## HW2. Due September 27

Exercise: Show that if $L$ is the Laplace operator on $\mathbb{R}^n$, then for $t > 0$, $\mathbf{P}_t f (x)=\frac{1}{(4\pi t)^{\frac{n}{2}}} \int_{\mathbb{R}^n} e^{-\frac{\|x-y\|^2}{4t} } f(y)dy.$

Exercise: Let $L$ be an essentially self-adjoint diffusion operator on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$. Show that if the constant function $1 \in \mathcal{D}(L)$ and if $L1=0$, then $\mathbf{P}_t 1=1.$

Exercise: Let $L$ be an essentially self-adjoint diffusion operator on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$

• Show that for every $\lambda > 0$, the range of the operator $\lambda \mathbf{Id}-L$ is dense in $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$.
• By using the spectral theorem, show that the following limit holds for the operator norm on $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, $\mathbf{P}_t=\lim_{n \to +\infty} \left( \mathbf{Id} -\frac{t}{n} L\right)^{-n}.$

Exercise: As usual, we denote by $\Delta$ the Laplace operator on $\mathbb{R}^n$. The Mac-Donald’s function with index $\nu \in \mathbb{R}$ is defined for $x \in \mathbb{R} \setminus \{ 0 \}$ by $K_\nu (x)=\frac{1}{2} \left( \frac{x}{2} \right)^\nu \int_0^{+\infty} \frac{e^{-\frac{x^2}{4t} -t}}{t^{1+\nu}} dt$.

• Show that for $\lambda \in \mathbb{R}^n$ and $\alpha > 0$, $\frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n} e^{i \langle \lambda , x \rangle} \left( \frac{\| x \| }{\sqrt{\alpha}} \right)^{1-\frac{n}{2}} K_{\frac{n}{2}-1} (\sqrt{\alpha} \| x \| ) dx=\frac{1}{\alpha +\| \lambda \|^2}.$
• Show that for $\nu \in \mathbb{R}$, $K_{-\nu}=K_\nu$.
• Show that $K_{1/2}(x)=\sqrt {\frac{\pi}{2x}} e^{-x}.$
• Prove that for $f \in \mathbf{L}^2 (\mathbb{R}^n,\mathbb{R})$ and $\alpha > 0$, $(\alpha\mathbf{Id}-\Delta)^{-1} f (x)=\int_{\mathbb{R}^n} G_\alpha (x-y) f(y) dy,$ where $G_\alpha(x)=\frac{1}{(2\pi)^{n/2}} \left( \frac{\| x \| }{\sqrt{\alpha}} \right)^{1-\frac{n}{2}} K_{\frac{n}{2}-1} (\sqrt{\alpha} \| x \| ).$ (You may use Fourier transform to solve the partial differential equation $\alpha g -\Delta g=f$).

Exercise: By using the previous exercise, prove that for $f \in \mathbf{L}^2 (\mathbb{R}^n,\mathbb{R})$, $\lim_{n \to + \infty} \left(\mathbf{Id} -\frac{t}{n} \Delta \right)^{-n} f =\frac{1}{(4\pi t)^{\frac{n}{2}}} \int_{\mathbb{R}^n} e^{-\frac{\|x-y\|^2}{4t} } f(y)dy,$ the limit being taken in $\mathbf{L}^2 (\mathbb{R}^n,\mathbb{R})$. Conclude that almost everywhere, $\mathbf{P}_t f (x)=\frac{1}{(4\pi t)^{\frac{n}{2}}} \int_{\mathbb{R}^n} e^{-\frac{\|x-y\|^2}{4t} } f(y)dy.$

## Lecture 5. Hormander’s type operators

For geometric purposes, it is often very useful to use the language of vector fields to study diffusion operators.
Let $\mathcal{O} \subset \mathbb{R}^n$ be a non-empty open set. A
smooth vector field  $V$ on $\mathcal{O}$ is a smooth map

$\begin{array}{llll} V: & \mathcal{O} & \rightarrow & \mathbb{R}^{n} \\ & x & \rightarrow & (v_{1}(x),...,v_{n}(x)). \end{array}$

We will often regard a vector field $V$ as a differential operator acting on the space of smooth functions $f: \mathcal{O} \rightarrow \mathbb{R}$ as follows:

$(Vf) (x)=\sum_{i=1}^n v_i (x) \frac{\partial f}{\partial x_i}.$
By the chain rule, we note that $V$ is a derivation, that is an operator on
$C^{\infty} (\mathcal{O} )$, linear over $\mathbb{R}$, satisfying for $f,g \in C^{\infty} (\mathcal{O} )$,

$V(fg)=(Vf)g +f (Vg).$
Conversely, it easily seen that any derivation on $C^{\infty} (\mathcal{O} )$ defines a vector field on $\mathcal{O}$ (Pick $x_0 \in \mathcal{O}$, observe that if $g$ is a smooth function such that $g(x_0)=0$ then $V((x-x_0)g)(x_0)=0$ and then, to compute $Vf(x_0)$ use a Taylor expansion around $x_0$).

