Timeline for Functional inequality?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 9, 2021 at 16:53 | comment | added | Ankoganit | Ah, true, I missed that completely. Shame, I was really hoping your approach is salvageable. | |
| Feb 9, 2021 at 16:52 | comment | added | nonuser | $g(y)=0$ only for $y<-1$ and not all $y$. | |
| Feb 9, 2021 at 16:51 | comment | added | Ankoganit | @Greedoid hm, what's wrong with it now? | |
| Feb 9, 2021 at 14:32 | comment | added | nonuser | It is still not working :( | |
| Feb 9, 2021 at 3:00 | comment | added | Ankoganit | @Greedoid I think the vector space argument can be skipped altogether. If f is a solution, then so is g(x)=f(x)-f(0)+f(0)x, and g satisfies g(0)=0. Then the rest of your very neat reasoning applies. | |
| Feb 8, 2021 at 17:45 | comment | added | nonuser | Do you think my method with Vector space can be saved? | |
| Feb 8, 2021 at 6:09 | comment | added | Ankoganit | The original proof had a small hole (while bounding $yf(x)$, I implicitly assumed $y$ is positive). I think it's fixed now. | |
| Feb 8, 2021 at 6:08 | history | edited | Ankoganit | CC BY-SA 4.0 |
added 224 characters in body
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| Feb 5, 2021 at 13:16 | vote | accept | Culver Kwan | ||
| Feb 5, 2021 at 13:16 | comment | added | Culver Kwan | Nice solution! (Perhaps mines is more complex than this) Checkmark incoming! | |
| Feb 5, 2021 at 12:48 | history | answered | Ankoganit | CC BY-SA 4.0 |