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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?

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There is a strongly related question of mine mathoverflow.net/questions/55235/… where Piero gave some arguments why in general it cannot be true. –  András Bátkai Apr 8 '11 at 10:46

3 Answers 3

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

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Thanks for this reply. But what would be the necessary/sufficient condition on the domain to make the eigenfunctions uniformly bounded? –  Denis Grebenkov Apr 9 '11 at 7:18

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.

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Thanks for this reply and helpful reference. But what would be the necessary/sufficient condition on the domain to make the eigenfunctions uniformly bounded? –  Denis Grebenkov Apr 9 '11 at 7:19

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.

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