Timeline for Is there a singular function that is Hölder continuous of every order less than $1$?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| May 24 at 23:03 | vote | accept | Nate River | ||
| May 24 at 1:56 | answer | added | Saúl RM | timeline score: 6 | |
| May 23 at 21:55 | history | edited | Nate River | CC BY-SA 4.0 |
edited title
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| May 23 at 21:23 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 13 characters in body
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| May 23 at 21:23 | comment | added | Nate River | @ChristianRemling I actually don’t know how to construct Cantor-like sets other than the fat/Smith-Volterra Cantor sets. What is the usual procedure? | |
| May 23 at 21:02 | comment | added | Christian Remling | The Cantor function is Holder continuous with exponent related to the Hausdorff dimension of $C$, so I think one can take a measure zero set of dimension $1$ to produce such an example. | |
| May 23 at 20:23 | comment | added | Thomas Kojar | I suppose you can try to use standard embedding results for Sobolev spaces and Holder. | |
| May 23 at 20:19 | history | asked | Nate River | CC BY-SA 4.0 |