Timeline for Lipschitz min implies Lipschitzian argmin?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 22, 2019 at 17:51 | comment | added | Dirk | And if it is convex, it is automatically locally Lipschitz (given that it is finite everywhere). | |
| Apr 21, 2019 at 6:05 | review | Close votes | |||
| Apr 25, 2019 at 18:55 | |||||
| Apr 20, 2019 at 21:07 | comment | added | Pietro Majer | Maybe you should add your definition of Lipschitz (maybe you mean "locally Lipschitz"?) and of supercoercive (is it $\lim_{\|x\|\to\infty}f(x)/\|x\|=+\infty$ what you mean?) because they are incompatible with each other in the common usage | |
| Apr 20, 2019 at 19:12 | comment | added | ABIM | Indeed, that's where the question stems from. However, in this setting, it is less clear to me if things are Lipschitz. | |
| Apr 20, 2019 at 15:41 | comment | added | Dirk | This reminds me of proximal mappings: there you have $f(x,y) = g(x) +\|x-y\|_2^2$ and the argmin over x is always 1-Lipschitz in y. | |
| S Apr 20, 2019 at 9:12 | history | suggested | CommunityBot | CC BY-SA 4.0 |
\operatorname*{}
|
| Apr 20, 2019 at 7:31 | comment | added | ABIM | Hmm..then is there any easy condition on $f$? | |
| Apr 20, 2019 at 6:41 | comment | added | Anthony Quas | I have in mind $g(x)=x^{2n}$ for example. | |
| Apr 20, 2019 at 6:19 | review | Suggested edits | |||
| S Apr 20, 2019 at 9:12 | |||||
| Apr 20, 2019 at 5:44 | comment | added | ABIM | If the map is strictly convex then this should rule this out no? | |
| Apr 19, 2019 at 21:41 | comment | added | Anthony Quas | Consider the case $X=\mathbb R$. If $g(x)$ is a conv x function with a very flat minimum at zero, then $f(x,y)=g(x)+xy$ has the property that the argmin is far from Lipschitz near $y=0$. | |
| Apr 19, 2019 at 19:43 | history | asked | ABIM | CC BY-SA 4.0 |