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Kirszbraun's theorem for $\mathbb{R}^2$ states the following:

Given any set $S\subset \mathbb{R}^2$ and any Lipschitz function $f:S\rightarrow \mathbb{R}^2$ with Lipschitz constant $k$, $0< k< \infty$, for any set $F$ which contains $S$ there exists a function $\tilde f:F\rightarrow C$ such that $Lip(\tilde{f})=k$, $\tilde{f}|_{S}=f$ and $C$ is any closed convex set containing $f(S)$.

I'm wondering if something similar could hold true for bi-Lipschitz functions, for example is the following claim true?

Given any two discrete sets of $n\ge 3$ linearly independent points $S_1,S_2\subset \mathbb{R}^2$, and any bi-Lipschitz function $f:S_1\rightarrow S_2$ with bi-Lipschitz constant $k$, $0<k<\infty$, there exists a function $\tilde{f}:C_1\rightarrow C_2$ which extends $f$ and has bi-Lipschitz constant $k$, where $C_i$ is the convex envelope of $S_i$.

This problem seems more complicated than what I expected since the only article I've found on this subject is this https://arxiv.org/pdf/1110.6124.pdf which computes the bi-Lipschitz extension only in the case $S$ is the border of the unit square and the extended function has the bi-Lipschitz constant multiplied by a "big" constant factor.

User YCor proved that my claim is false in general, but do you know any other reference on this problem? In particular is my claim true when $S$ is composed of 3 linearly independent points? (even this doesn't seem trivial to me)

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    $\begingroup$ Anyway it's certainly false. Consider an annulus of radius between $1/2$ and $2$ and the function $re^{i\theta}\mapsto r^{-1}e^{i\theta}$. You can restrict this to a large "dense enough" discrete set, say $r\in\{2^{k/n}:-n\le k\le n\}$ and $\theta\in \{2\pi k/n:0\le k\le n-1\}$ for large $n$. Then certainly any bilipschitz extension would have bilipschitz constant tending to $\infty$ (for $n\to\infty$), since otherwise letting $n$ tend to infinity and using compactness we could obtain a bilipschitz extension of the original map to the whole closed disk. $\endgroup$ – YCor May 5 '17 at 17:42
  • $\begingroup$ @YCor I'm sorry, but I don't understand your argument. Call $\Gamma_n$ the discrete set you defined for $n>>0$ and $f$ the function $f(re^{i\theta})=r^{-1}e^{i\theta}$, $f:\Gamma_n\rightarrow \Gamma_n$ has finite bilipschitz constant. Call $A$ the annulus of radius between $1/2$ and $2$, why are you saying that if there exists an extension $\tilde{f}:A\rightarrow A$ of $f$ with the same bilipschitz constant we can get an extension of $f$ to the whole disk with the same bilipschitz constant? Where's the contradiction? $\endgroup$ – user99087 May 5 '17 at 20:50
  • $\begingroup$ The extension to the convex hull would be a bilipschitz map on the closed disc of radius 2 that restricts to the inversion on the annulus... while a self-homeomorphism of the closed disc should always map the boundary to itself. $\endgroup$ – YCor May 5 '17 at 20:54
  • $\begingroup$ @YCor thank you very much, I get it now. Do you think that for a lower number of points it could work? I'm really interested in the case of 3 linearly independent points $\endgroup$ – user99087 May 5 '17 at 21:02
  • $\begingroup$ Would you first mind to add quantifiers to $k$? it's the whole point. $\endgroup$ – YCor May 5 '17 at 21:45
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It seems that you are asking the following:

Assume $f\colon\{x_1,x_2,x_3\}\to\mathbb R^2$ is a map with bi-Lipschitz constants $k$ and $1$ ($0<k<1$). Is it possible to extend $f$ to a bi-Lipschitz map $\bar f\colon\mathbb R^2\to\mathbb R^2$ with the same constants?

The answer in "no".

Set $x_i'=f(x_i)$. Assume $|x'_1-x'_3|=|x_1-x_3|$, $|x'_1-x'_2|=|x_1-x_2|$ and $|x'_2-x'_3|=k\cdot |x_2-x_3|$. Note that $$\ell=\frac{\measuredangle[x'_1\,^{x'_2}_{x'_3}]}{\measuredangle[x_1\,^{x_2}_{x_3}]}< k.$$

Assume $\bar f$ is a $1$-Lipschitz extension of $f$. Then $f$ sends the sides $[x_1,x_2]$ and $[x_1,x_3]$ isometrically to $[x'_1,x'_2]$ and $[x'_1,x'_3]$. It follows that the lower Lipschitz constant of $\bar f$ is at most $\ell$. Hence the statement follows.

(In fact one can adjust the values $|x_i-x_j|$ and $k>0$ so that $\ell=0$; in this case the triangle $[x_1'x_2'x_3']$ is degenerate).

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  • $\begingroup$ @user372511 I made an update in the answer, it should be clear now. $\endgroup$ – Anton Petrunin May 15 '17 at 18:59
  • $\begingroup$ I'm sorry to bother you again, but I realized I'm still not sure why the fact that $\overline{f}$ maps the two sides isometrically should imply that its lower Lipschitz constant is at most $l$ $\endgroup$ – user99087 May 29 '17 at 21:45
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    $\begingroup$ @user372511 Let $a$ be the area of a small sector of the hinge $[x_1\,^{x_2}_{x_3}]$ and $a'$ be the area of its image. Note that $\tfrac{a'}{a}$ gives an upper bound for the lower Lipschitz constant and $\tfrac{a'}{a}\le \ell$. $\endgroup$ – Anton Petrunin May 29 '17 at 21:56
  • $\begingroup$ @user372511, By Rademacher's theorem, any Lipschitz map is differentable almost everywhere. If the upper Lipschitz constant is 1 the lower gives the lower bound for the determinant of the Jacobian. $\endgroup$ – Anton Petrunin May 30 '17 at 15:55
  • $\begingroup$ @user372511 up to rescaling of the target space, $[k,\tfrac1k]$-bi-Lipschitz map is the same as $[k^2,1]$-bi-Lipschitz map. I prefer to fix upper constant, but after rescaling you may think of $[k,\tfrac1k]$-bi-Lipschitz if you wish. $\endgroup$ – Anton Petrunin May 31 '17 at 17:43

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