# Completeness of eigenfunctions of singular Sturm-Liouville problems

It is well known that eigenfunctions of a regular Sturm-Liouville problem form a complete orthogonal basis, but what about eigenfunctions of a singular Sturm-Liouville problem? Under what conditions are they complete? I'm primarily interested in the semi-infinite case $L_2(0,\infty)$ with well-behaved (continuously differentiable) coefficient and weight functions. I would be very grateful for proof sketches and references.

If $0$ is a regular endpoint for a Sturm-Liouville operator $Lf=-f''+qf$, and $\infty$ is a singular endpoint, then there are unique classical solutions $\varphi_{\lambda}$ of the eigenvalue problem with $$+\cos\alpha\varphi_{\lambda}(0)+\sin\alpha\varphi_{\lambda}'(0)=0 \\ -\sin\alpha\varphi_{\lambda}(0)+\cos\alpha\varphi_{\lambda}(0)=1.$$ Regardless of the possible condition (if any) that must be imposed at $\infty$ for a well-posed Sturm-Liouville problem, there is a measure $\mu_{\alpha}$ on $\mathbb{R}$ such that the following eigenfunction expansions hold, as well as the Parseval equality: $$f(x) = \int_{\mathbb{R}}\left(\int_{0}^{\infty}f(t)\varphi_{\lambda}(t)dt\right) \varphi_{\lambda}(x)d\mu_{\alpha}(\lambda), \\ Lf = \int_{\mathbb{R}}\left(\int_{0}^{\infty}f(t)\varphi_{\lambda}(t)dt\right)\lambda\varphi_{\lambda}(x)d\mu_{\alpha}(\lambda), \\ \|f\|^2 = \int_{\mathbb{R}}\left|\int_{0}^{\infty}f(t)\varphi_{\alpha}(t)dt\right|^2d\mu_{\alpha}(\lambda).$$ (Allowing for a weighted operator $L=\frac{1}{w}\left(-(pf')'+qf\right)$ requires working in the weighted space $L^2_w(0,\infty)$ instead of $L^2(0,\infty)$.) The spectral density measure $\mu_{\alpha}$ is purely discrete if the SL problem is in the limit circle case at $\infty$, which results in a countable orthonormal basis of eigenfunctions. However, in general, the spectral density measure may be discrete, continuous, singular continuous, or a combination of such components.