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.

$$\int Gf\ g \ dm = \int f \ Gg\ dm \quad\text{for every } f,g\in \mathcal D.$$

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).

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    $\begingroup$ I'm thinking about your problem, and hope to post an answer soon. In the meantime, I think you want to drop the condition $G({\cal D})\subset C_K$. This eliminates too many nice processes, and is not needed to make $\int Gf\, g\, dm$ finite. – Byron Schmuland $\endgroup$
    – user6096
    Feb 2 '12 at 16:12
  • $\begingroup$ Thanks for your comment. Yes, I agree, the assumption $G(\mathcal D)\subset C_K$ is superfluous for giving the statement and probably also for showing that it is true. On the other hand I don't find it very restrictive for the moment, so I would appreciate it if you can tell me also what kind of nice processes it eliminates. $\endgroup$
    – Hans
    Feb 2 '12 at 16:55

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).

  • $\begingroup$ @Byron Schmuland. Many thanks for your answer. I' m not sure I understood well, even in the compact case (for the moment I don't have access to your paper so I just comment on what you wrote above). Maybe it would help me, if you could tell me what goes wrong in the above argument when instead of taking $\tilde G$ (Friedrichs extension) I take another extension $\tilde{\tilde G}$ (sorry if I don't see something elementary). $\endgroup$
    – Hans
    Feb 4 '12 at 1:21
  • $\begingroup$ @Byron Schmuland. I thought bit more about your answer. It is ok for me in the compact case. I think that the compactness assumption on the state space implies that $G$ is essentially selfadjoint in $L^2(X;m)$, (because the range of $(\lambda-G)$ being dense in $C(X)$ is also dense in $L^2$), so the Friedrichs extension is the unique selfadjoint extension (it is the closure of $G$ in $L^2$). Moreover your argument works perfectly well, not only in the compact, but also in the locally compact case if one assumes esential selfadjointness. But if there is more than one selfadjoint extension? $\endgroup$
    – Hans
    Feb 4 '12 at 12:01
  • $\begingroup$ @Hans I'm still thinking about your problem from time to time, when I have moment to spare. I believe that a complete resolution of your problem can only come after we have clearly defined "the $L^2$ semigroup induced by $X$". Do you have a specific definition in mind? $\endgroup$
    – user6096
    Feb 7 '12 at 21:51
  • $\begingroup$ @Byron Schmuland. I reedited the question adding the definition of the $L^2$ semigroup induced by $X$. Also for this my reference is Fukushima/Oshima/Takeda. $\endgroup$
    – Hans
    Feb 8 '12 at 0:48
  • $\begingroup$ @Hans In that case, doesn't the Friedrichs extension of $(G,{\cal D})$ equal the $L^2$ generator of the process by definition? $\endgroup$
    – user6096
    Feb 8 '12 at 17:38

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.


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