1. James’ Theorem
1.1. Back to equivalence of superreflexivity and super RNP
Definition 1 Let
Here, sequences have length .
The equivalence of superreflexivity and super RNP can be formulated as follows.
Theorem 2 For 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 3 is 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
has a variant
This suggests the following
Definition 6 Say 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 7 is 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 1 There 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
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 8 1. Say a sequence is subsymmetric if for every sequence and ,
2. Say a sequence is additive if 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.
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 9 The modulus of convexity of a Banach space is
The modulus of smoothness of is
Say is uniformly convex if for all . Say is uniformly smooth if .
Remark 2 is 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 12 Say is -UC if such that for all , .
Say is -US if such that for all , .
Example 1 is -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 14 Given and , the following are equivalent:
- There exists a norm on such that and
- For every -valued dyadic martingale such that
2. implies the same statement for all martingales (but with a different ).
Example 2 Take . Then 1. is an equivalent definition of -UC.
Proof: Dyadic martingale means that and .
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.