Timeline for Is the composition of two nowhere differentiable functions still nowhere differentiable?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 4, 2019 at 2:59 | vote | accept | Liding Yao | ||
| May 27, 2019 at 18:05 | answer | added | Kostya_I | timeline score: 11 | |
| S May 27, 2019 at 16:57 | history | edited | David White | CC BY-SA 4.0 |
tried to improve spelling and formatting
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| S May 27, 2019 at 16:57 | history | suggested | Manfred Sauter | CC BY-SA 4.0 |
tried to improve spelling and formatting
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| May 27, 2019 at 16:42 | review | Suggested edits | |||
| S May 27, 2019 at 16:57 | |||||
| May 27, 2019 at 16:02 | comment | added | Manfred Sauter | A comment on your idea for a proof. The graph of $f$ is not purely unrectifiable. More precisely, a substantial part of the graph is contained in the graph of a $1$-Lipschitz function defined on the quadrant bisector. I would expect that therefore (using Lebesgue differentiation) there exists a set $E\subset\mathbb{R}$ of positive measure and $k>0$ such that for all $x_0\in E$ the set $\mathbb{R}\setminus I^k_{f,x_0}$ has positive lower density at $x_0$. So to me it seems that your idea might not work. | |
| May 26, 2019 at 8:56 | history | edited | Liding Yao | CC BY-SA 4.0 |
added 15 characters in body
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| May 26, 2019 at 8:55 | answer | added | Dominique Unruh | timeline score: 8 | |
| May 26, 2019 at 8:43 | history | asked | Liding Yao | CC BY-SA 4.0 |