With these notations, it is readily checked that if $V_0,V_1,\cdots, V_d$ are smooth vector fields on $\mathbb{R}^n$, then the second order differential operator

$L=V_0+\sum_{i=1}^d V_i^2$
is a diffusion operator. Here $V_i^2$ has to be understood as the operator $V_i$ composed with itself. Diffusion operators that may be written under the previous form are called Hormander’s type diffusion operators. It is easily seen that a Hormander’s type diffusion operator is elliptic in $\mathbb{R}^n$ if and only if for every $x \in \mathbb{R}^n$, the vectors $V_1(x),\cdots,V_d(x)$ form a basis of $\mathbb{R}^n$.
Proposition:  Let

$L=\sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial^2}{ \partial x_i \partial x_j} +\sum_{i=1}^n b_i (x)\frac{\partial}{\partial x_i},$

be a diffusion operator on $\mathbb{R}^n$ such that the $\sigma_{ij}$‘s and the $b_i$‘s are smooth functions. Let us assume that for every $x \in \mathbb{R}^n$, the rank of the matrix $(\sigma_{ij}(x))_{1 \le i,j \le n}$ is constant equal to $d$. Then, there exist smooth vector fields $V_0,V_1,\cdots, V_d$ on $\mathbb{R}^n$ such that $V_1,\cdots, V_d$ are linearly independent and

$L=V_0+\sum_{i=1}^d V_i^2.$

Proof:  We first assume $d=n$.  Since the matrix $(\sigma_{ij}(x))_{1 \le i,j \le n}$ is symmetric and positive, it admits a unique symmetric and positive square root $v(x)=(v_{ij}(x))_{1 \le i,j \le n}$. Let us assume for a moment that the $v_{ij}$‘s are smooth functions, in that case by denoting

$V_i=\sum_{j=1}^n v_{ij} \frac{\partial}{\partial x_j},$
the vector fields $V_1,\cdots,V_n$ are linearly independent and it is readily seen that the differential operator

$L-\sum_{i=1}^n V_i^2$
is actually a first order differential operator and thus a vector field. We therefore are let to prove that the $v_{ij}$‘s are smooth functions.  Let $\mathcal{O}$ be a bounded non empty set of $\mathbb{R}^n$ and $\Gamma$ be any contour in the half plane $\mathbf{Re} (z) >0$ that contains all the eigenvalues of $\sigma(x)$, $x \in \mathcal{O}$. We claim that

$v(x)=\frac{1}{2 \pi} \int_{\Gamma} \sqrt{z} (\sigma(x)-z\mathbf{I}_n)^{-1} dz, \quad x \in \mathcal{O}.$
Indeed, if $\Gamma'$ is another contour in the half plane $\mathbf{Re} (z) >0$ whose interior contains $\Gamma$, as a straightforward application of the Fubini’s theorem and Cauchy’s formula we have

$\begin{array}{ll} & \frac{1}{2 \pi} \int_{\Gamma} \sqrt{z} (\sigma(x)-z\mathbf{I}_n)^{-1} dz \times \frac{1}{2 \pi} \int_{\Gamma'} \sqrt{z} (\sigma(x)-z\mathbf{I}_n)^{-1} dz \\ =&\frac{1}{4 \pi^2} \int_{\Gamma'}\int_{\Gamma} \sqrt{z z'} (\sigma(x)-z\mathbf{I}_n)^{-1}(\sigma(x)-z'\mathbf{I}_n)^{-1} dz dz' \\ = & \frac{1}{4 \pi^2} \int_{\Gamma'}\int_{\Gamma} \sqrt{z z'} \frac{ (\sigma(x)-z\mathbf{I}_n)^{-1}-(\sigma(x)-z'\mathbf{I}_n)^{-1}}{z-z'} dz dz' \\ = & - \frac{1}{4 \pi^2} \int_{\Gamma}\int_{\Gamma'} \sqrt{z z'} \frac{ (\sigma(x)-z\mathbf{I}_n)^{-1}}{z'-z} dz' dz \\ =& - \frac{1}{4 \pi^2} \int_{\Gamma} (\sigma(x)-z\mathbf{I}_n)^{-1} \left( \int_{\Gamma'}\frac{ \sqrt{z z'} }{z'-z} dz' \right) dz \\ =& \frac{1}{2 i \pi} \int_{\Gamma} (\sigma(x)-z\mathbf{I}_n)^{-1} z dz \\ =& \frac{1}{2 i \pi} \int_{\Gamma} (\sigma(x)-z\mathbf{I}_n)^{-1} \sigma(x) dz. \end{array}$
In the last expression above, we may modify $\Gamma$ into a circle $\Gamma_R =\{ z, | z|=R \}$. Then by choosing $R$ big enough ($R > \sup_{x \in \mathcal{O}} \sqrt{ \| \sigma (x) \|}$), and expanding $\int_{\Gamma} (\sigma(x)-z\mathbf{I}_n)^{-1} dz$ in powers of $z$, we see that

