Simple curves with a positive lower bound for curvature

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 {\underset{\sim}{>\;<}}, we have finally arrived at

Theorem 1 (John Ma – Hon Leung – Wongting’s Theorem, 2011) Suppose {\gamma } is a simple plane curve with curvature {k\geq 1}, then {\gamma } is contained in a bounded region. If fact, it is contained in a square whose side is of length 4.

John Ma's example of a simple bounded curve with k>c>0

The key lemma is the following

Lemma 2 Suppose {\gamma(s)=(x(s), y(s)) } is a simple regular plane curve parametrized by arclength such that its curvature {k(s)\geq 1}. If {\gamma(0)=0, \gamma'(0)=e_1}, then {y(s)\leq 2} for all {s}.

Proof: To avoid unnecessary technicality, we assume that {\gamma(s)} has domain {\mathbb{R}}, i.e. it is a complete curve. Clearly we only have to show it for {s\geq 0}, as the case for {s\leq 0} follows immediately by considering the reflected curve.

Let {\theta(s)} be the angle from {e_1} to {\gamma'}, then as {k(s)=\frac{d\theta }{ds}\geq 1}, we can express {s=s(\theta)} and thus {\gamma} can be (re-)parametrized by {\theta}.

For {\theta \in [0,2\pi]}, observe that {y'(\theta)= y'(s)s'(\theta)= \sin \theta /k(\theta)>0} for {\theta \in(0,\pi)} and {y'(\theta)< 0} for {\theta \in (\pi, 2\pi)}, i.e. {y} is strictly increasing on {(0,\pi)} and strictly decreasing on {(\pi, 2\pi)}, and

\displaystyle  \begin{array}{rcl}  0<y(\pi)-y(0)=\displaystyle \int_0^\pi y'(\theta)d\theta =\int_0^\pi y'(s)s'(\theta)d\theta &= &\displaystyle \int_0^\pi \frac{\sin\theta}{k(\theta)}d\theta \\ &\leq& \displaystyle \int_0^\pi \sin \theta d\theta\\ &=&2. \end{array}

From this we see that {y(\theta)\leq 2} for {\theta\in [0,2\pi]}.

Suppose {y(2m\pi)\leq 0} for all integers {m\geq 1}, then the above argument shows that {y(\theta)\leq 2} for all {\theta\geq 0}.


Otherwise, there exists a smallest {m} such that {y(2m\pi)>0}.

We can assume w.l.o.g. that {m=1} (by the translation {\theta\rightarrow \theta-(m-1)2\pi}). We claim that there exists {\theta_0\in (0, \pi)} such that

\displaystyle  \begin{cases} y(\theta_0)=y(2\pi)\\ x(\theta_0)> x(2\pi). \end{cases}

First of all, note that {y'(\theta)>0} on {(0, \pi)}, so there exists a function {f:[y(0), y(\pi)]\rightarrow \mathbb{R}} such that {\gamma |_{[0, \pi]}} can be expressed as graph of the function {f} of {y}, i.e.

\displaystyle (f(y), y)= (f(y(\theta)), y(\theta))=(x(\theta), y(\theta))

for {\theta\in [0, \pi]}. As {\gamma'(\pi)//-e_1}, by Taylor-expanding {\gamma } around {\pi<\theta<2\pi}, we see that there is {\theta_1>\pi} such that {\gamma(\theta_1)} is below the graph of {f}, i.e.

\displaystyle  x(\theta_1)< f(y(\theta_1)).

As {\gamma} is simple, by continuity of {x(\theta)-f(y(\theta))},

\displaystyle  x(2\pi)< f(y(2\pi)). \ \ \ \ \ (1)

But {y(2\pi)\in (y(0), y(\pi))} as {y(0)\leq 0} and {y'(\theta)<0} for {\theta\in (\pi, 2\pi)}, so there exists {0<\theta_0<\pi} such that {y(\theta_0)= y(2\pi)}. So (1) implies

\displaystyle  \begin{cases} y(2\pi)= y(\theta_0)\\ x(2\pi)< f(y(2\pi))=f(y(\theta_0))= x(\theta_0). \end{cases}

This proves our claim.

Now, let {\Omega} be the domain bounded by {\gamma|_{[\theta_0, 2\pi]}} and the line segment {l:=(\gamma (2\pi), \gamma(\theta_0))}. Here we denote by {(p, q)} the line segment joining two points {p,q\in\mathbb{R}^2}, i.e. {(p, q)= \{(1-t)p+tq: t\in (0,1)\}}.

Note that by Taylor-expanding {\gamma} around {\theta= 2\pi}, we see that for all {\theta} slightly larger than {2\pi}, {\gamma(\theta)\subset \Omega}. We claim that {\gamma(\theta)\subset \Omega} for all {\theta> 2\pi}.

