What happens to continuous spectrum upon discretization? Excuse me for a bit of an vague question, but I haven't been able to find a definite answer for this for quite some time. My question is regarding (mostly non-normal )linear operators and their discretization. If we know that the operator has part of spectrum which is continuous, and we discretize the operator and obtain the eigenvalues numerically, what do we expect to see ? 
Could anyone explain the "main idea" behind how continuous spectrum will show up upon discretization of the operator itself ?
 A: Perhaps a concrete and simple example is helpful to develop intuition. Take the Laplacian $-d^2/dx^2$ on $\mathbb{R}$. The spectrum is continuous and unbounded, $E(k)=k^2$, $k\in\mathbb{R}$. 
Now discretize the $x$ variable on $\mathbb{Z}$. The spectrum remains continuous, but becomes bounded, $\tilde{E}(k)=2-2\cos k$, $k\in(-\pi,\pi]$. Notice that for small $k$ the two spectra coincide, $\tilde{E}(k)=E(k)+{\rm order}(k^4)$. Small $k$ means large wave lengths, and the discretization is not noticed if the wave length is larger than the spacing of the discretization.
You ask: "when we discretize the eigenvalues numerically, what do we expect to see?" The discrete spectrum appears not because of the discretization, but because numerically you will need to consider a finite system, rather than an infinite system. So you will restrict $x$ to some finite interval $(-N,N]$ of $\mathbb{Z}$ and you will need to introduce boundary conditions at the end points. Periodic boundary conditions are convenient, then the spectrum becomes $E_n=2-2\cos(n\pi/N)$, $-N<n\leq N$. For other boundary conditions the formula is different, but the spectrum remains discrete.
This is for a self-adjoint operator with a real spectrum, but I think the same applies to a non-self-adjoint operator: numerically you find a discrete spectrum not because of the discretization, but because of the finite system.
