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  "body": "\n## Statement\n\nLet $n \\ge 1$ and consider [[Lp spaces|SASP14MA1Z]] on $\\R^n$ with [[Lebesgue measure|2Z2QVWZ262]].\nLet $1 \\le p, q, r \\le \\infty$ with\n$$\\frac{1}{p} + \\frac{1}{q} = 1 + \\frac{1}{r}.$$\nLet $f \\in L^p(\\R^n)$ and $g \\in L^q(\\R^n)$.\nThen the convolution\n$$(f * g)(x) = \\int_{\\R^n} f(y) \\, g(x - y) \\, dy$$\nconverges absolutely for almost every $x \\in \\R^n$, the function $f * g$ lies in $L^r(\\R^n)$, and\n$$\\norm{f * g}_r \\le \\norm{f}_p \\norm{g}_q.$$\n\n## Remarks\n\nThe case $p = 1$ and $q = r$ follows from Minkowski's inequality for integrals: $f * g$ is an average of translates of $g$, each of norm $\\norm{g}_q$.\nThe case $r = \\infty$, with $p$ and $q$ conjugate, is Hölder's inequality applied for each fixed $x$; then $f * g$ is defined everywhere.\nThe general case follows from these two by the [[Riesz–Thorin theorem|AGRKRG1TMJ]], or directly by a three-factor Hölder inequality.\n\nExample: with $p = q = 1$, one gets $r = 1$ and $\\norm{f * g}_1 \\le \\norm{f}_1 \\norm{g}_1$.\nSo $L^1(\\R^n)$ is a Banach algebra under convolution.\n\nThe exponent relation is forced by scaling.\nIf $f_\\lambda(x) = f(\\lambda x)$, then $\\norm{f_\\lambda}_p = \\lambda^{-n/p} \\norm{f}_p$ and $f_\\lambda * g_\\lambda = \\lambda^{-n} (f * g)_\\lambda$.\nAn inequality $\\norm{f * g}_r \\le C \\norm{f}_p \\norm{g}_q$ for all $\\lambda \u003e 0$ therefore requires $1 + 1/r = 1/p + 1/q$.\n\nThe same inequality holds with the same proof on any unimodular locally compact group with Haar measure.\nThe constant $1$ is not sharp on $\\R^n$ when $1 \u003c p, q, r \u003c \\infty$; the sharp constant is known.\n\n## Explanations\n\n### L1 human\n\nConvolution smears one function out by another, like blurring a picture with a lens.\nYoung's inequality says how big the blurred result can be in terms of the sizes of the two ingredients.\nThe exponents must fit together in one specific way, which a change of scale explains.\n\n### L2 human\n\nYoung's inequality is the basic estimate for convolution operators $f \\mapsto f * g$ with a fixed kernel $g$.\nWith $g \\in L^1$ it shows that mollification $f * \\varphi_\\varepsilon$ is bounded on every $L^p$, which is the first step of approximation by smooth functions.\nWith $g$ a power such as $\\abs{x}^{-\\alpha}$, which is not in any $L^q(\\R^n)$, Young's inequality does not apply, and one needs the Hardy–Littlewood–Sobolev inequality, which uses weak $L^q$ instead.\n",
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