Research and Lecture notes

HW. Due October 25

Posted on October 18, 2016 by Fabrice Baudoin

Exercise 1. Let $L$ be a locally subelliptic and essentially self-adjoint diffusion operator. Let $P_t$ be the semigroup generated by $L$ . By using the maximum principle for parabolic pdes, prove that if $f \ge 0$ is in $L^2$ , then $P_t f \ge 0$ .

Exercise 2: Let $L$ be an essentially self-adjoint diffusion operator. Denote by $P_t^{(p)}$ the semigroup generated by $L$ in $L^p$ .

Show that for each $f \in \mathbf{L}_{\mu}^{1}$ , the $\mathbf{L}_{\mu}^{1}$ -valued map $t \rightarrow \mathbf{P}^{(1)}_t f$ is continuous.
Show that for each $f \in \mathbf{L}_{\mu}^{p}$ , $1 < p < 2$ , the $\mathbf{L}_{\mu}^{p}$ -valued map $t \rightarrow \mathbf{P}^{(p)}_t f$ is continuous.
Finally, by using the reflexivity of $\mathbf{L}_{\mu}^{p}$ , show that for each $f \in \mathbf{L}_{\mu}^{p}$ and every $p \ge 1$ , the $\mathbf{L}_{\mu}^{p}$ -valued map $t \rightarrow \mathbf{P}^{(p)}_t f$ is continuous.

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Lecture 10. The heat semigroup on the circle

Posted on October 18, 2016 by Fabrice Baudoin

In the next few lectures, we will show that the diffusion semigroups theory we developed may actually be extended without difficulties to a manifold setting. As a motivation, we start with a very simple example.

We first study the heat semigroup on the simplest (non Euclidean) Riemannian manifold: the circle $\mathbb{S}^1= \left\{ e^{i\theta}, \theta \in \mathbb{R} \right\}.$ The Laplace operator on $\mathbb{R}^n$ , $\Delta=\sum_{i=1}^n \frac{\partial^2}{\partial x_i^2}$ is the canonical diffusion operator on $\mathbb{R}^n$ . A natural question to be asked is: in the same way, is there a canonical diffusion operator on $\mathbb{S}^1$ . A first step, of course, is to understand what is a diffusion operator on $\mathbb{S}^1$ . We characterized diffusion operators as linear operators on the space of smooth functions that satisfy the maximum principle. Once a notion of smooth functions on $\mathbb{S}^1$ is understood, this maximum principle property can be taken as a definition. The circle $\mathbb{S}^1$ may be identified with the quotient space $\mathbb{R} / 2\pi \mathbb{Z}$ . More precisely, it is easily shown that a smooth function, $f :\mathbb{R} \rightarrow \mathbb{C}$ which is $2\pi$ periodic, i.e. $f(\theta+2\pi)=f(\theta),$ can be written as $f(\theta)=g\left( e^{i\theta} \right),$ for some function $g: \mathbb{S}^1 \rightarrow \mathbb{C}$ . Conversely, any function $g: \mathbb{S}^1 \rightarrow \mathbb{C}$ defines a $2\pi$ periodic function on $\mathbb{R}$ by setting $f(\theta)=g\left( e^{i\theta} \right).$ So, with this in mind, we simply say that $g: \mathbb{S}^1 \rightarrow \mathbb{C}$ is a smooth function if $f$ is. With this identification between the set of smooth $2\pi$ periodic functions on $\mathbb{R}$ and the set of smooth functions on $\mathbb{S}^1$ , it then immediate that the canonical diffusion operator $\Delta_{\mathbb{S}^1}$ on $\mathbb{S}^1$ should write, $\Delta_{\mathbb{S}^1} g ( e^{i\theta})=f''(\theta).$ The corresponding diffusion semigroup is also easily computed from the heat semigroup on $\mathbb{R}$ . Indeed, a natural computation leads to

$\left( e^{t \Delta_{\mathbb{S}^1} }g \right) ( e^{i\theta}) =\left( e^{t \frac{d^2}{d\theta^2} }f \right) ( \theta)$
$= \int_{-\infty}^{\infty} f(y+\theta) \frac{e^{-\frac{y^2}{4t} } }{\sqrt{4\pi t}} dy$
$=\sum_{k \in \mathbb{Z}} \int_{2k \pi }^{2k\pi +2\pi} f(y+\theta) \frac{e^{-\frac{y^2}{4t} } }{\sqrt{4\pi t}} dy$
$=\sum_{k \in \mathbb{Z}} \int_{0 }^{2\pi} f(y-2k\pi+\theta) \frac{e^{-\frac{(y-2k\pi)^2}{4t} } }{\sqrt{4\pi t}} dy$
$= \int_{0 }^{2\pi} f(y+\theta) p(t,y) dy$ ,
where $p(t,y)=\frac{1}{\sqrt{4\pi t}} \sum_{k \in \mathbb{Z}} e^{-\frac{(y-2k\pi)^2}{4t} }$ . This allows to define the heat semigroup on $\mathbb{S}^1$ as the family of operators defined by $\mathbf{P}_t g (e^{i\theta} )= \int_{0 }^{2\pi} g(e^{i\nu }) p(t,\theta-\nu) d\nu.$ The natural domain of this operator is $\mathbf{L}^2_\mu(\mathbb{S}^1,\mathbb{R})$ where $\mu$ is the measure on $\mathbb{S}^1$ which is defined through the property $\int_{\mathbb{S}^1} g d\mu = \int_0^{2\pi} f(\theta) d\theta.$ The reader may then check the following properties for this semigroup of operators:

(Semigroup property) $\mathbf{P}_{t+s} =\mathbf{P}_t \mathbf{P}_s$ ;
(Strong continuity) The map $t \rightarrow \mathbf{P}_t$ is continuous for the operator norm on $\mathbf{L}^2_\mu(\mathbb{S}^1,\mathbb{R})$ ;
(Contraction property) $\|\mathbf{P}_t g \|_{\mathbf{L}^2_\mu(\mathbb{S}^1,\mathbb{R}) } \le \| g \|_{\mathbf{L}^2_\mu(\mathbb{S}^1,\mathbb{R}) }$ ;
(Self-adjointness) For $g_1,g_2 \in \mathbf{L}^2_\mu(\mathbb{S}^1,\mathbb{R})$ , $\int_{\mathbb{S}^1} (\mathbf{P}_t g_1) g_2 d\mu=\int_{\mathbb{S}^1} g_1(\mathbf{P}_t g_2) d\mu$
(Markov property) If $g \in \mathbf{L}^2_\mu(\mathbb{S}^1,\mathbb{R})$ is such that $0 \le g \le 1$ , then $0 \le \mathbf{P}_t g \le 1$ .

Exercise.

