Consider an integral operator $(Kf)(x)=\int_{-1}^{1}K(x-y)f(y)dy$, where the kernel $K(-x)=K(x)$ is an even function.

Let $\lambda_n$ be the ordered eigenvalues of $K$ and $f_n(x)$ the corresponding eigenfunctions. The eigenfunctions $f_n(x)$ are either odd or even functions.

Is it generally true that they alternate, i.e. if $f_n(x)$ is even, then $f_{n+1}(x)$ is odd and vice versa?

Can one prove that the density of the eigenvalues corresponding to the even eigenfunctions is equal to the density of states corresponding to the odd eigenfunctions?

This post imported from StackExchange MathOverflow at 2015-02-26 12:34 (UTC), posted by SE-user Alex