# Can an integral operator with negative kernel have a positive eigenvalue?

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.

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Let $I_1 = [0,\frac{1}{2}], I_2 = (\frac{1}{2},1]$ and $A = -1_{I_1\times I_1}-2 \cdot 1_{I_1\times I_2}-1_{I_2\times I_2}-2\cdot 1_{I_2\times I_1}$.

Let $f = -1_{I_1}+1_{I_2}$. Then $Af = f$.

This is basically using the matrix $\begin{bmatrix} -1 & -2 \\ -2 & -1\end{bmatrix}$ which has eigenvalues $\{ -3, 1\}$.

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I think it can.

Take a different setup, where your Hilbert space is two-dimensional and consider $A=-\begin{pmatrix}0&1\\1&0\end{pmatrix}$. This has non-positive entries and eigenvalue $1$.

I think you can cook this up into an example for the Hilbert space $L^2(\mathbb{R})$. But I don't have time to write any details.

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True. My interest, in fact, is for an operator which has not finite rank. The closest possible case to a matrix that I want to allow is a compact operator. I should have added it. – Brian Dec 12 '12 at 18:34
How about tensoring with the identity or some other trick? I think the infinite rank condition is achievable too. – user108903 Dec 12 '12 at 18:36
That's clever too, but I would like to address the case of a kernel $A(x,y)$ that is not a finite sum of (negative) terms of the form $g(x)h(y)$. Say, $A(x,y)=-(x^2+y^2+1)^{-2}$. – Brian Dec 12 '12 at 18:42
Well, can't you just tensor with something more complicated? – user108903 Dec 12 '12 at 18:43
Alright, let me put the question in such a way that it reflects the "true" obstacle I see. Let me take $A(x,y)$ very nice in terms of both regularity and integrability. Say, $A(x,y)=-(1+exp(x^2+y^2))^{-1}$. Question: can $A$ have a positive eigenvalue? – Brian Dec 12 '12 at 18:47