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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 $f$?

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    I'm suggesting only a necessary condition, not a sufficient one. There are functions whose fixpoint sets are connected, and whose limits exist but are not continuous, such as (for $D=\mathbb R^2$, but in polar coordinates) $ (r,\theta)\mapsto (r,\theta+\sin \theta) $2012-08-29

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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 ...

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    I found an article: Arsove M.G., Some criteria for normality of families of continuous functions, Comm. Pure & Appl. Math., 9(3), 1956, 299-305. If the family consists of continuous functions if it is i. uniformly bounded over compact sets ii. equicontinuous over all compact sets. There are more results in that paper.2012-08-29