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...
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...
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.
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
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.