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. ${A_0}$

Henceforth, fix ${x,y \in X}$. We define ${A_0}$ to be the set of points in ${X}$ which are midway between ${x,y}$; this means that ${w \in A_0}$ iff $\displaystyle |w-y| = |w-x| = \frac{1}{2} |x-y|.$

Evidently the midpoint ${z}$ is in ${A_0}$. ${A_1}$

Now ${A_0}$ is symmetric with respect to ${z}$: ${2z - A_0 = A_0}$. This is easily checked, as follows: if ${w \in A_0}$, then $\displaystyle |(2z-w)-y| = |x-w| = \frac{1}{2}|x-y|, etc.$

Also ${A_0}$ is bounded, ${d_0:=\mathrm{diam}(A_0) <\infty}$. Then we can consider the set of points ${w' \in A_0}$ whose distance from any other point of ${A_0}$ is at most ${\frac{d}{2}}$. An example is ${z}$ ${A_0}$ is symmetric w.r.t. ${z}$. We can repeat this inductively; see below. ${A_n}$

With this we need a general lemma:

Lemma 2 Let ${A \subset X}$ be a bounded set symmetric with respect to ${z}$, with diameter ${d}$. Then the set ${A'}$ of points ${w \in A}$ such that for ${w' \in A}$ arbitrary, ${|w-w'| \leq \frac{d}{2}}$, satisfies the following properties:

1. ${\mathrm{diam}(A') \leq \frac{\mathrm{diam}(A)}{2}}$.
2. ${z \in A'}$.
3. ${A'}$ is symmetric with respect to ${z}$.

1 is easily checked from the definition of ${A'}$. 2 follows from symmetry: if ${w' \in A}$, then $\displaystyle 2|z - w'| = |2z - 2w'| = |(2z - w') - w'| \leq d.$

For 3, suppose ${w \in A', w \in A}$; then $\displaystyle | (2z - w') - w| = |(2z-w) - w'| \leq \frac{d}{2}$

since ${2z-w \in A}$.

So we have defined with this notation ${A_1 = (A_0)'}$, and inductively set ${A_n := (A_{n-1})'}$; each contains ${z}$ and is symmetric with respect to it. 1 in the lemma implies that ${\bigcap A_i = \{z\}}$, so we’ve completed the proof by the strategy discussion. Indeed, it is clear from the description that ${M}$ maps ${A_i}$ bijectively to the corresponding sets ${B_i}$ between ${M(x),M(y)}$.