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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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If $D$ is connected, then a necessary condition is that $\{x\mid f(x)=x\}$ must be connected, too. That disqualifies $t^\alpha$ for every $\alpha\notin\{0,1\}$. –  Henning Makholm Aug 29 '12 at 13:41
    
@HenningMakholm Thanks a lot! Indeed, sounds intuitively correct. But, is there a proof for that? –  Pantelis Sopasakis Aug 29 '12 at 13:46
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If $f$ is continuous, then every value of $\lim_n f^n(x)$ must be a fixpoint, so the limit (if it exists) maps $D$ surjectively to the set of fixpoints. (It's surjective because the limit function clearly maps every fixpoint to itself). Therefore $(\lim f)(D)=\{x\in D\mid f(x)=x\}$. But continuous functions map connected sets to connected sets. –  Henning Makholm Aug 29 '12 at 13:50
    
@HenningMakholm I only know that continuous functions map connected sets to connected ones. But does the converse hold as well? And if yes, is it sufficient to prove that $\lim f$ maps $D$ to a connected set? What if there is some other connected set $Z$ so that $(\lim f)(Z)$ is not connected? However, your suggestion serves as a criterion to test whether $\lim f$ is not continuous. –  Pantelis Sopasakis Aug 29 '12 at 14:04
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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) $$ (r,\theta)\mapsto (r,\theta+\sin \theta) $$ –  Henning Makholm Aug 29 '12 at 14:10
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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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On $\Re^n$ do we have a similar result for functions that are merely continuous. i.e. if $f^n$ is continuous and the family $\{f^n\}$ is uniformly bounded, is it also normal? –  Pantelis Sopasakis Aug 29 '12 at 16:25
    
No, certainly not. Uniformly bounded is not enough for normal. –  GEdgar Aug 29 '12 at 18:14
    
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. –  Pantelis Sopasakis Aug 29 '12 at 20:29
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