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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?

Asaf Shachar
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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.

Vogel
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  • @AsafShachar: I meant to say from $A(x_n)$ to Id (through the good set) – Vogel Jun 03 '19 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. – Vogel Jun 03 '19 at 15:28
  • I wanted to thank you again properly. Your answer is truly magnificent. Just one last question: You basically showed that we have continuous (non-singular) approximations. Do you think that we can "advance from there" to smooth approximations? I have edited the question to reflect my concern. Thanks again anyway, this is a cool answer. (Unfortunately, I have been sick the last couple of weeks, so I was only able to respond more properly at this time...). – Asaf Shachar Jun 16 '19 at 09:52