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  "body": "\n## Definition\n\nLet $L^2(0, \\infty)$ be the space of square-[[integrable|DWJF7ABPR1]] functions on $(0, \\infty)$ with the norm $\\norm{f}_2 = \\left( \\int_0^\\infty \\abs{f(t)}^2 \\, dt \\right)^{1/2}$.\nFor $f \\in L^2(0, \\infty)$, its [[Laplace transform|05XT5AVJBK]] $\\mathcal{L} f(s) = \\int_0^\\infty f(t) e^{-s t} \\, dt$ is defined for real $s \u003e 0$, and $\\mathcal{L}$ is a bounded linear operator from $L^2(0, \\infty)$ to $L^2(0, \\infty)$.\nThe *real Laplace inversion problem* is: given $F \\in L^2(0, \\infty)$, find $f \\in L^2(0, \\infty)$ with $\\mathcal{L} f = F$ on $(0, \\infty)$.\n*Ill-posedness* of this problem means that it is not [[well posed|FZ64956KX2]] in the sense of Hadamard.\nHere the solution is unique when it exists, but the inverse map $\\mathcal{L} f \\mapsto f$ is not continuous.\nThat is, there are $f_n$ with $\\norm{f_n}_2$ bounded away from $0$ and $\\norm{\\mathcal{L} f_n}_2 \\to 0$.\n\n## Remarks\n\nUniqueness holds by [[Lerch's uniqueness theorem|V3YTF5WJHY]].\nInstability, with the check: let $f_n(t) = e^{-t} \\sin(n t)$.\nThen $\\norm{f_n}_2^2 = \\int_0^\\infty e^{-2t} \\sin^2(n t) \\, dt = \\frac{n^2}{4(1 + n^2)} \\ge \\frac{1}{8}$ for $n \\ge 1$.\nIts transform is $\\mathcal{L} f_n(s) = \\frac{n}{(s + 1)^2 + n^2}$, and $\\norm{\\mathcal{L} f_n}_2^2 \\le \\int_{-\\infty}^{\\infty} \\frac{n^2}{(u^2 + n^2)^2} \\, du = \\frac{\\pi}{2 n} \\to 0$.\nSo no estimate $\\norm{f}_2 \\le C \\norm{\\mathcal{L} f}_2$ can hold.\nExistence also fails for most data: the range of $\\mathcal{L}$ consists of restrictions of holomorphic functions, so a generic $F \\in L^2(0, \\infty)$ is not a transform at all.\nContrast: inversion from values of $F$ on a vertical line in the complex plane, by the Bromwich integral, is a Fourier inversion and is stable in $L^2$.\n",
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