Uniform convexity and related metric geometric properties of Banach spaces
Recently, there has been a revival of type and cotype considerations, in nonlinear settings. This started with Enflo and Bourgain, and was revived by Naor and coauthors.
Our assignment is type and cotype. We split the contents into two parts: classical material (Maurey) and martingales (Pisier). I am preparing a book, parts of it will be posted on my webpage.
1. Martingale type versus type
Definition 1 Say a Banach space has martingale type if
has martingale cotype if
Type is , cotype is . Kwapien’s theorem states that only Hilbert space as typecotype. Classical type and cotype amounts to martingales with independant increments. The two notions differ. For instance,
has a martingale type is super-reflexive
has a type does not contain uniformly.
The best such that has a martingale type is denoted by . The best such that has a type is denoted by . There exists such that and .
1.1. Plan of course
- -valued martingales and RNP
- Uniform convexity and smoothness
- Property UMD: if it holds, all subtleties disappear.
- Nonlinear characterizations of superreflexifity
2. -valued martingales and RNP
Definition 2 Let be a Banach space, a measure space. A function is Bochner measurable if it is the pointwise limit of functions that take only finitely many values. Let
2.1. Elementary properties
Step functions, and the algebraic tensor product , are dense in .
Definition 3 Let be -algebras. Assume that .
A sequence of functions in is a martingale is for all , the conditional expectation
Conditional expectation is a contraction. So does . Indeed, for a.e. ,
Example 1 Let . Set . Then us a martingale, and converges to in .
Indeed, the union of is dense in .
2.3. Doob’s maximal inequalities
Proposition 4 (Doob) Let be nonnegative random variables, is -measurable. Assume that is a submartingale, i.e. for all ,
be the maximal function. Then
It follows that for all ,
Proof: Relies on stopping times. A stopping time is a -valued random variable such that for all , .
If is a martingale in ,
is again a martingale.
Let . Then
This implies that
Integrate this and use submartingale assumption.
Applying Proposition to yields
2.4. The fundamental property of martingales
Theorem 6 Let , . Then converges a.e. and in .
Proof: Apply Doob’s inequalities to the (shifted) martingale . Its -norm tends to so a.e. converges to .
2.5. Scalar case
Proof: By reflexivity, boundedness in implies existence of weakly converging subsequences, their weakly converges as well, so , weak limit.
In case , introduce the stopping time . A.e. and convergence hold on all . By Doob’s inequality, tends to , so .
2.6. Vector valued case
Theorem 7 does not extend to -valued martingales.
Example 2 , , coordinates, , . Then but is constant.
Example 3 , . Then but is constant.
Remark 1 bounded in implies that the norm converges a.e.
Indeed, the norm of is a submartingale. Nonnegative submartingales which are bounded in converge a.e.
3.1. RNP and convergence of martingales
The Radon-Nikodym Property solves this difficulty. It will be defined soon.
- has Radon-Nikodym Property.
- uniformly integrable in converges in and a.e.
- bounded in converges a.e.
- bounded in converges in and a.e.
- For all , the unit ball does not contain -separated infinite trees.
A -separated tree is a -valued martingale such that
- For all , takes finitely many values.
- For all and all , .
The martingale assumption merely means here that the set of values increases, each value being the average of values appearing in the next generation. The full picture is a tree where children are at least -far away from their father, and average to their father.
The issue here is wether -separates trees need to blow up or can be confined in a ball.
3.2. Radon-Nikodym Property
Definition 9 A valued measure on is a map such that there exists such that for all ,
Say is differentiable if there exists such that for all ,
Say that has RNP if every -valued measure is differentiable.
Example 4 , do not have RNP. Reflexive spaces, separable duals have RNP (Stegall).
But there exist RNP spaces which are not separable duals (Bourgain).
3.3. Proof of Theorem \ref
12: Given , define . Then is a -valued measure. RNP provides us with . For , , so .
41: ( admitted here). We can take standard, . Let be a -valued measure. Let be the dyadic filtration. Given , let be the length -interval containing , and set
Then the -limit of serves as a density for .
3.4. Open problem
It is known that RNP implies the Krein-Milman property (a closed convex set is the closed convex hull of its extreme points).
Question: Is the converse true, i.e. KMP implies RNP ?
Among many works on this question, here is a nice result. I like it because the proof uses martingales.
Theorem 10 (Edgar) Let be closed bounded convex in a separable space . Assume that has RNP. For all , there exists a function with values in the set of extremal points of , and whose expectation equals .
Here are facts used in the proof.
- There exists a continuous strictly convex function on .
- Convergence in holds for general directed index sets, not inly .
Proof: Assume is not extreme, . Set (resp. ) with probability . If is extreme, keep . If is not extreme, let pick at random… Use transfinite induction. For a limit ordinal , let . RNP garantees that the limit exists. Convergence show that there exists an ordinal beyong which become stationary. For every continuous strictly convex function on , the nonincreasing net converges, so for , . This implies that a.e. , i.e. takes its values in extremal points.