Timeline for Ekeland's standardness-property inheritable?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 21, 2022 at 17:44 | comment | added | Jochen Wengenroth | Thank you for this clarification. | |
| Sep 21, 2022 at 16:13 | history | edited | TaQ | CC BY-SA 4.0 |
just minor correction
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| Sep 21, 2022 at 16:10 | comment | added | TaQ | No, I consider the (polynomial) map $f:E\supseteq{\rm dom}\,f\to F=E$ given by $x\mapsto x+x^2$ for which I have not given the specific name "$f$". Here ${\rm dom}\,f$ is given by $\|x\|_0<\frac 14$ and $L:{\rm dom}\,f\to\mathcal L\,(F,E)$ is the family of (right) inverses for the derivative of $f$ occurring in Ekeland's theorem. It it rather "well-known" that this kind of maps (and even much more general ones) satisfy the Nash−Moser conditions. Does this clarify the state of matters? | |
| Sep 21, 2022 at 6:47 | comment | added | Jochen Wengenroth | Sorry, I don't get it. First you considered a fixed polynomial $x(s)=s+s^2$ (since you wrote $\|x\|_0=1/4$,it might rather be $s-s^2$?) and I first thought that the map $L:E\to E$ would be be $v\mapsto v/(1+2x)$, but then you plug in other functions $x$. What is the map $L:E\to E$, then? Do you claim that Nash-Moser applies to $L$ but Ekeland does not? | |
| Sep 21, 2022 at 5:09 | history | answered | TaQ | CC BY-SA 4.0 |