I'm interested in an integral equation with a symmetric difference (i.e. convolution) kernel of the form $h(|x-t|)$ on an interval $(-a,a)$, with an even free term $$\varphi(x)=f(|x|)+\lambda\int_{-a}^ah(|x-t|)\varphi(t)dt$$
When is the solution $\varphi(x)$ an even function? If I' m not mistaken, it would be sufficient if the resolvent was also a difference kernel. If the interval is the entire axis ($a=\infty$) and $h$ and $f$ are square integrable, then the resolvent kernel $r$ is a symmetric difference kernel $$\varphi(x)=f(|x|)+\lambda\int_{-\infty}^\infty r(|x-t|)f(|t|)dt$$ (see the first page in http://projecteuclid.org/euclid.jiea/1181075634), so I think that in this case an even free term implies that the solution is even. However, I wonder whether the solution would be always even if the interval was finite $(a < \infty)$?