Timeline for Does locally nilpotent imply nilpotent for continuous self-maps of intervals?
Current License: CC BY-SA 4.0
19 events
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| Nov 14, 2022 at 16:14 | comment | added | Dattier | Dédicace à Hybridex..... | |
| Sep 15, 2022 at 8:05 | history | edited | YCor |
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| Sep 15, 2022 at 6:58 | vote | accept | Dattier | ||
| Sep 14, 2022 at 22:33 | comment | added | Yemon Choi | @AlessandroDellaCorte The problem you link to uses the Baire Category Theorem in its solution, and to me that is the only resemblance/relevance. One might as well say that this is related to the Open Mapping Theorem for linear maps between Banach spaces, because that uses BCT... | |
| Sep 14, 2022 at 22:28 | answer | added | Christian Remling | timeline score: 7 | |
| Sep 14, 2022 at 22:21 | comment | added | Jochen Glueck | If $f$ is increasing on an interval $[0,d]$ for some $d > 0$, then due to $f(x) \le x$ for all $x \in [0,1]$ it follows that $f^n(x) \le f^n(d)$ for all $n \ge 0$ and all $x\in [0,d]$. So there exists $k$ such that $f^k = 0$ on $[0,d]$, and thus one can conclude that $f$ is nilpotent by the argument in @ChristianRemling's comment. So when trying to construct a counterexample, one apparently needs quite ugly behaviour of $f$ close to $0$. | |
| Sep 14, 2022 at 21:06 | history | edited | Sam Hopkins | CC BY-SA 4.0 |
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| Sep 14, 2022 at 21:05 | comment | added | Alessandro Della Corte | Lazy comment: I'm pretty sure that this is somewhat related: mathoverflow.net/q/34059/167834 | |
| Sep 14, 2022 at 20:33 | comment | added | Christian Remling | @NeilStrickland: I had similar thoughts, but I'm not sure we're getting much mileage out of an interval $(a,b)$ with $a>0$ and $f=0$ on $(a,b)$. The question might be if necessarily $f=0$ on $[0,d]$ for some $d>0$. (That would imply the claim immediately because we get closer to this set by a fixed amount during each iteration.) | |
| Sep 14, 2022 at 20:30 | history | edited | YCor | CC BY-SA 4.0 |
formatting, added tag
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| Sep 14, 2022 at 20:29 | comment | added | Neil Strickland | After replacing $f$ by $f^k$ for some $k$, we can assume that $f(0)=f(1)=0$. Now $f$ cannot have any fixed points apart from $x=0$, so we can apply the intermediate value theorem to $f(x)-x$ to see that $f(x)\leq x$ for all $x$, with equality only when $x=0$. The sets $Z_m=(f^m)^{-1}\{0\}$ are closed and their union is $[0,1]$, so some $Z_m$ must have nonempty interior by the Baire Category Theorem. I am not sure how much that helps. | |
| Sep 14, 2022 at 20:06 | comment | added | Christian Remling | @Malkoun: Yes, I misread the desired statement as $f\circ f=f$. | |
| Sep 14, 2022 at 19:52 | comment | added | Nicolast | @Christian: I don't understand your counterexample: doesn't it satisfy precisely $f\circ f=0$? | |
| Sep 14, 2022 at 19:48 | comment | added | Malkoun | I am confused. @ChristianRemling, isn't $f \circ f = 0$ in your example? I am probably tired and not thinking right... | |
| Sep 14, 2022 at 19:47 | comment | added | Christian Remling | What is true is that necessarily $f(0)=0$, by Sarkovski's theorem, because otherwise there would be other periodic orbits which obviously will not visit $x=0$. | |
| Sep 14, 2022 at 17:32 | comment | added | mathworker21 | why can't you just have shrinking triangles to $0$? | |
| Sep 14, 2022 at 17:09 | history | edited | Dattier | CC BY-SA 4.0 |
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| Sep 14, 2022 at 16:57 | review | Close votes | |||
| Sep 14, 2022 at 21:08 | |||||
| Sep 14, 2022 at 16:31 | history | asked | Dattier | CC BY-SA 4.0 |