Timeline for Is a locally invertible weak limit of injective maps injective almost everywhere?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 14, 2021 at 9:54 | vote | accept | Asaf Shachar | ||
| Oct 26, 2020 at 16:31 | answer | added | mlk | timeline score: 2 | |
| Oct 26, 2020 at 16:00 | comment | added | Leo Moos | @mlk You're of course correct, I went a bit too quickly there. The argument I was thinking of only gives only direction I believe. The correction you suggest seems sufficient. | |
| Oct 26, 2020 at 15:52 | comment | added | mlk | @LeoMoos I don't think strong $L^2$ alone gives you Hausdorff-convergence of the image. But you can get uniform convergence on an arbitrary large subset which gives you Hausdorff convergence of that image and then try to argue that the image of the leftover set is small via the weak $L^1$-convergence of the determinant. | |
| Oct 26, 2020 at 15:49 | comment | added | Leo Moos | @Asaf Sorry, I kept rewriting my comment in the hopes of making it more succinct. I don't think the argument requires thinking about the boundaries, but rather showing that the number of pre-images is constant equals $Q$. Have you tried something along the following lines: by the inverse function theorem, for all $y \in \mathrm{im}\, f$ there is a radius $\rho > 0$ so that the points in $B_\rho(y)$ have the same number of preimages; in other words it's locally constant. As the image of $f$ is connected, it must be constant globally. | |
| Oct 26, 2020 at 15:45 | comment | added | Leo Moos | Along the same lines as suggested above, from the weak $W^{1,2}$-convergence one finds that $Q \mathcal{H}^2(\mathrm{im} \, f) \leq \liminf_{n \to \infty} \mathcal{H}^2(\mathrm{im} \, f_n)$, where $f$ covers its image $Q \in \mathbf{Z}_{>0}$ times. One other hand, it seems to me that the strong $L^2$-convergence shows that $\mathrm{im} \, f_n \to \mathrm{im} \, f$ with respect to Hausdorff distance, and therefore $\mathcal{H}^2(\mathrm{im} \, f) = \lim_{n \to \infty} \mathcal{H}^2(\mathrm{im} \, f_n)$. | |
| Oct 26, 2020 at 15:43 | comment | added | Asaf Shachar | @LeoMoos Thanks, but I am not sure how do you continue from here. Also is it clear that $\int Jf$ is an integer multiple of the image's volume? (We have manifolds with boundary here; does that covering argument requires assuming that $f(\partial \Omega_1) \subseteq \partial \Omega_2$?) | |
| Oct 26, 2020 at 11:55 | comment | added | Asaf Shachar | Thanks, this sounds interesting. I know that the Jacobians $Jf_n$ converge weakly in $L^1(K)$ for $K \subset \subset \Omega_1$; Unfortunately, I am not sufficiently familiar with degree theory in the Sobolev context; it sounds like the right tool tough. I would be happy to see the details. | |
| Oct 26, 2020 at 11:24 | comment | added | mlk | I am pretty sure that what you ask is true as it is essentially just degree theory for Sobolev spaces in conjunction with the fact that non-negative determinants converge slightly better than one would expect. You don't even need to exclude the degenerate case of constants, as for those $|f^{-1}(y)|=0$ a.e. anyway. I'll try to give a full answer once I find the time. | |
| Oct 24, 2020 at 16:45 | comment | added | Asaf Shachar | I just mean that each $f_n$ is a Lipschitz map (with a Lipschitz constant which might depend on $n$) and that it's injective on $\Omega_1$. I am fine with replacing the injectivity assumption with the requirement $|f_n^{-1}(y)| \le 1$ for almost every $y \in \Omega_2$. And of course by $\det(df_n)>0$ I mean "almost everywhere", since $df$ might not be defined on all $\Omega_1$. | |
| Oct 24, 2020 at 13:27 | comment | added | Alexandre Eremenko | What does it mean "Lipschitz injective"? | |
| Oct 24, 2020 at 10:06 | history | asked | Asaf Shachar | CC BY-SA 4.0 |