AFAIK, a Green's function $G(x,\epsilon)$ has a singularity for $x=\epsilon$. This is clear in many analytical expressions for $G(x,\epsilon)$.
I do not understand, however, what happens to the singularity when $G(x, \epsilon)$ is expanded in terms of the linear operator eigenfunctions:
$G(x, \epsilon) = \sum_i \frac{u_i(x)u_i(\epsilon)}{\lambda_i}$.
For example, for rectangular-like domains and the dirichlet laplacian, eigenfunctions $u(x)$ are sines. How can I show that the series of $G(x, \epsilon)$ is still singular at $x=\epsilon$?
I found an answer in Courant & Hilbert's book pgs. 351-2.
The Green's function itself may be continuous and well-defined at $x=\epsilon$. However, the Green's function is singular in the sense that its first derivative does not exist at $x=\epsilon$.