[proofplan]
The boundary $M$ is a smooth real hypersurface in the complex manifold $X$, so its induced CR bundle consists exactly of the ambient antiholomorphic tangent directions that are tangent to $M$. Since $F$ is holomorphic on a neighbourhood of $M$, every ambient vector of type $(0,1)$ annihilates $F$. Applying this to each local section of $T^{0,1}M$ and then restricting to $M$ shows that every CR antiholomorphic vector field annihilates $u=F|_M$, which is precisely the CR equation for functions.
[/proofplan]
[step:Record the induced CR structure and the regularity of the boundary value]
Because $M=\partial\Omega$ is smooth, $M$ is a smooth real hypersurface of $X$. The induced antiholomorphic CR bundle on $M$ is the complex vector subbundle
\begin{align*}
T^{0,1}M = T^{0,1}X|_M \cap (TM\otimes_{\mathbb R}\mathbb C).
\end{align*}
Here $T^{0,1}X|_M$ denotes the restriction of the ambient antiholomorphic tangent bundle to $M$.
Since $F:U\to\mathbb C$ is holomorphic, it is smooth as a complex-valued function on the [open set](/page/Open%20Set) $U$. Therefore its restriction
$u:M\to\mathbb C$, defined by $u(x)=F(x)$ for $x\in M$, is smooth on $M$.
[/step]
[step:Test the boundary value against an arbitrary local CR antiholomorphic vector field]
Let $V\subset M$ be an open subset, and let
$\bar L:V\to T^{0,1}M$
be a smooth local section of the antiholomorphic CR bundle. For each $x\in V$, the vector $\bar L_x$ belongs to $T^{0,1}_xM$, hence by the definition of the induced CR structure it also belongs to the ambient space $T^{0,1}_xX$.
Holomorphicity of $F$ means that its ambient antiholomorphic differential vanishes:
\begin{align*}
\bar\partial F=0
\end{align*}
on $U$. Equivalently, every ambient tangent vector of type $(0,1)$ annihilates $F$. Applying this at the point $x\in V$ to the ambient vector $\bar L_x\in T^{0,1}_xX$ gives
\begin{align*}
\bar L_x(F)=0.
\end{align*}
Because $\bar L_x$ is tangent to $M$ and $u=F|_M$, the derivative of the restricted function $u$ along $\bar L_x$ is the restriction of the ambient derivative of $F$ along the same tangent vector. Hence
\begin{align*}
\bar L_x(u)=\bar L_x(F)=0.
\end{align*}
Since $x\in V$ was arbitrary, $\bar L u=0$ on $V$.
[guided]
We must prove the CR equation locally, so we test against an arbitrary local antiholomorphic CR vector field. Let $V\subset M$ be open, and let
$\bar L:V\to T^{0,1}M$
be a smooth local section. The point of using the induced CR structure is that it gives a direct inclusion into the ambient antiholomorphic directions: for every $x\in V$,
\begin{align*}
\bar L_x\in T^{0,1}_xM \subset T^{0,1}_xX.
\end{align*}
Now use the holomorphicity of $F:U\to\mathbb C$. For a [holomorphic function](/page/Holomorphic%20Function) on a complex manifold, the antiholomorphic differential vanishes:
\begin{align*}
\bar\partial F=0.
\end{align*}
This means that if $\xi\in T^{0,1}_xX$ is any ambient antiholomorphic tangent vector at a point $x\in U$, then $\xi(F)=0$. Since $x\in V\subset M\subset U$ and $\bar L_x\in T^{0,1}_xX$, we may substitute $\xi=\bar L_x$ and obtain
\begin{align*}
\bar L_x(F)=0.
\end{align*}
The final point is to pass from $F$ to its restriction $u=F|_M$. Because $\bar L_x$ lies in $T_xM\otimes_{\mathbb R}\mathbb C$, it differentiates functions on $M$ at $x$. The derivative of a restricted function along a tangent vector is computed by differentiating any smooth ambient extension along that same tangent vector. Here the ambient extension is precisely $F$, so
\begin{align*}
\bar L_x(u)=\bar L_x(F).
\end{align*}
Combining this identity with $\bar L_x(F)=0$ gives
\begin{align*}
\bar L_x(u)=0.
\end{align*}
Since $x\in V$ was arbitrary, $\bar L u=0$ on $V$.
[/guided]
[/step]
[step:Conclude that the restriction satisfies the tangential Cauchy Riemann equation]
The preceding step shows that every smooth local section $\bar L$ of $T^{0,1}M$ annihilates $u$. By the definition of the tangential Cauchy-Riemann operator on functions, this is exactly
\begin{align*}
\bar\partial_b u=0.
\end{align*}
Thus $u$ is a CR function on $M$.
[/step]
