MathOverflow is a question and answer site for professional mathematicians. 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

Let A be the unit square, $\{u_k\}$ is the set of all L2-normalized Laplacian eigenfunctions with Dirichlet boundary condition. Is it true that for any open subset V, $C_V = \inf\limits_k \int\limits_V dx |u_k(x)|^2 > 0$ ?

share|cite|improve this question

My offic mate and I believe this is true. By separation of variables the eigenfunctions are just $Csin(\pi kx)sin(\pi ly)$ for some fixed constant $C > 0$. Using the trig identity $\sin^2(x) = (1-\cos(2x))/2$, we see that

$\int_a^b \sin^2(kx)\ dx = (1/k)\int_{ak}^{bk} \sin^2(x)\ dx = (b-a)/2 - (1/2k)\int_{ak}^{bk}\cos(2x)\ dx \geq (b-a)/2 -1/2k$

The integral of $|u_k|^2$ on a small square is just

$\int_a^b\int_a^b \sin^2(\pi kx)\sin^2(\pi ly)\ dxdy$

so we can apply the previous line twice.

share|cite|improve this answer
These eigenfunctions are zero on the boundary. However, the problem is about arbitrary boundary conditions in which case the eigenfunctions are more complicated. – GH from MO Apr 8 '11 at 17:36
@GH: I usually understand "$L^2$ eigenfunctions with Dirichlet boundary conditions" to mean precisely the functions vanishing on the boundary... – Willie Wong Apr 8 '11 at 18:06
Willie, thanks. At any rate, what I suggested is a more interesting problem, I think. – GH from MO Apr 8 '11 at 22:34
Thanks, but it is not complete. The difficult case corresponds to degenerate eigenvalues, for which an eigenfunction is a linear combination of the products of sines. – Denis Grebenkov Apr 9 '11 at 7:14

If we take the usual trigonometric basis, it is indeed true as has been pointed out. However, there is a harder form of the question: Some of the eigenvalues are degenerate. If we allow arbritrary eigenfunctions, does the claim remain true?

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.