Extension of Homeomorphism of boundaries

Question

Let $A,B \subset \mathbb{R}^2$ be two open bounded sets with smooth boundary (boundary is a smooth closed curve).

Let us assume that we know $$\Phi \colon \partial A \to \partial B$$ is a homeomorphism. Is there a way to extend $\Phi$ to be a diffeomorphism $\tilde{\Phi} \colon A \to B$?

Answer 1

Here is one way to prove this.

1. Both $A, B$ are simply-connected, bounded by Jordan curves. Consider the (conformal) Riemann mappings $r_A: D^2\to A, r_B: D^2\to B$, where $D^2$ is the open unit disk in ${\mathbb C}$. By Caratheodory's extension theorem, both $r_A, r_B$ extend to homeomorphisms $R_A: \bar{D}^2\to \bar{A}, R_B: \bar{D}^2\to \bar{B}$ of the closures.

2. By composing, $f:= R_B^{-1}\circ \Phi\circ R_A$, we obtain a homeomorphism $f: S^1\to S^1$ of the boundary circle of the unit disk $D^2$.

3. Extend $f$ to a diffeomorphism $F: D^2\to D^2$ using, say, the Douady-Earle extension.

4. The composition $$ r_B\circ F \circ r_A^{-1}: A\to B $$ is the required diffeomorphism extending the homeomorphism $\Phi$ of the boundary circles.