Suppose that $T$ is a closed densely defined operator on a Hilbert space $H$ with $\rho(T) \neq \emptyset$. If $\lambda \in \sigma(T)$ is an isolated point then we know that $H = \mathcal{N}(E_o) \oplus \mathcal{R}(E_o)$ where $E_o$ is the associated spectral projection. Then it is known that $T$ is completely reduced by the pair of subspaces and if $T_1 = T|_{\mathcal{N}(E_o)}$ and $T_2 = T|_{\mathcal{R}(E_o)}$, then $\sigma(T_1) = \sigma(T) \setminus \{\lambda\}$ and $\sigma(T_2) = \{\lambda\}$. If we have the condition that $\mathcal{N}((T - \lambda I)^2) = \mathcal{N}(T - \lambda I)$, can it be proved that $(T - \lambda I)\mathcal{R}(E_o) = {0}$?. I am trying to prove that $\mathcal{R}(T - \lambda I) = \mathcal{N}(E_o) \oplus (T - \lambda I)\mathcal{R}(E_o) = \mathcal{N}(E_o) \oplus 0$.
Also $\mathcal{R}(E_o) = \{x: \|(T - \lambda)^nx\|^{1/n} \rightarrow 0\}$ is known for closed operators. Then $\mathcal{N}(T - \lambda I) \subseteq \mathcal{R}(E_o)$. Then converse needs to be proved.
In a previous problem of yours, I considered $L : L^2[0,1]\rightarrow L^2[0,1]$ given by $$ Lf = \int_{0}^{x}f(t)dt. $$ This is a quasinilpotent operator, i.e., $\sigma(L)=\{0\}$, and $\mathcal{N}(L)=\{0\}$. The range of $L$ is dense in $L^2[0,1]$. With this operator, let $$ H = L^2[0,1]\times L^2[0,1]. $$ and define $T : H \rightarrow H$ by $T(f,g)=(Lf,L^{-1}g)$ on $\mathcal{D}(T)=L^2 \times L(L^2)$. Then $T$ is unbounded, closed and densely-defined with $\sigma(T)=\{0\}$. Furthermore, $\mathcal{N}(T-0I)=\mathcal{N}((T-0I)^2)=\{0\}$.
The spectral projection $E_0$ associated with $\lambda=0$ is the orthogonal projection onto $L^2[0,1] \times \{0\}$. However $(T-0 I)E_0 \ne 0$.