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  "body": "\n## Definition\n\nLet $n \\ge 1$, let $\\mathcal{S}(\\R^n)$ be the Schwartz space, and let $\\mathcal{S}'(\\R^n)$ be the space of tempered distributions.\nFor $\\varphi \\in \\mathcal{S}(\\R^n)$, let\n$$\\mathcal{F}\\varphi(\\xi) = \\int_{\\R^n} \\varphi(x)\\, e^{-2\\pi i x \\cdot \\xi}\\, dx$$\nbe its Fourier transform.\nFor $T \\in \\mathcal{S}'(\\R^n)$, the *Fourier transform* $\\mathcal{F}T$ is the tempered distribution given by\n$$\\langle \\mathcal{F}T, \\varphi \\rangle = \\langle T, \\mathcal{F}\\varphi \\rangle \\quad \\text{for all } \\varphi \\in \\mathcal{S}(\\R^n).$$\n\n## Remarks\n\nThe definition makes sense because $\\mathcal{F}$ is a continuous linear bijection of $\\mathcal{S}(\\R^n)$ onto itself.\nFor an integrable function $f$, viewed as a tempered distribution, $\\mathcal{F}f$ is the usual Fourier transform; this follows from Fubini's theorem, $\\int (\\mathcal{F}f)\\varphi = \\int f (\\mathcal{F}\\varphi)$.\nOn $\\mathcal{S}'(\\R^n)$, $\\mathcal{F}$ is a linear bijection, and $\\mathcal{F}(\\partial_j T) = 2\\pi i \\xi_j\\, \\mathcal{F}T$.\n\nExample: for the [[Dirac delta|0KPHENZ5S7]] $\\delta$ at $0$, $\\langle \\mathcal{F}\\delta, \\varphi \\rangle = \\mathcal{F}\\varphi(0) = \\int \\varphi(x)\\, dx$.\nSo $\\mathcal{F}\\delta$ is the constant function $1$.\nBy the Fourier inversion formula, $\\mathcal{F}1 = \\delta$ in the same way.\n\nNon-example: on $\\R$, the function $e^{\\abs{x}}$ is locally integrable but does not define a tempered distribution.\nFor the Schwartz function $\\varphi(x) = 1/\\cosh(x/2)$, the integral $\\int e^{\\abs{x}} \\varphi(x)\\, dx$ diverges.\nSo the formula above does not define a Fourier transform of $e^{\\abs{x}}$ as a tempered distribution.\n",
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