Holomorphic function on disk with two fixed points is identity

Question

Let $f : \Bbb D \to \Bbb D$ a holomorphic function that has two fixed points : $\exists w_1, w_2$ such that $f(w_1)=w_1$ and $f(w_2)=w_2$. Show that $f(z)=z$ $\forall z \in \Bbb D$. I wanted to do something like : since $\lvert f(z) - z \rvert$ is continuous and differentiable then it atteins a local max by Rolle's theorem. By the maximum modulus principle, $f(z)-z$ is constant equal to $0$, so $f(z)=z$. I feel it is wrong because Rolle's theorem is one dimensional so the local max is reached only on the segment joining $w_1$ and $w_2$ and not in a neighborhood. How can I fix this, is there another way to do it ?

Answer 1

This is a standard (but very nice) application of Montel's theorem: define the family $\{g_n\}$ of functions as $g_0=\text{Id}, g_1=f; g_{n+1}=g_n\circ f$. Montel's theorem implies that there is a converging subsequence $g_{n_k}$ to a function $g$. Clearly this function cannot be constant ($g(w_1)\neq g(w_2)$). Now it is easy to see that $g\circ f=g$; the local invertibility principle implies that, locally, $f=\text{Id}$ and the identity principle does the rest.

Another way of proving the result is via Schwarz-Pick: an easy application of the theorem implies that $f$ is an automorphism of $\mathbb{D}$ and it is easy to see that the only one fixing two points is the identity.

The application of Rolle's theorem does not work since the existence of a maximum relative to a segment does not contradict the maximum principle: e.g., $-z^2$ has a maximum, relative to $[-1,1]$, at $0$.

Answer 2

Does this work?: consider a $P$ biholomorfism of $\Bbb D$ such that $P(0)=w1$, then consider the new function $g=P^{-1}\circ f\circ P$ which fixes the origin and some $z_1$ ,now by the max modullus principle $g/z$ is always less or equal to 1 and reaches 1 in $z_1$, so $g/z=1, g=z, f=z$.