Prove the Poisson summation formula: If $f:\mathbb{R} \rightarrow \mathbb{R}$ is a smooth and rapidly decreasing function, then $\sum_{m \in \mathbb{Z}} f(m) e^{im \theta}=\sum_{k \in \mathbb{Z}} \hat{f} (\theta -2k\pi).$
Deduce that the heat kernel on $\mathbb{S}^1$ may also be written $p(t,y)=\frac{1}{2\pi}\sum_{m \in \mathbb{Z}} e^{-m^2 t} e^{im y}.$

Exercise. From the previous exercise, the heat kernel on $\mathbb{S}^1$ is given by $p(t,y) =\frac{1}{2\pi}\sum_{m \in \mathbb{Z}} e^{-m^2 t} e^{im y} =\frac{1}{\sqrt{4\pi t}} \sum_{k \in \mathbb{Z}} e^{-\frac{(y 2k\pi)^2}{4t} }$ .

By using the subordination identity $e^{-\tau | \alpha | } =\frac{\tau}{2\sqrt{\pi}} \int_0^{+\infty} \frac{e^{-\frac{\tau^2}{4t}-t \alpha^2}}{t^{3/2}} dt, \quad \tau \neq 0, \alpha \in \mathbb{R},$ show that for $\tau > 0$ , $\frac{1+e^{-2\pi \tau}}{1-e^{-2\pi \tau}} =\frac{1}{2\pi} \sum_{k \in \mathbb{Z}} \frac{2\tau}{\tau^2+n^2}$
The Bernoulli numbers $B_k$ are defined via the series expansion $\frac{x}{e^x -1}=\sum_{k=0}^{+\infty} B_k \frac{x^k}{k!}.$ By using the previous identity show that for $k \in \mathbb{N}$ , $k \neq 0$ , $\sum_{n=1}^{+\infty} \frac{1}{n^{2k}} =(-1)^{k-1} \frac{(2\pi)^{2k} B_{2k} }{2(2k)!}.$

Exercise. Show that the heat kernel on the torus $\mathbb{T}^n=\mathbb{R}^n / (2 \pi \mathbb{Z})^n$ is given by $p(t,y) = \frac{1}{(4\pi t)^{n/2}} \sum_{k \in \mathbb{Z}^n} e^{-\frac{\|y+2k\pi\|^2}{4t} }=\frac{1}{(2\pi)^n} \sum_{l\in \mathbb{Z}^n} e^{i l \cdot y -\| l \|^2 t}.$

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Lecture 9. Diffusion semigroups in L^p

Posted on October 11, 2016 by Fabrice Baudoin

In the previous lectures, we have seen that if L is an essentially self-adjoint diffusion operator with respect to a measure, then by using the spectral theorem one can define a self-adjoint strongly continuous contraction semigroup on L2 with generator L. This semigroup is moreover positivity preserving and a contraction on the space of bounded square integrable functions. Our goal, in this lecture, is to define, for $1 \le p \le +\infty$ , the semigroup on $\mathbf{L}_{\mu}^p (\mathbb{R}^n,\mathbb{R})$ . This can be done by using the Riesz–Thorin interpolation theorem that we remind below. In this subsection, in order to simplify the notations we simply denote $\mathbf{L}_{\mu}^p (\mathbb{R}^n,\mathbb{R})$ by $\mathbf{L}_\mu^p$ . We first start with general comments about semigroups in Banach spaces.

Let $(B,\| \cdot \|)$ be a Banach space (which for us will be $\mathbf{L}_{\mu}^p$ , $1 \le p \le +\infty$ ).

We first have the following basic definition.

Definition: A family of bounded operators $(T_t)_{t \ge 0}$ on $B$ is called a contraction semigroup if:

$T_0 =\mathbf{Id}$ and for $s,t \ge 0$ , $T_{s+t}=T_s T_t$ ;
For each $x \in B$ and $t \ge 0$ , $\| T_t x \| \le \|x \|$ .

A contraction semigroup $(T_t)_{t \ge 0}$ on $B$ is moreover said to be strongly continuous if for each $x \in B$ , the map $t \to T_t x$ is continuous.

In this Lecture, we will prove the following result:

Theorem: let $L$ be an essentially self-adjoint diffusion operator. Denote by $(\mathbf{P}_t)_{t \ge 0}$ the self-adjoint strongly continuous semigroup associated to $L$ and constructed on $\mathbf{L}_{\mu}^2$ thanks to the spectral theorem. Let $1 \le p \le +\infty$ . On $\mathbf{L}_\mu^p$ , there exists a unique contraction semigroup $(\mathbf{P}^{(p)}_t)_{t \ge 0}$ such that for $f \in \mathbf{L}_\mu^p \cap \mathbf{L}_\mu^2$ , $\mathbf{P}^{(p)}_t f=\mathbf{P}_t f$ . The semigroup $(\mathbf{P}^{(p)}_t)_{t \ge 0}$ is moreover strongly continuous for $1 \le p < +\infty$ .

This theorem can be proved by using two sets of methods: Hille-Yosida theory on one hand and interpolation theory on the other hand. We shall use interpolation theory in $L^p$ space which takes advantage of the fact that $\mathbf{P}_t$ only needs to be constructed on $L^1$ and $L^\infty$ . The Hille-Yosida method starts from the generator and we sketch it below.

Definition:
Let $(T_t)_{t \ge 0}$ be a strongly continuous contraction semigroup on a Banach space $B$ . There exists a closed and densely defined operator $A: \mathcal{D}(A) \subset B \rightarrow B$ where $\mathcal{D}(A)=\left\{ x \in B,\quad \lim_{t \to 0} \frac{T_t x -x}{t} \text{ exists} \right\},$ such that for $x \in \mathcal{D}(A)$ , $\lim_{t \to 0} \left\| \frac{T_t x -x}{t} -Ax \right\|=0.$ The operator $A$ is called the generator of the semigroup $(T_t)_{t \ge 0}$ . We also say that $A$ generates $(T_t)_{t \ge 0}$ .

The following important theorem is due to Hille and Yosida and provides, through spectral properties, a characterization of closed operators that are generators of contraction semigroups.

Let $A: \mathcal{D}(A) \subset B \rightarrow B$ be a densely defined closed operator. A constant $\lambda \in \mathbb{R}$ is said to be in the spectrum of $A$ if the the operator $\lambda \mathbf{Id}-A$ is not bijective. In that case, it is a consequence of the closed graph theorem that if $\lambda$ is not in the spectrum of $A$ , then the operator $\lambda \mathbf{Id}-A$ has a bounded inverse. The spectrum of an operator $A$ shall be denoted $\rho(A)$ .

Theorem: A necessary and sufficient condition that a densely defined closed operator $A$ generates a strongly continuous contraction semigroup is that:

$\rho (A) \subset (-\infty,0]$ ;
$\| (\lambda \mathbf{Id} -A)^{-1} \| \le \frac{1}{\lambda}$ for all $\lambda > 0$ .

These two conditions are unfortunately difficult to directly check for diffusion operators.
We can bypass the study of the closure in $L^p$ of a diffusion operator by using interpolation theory.

