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I am reading a paper by Fukushima "On a stochastic calculus related to Dirichlet forms and distorted Brownian motions" and support it by a book "Dirichlet forms and symmetric Markov processes" by Fukushima, Oshima and Takeda.

Let me also remind that a dense subset $\tilde{C}$ of $C_0(X)$ is said to be a core of a Dirichlet form $\mathcal{E}$ if $\tilde{C}$ is $\mathcal{E}_1$-dense in $D[\mathcal{E}]$. A Dirichlet form possessing a core is called regular.

In the paper, there are the following 2 sentences:

"Let $\mathcal{E}$ be a Dirichlet form and $T_t$ the associated semigroup of Markovian symmetric operators on $L^2(X;m)$. If $\mathcal{E}$ is regular, then $T_t$ can be realized as \begin{equation} T_t\,f(x) = \int_X P_t(x, dy) f(y) \end{equation} by a transition function $P_t(x, E)$ on $X$ which is $m$-symmetric in the sense that $\int_XP_t\,f(x)\,g(x)\,m(dx) = \int_X P_tg(x)\,f(x)\,m(dx)$."

I could not find the above result in the book and I do not understand why we need the regularity assumption for that. Namely, we know that the space $C_0(X)$ of continuous functions with compact support is dense in $L^2(X,m)$, so we by Riesz-Markov-Kakutani there exists a transition function $P_t(x,dy)$ such that the above equation holds for $f\in C_0(X)$, and by density we can extend it to $L^2(X;m)$. Please tell me why this is wrong. Thank you very much.

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The result you quote is part of the main theorem of the book. I can look it up later today, but it's one part of the proof that every regular Dirichlet form has an associated Markov process.

The problem with your argument, in two words, is null sets. $T_t$ is an operator on $L^2$, and an element of $L^2$ isn't a function, it's an equivalence class mod a.e. equality. You would like to say "For each $x \in X$, consider the linear functional on $C_0(X)$ defined by $f \mapsto T_t f(x)$, and then use Riesz-Markov-Kakutani to find the measure associated to this functional and call it $P_t(x, \cdot)$." But $T_t f$ is only given as an element of $L^2$, and so $T_t f(x)$ isn't well defined; it depends on which representative of $T_t f$ you choose.

I think FOT has some counterexamples in the case that the form is not regular.

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  • $\begingroup$ Thank you very much, Nate. You made the idea very clear, you don't have to look up any notes. I have also seen one of your older questions on Dirichlet forms where you discussed a related example. Best wishes $\endgroup$ – tuko Sep 3 '16 at 15:36
  • $\begingroup$ Glad it helps. I think the "zero measure but positive capacity" issue is actually related to something else, so I took it out for now. $\endgroup$ – Nate Eldredge Sep 3 '16 at 15:39
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Although the regularity imposed in Fukushima's theory is quite natural, it is a bit of overkill if all one wants is a transition semigroup $(P_t)_{t\ge 0}$. There is work culminating in a short paper of S.E. Kuznetsov (1987-Prob. Theory and its Application; http://epubs.siam.org/doi/pdf/10.1137/1131031) that can be applied to a general Dirichlet form, the only extra condition being needed is that the state space is "nice" -- standard Borel is enough. Picking an initial distribution $\mu$ that is equivalent to the symmetry measure $m$, one can use the $L^2$ operators $T_t$ to define the finite dimensional distributions of a stochastic process, and then invoke Kolmogorov's theorem to gain the existence of such a process, call it $X=(X_t)$ on some probability space $(\Omega,\mathcal F,\Bbb P)$ such that $\Bbb P[X_0\in B]=\mu(B)$, $$ \Bbb P[X_s\in A,X_t\in B]=\int_X T_s(1_AT_{t-s}1_B)\,d\mu,\quad 0\le s<t, $$ etc. By the basic Dirichlet form theory, the semigroup $(T_t)$ is strongly continuous, so the process $X$ is necessarily stochastically continuous. This permits the application of Kuznetsov's result to obtain a (measurable in $(x,t)$) system $P_t(x,dy)$ of transition probabilities that satisfy the Chapman-Kolmogorov equation and represent $T_t$ as in the statement of your problem. The proof involves the Ray-Knight compactification, and builds on earlier work of J.B. Walsh dating to the early '70s.

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