I am reading about operators and their resolvents and I have the following question
If $A$ is a differential operator that commutes with translations, that is, $A\tau_h u=\tau_h Au$ then does the resolvent operator also commute?
Actualization.
in PERTURBATION OF NORMAL QUATERNIONIC OPERATORS (https://sweet.ua.pt/ukaehler/Webpage/Publications_files/tran7749_p1.pdf), I just read that "For example, in the complex case, given a linear operator $A$, any linear operator $B$ commuting with $A$ also commutes with the resolvent $(\lambda I−A)^{-1}$"
Given the above, my question would be true, but how can it be proved that if $B$ commutes with $A$ then the resolvent of $A$ also?
Actualization 2. I think I just checked with the newman series for the resolving operator.