Timeline for If the convolution of two functions $f\star g$ is equal to $g$, $f$ is even with compact support and $g$ is bounded, implies that $g$ is constant?

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May 26, 2021 at 20:12	comment	added	Alex M.		In order to use the machinery of the Fourier transform and deduce that $\hat f \hat g = \hat g$ from $f \star g = g$, you tacitly assume that $g$ is a function that admits a Fourier transform which is itself a function. But this excludes, among others, precisely the constant functions (the Fourier transforms of which are Dirac distributions).
Apr 25, 2020 at 5:06	vote	accept	Anton Sorokovskiy
Apr 24, 2020 at 22:39	comment	added	LSpice		Per Fourier inversion, we could take for $f$ the extension by $0$ of the function $x \mapsto \frac1{2\pi}(1 + 2\cos(x) + \cos(2x))$ on $[-\pi, \pi]$.
Apr 24, 2020 at 20:54	history	answered	Bruno Le Floch	CC BY-SA 4.0