Timeline for Averaged Parseval Relation for Sampling a Function on Integers
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 25, 2019 at 22:48 | vote | accept | kodlu | ||
| Jan 25, 2019 at 19:45 | answer | added | Willie Wong | timeline score: 1 | |
| Jan 25, 2019 at 18:01 | comment | added | kodlu | @WillieWong I would appreciate it if you wrote out the resulting estimate as an answer. Thanks. | |
| Jan 25, 2019 at 16:28 | comment | added | Willie Wong | but that doesn't matter for the $L^2$ estimates. | |
| Jan 24, 2019 at 22:43 | comment | added | kodlu | Sorry $|f|$ is almost constant, but $f$ is not. | |
| Jan 24, 2019 at 22:27 | comment | added | Willie Wong | If you assume $f$ is almost constant in magnitude, then pretty much all the $f(n)$ terms that appear in the summation can be replaced by $1$, no? Then the "combed" sum would be basically $\sum_{v = 1}^m N/v$ which shouldn't be too hard to estimate from below. | |
| Jan 24, 2019 at 20:13 | history | edited | kodlu | CC BY-SA 4.0 |
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| Jan 24, 2019 at 20:09 | comment | added | kodlu | We can assume $f$ takes values on the unit circle and even that $supp~ f=[1,N].$ | |
| Jan 24, 2019 at 15:55 | history | edited | Willie Wong |
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| Jan 24, 2019 at 15:53 | comment | added | Willie Wong | Isn't there a problem with your derivation near the end? The $L^2$ norm of the convolution $\widehat{\Phi_v} * \widehat{f}$ is not the integral you wrote. For $m \ll N$, if you take $f$ to be supported only at $f(1) = 1$ and $f(\mathrm{lcm}(\{1, \ldots, m\})+1) = 1$ and zero otherwise. The trivial lower bound of $A = 1$ seems to be sharp. So presumably you are more interested in when $m$ is large enough that $f_v$ has to be mostly non-trivial? | |
| Jan 24, 2019 at 10:07 | history | edited | kodlu | CC BY-SA 4.0 |
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| Jan 24, 2019 at 10:02 | history | asked | kodlu | CC BY-SA 4.0 |