Is there some sufficient condition on the kernel $K$ of a (say, finite rank, to simplify) integral operator $$ \mathcal K:f(x)\in L^2(\mathbb R)\mapsto \int K(x,y)f(y)dy $$ so that it has all its non-zero eigenvalues $\textbf{distinct}$ (i.e. with multiplicity one) ?
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6I can refer you to this paper. – Mhenni Benghorbal Aug 25 '12 at 12:34
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5A finite-rank operator on an infinite-dimensional Banach space has infinite-dimensional kernel. Are you talking about the nonzero eigenvalues? – Qiaochu Yuan Aug 25 '12 at 23:17