Consider two Markov semigroups $(T_t)_{t\geq 0}$ and $(S_t)_{t\geq 0}$ with infinitesimal generators $(A, D(A))$ and $(B, D(B))$. Suppose that some (bounded) operator $\Gamma$ is such that
$$
A\Gamma = \Gamma B.
$$
Does that imply that $T_t\Gamma = \Gamma S_t$?
This is immediately true if $A$ and $B$ are bounded as, in this case, $T_t = e^{tA}$ and $S_t = e^{tB}$ (converging absolutely), so that
$$
T_t\Gamma = \sum_{k\geq 0} \dfrac{t^k}{k!} A^k\Gamma = \sum_{k\geq 0} \dfrac{t^k}{k!} \Gamma B^k = \Gamma S_t.
$$
My first guess is that this should not be true in general. If, however, $T$ and $S$ are strongly continuous semigroups, then I have the feeling that the representation from Hille-Yosida should allow for this commutator relation. Unfortunately, I do not get a proof. It seems to me that the usual relations like
$$
T_t f – f = \int_0^t T_s A f ds
$$
are not helpful in this context. Instead, I would love to have a similar representation as in the bounded case. The closest I have found so far is the one from this answer. To conclude from there, one would need that
$$
\left(Iā\dfrac{t}{n}A\right)^{-n}\Gamma = \Gamma\left(I – \dfrac{t}{n}B\right)^{-n}
$$
which should follow from the fact that this negative power can be represented through
$$
\left( I – \dfrac{t}{n}A\right)^{-n} = \left(\left( I – \dfrac{t}{n}A\right)^n\right)^{-1} = \left(\sum_{k\geq 0} \dfrac{t^k}{n^k}A^k\right)^n,
$$
no?
But even if that argument works out, I would be interested in having a simpler proof that does not rely on this explicit expression of the semigroup in terms of the infinitesimal generator…
For those interested: The problem stems from looking into the Markov Mapping Theorem discussed in this paper as well as the one cited therein from Dynkin's book. Both can be interpreted as having the Markov semigroup satisfying some commutator relationship. Since the more recent work from Kurtz on Markov Mapping Theorems from the point of view of martingale problems, I was interested whether there is an easy argument to see that the initial theorem also works on the level of generators, at least in the case of strongly continuous generators. Otherwise, the result would be (mostly) purely theoretical as it is in most cases impossible to get information on the semigroup. There, $\Gamma f = f\circ \gamma$ for some measurable $\gamma$. Similarly, they would be looking at another operator $\alpha$ defined through $\alpha f(y) = \int f(x)\;\alpha(y,dx)$ for some Markov kernel $\alpha$. Both are bounded operators if I am not mistaken.
Edit: The spaces in the application are as follows: consider two measurable spaces $E$ and $B$, $\gamma: E\rightarrow B$ is measurable. Then $T$ is a semigroup on the space of bounded measurable functions $f:E\rightarrow \mathbb R$, in the strongly continuous case, we would look at $T$ as a semigroup on the space of bounded continuous functions (or continuous vanishing at infinity?). Similarly for $S$, exchangeing $E$ by $B$. To compute the infinitesimal generator, one would look at the bounded pointwise convergence resp.~uniform convergence. In this setting, as mentioned before, $\Gamma$ is the operator $f\mapsto f\circ \gamma$.
Best Answer
I will consider the case $(T_t)$ and $(S_t)$ are strongly continuous contraction semigroups. Also, I will assume that $\tilde{\mathcal{D}} = \mathcal{D}(A\Gamma)\cap \mathcal{D}(\Gamma B)$ is dense.
Let $t \geq 0$. By Hille-Yosida theorem, both $(I - tA)^{-1}$ and $(I - tB)^{-1}$ are contraction operators. Then for each $x \in \tilde{\mathcal{D}}$,
\begin{align*} (I - tA)^{-1}\Gamma(I - tB) x &= (I - tA)^{-1}(I - tA)\Gamma x = \Gamma x. \end{align*}
Since $\tilde{\mathcal{D}}$ is dense, it follows that $(I - tA)^{-1}\Gamma(I - tB)$ extends to a bounded operator that actually coincides with $\Gamma$. So, multiplying $(I - tB)^{-1}$ to the right of both sides, we get
$$ (I - tA)^{-1} \Gamma = \Gamma (I - tB)^{-1}. $$
From this, it follows that
$$ (I - tA)^{-n} \Gamma = \Gamma (I - tB)^{-n}. $$
Now replacing $t$ by $t/n$ and letting $n\to\infty$, we obtain $T_t \Gamma = \Gamma S_t$ as required.