**1. James’ Theorem **

** 1.1. Back to equivalence of superreflexivity and super RNP **

Definition 1Let

Here, sequences have length .

The equivalence of superreflexivity and super RNP can be formulated as follows.

Theorem 2For a Banach space , the following are equivalent:

is superreflexive.tends to as tends to .tends to as tends to .

*Proof:* Note that properties 2 and 3 are super properties. Go to ultraproducts where finite sequences become infinite sequences, and apply Ptak’s or Milman-Milman’s theorem.

Linial: is there a quantitative connection between and ? Answer: the proof gives something…

Corollary 3is superreflexive iff is superreflexive.

Indeed, property 2 is symmetric in and (up to a change in notation), . Since superreflexive implies reflexive, .

** 1.2. Variant of James’ Theorem **

James’ Theorem

Theorem 4 (James)In is non reflexive, then , there exist , such that .

has a variant

Theorem 5 (James)In is non reflexive, then , , there exist such that for all choices of signs of the form (call themadmissible), .

This suggests the following

Definition 6Say is -convex is there exists such that , there exist such that for all admissible choices of signs , .

This is a super property, so it leads to

Theorem 7is superreflexive iff is -convex.

*Proof:* If is -convex, then is reflexive. But -convexity is super, so is superreflexive. Conversely, assume that is not -convex. Then in some space finitely representable in , there exists a bounded sequence such that for all admissible choices of signs,

By Hahn-Banach, there exist unit functionals such that

This implies that or according wether or . This shows that is not reflexive.

Remark 1There exists finitely representable in which is the completion of equipped with a norm that satisfies

The unit ball of this space contains an infinite -separated dyadic tree.

** 1.3. Proof of James’ Theorems \ref **

and 5}

The proof we shall give is due to ergodic theorists Brunel and Sucheston around 1975. They realized that the following two properties clarify the proof a lot.

Definition 81. Say a sequence issubsymmetricif for every sequence and ,

2. Say a sequence isadditiveif for every ,

We are given a sequence such that

Call this a *James sequence*. We start with the end of the proof.

**Step 3**. Assuming that is subsymmetric and additive, we proceed to extracting squares.

Let

terms, which we symbolically denote by . Similarly, , i.e.

Let . Then , , so and do the job provided tends to infinity.

Using sequences , and , one gets 3 vectors such that all , , have norm close to (but not other combinations of signs). And it works the same for -tuples.

If stays bounded, set .

**Claim**: for all .

Indeed, when summing, signs create cancellations. It follows that behaves unconditionally, i.e. for arbitrary signs ,

So the closure of span is isomorphic to . So , and in particular, , is coarsely finitely representable in . This implies that is finitely representable in , i.e. is not superreflexive.

**Step 1**. We can upgrade the given sequence to be subsymmetric.

Consider all . By the James assumption, they stay bounded. In some ultraproduct, limits exist, leading to a norm on . The completion is a Banach space finitely representable in where the canonical basis is James and subsymmetric.

**Step 2**. We can upgrade the subsymmetric sequence to be subsymmetric and additive.

In the same manner, consider all . By subsymmetry, they form a nondecreasing family of norms on . The limiting Banach space is finitely representable in , and the canonical basis is James, subsymmetric and additive.

**2. Uniform convexity and smoothness **

Definition 9Themodulus of convexityof a Banach space is

Themodulus of smoothnessof is

Say isuniformly convexif for all . Say isuniformly smoothif .

Remark 2is UC iff is US.

** 2.1. UC versus superreflexivity **

Theorem 10 (David Milman)UC implies superreflexive.

*Proof:* Let . Let be a sequence (net) converging to . Then tends to . For the same reason, tends to . UC implies So UC implies reflexive. Since UC is super,

Note that UC is not isomorphism invariant. Answering a question of James, Enflo proved that

Theorem 11 (Enflo)is superreflexive iff is isomorphic to a UC space.

** 2.2. -uniform convexity **

We are heading towards a sharpening of Enflo’s theorem.

Definition 12Say is-UCif such that for all , .

Say is-USif such that for all , .

Example 1is -UC for , -UC for .

is -US for , -US for .

This is known as Clarkson inequality.

Theorem 13 (Pisier)The following are equivalent:

is superreflexive.such that is isomorphic to a -UC space.such that is isomorphic to a -US space., such that is isomorphic to a space which is both -UC and -US.

**Open problem**: can one have in 2 and 3 ?

Lemma 14Given and , the following are equivalent:

There exists a norm on such that andFor every -valued dyadic martingale such that

2. implies the same statement for all martingales (but with a different ).

Example 2Take . Then 1. is an equivalent definition of -UC.

*Proof:* Dyadic martingale means that and .

Define

infimum over all dyadic martingales starting at . To prove triangle inequality, given martingales starting at and , one uses the martingale starting at which jumps to and continues with the martingale.