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It can be shown that there are (smoothly bounded, Jordan) domains $E\subset \mathbb{R}^2$ which are $\textit{not}$ images of mappings $f$ from the unit disk (or any other planar domain), such that $\mathrm{d}f$ has fixed, constant singular values. That is, such mappings have 'limited transformation capability'.

On the other hand, I wonder if a characterization of all those domains onto which the unit disk (or more generally, any planar domain) can be transformed to by such $f$'s.

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  • $\begingroup$ Could you please give an example of a domain mentioned in the first sentence or give a proof/reference? Does "smoothly bounded" mean that the boundary is a $\mathbb C^\infty$ curve? Note that if $E$ is any domain in $\mathbb R^2$ then for any linear map $L$ we have $E=L(L^{-1} E)$, this seem to contradict to your first sentence... $\endgroup$
    – Dmitri Panov
    Commented May 14, 2021 at 14:58
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    $\begingroup$ @DmitriPanov Yes, the result I'm citing is from this* work: "For each smoothly bounded Jordan domain $D$, there exists a constant $C = C_D$ such that $\lambda_1(\partial E) \leq C \lambda_2 (E)/\iota(E)$ for every Jordan domain $E$ which the image of $D$ under a cps-homeomorphism" (In our language a cps-homeom. is a mapping with constant singular values. And $\iota(E)$ is the inradios, i.e., the radius of the greatest inscribed circle. $\lambda_1$ the length and $\lambda_2$ the area). Not every $E$ satisfies such inequality given $D$. *doi.org/10.1155/S0161171203204166 (theorem 3.1) $\endgroup$
    – Daniel Castro
    Commented May 14, 2021 at 17:30
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    $\begingroup$ @DmitriPanov What the previous inequality shows is that, for any constant singular values $\sigma_1$, $\sigma_2$, there are domains $E$ that cannot be obtained from an arbitrary $D$ (in particular, the unit disk). And in principle yes, I am restricted to $\mathbb{C}^\infty$ curves. Thank you ! $\endgroup$
    – Daniel Castro
    Commented May 14, 2021 at 17:37

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