Theorem: (Riesz-Thorin interpolation theorem) Let $1 \le p_0, p_1,q_0,q_1 \le \infty$ , and $\theta \in (0,1)$ . Define $1 \le p,q \le \infty$ by $\frac{1}{p}=\frac{1-\theta}{p_0} + \frac{\theta}{p_1}, \quad \frac{1}{q}=\frac{1-\theta}{q_0} + \frac{\theta}{q_1}.$ If $T$ is a linear map such that $T:\mathbf{L}_\mu^{p_0} \rightarrow \mathbf{L}_\mu^{q_0}, \quad \| T \|_{ \mathbf{L}_\mu^{p_0} \rightarrow \mathbf{L}_\mu^{q_0} } =M_0$ and $T:\mathbf{L}_\mu^{p_1} \rightarrow \mathbf{L}_\mu^{q_1}, \quad \| T \|_{ \mathbf{L}_\mu^{p_1} \rightarrow \mathbf{L}_\mu^{q_1} } =M_1,$ then, for every $f \in \mathbf{L}_\mu^{p_0} \cap \mathbf{L}_\mu^{p_1}$ , $\| T f \|_q \le M_0^{1-\theta} M_1^{\theta} \| f \|_p.$ Hence $T$ extends uniquely as a bounded map from $\mathbf{L}_\mu^{p}$ to $\mathbf{L}_\mu^{q}$ with $\| T \|_{ \mathbf{L}_\mu^{p} \rightarrow \mathbf{L}_\mu^{q} } \le M_0^{1-\theta} M_1^{\theta} .$

The statement that $T$ is a linear map such that $T:\mathbf{L}_\mu^{p_0} \rightarrow \mathbf{L}_\mu^{q_0}, \quad \| T \|_{ \mathbf{L}_\mu^{p_0} \rightarrow \mathbf{L}_\mu^{q_0} } =M_0$ and $T:\mathbf{L}_\mu^{p_1} \rightarrow \mathbf{L}_\mu^{q_1}, \quad \| T \|_{ \mathbf{L}_\mu^{p_1} \rightarrow \mathbf{L}_\mu^{q_1} } =M_1$ means that there exists a map $T: \mathbf{L}_\mu^{p_0} \cap \mathbf{L}_\mu^{p_1}\rightarrow \mathbf{L}_\mu^{q_0} \cap \mathbf{L}_\mu^{q_1}$ with $\sup_{ f \in \mathbf{L}_\mu^{p_0} \cap \mathbf{L}_\mu^{p_1} , \| f \|_{p_0} \le 1 } \| Tf \|_{q_0} =M_0$ and $\sup_{ f \in \mathbf{L}_\mu^{p_0} \cap \mathbf{L}_\mu^{p_1} , \| f \|_{p_1} \le 1 } \| Tf \|_{q_1} =M_1.$ In such a case, $T$ can be uniquely extended to bounded linear maps $T_0: \mathbf{L}_\mu^{p_0} \rightarrow \mathbf{L}_\mu^{q_0}$ , $T_1: \mathbf{L}_\mu^{p_1} \rightarrow \mathbf{L}_\mu^{q_1}$ . With a slight abuse of notation, these two maps are both denoted by $T$ in the theorem.

The proof of the theorem can be found in this post by Tao.

One of the (numerous) beautiful applications of the Riesz-Thorin theorem is to construct diffusion semigroups on $L^p$ by interpolation. More precisely, let $L$ be an essentially self-adjoint diffusion operator. We denote by $(\mathbf{P}_t)_{t \ge 0}$ the self-adjoint strongly continuous semigroup associated to $L$ constructed on $\mathbf{L}_{\mu}^2$ thanks to the spectral theorem. We recall that $(\mathbf{P}_t)_{t \ge 0}$ satisfies the submarkov property: That is, if $0 \le f \le 1$ is a function in $\mathbf{L}_{\mu}^2$ , then $0 \le \mathbf{P}_t f \le 1$ .

Theorem: The space $\mathbf{L}_{\mu}^1 \cap \mathbf{L}_{\mu}^{\infty}$ is invariant under $\mathbf{P}_t$ and $\mathbf{P}_t$ may be extended from $\mathbf{L}_{\mu}^1 \cap \mathbf{L}_{\mu}^{\infty}$ to a contraction semigroup $(\mathbf{P}^{(p)}_t)_{t \ge 0}$ on $\mathbf{L}_{\mu}^{p}$ for all $1 \le p \le \infty$ : For $f \in \mathbf{L}_{\mu}^p$ , $\| \mathbf{P}_t f \|_{ \mathbf{L}_{\mu}^p} \le \| f \|_{ \mathbf{L}_{\mu}^p}.$ These semigroups are consistent in the sense that for $f \in \mathbf{L}_{\mu}^p \cap \mathbf{L}_{\mu}^{q}$ , $\mathbf{P}^{(p)}_t f=\mathbf{P}^{(q)}_t f.$

Proof: If $f,g \in \mathbf{L}_{\mu}^1 \cap \mathbf{L}_{\mu}^{\infty}$ which is a subset of $\mathbf{L}_{\mu}^1 \cap \mathbf{L}_{\mu}^{\infty}$ , then $\left| \int_{\mathbb{R}^n} (\mathbf{P}_t f) g d\mu \right| = \left| \int_{\mathbb{R}^n} f(\mathbf{P}_t g) d\mu \right | \le \| f \|_{ \mathbf{L}_{\mu}^1} \| \mathbf{P}_t g \|_{ \mathbf{L}_{\mu}^\infty} \le \| f \|_{ \mathbf{L}_{\mu}^1} \| g \|_{ \mathbf{L}_{\mu}^\infty}.$ This implies $\| \mathbf{P}_t f \|_{ \mathbf{L}_{\mu}^1} \le \| f \|_{ \mathbf{L}_{1}^\infty}.$ The conclusion follows then from the Riesz-Thorin interpolation theorem $\square$

Exercise: Show that if $f \in \mathbf{L}_{\mu}^{p}$ and $g \in \mathbf{L}_{\mu}^{q}$ with $\frac{1}{p}+\frac{1}{q}=1$ then, $\int_{\mathbb{R}^n} f \mathbf{P}^{(q)}_t g d\mu=\int_{\mathbb{R}^n} g \mathbf{P}^{(p)}_t f d\mu.$

Exercise:

Show that for each $f \in \mathbf{L}_{\mu}^{1}$ , the $\mathbf{L}_{\mu}^{1}$ -valued map $t \rightarrow \mathbf{P}^{(1)}_t f$ is continuous.
Show that for each $f \in \mathbf{L}_{\mu}^{p}$ , $1 < p < 2$ , the $\mathbf{L}_{\mu}^{p}$ -valued map $t \rightarrow \mathbf{P}^{(p)}_t f$ is continuous.
Finally, by using the reflexivity of $\mathbf{L}_{\mu}^{p}$ , show that for each $f \in \mathbf{L}_{\mu}^{p}$ and every $p \ge 1$ , the $\mathbf{L}_{\mu}^{p}$ -valued map $t \rightarrow \mathbf{P}^{(p)}_t f$ is continuous.

We mention, that in general, the $\mathbf{L}_{\mu}^{\infty}$ valued map $t \rightarrow \mathbf{P}^{(\infty)}_t f$ is not continuous.

