Timeline for Can we approximate a vector field on the plane with non-vanishing vector fields in $W^{1,2}$?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Nov 20, 2019 at 9:26 | vote | accept | Asaf Shachar | ||
| Nov 20, 2019 at 9:26 | comment | added | Asaf Shachar | @Dap You are absolutely right...I didn't notice that Pietro's answer for my old question applies to this case as well. Thank you for mentioning this. | |
| Nov 20, 2019 at 5:23 | comment | added | Dap | I think this was already answered by Pietro Majer's answer here: mathoverflow.net/a/307832/112284. Set $p_n$ there to $V_n$ here. (And apply a cutoff function i.e. multiply the counterexample $p_n$ by a $C^\infty$ function that is $1$ on a neighborhood of $\mathbb D^2,$ and restrict $r_0$ to $(0,1).$) | |
| Nov 18, 2019 at 16:24 | comment | added | Willie Wong | @PietroMajer: I completely forgot about the $C^\infty$ part. Ok, I will undelete. | |
| Nov 18, 2019 at 15:06 | comment | added | Pietro Majer | Willie: Actually I think yours was a useful answer. Since Asaf's functions $V_n$ are assumed continuous, I'm not sure if my objection applies to your objection to his conjecture | |
| Nov 18, 2019 at 14:57 | comment | added | Willie Wong | Deleted my answer below because Pietro's comment made me doubt an essential part of my reasoning. For convenience, however, a survey of the VMO degree theory (which would apply in your case because $W^{1,2}$ embeds in $VMO$ in two dimensions) can be found here (Brezis, "New questions related to the topological degree".) | |
| Nov 18, 2019 at 14:45 | comment | added | Pietro Majer | No, the topological degree is well defined and continuous in $C^0$ situations. There are also extensions of the degree theory in different settings, not included in $C^0$, like VMO mappings: in that topology the degree is also locally constant, but maybe it is not a problem for you, because a nonzero degree of $f$ on $U$ wrto $p$ does not imply that $f(x)=p$ has a solution $x\in U$, like for the classical case (i.e. a continuous $f:\overline U\to\mathbb{R}^n$, $U\subset\mathbb{R}^n$ bounded open set, $p\notin f(\partial U)$) | |
| Nov 18, 2019 at 14:45 | answer | added | Willie Wong | timeline score: 3 | |
| Nov 18, 2019 at 14:31 | history | edited | Asaf Shachar | CC BY-SA 4.0 |
Fixed a mistake
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| Nov 18, 2019 at 14:26 | comment | added | Willie Wong | Minor side comment: compact support and isolated zeros are not compatible. Perhaps you meant that it only has isolated zeros on $\mathbb{D}^2$? | |
| Nov 18, 2019 at 13:56 | history | edited | Asaf Shachar | CC BY-SA 4.0 |
I have changed the required convergence rate
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| Nov 17, 2019 at 15:07 | comment | added | Pietro Majer | The strongest convergence you want must be strictly weaker of the uniform convergence, because if $V_n\to V$ uniformly on $\overline{\mathbb{D}^2}$, and $V$ has non-zero topological degree in $\mathbb{D}^2$ wrto $0$, the same holds eventually for the $V_n$, and they will vanish as well. But $W^{2,2}(\mathbb{D}^2,\mathbb{R}^2)$ is continuously embedded in $C^0(\mathbb{D}^2,\mathbb{R}^2)$: this convergence is too strong. | |
| Nov 17, 2019 at 14:36 | history | asked | Asaf Shachar | CC BY-SA 4.0 |