Timeline for For every monotonically-decreasing non-negative function $ f $, does there exist a function $ g $ so that $ f g $ is integrable? [closed]
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| S Jun 14, 2017 at 21:13 | history | suggested | LSpice | CC BY-SA 3.0 |
"decreasing non-decreasing" should be "decreasing non-negative"
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| Jun 14, 2017 at 20:58 | review | Suggested edits | |||
| S Jun 14, 2017 at 21:13 | |||||
| Apr 20, 2017 at 15:55 | vote | accept | Spinorbundle | ||
| Apr 20, 2017 at 2:25 | review | Reopen votes | |||
| Apr 20, 2017 at 8:17 | |||||
| S Apr 20, 2017 at 2:05 | history | suggested | Transcendental | CC BY-SA 3.0 |
Enhanced clarity of title and improved formatting.
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| Apr 20, 2017 at 1:43 | review | Suggested edits | |||
| S Apr 20, 2017 at 2:05 | |||||
| Apr 20, 2017 at 1:41 | history | closed |
Christian Remling Mikael de la Salle paul garrett R W Michael Renardy |
Not suitable for this site | |
| Apr 20, 2017 at 0:42 | answer | added | Hicham | timeline score: 0 | |
| Apr 19, 2017 at 21:10 | comment | added | Igor Rivin | A remark. If all you want is for the integral of $g$ not to converge, you just make $g([k, k+1]) = (-1)^k.$ | |
| Apr 19, 2017 at 20:42 | comment | added | paul garrett | A useful example, but/and better on Math Stack Exchange. | |
| Apr 19, 2017 at 20:33 | answer | added | Wlodek Kuperberg | timeline score: 10 | |
| Apr 19, 2017 at 18:15 | review | Close votes | |||
| Apr 19, 2017 at 22:29 | |||||
| Apr 19, 2017 at 17:59 | answer | added | Will Sawin | timeline score: 4 | |
| Apr 19, 2017 at 17:25 | history | asked | Spinorbundle | CC BY-SA 3.0 |