Hilbert-Schmidt integral operator with missing eigenfunctions

Question

I'm having some issues with the spectral decomposition of the integral operator \begin{equation} (Af)(x)=\int_0^1|x-y|f(y)dy,\text{ with $f\in L^2[0,1]$}. \end{equation} Since \begin{equation} \int_0^1\int_0^1|x-y|^2\,dx\,dy<\infty \text{ and } |x-y|=|y-x|, \end{equation} this is a self-adjoint Hilbert-Schmidt integral operator. Therefore the spectral theorem for compact self-adjoint operators guarantees the existence of an orthonormal basis of $L^2[0,1]$ of eigenfunctions of $A$. Since $Af$ is Lipshitz, if $Af=\lambda f$ with $\lambda\ne0$, $f$ must be twice continuously differentiable, and satisfy the o.d.e. \begin{align} f''(x)=\frac{2}{\lambda}\,f(x). \end{align} It follows that the only eigenfunctions associated with a strictly negative eigenvalue are of the form $f(x)=\cos((2n+1)\pi x)$, with $n\ge0$ an integer (the associated eigenvalue is $\lambda=\frac{-2}{\pi^2(2n+1)^2}$.) There are, however, no eigenfunctions associated with a $\lambda>0$. Furthermore we are able to prove there are no eigenfunctions associated with $\lambda=0$. So, since $f(x)=1$ is orthogonal to the eigenfunctions $\cos((2n+1)\pi x)$, with $n\ge0$, these are an incomplete base for $L^2[0,1]$. (Treating $A$ as a convolution and using a Fourier series decomposition yields the same form of eigenfunctions.)

I have two issues: (1) finite approximations of this operator imply a (single) positive eigenvalue (with the largest modulus out of all eigenvalues)---although $A$ doesn't have positive eigenvalues; and finally (2) the finite approximations also imply eigenfunctions that do not correspond to any of the functions $\cos((2n+1)\pi x)$, with $n\ge0$.

Answer 1

I figured a positive eigenvalue. For $\lambda>0$, if $f(x) := \alpha\exp(\tau x)+ \exp(-\tau x)$ with $\alpha$ a constant and $\tau=\sqrt{2/\lambda}$ then $Af-\lambda f=0$ if and only if $$\frac{e^{-\tau}-\alpha e^{\tau}-(\alpha-1)}{\tau}x+\frac{(-1-\tau) e^{-\tau}-\alpha(1-\tau)e^{\tau}-\alpha-1}{\tau^2}=0,\text{ for all }0\le x\le1.$$ To annihilate the linear term in x, you can solve for $\alpha$ in terms of $\tau$ to obtain $$\alpha=\frac{1+e^{-\tau}}{1+e^{\tau}}.$$ Substituting this value on the other term gives $$\frac{(-1-\tau) e^{-\tau}-\alpha(1-\tau)e^{\tau}-\alpha-1}{\tau^2}=\frac{(-\tau-2) e^{-\tau}-4+(\tau-2)e^{\tau}}{1+e^\tau},$$ which vanishes only when $$e^\tau=\frac{\tau+2}{\tau-2},$$ and a plot suggests that this is possible at a unique value of $\tau\approx 2.4$.