Show a non-constant, continuous function $f:\bar{D}\rightarrow \bar{D}$ is such that $f(\partial D)=\partial D$

Question

Suppose $\bar{D}= \{z:|z|\leq 1\}$ and assume we have a non-constant, continuous function $f:\bar{D}\rightarrow \bar{D}$, that is holomorphic on the interior of $\bar{D}$ and such that $f(\partial D)\subset \partial D$.

Show that $f(\partial D)=\partial D$.

Since $f$ is non-constant on $\bar{D}$, it is also non-constant on $\partial D$, because $\partial D$ is the closure of $\bar{D}$ and I think I have to use the fact that $|f|$ attains the max or min on the boundary $\partial D$. Could someone please help me to understand how to do that?

Answer 1

Hint. Let $\mathcal{B}$ be the set of non-constant, continuous functions $f:\bar{D}\rightarrow \bar{D}$, holomorphic in $D$ and such that $f(\partial D)\subset \partial D$ where $D=\{z: |z|<1\}$.

1) If $f\in \mathcal{B}$ then $0\in f(D)$.

If not $1/f\in \mathcal{B}$ and $1/|f|$ has a maximum in $\partial D$, which is a minimum for $|f|$. Therefore $|f|$ has a minimum AND a maximum in $\partial D$ where $|f|=1$. Hence $|f|=1$ in $\bar D$, which implies that $f$ is constant. Contradiction.

2) If $f\in \mathcal{B}$ then $f(D)=D$.

If not there is $a \in D$ such that $a\not \in f(D)$. Hence $M_a \circ f$ is never $0$ in $D$, where $M_a={{z-a} \over {1- \overline a z}}$. Since $M_a \circ f\in\mathcal{B}$ , by 1), we get a contradiction.

P.S. More generally, by Fatou's Theorem (see Theorem A), $f\in\mathcal{B}$ is a finite Blaschke product: $$f(z) = e^{i \theta} \prod_{k=1}^{n} {{z-a_k} \over {1- \overline a_kz}}$$ where, $n\in\mathbb{N}^+$, $\theta\in\mathbb{R}$, and $a_k \in D$ for $k=1,\dots,n$.

Answer 2

A non-constant holomorphic function is an open mapping, so $E = f(D) $ is open with $E \subset D$.

If $\partial E \subset \partial D$ then $$ D \cap \partial E \subset D \cap \partial D = \emptyset $$ and therefore (with $A^C$ denoting the complement of $A$ in $\Bbb C$) $$ D \setminus E = D \cap E^C = (D \cap \partial E) \cup (D \cap \overline E^C) = D \cap \overline E^C $$ is an open set, so that $$ D = E \cup (D \setminus E) $$ is a decomposition of the connected set $D$ into two disjoint open sets. It follows that one of the sets on the right-hand side must be empty, and thus $E = D$ and $\partial E = \partial D$.