Timeline for Can $C^1$ mappings with derivative of low rank be approximated by smooth maps?
Current License: CC BY-SA 3.0
23 events
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Jan 12, 2019 at 21:02 | history | edited | Piotr Hajlasz |
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Apr 23, 2018 at 19:30 | comment | added | Polatucha | @PiotrHajlasz I think it is done, green "ptaszek" marked. In the middle of May i should speak with Paweł about yor joint work, then this acceptance may become legally valid. | |
Apr 23, 2018 at 19:20 | vote | accept | Polatucha | ||
Apr 16, 2018 at 4:17 | comment | added | Piotr Hajlasz | @TarasBanakh The answer is in the negative, see below. | |
Apr 6, 2018 at 22:02 | review | Suggested edits | |||
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S Apr 6, 2018 at 21:59 | history | suggested | Piotr Hajlasz | CC BY-SA 3.0 |
I changed the title to a more appealing one.
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Apr 6, 2018 at 21:54 | review | Suggested edits | |||
S Apr 6, 2018 at 21:59 | |||||
Apr 6, 2018 at 21:01 | answer | added | Piotr Hajlasz | timeline score: 14 | |
Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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Nov 15, 2016 at 13:01 | comment | added | Polatucha | Let $U\overset{\text{open}}{\subset}\mathbb{R}^5$ and $f\in\mathcal{C}^1_{\leq 3}(U,\mathbb{R}^5)$ such that $\mathcal{H}^4(f(U))>0$. Now any smooth map with the same rank restriction, that is $f_\epsilon\in \mathcal{C}^\infty_{\leq 3}(U,\mathbb{R}^5)$ has at most $\mathcal{H}^m$ positive measure. And we can think that it is a problem because we have to approximate a set of dimension 4 by a set of dimension at most 3, but it is not clear to me that one can not do it because in examples images of such $f's$ are not a valid 4 manifolds but some kind of Cantor sets. | |
Nov 15, 2016 at 13:01 | comment | added | Polatucha | From this we have: | |
Nov 15, 2016 at 13:01 | comment | added | Polatucha | The powered up Sard theorem 3.4.3 from Federers book "Geometric measure theory" tells us that that the image under $f$ of the whole set U (it is equal to the set where $r(Df)\leq m$ since $f\in\mathcal{C}^1_{\leq m}(U,\mathbb{R}^n)$) can have positive $s-$dimensional Hausdroff measure for any $s<n$ because the optimal result is that $$\mathcal{H}^n(\lbrace x\:\vert\: dim\:im Df(x)\leq m\rbrace)=0$$ where $n=m-(n-m)/k$ and $k$ is how many times map $f$ is continuously differentiable. | |
Nov 15, 2016 at 13:01 | comment | added | Polatucha | Moreover, this is also true that mappings from $\mathcal{C}^1_{\leq m}(U,\mathbb{R}^n)$ can have much more "measure" in the image than smooth maps. | |
Nov 15, 2016 at 13:00 | comment | added | Polatucha | This is true, but I do not understand how not-density of small rank in whole $\mathcal{C}^\infty(U,\mathbb{R}^n)$ imply non-density for the subset $\mathcal{C}^1_{\leq m}(U,\mathbb{R}^n)$. As far as I understand identity mapping is far away from any map from the set I am interested in i.e. $$d_{\mathcal{C}^1}(\text{id},\mathcal{C}^1_{\leq m}(U,\mathbb{R}^n))>c$$ for some positive constant $c$ depending probably on the dimension $n$. | |
Oct 17, 2016 at 16:55 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
Sep 17, 2016 at 16:44 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
Aug 18, 2016 at 10:12 | comment | added | Taras Banakh | @Polatucha By Sard Theorem, any differentiable map $f:U\to R^n$ whose differential $Df$ has rank $<n$ at each point $x\in U$ has image $f(U)$ of zero Lebesgue measure. On the other hand, BFPT implies that any map $f:U\to R^n$ which is sufficiently close to the identity in $C^0$-topology has image $f(U)$ with non-empty interior. These two facts imply that maps with differentials of small rank cannot be dense in $C^\infty(U,R^n)$ even in the $C^0$-topology. | |
Aug 16, 2016 at 16:29 | comment | added | Polatucha | @TarasBanakh Thank you for the comment! Do you want to use BFPT to mapping with rank restriction and assume that domain U is a ball or apply this theorem to a ball in function space some transformation of mappings ? Would you please elaborate. | |
Aug 12, 2016 at 19:35 | comment | added | Taras Banakh | It seems that the answer is ``No'' due to the Brouwer Fixed Point Theorem and Sard Theorem. | |
Aug 4, 2016 at 14:37 | history | edited | Polatucha | CC BY-SA 3.0 |
added 72 characters in body
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Aug 4, 2016 at 14:35 | comment | added | Michael Albanese | You should link to your question on MSE. | |
Aug 4, 2016 at 14:22 | review | First posts | |||
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Aug 4, 2016 at 14:15 | history | asked | Polatucha | CC BY-SA 3.0 |