[proofplan]
The reduction has two parts. First, the local [Lewy extension principle](/theorems/9190) gives a holomorphic collar extension of the CR boundary function on the strictly pseudoconvex side, and a cutoff converts this collar-[holomorphic function](/page/Holomorphic%20Function) into a global smooth extension whose $\bar\partial$-error is supported away from the boundary. Second, the assumed compactly supported $\bar\partial$-solvability removes that error without changing the boundary trace, because the correction term is compactly supported in the interior.
[/proofplan]
[step:Extend the CR boundary value holomorphically into an interior collar]
Since $M=\partial\Omega$ is compact and $C^\infty$ strictly pseudoconvex, the local side of $M$ pointing into $\Omega$ is strictly pseudoconvex at every point of $M$. Applying the Lewy extension principle [citetheorem:9190] pointwise and using compactness of $M$ to pass to a finite collar, there exist an open neighbourhood $W\subset\mathbb C^n$ of $M$ and a function
\begin{align*}
H:W\cap\overline\Omega\to\mathbb C
\end{align*}
such that $H\in C^\infty(W\cap\overline\Omega;\mathbb C)$, $H$ is holomorphic on $W\cap\Omega$, and
\begin{align*}
H|_M=u.
\end{align*}
Here the CR hypothesis on $u$ is exactly the boundary compatibility condition required by the Lewy extension principle.
[/step]
[step:Cut off the collar extension without creating boundary error]
Choose open neighbourhoods $W_0,W_1\subset\mathbb C^n$ of $M$ such that
\begin{align*}
M\subset W_0\subset \overline{W_0}\subset W_1\subset \overline{W_1}\subset W.
\end{align*}
Choose a cutoff function
\begin{align*}
\eta:\overline\Omega\to[0,1]
\end{align*}
with $\eta\in C^\infty(\overline\Omega;\mathbb R)$, $\eta=1$ on $W_0\cap\overline\Omega$, and $\eta=0$ on $\overline\Omega\setminus W_1$. Define
\begin{align*}
\widetilde u:\overline\Omega\to\mathbb C
\end{align*}
by setting $\widetilde u=\eta H$ on $W\cap\overline\Omega$ and $\widetilde u=0$ on $\overline\Omega\setminus W_1$. This is well-defined and smooth because $\eta$ vanishes on a neighbourhood of $\overline\Omega\setminus W_1$ inside $W\cap\overline\Omega$. Since $\eta=1$ on $M$, we have
\begin{align*}
\widetilde u|_M=H|_M=u.
\end{align*}
Moreover, on $W_0\cap\Omega$ one has $\widetilde u=H$, so $\bar\partial\widetilde u=0$ there because $H$ is holomorphic.
[guided]
The point of the cutoff is to keep the boundary values and the holomorphicity near the boundary, while making the later $\bar\partial$-error live strictly inside $\Omega$. We first choose collar neighbourhoods $W_0$ and $W_1$ satisfying
\begin{align*}
M\subset W_0\subset \overline{W_0}\subset W_1\subset \overline{W_1}\subset W.
\end{align*}
The separation between $\overline{W_0}$ and $\mathbb C^n\setminus W_1$ lets us choose a smooth function
\begin{align*}
\eta:\overline\Omega\to[0,1]
\end{align*}
with $\eta=1$ on $W_0\cap\overline\Omega$ and $\eta=0$ on $\overline\Omega\setminus W_1$.
Now define
\begin{align*}
\widetilde u:\overline\Omega\to\mathbb C
\end{align*}
by $\widetilde u=\eta H$ where $H$ is defined, namely on $W\cap\overline\Omega$, and by $\widetilde u=0$ on $\overline\Omega\setminus W_1$. This patching is smooth because $\eta$ is already zero near the region where $H$ is no longer being used. Thus no compatibility condition is needed along the artificial cutoff interface.
On the boundary $M$, the cutoff equals $1$, so
\begin{align*}
\widetilde u|_M=(\eta H)|_M=H|_M=u.
\end{align*}
On the smaller collar $W_0\cap\Omega$, the same equality $\eta=1$ gives $\widetilde u=H$. Since $H$ is holomorphic on $W\cap\Omega$, it follows that
\begin{align*}
\bar\partial\widetilde u=0
\end{align*}
on $W_0\cap\Omega$. This actual vanishing on a collar, not merely infinite-order vanishing at $M$, is what will make the error form compactly supported in $\Omega$.
[/guided]
[/step]
[step:Show that the resulting $\bar\partial$-error is compactly supported and closed]
Define the $(0,1)$-form
\begin{align*}
f:=\bar\partial\widetilde u\in C^\infty(\Omega;\Lambda^{0,1}T^*\Omega).
\end{align*}
Since $f=0$ on $W_0\cap\Omega$, its support is contained in
\begin{align*}
\overline\Omega\setminus W_0.
\end{align*}
Since $\widetilde u=0$ on $\Omega\setminus W_1$, the support of $f$ is also contained in $\overline{W_1}\cap\overline\Omega$. Hence
\begin{align*}
\operatorname{supp} f\subset \overline{W_1}\cap(\overline\Omega\setminus W_0).
\end{align*}
The set on the right is compact and is contained in $\Omega$, because $M\subset W_0$ and $\overline\Omega$ is compact. Therefore
\begin{align*}
f\in C_c^\infty(\Omega;\Lambda^{0,1}T^*\Omega).
\end{align*}
Finally, the nilpotence identity for the Dolbeault operator gives
\begin{align*}
\bar\partial f=\bar\partial(\bar\partial\widetilde u)=0.
\end{align*}
[/step]
[step:Use compactly supported solvability to remove the $\bar\partial$-error]
By the assumed compactly supported solvability statement applied to the closed form $f$, there exists
\begin{align*}
v:\Omega\to\mathbb C
\end{align*}
with $v\in C_c^\infty(\Omega;\mathbb C)$ and
\begin{align*}
\bar\partial v=f.
\end{align*}
Define
\begin{align*}
F:\Omega\to\mathbb C
\end{align*}
by
\begin{align*}
F=\widetilde u-v.
\end{align*}
Since $\widetilde u$ is smooth on $\overline\Omega$ and $v$ is smooth on $\Omega$, the function $F$ is smooth on $\Omega$. Also,
\begin{align*}
\bar\partial F=\bar\partial\widetilde u-\bar\partial v=f-f=0.
\end{align*}
Thus $F$ is holomorphic on $\Omega$.
[/step]
[step:Check that the correction does not change the boundary trace]
Because $v\in C_c^\infty(\Omega;\mathbb C)$, the support $\operatorname{supp}v$ is a compact subset of $\Omega$. Hence there exists an open neighbourhood $W_2\subset\mathbb C^n$ of $M$ such that
\begin{align*}
W_2\cap\operatorname{supp}v=\varnothing.
\end{align*}
Therefore $v=0$ on $W_2\cap\Omega$. On this boundary collar,
\begin{align*}
F=\widetilde u.
\end{align*}
Since $\widetilde u\in C^\infty(\overline\Omega;\mathbb C)$ and $\widetilde u|_M=u$, the function $F$ extends smoothly to $\overline\Omega$ by the same boundary values and satisfies
\begin{align*}
F|_M=u.
\end{align*}
Thus $F\in\mathcal O(\Omega)\cap C^\infty(\overline\Omega;\mathbb C)$ is the required holomorphic extension of $u$.
[/step]
