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  "body": "\n## Definition\n\nLet $f, g \\colon \\R \\to \\C$ be Lebesgue measurable functions.\nLet $x \\in \\R$ be such that $\\int_\\R \\abs{f(y) \\, g(x - y)} \\, dy \u003c \\infty$.\nThe *convolution* of $f$ and $g$ at $x$ is the [[Lebesgue integral|DWJF7ABPR1]]\n$$(f * g)(x) = \\int_\\R f(y) \\, g(x - y) \\, dy.$$\nThe function $f * g$ is defined on the set of such $x$.\n\n## Remarks\n\nIf $f$ and $g$ lie in [[$L^1(\\R)$|SASP14MA1Z]], then $f * g$ is defined at almost every $x$, it lies in $L^1(\\R)$, and $\\norm{f * g}_1 \\le \\norm{f}_1 \\norm{g}_1$.\nIf $f \\in L^p(\\R)$ and $g \\in L^q(\\R)$ with $1/p + 1/q = 1$, then $f * g$ is defined everywhere and is bounded and uniformly continuous.\n\nExample: $f(x) = g(x) = e^{-x^2}$.\nCompleting the square gives $y^2 + (x - y)^2 = 2(y - x/2)^2 + x^2/2$, so\n$$(f * g)(x) = e^{-x^2/2} \\int_\\R e^{-2(y - x/2)^2} \\, dy = \\sqrt{\\pi/2} \\, e^{-x^2/2}.$$\nThe convolution of two Gaussians is again a Gaussian, wider than both.\n\nNon-example: $f = g = 1$ on $\\R$.\nThen $\\int_\\R \\abs{f(y) g(x - y)} \\, dy = \\int_\\R 1 \\, dy = \\infty$ for every $x$, so $f * g$ is defined nowhere.\n\nThe integral can also be written $\\int_\\R f(x - y) g(y) \\, dy$, by the substitution $y \\mapsto x - y$.\n",
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