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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

  1. $V$ is continuously embedded in $L^2(X,\xi)$
  2. If $C_0(X)$ is the space of continuous functions with compact support in $X$ then $V \cap C_0(X)$ is dense in $C_0(X)$
  3. 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:

  1. 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?

  2. 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).

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  • $\begingroup$ Can you make question 1 more precise? Is it something like this: "if we have two spaces $X,X'$ such that there is a Hilbert space isomorphism between $L^2(X), L^2(X')$ which maps $V$ to $V'$ and $a$ to $a'$, must $X,X'$ be homeomorphic?" $\endgroup$ – Nate Eldredge Aug 18 '17 at 16:28
  • $\begingroup$ @NateEldredge Well, usually one starts with a particular $a$ and $V$. Then to apply the theory developed by the author one needs to choose some $X$ that satisfies the conditions. If there are many such $X$ that fit the role then the results developed in the paper that use this framework are not well defined. I don't know how to specify it more mathematically, I'm sorry. $\endgroup$ – GuestUser Aug 20 '17 at 15:23
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For question 2, if you had $\Omega$ in place of $\bar{\Omega}$, the conditions you state would still be satisfied, but there are other interesting conditions that would not be. It would fail to be a regular Dirichlet space. In particular, you could not find a nice Markov process with state space $X = \Omega$ whose Dirichlet form was $a$. (Morally speaking, this process should be a Brownian motion in $\Omega$ that is killed at rate 1 inside $\Omega$, and reflected at the boundary. But in order to be reflected at the boundary, the process has to be able to reach the boundary, which you can't do with a process whose state space is only $\Omega$.)

(Incidentally, the author probably meant to say $V = H^1(\bar{\Omega})$.)

The standard modern reference on Dirichlet forms is Fukushima, Oshima and Takeda, Dirichlet Forms and Symmetric Markov Processes.

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  • $\begingroup$ Many thanks for the answer. You say that all conditions are satisfied whether $\Omega$ or $\bar\Omega$ is chosen, however the subsequent results in the paper read very differently depending on which set one picks. It seems strange then that the author of the paper is not more careful with the requirements. I had a very quick look at the book you recommend; it seems hard going but I will spend some time on it tomorrow at the office. Do you happen to know a text more aimed at PDE people? $\endgroup$ – GuestUser Aug 20 '17 at 15:17
  • $\begingroup$ You get a Dirichlet space in either case, but you get very different Dirichlet spaces. I don't know offhand of a more PDE text; the theory is probabilistic in origin and all the presentations I've seen are aimed at that point of view. $\endgroup$ – Nate Eldredge Aug 20 '17 at 18:29
  • $\begingroup$ Is it obvious that condition 3. in the OP implies the usual "unit contraction" property that defines a dirichlet form? $\endgroup$ – soup Aug 22 '17 at 13:20
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    $\begingroup$ @soup: Not to me. In fact, it looks more like the modern definition of "positivity preserving form". The paper cited by the OP is from 1976 so its definitions may not agree with those currently in use. $\endgroup$ – Nate Eldredge Aug 22 '17 at 15:51
  • $\begingroup$ Yes, in fact the reverse implication is true. $\endgroup$ – soup Aug 23 '17 at 9:40

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