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 http://math.stackexchange.com/q/1876303/357336

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