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  "body": "\n## Statement\n\nLet $n \\ge 1$, let $0 \u003c \\alpha \u003c n$, and let $I_\\alpha$ be the [[fractional integral|P1VFBJV1VN]] of order $\\alpha$ on $\\R^n$.\nLet $1 \\le p \u003c q \u003c \\infty$ with\n$$\\frac{1}{q} = \\frac{1}{p} - \\frac{\\alpha}{n}.$$\n\n1. If $p \u003e 1$, there is a constant $C = C(n, \\alpha, p)$ such that $\\norm{I_\\alpha f}_{L^q(\\R^n)} \\le C \\norm{f}_{L^p(\\R^n)}$ for all $f \\in L^p(\\R^n)$.\n2. If $p = 1$, so that $q = n/(n - \\alpha)$, then $I_\\alpha$ is of [[weak type|9534R6VMY0]] $(1, q)$: there is $C = C(n, \\alpha)$ with $\\abs{\\set{x : \\abs{I_\\alpha f(x)} \u003e \\lambda}} \\le \\left( C \\norm{f}_{L^1} / \\lambda \\right)^q$ for all $f \\in L^1(\\R^n)$ and $\\lambda \u003e 0$.\n\nHere $L^p(\\R^n)$ is the [[Lp space|SASP14MA1Z]] for Lebesgue measure.\n\n## Remarks\n\nA short proof is Hedberg's.\nFix $x$ and $R \u003e 0$, and split the integral defining $I_\\alpha \\abs{f}(x)$ into $\\abs{x - y} \u003c R$ and $\\abs{x - y} \\ge R$.\nThe near part is at most $C R^\\alpha Mf(x)$, where $M$ is the [[Hardy–Littlewood maximal function|VX1A56P5KQ]], by summing over dyadic shells.\nThe far part is at most $C R^{-n/q} \\norm{f}_p$, by [[Hölder's inequality|JQ9WXC729D]].\nChoosing $R$ to balance the two terms gives $\\abs{I_\\alpha f(x)} \\le C \\, Mf(x)^{p/q} \\norm{f}_p^{1 - p/q}$, and the [[maximal inequality|BF7T7VE57V]] for $M$ gives both parts.\n\nThe relation between $p$, $q$ and $\\alpha$ is forced by scaling.\nIf $f_\\delta(x) = f(\\delta x)$, then $I_\\alpha f_\\delta(x) = \\delta^{-\\alpha} I_\\alpha f(\\delta x)$, and an inequality $\\norm{I_\\alpha f}_q \\le C \\norm{f}_p$ for all $f$ and all $\\delta \u003e 0$ needs $\\frac{1}{q} = \\frac{1}{p} - \\frac{\\alpha}{n}$.\nAt $p = 1$ the strong bound fails: if $f \\ge 0$ is integrable with $\\int f \u003e 0$, then $I_\\alpha f(x)$ behaves like $c \\abs{x}^{\\alpha - n}$ at infinity, which is in weak $L^{q}$ but not in $L^{q}$ for $q = n/(n - \\alpha)$.\n\n## Explanations\n\n### L1 human\n\nFractional integration of order $\\alpha$ smooths a function by $\\alpha$ derivatives.\nThe inequality says that this smoothing also improves integrability: an $L^p$ function becomes an $L^q$ function with a larger exponent $q$.\nThe gain $\\frac{1}{p} - \\frac{1}{q} = \\frac{\\alpha}{n}$ is exactly what dimensional analysis allows.\n\n### L2 human\n\nThe inequality implies the [[Sobolev inequality|B03CA692K3]] for $1 \u003c p \u003c n$.\nA smooth function $u$ with compact support satisfies $\\abs{u(x)} \\le C \\, I_1(\\abs{\\nabla u})(x)$, and the case $\\alpha = 1$ then gives $\\norm{u}_{np/(n - p)} \\le C \\norm{\\nabla u}_p$.\nThere is also a bilinear form: $\\abs{\\iint f(x) g(y) \\abs{x - y}^{-\\lambda} \\, dx \\, dy} \\le C \\norm{f}_p \\norm{g}_r$ when $0 \u003c \\lambda \u003c n$, $1 \u003c p, r \u003c \\infty$ and $\\frac{1}{p} + \\frac{1}{r} + \\frac{\\lambda}{n} = 2$.\nLieb found the sharp constant of this bilinear form in the case $p = r$.\n",
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  "name": "Hardy–Littlewood–Sobolev inequality",
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