Timeline for Lavrentiev phenomenon between $C^1$ + Lipschitz derivative and $C^2$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 24, 2021 at 16:38 | vote | accept | Nate River | ||
| Jul 24, 2021 at 4:53 | comment | added | Nate River | Ah, you are right of course. | |
| Jul 24, 2021 at 4:24 | comment | added | Leo Moos | Let $u : [0,1] \to \mathbf{R}$ be a function in $C^{1,1} \setminus C^2$. It seems to me that modifying another answer by changing the sentence to 'define $f(x,y,\xi) = 0$ if $y = u(x)$ and $F(y)$ otherwise' is valid. The 'safe path' that this answer refers to would now correspond to the graph of $u$. As any $C^2$-regular function deviates from this, it seems to me that $\int_0^1 f(x,g(x)) \, \mathrm{d} x = + \infty$ for every $g \in C^2$. | |
| Jul 23, 2021 at 13:18 | answer | added | Alexandre Eremenko | timeline score: 1 | |
| Jul 23, 2021 at 5:24 | comment | added | Nate River | Ah, I intend for $f$ to be independent of $g’$. If we include $g’$ the answer is yes, but the independent case seems much more subtle. | |
| Jul 23, 2021 at 1:54 | comment | added | Alexandre Eremenko | Don't you want to include $g'$ in the arguments of f? | |
| Jul 23, 2021 at 0:49 | history | edited | Nate River | CC BY-SA 4.0 |
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| Jul 23, 2021 at 0:30 | history | edited | Nate River | CC BY-SA 4.0 |
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| Jul 23, 2021 at 0:24 | history | edited | Nate River | CC BY-SA 4.0 |
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| Jul 23, 2021 at 0:19 | history | asked | Nate River | CC BY-SA 4.0 |