 Solution

Let $L\varphi =\lambda \varphi$ with $\varphi \ne 0$ in the self-adjoint domain. By the boundary identity from part (i),Since its squared weighted norm is positive, $\boxed{\lambda \in \mathbb{R}}$.

For $L{\varphi }_n={\lambda }_n{\varphi }_n$ and $L{\varphi }_m={\lambda }_m{\varphi }_m$, the same identity givesThus ${\lambda }_n\ne {\lambda }_m$ impliesThis is orthogonality of Sturm-Liouville eigenfunctions. Degenerate eigenspaces can be given an orthonormal basis, although vectors in a degenerate eigenspace are not automatically orthogonal.