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  "body": "\n## Statement\n\nLet $(S, \\Sigma, \\mu)$ be a [[measure space|P15VQS5DJ9]].\nLet $p, q \\in (1, \\infty)$ with $\\frac{1}{p} + \\frac{1}{q} = 1$.\nLet $f, g \\colon S \\to \\R$ be [[measurable functions|YYNCJSR0PJ]].\nThen, with the [[integrals|DWJF7ABPR1]] taking values in $[0, \\infty]$ and the convention $0 \\cdot \\infty = 0$,\n$$\\int_S \\abs{f g} \\, d\\mu \\le \\Big( \\int_S \\abs{f}^p \\, d\\mu \\Big)^{1/p} \\Big( \\int_S \\abs{g}^q \\, d\\mu \\Big)^{1/q}.$$\n\n## Remarks\n\nProof idea: if a factor on the right is $0$ or $\\infty$ the claim is easy.\nOtherwise scale $f$ and $g$ so that both factors are $1$, apply [[Young's inequality|07VHQ31EYJ]] $ab \\le \\frac{a^p}{p} + \\frac{b^q}{q}$ pointwise to $\\abs{f(s)}$ and $\\abs{g(s)}$, and integrate.\nWhen the right side is finite, equality holds if and only if there are constants $\\alpha, \\beta \\ge 0$, not both $0$, with $\\alpha \\abs{f}^p = \\beta \\abs{g}^q$ almost everywhere.\n\nWith the counting measure on $\\set{1, \\dots, n}$ the statement becomes [[Hölder's inequality for sums|0BKDAGMGX9]].\nFor $p = q = 2$ it is the [[Cauchy–Schwarz inequality|0RFWW56Y22]] for integrals.\nThe same bound holds for $p = 1$ and $q = \\infty$, with the essential supremum of $\\abs{g}$ in place of the second factor.\n\n## Explanations\n\n### L1 human\n\nHölder's inequality bounds the integral of a product by the sizes of the two factors.\nThe size of $f$ is measured with the exponent $p$ and the size of $g$ with the matching exponent $q$.\nIt is the main tool for showing that a product of two functions is integrable.\n\n### L2 human\n\nIn the language of [[Lp spaces|SASP14MA1Z]], it says $\\norm{f g}_1 \\le \\norm{f}_p \\norm{g}_q$.\nSo each $g \\in L^q$ gives a bounded linear functional on $L^p$, which is the start of the duality between $L^p$ and $L^q$.\nIt is also the key step in the proof of the [[Minkowski inequality|N4QZS912MT]], the triangle inequality in $L^p$.\n",
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