Is a positive kernel positive definite?

Question

A symmetric function $k: \mathcal X \times \mathcal X \rightarrow \mathbb R$ is called a positive definite kernel if for any $x_1, \dots, x_n \in \mathcal X$ and $c_1, \dots, c_n \in \mathbb R$ we have \begin{equation} \sum_i \sum_j c_i c_j k(x_i,x_j) \geq 0. \end{equation}

I wonder whether $k(x,y) \geq 0$ is a sufficient condition for $k$ to be a kernel. I'm curious because I looked through multiple introductory scripts and this property didn't appear. But there have been more elaborate proofs that $k$ is a kernel for cases where $k(x,y) \geq 0$ is obvious, e.g. for the Gaussian kernel $k(x,y) = \exp\left(-\frac{\lVert x-y \rVert^2}{2\sigma^2}\right)$. But I can't think of a proof nor a counter example for the claim.

Is $k(x,y) \geq 0$ a sufficient condition for $k$ to be a kernel? If so, how could I see that? If not, what's a counter example?

Answer 1

It is not sufficient. For example, take any map for which $$ k(1,1) = k(2,2) = 1\\ k(1,2) = k(2,1) = 2 $$ Such as $k(x,y) = 1 + (x-y)^2$. Now, take $x = (1,2)$ $c = (1,-1)$ to find that $$ \sum_{i,j}c_ic_j k(x_i,x_j) = -2 $$