I'm looking for a solution to the following problem -
$\int_{-\infty}^{\infty} K(x-y) f(y) = \lambda f(x)$
Consider $K(x-y) = \left\{ \begin{array}{lr} e^{-(x-y)} & : x > y \\ 0 & : x < y \end{array} \right.$
In this case, we can actually find the eigenvalues and eigenvectors analytically, since $f(x) = e^{i \omega x}$ is an eigenvector:
$\int_{-\infty}^x e^{-(x-y)} e^{i \omega y} = \frac{e^{i \omega x}}{1 + i \omega}$
But now, if I try to do this numerically, i.e., if I discretize the integral: $\tilde{K}(x_i,y_j) = K(x_i - y_j)$, then I end up with a triangular matrix with 1's on the diagonal, and all the eigenvalues are thus 1.
I don't really see how to resolve this issue numerically, so any help would be much appreciated.