Jordan-Schönflies Curve Theorem states:
For any simple closed curve $\sigma$ in the plane, there is a homeomorphism $H$ of the plane which takes that curve into the standard circle.
Question: If $\sigma$ is a $C^k$ simple closed curve, can we choose $H$ to be a $C^k$ diffeomorphism?
I am trying to prove that if $Ran(\sigma)$ is the topology boundary of a bounded open set $U$, then $\overline{U}$ is a $C^k$ manifold with boundary whose manifold boundary is exactly $Ran(\sigma)$. (Edit: I have also found an alternative proof of it using the smooth Jordan-Brouwer Separation Theorem in this post: Differentiable Version of the Jordan-Brouwer Separation Theorem.)
Yes you can.
First you require an important extension of the Riemann Mapping Theorem (by Painlevè, Kellog, Warschawski—sorry, I don’t have references at hand but I believe they could be found easily online) that says that any biholomorphism of the open interior of a Jordan curve to the open unit disk centered at the origin extends continuously to a $C^k$ map of the boundary if and only if the boundary is a $C^k$ Jordan curve.
Apply the above extension first to the interior of the curve in $\mathbb{C}$ and then to the compactified exterior in the Riemann sphere $\widehat{\mathbb{C}}$ and you obtain a $C^k$-diffeomorphism as required.
$$\underline{Remarks}$$ The two biholomorphisms will differ on the $C^k$ simple closed curve by up to a diffeomorphism. But by (a smooth version of) Alexander’s Trick, that diffeomorphism can be extended to a diffeomorphism of the open unit disk, which upon composing appropriately with, say $H_1$ — the biholomorphism of the interior—will give a diffeomorphism that agrees on the boundary with that or $H_2$.