If this is true, then our proof would be complete, since {\Omega} lies below {\{y=2\}} by previous analysis. Suppose the claim is not true, then there exists a smallest {\theta_1> 2\pi} such that {\gamma(\theta_1)\not\in\Omega}. Clearly {\gamma(\theta_1)\in l}, as {\gamma } is simple. Let {p=\gamma(\theta_2)} be the point on {\gamma|_{[2\pi, \theta_1]}} such that {\displaystyle y(\theta_2)=\max_{\theta\in[2\pi, \theta_1]} y(\theta)}. Let {R} be the region bounded by {\gamma|_{[\theta_0, \theta_1]}\cup( \gamma(\theta_1),y( \theta_0))}. Then on {\gamma|_{[\theta_0, \theta_1]}}, the “curve normal” {n(\theta):= J\gamma'(\theta)}/|\gamma'(\theta)| of {\gamma} is the inward normal {N} of {R}, where {J} is the anticlockwise rotation by {\pi/2}. (This can be seen by looking at the point {\gamma (2\pi)}, and using the continuity of {\langle n, N\rangle\in \{\pm1\}} along the boundary {\partial R}. Although {N} may not be well-defined at {\gamma (\theta _1)} and {\gamma (\theta_0)}, this minor problem can be solved by smoothing {\partial R} near these two points without affecting {p}. ) Clearly {p\in \Omega}, so the inward normal {N} of {\partial R} at p is {e_2} (as {p+\varepsilon e_2\in R} for small {\varepsilon>0}), so {\gamma'(\theta_2)= e_1}, but then

\displaystyle k(\theta_2)= \frac{ \left|\begin{matrix} x' & x''\\ y' & y'' \end{matrix}\right| }{|\gamma'|^3}= \frac{\left|\begin{matrix} 1 & x''\\ 0 & y'' \end{matrix}\right| }{|\gamma'|^3}=\frac{ y'' }{|\gamma'|^3}\leq 0

as {y} attains a local maximum here, but this is a contradiction to the assumption that {k\geq 1}. \Box

The previous lemma can be strengthened to

Lemma 3 Suppose {\gamma(s)=(x(s), y(s)) } is a simple regular plane curve parametrized by arclength such that its curvature {k(s)\geq 1}. If {\gamma(0)=0, \gamma'(0)=e_1}, then {-2\leq y(s)\leq 2} for all {s}.

Proof: Again assume that the curve is complete. The previous lemma proves that {\gamma \subset \{y\leq 2\}}. So using {\theta } as the parameter again, and by the fact that {y (\theta)} is increasing on {(0, \pi)}, we see that {0<y(\pi)\leq 2} with {y'(\pi)//-e_1}. By translating the origin to {\gamma(\pi)} and applying a rotation by {\pi}, and using the previous lemma, it is easy to see that {-2<y(\theta)\leq 2} for all {\theta}. \Box

We can now prove the main theorem.

Proof: Again we can parametrize {\gamma} by {\theta \in I}. There are two cases: (i) {I} has length {\leq \pi}. (ii) {I} has length {>\pi} (can be infinite).

For the first case, we can assume {0} to be the midpoint of {I} and move {\gamma} so that {\gamma(0)=0} and {\gamma'(0)//e_1}. Then by Lemma 3, {\gamma} lies in {\{-2\leq y\leq 2\}}. For {0\leq \theta_0\leq \pi/2},

\displaystyle 0\leq x(\theta_0)-x(0)= \int_0^{\theta_0 }x'(\theta)d\theta=\int_0^{\theta_0} \frac{\cos \theta}{k(\theta)}d\theta\leq \int_0^{\pi/2} \cos \theta d\theta= 1.

Similarly for {-\pi/2\leq \theta_0<0},

\displaystyle  -1\leq x(\theta_0)\leq 0.

Thus {\gamma \subset \{-2\leq y\leq 2\}\cap \{-1\leq x\leq 1\}}.

For case (ii), we can assume that I\supset [0,\pi/2], we can then apply Lemma 3 to {\gamma (0)} and {\gamma(\pi/2)} respectively to show that {\gamma} is contained in a square of side 4. \Box

Remark 1 Clearly the theorem can be generalized to show that if {k\geq k_0>0}, then it is contained in a square of length {4/k_0}.

Question: Can we prove that {\gamma} is contained in a square of side {2} if {k\geq 1}?

Advertisements
This entry was posted in Geometry. Bookmark the permalink.

One Response to Simple curves with a positive lower bound for curvature

  1. Marco Barchiesi says:

    Nice theorem. Is it published somewhere?

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s