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  "body": "\n## Statement\n\nLet $(X, \\mu)$ and $(Y, \\nu)$ be $\\sigma$-finite [[measure spaces|P15VQS5DJ9]], let $1 \\le p \\le \\infty$ and $C \u003e 0$.\nLet $K \\colon X \\times Y \\to \\C$ be a [[measurable function|YYNCJSR0PJ]] such that\n$$\\int_Y \\abs{K(x, y)} \\, d\\nu(y) \\le C \\quad \\text{for almost every } x \\in X,$$\n$$\\int_X \\abs{K(x, y)} \\, d\\mu(x) \\le C \\quad \\text{for almost every } y \\in Y.$$\nLet $f \\in L^p(Y, \\nu)$.\nThen the integral\n$$(T f)(x) = \\int_Y K(x, y) \\, f(y) \\, d\\nu(y)$$\nconverges absolutely for almost every $x \\in X$, the function $T f$ lies in $L^p(X, \\mu)$, and\n$$\\norm{T f}_p \\le C \\norm{f}_p.$$\n\n## Remarks\n\nProof idea for $1 \u003c p \u003c \\infty$: let $p'$ be the conjugate exponent and write $\\abs{K} = \\abs{K}^{1/p'} \\abs{K}^{1/p}$.\nHölder's inequality in $y$ and the first hypothesis give $\\abs{T f(x)}^p \\le C^{p/p'} \\int_Y \\abs{K(x, y)} \\abs{f(y)}^p \\, d\\nu(y)$.\nIntegrating in $x$, Tonelli's theorem and the second hypothesis give $\\norm{T f}_p^p \\le C^{p/p'} \\cdot C \\norm{f}_p^p = C^p \\norm{f}_p^p$.\nThe case $p = \\infty$ uses only the first hypothesis, and the case $p = 1$ uses only the second.\n\nExample: for $X = Y = \\R^n$ with Lebesgue measure and $K(x, y) = g(x - y)$ with $g \\in L^1(\\R^n)$, both hypotheses hold with $C = \\norm{g}_1$.\nThe result is $\\norm{g * f}_p \\le \\norm{g}_1 \\norm{f}_p$, the case $q = 1$ of [[Young's convolution inequality|JXDKJCBXJ8]].\n\nBoth hypotheses are needed for the full range of $p$.\nOn $X = Y = (0, \\infty)$, the Hardy kernel $K(x, y) = 1/x$ for $y \u003c x$ and $K(x, y) = 0$ otherwise has $\\int_0^\\infty K(x, y) \\, dy = 1$ for every $x$, but $\\int_0^\\infty K(x, y) \\, dx = \\infty$ for every $y$.\nThe operator $T f(x) = \\frac{1}{x} \\int_0^x f(y) \\, dy$ is bounded on $L^p$ for $1 \u003c p \\le \\infty$ by [[Hardy's inequality|S7A8YFGG8X]], but not on $L^1$: for $f$ the indicator function of $(0, 1]$, $T f(x) = 1/x$ for $x \u003e 1$, which is not integrable.\n\n## Explanations\n\n### L1 human\n\nAn integral operator replaces a function by weighted averages of its values, with weights $K(x, y)$.\nSuppose that every row and every column of the weights has total mass at most $C$.\nThen the operator enlarges the $L^p$ size of a function by at most the factor $C$, for every $p$.\nThis is the continuous analogue of the fact that a matrix whose absolute row sums and column sums are at most $C$ has norm at most $C$ on every $\\ell^p$.\n\n### L2 human\n\nThis result is sometimes called the generalized Young inequality, since it contains the case $\\norm{g * f}_p \\le \\norm{g}_1 \\norm{f}_p$ of [[Young's convolution inequality|JXDKJCBXJ8]].\nIt is the quickest way to show that an integral operator with an integrable kernel acts boundedly on all $L^p$ spaces at once.\nFor $p = 2$ it is the [[Schur test|25HF1WPPKX]] with constant weights; the Schur test allows other weights and so applies to more kernels.\nIn the theory of [[integral equations|R3SG1SN7JQ]] it gives the bounds that the contraction principle needs for second-kind equations in $L^p$.\n",
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