While reading "Integral equations - a reference text" (Zabreiko et al. eds) I came up with this question I cannot answer:
Suppose $A:L^2(\mathbb{R})\to L^2(\mathbb{R})$ is a bounded linear integral operator with kernel $A(x,y)$ such that $A(x,y)\leqslant0$ $\forall x,y\in\mathbb{R}$. That is, $(Af)(x)=\int_{\mathbb{R}}A(x,y)f(y)dy$.
Suppose also (these additional assumptions might help, but I don't see the answer also under such a restriction) that $A(x,y)$ is as nice as needed, say, smooth and bounded, and, if needed, suppose also that $A$ is compact.
Can $A$ have a positive eigenvalue? Say, does the problem $Af=f$ have a solution in $L^2(\mathbb{R})$?
What I can trivially see is that since $A(x,y)$ is real then if $Af=f$ also $A\bar{f}=\bar{f}$, so if there are solutions to $Af=f$ then there are in particular real-valued solutions. And clearly such an $f$ cannot have a definite sign, because LHS and RHS of $Af=f$ would then have different sign. But apart from that, I don't see if this is the good beginning of a proof that a solution to $Af=f$ cannot exist, or if there is an obvious example of existence.