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Let $X$ be en compact metric set, and denote by $\mathcal{C}(X)$ the set of real continuous functions defined on $X$, endowed with the supremum norm $\|\cdot\|_{\infty}$. Let $\Omega$ be the generator of a Markov generator (for exemple with the definition of Liggett) generating the Markov process (toward its natural filtration) $\zeta=(\zeta_t)_{t\geq 0}$, having for initial law $\delta_u$ for a certain $u\in X$. Denote by $\mathcal{D}(\Omega)$ the domain of $\Omega$. Let $y,z\in\mathcal{D}(\Omega)$ such that, for any $f,g\in\mathcal{D}(\Omega)$ where $f$ is a function of $y$ and $g$ a function of $z$, then $fg\in\mathcal{D}(\Omega)$ and

$$\Omega(fg)=f\Omega g+g\Omega f.$$

Question: For $t\geq 0$, are $y(\zeta_t)$ and $z(\zeta_t)$ independents?

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    $\begingroup$ A trivial counter-example is when, say, $y$ is too rough, so that there are no non-constant functions of $y$ in the domain of the generator. (Say, if $\zeta$ is the Brownian motion and $y$ is nowhere differentiable.) I guess this is not the answer you are looking for, though. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Apr 9, 2021 at 21:35
  • $\begingroup$ (By the way, perhaps Liggett uses this notation, but I find it rather confusing: $\Omega$ is typically the probability space, not an operator. Also, quite often $\zeta$ denotes the lifetime of a Markov process rather than the Markov process, which is commonly denoted by $X_t$ or another capital letter.) $\endgroup$
    – Mateusz Kwaśnicki
    Commented Apr 9, 2021 at 21:39
  • $\begingroup$ Thank you for this remark, I edited it ($y,z\in\mathcal{D}(\Omega)$ rather than $\mathcal{C}(X)$) hoping that it is now more relevant. (Sorry for the choice of notations, indeed form Liggett.) $\endgroup$
    – G. Panel
    Commented Apr 9, 2021 at 22:57
  • $\begingroup$ This is still nitpicking, but now one can consider a very rough Markov process, so that again virtually no function of $y$ is in the domain of the generator. If I remember correctly, if $\zeta$ is the Brownian motion on, say, the Sierpiński triangle, and $y$ is a non-constant function in the domain of the generator, then $y^2$ is not in the domain of the generator. So I suppose one can craft $y$ in the domain in such a way that the only functions of $y$ which are in the domain of the generator are linear functions of $y$. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Apr 10, 2021 at 8:05

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