu ,\mu _{n-1})] ^{s-\sigma },
$$
and we conclude that
\begin{align*}
\| r_n\|_s & \leq (2\beta K_0+a_{3})
\widetilde{\varepsilon }[\max (\mu ,\mu _{n-1})]^{s-\sigma } \\
& \leq 9K_0\mu ^{5}\widetilde{\varepsilon }[\max (\mu ,\mu
_{n-1})] ^{s-\sigma }  \\
& =a_1\widetilde{\varepsilon }[\max (\mu ,\mu _{n-1})] ^{s-\sigma }.
\end{align*}
(f) \eqref{IV.19}$_n)$: By \eqref{IV.10} and \eqref{IV.11},
\begin{align*}
g_n& =S_{n-1}R_{n-1}-S_nR_n+(S_{n-1}-S_n)G(0)  \\
& =\underset{(4)}{\underbrace{(
S_{n-1}R_{n-1}-S_nR_{n-1})}}-\underset{(5)}{
\underbrace{S_nr_n}}+\underset{(6)}{\underbrace{(
S_{n-1}-S_n)G(0)}}.
\end{align*}
Case $s=0$: \eqref{II.6}, \eqref{IV.11} and
\eqref{IV.18}$_j$, $j\leq n-1$, imply
\begin{align*}
\| (4)\|_0&\leq \| (I-S_{n-1})R_{n-1}\|_0+\| (I-S_n)
R_{n-1}\|_0 \\
& \leq \beta \| R_{n-1}\|_{s_{\ast }-2}\mu _{n-1}^{2-s_{\ast
}}+\beta \mu _n^{2-s_{\ast }}\| R_{n-1}\|_{s_{\ast}-2} \\
& \leq (\beta a_1\widetilde{\varepsilon }\mu _{n-1}^{2-s_{\ast
}}+\beta a_1\widetilde{\varepsilon }\mu _n^{2-s_{\ast }})(
\mu ^{s_{\ast }-2-\sigma }+\overset{n-1}{\underset{j=2}{\sum }}\mu
_{j-1}^{s_{\ast }-2-\sigma }).
\end{align*}
Since $s_{\ast }-2>\sigma $ and $\beta \leq \mu $, then
$$
\| (4)\|_0 \leq \beta a_1\widetilde{
\varepsilon }(\mu _{n-1}^{2-s_{\ast }}+\mu _n^{2-s_{\ast }})
\mu _{n-1}^{s_{\ast }-2-\sigma }
\leq 2a_1\mu ^2\widetilde{\varepsilon }\mu _n^{-\sigma }.
$$
On the other hand, combining \eqref{II.4}, \eqref{IV.18}$_n$,
$\sigma <s_{\ast }-2$ and $\beta \leq \mu $, we obtain
$$
\| (5)\|_0\leq \beta \| r_n\|
_0\leq \beta a_1\widetilde{\varepsilon }[\max (\mu ,\mu
_{n-1})] ^{-\sigma }\leq a_1\mu ^2\widetilde{\varepsilon }
\mu _n^{-\sigma }.
$$
We also have by \eqref{I.5}, \eqref{II.3}, \eqref{II.6}. and $\sigma <s_{\ast }-2$,
\begin{align*}
\| (6)\|_0&\leq \| (I-S_{n-1})G(0)\|_0+\| (I-S_n)G(0)\|_0  \\
& \leq \beta \mu _{n-1}^{-\sigma }\| G(0)\|
_{\sigma }+\beta \mu _n^{-\sigma }\| G(0)\|_{\sigma }  \\
& \leq \beta ^2\mu _{n-1}^{-\sigma }|G(0)|_{s_{\ast }}+\beta ^2
\mu _n^{-\sigma }|G(0)|_{s_{\ast }}  \\
& \leq \beta ^2\widetilde{\varepsilon }\mu _n^{-\sigma }(\mu
^{\sigma }+1)\leq 2\mu ^{s_{\ast }}\widetilde{\varepsilon }\mu
_n^{-\sigma }.
\end{align*}
We finally get
$$
\| g_n\|_0\leq (2+3a_1)\mu ^{s_{\ast }}
\widetilde{\varepsilon }\mu _n^{-\sigma }.
$$
Case $s=s_{\ast }$: \eqref{II.5},
\eqref{IV.11}, \eqref{IV.18}$_j$, $1\leq j\leq n$, and
$\sigma <s_{\ast }-2$ show that
\begin{align*}
&\| (4)+(5)\|_{s_{\ast }}\\
& \leq \| S_{n-1}R_{n-1}\|_{s_{\ast }}+\| S_nR_n\|_{s_{\ast }} \\
& \leq \beta \mu _{n-1}^2\| R_{n-1}\|_{s_{\ast }-2}+\beta \mu
_n^2\| R_n\|_{s_{\ast }-2}  \\
&\leq \beta \mu _{n-1}^2a_1\widetilde{\varepsilon }(\mu
^{s_{\ast }-2-\sigma }+\underset{j=2}{\overset{n-1}{\sum }}\mu
_{j-1}^{s_{\ast }-2-\sigma })
+\beta \mu _n^2a_1\widetilde{\varepsilon }(\mu
^{s_{\ast }-2-\sigma }+\underset{j=2}{\overset{n}{\sum }}\mu _{j-1}^{s_{\ast
}-2-\sigma }) \\
& \leq \beta a_1\widetilde{\varepsilon }(\mu _{n-1}^2\mu
_{n-1}^{s_{\ast }-2-\sigma }+\mu _n^2\mu _n^{s_{\ast }-2-\sigma}) \\
& \leq 2\beta a_1\widetilde{\varepsilon }\mu _n^{s_{\ast }-\sigma }\leq
2\mu a_1\widetilde{\varepsilon }\mu _n^{s_{\ast }-\sigma }.
\end{align*}
Next,  by \eqref{I.5}, \eqref{II.5}, \eqref{II.3}, and $\beta \leq \mu $, we have
\begin{align*}
\| (6)\|_{s_{\ast }} & \leq \| S_nG(0)\|_{s_{\ast }}+\| S_{n-1}G(0)\|
_{s_{\ast }}  \\
& \leq \beta \mu _n^{s_{\ast }-\sigma }\| G(0)\|
_{\sigma }+\beta \mu _{n-1}^{s_{\ast }-\sigma }\| G(0)\|_{\sigma }  \\
& \leq 2\beta ^2\widetilde{\varepsilon }\mu _n^{s_{\ast }-\sigma }\leq
