Research and Lecture notes

Lecture 7. Diffusion semigroups in Lp

Posted on June 28, 2013 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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Lecture 6. The positivity preserving property of diffusion semigroups

Posted on June 27, 2013 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 5. The diffusion semigroup as a solution to the heat equation

Posted on June 25, 2013 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 $\frac{\partial u}{\partial t}= L u,\quad u (0,x)=0,$ . 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 $u=0$ $\square$

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Lecture 4. The heat kernel of a diffusion semigroup

Posted on June 24, 2013 by Fabrice Baudoin

The goal of this lecture is to prove that if a diffusion operator L is elliptic, then the semigroup it generates admits a smooth kernel. As a consequence, the semigroup generated by an elliptic diffusion operator is regularizing in the sense that it transforms any function into a smooth function. The key point is the following estimate that can be proved by using the theory of Sobolev spaces.

Proposition: Let $L$ be an elliptic diffusion operator with smooth coefficients on $\mathbb{R}^n$ which is symmetric with respect to a Borel measure $\mu$ . Let $u \in \mathbf{L}_\mu^2 (\mathbb{R}^n,\mathbb{R})$ such that $Lu,L^2u,\cdots, L^ku \in \mathbf{L}_\mu^2 (\mathbb{R}^n,\mathbb{R})$ , for some positive integer $k$ . If $k > \frac{n}{4}$ , then $u$ is a continuous function, moreover, for any bounded open set $\Omega \subset \mathbb{R}^n$ , and any compact set $K \subset \Omega$ , there exists a positive constant $C$ (independent of $u$ ) such that $\left(\sup_{x \in K} | u(x) | \right)^2 \le C \left( \sum_{j=0}^k \|L^j u \|^2_{ \mathbf{L}_\mu^2 (\Omega,\mathbb{R})} \right).$ More generally, if $k > \frac{m}{2}+\frac{n}{4}$ for some non negative integer $m$ , then $u \in \mathcal{C}^m(\mathbb{R}^n,\mathbb{R})$ and for any bounded open set $\Omega \subset \mathbb{R}^n$ , and any compact set $K \subset \Omega$ , there exists a positive constant $C$ (independent of $u$ ) such that $\left(\sup_{|\alpha| \le m} \sup_{x \in K} |\partial^\alpha u(x) | \right)^2 \le C \left( \sum_{j=0}^k \|L^j u \|^2_{ \mathbf{L}_\mu^2 (\Omega,\mathbb{R})} \right).$

We can explain by simple computations on the Laplace operator $\Delta$ how the $\frac{m}{2}+\frac{n}{4}$ comes into the play in the above proposition. Let $u$ be a smooth and rapidly decreasing function. By the inverse Fourier transform formula $\| \Delta^j u \|^2_{ \mathbf{L}^2 (\Omega,\mathbb{R})} =(2\pi)^{4j} \int_{\mathbb{R}^n} \| \lambda \|^{4j} | \hat{u} (\lambda) |^2 d\lambda,$ so that by Cauchy-Schwarz inequality we may bound
$\partial^\alpha u(x) = \int_{\mathbb{R}^n} (2i\pi\lambda)^{\alpha} e^{2i\pi \langle x, \lambda\rangle} \hat{u} (\lambda) d\lambda = \int_{\mathbb{R}^n} (2i\pi\lambda)^{\alpha} e^{2i\pi \langle x, \lambda\rangle} \sqrt{ \sum_{j=0}^k (2\pi)^{4j} \| \lambda \|^{4j} } \frac{\hat{u} (\lambda)}{\sqrt{ \sum_{j=0}^k (2\pi)^{4j} \| \lambda \|^{4j} } } d\lambda,$
by $\sum_{j=0}^k \|\Delta^j u \|^2_{ \mathbf{L}^2 (\mathbb{R}^n,\mathbb{R})}$ only when $\int_{\mathbb{R}^n} \frac{\|\lambda\|^{2\alpha} d\lambda}{ \sum_{j=0}^k (2\pi)^{4j} \| \lambda \|^{4j}} < \infty$ that is if $k > \frac{m}{2}+\frac{n}{4}$ . We are now in position to prove the following regularization estimate for the semigroup associated with an elliptic operator.

Proposition: Let $L$ be an elliptic 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 $\frac{n}{4}$ .

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 previous proposition, 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.

