Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In my computations I have obtained a sequence of eigenvalues $\lambda_k, \; k\in \mathbb{N}$ of double multiplicity. Thus, the basis for the eigenspace of $\lambda_k$ is given by $\psi_k(x) = \left\{e^{\frac{i \pi k}{l}x}, e^{-\frac{i \pi k}{l}x}\right\}$, where $l$ is the length of the interval in which the boundary value problem was considered. My question is: what should I do in order to show that the set $\{\psi_k\}_{k\in\mathbb{N}}$ is complete in $L^2(0,l)$? Basically, I have a problem with $\psi_k$ being given by two functions, instead of one. :) Thank you in advance!

share|cite|improve this question
The eigenvectors you gave are all constant functions. So altogether, they span the one-diensional space of constant functions. It would help if you said what the operator is, and maybe the details of your calculations. – 1015 May 13 '13 at 20:46
Sorry, that was a typo, I edited it. I should pay more attention! The original problem had the form \begin{equation} d^2 u^{iv} + (\lambda \beta_2 - P_1)u'' + (P_2 - \lambda\beta_1) u = 0 \end{equation} with the condition on the boundary $u''(0) = u''(l) = u(0) = u(l) = 0$. I posted it in one of my questions earlier. – Dina May 13 '13 at 21:01
Then with all these eigenvectors, you have recovered the canonical orthonormal (well, maybe not length one, depending on the inner product normalization) basis of $\{e^{ik\pi x/l}\;;\;k\in\mathbb{Z}\}$ of $L^2$. Your operator is even diagonal in the canonical basis. It does not make sense to say that $\psi_k(x)$ is equal to these sets. What you mean is that $\psi_k$ is either one. – 1015 May 13 '13 at 21:03
Ah, I see it now, with $k\in\mathbb{Z}$. So simple! Thank you, everything is clearer now! – Dina May 13 '13 at 21:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.