Due to the consistency property, we always remove the subscript $p$ from $\mathbf{P}^{(p)}_t$ and only use the notation $\mathbf{P}_t$ .

To finish this Lecture, we finally connect the heat semigroup in $L^p$ to $L^p$ solutions of the heat equation.

Proposition: Let $f \in L^p_\mu(\mathbb{R}^n)$ , $1 \le p \le \infty$ , and let $u (t,x)= P_t f (x), \quad t \ge 0, x\in \mathbb{R}^n.$ Then, if $L$ is elliptic with smooth coefficients, $u$ is smooth on $(0,+\infty)\times \mathbb{R}^n$ and is a strong solution of the Cauchy problem $\frac{\partial u}{\partial t}= L u,\quad u (0,x)=f(x).$

Proof: The proof is identical to the $L^2$ case. For $\phi \in C^\infty_0 ((0,+\infty) \times \mathbb{R}^n)$ , we have
$\int_{\mathbb{R}^n \times \mathbb{R}} \left( \left( -\frac{\partial}{\partial t} -L \right) \phi (t,x) \right) u(t,x) d\mu(x) dt$
$=\int_{\mathbb{R}} \int_{\mathbb{R}^n} \left( \left( -\frac{\partial}{\partial t} -L \right) \phi (t,x) \right) P_t f (x) dx dt$
$= \int_{\mathbb{R}} \int_{\mathbb{R}^n} P_t \left( \left( -\frac{\partial}{\partial t} -L \right) \phi (t,x) \right) f (x) dx dt$
$= \int_{\mathbb{R}} \int_{\mathbb{R}^n} -\frac{\partial}{\partial t} \left( P_t \phi (t,x) f(x) \right) dx dt =0.$
Therefore $u$ is a weak solution of the equation $\frac{\partial u}{\partial t}= L u$ . Since $u$ is smooth it is also a strong solution $\square$ .

We now address the uniqueness of solutions. As in the $L^2$ case, we assume that $L$ is elliptic with smooth coefficients and that there is a sequence $h_n\in C_0^\infty(\mathbb{R}^n)$ , $0 \le h_n \le 1$ , such that $h_n\nearrow 1$ on $\mathbb{R}^n$ , and $||\Gamma (h_n,h_n)||_{\infty} \to 0$ , as $n\to \infty$ .

Proposition: Let $v(x,t)$ be a non negative function such that $\frac{\partial v}{\partial t} \le L v,\quad v(x,0)=0,$ and such that for every $t > 0$ , $\| v ( \cdot,t) \|_{L^p_\mu(\mathbb{R}^n)} <+\infty$ , where $1 < p < +\infty$ . Then $v(x,t)=0$ .

Proof: Let $x_0 \in \mathbb{R}^n$ and $h \in C_0^\infty(\mathbb{R}^n)$ . Since $u$ is a subsolution with the zero initial data, for any $\tau\in (0,T)$ ,
$\int_0^\tau \int_{\mathbb{R}^n} h^2(x) v^{p-1}(x,t) L v(x,t) d\mu(x) dt$
$\geq \int_0^\tau \int_{\mathbb{R}^n} h^2(x) v^{p-1} \frac{\partial v}{\partial t} d\mu(x) dt$
$= \frac{1}{p} \int_0^\tau \frac{\partial }{\partial t} \left( \int_{\mathbb{R}^n} h^2(x) v^{p} d\mu(x)\right) dt$
$= \frac{1}{p}\int_{\mathbb{R}^n} h^2(x) v^{p}(x,\tau) d\mu(x).$
On the other hand, integrating by parts yields
$\int_0^\tau \int_{\mathbb{R}^n} h^2(x) v^{p-1}(x,t) L v(x,t) d\mu(x) dt = - \int_0^\tau \int_{\mathbb{R}^n} 2h v^{p-1} \Gamma(h,v) d\mu dt - \int_0^\tau \int_{\mathbb{R}^n} h^2 (p-1) v^{p-2} \Gamma(v) d\mu dt .$

Observing that
$0\leq \left(\sqrt{\frac{2}{p-1}\Gamma(h)}v - \sqrt{\frac{p-1}{2}\Gamma(v)}h \right)^2 \leq \frac{2}{p-1}\Gamma(h)v^2 + 2 \Gamma(h,v) h v +\frac{p-1}{2}\Gamma(v)h^2 ,$
we obtain the following estimate.
$\int_0^\tau \int_{\mathbb{R}^n} h^2(x) v^{p-1}(x,t) L v(x,t) d\mu(x) dt$
$\leq \int_0^\tau \int_{\mathbb{R}^n} \frac{2}{p-1} \Gamma(h) v^p d\mu dt - \int_0^\tau \int_{\mathbb{R}^n} \frac{p-1}{2}h^2 v^{p-2} \Gamma(v) d\mu dt$
$= \int_0^\tau \int_{\mathbb{R}^n} \frac{2}{p-1} \Gamma(h) v^p d\mu dt - \frac{2(p-1)}{p^2} \int_0^\tau \int_{\mathbb{R}^n} h^2 \Gamma(v^{p/2}) d\mu dt$

Combining with the previous conclusion we obtain ,
$\int_{\mathbb{R}^n} h^2(x) v^{p}(x,\tau) d\mu(x) + \frac{2(p-1)}{p} \int_0^\tau \int_{\mathbb{R}^n} h^2 \Gamma(v^{p/2}) d\mu dt \leq \frac{2 p}{(p-1) } \| \Gamma(h) \|_\infty^2 \int_0^\tau \int_{\mathbb{R}^n} v^p d\mu dt.$
By using the previous inequality with an increasing sequence $h_n\in C_0^\infty(\mathbb{R}^n)$ , $0 \le h_n \le 1$ , such that $h_n\nearrow 1$ on $\mathbb{R}^n$ , and $||\Gamma (h_n,h_n)||_{\infty} \to 0$ , as $n\to \infty$ , and letting $n \to +\infty$ , we obtain
$\int_{\mathbb{R}^n} v^{p}(x,\tau) d\mu(x)=0$ thus $v=0$ $\square$ .

As a consequence of this result, any solution in $L^p_\mu(\mathbb{R}^n)$ , $1 < p < +\infty$ of the heat equation $\frac{\partial u}{\partial t}= L u$ is uniquely determined by its initial condition, and is therefore of the form $u(t,x)=P_tf(x)$ . We stress that without further conditions, this result fails when $p=1$ or $p=+\infty$ .

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HW due October 11

Posted on October 6, 2016 by Fabrice Baudoin

Exercise: Let $L$ be an elliptic diffusion operator with smooth coefficients that is essentially self-adjoint with respect to a measure $\mu$ . Let $\alpha$ be a multi-index. If $K$ is a compact set of $\mathbb{R}^n$ , show that there exists a positive constant $C$ such that for $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ , $\sup_{x \in K} |\partial^{\alpha} \mathbf{P}_t f(x)| \le C \left( 1 +\frac{1}{t^{|\alpha|+\kappa}} \right) \| f \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})},$ where $\kappa$ is the smallest integer larger than $\frac{n}{4}$

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Lecture 8. Positivity preserving property of diffusion semigroups

Posted on October 6, 2016 by Fabrice Baudoin

In the previous lectures, we have proved that if L is a diffusion operator that is essentially self-adjoint then, by using the spectral theorem, we can define a self-adjoint strongly continuous contraction semigroup with generator L and this semigroup is unique. A remarkable property of the semigroup is that it preserves the positivity of functions.