$\lim_{R \to +\infty} \frac{1}{2 i \pi} \int_{\Gamma_R} (\sigma(x)-z\mathbf{I}_n)^{-1} dz= \mathbf{Id} .$
As a conclusion,

$\left( \frac{1}{2 \pi} \int_{\Gamma} \sqrt{z} (\sigma(x)-z\mathbf{I}_n)^{-1} dz \right)^2 = \sigma(x),$
so that, as we claimed it,

$v(x)=\frac{1}{2 \pi} \int_{\Gamma} \sqrt{z} (\sigma(x)-z\mathbf{I}_n)^{-1} dz, \quad x \in \mathcal{O}.$
This expression of the square root of $\sigma$ clearly shows that the $v_{ij}$‘s are smooth functions. By putting things together, we therefore proved the proposition in the case $d=n$.

Let us now turn to the case $d < n$. By smoothly choosing an orthonormal basis of $\mathbb{R}^n$ which is adapted to the orthogonal decomposition

$\mathbb{R}^n=\mathbf{Ker} (\sigma(x) ) \oplus \mathbf{Ker} (\sigma(x) )^\perp,$
we get a decomposition

$\sigma(x)=M(x)\left( \begin{array}{ll} \mathcal{V}(x) & 0 \\ 0 & 0 \end{array} \right)M(x)^{-1},$
where $M(x)$ is an orthogonal matrix with smooth coefficients and where $\mathcal{V}(x)$ is a $d \times d$ symmetric and positive matrix with smooth coefficients. We may now apply the first part of the proof to the matrix $\mathcal{V}(x)$ and easily conclude $\square$

## Lecture 4. Diffusion semigroups as solutions of heat equations

In this lecture, we show that the diffusion semigroup that was constructed in the previous lectures appears as the solution of a parabolic Cauchy problem. Under an ellipticity and completeness assumption, it is moreover the unique square integrable solution.

Proposition: Let $L$ be an essentially self-adjoint diffusion operator and let $(\mathbf{P}_t)_{t \ge 0}$ be the corresponding diffusion semigroup. Let $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, and let $u (t,x)= \mathbf{P}_t f (x), \quad t \ge 0, x\in \mathbb{R}^n.$ Then $u$ is a weak solution of the Cauchy problem
$\frac{\partial u}{\partial t}= L u,\quad u (0,x)=f(x).$

Proof: For $\phi \in \mathcal{C}_c ((0,+\infty) \times \mathbb{R}^n,\mathbb{R})$, we have
$\int_{\mathbb{R}^{n+1}} \left( \left( -\frac{\partial}{\partial t} -L \right) \phi (t,x) \right) u(t,x) d\mu(x) dt$
$=\int_{\mathbb{R}} \int_{\mathbb{R}^{n}} \left( \left( -\frac{\partial}{\partial t} -L \right) \phi (t,x) \right) \mathbf{P}_t f (x) d\mu(x) dt$
$= \int_{\mathbb{R}} \int_{\mathbb{R}^{n}} \mathbf{P}_t \left( \left( -\frac{\partial}{\partial t} -L \right) \phi (t,x) \right) f (x) d\mu(x) dt$
$= \int_{\mathbb{R}} \int_{\mathbb{R}^{n}} -\frac{\partial}{\partial t} \left( \mathbf{P}_t \phi (t,x) f(x) \right) d\mu(x) dt =0.$

If the operator $L$ is furthermore assumed to be elliptic, then as we have seen in the previous lecture, the map $(t,x) \to \mathbf{P}_t f(x)$ is smooth and therefore, the above solution is also strong.

We now address uniqueness questions. We need further assumptions that already have been met before. We consider an elliptic diffusion operator $L$ with smooth coefficients on $\mathbb{R}^n$ such that:

• There is a Borel measure $\mu$, symmetric and invariant for $L$ on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$;
• 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 $n\to \infty$.

Under these assumptions we already know that $L$ is essentially self-adjoint. The next proposition implies that $(t,x) \to \mathbf{P}_t f(x)$ is the unique solution of the parabolic Cauchy problem.

