Let $\Omega \subseteq \mathbb{R}^n$ be an open bounded domain with a smooth boundary. Fix $1<p<n$.


Let $A \in W^{1,p}(\Omega;\text{End}(\mathbb{R}^n)) \cap C(\Omega;\text{End}(\mathbb{R}^n))$ with $\det A > 0$ a.e.



> 
> Do there exist $u\_n \in C^{\infty}\big(\Omega,\text{GL}(\mathbb{R}^n)\big)$ such that $u\_n \to A$ in $W^{1,p}\_{loc}$?
> 
> 
> I am also interested in a weaker result: Are there $u\_n \in C^{\infty}\big(\Omega,\text{End}( \mathbb{R}^n)\big)$, such that $u\_n(x) \in \text{GL}(\mathbb{R}^n)$ a.e. **and**
> $u\_n \to A$ in $W^{1,p}\_{loc}$?
> 
> 
> 


I don't really need the $u\_n$ to be defined on all $\Omega$. It suffices that for **every** arbitrarily small ball in $\Omega$, there would be a neighbourhood where such a sequence $u\_n$ would be defined.


The problem is that *it is not always true* that $A\_x \in \text{GL}( \mathbb{R}^n)$ for *every* $x \in \Omega$. The rank can fall on a subset of measure zero.


**If we knew $A(x) \in \text{GL}(\mathbb{R}^n)$ everywhere then the answer would be positive.** This follows from the facts that "being invertible" is an open condition, and that continuous Sobolev maps can be *approximated uniformly* by smooth maps over compact subsets.




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In more detail, let $K \subseteq \Omega$ be compact. Since we assumed $A \in C\big(\Omega, \text{GL}(\mathbb{R}^n) \big)$, the map $\psi:x \to A\_x$, considered as a map $K \to \text{GL}( \mathbb{R}^n)$, is continuous. Thus $\psi(K) $ is compact and $\text{dist}\big(\psi(K),\partial \text{GL}(\mathbb{R}^n)\big)>0$.


Now consider each component of $\psi(x)=A\_x \in \text{End}(\mathbb{R}^n) $. We can approximate each component of $\psi$ using mollification on an open subset of $\Omega$ containing $K$. Since each component is a continuous function, the mollifications *converge uniformly* on $K$. This implies that from a certain point in the mollified sequence, $\text{dist}(u\_n,A)<\text{dist}\big(\psi(K),\partial \text{GL}(\mathbb{R}^n)\big)$, so the $u\_n$ are invertible.



Can we stay invertible while approximating linear maps in Sobolev spaces?
I will recycle the answer I've been writing into some words of explanation on Alex Gavrilov's example, which is more simple and elegant. We can focus on the first column $p\_n$ of an approximating sequence $u\_n:=[p\_n,q\_n]$ of $A(x,y):=\begin{bmatrix}x&-y\\y&x\end{bmatrix}$. If $p\_n\in C^0\cap W^{1,1}\_{loc}(\mathbb{R}^2,\mathbb{R}^2)$ converges in $W^{1,1}\_{loc}(\mathbb{R}^2,\mathbb{R}^2)$, then for at least one $r\_0>0$ (actually, for a.e. $r>0$) the sequence restricted to the boundary, $p\_{n|\partial B(0,r\_0)}$ is in $W^{1,1}(\partial B(0,r\_0))$ and converges there, therefore also uniformly. Then for some $n\_0$, $$\|p\_{n\_0|\partial B(0,r\_0)}-{\operatorname{id}\_{\partial B(0,r\_0)}}\|\_{\infty,\partial B(0,r\_0) }<r\_0$$ that implies that $ p\_{n\_0|B(0,r\_0)} $ has degree $1$ and $p\_n(x,y)$ vanishes in some $(x\_0,y\_0)\in B(0,r\_0)$, so that $u\_{n\_0}(x\_0,y\_0)$ is not invertible. 


For the weaker version, it is sufficient to approximate $A$ in $W^{1,p}\_{loc}$ with a sequence of polynomial maps $u\_n$, since the polynomial $\det(u\_n)$ is automatically non-vanishing a.e. as soon as it is not identically zero (which of course is eventually true, since $\det(u\_n(x\_0))\to\det(A(x\_0))>0$ holds at some point $x\_0)$ )


