Let $ f $ be a holomorphic function which is also bounded on the closed unit disk. In addition, assume that $ f $ accepts real values on the unit circle. I want to prove that $ f $ is constant.
So, I think the natural way would be to extend $ f $ so that $ f $ would be entire and then use Liouville's theorem.
Im not sure what is the correct way to extend an analytic function that defined on the closed unit disk, I thought about something of the form:
$ f\left(z\right)=\begin{cases}
f\left(z\right) & z\in\overline{D}\left(0,1\right)\\
f\left(\frac{z}{|z|}\right) & z\notin\overline{D}\left(0,1\right)
\end{cases} $
I assume that this is not an acceptable continuation because it dosent really use the fact the $ f $ accepts real values on the unit circle.
What would be the correct way to extend the function so it would be entire?
Thanks in advance.
Best Answer
The Möbius transform $$\Phi: \overline{\mathbb{H}_+} \rightarrow D(0,1),\; z \mapsto \frac{z-i}{z+i}$$ maps the closed upper half plane $\overline{\mathbb{H}_+} = \{z \in \mathbb{C}: \Re \, z \geq 0\}$ holomorphically onto the unit disk (see Mapping unit disc onto upper half plane). In particular, it maps the real axis onto the boundary of the unit disk.
Now, the map $$f \circ \Phi: \mathbb{H}_+ \rightarrow \mathbb{C}$$ is holomorphic and accepts real values on the real axis since f accepts real values on the boundary of the unit disk. Using the Schwarz Reflection Principle (see https://en.wikipedia.org/wiki/Schwarz_reflection_principle) we can extend $f \circ \Phi$ to an entire function which is bounded since f is bounded. Then, Liouville gives that f is constant.