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  "body": "\n## Definition\n\nLet $n \\ge 1$.\nA *Calderón–Zygmund operator* is a linear operator $T \\colon L^2(\\R^n) \\to L^2(\\R^n)$ such that:\n\n1. $T$ is bounded on the [[$L^2$ space|SASP14MA1Z]], that is $\\norm{Tf}_{L^2} \\le C\\norm{f}_{L^2}$ for some constant $C$ and all $f \\in L^2(\\R^n)$;\n2. there is a [[Calderón–Zygmund kernel|78N6AXP9VV]] $K$ such that, for every $f \\in L^2(\\R^n)$ with compact support,\n$$Tf(x) = \\int_{\\R^n} K(x, y) f(y) \\, dy \\quad \\text{for almost every } x \\notin \\operatorname{supp} f.$$\n\n## Remarks\n\nExample: the Hilbert transform $H$ on $\\R$ is a Calderón–Zygmund operator.\nIt is bounded on $L^2(\\R)$ with norm $1$, since its multiplier $-i\\sgn\\xi$ has modulus $1$.\nFor $x$ 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.\n\nNon-example: on $\\R$, let $Tf = f * \\mathbf{1}_{[0, 1]}$, so $Tf(x) = \\int \\mathbf{1}_{[0, 1]}(x - y) f(y) \\, dy$.\nIt is bounded on $L^2(\\R)$ by Young's inequality, so clause 1 holds.\nClause 2 fails, because its kernel $\\mathbf{1}_{[0, 1]}(x - y)$ jumps at $x - y = 1$.\nFor $y = 0$, $x = 1 - \\eta$ 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 $\\eta \\to 0$.\n\nThe kernel does not determine the operator: the identity and the zero operator both have kernel $0$ off the diagonal.\nClause 2 is only required off the support of $f$, since near the diagonal the integral need not converge absolutely.\nThe [[singular integral operators|HTMCEARGDD]] with smooth mean-zero angular part are Calderón–Zygmund operators.\n",
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