Suppose $\Omega$ is a simply connected domain surrounded by a piecewise smooth simple closed curve, i.e. $\partial\Omega$ is a piecewise smooth simple closed curve. $f\in H(\Omega)$ is a bounded holomorphic function on $\Omega$. Can we always extend $f$ analytically past $\partial\Omega$? If it's not right, what about assuming that $f$ is injective?
In Ahlfors' Complex Analysis, while proving Schwarz-Christoffel formula, he proves many propositions on the boundary behavior. Approximately, they follow the paradigm:
If $\Omega$ is a simply connected domain and $\partial\Omega$ is well-behaved enough, $f\colon\Omega\to\mathbb D$ is a biholomorphism onto the unit disc, then $f$ could be extended analytically past $\partial\Omega$.
For example, if $\partial\Omega$ is
a polygon or(EDIT: I've reread that paragraph and found that he didn't claim such a proposition for the polygon, though in the proof he did some analytic extension, which couldn't be pullbacked in a transformation) a real analytic simple closed curve, then the extension is possible. I don't want to repeat the argument, which is quite lengthy, but only point out the key idea. We use a good function (at least, locally biholomorphic) $\varphi$ to parametrize a neighborhood of $z_0\in\partial\Omega$ locally, then consider $F(\zeta)=\log f(\phi(\zeta))$, where $\log$ is an appropriate branch of the logarithm function. When $\phi(\zeta)$ tends to the boundary, $f(\phi(\zeta))$ tends to the unit circle and the imaginary part of $F(\zeta)$ tends to $0$, then we can apply Schwarz reflection principle to extend $F$, therefore $f$ to some degree.
Apparently, the preceding argument relies on the analyticity of $\partial\Omega$, and the bijectivity of $f$. It seems to me that non-extendability happens at unbounded points, just like Riemann's theorem on removable singularities, or Casorati-Weierstrass theorem for essential singularities. I need to see whether it's generally true for good boundaries, so I pay attention on smooth or piecewise smooth boundaries, and no matter whether you define smooth as $\mathcal C^1$ or $\mathcal C^\infty$.
Any idea? Thanks.