TITLE: Continuity of $f\circ f \circ \cdots$
QUESTION [5 upvotes]: Let $f:D\to D$, where $D\subseteq \mathbb{R}^n$, be a continuous function. Under what conditions is $f\circ f \circ \cdots$ continuous? Here, $\circ$ stands for the composition operator and sometimes the notation $f^2=f\circ f$ is used. So in this notation, when is $\lim_n f^n$ continuous?
There is a counterexample I can think of: $f(t)=t^\alpha$ over $D=[0,1]$ for $\alpha>1$. Then $f^n(t)=t^{\alpha^n}$ and the point-wise limit of $f^n$ is $\left(\lim_n f^n\right)(t)=0$ for $t\in[0,1)$ and $\left(\lim_n f^n\right)(1)=1$ which is not continuous.
One think one could suggest is that $f^n$ be a uniformly convergent sequence of functions. But what should this imply for $f$?

REPLY [2 votes]: In complex analysis, such things are studied.  Going back to Fatou and Julia.  Say $f$ is analytic on a domain in the complex plane.  The most important condition is that the set $\{f^n : n \ge 1 \}$ of iterates should be a normal family ... this is what you need for the limit to be an analytic function.  
Link ... normal family ...