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Does it true that there exists a compactly supported eigenfunction corresponding to the first positive eigenvalue $\lambda_1$ of hyperbolic Laplacian operator $\Delta$ on $L^2(S)$, $S$ is a hyperbolic surface? It's not compact but finite volume. Thanks.

Amateur
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The Laplacian is an elliptic operator so the eigenfunctions are real analytic. They cannot be zero on an open subset unless they are zero on the whole (connected) manifold.

orangeskid
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  • If $f$ is eigenfunction then $g=f$ on a compact bounded subset and otherwise $g=0$. Then $g$ is still an eigenfunction. Right? – Amateur Sep 12 '14 at 10:08
  • No @Amateur, that is not right. There are plenty of problems with your construction, for example, regularity. Moreover, your construction also does not satisfies the unique property continuation, which is what this answer tells you a eigenfunction must satisfies. – Tomás Sep 12 '14 at 10:22
  • Eigenfunction for the laplacian means $\Delta f - \lambda f = 0$ for the eigenvalue ($\pm$) $\lambda$. Compactly supported means $f= 0$ on the complement of some compact subset.OK, if that is what you want from $g$ then you can multiply $f$ by some smooth function equal $1$ on some compact subset and $0$ outside some larger compact subset. But $g$ will not be an eigenfunction anymore. The problem is on the transition from $f$ to $0$ which cannot be done abruptly but only smoothly, one cannot expect some relationship of the form $\delta g = \mathrm{const}\cdot g$ be preserved. – orangeskid Sep 12 '14 at 10:28