**Talagrand’s inequality and hypercontractivity**

**1. Talagrand’s inequality **

Theorem 1 (Talagrand)Let . Let be the uniform measure on .

This improves on the KKL theorem. Indeed, it implies

Theorem 2 (Kahn, Kalai, Linial)Let . Then there exists such that

Indeed, for any ,

Today, I will prove Talagrand’s inequality in a slightly extended context.

**2. Markov chains **

A *Markov kernel* on a finite set is a function such that for all . Say is *reversible* with respect to *invariant* probability measure if

and

The corresponding *Dirichlet form* is .

The *spectral gap* is the best constant such that .

The *logarithmic Sobolev constant* is the best constant in

Here,

It is stronger than spectral gap, (with our choice of normalization).

Spectral gap is equivalent to exponential decay under heat flow.

Logarithmic Sobolev inequality is equivalent to hypercontractivity of heat flow,

with .

On a product space , there is a product kernel, and

Example 1Let .

In other words, , , .

Example 2Product of copies of previous example.

Then

For this example, Talagrand’s inequality states that

On the biased cube, i.e. , , , (Bonami-Beckner inequality).

** 2.1. Proof of Talagrand’s inequality for products **

Now I give the proof.

Assume . Write

We treat separately small and large values of . Write

Since

for ,

Commutation: commutes with the ‘s, so

by hypercontractivity and Hölder, given by

Finally,

gives Talagrand’s inequality.

The main ingredients of the proof were

- Expression of energy as a sum of terms.
- Commutation.
- Global hypercontractivity.

This carries over to certain non product situations.

** 2.2. The symmetric group **

Let be the symmetric group, equipped with the kernel corresponding to the random walk associated to the generating set consisting of all transpositions, i.e.

Commutation holds: .

The spectral gap is due to Diaconis-Shahshahani in 1981.

Hypercontractivity was proven by Diaconis and Saloff-Coste (1996) and Lee-Yau (1998).

The above proof automatically gives the following Talagrand inequality.

In terms of influences of Boolean functions, there exists a transposition such that

It is not that exciting since it follows from spectral gap. For Talagrand’s inequality to improve on spectral gap, must be substantially larger than , i.e.

**3. Continuous setting **

** 3.1. Gaussian case **

From now on, is the Gaussian measure on . Then Talagrand’s inequality reads

The proof goes along similar lines. is replaced by the Ornstein-Uhlenbeck semigroup

It commutes with partial derivatives. Hypercontractivity holds with .

** 3.2. Extensions **

On , use measure . Assume a lower bound on the Hessian (the correct formulation involves the Ricci curvature of the semigroup). Then commutation holds up to a factor :

And one gets a Talagrand inequality.

** 3.3. The round sphere **

Let . Write

Commutation holds. . This gives a Talagrand inequality

**4. Talagrand inequality **

** 4.1. influences **

Apply Talagrand’s inequality to characteristic functions of sets. The -influence makes sense (Keller, Mossel, Sen 2010).

In the Gaussian case, it has the following geometric meaning. Given , let

Then define

Then

Theorem 3 (Keller, Mossel, Sen 2010)There exists such that

This is optimal.

** 4.2. Proof of the -influence bound **

I show that a slight modification of above proof gives Keller, Mossel and Sen’s result.

Change into :

Use spectral gap,

by hypercontractivity.

Trick: If , is Lipschitz,

This easily follows from the integral representation of the Ornstein-Uhlenbeck semigroup. Estimate

Get

Talgrand’s inequality for norms reads

From there, the application to influences is immediate.

** 4.3. Connections with isoperimetry **

On , use Gaussian measure again. The Talagrand inequality reads

In terms of influences, there exists such that

In case one only one factor, this says that

We recognize a quantitative form of the Gaussian isoperimetric inequality

Note that

as tends to . So we merely lost a constant factor.