Proposition Let $L$ be a diffusion operator that satisfies the above assumptions. Let $u(t,x)$ be a smooth solution of the Cauchy problem $\frac{\partial u}{\partial t}= L u,\quad u (0,x)=0,$ where $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$. Assume that $\| u (t , \cdot) \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})} < +\infty$. Then $u(t,x)=0$

Proof: Let $h_n$ be as above. On one hand, we have $\int_0^\tau \int_{\mathbb{R}^n} h_n^2 u Lu d\mu dt =\frac{1}{2} \int_0^\tau \frac{\partial}{\partial t} \int_{\mathbb{R}^n} h_n^2 u^2 (t,x) d\mu dt = \int_{\mathbb{R}^n} h_n^2 u^2 (\tau,x) d\mu$.
On the other hand, we have $\int_{\mathbb{R}^n} h_n^2 u Lu d\mu =-\int_{\mathbb{R}^n} \Gamma(h_n^2 u, u) d\mu =-\int_{\mathbb{R}^n} h_n^2 \Gamma(u) d\mu -2 \int_{\mathbb{R}^n} uh_n \Gamma(u,h_n)d\mu$.
From Cauchy-Schwarz inequality, we now have $\int_{\mathbb{R}^n} uh_n \Gamma(u,h_n)d\mu \ge - \left( \int_{\mathbb{R}^n} u^2\Gamma(h_n) d\mu\right)^{1/2}\left( \int_{\mathbb{R}^n} h_n^2\Gamma(u) d\mu\right)^{1/2}$
$\ge -\frac{1}{2} \int_{\mathbb{R}^n} u^2\Gamma(h_n) d\mu-\frac{1}{2} \int_{\mathbb{R}^n} h_n^2\Gamma(u) d\mu$.

We deduce that $\int_{\mathbb{R}^n} h_n^2 u Lu d\mu \le \int_{\mathbb{R}^n} u^2 \Gamma(h_n) d\mu$. As a conclusion we obtain that $\int_{\mathbb{R}^n} h_n^2 u^2 (\tau,x) d\mu \le \int_0^\tau \int_{\mathbb{R}^n} u^2 \Gamma(h_n) d\mu dt$. Letting $n \to \infty$, yields $\int_{\mathbb{R}^n} u^2 d\mu \le 0$ and thus $u=0$ $\square$

## Homework 1. MA5016. Due September 13 in class

Exercise 1. Let $L: \mathcal{C}^{\infty} (\mathbb{R}^n) \rightarrow \mathcal{C} (\mathbb{R}^n)$ be a linear operator such that:

• $L$ is a local operator, that is if $f=g$ on a neighborhood of $x$ then $Lf(x)=Lg(x)$;
• If $f \in \mathcal{C}^{\infty} (\mathbb{R}^n)$ has a global maximum at $x$ with $f(x)\ge 0$ then $Lf (x) \le 0$.

Show that for $f \in \mathcal{C}^{\infty} (\mathbb{R}^n)$ and $x \in \mathbb{R}^n$,

$Lf(x)=\sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial^2 f}{ \partial x_i \partial x_j} +\sum_{i=1}^n b_i (x)\frac{\partial f}{\partial x_i} -c(x)f(x),$

where $b_i$, $c$ and $\sigma_{ij}$ are continuous functions on $\mathbb{R}^n$ such that for every $x \in \mathbb{R}^n$, $c(x) \ge 0$ and the matrix $(\sigma_{ij}(x))_{1\le i,j\le n}$ is symmetric and nonnegative.

Exercise 2.

• 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 $C^2$,
$L \phi (f)=\phi'(f) Lf+\phi''(f) \Gamma(f,f).$

Exercise 3 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 4 (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.

Exercise 5: 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 $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$

## Lecture 3. Semigroups generated by diffusion operators

In this lecture, we consider a diffusion operator L which is essentially self-adjoint. Its Friedrichs extension is still denoted by L.

The fact that we are now dealing with a non negative self-adjoint operator allows us to use spectral theory in order to define the semigroup generated by L. We recall the following so-called spectral theorem.

Theorem: Let $A$ be a non negative self-adjoint operator on a separable Hilbert space $\mathcal{H}$. There exist a measure space $(\Omega, \nu)$, a unitary map $U: \mathbf{L}_{\nu}^2 (\Omega,\mathbb{R}) \rightarrow \mathcal{H}$ and a non negative real valued measurable function $\lambda$ on $\Omega$ such that $U^{-1} A U f (x)=\lambda(x) f(x),$ for $x \in \Omega$, $Uf \in \mathcal{D}(A)$. Moreover, given $f \in \mathbf{L}_{\nu}^2 (\Omega,\mathbb{R})$, $Uf$ belongs to $\mathcal{D}(A)$ if only if $\int_{\Omega} \lambda^2 f^2 d\nu < +\infty$.

