Theorem 1 Let have type 2 and cotype . Let be a subspace of . Every factors through a Hilbert space.
Corollary 2 One recovers Kwapien’s theorem: if has type and cotype , then is isomorphic to a Hilbert space.
Corollary 3 Let be a Hilbert space isometrically embedded in . Assume has type . Then is complemented in .
The proof of the factorization theorem relies on the following Lemma, proved yesterday.
Lemma 4 There exists a measure on the unit sphere of such that for all ,
Proof: Let be the set of finite dimensional subspaces of . For , let be equal to on and to elsewhere. Then has cotype , so there is a corresponding measure on the unit sphere of . Define
For , there is a unique such that . Set . The ultraproduct is again a Hilbert space. The collection of maps defines a map . Let be the closure of the image .
Rosenthal 1972, Nikishin 1970, 1972, Maurey, revisited by Pisier in 1985. Rosenthal and Maurey look for factorizations of maps through multiplication operators . Nikishin considers maps , removes a set of small measure so that map factors to weak , .
Definition 5 Weak , denoted by , is the set of functions such that
The best is .
If , a function in belongs to ,
Therefore, for , an alternative norm on is
More generally, the Lorentz space is the set of functions whose rearrangement satisfies
Theorem 6 (Pisier) For , the following are equivalent:
- There exists a probability density such that for , the set is bounded in .
- such that for every and ,
Proof: For every density , let be the set where is achieved by . Estimate
sum up and use Hölder inequality and the fact that is a probability measure.
Conversely, assume for simplicity that . We shall maximize the integral functional on the set ,
Assume that achieves its maximum at some and (rescale if necessary) that the maximum equals . Let , , ,
this show that the set is bounded in .
What if the maximum is not achieved ? Use Ekeland’s -minimization. If is a complete metric space, and bounded below. Then, , there exists an such that for all ,
Proof is by transfinite induction.
Pick an -maximum . Then for all , , ,
and proof continues as before…
1.2. Sublinear maps
Say an operator is sublinear if and .
Example 1 Given , has this property.
The previous theorem implies
Proposition 7 Let be sublinear and satisfy . The following are equivalent:
- factors through where the second arrow is multiplication by , .
- such that for every ,
Theorem 8 (Pisier 1985) Let . Let be sublinear and bounded. If has type , then factors through .
Proof: One expresses in terms of max.
1.3. Back to linear maps
In the linear case, one case dualize. Maps dualize to . If is separable, the image is contained in some .
Proposition 9 Let be bounded and linear. The following are equivalent:
- factors through for some probability measure .
- For all , .
Theorem 10 If has cotype , and is bounded and linear, then factors through for some probability measure .
This is what motivated Rosenthal in 1972.
Corollary 11 A reflexive subspace of cannot contain for all .
If -embeds in , then the basis functions have nearly disjoint supports (there are disjoint subsets such that . According to factorization theorem, up to a change of measure, for some .
Let be the circle. Let be a family of characters. This defines a set which is translation invariant. Since the change of measure should be translation invariant, there is no change of measure. This such that (Ebenstein-Bachelis).
1.5. Comparison of Gaussian and Bernoulli averages
Theorem 12 Let (resp. ) be independant Gaussian (resp. Bernoulli) variables. The following are equivalent:
- such that for all , .
- has non trivial cotype.
Note that the reverse inequality always holds:
Proof: If has trivial cotype, then it -contains for which comparison fails. Conversely, factorize the map , .
On monday, -convexity.