2\mu ^2\widetilde{\varepsilon }\mu _n^{s_{\ast }-\sigma }.
\end{align*}
Therefore,
$$
\| g_n\|_{s_{\ast }}\leq 2\mu (a_1+\mu )
\widetilde{\varepsilon }\mu _n^{s_{\ast }-\sigma }.
$$
We can finally conclude using \eqref{IV.4} and $\mu \leq a_1$, that
$$
\| g_n\|_{s_{\ast }}\leq 4a_1\mu ^2\widetilde{\varepsilon
}\mu _n^{s_{\ast }-\sigma }\leq a_{2}\widetilde{\varepsilon }\mu
_n^{s_{\ast }-\sigma }.
$$
(g) \eqref{IV.20}$_n$ : By \eqref{IV.9}, we have
$$
\theta _n=|G(\widetilde{w}_n)|_0\leq |G(w_n)-G(\widetilde{w}_n)|_0
+|G(w_n)|_0
$$
Using \eqref{IV.22}:
$$
G(w_n)=(I-S_{n-1})R_{n-1}+(I-S_{n-1})G(0)+r_n.
$$
Then
$$
\theta _n\leq \underset{(7)}{\underbrace{|G(
w_n)-G(\widetilde{w}_n)|_0}}+\underset{
(8)}{\underbrace{|(I-S_{n-1})R_{n-1}|
_0}}+\underset{(9)}{\underbrace{|(
I-S_{n-1})G(0)|_0}}+\underset{(10)}
{\underbrace{|r_n|_0}}.
$$
Since we proved that $|w_n|_{2}\leq 1$ and $|\widetilde{
w}_n|_{2}\leq 1$, we can apply Proposition  \ref{VI.1} to get
$$
(7)\leq \beta K_0\| w_n-\widetilde{w}_n\|
_{2+n_{\ast }}(\| \varphi \|_{2+n_{\ast }}+\|
w_n\|_{2+n_{\ast }}+\| \widetilde{w}_n\|_{2+n_{\ast}}+1).
$$
Equations \eqref{II.4}, \eqref{IV.15}$_n$, \eqref{IV.17}$_n$, and
$3+n_{\ast }\leq 4+2n_{\ast }-\tau \leq \sigma -\tau $
imply
$$
(7)\leq 2\beta ^2K_0\sqrt{\widetilde{\varepsilon }}\mu
_n^{2+n_{\ast }-\sigma }(\mu ^2+2\sqrt{\widetilde{\varepsilon }}
+2\beta \sqrt{\widetilde{\varepsilon }}+1).
$$
Since $\widetilde{\varepsilon }\leq \frac{1}{(6\beta ^2)^2}$, 
$\beta \leq \mu $ and $4+n_{\ast }-\sigma \leq 4+2n_{\ast }-\sigma \leq 0$
then
$$
(7)\leq 4\mu ^{5}K_0\sqrt{\widetilde{\varepsilon }}\mu_n^{-2}.
$$
In the case $n=1$, $(8)=0$. For $n\geq 2$, since $\beta \leq\mu $,
 $n_{\ast }-\sigma \leq -2$ and $\mu ^{4}a_1\sqrt{\widetilde{
\varepsilon }}\leq a_{2}\sqrt{\widetilde{\varepsilon }}\leq 1$, combining
\eqref{II.1}, \eqref{II.6}, \eqref{IV.11},  and \eqref{IV.18}$_j$,
$j\leq n-1$, we obtain
\begin{align*}
(8) & \leq \beta \| (I-S_{n-1}) R_{n-1}\|_{n_{\ast }}  \\
& \leq \beta ^2\mu _{n-1}^{n_{\ast }-s_{\ast }+2}a_1\widetilde{
\varepsilon }\big(\mu ^{s_{\ast }-2-\sigma }+
\sum_{j=2}^{n-1} \mu _{j-1}^{s_{\ast }-2-\sigma }\big)  \\
& \leq \beta ^2a_1\widetilde{\varepsilon }\mu _{n-1}^{s_{\ast }-\sigma
}\leq \sqrt{\widetilde{\varepsilon }}\mu _n^{-2}.
\end{align*}
Equations \eqref{I.5}, \eqref{II.1}, \eqref{II.3}, \eqref{II.6}, \eqref{IV.5}, and
$\beta \leq \mu $ imply
\begin{align*}
(9)& \leq \beta \| (I-S_{n-1})G(0)\|_{n_{\ast }}\leq \beta ^2
\mu _{n-1}^{n_{\ast }-s_{\ast}}\| G(0)\|_{s_{\ast }} \\
& \leq \beta ^{3}\mu _{n-1}^{-2}\widetilde{\varepsilon }\leq \beta ^{3}\mu
^2\widetilde{\varepsilon }\mu _n^{-2}\leq \sqrt{\widetilde{\varepsilon }}
\mu _n^{-2}.
\end{align*}
Finally,  by \eqref{II.1} and \eqref{IV.18}$_n$,
\begin{align*}
(10)\leq \beta \| r_n\|_{n_{\ast }}
& \leq\beta a_1\widetilde{\varepsilon }[\max (\mu ,\mu _{n-1})
] ^{n_{\ast }-\sigma }  \\
& \leq \mu a_1\widetilde{\varepsilon }[\max (\mu ,\mu
_{n-1})] ^{-2}\leq \sqrt{\widetilde{\varepsilon }}\mu
_n^{-2}.
\end{align*}
Thus, we conclude that
$$
\theta _n\leq 7K_0\mu ^{5}\sqrt{\widetilde{\varepsilon }}\mu
_n^{-2}=a_{3}\sqrt{\widetilde{\varepsilon }}\mu _n^{-2}\leq 1.
$$
(h) \eqref{IV.21}: We have
$$
A_n(2)\leq \max \Big(1,|\frac{\partial g}{\partial u}
(\varphi +\widetilde{\varepsilon }\widetilde{w}_n)|
_{2},\max_{1\leq i,j\leq n} |\frac{\partial F}{\partial u_{i
\overline{j}}}(\varphi _{k\overline{l}}+\widetilde{\varepsilon }(
\widetilde{w}_n)_{k\overline{l}})|_{2}+\theta
_n\Big).
$$
Using \eqref{II.9}, \eqref{IV.1}, \eqref{IV.16}$_n$ and \eqref{IV.20}$_n$,
 we get $A_n(2)\leq M_0$.
\end{proof}


