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A unimodal map is a continuous map $f:[0,1]\longrightarrow [0,1]$ such that there is only one turning point (critical point), denoted by $c$, and $f(0)=f(1)=0$.

Unimodal map is related to kneading invariant and kneading map, constructed by Hofbauer and Keller.

One assumption is always being made by several books in dynamics. They all assume $$f^{2}(c)<c<f(c),$$ where $f^{n}(x)$ is the $n-$times iteration of $f$ at $x$.

They all use an one-line explanation:

If this condition does not hold, then $f$ has no interesting dynamics.

I have drew several pictures but I cannot see why. Even though we have $c<f^{2}(c)<f(c)$, the dynamics is still complicated, since I cannot see a clear rule about where $f^{n}(c)$ will go for $n\geq 3$.

So what does it mean by "interesting" here? and why does this assumption make the dynamics "interesting"?

Thank you!

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    $\begingroup$ In this case, the interval $I=[c,f(c)]$ is mapped into itself and the restriction of $f$ to $I$ is decreasing. This means that every orbit in $I$ converges either to a fixed point or a period two orbit. For points in $[0,c]$, the orbit is either monotonic (and so converges to a fixed point), or it eventually leaves $[0,c]$, whereupon it enters $I$. For points in $[f(c),1]$, the image is in $[0,f^2(c)]\subset [0,c]\cup I$, so that we have shown that every orbit approaches a fixed point or period 2 orbit. Pretty dull... $\endgroup$
    – Anthony Quas
    Commented Jun 26, 2020 at 2:34
  • $\begingroup$ @AnthonyQuas brilliant! Post an answer so that I can accept and vote? $\endgroup$
    – JacobsonRadical
    Commented Jun 26, 2020 at 13:51
  • $\begingroup$ @AnthonyQuas could you please elaborate a little bit why every orbit in $I$ converges either to a fixed point or a period two orbit? $\endgroup$
    – JacobsonRadical
    Commented Jun 26, 2020 at 14:00
  • $\begingroup$ The point is $f$ is a decreasing map from $I$ to $I$, so that $f^2$ is an increasing map from $I$ to $I$. It's pretty easy to see that if you have an increasing map, everything converges to a fixed point. Probably math.stackexchange.com would be a better place if you want to ask about that. $\endgroup$
    – Anthony Quas
    Commented Jun 26, 2020 at 16:05

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