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Let $G\subseteq \mathbb{C}$ be a region that contains the closed unit disc and let $g: G\rightarrow \mathbb{C}$ be an analytic function.

Further assume that $|f(z)|=1$ for $|z|=1$.

What can you say about $f$?

Not sure how Schwarz lemma can be applied here.

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Answer.

Case I. If $f$ does not have zeros inside the open unit disc $D$ centered at the origin, then $f$ is constant.

Case II. If $f$ has $n$ roots inside the open unit disc, then $f$, then $$ f(z)=\lambda\prod_{j=1}^n \varphi_{a_j}(z), $$ where $a_j\in D$, and $\varphi_{a_j}(z)=\dfrac{z-a_j}{1-\overline{a_j}z}$, and $\lvert\lambda\rvert=1$.

Explanation. If no roots, then $1/f$ is also holomorphic, and both $f$ and $1/f$ are bounded by $1$ in $\overline{D}$, and hence $f$ is constant.

If $n$ roots, and $f(a)=0$, for $\lvert a\rvert<1$, then $$ g(z)=\frac{f(z)}{\frac{z-a}{1-\overline{a}z}} $$ has $n-1$ roots in $D$, and $\lvert g(z)\rvert=1$ as well, since $$ \left|\frac{z-a}{1-\overline{a}z}\right|=1, $$ for $\lvert z\rvert=1$.

Next, use induction.

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    $\begingroup$ Any reasoning behind this? $\endgroup$ – Curious Droid Nov 3 '14 at 20:42
  • $\begingroup$ See my updated answer. $\endgroup$ – Yiorgos S. Smyrlis Nov 3 '14 at 20:55
  • $\begingroup$ Why is this true: If no roots, then $1/f$ is also holomorphic, and both $f$ and $1/f$ are bounded by 1 in $\overline{D}$, and hence f is constant. $\endgroup$ – Curious Droid Nov 3 '14 at 21:08
  • $\begingroup$ Blaschke product. +1. $\endgroup$ – orangeskid Nov 3 '14 at 21:08

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