Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 2 / 10E / b / Solution

The strict distance inequality implies $d(fx,fy)\le d(x,y)$ for all points, so $f$ is Lipschitz continuous. Henceis a continuous function, since $|g(x)-g(y)|\le d(fx,fy)+d(x,y)\le 2d(x,y)$. On the nonempty compact metric space, $g$ attains its minimum at some $p$.

If $fp\ne p$, the strict distance inequality applied to $fp,p$ givesa contradiction. Thus $fp=p$. If $p,q$ were different fixed points, then $d(p,q)=d(fp,fq)<d(p,q)$, again impossible. There is exactly one fixed point. This proves strict distance decrease on a compact metric space without assuming a uniform contraction constant.