## Lecture 2. Essentially self-adjoint diffusion operators

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
$L=\sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial^2}{ \partial x_i \partial x_j} +\sum_{i=1}^n b_i(x)\frac{\partial}{\partial x_i},$
where $b_i$ and $\sigma_{ij}$ are continuous functions on $\mathbb{R}^n$ and for every $x \in \mathbb{R}^n$, the matrix $(\sigma_{ij}(x))_{1\le i,j\le n}$ is a symmetric and non negative matrix.

We will assume that $L$ is symmetric with respect to a measure $\mu$ which is equivalent to the Lebesgue measure, that is, for every smooth and compactly supported functions $f,g : \mathbb{R}^n \rightarrow \mathbb{R} \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$, $\int_{\mathbb{R}^n} g Lf d\mu= \int_{\mathbb{R}^n} fLg d\mu.$

ExerciseShow that if $L$ is symmetric with respect to $\mu$ then, in the sense of distributions $L'\mu=0,$ where $L'$ is the adjoint of $L$ in the distribution sense.

Exercise Show that if $f : \mathbb{R}^n \rightarrow \mathbb{R}$ is a smooth function and if $g \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$, then we still have the formula $\int_{\mathbb{R}^n} f Lg d\mu =\int_{\mathbb{R}^n} gLf d\mu.$

Exercise On $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$, let us consider the diffusion operator $L=\Delta +\langle \nabla U, \nabla \cdot \rangle,$ where $U: \mathbb{R}^n \rightarrow \mathbb{R}$ is a $C^1$ function. Show that $L$ is symmetric with respect to the measure $\mu (dx)=e^{U(x)} dx$.

Exercise (Divergence form operator). On $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$, let us consider the operator $Lf=\mathbf{div} (\sigma \nabla f),$ where $\mathbf{div}$ is the divergence operator defined on a $C^1$ function $\phi: \mathbb{R}^n \rightarrow \mathbb{R}^n$ by
$\mathbf{div} \text{ } \phi=\sum_{i=1}^n \frac{\partial \phi_i}{\partial x_i}$
and where $\sigma$ is a $C^1$ field of non negative and symmetric matrices. Show that $L$ is a diffusion operator which is symmetric with respect to the Lebesgue measure.

For every smooth functions $f,g: \mathbb{R}^n \rightarrow \mathbb{R}$, let us define the so-called carre du champ, which is the symmetric first-order differential form defined by $\Gamma (f,g) =\frac{1}{2} \left( L(fg)-fLg-gLf \right).$ A straightforward computation shows that $\Gamma (f,g)=\sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial f}{\partial x_i} \frac{\partial g}{\partial x_j},$ so that for every smooth function $f$, $\Gamma(f,f) \ge 0.$

Exercise.

• Show that if $f,g :\mathbb{R}^n \rightarrow \mathbb{R}$ are $C^1$ functions and $\phi_1,\phi_2: \mathbb{R} \rightarrow \mathbb{R}$ are also $C^1$ then,
$\Gamma (\phi_1 (f), \phi_2 (g))=\phi'_1 (f) \phi_2'(g) \Gamma(f,g).$
• Show that if $f:\mathbb{R}^n \rightarrow \mathbb{R}$ is a $C^2$ function and $\phi: \mathbb{R}\rightarrow \mathbb{R}$ is also $C^2$,
$L \phi (f)=\phi'(f) Lf+\phi''(f) \Gamma(f,f).$

The bilinear form we consider is given for $f,g \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ by $\mathcal{E} (f,g)=\int_{\mathbb{R}^n} \Gamma (f,g) d\mu.$
This is the energy functional (or Dirichlet form) associated to $L$. It is readily checked that $\mathcal{E}$ is symmetric $\mathcal{E} (f,g)=\mathcal{E} (g,f)$,
and non negative $\mathcal{E} (f,f) \ge 0.$ It is easy to see that
$\mathcal{E}(f,g)=-\int_{\mathbb{R}^n} fLg d\mu=-\int_{\mathbb{R}^n} gLf d\mu.$

The operator $L$ on the domain $\mathcal{D}(L)= \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ is a densely defined non positive symmetric operator on the Hilbert space $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$. However, in general, it is not self-adjoint, indeed we easily see that
$\left\{ f \in \mathcal{C}^\infty (\mathbb{R}^n,\mathbb{R}), \| f \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})} +\|Lf \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})} < \infty \right\} \subset \mathcal{D}(L^*).$

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 $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, there exists a densely defined non positive self-adjoint extension of $L$.

Proof: The idea is to work with a Sobolev type norm associated to the energy form $\mathcal{E}$. On $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$, let us consider the following norm
$\| f\|^2_{\mathcal{E}}=\| f \|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }+\mathcal{E}(f,f).$
By completing $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ with respect to this norm, we get a Hilbert space $(\mathcal{H},\langle \cdot , \cdot \rangle_{\mathcal{E}})$. Since for $f \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$, $\| f \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) } \le \| f\|_{\mathcal{E}}$, the injection map $\iota : ( \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), \| \cdot \|_{\mathcal{E}}) \rightarrow (\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}), \| \cdot \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) })$ is continuous and it may therefore be extended into a continuous map $\bar{\iota}: (\mathcal{H}, \| \cdot \|_{\mathcal{E}}) \rightarrow (\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}), \| \cdot \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) })$. Let us show that $\bar{\iota}$ is injective so that $\mathcal{H}$ may be identified with a subspace of $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$. So, let $f \in \mathcal{H}$ such that $\bar{\iota} (f)=0$. We can find a sequence $f_n \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$, such that $\| f_n -f \|_{\mathcal{E}} \to 0$ and $\| f_n \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) } \to 0$. We have
$\| f \|_{\mathcal{E}} =\lim_{m,n \to + \infty} \langle f_n, f_m \rangle_{\mathcal{E}}$
$=\lim_{m \to + \infty} \lim_{n \to + \infty} \langle f_n,f_m \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }+\mathcal{E}(f_n,f_m)$
$=\lim_{m \to + \infty} \lim_{n \to + \infty} \langle f_n,f_m \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }- \langle f_n,Lf_m \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }=0$
thus $f=0$ and $\bar{\iota}$ is injective. Let us now consider the map
$B=\bar{\iota} \cdot \bar{\iota}^* : \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) \rightarrow \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) .$
It is well defined due to the fact that since $\bar{\iota}$ is bounded, it is easily checked that $\mathcal{D}(\bar{\iota}^*)= \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}).$

Moreover, $B$ is easily seen to be symmetric, and thus self-adjoint because its domain is equal to $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$. Also, it is readily checked that the injectivity of $\bar{\iota}$ implies the injectivity of $B$. Therefore, we deduce that the inverse
$A=B^{-1}: \mathcal{R} (\bar{\iota} \cdot \bar{\iota}^*) \subset\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) \rightarrow \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$
is a densely defined self-adjoint operator on $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$. Now, we observe that for $f,g \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$,
$\langle f,g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}-\langle Lf,g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}$
$= \langle \bar{i}^{-1}(f),\bar{i}^{-1}(g) \rangle_\mathcal{E}$
$= \langle (\bar{i}^{-1})^* \bar{i}^{-1} f,g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}$
$= \langle (\bar{i} \bar{i}^*)^{-1} f,g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}$
Thus $A$ and $\mathbf{Id}-L$ coincide on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$.
By defining, $-\bar{L}=A-\mathbf{Id},$ we get the required self-adjoint extension of $-L$ $\square$

The operator $\bar{L}$, as constructed above, is called the Friedrichs extension of $L$.

Definition: If $\bar{L}$ is the unique non positive self-adjoint extension of $L$, then the operator $L$ is said to be essentially self-adjoint on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$. In that case, there is no ambiguity and we shall denote $\bar{L}=L$.

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

Lemma: If for some $\lambda > 0$, $\mathbf{Ker} (-L^* +\lambda \mathbf{Id} )= \{ 0 \},$ then the operator $L$ is essentially self-adjoint on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$

Proof: We make the proof for $\lambda=1$ and let the reader adapt it for $\lambda \neq 0$. Let $-\tilde{L}$ be a non negative self-adjoint extension of $-L$. We want to prove that actually, $-\tilde{L}=-\bar{L}$. The assumption $\mathbf{Ker} (-L^* + \mathbf{Id} )= \{ 0 \}$ implies that $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ is dense in $\mathcal{D}(-L^*)$ for the norm
$\| f \|^2_{\mathcal{E}}=\| f \|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) } -\langle f , L^* f \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }.$
Since, $-\tilde{L}$ is a non negative self-adjoint extension of $-L$, we have
$\mathcal{D}(-\tilde{L}) \subset \mathcal{D}(-L^*).$
The space $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ is therefore dense in $\mathcal{D}(-\tilde{L})$ for the norm $\| \cdot \|_{\mathcal{E}}$.

At that point, we use some notations introduced in the proof of the Friedrichs extension theorem. Since $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ is dense in $\mathcal{D}(-\tilde{L})$ for the norm $\| \cdot \|_{\mathcal{E}}$, we deduce that the equality
$\langle f,g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}-\langle \tilde{L}f,g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}= \langle \bar{i}^{-1}(f),\bar{i}^{-1}(g) \rangle_\mathcal{E} ,$
which is obviously satisfied for $f,g \in \mathcal{C}_c(\mathbb{R}^n,\mathbb{R})$ actually also holds for $f,g \in \mathcal{D}(\tilde{L})$. From the definition of the Friedrichs extension, we deduce that $\bar{L}$ and $\tilde{L}$ coincide on $\mathcal{D}(\tilde{L})$. Finally, since these two operators are self adjoint we conclude $\bar{L}=\tilde{L}$ $\square$

Given the fact that $-L$ is given here with the domain $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$, the condition $\mathbf{Ker} (-L^* +\lambda \mathbf{Id} )= \{ 0 \},$ is equivalent to the fact that if $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ is a function that satisfies in the sense of distributions $-Lf+\lambda f=0,$ then $f=0$.

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 $L$ is elliptic with smooth coefficients and if there exists an increasing sequence $h_n\in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$, $0 \le h_n \le 1$, such that $h_n\nearrow 1$ on
$\mathbb{R}^n$, and $||\Gamma (h_n,h_n)||_{\infty} \to 0$, as $n\to \infty$, then the operator $L$ is essentially self-adjoint on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$

Proof: Let $\lambda > 0$. According to the previous lemma, it is enough to check that if $L^* f=\lambda f$ with $\lambda > 0$, then $f=0$. As it was observed above, $L^* f=\lambda f$ is equivalent to the fact that, in the sense of distributions, $Lf =\lambda f$. From the hypoellipticity of $L$, we deduce therefore that $f$ is a smooth function. Now, for $h \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$,
$\int_{\mathbb{R}^n} \Gamma( f, h^2f) d\mu =-\langle f, L(h^2f)\rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}$
$=-\langle L^*f ,h^2f \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}$
$=-\lambda \langle f,h^2f\rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}$
$=-\lambda \langle f^2,h^2 \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})} \le 0.$
Since $\Gamma( f, h^2f)=h^2 \Gamma (f,f)+2 fh \Gamma(f,h)$, we deduce that
$\langle h^2, \Gamma (f,f) \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}+2 \langle fh, \Gamma(f,h)\rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})} \le 0.$
Therefore, by Cauchy-Schwarz inequality
$\langle h^2, \Gamma (f,f) \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})} \le 4 \| f |_2^2 \| \Gamma (h,h) \|_\infty.$
If we now use the sequence $h_n$ and let $n \to \infty$, we obtain $\Gamma(f,f)=0$ and therefore $f=0$, as desired $\square$

The assumption on the existence of the sequence $h_n$ 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 $L$ is complete, or in other words that the balls of the diffusion operator $L$ are compact.

Exercise: Let $L$ be an elliptic diffusion operator with smooth coefficients. We assume that $L$ defined on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ is symmetric with respect to the measure $\mu$. Let $\Omega \subset \mathbb{R}^n$ be a non empty set whose closure $\bar{\Omega}$ is compact. Show that the operator $L$ is essentially self-adjoint on
$\{ u :\bar{\Omega} \to \mathbb{R},\text{ u smooth}, \text{ } u=0 \text{ on } \partial\Omega \}.$

Exercise:Let
$L=\Delta +\langle \nabla U, \nabla \cdot\rangle,$
where $U$ is a smooth function on $\mathbb{R}^n$. Show that with respect to the measure $\mu(dx)=e^{U(x)} dx$, the operator $L$ is essentially self-adjoint on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$.

Exercise: On $\mathbb{R}^n$, we consider the divergence form operator
$Lf=\mathbf{div} (\sigma \nabla f),$
where $\sigma$ is a smooth field of positive and symmetric matrices that satisfies
$a \|x \|^2 \le \langle x , \sigma x \rangle \le b \|x \|^2, \quad x \in \mathbb{R}^n,$
for some constant $0 < a \le b$. Show that with respect to the Lebesgue measure, the operator $L$ is essentially self-adjoint on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$

Exercise: On $\mathbb{R}^n$, we consider the Schrodinger type operator $H=L-V$, where $L$ is a diffusion operator and $V:\mathbb{R}^n \rightarrow \mathbb{R}$ is a smooth function. We denote
$\Gamma (f,g) =\frac{1}{2} \left( L(fg)-fLg-gLf \right).$
Show that if there exists an increasing sequence $h_n\in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$, $0 \le h_n \le 1$, such that $h_n\nearrow 1$ on
$\mathbb{R}^n$, and $||\Gamma (h_n,h_n)||_{\infty} \to 0$, as $n\to \infty$ and that if $V$ is bounded from below then $H$ is essentially self-adjoint on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$.

This entry was posted in Diffusions on manifolds. Bookmark the permalink.