Homeomorphism that maps one ball to another

Question

Let $D$ be a domain in $\mathbb R^n$. There are two balls $B_1$ and $B_2$ in $D$ with the same small radius $\varepsilon$. How to construct a bi-Lipschitz homeomorphism $f$ of $D$ onto itself that maps $B_1$ onto $B_2$?

Answer 1

I'm assuming $D$ is open and connected. Then there is a path $$\gamma:\quad s\mapsto \gamma(s)\in D\qquad(0\leq s\leq L)\ ,$$ parametrized by arc length, with $\gamma(5\epsilon)={\bf p}$, $\gamma(L-5\epsilon)={\bf q}$, where ${\bf p}$ and ${\bf q}$ are the centers of the two balls. We may even assume that the ends of $\gamma$ are straight within the last $10\epsilon$. Use this path to construct a tube $T\subset D$ of thickness $10\epsilon$ that contains the two balls in its interior.

This tube is the image of a cylinder $C$ of length $L$ along the $x_1$-axis in an auxiliary space; say $T=\phi(C)$. Construct a homeomorphism $g:\>C\to C$ that fixes the boundary $\partial C$ pointwise, and maps the copy $\hat B_1:=\phi^{-1}(B_1)\subset C$ onto $\hat B_2\subset C$. We now "transport" $g$ to $T$ via the parametrization $\phi:\>C\to T$ of the tube, and define $$f(x)=x \quad(x\in D\setminus T), \qquad f(x):=\phi\bigl( g\bigl(\phi^{-1}(x)\bigr)\bigr)\quad(x\in T)\ .$$ This $f$ will be Lipschitz with a Lipschitz constant of order of magnitude ${L\over\epsilon}$.