\section{Appendix 2}

In the rest of this paper, we  prove  estimates \eqref{III.13}--\eqref{III.15}
for $L_{\nu }$. We shall need the following result.

\begin{proposition} \label{propVII.1}
The operator
$$
 P=\sum_{i,j=1}^n \frac{\partial F}{
\partial u_{z_i\overline{z_j}}}(u_{z_i\overline{z_j}})
\partial _{z_i}\partial _{\overline{z_j}},
$$
 where $u\in C^{3}(\overline{\Omega })$, is formally self-adjoint.
\end{proposition}

\begin{proof}
Let  $\Sigma =\{z\in \Omega /F(u_{z_i\overline{z_j}})(z)=0\}$.
Since
$$
 P=\sum_{i=1}^n \partial _{z_i}
\Big(\overset{n}{\underset{j=1}{\sum }}\frac{\partial F}{\partial u_{z_i
\overline{z_j}}}\partial _{\overline{z_j}}\Big)
-\sum_{i,j=1}^n \partial _{z_i}\big(\frac{\partial F}{\partial u_{z_i\overline{z_j}}}
(u_{i\overline{j}})\big)\partial _{\overline{z_j}},
$$
it is sufficient to prove that for $j=1,\dots ,n$ and $z\in \overline{\Omega }$,
\begin{align*}
A_j(z) & =\sum_{i=1}^n \partial_{z_i}
\big(\frac{\partial F}{\partial u_{z_i\overline{z_j}}}(
u_{z_i\overline{z_j}})(z)\big) \\
&  =\overset{n}{\underset{i,p,q=1}{\sum }}\frac{\partial ^2F}{\partial
u_{z_i\overline{z_j}}\partial u_{z_p\overline{z_{q}}}}(u_{z_i
\overline{z_j}})u_{z_iz_p\overline{z_{q}}}(z)=0.
\end{align*}
Using the relation \eqref{II.10}, we get $A_j(z)=0$ for any
$z\notin \Sigma $. The continuity of the determinant
function allow as to have the  conclusion when $z\in \Sigma $.
\end{proof}

