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  "body": "\n## Statement\n\nLet $n \\ge 2$, let $\\Omega \\subseteq \\R^n$ be open, and let $u \\colon \\Omega \\to [-\\infty, \\infty)$ be [[upper semicontinuous|RN4G8SGP6P]].\nFor $a \\in \\R^n$ and $r \u003e 0$ let $\\sigma_{a,r}$ be the normalised surface measure on the sphere $S(a, r)$, and let $\\lambda$ be Lebesgue measure.\nAssume that each $a \\in \\Omega$ has some $\\rho(a) \u003e 0$ such that $u(a) \\le \\int u \\, d\\sigma_{a,r}$ for all $0 \u003c r \u003c \\rho(a)$, that is, $u$ is [[subharmonic|Z1MHMGFNFW]] in the local sense.\nThen for every $a \\in \\Omega$ and every $r \u003e 0$ with $\\overline{B(a, r)} \\subseteq \\Omega$:\n\n1. $u(a) \\le \\displaystyle\\int_{S(a, r)} u \\, d\\sigma_{a,r}$;\n2. $u(a) \\le \\dfrac{1}{\\lambda(B(a, r))} \\displaystyle\\int_{B(a, r)} u \\, d\\lambda$.\n\n## Remarks\n\nProof idea: since $u$ is upper semicontinuous, there are continuous functions $\\varphi_k$ on $S(a, r)$ that decrease to $u$ there.\nLet $h_k$ be the harmonic function in $B(a, r)$ with boundary values $\\varphi_k$, given by the [[Poisson integral formula|PCZBE7EX01]].\nThe function $u - h_k$ again satisfies the local sub-mean inequality, so the [[maximum principle for subharmonic functions|9XJ5FBQD7M]] gives $u \\le h_k$ in the ball.\nHence $u(a) \\le h_k(a) = \\int \\varphi_k \\, d\\sigma_{a,r}$, and monotone convergence gives part 1.\nPart 2 follows by integrating part 1 over the radii from $0$ to $r$.\n\nSo a locally defined subharmonic function satisfies the inequality on every ball whose closure lies in $\\Omega$, not only on small ones.\nExample: $u(x) = \\abs{x}^2$ on $\\R^n$ has sphere mean $\\abs{a}^2 + r^2$ and ball mean $\\abs{a}^2 + \\frac{n}{n+2} r^2$, both at least $\\abs{a}^2$.\nFor a harmonic function both inequalities become equalities, which is the [[mean value property|0HYPR64JM2]].\nThe ball $\\overline{B(a, r)}$ must lie in $\\Omega$: for $u(x) = -\\log\\abs{x}$ on $\\R^2 \\setminus \\set{0}$, which is harmonic there, the circle mean about $a = (1, 0)$ with $r = 2$ is $-\\log 2 \u003c 0 = u(a)$.\n\n## Explanations\n\n### L1 human\n\nA subharmonic function at a point is never larger than its average over a sphere or a ball around that point.\nThe definition only asks this for tiny spheres, and this result says it then holds for every sphere that fits in the domain.\n\n### L2 human\n\nThe global inequality is what makes subharmonic functions useful: it gives the maximum principle, local integrability, and comparison with harmonic functions on balls.\nThe averages over spheres of radius $r$ are increasing in $r$ and tend to $u(a)$ as $r \\to 0$.\nFor $C^2$ functions the condition is equivalent to $\\Delta u \\ge 0$.\n",
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