More precisely, we are going to prove that if $L$ is am essentially self-adjoint operator with respect to a measure $\mu$ then, by denoting $(P_t)_{t \ge 0}$ the semigroup generated by $L$ : If $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ satisfies $0 \le f \le 1$ , then $0 \le \mathbf{P}_t f \le 1$ , $t \ge 0$ . This property is called the submarkov property of the semigroup $(\mathbf{P}_t)_{t \ge 0}$ . The terminology stems from the connection with probability theory where $(\mathbf{P}_t)_{t \ge 0}$ is interpreted as the transition semigroup of a sub-Markov process.

As a first step, we prove the positivity preserving property, which is a consequence of the following functional inequality satisfied by diffusion operators:

Lemma: (Kato’s inequality for diffusion operators). Let $L$ be a diffusion operator on $\mathbb{R}^n$ which is symmetric with respect to a measure $\mu$ . Let $u \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ . Define $\mathbf{sgn} \text{ }u(x)=0 \quad \text{ if } u(x)=0$ and $\mathbf{sgn} \text{ }u (x)=\frac{u(x)}{|u(x)|} \quad \text{ if } u(x)\neq 0.$ In the sense of distributions, we have the following inequality $L |u | \ge ( \mathbf{sgn} \text{ }u ) Lu.$

Proof: If $\phi$ is a smooth and convex function and if $u$ is assumed to be smooth, it is readily checked that $L \phi(u)=\phi'(u) Lu +\phi''(u) \Gamma(u,u) \ge \phi'(u)Lu.$ By choosing for $\phi$ the function $\phi_\varepsilon(x)=\sqrt{x^2 +\varepsilon^2}, \quad \varepsilon > 0$ , we deduce that for every smooth function $u \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ , $L\phi_\varepsilon (u) \ge \frac{u}{\sqrt{x^2 +\varepsilon^2}} Lu.$ As a consequence this inequality holds in the sense of distributions, that is for every $f \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ , $f \ge 0$ , $\int_{\mathbb{R}^n} f L\phi_\varepsilon (u) d\mu \ge \int_{\mathbb{R}^n} f \frac{u}{\sqrt{u^2 +\varepsilon^2}} Lu d\mu$ . Letting $\varepsilon \to 0$ gives the expected result $\square$

We are now in position to state and prove the positivity preserving theorem.

Proposition: Let $L$ be an essentially self-adjoint diffusion operator on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ . If $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ is almost surely nonnegative $(f \ge 0)$ , then we have for every $t \ge 0$ , $\mathbf{P}_t f \ge 0$ almost surely.

Proof: The main idea is to prove that for $\lambda > 0$ , the resolvent operator $(\lambda \mathbf{Id}-L)^{-1}$ which is well defined due to essential self adjointness preserves the positivity of function. Then, we may conclude by the fact that, as it is seen from spectral theorem, $\mathbf{P}_t=\lim_{n \to +\infty} \left( \mathbf{Id} -\frac{t}{n} L\right)^{-n}$ .

We first extend Kato’s inequality to a larger class of functions.

Let $\lambda > 0$ . We consider on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ the norm $\| f \|^2_{\lambda} =\| f \|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }+\lambda\mathcal{E}(f,f) =\| f \|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }+\lambda \int_{\mathbb{R}^n} \Gamma (f,f) d\mu$ and denote by $\mathcal{H}_\lambda$ the completion of $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ . Our goal will be to show that the Kato’s inequality is also satisfied for $u \in \mathcal{H}_\lambda$ . As in the proof of Kato’s inequality, we first consider smooth approximations of the absolute value. For $\varepsilon > 0$ we introduce the function $\phi_\varepsilon(x)=\sqrt{x^2 +\varepsilon^2}-\varepsilon, \quad \varepsilon > 0.$ It is easily seen that for $u \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ , $\phi_\varepsilon(u) \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ . Let now $u_n \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ be a Cauchy sequence for the norm $\| \cdot \|_{\lambda}$ . We claim that the sequence $\phi_\varepsilon(u_n)$ is also a Cauchy sequence. Indeed, since $| \phi_\varepsilon(x)-\phi_\varepsilon(y) | \le |x-y|,$ we have, on one hand $\| \phi_\varepsilon(u_n)-\phi_\varepsilon(u_m) \|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) } \le \| \phi_\varepsilon(u_n)-\phi_\varepsilon(u_m) \|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}.$ Now, keeping in mind that $\mathcal{E}$ is a nonnegative bilinear form and thus satisfies Cauchy-Schwarz-inequality, we have on the other hand
$\sqrt{\mathcal{E}(\phi_\varepsilon(u_n)-\phi_\varepsilon(u_m),\phi_\varepsilon(u_n)-\phi_\varepsilon(u_m)) }$
$\le \|\phi'_\varepsilon(u_n)\|_\infty \sqrt{\mathcal{E}(u_n-u_m,u_n-u_m) }+\sqrt{\| \Gamma(u_m,u_m)\|_\infty } \|\phi'_\varepsilon(u_n)-\phi'_\varepsilon(u_m) \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}$
$\le \sqrt{\mathcal{E}(u_n-u_m,u_n-u_m) }+\frac{1}{\varepsilon} \sqrt{\| \Gamma(u_m,u_m)\|_\infty } \|u_n-u_m \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}.$
As a consequence, $\phi_\varepsilon(u_n)$ is a Cauchy sequence and thus converges toward an element of $\mathcal{H}_\lambda$ . If $u$ denotes the limit of $u_n$ in $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ , the limit in $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ of $\phi_\varepsilon(u_n)$ is $\phi_\varepsilon(u)$ . As a conclusion, if $u \in \mathcal{H}_\lambda$ then $\phi_{\varepsilon}(u) \in \mathcal{H}_\lambda$ .

From the proof of Kato’s inequality, if $u \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ then for every $f \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ , $f \ge 0$ , $\int_{\mathbb{R}^n} f L\phi_\varepsilon (u) d\mu \ge \int_{\mathbb{R}^n} f \frac{u}{\sqrt{u^2 +\varepsilon^2}} Lu d\mu$ . This may be rewritten as $- \int_{\mathbb{R}^n} f \frac{u} {\sqrt{u^2 +\varepsilon^2}} Lu d\mu \ge \mathcal{E} (f,\phi_\varepsilon (u)).$
Let $f \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ and $u \in \mathcal{D}(L)\subset \mathcal{H}_\lambda$ . We consider a sequence $u_n \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ such that $u_n \to u$ for the norm $\| \cdot \|_\lambda$ . In particular $u_n \to u$ for the norm $\| \cdot \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}$ , so that by passing to a subsequence we can suppose that $u_n(x) \to u(x)$ pointwise almost surely. Applying the inequality to $u_n$ and letting $n \to +\infty$ leads to the conclusion that the inequality also holds for $u \in \mathcal{D}(L)$ and $f \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ .

