In fact it is about holomorphic functions:

Does every holomorphic function $f:\mathbb{D}\longrightarrow \mathbb{D}$ have a fixed point?Here $\mathbb{D}$ is the open unit disk centered around $0$ on the complex plane$\mathbb{C}$.

It is an exercise from Stein's book,COMPLEX ANALYSIS,exercise 12 to Chapter 8.

My idea is as follows:first extend $f$ to the closure of the unit disk $\bar{\mathbb{D}}$,i.e.$\tilde{f}:\bar{\mathbb{D}}\longrightarrow\bar{\mathbb{D}}$.$\bar{f}$is continuous.Then apply the method of proving the classical fixed point theory on the unit disk.Yet this way we can at most get that $\bar{f}$ has fixed points,not $f$.

Are there any flaws in the above sketch of proof?

Or would someone be kind enough to give some hints on this problem?Thank you very much!

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    $\begingroup$ The open unit disk is biholomorphic to another open subset of $\mathbb{C}$ that might be easier to work with... $\endgroup$ – Qiaochu Yuan Sep 23 '11 at 15:01
  • $\begingroup$ If you use the hyperbolic metric, you have that holomorphic maps $F:\mathbb{D}\to\mathbb{D}$ do not increase distance, and because of Schwarz's Lemma, you have that a holomorphic map is either an isometry or a contraction (not necessarily strict) with this metric (you can consult Complex Analysis in the Spirit of Lipman Bers, for example). $\endgroup$ – rfauffar Sep 23 '11 at 16:29
  • $\begingroup$ If the contraction is strict, then you have Banach's fixed point theorem. If your function is an isometry, then it is a Moebius transformation, and you can study its fixed points in the disk. If not, then you can reduce your problem to seeing contractions that are not strict. $\endgroup$ – rfauffar Sep 23 '11 at 16:32
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    $\begingroup$ One flaw in our reasoning is that it is not clear that $f$ can be extended continuously to the closed disk. $\endgroup$ – lhf Sep 23 '11 at 16:38
  • $\begingroup$ Also consider the example $f(z) = (z+2)/3$ $\endgroup$ – GEdgar Sep 23 '11 at 17:30

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