Actually I was having a look at this problem Uniform convergence of a sequence of functions and was wondering if the result still holds hood if the boundedness condition on the domain is dropped. Thanks for any help.
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If $D \subset \mathbb C$ is any domain such that $\mathbb C \backslash D$ contains at least two points, then by Montel's theorem the analytic functions from $D \to D$ form a normal family. If $f$ is such a function, and $f(p) = p$ and $f'(p) = 0$ for some $p \in D$, then the iterates $F_n$ of $f$ have a subsequence converging uniformly on compact sets to an analytic function $g$, and it is easy to show that $g$ is constant.
Robert Israel
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The simple counterexample is easy: Take $f(z) = z^2$ on $D = \mathbb{C}$; then $f(0) = 0$, $f^{\prime}(0) = 0$, and $f$ is certainly analytic as a map from $\mathbb{C}$ to itself. Yet clearly the $n$th iterate of $f$ at $z = 2$ will be $2^{(2^n)}$, which clearly diverges, and in general for $\vert z \vert > 1$, $F_n(z) = f \circ f \circ \dotsb \circ f(z)$ will diverge in general.
The issue is that $f: D \to D$ was what forced a very strong boundedness when $D$ was bounded (in particular, why you couldn't just use $z \to z^2$ on a disk of radius $> 1$ to get an easy counterexample in the original problem) and you don't get that when $D$ is unbounded, at least not immediately.
I wonder if the formula would work on a strange domain in $\mathbb{C}$; it is not immediately obvious to me.
Charles Baker
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I think that it will work on a simply connected proper subset of C as then we can use the Riemann Mapping Theorem. – Ester Jan 05 '15 at 19:09