## A theorem of Mazur-Ulam on isometric maps of vector spaces November 22, 2009

Posted by Akhil Mathew in analysis, functional analysis, MaBloWriMo.
Tags: ,

I first posted this entry at Climbing Mount Bourbaki, where I have continued the MaBloWriMo series into topics in Riemannian geometry such as the Cartan-Hadamard theorem.  This particular material came up as part of the proof that distance-preserving maps between Riemannian manifolds are actually isometries.  However, the style of the entry seemed appropriate for this blog, so I’m placing it here as well.

The result in question is:

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.

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 ${M(x),M(y)}$.