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  "body": "\n## Definition\n\nLet $\\mathcal{S}(\\R^n)$ be the [[Schwartz space|GE020MY43M]] and $\\mathcal{S}'(\\R^n)$ the space of [[tempered distributions|8TK1JNG87X]], with pairing $\\langle u, \\varphi \\rangle$.\nLet $K \\colon \\R^n \\times \\R^n \\to \\C$ be a measurable kernel and let $T$ be the [[integral transform|YCK709QB06]] $(T\\varphi)(x) = \\int_{\\R^n} K(x, y)\\, \\varphi(y)\\, dy$.\nLet $T^t$ be the transform with the transposed kernel, $(T^t\\varphi)(y) = \\int_{\\R^n} K(x, y)\\, \\varphi(x)\\, dx$.\nAssume that $T$ and $T^t$ map $\\mathcal{S}(\\R^n)$ continuously into $\\mathcal{S}(\\R^n)$.\nThe *transform* of $u \\in \\mathcal{S}'(\\R^n)$ is the tempered distribution $Tu$ given by\n$$\\langle Tu, \\varphi \\rangle = \\langle u, T^t \\varphi \\rangle \\quad (\\varphi \\in \\mathcal{S}(\\R^n)).$$\n\n## Remarks\n\nThe definition agrees with the integral on Schwartz functions: for $f, \\varphi \\in \\mathcal{S}(\\R^n)$, [[Fubini's theorem|0R0T1VWH5X]] gives $\\int (Tf)\\varphi = \\iint K(x, y) f(y) \\varphi(x)\\, dy\\, dx = \\int f\\, (T^t\\varphi)$, when the double integral converges absolutely.\nExample: the [[Fourier transform|6V59T5KJBX]] has the symmetric kernel $K(x, y) = e^{-2\\pi i x \\cdot y}$, so $T^t = T$ maps $\\mathcal{S}$ onto $\\mathcal{S}$.\nFor the Dirac delta $\\delta$ at $0$, $\\langle T\\delta, \\varphi \\rangle = (T\\varphi)(0) = \\int \\varphi$, so $T\\delta = 1$.\nNon-example: the kernel $K(x, y) = e^{xy}$ on $\\R$ does not qualify, since for the Schwartz function $\\varphi(y) = e^{-\\sqrt{1 + y^2}}$ the integral $\\int_\\R e^{2y} \\varphi(y)\\, dy$ diverges, so $T\\varphi$ is not defined at $x = 2$.\n",
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