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Let $0<\sigma_1<\sigma_2$, and let $D \subseteq \mathbb{R}^2$ be the closed unit disk.

Let $f \in W^{1,\infty}(D,\mathbb{R}^2)$, and suppose that the singular values of $df$ are a.e. equal to $\sigma_1,\sigma_2$.

Do there exist $f_n \in C^{\infty}(D^o,\mathbb{R}^2)$ such that $f_n \to f$ in $W^{1,2}$ and the singular values of $df_n$ are everywhere equal to the $\sigma_i$?

Can we also get the $f_n$ to converge uniformly to $f$? ($f$ is continuous, since it is in $W^{1,\infty}$).

The standard mollification process won't work here-it is an averaging process, and averaging matrices with given singular values reduces the norm. (all these matrices lie on Euclidean sphere of radius $\sqrt{\sigma_1^2+\sigma_2^2}$, which is strictly convex).


Here is a concrete example:

Suppose that $\frac{\sigma_2}{\sigma_1}=n$ is a natural number.

The map $f:re^{i \theta} \to \sigma_1re^{i(\sigma_2/\sigma_1) \theta}$ (which in general is smooth on the disk only after removing a ray) has constant singular values $\sigma_i$.

Since we assumed $\frac{\sigma_2}{\sigma_1}=n$, $$ f(z)=f(re^{i\theta})= re^{in\theta}=\frac{z^n}{|z|^{n-1}}, $$ is smooth on the entire disk without the origin, and is in $W^{1,\infty}(D,\mathbb{R}^2)$.

Can we approximate $f$ in $W^{1,2}$ with smooth maps having the fixed singular values $\sigma_i$?

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