Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 3 / 11G / Solution

A map $T:X\to Y$ between metric spaces is a contraction mapping if some $q<1$ satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$.

The Banach fixed-point theorem states that a contraction from a nonempty complete metric space to itself has a unique fixed point. Indeed, for $x_{n+1}=T{x}_n$,so the geometric-series estimate makes $(x_n)$ Cauchy. Completeness gives a limit $x_{\ast }$, continuity gives $T{x}_{\ast }=x_{\ast }$, andproves uniqueness.

Every solution of $x=\cos x$ lies in $[-1,1]$. The cosine maps this complete interval into itself and, by the mean value theorem, has Lipschitz constant at most $\sin 1<1$ there. It therefore has exactly one real fixed point.

The mean value inequality says that on a convex domain, a uniform derivative bound $\Vert Df\Vert \le M$ implies $\Vert f(x)-f(y)\Vert \le M\Vert x-y\Vert$. Equip ${\mathbb{R}}^2$ with the maximum norm and takeFor $(x,y)\in D$, elementary cosine bounds giveso $f(D)\subseteq D$. The maximum absolute row sum ofis at most $2\sin (1/2)<1$. The mean value inequality makes $f$ a contraction on the complete set $D$, so it has a fixed point.