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Do eigenfunctions of inverse of elliptic operator (eg: Laplacian) form basis of $L^P(\Omega)$ ? For p=2 we know the answer is yes, I am looking for p>2. More generally, is it true that eigenfunctions of any compact on separable Banach Space form a basis for that?

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  • $\begingroup$ What do you need this basis for? $\endgroup$
    – András Bátkai
    Commented Jan 8, 2014 at 23:17

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The operator which maps the function f(x) to $\int_0^x f(y)dy$ is compact, but it has no eigenfunctions. So the answer to your second question is no, even for Hilbert space. Also, if we are not talking about orthonormal bases in Hilbert space, there are different notions of basis (Schauder, Riesz). The inverse of Laplacian in $L^p$ is rather well behaved, but you should not expect too many sweeping abstract generalizations.

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