**1. Super-reflexivity **

** 1.1. Uniform convexity **

A Banach space is *uniformly convex* if the midpoint of two points on the unit sphere which are sufficiently far apart is deep inside the ball.

**Example**. Hilbert spaces, spaces with are uniformly convex.

Uniform convexity carries to direct sums .

** 1.2. Super-reflexivity **

Uniform convexity is not stable under renorming.

Definition 1A Banach space is super-reflexive if it admits an equivalent norm which is uniformly convex.

Here is an equivalent definition.

Theorem 2 (Enflo)A Banach space is super-reflexive iff every one of its ultrapowers is reflexive.

In other words, every space which is finitely representable in it is reflexive.

Recall that, given an ultrafilter on an index set , the *ultraproduct* of a family of Banach spaces is the quotient of by the subspace of sequences that tend to 0 along . This is a Banach space. If all , the ultraproduct is called an ultrapower.

**2. Pestov’s theorem **

Here is Pestov’s main theorem.

Theorem 3 (Pestov)Let be a metric space. Let be a locally finite group acting on by isometries. Assume that admits a coarse/uniform embedding into a Banach space . Then there is a coarse/uniform embedding from to some ultrapower of some , and the action of on extends to an action by affine isometries on the affine span of .

** 2.1. Proof **

Let be the initial coarse embedding. Start with a finite subgroup of . Define

Then has the same expansion and compression moduli as . acts on via left translations, and is equivariant for this action.

Next we glue all maps together. Let denote the set of all finite subgroups of . Because is locally finite, there exists an ultrafilter on such that for all , the set

is in . Indeed, the intersection of two such subsets

and the subgroup generated by two finite subgroups is finite again. So we are dealing with a filter basis.

Fix an origin . Let be the ultraproduct over of spaces , based at instead of 0 (this makes an affine space instead of a vectorspace, and so much for the ultraproduct). Define as the sequence . It is well defined and has the same expansion and compression moduli as .

Given , observe that -almost surely,

is well defined, and this provides an isometry of . Isometries of subsets of Hilbert spaces extend to affine hulls, so the action extends.

As defined, is not quite an ultrapower. For every , embeds non canonically into , so

** 2.2. Application to uniform embeddability of Urysohn space **

Let be a uniform embedding. Let be a dense locally finite subgroup of . Pestov’s theorem implies that there is a uniform embedding of to a reflexive Banach space , equivariant with respect to an affine isometric action of . is a homeo onto . The topologies of pointwise convergence induced on by the two actions of coincide, so, by density, the action on the affine span of extends to . I.e. we get a continuous group homomorphism

Since there is an injective equivariant map, it is injective.

Now we use the fact that contains a subgroup isomorphic to . So we get a linear isometric representation of on a reflexive Banach space. According to Megerelishvili’s theorem, this must be the trivial representation. So acts faithfully by translations. Contradiction, since is non abelian.