Bi-Lipschitz extension

Question

Given a bi-Lipschitz homeomorphism $\Phi:\mathbb{B}^n(0,1)\to\mathbb{R}^n$, (that is a bi-Lipschitz map onto the image), can one find a bi-Lipschitz homeomorphism $\Psi:\mathbb{R}^n\to\mathbb{R}^n$ such that $$ \Psi|_{\mathbb{B}^n(0,\frac{1}{2})}=\Phi|_{\mathbb{B}^n(0,\frac{1}{2})}\ \ ? $$

It follows from the planar bi-Lipschitz Schoenflies theorem [T] that when $n=2$, the answer is yes since you can actually extend $\Phi$ from $\mathbb{B}^2(0,1)$ to $\mathbb{R}^2$ in a bi-Lipschitz manner, but in higher dimensions it is not possible to extend from $\mathbb{B}^n(0,1)$, and this is why I am asking about extension from a smaller ball.

Thickening of the Fox-Artin arc creates a domain $\Omega$ in $\mathbb{R}^3$ that is bi-Lipschitz homeomorphic to $\mathbb{B}^3(0,1)$, but no bi-Lipschitz homeomorphism $\Phi:\mathbb{B}^3\to\Omega$ can be extended to a homeomorphism of $\mathbb{R}^3$. This was observed by Fred Gehring in 1968, but details were not provided. Details are provided in Theorem 3.7 in [M].

[M] Martin, G.: Quasiconformal and bi-Lipschitz homeomorphisms, uniform domains and the quasihyperbolic metric. Trans. Amer. Math. Soc. 292 (1985), 169–191.

[T] Tukia, P.: The planar Schönflies theorem for Lipschitz maps. Ann. Acad. Sci. Fenn. Ser. A I Math. 5 (1980), no. 1, 49–72.

Answer 1

Let me describe a baby case of Sulivan's construction which is beautiful.

Assume I have a homeomorphism $h$ from pentagon to itself that sends each side to itself.

Let us assume that the pentagon has right angles in the conformal disc model of Lobachevsky plane. Extend the map $h$ to the disc applying reflections in the sides of pentagon, so the obtained map $\tilde h$ commutes with and any reflection in a side of pentagon.

Note that $\tilde h$ is bi-Lipschitz and it is identity on the boundary of the disc. So we can extend $\tilde h$ by identity map outside of disc.

It is problematic to do the same for general $h$ and higher dimensions, but all this seem to be solved by Sullivan.

Answer 2

The following result answers the question by providing a locally bi-Lipschitz extension.

Theorem. Given a bi-Lipschitz homeomorphism $\Phi:\mathbb{B}^n(0,1)\to\mathbb{R}^n$, (that is a bi-Lipschitz map onto the image), one can find a locally bi-Lipschitz homeomorphism $\Psi:\mathbb{R}^n\to\mathbb{R}^n$ such that $$ \Psi|_{\mathbb{B}^n(0,\frac{1}{2})}=\Phi|_{\mathbb{B}^n(0,\frac{1}{2})}. $$

This result is a straightforward consequence of the Theorem 5.10 in [TV]. If we regard $\mathbb{B}^n(0,1)$ as a subset of $\mathbb{S}^n$ (stereographic projection + one point compactification of $\mathbb{R}^n$), then $\Phi$ can be extended from $\mathbb{B}^n(0,\frac{1}{2})$ to a bi-Lipschitz map of $\mathbb{S}^n$. Such a map gives a locally bi-Lipschitz map of $\mathbb{R}^n$. Theorem 5.10 is a version of the generalized Schoenflies theorem for bi-Lipschitz maps.

Remark. I think it is possible to find not only a locally bi-Lipschitz, but a bi-Lipschitz extension. Argument would be the same as above, with the only difference that we would place at the north poles (from which we have stereographic projections) points there the extension and its inverse are differentiable. But, I have to think about this argument a bit more.

[TV] Tukia, P.; Väisälä, J.: Lipschitz and quasiconformal approximation and extension. Ann. Acad. Sci. Fenn. Ser. A I Math. 6 (1981), no. 2, 303–342 (1982).