# Can we approximate a.e. invertible matrices with everywhere invertible matrices in $L^2$ sense?

Let $$\mathbb{D}^n=\{ x \in \mathbb{R}^n \, | \, |x| \le 1\}$$ be the closed unit ball, and let $$A:\mathbb{D}^n \to \mathbb{R}^{n^2}$$ be real-analytic on the interior $$(\mathbb{D}^n)^o$$ and smooth on the entire closed ball $$\mathbb{D}^n$$. Suppose that $$n \ge 2$$, and that $$\det A >0$$ a.e. on $$\mathbb{D}^n$$.

Are there smooth maps $$A_k: \mathbb{D}^n \to \mathbb{R}^{n^2}$$, such that $$A_k \to A$$ in $$L^2(\mathbb{D}^n , \mathbb{R}^{n^2})$$ and $$\det A_k >0$$ everywhere on $$\mathbb{D}^n$$?

Edit:

In Vogel's elegant answer, it is proved that we can approximate $$A$$ via continuous maps. Can we approximate it with smooth maps?

I think the answer should be positive, but I am having trouble with the details:

Using mollifiers, we can approximate any continuous $$A_k \in L^2(\mathbb{D}^n , \mathbb{R}^{n^2})$$ with a smooth version $$\tilde A_k$$ in such a way to ensure that $$\tilde A_k$$ will converge to $$A_k$$ uniformly on compact subsets of $$(\mathbb{D}^n)^o$$. Since $$s_k=\min_{x \in \mathbb{D}^n}\text{dist}(A_k(x), {\det}^{-1}(0))>0,$$ then if $$\| \tilde A_k(y) -A_k(y)\| < s_k$$ we have $$\det(\tilde A_k(y))>0$$ as we wanted. The problem is that we only have uniform convergence $$\tilde A_k \to A_k$$ on compact subsets of the interior of $$\mathbb{D}^n$$, so I think we might have a problem on the boundary...

Any ideas about how to finish this reduction?

• $\mathbb D^n$ is the closed unit disc? Do you mean poly disc?
– zhw.
May 24, 2019 at 16:48
• It is the closed unit ball. I have edited the question to make this clear. May 24, 2019 at 19:11

Apply Vitali’s covering theorem to get a sequence of disjoint closed balls $$B_n\subset\{x|det A(x)>0\}$$ that cover everything but a null set. We will take $$A_k$$ to be equal to the identity matrix outside $$B_1,\cdots,B_k.$$ Inside each ball $$B_n=B(x_n,r_n)$$ with $$n\leq k$$ define $$A_k$$ by:
• $$A_k(x_n+tr_ny)=A(x_n+g(t)r_ny)$$ for $$0\leq t\leq 1-1/4k$$ and unit vectors $$y,$$ where $$g(0)=g(1-1/4k)=0$$ and $$g(1-1/2k)=1-1/2k,$$ with linear interpolation between these points
• $$A_k(x_n+tr_ny)=\gamma_n(4k(t-1+1/4k))$$ for $$1-1/4k\leq t\leq 1$$ and unit vectors $$y,$$ where $$\gamma_n$$ is a choice of path from $$A(x_n)$$ to the identity matrix
This should be continuous and give $$A_k\to A$$ in measure, which easily implies $$L^2$$ convergence since everything is bounded.
• @AsafShachar: I meant to say from $A(x_n)$ to Id (through the good set) Jun 3, 2019 at 10:01
• @AsafShachar: applying $A$ to $\gamma_n$ was a mistake. Sorry. I am not assuming $A$ is the identity matrix at any point. Jun 3, 2019 at 15:28