Let $X$ be a Banach space and $\Phi : X \rightarrow X$ a bounded operator.
Let also $R(\cdot,\Phi) : \rho(\Phi) \rightarrow B(X)$ the resolvent operator of $\Phi$ defined in the usual way as $R(\lambda,\Phi) := (\lambda I -\Phi)^{-1}$ for $\lambda \in \rho(\Phi)$.
Is it always true, or under which assumptions, does $\Phi$ commute with its resolvent?
And if yes, how to show it?
I think that maybe the Neumann series plays a role in an eventual proof.
Best Answer
You have $$ I=R(\lambda,\Phi)\,(\lambda I - \Phi)=\lambda\,R(\lambda,\Phi) - R(\lambda,\Phi)\Phi. $$You also have $$ I=(\lambda I - \Phi)\,R(\lambda,\Phi)=\lambda\,R(\lambda,\Phi) - \Phi\,R(\lambda,\Phi). $$ Subtract, and you get $$ R(\lambda,\Phi)\Phi=\Phi\,R(\lambda,\Phi). $$ In general, any algebraic expression on $\Phi$ will commute with $\Phi$. And then you can move up from there to functional calculus (which would give a different prove of the equality).