# Hilbert-Schmidt integral operator with missing eigenfunctions

I'm having some issues with the spectral decomposition of the integral operator $$$$(Af)(x)=\int_0^1|x-y|f(y)dy,\text{ with f\in L^2[0,1]}.$$$$ Since $$$$\int_0^1\int_0^1|x-y|^2\,dx\,dy<\infty \text{ and } |x-y|=|y-x|,$$$$ this is a self-adjoint Hilbert-Schmidt integral operator. Therefore the spectral theorem for compact self-adjoint operators guarantees the existence of an orthonormal basis of $$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 $$n\ge0$$.

• Fixed, thanks. The boundary conditions can be obtained by splitting the domain of the integral operator to remove the absolute value and differentiating. They are $\lambda(f(0)+f(1))=\int_0^1f(y)dy$ and $f'(0)+f'(1)=0$ Commented Jun 21, 2021 at 17:12
• Form observation (1), it seems the most likely explanation is missing a positive eigenvalue which should be there en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem Commented Jun 21, 2021 at 17:58
• I don't obtain that for the boundary conditions. Please check. You have a positive compact operator, the Krein-Rutman Theorem applies (the infinite version of Perron-Frobenius). There is a positive eigenvalue, correponding to the only non sign changing eigensolution. Commented Jun 21, 2021 at 18:22
• \begin{align*} &(\int_0^x(x-y)f(y)dy+\int_x^1(y-x)f(y)dy)^\prime \\ =&\int_0^x f(y)dy +x(f(x)-f(0)) -xf(x)-\int_x^1f(y)dy \\&- x(f(1)-f(x))+f(1)-xf(x)\\ &= \int_0^x f(y)dy -\int_x^1f(y)dy - x(f(1)+f(0)) +f(1) \end{align*} Commented Jun 21, 2021 at 19:47
• @EvanGorman: Actually, they do. The antiperiodic eigenfunctions (that is $f(1)=-f(0)$, $f'(1)=-f'(0)$) that don't satisfy your bc's are the sine functions, but that too messes things up (the antiperiodic eigenfunctions would have spanned the whole space, being the eigenfunctions of a self-adjoint operator). Commented Jun 21, 2021 at 19:53

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 $$\tau\approx 2.4$$.
• @EvanGorman: There may be other positive eigenvalues. The one above corresponds to searching for one with an eigenfunction of the form $A e^{\tau}+B e^{-\tau x}$ with $B\ne0$. Commented Jun 22, 2021 at 2:35