Timeline for Is a locally invertible weak limit of injective maps injective almost everywhere?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 14, 2021 at 9:54 | vote | accept | Asaf Shachar | ||
| Oct 26, 2020 at 18:59 | history | edited | mlk | CC BY-SA 4.0 |
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| Oct 26, 2020 at 18:51 | comment | added | mlk | @AsafShachar Ah, yeah I forgot that detail. However the fix is easy. Since $\Omega_1$ has $C^1$ boundary there is an extension operator for $W^{1,2}$. So we can extend $f_n$ to $\Omega \supset \Omega_1$ with the same convergence and then $\overline{\Omega_1}$ is compact within $\Omega$. | |
| Oct 26, 2020 at 17:42 | comment | added | Asaf Shachar | I don't think that it's trivial to guarantee this, since even though $f \in C^1$, we don't know that it's injective (yet), so $A$ may well have some weird pre-images which are close to the boundary of $\Omega_1$ | |
| Oct 26, 2020 at 17:40 | comment | added | Asaf Shachar | Thanks, that looks like a nice solution! I have one question: Müller's result gives weak convergene of the Jacobians on $L^1(K)$ for compactly contained subsets $K \subset \subset \Omega_1$. So, for your argument to work you need to make sure that $B=f^{-1}(A)$ is compactly contained in the interior $\Omega_1$. (I actually think that you need it also for the first $\liminf$ inequality $\int_B \det df dx \leq \liminf_{n\to\infty} \int_B \det df_n dx $.) | |
| Oct 26, 2020 at 16:38 | history | edited | mlk | CC BY-SA 4.0 |
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| Oct 26, 2020 at 16:31 | history | answered | mlk | CC BY-SA 4.0 |