Is the resolvent of a local operator local?

Question

Let $A$ denote a bounded linear operator on the Hilbert space $l^2(\mathbb{Z})$. We call $A$ a local operator if and only if there exists a $C \geq 0$ such that $\langle e_x | A | e_y \rangle = 0 $ if $|x-y| \geq C$. Here, $e_x \in l^2(\mathbb{Z})$ acts such that $e_x(y) = 1 \iff x=y$ and is $0$ for all other $y \in \mathbb{Z}$. The family $(e_x)_{x \in \mathbb{Z}}$ forms an orthonormal basis of $l^2(\mathbb{Z})$.

I want to find out whether there exists any relation between $A$ being local and its resolvent being local. I can informally show the following for any z in the resolvent set of $A$:

\begin{align*} \langle e_x | (A-z)^{-1} | e_y \rangle \neq 0 &\iff (A-z)^{-1} \text{ maps a part of } e_y \text{ into } e_x \\ &\iff A-z \text{ maps a part of } e_x \text{ into } e_y \\ &\iff \langle e_y | A-z | e_x \rangle \neq 0 \end{align*} Now, if $|x-y| \geq C$, then \begin{align*} \langle e_y | A-z | e_x \rangle = \langle e_y | A | e_x \rangle - z \langle e_y | e_x \rangle = 0 \end{align*} From this I would obtain \begin{align*} A \text{ is local } \iff (A-z)^{-1} \text{ is local for some } z \text{ in the resolvent set of } A \end{align*} On top of that, the range $C$ of $A$ would be the same for all resolvents $(A-z)^{-1}$.

I am not entirely convinced of my argument, but the only point that I could see being wrong is the equivalence of $\langle e_x | (A-z)^{-1} | e_y \rangle \neq 0$ and $\langle e_y | A-z | e_x \rangle \neq 0$. If my informal argument is correct, there must be a rigurous way to show this. Can you help me make my argument more precise or show me where my error lies?

EDIT: I have changed my notation to avoid misunderstandings. Before that I had denoted by $x$ both the integer $x \in \mathbb{Z}$ and also the element $e_x \in l^2(\mathbb{Z})$.

Answer 1

For each $n\in\mathbb N$ and $z\in\mathbb C$, let $J_n(z)$ denote the $n\times n$ Jordan block, $$ J_n(z)=\begin{bmatrix} z &1&0&0&\cdots&0\\ 0&z &1&0&\cdots&0\\ \vdots&\ddots&\ddots&\ddots&\cdots&\vdots\\ 0&\cdots&0&z &1&0\\ 0&\cdots&\cdots&0&z &1\\ 0&\cdots&\cdots&\cdots&0&z \\ \end{bmatrix} $$ Then, for nonzero $z$, $$ J_n(z)^{-1}=\begin{bmatrix} z^{-1} &-z^{-2} &z^{-3}&-z^ {-4}&\cdots&(-1)^{n+1}z^{-n}\\ 0&z^{-1} &-z^{-2}&z^{-3}&\cdots&(-1)^{n}z^{n-1}\\ \vdots&\ddots&\ddots&\ddots&\cdots&\vdots\\ 0&\cdots&0&z^{-1} &-z^{-2}&z ^{-3}\\ 0&\cdots&\cdots&0&z^{-1} &-z^{-2}\\ 0&\cdots&\cdots&\cdots&0&z^{-1} \\ \end{bmatrix} $$ Now form $$ A=\bigoplus_{n=1}^\infty J_n(0)=\begin{bmatrix}J _1(0)\\& J _2(0)\\&&\ddots\end{bmatrix}. $$ Then for all $z$ we have that $A-z$ is local with $C=2$. But for all $z\ne0$, $$ (A-z)^{-1}=\begin{bmatrix}J _1(0)^{-1}\\& J _2(0)^{-1}\\&&\ddots\end{bmatrix} $$ and so if $x$ denotes the first row of $J_n(z)^{-1}$ in $A$, then $$ \langle e_x|(A-z)^{-1}|e_{x+n}\rangle =(-1)^{n+1}z ^{-n}\ne0. $$ Thus $(A-z)^{-1}$ is not local.

Answer 2

I have stumbled upon another way to see that the resolvent of a local operator is not necessarily local, by using the Neumann series. Informally, we have \begin{align*} (A-z)^{-1} = -\frac{1}{z} \left(I - \frac{A}{z} \right)^{-1} = - \frac{1}{z} \sum_{k=o}^\infty \left( \frac{A}{z} \right)^k \end{align*} Unless $A$ is a diagonal matrix (i.e. has range $0$), the terms $A^k$ will have longer and longer ranges. Thus, the resolvent can not be local.