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Given a piecewise $C^1$ Jordan curve $\sigma$, there is a region $A$ bounded by $\sigma$. Is it true that $\overline A$ is $C^1$-diffeomorphic to a closed polygon? (I believe this is the same as claiming $\overline A$ has a $C^1$ triangulation.)

A curve $\phi: [0,1]\to \mathbb R^2$ is said to be piecewise $C^1$ if there exist finitely many $a_i$ that $0 = a_0 < a_1 < \cdots < a_{n-1} < a_n = 1$, and $\phi$ is $C^1$ on each $[a_i, a_{i+1}]$.

The claim holds if $\sigma$ is $C^1$ (not just piecewise $C^1$) as is shown in this question.

I am looking for a rigorous proof of Green's theorem which is said to hold for piecewise $C^1$ Jordan curves.

Zhang Yuhan
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  • Please, add your definition of a piecewise $C^1$ curve. – Moishe Kohan Dec 03 '20 at 13:35
  • OK, I have edited the question. – Zhang Yuhan Dec 03 '20 at 14:50
  • Then the claim is false. – Moishe Kohan Dec 03 '20 at 14:54
  • OK, is there any reference or counterexample? Is it still false if the direction of one-sided limits of $\frac {d\phi} {dt}$ at $a_i$ are not opposite? – Zhang Yuhan Dec 03 '20 at 15:22
  • An example would be a curve which contains a small neighborhood of $(0,0)$ in ${(x,y): x^2=y^3}$. But if you make the assumption about derivative as in your comment then such $C^1$-diffeomorphism probably exists. This essentially a local question, I will have to think more about it. In any case, you do not really need this for Green's Theorem, see here. – Moishe Kohan Dec 03 '20 at 20:13

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