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  "body": "\n## Statement\n\nLet $n \\ge 1$ and let $K$ be a locally integrable function on $\\R^n \\setminus \\set{0}$.\nAssume that there is a constant $A$ such that:\n\n1. $\\abs{K(x)} \\le A\\abs{x}^{-n}$ for all $x \\ne 0$;\n2. $\\int_{\\abs{x} \\ge 2\\abs{y}} \\abs{K(x - y) - K(x)} \\, dx \\le A$ for all $y \\ne 0$;\n3. $\\int_{r \u003c \\abs{x} \u003c R} K(x) \\, dx = 0$ for all $0 \u003c r \u003c R$.\n\nFor $\\varepsilon \u003e 0$, let $T_\\varepsilon f(x) = \\int_{\\abs{y} \u003e \\varepsilon} K(y) f(x - y) \\, dy$.\nThen for each $1 \u003c p \u003c \\infty$ there is a constant $C$, depending only on $n$ and $p$, such that\n$$\\norm{T_\\varepsilon f}_{L^p(\\R^n)} \\le C A \\norm{f}_{L^p(\\R^n)} \\quad \\text{for all } f \\in L^p(\\R^n) \\text{ and } \\varepsilon \u003e 0.$$\nMoreover, the operators $T_\\varepsilon$ are of weak type $(1, 1)$, uniformly in $\\varepsilon$.\n\n## Remarks\n\nThe proof has three steps.\nThe three conditions together give a uniform bound on the Fourier transforms of the truncated kernels, hence a uniform $L^2$ bound by the Plancherel theorem.\nThe [[Calderón–Zygmund lemma|K1YKXVPVDG]] splits $f$ into a bounded good part and a bad part with mean zero on disjoint cubes.\nCondition 2 controls the operator on the bad part away from the cubes, and this gives the weak type $(1, 1)$ bound.\nThe [[Marcinkiewicz interpolation theorem|R1PHMVBJ9J]] then gives $1 \u003c p \\le 2$, and duality gives $2 \\le p \u003c \\infty$.\n\nExample: on $\\R$, $K(x) = \\frac{1}{\\pi x}$ satisfies condition 1 with $A = 1/\\pi$, and condition 3 because $K$ is odd.\nFor condition 2, $\\abs{K(x - y) - K(x)} = \\frac{\\abs{y}}{\\pi \\abs{x}\\abs{x - y}} \\le \\frac{2\\abs{y}}{\\pi x^2}$ when $\\abs{x} \\ge 2\\abs{y}$, whose integral is $2/\\pi$.\nSo the truncated Hilbert transforms are bounded on $L^p(\\R)$ uniformly in $\\varepsilon$, which recovers the [[M. Riesz theorem|PJFH51Z48M]].\n\nThe cancellation condition 3 is needed.\nFor $K(x) = \\abs{x}^{-n}$, conditions 1 and 2 hold, but $T_\\varepsilon f(x)$ grows like $\\log(1/\\varepsilon)$ for $f$ equal to $1$ near $x$.\nThe conclusion fails for $p = 1$ and $p = \\infty$, as already for the Hilbert transform.\n\n## Explanations\n\n### L1 human\n\nIntegrating a function against a kernel of size $\\abs{x}^{-n}$ is dangerous, since this kernel is not integrable near $0$ or near infinity.\nSuppose the kernel is smooth on average, and its positive and negative values balance on every shell.\nThe theorem says that the integral then gives an operator that preserves $L^p$ size, for $1 \u003c p \u003c \\infty$.\nThe Hilbert transform and the Riesz transforms are the main examples.\n\n### L2 human\n\nThis is the central theorem of the theory of [[singular integral operators|HTMCEARGDD]].\nIts method, an $L^2$ bound, the Calderón–Zygmund decomposition and interpolation, also proves the Hörmander–Mikhlin multiplier theorem and the bounds for maximal functions.\nIt extends to non-convolution operators, the [[Calderón–Zygmund operators|3SCXCW4AH3]], once an $L^2$ bound is known; the [[T(1) theorem|3FBF628XEF]] gives a test for that $L^2$ bound.\n",
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