\subsection{Estimates in the elliptic Zone of $L$}
Let $Q=\overset{2n}{\underset{i,j=1}{\sum }}b^{ij}D_{x_i}D_{x_j}+b$ be a
degenerate elliptic operator with real coefficients $b$, $b^{ij}=b^{ji}\in
C^{s_{\ast },\tau }(\overline{\Omega })$. Assume that there is
a continuous function $\lambda (x)\geq 0$ defined in $\overline{
\Omega }$ such that
$$
\sum^{2n}_{i,j=1} b^{ij}\xi _i\xi _j\geq \lambda (x)|\xi |^2.
$$
Let $S$ be a subset of $\overline{\Omega }$ satisfying
$\{x\in \overline{\Omega }: \lambda (x)=0\} \subset S$.

\begin{lemma} \label{lmVII.2}
Assume that $Q$ is uniformly elliptic in $\overline{\Omega }$;
that is
$\lambda (x)\geq \lambda _0$, $\lambda _0$ is a positive constant
Then for any integer $1\leq s\leq s_{\ast }$ there exists a constant
$C_s'$ depending only on $s,\lambda _0$ and $A(0)$ such that for
any real function $u\in C^{s_{\ast },\tau }(\Omega )\cap H_0^{1}(\Omega )$,
\begin{gather}
\| u\|_1\leq C_1'(\| Qu\|_0+A(2)\| u\|_0),\label{VII.1} \\
\| u\|_s\leq C_s'(\| Qu\|_{s-1}+\sum_{i\leq s-2,\, i+j\leq s-1}
A(i+2)\| u\|_j),\; s\geq 2.  \label{VII.2}
\end{gather}
\end{lemma}

It is not difficult to prove \eqref{VII.1}. In fact, we need
only to apply well-known standard techniques to the linear elliptic operator
$Q$ and to calculate several constants precisely. By induction with respect
to $s$ and patient calculation, \eqref{VII.2} follows from
\eqref{VII.1}.

For $\delta >0$, we define the set $S_{\delta }$ by
$$S_{\delta }=\{x\in \overline{\Omega },d(x,S)<\delta\}.
$$

\begin{lemma} \label{lmVII.3}
Assume that $S$ is a compact $C^{\infty }$ submanifold of $\Omega $\ and $
\Omega \backslash S$ is connected. Then there exists a function $\mu \in
L^{\infty }(\Omega )$ and a constant $C>0$ such that $\mu =0$
on $S$, $m_{\delta }=\inf_{\overline{\Omega }\backslash S_{\delta }}\mu >0$
for any sufficiently small $\delta $ and
\begin{equation}
\int_{\Omega }\mu u^2dx\leq C\big\{\| Qu\|_0\|
u\|_0+\frac{1}{2}\sup [b_{ij}^{ij}-2b] \|u\|_0^2\big\},  \label{VII.3}
\end{equation}
for $u\in C^{s_{\ast },\tau }(\Omega )\cap H_0^{1}(\Omega )$.
\end{lemma}

\begin{proof}
Standard techniques of elliptic operators give
$$
\int \lambda |Du|^2dx\leq C\big\{\| Qu\|_0\| u\|_0
+\frac{1}{2}\sup [b_{ij}^{ij}-2b]\| u\|_0^2\big\}.
$$
Hence, it suffices to show that
$\int \mu u^2dx\leq \int \lambda |Du|^2dx$.
First, let us fix a point $p\in \overline{\Omega \backslash S}$ arbitrarily.

By virtue of the fundamental theorem of ordinary differential equations, we
can construct a family of curves $c(t,x)\in C^{\infty }(
[0,T_p] \times U_p)$ such that
$c(0,x)=x$, $c(t,x)\notin S$ for $0<t<T_p$ when
$x\in \overline{\Omega \backslash S}$,
$c(T_p,x)\notin \overline{\Omega }$,
$|\dot{c}(t,x)|\equiv 1$, $\sup_{x\in U_p}\tau_x<\infty $,
and $c(t,.)$ is a local $C^{\infty }$ diffeomorphism defined
in $U_p$ for any fixed $t$.

Here, $T_p$ is a positive constant, $U_p$ is a sufficiently small open
neighborhood of $p$, and, $\tau _x=\inf \{t\geq 0:c(t,x)
\notin \Omega \text{ }\}$
We define a function $\mu _p(x)$ by
$$
\mu _p(x)=\inf \{\lambda (c(t,x)):0\leq t\leq \text{ }\tau _x\}.
$$
For $u\in C^{1}(\overline{\Omega })$ satisfying $u\big|_{\partial \Omega }=0$,
since
$$
u(x)=u(c(0,x))-u(c(\tau_x,x))=-\int_0^{\tau _x}Du(c(t,x)).
\dot{c}(t,x)dt,
$$
we have
$$
|u(x)|^2\leq C\int_0^{\tau_x}|Du(c(t,x))|^2dt.
$$
Multiplying this inequality by $\mu _p$ and using its definition,
we obtain
$$
\mu _p(x)|u(x)|^2\leq C\int_0^{\tau _x}\lambda (c(t,x))|
Du(c(t,x))|^2dt,
$$
which implies
$$
\int_{U_p}\mu _p|u|^2\leq C\int_{\Omega }\lambda |Du|^2dt.
$$
Secondly, we note that the above argument ensures the existence of a finite
number of points $p_1,\dots ,p_N$ such that
$\overline{\Omega \backslash S}\subset \cup_{i=1}^{N} U_{p_i}$ and
$$
\int_{U_{p_i}}\mu _{p_i}|u|^2\leq C\int_{\Omega }\lambda
|Du|^2dt.
$$
Therefore, we have only to define $\mu $ by
$$
\mu (x)=\begin{cases}
\min \{\mu _{p_i}(x):x\in U_{p_i},1\leq i\leq n\}, &\mbox{if }
 x\in \Omega \backslash S, \\
0, &\mbox{if } x\in S.
\end{cases}
$$
\end{proof}

