Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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) ?

share|cite|improve this question
3  
I can refer you to this paper. – Mhenni Benghorbal Aug 25 '12 at 12:34
4  
A 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.