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  "body": "\n## Definition\n\nLet $n \\ge 1$ and $1 \\le p \\le \\infty$.\nLet $\\mathcal{S}(\\R^n)$ be the [[Schwartz space|CG9JEATW3Q]], let $\\hat f$ be the [[Fourier transform|DPVM34ADZV]] of $f$, and let $g^\\vee$ be the inverse Fourier transform of $g$.\nLet $m \\colon \\R^n \\to \\C$ be a bounded measurable function, and let $T_m f = (m \\hat f)^\\vee$ for $f \\in \\mathcal{S}(\\R^n)$.\nThe function $m$ is an *$L^p$ multiplier* if there is a constant $C$ such that\n$$\\norm{T_m f}_{L^p(\\R^n)} \\le C \\norm{f}_{L^p(\\R^n)} \\quad \\text{for all } f \\in \\mathcal{S}(\\R^n).$$\nThe smallest such $C$ is the multiplier norm of $m$, and $\\mathcal{M}_p$ denotes the space of $L^p$ multipliers.\n\n## Remarks\n\nExample: for $h \\in \\R^n$, let $m(\\xi) = e^{2\\pi i \\xi \\cdot h}$, with the convention $\\hat f(\\xi) = \\int f(x) e^{-2\\pi i x \\cdot \\xi} \\, dx$.\nThen $T_m f(x) = f(x + h)$, so $\\norm{T_m f}_{L^p} = \\norm{f}_{L^p}$ and $m \\in \\mathcal{M}_p$ with norm $1$ for every $p$.\n\nNon-example: in $\\R^1$, $m(\\xi) = \\xi^2$ is not bounded, so it is not an $L^p$ multiplier.\nIndeed, for $p = 2$ the Plancherel theorem gives $\\norm{T_m f}_{L^2} = \\norm{\\xi^2 \\hat f}_{L^2}$, and its ratio to $\\norm{f}_{L^2} = \\norm{\\hat f}_{L^2}$ is unbounded when $\\hat f$ is a fixed bump moved far out.\n\nBy the Plancherel theorem, $\\mathcal{M}_2 = L^\\infty(\\R^n)$ with equal norms.\nBy duality, $\\mathcal{M}_p = \\mathcal{M}_{p'}$ when $1/p + 1/p' = 1$.\nThe space $\\mathcal{M}_1 = \\mathcal{M}_\\infty$ consists of the Fourier transforms of finite Borel measures.\nBy the [[Riesz–Thorin theorem|AGRKRG1TMJ]], $\\mathcal{M}_p \\subseteq \\mathcal{M}_q$ when $q$ lies between $p$ and $2$.\n",
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