Order of the eigenvalues of Sturm-Liouville operator Suppose we are working on a bounded interval $[a,b]$ with Sturm-Liouville operator $L$ given by
$$
     Lf = \frac{1}{w(x)}\left[-\frac{d}{dx}\left(p(x)\frac{df}{dx}\right)+q(x)f\right].
$$
How can I prove that the eigenvalues of the operator can be ordered as an increasing sequence such that : $\lambda_0 < \lambda_1 < \lambda_2... < \lambda_n < ... \to + \infty$
 A: What's your knowledge in basic Functional Analysis? If enough, note that it is consequence of the spectral theorem for compact operators (having in mind that eigenvalues for Sturm-Liouville-type operators are not defined in the classical way, but inversed).
See https://en.wikipedia.org/wiki/Compact_operator_on_Hilbert_space#Spectral_theorem.
A: The problem needs to be a regular problem on $[a,b]$; otherwise what you say may not be true at all. If the problem is regular, and if you have endpoint conditions
$$
                 \cos\alpha f(a)+\sin\alpha f'(a) = 0 \\
                 \cos\beta f(b)+\sin\beta f'(b) = 0,
$$
then you basically get what you want. One way to argue is by looking at the classical solutions of the Sturm-Liouville equation $Lf=\lambda f$ with endpoint conditions
$$
                    f(a) = -\sin\alpha,\;\;\; f'(a)=\cos\alpha.
$$
Then $f(x,\lambda)$ and $f'(x,\lambda)$ will depend holomorphically on $\lambda$, which means that the solutions where $\cos\beta f(b,\lambda)+\sin\beta f'(b,\lambda)=0$ will correspond to the zeroes of a holomorphic function of $\lambda$. Then some simple estimates of $\langle Lf,f\rangle$ will show that this form is bounded below, and, therefore, the eigenvalues are bounded below. Knowing that the eigenvalues are the zeros of a power series in $\lambda$ that is not identically $0$ tells you that the set of eigenvalues do not cluster and, therefore, can be ordered.
