## 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.

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

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 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 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 ${k\geq 1}$?