The goal of the next few lectures will be to introduce the semigroup generated by a diffusion operator. The construction of the semigroup is non trivial because diffusion operators are unbounded operators.

We consider a diffusion operator

where and are continuous functions on and for every , the matrix is a symmetric and non negative matrix.

We will assume that is symmetric with respect to a measure which is equivalent to the Lebesgue measure, that is, for every smooth and compactly supported functions ,

**Exercise***Show that if is symmetric with respect to then, in the sense of distributions where is the adjoint of in the distribution sense.*

**Exercise*** Show that if is a smooth function and if , then we still have the formula *

**Exercise*** On , let us consider the diffusion operator where is a function. Show that is symmetric with respect to the measure .*

**Exercise*** (Divergence form operator). On , let us consider the operator where is the divergence operator defined on a function by*

and where is a field of non negative and symmetric matrices. Show that is a diffusion operator which is symmetric with respect to the Lebesgue measure.

For every smooth functions , let us define the so-called *carre du champ*, which is the symmetric first-order differential form defined by A straightforward computation shows that so that for every smooth function ,

**Exercise.**

- Show that if are functions and are also then,

- Show that if is a function and is also ,

The bilinear form we consider is given for by

This is the energy functional (or Dirichlet form) associated to . It is readily checked that is symmetric ,

and non negative It is easy to see that

The operator on the domain is a densely defined non positive symmetric operator on the Hilbert space . However, in general, it is not self-adjoint, indeed we easily see that

A famous theorem of Von Neumann asserts that any non negative and symmetric operator may be extended into a self-adjoint operator. The following construction, due to Friedrich, provides a canonical non negative self-adjoint extension.

**Theorem:***(Friedrichs extension) On the Hilbert space , there exists a densely defined non positive self-adjoint extension of .*

**Proof:** The idea is to work with a Sobolev type norm associated to the energy form . On , let us consider the following norm

By completing with respect to this norm, we get a Hilbert space . Since for , , the injection map is continuous and it may therefore be extended into a continuous map . Let us show that is injective so that may be identified with a subspace of . So, let such that . We can find a sequence , such that and . We have

thus and is injective. Let us now consider the map

It is well defined due to the fact that since is bounded, it is easily checked that

Moreover, is easily seen to be symmetric, and thus self-adjoint because its domain is equal to . Also, it is readily checked that the injectivity of implies the injectivity of . Therefore, we deduce that the inverse

is a densely defined self-adjoint operator on . Now, we observe that for ,

Thus and coincide on .

By defining, we get the required self-adjoint extension of

The operator , as constructed above, is called the Friedrichs extension of .

**Definition:** *If is the unique non positive self-adjoint extension of , then the operator is said to be essentially self-adjoint on . In that case, there is no ambiguity and we shall denote .*

We have the following first criterion for essential self-adjointness.

**Lemma:*** If for some , then the operator is essentially self-adjoint on . *

**Proof:** We make the proof for and let the reader adapt it for . Let be a non negative self-adjoint extension of . We want to prove that actually, . The assumption implies that is dense in for the norm

Since, is a non negative self-adjoint extension of , we have

The space is therefore dense in for the norm .

At that point, we use some notations introduced in the proof of the Friedrichs extension theorem. Since is dense in for the norm , we deduce that the equality

which is obviously satisfied for actually also holds for . From the definition of the Friedrichs extension, we deduce that and coincide on . Finally, since these two operators are self adjoint we conclude

Given the fact that is given here with the domain , the condition is equivalent to the fact that if is a function that satisfies in the sense of distributions then .

As a corollary of the previous lemma, the following proposition provides a useful sufficient condition for essential self-adjointness that is easy to check for several diffusion operators (including Laplace-Beltrami operators on complete Riemannian manifolds).

**Proposition:*** If the diffusion operator is elliptic with smooth coefficients and if there exists an increasing sequence , , such that on*

, and , as , then the operator is essentially self-adjoint on .

**Proof:** Let . According to the previous lemma, it is enough to check that if with , then . As it was observed above, is equivalent to the fact that, in the sense of distributions, . From the hypoellipticity of , we deduce therefore that is a smooth function. Now, for ,

Since , we deduce that

Therefore, by Cauchy-Schwarz inequality

If we now use the sequence and let , we obtain and therefore , as desired

The assumption on the existence of the sequence will be met several times in this course. We will see later that, from a geometric point of view, it says that the intrinsic metric associated to is complete, or in other words that the balls of the diffusion operator are compact.

**Exercise:*** Let be an elliptic diffusion operator with smooth coefficients. We assume that defined on is symmetric with respect to the measure . Let be a non empty set whose closure is compact. Show that the operator is essentially self-adjoint on*

**Exercise:***Let*

where is a smooth function on . Show that with respect to the measure , the operator is essentially self-adjoint on .

**Exercise:*** On , we consider the divergence form operator*

where is a smooth field of positive and symmetric matrices that satisfies

for some constant . Show that with respect to the Lebesgue measure, the operator is essentially self-adjoint on

**Exercise:*** On , we consider the Schrodinger type operator , where is a diffusion operator and is a smooth function. We denote*

Show that if there exists an increasing sequence , , such that on

, and , as and that if is bounded from below then is essentially self-adjoint on .