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  "body": "\n## Statement\n\nLet $n \\ge 1$, let $f \\in L^1(\\R^n)$, and let $\\hat f(\\xi) = \\int_{\\R^n} f(x) e^{-2\\pi i x \\cdot \\xi}\\, dx$ be its [[Fourier transform|DPVM34ADZV]].\nFix an index $j \\in \\set{1, \\dots, n}$.\n\n1. If the function $x \\mapsto x_j f(x)$ lies in [[$L^1(\\R^n)$|SASP14MA1Z]], then $\\hat f$ has a continuous partial derivative in $\\xi_j$, and $\\partial_{\\xi_j} \\hat f(\\xi)$ is the Fourier transform of $-2\\pi i x_j f(x)$.\n2. If $f$ is continuously differentiable and $\\partial_j f$ lies in $L^1(\\R^n)$, then the Fourier transform of $\\partial_j f$ is $\\xi \\mapsto 2\\pi i \\xi_j \\hat f(\\xi)$.\n\n## Remarks\n\nFor part 1, differentiate under the integral sign.\nThe difference quotients of $e^{-2\\pi i x \\cdot \\xi}$ in $\\xi_j$ are bounded by $2\\pi \\abs{x_j}$, so [[dominated convergence|BYNV84PF86]] applies with dominating function $2\\pi \\abs{x_j f(x)}$.\nFor part 2, integrate by parts in $x_j$.\nOn almost every line parallel to the $j$-th axis, $f$ and $\\partial_j f$ are integrable, so $f$ has limits at both ends, and these limits are $0$; the boundary terms vanish.\n\nExample: for $f(x) = e^{-\\pi x^2}$ on $\\R$, part 2 gives $\\widehat{f'}(\\xi) = 2\\pi i \\xi e^{-\\pi \\xi^2}$, and part 1 gives $\\frac{d}{d\\xi} e^{-\\pi \\xi^2} = -2\\pi \\xi e^{-\\pi \\xi^2}$; both agree with $f' = -2\\pi x f$.\n\nIterating, if $f$ lies in the Schwartz space then $\\widehat{\\partial^\\alpha f}(\\xi) = (2\\pi i \\xi)^\\alpha \\hat f(\\xi)$ and $\\partial^\\alpha \\hat f$ is the transform of $(-2\\pi i x)^\\alpha f$ for every multi-index $\\alpha$.\nThe integrability hypotheses cannot be dropped: if $x_j f$ is not integrable, $\\hat f$ need not be differentiable, as for $f(x) = 1/(1 + x^2)$ on $\\R$, whose transform $\\pi e^{-2\\pi \\abs{\\xi}}$ has a corner at $0$.\n\n## Explanations\n\n### L1 human\n\nThe Fourier transform swaps two operations.\nTaking a derivative of $f$ becomes multiplying $\\hat f$ by $2\\pi i \\xi$.\nMultiplying $f$ by $x$ becomes, up to a constant, taking a derivative of $\\hat f$.\n\n### L2 human\n\nThese rules turn a linear differential operator with constant coefficients $P(\\partial)$ into multiplication by the polynomial $P(2\\pi i \\xi)$.\nThis is how the transform solves the heat, wave and Schrödinger equations on $\\R^n$.\nThey also give the duality between smoothness and decay: the more derivatives $f$ has in $L^1$, the faster $\\hat f$ decays, and the faster $f$ decays, the smoother $\\hat f$ is.\nThis duality is the reason the Schwartz space is mapped onto itself.\n",
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