My professor in the course "Math for Physicists" mentioned that the eigenfunction of a Hermitian operator or a Sturm-Liouville operator is an orthonormal basis for the function space on which the operator is defined, that the eigenvalues are real, and that the eigenfunctions are orthogonal.
I proved that the eigenvalues of the operators are real and that the eigenfunctions are orthogonal, but I haven't managed to prove that they span the function space (I assume that he meant a complete system; please correct me if I'm wrong). The professor said that he didn't want to deal with proving the completeness of the system because it would take him too long, and not because of a lack of our (the students) prerequisite knowledge.
I haven't learned functional analysis or measure theory, which seemed relevant to this question (in addition to the spectral theorem for the infinite-dimensional case, which we also haven't been introduced to; we were presented only with the finite-dimensional case), so I wonder if there is a way to prove this theorem that could have been presented to us without those concepts (even an informal one).
Additionally, I would be thankful if someone could provide a reference that introduces and proves this theorem.
Thank you for any information!
Best Answer
Suppose $\mu$ is a probability measure on $[0,1]$, and consider $Mf=xf(x)$. If $\mu$ has an atom at $x_0\in[0,1]$, then $M\chi_{\{x_0\}}=x_0\chi_{\{x_0\}}$, which means that $M$ has an eigenvector $\chi_{\{x_0\}}$ with eigenvalue $x_0$.
If $m$ is Lebesgue measure on $[0,1]$, then $M$ has only continuous spectrum, without any eigenvalues. Continuous spectrum gives you "approximate" eigenvectors in the sense that \begin{align} \|(M-\lambda I)\chi_{[\lambda-\epsilon,\lambda+\epsilon]}\|_{L^2[0,1]}^2 &= \int_{\lambda-\epsilon}^{\lambda+\epsilon}(x-\lambda)^2dm(x) \\ &=\left.\frac{(x-\lambda)^3}{3}\right|_{x=\lambda-\epsilon}^{x=\lambda+\epsilon}=\frac{2\epsilon^2}{3} \\ \|\chi_{[\lambda-\epsilon,\lambda+\epsilon]}\|^2 &=\int_{\lambda-\epsilon}^{\lambda+\epsilon}dm(x)=2\epsilon, \end{align} which implies that $$ \frac{\|(M-\lambda I)\chi_{[\lambda-\epsilon,\lambda+\epsilon]}\|}{\|\chi_{[\lambda-\epsilon,\lambda+\epsilon]}\|}=\frac{\epsilon}{\sqrt{3}} \rightarrow 0 \mbox{ as $\epsilon\rightarrow 0$}. $$ So the following is an approximate unit eigenvector sequence, with approximate eigenvalue $\lambda$, as $\epsilon\downarrow 0$: $$ \varphi_{\lambda,\epsilon}=\frac{\chi_{[\lambda-\epsilon,\lambda+\epsilon]}}{\|\chi_{[\lambda-\epsilon,\lambda+\epsilon]}\|}. $$ Every $\lambda\in [0,1]$ is an approximate eigenvalue. There are no actual eigenvalues for $M$. $M$ has only continuous spectrum, with $\sigma(M)=[0,1]$. With that mind, you're not going to prove that $M$ has a complete set of eigenvectors because it has no eigenvectors.