We may apply the spectral theorem to the self-adjoint operator $-L$ in order to define $e^{tL}$. More generally, given a Borel function $g :\mathbb{R}_{\ge 0} \to \mathbb{R}$ and the spectral decomposition of $-L$, $U^{-1} L U f (x)=-\lambda(x) f(x)$, we may always define an operator $g(-L)$ as being the unique operator that satisfies $U^{-1} g(-L) U f (x)= g\circ \lambda (x) f(x).$ We may observe that $g(-L)$ is a bounded operator if $g$ is a bounded function.

As a particular case, we define the diffusion semigroup $(\mathbf{P}_t)_{t \ge 0}$ on $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ by the requirement $U^{-1} \mathbf{P}_t U f (x)=e^{-t \lambda (x)} f(x).$

This defines a family of bounded operators $\mathbf{P}_t : \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) \rightarrow \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ whose following properties are readily checked from the spectral decomposition:

• For $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, $\| \mathbf{P}_t f \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})} \le \| f \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}.$
• $\mathbf{P}_0=\mathbf{Id}$ and for $s,t \ge 0$, $\mathbf{P}_s \mathbf{P}_t =\mathbf{P}_{s+t}$.
• For $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, the map $t \to \mathbf{P}_t f$ is continuous in $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$.
• For $f,g \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, $\int_{\mathbb{R}^n} (\mathbf{P}_t f) g d\mu= \int_{\mathbb{R}^n} f(\mathbf{P}_t g) d\mu$

We summarize the above properties by saying that $(\mathbf{P}_t)_{t \ge 0}$ is a self-adjoint strongly continuous contraction semigroup on $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$.

From the spectral decomposition, it is also easily checked that the operator $L$ is furthermore the generator of this semigroup, that is for $f \in \mathcal{D}(L)$, $\lim_{t \to 0} \left\| \frac{\mathbf{P}_t f -f}{t} -Lf \right\|_{ \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }=0.$ From the semigroup property, it implies that for $t \ge 0$, $\mathbf{P}_t \mathcal{D}(L) \subset \mathcal{D}(L)$, and that for $f \in \mathcal{D}(L)$, $\frac{d}{dt}\mathbf{P}_t f= \mathbf{P}_t Lf=L \mathbf{P}_t f,$ the derivative on the left hand side of the above equality being taken in $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$.

It is easily seen that the semigroup $(\mathbf{P}_t)_{t \ge 0}$ is actually unique in the followings sense:

Proposition: Let $\mathbf{Q}_t : \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) \rightarrow \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, $t \ge 0$, be a family of bounded operators such that:

• For $s,t \ge 0$, $\mathbf{Q}_s \mathbf{Q}_t =\mathbf{Q}_{s+t}$.
• For $f \in \mathcal{D}(L)$, $\lim_{t \to 0} \left\| \frac{\mathbf{Q}_t f -f}{t} -Lf \right\|_{ \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }=0,$

then for every $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ and $t \ge 0$, $\mathbf{P}_t f=\mathbf{Q}_t f$.

Exercise: Show that if $L$ is the Laplace operator on $\mathbb{R}^n$, then for $t > 0$, $\mathbf{P}_t f (x)=\frac{1}{(4\pi t)^{\frac{n}{2}}} \int_{\mathbb{R}^n} e^{-\frac{\|x-y\|^2}{4t} } f(y)dy.$

Exercise: Let $L$ be an essentially self-adjoint diffusion operator on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$. Show that if the constant function $1 \in \mathcal{D}(L)$ and if $L1=0$, then $\mathbf{P}_t 1=1.$

Exercise: Let $L$ be an essentially self-adjoint diffusion operator on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$

• Show that for every $\lambda > 0$, the range of the operator $\lambda \mathbf{Id}-L$ is dense in $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$.
• By using the spectral theorem, show that the following limit holds for the operator norm on $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, $\mathbf{P}_t=\lim_{n \to +\infty} \left( \mathbf{Id} -\frac{t}{n} L\right)^{-n}.$

Exercise: As usual, we denote by $\Delta$ the Laplace operator on $\mathbb{R}^n$. The Mac-Donald’s function with index $\nu \in \mathbb{R}$ is defined for $x \in \mathbb{R} \setminus \{ 0 \}$ by $K_\nu (x)=\frac{1}{2} \left( \frac{x}{2} \right)^\nu \int_0^{+\infty} \frac{e^{-\frac{x^2}{4t} -t}}{t^{1+\nu}} dt$.

