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Let $\gamma:[0,L]\rightarrow \mathbb{R}^2$ be a $C^\infty$ curve parameterized by arc length. We suppose that $\gamma$ is a simple closed curve that bounds a bounded domain in $\mathbb{R}^2$.

We denote by $\nu(t)$ the inward unit normal at $\gamma(t)$. For any $\epsilon>0$ sufficiently small, $t\in[0,L]\mapsto \gamma_\epsilon(t)=\gamma(t)+\epsilon \nu(t)$ is a still a smooth curve.

Question. Is it true that the length of the curve $\gamma_\epsilon$ is less than or equal to the length of $\gamma$?

Kosh
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  • Seems unlikely to me. – copper.hat Dec 22 '19 at 23:10
  • @Arnaud answer is fine. – Kosh Dec 22 '19 at 23:26
  • Cool. Just not clear how $\epsilon$ factors into his answer. – copper.hat Dec 22 '19 at 23:29
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    If you compute the norm to the square of $\gamma_\epsilon'(t)$ you get $|\gamma_\epsilon'(t)|^2= (1+ \epsilon\cdot c(t))^2$. That is, $|\gamma_\epsilon'(t)|= |1+ \epsilon\cdot c(t)|$. But the curvature is bounded, and therefore for $\epsilon$ sufficiently small you can remove the absolute value and then use the Hopf theorem which links the total curvature to the winding number. – Kosh Dec 22 '19 at 23:35
  • Thanks ${}{}{}$ – copper.hat Dec 22 '19 at 23:38

1 Answers1

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Yes: use the fact that $d\nu/dt = -c(t) d\gamma/dt$ where $c(t) = d\arg \gamma / dt$ is the curvature, and that $\int c(t)dt = 2\pi$.

Arnaud
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