A kernel $K(x,y)$ is SR(2) if it is sign-regular up to order 2, as defined by Samuel Karlin based on work of I. J. Schoenberg on variation diminishing kernels. One criterion for sufficiently differentiable kernels is: $\frac{d^2}{dx\hspace{1pt}dy}\, \log K(x,y)$ is non-negative or non-positive on the domain of $K$. A consequence of sign-regularity is order preservation or reversal, e.g. monotonic functions map to monotonic functions under the linear mapping of the kernel (under certain conditions).
I can show, using an integral representation and tedious consideration of many possible combinations of ranges for the variables, that the kernel:
$\quad \quad K(\phi,\theta) = f(\cos(\phi-\theta)) - f(\cos(\phi+\theta))$, $\quad 0 < \phi,\hspace{1pt} \theta < \pi/2$
is order preserving (reversing) for monotonic functions on $(0,\pi/2)$ when $f(x)=g(x^2)$ with $g$ non-decreasing and convex (concave) on $(0,\infty)$. But the kernel is apparently sign regular as well under these conditions, and I would ideally like to use sign-regularity to show that the linear transform does not increase the number of zeros of a function on $(0,\pi/2)$ (is variation diminishing).
The sign regularity condition can be formulated as the non-negativity (non-positivity) of the determinant of the 2x2 matrix with kernel evaluated at four points at the corners of a rectangle in $(0,\pi/2)\times (0,\pi/2)$.
I was hoping someone familiar with sign regularity (and total positivity) might know of a way to prove that this kernel is sign regular. Thanks.
