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Sturm-Liouville Problem

How could one prove that there are at most countably many eigenvalues of the Sturm-Liouville problem $−Lu=ju$, $j$ = eigenvalue, and $u$ is in $C^2[a,b]$?

I have put some more thought and I am still confused about this problem. Suppose I have $-Lu=\lambda u$ where $-Lu=(pu')'+qu.$ How can you justify that if $\left\{ c_{\alpha}\right\} _{\alpha\in\Gamma}$ is a set of non-zero real numbers such that $\mathbb{\sum}_{j=1}^{N}c_{\alpha_{j}}^{2}\leq1$ for every choice of $N\geq1$ and every choice of distinct $\alpha.$ Then $\Gamma$ is a countable set. How to proceed from there? Is there an easy solution? Thank you very much for help.

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marked as duplicate by Grigory M, t.b., Asaf Karagila, JavaMan, Jonas Teuwen Nov 8 '11 at 0:13

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I'm familiar with the line of proof that Wikipedia mentions, which is: show that the Sturm-Liouville operator is compact and apply the spectral theorem for compact operators (which is considerably easier to prove than the general one). I take it that you are looking for a more direct way? –  Tim van Beek May 20 '11 at 8:33

1 Answer 1

$\sum_{j=1}^N c_{\alpha_j}^2\le 1$ implies $0< c_\alpha^2\le 1$ for each $\alpha\in\Gamma$.
Let $A_n$, $n\ge 1$, be the subset of $S=\{c_\alpha\}_{\alpha\in\Gamma}$ defined by $$A_n = \left\{c_\alpha\in S : \frac {1}{n+1} < c_\alpha^2 \le \frac 1 n\right\}$$ $A_n$ must contain $n$ elements at most, otherwise $$\sum_{c_\alpha\in A_n} c_\alpha^2 > \sum_{c_\alpha\in A_n} \frac{1}{n+1} = \frac{\text{Card}(A_n)}{n+1} \ge 1$$ Evidently $\cup_n A_n = S$; $S$ is a countable union of finite sets, so it is countable.

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