I am looking for either an obvious answer or some assessment of how difficult this is.
Consider a compact operator $K$ from $L^2([a,b])\to L^2([a,b])$ with an eigenvalue at $\lambda_0\neq 0.$ Consider the (not closed) subspace $A=C([a,b])$ with the $L^2$ inner product. Suppose we know that in $A,$ the geometric multiplicity of $\lambda_0$ is $n$. Given that $A$ is dense, due to Lusins Theorem, it's hard to picture there being additional eigenfunctions in $L^2$ that are not in $A.$ Is my intuition correct or does it depend on $K$? I guess I'm looking for a version of the Fredholm alternative on dense subspaces. In general, it seems harder to calculate the dimension of the eigenspace in $L^2$ than $C$ since pointwise arguments can be made on $C.$
The vectors $v \in L^2$ such that $Kv = \lambda_0 v$ form a finite-dimensional subspace $W$ of $L^2$. The intersection of that with $C([a,b])$ is a subspace of dimension $n$. But the dimension of $W$ could be any integer $\ge n$.
The fact that continuous functions are dense is irrelevant: the continuous approximations of a discontinuous function in $W$ won't be eigenfunctions for eigenvalue $\lambda_0$.
It's easy to cook up examples of this: just take $n$ linearly independent continuous functions and $m$ functions which each have a single discontinuity at one point (different for each function), and let $K$ be a projection on the linear span of these $m+n$ functions.