---
tag: 8VB096R12H
kind: definition
name: Singular integral operator
msc: [42B20, 47G10]
defines: HTMCEARGDD
uses: [DWJF7ABPR1, 05M5SWSZZD]
sources:
  - url: https://en.wikipedia.org/wiki/Singular_integral
---

## Definition

Let $n \ge 1$, let $S^{n-1}$ be the unit sphere of $\R^n$ with its surface measure $d\sigma$, and let $\Omega \colon S^{n-1} \to \C$ be an integrable function with
$$\int_{S^{n-1}} \Omega(\omega) \, d\sigma(\omega) = 0.$$
Let $K(x) = \Omega(x/\abs{x})/\abs{x}^n$ for $x \ne 0$.
The *singular integral operator* with kernel $K$ is
$$Tf(x) = \lim_{\varepsilon \to 0^+} \int_{\abs{y} > \varepsilon} K(y) f(x - y) \, dy,$$
defined first for smooth functions $f$ with compact support, where the [[limit|05M5SWSZZD]] exists for every $x$.

## 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.