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  "body": "\n## Definition\n\nLet $n \\ge 1$, let $\\delta \\ge 0$ and let $R \u003e 0$.\nLet $\\hat f(\\xi) = \\int_{\\R^n} f(x) e^{-2\\pi i x \\cdot \\xi} \\, dx$ be the [[Fourier transform|DPVM34ADZV]] of a [[Schwartz function|CG9JEATW3Q]] $f$ on $\\R^n$, and let $t_+ = \\max(t, 0)$, with the convention $t_+^0 = \\mathbf{1}_{t \u003e 0}$.\nThe *Bochner–Riesz mean* of order $\\delta$ and radius $R$ of $f$ is\n$$S^\\delta_R f(x) = \\int_{\\R^n} \\Big(1 - \\frac{\\abs{\\xi}^2}{R^2}\\Big)_+^\\delta \\hat f(\\xi) \\, e^{2\\pi i x \\cdot \\xi} \\, d\\xi.$$\nThe operator $S^\\delta = S^\\delta_1$ is the Bochner–Riesz operator of order $\\delta$.\n\n## Remarks\n\nExample: for $\\delta = 0$, $S^0_R f$ is the spherical partial sum, the inverse Fourier transform of $\\hat f$ restricted to the open ball of radius $R$.\nFor $n = 1$ and $\\delta = 1$, the multiplier is $(1 - \\xi^2/R^2)_+$, which is a continuous function with compact support.\n\nThe multiplier of $S^\\delta_R$ is the dilate of the multiplier of $S^\\delta$ by $R$, and dilations do not change $L^p$ operator norms.\nSo $S^\\delta_R$ is bounded on $L^p(\\R^n)$ uniformly in $R$ if and only if $(1 - \\abs{\\xi}^2)_+^\\delta$ is an [[$L^p$ multiplier|RAGR38Q40B]].\nIn that case, $S^\\delta_R f \\to f$ in $L^p(\\R^n)$ as $R \\to \\infty$ for every $f \\in L^p(\\R^n)$, $1 \\le p \u003c \\infty$.\n\nNon-example of convergence: for $\\delta = 0$, $n \\ge 2$ and $p \\ne 2$, the ball multiplier is not an $L^p$ multiplier, by [[Fefferman's ball multiplier theorem|4A5MJY4XPV]].\n\nFor $\\delta \u003e (n - 1)/2$, the kernel of $S^\\delta$ is integrable, so $S^\\delta$ is bounded on $L^p(\\R^n)$ for all $1 \\le p \\le \\infty$.\nThe value $(n - 1)/2$ is the critical index.\n",
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