Can closed curves have small curvature?

Question

Let $\gamma$ be a smooth curve in Euclidean space of length $2\pi$ whose curvature function satisfies $-1 < k(t) < 1$. Can $\gamma$ be closed?

This seems like it should be an easy exercise, at least in the plane, but I'm stuck...

Answer 1

do Carmo's Differential Geometry of Curves and Surfaces mentions Fenchel's theorem on page 399:

The total curvature of a simple closed curve is at least $2\pi$

(plus something about when equality holds; this statements differs from the wikipedia page for Fenchel's theorem.)

In your case, you have that the total curvature $\int |k(s)| ds$ is $< 2\pi$. So the curve cannot be closed (if it is simple).

There is a generalization to piecewise regular closed curves in $\mathbb R^n$, known as the Fenchel-Borsuk theorem, though I can't find a handy reference right now.

Answer 2

If the curve was closed, then the Gauss-Bonnet formula for curves would apply, that is, $\int_0^{2\pi} k(t) dt= ±2\pi$ (where the curve is parameterized by arc length), but by the inequality $|k(t)|<1$ we get $\int_0^{2\pi} k(t) dt<\int_0^{2\pi}1dt = 2\pi$ and $\int_0^{2\pi} k(t) dt> -2\pi$.

edit : We might need to add the hypothesis that the curve is simple for this method to work, so this might not be a good solution.

edit 2 : I think that if the curve is not simple, then $\int_0^{2\pi} k(t) dt= 2k\pi$ where k is the winding number of the curve, which would fix this solution. See this link

Answer 3

Take the mid-point and assume that the unit tangent vector at that point is looking in the direction of the $x$-axis. Now estimate the $x$-projection of the unit tangent vector at the distance $t$ from the midpoint. It is greater than $\cos t$ because the angle of rotation is less than $t$ due to the curvature restriction. But $\int_{-\pi}^\pi\cos t\,dt=0$, so you finish to the right of where you started. This works in all dimensions.