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  "body": "\n## Definition\n\nLet $n \\ge 1$, 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\n$$\\int_{S^{n-1}} \\Omega(\\omega) \\, d\\sigma(\\omega) = 0.$$\nLet $K(x) = \\Omega(x/\\abs{x})/\\abs{x}^n$ for $x \\ne 0$.\nThe *singular integral operator* with kernel $K$ is\n$$Tf(x) = \\lim_{\\varepsilon \\to 0^+} \\int_{\\abs{y} \u003e \\varepsilon} K(y) f(x - y) \\, dy,$$\ndefined first for smooth functions $f$ with compact support, where the [[limit|05M5SWSZZD]] exists for every $x$.\n\n## Remarks\n\nThe limit is a principal value: the mean zero condition makes $\\int_{\\varepsilon \u003c \\abs{y} \u003c 1} K(y) \\, dy = 0$, so $f(x - y)$ may be replaced by $f(x - y) - f(x) = O(\\abs{y})$ near $y = 0$, which is integrable against $\\abs{y}^{-n}$.\nWhen $\\Omega$ is Lipschitz continuous, $T$ extends to a bounded operator on $L^p(\\R^n)$ for $1 \u003c p \u003c \\infty$, and it is of weak type $(1, 1)$.\nThe more general Calderón–Zygmund operators have kernels $K(x, y)$ that need not be of convolution type.\n\nExample: for $n = 1$, $S^0 = \\set{-1, 1}$, and $\\Omega(\\pm 1) = \\pm 1/\\pi$ has mean zero.\nThen $K(x) = 1/(\\pi x)$ and $T$ is the Hilbert transform.\nFor $n \\ge 2$, $\\Omega_j(\\omega) = c_n\\omega_j$ has mean zero by symmetry, and gives the Riesz transforms.\n\nNon-example: $\\Omega \\equiv 1$ fails the mean zero condition.\nFor $f$ equal to $1$ near $x$, the integral $\\int_{\\varepsilon \u003c \\abs{y} \u003c 1} \\abs{y}^{-n} \\, dy$ is a constant times $\\log(1/\\varepsilon)$, so the limit does not exist.\n",
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