Let $\mathcal L$ be the infinitesimal Generator of a Markov Prozess. Assume we know, that applied to some suitable, given function $f$ we know that we get the function $a(x)$ i.e $$\mathcal L f(x)=a(x)$$ Is it possible to directly conclude $e^{t \mathcal L} f(x)=?$
I know, that $e^{t \mathcal L}$ could be replaced by the transition semigroup and we could compute that, but is there a way to use our previous knowledge of $a(x)$ to arrive there aswell?
Example: The Markov process is standard $1$-dimensional Brownian motion. Consider $f_0(x):=1$ and $f_1(x)=x$. Then $\mathcal L f_0(x) =\mathcal L f_1(x) =0$ for all $x$, but $e^{t\mathcal L}f_0(x)=f_0(x)$ while $e^{t\mathcal L}f_1(x)=f_1(x)$ fo all $x$.