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?