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  "body": "\n## Statement\n\nLet $n \\ge 1$ and let $\\mathcal{S}(\\R^n)$ be the [[Schwartz space|CG9JEATW3Q]] with its Fréchet topology.\nLet $\\mathcal{F} f(\\xi) = \\int_{\\R^n} f(x) e^{-2\\pi i x \\cdot \\xi}\\, dx$ be the [[Fourier transform|DPVM34ADZV]].\nThen $\\mathcal{F}$ maps $\\mathcal{S}(\\R^n)$ onto $\\mathcal{S}(\\R^n)$, it is a linear bijection, and both $\\mathcal{F}$ and its inverse are continuous.\nThe inverse is the [[inverse Fourier transform|JB8QMT8DCH]]\n$$\\mathcal{F}^{-1} g(x) = \\int_{\\R^n} g(\\xi)\\, e^{2\\pi i x \\cdot \\xi}\\, d\\xi.$$\n\n## Remarks\n\nBy the [[rules for differentiation and multiplication|HHNR31W3WD]], $\\xi^\\alpha \\partial^\\beta \\hat f$ is, up to constants, the transform of $\\partial^\\alpha (x^\\beta f)$.\nThat function is in $\\mathcal{S}(\\R^n) \\subset L^1(\\R^n)$, so its transform is bounded, which shows $\\hat f \\in \\mathcal{S}(\\R^n)$.\nThe same estimates, with a factor $(1 + \\abs{x})^{-n-1}$ to control the integral, give continuity.\nBijectivity is the inversion formula $\\mathcal{F}^{-1} \\mathcal{F} f = f$, proved by pairing with a Gaussian of shrinking width.\nAlso $\\mathcal{F}^4 f = f$, since $\\mathcal{F}^2 f(x) = f(-x)$.\n\nExample: the Gaussian $e^{-\\pi \\abs{x}^2}$ is a Schwartz function and is fixed by $\\mathcal{F}$.\n\nOn $\\mathcal{S}(\\R^n)$ the transform also preserves the $L^2$ norm, $\\int \\abs{\\hat f}^2 = \\int \\abs{f}^2$.\nSince $\\mathcal{S}(\\R^n)$ is dense in $L^2(\\R^n)$, this is how the transform is extended to a unitary map of $L^2(\\R^n)$.\nBy duality it also extends to tempered distributions.\nThe space $L^1(\\R^n)$ has no such property: the transform of an integrable function need not be integrable.\n\n## Explanations\n\n### L1 human\n\nSchwartz functions are smooth and decay fast, and the Fourier transform turns smoothness into decay and decay into smoothness.\nSo it sends this space to itself, and nothing is lost: every Schwartz function is the transform of exactly one other.\n\n### L2 human\n\nThis is the cleanest setting for Fourier analysis on $\\R^n$: all formulas (inversion, Plancherel, the convolution theorem, differentiation rules) hold on $\\mathcal{S}(\\R^n)$ without any conditions.\nResults on $L^p$ spaces are then obtained by density, and results on tempered distributions by duality.\n",
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