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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.

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.

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 $$ \Psi|_{\mathbb{B}^n(0,\frac{1}{2})}=\Phi|_{\mathbb{B}^n(0,\frac{1}{2})}\ \ ? $$

When $n=2$ the answer is yes and 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 $\mathbb{R}^3$ and a bi-Lipschitz homeomorphism $\Phi:\mathbb{B}^3\to\Omega$ such that $f$ cannot 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:

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

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.

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 $$ \Psi|_{\mathbb{B}^n(0,\rfrac{1}{2})}=\Phi|_{\mathbb{B}^n(0,\rfrac{1}{2})}\ \ ? $$ When $n=2$ the answer is yes and 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 and this is why I am asking about extension from a smaller ball.

Thickening of the Fox-Artin arc created a domain $\Omega$ in $\mathbb{R}^2$ and a bi-Lipschitz homeomorphism $\Phi:\mathbb{R}^3\to\Omega$ such that $f$ cannot 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:

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

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 $$ \Psi|_{\mathbb{B}^n(0,\frac{1}{2})}=\Phi|_{\mathbb{B}^n(0,\frac{1}{2})}\ \ ? $$

When $n=2$ the answer is yes and 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 $\mathbb{R}^3$ and a bi-Lipschitz homeomorphism $\Phi:\mathbb{B}^3\to\Omega$ such that $f$ cannot 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:

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