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  "body": "\n## Statement\n\nLet $f \\in L^2(\\R)$, an element of the [[$L^2$ space|SASP14MA1Z]], and let $F = Wf$ be its [[Weierstrass transform|5HYB6DDZH8]].\nLet $\\hat g(\\xi) = \\int_\\R g(x)\\, e^{-i x \\xi}\\, dx$ denote the [[Fourier transform|DPVM34ADZV]], extended to $L^2(\\R)$.\nThen $F \\in L^2(\\R)$, $\\hat F(\\xi) = e^{-\\xi^2} \\hat f(\\xi)$ for almost every $\\xi$, and $f$ is recovered from $F$ as the function in $L^2(\\R)$ whose Fourier transform is\n$$\\hat f(\\xi) = e^{\\xi^2}\\, \\hat F(\\xi).$$\nIn particular $W$ is injective on $L^2(\\R)$.\n\n## Remarks\n\nThe kernel $k(u) = \\frac{1}{\\sqrt{4\\pi}} e^{-u^2/4}$ is integrable with $\\hat k(\\xi) = e^{-\\xi^2}$, and $F = k * f$.\nThe convolution theorem on $L^1 * L^2$ gives $\\hat F = \\hat k \\hat f$, and the [[Plancherel theorem|HPZHRZ0502]] makes the Fourier transform a bijection of $L^2(\\R)$ up to the factor $2\\pi$, which yields the formula.\nFormally $W = e^{D^2}$ with $D = d/dx$, and the inverse is $e^{-D^2}$; for example, $W[e^{ax}] = e^{a^2} e^{ax}$ and $e^{-D^2} e^{ax} = e^{-a^2} e^{ax}$.\nThe inversion is unstable: the factor $e^{\\xi^2}$ grows very fast, so a small change of $F$ at high frequencies can change $f$ a lot.\nNot every function in $L^2$ is a Weierstrass transform: $F$ must satisfy $e^{\\xi^2} \\hat F \\in L^2$, and every such $F$ extends to an entire function.\n\n## Explanations\n\n### L1 human\n\nThe Weierstrass transform blurs a function as heat flow blurs a temperature profile.\nThis theorem says the blur loses no information: the original profile can be computed back from the blurred one.\nIn practice the computation is very sensitive to small errors, as running heat flow backwards in time always is.\n\n### L2 human\n\nThe inversion is the uniqueness theorem for the backward heat equation at a fixed time, in the class of square-integrable data.\nOn the Fourier side it is division by the Gaussian $e^{-\\xi^2}$, which never vanishes; this is why the transform is injective.\nThe same reasoning applies to any convolution with an integrable kernel whose Fourier transform has no zeros.\nThe operator notation $e^{D^2}$ also explains how $W$ acts on polynomials: it maps the Hermite polynomial $H_n(x/2)$ to $x^n$, since $e^{-D^2} x^n = H_n(x/2)$ for the physicists' Hermite polynomials $H_n$.\n",
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