Timeline for Could we assume without loss of generality that all coefficients are positive?
Current License: CC BY-SA 4.0
5 events
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| May 28, 2019 at 13:56 | comment | added | Mikael de la Salle | Because the sum of the coefficients (=Fourier coefficients here) of $\alpha |\beta|^2$ is the value of $\alpha |\beta|^2$ at $z=1$, so is $0$. | |
| May 28, 2019 at 13:40 | comment | added | Mikael de la Salle | Seen as a function on the circle, $\alpha(z) = |z-1|^2$. Or any positive trigonometric polynomial vanishing at $z=1$. | |
| May 28, 2019 at 13:38 | comment | added | MSMalekan | @MikaeldelaSalle: Could you please give me one of such counterexamples? | |
| May 28, 2019 at 13:23 | comment | added | Mikael de la Salle | When $G$ is commutative (for example $\mathbf Z$), you have the Fourier transform that translates your question to an elementary question on functions on the circle. You will easily find counterexamples in that case. | |
| May 28, 2019 at 12:27 | history | asked | MSMalekan | CC BY-SA 4.0 |