Look at the Hilbert space $l^2( \mathbb{Z}) $ and let $A$ be a translation invariant band operator. I.e. if $\{ e_n \}_{n \in \mathbb Z} $ is the standard basis for $l^2( \mathbb{Z}) $ then it holds that \begin{align*} \langle e_n, A e_m \rangle = 0 \end{align*} for $\vert n-m \vert \geq K$ for some $K$. And translation invariance gives \begin{align*} \langle e_n, A e_m \rangle = \langle e_{n+k}, A e_{m+k} \rangle \end{align*} for all $k \in \mathbb{Z}$.
The eigenvalue equation reads
\begin{align*} A \psi = \lambda \psi \end{align*}
for some function $\psi \in l^2 ( \mathbb{Z}) $.
Since $A$ is a band matrix we could extend this equation for functions $\phi: \mathbb{Z} \to \mathbb{C}$ which we do not require to be square-summable and ask for solutions to this equation.
Is it now true that $\lambda \in \sigma(A)$ if and only if there exists such a solution $\psi$ which is bounded by a polynomial? Maybe one needs to assume translation invariance.
Example: For the discrete Laplacian \begin{align*} \langle e_n, A e_m \rangle = -2 \delta_{n,m} + \delta_{n+1,m} + \delta_{n,m+1} \end{align*} we get solutions of the form $\psi(x) =e^{ipx}$ which are not square integrable, but are still solutions of the finite difference equation. Notice that one can truncate these to construct af Weyl-sequence and it seems that this could prove at least one direction of the question.
The issue is discussed in chapter 7.1 https://arxiv.org/pdf/0709.3707.pdf here for Hamiltonians $A$, but I am particularly interested in the non-normal case.
EDIT: (Partially progress using translation invariance) By translation invariance the eigenvalue equation can be written as \begin{align*} \sum_{i = - K}^K a_i \psi_{i+n} = \lambda \psi_n \end{align*} which is just linear recursion. So one can find the solutions using the characteristic equation and then they are either constant or exponentially increasing/decreasing. How can we show that the exponentially increasing solutions do not correspond to elements in the spectrum of $A$? Edit2: In this case we can prove the result using Fourier Transformation, but what if the operator is not translation invariant, but translation invariant plus a compact pertubation?
