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  "body": "\n## Statement\n\nLet $(X, \\mu)$ and $(Y, \\nu)$ be $\\sigma$-finite [[measure spaces|P15VQS5DJ9]], and let $K \\colon X \\times Y \\to \\C$ be a [[measurable function|YYNCJSR0PJ]].\nAssume that there are measurable functions $p \\colon X \\to (0, \\infty)$ and $q \\colon Y \\to (0, \\infty)$ and constants $\\alpha, \\beta \u003e 0$ such that\n$$\\int_Y \\abs{K(x, y)} \\, q(y) \\, d\\nu(y) \\le \\alpha \\, p(x) \\quad \\text{for almost every } x \\in X,$$\n$$\\int_X \\abs{K(x, y)} \\, p(x) \\, d\\mu(x) \\le \\beta \\, q(y) \\quad \\text{for almost every } y \\in Y.$$\nThen for every $f \\in L^2(Y, \\nu)$ the integral\n$$(T f)(x) = \\int_Y K(x, y) \\, f(y) \\, d\\nu(y)$$\nconverges absolutely for almost every $x$.\nMoreover, $T$ is a [[bounded linear operator|03TM135G0N]] from $L^2(Y, \\nu)$ to $L^2(X, \\mu)$ with operator norm $\\norm{T} \\le \\sqrt{\\alpha \\beta}$.\n\n## Remarks\n\nProof idea: write $\\abs{K} \\abs{f} = \\big(\\abs{K} q\\big)^{1/2} \\big(\\abs{K} \\abs{f}^2 / q\\big)^{1/2}$ and apply the Cauchy–Schwarz inequality in $y$.\nThe first hypothesis gives $\\abs{T f(x)}^2 \\le \\alpha \\, p(x) \\int_Y \\abs{K(x, y)} \\abs{f(y)}^2 q(y)^{-1} \\, d\\nu(y)$.\nIntegrating in $x$ and exchanging the integrals by Tonelli's theorem, the second hypothesis gives $\\norm{T f}_2^2 \\le \\alpha \\beta \\norm{f}_2^2$.\n\nWith $p = q = 1$ the test says: if $\\int_Y \\abs{K(x, y)} \\, d\\nu(y) \\le \\alpha$ for almost every $x$ and $\\int_X \\abs{K(x, y)} \\, d\\mu(x) \\le \\beta$ for almost every $y$, then $\\norm{T} \\le \\sqrt{\\alpha \\beta}$ on $L^2$.\nFor $\\alpha = \\beta$ this is the case $p = 2$ of [[Young's inequality for integral operators|48NS4ZCSR3]].\nThe freedom to choose the weights $p$ and $q$ is what makes the test useful for kernels such as $1/(x + y)$, which have no finite row or column integrals.\n\nExample: let $X = Y = (0, \\infty)$ with Lebesgue measure and $K(x, y) = 1/(x + y)$.\nTake $p(x) = x^{-1/2}$ and $q(y) = y^{-1/2}$.\nThe substitution $y = x u$ gives $\\int_0^\\infty \\frac{y^{-1/2}}{x + y} \\, dy = x^{-1/2} \\int_0^\\infty \\frac{u^{-1/2}}{1 + u} \\, du = \\pi x^{-1/2}$, and by symmetry the same holds for the second hypothesis.\nSo $\\alpha = \\beta = \\pi$ and $\\norm{T} \\le \\pi$, which is the continuous form of [[Hilbert's inequality|G4R4MWWP3C]].\n\n## Explanations\n\n### L1 human\n\nAn integral operator acts on a function by averaging it against a kernel $K(x, y)$.\nThe Schur test gives a simple way to see that such an operator does not blow up the size of functions, measured in $L^2$.\nOne only has to find one positive test function for the rows and one for the columns of the kernel, and check two integral inequalities.\n\n### L2 human\n\nThe Schur test is the main elementary tool to prove that an integral operator is bounded on $L^2$ when its kernel is not square-integrable.\nSuch an operator is not [[Hilbert–Schmidt|W3BX66Y47G]].\nIt is used for kernels with a singularity or slow decay, such as homogeneous kernels on $(0, \\infty)$ and kernels in the study of Bergman projections.\nIt is the continuous analogue of the bound of the norm of a matrix by the geometric mean of the largest row sum and the largest column sum.\nFor $L^p$ with $p \\ne 2$, the same idea with Hölder's inequality in place of Cauchy–Schwarz gives a version with weights.\n",
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