5
$\begingroup$

Let $A$ be a bounded domain in $\mathbb R^d$, $d>1$, and $\{u_k\}$ is the set of all $L^2$-normalized Laplacian eigenfunctions on $A$ with Dirichlet boundary condition (i.e., $\|u_k\|_2 = 1$).

Is it true that these eigenfunctions are uniformly bounded, i.e., $sup_k \|u_k\|_\infty < \infty$, where $\|.\|_\infty$ is the $L^\infty$-norm (the maximum)? In other words, does there exist a constant $C_A$ such that for any $k$ and any $x\in A$, $|u_k(x)| < C_A$?

If the answer is positive, please provide a reference or a proof.

If the answer is negative, please provide a counter-example. In that case, what are the conditions on the domain $A$ to make this statement true?

$\endgroup$
1

3 Answers 3

7
$\begingroup$

The answer is no. The following reference specifically discusses the case of the two-dimensional disk: http://www.staff.uni-oldenburg.de/daniel.grieser/wwwpapers/diss.pdf

$\endgroup$
1
  • $\begingroup$ Thanks for this reply. But what would be the necessary/sufficient condition on the domain to make the eigenfunctions uniformly bounded? $\endgroup$ Apr 9, 2011 at 7:18
4
$\begingroup$

There is a paper on this subject (containing also further references) by C. D. Sogge, Eigenfunction and Bochner-Riesz estimates on manifolds with boundary. Math. Res. Lett. 9, No.2-3, 205-216 (2002), ArXiv: math/0202032. It is stated that typically there is some growth in $L_\infty$ metric, not uniform boundedness.

$\endgroup$
1
  • $\begingroup$ Thanks for this reply and helpful reference. But what would be the necessary/sufficient condition on the domain to make the eigenfunctions uniformly bounded? $\endgroup$ Apr 9, 2011 at 7:19
1
$\begingroup$

I don't have a general answer (I guess it is yes, there are uniformy bounded, at least when $A$ is a smooth bounded domain). At least, let me mention the case of the torus $\mathbb T^d=\mathbb R^d/\mathbb Z^d$. The eigenfunctions are the exponentials $\exp(2i\pi m\cdot x)$, where $m\in\mathbb N^d$. They are uniformly bounded and this fact is crucial in the Riesz-Thorin interpolation theorem that the Fourier series of an $L^p$-function $f$ belongs to $\ell^{p'}$ whenever $1\le p\le2$ and $p'$ is the conjugate exponent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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