TITLE: Riemann mapping theorem and smoothness on the boundary
QUESTION [10 upvotes]: 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 $\mathbb D$?

REPLY [2 votes]: 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.