accumulation point of a bounded linear operator

Question

Let $T$ be a bounded linear operator acting on a complex Banach space $B.$ Assume that there exists $\epsilon>0$ such that $\lambda I - T$ is one-to-one with a dense image for all $0<|\lambda|<\epsilon.$ In this case, $0$ is not an accumulation point of the spectrum $\sigma(T)?$

Answer 1

If the range of $\lambda I - T$ is dense but not all of $B$, then $\lambda$ is in the spectrum.

For example, let $T$ be multiplication by the variable $x$ on $L^2[0,1]$. Then for every $\lambda \in \sigma(T) = [0,1]$, $\lambda I - T$ is one-to-one with dense range.

EDIT: To see that the range is dense, note that the range includes all $L^2$ functions that are $0$ in a neighbourhood of $\lambda$.