I have announced that superreflexivity is equivalent to being isomorphic to -UC for some . It implies that superreflexivity is equivalent to being isomorphic to a space which is uniformly non square. Indeed, James’ theorem states that non reflexive implies uniformly non square. The latter is a super property, the former is isomorphism invariant, so isomorphic to a space which is uniformly non square implies superreflexive. Conversely, uniform convexity implies non square.

**1. End of equivalence of superreflexivity and being isomorphic to UC/US **

There remains to prove that superreflexive implies

(*) , such that for all dyadic martingales convergent in ,

This relies on ideas due to James and independently to Gurarii-Gurarii. The keyword is *basic sequence*. A sequence in a Banach space is -basic if for all , for all sequences of coeffcients ,

For instance, a martingale is a -basic sequence.

** 1.1. Step 1 **

Theorem 1 (Gurarii-Gurarii, James)The following are equivalent:

is superreflexive., , such that for all , all -basic sequences in (and any quotient of ) such that, , such that for all , all -basic sequences in (and any quotient of ) such that

*Proof:* 12. Blocking argument. -basic is stable under blocking, i.e. grouping consecutive terms of the sequence in blocks. Thus the best constant in the wanted inequality (for -basic sequences) is submultiplicative. Either this constant decays polynomially (and we are done) or it stays . In that case, we have a sequence with and . One can choose a biorthogonal sequence (i.e. ) such that . Set . Then or depending wether or not, and , this contradicts superreflexivity.

Observe that if is a closed subspace, . So the result holds also in quotients of .

23. By duality.

31. If is non reflexive, there exists sequences , such that or depending wether or not. We can assume that is -basic. Thus

cannot be bounded by , which is .

** 1.2. Step 2 **

Prove that superreflexive implies superreflexive.

We already know hat superreflexive implies -convex. This easily implies that the average over admissible

(since the constant is nearly , fixed), which passes through integration.

** 1.3. Step 3 **

Apply Step 1 to .

** 1.4. Step 4 **

See the notes.

**2. Relationship between UC and type, cotype **

Recall that -UC implies cotype . Converse fails ( and have cotype ). We shall see that the converse holds in the family of UMD spaces.

** 2.1. Interpolation **

Theorem 2 (Pisier, Xu)Let . There exists a Banach space such that , , isomorphic to , and has the same type and cotype as , except if .

We use (Lions-Peetre) real interpolation, , with .

Definition 3Let , be Banach spaces with continuous inclusions in a third Banach space . Define

On , put the norm

and complete to get .

Example 1is aLorentzspace.

If , coincides with the ordinary Lebesgue space.

If , is the weak space.

Indeed, , where is the increasing rearrangement of , and by definition,

If ,

by Hardy inequality applied to .

Example 2Similarly, is aLorentz sequence space.

If , coincides with the ordinary sequence space.

If , is the weak space.

Proposition 4More generally, if . Also, for all ,

if ,

if .

This follows from the Hölder-Minkowski inequality if .

** 2.2. Comparison spaces **

Here is a -parameter family

Definition 5For a sequence of real numbers, let

Replacing the norm in the definition by , we get .

was discovered by James. This was the first example of a Banach space such that . contains ‘s uniformly, so has trivial type and cotype. However, the dual has cotype or depending wether or .

Lemma 6

More generally, .

*Proof:* The trick for is a property of for integer (this suffices for sequence spaces). For ,

The main point of interpolation theory is the continuity of operators. If and has norm , then so does .

Apply this to , where . This gives

i.e. .

** 2.3. Reiteration **

This is the abstract version of Marcinkiewicz’ theorem in harmonic analysis.

Proposition 7Let , . Then

where does not depend on the ‘s.

Corollary 8Let . Then where .

*Proof:* Apply reiteration lemma to inclusions . Get

i.e.

Lemma 9Let . The family is -convex in the following sense: . It is -concave in the following sense: .

*Proof:* By definition, . Since sup of integral is less than integral of sup, , . Interpolate and apply Hölder-Minkowski.

Concavity is by duality, since .

** 2.4. Proof that has type for all **

Let . Consider thet map , . For ,

Khintchin implies that is bounded. Up to constant,

Taking expectation gives the type inequality.