Singular integral operator

Remarks

The limit is a principal value: the mean zero condition makes $\int_{\varepsilon < \abs{y} < 1} K(y) \, dy = 0$, so $f(x - y)$ may be replaced by $f(x - y) - f(x) = O(\abs{y})$ near $y = 0$, which is integrable against $\abs{y}^{-n}$. When $\Omega$ is Lipschitz continuous, $T$ extends to a bounded operator on $L^p(\R^n)$ for $1 < p < \infty$, and it is of weak type $(1, 1)$. The more general Calderón–Zygmund operators have kernels $K(x, y)$ that need not be of convolution type.

Example: for $n = 1$, $S^0 = \set{-1, 1}$, and $\Omega(\pm 1) = \pm 1/\pi$ has mean zero. Then $K(x) = 1/(\pi x)$ and $T$ is the Hilbert transform. For $n \ge 2$, $\Omega_j(\omega) = c_n\omega_j$ has mean zero by symmetry, and gives the Riesz transforms.

Non-example: $\Omega \equiv 1$ fails the mean zero condition. For $f$ equal to $1$ near $x$, the integral $\int_{\varepsilon < \abs{y} < 1} \abs{y}^{-n} \, dy$ is a constant times $\log(1/\varepsilon)$, so the limit does not exist.