Calderón–Zygmund operator

Remarks

Example: the Hilbert transform $H$ on $\R$ is a Calderón–Zygmund operator. It is bounded on $L^2(\R)$ with norm $1$, since its multiplier $-i\sgn\xi$ has modulus $1$. For $x$ outside the support of $f$, the principal value is an ordinary integral, so $Hf(x) = \int \frac{f(y)}{\pi(x - y)} \, dy$, and $\frac{1}{\pi(x - y)}$ is a Calderón–Zygmund kernel.

Non-example: on $\R$, let $Tf = f * \mathbf{1}_{[0, 1]}$, so $Tf(x) = \int \mathbf{1}_{[0, 1]}(x - y) f(y) \, dy$. It is bounded on $L^2(\R)$ by Young's inequality, so clause 1 holds. Clause 2 fails, because its kernel $\mathbf{1}_{[0, 1]}(x - y)$ jumps at $x - y = 1$. For $y = 0$, $x = 1 - \eta$ and $x' = 1 + \eta$, the difference of the kernel values is $1$, while the bound $A(2\eta)^\delta (1 - \eta)^{-1-\delta}$ tends to $0$ as $\eta \to 0$.

The kernel does not determine the operator: the identity and the zero operator both have kernel $0$ off the diagonal. Clause 2 is only required off the support of $f$, since near the diagonal the integral need not converge absolutely. The singular integral operators with smooth mean-zero angular part are Calderón–Zygmund operators.