Suppose we have a bounded operator T on a Banach space X, for which it holds that $ 0$ is an isolated point of the spectrum, and $ \cup_{n \in \mathbb{N}}({\rm Ker}(T^n))={\rm Ker}(T)$ is one-dimensional. Suppose further that T has no nontrivial, closed, invariant subspaces. Is it true that $ 0 $ must be a pole of the resolvent?
It is quite easy to construct counterexamples when the assumption that $0$ is an eigenvalue is dropped (see for instance this post), but I have been struggling to come up with counterexamples in the case outlined above. Thanks!