Show that $f(x) :=\cos(\cos(x))$ is a contraction mapping. Find the fixed point. 
Show that $f(x) := \cos(\cos(x))$ is a contraction mapping. Find the fixed point.

I know that $|\cos(x)|\leq 1$ which means that $|\cos(\cos(x))| \in [\cos 0,\cos 1]$
I need to find $\alpha <1$ s.t.
$$|\cos(\cos(x)) - \cos(\cos(y))| \leq \alpha d|x-y| .$$
 A: In the interval $X:=\bigl[0,{\pi\over2}\bigr]$ the function $\cos$ decreases monotonically from $1$ to $0$. Therefore $f:=\cos\circ\cos$ increases from $\cos 1>\cos{\pi\over3}={1\over2}$ to $\cos0=1$. It follows that we can consider  the restriction  $f\restriction X$ as a selfmap of the complete metric space $X$.
Furthermore we have
$$0\leq f'(x)=\sin(\cos x)\cdot\sin x\leq\sin 1\cdot1<\sin{\pi\over3}={\sqrt{3}\over2}\qquad(x\in X)\ ,$$
and this shows that $f$ is a contraction of $X$. According to the fixed point theorem we therefore have a unique fixed point $\xi$ of the restriction $f\restriction X$.
It remains to show that there are no fixed points of $f$ outside of $X$: From $$f({\mathbb R})=\cos\bigl([{-1},1]\bigr)\subset\left[{1\over2},1\right]$$ it immediately follows that there can be no fixed points $\xi<0$ or $\xi>{\pi\over2}$.
A: Hint Computing gives $$f'(x) = \sin \cos x \sin x ,$$ and so $$|f'(x)| = |\sin \cos x| \, |\sin x|.$$
Thus, $|f'(x)| \leq 1$, and if equality holds for some $a$ we would have $|\sin \cos a| = 1$ and $|\sin a| = 1$. (It may also be useful to observe that $f$, and hence $f'$, is periodic.)
