# On closed simple curve with curvature at most 1

I am looking for the reference to the following theorem.

I have to apply a similar statement, and it would be nice to trace the source. Please note, I know few proofs in fact it is Problem 3 in my collection of exercises.

Theorem. Let $\gamma$ be a closed simple plane curve with curvature at most 1. Then the disc bounded by $\gamma$ contains a circle of radius $1$.

Here is an illustration for those who are lazy to read.

P.S. So, do you think I can claim it as mine?

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Are you aware of the paper "Optimal Isoperimetric Inequalities" by F. J. Almgren? A generalization (the "Area mean curvature characterization of spheres") of a weaker form of your question is an essential ingredient. Specifically, the fact that a closed simple curve with curvature at most one bounds a region of area at least $\pi$ with equality only for a circle. The paper claims this was new, though maybe it was known in the one-dimensional case earlier... –  Rbega Mar 31 '13 at 20:47
I believe that this very question was asked on a written Ph.D. qualifying exam at Harvard while I was there (it might have even been the one I took), but after the exam the faculty member realized that he did not know the answer himself, even though it seemed obvious and elementary. I believe it took year or two before an entering graduate student was able to provide a proof. My guess is that if this ever appeared in a paper, it was stated as only a lemma. So it would be difficult to figure out who did it first. –  Deane Yang Mar 31 '13 at 23:20
Nice picture! I think the question on the Harvard qualifying exam assumed convexity. –  Deane Yang Mar 31 '13 at 23:21
@Anton it is Theorem (3) on page 452 –  Rbega Apr 1 '13 at 8:44
@Rbega, thank you, sorry, I did not read your comment carefully; I thought Almgren claims that it IS new even in one-dimensional case. –  Anton Petrunin Apr 1 '13 at 14:23

I would check work from the 1980s on the cut locus being a tree on surfaces. Consider the inward normal exponential map of the simple curve, take its cut locus, the latter is a tree, take a leaf of the tree. This point will be at least distance 1 away from the boundary if the (signed) curvature of the curve is less than 1. This works for all curves (convex or not).

For the cut locus, see for example Itoh, Jin-ichi, Essential cut locus on a surface. Proceedings of the Fifth Pacific Rim Geometry Conference (Sendai, 2000), 53–59, Tohoku Math. Publ., 20, Tohoku Univ., Sendai, 2001.

See also Zamfirescu, Tudor, On the critical points of a Riemannian surface. Adv. Geom. 6 (2006), no. 4, 493–500.

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This is not an answer, since I don't know a reference. But I now remember the proof when the curve is known to be convex. Here's a sketch:

The trick is to put the disk of radius 1 so that it osculates the curve at a point where the curvature is largest. Now parameterize the curve as a function of the angle $\theta$ that the outer unit normal makes with the positive $x$-axis with the origin placed at the center of the disk. Define the support function $h$ as a function of theta to be the dot product of the corresponding point on the curve with the outer unit normal. First, note that the disk is inside the curve if and only if $h \ge 1$ for all $\theta$. It can be shown that $h$ satisfies the ODE $$h'' + h = \rho$$ where $\rho$ is the reciprocal of curvature. The result now follows by the Sturm comparison theorem. This proof works nicely in the sense that it can be generalized to higher dimensions using the second fundamental form.

This all uses standard stuff from convex geometry, so I think it's likely it was known to Blaschke if not even earlier than that. I imagine that the proof above could be adapted to nonconvex curves, but I haven't tried. If so, I would imagine that this was also known a long time ago.

Finally, the Harvard graduate student who figured out the proof somehow knew about the support function and that it satisfies the ODE above. The rest of us knew only the more usual differential geometric definitions of curvature, which were useless for this question.

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In the nonconvex case, the proof does not work directly, at least the disk of radius 1 so that which osculates the curve at a point where the curvature is largest may not lie in the domain. In a proof I know, you look at the domains $\Omega_t$ on distance $t\le 1$ inside the curve and check that one of the connected component is a "$t$-tear"; i.e., it is bounded by a smooth closed curve with curvature $\le \tfrac1t$ and with at most one corner. It follows that $\Omega_1$; hence the result follows. –  Anton Petrunin Apr 1 '13 at 15:08
Anton, thanks! I now see your point (which is shown very clearly in your picture). I don't fully understand your proof, but that's because I'm not particularly good at geometric arguments like this. I will have to study it more carefully. –  Deane Yang Apr 1 '13 at 16:10