Timeline for $C^1$ extension with compact support
Current License: CC BY-SA 4.0
9 events
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| Jul 22, 2021 at 17:34 | comment | added | Iosif Pinelis | @Bogdan : I think this is impossible in general. Consider e.g. $\phi_0(x_1,x_2):=x_1$ for $(x_1,x_2)\in\overline\omega$, where $\omega$ is the open unit disk. Then $\sup_{\overline\omega}|\phi_0|=1$, but $f(x_1,0)>1$ for all $x_1>1$ close enough to $1$. Perhaps, you misunderstood the proposition. | |
| Jul 22, 2021 at 17:09 | comment | added | Bogdan | Sorry. I mean in $|f|\leq sup_{x\in\overline{\omega}} |\phi_0|$ in $\Omega$. I found an article which says something like this but I do not understand exactly. Here it is: core.ac.uk/download/pdf/82133103.pdf (the proposition at page 326). Is it indeed true that $f$ can be chosen that way? | |
| Jul 22, 2021 at 15:42 | comment | added | Iosif Pinelis | @Bogdan : Since $f$ and $\phi_0$ are continuous and $f=\phi_0$ on $\omega$, we have $f=\phi_0$ on $\overline\omega$. | |
| Jul 22, 2021 at 15:14 | comment | added | Bogdan | Is it true that the Whitney Extension of $\phi_0$ satisfies the inequality $|f|\leq |\phi_0|$ on $\overline{\omega}$? | |
| Jul 22, 2021 at 12:56 | vote | accept | Bogdan | ||
| Jul 22, 2021 at 12:44 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 22, 2021 at 12:35 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 22, 2021 at 12:29 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 22, 2021 at 12:13 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |