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

*Posted by Akhil Mathew in analysis, functional analysis, MaBloWriMo.*

Tags: isometries, linear maps

trackback

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

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 , thenAlso 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 .

## Comments»

No comments yet — be the first.