Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I would like to prove that the Legendre polynomials form a complete basis on the interval [-1, 1] using functional analysis.

Here is what I came up with so far. Legendre polynomials $P_n(x)$ are solutions of the following Sturm–Liouville problem:

$$-\frac{d}{dx}\left[p(x)\frac{du}{ dx}\right]+q(x)u=\lambda w(x)u$$

with $p(x)=1-x^2$, $q(x)=0$, $w(x)=1$ and $\lambda=n(n+1)$. This can be written as:

$$L u = \lambda u$$


$$L u = {1 \over w(x)} \left(-{d\over dx}\left[p(x){du\over dx}\right]+q(x)u \right)$$

Using proper boundary conditions, it can be proven that $L$ is self-adjoint. The eigenvalues are $n(n+1)$. They are discrete. One can then easily prove that the eigenvectors ($P_n(x)$) are orthogonal.


1) What else do I need to prove to show that the spectrum is discrete?

2) What exactly do I need to prove to show that these eigenvectors form a complete basis? What mathematical theorem/theorems can be used?

3) How would 2) change if some part of the spectrum was continuous?

share|cite|improve this question
For 1 (and then 3 as well): what happens if $\lambda$ is not of the form $n(n+1)$? –  Guess who it is. May 9 '12 at 2:19
@J.M., for non-integer $n$ (i.e. a general $\lambda$), the solutions are Legendre functions ( So it would seem that all $\lambda$ are actually allowed, thus the spectrum being continuous... –  Ondřej Čertík May 9 '12 at 4:01
Yes. However, note that the Legendre functions in general are singular at the ends of the interval $(-1,1)$; that's what makes the Legendre polynomials special... –  Guess who it is. May 9 '12 at 4:04
Ah -- I forgot about the boundary condition -- so requiring the solutions to satisfy some conditions (for example being non-singular, or I can even require something stronger, like being $\pm 1$ at the endpoints), then the only allowed $\lambda$ are of the form $n(n+1)$, thus proving that the spectrum is discrete. If we allow Legendre functions, then the spectrum is continuous --- because I think that the Legendre functions can be normalized to (Dirac) delta function (using physics terminology). What about 2? –  Ondřej Čertík May 9 '12 at 4:11

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.