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  "body": "\n## Statement\n\nLet $\\mathcal{C}$ be the set of continuous functions $[0, \\infty) \\to \\C$, with pointwise addition and the [[convolution on the half-line|FCKX8KFVGN]]\n$$(f * g)(t) = \\int_0^t f(\\tau) \\, g(t - \\tau) \\, d\\tau$$\nas multiplication.\nThen $\\mathcal{C}$ is a commutative [[rng|MBK7NFE74S]], that is, a ring that need not have a unit.\nIt has no [[zero divisors|0NNRDW10RZ]]: if $f, g \\in \\mathcal{C}$ and $f * g = 0$, then $f = 0$ or $g = 0$.\n\n## Remarks\n\nCommutativity and associativity follow from the corresponding facts on the line, and distributivity is linearity of the integral.\nFor the absence of zero divisors, suppose $f * g = 0$ and $f(t_0) \\ne 0$.\nBy continuity $f$ is nonzero near $t_0$, so $f$ does not vanish almost everywhere on $(0, \\lambda)$ for any $\\lambda \u003e t_0$.\nFor each $T \u003e t_0$, the [[Titchmarsh convolution theorem|WGY75H4VWW]] then gives $g = 0$ almost everywhere on $(0, T - t_0)$; letting $T \\to \\infty$ and using continuity gives $g = 0$.\n\n$\\mathcal{C}$ has no unit element.\nIf $e * f = f$ for all $f$, then $e * 1 = 1$, but $(e * 1)(0) = 0 \\ne 1$.\n\nExample: $1 * 1 = t$ and $t * t = t^3/6$, and in general no product of nonzero polynomials in $\\mathcal{C}$ is $0$, in line with the formula $t^m * t^n = \\frac{m!\\,n!}{(m+n+1)!} t^{m+n+1}$.\n\n## Explanations\n\n### L1 human\n\nIn ordinary arithmetic, a product is zero only when one factor is zero.\nThe same rule holds when continuous functions on $[0, \\infty)$ are \"multiplied\" by convolution.\nThis is far from obvious, because convolution mixes values of the two functions in a complicated way.\n\n### L2 human\n\nA commutative rng without zero divisors has a field of fractions, built from pairs $f/g$ as for the rational numbers.\nFor $\\mathcal{C}$ this is the [[Mikusiński field of operators|82YAXZMMQ1]], the setting of the Mikusiński operational calculus.\nThe same conclusion holds for locally integrable functions, with equality almost everywhere.\n",
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