I have the following question which I assume might be discussed in the literature under a terminology I am unaware of. Perhaps the notion doesn't even make sense and someone can explain why. I define the following notion:
Definition: Let $A\in \mathbb{C}_{n\times n}$ be a Hermitian matrix. I call a number $\lambda\in \mathbb{C}$ is a $\delta$-eigenvalue if there exists a normalized eigenfunction $\psi \in \mathbb{C}^n$, $\Vert \psi\Vert=1$, such that $\Vert A\psi-\lambda \psi \Vert\leq\delta$.
I am interested in whether there is a way to compute all $\delta$-eigenvalue of such a matrix? I'm pretty sure that it follows by a simple argument that $\big\{ \lambda\in \mathbb{C}: \text{dist}\big(\lambda,\sigma(A)\big)\leq \delta \big\}$ are all $\delta$-eigenvalues of $A$, but is it possible to have more $\delta$-eigenvalues.
My motivation is studying a self-adjoint operator $H:\ell^2(\mathbb{Z}^d)\to \ell^2(\mathbb{Z}^d)$ with purely essential spectrum. I want to find a Weyl sequence by restricting to finite subspaces and look for $\delta$-eigenvalues, which correspond to compactly supported functions in a Weyl sequence.
I would appreciate any insight that anyone has on the matter.
As $A$ is self-adjoint there is an orthonormal basis $\{e_k\}_{k=1}^n$ of eigenvectors. Let $\lambda_k$ denote the corresponding eigenvalues. For $x\in \mathbb{C}^n,$ let $x_k=\langle x,e_k\rangle.$ Then for $\|x\|=1$ we get $$\|(A-\lambda I)x\|^2=\sum_{k=1}^n|\lambda_k-\lambda|^2|x_k|^2\ge \min_k|\lambda_k-\lambda|^2=d(\lambda,\sigma(A))^2$$ Therefore if $\|x\|=1$ and $\|(A-\lambda I)x\|\le \delta,$ then $d(\lambda,\sigma(A))\le \delta.$
Conversely if $d(\lambda,\sigma(A))le \delta$ then $\lambda$ is a $\delta$-eigenvalue. Indeed, there exists $\lambda_0\in \sigma(A)$ such that $|\lambda-\lambda_0|\le \delta.$ Let $\|\varphi \|=1$ and $A\varphi=\lambda_0\varphi.$ Then $$\|(A-\lambda I)\varphi\|\le \|(A-\lambda_0I)\varphi\|+|\lambda-\lambda_0|\le \delta$$ Summarizing $\lambda$ is a $\delta$-eigenvalue iff $d(\lambda,\sigma(A))\le \delta.$
A similar argument can be used for infinite dimensional space, by applying the resolution of the identity. Namely if $$A=\int\limits_{\sigma(A)}t\,dE(t)$$ then for $\|x\|=1$ we get $$\|(A-\lambda I)x\|^2=\int\limits_{\sigma(A)}|t-\lambda|^2\,d\langle E(t)x,x\rangle \\ \ge d(\lambda,\sigma(A))^2\int\limits_{\sigma(A)}d\langle E(t)x,x\rangle =d(\lambda,\sigma(A))^2$$