Scribe: Yanqi Qiu.

**1. -convexity **

** 1.1. Definition **

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

**2. Hypercontractivity **

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,

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## About metric2011

metric2011 is a program of Centre Emile Borel, an activity of Institut Henri Poincaré, 11 rue Pierre et Marie Curie, 75005 Paris, France. See
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