Let $f:\ [0,1] \to [0,1]$ be continuous and increasing. Also let $p \in [0,1]$. Define $$x_n = f^n(p) = f(f(...f(p))...)$$ Prove that either $p$ is a fixed point of $f$ or $x_n$ converges to a fixed point of $f.$

What I have so far:

We know that there are exactly 3 possible cases when $p \in [0,1]$.

  1. $f(p) = p$

In this case, we are done as $p$ is a fixed point of $f$.

  1. $f(p) > p$

In this case we have $$p < f(p) \leq f^2(p) \leq f^3(p) \leq ...$$

  1. $f(p) < p$

Here we have that $$...f^3(p) \leq f^2(p) \leq f(p) < p$$

I suppose I am struggling to see how to show that $x_n$ conveges to a fixed point of $f$. Any help would be greatly appreciated!

  • $\begingroup$ The book "A Mathematician's Miscellany" by John E. Littlewood contains a visual proof of this result. $\endgroup$ – Shahab Oct 31 '18 at 15:08

In case $2, (x_n)$ is an increasing sequence bounded above by $1,$ hence it converges to its supremum, say $x.$

Since $f$ is continuous, $$x_n \to x \implies f(x_n)\to f(x).$$ That is $f^{n+1}(p) \to f(p).$ But $f^{n+1}(p)=x_{n+1} \to x.$ Therefore $f(x)=x.$ Hence, $x$ is a fixed point.

Similarly for case $3.$

  • $\begingroup$ Thank you this makes sense now, knew I was right there! $\endgroup$ – User1996424 Dec 13 '17 at 3:47

Assume the second case is valid. Then $\{f^{n}(p)\}_{n}$ is an increasing sequence on $[0,1]$, of course, it is bounded (by $1$, for example). So $\lim_{n}f^{n}(p)=x_{0}$ for some $x_{0}\in[0,1]$. Now continuity of $f$ at $x=x_{0}$ implies that $f(x_{0})=f(\lim_{n}f^{n}(p))=\lim_{n}f(f^{n}(p))=\lim_{n}f^{n+1}(p)=x_{0}$.


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