Topology without tears:
I am asked to show using exercise 2 and 3 that $f:\mathbb{R}\to \mathbb{R}$ defined as $f(x)=\cos x$ does not satisfy the contraction mapping theorem but does have a fixed point.
Contraction mapping theorem:Let $(X,d)$ be a complete metric space and $f$ a contraction mapping of $(X,d)$ into itself.Then $f$ has precisely one fixed point.
Exercise 2:Extend the contraction mapping theorem by showing that if $f$ is a mapping of a complete metric space $(X,d)$ into itself and $f^N$ is a contraction mapping for some positive integer $N$, then $f$ has precisely one point.
Exercise 3:Let $f:[a,b]\to[a,b]$ be differentiable.Then $f$ is a contraction if and only if $f'(x)\leq r<1,\forall x\in[a,b]$
To say $f$ does not satisfy Contraction Mapping Theorem,I need to show $f$ is not a contraction map. $f'(x)=-\sin x$. If I use exercise 3,since $-1 \leq -\sin x\leq1,\forall x\in \mathbb{R}$ can I say $\nexists r\in(0,1)$ such that $f'(x)\leq r< 1$ So $f$ is not a contraction?
How should I proceed?How to find $N$ such that $f^{N}$ is a contraction?