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I am trying to solve the following problem :

Let $f:[a,b]→[a,b]$ be a continuous function. Suppose that $f$ admits two successive fixed unstable points $r_1$ and $r_2$ such that $r_1<r_2$ and there is no fixed point in the interval $[r_1, r_2]$.

Show that there is a $2$-periodic point $s\in[r_1,r_2].$

Hint: study the function defined by $g(x)=f^2(x)-x.$

My answer : As $r_1$ is unstable then $\exists x>r_1:f^2(x)>x>r_1.$

As $r_2$ is unstable then $\exists y<r_2:f^2(y)<y<r_2.$

As $f^2(x)-x>0$ and $f^2(y)-y<0$ then by the Intermediate value theorem there exist a point $s\in [x,y]$ such that $f^2(s)=s.$

Hence there is a $2$-periodic point.

I have two questions:

$(1)$ Is my answer correct ?

$(2)$ If so, we could use the same argument to show that there is periodic points of all periods

In particular as the system will have $3$-periodic points so the system is chaotic?

Stephen Donovan
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kiki
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  • Why is $f^2$ monotonically increasing? One could get increasing distance also in the other direction. // I do not think that you get the necessary monotonic behavior for $f^3$, and the fixed point you get for $f^4$ could be a point from the 2-cycle. – Lutz Lehmann Apr 18 '21 at 10:46
  • Unstable here stands for the two directions , sorry for not mentioning that. I think that if iterates of points starting at the left of $r_2$ will move away from $r_2$ and if iterates of points starting at the right of $r_1$ will also move away from $r_1$ then we should have a 3-periodic point also which in turn means that we have every period. – kiki Apr 18 '21 at 11:55

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