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I'd like to know whether there's a general condition on an operator for it to have an eigenfunction. For example, differential operator has eigenfunction $f_k (x)=e^{kx}$ , and differential operator have many properties such as linear and obeys the shift theorem. The condition I'm asking about maybe similar to these kinds of properties.

To be more specific, I'm interested in an integral operator $(Kf)(x)=∫ r(x,y)f(y)dy$ where r(x,y) is a square integratable function and f(x) is a probability density function. Thank you!

iridium
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    Ordinary differential operators $Lf=f^{(n)}+a_{n-1}f^{(n-1)}+\cdots+a_{1}f^{(1)}+a_{0}f$ have eigenfunctions on an interval $[a,b]$ if the coefficients satisfy some minimal properties. That's because $Lf=\lambda f$ is just another ODE for which theorems about existence and uniqueness of solutions apply. In fact, you'll have $n$ linearly-independent solutions in this case because the order of the ODE is $n$ (n=1 in your case.) Once you begin imposing conditions at two endpoints of an interval, then it becomes trickier, but that's a different discussion. – Disintegrating By Parts Jun 08 '14 at 15:29
  • Thank you for the reply! Are you also familiar with other kinds of operators such as an integral operator? – iridium Jun 08 '14 at 15:38
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    If $r(x,y)$ is square integrable, i.e., $\int\int |r(x,y)|^{2}dx,dy < \infty$, then $K$ is a classical operator type known as Hilbert-Schmidt. http://en.wikipedia.org/wiki/Hilbert%E2%80%93Schmidt_integral_operator A lot can be said about eigenfunctions of such operators http://en.wikipedia.org/wiki/Compact_operator . The F. Riesz Theory of compact operators is refined, clean, and almost unchanged for a hundred years now. – Disintegrating By Parts Jun 08 '14 at 15:44

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