Continuous kernel

Remarks

Example: $K(x, s) = \min(x, s)$ on $[0, 1]$. On the closed triangle $s \le x$ it equals $s$, and on the closed triangle $x \le s$ it equals $x$. Both formulas are continuous and agree on the diagonal, so $K$ is continuous on the square.

Non-example: $K(x, s) = \abs{x - s}^{-1/2}$ for $x \ne s$, with any value on the diagonal, is not continuous on $[0, 1]^2$, since it is unbounded near the diagonal. It is a weakly singular kernel.

A continuous kernel is bounded, $\abs{K(x, s)} \le M$, and uniformly continuous on the compact square. So $x \mapsto \int_a^b K(x, s) \varphi(s) \, ds$ is continuous for every continuous $\varphi$, and it is bounded by $M (b - a) \max \abs{\varphi}$. The same definition is used with $[a, b]$ replaced by a compact subset of $\R^m$.