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Let $\mathbb{D}^2=\{ x \in \mathbb{R}^2 \, | \, |x| \le 1\}$ be the closed unit disk, and set $$\begin{equation*} A(x,y)=\left( \begin{array}{cc} x & -y \\ y & x \end{array} \right) \end{equation*}. $$

Are there smooth maps $A_n: \mathbb{D}^2 \to \mathbb{R}^{4}=M_2(\mathbb{R})$, such that $A_n \to A$ in $L^2(\mathbb{D}^2 , \mathbb{R}^{4})$ and $\det A_n >0$ everywhere on $\mathbb{D}^2$?

Note that it is impossible to approximate $A$ in $W^{1,2}$ by invertible matrices.

Also, a radially symmetric approximation does not exist: Write $A=rR_{\theta}$ in polar coordinates . Then $A_n=f_n(r)R_{\theta}$ is continuous at the origin only if $\lim_{r \to 0}f_n(r)=0$, which implies $A_n(0,0)=0$. So, any non-degenerate continuous approximation must break the radial symmetry of $A$.

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    $\begingroup$ Just push the disk at some boundary point with some sufficiently thin object all the way to $0$, so that the image contains the origin no more and take the composition of $A$ and the resulting mapping. $\endgroup$
    – fedja
    Commented May 28, 2019 at 11:34
  • $\begingroup$ Thanks, I think I understand visually what you are saying: you "push inwards" (inside) some thin neck of the disk, starting on the boundary, and moving pass the origin. So you are "left with" only a part of the disk which does not contain the origin. So, the point is that as this neck gets thinner and thinner, the integral of the difference gets smaller (since all the quantities here are bounded, only the measure of the removed thin domain plays a key role). Does that sound like a reasonable description of your suggestion? $\endgroup$
    – Asaf Shachar
    Commented May 28, 2019 at 13:34
  • $\begingroup$ I guess that writing down explicitly such a "pushing map" is not entirely trivial, but I am rather "visually convinced" that it can be done...Finally, this question was a "toy model" of this question, which may interest you. $\endgroup$
    – Asaf Shachar
    Commented May 28, 2019 at 13:34

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