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 $\frac{n}{4}$ .

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

Theorem: Let $L$ be an elliptic 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 3. Semigroup generated by a symmetric diffusion operator

Posted on June 24, 2013 by Fabrice Baudoin

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 $q(t,x)=\frac{\Gamma\left( \frac{n+1}{2} \right)}{\pi^{\frac{n+1}{2}} } \frac{t}{(t^2+\| x \|^2 )^{\frac{n+1}{2}} }.$

Posted in Curvature dimension inequalities | 3 Comments

Lecture 2. Essentially self-adjoint diffusion operators

Posted on June 22, 2013 by Fabrice Baudoin

The goal of the next few lectures will be to introduce the semigroup generated by a diffusion operator. This semigroup will play pervasive role throughout these lectures and is the main tool associated to the curvature dimension inequalities. 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 $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ .

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Lecture 1. Diffusion operators

Posted on June 21, 2013 by Fabrice Baudoin

In this first lecture we introduce the main characters of this course: The 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 three 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 $Lf (x) \ge 0$ .

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 $Af (x) \ge 0$ .

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.

Posted in Curvature dimension inequalities | 4 Comments

MA 696. Curvature dimension inequalities

Posted on June 13, 2013 by Fabrice Baudoin

Next Fall, I will teach a graduate course on curvature dimension inequalities, and, as usual, the Lectures will be posted on this blog.

The theory of curvature dimension inequalities and of their applications to the geometric analysis of manifolds is, at that time, my main research interest. I am therefore quite enthusiastic about the perspective of preparing the set of Lecture notes.

Curvature dimension inequalities were first introduced and extensively used by Dominique Bakry. The original motivation was to study hypercontractivity criteria for diffusion semigroups and boundedness properties of Riesz transforms. They nowadays play a crucial role in the geometric analysis of manifolds, because they provide a very robust and synthetic way to analyse the impact of curvature bounds on the global geometry of a space.

We will mainly focus on curvature dimension inequalities in Riemannian geometry and, at the end of the course will cover more recent applications and generalizations to sub-Riemannian manifolds. Only very basic notions of differential geometry will be required.

The course will cover the following topics:

1. Symmetric diffusion semigroups;
2. The Laplace-Beltrami operator on a Riemannian manifold, Bochner’s formula;
2. The heat semigroup on a Riemannian manifold;
3. Li-Yau type inequalities and Harnack estimates;
4. The heat kernel proof of Bonnet-Myers theorem;
5. Sub-Riemannian manifolds with transverse symmetries;
6. Sub-Riemannian Li-Yau inequalities;
7. Open problems and recent developments: Geometric analysis of contact manifolds.

Posted in Curvature dimension inequalities | Leave a comment

Lecture notes on rough paths theory

Posted on April 10, 2013 by Fabrice Baudoin

For those who are interested, here are the notes corresponding to the lectures posted on this blog. All comments are welcome.

Posted in Rough paths theory | 2 Comments

Lecture 30. The Stroock-Varadhan support theorem

Posted on April 10, 2013 by Fabrice Baudoin

To conclude this course, we are going to provide an elementary proof of the Stroock–Varadhan support theorem which is based on rough paths theory. We first remind that the support of a random variable $X$ which defined on a metric space $X$ is the smallest closed $F$ such that $\mathbb{P}(X\in F)=1$ . In particular $x \in F$ if and only if for every open ball $\mathbf{B}(x, \varepsilon)$ , $\mathbb{P}( X \in \mathbf{B}(x, \varepsilon)) > 0$ .

Let $(B_t)_{0 \le t \le T}$ be a $d$ -dimensional Brownian motion. We can see $B$ as a random variable that takes its values in $C^{p-var}([0,T],\mathbb{R}^d)$ , $p > 2$ . The following theorem describes the support of this random variable.

Proposition: Let $p > 2$ . The support of $B$ in $C^{p-var}([0,T],\mathbb{R}^d)$ is $C_0^{0,p-var}([0,T],\mathbb{R}^d)$ , that is the closure for the $p$ -variation distance of the set of smooth paths starting at 0.

Proof: The key argument is a clever application of the Cameron-Martin theorem. Let us recall that this theorem says that if
$h \in \mathbb{W}_0^{1,2} =\left\{ h : [0,T] \to \mathbb{R}^d, \exists k \in L^2([0,T], \mathbb{R}^d), h(t)=\int_0^t k (s) ds \right\},$
then the distribution of $B+h$ is equivalent to the distribution of $B$ .

