Let $(G, \mathcal D)$ be a densely defined operator on $C_0$ (continuous functions vanishing at infinity on some nice topological space) whose closure $\bar G$ generates a Feller semigroup and let $X$ be a Markov process corresponding to it.
Assume that $\mathcal D\subset C_K$ (continuous functions with compact support), that $G(\mathcal D)\subset C_K$ and that $G$ is symmetric with respect to a Radon measure $m$(Edit: with full support, but not necessarily finite), i.e.
$\int Gf\ g \ dm = \int f \ Gg\ dm$ for every $f,g\in \mathcal D$.
I guess that the Dirichlet form $\mathcal E$ of $X$ (defined as in the book of Fukushima/Oshima/Takeda by using the transition kernel, see (1.4.13) on p.30 in the last edition) is given by the closure of
$\mathcal D\ni f,g \mapsto \int Gf\ g dm$.
In other terms the Friedrichs extension of $G$ in $L^2(dm)$ should be the generator of the $L^2$ semigroup induced by $X$. (Edit: by $L^2$ semigroup induced by $X$ I mean the semigroup corresponding to the Dirichlet form $\mathcal E$ )
Is this true? I didn't find a reference nor a simple argument for showing this.
Or is it possible that a selfadjoint extension other than the Friedrichs one generates the $L^2$ semigroup induced by $X$?
Edit: From the answer of Byron Schmuland it is clear to me that the guess is true if $m$ is finite (this happens for example when the state space is compact)compact. Observe that in this case $G$ is essentially selfadjoint in $L^2$, so the Friedrichs extension is just the closure of $G$ and there are no other selfadjoint extensions. I'm still confused about the case of $m$ being not finitenoncompact state space. I would also appreciate partial answers which work for some concrete example of $G$ (say elliptic partial differential operators, or discrete operators).
