Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 2 / 12F / a / Solution

A map $f:A\to \mathbb{R}$ is Lipschitz continuous with Lipschitz constant $L$ whenfor all $y,z\in A$.

Fix $y_0\in A$. The Lipschitz inequality and the triangle inequality implywhile choosing $y=y_0$ gives a finite upper bound. Hence the infimum defining $F(x)$ is a real number.

If $x\in A$, then $f(x)\le f(y)+Ld(x,y)$ for every $y\in A$, while $y=x$ gives the reverse inequality after taking the infimum. Thus $F(x)=f(x)$. For arbitrary $x,z\in X$,Taking infima gives $F(x)\le F(z)+Ld(x,z)$; interchanging $x,z$ givesThis is the McShane extension theorem: $F$ extends $f$ without increasing its Lipschitz constant.