How to prove Schwarz reflection principle?

Question

Suppose that $f$ is non-vanishing and continuous on a closed unit disk that is holomorphic in the interior $D$. Show that if

$\lvert f(z) \rvert = 1$ whenever $\lvert z \rvert = 1$

then $f$ is constant.

I'm not sure where to start with this one or how to go about this proof at all. Any help would be appreciated.

How is this question related to the Schwarz reflection principle? — lhf, Feb 26 '13 at 11:34

Guillermo · Answer 1 · 2013-02-26T11:31:41.203

2

Since $|f|$ is non-vanishing, there exists $M>0$ such that $|f(z)|>M$ for all $z \in \bar D$.

Define $f(z)=\frac{1}{f(\frac{1}{\bar z})}$ for $z \in \mathbb{C}-D$. Note that if $z \in \mathbb{C}-D$, then $\frac{1}{z} \in D$.

We claim that $f$ is bounded and entire. then it is a constant function.

You can easily check that $|f|$ is bounded, and $f$ is holomorphic in D and $\mathbb{C}-D$

It suffices to show that $f$ is holomorphic on boundary ($|z|=1$)

Apply Morera's theorem !

edited Feb 26 '13 at 11:31

answered Feb 26 '13 at 11:25

Guillermo

You can't just redefine $f$ like you do... did you mean using a new name $g$ from line 2 on? – vonbrand Feb 26 '13 at 12:38
@vonbrand $f$ is originally not defined for $z$ outside the unit disc. Ryuichi is defining an extension of the given $f$. It's common abuse of notation to keep the same name for the extended function. – mrf Feb 26 '13 at 13:16
@mrf, no coffee yet. Sorry. – vonbrand Feb 26 '13 at 13:43

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