Finally, by using the same type of arguments as above, it is shown for $u \in \mathcal{H}_\lambda$ , when $\varepsilon \to 0$ , $\varepsilon (u)$ tends to $|u|$ in the norm $\| \cdot \|_\lambda$ . Thus, if $u \in \mathcal{H}_\lambda$ , $|u| \in \mathcal{H}_\lambda$ .

As a consequence of all this, if $u \in \mathcal{D}(L)$ , $|u| \in \mathcal{H}_\lambda$ , and moreover for $f \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ , $- \int_{\mathbb{R}^n} f (\mathbf{sgn} \text{ }u )Lu d\mu \ge \mathcal{E} (f,|u|).$
And this last inequality is easily extended to $f \in \mathcal{H}_\lambda$ by density of $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ in $\mathcal{H}_\lambda$ for the norm $\| \cdot \|_\lambda$ . In particular when $f=|u|$ we get that for $u \in \mathcal{D}(L)$ , $\mathcal{E}(|u|, |u|) \le \mathcal{E}(u, u).$ Again by density, this inequality can be extended to every $u \in \mathcal{H}_\lambda$ . Since $L$ is essentially self-adjoint we can consider the bounded operator $\mathbf{R}_\lambda=( \mathbf{Id}-\lambda L)^{-1}$ that goes from $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ to $\mathcal{D}(L)\subset \mathcal{H}_\lambda$ . For $f \in \mathcal{H}_\lambda$ and $g \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ with $g \ge 0$ , we have
$\langle | f | , \mathbf{R}_\lambda g \rangle_\lambda = \langle | f | , \mathbf{R}_\lambda g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}-\lambda \langle |f| , L\mathbf{R}_\lambda g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}$
$=\langle |f|, (\mathbf{Id}-\lambda L) \mathbf{R}_\lambda g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}$
$=\langle |f|, g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}$
$\ge | \langle f, g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}|$
$\ge |\langle f , \mathbf{R}_\lambda g \rangle_\lambda|.$
Moreover, from the previous inequality, for $f \in \mathcal{H}_\lambda$ ,
$\|\text{ } | f|\text{ } \|_\lambda^2 =\| \text{ }| f |\text{ } \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}^2+\lambda \mathcal{E}(|f|,|f|)$
$\ge \| f \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}^2+\lambda \mathcal{E}(f,f) \ge \| f \|_\lambda^2.$
By taking $f= \mathbf{R}_\lambda g$ in the two above sets of inequalities, we draw the conclusion
$|\langle \mathbf{R}_\lambda g , \mathbf{R}_\lambda g \rangle_\lambda| \le \langle | \mathbf{R}_\lambda g | , \mathbf{R}_\lambda g \rangle_\lambda \le \|\text{ } | \mathbf{R}_\lambda g|\text{ } \|_\lambda \|\mathbf{R}_\lambda g\|_\lambda\le |\langle \mathbf{R}_\lambda g , \mathbf{R}_\lambda g \rangle_\lambda|.$
The above inequalities are therefore equalities which implies $\mathbf{R}_\lambda g = | \mathbf{R}_\lambda g|.$ As a conclusion if $g \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ is $\ge 0$ , then for every $\lambda > 0$ , $( \mathbf{Id}-\lambda L)^{-1} g \ge 0$ . Thanks to spectral theorem, in $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ , $\mathbf{P}_t g=\lim_{n \to +\infty} \left( \mathbf{Id} -\frac{t}{n} L\right)^{-n}g.$ By passing to a subsequence that converges pointwise almost surely, we deduce that $\mathbf{P}_t g \ge 0$ almost surely $\square$

Exercise: Let $L$ be an elliptic diffusion operator with smooth coefficients that is essentially self-adjoint. Denote by $p(t,x,y)$ the heat kernel of $\mathbf{P}_t$ . Show that $p(t,x,y) \ge 0$ . (Remark: It actually possible to prove that $p(t,x,y) > 0$ ).

Besides the positivity preserving property, the semigroup is a contraction on $\mathbf{L}_{\mu}^\infty (\mathbb{R}^n,\mathbb{R})$ . More precisely,

Proposition: Let $L$ be an essentially self adjoint diffusion operator on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ . If $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) \cap \mathbf{L}_{\mu}^\infty (\mathbb{R}^n,\mathbb{R})$ , then $\mathbf{P}_t f \in \mathbf{L}_{\mu}^\infty (\mathbb{R}^n,\mathbb{R})$ and $\| \mathbf{P}_t f \|_\infty \le \| f \|_\infty.$

Proof: The proof is close and relies on the same ideas as the proof of the positivity preserving property. So, we only list below the main steps and let the reader fills the details.

As before, for $\lambda > 0$ , we consider on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ the norm $\| f \|^2_{\lambda} =\| f \|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }+\lambda\mathcal{E}(f,f)$ and denote by $\mathcal{H}_\lambda$ the completion of $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ .

The first step is to show that if $0 \le f \in \mathcal{H}_\lambda$ , then $1 \wedge f$ (minimum between $1$ and $f$ ) also belongs to $\mathcal{H}_\lambda$ and moreover $\mathcal{E}( 1 \wedge f, 1 \wedge f) \le \mathcal{E}(f,f).$

Let $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ satisfy $0 \le f \le 1$ and put $g=\mathbf{R}_\lambda f=( \mathbf{Id}-\lambda L)^{-1} f \in \mathcal{H}_\lambda$ and $h=1 \wedge g$ . According to the first step, $h \in \mathcal{H}_\lambda$ and $\mathcal{E}( h, h ) \le \mathcal{E}(g,g)$ . Now, we observe that:
$\| g-h \|_\lambda^2 =\| g \|_\lambda^2 -2 \langle g,h \rangle_\lambda +\| h \|_\lambda^2$
$=\langle \mathbf{R}_\lambda f,f \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }-2 \langle f,h \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) } +\| h\|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) } +\lambda \mathcal{E}(h,h)$
$= \langle \mathbf{R}_\lambda f,f \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }-\| f\|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }+\| f-h\|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }+\lambda \mathcal{E}(h,h)$
$\le \langle \mathbf{R}_\lambda f,f \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }-\| f\|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }+\| f-g\|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }+\lambda \mathcal{E}(g,g)=0$
As a consequence $g=h$ , that is $0\le g \le 1$ .

The previous step shows that if $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ satisfies $0 \le f \le 1$ then for every $\lambda > 0$ , $0 \le ( \mathbf{Id}-\lambda L)^{-1} f \le 1$ . As in the previous proposition, we deduce that $0 \le \mathbf{P}_t f \le 1$ almost surely $\square$

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Lecture 7. Heat kernels of subelliptic semigroups

Posted on September 29, 2016 by Fabrice Baudoin

In this Lecture, we use the local regularity theory of subelliptic operators, to prove the existence of heat kernels.

