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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.

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

2 Answers 2

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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.

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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 $z\in\mathbb{C}$.

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    $\begingroup$ 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\} $ $\endgroup$
    – FreeZe
    Aug 8, 2021 at 9:30
  • $\begingroup$ Sorry, I corrected a grammatical error. $\endgroup$ Aug 8, 2021 at 9:39
  • $\begingroup$ Can you tell what is wrong with my definition? is it not holomorphic? $\endgroup$
    – FreeZe
    Aug 8, 2021 at 11:06
  • $\begingroup$ $f\left(z\right)=\begin{cases} f\left(z\right) & z\in\overline{D}\left(0,1\right)\\ f\left(\frac{z}{|z|}\right) & z\notin\overline{D}\left(0,1\right) \end{cases}=\begin{cases} f\left(z\right) & z\in\overline{D}\left(0,1\right)\\ f\left(\sqrt{\frac{z}{\bar z}}\right) & z\notin\overline{D}\left(0,1\right) \end{cases}$ may be not holomorphic function $\endgroup$ Aug 8, 2021 at 11:23

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