Timeline for Smooth approximation of Hölder functions "from below"
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Apr 17, 2023 at 13:19 | comment | added | António Borges Santos | @IosifPinelis please excuse the delay. I just wanted to wait a bit, as I thought fedja might have something to add. As you pointed out, your nice answer in that was rather intricate in the end and not as simple as the argument that fedja seemed to have initially thought about... | |
| Apr 17, 2023 at 13:19 | vote | accept | António Borges Santos | ||
| Apr 14, 2023 at 14:32 | comment | added | Iosif Pinelis | @AntónioBorgesSantos : Do you have a response to the answer below? | |
| Apr 12, 2023 at 17:51 | comment | added | Iosif Pinelis | @fedja : Looking at the answer below, it seems one needs to sort out small intervals, to avoid their clustering. Also, it seems one needs to sort not only the values of $f$ but also the values of its argument. | |
| Apr 12, 2023 at 11:37 | comment | added | Giorgio Metafune | @IosifPinelis No, sorry again, since I truncate with $f$ everything is lost...I mixed all things | |
| Apr 12, 2023 at 11:15 | comment | added | Giorgio Metafune | @IosifPinelis You are right...this is only Lipschitz approximation | |
| Apr 12, 2023 at 11:12 | comment | added | Iosif Pinelis | @GiorgioMetafune : But when you truncate, can't you lose the smoothness? | |
| Apr 12, 2023 at 11:10 | comment | added | Giorgio Metafune | @IosifPinelis Sorry for being unclear. Having $f_n$, use $(f \wedge f_n) \vee (-f)$. | |
| Apr 12, 2023 at 11:05 | comment | added | Iosif Pinelis | @GiorgioMetafune : Can you explain what you mean by "so on"? | |
| Apr 12, 2023 at 11:04 | comment | added | Iosif Pinelis | @fedja : I am afraid that where $f$ was $0$ (or very close to $0$) (and remained so after the composition with $G$) it might be not necessary that the (near) vanishing property is preserved after the convolution. | |
| Apr 12, 2023 at 10:44 | comment | added | Giorgio Metafune | As @fedja suggested or also: first approximate $f$ with smooth $f_n$ in $C^\gamma$ with $\beta<\gamma<\alpha$ and then take $f \wedge f_n$ and so on. I am using the fact that a Lipschitz map $F$ defines a continuous map $u \mapsto F(u)$ from $C^\gamma \to C^\beta$ (if you want $\beta=\gamma$ you need $F \in C^1$). You need also to write $f \wedge g=(f-g)^-+f$. | |
| Apr 12, 2023 at 5:31 | history | edited | Jukka Kohonen | CC BY-SA 4.0 |
Giving Hölder his dots
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| Apr 12, 2023 at 3:33 | answer | added | Iosif Pinelis | timeline score: 4 | |
| Apr 12, 2023 at 2:19 | comment | added | fedja | Looks like you can get away with first taking the composition $G\circ f$ where $G(x)=0$ of $[-\delta,\delta]$, $x-\delta$ for $x>\delta$ and $x+\delta$ for $x<-\delta$ and then convolving this composition with a smooth narrow bump, cannot you? | |
| Apr 11, 2023 at 20:25 | history | asked | António Borges Santos | CC BY-SA 4.0 |