I was reading a proof by Beardon of the Wolff-Denjoy Theorem in Complex Dynamics. In the proof, a family of maps $$f_\epsilon=(1-\epsilon)f(z)$$ is used, where $f(z)$ is an analytic map from the unit disk $\Delta$ to itself.
Thus, for all $\epsilon >0$ the maps $f_\epsilon$ map the unit disk to a compact subset of itself. Now it is claimed, that each of these $f_\epsilon$ has a fixed point in $\Delta$.
Is it true that any analytic function mapping the unit disk to a compact subset of itself has a fixed point? (Otherwise this must follow for the $f_\epsilon$ from some other properties of $f$, which I did not mention here...)
I thought about using Brouwer fixed point theorem, but as the unit disk is not compact and $f$ may not extend continuously to $\partial D$, I do not see how this may work.