A positive-semidefinite, symmetric convolution kernel on the circle $\mathbb{T}^1$ is a function $k:\mathbb{T}^1\to\mathbb{R}$ such that $k(x)=k(-x)$, and $\sum_{i=1}^n\sum_{j=1}^n k(x_i-x_j)c_i c_j\geq 0$ for all sequences $x_1 \ldots x_n \in \mathbb{T}^1$ and $c_1 \ldots c_n\in\mathbb{R}$.
Question: does there exist a positive-semidefinite convolution kernel, that is continuous at $0$ but discontinuous somewhere else?
Motivation: Mercer's theorem tells us that a symmetric continuous positive-semidefinite kernel has a basis of eigenfunctions, whose eigenvalues have finite sum. I am looking for examples of discontinuous functions to show how these predictions fail. Convolution kernels on the circle are simple place to look, as their spectrum can be computed by Fourier analysis.
Thus the problem is equivalent to asking if there is a cosine series with non-negative terms that is continuous at 0 but not somewhere else.
Attempts so far:
A first thought would be a step function: $k(x) = I(|x|<\Theta)$. But this is not positive semidefinite.
A second try goes in reverse: take $k(x) = \sum_{n=1}^{\infty} {\cos(nx) \over n}$. The eigenvalues are 1/n, so the conclusions of Mercer's theorem cannot hold. But also see that $k(0) = \infty$. This kernel is discontinuous in a very specific way: it's value tends to infinity at $x=0$.
A third try would be $k(x) = \sum_{n=1}^{\infty} {\cos(n^2x) \over n^2}$. This converges absolutely for all $x$. But it turns out to be "Riemann's example" - a classic pathological function that is continuous everywhere but differentiable almost nowhere.
Are all discontinuous positive-semidefinite convolution kernels discontinuous at $x=0$? Or are there some which are well behaved at $0$ with other types of discontinuity, such as a jump at a point $x\neq0$?
