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The precise domain of the generator $A$ of an Itō diffusion on a Hilbert space $H$ (assume $H=\mathbb R^d$, if that's easier for you to work with) can usually not be determined explicitly$^1$. Usually, we can only show that $C_c^2(H)$ is contained in $\mathcal D(A)$.

Now, we know that if $\mu$ is a measure on $H$ with a density $p$ with respect to a reference measure $\lambda$ on $H$, then $\mu$ is invariant with respect to the diffusion process iff $$A^\ast p=0\tag2,$$ where $A^\ast$ denotes the adjoint of $A$ with respect to the bracket $$\langle\;\cdot\;,\;\cdot\;\rangle:\mathcal B_b(H)\times\mathcal L^1(\lambda)\;,\;\;\;(f,g)\mapsto\int fg\:{\rm d}\lambda\tag3,$$ where $\mathcal B_b(H)$ denotes the space of bounded Borel measurable real-valued functions on $H$.

(If you want to think of an example: Take standard Brownian motion; then $A=\frac12\Delta$ and $\mathcal D(A)$ is the closure of $C_c^\infty(\mathbb R^d)$ with respect to the graph norm of the Laplacian. This operator is self-adjoint; so $A=A^\ast$.)

Question: From this result, it seems like there is no chance to construct a diffusion process for a given $\mu$ with a non-differentiable or even non-continuous density $p$. Is that really the whole story? Nothing we can do about?

EDIT: In case it is not clear what I'm asking: Say we have a probability measure $\mu$ with non-differentiable density $p$ with respect to the Lebesgue measure on $\mathbb R^d$. Can we choose the drift and diffusion coefficients $b$ and $\sigma$ of an SDE $${\rm d}X_t=b(X_t){\rm d}t+\sigma(X_t){\rm d}W_t\tag4$$ such that $\mu$ is invariant with respect to the transition kernel of $(X_t)_{t\ge0}$?

Remark: The reason for this question is the following observation: No matter how smooth $p$ is, a Metropolis-Hastings kernel $\kappa$ with target distribution $\mu$ will always be $\mu$-reversible and hence $\mu$ will always be $\kappa$-invariant. Assume that we are using a Gaussian proposal kernel (this is usually called "Random Walk Metropolis"). Now, by specifying a constant holding rate, we can easily embed $\kappa$ into continuous-time. The resulting process, doesn't seem to be far from a (discretized) Brownian motion. Though, clearly, there is acceptance-rejection occurring. In that sense, we maybe can do something similar by running a suitably modified simulation of a diffusion process.

$^1$ Please correct me, if that's wrong, but it won't be important for this question.

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    $\begingroup$ Maybe I am misunderstanding, but are you starting with some general measure $\mu$ that has poor regularity and trying to construct a diffusion process that converges to it in law? The closest think I've seen in such a generality is using Dirichlet forms eg. see here "Construction of diffusion processes on fractals, d-sets, and general metric measure spaces" projecteuclid.org/journals/kyoto-journal-of-mathematics/… $\endgroup$
    – Thomas Kojar
    Commented Jun 30, 2023 at 20:27
  • $\begingroup$ In SPDEs, I've seen the use of Stochastic-quantization where they start from some very singular measure eg. that with distributional density (Euclidean free field), and construct an SPDE converging to it up to renormalizations. $\endgroup$
    – Thomas Kojar
    Commented Jun 30, 2023 at 20:29
  • $\begingroup$ It seems you're mixing up $C^2_c(H)\subset \mathcal{D}(A)$ with $\mathcal{D}(A)\subset C^2_c(H)$. Otherwise, I don't see how you would arrive to the conclusion that $p$ has to be differentiable (which is not true). $\endgroup$
    – m7e
    Commented Jul 1, 2023 at 13:21
  • $\begingroup$ @m7e Sorry, but I cannot follow. We have $\mathcal D(A)\subseteq C^2(H)$. On the other hand, if $\mu$ is invariant, then $p$ must belong to $\mathcal D(A^\ast)$. If, for example, $A$ is self-adjoint, then this means $p\in C^2(H)$. What am I missing? $\endgroup$
    – 0xbadf00d
    Commented Jul 1, 2023 at 13:37
  • $\begingroup$ No, $\mathcal{D}(A)\subset C^2(H)$ does not necessarily hold. In fact, there are examples (if the drift coefficient of the diffusion is a distribution) such that $\mathcal{D}(A)$ and $C^2(H)$ are disjoint. $\endgroup$
    – m7e
    Commented Jul 1, 2023 at 19:42

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