# Analytic extension of holomorphic function defined on the unit disk

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?

• Do you mean that $f$ takes real values on the unit circle, that is $f(x) \in \mathbb R$ for $x \in D(0, 1)$? Aug 8, 2021 at 8:39
• Try using that the unit disc can be mapped biholomorphically onto the upper half plane. Aug 8, 2021 at 8:46
• @Ramanujan Unit circle is $C(0,1 )$ that is the circle $|z|=1$. With $D(0,1)$ I denote the unit disk, that is $|z|<1$ Aug 8, 2021 at 8:52
• I meant to say $C(0,1)$. Aug 8, 2021 at 8:56

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.
If $$f(z)$$ accepts real values on the unit circle i.e. if $$f(z)=u(x,y)+i\cdot0$$ for all $$(x,y)\in\{x^2+y^2= 1\}$$. Then, by the Cauchy-Riemann condition, we have the following equation $$\begin{cases} \frac{\partial u}{\partial x}=0 \\ \frac{\partial u}{\partial y}=0, \end{cases}$$ for all $$(x,y)\in\{x^2+y^2= 1\}$$. From here $$f(z)=u(x,y)=Const$$ for $$z\in S(0,1)$$. Then, by the theorem, the uniqueness $$f(z)=Const$$ for all $$z\in\mathbb{C}$$.
• The unit circle is not $\left\{ \left(x,y\right):x^{2}+y^{2}\leq1\right\}$. It is $\left\{ \left(x,y\right):x^{2}+y^{2}=1\right\}$ Aug 8, 2021 at 9:30
• $f\left(z\right)=\begin{cases} f\left(z\right) & z\in\overline{D}\left(0,1\right)\\ f\left(\frac{z}{|z|}\right) & z\notin\overline{D}\left(0,1\right) \end{cases}=\begin{cases} f\left(z\right) & z\in\overline{D}\left(0,1\right)\\ f\left(\sqrt{\frac{z}{\bar z}}\right) & z\notin\overline{D}\left(0,1\right) \end{cases}$ may be not holomorphic function Aug 8, 2021 at 11:23