I want to ask a question that arises from reading this paper.

Let $X$ be a locally compact space which is countable at infinity and let $\xi$ be a Radon measure on $X$. Suppose $V$ is a Hilbert space and there is a continuous and coercive bilinear form $a:V \times V \to \mathbb{R}$ such that

- $V$ is continuously embedded in $L^2(X,\xi)$
- If $C_0(X)$ is the space of continuous functions with compact support in $X$ then $V \cap C_0(X)$ is dense in $C_0(X)$
- For all $v \in V$, $|v| \in V$ and $a(v^+,v^-) \leq 0$.
We call $(V,a)$ a Dirichlet space.

(The $v^+$, $v^-$ makes sense if we consider $V$ as a sublattice of $L^2(X,\xi)$ so that $v_1 \geq v_2$ means $v_1 \geq v_2$ $\xi$-a.e. in $X$.)

The author then gives two examples, both on bounded regular domain $\Omega$ in $\mathbb{R}^n$ with $\xi$ chosen to be the Lebesgue measure.

- (Ex 1) $X=\Omega$, $V=H^1_0(\Omega)$, $$a(u,v) = \int_\Omega \nabla u\cdot \nabla v + uv$$ So far so ok.
- (Ex 2) $X=\bar \Omega$ (closure of $\Omega$), $V=H^1(\Omega)$, $a(u,v)$ same as above.

I have two questions:

Given a bilinear form $a$ and a space $V$ (let's say some Sobolev space on $\Omega$) and given the conditions that $X$ must satisfy, is this information enough to determine uniquely what $X$ should be?

Consider the two examples: why would I think to choose $X=\bar\Omega$ in the second example? I don't see what goes wrong if I choose plain $\Omega$ for $X$.

The author cites a thesis of Ancona for related reading, but this thesis is impossible to find (for me).