Proposition: Let $L$ be a locally subelliptic diffusion operator with smooth coefficients that is essentially self-adjoint with respect to a measure $\mu$ . Denote by $(\mathbf{P}_t)_{t \ge 0}$ the corresponding semigroup on $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ .

If $K$ is a compact set of $\mathbb{R}^n$ , there exists a positive constant $C$ such that for $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ , $\sup_{x \in K} |\mathbf{P}_t f(x)| \le C \left( 1 +\frac{1}{t^{\kappa}} \right) \| f \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})},$ where $\kappa$ is the smallest integer larger than $\frac{n}{4\varepsilon}$ and $\varepsilon >0$ .
For $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ , the function $(t,x)\rightarrow \mathbf{P}_tf (x)$ is smooth on $(0,+\infty)\times \mathbb{R}^n$ .

Proof: Let us first observe that from the spectral theorem that if $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ then for every $k \ge 0$ , $L^k \mathbf{P}_t f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ and $\|L^k \mathbf{P}_t f \|_{ \mathbf{L}_\mu^2 (\Omega,\mathbb{R})} \le \left(\sup_{\lambda \ge 0} \lambda^k e^{-\lambda t}\right) \|f \|_{ \mathbf{L}_\mu^2 (\Omega,\mathbb{R})}.$ Now, let $K$ be a compact set of $\mathbb{R}^n$ . From the proposition in, there exists therefore a positive constant $C$ such that $\left(\sup_{x \in K} | \mathbf{P}_t f(x) | \right)^2 \le C \left( \sum_{k=0}^{\kappa} \|L^k \mathbf{P}_t f \|^2_{ \mathbf{L}_\mu^2 (\Omega,\mathbb{R})} \right).$ Since it is immediately checked that $\sup_{\lambda \ge 0} \lambda^k e^{-\lambda t}=\left( \frac{k}{t}\right)^k e^{-k},$ the bound $\sup_{x \in K} |\mathbf{P}_t f(x)| \le C \left( 1 +\frac{1}{t^{\kappa}} \right) \| f \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}$ easily follows. We now turn to the second part. Let $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ . First, we fix $t > 0$ . As above, from the spectral theorem, for every $k \ge 0$ , $L^k \mathbf{P}_t f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ , for any bounded open set $\Omega$ . By hypoellipticity of $L$ , we deduce therefore that $\mathbf{P}_t f$ is a smooth function.

Next, we prove joint continuity in the variables $(t,x)\in (0,+\infty)\times \mathbb{R}^n$ . It is enough to prove that if $t_0 >0$ and if $K$ is a compact set in $\mathbb{R}^n$ , $\sup_{x \in K} | \mathbf{P}_{t} f(x) - \mathbf{P}_{t_0} f(x) | \rightarrow_{t \to t_0} 0.$ From the previous proposition, there exists a positive constant $C$ such that $\sup_{x \in K} | \mathbf{P}_{t} f(x) - \mathbf{P}_{t_0} f(x) | \le C \left( \sum_{k=0}^{\kappa} \|L^k \mathbf{P}_t f-L^k \mathbf{P}_{t_0} f \|^2_{ \mathbf{L}_\mu^2 (\Omega,\mathbb{R})} \right).$ Now, again from the spectral theorem, it is checked that $\lim_{t \to t_0} \sum_{k=0}^{\kappa} \|L^k \mathbf{P}_t f-L^k \mathbf{P}_{t_0} f \|^2_{ \mathbf{L}_\mu^2 (\Omega,\mathbb{R})}=0.$ This gives the expected joint continuity in $(t,x)$ . The joint smoothness in $(t,x)$ is a consequence of the second part of the previous proposition and the details are let to the reader $\square$

Remark: If the bound $\sup_{x \in K} |\mathbf{P}_t f(x)| \le C(t) \| f \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}$ uniformly holds on $\mathbb{R}^n$ , that is if $\| \mathbf{P}_t \|_{ \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) \rightarrow \mathbf{L}_{\mu}^\infty (\mathbb{R}^n,\mathbb{R})} < \infty,$ then the semigroup $(\mathbf{P}_t)_{t \ge 0}$ is said to be ultracontractive.

We are now in position to prove the following fundamental theorem:

Theorem: Let $L$ be a locally subelliptic and essentially self-adjoint diffusion operator. Denote by $(\mathbf{P}_t)_{t \ge 0}$ the corresponding semigroup on $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ . There is a smooth function $p(t,x,y)$ , $t \in (0,+\infty), x,y \in \mathbb{R}^n$ , such that for every $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ and $x \in \mathbb{R}^n$ , $\mathbf{P}_t f (x)=\int_{\mathbb{R}^n} p(t,x,y) f(y) d\mu (y).$ The function $p(t,x,y)$ is called the heat kernel associated to $(\mathbf{P}_t)_{t \ge 0}$ . It satisfies furthermore:

(Symmetry) $p(t,x,y)=p(t,y,x)$ ;
(Chapman-Kolmogorov relation) $p(t+s,x,y)=\int_{\mathbb{R}^n} p(t,x,z)p(s,z,y)d\mu(z)$ .

Proof: Let $x\in \mathbb{R}^n$ and $t > 0$ . From the previous proposition, the linear form $f \rightarrow \mathbf{P}_t f (x)$ is continuous on $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ , therefore from the Riesz representation theorem, there is a function $p(t,x,\cdot)\in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ , such that for $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ , $\mathbf{P}_t f (x)=\int_{\mathbb{R}^n} p(t,x,y) f(y) d\mu (y).$ From the fact that $\mathbf{P}_t$ is self-adjoint on $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ , $\int_{\mathbb{R}^n} (\mathbf{P}_t f) g d\mu=\int_{\mathbb{R}^n} f(\mathbf{P}_t g) d\mu,$ we easily deduce the symmetry property $p(t,x,y)=p(t,y,x).$ And the Chapman-Kolmogorov relation $p(t+s,x,y)=\int_{\mathbb{R}^n} p(t,x,z)p(s,z,y)d\mu(z)$ stems from the semigroup property $\mathbf{P}_{t+s} =\mathbf{P}_t \mathbf{P}_s$ . Finally, from the previous proposition the map $(t,x) \rightarrow p(t,x,\cdot) \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ is smooth on $(0,+\infty) \times \mathbb{R}^n$ for the weak topology of $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ . This implies that it is also smooth on $(0,+\infty) \times \mathbb{R}^n$ for the norm topology. Since, from the Chapman-Kolmogorov relation $p(t,x,y)=\langle p(t/2,x,\cdot), p(t/2,y.\cdot) \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) },$ we conclude that $(t,x,y)\rightarrow p(t,x,y)$ is smooth on $(0,+\infty) \times \mathbb{R}^n \times \mathbb{R}^n$ $\square$

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Lecture 6. Subelliptic diffusion operators

