Timeline for Prove that ..., f(x-2), f(x-1), f(x), f(x+1), f(x+2),... is algebraically linearly independent without the Fourier transform
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 31, 2014 at 15:48 | vote | accept | Peter Luthy | ||
| Jul 29, 2014 at 21:34 | comment | added | Aaron Meyerowitz | I wonder if anything could be done with the bi-infinite sequence $\langle a \rangle$ defined by $a_i=\sqrt{\int_i^{i+1}f^2(x)dx}$ so $|| f ||=\sum_{-\infty}^{\infty}a_i^2.$ Dependence of shifts of $f$ do not correspond to dependence of shifts of $\langle a \rangle$ (i.e. a linear recurrence) but maybe some inequality can be brough to bear. | |
| Jul 29, 2014 at 19:05 | comment | added | bartgol | Oh, of course. I read the question a little too fast. Silly me. =P | |
| Jul 29, 2014 at 18:56 | comment | added | J. E. Pascoe | Bartgol, he's talking about the functions $f(x-k) \in L^2,$ not the values of $f$ at the integers which wouldn't be well defined anyway. | |
| Jul 29, 2014 at 18:51 | comment | added | bartgol | Actually, I'm pretty sure I can build a function $f\in L^2$ that is equal to 1 on every integer. You just need to make the spikes narrower and narrower as you get to infinity. That function would not be in $H^1$, of course, but that's another story. | |
| Jul 29, 2014 at 18:44 | comment | added | bartgol | That set is not linearly independent (what if $f$ is odd?). Perhaps you meant that there is an infinite subset of $V$ that is linearly independent? | |
| Jul 29, 2014 at 17:56 | answer | added | J. E. Pascoe | timeline score: 11 | |
| Jul 29, 2014 at 17:38 | history | asked | Peter Luthy | CC BY-SA 3.0 |