Analytic continuation of a conformal map The following problem is stated in Greene and Krantz, Problem 15, page 413:

Let $U$ be a bounded simply connected domain in $\mathbb{C}$ with real analytic boundary (i.e. the boundary is locally the graph of real function that is analytic). By the Riemann mapping theorem, there exists a conformal map $f$ from $\mathbb{D}$ to $U$. Show that $f$ can be analytically continued to some open neighborhood of $\overline{\mathbb{D}}$.

The problem comes with a hint: It is important that the map take the boundary to the boundary, then one can reduce to the case where the boundary is flat via a change of variable. Then apply Schwartz reflection.
I have no idea how to make this hint work out.
 A: The point is that you can reflect in analytic arcs, not only in straight lines. 
Let $C$ be an analytic arc, that is, there is a real-analytic injective $\varphi \colon [0,1] \to \mathbb{C}$ with non-vanishing derivative such that $C$ is the image of $\varphi$. The power series expansions of $\varphi$ give a holomorphic continuation $\tilde{\varphi} \colon (-\varepsilon,1+\varepsilon)\times (-i\varepsilon,i\varepsilon) \to \mathbb{C}$ for some $\varepsilon > 0$, and if $\varepsilon$ is chosen small enough, $\tilde{\varphi}$ is a biholomorphism to its image $V$. Then on $V$ you have the antiholomorphic map
$$\sigma_C \colon z \mapsto \tilde{\varphi}\left(\overline{\tilde{\varphi}^{-1}(z)}\right),$$
the reflection in $C$. Evidently, $\sigma_C$ is an involution that fixes the points of $C$.
If $h \colon V\to W$ is a holomorphic map, $C_1$ is a one-sided analytic boundary arc of $V$, and $C_2$ a one-sided analytic boundary arc of $W$, and we have $h(z) \to C_2$ whenever $z \to C_1$ in $V$, then $\sigma_{C_2}\circ h \circ \sigma_{C_1}$ gives a holomorphic continuation of $h$ across $C_1$ (it may not be immediately obvious that $h$ extends continuously to $C_1$ with values in $C_2$, but the classical reflection principle for straight lines, which here is wrapped in the parametrisations of the analytic arcs yields that).
Since every proper arc of the unit circle is easily seen to be a one-sided analytic boundary arc of the unit disk, and $U$ has by assumption also an analytic boundary, it remains to be seen that for every small enough boundary arc $C_1$ of the unit disk you can choose a boundary arc $C_2$ of $U$ such that $z \to C_1$ implies $f(z)\to C_2$, and that the local continuations thus obtained fit together to yield a continuation of $f$ to a neighbourhood of $\overline{\mathbb{D}}$.
However, I find it easier to work in the other direction and consider $g = f^{-1} \colon U \to \mathbb{D}$. Since $w \to \partial U \implies g(w) \to \partial\mathbb{D}$, the real-valued harmonic function $\lambda(w) = \log \lvert g(w)\rvert$ on $U\setminus\{f(0)\}$ satisfies $\lambda(w)\to 0$ for $w\to \partial U$ (we have no need to choose a boundary arc of $\mathbb{D}$ this way), and can be reflected across each boundary arc of $U$. The local reflections agree in a neighbourhood of $\partial U$ by the identity theorem for harmonic functions, so we obtain a global continuation of $\lambda$ to a neighbourhood of $\partial U$. Locally, $\lambda$ is the real part of a holomorphic function $h$, and the imaginary part can be chosen so that $e^h \equiv g$ in $U$. The local holomorphic continuations of $g$ across the boundary again fit together to give a holomorphic continuation $\tilde{g}$ of $g$ to a neighbourhood $\tilde{U}$ of $\overline{U}$. If the neighbourhood is small enough, $\tilde{g}$ is a biholomorphism and $\tilde{f} = \tilde{g}^{-1}$ is the desired analytic continuation of $f$.
