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  "body": "\n## Definition\n\nLet $(X, \\mu)$ and $(Y, \\nu)$ be [[measure spaces|56MJSVTHG0]] whose measures are [[sigma-finite|CW5CEF84T2]].\nLet $K \\colon X \\times Y \\to \\C$ be a measurable function, the *kernel*.\nThe *integral operator* with kernel $K$ is the [[linear map|0RJF668P1T]] $T$ given by\n$$(Tf)(x) = \\int_Y K(x, y)\\, f(y)\\, d\\nu(y).$$\nIts domain is the set of measurable functions $f \\colon Y \\to \\C$ for which the [[Lebesgue integral|XVEWR93EMV]] $\\int_Y \\abs{K(x, y) f(y)}\\, d\\nu(y)$ is finite for $\\mu$-almost every $x \\in X$.\n\n## Remarks\n\nExample: on $X = Y = [0, 1]$ with Lebesgue measure, the kernel $K(x, y) = 1$ for $y \u003c x$ and $K(x, y) = 0$ for $y \\ge x$ gives the Volterra operator $(Vf)(x) = \\int_0^x f(y)\\, dy$.\nIt is defined for every integrable $f$, because $\\int_0^1 \\abs{K(x, y) f(y)}\\, dy \\le \\int_0^1 \\abs{f}$.\nNon-example: the identity map of $L^2([0,1])$ is linear but is not an integral operator with a function kernel.\nSuch a kernel would give $\\int_A K(x, y)\\, dy = 1$ for almost every $x \\in A$ and $0$ for almost every $x \\notin A$, for every interval $A$.\nThis would force the measure $K(x, y)\\, dy$ to be a point mass at $x$, which is impossible.\nIf $K$ is in $L^2(X \\times Y)$, the operator maps the [[Lebesgue space|7KGF78FGEF]] $L^2(Y)$ to $L^2(X)$ and is bounded with norm at most $\\norm{K}_{L^2}$.\nSuch operators are called Hilbert–Schmidt integral operators, and they are compact.\n",
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