I want to prove the followig: let $T(X\to X)$ be a closed densely defined linear operator in a complex Banach space $H$. Let $D_0\subset D(T)$ and suppose that for some $\lambda\in\rho(T)$ we have $(T-\lambda)D_0$ is dense in $X$ then $(T-\mu)D_0$ is dense for any $\mu\in\rho(T)$.
I think it is posible to use some resolvent identity or, if $y\in X$, there exists $(x_n)_n\subset D_0$ such that $$y=\lim\limits_{n\to\infty}(T-\lambda)x_n.$$ Thus we can take $y_n:=(T-\mu)^{-1}(T-\lambda)x_n$ and clearly $(T-\mu)y_n$ converges to $y$. However, $(y_n)_n$ not necesarily lie in $D_0$.
Thanks for your ideas!
Here: $\rho(T)$ denotes the resolvent set of $T$.