Why does the set of an hermitian operator's eigenfunctions spans the functions space

Question

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.

Answer 1

All these results are called "spectral theorems". They depend on what hypotheses you adjoin to the problem other than just being self-adjoint.

The "nice" case is for compact self-adjoint operators. Here the statement of the spectral theorem is essentially the same as in the finite dimensional case: there is an orthonormal basis of the space made up of eigenvectors, and all eigenvalues are real. Everything here is the same except that the basis in question is a Schauder basis (i.e. basis representations are infinite sums rather than finite sums, so there is a limit process going on when we write a basis representation).

Probably the more important case for many physical applications (for example in quantum mechanics) is for bounded self-adjoint operators. Here the work is harder, and the resulting "diagonal basis" is not just a countable version of what we see in finite dimensions.

Instead we get a measure space $(X,\Sigma,\mu)$, a function $f \in L^\infty(X,\mu)$, and a unitary similarity transformation defined by $U : H \to L^2(X,\mu)$. This similarity transformation turns our given bounded self-adjoint operator into the multiplication operator $T : L^2(X,\mu) \to L^2(X,\mu)$ defined by $(T \varphi)(x) = f(x) \varphi(x)$.

Things are more complicated still in the unbounded case. I won't try to go there, but there are hypotheses that can be used in this case as well.

Answer 2

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.

Answer 3

The correct rigorous version of this is the Spectral Theorem.

I don't know which "certain set of conditions" the professor had in mind. In general self-adjoint operators on an infinite-dimensional Hilbert space might have continuous spectrum, in which case there are no eigenfunctions at all (but a physicist might speak of "generalized eigenfunctions" that are not members of the Hilbert space, and the "sum" will become an integral).