Let X be a complete metric space and let $S: X \to X$ be a map. Assume that there exists $m\geq 1$, so: $S^m=\underbrace{S∘S∘··∘S}_{m\text{}}$ is a contraction.
1) Show that S has a unique fixed point.
2) Show that for $m=2$ we can take $S=\cos:[0,\pi/2]→[\pi/2]$
Definition:
Let $X = (X,d)$ be a metric space.
A map $S:T\to T$ is a contraction if there exists a number $0≤\beta<1$, so:
$d(Sx,Sy)\leq \beta d(x,y)$ for all $x,y\in X$
My solutions:
1) First we have to show that $S$ has a fixed point by showing that $S^m$ has a fixed point. But since $S^m$ is a contraction on a complete metric space then it follows from Banach's theorem that $S^m$ has a fixed point (and it's unique).
If $x$ is a fixed point for $S^m$ then $S^mx=x$ and $S^{m+1}x=Sx$.
Now we have to show that $Sx=x$.
$Sx = S(S^mx)= S^m(Sx) \implies Sx = x$
2) This part I'm not sure about how to understand the question. Is there anyone who can help me with this part? I'll appreciate any hint.
$1.$ As it is right now, you have (correctly) proved that $S$ has a fixed point, but it remains to show that it is unique. For that matter, let $y$ be another fixed point of $S$, then recursively, one has $S^my=y$, but $x$ is the unique fixed point of $S^m$, so that $x=y$.
$2.$ You have to prove that $\cos\circ\cos$ is a contraction of $[0,\pi/2]$, for example see the answer here.