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  "body": "\n## Definition\n\nLet $k \\in L^1(\\R)$, let $f \\in L^2(\\R)$, and let $\\lambda \\in \\C$.\nWrite $\\hat g(\\xi) = \\int_{-\\infty}^{\\infty} g(x) e^{-i x \\xi} \\, dx$ for the [[Fourier transform|DPVM34ADZV]] of $g$.\nThe *Fourier transform method* for the [[convolution integral equation|EY75YD988Z]]\n$$\\varphi(x) - \\lambda \\int_{-\\infty}^{\\infty} k(x - t) \\, \\varphi(t) \\, dt = f(x), \\quad x \\in \\R,$$\nwith unknown $\\varphi \\in L^2(\\R)$ consists of the following steps.\n\n1. Apply the Fourier transform to both sides; by the convolution theorem the equation becomes $(1 - \\lambda \\hat k(\\xi)) \\, \\hat\\varphi(\\xi) = \\hat f(\\xi)$ for almost every $\\xi \\in \\R$.\n2. Solve for $\\hat\\varphi(\\xi) = \\hat f(\\xi) / (1 - \\lambda \\hat k(\\xi))$.\n3. Recover $\\varphi$ from $\\hat\\varphi$ with the inverse Fourier transform.\n\nThe function $\\xi \\mapsto 1 - \\lambda \\hat k(\\xi)$ is the *symbol* of the equation.\n\n## Remarks\n\nIf $1 - \\lambda \\hat k(\\xi) \\ne 0$ for every $\\xi \\in \\R$, the method gives the unique solution in $L^2(\\R)$.\nIndeed $\\hat k$ is continuous and tends to $0$ at infinity by the [[Riemann–Lebesgue lemma|WVKDC011KP]], so $\\abs{1 - \\lambda \\hat k}$ is bounded below by a positive constant.\nThen $\\hat f / (1 - \\lambda \\hat k)$ lies in $L^2(\\R)$, and the [[Plancherel theorem|HPZHRZ0502]] makes every step reversible.\n\nExample: $k(s) = e^{-\\abs{s}}$, $\\lambda = \\frac14$, $f(x) = \\frac12 e^{-\\abs{x}}$.\nHere $\\hat k(\\xi) = \\frac{2}{1 + \\xi^2}$, the symbol $1 - \\frac{1}{2(1 + \\xi^2)}$ is at least $\\frac12$, and step 2 gives $\\hat\\varphi(\\xi) = \\frac{1}{1/2 + \\xi^2}$, so $\\varphi(x) = \\frac{1}{\\sqrt 2} e^{-\\abs{x}/\\sqrt 2}$.\n\nNon-example: for $k(s) = \\frac12 e^{-\\abs{s}}$ and $\\lambda = 1$ the symbol $\\frac{\\xi^2}{1 + \\xi^2}$ vanishes at $\\xi = 0$.\nStep 2 then gives a function $\\hat f(\\xi) (1 + \\xi^2)/\\xi^2$ that need not be in $L^2(\\R)$, so the method does not produce a solution for every $f \\in L^2(\\R)$.\n\nThe method needs the integral to run over the whole group $\\R$.\nOn a half-line one gets a [[Wiener–Hopf equation|QYY4NMDY5H]], which needs the [[Wiener–Hopf method|H4NXEY6QTZ]] instead.\nFor equations on $[0, \\infty)$ of Volterra type, the [[Laplace transform method|EHGECREYB4]] plays the same role.\n",
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