Timeline for Root of positive function in Fourier algebra
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 5, 2015 at 9:16 | vote | accept | Hannes Thiel | ||
| Mar 4, 2015 at 22:18 | answer | added | Yemon Choi | timeline score: 5 | |
| Mar 4, 2015 at 17:24 | answer | added | Christian Remling | timeline score: 1 | |
| Mar 4, 2015 at 16:21 | comment | added | Hannes Thiel | @NoamD.Elkies: Yes, that's what I mean by $C_0(\mathbb{R})$. | |
| Mar 4, 2015 at 16:18 | comment | added | Noam D. Elkies | Thanks for this eplanation/translation. So $C_0({\mathbb R})$ means continuous and vanishing at infinity? | |
| Mar 4, 2015 at 15:36 | comment | added | Hannes Thiel | @Noam: For $G=\mathbb{R}$, my question comes down to: If $f\in C_0(\mathbb{R})$ is the Fourier transform of some function in $L^1(\mathbb{R})$, is then $|f|$ the Fourier transform of some other function in $L^1(\mathbb{R})$? | |
| Mar 4, 2015 at 15:34 | comment | added | Hannes Thiel | Yes, for every locally compact abelian group $G$, the Fourier algebra $A(G)$ is naturally isometrically isomorphic to $L^1(\widehat{G})$ via the Fourier transform. In particular, since $\widehat{\mathbb{R}}\cong\mathbb{R}$, we have $A(\mathbb{R})=\{\widehat{f} : f\in L^1(\mathbb{R})\}$, with the norm coming from $L^1(\mathbb{R})$. | |
| Mar 4, 2015 at 15:04 | comment | added | Noam D. Elkies | That definition is what I'm not sure of. Are you saying then that the two $F$'s (sorry for using the same letter above; it's too late now to fix) are to be $L^1$? | |
| Mar 4, 2015 at 14:56 | comment | added | Noam D. Elkies | So for example when $G = \bf R$ you're hoping that if $f$ is the Fourier transform of a continuous function $F$ (with $F(x) \rightarrow 0$ as $x \rightarrow \pm \infty$) then $|\,f\,|$ is also the Fourier transform of some continuous function $F\,$? | |
| Mar 4, 2015 at 14:26 | history | asked | Hannes Thiel | CC BY-SA 3.0 |