TITLE: Why does the set of an hermitian operator's eigenfunctions spans the functions space
QUESTION [10 upvotes]: During a discussion about linear hermitian operators, my professor claimed that if a linear operator $M$ is hermitian under a certian set of conditions, then genrally any function that fulfills these conditions can be expressed as an infinite sum of the operator's eigenfunctions.
However, he did not prove it, as it is a "math for physicists" course which is quite informal. I am intrested in the proof.
We proved a similar theorem once in a linear algebra course (the set of an hermitian $(n\times n)$ matrix's eigenvectors spans the space), but the proof was based on mathematical induction on the space's dimensions, and here we have infinite dimensions.

REPLY [4 votes]: In addition to Ian's and Bob Israel's points, there is another useful class: (suitable) unbounded operators whose inverse (or resolvent) is compact. Abstractly, it is not clear why this is an interesting class, except for the classical fact that the spectra are in correspondence by $\lambda\longleftrightarrow 1/\lambda$. It is only when we somehow know/prove things akin to the fact that Laplace-Beltrami operators on compact Riemannian manifolds have compact resolvents that this looms large.
But/and, no, generally, there is no simple (and true) assertion about eigenvectors, because the literal eigenvectors do not span the whole space in any sense. That is, there is "continuous spectrum", and/or a need for "generalized eigenvectors".
The simplest case of the latter is the spectral decomposition of $L^2(\mathbb R)$ with respect to the Laplacian. Fourier inversion expresses nice functions as superpositions of exponentials $e^{ix\xi}$, which are not in $L^2$, etc.