# Showing $f_{n}$ is a normal family, where $f_{n}$ is the $n^{th}$ iterate of $f$

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.

-

If $|f'(0)|<1$, then there is a constant $m$, $0<m<1$, 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.
Thank you for the help. I fully understand the $|f^{'}(0)|>1$ answer. How exactly is it in the first sentence you deduce $|f^{'}(z)| \leq m$ for $z \in D$? – pel May 14 '11 at 3:32
You are simply using continuity of the function $f^{'}$ in the first sentence. – pel May 14 '11 at 3:36
@pel: Exactly. Just take $m$ with $|f'(0)|<m<1$ and apply continuity of $f'$. (And you're welcome.) – Jonas Meyer May 14 '11 at 3:37