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  "body": "\n## Statement\n\nLet $n \\ge 1$, and let $f \\colon \\R^n \\to \\C$ be locally integrable, that is, integrable on every compact set.\nLet $g \\colon \\R^n \\to \\C$ be a [[continuously differentiable function|0VW90BBSJ4]] with compact support.\nThen the [[convolution|2Z29JN3VES]] $f * g$ is defined everywhere and is continuously differentiable, and for $j = 1, \\dots, n$\n$$\\frac{\\partial}{\\partial x_j} (f * g) = f * \\frac{\\partial g}{\\partial x_j}.$$\n\n## Remarks\n\nWrite $(f * g)(x) = \\int f(y) g(x - y) \\, dy$ and [[differentiate under the integral sign|0AR06W1E1B]].\nLet $x$ range over a bounded set and let $0 \u003c \\abs{t} \\le 1$.\nBy the mean value theorem, the difference quotients $(g(x + t e_j - y) - g(x - y))/t$ are bounded by $\\sup \\abs{\\partial_j g}$.\nThey vanish for $y$ outside a fixed compact set $K$.\nSince $f$ is integrable on $K$, the [[dominated convergence theorem|BYNV84PF86]] lets the limit pass inside the integral.\nContinuity of $f * \\partial_j g$ follows in the same way from the uniform continuity of $\\partial_j g$.\n\nExample: on $\\R$, take $f = \\mathbf{1}_{[0, \\infty)}$ and $g$ a continuously differentiable function with compact support.\nThen $(f * g)(x) = \\int_{-\\infty}^x g(u) \\, du$, and the formula gives $(f * g)' = f * g' = g$, as the fundamental theorem of calculus predicts.\n\nThe derivative falls on the smooth factor; $f$ itself need not be differentiable or even continuous.\nWithout compact support (or some decay) of $g$ the integral need not converge: for $f = 1$ and $g(x) = 1 + x^2$ the convolution is defined nowhere.\n\n## Explanations\n\n### L1 human\n\nTo differentiate a convolution, it is enough to differentiate one of the two factors.\nYou may choose the nicer one.\nSo a rough function convolved with a smooth one has a derivative, even if the rough function does not.\n\n### L2 human\n\nThis is the basic step behind [[smoothing by convolution|4P7KPA3Y7R]]: by induction, $f * g$ is as differentiable as $g$.\nIt also explains why convolution commutes with constant-coefficient differential operators $P(\\partial)$: $P(\\partial)(f * g) = f * P(\\partial) g$.\nIn the theory of distributions the same formula holds with $\\partial(f * g) = (\\partial f) * g = f * (\\partial g)$ for any distributions $f, g$ one of which has compact support.\n",
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