Today’s will be a relatively short post, and primarily algebraic. I had mentioned a couple of days back the following result:

Theorem 1 (Mazur-Ulam) An isometry ${M: X \rightarrow X'}$ of a normed linear space ${X}$ onto another normed linear space ${X'}$ with ${M(0)=0}$ is linear.

We needed the theorem to show that a distance-preserving map between Riemannian manifolds is an isometry.   Recall that $M$‘s being an isometry means that it preserves distances between points.

Apparently the theorem has no relation to Barry Mazur, but to another Mazur.

I will follow Lax’s Functional Analysis in the proof.

Reduction

It is enough to prove that

$\displaystyle M\left( \frac{x+y}{2}\right) = \frac{1}{2}(M(x) + M(y))\ \ \ \ \ (1)$

for all ${x,y \in X}$.

Indeed, if this is the case, then by ${M(0)=0}$, we get ${M\left( \frac{1}{2^n} x\right) = \frac{1}{2^n} M(x)}$ by induction. So

$\displaystyle M\left( \frac{k}{2^n} x\right) = \frac{k}{2^n} M(x)$

for all ${k}$. By density of the dyadic fractions and continuity, we find ${M(kx)=kM(x)}$ for all ${k \in \mathbb{R}^+}$. Also (1) and what’s already proved imply ${M(x+y) = M(x)+M(y)}$, so ${M(-x)=-M(x)}$, which proves linearity.

Strategy of the proof

The idea is to use a purely norm-theoretic way of describing the midpoint of ${x,y}$. This must be reflected by ${M}$, so it will prove (1). In particular, given ${x,y \in X}$, we will define sets ${A_0, A_1, \dots}$ with ${\bigcap A_i = \{ \frac{1}{2}(x+y)\}}$, with the sets defined solely in terms of the norm structure on ${X}$. It will thus follow that ${M}$ maps these sets onto their analogs on ${X'}$—which will prove (1) and the theorem. (more…)