This is a sequel to the post about An unbounded curve with a positive lower bound for curvature. In that post we give an example of an unbounded curve whose curvature has a positive lower bound. This curve is not simple. It was conjectured by Wongting that a simple plane curve which has a positive lower bound for curvature should be bounded. After discussing with infinitely many people and a number of failed attempts , we have finally arrived at
Theorem 1 (John Ma – Hon Leung – Wongting’s Theorem, 2011) Suppose is a simple plane curve with curvature , then is contained in a bounded region. If fact, it is contained in a square whose side is of length 4.
The key lemma is the following
Lemma 2 Suppose is a simple regular plane curve parametrized by arclength such that its curvature . If , then for all .
Proof: To avoid unnecessary technicality, we assume that has domain , i.e. it is a complete curve. Clearly we only have to show it for , as the case for follows immediately by considering the reflected curve.
Let be the angle from to , then as , we can express and thus can be (re-)parametrized by .
For , observe that for and for , i.e. is strictly increasing on and strictly decreasing on , and
From this we see that for .
Suppose for all integers , then the above argument shows that for all .
We can assume w.l.o.g. that (by the translation ). We claim that there exists such that
First of all, note that on , so there exists a function such that can be expressed as graph of the function of , i.e.
for . As , by Taylor-expanding around , we see that there is such that is below the graph of , i.e.
But as and for , so there exists such that . So (1) implies
This proves our claim.
Now, let be the domain bounded by and the line segment . Here we denote by the line segment joining two points , i.e. .
Note that by Taylor-expanding around , we see that for all slightly larger than , . We claim that for all .
If this is true, then our proof would be complete, since lies below by previous analysis. Suppose the claim is not true, then there exists a smallest such that . Clearly , as is simple. Let be the point on such that . Let be the region bounded by . Then on , the “curve normal” of is the inward normal of , where is the anticlockwise rotation by . (This can be seen by looking at the point , and using the continuity of along the boundary . Although may not be well-defined at and , this minor problem can be solved by smoothing near these two points without affecting . ) Clearly , so the inward normal of at is (as for small ), so , but then
as attains a local maximum here, but this is a contradiction to the assumption that .
The previous lemma can be strengthened to
Proof: Again assume that the curve is complete. The previous lemma proves that . So using as the parameter again, and by the fact that is increasing on , we see that with . By translating the origin to and applying a rotation by , and using the previous lemma, it is easy to see that for all .
We can now prove the main theorem.
Proof: Again we can parametrize by . There are two cases: (i) has length . (ii) has length (can be infinite).
For the first case, we can assume to be the midpoint of and move so that and . Then by Lemma 3, lies in . For ,
Similarly for ,
For case (ii), we can assume that , we can then apply Lemma 3 to and respectively to show that is contained in a square of side 4.
Remark 1 Clearly the theorem can be generalized to show that if , then it is contained in a square of length .
Question: Can we prove that is contained in a square of side if ?