Calderón–Zygmund kernel

Remarks

Example: $n = 1$ and $K(x, y) = \frac{1}{\pi(x - y)}$, the kernel of the Hilbert transform. Clause 1 holds with $A = 1/\pi$. For clause 2, if $\abs{x - x'} \le \frac{1}{2}\abs{x - y}$ then $\abs{x' - y} \ge \frac{1}{2}\abs{x - y}$, and $$\abs{K(x, y) - K(x', y)} = \frac{\abs{x - x'}}{\pi \abs{x - y}\abs{x' - y}} \le \frac{2}{\pi}\frac{\abs{x - x'}}{\abs{x - y}^2},$$ so clause 2 holds with $\delta = 1$; clause 3 is the same by symmetry.

Non-example: $K(x, y) = \abs{x - y}^{-n - 1/2}$ fails clause 1, because it is too large near the diagonal.

A convolution kernel $K(x, y) = k(x - y)$ is a Calderón–Zygmund kernel when $\abs{k(x)} \le A\abs{x}^{-n}$ and $\abs{\nabla k(x)} \le A\abs{x}^{-n-1}$ for $x \ne 0$, by the mean value theorem. A weaker smoothness condition often used for convolution kernels is the Hörmander condition $\int_{\abs{x} > 2\abs{y}} \abs{k(x - y) - k(x)} \, dx \le A$ for all $y \ne 0$. The size and smoothness conditions alone do not give an $L^2$ bound; one also needs cancellation, as for the singular integral operators with mean-zero angular part.