\begin{lemma}  \label{lmVII.4}
  For $u\in C_0^{1}(\Omega )$,
\begin{gather}
\sum_{k} \| [\partial _{k},Q] u\|_0^2 \leq
 C(A(2)\| Qu\|_1\|u\|_1+A(2)^2\| u\|_1^2),
\label{VII.4} \\
\sum_{k}\| [\partial _{k},Q] u\|_s^2
\leq C(A(2)\| Qu\|_{s+1}\|u\|_{s+1}
+\sum_{(i,j)\in \Lambda _{s+1}}A(i+2)^2\| u\|_j^2)\,\; s\geq 1.
\label{VII.5}
\end{gather}
\end{lemma}

\begin{proof} \cite[Lemma 1.7.1]{o1} shows that
$$
(b_{k}^{ij}u_{ij})^2\leq CA(2) b^{ij}u_{li}u_{lj},
$$
which implies
\begin{align*}
\sum_{k}\| [\partial _{k},Q] u\|_0^2
&\leq C\sum_{k} \int \{(b_{k}^{ij}u_{ij})^2+(b_{k}u)^2\} \\
& \leq CA(2)\sum_k \int b^{ij}u_{li}u_{lj}+CA(1)^2\| u\|_1^2.
\end{align*}
Integrating by parts
$$
\int b^{ij}u_{li}u_{lj}=-\langle (Qu)_{l},u_{l}\rangle +\langle [\partial _{l},Q]
u,u_{l}\rangle +\frac{1}{2}\langle (b_{ij}^{ij}-2b)u_{l},u_{l}\rangle ,
$$
which implies
$$
\int b^{ij}u_{li}u_{lj}\leq C\big(\| Qu\|_1\|
u\|_1+\sum_k \| [\partial _{k},Q]
u\|_0\| u\|_1+A(2)\| u\|_1^2\big).
$$
 From these inequalities, and using the inequality
$\alpha \beta \leq \varepsilon \alpha ^2+\frac{1}{\varepsilon }\beta ^2$
it follows that
$$
\sum_k \| [\partial _{k},Q] u\|_0^2\leq C(A(2)\| Qu\|_{s+1}
\|u\|_{s+1}+A(2)^2\| u\|_1^2).
$$
 For $s\geq 1$, \eqref{VI.5} is proved by recursion on $s$ using
\eqref{VI.4}.
\end{proof}

\begin{lemma} \label{lmVII.5}
Let $\chi \in C^{\infty }$ satisfy
$\mathop{\rm supp} \nabla \chi \subset \Omega $. For any integer
$0\leq s\leq s_{\ast }$, there
exists a constant $C_s>0$ such that for all $u\in C^{s_{\ast },\tau}(\Omega )$,
\begin{equation}
\| [\chi ,Q] u\|_s^2\leq C_s\big(A(
2)\| Qu\|_s\| u\|_s+\sum_{(i,j)\in \Lambda _s} A(i+2)^2\| u\|_j^2\big).  \label{VII.6}
\end{equation}
\end{lemma}

\begin{proof}
Let us consider a cut-off function
$\widetilde{\chi }\in C_0^{\infty}(\Omega )$ satisfying
$0\leq \widetilde{\chi }\leq 1$ and
$\widetilde{\chi }=1$ on $\cup_i\mathop{\rm supp}\partial _i\chi$,
and define an operator
$\widetilde{Q}=\widetilde{b}{}^{ij}D_{x_i}D_{x_j}+
\widetilde{b}$ by $\widetilde{Q}=\widetilde{\chi }Q$.
Since $[\chi ,\widetilde{Q}] u=[\chi ,Q] u$ and
$\| \widetilde{Q}u\|_s\leq C\| Qu\|_s$, it will suffice to prove
\eqref{VII.6} for $\widetilde{Q}$.

