Existence of a period three cycle for a continuous unimodal interval map (A question related to Li-Yorke theorem)

Question

Let $f$ be a continuous unimodal map from a closed interval $I$ to itself. Suppose that there exist $x_1$, $x_2$, $x_3$, $x_4\in I$ such that $f(x_1)=x_2$, $f(x_2)=x_3$, $f(x_3)=x_4$, and $x_4<x_1<x_2<x_3$. Then is it true that $f$ has a period three cycle? Some posts (and corresponding answers) suggested to use the intermediate value theorem for $g(x)=f^3(x)-x$. However, a zero of $g$ is either a point of period 3 for $f$ or a fixed point of $f$. I do not see how to proceed if we get a fixed point.

Answer 1

Let's check consequences on intervals:

• The maximum of $f$ is after $x_2$, so $f([x_1,x_2])=[x_2,x_3]$.

• The image of an interval contains at least the interval between the values at the endpoints, so $f([x_2,x_3])\supset [x_4,x_3]\supset [x_1,x_3]$.

• The maximizer $c$, $f(c)>x_3$, is inside $[x_2,x_3]$, otherwise $f(x_3)=x_4<x_3=f(x_2)$ is not possible (remember unimodal).

As a consequence there is an interval $J\subset [c,x_3]$ with $$ f(J)=[x_1,x_2], ~~ f^2(J)=[x_2,x_3], ~~ f^3(J)\supset J. $$ and $f^3$ restricted to $J$ is falling. This is sufficient to find a fixed point $x_*$. As $x_*\in[c,x_3]$ and $f(x_*)\in[x_1,x_2]$, this is not a fixed point of $f$, thus a true period-3 point.

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To be a bit more correct and independent of the location of the maximum, the interval $[x_2,x_3]$ contains a segment where $f$ is falling. The image of that segment contains the interval $[x_1,x_2]$. Let $a<b$ inside that segment be the points with $f(a)=x_2$ and $f(b)=x_1$. Then $$ f^2(a)=x_3, ~~ f^3(a)=x_4<x_2<a\\ f^2(b)=x_2, ~~ f^3(b)=x_3>b $$ The interval $[a,b]$ satisfies the assumptions of the intermediate value theorem for $g$.