Density of odd and even eigenstates of an integral operator

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?

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I do not get the definition of your operator $K$. What is $x$ on the left hand side, and where is ist on the right hand side? It would be also better if you used different symbols for the operator and for the kernel, but that is not that big of a problem. – András Bátkai Oct 31 '13 at 12:16
Sorry, it was a misprint in the definition of $K$. – Alex Oct 31 '13 at 12:50
Is $f$ a function (real, complex?) on line, or on interval? – Piotr Migdal Nov 7 '13 at 13:37