I have no doubt that the following observation is quite well known. Let $\varphi:[0,1]\to [0,1]$ be a continuous map. Assume that the iterates $\varphi^n$ converge pointwise to some continuous map $\varphi_\infty$. Then the convergence is in fact uniform. However, I was unable to locate a reference. Does anybody know where I can find it written? Thanks in advance.

1$\begingroup$ I think your need monotonicity w.r.t n, then it becomes Dini's theorem. $\endgroup$– Piyush GroverMar 30, 2013 at 21:40

$\begingroup$ Counterexample if not monotonic in this file: math.ubc.ca/~feldman/m321/dini.pdf $\endgroup$– Piyush GroverMar 30, 2013 at 21:45

1$\begingroup$ Your wording suggests that you have a proof and only seek a reference. It this correct? $\endgroup$– Sergei IvanovMar 31, 2013 at 8:29

1$\begingroup$ @Sergei Ivanov. Yes, exactly. $\endgroup$– EtienneMar 31, 2013 at 8:55

1$\begingroup$ There is a tag 'referencerequest'. $\endgroup$– Sergei IvanovMar 31, 2013 at 9:06
2 Answers
I don't know a reference but maybe the following proof is shorter than yours.
By continuity, $\varphi\circ\varphi_\infty=\varphi_\infty$. Hence $\varphi$ is identity on the set $I:=\varphi_\infty([0,1])$. Hence $I$ is the set of fixed points of $\varphi$. And it is compact and connected. Then there are two cases: either $I$ is a nontrivial interval $[a,b]$ or it is a single point $\{a\}$.
In the first case, we have $f(x)<x$ for all $x>b$ and $f(x)>x$ for all $x<a$, otherwise there is a fixed point outside $I$. There is an $\varepsilon>0$ such that $f(x)>a$ for all $x\in(b,b+\varepsilon)$ and $f(x)<b$ for all $x\in(a\varepsilon,a)$. Let $U=(a\varepsilon,b+\varepsilon)$. Then $\varphi(U)\subset U$ and for every $x\in U$ we have $$ \varphi^n(x)x<\delta(\varepsilon) $$ where $\delta(\varepsilon)\to $ as $\varepsilon\to 0$ ($\delta(\varepsilon)$ is the maximum of the diameters of the sets $\varphi([a\varepsilon,a])$ and $\varphi([b,b+\varepsilon])$) For every $x\in[0,1]$, the iterations $\varphi^n(x)$ eventually get into $U$, hence $$ \bigcup_n \varphi^{n}(U) = [0,1] . $$ By compactness, this implies that $[0,1]$ is covered by finitely many sets $f^{n}(U)$ and hence there exist $n$ such that $\varphi^n([0,1])\subset U$. After this $n$, all iterations stay within the distance $\delta(\varepsilon)$ from the limit. Since $\varepsilon$ can be arbitrarily small, this implies uniform convergence.
In the second case (when $I$ is a single point $a$), similarly observe that $f(x)<x$ for $x>a$ and $f(x)>x$ for $x<a$, and the same holds for $\varphi^2$, $\varphi^3$, etc. This implies that if e.g. $x>a$ and $f(x)<a$ then all iterations $\varphi^n(x)$ lie between $x$ and $f(x)$. This easily implies that there is an arbitrarily small neighborhood $U$ of $a$ such that $f(U)\subset U$. Then uniform convergence follows similarly to the first case.

$\begingroup$ Thanks for this nice proof, which is quite different from (and perhaps shorter than) the one I knew. I still cannot believe that the result has never been stated (and proved) in some book and/or paper, but a google search led to nothing. $\endgroup$– EtienneMar 31, 2013 at 10:35
This result seems to be a consequence of Dini's theorem (as noted by nonlinearism):
If $\varphi^n$ converges to a continuous function $\varphi_\infty$, we have $\varphi_\infty = 0$. Hence, $\varphi < 1$ on $[0,1]$. By compactness, we have $\varphi < p$ for some $p < 1$. Therefore, the convergence is monotonic and we can apply Dini's theorem.
This should work for all $\varphi : K \to [1,1]$, where $K$ is a compact metric space (for $\varphi < 0$, one has to adopt the proof in an obvious manner).

1$\begingroup$ Why $\varphi_\infty=0$? What is $\varphi(x)=x$? $\endgroup$ Mar 31, 2013 at 8:27

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$\begingroup$ Sorry, I misunderstood your "iterate" (I thought of "power"). $\endgroup$– gerwMar 31, 2013 at 18:56