This is closely related to the question here. The setup is that $U\subset\mathbb{C}$ is an open bounded simply connected domain with $C^\infty$ boundary. If $\phi:U\rightarrow\mathbb{D}$ is a biholomorphic mapping from $U$ to the unit disk, it is known that $\phi$ extends to a smooth map $\overline{U}\rightarrow\overline{\mathbb{D}}$. However, the question that is not addressed is whether this extension is actually a diffeomorphism. Now a comment by the OP on the linked question states that Graeme Segal showed something like this in some unpublished notes, which I can't find anywhere. So let me now ask this precisely:

Does $\phi$ as above extend to a diffeomorphism? The answer seems to be yes, but I would appreciate any reference that confirms this.

Incidentally, I'm also interested in analogous $C^k$-boundary and boundary-with-corners statements, but those seem even more exotic.


1 Answer 1


The main reference on this topic is the book "Boundary behavior of conformal maps" by Pommerenke.

If the curve is $C^\infty$, then the biholomorphic mapping extends to a smooth map on the closure of the unit disk, and the derivative is non-vanishing. This is implicit in Theorem 3.2 of Pommerenke's book, see also Theorem 3.5.

For boundaries with corners, you might be interested in Section 3.4.

  • $\begingroup$ Very interesting, the book explains that a mapping $\psi:\mathbb{D}\rightarrow U$ does not necessarily extend to a diffeomorphism since the derivative may not have a continuous extension up to the boundary! However, if the boundary is assumed to be "Dini-smooth", then $\psi'$ does extend continuously to the boundary. Incidentally, let me mention that the Internet Archive has a copy of the book available to borrow. $\endgroup$
    – J_P
    Jun 8 at 19:34
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    $\begingroup$ Whoops, I made a mistake... The book uses "smooth curve" to mean $C^1$ curve. For $C^\infty$ boundaries the result does seem to hold as I thought. This is apparently called the Kellogg-Warschawski theorem. $\endgroup$
    – J_P
    Jun 8 at 19:43
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    $\begingroup$ @J_P Yes, indeed, the term "smooth curve" is a bit misleading here. Glad this helped. $\endgroup$ Jun 9 at 18:32

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