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…)