Timeline for An square root of the multiplicative operator on $\ell^1(\mathbb{Z}_n)$
Current License: CC BY-SA 4.0
3 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 16, 2022 at 12:47 | comment | added | Gro-Tsen | I suppose $\mathbb{Z}_n$ means $\mathbb{Z}/n\mathbb{Z}$, not $n$-adics. But then why do you write $\ell^1(\mathbb{Z}_n)$ for what I suppose is just $\mathbb{C}^{\mathbb{Z}/n\mathbb{Z}}$ (the $1$ is purely irrelevant, right?)? Doesn't Fourier analysis immediately show the answer to be “convolution by the inverse Fourier transform of any square root of the Fourier transform of $x$”? Is there some subtlety that I missed? | |
| Jun 16, 2022 at 12:35 | history | edited | ABB |
edited tags
|
|
| Jun 16, 2022 at 11:47 | history | asked | ABB | CC BY-SA 4.0 |