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May 14, 2018 at 17:02 history edited Robert Israel CC BY-SA 4.0
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May 14, 2018 at 16:14 comment added Yemon Choi @MeisamSoleimaniMalekan No, it doesn't, but that is not what Robert is claiming. $\widehat{f}$ is a bounded continuous function on $\widehat{G}$, since $\Vert f \Vert_1 < \infty$.
May 14, 2018 at 10:40 comment added MSMalekan I do not understand the last line of your answer. Does Fourier transform image $C_c(G)$ on $C_c(\hat G)$?
May 14, 2018 at 8:28 history edited Robert Israel CC BY-SA 4.0
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May 14, 2018 at 8:14 history answered Robert Israel CC BY-SA 4.0