If $f$ is analytic defined on $D:=\{z:|z|<1\}$ and $|f(z)|\le 1$, can $f$ always be extended analytically to $\overline D$? 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 $f$ (a) exists, (b) is conti...

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,...

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, $|f(e^{i\theta})|=1$ and thus $f(z)$ tends ...

When doing this problem, I really wanted the condition that $\hat{f}(\xi) = 0\; \forall\; x \in \mathbb{R}$. Indeed, by pulling out the textbook, this is the condition given for the problem. First, we get agreement of $A$ and $B$ on the real axis since $\hat{f}(\xi) = 0\; \forall \xi \in \mathb...