When the point spectrum is discrete? Are there some criteria to tell when the point spectrum of a linear operator is discrete? In general it is not the same (take the spectrum of the "annihilation" operator). More specifically, what are the conditions that should be satisfied by a symmetric (or even self-adjoint) operator to have "point spectrum" = "discrete spectrum"?
 A: When dealing with differential operators, the most common condition is where the resolvent operator is compact. Resolvents of differential operators are often compact on bounded domains because they map uniformly bounded sets of continuous functions to sets of functions with uniformly bounded derivatives, which means the image is an equicontinuous set of functions. Some argument of that type is usually given to prove a resolvent is compact. Showing compactness is trickier on infinite domains, and the resolvent need not be compact, which is the case for the simplest differential operator, the momentum operator.
Not all operators with point spectrum have some compact resolvent, though. When considered on $L^{2}$, the Fourier transform has 4 eigenvalues and a complete orthonormal basis of eigenfunctions (the Hermite functions.) If the dimensions of any of the eigenspaces is infinite, then an operator cannot be compact, unless the eigenvalue is $0$. Or, if the eigenvalues cluster somewhere other than at $0$, the operator cannot be compact.
If the resolvent $R(\lambda) = (A-\lambda I)^{-1}$ of an operator $A : \mathcal{D}(A)\subset \mathcal{H}\rightarrow\mathcal{H}$ on a separable infinite-dimensional Hilbert space $\mathcal{H}$ is compact for some $\lambda$, then the eigenvalues cluster only at $0$, which forces the eigenvalues of $A$ to cluster only at $\infty$.
