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.


If $f\colon\Omega\to\mathbb{D}$ is biholomorphic and has an analytic continuation across $\partial\Omega$, it must map $\partial\Omega$ to $\partial\mathbb{D}$. Since $f'$ can have only isolated zeros on $\partial\Omega$, $f$ is locally biholomorphic at almost all points of $\partial\Omega$, and thus $\partial\Omega\cap V = f^{-1}(\partial\mathbb{D}\cap U)$ must be an analytic arc. So that $\partial\Omega$ consists of analytic arcs is essential for the existence of a continuation across the boundary at least if $f$ is injective and the image of $f$ has a boundary consisting of analytic arcs. Sometimes it is of course possible to continue $f$ across the boundary if $\partial\Omega$ is less regular, but a general theorem asserting the existence of a continuation across the boundary needs strong premises.

It seems to me that non-extendability happens at unbounded points

Not only there. If $(\alpha_n)$ is a sequence in $\mathbb{D}\setminus\{0\}$ with

$$\sum_{n=1}^\infty 1 - \lvert\alpha_n\rvert < \infty,$$

the Blaschke product

$$B_{k,\alpha}(z) = z^k\cdot\prod_{n=1}^\infty \frac{\lvert \alpha_n\rvert}{\alpha_n}\frac{z-\alpha_n}{1-\overline{\alpha_n}z}$$

is, for every $k\in\mathbb{N}$ a bounded holomorphic function on $\mathbb{D}$, and it cannot be analytically continued across every part of the boundary containing a limit point of $(\alpha_n)$. Choosing $(\alpha_n)$ so that all points on $\partial\mathbb{D}$ are limit points, you get a bounded holomorphic function whose natural boundary is $\partial\mathbb{D}$.

  • $\begingroup$ Since I'm a newbie, I need time to check the details. Thanks. $\endgroup$ – Yai0Phah Jan 30 '14 at 19:18
  • $\begingroup$ Blaschke product seems a well-known result. If boundaries of both domain and the image are simple, closed and analytic, then the biholomorphism $f$ could be extended analytically across the boundary, right? For Carathéodory's theorem, it could be extended homeomorphically to the boundary, then for each boundary point $z_0$ and $w_0=f(z_0)$, we could analytically rectify the boundaries of the domain and the image to a straight line locally, then apply Schwarz reflection principle and pullback to the original map? $\endgroup$ – Yai0Phah Jan 31 '14 at 8:37
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    $\begingroup$ Yes, you can do that except where the boundary (of either domain) has a corner. In general, you can only continue across the arcs of the boundary, not where two arcs meet. $\endgroup$ – Daniel Fischer Jan 31 '14 at 11:25

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