Properties of an integral operator: $(Au)(x)=\int_{\bf R}e^{-|x|-|y|}u(y)dy$

Question

My questions are motivated by the following exercise:

Consider the eigenvalue problem $$ \int_{-\infty}^{+\infty}e^{-|x|-|y|}u(y)dy=\lambda u(x), x\in{\Bbb R}.\tag{*} $$ Show that the spectrum consists purely of eigenvalues.

Let $A:L^2({\Bbb R})\to L^2({\Bbb R})$ be a linear operator such that $$(Au)(x)=\int_{\Bbb R}k(x,y)u(y)dy$$ where $k(x,y)=e^{-|x|-|y|}$. Then $A$ is self-adjoint, since $\overline{k(x,y)}=k(y,x)$. Thus $\sigma(A)\subset{\Bbb R}$.

My first question: is $A$ invertible?

$A$ is a Hilbert-Schmidt operator since $k\in L^2(\Bbb{R}^2)$ and thus $A$ is compact. Then the answer should be NO since $L^2({\Bbb R})$ is an infinite-dimensional Hilbert space. It follows that $\lambda=0$ must be an eigenvalue of $A$ according to the conclusion in the exercise. But $\operatorname{ker}(A)={0}$ which implies that $\lambda=0$ is not an eigenvalue of $A$.

My second question: what mistake do I make above?

Answer 1

You miscomputed $\ker A$: it's the set of (equivalence classes of) functions of $L^2$ such that $$\int_{\Bbb R}e^{-|y|}u(y)dy=0,$$ which is not reduced to $0$.

Answer 2

I just found that $\operatorname{ker}(A)$ is not $\{0\}$ since $Af=0$ for $f(x)=x\cdot 1_{[-1,1]}(x)$.

Since the spectrum of compact self-adjoint operators consists entirely of eigenvalues, with the possible exception of $\lambda=0$ and the example above shows that $0$ is an eigenvalue of $A$, we have a proof for the exercise.