Let $\Omega$ be a domain (i.e. open and connected subset) in the complex plane $\mathbb C$. Let $b\Omega$ be its boundary and $\overline{\Omega}=\Omega\cup b\Omega$. Let $C^\infty(b\Omega)$, $C^\infty(\Omega)$ and $C^\infty(\overline{\Omega})$ be the set of smooth (i.e.continuous partial derivatives of all orders) functions on $b\Omega$, $\Omega$ and $\overline{\Omega}$ respectively. Finally, let $H(\Omega)$ be the set of holomorphic functions on $\Omega$.
I read the following theorem and it has raised me some questions about boundary values of holomorphic functions and extensions of such functions to the boundary. Let me state the theorem and I' ll make the questions after.
Suppose that h is a holomorphic function $\Omega$ that extends to be a continuous function on $\overline{\Omega}$. If the boundary values of h are in $C^\infty(b\Omega)$, then $h\in C^\infty(\overline\Omega)$.
In other words, there exists a function $\tilde{h}$ such that $\tilde{h}\in C(\overline{\Omega})$ and such that $\tilde{h}=h$ on $\Omega$. The conclusion of the theorem can be stated like this: "then $\tilde{h}\in C^\infty(\overline\Omega)$". Since $\tilde{h}=h$ on $\Omega$ (from the hypothesis), it is obvious that $h\in C^\infty(\Omega)$, since then $\tilde{h}\in H(\Omega)\subset C^\infty(\Omega)$. Also, from the hypothesis, we have that the boundary values of h, from now noted $b.v.h$, are in $C^\infty(b\Omega)$.
Now the question(s):
- If we haven't prove the theorem. Can we say that $\tilde{h}=b.v.h$ on $b\Omega$?
- Does the conclusion of the theorem imply that $\tilde{h}=b.v.h$ on $b\Omega$?
- Can the function $b.v.h$ be different from the continuous extension of h on $b\Omega$?
If I understand this correctly, the $b.v.h$, even if it is smooth, can be different from the extension of the $h\in H(\Omega)$ to the boundary of $\Omega$, even if the extension is continuous?