Nilpotency of $(A-z)P_z$ imply $z$ is an Eigenvalue?

Question

Let $z$ be an isolated point in the spectrum of some possibly unbounded operator $A:D\subset X \rightarrow X$ on a Banach space $X$. Where the spectrum $\sigma(A)$ without $z$ is closed. Let $P_z$ be the Riesz projector corresponding to $z$. Assume $((A-z)P_z)^m=0$ for some $m$. Then $z$ is an Eigenvalue of $A$. But why?

Answer 1

Intuition

The proof below is based on two observations. First, the converse of Cauchy's integral theorem implies that $P_z$ has non-zero range $\mathcal{R}$. Second, if $(A-z)P_z$ is nilpotent then starting with some non-zero vector in $\mathcal{R}$ we can construct a finite sequence of vectors $$ u, (A-z)P_zu,\dots, ((A-z)P_z)^{k-1}u, ((A-z)P_z)^ku=0\tag1 $$ all of which are non-zero except the last one. Taking the penultimate element of the sequence we obtain an eigenvector.

Proof

Let $\rho(A)$ and $\sigma(A)$ denote the resolvent set and the spectrum of $A$, respectively. Let $\Gamma_z\subset\rho(A)\cup\{z\}$ denote a simply connected open subset of $\mathbb{C}$ such that $z\in\Gamma_z$. Let $\gamma_z:[0,1]\to\Gamma_z$ denote a simple rectifiable loop that lies witin $\Gamma_z$ and encloses $z$. Recall the definition $$ P_z=-\frac{1}{2\pi i}\oint_{\gamma_z}(A-wI)^{-1}dw\tag2 $$ which is independent of the precise choice of $\gamma_z$ as long as it meets the requirements above. Note that $\Gamma_z$ is simply connected, so we can use the converse of Cauchy's integral theorem to rule out the possibility that $P_z=0$. Let then $\mathcal{R}$ denote the range of $P_z$ and let $u$ be a non-zero vector in $\mathcal{R}$. Further, let $k$ be the smallest positive integer such that $((A-z)P_z)^ku=0$ which is well-defined by assumption. In fact, $1\leq k \leq m$. Define $v:=((A-z)P_z)^{k-1}u$ and note that $v\ne 0$ by definition of $k$. Moreover, $v\in \mathcal{R}$ because $\mathcal{R}$ is invariant under $P_z$ and $A-z$. Finally, $(A-z)P_zv=0$, so $Av=zv$.