Timeline for When is $|\int_a^b \exp(-izx)f(x) \, \mathrm{d}x| \leq |\int_a^b f(x) \, \mathrm{d}x|$ for a general $f$? [closed]
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 8, 2017 at 20:39 | review | Reopen votes | |||
| Aug 8, 2017 at 21:58 | |||||
| Aug 8, 2017 at 20:20 | history | edited | AD500712838 | CC BY-SA 3.0 |
added 5 characters in body
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| Aug 8, 2017 at 20:20 | comment | added | AD500712838 | Thank you for pointing that out. It is assumed that $z$ is real. I edited the question. | |
| Aug 8, 2017 at 20:12 | history | closed |
Michael Renardy Henry.L Yoav Kallus Peter Humphries Stefan Kohl♦ |
Needs details or clarity | |
| Aug 8, 2017 at 19:02 | comment | added | Gerald Edgar | You say "all $z$" ... do you include complex $z$? I think you need the restriction $z$ real even for the case where $f(z) \ge 0$. | |
| Aug 8, 2017 at 18:59 | review | Close votes | |||
| Aug 8, 2017 at 20:12 | |||||
| Aug 8, 2017 at 18:28 | comment | added | Yemon Choi | Basically, if you want something like this to be true, you need a lot more control over $f$ | |
| Aug 8, 2017 at 18:26 | comment | added | Yemon Choi | How general do you want $f$ to be? I mean, any $f$ which has mean zero over the interval is very unlikely to satisfy this inequality | |
| Aug 8, 2017 at 18:04 | history | asked | AD500712838 | CC BY-SA 3.0 |