# Localization of Laplacian eigenfunction on the unit square?

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, $C_V = \inf\limits_k \int\limits_V dx |u_k(x)|^2 > 0$ ?

-

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.

-
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?

-