Lecture 5. Dirichlet forms

As in the previous lecture, let $(X, \mathcal{B}, \mu)$ be a good measurable space equipped with a $\sigma$ -finite measure $\mu$ .

Definition: A function $v$ on $X$ is called a normal contraction of the function $u$ if for almost every $x,y \in X$ ,
$| v(x)-v(y)| \le |u(x) -u(y)| \text{ and } |v(x)| \le |u(x)|.$

Definition: Let $(\mathcal{E},\mathcal{F}=\mathbf{dom}(\mathcal{E}))$ be a densely defined closed symmetric form on $L^2(X,\mu)$ .
The form $\mathcal{E}$ is called a Dirichlet form if it is Markovian, that is, has the property that if $u \in \mathcal{F}$ and $v$ is a normal contraction of $u$ then $v \in \mathcal{F}$ and $\mathcal{E}(v,v) \le \mathcal{E} (u,u).$

The main theorem is the following.

Theorem: Let $(P_t)_{t \ge 0}$ be a strongly continuous self-adjoint contraction semigroup on $L^2(X,\mu)$ . Then, $(P_t)_{t \ge 0}$ is a Markovian semigroup if and only if the associated closed symmetric form on $L^2(X,\mu)$ is a Dirichlet form.

Proof: Let $(P_t)_{t \ge 0}$ be a strongly continuous self-adjoint contraction Markovian semigroup on $L^2(X,\mu)$ . There exists a transition function $\{p_t,t \geq 0 \}$ on $X$ such that for every $u \in L^\infty(X,\mu)$ and a.e. $x \in X$
$P_tu (x)=\int_X u(y) p_t(x,dy), \quad t > 0.$
Denote
$k_t(x)= P_t1 (x)= \int_X p_t(x,dy).$
We observe that from the Markovian property of $P_t$ , we have $0 \le k_t \le 1$ a.e.
We have then
$\frac{1}{2} \int_X \int_X (u(x)-u(y))^2 p_t(x,dy) d\mu(x)=\int_X u(x)^2 k_t(x)d\mu(x)-\int_X u(x) P_t u (x) d\mu(x).$
Therefore,
$\langle u-P_t u,u \rangle=\frac{1}{2} \int_X \int_X (u(x)-u(y))^2 p_t(x,dy) d\mu(x)+\int_X u(x)^2 (1-k_t(x))d\mu(x).$
Let us now assume that $u \in \mathcal{F}$ and that $v$ is a normal contraction of $u$ . One has
$\int_X \int_X (v(x)-v(y))^2 p_t(x,dy) d\mu(x) \le \int_X \int_X (u(x)-u(y))^2 p_t(x,dy) d\mu(x)$
and
$\int_X v(x)^2 (1-k_t(x))d\mu(x) \le \int_X u(x)^2 (1-k_t(x))d\mu(x).$
Therefore,
$\langle v-P_t v,v \rangle \le \langle u-P_t u,u \rangle$
Since $u \in \mathcal{F}$ , one knows that $\frac{1}{t}\langle u-P_t u,u \rangle$ converges to $\mathcal{E}(u)$ when $t \to 0$ . Since $\frac{1}{t} \langle v-P_t v,v \rangle$ is non-increasing and bounded it does converge when $t \to 0$ . Thus $v \in \mathcal{F}$ and
$\mathcal{E}(v) \le \mathcal{E}(u).$
One concludes that $\mathcal E$ is Markovian.

Now, consider a Dirichlet form $\mathcal E$ and denote by $P_t$ the associated semigroup in $L^2(X,\mu)$ and by $A$ its generator.
The main idea is to first prove that for $\lambda >0$ , the resolvent operator $(\lambda \mathbf{Id}-A)^{-1}$ preserves the positivity of function. Then, we may conclude by the fact that for $f \in L^2(X,\mu)$ , in the $L^2(X,\mu)$ sense
$P_tf=\lim_{n \to +\infty} \left( \mathbf{Id} -\frac{t}{n} L\right)^{-n}f.$

Let $\lambda >0$ . We consider on $\mathcal{F}$ the norm
$\| f \|^2_{\lambda} =\| f \|^2_{L^2(X,\mu) }+\lambda\mathcal{E}(f,f)$
From the Markovian property of $\mathcal E$ , if $u \in \mathcal{F}$ , then $|u| \in \mathcal{F}$ and
$\mathcal{E}(|u|, |u|) \le \mathcal{E}(u, u).$

We consider the bounded operator
$\mathbf{R}_\lambda=( \mathbf{Id}-\lambda A)^{-1}$
that goes from $L^2(X,\mu)$ to $\mathcal{D}(A)\subset \mathcal{F}$ .
For $f \in \mathcal{F}$ and $g \in L^2(X,\mu)$ with $g \ge 0$ , we have
$\langle | f | , \mathbf{R}_\lambda g \rangle_\lambda= \langle | f | , \mathbf{R}_\lambda g \rangle_{L^2(X,\mu)}-\lambda \langle |f| , A\mathbf{R}_\lambda g \rangle_{L^2(X,\mu)}$
$=\langle |f|, (\mathbf{Id}-\lambda A) \mathbf{R}_\lambda g \rangle_{L^2(X,\mu)}$
$=\langle |f|, g \rangle_{L^2(X,\mu)}$
$\ge | \langle f, g \rangle_{L^2(X,\mu)}|$
$\ge |\langle f , \mathbf{R}_\lambda g \rangle_\lambda|.$

Moreover, from previous inequality , for $f \in \mathcal{F}$ ,
$\|\text{ } | f|\text{ } \|_\lambda^2 =\| \text{ }| f |\text{ } \|_{L^2(X,\mu)}^2+\lambda \mathcal{E}(|f|,|f|)$
$\le \| f \|_{L^2(X,\mu)}^2+\lambda \mathcal{E}(f,f)$
$\le \| f \|_\lambda^2.$

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 $g \in L^2(X,\mu)$ is a.e. $\ge 0$ , then for every $\lambda >0$ , $( \mathbf{Id}-\lambda A)^{-1} g \ge 0$ a.e.. Thanks to the spectral theorem, in $L^2(X,\mu)$ ,
$P_t g=\lim_{n \to +\infty} \left( \mathbf{Id} -\frac{t}{n} A\right)^{-n}g.$
By passing to a subsequence that converges pointwise almost surely, we deduce that $P_t g \ge 0$ almost surely.
The proof of
$g \le 1, \text{ a.e } \implies P_t g \le 1, \text{ a.e }.$
follows the same lines and is let as an exercise to the reader. $\square$ .