For $s=0$: The corollary to Lemma 1.7.1 in \cite{o1} shows that
$$
\big(\sum_{i,j} \widetilde{b}{}^{ij}u_j\big)^2\leq
2A(0)\widetilde{b}{}^{ij}u_iu_j.
$$
which gives
$$
\| [\chi ,\widetilde{Q}] u\|_0^2\leq CA(0)\int \widetilde{b}{}^{ij}u_iu_j
+CA(0)^2\|u\|_0^2,
$$
Integrating by parts we have
$$
\int \widetilde{b}{}^{ij}u_iu_j
=-\langle \widetilde{Q}u,u\rangle
+\frac{1}{2}\langle (\widetilde{b}_{ij}^{ij}-2
\widetilde{b})u,u\rangle
 \leq \| \widetilde{Q}u\|_0\| u\|_0+CA(2)\| u\|_0^2,
$$
which implies \eqref{VII.6}$_0$.

Note that \eqref{VII.6}$_{s\geq 1}$ follows from \eqref{VII.6}$_0$ by induction
with respect to $s$
\end{proof}

\subsection{Estimates near the degenerate points of $L$}
For $t\geq t_0\geq 1$, we define
$$
V_{t}(0)=\{x\in \Omega ,|x_n|<\frac{1}{t}
\}\cap B(0,\delta _1).
$$

\begin{proposition} \label{propVII.6}
For any integer $0\leq s\leq s_{\ast }$ and any function $u\in
C_0^{s_{\ast },\tau }(V_{t}(0))$, there exists a
constant $C_s''=C_s''(n,\Omega,\varphi ,\delta _1)>0$ such that
\begin{gather}
\| u\|_0\leq C_0^{''}t^{-1}\| L_{\nu}u\|_0,  \label{VII.7} \\
\| u\|_s\leq C_s''t^{-1}(\| L_{\nu
}u\|_s+\underset{(i,j)\in \Lambda _s}{\sum }A(
i+2)\| u\|_j),\quad s\geq 1,  \label{VII.8}
\end{gather}
where $\delta _1$ is as in Lemma \ref{lmIII.1}.
\end{proposition}

\begin{proof}
Let $v=(T-e^{tx_n})^{-1}u$, and  $T>5e$ a constant.
A direct computation gives
\begin{gather*}
Qu=(T-e^{tx_n})Qv-te^{tx_n}\{2b^{nj}v_j+tb^{nn}v\}, \\
\int (T-e^{tx_n})^{-1}Qu.v=-I+II-III-IV,
\end{gather*}
where
\begin{gather*}
I= \int b^{ij}v_iv_j, \quad
II=\frac{1}{2} \int \{b_{ij}^{ij}-2b\}v^2, \\
III=t^2 \int e^{tx_n}b^{nn}(T-e^{tx_n})^{-1}v^2, \quad
IV=2t \int e^{tx_n}(T-e^{tx_n})^{-1}vb^{nj}v_j.
\end{gather*}
Using the Cauchy-Schwartz inequality, we get
$$
|IV|\leq  \int b^{ij}v_iv_j+4t^2
\int e^{2tx_n}(T-e^{tx_n})^{-2}b^{nn}v^2.
$$
Since
$$
e^{tx_n}(T-e^{tx_n})^{-1}-4e^{2tx_n}(
T-e^{tx_n})^{-2}=e^{tx_n}(T-e^{tx_n})^{-2}(T-5e^{tx_n}),
$$
it follows that
$$
t^2 \int e^{2tx_n}(T-e^{tx_n})^{-4}(
T-5e^{tx_n})b^{nn}u^2\leq - \int (T-e^{tx_n})^{-2}Qu.u-II.
$$
Also
$$
e^{-1}\leq e^{tx_n}\leq e, \quad
(T-e^{-1})^{-1}\leq (T-e^{tx_n})^{-1}\leq (T-e)^{-1};
$$
therefore,
\begin{equation}
C_0t^2\inf_{V_{t}(0)} (b^{nn})\| u\|_0^2
\leq C\big\{\| Qu\|_0\|
u\|_0+\frac{1}{2}\underset{V_{t}(0)}{\sup }[
b_{ij}^{ij}-2b] \| u\|_0^2\big\}.  \label{VII.9}
\end{equation}
To prove \eqref{VII.7}, we apply \eqref{VII.8}. So,  for
$u\in C_0^{s_{\ast },\tau }(V_{t}(0))$, we can write
$$
t\big\{tC_0\underset{V_{t}(0)}{\inf }(b^{nn})-
\frac{C}{2}\sup_{V_{t}(0)}|b_{ij}^{ij}-2b|\big\}\| u\|_0^2\leq C\|
Qu\|_0\| u\|_0,
$$
with $Q=L_{\nu }$ and
$b^{nn}=(\Phi ^{nn}+4(\theta +\nu ))$.
If $|w|_{3,\tau }\leq 1$, $|x|\leq \delta _0$
and $\varepsilon \leq \varepsilon _1$, we have
$$
\Phi ^{nn}\geq \overset{n-1}{\underset{i=1}{\prod }}\sigma _i-M\delta
_1-M\varepsilon _1=\alpha >0.
$$
Taking
$t\geq t_0=\max (\frac{4(C+1)A(2)}{\alpha C_0},1)$,
\eqref{VII.7} is proved.
To prove \eqref{VII.8}, we use \eqref{VII.7} and recursion on $s$.
We now  estimate  $\| \chi u\|_s$.
\end{proof}

