**1. Factorization **

Theorem 1Let have type 2 and cotype . Let be a subspace of . Every factors through a Hilbert space.

Corollary 2One recovers Kwapien’s theorem: if has type and cotype , then is isomorphic to a Hilbert space.

Corollary 3Let 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 4There 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 .

** 1.1. **

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 5Weak , 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 1Given , has this property.

The previous theorem implies

Proposition 7Let 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 9Let be bounded and linear. The following are equivalent:

factors through for some probability measure .For all , .

Theorem 10If has cotype , and is bounded and linear, then factors through for some probability measure .

** 1.4. Applications **

This is what motivated Rosenthal in 1972.

Corollary 11A 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 12Let (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.

** References **