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  "body": "\n## Statement\n\nLet $p \u003e 1$ be a real number.\n\n1. (Discrete form.) Let $a_1, a_2, \\dots$ be nonnegative real numbers. Then\n$$\\sum_{n=1}^{\\infty} \\Big( \\frac{a_1 + a_2 + \\dots + a_n}{n} \\Big)^{p} \\le \\Big( \\frac{p}{p - 1} \\Big)^{p} \\sum_{n=1}^{\\infty} a_n^{p}.$$\n2. (Integral form.) Let $f \\colon (0, \\infty) \\to [0, \\infty)$ be a [[measurable function|YYNCJSR0PJ]], and let $F(x) = \\frac{1}{x} \\int_0^x f(t) \\, dt$ for $x \u003e 0$. Then\n$$\\int_0^{\\infty} F(x)^{p} \\, dx \\le \\Big( \\frac{p}{p - 1} \\Big)^{p} \\int_0^{\\infty} f(x)^{p} \\, dx.$$\n\nHere the [[series|0CVVKY7GY2]] and [[integrals|DWJF7ABPR1]] take values in $[0, \\infty]$.\n\n## Remarks\n\nProof idea for the integral form: substitute $t = s x$ to get $F(x) = \\int_0^1 f(s x) \\, ds$, an average of the dilates of $f$.\nBy Minkowski's integral inequality, the continuous form of the [[Minkowski inequality|N4QZS912MT]], $\\norm{F}_p \\le \\int_0^1 \\norm{f(s \\, \\cdot)}_p \\, ds$.\nSince $\\norm{f(s \\, \\cdot)}_p = s^{-1/p} \\norm{f}_p$ and $\\int_0^1 s^{-1/p} \\, ds = \\frac{p}{p-1}$, the result follows.\n\nThe constant $\\big(\\frac{p}{p-1}\\big)^p$ is the best possible in both forms.\n\nThe hypothesis $p \u003e 1$ is needed.\nFor $p = 1$ and $f$ the indicator function of $(0, 1]$, one gets $F(x) = 1/x$ for $x \u003e 1$, so the left side is infinite while the right side is $1$.\n\n## Explanations\n\n### L1 human\n\nTake a sequence of nonnegative numbers and replace each term by the average of the terms up to it.\nHardy's inequality says the new sequence is not much larger than the old one, when size is measured by the sum of $p$-th powers.\nThe loss is at most the factor $\\big(\\frac{p}{p-1}\\big)^p$, which is $4$ for $p = 2$.\n\n### L2 human\n\nIn operator language, the averaging operator $f \\mapsto F$ is bounded on the [[Lp space|SASP14MA1Z]] $L^p(0, \\infty)$ for $1 \u003c p \u003c \\infty$, with norm $\\frac{p}{p-1}$.\nIt is a basic example of a bounded integral operator, and it is used in Sobolev-type estimates and in the theory of weighted inequalities.\nLetting $p \\to \\infty$ after replacing $a_n$ by $a_n^{1/p}$ leads to [[Carleman's inequality|B54HPY6C5Z]].\n",
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