Timeline for Is there a singular function that is Hölder continuous of every order less than $1$?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 24 at 23:04 | comment | added | Nate River | Just read it! Great solution, and very natural idea despite the messiness. | |
| May 24 at 23:03 | vote | accept | Nate River | ||
| May 24 at 2:59 | comment | added | Saúl RM | Great. I just adapted the construction of the Cantor step function to a Cantor set of dimension 1 but measure $0$ (I already had an idea of how to construct that due to the question the other day). The proof turned out a bit ugly, but that seems difficult to avoid due to the intrinsic ugliness of decimal representations. It would be nice if there is a more elegant construction though | |
| May 24 at 2:42 | comment | added | Nate River | Man I have absolutely no idea how you come up with these crazy constructions on the fly. I will read this and get back to you when I understand. | |
| May 24 at 1:56 | history | answered | Saúl RM | CC BY-SA 4.0 |