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


2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – Andreas132
    Commented Jul 4 at 10:15
  • $\begingroup$ I have edited the answer, please look at the example now. $\endgroup$ Commented Jul 4 at 22:43
  • $\begingroup$ 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. $\endgroup$
    – Andreas132
    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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .