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Ok so the question is very simple, if you have the Hilbert Schmidt operator: $$Kf[x]=\int_a^b k(x,y)f(y)dy,$$ with $f\in L^2(a,b)$, how can you find his eigenvalues(i.e, $Kf_n=\lambda_n f_n$)? You need to solve some integral equations? Thanks

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  • $\begingroup$ Is $k(x,y)=\overline{k(y,x)}$? $\endgroup$ – Alex R. Nov 21 '15 at 1:15
  • $\begingroup$ Actually I think it is $\endgroup$ – Carlos Vázquez Monzón Nov 21 '15 at 1:22
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If your problem comes from a more strongest formulation like, for example, this ODE $-u''(x)+\mu u(x)=f(x)$ with some boundary conditions try to use the condition of eigenvalue there.

Continuing with the example, you know that $f \neq 0$ will be an eigenvector with eigenvalue $\lambda$ if and only if $-\lambda f''(x)+\mu \lambda f(x)=\lambda f(x)$. So you are looking for the values of $\lambda$ that make that $-\lambda f''(x)+\mu \lambda f(x)=\lambda f(x)$ has more than one solution. You just need to solve the equation and use the boundary conditions to check if there is any non-trivial solution.

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  • $\begingroup$ My ODE is more simple: $u''(x)=-f(x)$. The problem is that im looking for the solution of an arbitrary $f$, so I need a general solution $\endgroup$ – Carlos Vázquez Monzón Nov 21 '15 at 1:39
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The explicit formulas of the eigenfunctions will depend on the kernel $k$ but to show existence you can use the so-called Hilbert-Schmidt Theorem (which is an analogue of the spectral theorem):

Theorem (Hilbert-Schmidt): Let $H$ be a separable $\Bbb{C}$-Hilbert space, $K\in\operatorname{End}(H)$ be compact and self-adjoint. Then there is a countable orthonormal basis of $H$ consisting of the eigenfunctions of $K$.

The conditions on $k$ guarantees that $K$ satisfies the conditions of the theorem (you're going to need $k$ to be (at the very least) square integrable with respect to the product measure as well).

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