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I think of a theorem about a differentiable simple closed curve in 2D that I would like to prove. Here it is:

Let $C$ be a differentiable, regular, simple, closed curve in $\mathbb{R}^2$ parametrized by $\gamma:[a,b]\rightarrow\mathbb{R}^2$. Let $\gamma(t)=\left(x(t),y(t)\right)$. Then, there exist distinct points $p,q \in [a,b]$ such that $y'(p)=y'(q)=0$ and $x'(p)<0$, $x'(q)>0$.

I try on several random curves and it works, so I am certain this is true. I try to find a counterexample, but none comes to mind. However, I don't study much about Differential Geometry yet, so I am stuck on finding suitable tools to use. I try to look into some books and sites, and none of them give me anything related to that.

Does anyone have idea on proving it, or suitable tools that I can use? Or maybe suitable reference?

Joel
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  • I suppose Bolzano's theorem should be of help here. – Vincenzo Tibullo Aug 20 '14 at 12:10
  • What about the following curve $\gamma(t)=(\cos(2t)\cos(t),\cos(2t)\sin(t))$ defined on $\left[\frac{5\pi}{4},\frac{7\pi}{4}\right]$. It is a differentiable, regular, simple closed curve but the derivative of $y(t)$ vanishes only once. May be you should ask for more regularity on $\gamma$, for example $\mathcal C^1$ on $[a,b]$ and such that $\gamma'(a)=\gamma'(b)$. – Bebop Aug 20 '14 at 12:38

1 Answers1

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Here is a straight forward argument:

Without loss of generality we may assume that $y(t)$ attains its maximum at $t = a$. Choose $v \in ]a,b[$ such that $\gamma(t)$ attains its minimum at $v$. Then clearly $y'(a) = y'(v) = 0$. By regularity $x'(a) \neq 0$ and $x'(v) \neq 0$.

Now assume your statement is wrong. Then we may without loss of generality assume that $x'(a) > 0$ and $x'(v) > 0$. Then, since $x'(v) > 0$, we find that $x(v + t) > x(v - s)$ for all $t,s, \in ]0,\epsilon[$ for some small positive $\epsilon$.

We claim that for all $r \in ]a,v[$ and $\tilde r \in ]v,b[$ with $y(r) = y(\tilde r)$ we have $x(r) < x(\tilde r)$. In fact this is clear from continuity, since $\gamma$ is simple closed.

As above we find that $x(a + t) > x(a)$ for all sufficiently small $t > 0$ since $x'(a) > 0$. From the above claim we thus conclude that $x(b - t) > x(a) = x(b)$ for all sufficiently small $t > 0$. Therefore $x'(b) = \lim_{t \searrow 0} \frac{x(b) - x(b - t)}{t} < 0$, a contradiction.

Wolfgang Spindeler
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