• Show that for $\lambda \in \mathbb{R}^n$ and $\alpha > 0$, $\frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n} e^{i \langle \lambda , x \rangle} \left( \frac{\| x \| }{\sqrt{\alpha}} \right)^{1-\frac{n}{2}} K_{\frac{n}{2}-1} (\sqrt{\alpha} \| x \| ) dx=\frac{1}{\alpha +\| \lambda \|^2}.$
• Show that for $\nu \in \mathbb{R}$, $K_{-\nu}=K_\nu$.
• Show that $K_{1/2}(x)=\sqrt {\frac{\pi}{2x}} e^{-x}.$
• Prove that for $f \in \mathbf{L}^2 (\mathbb{R}^n,\mathbb{R})$ and $\alpha > 0$, $(\alpha\mathbf{Id}-\Delta)^{-1} f (x)=\int_{\mathbb{R}^n} G_\alpha (x-y) f(y) dy,$ where $G_\alpha(x)=\frac{1}{(2\pi)^{n/2}} \left( \frac{\| x \| }{\sqrt{\alpha}} \right)^{1-\frac{n}{2}} K_{\frac{n}{2}-1} (\sqrt{\alpha} \| x \| ).$ (You may use Fourier transform to solve the partial differential equation $\alpha g -\Delta g=f$).

Exercise: By using the previous exercise, prove that for $f \in \mathbf{L}^2 (\mathbb{R}^n,\mathbb{R})$, $\lim_{n \to + \infty} \left(\mathbf{Id} -\frac{t}{n} \Delta \right)^{-n} f =\frac{1}{(4\pi t)^{\frac{n}{2}}} \int_{\mathbb{R}^n} e^{-\frac{\|x-y\|^2}{4t} } f(y)dy,$ the limit being taken in $\mathbf{L}^2 (\mathbb{R}^n,\mathbb{R})$. Conclude that almost everywhere, $\mathbf{P}_t f (x)=\frac{1}{(4\pi t)^{\frac{n}{2}}} \int_{\mathbb{R}^n} e^{-\frac{\|x-y\|^2}{4t} } f(y)dy.$

Exercise:

• Show the subordination identity $e^{-y | \alpha | } =\frac{y}{2\sqrt{\pi}} \int_0^{+\infty} \frac{e^{-\frac{y^2}{4t}-t \alpha^2}}{t^{3/2}} dt, \quad y > 0, \alpha \in \mathbb{R}.$
• The Cauchy’s semigroup on $\mathbb{R}^n$ is defined as $\mathbf{Q}_t=e^{-t \sqrt{-\Delta}}$. By using the subordination identity and the heat semigroup on $\mathbb{R}^n$, show that for $f \in \mathbf{L}^2 (\mathbb{R}^n,\mathbb{R})$, $\mathbf{Q}_tf (x)=\int_{\mathbb{R}^n} q(t,x-y) f(y) dy,$ where $q(t,x)=\frac{\Gamma\left( \frac{n+1}{2} \right)}{\pi^{\frac{n+1}{2}} } \frac{t}{(t^2+\| x \|^2 )^{\frac{n+1}{2}} }.$

## 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 $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$.

## Lecture 1. Diffusion operators

Definition: A differential operator $L$ on $\mathbb{R}^n$, is called a diffusion operator if it can be written

$L=\sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial^2}{ \partial x_i \partial x_j} +\sum_{i=1}^n b_i (x)\frac{\partial}{\partial x_i},$

where $b_i$ and $\sigma_{ij}$ are continuous functions on $\mathbb{R}^n$ and if for every $x \in \mathbb{R}^n$, the matrix $(\sigma_{ij}(x))_{1\le i,j\le n}$ is a symmetric and nonnegative matrix.

If for every $x \in \mathbb{R}^n$ the matrix $(\sigma_{ij}(x))_{1\le i,j\le n}$ is positive definite, then the operator $L$ is said to be elliptic. The first example of a diffusion operator is the Laplace operator on $\mathbb{R}^n$:

$\Delta=\sum_{i=1}^n \frac{\partial^2}{\partial x_i^2}.$

It is of course an elliptic operator.

One of the first property of diffusion operators is that they satisfy a maximum principle. Before we state this principle let us recall the following simple result from linear algebra.

Lemma. Let $A$ and $B$ be two symmetric and nonnegative matrices, then $\mathbf{tr} (AB) \ge 0.$

Proof:

Since $A$ is symmetric and non negative, there exists a symmetric and non negative matrix $S$ such that $S^2=A$. We have then

$\mathbf{tr} ( AB)=\mathbf{tr} ( S^2 B)=\mathbf{tr} ( S B S)=\mathbf{tr} ( S^* BS).$

The matrix $S^* BS$ is seen to be symmetric and nonnegative and thus has a non negative trace.$\square$

Proposition (Maximum principle for diffusion operators). Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be a smooth function that attains a local minimum at $x$. If $L$ is a diffusion operator then $Lf(x) \ge 0$.

