Show that there is $x\in[-1;1]$ such as the set of limit point of $u_n(x)$ is $[-1;1]$ where $u_0(x) = x, u_{n+1}(x) = 2u_n(x)^2 - 1$

Question

Let be $f$ defined by $f(x) = 2x^2 -1$.

Let be $(u_n(x))$ the sequence defined by: $u_0(x) = x, u_{n+1}(x) = f(u_n(x))$

Show that there is $x\in[-1;1]$ such as the set of limit point of $u_n(x)$ is $[-1;1]$.

I started with studying $f$ to find stable intervals, potential limits etc... I've only found that $1$ and $-1/2$ are fixed point for $f$. I don't know what to do next...

Answer 1

Hint: $f\left(\left[-1,1\right]\right)=\left[-1,1\right]$.

Update: You can actually solve this recurrence. Note that $2\cos^{2}x-1=\cos2x$, and therefore it is easy to check that

$$u_{n}=\cos\left(2^{n}C\right)$$

is a solution, where $C$ is an arbitrary constant. In particular, if $u_{0}=x$ then $C=\arccos x$.