# Can we stay invertible while approximating linear maps in Sobolev spaces?

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.

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.

## 2 Answers

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 $x_0)$ )

• Thanks, this is interesting. I hope you will be willing to clarify two points for me: (1) Regarding your last point, how do we rule out the possibility $\det(u_n)=0$ a.e. in general? It is not clear to me that the given assumptions $\det A > 0$ a.e. and $x \to A_x$ is continuous imply the existence of a compact subset (locally around every point) where $\det A \neq 0$. This holds in the example considered above. (2) Can you give me any reference for the phenomena that "Sobolev convergence" passes to the boundary? (I am referring to the restrictions $p_{n|\partial B(0,r_0)}$). Thanks again. – Asaf Shachar Aug 8 '18 at 14:28
• On a second thought, I also don't see how can you approximate $A$ with polynomials in $W^{1,p}$. (You can certainly approximate $A$ uniformly, and in particular in $L^p$, but how do you make sure the derivatives of the polynomials approximate the derivatives of $A$?) – Asaf Shachar Aug 8 '18 at 17:04
• I added a rmk for (1). The fact (2) about traces comes from the embedding $W^{1,1}(I\times J)$ in the Bochner space $L^1(I, W^{1,1}(J))$ which is essentially Fubini-Tonelli theorem. – Pietro Majer Aug 9 '18 at 6:06
• On the last point: you can first approximate in $W^{1,p}_{loc}$ by a map in $C^\infty(\mathbb{R}^n)$, which can be approximated $C^\infty_{loc}$ by polynomials. (Alternatively, one can do the argument of the last point with real analytic maps as well, instead of polynomial maps. To approximate a $W^{1,p}_{loc}$ function with a real analytic function, multiply by a cut-off function, then mollify by the Weierstrass kernel). – Pietro Majer Aug 9 '18 at 6:07

Consider the case $n=2$,

\begin{equation*} A(x,y)=\left( \begin{array}{cc} x & -y \\ y & x \end{array} \right) \end{equation*}

This function is not just in a Sobolev space, it is analytic. Obviously $\det A>0$ a.e, that is, except the origin. But you cannot approximate it by invertible matrices very well due to topological obstacles. (Not in $W^{1,p}$, at least.)

Your second question is completely different. I presume that the answer to it is positive (although I did not think about the details). You simply approximate $A$ by a smooth $u_n$ somehow, and if $\det u_n=0$ on a set of positive measure, then you tweak $u_n$ a bit.

• Thanks. Can you please elaborate on these "topological obstructions"? (Why approximation by invertible matrices in $W^{1,p}$ is not possible?) – Asaf Shachar Aug 8 '18 at 12:16
• Because this is the operator of multiplication by $x+iy$, and it is not possible to approximate $f(z)=z$ near the origin by a function which does not vanish. (This is not exactly the same thing, but the conclusion is the same.) – Alex Gavrilov Aug 8 '18 at 12:20