Timeline for Which functions $f: \mathbb{R} \to \mathbb{R}$ is injective over some subinterval of $(x,y)$ whenever $x<y$ and $f(x) \ne f(y)$?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| May 9, 2017 at 11:32 | comment | added | Aditya Guha Roy | For the step function see that it is (nearly trivial) devil that prevents us from having differentiability 'the-almost' everywhere, though yes I think the total differentiability may do a clean work. Also if anyone has a proof about any claims on differentiable functions / absolutely continuous functions etc . , you can post it here. | |
| May 9, 2017 at 11:29 | comment | added | Aditya Guha Roy | Ok if you (anyone) suspects the source : I say I never read it being true. I actually was thinking about it because I felt that if this was true for certain (though peculiarly) general functions then many things were true for many general categories of functions (one such thing is : mathoverflow.net/questions/269064/… ) , however unfortunately it is not so general that it holds with the sufficiency of continuity, but I don't know what to tell about absolute continuity. | |
| May 7, 2017 at 13:54 | comment | added | Aditya Guha Roy | In general, I think absolute continuity is enough, though it may not be the strongest and the best solution to it. Also it may have some lacks. | |
| May 7, 2017 at 13:53 | comment | added | Aditya Guha Roy | @ClaudioGorodski I think differentiability over the whole domain is necessary (just think about the step functions). | |
| May 7, 2017 at 12:57 | comment | added | Claudio Gorodski | Differentiablity is of course enough. | |
| May 7, 2017 at 3:59 | comment | added | Aditya Guha Roy | Unfortunately continuity is not enough : (The devil here is) the Weierstrass function. | |
| May 7, 2017 at 3:56 | history | asked | Aditya Guha Roy | CC BY-SA 3.0 |