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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.
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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 21:01 answer added Piotr Hajlasz timeline score: 14
Apr 13, 2017 at 12:19 history edited CommunityBot
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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
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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