The Fourier transform of convolution is: $F (f \ast g)= F (f) \cdot F(g)$, where $F$ represents Fourier map. I wonder if there exists similar statement for inverse Fourier map i.e. does it hold that $F ^{-1}(f \ast g)= F^{-1} (f) \cdot F^{-1}(g)$. I haven't found anywhere such statement (so i think that is not true), but I would like to be sure that it doesn't holds. Thanks a lot in advance!
2026-04-09 16:57:08.1775753828
Does there exist Convolution theorem for Inverse Fourier map?
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With your convention, we have $F^{-1}(f)(x) = {1\over 2\pi}F(f)(-x)$. As a result, $$ F^{-1}(f*g)(x) = {1 \over 2\pi} F(f*g)(-x) = {1 \over 2\pi}\ F(f)(-x)\ F(g)(-x) = 2\pi\ F^{-1}(f)(x)\ F^{-1}(g)(x). $$
We end up with $$ F^{-1}(f*g) = 2\pi\ F^{-1}(f)\ F^{-1}(g). $$