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  "body": "\n## Definition\n\nLet $K \\colon (0, \\infty) \\times (0, \\infty) \\to \\C$ be the [[kernel|9M0NNNFJGM]] of an [[integral transform|YCK709QB06]]\n$$(Tf)(x) = \\int_0^\\infty K(x, y)\\, f(y)\\, dy.$$\nLet $A$ be a class of functions on which $T$ is defined, such as the [[Lebesgue space|7KGF78FGEF]] $L^2(0, \\infty)$, with $T(A) \\subseteq A$.\n\n1. $K$ is *symmetric* if $K(x, y) = K(y, x)$ for all $x, y \u003e 0$.\n2. $K$ is *self-reciprocal* on $A$ (a *Fourier kernel*) if $T(Tf) = f$ for every $f \\in A$; that is, $g = Tf$ implies $f = Tg$.\n\n## Remarks\n\nExample: $K(x, y) = \\sqrt{2/\\pi}\\, \\cos(xy)$ is symmetric, since $\\cos(xy) = \\cos(yx)$.\nIt is self-reciprocal on $L^2(0, \\infty)$: this is the inversion theorem for the Fourier cosine transform, which follows from the Fourier inversion theorem applied to the even extension of $f$.\nThe sine kernel $\\sqrt{2/\\pi}\\, \\sin(xy)$ and the Hankel kernel $\\sqrt{xy}\\, J_\\nu(xy)$ for $\\nu \\ge -1/2$, with $J_\\nu$ the Bessel function of the first kind, are also symmetric and self-reciprocal.\nNon-example: the Laplace kernel $e^{-xy}$ is symmetric but not self-reciprocal.\nFor $f(y) = e^{-y}$, $(Tf)(x) = 1/(1 + x)$, and $T(Tf)(x) = \\int_0^\\infty e^{-xy}/(1 + y)\\, dy$ tends to $+\\infty$ as $x \\to 0^+$, while $f(x) \\to 1$.\n",
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