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Asked once on SE-mathematics.

Let $U$ be an open subset in $\mathbb{R}^n$, $m\in\mathbb{N}$, $1\leq m<n$ and let $$\mathcal{C}^k_{\leq m}(U,\mathbb{R}^n):=\lbrace g\in\mathcal{C}^k(U,\mathbb{R}^n)\mid\dim \operatorname{im} Df(x)\leq m\:\forall x\in U\rbrace,$$ where $\mathcal{C}^k(U,\mathbb{R}^n)$ mean $k-$times continuously differentiable mappings from U to $\mathbb{R}^n$. Is it true that $$\mathcal{C}^\infty_{\leq m}(U,\mathbb{R}^n)\overset{\text{dense}}{\subset}\mathcal{C}^1_{\leq m}(U,\mathbb{R}^n),$$ with the usual $\left(\mathcal{C}^1,d(\cdot,\cdot)_{\mathcal{C}^1}\right)$ distance $$d(f,g)_{\mathcal{C}^1}=\sup\limits_{x\in U}\left|f(x)-g(x)\right|+ \sup\limits_{x\in U}\left\|Df(x)-Dg(x)\right\|.$$

$|\cdot|$ is length of a vector from $\mathbb{R}^n$ and $\|\cdot\|$ is length of vector from $\mathbb{R}^{n^2}$. Link to mathSE question https://math.stackexchange.com/q/1876303/357336

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    $\begingroup$ You should link to your question on MSE. $\endgroup$
    – Michael Albanese
    Commented Aug 4, 2016 at 14:35
  • $\begingroup$ It seems that the answer is ``No'' due to the Brouwer Fixed Point Theorem and Sard Theorem. $\endgroup$
    – Taras Banakh
    Commented Aug 12, 2016 at 19:35
  • $\begingroup$ @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. $\endgroup$
    – Polatucha
    Commented Aug 16, 2016 at 16:29
  • $\begingroup$ @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. $\endgroup$
    – Taras Banakh
    Commented Aug 18, 2016 at 10:12
  • $\begingroup$ 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 $n$. $\endgroup$
    – Polatucha
    Commented Nov 15, 2016 at 13:00

1 Answer 1

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There is a counterexample.

Example. There is $f\in C^1(\mathbb{R}^5,\mathbb{R}^5)$ with ${\rm rank}\, Df\leq 3$ that cannot be approximated in the supremum norm by mappings $g\in C^2(\mathbb{R}^5,\mathbb{R}^5)$ satisfying ${\rm rank}\, Dg\leq 3$.

Example. There is $f\in C^1(\mathbb{R}^7,\mathbb{R}^7)$, ${\rm rank}\, Df\leq 4$, that cannot be approximated in the supremum norm by mappings $g\in C^3(\mathbb{R}^7,\mathbb{R}^7)$ satisfying ${\rm rank}\, Dg\leq 4$.

Both examples are special cases of the following result proved in [1]:

Theorem. Suppose that $m+1\leq k<2m-1$, $\ell\geq k+1$, $r\geq m+1$, and the homotopy group $\pi_k(S^m)$ is non-trivial. Then there is a map $f\in C^1(\mathbb{R}^\ell, \mathbb{R}^r)$ with ${\rm rank}\, Df\leq m $ in $\mathbb{R}^\ell$ that cannot be approximated by maps of class $ C^{k-m+1}$.

You can find infinitely many more examples.

[1] P. Goldstein, P. Hajłasz, $C^1$ mappings in $\mathbb{R}^5$ with derivative of rank at most $3$ cannot be uniformly approximated by $C^2$ mappings with derivative of rank at most $3$. . J. Math. Anal. Appl. 468 (2018), 1108–1114. (arXiv:1804.08289).

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