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  "body": "\n## Statement\n\nLet $n \\ge 1$, and let $f, g \\colon \\R^n \\to \\C$ be Lebesgue measurable.\nLet $x \\in \\R^n$.\nThen $\\int_{\\R^n} \\abs{f(y) g(x - y)} \\, dy \u003c \\infty$ if and only if $\\int_{\\R^n} \\abs{g(y) f(x - y)} \\, dy \u003c \\infty$.\nIn that case $(f * g)(x) = (g * f)(x)$, where $*$ is the [[convolution|2Z29JN3VES]].\nIn particular $f * g = g * f$ for all $f, g \\in L^1(\\R^n)$.\n\n## Remarks\n\nThe proof is the substitution $y = x - z$.\n[[Lebesgue measure|2Z2QVWZ262]] is invariant under the reflection $z \\mapsto -z$ and under translations, so\n$$\\int_{\\R^n} f(y) g(x - y) \\, dy = \\int_{\\R^n} f(x - z) g(z) \\, dz = (g * f)(x),$$\nand the same computation with absolute values gives the equivalence of the two convergence conditions.\n\nExample: on $\\R$, with $f = \\mathbf{1}_{[0, 1]}$ and $g = \\mathbf{1}_{[0, 2]}$, both $f * g$ and $g * f$ give at $x$ the length of $[0, 1] \\cap [x - 2, x]$ and of $[0, 2] \\cap [x - 1, x]$, which are equal: both are $\\min(x, 1)$ for $0 \\le x \\le 2$.\n\nThe statement uses that the group $\\R^n$ is abelian.\nOn a non-abelian locally compact group, convolution $f * g$ with $(f * g)(x) = \\int f(y) g(y^{-1} x) \\, dy$ is in general not commutative.\n\n## Explanations\n\n### L1 human\n\nWhen you convolve two functions, it does not matter which one you slide and which one stays put.\nThe answer is the same either way.\nThis is like the product of two polynomials, which does not depend on the order of the factors.\n\n### L2 human\n\nCommutativity, together with [[associativity|HZ5J8GGS16]] and bilinearity, makes $L^1(\\R^n)$ a commutative algebra under convolution.\nIt is the reason why the [[convolution on the line|ERP5737VZ1]] can be written with either function shifted.\nUnder the Fourier transform it becomes the commutativity of the pointwise product $\\hat f \\hat g = \\hat g \\hat f$.\n",
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