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



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> 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 $f(S)$. 
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I'm wondering if something similar could hold true for bi-Lipschitz functions, for example is the following claim true?



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> 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 $S\_i$.
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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)



Bi-Lipschitz version of Kirszbraun's extension theorem
It seems that you are asking the following:



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> 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?
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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).


