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  "body": "\n## Definition\n\nLet $f \\colon \\R \\to \\C$ be measurable with $\\int_\\R \\abs{f(y)}\\, e^{-c y^2}\\, dy \u003c \\infty$ for every $c \u003e 0$.\nThe *Weierstrass transform* of $f$ is the function $Wf \\colon \\R \\to \\C$ given by\n$$(Wf)(x) = \\frac{1}{\\sqrt{4\\pi}} \\int_\\R f(y)\\, e^{-(x - y)^2/4}\\, dy.$$\n\n## Remarks\n\nThe kernel $\\frac{1}{\\sqrt{4\\pi}} e^{-u^2/4}$ is the density of the [[normal distribution|05AQE0QR0D]] with mean $0$ and variance $2$.\nSo $Wf(x) = \\mathbb{E}[f(x + Y)]$ for a random variable $Y$ with that law.\nIt is the [[heat kernel|KD404MF042]] $\\frac{1}{\\sqrt{4\\pi t}} e^{-u^2/(4t)}$ at $t = 1$, so $Wf = u(\\cdot, 1)$ where $u$ solves the [[heat equation|VP2HN86VAG]] $u_t = u_{xx}$ with $u(\\cdot, 0) = f$.\nFor integrable $f$ and $\\hat f(\\xi) = \\int f(x) e^{-i x \\xi}\\, dx$, one has $\\widehat{Wf}(\\xi) = e^{-\\xi^2} \\hat f(\\xi)$.\nThe growth condition can be weakened to $\\int \\abs{f(y)} e^{-c y^2}\\, dy \u003c \\infty$ for one $c \u003c \\frac14$, which is enough for the integral to converge for every $x$.\n\nExample: $f(x) = x^2$.\nWith $Y$ of mean $0$ and variance $2$, $Wf(x) = \\mathbb{E}[(x + Y)^2] = x^2 + 2x\\, \\mathbb{E}[Y] + \\mathbb{E}[Y^2] = x^2 + 2$.\nLikewise $W[e^{a x}](x) = e^{a x} \\mathbb{E}[e^{aY}] = e^{a^2} e^{a x}$ for every real $a$.\n\nNon-example: $f(x) = e^{x^2/4}$ fails the condition for $c \u003c \\frac14$, and indeed $\\int_\\R e^{y^2/4} e^{-(x-y)^2/4}\\, dy = e^{-x^2/4} \\int_\\R e^{x y/2}\\, dy$ diverges for every $x$.\n",
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