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Let $U\subset \mathbb C$ be open, bounded, simply connected, with $C^\infty$ boundary. Apply the Riemann mapping theorem to get a bilolomorphic isomorphism $$ f:U\to \mathbb D $$ between $U$ and the unit disc $\mathbb D:=\{z\in \mathbb C:|z|<1\}$.

How can I see that $f$ extends to a $C^\infty$ map from the closure of $U$ to the closure of $\mathbb D$?

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  • $\begingroup$ You might already know this: there is a theorem of Caratheodory which implies that $f$ extends to a continuous map. Perhaps examining the proof will show that the extension is $C^{\infty}$ if the boundary of $U$ is so. $\endgroup$
    – naf
    Commented Dec 4, 2011 at 12:00
  • $\begingroup$ Can you give a reference? Thanks！ $\endgroup$
    – MZWang
    Commented Dec 5, 2011 at 4:55

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Another answer, since it is different from the previous one:

The result you want was proved by Painleve in 1887, long BEFORE Cartheodory's theorem. The proof is given in the very nice survey article:

http://www.ams.org/journals/bull/1990-22-02/S0273-0979-1990-15879-3/S0273-0979-1990-15879-3.pdf (page 238). The paper is:

S. Bell MAPPING PROBLEMS IN COMPLEX ANALYSIS AND THE D-BAR problem (bull AMS, 1990). (it still uses elliptic regularity, but he refers to other proofs which are different).

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    $\begingroup$ I also found another proof of that same result in an unpublished note by Graeme Segal called "Sewing Riemann surfaces together". G. Segal also addresses a subtle point that is somehow omitted in Bell's paper: the non-vanishing of the derivative at the boundary. Quoting G. Segal: "it is easy to show that [the derivative] cannot vanish to finite order, but to show that it cannot vanish to infinite order at a boundary point"... and then there is a little lemma that's not too difficult, but that I won't reproduce here. $\endgroup$
    – André Henriques
    Commented May 20, 2012 at 21:17
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See http://maths.sogang.ac.kr/shcho/pdf/P18.pdf, theorem 1.3 (which is the same as theorem 3.4).

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  • $\begingroup$ After having a closer look at that article, I'm not longer very satisfied by it. The argument I care about is entirely contained in the last page (one can safely disregard everything that comes before). The assumption that the boundary is $C^\infty$ is only used implicitly in the sentence "by the regularity theory of the Laplace operator". Also, I can't follow the last argument (last line of the paper): if you know that $|f(z)|^2$ is smooth, how do you conclude that $f(z)$ is smooth? $\endgroup$
    – André Henriques
    Commented Dec 9, 2011 at 10:28
  • $\begingroup$ I will try to find another reference (though I suspect the "regularity theory for the Laplace operator" cannot be escaped entirely, since this result is clearly a part of it...) $\endgroup$
    – Igor Rivin
    Commented Dec 9, 2011 at 10:45
  • $\begingroup$ I'm perfectly happy with an argument that uses the Laplace operator. But why not use the $\bar \partial$ operator directly? Isn't this regularity thing a common feature of all elliptic differential operators? $\endgroup$
    – André Henriques
    Commented Dec 10, 2011 at 10:24
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This is well-known result by Kellogg (O. Kellogg: On the derivatives of harmonic functions on the boundary, Trans. Amer.Math. Soc. 33 (1931), 689-692.), and Warschawski (On the higher derivatives at the boundary in conformal mapping,} Trans. Amer. Math. Soc, {\bf 38}, No. 2 (1935), 310-340.), where they prove even more, that the if the boundary is C^{n,\alpha}, then the conformal parametrization is C^{n,\alpha} up to the boundary.

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By a theorem of Caratheodory it extends to a homeomorphism.

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