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 of a normed linear space onto another normed linear space with is linear. *

We needed the theorem to show that a distance-preserving map between Riemannian manifolds is an isometry. Recall that ‘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

for all .

Indeed, if this is the case, then by , we get by induction. So

for all . By density of the dyadic fractions and continuity, we find for all . Also (1) and what’s already proved imply , so , which proves linearity.

**Strategy of the proof **

The idea is to use a purely norm-theoretic way of describing the midpoint of . This must be reflected by , so it will prove (1). In particular, given , we will define sets with , with the sets defined solely in terms of the norm structure on . It will thus follow that maps these sets onto their analogs on —which will prove (1) and the theorem.

** **

Henceforth, fix . We define to be the set of points in which are **midway between **; this means that iff

Evidently the midpoint is in .

** **

Now is **symmetric with respect to **: . This is easily checked, as follows: if , then

Also is bounded, . Then we can consider the set of points whose distance from any other point of is at most . An example is — is symmetric w.r.t. . We can repeat this inductively; see below.

** **

With this we need a general lemma:

Lemma 2Let be a bounded set symmetric with respect to , with diameter . Then the set of points such that for arbitrary, , satisfies the following properties:

- .
- .
- is symmetric with respect to .

1 is easily checked from the definition of . 2 follows from symmetry: if , then

For 3, suppose ; then

since .

So we have defined with this notation , and inductively set ; each contains and is symmetric with respect to it. 1 in the lemma implies that , so we’ve completed the proof by the strategy discussion. Indeed, it is clear from the description that maps bijectively to the corresponding sets between .

November 22, 2009 at 1:18 pm

[…] functional analysis. Tags: isometries, linear maps trackback I first posted this entry at Climbing Mount Bourbaki, where I have continued the MaBloWriMo series into topics in Riemannian geometry such as the […]

November 11, 2011 at 10:56 am

Nice post! Just a couple of didactic remarks: it is crucial that be surjective for this proof to work, because one can cook up non-surjective isometries which are not affine. For instance, the map defined by is an isometry if is equipped with the max (or ) norm. Also, the proof simplifies greatly if the codomain is strictly convex (no sphere contains a line segment), because already consists of a single point. In this case, surjectivity is not even needed! As an example, the spaces for are strictly convex.

November 11, 2011 at 6:38 pm

Thanks! These are interesting comments; I haven’t thought about this material in a while, and may use this in my section for the analysis class I’m TAing.

November 11, 2011 at 11:02 am

Also — you probably know this, but since you brought it up — the Mazur of this post is this guy. I’ll bet he and Ulam thought up this proof during a session at the Scottish Café in Lwow, in between bouts of drinking strong coffee and brandy.