Posted on September 27, 2016 by Fabrice Baudoin

This lecture is an introduction to the regularity theory of diffusion operators. Most of the statements will be given without proofs. For a good and easy introduction to the theory in the elliptic case, we refer to the book : Introduction to partial differential equations by Folland. For the proof of Hormander’s theorem, we refer to Hairer’s lecture notes and the references therein.
Definition: Let $L$ be a diffusion operator with smooth coefficients which is defined on an open set $\Omega \subset \mathbb{R}^n$ . We say that $L$ is subelliptic on $\Omega$ , if for every compact set $K \subset \Omega$ , there exist a constant $C$ and $\varepsilon >0$ such that for every $u \in C^\infty_0(K)$ ,
$\| u \|^2_{(2\varepsilon)} \le C \left( \| Lu \|_2^2+\|u\|^2_2\right).$

In the above definition, we denoted for $s\in \mathbb{R}$ , the Sobolev norm

$\|f \|^2_{(s)}=\int_{\mathbb{R}^n} | \hat{f} (\xi) |^2 (1+\| \xi \|^2)^s d\xi <+\infty ,$

where $\hat{f} (\xi)$ is the Fourier transform of $f$ , and $\| \cdot \|_2$ is the classical $L^2$ norm. It is well-known (Weyl’s theorem) that elliptic operators are subelliptic in the sense of the previous definition with $\varepsilon=1$ . There are many interesting examples of diffusion operators which are subelliptic but not elliptic. Let, for instance,

$L=\sum_{i=1}^d V_i^2 +V_0$
where $V_0,V_1,\cdots,V_d$ are smooth vector fields defined on an open set $\Omega$ . We denote by $\mathfrak{V}$ the Lie algebra generated by the $V_i$ ‘s, $1 \le i \le d$ , and for $x \in \Omega$ ,

$\mathfrak{V}(x)=\{ V(x), V \in \mathfrak{V} \}.$
The celebrated Hormander’s theorem states that if for every $x \in \Omega$ , $\mathfrak{V}(x)=\mathbb{R}^n$ , then $L$ is a subelliptic operator. In that case $\varepsilon$ is $1/d$ , where $d$ is the maximal length of the brackets that are needed to generate $\mathbb{R}^n$ .

If $L$ is a subelliptic diffusion operator, using the theory of pseudo-differential operators, it can be proved that the subellipticity defining inequality self-improves into a family of inequalities of the type

$\| u \|^2_{(2\varepsilon+s)} \le C \left( \| Lu \|_{(s)}^2+\|u\|^2_{(s)}\right), \quad u \in C^\infty_0(K),$
where $s\in \mathbb{R}$ and the constant $C$ only depends on $K$ and $s$ . This implies, in particular, by a usual bootstrap argument and Sobolev lemma that subelliptic operators are hypoelliptic. Iterating the latter inequality also leads to

$\| u \|^2_{(2k\varepsilon)} \le C \sum_{j=0}^k \| L^j u\|^2_2, \quad u \in C^\infty_0(K),$
where $k \ge 0$ . This may be used to bound derivatives of $u$ in terms of $L^2$ norms to iterated powers of $u$ . Indeed, if $\alpha$ is a multi-index and $k$ is such that $4k\varepsilon> 2 | \alpha | +n$ , then we get $\sup_{x \in K} | \partial^\alpha u (x) |^2 \le C\| u \|^2_{(2k\varepsilon)}$ and therefore

$\sup_{x \in K} | \partial^\alpha u (x) |^2 \le C' \sum_{j=0}^k \| L^j u\|^2_2.$
Along the same lines, we also get the following result.

Proposition: Let $L$ be a subelliptic diffusion operator with smooth coefficients on an open set $\Omega \subset \mathbb{R}^n$ . Let $u \in L^2(\Omega)$ such that, in the sense of distributions,

$Lu,L^2u,\cdots, L^ku \in L^2(\Omega),$
for some positive integer $k$ . Let $K$ be a compact subset of $\Omega$ and denote by $\varepsilon$ the subellipticity constant. If $k>\frac{n}{4 \varepsilon}$ , then $u$ is a continuous function on the interior of $K$ and there exists a positive constant $C$ such that

$\sup_{x \in K} | u(x) |^2 \le C \sum_{j=0}^k \|L^j u \|^2_{ L^2(\Omega)}.$
More generally, if $k>\frac{m}{2\varepsilon}+\frac{n}{4\varepsilon}$ for some non negative integer $m$ , then $u$ is $m$ -times continuously differentiable in the interior of $K$ and there exists a positive constant $C$ such that

$\sup_{|\alpha| \le m} \sup_{x \in K} |\partial^\alpha u(x) |^2 \le C \sum_{j=0}^k \|L^j u \|^2_{ L^2(\Omega)} .$

As a consequence of the previous result, we see in particular that
$\bigcap_{k \ge 0} \mathcal{D}(L^k) \subset C^\infty(\mathbb{R}^n).$

We can define subelliptic operators on a manifold by using charts:

Definition: Let $L$ be a diffusion operator on a manifold $\mathbb{M}$ . We say that $L$ is subelliptic on $\mathbb{M}$ if it is in any local chart.

The previous Proposition can then be extended to the manifold case:
Proposition: Let $\mathbb{M}$ be a manifold endowed with a smooth positive measure $\mu$ , and let $L$ be a subelliptic diffusion operator with smooth coefficients on an open set $\Omega \subset \mathbb{M}$ . Let $u \in L_\mu^2(\Omega)$ such that, in the sense of distributions,

$Lu,L^2u,\cdots, L^ku \in L_\mu^2(\Omega),$
for some positive integer $k$ . Let $K$ be a compact subset of $\Omega$ . There exists a constant $\varepsilon >0$ such that If $k>\frac{n}{4 \varepsilon}$ , then $u$ is a continuous function on the interior of $K$ and there exists a positive constant $C$ such that

$\sup_{x \in K} | u(x) |^2 \le C \sum_{j=0}^k \|L^j u \|^2_{ L_\mu^2(\Omega)} .$
More generally, if $k>\frac{m}{2\varepsilon}+\frac{n}{4\varepsilon}$ for some non negative integer $m$ , then $u$ is $m$ -times continuously differentiable in the interior of $K$ and there exists a positive constant $C$ such that

$\sup_{|\alpha| \le m} \sup_{x \in K} |\partial^\alpha u(x) |^2 \le C \sum_{j=0}^k \|L^j u \|^2_{ L_\mu^2(\Omega)} .$

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HW2. Due September 27

Posted on September 20, 2016 by Fabrice Baudoin

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, $\mathbf{P}_t f (x)=\frac{1}{(4\pi t)^{\frac{n}{2}}} \int_{\mathbb{R}^n} e^{-\frac{\|x-y\|^2}{4t} } f(y)dy.$

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Lecture 5. Hormander’s type operators

Posted on September 20, 2016 by Fabrice Baudoin

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 $\square$

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Lecture 4. Diffusion semigroups as solutions of heat equations

Posted on September 8, 2016 by Fabrice Baudoin

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 $u=0$ $\square$