If self-composition of an analytic function on the disk gives a linear fractional map, is the original function linear fractional?

Question

Suppose $f: \mathbb{D} \rightarrow \mathbb{D}$ is analytic. Furthermore, suppose $f \circ f = g$, and $g$ is a linear fractional map. Does this guarantee that $f$ is linear fractional? I know it would be true if the domain was the sphere instead of the disk, or if $g$ was onto the disk (as we already have that $f$ is injective because $g$ is). But I want this to be true in general. Any insight?

Answer 1

Yes. Consider the sequence of open disks $$\mathbb D\subset g^{-1}(\mathbb D)\subset g^{-2}(\mathbb D)\subset \dots . $$ Let $\mathbb D'$ be the union. This is either an open disk or the complement of a point in the Riemann sphere. You can extend $f$ to $\mathbb D'$ by defining $f(z)=g^{-n}(f(g^n(z)))$ for sufficently large $n$. Now $f$ maps $\mathbb D'$ to itself in a one to one and onto fashion, so it is a linear fractional map.