\begin{proposition}  \label{propVII.7}
For any cut-off function $\chi \in C_0^{\infty }(V_{t}(
0))$, $u\in C^{s_{\ast},\tau }(\Omega )\cap H_0^{1}(\Omega )$
and $1\leq s\leq s_{\ast }$,
\begin{equation} \label{VII.10}
\| \chi u\|_s\leq
2C_s''(\| L_{\nu}u\|_s+\| [\chi ,L_{\nu }] u\|_s
+\sum_{j<s,\, (i,j)\in \Lambda _s}
(|\varphi +\varepsilon w|_{i+4,\tau }+1)\| u\|_j).
\end{equation}
\end{proposition}

\begin{proof}
Let us consider a cut-off function $\chi \in C_0^{\infty }(V_{t}(0))$.
For $u\in C^{s_{\ast },\tau }\cap H_0^{1}(\Omega )$, since
$\mathop{\rm supp}\chi \subset V_{t}(0)$, we have by \eqref{VII.9} for any
$1\leq s\leq s_{\ast }$,
\begin{align*}
\| \chi u\|_s&\leq  C_s''t^{-1}\big(
\| \chi L_{\nu }u\|_s+\| [\chi ,L_{\nu }]
u\|_s+\sum_{j<s,\, (i,j)\in \Lambda _s} A(i+2)\| u\|_j\big) \\
&\quad +C_s''t^{-1}A(2)\| \chi u\|_s.
\end{align*}
We have $A(2)\leq M_0$. We fix $t\geq t_0$ such that for
 $1\leq s\leq s_{\ast }$,
$C_s''t^{-1}A(2)\leq \frac{1}{2}$.
On the other hand,
$$
A(i+2)=\max \big(1,|\frac{\partial g}{\partial u}
(\varphi +\varepsilon w)|_{i+2},\underset{1\leq p,q\leq n
}{\max }|\frac{\partial F}{\partial u_{p\overline{q}}}(\varphi
_{k\overline{l}}+\varepsilon w_{k\overline{l}})|_{i+2}+\theta\big).
$$
But, for $k\in \{0,1,2\}$,
$|\partial ^{k}\varphi+\varepsilon \partial ^{k}w|_0\leq |\varphi |_{2}+1$,
then by \eqref{II.9}, since $\theta \leq 1$, we get, for
$0\leq i\leq s_{\ast }-2$,
\begin{equation}
A(i+2)\leq C(\varphi )\big(|\varphi
+\varepsilon w|_{i+4,\tau }+1\big).  \label{VII.11}
\end{equation}
and we deduce \eqref{VII.10}.
\end{proof}

\subsection{Proof of the estimates \eqref{III.13}--\eqref{III.15}
for $L_{\nu }$}.
Since $\| u\|_s\leq \| (1-\chi )u\|
_s+\| \chi u\|_s$, it will suffice to estimate $\|
(1-\chi )u\|_s\ $and $\| \chi u\|_s$.

\begin{proof}[Proof of \eqref{III.13}]
Since $\chi =1$ in a neighborhood of zero in $V$, then, there exists
$\delta >0$ such that $\mathop{\rm Supp}(1-\chi )\subset \overline{\Omega }
\backslash B(0,\delta )$.

Let us consider the cut-off functions: $\widetilde{\chi }$,
$\widetilde{\widetilde{\chi }}\in C_0^{\infty }(\overline{\Omega }
\backslash S)$,
$0\leq \widetilde{\chi },\widetilde{\widetilde{\chi}}\leq 1$
and such that $\widetilde{\chi }=1$ on
$\mathop{\rm supp}\partial_i\chi $ and
$\widetilde{\widetilde{\chi }}=1$ on $\mathop{\rm supp}\widetilde{\chi }$.
Let $\mu $ be the function given by Lemma \ref{lmVII.3} ($m_{\delta }$
depends only on $\varphi ,\Omega ,n$).

