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.

  • $\begingroup$ Do you mean any analytic map. Otherwise, you can map the whole disk to two points. $\endgroup$ – sqtrat Mar 3 '16 at 13:57
  • $\begingroup$ yes, I mean that, sorry $\endgroup$ – TheAbelian Mar 3 '16 at 13:58

If $K \subset \Delta$ is compact, there is an $r < 1$ with $K \subset \overline{D_r(0)} = \{ z : \lvert z\rvert \leqslant r\}$. Now you can apply Brouwer's fixed point theorem to ${f_{\epsilon}}\lvert_{\overline{D_r(0)}} \colon \overline{D_r(0)} \to K \subset \overline{D_r(0)}$ to conclude that $f_\epsilon$ has a fixed point in $\overline{D_r(0)}$, and a fortiori in $\Delta$.


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