Let us denote by $F$ the support of $B$ . It is clear that $F \subset C_0^{0,p-var}([0,T],\mathbb{R}^d)$ , because the paths of $B$ have bounded $q$ variation for $2 < q < p$ .
Let now $x \in F$ . We have for $\varepsilon > 0$ , $\mathbb{P}( d_{p-var}( B, x) < \varepsilon) > 0$ . From the Cameron Martin theorem, we deduce then for $h \in \mathbb{W}_0^{1,2}$ , $\mathbb{P}( d_{p-var}( B+h, x) < \varepsilon) > 0$ . This shows that $x-h \in F$ . We can find a sequence of smooth $x_n$ that converges to $x$ in $p$ -variation. From the previous argument $x-x_n\in F$ and converges to 0. Thus $0 \in F$ and using the same argument shows then that $\mathbb{W}_0^{1,2}$ is included in $F$ . This proves that $F=C_0^{0,p-var}([0,T],\mathbb{R}^d)$ $\square$

The following theorem due to Stroock and Varadhan describes the support of solutions of stochastic differential equations. As in the previous proof, we denote
$\mathbb{W}_0^{1,2} =\left\{ h : [0,T] \to \mathbb{R}^d, \exists k \in L^2([0,T], \mathbb{R}^d), h(t)=\int_0^t k (s) ds \right\}.$

Theorem: Let $\gamma > 2$ and let $V_1,\cdots,V_d$ be $\gamma$ -Lipschitz vector fields on $\mathbb{R}^n$ . Let $x_0 \in \mathbb{R}^n$ . Let $(X_t)_{t \ge 0}$ be the solution of the Stratonovitch differential equation:
$X_t=x_0 + \sum_{i=1}^d \int_0^t V_i (X_s) \circ dB^i_s.$
Let $p > 2$ . The support of $X$ in $C^{p-var}([0,T],\mathbb{R}^d)$ is the closure in the $p$ -variation topology of the set:
$\left\{ x^h, h \in \mathbb{W}_0^{1,2} \right\},$
where $x^h$ is the solution of the ordinary differential equation
$x_t^h=x_0 + \sum_{i=1}^d \int_0^t V_i (x^h_s) dh^i_s$ .

Proof: We denote by $B^n$ the piecewise linear process which is obtained from $B$ by interpolation along a subdivision $D_n$ which is such that $D_{n+1} \subset D_n$ and whose mesh goes to 0. We know that $B^n \in \mathbb{W}_0^{1,2}$ and that $x^{B^n}$ almost surely converges in $p$ -variation to $X$ . As a consequence $B$ almost surely takes its values in the closure of:
$\left\{ x^h, h \in \mathbb{W}_0^{1,2} \right\}.$
This shows that the support of $B$ is included in the closure of $\left\{ x^h, h \in \mathbb{W}_0^{1,2} \right\}$ . The converse inclusion is a little more difficult and relies on the Lyons’ continuity theorem. It can be proved by using similar arguments as for $B$ (details are let to the reader) that the support of $S_2(B)$ is the is the closure in the $p$ -variation topology of the set:
$\left\{ S_2(h), h \in \mathbb{W}_0^{1,2} \right\},$
where $S_2$ denotes, as usual, the lift in the Carnot group of step 2. Take $h \in \mathbb{W}_0^{1,2}$ and $\varepsilon > 0$ . By the Lyons’ continuity theorem, there exists therefore $\eta > 0$ such that $d_{p-var} (S_2(h),S_2(B)) < \eta$ implies $\| X- x^h \|_{p-var} < \varepsilon$ . Therefore
$0< \mathbb{P} \left( d_{p-var} (S_2(h),S_2(B)) < \eta \right) \le \mathbb{P} \left( \| X- x^h \|_{p-var} < \varepsilon\right).$
In particular, we have $\mathbb{P} \left( \| X- x^h \|_{p-var} < \varepsilon\right) > 0$ . This proves that $x^h$ is in the support of $X$ . So, the proof now boils down to the statement that the support of $S_2(B)$ is the closure in the $p$ -variation topology of the set:
$\left\{ S_2(h), h \in \mathbb{W}_0^{1,2} \right\} \square$