Lecture 7. Diffusion operators as Markovian operators

Posted on February 16, 2019 by Fabrice Baudoin

In this section, 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. Our goal is to prove that if L is essentially self-adjoint, then the semigroup it generates is Markovian. We will also prove that this semigroup is solution of the heat equation associated to L.

As before, we will assume that there is Borel measure \mu which is equivalent to the Lebesgue measure and that symmetrizes L in the sense that for every smooth and compactly supported functions f,g : \mathbb{R}^n \rightarrow \mathbb{R},
\int_{\mathbb{R}^n} g Lf d\mu= \int_{\mathbb{R}^n} fLg d\mu.

Our first goal will be to prove that if L is essentially self-adjoint, then the semigroup it generates in L^2(\mathbb{R}^n, \mu) is Markovian. The key lemma is the so-called Kato inequality:

Lemma: Let L be a diffusion operator on \mathbb{R}^n with symmetric and invariant measure \mu. Let u \in C^\infty_0 (\mathbb{R}^n). Define
\mathbf{sgn} \text{ }u=0 \quad \text{ if } u(x)=0,
                                   =\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 C_0^\infty (\mathbb{R}^n),
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}_0^\infty  (\mathbb{R}^n), 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

From Kato inequality, it is relatively easy to see that if L is an essentially self-adjoint diffusion operator, then the associated quadratic form is Markovian. As a consequence, we deduce the following theorem.

Proposition: Let L be a diffusion operator on \mathbb{R}^n with symmetric and invariant measure \mu. Assume that L is essentially self-adjoint, then the semigroup it generates is Markovian.

Next, we connect the semigroup associated to a diffusion operator L to the parabolic following Cauchy problem:
\frac{\partial u}{\partial t}= L u, \quad u (0,x)=f(x).

In the remainder of the section, we assume that the diffusion operator L is elliptic with smooth coefficients and that there exists 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. In particular, we know from this assumption that the operator L is essentially self-adjoint.

Proposition:  Let f \in L^p(\mathbb{R}^n,\mu), 1 \le p \le \infty, and let
u (t,x)= P_t f (x), \quad t \ge 0, x\in \mathbb{R}^n.
Then 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: 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.

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(\mathbb{R}^n,\mu)} <+\infty,
where 1 <p <+\infty. Then v(x,t)=0.

Proof: Let x_0 \in X 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_X 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_X 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_X v^{p}(x,\tau) d\mu(x)=0 thus v=0. \square

As a consequence of this result, any solution in L^p({\mathbb{R}^n},\mu), 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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