Proof:

Let
$L=\sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial^2}{ \partial x_i \partial x_j} +\sum_{i=1}^n b_i(x)\frac{\partial}{\partial x_i},$

and let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be a smooth function that attains a local minimum at $x$. We have

$Lf(x) =\sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial^2 f}{ \partial x_i \partial x_j} (x)$
$=\mathbf{tr} \left( \sigma (x) \mathbf{Hess } f (x) \right),$

where $\sigma(x)$ is the symmetric and non negative matrix with coefficients $\sigma_{ij}(x)$ and $\mathbf{Hess } f (x)$ is the Hessian matrix of $f$, that is the symmetric matrix with coefficients $\frac{\partial^2 f}{ \partial x_i \partial x_j} (x)$. Since $x$ is a local minimum of $f$, $\mathbf{Hess } f (x)$ is a non negative matrix. We can now use the previous lemma to get the expected result $\square$

It is remarkable that, together with the linearity, this maximum principle characterizes the diffusion operators:

Theorem. Let $\mathcal{C}^{\infty} (\mathbb{R}^n)$ be the set of smooth functions $\mathbb{R}^n \rightarrow \mathbb{R}$ and let $\mathcal{C} (\mathbb{R}^n)$ be the set of continuous functions $\mathbb{R}^n \rightarrow \mathbb{R}$. Let now $L: \mathcal{C}^{\infty} (\mathbb{R}^n) \rightarrow \mathcal{C} (\mathbb{R}^n)$ be an operator such that:

• $L$ is linear;
• If $f \in \mathcal{C}^{\infty} (\mathbb{R}^n)$ has a local minimum at $x$, $Lf (x) \ge 0$.

Then $L$ is a diffusion operator.

Proof:

Let us consider an operator $L$ that satisfies the two above properties. As a first observation, it is readily seen from the third point that $L$ transforms constant functions into the zero function. Let now $y \in \mathbb{R}^n$ be fixed in the following argument. We are going to show that if $g$ is a smooth function, then

$L( \| x-y \|^3 g) (y) =0.$

The idea will then be to use the Taylor expansion formula. For $\varepsilon >0$, the function

$x \rightarrow \| x-y \|^3 g(x) +\varepsilon \| x -y \|^2$

admits a local minimum at $y$, thus

$L( \| x-y \|^3 g) (y) \ge - \varepsilon L ( \| x -y \|^2)(y).$

By letting $\varepsilon \to 0$, we therefore obtain

$L( \| x-y \|^3 g) (y) \ge 0.$

By considering now the function

$x \rightarrow \| x-y \|^3 g(x) -\varepsilon \| x -y \|^2,$

we show in the very same way that

$L( \| x-y \|^3 g) (y) \le 0.$

As a conclusion

$L( \| x-y \|^3 g) (y) = 0.$

Let now $f$ be a smooth function. By the Taylor expansion formula, there exists a smooth function $g$ such that in a neighborhood of $y$

$f(x)$
$=f(y)+\sum_{i=1}^n (x_i -y_i) \frac{\partial f}{\partial x_i}(y)+\frac{1}{2} \sum_{i,j=1}^n (x_i-y_i)(x_j-y_j) \frac{\partial^2 f}{\partial x_i \partial x_j} (y) + \| x-y \|^3 g(x).$

By applying the operator $L$ to the previous equality, and by taking account the previous observations we obtain

$Lf(y)=\sum_{i=1}^n L(x_i -y_i)(y) \frac{\partial f}{\partial x_i}(y)+\frac{1}{2} \sum_{i,j=1}^n L((x_i-y_i)(x_j-y_j))(y) \frac{\partial^2 f}{\partial x_i \partial x_j} (y).$

By denoting now,

$b_i(y)= L(x_i -y_i)(y),$

and

$\sigma_{ij} (y)=\frac{1}{2} L((x_i-y_i)(x_j-y_j))(y) ,$

we reach the conclusion

$Lf(y)=\sum_{i,j=1}^n \sigma_{ij} (y) \frac{\partial^2 f}{ \partial x_i \partial x_j}(y) +\sum_{i=1}^n b_i (y)\frac{\partial f}{\partial x_i}(y).$

The matrix, $(\sigma_{ij}(y))_{1\le i,j \le n}$ is seen to be non negative, because for every $\lambda \in \mathbb{R}^n$,