By \eqref{VII.3}, there exists $C_0=C_0(\varphi,\Omega ,n)>0$ such that
$$
\| (1-\chi )u\|_0^2= \int_{\overline{\Omega }
\backslash B(0,\delta )}u^2dx
\leq \frac{1}{m_{\delta }} \int \mu u^2dx
\leq C_0(\| u\|_0\| L_{\nu }u\|_0+B\| u\|_0^2),
$$
where $B=\frac{1}{2}\sup [b_{ij}^{ij}-2b] $.
By proposition  \ref{VII.1}, $\sum_{ij}b_{ij}^{ij}=0$, and the
hypothesis (A2) imply that $-2b\leq \varrho $. So,
$B\leq \varrho $ and we have
$$
\| (1-\chi )u\|_0^2\leq C_1(\|u\|_0\| L_{\nu }u\|_0
+\varrho \| u\|_0^2).
$$
Since $\mathop{\rm Supp}\widetilde{\widetilde{\chi }}\subset
 \overline{\Omega }\backslash\{0\}$, we also have by the same way,
$$
\| \widetilde{\widetilde{\chi }}u\|_0^2\leq C_1(
\| u\|_0\| L_{\nu }u\|_0+\varrho \|u\|_0^2).
$$
On the other hand, by \eqref{VII.8},
$$
\| \chi u\|_0^2\leq C_{2}\| L_{\nu }\chi u\|_0^2
\leq C_{2}(\| L_{\nu }u\|_0^2+\| [\chi ,L_{\nu }] u\|_0^2),
$$
but $\widetilde{\chi }L_{\nu }\widetilde{\widetilde{\chi }}u
=\widetilde{\chi}L_{\nu }u$ and
$[\chi ,L_{\nu }] u=[\chi ,\widetilde{\chi }L_{\nu }]
\widetilde{\widetilde{\chi }}u$. Since
$A(2)\leq M_0$ and $\nu \leq 1$, using Lemma \ref{lmVII.5}, we get
\begin{align*}
\| [\chi ,L_{\nu }] u\|_0^2=\| [\chi
,\widetilde{\chi }L_{\nu }] \widetilde{\widetilde{\chi }}u\|_0^2
&\leq C\big[\| \widetilde{\chi }L_{\nu }\widetilde{
\widetilde{\chi }}u\|_0\| \widetilde{\widetilde{\chi }}
u\|_0+(M_0+1)^2\| \widetilde{\widetilde{\chi }}u\|_0^2\big]  \\
& \leq C'\big(\| L_{\nu }u\|_0\| \widetilde{\widetilde{\chi }}u\|_0
+\| \widetilde{\widetilde{\chi }}u\|_0^2\big).
\end{align*}
Combining these inequalities with the fact that $\varrho <<1$, and using the
inequality
$\alpha \beta \leq \varepsilon \alpha ^2+\frac{1}{\varepsilon }\beta ^2$,
we get \eqref{III.13}
\end{proof}

\begin{proof}[Proof of \eqref{III.14}]
We have $\mathop{\rm supp}(1-\chi)\subset \overline{\Omega }\backslash B(0,\delta )$.
Or $\varphi $ is strictly plurisubharmonic on $E=\mathop{\rm supp}(1-\chi )$,
then for $\varepsilon \leq \varepsilon _{4}$ small enough, $L$ is uniformly
elliptic on $E$. Using \eqref{VII.1} and the estimation $A(2)\leq M_0$,
we have
$$
\| (1-\chi )u\|_1\leq C_1'(\| L_{\nu }u\|_0+(M_0+1)\| u\|_0
+\| [\chi ,L_{\nu }] u\|_0).
$$
Applying Lemma \ref{lmVII.5}, we get
$$
\| [\chi ,L_{\nu }] u\|_0\leq C_0(\|L_{\nu }u\|_0+(M_0+1)\| u\|_0),
$$
therefore,
$$
\| (1-\chi )u\|_1\leq C_1(M_0)(\| L_{\nu }u\|_0+\| u\|_0).
$$
On the other hand, since $A(2)\leq M_0$, we get using \eqref{VII.10},
$$
\| \chi u\|_1\leq C_1(M_0)(\|
L_{\nu }u\|_1+\| [\chi ,L_{\nu }] u\|_1+\| u\|_0)\,.
$$
But $\widetilde{\chi }L_{\nu }\widetilde{\widetilde{\chi }}u=\widetilde{\chi
}L_{\nu }u$
and $[\chi ,L_{\nu }] u=[\chi ,\widetilde{\chi }L_{\nu }]
\widetilde{\widetilde{\chi }}u$, so, since $A(2)\leq M_0$, Lemma \ref{lmVII.5}
gives
\begin{align*}
\| [\chi ,L_{\nu }] u\|_1& \leq C_1(
\| \widetilde{\chi }L_{\nu }\widetilde{\widetilde{\chi }}u\|_1
+(1+M_0)\| \widetilde{\widetilde{\chi }}u\|_1) \\
& \leq C_1(\| L_{\nu }u\|_1+(1+M_0)
\| \widetilde{\widetilde{\chi }}u\|_1).
\end{align*}
Since $L_{\nu }$ is uniformly elliptic on
$\mathop{\rm supp}\widetilde{\widetilde{\chi }}$
and $A(2)\leq M_0$, then we have by \eqref{VII.1},
$$
\| \widetilde{\widetilde{\chi }}u\|_1
\leq C_1'(\| L_{\nu }u\|_0+(M_0+1)\|u\|_0
+\| [\widetilde{\widetilde{\chi }},L_{\nu }]u\|_0),
$$
which using \eqref{VII.6} gives
$$
\| \widetilde{\widetilde{\chi }}u\|_1\leq C_1(
M_0)(\| L_{\nu }u\|_1+\| u\|_0).
$$
Combining these inequalities, we get \eqref{III.14}.
\end{proof}
The proof of \eqref{III.15} is identical to that of
\eqref{III.14} using the inequalities \eqref{VII.1},
\eqref{VII.2}, \eqref{VII.6}, and \eqref{VII.10}.

\subsection*{Acknowledgments}
The author wishes to thank  the anonymous referee for his or her helpful
comments.


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\end{document}