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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.)

Zhang Yuhan
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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 $H_2$.

Jack LeGrüß
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    On the Higher Derivatives at the Boundary in Conformal Mapping I have found this paper of Warschawski.

    We need to use the extension theorem to both the biholomorphim $H_1$ from the interior region to ${|z|<1}$ and $H_2$ from the exterior region to ${|z|>1}$, but how can we guarantee the extended values and derivatives of $H_1$ and $H_2$ on the curve to be the same?

     – Zhang Yuhan Oct 07 '20 at 02:46
  • @Zhang: I have provided remarks on how to ensure the values and derivatives are agreed on the boundary. – Jack LeGrüß Oct 07 '20 at 10:22
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    If the extended derivatives of $H_1$ and $H_2$ on the curve are nonzero, I can use Whitney Extension Theorem and Inverse Function Theorem to prove $H_1$ and $H_2$ differ on the curve by up to a diffeomorphism. However, since a diffeomorphism must have nonzero Jacobian determinant, it would be troublesome if the derivative of $H_1$ or $H_2$ vanishes on the curve. Is there anyway to avoid it? – Zhang Yuhan Oct 07 '20 at 10:57
  • Oh no, you don’t need to go through that trouble. The ‘difference’ in diffeomorphism on the curve is simply the composition $H_1^{-1}\circ H_2$ (modulo ordering and\or inversion). – Jack LeGrüß Oct 07 '20 at 11:36
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    But the final diffeomorphism $H$ on $\mathbb{C}$ must have nonzero Jacobian determinant at every point. If, let's say, the extended derivative of $H_2$ is zero at the point $z_0$ on the curve, it would be impossible for the Jacobian determinant of $H$ at $z_0$ to be nonzero. – Zhang Yuhan Oct 07 '20 at 11:51
  • Umm, I think that is not going to happen. Boundary behavior of biholomorphisms in this case will prevent that (Note that the derivative of inverse is $(f^{-1})’(f(z))=\frac{1}{f’(z)}$, so they’d be nonzero on the boundary). – Jack LeGrüß Oct 07 '20 at 17:31
  • Thanks. I would continue going through the details for a while, but I think this is the right direction. – Zhang Yuhan Oct 08 '20 at 04:41