Let $\gamma: (0,1) \to \mathbb{R}^2$ be a $C^\infty$ regular plane curve, and suppose that its curvature is at least $k_0$ everywhere. Is the image of $\gamma$ contained in disk of radius $\frac{1}{k_0}$?

Intuitively, it seems likely to be true. If we choose $p$ at a distance of $\frac{1}{k_0}$ from and perpendicular to $\gamma(0)$, then it seems like the curve will be inching towards $p$. However, I can't seem to prove that this is the case.

If the image of $\gamma$ is in $\mathbb{R}^3$ (or $\mathbb{R}^n$ for any $n > 2$), then I believe the result is false, since we can just take tightly wound helix that travels upwards very high. Therefore, if the result is true in $\mathbb{R}^2$, then I think it really uses the $2$-dimensional-ness of the space.

One thing I know is that if $|\gamma(t)|$ is maximized when $t = t_0$, then $|k(t_0)| \ge \frac{1}{|\gamma(t_0)|}$. Maybe we can translate the curve so that the $p$ defined above is the origin, and then use something like this somehow? The inequality goes in the wrong direction, so we probably need some sort of opposite result, but I'm not sure what that result would be.

  • $\begingroup$ I think the analogous statement would be that the curve stays within the $n$-sphere of radius $1/k_0$, which is probably also true. $\endgroup$ – Elliot G Mar 16 '18 at 6:59
  • $\begingroup$ I agree that's the analogous statement, but it seems to me that a helix would still be a counter-example, would it not? It could go up higher than $1/k_0$. $\endgroup$ – user541304 Mar 16 '18 at 7:01

This is not true: the "helical" behaviour can manifest - even in the plane - so long as you allow self-intersections. For example, a suitable reparametrization of $$\gamma:\mathbb R \to \mathbb R^2:\gamma(t)=(t+2 \sin t,2 \cos t)$$ produces an unbounded curve looking something like this:

projection of helix

Hopefully just from the picture you believe this curve has curvature bounded below - otherwise, you can do a manual calculation to verify that $\gamma'\times\gamma''$ stays away from zero. Since it is unbounded, it leaves every disc, and thus is a counterexample.

The trick we're pulling here is essentially the same as the one required to drive a car that can only turn left - if you want to turn right, just turn left three times.

If you assume $\gamma$ is a simple closed curve then my intuition is that your implication holds - the fact that the curvature has constant sign will then tell you that $\gamma$ is convex, so writing it as a polar graph $r(\theta)$ will probably be fruitful.

  • $\begingroup$ Thanks! What if - even stronger than being simple - we assume $\gamma$ is the graph of a function over the interval $(-1, 1)$. Do you know if the result is easier then? $\endgroup$ – user541304 Mar 16 '18 at 12:49
  • $\begingroup$ A simple closed curve is compact so must be contained in a disk of finite radius in any event! $\endgroup$ – Robert Lewis Mar 16 '18 at 16:17
  • $\begingroup$ I wonder what happens if we assume the curvature is increasing, perhaps at a certain minimum rate? $\endgroup$ – Robert Lewis Mar 16 '18 at 16:18
  • $\begingroup$ And I wonder what happens if self-intersections are forbidden. Anyway, thanks for a really nice answer! $\endgroup$ – Robert Lewis Mar 16 '18 at 16:22
  • $\begingroup$ @RobertLewis: yes, of course the nontrivial part is just the explicit estimate. Just disallowing self-intersections makes it a very interesting question - my intuition is that it should be enough, but I have no idea what a proof would look like: injectivity is a tough condition to leverage geometrically. Perhaps some kind of two-point estimate (cf. this argument in the parabolic case). $\endgroup$ – Anthony Carapetis Mar 16 '18 at 22:12

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