Weakly singular kernel

Remarks

Example: $K(x, s) = \abs{x - s}^{-1/2}$ on $\Omega = [0, 1]$, with $m = 1$, $M = 1$ and $\alpha = 1/2 < 1$. It is continuous off the diagonal.

Non-example: $K(x, s) = 1/(x - s)$ on $[0, 1]$. It satisfies the bound only with $\alpha = 1$, which is not less than $m = 1$; the integral $\int_0^1 \abs{x - s}^{-1} \, ds$ diverges.

Since $\alpha < m$, the function $s \mapsto \abs{x - s}^{-\alpha}$ is integrable over $\Omega$, so $\int_\Omega K(x, s) \varphi(s) \, ds$ exists for bounded continuous $\varphi$. Assume also that $\Omega$ is Jordan measurable, for example a closed interval or a closed ball. Then the integral operator of a weakly singular kernel maps continuous functions on $\Omega$ to continuous functions and is compact for the maximum norm. The kernel $\log \abs{x - s}$ is weakly singular on a bounded set: it satisfies the bound for every $\alpha \in (0, m)$.