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  "body": "\n## Definition\n\nLet $(X, \\mu)$ be a $\\sigma$-finite [[measure space|P15VQS5DJ9]], and let $L^2(X)$ be the [[$L^2$ space|SASP14MA1Z]] of $\\mu$.\nLet $K \\colon X \\times X \\to \\C$ be a measurable function with\n$$\\int_X \\int_X \\abs{K(x, y)}^2 \\, d\\mu(x) \\, d\\mu(y) \u003c \\infty.$$\nThe *Hilbert–Schmidt integral operator* with kernel $K$ is the operator $T_K \\colon L^2(X) \\to L^2(X)$ given by\n$$(T_K f)(x) = \\int_X K(x, y) \\, f(y) \\, d\\mu(y).$$\n\n## Remarks\n\nThe operator is well defined: for almost every $x$, $K(x, \\cdot) \\in L^2(X)$, and the Cauchy–Schwarz inequality gives $\\norm{T_K f} \\le \\norm{K}_{L^2(X \\times X)} \\norm{f}$.\nIts Hilbert–Schmidt norm $\\left(\\sum_n \\norm{T_K e_n}^2\\right)^{1/2}$, for any orthonormal basis $(e_n)$, equals $\\norm{K}_{L^2(X \\times X)}$.\n\nExample: on $X = [0, 1]$ with Lebesgue measure, $K(x, y) = \\min(x, y)$ satisfies $\\int_0^1 \\int_0^1 \\min(x, y)^2 \\, dx \\, dy = 1/6 \u003c \\infty$, so $T_K$ is a Hilbert–Schmidt integral operator.\n\nNon-example: on $[0, 1]$, $K(x, y) = \\abs{x - y}^{-1/2}$ fails the condition, since $\\int_0^1 \\int_0^1 \\abs{x - y}^{-1} \\, dx \\, dy = \\infty$.\nIts operator is still bounded and compact on $L^2[0, 1]$, but it is not a Hilbert–Schmidt operator.\n\nKernels with the condition above are the [[square-integrable kernels|WWMWX7AD4V]].\n",
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