Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 4 / 1A / Solution

The strict distance inequality also gives $d(fx,fy)\le d(x,y)$ when $x=y$, so $f$ is Lipschitz continuous with constant at most one, hence continuous. Put $g(x)=d(x,f(x))$. The triangle inequality givesThus $g$ is also Lipschitz continuous and continuous.

Assume the compact metric space is nonempty, as needed for existence of a point. By the extreme value theorem, $g$ attains its infimum at some $p$. If $h=g(p)>0$, then $p\ne f(p)$, and strict distance decrease givescontradicting minimality. Since distances are nonnegative, $h=0$ and $f(p)=p$. If two different fixed points $p,q$ existed, then $d(p,q)=d(f(p),f(q))<d(p,q)$, another contradiction. Therefore there is exactly one fixed point. This is strict distance decrease on a compact metric space; no uniform contraction factor less than one was assumed.