Symmetric Feller processes and Dirichlet Forms - MathOverflow

Symmetric Feller processes and Dirichlet Forms

2012-02-01T18:45:28Z

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 the state space is 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 noncompact state space. I would also appreciate partial answers which work for some concrete example of $G$ (say elliptic partial differential operators, or discrete operators).

Answer by Byron Schmuland for Symmetric Feller processes and Dirichlet Forms

2012-02-03T19:24:51Z

I've decided to post an incomplete preliminary answer.

I ran into your problem when I was writing [1]. On page 258 you will see my resolution.

I should point out that in my case, the underlying space $X$ was compact, and that $m$ was a finite measure with full support. Thus, $C(X)$ embeds into $L^2(X;m)$ with a continuous, linear injection in the obvious way. This may not hold in the locally compact case, and I'm not sure how serious a problem that is.

Translated into your notation, and letting $\tilde G$ be the Friedrichs extension we note that $\bar G$ and $\tilde G$ agree on $\cal D$ and so the resolvent operators $\bar R_\lambda$ and $\tilde R_\lambda$ agree on $(\lambda-G)({\cal D})$. We deduce that $\bar R_\lambda= \tilde R_\lambda$ on $C(X)$ and using the Yosida approximation conclude the same about the semigroup operators $\bar T_t$ and $\tilde T_t$.

I hope this is of some help. If anything is unclear, let me know.

[1] A result on the infinitely many neutral alleles diffusion model. Journal of Applied Probability 28, 253-267 (1991).

Answer by Hans for Symmetric Feller processes and Dirichlet Forms

2012-02-08T22:01:53Z

I think that the guess is true under the general assumptions I made, by following Byron Schmuland's reasoning. Let me spell it out the way I understood it.

I denote by $T$ the $L^2(dm)$ semigroup induced by $X$, by $\bar T$ the Feller semigroup generated by the closure of $G$ in $C_0$ and by $\tilde T$ the $L^2(dm)$ semigroup generated by the Friedrichs extension of $G$.

The semigroup $T$ is characterized by

$T=\bar T$ on $L^2(dm)\cap C_0$ (I take this as definition of $T$ as in Fukushima et al.)

So it is enough to show that $\tilde T = \bar T$ on $C_K$ (which is both dense in $L^2(dm)$ and $C_0$).

By Yosida approximation it is enough to show that the corresponding resolvents satisfy for $\lambda>0$

$\tilde R_\lambda = \bar R_\lambda$ on $C_K$

By definition of resolvent $\tilde R_\lambda= \bar R_\lambda$ on $\mathcal F:=(\lambda-G)(\mathcal D)$. The $C_0$-closure of $\mathcal F$ is $C_0$ since $G$ generates, in particular it contains $C_K$. It follows that also the $L^2$ closure of $\mathcal F$ contains $C_K$, so we are done.

Observe that under the assumptions I made in the question, $G$ is automatically essentially selfadjoint, so there is no other selfadjoint extension other than the Friedrichs one. (a criterion for essential selfadjointness is that $(\lambda -G)(\mathcal D)$ is dense in $L^2$ which we have shown above). So the interesting case I was wondering about actually doesn't happen.

Let me know if there is a flaw in what I wrote.