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  "body": "\n## Definition\n\nLet $V$ be a complex vector space of functions, and let $\\Omega$ be a set.\nLet $T$ be an [[integral transform|YCK709QB06]], a linear map from $V$ to the space of complex functions on $\\Omega$.\nA *convolution* of $T$ is a bilinear map $\\circledast \\colon V \\times V \\to V$ such that\n$$T(f \\circledast g)(\\omega) = (Tf)(\\omega) \\, (Tg)(\\omega)$$\nfor all $f, g \\in V$ and all $\\omega \\in \\Omega$.\n\n## Remarks\n\nThe identity $T(f \\circledast g) = Tf \\cdot Tg$ is called the factorization identity or convolution theorem of $T$.\nIf $T$ is injective, then $T$ has at most one convolution, namely $f \\circledast g = T^{-1}(Tf \\cdot Tg)$, and then $\\circledast$ is automatically commutative and associative.\nVariants allow a fixed weight function $\\gamma$ on $\\Omega$, with $T(f \\circledast g) = \\gamma \\cdot Tf \\cdot Tg$, or mix several transforms.\n\nExample: for the Mellin transform $(\\mathcal{M} f)(s) = \\int_0^\\infty x^{s-1} f(x) \\, dx$, the [[Mellin convolution|A2PMJG7J68]]\n$$(f \\circledast g)(x) = \\int_0^\\infty f(x/y) \\, g(y) \\, \\frac{dy}{y}$$\nis a convolution.\nIndeed, the substitution $x = uy$ gives $\\int_0^\\infty x^{s-1} \\int_0^\\infty f(x/y) g(y) \\frac{dy}{y} dx = \\int_0^\\infty u^{s-1} f(u) \\, du \\int_0^\\infty y^{s-1} g(y) \\, dy$ when the integrals converge absolutely.\nThe [[convolution on the line|ERP5737VZ1]] is a convolution of the Fourier transform, and the [[convolution on the half-line|FCKX8KFVGN]] is one of the Laplace transform.\n\nNon-example: the pointwise product $fg$ is not a convolution of the Fourier transform $\\hat f(\\xi) = \\int_\\R f(x) e^{-i x \\xi} \\, dx$ on $L^1(\\R) \\cap L^2(\\R)$.\nFor $f = g = \\mathbf{1}_{[0,1]}$, the product is $\\mathbf{1}_{[0,1]}$ with $\\hat{f}(0) = 1$, but the factorization identity would need $\\widehat{fg}(\\xi) = \\hat f(\\xi)^2$, and $\\hat f(\\xi)^2 \\ne \\hat f(\\xi)$ at $\\xi = \\pi$, where $\\hat f(\\pi) = 2/(i\\pi) \\ne 0$.\n",
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