If $f$ is analytic defined on $D:=\{z:|z|<1\}$ and $|f(z)|\le 1$, can $f$ always be extended continuously to $\overline D$? In other words, can all analytic functions $f:D\to\overline D$ be extended to continuous functions $\tilde f:\overline D\to\overline D$ (such that $\tilde f|_D=f$)?
Ideally, for $|z_0|=1$, we could simply define $f(z_0)=\lim_{z\to z_0}f(z)$. It seems like a reasonable assumption that this extended version of $f$ exists and is continuous, but it's not obvious to me that these limits must exist. This feels like it should be a standard result in complex analysis, but I haven't been able to find a good reference.
(Unfortunately, the hope that $f$ can be extended to an analytic function is dashed by the example of $\sqrt{z-1}$.)
The answer is no. Consider $f(z) = e^{(z+1)/(z-1)}$. The map $z\mapsto \frac{z+1}{z-1}$ takes the unit circle to the imaginary axis and the interior of the unit circle to the left half-plane; therefore $f(z)$ is indeed a map from $D$ to itself. However, $f(x)$ tends to $0$ as $x$ tends to $1$ from below, but $|f(e^{i\theta})|=1$ and thus $f(z)$ cannot tend to $0$ as $\theta\to0$.