Today, I want to prove equivalence of superreflexibility and super RNP. Tomorrow, I will continue with the equivalence of these properties with the following statement.

*There exist , , such that for all square integrable martingales with increments . *

* *

**1. Super properties **

Definition 1Let , be Banach spaces. Say isfinitely representable inif , every finite dimensional subspace can be embedded in with distorsion .

Example 1is finitely representable in , whereas these spaces

Definition 2Let be a property of Banach spaces. We say that hassuper-if every space finitely representable in has .

Say is a super property if super-.

Example 2Type (and cotype ) are super properties.

** 1.1. Ultraproducts **

Definition 3Let be a family of Banach spaces. Let be an ultrafilter on . Theultraproduct

where in the restricted product one considers bounded sequences , and one mods out sequences such that . defines a norm on it.

When are all the same, one speaks of the *ultrapower* .

Proposition 4is finitely representable in iff there exist an index set and an ultra embeds isometrically in the ultraproduct .

Ultraproducts are nice for clearing out ‘s from definitions and statements. They hide estimates, which may be a dommage since estimates are at the heart of functional analysis and its applications.

Proof requires two lemmas. The first one is easy.

Lemma 5Let be a finite dimensional Banach space. Let , . Let be a linear map. Assume is -isometric on an -net of the unit sphere of . Then is -isometric on the whole of .

Lemma 6is finitely representable in .

*Proof:* Let be finite dimensional. Pick a finite -net in its unit sphere. Let be a basis of . View as the limit of a bounded sequence . By definition, for all ,

so for large enough , is an -isometric embedding on , so is nearly isometric on .

*Proof:* of Proposition 4. In one direction, by transitivity of finite representability. Conversely, if is finitely representable in , take the set of pairs where runs through finite dimensional subspaces of . For each , choose a -isometric embedding of in . Then for every , the sequence represents a point , and is isometric.

Proposition 7If a property is stable under isomorphisms and subspaces, so does super-.

*Proof:* Let have super-, let be isomorphic to . Then is isomorphic to . is finitely representable in , so it has . So does . Let be finitely representable in . Then embeds isometrically in , so it has . This proves that has super-.

** 1.2. Equivalence of super reflexivity and super RNP **

The following theorem describes non reflexivity as a gap between bounded variation and boundedness.

Theorem 8 (Ptak, Milman-Milman)For a Banach space, the following are equivalent:

is not reflexive.For all , there exists sequences and such thatThere exists , there exists sequences and such thatFor all , there exists a sequence such that for all finitely supported sequences ,There exists , there exists a sequence such that for all finitely supported sequences ,For all , there exists a sequence such that for all finitely supported sequences ,There exists , there exists a sequence such that for all finitely supported sequences ,The inclusion factors through .

*Proof:* 68: Extend each functional

to by Hahn-Banach. This gives a map .

46: Abel summation.

24: Because .

12: If is reflexive, one can assume that tends to in and tends to in , then is the limit of and , contradiction.

12 is the serious part. We use the following fact (local reflexivity, related to the fact that is finitely representable in for all ): For all , for all , for all , there exists such that and for all , .

Assume is not reflexive. There exists with norm and . Since , there exists such that . Then above fact yields such that and . Since , by Hahn-Banach, there exists such that and . Again, there is such that and , and so on… Observe that in addition, we can require that for in any prescribed finite set (replace by , and so on). It follows that, given , we can require that for all sequences ,

Indeed, bad choices would live in a finite dimensional space, the unit ball of which can be discretized. So the argument gives 4. as well.

Proposition 9Super reflexivity is equivalent to super RNP.

*Proof:* Since reflexivity implies RNP, super reflexivity implies RNP. Conversely, need a lemma.

Lemma 10Assume is not reflexive. Then there exists a space , finitely representable in , such that is the completion of with respect to a norm such that

*Proof:* of Lemma. The finite dimensional analogue of this would be

which is one of the above equivalent properties. Now take an ultralimit in which finite sums converge to integrals. By construction, is finitely representable in .

Now any as in the Lemma fails to have RNP. Indeed, the following martingale diverges. Given an interval , let and be its left and right halves. Given a sequence , define recursively . Start with , set . Then . is supported on , so .

In Lemma 10, one can even require (with a different )

We shall need this tomorrow when proving the following

Theorem 11 (James)non reflexive implies that contains squares, i.e. for all , there exist unit vectors , such that .

It follows that span is -isometric to . One cannot go beyond dimension . Instead, existence of higher-dimensional cubes has to do with type, see Maurey’s course.

** References **