Timeline for Can a differentiable function be nowhere locally $\alpha$-Hölder for all $\alpha > 0$?
Current License: CC BY-SA 4.0
11 events
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| Jun 9 at 8:07 | comment | added | Pietro Majer | btw the existence of a nested sequence of intervals with unbounded incremental ratio is also what one gets assuming f′≤0 a.e. and f not decreasing. mathoverflow.net/questions/471848/… | |
| Jun 9 at 5:34 | comment | added | Pietro Majer | Yes, and multiplying everything by $dist(x,C)^2$ so that f'=0 on C. | |
| Jun 9 at 2:17 | comment | added | Nate River | @PietroMajer and of course on the complement, which is a union of open intervals, it is Lipschitz with fixed constant on each open interval. This is a pretty neat decomposition result. | |
| Jun 8 at 20:24 | comment | added | Saúl RM | @PietroMajer yes, that is a nice statement. I guess you can obtain such $f$ by introducing one small copy of the function $x^2\sin(1/x^3)$ in each of the extremes of the (countably many) gaps of the set $C$ | |
| Jun 8 at 20:12 | comment | added | Pietro Majer | So the set of points $x\in[0,1]$ such that $f’$ is not locally bounded at $x$ is a closed set with empty interior, and in fact any closed set with empty interior $C\subset[0,1]$ is the set of points $x\in[0,1]$ such that $f’$ is not locally bounded at $x$ for some everywhere differentiable $f:[01,\to\mathbb R$ | |
| Jun 8 at 14:31 | vote | accept | Nate River | ||
| Jun 8 at 14:29 | comment | added | Nate River | Ahh understood. | |
| Jun 8 at 14:20 | history | edited | Saúl RM | CC BY-SA 4.0 |
added 13 characters in body
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| Jun 8 at 14:19 | comment | added | Saúl RM | Indeed, $x^2\sin(1/x^3)$ has unbounded derivative near $0$. In case that is the confusion, by "locally unbounded" I mean "unbounded in every open interval" | |
| Jun 8 at 14:19 | comment | added | Nate River | Hmm, doesn’t $x^2 \sin (1/x^3)$ have unbounded derivative near 0? | |
| Jun 8 at 11:43 | history | answered | Saúl RM | CC BY-SA 4.0 |