Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.