Let $\alpha :[0, l] \rightarrow \mathbb{R}^{2}$ parameterize a simple closed curve by arc length. Suppose that there is a constant $R$ such that the curvature of $\alpha$ satisfies $0<\kappa_{\sigma}(s) \leq R$. Prove that $\text{length}({\alpha}) \geq \frac{2 \pi}{R}.$

Hint: Note that the hypothesis say that the curve is curved less than the circle of radius $\frac{1}{R},$ then its length is greater than the length of the circle.

That's the problem. I was thinking about using the theorem which says that the length of a curve C in $\mathbb{R}^2$ has length $$\frac{1}{4}\int_{0}^{2\pi}length(P_\sigma(C))d\sigma,$$ where $P_\sigma(C)$ is the projection of the curve. The problem is that I don't know how to get a relation between the curvature and the projection. Maybe I am completely wrong and there is another way to solve the problem. Any ideas?

  • $\begingroup$ Hint: $\kappa_{\alpha}(s) = \left| \frac{d\theta}{ds}\right|$ where $\theta$ is the angle between the tangent vector at $\alpha(s)$ and $x$-axis. What is $\int_0^l \frac{d\theta}{ds} ds$? $\endgroup$ – achille hui Sep 14 '19 at 4:22
  • $\begingroup$ Awesome, now it's done Thank you so much, really. I will post the completely answer just in case someone need it one day $\endgroup$ – vicase98 Sep 14 '19 at 13:59

Here is a different solution: By the Hopf Umlaufsatz, $\int_C\kappa\,ds = 2\pi$. Now use $\kappa\le R$ to get $2\pi \le R\,\text{length}(\alpha)$.


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