# Is there a positive-semidefinite convolution kernel, that is continuous at $0$ but discontinuous elsewhere?

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 $x\neq0$?

• If there are counterexamples, the theorem is not valid. Please specify the additional conditions that are crucial for the theorem. – Peter Aug 24 '18 at 11:35
• Thanks - the title wasn't phrased right. The theorem is certainly valid, I'm looking for examples that don't fulfill the conditions of the theorem. Title changed accordingly. – Neuromath Aug 24 '18 at 12:41

## 1 Answer

The answer is no: if the kernel is discontinuous, it cannot be finite at $x=0$. The trick came from a related question. (At least for kernels definable by Fourier series, and any others will be pretty pathological.)

If the kernel function can be expressed as a Fourier series, this series will have only cosines, and they will have only non-negative coefficients. If such a Fourier series converges at 0, the coefficients must be in $l^1$, which means the kernel function is continuous everywhere.