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I am interested in the following question. Let $\Omega \subset \mathbb C$ be a bounded simply-connected domain and let $f: \Omega \rightarrow \Delta$ be a biholomorphism onto the unit disk. Does $f$ extend continuously to the boundary?

If this is difficult, consider the following example. Let $\Omega = \Delta \setminus [0, 1)$, and let $f: \Omega \rightarrow \Delta$ be a biholomorphism. Is it true that for $x \in [0, 1)$ we have that $$\lim_{\varepsilon \rightarrow 0^+}f(x + i \varepsilon) = \lim_{\varepsilon \rightarrow 0^+}f(x - i \varepsilon)? $$

Levi
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  • First of all: Convince yourself that the open mapping theorem forces any continuous extension to $\overline{\Omega}$ to be injective. Hence, if the inverse map $\Delta\to \Omega$ extends continuously, it is automatically a homeomorphism (being a continuous bijection from a compact space to a Hausdorff space).

    There is a very deep theory on this subject, which contains the general result that, in a suitable sense, the maps always extend to the boundary. In the usual sense, though, there is a continuous extension to the boundary of $\Delta$ whenever the boundary of $\mathbb{C}\setminus\Omega$

     – WoolierThanThou Oct 31 '19 at 09:33
  • is locally connected. – WoolierThanThou Oct 31 '19 at 09:34
  • I realise this is in the opposite order of what you're writing, but the theory usually deals with the maps from the unit disk as opposed to into the unit disk. – WoolierThanThou Oct 31 '19 at 09:34
  • @WoolierThanThou What do you mean by "the theory"? Do you have any reference? Also, I am not a big fan of considering the other direction since the Riemann mapping theorem works conceptually much nicer in the direction my question goes. Also the example $\Omega = \Delta \setminus [0, 1)$ can not be studied using your framework because the map $\Omega \rightarrow \Delta$ extends, but $\Delta \rightarrow \Omega$ does not. – Levi Oct 31 '19 at 09:39
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    Rao's "Complex Analysis: An Invitation" contains an introduction to prime ends. You should probably also be able to find it in in Ahlfors. – WoolierThanThou Oct 31 '19 at 09:40
  • And it's not true that the Riemann map $f:\Omega\to \Delta$ extends. Note that $\lim_{\varepsilon\to 0^+} f(t+i\varepsilon)\neq \lim_{\varepsilon\to 0^-} f(t-i\varepsilon)$ for any $t\in (0,1)$. – WoolierThanThou Oct 31 '19 at 09:41
  • What do you mean with $\lim_{\varepsilon \rightarrow 0^+} f(t + \varepsilon)$? – Levi Oct 31 '19 at 09:43
  • Do you mean $f(t + i \varepsilon)$? – Levi Oct 31 '19 at 09:44
  • Yeah, sorry. My bad. – WoolierThanThou Oct 31 '19 at 09:44
  • I don't see why they are not equal. Mind writing an answer? :) – Levi Oct 31 '19 at 09:46

2 Answers2

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The answer for your question is very simple:

$f: \Omega \to \Delta$ bi-holomorphism extends continuously to the boundary iff $\Omega$ is a Jordan domain

(here we assume $\Omega$ bounded as in the question - otherwise the result holds only with obvious modifications to allow for the point at infinity to "complete" the boundary to a simple curve in $\hat C$ as the upper plane and the usual Mobius transform to the unit disc shows)

Note that the converse is false and as noted in the comments, for bi-holomorphisms $g:\Delta \to \Omega$ we have continuous extension iff $\partial \Omega$ is locally connected and continuous injective extension (hence topological homeomorphism from the closed unit disc) extension iff $\Omega$ is a Jordan domain (and then lots of results about the curve bounding $\Omega$ if we impose stricter conditions on the boundary map like real differentiability or even Lipschitz, complex differentiability etc

The proof can be found for example in Conway, Functions of Complex Variables II, page 53-54 and it follows from three facts: extension of $f: \Omega \to \Delta$ bi-holomorphism to one boundary point implies the boundary point is simple and the extension is injective at simple boundary points; in our case the extension is then injective everywhere; the image of $\partial \Omega$ is a compact subset of the unit circle and as it follows that all points with radial limits for $f^{-1}$ on the unit circle are in the image, the image must be also surjective as those have full lebesgue measure. Hence the extension is a homeomorphism of the boundaries.

Conrad
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    the only non-trivial part (up to classical stuff about radial limits etc which are part of general theory, eg Hardy spaces) is the injectivity of the extension at simple boundary points - Conway proves it directly but it follows from a general result which states that curves in $\Omega$ that end on boundary points are sent by $f$ (no extension condition needed here, just conformal isomorphism of the open domains) to curves that end on boundary points of $\Delta$ and different end points on $\Omega$ go to different endpoints on $\Delta$; this is obviously what breaks down for the converse – Conrad Oct 31 '19 at 19:41
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These properties are equivalent (if $\Omega$ is bounded and simply connected):

  1. A Riemann map $\tau\colon\Delta\to\Omega$ extends homeomorphically to a map $\bar\Delta\to\bar\Omega$
  2. $\partial\Omega\cong S^1$ (that is, a Jordan curve)
  3. Every point of $\partial\Omega$ is simple, which means that for every $\alpha_n\to\beta\in\partial\Omega$ there exist $\gamma\colon [0,1)\to\Omega$ and a sequence $t_n\in [0,1)$ such that $\gamma(t_n)=\alpha_n$.

A (partial) proof can be found in Rudin's Real and Complex Analysis. In addition, a theorem of Painlevé says that this extension is $C^\infty$ if the boundary of $\Omega$ has the same regularity.

Lorenzo Cecchi
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