# For a closed plane curve, showing some inequalities.

I have a problem following :

Let $\gamma:[0,T]→\mathbb{R}^2$ be a closed plane curve, i.e., a regular parametrized curve such that $\gamma$ and all its derivatives agree at 0 and $T$. For convenience of formulation, we assume that $\gamma$ is parametrized by arc-length

(1) Show that there exist $t_1, t_2 \in [0,T]$ such that $\gamma'(t_1)=-\gamma'(t_2)$.

(2) Let $\kappa(t)$ denote the curvature of $\gamma$ at t. Show that $$\int_0^{T} |\kappa(t)|dt \ge 2\pi$$

I tried to solve this problems. Since $\gamma$ is parametrized by arc-length, it is unit-speed curve. So, I know the part(1) geometrically, but I don't know the perfect answer. My tutor gave me a hint 'consider the integral $$\int_0^{T} \gamma'(t)dt$$

I know the value of the integral is zero because $\gamma$ is a closed curve. But I don't proceed more..

Also, for part (2), I hope to apply Gauss-Bonnet Theorem.

Since it is a plane curve, the Gaussian curvature is 0, and the curvature $\kappa$ is equal to the geodesic curvature $\kappa_g$. But I don't know why does the inequality in (2) hold.

• The value of the second integral is $T$, not zero. – Michael Hoppe Oct 19 '13 at 13:27
• Oh my mistake.. I edited it – NNNN Oct 19 '13 at 13:39
• @Landscape Thank you for your attention. The conditions that I wrote are all. So we may assume that it is not simple, generally. – NNNN Oct 22 '13 at 9:58

For convenience, let us extend the domain of $\gamma$ to $\Bbb R$ by defining $\gamma(t+nT)=\gamma(t)$, $\forall n\in\Bbb Z$.

(1) Denote the unit circle in $\Bbb R^2$ by $S^1$, i.e. $$S^1=\{(\cos \theta, \sin \theta)\in \Bbb R^2\mid \theta\in \Bbb R\}.$$

Since $\gamma$ is parameterized by arc-length, $\gamma'$ can be considered as a continuous and $T$-periodic map from $\Bbb R$ to $S^1$. Therefore, $\gamma'(\Bbb R)=\gamma'([0,T])$ must be a closed and connected subset of $S^1$, i.e. there exist $a\le b$, such that $$\gamma'(\Bbb R)=\{(\cos \theta, \sin \theta)\in \Bbb R^2\mid \theta\in [a,b]\}.$$ To prove (1), it suffices to show that $b-a\ge\pi$. Denote $\theta=\frac{a+b}{2}$ and $v=(\cos\theta, \sin\theta)$. $$\int_0^T\langle v,\gamma'(t)\rangle dt= \langle v,\int_0^T\gamma'(t) dt\rangle =0\Longrightarrow \exists t\in[0,T],\, \langle v,\gamma'(t)\rangle\le 0\Longrightarrow b-a\ge \pi,$$ which completes the proof.

(2) To apply Gauss–Bonnet theorem, let us follow the statements and notations in this page, and apply Equation $(3)$ there.

If $\gamma$ is a simple closed curve, we may choose $M$ as the closure of the bounded component of $\Bbb R^2\setminus \gamma([0,T])$. Then $\partial M=\gamma([0,T])$, $\chi(M)=1$, $K=0$, $|\kappa_g|=|\kappa|$, and there is no jump angle, so by Gauss–Bonnet theorem, $$\int_0^T|\kappa(t)|dt\ge |\int_0^T\kappa(t)dt|=2\pi.$$

Otherwise, $\gamma$ is self-intersecting. Let $$t_1:=\sup \{t\in[0,T]: \gamma|_{[0,t]} \text{ is a simple curve} \}.$$ Then $0<t_1<T$ and there exists $s_1\in [0,t_1)$, such that $\gamma|_{[s_1,t_1]}$ is a simple closed curve, which has only one posible jump angle $\alpha_1$ at $\gamma(s_1)=\gamma(t_1)$ with $|\alpha_1|\le \pi$. Similar to the simple closed curve case, we may choose $M$ as the closure of the bounded component of $\Bbb R^2\setminus \gamma([s_1,t_1])$ with $\partial M=\gamma([s_1,t_1])$. By Gauss–Bonnet theorem, $$\int_{s_1}^{t_1}|\kappa(t)|dt\ge |\int_{s_1}^{t_1}\kappa(t)dt|=|2\pi-\alpha_1|\ge \pi.$$ Similarly, let $$t_2:=\sup \{t\in[t_1,s_1+T]: \gamma|_{[t_1,t]} \text{ is a simple curve} \}.$$

Then $t_2>t_1$ and there exists $s_2\in [t_1,t_2)$, such that $\gamma|_{[s_2,t_2]}$ is a simple closed curve, which has only one possible jump angle $\alpha_2$ at $\gamma(s_2)=\gamma(t_2)$ with $|\alpha_2|\le \pi$. Then we may choose $M$ as the closure of the bounded component of $\Bbb R^2\setminus \gamma([s_2,t_2])$ with $\partial M=\gamma([s_2,t_2])$. By Gauss–Bonnet theorem, $$\int_{s_2}^{t_2}|\kappa(t)|dt\ge |\int_{s_2}^{t_2}\kappa(t)dt|=|2\pi-\alpha_2|\ge \pi.$$

Since $s_1<t_1\le s_2<t_2\le s_1+T$,

$$\int_0^T|\kappa(t)|dt=\int_{s_1}^{s_1+T}|\kappa(t)|dt\ge\int_{s_1}^{t_1}|\kappa(t)|dt+\int_{s_2}^{t_2}|\kappa(t)|dt\ge 2\pi.$$

• Thank you very much!!! By the way, I don't understand one in your answer. In the equality $$\int_0^T\langle v,\gamma'(t)\rangle dt= \langle v,\int_0^T\gamma'(t) dt\rangle =0$$, why do you assert that the first equality holds? Could you explain that more precisely? – NNNN Oct 22 '13 at 12:24
• @JeongNam-ho: It's just the linearity of integral. To see this more clearly, you may write write $v=(v_1,v_2)$, $\gamma'(t)=(x'(t),y'(t))$, and evaluate both sides of the equality. – 23rd Oct 22 '13 at 12:33
• @JeongNam-ho: When I looked back at this post incidentally, I found that the remark $\gamma'(\Bbb R)=S^1$ in the last version of my answer is incorrect in general, so I removed it. (The statement is correct when $\gamma$ is a simple closed curve and when I made the remark, I overlooked the self-intersecting case.) Here is a counter-example of the remark for self-intersecting case. Let $\gamma$ be the arc-length reparametrization of the curve $t\mapsto (2\sin t, \sin 2t)$. Then it's easy to check that $(0,1)\notin \gamma'(\Bbb R)$. – 23rd Nov 6 '13 at 11:15
• @JeongNam-ho: Moreover, when I checked my answer, I found I had given a wrong link to Gauss–Bonnet theorem. I cannot believe how that could happen! Anyway, the link is corrected now. I sincerely apologize for these mistakes. – 23rd Nov 6 '13 at 11:16
• Thanks a lot! But I understood your original mean, so I had no uncomfortablity. Anyway, thank your attention:) – NNNN Nov 13 '13 at 9:46