Timeline for Convergence of a sequence by iteration
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Jul 5, 2018 at 19:27 | comment | added | Mateusz Kwaśnicki | @MB2009: But this does not imply continuity of $\nabla F$. Anyway, apparently a smooth counterexample exists, as I attempt to explain in my answer. (The example discussed there is only $C^{1,1-\epsilon}$ for any $\epsilon > 0$, but you can modify it slightly to get $C^\infty$, if you like). | |
| Jul 5, 2018 at 19:19 | answer | added | Mateusz Kwaśnicki | timeline score: 4 | |
| Jul 5, 2018 at 17:24 | comment | added | user111097 | @PiotrHajlasz Yes. Actually the statement on $\nabla F$ implies implicitly that $F$ is differentiable. | |
| Jul 5, 2018 at 17:17 | comment | added | Piotr Hajlasz | A convex function need not be differentiable everywhere. If $x_n$ is a point where $F$ is not differentiable, then $x_{n+1}$ is not even defined. Do you assume that $F\in C^1$? | |
| Jul 5, 2018 at 14:38 | history | asked | user111097 | CC BY-SA 4.0 |