Dirichlet form

Remarks

Example: on $L^2(\R^n)$ with Lebesgue measure, take $\mathcal{F}$ to be the Sobolev space $W^{1,2}(\R^n)$ and $\mathcal{E}(u, v) = \int_{\R^n} \nabla u \cdot \nabla v \, dx$. It is symmetric and nonnegative, and $\norm{\cdot}_1$ is the Sobolev norm, which is complete. The truncation $v = \min(\max(u, 0), 1)$ of $u \in W^{1,2}$ lies in $W^{1,2}$, with $\nabla v = \nabla u$ almost everywhere on $\set{0 < u < 1}$ and $\nabla v = 0$ almost everywhere elsewhere, so $\mathcal{E}(v, v) \le \mathcal{E}(u, u)$.

Non-example: on $L^2(\R)$ take $\mathcal{F} = W^{2,2}(\R)$ and $\mathcal{E}(u, v) = \int u'' v'' \, dx$. Clauses 1 to 3 hold, but clause 4 fails. If $u \in W^{2,2}(\R)$ is smooth and crosses the level $1$ with nonzero slope, its truncation has a corner there, so it is not in $W^{2,2}(\R)$.

A closed, symmetric, nonnegative, densely defined form corresponds to a nonnegative self-adjoint operator $A$ with $\mathcal{E}(u, v) = (A^{1/2} u, A^{1/2} v)$ and $\mathcal{F}$ the domain of $A^{1/2}$. Clause 4 holds if and only if the semigroup $T_t = e^{-tA}$ is Markovian: $0 \le f \le 1$ almost everywhere implies $0 \le T_t f \le 1$ almost everywhere, for every $t > 0$. Let $X$ be a locally compact separable metric space and $m$ a Radon measure of full support. The form is then regular if $\mathcal{F} \cap C_c(X)$ is dense in $\mathcal{F}$ for $\norm{\cdot}_1$ and dense in $C_c(X)$ for the uniform norm.