# Is the resolvent of a local operator local?

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 $$e_x \in l^2(\mathbb{Z})$$.

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.
• I think there is a misunderstanding due to my abuse of notation: I denote by $x$ both the integer $x \in \mathbb{Z}$ and also the element $|x \rangle \in l^2(\mathbb{Z})$ such that $| x \rangle (y) = 1 \iff x=y$ and $0$ otherwise for all $y \in \mathbb{Z}$. If we instead denote the corresponding element of $l^2(\mathbb{Z})$ by $e_x$, we avoid this problem. Then, if we multiply both sides by $a$, this does not affect the distance between the elements $a e_x$ and $a e_y$ and we can not conclude that $\langle a e_x |A | a e_y \rangle =0$. I will change the notation in the original question. Commented Jul 4 at 10:15
• Thanks for the answer, that solves my question. We can also see from your example where my error lies, it is in the informal equality of $(A-z)^{-1}$ maps a part of $e_y$ into $e_x$ and $A-z$ maps a part of $e_x$ into $e_y$. As we can already see in the inverse of $J_n(z)$, this is wrong. Commented Jul 5 at 14:12
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.