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Assume that $f$ is an entire function with $f(0)=0$. Consider the family of functions {$f_{n}$} where $f_{n}$ is the $n^{th}$ iterate of $f$, i.e. $f_{n}=f \circ f \circ \ldots \circ f$ ($n$ times).

I'm trying to show that if |f^{'}(0)|<1, then there is an open set $U$ containing zero so that the family {$f_{n}$} is normal on $U$. I have been able to show |f_{n}^{'}(0)|<1 and not much more.

Lastly if we take |f^{'}(0)|>1 then there does not exist an open set $U$ containing zero such that {$f_{n}$} is normal, but I'm not sure how to show this either. I'm pretty sure we can do this problem without relying on Fundamental Normality Test. Thank you for the help and guidance.

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If |f'(0)|<1, then there is a constant $m$, $0, and a disk $D$ centered at $0$ such that |f'(z)|\leq m for all $z\in D$. Then for all $z\in D$, |f(z)|=|\int_{[0,z]}f'(w)dw|\leq m|z|. Inductively you get that $f_n(D)\subseteq m^n D$, so $(f_n)_n$ converges uniformly to $0$ on $D$.

If |f'(0)|>1, then f_n'(0)\to\infty by the chain rule, while $f_n(0)=0$ for all $n$, which implies that local uniform convergence of a subsequence is impossible (even if you allow $\infty$). If the sequence were normal on a neighborhood of $0$, then some subsequence $(f_{n_k})_k$ would converge uniformly on a neighborhood of $0$ to a holomorphic function $g$ (with $g(0)=0$). But this would imply that f_{n_k}'(0)\to g'(0)\neq\infty as a consequence of Cauchy's integral formula.

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    @pel: Exactly. Just take $m$ with |f'(0)| and apply continuity of $f'$. (And you're welcome.)2011-05-14