Timeline for Differential equation changing sign almost everywhere
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 14, 2018 at 7:17 | comment | added | Our | @MateuszKwaśnicki Ok, I was confused because I thought you were using the fact that it takes the values $-1,1$ almost everywhere somewhere else. Thanks for the response, by the way. | |
| Aug 13, 2018 at 22:46 | comment | added | Mateusz Kwaśnicki | @onurcanbektas: I am sorry, I could not figure out which part of the proof is puzzling you. If $g'$ is constant everywhere, then $f(x) = \pm 1 + k e^x$ everywhere. Otherwise, $g'$ takes at least two values, and Clarkson's result allows us to conclude that $g'$ cannot belong to $\{-1, 1\}$ a.e. | |
| Aug 13, 2018 at 10:52 | comment | added | Our | @MateuszKwaśnicki Sir, I'm little confused; if $g(x) \in \{-1, 1\}$ almost everywhere, then $f(x) = \pm 1 + ke^x$ almost everywhere, so with the initial condition, $f(x) = \pm 1$ almost everywhere; but this is valid not all of $f's$ domain, so $f$ does not have to be a constant function (maybe constant almost everywhere ?) | |
| Jul 26, 2018 at 21:02 | comment | added | Paul | The conjecture becomes false if on assuming if we assume f continous everywhere and differentiable almost everywhere | |
| Jul 24, 2018 at 8:44 | comment | added | Paul | To avoid this problem, one can use a direct reasoning: With the result of J.A. Clarkson, one can prove if $g :\mathbb{R}→\mathbb{R}$ everywhere differentiable, and g′(x)∈{−1,1} almost everywhere then $g '$ must be constant. we apply it to $g = \int f + f$ and thus $g '= f + f'$ is constant ie $f + f '= 1$ everywhere or $f + f' = -1$everywhere and so $f (x) = ke ^ {- x} \pm 1$ but condition $f'(0)=0$ implie $k=0$ so f is the constant function 1 or the constant function -1 | |
| Jul 24, 2018 at 8:43 | comment | added | Mateusz Kwaśnicki | @MeisamSoleimaniMalekan: I am sorry, I did not get your point. Can you add some details? | |
| Jul 24, 2018 at 8:42 | comment | added | Mateusz Kwaśnicki | @Tina: Thanks, I updated the answer and added some details. | |
| Jul 24, 2018 at 8:41 | history | edited | Mateusz Kwaśnicki | CC BY-SA 4.0 |
added 400 characters in body
|
| Jul 24, 2018 at 2:21 | comment | added | MSMalekan | The set $E(-1,1)+f(0)$ should be considered instead of $E(-1,1)$. This set is non empty because of your condition $f'(0)=0$, and not by virtue of Darboux's theorem. | |
| Jul 23, 2018 at 23:24 | vote | accept | Paul | ||
| Jul 23, 2018 at 23:19 | comment | added | Paul | @ Mateusz Kwaśnicki Thank you for ur answer and especially the result of J.A. Clarkson. I have a small remark about your reasoning, it does not explain the relevance of the hypothesis f '(0) = 0 because without it the conjecture is wrong | |
| S Jul 23, 2018 at 15:57 | history | suggested | psmears | CC BY-SA 4.0 |
Improve wording and grammar
|
| Jul 23, 2018 at 14:48 | review | Suggested edits | |||
| S Jul 23, 2018 at 15:57 | |||||
| Jul 23, 2018 at 12:38 | history | answered | Mateusz Kwaśnicki | CC BY-SA 4.0 |