Suppose $f$ is analytic in the unit disc $D(0,1)$ and maps the unit circle into itself. Show then that $f$ maps the entire disc onto itself.
So the outline wants us to use the Max Modulus Theorem to show that $f$ maps $D(0,1)$ into itself. Then, use the fact that we proved that if $f:S \to T$, $f$ non-constant and analytic on $S$, and if $f(z)$ is a boundary point, $z$ is a boundary of $S$ to show that the mapping is onto.
I'm not sure if mapping the unit circle into itself means that $|f|=1$ on the unit circle. Also is the unit disc compact? Thanks!
[Math] Show that $f$ maps the entire unit disc onto itself.
complex-analysis
Best Answer
If $w \in D$ and $w \not \in f(D)$ then $$ z \mapsto \frac{1}{f(z)-w} $$ is holomorphic on $\overline{D}$. By the maximum modulus principle, for any $z \in D$ $$ \left|\frac{1}{f(z)-w}\right| \leq \max_{\omega \in \partial D} \left| \frac{1}{f(\omega)-w} \right| \leq \max_{\omega \in \partial D}\frac{1}{\left||f(\omega)|-|w|\right|} = \frac{1}{1-|w|} $$ so $|f(z) - w| \geq 1-|w|$. This means that $D \setminus f(D)$ is open. By the open mapping theorem either $f$ is constant or $f(D)$ is open. The latter implies that $f(D)=D$ since $D$ is connected. Moreover, the image of a compact set under a continuous function is compact. Therefore if $f$ is not constant then $f(\overline{D}) = \overline{D}$.
Or, alternatively, this bound shows that $w$ can be moved in a straight line towards $0$ while the radius of the "image free" disc around it increases. For $w=0$ the bound becomes $|f(z)| \geq 1$ so that $f(D) \cap D = \emptyset$. So either $f(D)$ contains all of $D$ or avoids it entirely. In the latter case $f$ maps into the unit circle. This would mean that $\overline{f} = f^{-1}$ but $\overline{f}$ can be holomorphic (complex differentiable) only if $f'$ vanishes identically. The conclusion is that either $f(D)=D$ or $f$ is constant.