Scribe: Yanqi Qiu.
Let , viewed as a group and a measure space. Let denote the -th coordinate.
Definition 1 Define the Rademacher subspace as the linear span of functions of the form , , .
Note that has cotype iff the natural map is continuous. Or dually, that the map is continuous. This gives an elegant proof of the fact that has cotype .
Definition 2 Say that is -convex if there exists such that for all , the orthogonal projection has norm .
Thanks to Khintchin’s inequality, one could replace with in this definition.
Example 1 If , is -convex.
Theorem 3 (Pisier) If has type , then is -convex.
1.2. Proof of Pisier’s theorem
One uses Fourier analysis on . Every function can be written uniquely
where is the Fourier-Walsh basis, . The orthogonal projection of onto is
It can be recovered, by a contour integration, from the holomorphic function
Thus one needs estimates on this function or on its semi group version
Note that is a convex combination of contractive commuting projections.
Lemma 4 Let . Assume that is a convex combination of contractive commuting projections. Let be the best constant in the inequality
Then for all ,
Note that . Applying the Lemma to leads to the following estimate on the generator of the semi-group (),
The contour integral
gives the estimate
Let us take the opportunity to discuss a related inequality. It is based again on analysis on the discrete cube.
Let . Then is a nonnegative function on . Let denote convolution with . Then .
Theorem 5 (Bonami, Beckner) If and , then
This boils down to the inequality
which is proven by brute force, see Johnson and Lindenstrauss’ article in the Handbook in Geometry of Banach Spaces.
This property tensorizes: Applying BB inequality times (combining it with Hölder-Minkowski’s inequality) gives a BB inequality on . Since ,
for every Banach space , so BB inequality is valid .
2.1. Proof of Kahane’s inequality
Here is a simple proof (due to Christer Borrell) of Kahane’s inequality from BB. Let . Then , so BB implies (in particular)
with . The constant in Kahane’s inequality is . This implies a sharpening of a result of Landau-Shepp-Fernique,