$\sum_{i,j=1}^n \lambda_i \lambda_j \sigma_{ij} (y)$
$= \frac{1}{2}\sum_{i,j=1}^n \lambda_i \lambda_j L((x_i-y_i)(x_j-y_j))(y)$
$=\frac{1}{2} L( \langle \lambda , x-y \rangle^2)(y).$

Now, the function $x \rightarrow \langle \lambda , x-y \rangle^2$ is seen to attain a local minimum at $y$, so that from the maximum principle

$L( \langle \lambda , x-y \rangle^2)(y) \ge 0.$

Finally, since $L$ transforms smooth into continuous functions, the functions $b_i$‘s and $\sigma_{ij}$‘s are seen to be continuous $\square$

Exercise. Let $L: \mathcal{C}^{\infty} (\mathbb{R}^n) \rightarrow \mathcal{C} (\mathbb{R}^n)$ be a linear operator such that:

• $L$ is a local operator, that is if $f=g$ on a neighborhood of $x$ then $Lf(x)=Lg(x)$;
• If $f \in \mathcal{C}^{\infty} (\mathbb{R}^n)$ has a global maximum at $x$ with $f(x)\ge 0$ then $Lf (x) \le 0$.

Show that for $f \in \mathcal{C}^{\infty} (\mathbb{R}^n)$ and $x \in \mathbb{R}^n$,

$Lf(x)=\sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial^2 f}{ \partial x_i \partial x_j} +\sum_{i=1}^n b_i (x)\frac{\partial f}{\partial x_i} -c(x)f(x),$

where $b_i$, $c$ and $\sigma_{ij}$ are continuous functions on $\mathbb{R}^n$ such that for every $x \in \mathbb{R}^n$, $c(x) \ge 0$ and the matrix $(\sigma_{ij}(x))_{1\le i,j\le n}$ is symmetric and nonnegative.

The previous theorem is actually a special case of a beautiful theorem that is due to Courrège that classifies the operators satisfying the positive global maximum principle. We mention this theorem without proof because the result will not be needed in the following. A complete proof may be found in the original article by Courrège.

We denote by $\mathcal{C}_c^{\infty} (\mathbb{R}^n)$ the space of smooth and compactly supported functions $\mathbb{R}^n \rightarrow \mathbb{R}$. A linear operator $A:\mathcal{C}_c^{\infty} (\mathbb{R}^n) \rightarrow \mathcal{C} (\mathbb{R}^n)$ is said to satisfy the positive maximum principle if for every function $f \in \mathcal{C}_c^{\infty}(\mathbb{R}^n)$ that has a global maximum at $x$ with $f(x)\ge 0$ then $Af (x) \le 0$.

In the following statement $\mathcal{B}(\mathbb{R}^n)$ denotes the set of Borel sets on $\mathbb{R}^n$ and a kernel $\mu$ on $\mathbb{R}^n \times \mathcal{B}(\mathbb{R}^n)$ is a family $\left\{ \mu(x,\cdot), x \in \mathbb{R}^n\right\}$ of Borel measures.

Theorem (Courrège’s theorem) Let $A:\mathcal{C}_c^{\infty} (\mathbb{R}^n) \rightarrow \mathcal{C} (\mathbb{R}^n)$ be a linear operator. Then $A$ satisfies the positive maximum principle if and only if there exist functions $(\sigma_{ij}(x))_{1\le i,j\le n}$, $b_i$, $c:\mathbb{R}^n \rightarrow \mathbb{R}$ and a kernel $\mu$ on $\mathbb{R}^n \times \mathcal{B}(\mathbb{R}^n)$ such that for every $f \in \mathcal{C}_c^{\infty} (\mathbb{R}^n)$ and $x \in \mathbb{R}^n$,

$Af(x) = \sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial^2 f}{ \partial x_i \partial x_j} +\sum_{i=1}^n b_i (x)\frac{\partial f}{\partial x_i} -c(x)f(x)$
$+\int_{\mathbb{R}^n} \left( u(y) -\chi(y-x)u(x)-\sum_{j=1}^n\frac{\partial u}{\partial x_j}(x)\chi(y-x)(y_j-x_j) \right)\mu(x,dy),$

where $\chi \in \mathcal{C}_c^{\infty} (\mathbb{R}^n)$, with $0 \le \chi \le 1$ takes the constant value 1 on the ball $\mathbf{B}(0,1)$. In addition, for every $x \in \mathbb{R}^n$, $c(x) \ge 0$ and the matrix $(\sigma_{ij}(x))_{1\le i,j\le n}$ is a symmetric and nonnegative matrix. The functions $b_j$‘s and $c$ are continuous. Moreover for every $y \in \mathbb{R}^n$, the function $x \to \sum_{i,j} \sigma_{ij}(x) y_i y_j$ is upper semicontinuous.

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