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  "body": "\n## Statement\n\nLet $n \\ge 1$, and let $f, g \\colon \\R^n \\to \\C$ be continuous functions with compact supports $\\operatorname{supp} f$ and $\\operatorname{supp} g$.\nHere the support of a function is the closure of the set where it is not $0$.\nThen the [[convolution|2Z29JN3VES]] $f * g$ is continuous, and\n$$\\operatorname{supp}(f * g) \\subseteq \\operatorname{supp} f + \\operatorname{supp} g = \\set{a + b : a \\in \\operatorname{supp} f, \\ b \\in \\operatorname{supp} g}.$$\nIn particular $f * g$ has compact support.\n\n## Remarks\n\nIf $(f * g)(x) = \\int f(y) g(x - y) \\, dy \\ne 0$, then $f(y) g(x - y) \\ne 0$ for some $y$.\nSo $y \\in \\operatorname{supp} f$ and $x - y \\in \\operatorname{supp} g$, and $x = y + (x - y)$ lies in the [[sumset|XNBN31HQWB]] $\\operatorname{supp} f + \\operatorname{supp} g$.\nThis sumset is the image of the [[compact set|0R6EWFKRCF]] $\\operatorname{supp} f \\times \\operatorname{supp} g$ under the continuous map $(a, b) \\mapsto a + b$, so it is compact and hence closed, and it contains the closure of $\\set{f * g \\ne 0}$.\n\nExample: if $f$ and $g$ are continuous, positive on $(0, 1)$ and zero elsewhere, then $f * g$ is positive exactly on $(0, 2)$, and $\\operatorname{supp}(f * g) = [0, 2] = [0, 1] + [0, 1]$.\n\nThe inclusion can be strict, because of cancellation in the integral.\nLet $\\varphi$ be continuous, positive on $(0, 1)$ and zero elsewhere, and put $\\psi = \\varphi * \\varphi$, supported on $[0, 2]$.\nFor $f(x) = \\varphi(x) - \\varphi(x - 2)$ and $g(x) = \\varphi(x) + \\varphi(x - 2)$ the cross terms cancel and $f * g(x) = \\psi(x) - \\psi(x - 4)$.\nSo $\\operatorname{supp}(f * g) = [0, 2] \\cup [4, 6]$, while $\\operatorname{supp} f + \\operatorname{supp} g = [0, 6]$.\nFor integrable functions the same statement holds with essential supports, in the form $\\operatorname{supp}(f * g) \\subseteq \\overline{\\operatorname{supp} f + \\operatorname{supp} g}$.\n\n## Explanations\n\n### L1 human\n\nIf $f$ lives on a set $A$ and $g$ lives on a set $B$, then $f * g$ lives on the set of all sums $a + b$.\nFor instance, two bumps on $[0, 1]$ give a bump on $[0, 2]$.\nSo convolution can spread a function out, but only by a controlled amount.\n\n### L2 human\n\nThis is why convolution with a [[mollifier|CQMK4YPE1J]] supported in a ball of radius $\\varepsilon$ moves the support of a function by at most $\\varepsilon$.\nTogether with [[smoothing by convolution|4P7KPA3Y7R]] it shows that smooth functions with compact support are dense in $L^p$ for $1 \\le p \u003c \\infty$.\nOn the line, the [[Titchmarsh convolution theorem|WGY75H4VWW]] gives an equality for the convex hulls of the supports.\n",
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