Timeline for Smooth approximation of Hölder functions "from below"
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 17, 2023 at 13:19 | vote | accept | António Borges Santos | ||
| Apr 13, 2023 at 13:57 | comment | added | Iosif Pinelis | @GiorgioMetafune : Nice argument! | |
| Apr 13, 2023 at 7:03 | comment | added | Giorgio Metafune | For continuous $f \geq 0$ this simple construction works. If $M$ is the maximum of $f$, $K=\{f \geq M/2\}$, $V=\{f >M/4\}$ let $g$ be smooth with compact support in $V$ with $0 \leq g \leq M/4$ and $g=M/4$ in $K$. Then $0 \leq g \leq f$ and $f-g \leq (3/4)M$. Iterating this argument one finds $0 \leq f-(g_1+\dots g_n) \leq (3/4)^n M$ (in the previous comment I forgot to say that $f-g$ should be small). | |
| Apr 12, 2023 at 18:49 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
edited body
|
| Apr 12, 2023 at 18:15 | comment | added | Iosif Pinelis | @GiorgioMetafune : I don't think that this proof implies the result you mention, for continuous (rather than Hölder-continuous) functions. However, I think the just-continuous case should be somewhat simpler. | |
| Apr 12, 2023 at 17:49 | comment | added | Giorgio Metafune | Does your proof give that for any continuous and positive $f$, there exists a smooth $g$ such that $0 \leq g \leq f$ poitwise? Also that does not look obvious to me (unless $f$ is strictly positive). | |
| Apr 12, 2023 at 16:22 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 84 characters in body
|
| Apr 12, 2023 at 16:03 | history | undeleted | Iosif Pinelis | ||
| Apr 12, 2023 at 16:03 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 1727 characters in body
|
| Apr 12, 2023 at 7:11 | history | deleted | Iosif Pinelis | via Vote | |
| Apr 12, 2023 at 3:33 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |